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Rewrite $\frac{125}{\left(\frac{1}{625}\right)^{-x-3}}=5^3$ in a common base then solve for $x$ I am to rewrite $\frac{125}{(\frac{1}{625})^{-x-3}}=5^3$ and then solve for x. My textbooks solutions section says the solution is -3. I gave it a shot and got 3.25. Here is my working:
$$\frac{125}{\left(\frac{1}{625}\right)^{-x-3}}=5^3$$
$$\frac{5^3}{\left(\frac{1}{5^4}\right)^{-x-3}}=5^3$$
(Here's where I get pretty confused. I think I can pull the denominator component $\frac{1}{5^4}$ up above and transform from a fraction to a full number, but I'm unsure how to handle the exponent $-x-3$)
$$5^3 \times 5^{4(x-3)}=5^3$$
$$5^{12x-36}=5^3$$
$$12x-36=3$$
$$12x=39$$
$$x=\frac{39}{12}=3.25$$
Where did I go wrong and how can I arrive at 3? I suspect I went stray between the second and 3rd lines. If this is so, rather than just provide a solution please please do also explain the logic explicitly... how to I 'pull up' $\left(\frac{1}{5^4}\right)^{-x-3}$, what are the rules for each component?
| Your first mistake starts here:
$$5^3 \times 5^{4(x-3)}=5^3$$
This is not correct. Because,
$$\begin{align}\left(\frac{1}{625}\right)^{-x-3}&=\left(5^{-4}\right)^{-x-3}\\
&=5^{-4\times (-x-3)}\\
&=5^{4(x+3)}.\end{align}$$
Your second mistake is:
$$5^3 \times 5^{4(x-3)}=5^3$$
$$5^{12x-36}=5^3$$
The correct one is as follows:
$$5^{3+4(x-3)}=5^3$$
But, we don't need this. Because there is a simpler way:
$$\begin{align}5^3 \times 5^{4(x-3)}=5^3 &\iff 5^{4(x-3)}=5^0\\ &\iff 4(x-3)=0\end{align}$$
However, even the last one we wrote is not valid. Because, we have
$$\begin{align}5^3 \times 5^{4(x+3)}=5^3 &\iff 5^{4(x+3)}=5^0\\ &\iff 4(x+3)=0 \\ &\iff x=-3.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determine $\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$ Determine $$\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$$ | Hint:
For small enough $x,y$ such that the ordered pair $(x,y)\ne (0,0)$, we have $|\sin(xy^2)|\lt |xy^2|$
$|y|=\sqrt{y^2}\le\sqrt{y^2+x^2}$ and so by similar arguments, we have:
$\big|\frac{x^2 y\sin(xy^2)}{(x^4+y^4)\sqrt{x^2+y^2}}-0\big|\le \frac{|x^3|y^2}{x^4+y^4}\le \frac{|x|x^2}{\sqrt{x^4+y^4}}\le |x|\le \sqrt{x^2+y^2}$
Can you fill in the gaps and finish using definition of limit?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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If $P_n(x)=\sum_{k=0}^{[(n-1)/2]}(-1)^k\ _nC_{2k+1}\ x^k$, find $A(x)$ and $B(x)$ such that $P_{n+2}(x)=A(x)P_{n+1}(x)+B(x)P_n(x)$ A high school math problem:
For $n \in N$ (the natural number set), let us define the polynomial $P_n(x)$ as follows:
$$P_n(x) = \sum_{k=0}^{[(n-1)/2]}(-1)^k \ {}_nC_{2k+1} \ x^k$$
where $[(n-1)/2]$ represents the greatest integer that does not exceed $(n-1)/2$.
Find the expressions for the polynomials $A(x)$ and $B(x)$ such that for all $n \in N$
$$P_{n+2}(x) = A(x)P_{n+1}(x) + B(x)P_n(x)$$
Someone has said it should start from rearranging the formula and stating that
$$P_n(x) = \frac{(1+i \sqrt{x})^n - (1-i \sqrt{x})^n }{2i\sqrt{x}}$$
And from here I am able to do the rest of the steps. But I do not see intuitively how I can reach this expression.
| We obtain
\begin{align*}
\color{blue}{P_n(x)}&\color{blue}{=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(-1)^k\binom{n}{2k+1}x^k}\tag{1}\\
&=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}i^{2k}\binom{n}{2k+1}\left(\sqrt{x}\right)^{2k}\tag{2}\\
&=\frac{i\sqrt{x}}{i\sqrt{x}}\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}i^{2k}\binom{n}{2k+1}\left(\sqrt{x}\right)^{2k}\\
&=\frac{1}{i\sqrt{x}}\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\binom{n}{2k+1}\left(i\sqrt{x}\right)^{2k+1}\\
&=\frac{1}{2i\sqrt{x}}\left[\sum_{k=0}^{n}\binom{n}{k}\left(i\sqrt{x}\right)^{k}-\sum_{k=0}^{n}(-1)^k\binom{n}{k}\left(i\sqrt{x}\right)^{k}\right]\tag{3}\\
&\,\,\color{blue}{=\frac{1}{2i\sqrt{x}}\left[\left(1+i\sqrt{x}\right)^n-\left(1-i\sqrt{x}\right)^n\right]}\tag{4}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we use for convenient notation $\binom{n}{k}:=_nC_{k}$.
*In (2) we use the identities $i^2=-1$ and $x=\left(\sqrt{x}\right)^2$.
*In (3) we represent the terms with odd index $2k+1$ as sum of all terms minus the terms with even index $2k$. Note, we have to divide by $2$ since we add the odd terms twice.
*In (4) we apply the binomial theorem twice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4135453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ to prove $3^{n+2}$ does not divide $2^{3^n}+1$ Prove for all natural numbers $n$ such that $2^{3^n}+1$ is not divisible by $3^{n+2}$.
My working for not divisible: Induction proof
Base case: n = 0
$$3^{0+2}=9$$
and
$$2^{3^0}+1=3$$
9 cannot divide 3, so base case is true.
Assume n=k is true. That means $B=3^{k+2}$ does not divide $A=2^{3^k}+1$. Now prove $3^{k+3}$ does not divide $2^{3^{k+1}}+1$.
Using $a^3+b^3=(a+b)(a^2-ab+b^2)$:
$$(2^{3^k})^3+1=(2^{3^k}+1)((2^{3^k})^2-2^{3^k}+1)$$
$$=A((2^{3^k})^2-A+2)$$
$$3^{k+3}=3\times 3^{k+2}$$
$$=3B$$
That's where I need help!
I want to show that $A((2^{3^k})^2-A+2)$ is not divisible by $3B$ by using hypothesis "$A$ is not divisible by $B$".
| Hint: It is easy to see that $$3~|~(2^{3^k})^2-2^{3^k}+1$$ but $$9\nmid(2^{3^k})^2-2^{3^k}+1.$$
Note: To complete the proof, use the Unique Factorization Theorem of integers and count the multiplicity of the prime factor $3$. (This theorem requires only the Division Algorithm to prove.)
An alternative formal way to show that $3B\nmid A\cdot((2^{3^k})^2-A+2)$ is to use the 3-adic order: Compare the $3$-adic order $\nu_3$ of the two integers, one has $$\nu_3(3B)=\nu_3(3)+\nu_3(B)=1+\nu_3(B)$$ and $$\nu_3(A\cdot ((2^{3^k})^2-A+2))=\nu_3(A)+\nu_3((2^{3^k})^2-A+2)=\nu_3(A)+1,$$ where the latter equality follows from the above suggestions and comments.
Now the induction hypothesis $B\nmid A$ shows that $$\nu_3(B)>\nu_3(A)$$
$$\Rightarrow \nu_3(B)+1>\nu_3(A)+1$$
$$\Rightarrow 3B\nmid A\cdot((2^{3^k})^2-A+2),$$ completing the case $n=k+1$ and the proof by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying the sequence defined by $A_0=6$ and $A_n=\frac{8}{9}A_{n-1}+\frac{6}{5}(\frac{20}{9})^n$, using sigma notation
I have $A_0 = 6$ and
$$A_n = \left(\frac{8}{9}\right)A_{n-1} + \left(\frac{6}{5}\right) \left(\frac{20}{9}\right)^n$$
and I want to simplify it with sigma notation.
I have it up to
$$\begin{align}A_4 = &6 \left(\frac{8}{9}\right)^4 + {\left(\frac{8}{9}\right)^3} \left(\frac{6}{5}\right) {\left(\frac{20}{9}\right)^{1}} + {\left(\frac{8}{9}\right)^2} \left(\frac{6}{5}\right){\left(\frac{20}{9}\right)^{2}} \\
&+ {\left(\frac{8}{9}\right)^1}\left(\frac{6}{5}\right){{\left(\frac{20}{9}\right)^{3}}} + \left(\frac{6}{5}\right) {\left(\frac{20}{9}\right)^4}
\end{align}$$ because I wanted to find the pattern
Apparently the answer is
$$4\left(\frac{8}{9}\right)^n + 2\left(\frac{20}{9}\right)^n$$ and the simplification into sigma notation is $$6\left(\frac{8}{9}\right)^n + \sum_{k=0}^{n}\left(\frac{8}{9}\right)^{n-k}\left(\frac{6}{5}\right)\left(\frac{20}{9}\right)^k$$
but I can seem to can seem to justify that. It's has to be some sort of algebraic manipulation that I did wrong but I don't see how the $k=0$ term works? To me, it seems that the sigma notation adds an extra term when $k=0$
(btw, this is based on the second-last and last pages of the PowerPoint presentation "Mathematics Geometry: Menger Sponge" (PPT link via ubc.ca) by the University of British Columbia Science and Mathematics Education Research Group.)
--
Personally, I think it's a typo since on Wolfram Alpha, if I use my $k=1$ then I get the right result but with $k=0$ it doesn't work
| (I have done this type of problem
so often,
I could almost do it in my sleep.
However,
each time I do it a little differently
and from scratch,
so here is the
latest version.)
If
$a_n
=ua_{n-1}+b_n
$
then,
dividing by $u^n$,
$\dfrac{a_n}{u^n}
=\dfrac{ua_{n-1}}{u^n}+\dfrac{b_n}{u^n}
=\dfrac{a_{n-1}}{u^{n-1}}+\dfrac{b_n}{u^n}
$.
Letting
$c_n = \dfrac{a_n}{u^n}$,
this becomes
$c_n
=c_{n-1}+\dfrac{b_n}{u^n}
$
so
$c_n-c_{n-1}
=\dfrac{b_n}{u^n}
$.
The left side telescopes
when summed so that
$c_m-c_0
=\sum_{n=1}^m (c_n-c_{n-1})
=\sum_{n=1}^m \dfrac{b_n}{u^n}
$
or
$\dfrac{a_m}{u^m}
=a_0+\sum_{n=1}^m \dfrac{b_n}{u^n}
$
or
$a_m
=u^ma_0+\sum_{n=1}^m u^{m-n}b_n
$.
If
$b_n=r s^n$
(as in this problem),
$\begin{array}\\
a_m
&=u^ma_0+\sum_{n=1}^m u^{m-n}rs^n\\
&=u^ma_0+ru^m\sum_{n=1}^m (s/u)^n\\
&=u^ma_0+ru^m\sum_{n=1}^m t^n
\qquad t = s/u\\
&=u^ma_0+ru^m\dfrac{t-t^{m+1}}{1-t}
\qquad\text{if } t \ne 1\\
&=u^ma_0+ru^m\dfrac{s/u-(s/u)^{m+1}}{1-s/u}\\
&=u^ma_0+r\dfrac{su^{m-1}-s^{m+1}/u}{1-s/u}\\
&=u^ma_0+r\dfrac{su^{m}-s^{m+1}}{u-s}\\
&=u^ma_0+rs\dfrac{u^{m}-s^{m}}{u-s}\\
&=u^m(a_0+\dfrac{rs}{u-s})-rs\dfrac{s^{m}}{u-s}\\
&=u^m(a_0+v)-vs^m
\qquad v=\dfrac{rs}{u-s}\\
\\
&=u^ma_0+ru^mm
\qquad\text{if } t = 1\\
&=u^m(a_0+rm)\\
\end{array}
$
If
$a_0=6,
u=\dfrac89,
r=\dfrac65,
s=\dfrac{20}{9}
$
then
$u-s
=-\dfrac{12}{9}
=-\dfrac43
$,
$rs
=\dfrac83
$,
$v
=-2
$
so that
$\begin{array}\\
a_n
&=u^m(a_0+v)-vs^m\\
&=(8/9)^m(6-2)+2(20/9)^m\\
&=\dfrac{4\cdot 8^n+2\cdot 20^n}{9^n}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to calculate the limit $\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$ $$I=\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$$
I tried $$\frac{n}{\sqrt{n^2+n+n}}\leq\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\leq\frac{n}{\sqrt{n^2+n+1}}$$
But$$\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^2+n+n}}\right)^n=\lim_{n\to\infty}\left(1+\frac2n\right)^{\displaystyle-\frac{n}{2}}=e^{\displaystyle \lim_{n\to\infty}-\frac n2\ln(1+\frac2n)}=\frac1e$$
And in the same way,i got $$\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^2+n+1}}\right)^n=\lim_{n\to\infty}\left(1+\frac1n+\frac{1}{n^2}\right)^{\displaystyle-\frac{n}{2}}=e^{\displaystyle \lim_{n\to\infty}-\frac n2\ln(1+\frac1n+\frac{1}{n^2})}=\frac{1}{\sqrt e}$$
So i only got $$\frac1e\leq I\leq\frac{1}{\sqrt e}$$
Could someone help me get the value of $I$. Thanks!
| If you like generalized harmonic numbers
$$S_n=\sum_{k=1}^n \frac{1}{\sqrt{n^2+n+k}}=H_{n^2+2 n}^{\left(\frac{1}{2}\right)}-H_{n^2+n}^{\left(\frac{1}{2}\right)}$$
Using twice the asymptotics
$$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2 p^{1/2}}-\frac{1}{24 p^{3/2}}+O\left(\frac{1}{p^{5/2}}\right)$$ and continuing with Taylor expansions
$$S_n=1-\frac{3}{4 n}+\frac{5}{8 n^2}+O\left(\frac{1}{n^3}\right)$$
$$n \log(S_n)=-\frac{3}{4}+\frac{11}{32 n}+O\left(\frac{1}{n^2}\right)$$
$$S_n^n=e^{n \log(S_n)}=e^{-3/4}\left(1+ \frac{11}{32 n}\right)+O\left(\frac{1}{n^2}\right)$$
Trying for $n=10$ the "exact" value is $0.487638$ while the truncated expansion gives $0.488604$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
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Nonhomogeneous First order differential equation I'm trying to understand what is wrong with this solution, since I'm not getting the same answer in Matlab
$y'-xy=xy^{3/2}\, ,y(1)=4$
\begin{align*}
y'-xy=&xy^{3/2}&\\
\dfrac{dy}{dx}=&x(y+y^{3/2})&\\
\dfrac{dy}{(y+y^{3/2})}=&x\, dx&\\
-2\ln \dfrac{1+\sqrt{y}}{\sqrt{y}}=&\dfrac{x^2}{2}+c&\\
y=&\dfrac{-1}{(1-\mathrm{e}^{\tfrac{-x^2}{4}+c})^2}&\\
4=&\dfrac{-1}{(1-\mathrm{e}^{\tfrac{-1}{4}+c})^2}&\\
c=&-0.655&\\
y=&\dfrac{-1}{(1-\mathrm{e}^{\tfrac{-x^2}{4}-0.655})^2}&
\end{align*}
This is what I'm getting in Matlab
$ \left(\begin{array}{c}
\frac{{{\left(\textrm{tanh}\left(\frac{x^2 }{8}+\textrm{atanh}\left(3\right)-\frac{1}{8}\right)+1\right)}}^2 }{4}\\
\frac{{{\left(\textrm{tanh}\left(-\frac{x^2 }{8}+\textrm{atanh}\left(5\right)+\frac{1}{8}\right)-1\right)}}^2 }{4}
\end{array}\right)$
| I think $$y=\frac {1}{(1-e^{-\frac {x^2}{4}+c})^2}$$ instead of what you wrote. $y$ cannot be negative anyway, as $\sqrt y$ is present in the equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
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Perfect Square With Two Integer Variables I am trying to solve a number theory problem in general form. However, I got stuck in the following step:
$a,b,n \in \mathbb Z^{+}$ for which values of $n$, this equation is solvable $\frac{(n+1)(n+2a)}{2} = b^2$ ?
Can we make a general statement about $n$ ? By the way I have tried to apply some inequalities with cases $n>a$, $n<a$ and $n=a$ . Also I think modular arithmetic doesn't work so good for this problem.
| Clearing denominators and expanding, the question is to find all $n\in\Bbb{Z}$ for which there exist $a,b\in\Bbb{Z}$ such that
$$2b^2=(n+1)(n+2a).$$
So let $n\in\Bbb{Z}$ be given and suppose $a,b\in\Bbb{Z}$ satisfy the above.
Clearly $2b^2$ is divisible by $n+1$, say
$$2b^2=m(n+1),$$
for some $m\in\Bbb{Z}$. It follows that
$$2a=\frac{2b^2}{n+1}-n=m-n,$$
where in particular $m\equiv n\pmod{2}$. With this in mind, let's see for which $n\in\Bbb{Z}$ we can find appropriate $a,b\in\Bbb{Z}$.
*
*If $n$ is even, say $n=2k$, we can take $b=n+1$ to get
$$2b^2=2(n+1)^2,$$
corresponding to $m=2(n+1)$, and correspondingly $a=\frac{m-n}{2}=k+1$, to get
$$\frac{(n+1)(n+2a)}{2}=\frac{(2k+1)(2k+2k+2)}{2}=(2k+1)^2=(n+1)^2=b^2.$$
*If $n$ is odd then $n+1$ is even, so $n+1=2^ck$ for some $c,k\in\Bbb{Z}$ with $c\geq1$ and $k$ odd. If $c$ is odd, say $c=2d+1$, then we may take $b=2^dk$ to get
$$2b^2=2(2^dk)^2=2^{2d+1}k^2=2^ck^2=k(n+1),$$
corresponding to $m=k$, and correspondingly $a=\frac{m-n}{2}=\frac{k+1}{2}-2^{2d}k$, to get
$$\frac{(n+1)(n+2a)}{2}=\frac{2^ck\cdot k}{2}=(2^dk)^2=b^2.$$
*Finally, if $n$ is odd and $n+1=2^ck$ with $c$ is even, say $c=2d$, then
$$2b^2=m(n+1)=2^cmk=(2^d)^2mk,$$
which shows that $m$ is divisible by $2$, i.e. $m\equiv0\pmod{2}$. But this contradicts $m\equiv n\pmod{2}$. So for $n$ of the form $n=4^dk-1$ with $k$ odd and $d\geq1$ there are no such $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int_{0}^{\infty} \frac{\left(\frac{\pi x}{2}-\log (x)\right)^3}{\left(x^2+1\right)^2 \left(\log^2(x)+\frac{\pi ^2}{4}\right)} = \pi$ proof Today @integralsbot at twitter posted an interesting integral relation for complicated integral that gives $\pi$ as a result. I don't know how this bot works or where does it take the relation it posts from, but some of the problems are really difficult to prove and interesting. Such as this example.
$$\int_{0}^{\infty} \frac{\left(\frac{\pi x}{2}-\log (x)\right)^3}{\left(x^2+1\right)^2 \left(\log^2(x)+\frac{\pi ^2}{4}\right)} = \pi \tag{1}\label{eq1}$$
I've tried to calculate this and it appears to be very difficult (for starters wolfram alpha can't calculate it). The plot of the function under the integral doesn't look very extraordinarily:
Using partial fraction decomposition the integral can be transformed to the following form:
$$\int_{0}^{\infty} \left( \frac{3 \pi x}{2 \left(x^2+1\right)^2}-\frac{\log (x)}{\left(x^2+1\right)^2}+\frac{\pi^3 x^3-6 \pi ^2 x^2 \log (x)-3 \pi ^3 x+2 \pi ^2 \log (x)}{2 \left(x^2+1\right)^2 \left(4\log ^2(x)+\pi ^2\right)} \right)$$
The first two terms of the integral can be more
$$ \int_{0}^{\infty} \frac{3 \pi x}{2 \left(x^2+1\right)^2} = \frac{3 \pi}{4} $$
or less easily solved
$$ \int_{0}^{\infty} -\frac{\log (x)}{\left(x^2+1\right)^2} = \frac{\pi}{4} $$
to yield:
$$ \int_{0}^{\infty} \left( \frac{3 \pi x}{2 \left(x^2+1\right)^2} - \frac{\log (x)}{\left(x^2+1\right)^2} \right) = \pi \tag{2}\label{eq2} .$$
This is interesting by itself, since plot of the integrated function has interesting shape with two inflection points of the first derivative, which I wouldn't expect to be equal to $\pi$ (of course the algebra clearly shows that it is)
What's however even more interesting is that, if I'm not mistaken, it follows from \eqref{eq1} and \eqref{eq2} that
$$ \int_{0}^{\infty} \frac{\pi^3 x^3-6 \pi ^2 x^2 \log (x)-3 \pi ^3 x+2 \pi ^2 \log (x)}{2 \left(x^2+1\right)^2 \left(4\log ^2(x)+\pi ^2\right)} = 0 \tag{3}\label{eq3}.$$
This seems to be correct when checked with numerical integration (I've checked in relatively high precision of ~100 decimal digits) but seem to be very hard to prove. Plot of the function under the integral \eqref{eq3} doesn't appear (to me) to suggest any trivial solution:
The most obvious way to transform integral \eqref{eq3} is to expand the numerator which gives:
$$
\int_{0}^{\infty} \left( -\frac{3 \pi ^2 x^2 \log (x)}{\left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}-\frac{3
\pi ^3 x}{2 \left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}+\frac{\pi ^2 \log
(x)}{\left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}+\frac{\pi ^3 x^3}{2
\left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)} \right), $$
but the resulting integrals are also very difficult, perhaps too difficult to directly attack the problem in this way (I've tried to solve the second one, since it seems to have the simplest numerator, but didn't managed to get any results yet) and neither their numerical values (-1.93789..., 3.229820..., -0.322982..., -0.968946...) nor the plots indicate any obvious cancellations. Maybe there is some indirect way? Perhaps the integral \eqref{eq3} can be shown to be equal to $0$ without directly solving the integral at all?
As I mentioned in the beginning of this post, I don't know where does this \eqref{eq1} relation come from but it seems to be true and interesting and must've been found somehow. Therefore I believe some solution of this problem exists. I post it as an interesting challenge/puzzle. Maybe someone knows some of the relations, can solve some of the unsolved integrals, sees clever substitution or can provide any interesting insight into this problem.
So, to finally pose the question.
How to prove/solve either \eqref{eq1} or \eqref{eq3}?
| Let
$$f(x)= \frac{\pi^3 x^3-6 \pi ^2 x^2 \log (x)-3 \pi ^3 x+2 \pi ^2 \log (x)}{2 \left(x^2+1\right)^2 \left(4\log ^2(x)+\pi ^2\right)} $$ that is to say
$$f(x)=\frac{\pi ^2 \left(\pi x \left(x^2-3\right)-2\left(3x^2-1\right) \log
(x)\right)}{2 \left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}$$
and write
$$I=\int_0^\infty f(x)\,dx=\int_0^1 f(x)\,dx+\int_1^\infty f(x)\,dx$$ For the second integral, let $x=\frac 1 x$. So, now we have
$$I=\int_0^1 f(x) \,dx-\int_0^1 g(x) \,dx=\int_0^1 [f(x)- g(x)] \,dx$$ with
$$g(x)=\frac{\pi ^2 \left(\pi \left(3 x^2-1\right)+2 x \left(x^2-3\right) \log
(x)\right)}{2 x \left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}$$
$$f(x)-g(x)=\frac{\pi ^2 \left(\pi \left(x^4-6 x^2+1\right)-8 x \left(x^2-1\right) \log
(x)\right)}{2 x \left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}$$ and, numerically,
$$\int_0^1 [f(x)- g(x)] \,dx=0$$ which still needs to be proved.
In fact, what happens if that
$$\int_0^a [f(x)- g(x)] \,dx=-\int_a^1 [f(x)- g(x)] \,dx$$ where $a$ is the solution of
$$\pi \left(x^4-6 x^2+1\right)-8 x \left(x^2-1\right) \log (x)=0$$
$$a=0.1928617994587428536765548799193482174970163765555\cdots$$ which is not recognized by inverse symbolic calculators.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4142470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
How do I find all values $x$ such that a vector is a linear combination of a nonempty set of vectors in vector space $\mathbb{R^3}$? In vector space $\mathbb{R^3}$.
Find all values $x$ such that $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}$ $\in$ span $\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} \}$.
My solution:
I used the equation:
$r_1\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}$ + $r_2\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$ + $r_3\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$ = $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}$
which can be represented with the augmented matrix:
$$ \left[
\begin{array}{ccc|c}
1&0&1&2x^2\\
1&1&2&-3x\\
3&1&4&1\\
\end{array}
\right] $$
which has an RREF of:
$$ \left[
\begin{array}{ccc|c}
1&0&1&2x^2\\
0&1&1&-3x-2x^2\\
0&0&0&1-4x^2+3x\\
\end{array}
\right] $$
I understood this as the system could only be consistent iff:
$1-4x^2+3x = 0$
So then, I solved for $x$ which yields $x = 1,\frac {-1}{4} $.
Is it correct to say, then, that $1$ and $\frac {-1}{4}$ are all the values of $x$ that will make $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}$ $\in$ span $\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} \}$?
| Here's another approach. Just for your own entertainment.
Using techniques from multivariable calculus you can write $$\text{span}\Bigg\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}\Bigg\}=\big\{(x,y,z)\in \mathbb{R}^3:2x+y-z=0\big\}$$ Therefore $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}\in \text{span}\Bigg\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}\Bigg\}$ if and only if $$2(2x^2)-3x-1=0 \iff (4x+1)(x-1)=0$$ So $x=-1/4,1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Summation to infinity with a general pattern $$
\sum_{n=1}^{\infty} \frac{(n-1)!}{(x+1)(x+2)..... (x+n)}.
$$
I tried solving it by inserting some values
$\frac{1}{x+1}$ +$\frac{1}{(x+1)(x+2)}$+$\frac{2}{(x+1)(x+2)(x+3)}$
$\frac{1}{x+1}$+$\frac{1}{x+1}$-$\frac{1}{x+2}$+$\frac{1}{x+1}$-$\frac{1}{x+2}$+$\frac{1}{x+3}$-$\frac{1}{x+2}$
This gives me a hint that
$\frac{1}{x+1}$ this comes n times
-$\frac{1}{x+2}$ comes n-1times and so on
But since the terms tend to infinity I dont find it useful so any hint?
Answer is $\frac{1}{x}$
| A more general result is true.
$\textbf{Lemma:}$ Let $(a_n)_{n\ge 1}$ be a positive sequence. Then for $x \ne 0$,
$$
\sum_{n=1}^{\infty} \frac{a_1a_2\cdots a_{n-1}}{(x+a_1)(x+a_2)\cdots(x+a_n)} = \frac{1}{x}
$$
Proof. Define $A_0 = \frac{1}{x}$ and $$A_N = \frac{a_1a_2\cdots a_{N}}{x(x+a_1)(x+a_2)\cdots(x+a_N)}$$
Note that $$A_{n-1} - A_n = \frac{a_1a_2\cdots a_{n-1}}{(x+a_1)(x+a_2)\cdots(x+a_n)}$$
So,
$$
\begin{align*}
\sum_{k=1}^{N} \frac{a_1a_2\cdots a_{n-1}}{(x+a_1)(x+a_2)\cdots(x+a_n)} &= \sum_{k=1}^{N} (A_{n-1} - A_n) \\ &= A_0 - A_N \\
&= \frac{1}{x} - \frac{a_1a_2\cdots a_{N}}{x(x+a_1)(x+a_2)\cdots(x+a_N)}
\end{align*}
$$
Letting $N \to \infty$, we get the desired result.
In your case, $a_n = n$.
| {
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"url": "https://math.stackexchange.com/questions/4147826",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
Prove that $\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}$
My question:
Let $a,b,c$ be positive real numbers satisfy $a+b+c=3.$
Prove that $$\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}.$$
I have tried to change the LHS to $$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2}$$
And using Cauchy–Schwarz inequality for it
$$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2}\geq \frac{(a+b+c)^2}{ab+bc+ca+ac^2+ba^2+cb^2}$$
Then because $$ab+ca+ca\leq \frac{(a+b+c)^2}{3}=\frac{3^2}{3}=3,$$
$$\frac{(a+b+c)^2}{ab+bc+ca+ac^2+ba^2+cb^2}\geq \frac{9}{3+ac^2+ba^2+cb^2}$$
Finally, I can't prove $ac^2+ba^2+cb^2\leq 3$ $$ $$
I look forward to your help, thank you!
| Partial answer hint :
For $x,y,z\in[0.5,2]$ we have the inequality :
$$\frac{x}{y+z^{2}}-\frac{1}{x+y+z}\frac{x}{y+z}\ge0$$
Summing and using the constraint gives the inequality .
Now we have one variable less than $1/2$ come back further .
Extended comment :
I have almost if finish if $\max({a,b,c})=a\ge 2$ the inequality is obvious . Next we have the following inequalities for $2\geq a\geq 1.5\geq c\ge 0.5\geq b$: $$\frac{a}{b+c^{2}}+\frac{c}{a+b^{2}}+\min\left(\frac{1}{8}b,\frac{1}{8}c\right)-\left(\frac{a}{b+c^{2}}+\max\left(\frac{1}{2}b,\frac{1}{2}c\right)\right)\ge 0$$ And $$\frac{a}{b+c^{2}}+\frac{b}{c+a^{2}}+\frac{c}{a+b^{2}}-\left(\frac{a}{b+c^{2}}+\frac{c}{a+b^{2}}+\min\left(\frac{1}{8}b,\frac{1}{8}c\right)\right)\ge0$$ And finally for $2\ge a\ge 1.6\geq c\ge 0.5\ge b$ : $$\frac{a}{b+c^2}+\max(0.5b,0.5c)\ge 1.5$$ and $a+b+c=3$ and $a,b,c\geq 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149973",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations and worked out examples convince me that this inequality indeed holds true. I have tried to solve this problem by induction. Clearly, for $N=1$ we have,
$$\mathcal{P}(1) = \left(\frac{2}{3}\right)^{\tfrac{3}{4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{4}} \approx 0.8774 < 1.$$
Now assume the inequality holds for $N$, then for $N+1$ we have,
\begin{align}
\mathcal{P}(N+1) &=\left(\frac{2N+2}{2N+3}\right)^{\tfrac{2N+3}{2N+4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{2N+4}} \\
&= \left(\left(\frac{2N}{2N+1}\right)\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)\right)^{\left(\frac{2N+1}{2N+2}\right)\left(\frac{2N+2}{2N+1}\cdot\frac{2N+3}{2N+4}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\frac{2N+2}{2N+4}}\\[1em]
&= \left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)\left(1 + \frac{1}{2(N+1/2)(N+2)}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\left(1 - \frac{1}{N+2}\right)}\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \small\left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}}\cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \mathcal{P}(N) \cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \cdot\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}
\end{align}
Now from here we know that the first three terms are all smaller than 1 ($\mathcal{P}(N) < 1$ by induction hypothesis). However the last term is larger than one. For the proof by induction to work out, we need that this last term cancels against,
$$\left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}}.$$ But I do not see how it does. Any help is greatly appreciated.
| We have
$$
(\mathcal{P}(N))^{2N + 2} = \left( {\frac{{2N}}{{2N + 1}}} \right)^{2N + 1} 2 = \frac{1}{{\left( {1 + \frac{1}{{2N}}} \right)^{2N} }}\frac{2}{{1 + \frac{1}{{2N}}}} < \frac{4}{9} \cdot 2 < 1,
$$
since $
{\left( {1 + \frac{1}{n}} \right)^n }
$ is increasing monotonically to $e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Differential equation with really strange solution... I wish to solve the following differential equation $$\frac{dy}{dx}=\frac{\sqrt{y+7}}{(x+3)(x+8)}$$ with initial condition $y(0)=2$. I think my method is correct but my final solution seems like a really strange answer... This is what I got:
$$\Rightarrow y=2+\frac{1}{100}\left(\ln\left|\frac{3(x+3)}{8(x+8)}\right|\right)\ln\left(\left|\frac{8(x+3)}{3(x+8)}\right|\right)$$
I’m not sure how to check if this is correct; is it right? Thanks!!
This is my working:
We have the differential equation $\frac{dy}{dx}=\frac{\sqrt{(y+7)}}{(x+3)(x+8)}$. This is clearly separable, so we rearrange to get the following:
$$\frac{1}{\sqrt{y+7}}dy=\int\frac{1}{(x+8)(x+3)}dx$$
$$\Rightarrow 2\sqrt{y+7}=\frac{1}{5}\ln\left|\frac{x+3}{x+8}\right|$$
$$\Rightarrow \sqrt{y+7}=\frac{1}{10}\ln\left|\frac{x+3}{x+8}\right|$$
$$\Rightarrow y+7=\frac{1}{100}\left(\ln\left|\frac{x+3}{x+8}\right|\right)^2$$
$$\Rightarrow y=-7+\frac{1}{100}\left(\ln\left|\frac{x+3}{x+8}\right|\right)^2+C$$
Now set $y(0)=2$:
$$2=-7+\frac{1}{100}\left(\ln\left|\frac{0+3}{0+8}\right|\right)^2+C$$
$$\Rightarrow 9=\frac{1}{100}\left(\ln\left|\frac{3}{8}\right|\right)^2+C$$
$$\Rightarrow 900=\left(\ln\frac{3}{8}\right)^2+100C$$
$$\Rightarrow 100C=900-\ln\left(\frac{3}{8}\right)^2$$
$$\Rightarrow C=9-\frac{1}{100}\ln\left(\frac{3}{8}\right)^2$$
Substituting this value of $C$ into the general solution yields the following:
$$y=-7+\frac{1}{100}\left(\ln\left|\frac{x+3}{x+8}\right|\right)^2+9-\frac{1}{100}\left(\ln\frac{3}{8}\right)^2$$
$$\Rightarrow y=2+\frac{1}{100}\left[\left(\ln\frac{x+3}{x+8}\right)^2-\ln\left(\frac{3}{8}\right)^2\right]$$
$$\Rightarrow y=2+\frac{1}{100}\left[\ln\left|\frac{x+3}{x+8}\right|+\ln\left(\frac{3}{8}\right)\right]\left[\ln\left|\frac{x+3}{x+8}\right|-\ln\frac{3}{8}\right],$$ by the difference of two squares identity.
$$\Rightarrow y=2+\frac{1}{100}\left(\ln\left|\frac{3(x+3)}{8(x+8)}\right|\right)\left(\left|\frac{8(x+3)}{3(x+8)}\right|\right)$$
| You should have added $C$, the constant, before squaring. That made all the difference.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the perimeter of triangle $ABC$. In triangle $ABC$, point $M$ is the midpoint of side $BC$. Suppose that the point
$D$ and $E$ on the $CA$ side so $CD = DE$ with $D$ between $C$ and $E$.
Suppose also points $F$ and $G$ on the side $AB$ so that $AM$, $BD$, and $CF$ meet at one point. Likewise with $AM$, $BE$, and $CG$. If, $AF + BM + CD = 17$ and $AG + BM + CE = 23$. Find the perimeter of triangle $ABC$.
With Ceva's Theorem obtained
$-$ $EG || DF || CB$
$-$ $\Delta ABC$ ~ $\Delta AEG$ ~ $\Delta ADF$
$GF=BF$
Let $AG = ax$, $AE = bx$, $GF = ay$, $ED = by$, $BM = z$
$ax + ay + z + by = 17$
$ax + z + 2by = 23$,
We get $y (b-a) = 6$
Perimeter $\Delta ABC$ = $(\frac{x}{y}-1)+40$
But i can find the value of $x/y$. if someone can help that great!
| Let $AG/AB = AE/AC = r$. Then because $F$ is the midpoint of $GB$, we have $$\frac{AF}{AB} = \frac{AG + (AB - AG)/2}{AB} = \frac{1+r}{2}.$$ Similarly, $$\frac{CE}{AC} = \frac{AC - AE}{AC} = 1 - r, \quad \frac{DC}{AC} = 1 - \frac{1+r}{2} = \frac{1-r}{2}.$$ Therefore, $$17 = AF + BM + CD = \frac{1+r}{2} c + \frac{a}{2} + \frac{1-r}{2} b, \\ 23 = AG + BM + CE = r c + \frac{a}{2} + (1-r) b,$$ where $a, b, c$ are the respective sides of $\triangle ABC$. Then multiplying the first equation by $2(1-2r)/(1-r)$ and the second by $2r/(1-r)$, we get $$a+b+c = 22 + \frac{12}{1-r}.$$ This suggests that the answer depends on $r$.
To prove that the triangle is not uniquely determined, we can illustrate two distinct examples.
Example 1. Choose $r = 1/4$; we get $a+b+c = 38$. Then choose $a = b = 18$, $c = 2$. Consequently $AG = 1/2$, $BM = 9$, $CE = 27/2$, $AF = 5/4$, and $CD = 27/4$.
Example 2. Choose $r = 1/13$; we get $a+b+c = 35$. Then choose $a = 14$, $b = 17$, $c = 4$. Consequently $AG = 4/13$, $BM = 7$, $CE = 204/13$, $AF = 28/13$, and $CD = 102/13$.
It is worth noting that we must have $0 < r < 5/11$, otherwise the triangle inequality is not met.
| {
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Consider hyperboloid $z^2 = x^2 − 4y^2 + 5$. Find an equation of the line which belongs to this hyperboloid and the point $(0, 1, 1)$. Consider hyperboloid $z^2 = x^2 − 4y^2 + 5$. Find an equation of the line which belongs to this hyperboloid and the point $(0, 1, 1)$.
This is what I have so far:
$I(s) = OP + sv$
Letting $v = (a, b, c) $
$I(s) = (0, 1, 1) + s(a, b, c)$
$x = as, y = bs + 1, z = cs + 1$
I then plugged the parameterization into the equation $z^2 = x^2 − 4y^2 + 5$, giving me $s^2(a^2-4b^2-c^2)-2s(4b+c)=0$.
However, how do I proceed from here to find $(a, b, c)$?
| We'll assume $s \neq 0$ to simplify things since we know $I(0) = (0,1,1)$. Notice then that
\begin{align*}
(cs+1)^2=(as)^2 -4(bs+1)^2+5 &\implies s \left(sa^2 - 4 b^2 s - 8 b - c^2 s - 2 c\right) = 0\\
& \implies s\left(a^2 -4b^2 -c^2\right) = 8b+2c \tag{1}
\end{align*}
From here, since it should hold that $I(s)$ is in the hyperboloid for any $s$, choosing some arbitrary $s_1, s_2 \in \mathbb{R} \setminus \{0\}$ we get that
$$
(s_1-s_2) \left(a^2 -4b^2 -c^2\right) = 0
$$
but since $s_1$ need not be equal to $s_2$, to guarantee that the above equation holds in general it must be the case that
$a^2 -4b^2 -c^2 = 0$. Plugging the previous condition in $(1)$ we get $c = - 4b$. So combining the last two conditions we get that
\begin{align}
a^2 = 4\left(- \frac{c}{4} \right)^2 +c^2 \implies c = \pm \frac{2a}{\sqrt{5}}\\
a^2 = 4b^2 +(-4b)^2 \implies b = \mp \frac{a}{2\sqrt{5}}\\
\end{align}
noticing that $c = - 4b$ imposes $b$ and $c$ to have different signs. So your line has a free parameter $a$, and choosing $a=1$ for simplicity you get your solutions to be
$$
\boxed{I(s) = (0,1,1) + s\left(1, \pm \frac{1}{2\sqrt{5}}, \mp \frac{2}{\sqrt{5}}\right)}
$$
Here's a graph of how the solutions look like:
where the dotted line is the solution with the $y$ coordinate $+ \frac{1}{2\sqrt{5}}$ and the filled-out line is the solution with the $y$ coordinate $- \frac{1}{2\sqrt{5}}$.
| {
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Looking for other approaches to find the height $AH$ in triangle $ ABC $ where $A(1,5)$ , $B(7,3)$, $C(2,-2)$
We have a triangle with vertices $A(1,5)$ , $B(7,3)$, $C(2,-2)$. What
is the length of the height $AH$ in the triangle $ABC$ ?
$1)4\qquad\qquad2)3\sqrt2\qquad\qquad3)5\qquad\qquad4)4\sqrt2$
This is a problem from a timed exam.
Here is my approach:
The equation of the line passes through $B$ and $C$ is $y_=x-4$ so slope of $AH$ is $-1$ and its a line passes through $A(1,5)$ so the equation for this line becomes $y=-x+6$. in order to find coordinate of $H$ we should equate the formulas of those lines:
$$x-4=-x+6\quad\to\quad x=5\quad\text{and $\quad y=1\quad$ So$\quad H (5,1)$}$$
$$\text{Distance from $A$ to $H$:$\qquad$}AH=\sqrt{(5-1)^2+(1-5)^2}=4\sqrt2$$
Although I believe my answer is quick but I'm looking for other ideas to solve this problem.
| Area of triangle is \begin{align}\frac12|x_1y_2+x_2y_3+x_3y_1-x_2y_1-x_3y_2-x_1y_3|&=\frac12|3-14+10-35-6+2|\\
&=20 \end{align}
Hence
$$AH \times BC=40$$
$$AH \times \sqrt{5^2+5^2}=40$$
$$5\sqrt{2} AH = 40$$
$$AH=\frac{8}{\sqrt2}=4\sqrt2$$
| {
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$\measuredangle C=120^\circ$ and two altitudes $AH$ and $BD$ are altitudes of $\triangle ABC$ and $\measuredangle ACB=120^\circ$. If $S_{\triangle HCD}=\dfrac{15\sqrt3}{4},$ find the area of $\triangle ABC$.
$$S_{\triangle HCD}=\dfrac12\cdot CH\cdot CD\cdot\sin\measuredangle HCD=\dfrac{\sqrt3}{4}CH\cdot CD=\dfrac{15\sqrt{3}}{4}\ \implies CH\cdot CD=15$$
On the other hand $$S_{\triangle ABC}=\dfrac12\cdot AC\cdot BC\cdot\sin\measuredangle ACB=\dfrac{\sqrt3}{4}AC\cdot BC=?$$
I noted that $ABDH$ is inscribed, because $\measuredangle ADB=\measuredangle AHB=90^\circ$, so $$AC\cdot CD=BC\cdot CH.$$ I am stuck here. Thank you in advance!
| As $\angle BCD = 60^0, BD = CD \tan 60^0 = CD \sqrt3$
$CH = AC \cos 60^0 \implies AC = 2 \cdot CH$ as $\angle ACH = 60^0$
Area $\triangle ABC = \dfrac{1}{2} \cdot AC \cdot BD = \dfrac{1}{2} \cdot 2 \cdot CH \cdot CD \sqrt3 = 15 \sqrt3$
(as you already found that $CH \cdot CD = 15$)
| {
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Computing probabilities for each value of a random variable Assume you have a bag of 6 marbles, 2 of which are red and 4 of which are blue, and let the random variable Y represent the number of red marbles drawn after 3 are taken from the bag.
I'm trying to compute the probability for each value of Y.
What I've attempted:
Total number of combinations: ${}_6 \mathrm{ C }_3 = {6 \choose 3} = \frac{6!}{3!(6-3)!} = 20$
$P(Y = 0) = \frac{4 \choose 3}{20} = \frac{4}{20}$ (because there are four blue marbles in the bag and we are looking for the combinations of drawing three of these four)
$P(Y = 1) = 2\times\frac{4 \choose 2}{20} = \frac{12}{20}$ (because if you draw one red marble you have to draw two blue marbles from the bag to get Y = 1)
$P(Y = 2) = \frac{4 \choose 1}{20} = \frac{4}{20}$ (because if you draw two red marbles you have to draw one blue marble from the four in the bag to get Y = 2)
Have I done this correctly? I would appreciate any help.
| I have a slightly different way of doing it, and my results are the same as yours.
$P(Y=0)=\frac{4\times 3\times 2}{6\times 5\times 4}=\frac{1}{5}$
$P(Y=1)=3\times \frac{2\times 4\times 3}{6\times 5\times 4}=\frac{3}{5}$
$P(Y=2)=3\times \frac{2\times 1\times 4}{6\times 5\times 4}=\frac{1}{5}$
| {
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If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac {x}2$?
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac
{x}2$?
$1)-3\qquad\qquad2)2\qquad\qquad3)-2\qquad\qquad4)3$
Here is my method:
$$1-\sin x=4+4\sin x\quad\Rightarrow\sin x=-\frac{3}5$$
We have $\quad\sin x=\dfrac{2\tan(\frac x2)}{1+\tan^2(\frac{x}2)}=-\frac35\quad$. by testing the options we can find out $\tan(\frac x2)=-3$ works (although by solving the quadratic I get $\tan(\frac x2)=-\frac13$ too. $-3$ isn't the only possible value.)
I wonder is it possible to solve the question with other (quick) approaches?
| Hint:
$ \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$
$1=\sin^2 x + \cos^2 x$
Complete the square... and divide by something... then you should be able to use tangent formula
| {
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"source": "stackexchange",
"question_score": "3",
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Proving upper bound on a term involving binomial coefficients I am trying to show:
$$\binom{n}{k}\left(1-\frac{k}{n}\right)^{2n-2k+2} \left(\frac{k-1}{n}\right)^{2k}\leq \frac{1}{n^2}$$ for $k\in \{2,3,\cdots,n-1\}$ for all $n$
I observe that numerically that is true but analytically, I am not able to look at the term on LHS as a term of a binomial expansion or something like that to get an upper bound. Any ideas?
| This is not a proof.
$$a_k=\binom{n}{k}\left(1-\frac{k}{n}\right)^{2n-2k+2} \left(\frac{k-1}{n}\right)^{2k}$$
Expanding as series for large $n$ and keeping the first term only, we have
$$a_2=\frac{1}{2 e^4 n^2}+O\left(\frac{1}{n^3}\right)\qquad \qquad
a_3=\frac{32}{3 e^6 n^3}+O\left(\frac{1}{n^5}\right)$$
$$a_4=\frac{2187}{8 e^8 n^4}+O\left(\frac{1}{n^5}\right)\qquad \qquad
a_5=\frac{131072}{15 e^{10} n^5}+O\left(\frac{1}{n^6}\right)$$
$$a_{n-4}=\frac{131072}{3 e^{10} n^6}+O\left(\frac{1}{n^7}\right)\qquad \qquad
a_{n-3}=\frac{2187}{2 e^8 n^5}+O\left(\frac{1}{n^6}\right)$$
$$a_{n-2}=\frac{32}{e^6 n^4}+O\left(\frac{1}{n^5}\right)\qquad \qquad
a_{n-1}=\frac{1}{e^4 n^3}+O\left(\frac{1}{n^5}\right)$$
| {
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Smallest $s$ such that $\sum_i^n (1-\frac{1}{i})^s<\epsilon$ Take $f(n)$ defined as follows
$$\begin{array}{lll}
g(s,n)&=&\frac{1}{n}\sum_{i=1}^n \left(1-\frac{1}{i}\right)^s\\
f(n)&=&\text{smallest } s \text{ such that } g(s,n)<\epsilon
\end{array}$$
Empirically, $f(n)$ grows linearly in $n$, and in $-\log\epsilon$, is there a good way to see this analytically?
notebook
| First, suppose $s \ge n\log\dfrac{1}{\epsilon}$. Then since $1-\dfrac{1}{k} \le 1-\dfrac{1}{n}$ for $1 \le k \le n$, and $\left(1-\dfrac{1}{n}\right)^n \le \dfrac{1}{e}$ for all $n \ge 1$, we have
\begin{align*}
g(s,n) &= \dfrac{1}{n}\sum_{k = 1}^{n}\left(1-\dfrac{1}{k}\right)^s
\\
&\le \dfrac{1}{n}\sum_{k = 1}^{n}\left(1-\dfrac{1}{n}\right)^s
\\
&= \left(1-\dfrac{1}{n}\right)^s
\\
&\le \left(1-\dfrac{1}{n}\right)^{n\log\tfrac{1}{\epsilon}}
\\
&\le e^{-\log\tfrac{1}{\epsilon}}
\\
&= \epsilon.
\end{align*}
Now, suppose $n \ge 4$ and $s \le \dfrac{1}{2\ln 4}n\log\dfrac{1}{2\epsilon}$. Then since $1-\dfrac{1}{k} \ge 1-\dfrac{2}{n}$ for $\left\lceil\tfrac{n}{2}\right\rceil \le k \le n$, and $\left(1-\dfrac{2}{n}\right)^{n/2} \ge \dfrac{1}{4}$ for all $n \ge 4$, we have
\begin{align*}
g(s,n) &= \dfrac{1}{n}\sum_{k = 1}^{n}\left(1-\dfrac{1}{k}\right)^s
\\
&\ge \dfrac{1}{n}\sum_{k = \left\lceil\tfrac{n}{2}\right\rceil}^{n}\left(1-\dfrac{1}{k}\right)^s
\\
&\ge \dfrac{1}{n}\sum_{k = \left\lceil\tfrac{n}{2}\right\rceil}^{n}\left(1-\dfrac{2}{n}\right)^s
\\
&\ge \dfrac{1}{n}\left(n-\left\lceil\tfrac{n}{2}\right\rceil+1\right)\left(1-\dfrac{2}{n}\right)^s
\\
&\ge \dfrac{1}{2}\left(1-\dfrac{2}{n}\right)^s
\\
&\ge \dfrac{1}{2}\left(1-\dfrac{2}{n}\right)^{\tfrac{1}{2\ln 4}n\log\tfrac{1}{2\epsilon}}
\\
&\ge \dfrac{1}{2}\cdot 4^{-\tfrac{1}{\ln 4}\log\tfrac{1}{2\epsilon}}
\\
&= \epsilon.
\end{align*}
This proves that $\dfrac{1}{2\ln 4}n\log\dfrac{1}{2\epsilon}+1 \le f(n,\epsilon) \le n\log\dfrac{1}{\epsilon}$ for $n \ge 4$. So $f(n,\epsilon)$ grows linearly in $n$ and linearly in $\log\dfrac{1}{\epsilon}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the area of the part of the sphere $x^2+y^2+z^2=2$ that lies inside the cylinder $y^2+z^2=1$ This is what I have tried so far. I am just not sure whether I got it right or not.
We have $y^2+z^2=1$, then $r=1,\quad x=\sqrt{2-y^2-z^2}\quad\text{and}\quad x=\sqrt{2-r^2}$
$$\frac{\partial{x}}{\partial{y}}=-\frac{y}{\sqrt{2-y^2-z^2}}\quad\text{and}\quad
\frac{\partial{x}}{\partial{z}}=-\frac{z}{\sqrt{2-y^2-z^2}}
$$
$$\begin{align}
A(S)
&=2\iint\limits_{R}\sqrt{1+\left(-\frac{y}{\sqrt{2-y^2-z^2}}\right)^2+\left(-\frac{z}{\sqrt{2-y^2-z^2}}\right)^2}\,\mathrm dA\\
&=2\iint\limits_{R}\sqrt{1+\frac{y^2+z^2}{2-y^2-z^2}}\,\mathrm dA\\&=2\sqrt{2}\iint\limits_{R}\frac{1}{\sqrt{2-y^2-z^2}}\,\mathrm dA
\end{align}
$$
Converting to polar coordinate $R=\{(r, \theta)\mid0\leq r \leq 1, 0\leq \theta \leq 2\pi\}$
$$A(S)= 2\sqrt{2}\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\frac{r}{\sqrt{2-r^2}}\,\mathrm dr\,\mathrm d\theta=4\pi(2-\sqrt{2})$$
| Yes it is correct.
Another method, that gives the answer faster id to use the spherical coordinates $(r,\theta,\phi)$ oriented not around $z$ axis as usual, but around $x$ axis (so that $y^2+z^2 = r^2 \sin^2\theta$).
The sphere is given by the condition $r=\sqrt{2}$. The element of the area of a sphere is $dA = r^2\sin\theta\, d\theta\, d\phi = 2 \sin\theta\, d\theta\, d\phi$.
The inside of the cylinder is given by the condition $r^2\sin^2\theta \le 1$, which with the condition $r=\sqrt{2}$ gives the condition $$\sin^2 \theta < \frac{1}{2} $$
that is (remembering that for spherical coordinates $\theta\in[0,\pi]$)
$$\theta\in[0,{\pi/ 4}] \cup [{3\pi/ 4},\pi] =: I$$
We have then
\begin{align} A(S) &= \int_{\theta\in I} \int_{\phi\in[0,2\pi]} dA = 2\left(\int_0^{\pi/4}\sin\theta d\theta + \int_{3\pi/4}^{\pi}\sin\theta d\theta\right) \int_0^{2\pi} d\phi = \\
&= 2\cdot\big((1-\frac{\sqrt{2}}{2})+(1-\frac{\sqrt{2}}{2})\big)\cdot 2\pi = 4\pi(2-\sqrt{2})\end{align}
| {
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Interesting Inequality using AM-GM and other identities. Let $a, b, c > 0$. Prove that $$\sqrt{a^2-ab+b^2} + \sqrt{b^2 - bc + c^2} + \sqrt{c^2 - ca + a^2} \le \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b}.$$ This should be solvable with AM-GM and a few other inequalities, but I am a little stuck on this problem.
My idea was to remove the radical. $\sqrt{a^2-ab+b^2} \le \frac{a^2-ab+b^2}{a+b} + \frac{a+b}{4}$ by AM-GM. Adding this up cyclically, it suffices to show the inequality
$$\frac{5}{2}\left(a+b+c\right)-3\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\right) \le \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b},$$
which I'm pretty sure is true, but I have no clue how to prove.
This inequality resembles https://artofproblemsolving.com/community/c6h1288310p6804993 and https://artofproblemsolving.com/community/q2h1817483p12130020, the latter of which is a weaker version of this inequality.
| The inequality
$$\frac{5}{2}\left(a+b+c\right)-3\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\right) \le \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b},$$
doesn't true, example $a = \frac 13,\, b = \frac 1 4,\,c = 1.$
Let $a = \frac{1}{x},\,b = \frac{1}{y},\,c = \frac{1}{z},$ the inequality become
$$x \sqrt{y^{2}-y z+z^{2}}+y \sqrt{z^{2}-zx+x^{2}}+z \sqrt{x^{2}-x y+y^{2}} \leqslant x^2+y^2+z^2.$$
By the Cauchy-Schwarz inequality we have
$$\left(\sum x \sqrt{y^{2}-y z+z^{2}}\right) ^2 \leqslant (x+y+z) \sum x(y^2-yz+z^2).$$
It's remain to prove that
$$(x+y+z) \sum x(y^2-yz+z^2) \leqslant (x^2+y^2+z^2)^2,$$
equivalent to
$$x^4+y^4+z^4+xyz(x+y+z) \geqslant \sum xy(x^2+y^2).$$
Which is the Schur inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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} |
Computing $\int_{-2}^{2}\frac{1+x^2}{1+2^x} dx$ I am trying to compute the following integral by different methods, but I have not been able to come up with the result analytically.
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx$$
First I tried something like: $2^{x}=e^{x\ln{2}}\Rightarrow u=x\ln{2} \iff x=\frac{u}{\ln{2}}$ $\Rightarrow$ $\frac{du}{\ln{2}}=dx$.
On the other hand, $1+x^{2}=(x-1)^{2}-2x$
Replacing
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx=\int_{-2}^{2}\frac{(x-1)^{2}-2x}{1+e^{x\ln{2}}}dx=\int_{-2}^{2}\frac{(x-1)^{2}-2x}{1-(-e^{x\ln{2}})}dx$$
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx=\int_{-2}^{2}((x-1)^{2}-2x)\sum_{n=0}^{\infty}(-e^{x\ln{2}})^{n}dx$$
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx=\int_{-2}^{2}((x-1)^{2}-2x)\sum_{n=0}^{\infty}((-1)^{n}e^{nx\ln{2}})dx=\int_{-2}^{2}((x-1)^{2}-2x)\sum_{n=0}^{\infty}\frac{((-1)^{n}n^{n}x^{n}\ln^{n}{2})}{n!}dx$$
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx=\sum_{n=0}^{\infty}\frac{((-1)^{n}n^{n}\ln^{n}{2})}{n!}\int_{-2}^{2}(1+x^2)x^{n}dx$$
I do not know if the reasoning is correct. I hope someone can help me.
Note: By symmetry the integral can be reduced to $f(-x)=2^{x}f(x)$ so $2I=I+I=\int_{-2}^{2}(1+2^{x})\frac{1+x^2}{1+2^x}dx=\frac{14}{3}$
| Note that $\frac{1}{1+2^x}= \frac12 - \frac12 \tanh (\frac x2\ln2)
$,
where the odd function $\tanh(\cdot)$ vanishes under integration. Thus
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx= \int_{-2}^{2}\frac12(1+x^2)dx=\frac{14}3
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to graph and solve this equation? I'am trying to solve this equation.
\begin{equation}
x^{8}+(x+2)^{8}=2
\end{equation}
What I tried:
\begin{equation}g(x)=x^{8}+(x+2)^{8}\end{equation}
\begin{equation}
\begin{array}{l}
\text { }\\
y=x+1
\end{array}
\end{equation}
\begin{equation}
(y+1)^{8}+(y-1)^{8}=2
\end{equation}
\begin{equation}
(y+1)^{8}=y^{8}+a_{7} y^{7}+a_{6} y^{6}+\cdots+a_{1} y+1
\end{equation}
\begin{equation}
(y-1)^{8}=y^{8}-a_{7} y^{7}+a_{6} y^{6}+\cdots-a_{1} y+1
\end{equation}
\begin{equation}
\begin{array}{l}
2\left(y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}+1\right)=2 \Leftrightarrow \\
y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}=0 \Leftrightarrow y^{2}=0 \Leftrightarrow y=0
\end{array}
\end{equation}
y=0, x= -1
Maybe there is another way of solving this...
| OK then, let's use calculus.
Consider the function $f(x) = x^8 + (2 + x)^8$. Note that $f(x) \to \infty$ as $x \to \pm \infty$, so the function must achieve a global minimum at a stationary point, i.e. where $f'(x) = 0$.
We have,
$$f'(x) = 8x^7 + 8(x + 2)^7.$$
This is $0$ if and only if
$$8x^7 + 8(x + 2)^7 = 0 \iff (x + 2)^7 = -x^7 \iff \left(\frac{x + 2}{x}\right)^7 = -1.$$
Note, for the above to work, we must observe that $x = 0$ is definitely not a solution. As we are considering only the reals, we can take the unique seventh root of both sides to obtain
$$\frac{x + 2}{x} = -1 \iff x + 2 = -x \iff x = -1.$$
That is, there exists one and only one stationary point for the whole function: at $x = -1$. Therefore, the global minimum of $f$ must be achieved only at this point. That is,
$$2 = f(-1) \le f(x)$$
for all $x \in \Bbb{R}$, with equality if and only if $x = -1$. This implies $f(x) = 2$ if and only if $x = -1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse.
An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse.
Here is a procedure how to do it analiticaly.
*
*If $T(x_0,y_0)$ is a touching point, then $x_0+y_0=5$
*The equation of ellipse is $${(x_0-{3\over 2})^2\over a^2} +{(y_0+1)^2 \over b^2}=1$$
*Since $2e=1$ we have $a^2-b^2 = {1\over 4}$
*Since the slope of tangent is $-1$ we have $${2(x_0-{3\over 2})\over a^2} -{2(y_0+1)\over b^2}=0$$
And now we have to solve this tedious system. How to do it more geometrical?
| The ellipse divides the plane into three regions.
*
*the exterior, where the sum of the distances to the foci is greater than $c$.
*the ellipse itself, where the sum of the distances to the foci is exactly $c$.
*the interior, where the sum of the distances to the foci is smaller than $c$.
Parametrize the tangent as $x=t,y=5-t$. Thus the sum of the distances to the foci along the tangent has unique global minimum at the tangency point.
Explicitly, the function
\begin{align}
f(t):={}&\sqrt{(t-1)^2+(5-t+1)^2}+
\sqrt{(t-2)^2+(5-t+1)^2}
\\
={}&
\sqrt{(t-1)^2+(6-t)^2}+
\sqrt{(t-2)^2+(6-t)^2}
\end{align}
has global minimum at $t=\frac {34}9$, yielding the tangency point $(\frac {34}9,\frac{11}9)$. This can be found by differentiating $f(t)$ and imposing $f'(t)=0$, which gives
$$
\frac{4 \sqrt{2} (t-4)}{\sqrt{(t-8) t+20}}=-\frac{8 t-28}{\sqrt{2 (t-7) t+37}}
$$
which can be squared to a quadratic equation
$$
9 t^2-88 t+204=0\Rightarrow (t-6 ) ( 9 t-34)=0,
$$
with roots $6,\frac {34}9$, the latter being the actual solution of the original equation before squaring.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction that $3 \mid n^4-n^2 \forall n \in \mathbb{Z^+}, n \ge 2$. Proposition: $3 \mid n^4-n^2$ for all $n \in \mathbb{Z^+}, n \ge 2$
My attempt
Lemma: $3 \mid (m-1)m(m+1)$
Proof.
Suppose $m \in \mathbb{Z}$. By the QRT, we have $m=3q+r$ $\,\ni\, r \in \{0,1,2\}$, and $q=\lfloor{\frac{m}{3}}\rfloor$.
We want to show that $(m-1)m(m+1)$ is divisible by 3.
Clearly, if any one of the three factors in $(m-1)m(m+1)$ is divisible by 3, then the product must also be divisible by 3.
$\cdot$ Case 1 ($r=0$): If $r=0$, then $m=3q$ is divisible by 3.
$\cdot$ Case 2 ($r=1$): If $r=1$, then $m-1=(3q+1)-1=3q$ is divisible by 3.
$\cdot$ Case 3 ($r=2$): If $r=2$, then $m+1=(3q+2)+1=(3q+3)=3(q+1)$ is divisible by 3.
In either case, one of the three factors in $(m-1)m(m+1)$ is divisible by 3. Thus $3 \mid (m-1)m(m+1)$. Therefore, the product of any 3 consecutive integers is divisible by 3.
Proof.
Base case: Let $n=2$. Then $\exists k \in \mathbb{Z}$ such that $12=3k$, namely, $k=4$. Thus $3 \mid 12$ and hence the claim holds for the base case.
Assume for positive integer $m \ge 2$ that $m^4-m^2=3k$ for some $k \in \mathbb{Z}$. We need only to show that this implies that $3 \mid (m+1)^4-(m+1)^2$.
$(m+1)^4-(m+1)^2= \sum_{i=0}^{4} {4\choose i} m^i \cdot 1^{4-i}-(m+1)^2=(m^4-m^2)+4m^3+6m^2+2m=(m^4-m^2)+3(m^3+2m^2+m)+(m^3-m)$.
By the inductive hypothesis, $3 \mid(m^4-m^2)$. And since $(m^3+2m^2+m) \in \mathbb{Z}$, it follows that $3 \mid 3(m^3+2m^2+m)$. Note that $(m^3-m)=m(m^2-1)=(m-1)m(m+1)$. By the Lemma, $3 \mid (m^3-m)$.
Thus we have $3 \mid (m^4-m^2)$, $3 \mid 3(m^3+2m^2+m)$, and $3 \mid (m^3-m)$.
To show that their sum is divisible by 3, let $k_1,k_2,k_3 \in \mathbb{Z} \,\ni\, m^4-m^2=3k_1, 3(m^3+2m^2+m)=3k_2$, and $(m^3-m)=3k_3$. Thus $3k_1+3k_2+3k_2=3(k_1+k_2+k_3)=m^4-m^2+3(m^3+2m^2+3m)+(m^3-m)$, where $(k_1+k_2+k_3) \in \mathbb{Z}$.
Hence $3 \mid (m+1)^4-(m+1)^2$.
Therefore, by induction, $3 \mid n^4-n^2, \forall n \in \mathbb{Z^+}, n \ge 2$.
| Your proof seems correct but messy.
There’s a much easier inductive style proof. You can show it for $0,1,2$ by calculation and then note that $3| (n+3)^4-(n+3)^2-n^4+n^2$ by basic modular arithmetic.
| {
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Find the largest number that $ n(n^2-1)(5n+2) $ is always divisible by? My Solution:
$$ n(n^2-1)(5n+2) = (n-1)n(n+1)(5n+2) $$
*
*This number is divisible by 6 (as at least one of 2 consecutive integers is divisible by 2 and one of 3 consecutive integers is divisible by 3.
*$ 5n+2 \equiv 5n \equiv n \mod 2 $ then $n$ and $5n+2$ have the same pairness and at least one of $n+1$ and $5n+2$ is divisible by 2.
*$ n \equiv 5n \equiv 5n+4 \mod 4 \to $
if $ 2\ | \ n+1 \to n - 1 $ or $ n + 1 $ is divisible by 4
if $ 2\ | \ 5n+2 \to n $ or $ 5n + 2 $ is divisible by 4
The expression is divisible by 6 and has 2 even integers and one of them is divisible by 4 $\to$ is divible by 24.
| Alternatively.
Let $p$ be an odd prime. If $p|n$ but $p^2 \not \mid n$ then we have $p\not \mid n-1, n-2, 5n + 2$ so we need never have any odd prime $p^2|n(n^2-1)(5n+2)$.
If $p=3$ we must have one of $n-1, n, n+1$ be divisible by $3$ so we must have $3|n(n^2-1)(5n+2)$ but we need not have $3^2|n(n^2-1)(5n+2)$.
If $p=5$ we will not have $5|5n+2$ and as we have $5$ options for $n \pmod 5$ we can avoid $n \equiv 0, 1, 4$ (by having $n\equiv 2,3\pmod 5$ and thus have $n, n+1, n-1\not \equiv 0\pmod 5$ and so $5\not \mid n(n^2-1)(5n+2)$.
If $p > 5$ we can have more than $4$ options for $n\pmod 5$ and so we can have $n \not \equiv 0, 1,-1, -2\cdot 5^{-1} \pmod p$. ANd that way we can avoid $p|n(n^2-1)(5n+2)$.
So no odd prime other than $3$ need divide $n(n^2-1)(5n+2)$ and $3^2$ need not divide it.
And as what power of $2$ must divide $n(n^2-1)(5n+2)$.
One of $n,n+1$ must be even so $2$ must divide. If $n$ is odd then $n-1$ and $n+1$ is even and so $4$ must divide. But one of $n-1$ or $n+1$ must be divisible by $4$ so $8$ must divide.
If $n$ is even then $5n+2$ is even so $4$ must divide. If $n$ is not divisible by $4$ then $n = 4k + 2$ for some $k$ and $5n+2 = 20k + 12$ is divisible by $4$ and so $8$ must divide. And if $n$ is divisible by $4$ then as $5n + 2$ is even $8$ must divide.
That exhausts all cases so $8$ must divide $n(n^2-1)(5n+2)$.
Now no higher power of $2$ must divide as we can chose $n$ odd so that $4$ but not $8$ divides one of $n\pm 1$. For example if $n = 5$ and so $n+1 = 6$ and $n-1 =4$ and only $8$ need divide. Or we can choose an even $k$ where $n = 4k + 2$ and $5n+2 = 20k + 12= 4(5k + 3)$ is divisible by $4$ but not by $8$.
So we must have $3$ and $8$ divide $n(n^2-1)(5n+2)$ but no other prime or higher power of $3$ or higher power of $2$ need divide.
SO $24$ is the largest number that must divide.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given an equation with trigonometric roots $\tan a$ and $\cot a$, determine $\sin a + \cos a$. Question: Given that $\tan a$ and $\cot a$ are two real roots of the equation $x^2 + k^2 - kx - 3 = 0$, and $3\pi < a < \frac{7\pi}{2}$, find the value of $\sin a + \cos a$.
My solution can be found below in the answers section.
| My solution:
First realize that the range for $a$ can be simplified to $\pi < a < \frac{3\pi}{2}$.
Since $\tan a$ and $\cot a$ have a product of $1$, using Vieta's formula, $k^2 - 3 = 1 \Longrightarrow k = \pm 2$.
So, either $x^2 + 2x + 1 = 0$ or $x^2 - 2x + 1 = 0$.
Hence, $x = \pm 1$.
So, using the fact that $\tan a$ and $\cot a$ have a product of $1$, either $\tan a = \cot a = 1$ or $\tan a = \cot a = -1$.
If $\tan a = \cot a = 1$ $\Longrightarrow \sin a = \cos a.$ This would make $a = \frac{\pi}{4}$. However, this does not fit within the range for $a$.
So, $\tan a = \cot a = -1 \Longrightarrow \sin a = -\cos a.$ This would make $a = \frac{5\pi}{4}$. This fits within the range for $a$.
However, we must read the problem carefully as it asks for the value of $\sin a + \cos a$ and not just the value of $a$ itself.
So, the answer is $-\frac{\sqrt{2}}{2} + -\frac{\sqrt{2}}{2} = \boxed{-\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given a general region, find the double integral bounded between $y = x$ and $y=3x-x^2 $ $$ J = \iint_R (x^2-xy)\,dx \,dy, $$
Suppose region R is bounded between $y = x$ and $y=3x-x^2 $
My attempt using vertical integration:
$$ \int^{x=2}_{x=0} \int^{y=3x-x^2}_{y=x} \left({x^2-xy}\right)dy\ dx$$
$$\int^2_0 \left[x^2y-x\frac{y^2}{2}\right]^{3x-x^2}_{x}\, dx$$
$$\int^2_0 \frac{-x^5+4x^4-4x^3}{2} \,dx $$
$$\boxed{J = -\frac{8}{15}}$$
My attempt using horizontal integration :
$$ \int^{y=2}_{y=0} \int^{x=y}_{x=3\,\pm \sqrt{9-y}} \left({x^2-xy}\right)dx\ dy$$
For $ x = 3+\sqrt{9-y}$
$$ \int^2_0 \left[\frac{x^3}{3}-\frac{x^2}{2}y\right]^y_{3+\sqrt{9-y} }\,dy$$
For $ x = 3-\sqrt{9-y}$
$$ \int^2_0 \left[\frac{x^3}{3}-\frac{x^2}{2}y\right]^y_{3-\sqrt{9-y} }\,dy$$
My doubts :
1.) How do I set my limit of integration for horizontal integration, if there is $\pm$ to be considered ?
2.) the answer as negative what does that imply in questions related to double integrals?
Could you guys please help
| Note that in your second case, your region is divided into two (under and above the green line). When you find the inverse functions of $y(x)$'s you have $$x(y) = y$$
for $y = x$ and $$x(y) = \frac{3 \pm \sqrt{9-4y}}{2}$$ (separate them into two at $x = 3/2)$ for $y = 3x-x^2$
Now, move your pen from left to right to notice what should be the bounds for integration over each region.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate $\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}$ As seen in the title I'm interested in a way to evaluate
$$\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}$$
But I'm not sure what to do, I did attempt something but ended up with a wrong result
$$\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k-\sqrt{2}n\right)\left(k+\sqrt{2}n\right)}$$
then applying partial fraction decomposition
\begin{align*}
&=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3\left(k-\sqrt{2}n\right)}+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3\left(k+\sqrt{2}n\right)}+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^2n^2}\\
&=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3}\int _0^1x^{k-\sqrt{2}n-1}\:\mathrm{d}x+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3}\int _0^1x^{k+\sqrt{2}n-1}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=-\sum _{n=1}^{\infty }\int _0^1\operatorname{Li}_3\left(-x\right)x^{-\sqrt{2}n-1}\:\mathrm{d}x-\sum _{n=1}^{\infty }\int _0^1\operatorname{Li}_3\left(-x\right)x^{\sqrt{2}n-1}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=\int _0^1\frac{\operatorname{Li}_3\left(-x\right)}{x\left(1-x^{\sqrt{2}}\right)}\:\mathrm{d}x-\int _0^1\frac{x^{\sqrt{2}}\operatorname{Li}_3\left(-x\right)}{x\left(1-x^{\sqrt{2}}\right)}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=\int _0^1\frac{\operatorname{Li}_3\left(-x\right)}{x}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=-\frac{7}{8}\zeta (4)+\frac{5}{4}\zeta (4)\\
&=\frac{3}{8}\zeta (4)
\end{align*}
Which seems to be a wrong answer due to numerical methods and approximations, what did I do wrong? how else could I approach this problem?
| Let $S_1(n)=\sum\limits_{k=1}^{\infty } \frac{\left ( -1 \right )^{k-1}}{ k^2-2n^2} $, then $S=\sum\limits_{n=1}^{\infty }\left (\sum\limits_{k=1}^{\infty } \frac{\left ( -1 \right )^{k-1}}{n^2\left ( k^2-2n^2 \right )} \right )=\sum\limits_{n=1}^{\infty }\frac{S_1(n)}{n^2}$
Let's consider the integral in the complex plane along a big circle of radius $R$ (counter-clockwise) with the center at $(0;0)$:
$\oint=\oint_C\frac{\pi}{\sin\pi z}\frac{1}{z^2-2n^2}dz=2\pi i \sum Res\frac{\pi}{\sin\pi z}\frac{1}{z^2-2n^2}$ inside the circle
It can be shown that $\oint\to0$ as $R\to\infty\,\,\, \Rightarrow$
$\sum Res\frac{\pi}{\sin\pi z}\frac{1}{z^2-2n^2}=\sum\limits_{k=-\infty}^{\infty } \frac{(-1)^k}{k^2-2n^2} +\frac{\pi}{\sqrt2 n\sin(\sqrt2 \pi n)}=0$
$-\sum\limits_{k=-\infty}^{\infty } \frac{(-1)^k}{k^2-2n^2}=2S_1(n)+\frac{1}{2n^2}=\frac{\pi}{\sqrt2 n\sin(\sqrt2 \pi n)}\,\,\,\Rightarrow\,\,\,S_1(n)=\frac{\pi}{2\sqrt2 n\sin(\sqrt2 \pi n)}-\frac{1}{4n^2}\,\,\,\Rightarrow$
$$S=\sum\limits_{n=1}^{\infty }\frac{\pi}{2\sqrt2 n^3\sin(\sqrt2 \pi n)}-\frac{1}{4}\zeta(4)$$
The evaluation of $\sum\limits_{n=1}^{\infty }\frac{1}{n^3\sin(\sqrt2 \pi n)}=-\frac{13 \pi^3}{360 \sqrt{2}}$ is done here: Compute the series $\sum_{n=1}^{+\infty} \frac{1}{n^3\sin(n\pi\sqrt{2})}.$
NB The approach is the same: integration along the closed contour in the complex plane with the special integrand $\biggl(\frac{1}{z^3\sin(\pi z)\sin(\sqrt2\pi z-\pi z)}\biggr)$
Finally,
$$S=-\frac{13\pi^4}{360*4}-\frac{\pi^4}{360}=-\frac{17\pi^4}{1440}$$
| {
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"url": "https://math.stackexchange.com/questions/4190469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to evaluate $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$?
Evaluate: $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$.
I've just started learning limits and calculus and this is an exercise problem from my textbook.
To solve the problem, I tried factorizing the numerator and denominator of the fraction. The numerator can be factorized as $(x-1)(x+1)(2x^2-1)$ and the numerator can be factorized as $x(6x^3+x^2-3x)$. So, we can rewrite the problem as follows: $$\lim\limits_{x\to\infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}=\lim\limits_{x\to\infty}\frac {(x-1)(x+1)(2x^2-1)}{x(6x^3+x^2-3)}$$ But this doesn't help as there is no common factor in the numerator and denominator. I've also tried the following: $$\lim\limits_{x\to\infty}\frac {(x-1)(x+1)(2x^2-1)}{x(6x^3+x^2-3)}=\lim\limits_{x\to\infty}\frac{x-1}{x}\cdot \lim\limits_{x\to\infty}\frac {(x+1)(2x^2-1)}{6x^3+x^2-3}=1\cdot \lim\limits_{x\to\infty}\frac {(x+1)(2x^2-1)}{6x^3+x^2-3}$$ Here I used $\frac{x-1} x=1$ as $x$ approaches infinity. Yet this does not help.
The answer is $\frac 1 3$ in the book but the book does not include the solution.
So, how to solve the problem?
| When $x$ is very large the dominant term in Num is $2x^4$ and in den it is $6x^4$. So the required limit is $$\lim_{x \to \infty} \frac{2x^4}{6x^4}=\frac{1}{3}.$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Sum a product of Bernoulli numbers and binomial coefficients Context: I am interested in developing the large-$x$ asymptotic series of the digamma function
$$\psi\Big(\frac{1}{2}+ix\Big)$$
for real positive $x$.
For this I am using the known asymptotic expansion of $\psi(x)$:
$$\psi(z) \sim \ln z - \frac{1}{2z} - \sum_{n=1}^\infty \frac{B_{2n}}{2n z^{2n}}, \qquad z\rightarrow \infty$$
where $B_{2n}$ are the Bernoulli numbers, in conjunction with the generalized binomial expansion for negative powers:
$$\frac{1}{(1+y)^p} = \sum_{n=0}^\infty \begin{pmatrix}-p\\n\end{pmatrix} y^n$$.
Question: In the process of applying the binomial expansions to the third term of the large $x$ behavior of the digamma function, I end up with the following double sum:
$$S = -\sum_{d=2}^\infty \Big(\sum_{n=1}^{\lfloor d/2 \rfloor} \frac{B_{2n}}{2n} \begin{pmatrix}-2n\\d-2n\end{pmatrix}2^{2n}\Big)\big(\frac{-i}{2x}\big)^d$$
What is the closed-form formula for the inner $n$-sum, for $d$ odd and for $d$ even? I believe this should be possible since the odd $d$ case leads to a series that is purely imaginary which (I observe by explicit computation of the first few terms) exactly cancels the imaginary part arising from the expansion of $\ln(z) - \frac{1}{2z}$. I am therefore able to guess
$$-\sum_{n=1}^{\lfloor d/2 \rfloor} \frac{B_{2n}}{2n} \begin{pmatrix}-2n\\d-2n\end{pmatrix}2^{2n} = 1-\frac{1}{d},\qquad d = \{3,5,7,\ldots\}$$
*
*How can I prove this relation? And what is the closed-form result for $d$ even and positive?
*Bonus: what is the large $x$ behavior of $\psi(\frac{1}{2}+ix)$?
| We seek to evaluate
$$\sum_{q=1}^{\lfloor n/2\rfloor} \frac{B_{2q}}{2q}
{-2q\choose n-2q} 2^{2q}.$$
Taking advantage of the odd index Bernoulli numbers being zero except
$B_1$ this is
$$(-1)^{n+1} + \sum_{q=1}^n \frac{B_{q}}{q}
{-q\choose n-q} 2^q.$$
Now with the binomial coefficient we get
$$(-q)^{\underline{n-q}}/(n-q)! =
(-1)^{n-q} (n-1)^{\underline{n-q}}/(n-q)! =
(-1)^{n-q} {n-1\choose n-q}.$$
Continuing,
$$(-1)^{n+1} +
\sum_{q=1}^n \frac{B_q}{q} (-1)^{n-q} {n-1\choose q-1} 2^q
= (-1)^{n+1} +
\frac{1}{n} \sum_{q=1}^n B_q (-1)^{n-q} {n\choose q} 2^q
\\ = (-1)^{n+1} + \frac{1}{n} (-1)^{n+1}
+ \frac{1}{n} \sum_{q=0}^n B_q (-1)^{n-q} {n\choose q} 2^q
\\ = (-1)^{n+1} \frac{n+1}{n}
+ (n-1)! [z^n] \exp(-z) \frac{2z}{\exp(2z)-1}.$$
Now we have
$$\frac{2}{\exp(2z)-1} =
\frac{1}{\exp(z)-1} - \frac{1}{\exp(z)+1}$$
as well as
$$\frac{1}{\exp(z)} \frac{1}{\exp(z)-1}
= \frac{1}{\exp(z)-1} - \frac{1}{\exp(z)}$$
and
$$\frac{1}{\exp(z)} \frac{1}{\exp(z)+1}
= - \frac{1}{\exp(z)+1} + \frac{1}{\exp(z)}$$
which yields for the sum
$$(-1)^{n+1} \frac{n+1}{n}
+ (n-1)! [z^n]
\left[\frac{z}{\exp(z)-1} + \frac{z}{\exp(z)+1}
- \frac{2z}{\exp(z)} \right]$$
or
$$(-1)^{n+1} \frac{1-n}{n}
+ \frac{1}{n} B_n
+ (n-1)! [z^n] \frac{z}{\exp(z)+1}.$$
Observe that
$$\frac{z}{\exp(z)+1}
= \frac{z}{\exp(z)-1} - \frac{2z}{\exp(2z)-1}$$
This will produce at last
$$(-1)^{n+1} \frac{1-n}{n}
+ \frac{1}{n} B_n
+ \frac{1}{n} B_n - \frac{1}{n} 2^n B_n$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{
(-1)^{n+1} \frac{1-n}{n} +
\frac{1}{n} (2-2^n) B_n.}$$
Again with the Bernoulli numbers at odd indices being zero we get for
$n\ge 3$, $n$ odd the closed form
$$\bbox[5px,border:2px solid #00A000]{
\frac{1-n}{n}.}$$
Here we have consulted OEIS A036968, the
Genocchi numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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How to make use of angle sum and difference identities to find the value of sine and cosine?
Calculate: $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$
What is the easiest way to find $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$ (without a calculator)? If I know that $\frac{5\pi }{12}=\frac{\pi }{4}+\frac{\pi }{6}$ and $\frac{\pi }{12}=\frac{\pi }{3}-\frac{\pi }{4}$, then I can apply angle sum and difference identities. But how do I know $\frac{5\pi }{12}= \frac{\pi }{4}+\frac{\pi }{6}$ and $\frac{\pi }{12}= \frac{\pi }{3}-\frac{\pi }{4}$ in the first place. I know $ \frac{\pi }{4}+\frac{\pi }{6} = \frac{5\pi }{12}$ and $ \frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}$ but I can't go the other way round.
I gave $\frac{5\pi}{12}$ and $\frac{\pi}{12}$ as an example, I want the general solution for any value in pi rational form $\frac{\pi p}{q}$.
| After practicing a lot of problems in trigonometry, I realized that I need to know the values of sine and cosine only at $\{ \pi/6, \pi/4, \pi/3\}$, and the double angle, half angle and sum and difference identities. @marwalix et al. answered the same.
By applying the double angle formula (i.e. $\sin \left( 2\theta \right) =2\sin \theta \cos \theta$), I found the value of $\{\pi/5, 2\pi/5\}$:
$$\dfrac{\pi }{5}=2\cdot \dfrac{\pi }{10}$$
$$\dfrac{2\pi }{5}=2\cdot \dfrac{\pi }{5}$$
By applying the half angle formula (i.e. $\sin \dfrac{\theta }{2}=\sqrt{\dfrac{1-\cos \theta }{2}}$, $\cos \dfrac{\theta }{2}=\sqrt{\dfrac{1+\cos \theta }{2}}$), I found the value of $\{ \pi/12, \pi/8\}$:
$$\dfrac{\pi }{12}=\dfrac{1}{2}\cdot \dfrac{\pi }{6}$$
$$\dfrac{\pi }{8}=\dfrac{1}{2}\cdot \dfrac{\pi }{4}$$
And finally, by applying $\sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta$, and $\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$, I found the value of $\{ \pi/12, 3\pi/10, 3\pi/8, 5\pi/12\}$:
$$\dfrac{\pi }{12}=\dfrac{\pi }{3}-\dfrac{\pi }{4}=\dfrac{\pi }{4}-\dfrac{\pi }{6}$$
$$\dfrac{3\pi }{10}= \dfrac{\pi }{10}+\dfrac{\pi }{5}=\dfrac{2\pi }{5}-\dfrac{\pi }{10}$$
$$\dfrac{3\pi }{8}=\dfrac{\pi }{4}+\dfrac{\pi }{8}, \dfrac{5\pi }{12}=\dfrac{\pi }{4}+\dfrac{\pi }{6}$$
The values that I found at $\{0, \pi/12, \pi/10, \pi/6, \pi/5, \pi/4, 3\pi/10, \pi/3, 3\pi/8, 2\pi/5, 5\pi/12, \pi/10\}$ are as follows:
| {
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"url": "https://math.stackexchange.com/questions/4191686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the maximum of two variable Question :
If $\sqrt x + \sqrt y = \sqrt {135}$ for $x,y \in Z_{\ge 0}$
Find $max ~(xy)$
My first approach :
suppose $a = \sqrt{x}$ , $b = \sqrt{y}$
$ ab = \left(xy\right)^\frac{1}{2}, ~ a+b = \left(135\right)^\frac{1}{2}$
usage of AM-GM , equality hold if (and only if ) $a=b$:
$$ 0 \le \sqrt{ab} \le \frac{a+b}{2}$$
$$ 0 \le \left(xy\right)^{\frac{1}{2} \cdot \frac{1}{2}= \frac{1}{4}} \le \frac{135^{\frac{1}{2}}}{2}$$
$$ max~(xy) = \frac{135^2}{2^4}$$
My second approach:
$$\sqrt{y} = \sqrt{135}-\sqrt{x}$$
Square both sides :
$$ y = 135 -2\sqrt{135x}+x $$
Maximizing the function :
$$f(x) = (x)(135-2\sqrt{135x}+x)$$
$$f’(x) = \frac{d}{dx}(x)[135-2\sqrt{135x}+x] + [x] \frac{d}{dx}(135-2\sqrt{135x}+x) $$
$$ f’(x) = 135+2x-3\sqrt{135x}$$
Solving for critical points :
$$ 2x-3\sqrt{135x}+135 = 0$$
$$ x = \frac{135}{4} , x= {135}$$
$$ f \left(\frac{135}{4}\right) = \frac{135^2}{4^2} = \frac{135^2}{2^4} , ~~ f\left(135\right) = 0 $$
Hence,
$$ max~(xy) = \frac{135^2}{2^4}$$
Though both approaches gave the same answer, I still want to validate my approaches. Please guys if you could validate the solution
| $$y = 135 -2\sqrt{135x}+x$$
So $x=15k^2$ ( where $k$ is a non negative integer) because $135x$ is a perfect square
$$\sqrt{y} = \sqrt{135}-\sqrt{x}=(3-k)\sqrt{15}$$
$$\implies y=15(3-k)^2$$
(Note: $k\le3$ otherwise $\sqrt{y}$ will be negative)
So the general solution is $(x,y)=(15k^2,15(3-k)^2)$
$$\implies xy=225(3k-k^2)^2$$
$0\le k\le3$, so $(3k-k^2)^2$ is maximum when $k=1$ or $2$.
Therefore $\max(xy)=225\cdot4=900$
| {
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"timestamp": "2023-03-29T00:00:00",
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can $x^2+a=y^2$ where a is prime Let a sequence of the sum of prime numbers be $S_1=2$, $S_2=5$, $S_3=10$, $S_4=17...$
Can $S_n$ and $S_{n+1}$ both be square numbers $x^2$ and $y^2$ respectively?
I have already made the observation that $x^2+a=y^2$
$a=(y+x)(y-x)$
and therefore $y-x=1$ and $x+y=a$ for this to hold.
| If $S_n=x^2$, and $S_{n+1}=y^2,$ then, as you’ve noticed, $y+x=p_{n+1}, y-x=1.$ Solving these equations:
$$y=\frac{p_{n+1}+1}2,x=\frac{p_{n+1}-1}2$$
This also means that you need $$S_n=\left(\frac{p_{n+1}-1}2\right)^2.$$
Now for $n\geq 5:$
$$\begin{align}S_n&<1+3+5+7+9+11+\cdots +p_n\\&= \left(\frac{p_n+1}2\right)^2\\&\leq\left(\frac{p_{n+1}-1}2\right)^2\end{align}$$
(Where the first sum is the sum of the odd numbers from $1$ to $p_n,$ where the sum of the first $k$ odd numbers is known to be $k^2.$)
So there are no example $n\geq 5.$
You just need to check $n=1,2,3,4$ by hand.
| {
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Proving an inequality given $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\le 1$ Given that $a,b,c > 0$ are real numbers such that $$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\le 1,$$ prove that $$\frac{1}{b+c+1}+\frac{1}{c+a+1}+\frac{1}{a+b+1}\ge 1.$$
I first rewrote $$\frac{1}{a+b+1} = 1 - \frac{a+b}{a+b+1},$$ so the second inequality can be rewritten as $$\frac{b+c}{b+c+1} + \frac{c+a}{c+a+1} + \frac{a+b}{a+b+1} \le 2.$$ Cauchy-Schwarz gives us $$\sum \frac{a+b}{a+b+1} \geq \frac{(\sum \sqrt{a+b})^2}{\sum a+ b+ 1}.$$ That can be rewritten as $$\frac{2(a+b+c) + 2\sum \sqrt{(a+b)(a+c)}}{2(a+b+c) + 3},$$ which is greater than or equal to $$\frac{2(a+b+c) + 2 \sum(a + \sqrt{bc})}{2(a+b+c) + 3} = \frac{4(a+b+c) + 2 \sum \sqrt{bc}}{2(a+b+c) + 3} \geq 2,$$ which is the opposite of what I want. Additionally, I'm unsure of how to proceed from here.
| Since the two inequalities are completely symmetric in $a,b,c$. WLOG, we only need to study the case $a \ge b \ge c$.
Let $\Lambda = a + b + c + 1$. The two inequalities can be rewritten
as
$$\sum_{cyc}\frac{a}{\Lambda -a } \stackrel{def}{=}\frac{a}{b+c+1} + \frac{b}{c+a+1} + \frac{c}{a+b+1} \le 1\\
\sum_{cyc}\frac{1}{\Lambda-a}\stackrel{def}{=}\frac{1}{b+c+1}
+ \frac{1}{c+a+1} + \frac{1}{a+b+1} \stackrel{?}{\ge} 1
$$
Notice for $x \in (0,\Lambda)$, the map $x \mapsto \frac{1}{\Lambda - x}$ is increasing. We have
$$a \ge b \ge c \quad\implies\quad \frac{1}{\Lambda - a} \ge \frac{1}{\Lambda - b} \ge \frac{1}{\Lambda - c}$$
By Rearrangement inequality, we have
$$\begin{align}
1 \ge \sum_{cyc}\frac{a}{\Lambda-a} \ge \sum_{cyc}\frac{b}{\Lambda-a}
= \frac{b}{\Lambda-a} +\frac{c}{\Lambda-b} + \frac{a}{\Lambda-c}
\\
1 \ge \sum_{cyc}\frac{a}{\Lambda-a} \ge \sum_{cyc}\frac{c}{\Lambda-a}
= \frac{c}{\Lambda-a} +\frac{a}{\Lambda-b} + \frac{b}{\Lambda-c}
\end{align}
$$
From these, using the decomposition you have, one obtain:
$$\sum_{cyc}\frac{1}{\Lambda-a}
= \sum_{cyc}\left(1 - \frac{b+c}{\Lambda-a}\right) = 3 -
\sum_{cyc}\frac{b+c}{\Lambda - a}
\ge 3 - (1+1) = 1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Committee with 4 members
My school's Future Mathematicians of America club has 16 members, 7 boys and 9 girls. A president and a 3-person executive committee are chosen (where the president cannot serve on the committee). What is the probability that the president is the same gender as the majority of the committee?
I ended up getting:$${{7\left(9\binom{6}{2} + \binom{6}{3}\right) + 9\left(7\binom{8}{2} + \binom{8}{3}\right)}\over{\binom{16}{4}}} = {{479}\over{260}},$$which I know can't be right since that's greater than $1$. What did I do wrong?
EDIT: Okay, so with the helpful comment by JMoravitz in the comments section, maybe it should be the following instead?
$${{7\left(9\binom{6}{2} + \binom{6}{3}\right) + 9\left(7\binom{8}{2} + \binom{8}{3}\right)}\over{16 \binom{15}{3}}} = {{479}\over{1040}}$$
| Here's a different approach that yields the same result. First choose the four boys and girls, and then choose a president from the majority:
$$\frac{\binom{7}{0}\binom{9}{4}4+\binom{7}{1}\binom{9}{3}3+\binom{7}{3}\binom{9}{1}3+\binom{7}{4}\binom{9}{0}4}{\binom{16}{4}4}=\frac{479}{1040}$$
Note the absence of $\binom{7}{2}\binom{9}{2}$ in the numerator.
| {
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"source": "stackexchange",
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Pardon my mistakes in asking t I have problems solving equations of LIMITS. To be specific the ones I have to rationalize. https://www.ck12.org/book/ck-12-precalculus-concepts/section/14.5/
Example 5.
The last example is what's giving me issues. The denominator at the $4$-th equation to be precise. I don't know how it was manipulated.
Am an undergraduate in the University of Calabar in Nigeria. Am studying for my exams which will commence on 22nd of this month. I believe the answer will help me pass maths exams with good grades and the answer will also help others who have same issue. And I'll definitely reffer my friends to this website. Admin please kindly approve as the responses will help me in no small way. Thanks
Evaluate the following limit:
$$\lim{x\to 0}\left(\frac{3}{x\sqrt{9−x}}-\frac1x\right)$$
\begin{align}
\lim_{x \to 0}\left(\frac{3}{x\sqrt{9−x}}-\frac1x\right) &= \lim_{x \to 0}\left(\frac{3}{x\sqrt{9−x}}-\frac{\sqrt{9-x}}{x\sqrt{9-x}}\right) \\
&=\lim_{x \to 0} \left(\frac{3-\sqrt{9-x}}{x\sqrt{9-x}} \right)\\
&= \lim_{x \to 0} \left( \frac{3-\sqrt{9-x}}{x\sqrt{9-x}}\cdot \frac{3+\sqrt{9-x}}{3+\sqrt{9-x}}\right)\\
&= \lim_{x \to 0} \left( \frac{9-(9-x)}{x\sqrt{9-x}}\right)\\
&= \lim_{x \to 0} \frac{x}{x\sqrt{9-x}}\\
&= \lim_{x \to 0} \frac1{\sqrt{9-x}}\\
&= \frac1{\sqrt{9-0}}\\
&= \frac13
\end{align}
| The answer is indeed wrong.
\begin{align}
\lim_{x \to 0} \left( \frac{(3-\sqrt{9-x}}{x\sqrt{9-x}} \cdot \frac{3+\sqrt{9-x}}{3+\sqrt{9-x}} \right) &= \lim_{x \to 0} \frac{1}{3+\sqrt{9-x}}\lim_{x \to 0}\frac{9-(9-x)}{x\sqrt{9-x}}\\
&= \frac16 \cdot \lim_{x \to 0}\frac{x}{x\sqrt{9-x}}\\
&=\frac16 \cdot \lim_{x \to 0}\frac1{\sqrt{9-x}}\\
&=\frac1{18}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4198756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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With $x^2+y^2=1$ find Minimum and Maximum of $x^5+y^5$ (do not use derivative) It's easy to see that the minimum is $-1$ and maximum is $1$.
My idea is put $x=\cos(a)$, $y=\sin(a)$ and $t=x+y$, so I have $-\sqrt{2}\le t \le \sqrt{2}$ then $0 \le t^2 \le 2$
When $t=x+y$ then $(x+y)^2=1+2xy$.
Then
\begin{aligned}
x^5+y^5&=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)\\
&=x^3+y^3-x^2y^2(x+y)\\
&=(x+y)(x^2-xy+y^2)-x^2y^2(x+y)\\
&=(x+y)(1-xy-x^2y^2)\\
&=\left( {x + y} \right)\left( {1+\frac{1}{2} - {{\left( {x + y} \right)}^2} - \frac{{{{\left( {x + y} \right)}^4}}}{4} + \frac{{{{\left( {x + y} \right)}^2}}}{2} - \frac{1}{4}} \right)\\
&=t\left( {\frac{5}{4} - \frac{{{t^4}}}{4}} \right)\\
\end{aligned}
This problem is for those who have not studied derivatives so I dont have any idea for next step. Anyone can help me for the hint or other solutions? Very Thanks
| $x^2+y^2=1$ implies $x \in [-1,1]$ and $y \in [-1,1]$
(You can verify this by the fact that all (x,y) lie on the unit circle of radius 1 so x and y must lie in the interval [-1,1])
Now, $x^2 \in [0,1]$ and $y^2 \in [0,1]$
Also $x^3 \in [-1,1]$ (since $x \in [-1,1]$)
and $y^3 \in [-1,1]$ (since $y \in [-1,1]$)
Now $$-1 \leq x^3 \leq 1 \implies -x^2 \leq x^5 \leq x^2 \tag{i}$$ (multiplied $x^2\geq0$ throughout the inequality)
Similarly $$-1 \leq y^3 \leq 1 \implies -y^2 \leq y^5 \leq y^2 \tag{ii}$$ (multiplied $y^2\geq0$ throughout the inequality)
From inequalities (i) and (ii)
$$-x^2-y^2 \leq x^5+y^5 \leq x^2+y^2$$
$$ -1\leq x^5+y^5 \leq 1$$
Since $x^5+y^5$ actually takes values $-1$ and $1$ for atleast one $(x,y)$ such that $x^2+y^2=1$ (for instance $x^5+y^5=-1$ for $x=-1, y=0$ and $x^5+y^5=1$ for $x=1, y=0$)
Therefore -1 and 1 are minimum and maximum values of $x^5+y^5$ respectively.
| {
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"url": "https://math.stackexchange.com/questions/4199197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Project parabola onto another plane (edited) I am starting with a paraboloid in the form
$$A x^{2} + B x y + C y^{2} + D x + E y + F = z$$
and a plane
$$a x + b y + c z + d = 0$$
By equating the z terms in the two equations, I have the equation of the resulting 2nd degree curve in the world coordinate frame as
$$
f = \begin{cases} A_1 x^{2} + B_1 x y + C_1 y^{2} + D_1 x + E_1 y + F_1 = 0 \\
z = -\frac{a x + b y + d}{c}
\end{cases}
$$
The first part of "f" is the projection of the curve of intersection along the z-axis onto the x-y plane.
For the case where this curve "f" is a parabola, I am trying to find its equation as associated with a coordinate frame attached to the plane of intersection. This new frame has its z axis aligned with the plane normal, but I can choose the origin and the orientation of x axis on the plane to make the equation simpler.
Example scenario
shows a parabloid with the equation:
$$ -0.05x^2-1.1y^2+0.5xy+0.5x+5=z$$ the plane is
$$0.485x-0.485y+0.728z-0.97 = 0$$
Th equation of the resultant parabola(cyan) on the x-y plane is
$$-0.036x^2+0.36xy-0.8y^2+0.85x-0.48y+2.67=0$$
I am trying to find the equation of the white parabola. I have tried projecting 4 points onto the intersection plane and fitting a parabola to it. Those two possible solutions for that are shown in green and red. As can be seen, the red one, which is the closest, doesn't quite match up to the white. I can of course pick a lot more points and try a least square solution, but I was wondering if there is a smarter way to do this.
| First, define $\mathbf{r} = [x, y, z]^T $ as the position vector. Then the equation
$ z= A x^2 + B x y + C y^2 + D x + E y + F $
can be written as
$ \mathbf{r}^T \mathbf{Q r} + \mathbf{b}^T \mathbf{r} + c = 0 \hspace{12pt}(1)$
where
$\mathbf{Q} = \begin{bmatrix} A && \frac{1}{2} B && 0 \\ \frac{1}{2} B && C && 0 \\ 0 && 0 && 0 \end{bmatrix}$
$\mathbf{b} = \begin{bmatrix} D , E , -1 \end{bmatrix}^T $
$c = F $
And the intersecting plane can be written as $ \mathbf{n}^T (\mathbf{r} - \mathbf{r}_0) = 0 $ where
$\mathbf{n} = \begin{bmatrix} a, b, c \end{bmatrix}^T$, and $\mathbf{r_0}$ is any point on the plane, satisfying $\mathbf{n}^T \mathbf{r_0} = - d $
Next, write the vector equation of the plane,
$ \mathbf{r} = \mathbf{r_0} + u_1 \mathbf{v_1} + u_2 \mathbf{v_2}\hspace{12pt} (2) $
where $\mathbf{v_1, v_2}$ are two unit mutually orthogonal vectors are orthogonal to vector $\mathbf{n}$.
Equation (2) can be written compactly as
$ \mathbf{r} = \mathbf{r_0} + \mathbf{V u} \hspace{12pt} (3) $
where $\mathbf{V} = [ \mathbf{v_1}, \mathbf{v_2} ] $ and $ \mathbf{u} = [u_1, u_2]^T$
Now substitute (3) into (1), to obtain,
$ ( \mathbf{r_0} + \mathbf{V u} )^T \mathbf{Q}(\mathbf{r_0} + \mathbf{V u}) + \mathbf{b}^T (\mathbf{r_0} + \mathbf{V u}) + c = 0 $
Expanding,
$ \mathbf{u}^T \mathbf{V}^T \mathbf{Q} \mathbf{V u} + \mathbf{u}^T (2 \mathbf{V}^T \mathbf{Q r_0} + \mathbf{V}^T \mathbf{b} ) + \mathbf{r_0}^T \mathbf{Q r_0} + \mathbf{b}^T\mathbf{r_0}+ c = 0 \hspace{12pt} (4)$
This is the equation of the intersection between the quadric and the plane.
It specifies the relation between the coordinates $u_1$ and $u_2$ in the $\mathbf{r_0 ,V}$ coordinate frame.
The next step is to diagonalize $\mathbf{V}^T \mathbf{Q V} $ so that,
$\mathbf{V}^T \mathbf{Q V} = \mathbf{ R D R}^T $
where $R$ is a $2 \times 2$ rotation matrix and $D $ is a $2 \times 2 $ diagonal matrix.
The above intersection will be a parabola if and only if $\mathbf{D}$ has exactly one diagonal entry equal to $0$. And we can assume that this zero entry is $D_{22}$.
Define the vector $\mathbf{w} = \mathbf{R}^T \mathbf{u} $, then equation (4) in terms of $w$ becomes
$ \mathbf{w}^T D \mathbf{w} + \mathbf{w}^T \mathbf{g} + h = 0 \hspace{12pt} (5)$
where $ \mathbf{g} = \mathbf{R}^T ( 2 \mathbf{V}^T \mathbf{Q r_0} + \mathbf{V}^T \mathbf{b} ) $ and $ h = \mathbf{r_0}^T \mathbf{Q r_0} +\mathbf{b}^T \mathbf{r_0}+ c $
Remember that $D_{22} = 0 $, and let $D_{11} = a , \mathbf{g} = [g_1, g_2]^T$ and $\mathbf{w} = [x, y]^T $ (not to be confused with the $x$ and $y$ in the world coordinate frame $Oxyz$ ), then equation (5) reads,
$ a x^2 + g_1 x + g_2 y + h = 0 $
For this to be a parabola, we can assume that $g_2 \ne 0$ and $a \ne 0$, then by completing the square, we have
$ a (x + \dfrac{g_1}{2 a} )^2 + g_2 y + h - \dfrac{g_1^2}{4 a} = 0 $
so that,
$ y = -\dfrac{a}{g_2} (x + \dfrac{g_1}{2a} )^2 + \dfrac{1}{g_2} ( \dfrac{g_1^2}{4a} - h ) $
And this specifies the $y$ coordinate as a function of $x$ coordinate in the $\mathbf{r_0},\mathbf{W}$ coordinate frame.
To summarize, at the end we will have two axes given by the columns of the matrix $\mathbf{W} =\mathbf{VR}$. Let these two columns be $\mathbf{w_1} $ and $\mathbf{w_2} $, then the parabola of intersection is given explicitly (and parametrically) by
$\mathbf{r} =\mathbf{r_0} + x \mathbf{w_1} + y \mathbf{w_2} = \mathbf{r_1} + t \mathbf{w_1} + \alpha t^2 \mathbf{w_2} \hspace{12pt} (6) $
where the $\alpha = -\dfrac{a}{g_2}$ and $\mathbf{r_1} $ is the vertex of the parabola, given by,
$ \mathbf{r_1} = \mathbf{r_0} + x_0 \mathbf{w_1} + y_0 \mathbf{w_2} \hspace{12pt} (7)$
with $x_0 = - \dfrac{g_1}{2a}, y_0 = \dfrac{1}{g_2} ( \dfrac{g_1^2}{4a} - h ) $
| {
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"source": "stackexchange",
"question_score": "2",
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How do i approach ahead in this question
Let $a,b \in\Bbb N$, $a$ is not equal to $b$ and the quadratic equations $(a-1)x^2 -(a^2 + 2)x +a^2 +2a=0$ and $(b-1)x^2 -(b^2 + 2)x +b^2 +2b=0$ have a common root, then the value of $ab/5$ is
So what I did was,
I subtracted the two equations and got
$x=(a+b+2)/(a+b)$
I tried putting it in equation,it didn’t work,then i tried adding the equations and then put the value,still didn’t work.
I cant seem to figure out how to approach this problem.Can anybody help out?
| If you like using Vieta's formulas, here is the possible solution based on these formulas:
Assuming $x$ as a coefficient, you can observe that, $a$ and $b$ are root of the quadratic respect to the variable $u:$
$$u^2(1-x)+u(x^2+2)-(x^2+2x)=0, ~ x≠1$$
Then, using the Vieta's formulas, we have
$$\begin{align}&ab=\frac{x^2+2x}{x-1}\\
\implies &x^2+x(2-ab)+ab=0 \\
\implies &x_1x_2=ab\end{align}$$
Then, we also have a quadratic respect to the variable $x:$
$$(a-1)x^2 -(a^2 + 2)x +a^2 +2a=0, ~ a≠1$$
Since $a≠0$, applying Vieta's formulas again, we get
$$\begin{align}&\frac{a^2+2a}{a-1}=ab\\
\implies &b=\frac{a+2}{a-1}\\
\implies &b=1+\frac{3}{a-1}\\
\implies &a\in\left\{2,4\right\}\\
\implies &ab=8 \\
\implies &\frac{ab}{5}=\frac 85. \end{align}$$
| {
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If $x^3 + \frac{1}{x^3} = 2\sqrt{5}$, then find $x^2 + \frac{1}{x^2}$. Question: If $x$ is a real number satisfying $x^3 + \frac{1}{x^3} = 2\sqrt{5}$, determine the exact value of $x^2 + \frac{1}{x^2}$.
My partial solution: $(x + \frac{1}{x})^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}.$
We know that $x^3 + \frac{1}{x^3} = 2\sqrt5 \Longrightarrow (x + \frac{1}{x})^3 = 2\sqrt5 + 3x + \frac{3}{x}.$
So, $(x + \frac{1}{x})^3 = 2\sqrt5 + 3(x + \frac{1}{x})$. Let $y = x + \frac{1}{x}$.
Then, $y^3 = 2\sqrt5 + 3y$. We want to solve for $x^2 + \frac{1}{x^2}$. $(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2$.
So, we want to solve for $y^2 - 2$. But, I'm confused about how to solve the equation $y^3 = 2\sqrt5 + 3y$, should I check random values?
Any help is appreciated, thanks in advance! (Thanks for the suggestions in the comments :)
| From $x^3+\frac{1}{x^3}=2\sqrt{5}$ we get $\left(x^3+\frac{1}{x^3}\right)^2=20$ or $x^6+\frac{1}{x^6}=18$
Let $y=x^2+\frac{1}{x^2}$. Then $y^3 = 3y + x^6+\frac{1}{x^6} = 3y +18.$
The only real solution of $y^3-3y-18=0$ is $y=3$. Therefore $x^2+\frac{1}{x^2}=3.$
By the way, $x = \frac{\sqrt{5}\pm 1}{2}$ can be used to see that this is actually the solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real. I want to solve this problem using only the AM-GM inequality. Can someone give me the softest possible hint? Thanks.
Useless fact: from equality we can conclude $abc \le 1$.
Attempt 1:
Adding $(ab + bc + ca)$ to both sides of inequality and using the equality leaves me to prove: $ab + bc + ca \le 3$.
Final edit: I found a easy way to prove above.
$18 = 2(a+b+c)^2 = (a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) + 4ab + 4bc + 4ca \ge 6(ab + bc + ca) \implies ab + bc + ca \le 3$
(please let me know if there is a mistake in above).
Attempt 2: multiplying both sides of inequality by $2$, we get:
$(a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$. By substituting $x = a+b, y = b+c, z = c+a$ and using $x+y+z = 6$ we will need to show:
$x^2 + y^2 + z^2 \ge 12$. This doesnt seem trivial either based on am-gm.
Edit: This becomes trivial by C-S.
$(a+b).1 + (b+c).1 + (c+a).1 = 6 \Rightarrow \sqrt{((a+b)^2 + (b+c)^2 + (c+a)^2)(1 + 1 + 1)} \ge 6 \implies (a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$
Attempt 3:
$x = 1-t-u$, $y = 1+t-u$, $z = 1 + 2u$
$(1-u-t)^2 + (1-u+t)^2 + (1+2u)^2 + (1-u-t)(1-u+t) + (1+t-u)(1+2u) + (1-t-u)(1+2u)$
$ = 2(1-u)^2 + 2t^2 + (1 + 2u)^2 + (1-u)^2 - t^2 + 2(1+2u)$
expanding we get:
$ = 3(1 + u^2 -2u) + t^2 + 1 + 4u^2 + 4u + 2 + 4u = 6 + 7u^2 + t^2 + 2u\ge 6$.
Yes, this works.. (not using am-gm or any such thing).
| Second approach : using only AM-GM
The second approach (which is correct thus far) does work out using only AM-GM. It works using CS quite clearly, but it comes out through AM-GM as well, although you have to sum a few instances of the two variable AM-GM.
We want to prove that $x^2+y^2+z^2 \geq 12$ for $x+y+z =6$ (Note : this itself was a hint in the comments!). Apply AM-GM to $\frac{x^2}{x^2+y^2+z^2}$ and $\frac 13$, this gives you some inequality, which I'll hide :
$\frac{x}{\sqrt{3(x^2+y^2+z^2)}}\leq \frac{x^2}{2(x^2+y^2+z^2)} + \frac 16$.
Derive similar inequalities using AM-GM for $y,z$, and add them up and rearrange to get the result.
(I wanted to provide a summary of the third approach, but I guess I'll avoid it since it's technical)
| {
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Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution
Ok, so it's been a really long time since I've done a problem like this but after doing a little bit o studying, this is how far I've gotten.
$$\int\frac{x^4}{\sqrt{a^2-x^2}}\,dx$$
Focusing on the denominator and using a triangle, I found that $\sin\theta=\frac{x}{a}$, therefore $x=a\sin\theta$ and $dx=a\cos\theta \, d\theta$
Plugging this into the given problem, I got $\int\frac{x^4}{\sqrt{a^2-x^2}}\, dx$ ${}= \int\frac{a^4 \sin^4\theta}{a\cos\theta}\cdot a\cos\theta \, d \theta $= $a^4\int (\sin^4 \theta ) \, d \theta$
Since $\sin^2\theta=\frac{1-\cos2\theta}{2}$, I can say
$$a^4\int (\sin^4 \theta \,d \theta =a^4\int\frac{1-2\cos2\theta+\cos^22\theta}{4}\,)d\theta$$ Pulling out the 4 from the denominator and substituting $\cos^22\theta=1-\sin^22\theta$, I get
$$\frac{a^4}{4} \int({1-2\cos2\theta+1-\sin^22\theta})\, d\theta=\frac{a^4}{4}\int({2-2\cos2\theta-\sin^22\theta}) \, d\theta$$
Substituting $\sin^22\theta for \frac{1-\cos4\theta}{2}$ gets me
$$\frac{a^4}{4}\int(2-2\cos2\theta-\frac{1-\cos4\theta}{2})d\theta=\frac{a^4}{8}\int(4-4\cos2\theta-1+\cos4\theta) \, d\theta=\frac{a^4}{8}\int(3-4 \cos2\theta + \cos4\theta)\, d\theta$$
$$=\frac{a^4}{8}[3\theta-\frac{4\sin2\theta}{2}+\frac{\sin4\theta}{4}]+C$$
From here, there are some substitutions that we must do.
We know that $sin\theta=\frac{x}{a}$ So, $\theta=sin^{-1}(\frac{x}{a})$
Also, $sin2\theta=2sin\theta cos\theta$=$2*\frac{x}{a}*\frac{\sqrt{x^2-a^2}}{a}$ and $sin4\theta=4cos^3\theta sin\theta-4sin^3\theta cos\theta=4*({\frac{\sqrt{a^2-x^2}}{a}})^3*\frac{x}{a}-4*(\frac{x}{a})^3*\frac{\sqrt{a^2-x^2}}{a}$
Substituting this gives,
$\frac{a^4}{8}[3sin^{-1}\frac{x}{a}-\frac{8x\sqrt{a^2-x^2}}{a^2}+\frac{4x(\sqrt{a^2-x^2}^3}{a^2}-\frac{4x^3\sqrt{a^2-x^2}}{a^2}]+C$
And then finally, $\frac{3a^4}{8}sin^{-1}\frac{x}{a}-\frac{a^2x\sqrt{a^2-x^2}}{2}+\frac{a^2x(\sqrt{a^2-x^2}^3}{2}-\frac{ax^3\sqrt{a^2-x^2}}{2}]+C$
Is this correct? Please critique, comment, and help clarify. I really need to understand this problem.
| This is a binomial differential integral of the type:
$I=\int x^{m}(A+B x^{n})^{q/r}dx$.
Chebiceff showed that:"such an integral can be traced back to an integral of rational function only when
$\frac{m+1}{n}$ or $\frac{m+1}{n}+\frac{q}{r}$,
are integers.
The second case is the one that interests us.
Let it be
$a^{2}-x^{2}=t^{2} x^{2}$,
$x=\Big(\frac{a^{2}}{t^{2}+1}\Big)^{(1/2)}$,
$dx=\frac{-t a}{(t^{2}+1)^{3/2}}dt$,
$ a^{2}-x^{2}=\frac{a^{2}t^{2}}{t^{2}+1}$,
$I=a^{4}\int \frac{1}{(t^{2}+1)^{3}}dt$,
then
$I=\frac{3a^{4}asin(\frac{x}{a})}{8}-\frac{x\sqrt{a^{2}-x^{2}}(2x^{2}+3a^{2})}{8}$, (Maple).
| {
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"url": "https://math.stackexchange.com/questions/4209191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving the equation $3\cos x+\frac{1}{2}=\cos ^{2}x\left(1+2\cos x\left(1+4\sin ^2x\right)\right)$ Solve the equation:$$3\cos x+\frac{1}{2}=\cos ^{2}x\left(1+2\cos x\left(1+4\sin ^2x\right)\right)$$
To solve it, I tried writing the equation in term of $\cos x$. ( I denote $\cos x$ by $c$):
$$3c+\frac12=c^2(1+2c(5-4c^2))$$
$$3c+\frac12=c^2+10c^3-8c^5$$
$$16c^5-20c^3-2c^2+6c+1=0 \qquad\text{Where $c\in[-1,1]$}$$
I tried $c=\pm1,\pm\frac12,0$, but neither of them satisfied the equation. So I don't know how to find $\cos x$.
| Your calculations are right. We have $c=\pm\frac 12\sqrt 2$ as two of its roots. Hence, we get $x\in\{\pm\frac{\pi}{4}+k\pi\mid k\in\mathbb Z\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this proof regarding the nonexistence of odd perfect numbers correct? Preamble: This post is an offshoot of this earlier question, which was not so well-received in MathOverflow.
Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
If $m$ is odd and $\sigma(m)=2m$, then $m$ is called an odd perfect number. Euler proved that a hypothetical odd perfect number must necessarily have the form
$$m = q^k n^2$$
where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
As proved in MO, we have the following equation:
$$I(n^2) - \frac{2(q - 1)}{q} = \frac{1}{q^{k+1}}\cdot{I(n^2)},$$
whereupon, starting from the lower bound
$$I(n^2) > \frac{2(q - 1)}{q}$$
we get the recursive estimates
$$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\cdot\frac{2(q - 1)}{q}$$
$$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\bigg(\frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\cdot\frac{2(q - 1)}{q}\bigg)$$
$$= \frac{2(q - 1)}{q}\cdot\Bigg(1 + \frac{1}{q^{k+1}} + \bigg(\frac{1}{q^{k+1}}\bigg)^2\Bigg)$$
$$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\Bigg(\frac{2(q - 1)}{q}\cdot\Bigg(1 + \frac{1}{q^{k+1}} + \bigg(\frac{1}{q^{k+1}}\bigg)^2\Bigg)\Bigg)$$
$$= \frac{2(q - 1)}{q}\cdot\Bigg(1 + \frac{1}{q^{k+1}} + \bigg(\frac{1}{q^{k+1}}\bigg)^2 + \bigg(\frac{1}{q^{k+1}}\bigg)^3\Bigg)$$
$$\ldots$$
$$\ldots$$
$$\ldots$$
Repeating the process ad infinitum, we get:
$$I(n^2) > \frac{2(q - 1)}{q}\cdot\Bigg(\sum_{i=0}^{\infty}{\bigg(\frac{1}{q^{k+1}}\bigg)^i}\Bigg).$$
But
$$\sum_{i=0}^{\infty}{\bigg(\frac{1}{q^{k+1}}\bigg)^i}$$
is an infinite geometric series, with sum
$$\frac{a_0}{1 - r}$$
where the first term $a_0 = 1$ and the common ratio
$$r = \frac{1}{q^{k+1}}.$$
Hence, we obtain
$$\sum_{i=0}^{\infty}{\bigg(\frac{1}{q^{k+1}}\bigg)^i} = \frac{1}{1 - \frac{1}{q^{k+1}}},$$
from which we finally get
$$I(n^2) > \frac{2(q - 1)}{q}\cdot\frac{q^{k+1}}{q^{k+1} - 1}.$$
But we can simplify the RHS of the last inequality as follows:
$$\frac{2(q - 1)}{q}\cdot\frac{q^{k+1}}{q^{k+1} - 1} = \frac{2q^k (q - 1)}{q^{k+1} - 1} = \frac{2q^k}{\sigma(q^k)} = \frac{2}{I(q^k)} = I(n^2).$$
We have therefore finally arrived at the contradiction
$$I(n^2) > I(n^2).$$
We therefore conclude that there cannot be any odd perfect numbers.
Here is my:
QUESTION: Does this proof hold water? If it does not, where is the error in the argument, and can it be mended so as to produce a valid proof?
| Assuming your statements about the recursive estimates for $I(n^2)$ are correct, $I(n^2)$ may still not be greater than the limit of the recursive estimates. If we know $a>x_n$ for all $n$, and $x_n\to x$ as $n\to\infty$, we can only deduce that $a\ge x$.
| {
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Expected value of $\overline{ABC} \times \overline{DEF}$ Here is a question from HMMT:
https://hmmt-archive.s3.amazonaws.com/tournaments/2013/nov/team/solutions.pdf
The digits $1$, $2$, $3$, $4$, $5$, $6$ are randomly chosen (without replacement) to form the three-digit numbers $M = \overline{ABC}$ and $N = \overline{DEF}$. For example, we could have $M = 413$ and $N = 256$. Find the expected value of $M \cdot N$.
Here's what I did. Each digit on average is going to be $(1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5$, so the expected value is $(100(3.5) + 10(3.5) + 3.5)^2 = 150932.25$. However, the answer at the link above is $143745$. What did I do wrong? Did I overcount something?
| The correct answer is 143745.
$$
\frac{1}{6}\cdot\frac{1\cdot(2+3+4+5+6)+2\cdot(1+3+4+5+6)+...+6\cdot(1+2+3+4+5)}{5}\cdot111\cdot111
$$
Your answer is incorrect because you average the numbers evenly
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find minimum value of $\frac{x^3}{\sqrt{2(y^4+1)}}+\frac{y^3}{\sqrt{2(z^4+1)}}+\frac{z^3}{\sqrt{2(x^4+1)}}$
Let $x,y,z>$ and $x+y+z=xy+yz+zx$
. Find the minimum value of $$P=\frac{x^3}{\sqrt{2(y^4+1)}}+\frac{y^3}{\sqrt{2(z^4+1)}}+\frac{z^3}{\sqrt{2(x^4+1)}}$$
My solution: I know the minimum value is $\frac{3}{2}$ when $x=y=z=1$
So $$P=\frac{x^4}{\sqrt{x^2.2(y^4+1)}}+\frac{y^4}{\sqrt{y^2.2(z^4+1)}}+\frac{z^4}{\sqrt{z^2.2(x^4+1)}}$$
$$\ge\frac{(x^2+y^2+z^2)^2}{\sqrt{2x^2(y^4+1)}+\sqrt{2y^2(z^4+1)}+\sqrt{2z^2(x^4+1)}}$$
$$\ge\frac{(x^2+y^2+z^2)^2}{\sqrt{2(x^2+y^2+z^2)(y^4+1+z^4+1+x^4+1)}}$$
$$\ge\sqrt{\frac{(x+y+z)^3}{2(x^4+y^4+z^4+3)}}$$ because $(x^2+y^2+z^2)\ge(x+y+z)$
So now i need prove $x^4+y^4+z^4+3 \le \frac{2}{9}(x+y+z)^3$ but i stuck here for a hour. So please help me, thnank
| The inequality, which you got is wrong. Try $x=y=2$ and $z\rightarrow0^+$.
For $x=y=z=1$ we obtain a value $\frac{3}{2}$.
We'll prove that it's a minimal value.
Indeed, by Holder $$\sum_{cyc}\frac{x^3}{\sqrt{y^4+1}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x^3}{\sqrt{y^4+1}}\right)^2\sum\limits_{cyc}x^3(y^4+1)}{\sum\limits_{cyc}x^3(y^4+1)}}\geq\sqrt{\frac{(x^3+y^3+z^3)^3}{\sum\limits_{cyc}x^3(y^4+1)}}$$ and it's enough to prove that:
$$2(x^3+y^3+z^3)^3\geq9\sum_{cyc}(x^4z^3+x^3),$$ which is true because
$$(x^3+y^3+z^3)^3\geq9\sum_{cyc}x^4z^3$$ and $$(x^3+y^3+z^3)^2\geq9.$$
Can you end it now?
| {
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Solution Verification: Show that $\sum_{i=0}^{n}{\binom{n+a}{n-i}\binom{i+a+1}{i}}=2^{n}\binom{n+a}{a}+2^{n-1}\binom{n+a}{a+1}$ for $n\geq 1$ I solved it by counting the same object using different methods. Say that there are $n+a$ boys and $1$ girl. I want to form a team of $a+1$ kids with some reserve players, but I want the girl to be either in the team or at least be a reserve player.
Method 1
I first select $i+a$ boys, then from these and the girl ($i+a+1$ kids) I select $a+1$ to be in the team while the rest ($i$ kids) will be the reserve players. The number of possibilities is;
$\sum_{i=0}^{n}{\binom{n+a}{i+a}\binom{i+a+1}{a+1}}= \sum_{i=0}^{n}{\binom{n+a}{n-i}\binom{i+a+1}{i}} $
Method 2
If the girl is in the team, then I need to select $a$ boys to complete the team while the remaining $n$ boys can be reserve players or not. The number of possibilities is;
$2^{n}\binom{n+a}{a}$
If the girl is only a reserve player, I need to select $a+1$ boys to be in the team while the remaining $n-1$ boys can be reserve players or not. The number of possibilities is;
$2^{n-1}\binom{n+a}{a+1}$
Conclusion
Since both method count the same objects then I can conclude that;
$\sum_{i=0}^{n}{\binom{n+a}{n-i}\binom{i+a+1}{i}}=2^{n}\binom{n+a}{a}+2^{n-1}\binom{n+a}{a+1}$
I want to know if my solution is correct and if there’s any different solution out there.
| OP asks for an alternate evaluation of
$$\sum_{q=0}^n {n+a\choose n-q} {a+1+q\choose q}.$$
We have by inspection that this is
$$[z^n] (1+z)^{n+a} \frac{1}{(1-z)^{a+2}}$$
which is in turn
$$\underset{z}{\mathrm{res}}\;
\frac{1}{z^{n+1}} (1+z)^{n+a} \frac{1}{(1-z)^{a+2}}.$$
Now we put $z/(1+z) = w$ so that $z=w/(1-w)$ and $dz = 1/(1-w)^2 \; dw$
to get
$$\underset{w}{\mathrm{res}}\;
\frac{1}{w^n} \frac{1-w}{w}
\frac{1}{(1-w)^a}
\frac{(1-w)^{a+2}}{(1-2w)^{a+2}}
\frac{1}{(1-w)^2}
\\ = \underset{w}{\mathrm{res}}\;
\frac{1-w}{w^{n+1}}
\frac{1}{(1-2w)^{a+2}}.$$
This is
$$[w^n] (1-w) \frac{1}{(1-2w)^{a+2}}
= 2^n {n+a+1\choose a+1} - 2^{n-1} {n+a\choose a+1}
\\ = 2^n \frac{n+a+1}{a+1} {n+a\choose a}
- 2^{n-1} {n+a\choose a+1}
\\ = 2^n {n+a\choose a}
+ 2^n \frac{n}{a+1} {n+a\choose a}
- 2^{n-1} {n+a\choose a+1}
\\ = 2^n {n+a\choose a} + 2^{n-1} {n+a\choose a+1}$$
which is the claim.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of ways to sample $x$ beads from 3 red beads, 3 green beads, and 3 blue beads Suppose I wanted to sample $x$ beads from 3 green, 3 blue, and 3 red beads (apart from colour, the beads are not distinct). I've been trying to find one equation I can use to solve for something like this (likely involving combinatorics), but I can't seem to figure it out. For $x=1$ the answer is just $3\choose1$. For $x=2$, if the beads are of different colours there are 3 ways, and if the beads are of the same colour there is also 3 choices to choose, so I get a total of 6. But I can't seem to find a pattern here to come up with one unified equation.
| we will use generating functions Then , if the same color of beads are different from each other :
Generating function for different red beads = $$1 + C(3,1)x + C(3,2)x^2 + C(3,3)x^3 = 1+3x +3x^2 +x^3$$
Generating function for different blue beads = $$1 + C(3,1)x + C(3,2)x^2 + C(3,3)x^3 = 1+3x +3x^2 +x^3$$
Generating function for different green beads = $$1 + C(3,1)x + C(3,2)x^2 + C(3,3)x^3 = 1+3x +3x^2 +x^3$$
Then find the coefficient of $[x^k]$ in the expansion of $$( 1+3x +3x^2 +x^3)^3$$
If the same color beads are not distinct then :
Generating function for red beads = $$ 1+x +x^2 +x^3$$
Generating function for blue beads = $$ 1+x +x^2 +x^3$$
Generating function for green beads = $$ 1+x +x^2 +x^3$$
Then find the coefficient of $[x^k]$ in the expansion of $$( 1+x +x^2 +x^3)^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I find the value of $f(0)$? Let $f$ be a polynomial such that $f(0)>0$ and $f(f(x))=4x+1$ for all $x\in R$, then $f(0)$ is $?$
So this is what I've tried so far,
$f(f(0))=1$ so $f(f(0))=4f(0)+1$ (*)
Also, $f(f(1))=5$,this implies $f(5)=16f(0)+5$ Now how should I find the value of $f(5)$ in terms of $f(1)$ so that I can solve (*)? Any help is appreciated .
The answer given is $1/2$
| $f(x) = ax + b$ $(*)$
$f(f(x)) = a^2x + ab + b = 4x + 1$
Comparing coefficients,
$b(a+1) = 1$
$a^2 = 4$
$\implies a = 2, b = \frac{1}{3}$
($a$ is not $-2$ by the fact that $f(0) > 0$, $a = -2$ would result in $b = -\frac{1}{3} \implies f(0) = -\frac{1}{3}$)
Therefore $f(x) = 2x + \frac{1}{3}$ and $f(0) = \frac{1}{3}$
(($*$) comes from the fact that if a polynomial $f(x)$ is degree $n$, then the polynomial $f(f(x))$ is of degree $n^2$)
| {
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"timestamp": "2023-03-29T00:00:00",
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Interesting inequality with three positive reals which seems trivial Let $a,b,c$ be positive real numbers, such that $(ab)^2 + (bc)^2 + (ca)^2 = 3$. Prove that
$(a^2 - a + 1)(b^2 - b + 1)(c^2 - c + 1) \geq 1$.
First I multiplied both sides by $(a+1)(b+1)(c+1)$ and expand the inequality:
$$a^3b^3c^3+a^3b^3+b^3c^3+c^3a^3\geq abc+ab+bc+ca$$
But stuck at this point.
After that i tried to use this inequality:
$$(a^2-a+1)(b^2-b+1) \geq \frac{a^2+b^2}{2}$$
Let $a^2=x$, $b^2=y$, and $c^2=z$.
$xy+yz+zx=3$
Then the statement is equivalent to:
$$(x+y)(y+z)(z+x) \geq 8$$
| Now, since $$\prod_{cyc}(x+y)\geq\frac{8}{9}(x+y+z)(xy+xz+yz),$$ it's enough to prove that $$x+y+z\geq3,$$ which is for you.
| {
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Evaluating $\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}$?
Calculate this limit
$$\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}$$
My attempt: using the limit development : we find
$$\exp(\sin(x)-x)=\exp(x-\frac{x^{3}}{3!}+o (x^{3})-x)=\exp(-\frac{x^3}{3!}+o(x^3))=1-\frac{x^3}{3!}+o(x^3)$$
So:
\begin{align}
\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}
&=\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)}\left(\frac{1}{\frac{x^3}{3!}+o(x^3)}\right)\\\
& \sim \lim\limits_{x\rightarrow 0}\left(\frac{\sin(x)}{x}\right)\left(\frac{3!}{x^{2}\exp(x)}\right)=\pm\infty.
\end{align}
Is this correct?
| You don't really need to factor like that. You can just say $\sin(x)=x-x^3/6+o(x^3)$ so
$$\exp(\sin(x))=1+\sin(x)+\sin(x)^2/2+\sin(x)^3/6+o(\sin(x)^3) \\
= 1 + (x-x^3/6) + (x-x^3/6)^2/2 + (x-x^3/6)^3/6 + o(x^3) \\
= 1 + x + x^2/2 - x^3/6 + x^3/6 + o(x^3) \\
= 1 + x + x^2/2 + o(x^3)$$
so $\exp(x)-\exp(\sin(x))=x^3/6 + o(x^3)$.
Your method is a little bit easier in this particular problem but it is pretty limited in its scope compared to this method.
The one actual issue in your calculation is that $\sin(x)/x \to 1$ from either side and $1/x^2 \to +\infty$ from either side, so you have $+\infty$ on both sides.
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum value of a expression Problem: If $\alpha+\beta+\gamma=20$, then what is $\max(\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5})$?
My attempt: Assume $\alpha \geq \beta \geq \gamma$. Then $\alpha+\beta+\gamma \leq 3\alpha$ so $25 \leq 3\alpha+5$.
Also $\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5}\leq 3\sqrt{3\alpha+5}$
At this point, I don't have idea what to do next. What should I do next?
Am I doing it incorrectly?
| Let $u=\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5}$.
Denote:
$$\begin{cases}
3\alpha +5=a^2\\
3\beta+5=b^2\\
3\gamma +5=c^2
\end{cases}$$
Then, the problem becomes:
$$v=a+b+c\to max \ \ s.t.\ \ a^2+b^2+c^2=75$$and:
$$v^2=a^2+b^2+c^2+2(ab+bc+ca)=75+2(ab+bc+ca)\le \\
75+2(a^2+b^2+c^2)=225,$$
the equality occurs for $a=b=c=5$.
Hence: $v(5,5,5)=15$ or $u(20/3,20/3,20/3)=15$ is max.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving an integral equivalence involving floor and ceiling functions During the course of looking at the Euler–Mascheroni constant I have run across the following result:
$$\gamma
= \int \limits_1^\infty \Bigg( \frac{1}{\lfloor x \rfloor} - \frac{1}{x} \Bigg) \ dx
= \int \limits_1^\infty \Bigg( \frac{\lceil x \rceil}{x^2} - \frac{1}{x} \Bigg) \ dx.$$
Representation of this constant using the first integral is a well-known result. What is the simplest way to prove the equivalence of the two integrals?
| Applying the suggestion (in comments) from Paramanand Singh we get:
$$\begin{align}
\int \limits_1^\infty \Bigg( \frac{\lceil x \rceil}{x^2} - \frac{1}{\lfloor x \rfloor} \Bigg) \ dx
&= \sum_{n=1}^\infty \int \limits_n^{n+1} \Bigg( \frac{\lceil x \rceil}{x^2} - \frac{1}{\lfloor x \rfloor} \Bigg) \ dx \\[6pt]
&= \sum_{n=1}^\infty \int \limits_n^{n+1} \Bigg( \frac{n+1}{x^2} - \frac{1}{n} \Bigg) \ dx \\[6pt]
&= \sum_{n=1}^\infty \Bigg[ - \frac{n+1}{x} - \frac{x}{n} \Bigg]_{x=n}^{x=n+1} \\[6pt]
&= \sum_{n=1}^\infty \Bigg[ \Bigg( - \frac{n+1}{n+1} - \frac{n+1}{n} \Bigg) - \Bigg( - \frac{n+1}{n} - \frac{n}{n} \Bigg) \Bigg] \\[6pt]
&= \sum_{n=1}^\infty \Bigg[ \Bigg( - 2 - \frac{1}{n} \Bigg) - \Bigg( - 2 - \frac{1}{n} \Bigg) \Bigg] \\[6pt]
&= \sum_{n=1}^\infty 0 = 0. \\[6pt]
\end{align}$$
The equivalence of the two integrals in question follows simply from here, which establishes the latter representation of the Euler-Mascheroni constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution verification for limit problem $\lim\limits_{x\to 0}\dfrac{\sin(4x)\tan^2(3x)+6x^2}{2x^2+\sin(3x)\cos(2x)}.$ Find the value of
$$\lim\limits_{x\to 0}\dfrac{\sin(4x)\tan^2(3x)+6x^2}{2x^2+\sin(3x)\cos(2x)}.$$
I try as below.
\begin{align}
\lim\limits_{x\to 0}\dfrac{\sin(4x)\tan^2(3x)+6x^2}{2x^2+\sin(3x)\cos(2x)}&=\lim\limits_{x\to 0}\dfrac{6x^2\left(\dfrac{\sin(4x)\tan^2(3x)}{6x^2}+1\right)}{2x^2\left(1+\dfrac{\sin(3x)\cos(2x)}{2x^2}\right)}\\
&=3\lim\limits_{x\to 0}\dfrac{\left(\dfrac{\sin(4x)}{4x}\cdot\left(\dfrac{\tan(3x)}{3x}\right)^2 6x+1\right)}{\left(1+\dfrac{\dfrac{1}{2}\left(\sin(5x)+\sin(x)\right)}{2x^2}\right)}\\
&=3\lim\limits_{x\to 0}\dfrac{\left(\dfrac{\sin(4x)}{4x}\cdot\left(\dfrac{\tan(3x)}{3x}\right)^2 6x+1\right)}{\left(1+\dfrac{\sin(5x)}{4x^2}+\dfrac{\sin(x)}{4x^2}\right)}\\
&=3\lim\limits_{x\to 0}\dfrac{\left(\dfrac{\sin(4x)}{4x}\cdot\left(\dfrac{\tan(3x)}{3x}\right)^2 6x+1\right)}{\dfrac{1}{x}\left(x+\dfrac{5}{4}\dfrac{\sin(5x)}{5x}+\dfrac{1}{4}\dfrac{\sin(x)}{x}\right)}\\
&=3\lim\limits_{x\to 0}\dfrac{\left(\dfrac{\sin(4x)}{4x}\cdot\left(\dfrac{\tan(3x)}{3x}\right)^2 6x^2+x\right)}{\left(x+\dfrac{5}{4}\dfrac{\sin(5x)}{5x}+\dfrac{1}{4}\dfrac{\sin(x)}{x}\right)}\\
&=3\dfrac{\left(\lim\limits_{x\to 0}\dfrac{\sin(4x)}{4x}\left(\lim\limits_{x\to 0}\dfrac{\tan(3x)}{3x}\right)^2 \lim\limits_{x\to 0} 6x^2+\lim\limits_{x\to 0} x\right)}{\left(\lim\limits_{x\to 0} x+\dfrac{5}{4}\lim\limits_{x\to 0} \dfrac{\sin(5x)}{5x}+\dfrac{1}{4}\lim\limits_{x\to 0} \dfrac{\sin(x)}{x}\right)}\\
&=3\cdot\dfrac{1\cdot 1^2\cdot 0+0}{0+\dfrac{5}{4}+\dfrac{1}{4}}\\
&=0.
\end{align}
I'm not sure with my answer. Does my answer correct?
| More simply we can proceed as follows
$$\dfrac{\sin(4x)\tan^2(3x)+6x^2}{2x^2+\sin(3x)\cos(2x)}=\frac 4 3 x\cdot\dfrac{\frac{\sin(4x)}{4x}\frac{\tan^2(3x)}{(3x)^2}9x+\frac 3 2}{\frac 2 3x+\frac{\sin(3x)}{3x}\cos(2x)}\to 0\cdot \frac{1\cdot 1\cdot 0+\frac32}{0+1\cdot1}=0$$
| {
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Why is integral of $(x+y)^2 \,dx$ different than integral of $(x^2+ 2xy + y^2) \,dx$? I calculated the integral of $(x + y)^2 \,dx$ using the substitution method and got $\frac{1}{3}(x+y)^3$ as result.
Then I applied the distributive property to $(x + y)^2$, calculated the integral of $(x^2+2xy+y^2) \,dx$ and I got $\frac{x^3}{3}+x^2y+xy^2$ as result, which is not equal to $\frac{1}{3}(x+y)^3$.
But, knowing that $(x+y)^2$ = $x^2+2xy+y^2$, why aren't the results of these integrals the same?
| Their difference is constant (it's $\frac{y^3}3$). So, you got the same answer by both methods.
To be more precise,$$\frac13(x+y)^3=\left(\frac13x^3+x^2y+xy^2\right)+\frac{y^3}3,$$and therefore\begin{align}\frac{\mathrm d}{\mathrm dx}\left(\frac13(x+y)^2\right)&=\frac{\mathrm d}{\mathrm dx}\left(\left(\frac13x^3+x^2y+xy^2\right)+\frac{y^3}3\right)\\&=\frac{\mathrm d}{\mathrm dx}\left(\frac13x^3+x^2y+xy^2\right).\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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evaluate $\int_0^{\pi/2} x^2\log(\sin x)\,dx$ I am a high school student , I know how to evaluate $\int_0^{\pi/2} x\log(\sin x)\,dx$.
It would be great if someone can help me evaluating $\int_0^{\pi/2} x^2\log(\sin x)\,dx$ and tell me if this integral is elementary or non elementary .
I tried using the "a-x" property but it resulted in $0=0$
| Another possibility is to write
$$\log[\sin(x)]=\sum_{n=1}^\infty (-1)^n \frac{2^{2 n-3}\,\, (E_{2 n-1}(1)-E_{2 n-1}(0))}{n \,(2 n-1)!}\left(x-\frac{\pi }{2}\right)^{2 n}$$ where appear the Euler polynomial and use
$$\int_0^{\frac \pi 2}x^2\left(x-\frac{\pi }{2}\right)^{2 n}\,dx=\frac{\left(\frac{\pi }2\right)^{2 n+3}}{(n+1) (2 n+1) (2 n+3)}$$ This gives
$$I_2=\frac 1{16}\sum_{n=1}^\infty (-1)^n\frac{\pi ^{2 n+3} (E_{2 n-1}(1)-E_{2 n-1}(0))}{ (2 n+3)!}$$
which, after a series of tedious manipulations, leads to the result already given by @Tuvasbien.
Numerically, this summation does not converge quite fast. Considering the partial (from $n=1$ to $n=p$)
$$\left(
\begin{array}{cc}
p & \Sigma_p \\
1 & -0.159385 \\
2 & -0.178112 \\
3 & -0.183246 \\
4 & -0.185204 \\
5 & -0.186108 \\
6 & -0.186581 \\
7 & -0.186852 \\
8 & -0.187019 \\
9 & -0.187127 \\
10 & -0.187200 \\
\cdots & \cdots \\
\infty & -0.187426
\end{array}
\right)$$
You could easily generalized the result for
$$I_p=\int_0^{\frac \pi 2}x^p\log[\sin(x)]\,dx$$ since
$$\int_0^{\frac \pi 2}x^p\left(x-\frac{\pi }{2}\right)^{2 n}\,dx=\left(\frac{\pi }{2}\right)^{2 n+p+1}\frac{(2 n)!\,\,p!}{(2 n+p+1)!}$$
Edit
Numerically, if you look at this question of mine, we could have very good approximations writing
$$I_2\sim\frac 1{16}\sum_{n=1}^p (-1)^n\frac{\pi ^{2 n+3} (E_{2 n-1}(1)-E_{2 n-1}(0))}{ (2 n+3)!}-\frac {\pi^3}8\sum_{n=p+1}^\infty \frac{1}{n (n+1) (2 n+1) (2 n+3)}$$ and
$$\sum_{n=p+1}^\infty \frac{1}{n (n+1) (2 n+1) (2 n+3)}=\frac{1}{3} \left(H_{p+\frac{3}{2}}-H_p\right)-\left(H_{p+1}-H_{p+\frac{1}{2}}\right)$$
Using $p=5$ gives an absolute error equal to $9.54\times 10^{-10}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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square root of $1 \pmod m$ for prime $m$ and non prime $m$ While doing square root $1$ moduli i.e. $x ^2 \equiv 1 \pmod m$ I noticed the following:
If the moduli $m$ is prime it seems that the square is always $1$ and $m - 1$
But if it is not prime, it is not the case and I am not sure if there is a general formula for that.
Examples:
$1 \pmod 3$: $1$ and $2$
$1 \pmod 4$: $1$ and $3$
$1 \pmod 5$: $1$ and $4$
$1 \pmod 6$: $1$ and $5$
$1 \pmod 7$: $1$ and $6$
$1 \pmod 8$: $1$ and $3$ and $5$ and $7$
$1 \pmod 9$: $1$ and $8$
$1 \pmod {11}$: $1$ and $10$
Here we see that for prime modulo we have $1$ and $m - 1$ while for non prime it varies.
Am I right to generalize that for prime it always holds? What is the property that causes this and is there a general conclusion for non prime?
| Outline for an answer.
Let $g(m)$ be the number of distinct roots to $x^2\equiv1\pmod m.$
If $m,n$ are relatively prime, then show $g(mn)=g(m)g(n).$ (Hint: use Chinese remainder theorem.) (This property is called being “multiplicative.”)
This is true for any polynomial congruence in one variable. If $h(m)$ is the number of solutions to $x^3+x\equiv 1\pmod m,$ then $h$ is multiplicative.
So we only need to comoute $g(p^k),$ where $p$ is a prime.
If $p$ is an odd prime, then show $g(p^k)=2$ for any $k>0.$
This is because $x^2\equiv 1\pmod {p^k},$ means $p^k\mid x^2-1=(x-1)(x+1).$ But $x-1$ and $x+1$ cannot both be divisible by $p,$ so this means $p^k\mid x-1$ or $p^k\mid +1,$ and thus $x\equiv\pm 1\pmod{p^k}.$
If $p=2,$ you have the problem that $x-1$ and $x+1$ can both be divisible by $2.$
Show that:
$$g(2^k)=\begin{cases}1&k=0,1\\2&k=2\\4&k>2 \end{cases}$$
Finally, if $$m=2^kp_1^{k_1}p_2^{k_2}\cdots p_{n}^{k_n}$$ with the $p_i$ distinct odd primes, then:
$$g(m)=g(2^k)2^n$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complex number related problem Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______
My solution is as follow
${z_1} = 2{e^{i{\theta _1}}};{z_2} = 2{e^{i{\theta _2}}};{z_3} = 2{e^{i{\theta _3}}}$ & $Z = {z_1} + {z_2} + {z_3} = 2\left( {{e^{i{\theta _1}}} + {e^{i{\theta _2}}} + {e^{i{\theta _3}}}} \right)$
$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {{e^{i{\theta _1}}} - {e^{i{\theta _2}}}} \right| = \left| {{e^{i{\theta _1}}} - {e^{i{\theta _3}}}} \right|$
Let ${\theta _1} = 0$
$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {1 - \left( {\cos {\theta _2} + i\sin {\theta _2}} \right)} \right| = \left| {1 - \left( {\cos {\theta _3} + i\sin {\theta _3}} \right)} \right|$
$ \Rightarrow \left| {1 - \cos {\theta _2} - i\sin {\theta _2}} \right| = \left| {1 - \cos {\theta _3} - i\sin {\theta _3}} \right| \Rightarrow \left| {2{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| {2{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right|$
$\Rightarrow \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right| \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {\cos \frac{{{\theta _2}}}{2} + i\sin \frac{{{\theta _2}}}{2}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {\cos \frac{{{\theta _3}}}{2} + i\sin \frac{{{\theta _3}}}{2}} \right)} \right|$
$ \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right|$
$ \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right|$
$\Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|$
${\theta _2} \ne {\theta _3}$
How do I proceed further?
| Algebra + Geometry approach
Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______
Let's try interpret the problem geometrically.
$|z_1| = |z_2 | = |z_3|$ means that the three points lie on a circle.
$|z_1 - z_2 | = |z_1 - z_3|$ means that the chord substended by $z_1$ and $z_2$ is equal to the chord subtended by $z_1$ and $z_3$. This means that $z_1$ is the mid point of the arc on the circle starting at $z_3 $ and ending at $z_2$.
It is also given that $|z_1 + z_2 + z_3 |= 2$, we can rearrange this to give: $|1 + \frac{z_2}{z_1} +\frac{z_3}{z_1} | = 1$ but we know that $\frac{z_2}{z_1} = e^{i \theta}$ and $\frac{z_3}{z_1} = e^{- i \theta}$, this leads to:
$$| 1+ 2 \cos \theta | = 1$$
It is clear the the only two principle solutions for $\theta$ are $\theta = \{\pi, \frac{\pi}{2} \}$... but if $\theta=\pi$ then $z_2 = z_3$ violating the condition in question (draw a picture!)
We go back to the question : $|z_1 + z_2 | |z_1 + z_3| = \frac{|z_1^2 - z_2^2| | z_1^2 - z_3^2|}{|z_1 - z_2| |z_1 - z_3|}$
We know when we square a complex number, we square the magnitude and double the angle. Hence, $z_1^2 - z_2^2$ represents a chord having twice the angle as $z_1 - z_2$ and on a circle with squared the radius and similarly does $|z_1 - z_3|^2$.
Ok, but what is the expression for the length of chord which subtends an angle $\theta$ in a circle of radius $r$? By the law of cosines, I find it as $ 2r \sin \frac{\theta}{2}$... now we begin the bash:
$$|z_1 + z_2 | |z_1 + z_3| = \frac{|z_1^2 - z_2^2| | z_1^2 - z_3^2|}{|z_1 - z_2| |z_1 - z_3|} = \frac{ (2r^2 \sin \theta)^2}{(2r \sin \frac{\theta}{2})^2}=4r^2 \cos^2 \frac{\theta}{2}$$
We know $r=2$ and $ \theta= \frac{\pi}{2}$, which gives the final answer as $8$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate maximum and minimum when second partial derivative test fail
Calculate the maximum and minimum of function $z = f(x,y) = x^2 - y^2 +2$ subjected to the inequality constraint $D=\{(x,y)| x^2 + \frac{y^2}{4} \leq 1\}$.
My solution:
First form the function
$$
g(x,y) = x^2 + \frac{y^2}{4} - c, ~0 \leq c \leq 1.
$$
Then form the Lagrangian function
$$
L(x,y,\lambda) = x^2 - y^2 + 2 + \lambda\left(x^2 + \frac{y^2}{4} - c\right).
$$
Therefore we have
$$
\left \{
\begin{array}{ll}
L_x' = 2x +2\lambda x = 0\\
L_y' = -2y + \frac{\lambda}{2}y = 0 \\
L_\lambda' = x^2 + \frac{y^2}{4} - c
\end{array} \right.
$$
After solving above equations, so we can got saddle point like $(\varphi(c),\psi(c))$.
*
*when $c = 0$, we have $x = y = 0$.
*when $c \neq 0$, there are two kind of solutions: (1 $x = 0$, we have $y = \pm 2\sqrt{c}$; (2 $y=0$, we have $ x = \pm \sqrt{c}$.
My problem is that i cant use second partial derivative test for judging $(0,\pm 2\sqrt{c})$ and $(\pm \sqrt{c},0)$ are maximum or minimum, obviously $AC-B^2 = 0$. How can i do next? Thanks in advance!
| HINT: It is an absolute maximum (and minimum). It is easy to see that it is obtained on the boundary. Then you will have a finite number of possible points. Simply compare them.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all analytic functions in the disk $|z-2|<2$ such that $f \left(2+\frac{1}{n} \right) = \frac{1-n}{5n+3} + \sin{\left(\frac{n \pi}{2} \right)}$ Find all analytic functions (or prove that no such exist) inside the disk $|z-2|<2$ that satisfy the following condition:
$$f \left(2+\frac{1}{n} \right) = \frac{1-n}{5n+3} + \sin{\left(\frac{n \pi}{2} \right)}, \ n \in \mathbb{N}$$
For $n=2k$ the expression simplifies quite a bit since $\sin{\left(\frac{n \pi}{2} \right)}= \sin{\left(k \pi \right)} = 0$ so we're left with $$f \left(\frac{4k+1}{2k}\right) = \frac{1-2k}{10k+3}$$
Similarly for $n=4k-3$ and $n=4n-1$ we can simplify the sinus expression to $1$ and $-1$. So we're getting $3$ different expressions for $f\left(2+\frac{1}{n}\right)$ based on whether $n=2k$ , $n=4k-3$ or $n=4k-1$. My idea is to use the identity theorem for any of these two expressions and prove that no such function exists but I can't quite tie it all together.
| If $n$ is a multiple of $4$, then\begin{align}f\left(2+\frac1n\right)&=\frac{1-n}{5n+3}\\&=\frac{1/n-1}{5+3/n}\\&=\frac{2+1/n-3}{-1+3(2+1/n)}.\end{align}Now, for each $z\in D_2(2)$, let $g(z)=\frac{z-3}{-1+3z}$. Then, by the identity theorem, $g=f$. But that's impossible, since $g(3)\ne f(3)$: $g(3)=0$, whereas$$f(3)=f\left(2+\frac11\right)=1.$$So, no such function $f$ exists.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by contradiction that if $n^3$ is a multiple of $3$, then $n$ is a multiple of $3$ Problem statement:
Using proof by contradiction, prove that if $n^3$ is a multiple of $3$ , then $n$ is a multiple of $3.$
Attempt 1:
Assume that there is exist $n$ which is a multiple of $3$ such that $n^3$ is not a multiple of $3.$
Then $n = 3k $ , $n^3 = 27 k^3 $ which a multiple of $3$, which contradicts the assumption that $n^3$ is a multiple of $3.$
Attempt 2:
Assume that there exist $n^3$ which is a multiple of $3$ such that $n$ is not a multiple of $3.$
Then $n = k+ 1 $ or $n= k+2.$
Then $n^3 = k^3 + 3k^2 + 3k + 1$
or $n^3 = k^3 + 6 k^2 + 12 k + 8.$
Both cases contradict the assumption that $n^3 $ is a multiple of $3.$
Which solution is correct ? Or do both work ?
| If $P,Q$ are statements then $P \implies Q$ is same as $ \neg Q \implies \neg P $.
Here $P : n^3$ is a multiple of $3$ and $Q: n$ is a multiple of $3$.
So if you want to prove it by contradiction you have to assume that $n^3$ is multiple of $3$ but not $n$ and then get a contradiction.
You can do this by noting that every number is in the form $3k, 3k+1$ or $3k +2$. If $n$ is a not a multiple of $3$ then it has to be in the form of $3k+1$ or $3k +2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral using the definition of integral is returning double the real answer? I'm trying to evaluate an integral using the definition of the definite integral but for some reason my answer is 12 when it should be 6. I assumed that I'd missed a 2 somewhere that would go in the denominator but I've tried this exercise multiple times and I keep getting the same answer.
My steps:
$$ \int_0^3 (2x-1)dx = \lim_{n\to \infty} \sum_{k=1}^n f\left(\frac{k(3-0)}{n}\right) \frac{3}{n} $$
First I solve the function that is inside the summation.
$$ f\left(\frac{k(3-0)}{n}\right) = f\left(\frac{3k}{n}\right) = 2\left(\frac{3k}{n}\right) - 1 = \frac{6k}{n} - 1 $$
Then I solve the summation.
$$ \sum_{k=1}^{n} \left( \frac{6k}{n} - 1 \right) \frac{3}{n} = \frac{3}{n}\sum_{k=1}^{n} \left( \frac{6k}{n} - 1 \right) = \frac{3}{n}\left(\sum_{k=1}^{n} \frac{6k}{n} - \sum_{k=1}^{n} (-1) \right) = \frac{3}{n}\left( \left(\sum_{k=1}^{n} \frac{6}{n}k \right) - (-n) \right) = \frac{3}{n}\left(\frac{6}{n}\left(\sum_{k=1}^{n}k \right) + n \right) = \frac{3}{n} \left( \frac{6}{n} \cdot \frac{n(n+1)}{2} + n \right) = \frac{3}{n} \left( 3(n+1) + n \right) = \frac{3}{n}(4n + 3) = \frac{12n + 9}{n} $$
Then I find the limit of the solved summation as $n \to \infty$.
$$ \lim_{n\to\infty} \frac{12n + 9}{n} = 12 $$
However, the answer should be 6. What am I doing wrong here?
| As Michael said, it's just a sign error:
$$\cdots= \frac{3}{n} \left( \frac{6}{n} \cdot \frac{n(n+1)}{2} \color{red}{-} n \right) = \frac{3}{n} \left( 3(n+1) \color{red}{-} n \right) = \frac{3}{n}(\color{red}{2}n + 3) = \frac{\color{red}{6}n + 9}{n}$$
| {
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Ivan has 3 red blocks, 4 blue blocks and 2 green blocks. He builds the tower with randomly selected blocks but only stops when the tower consists of all three colours.
*
*What is the probability that the tower is 4 blocks tall?
My approach is the following but I am not sure at all:
to make it 3 colours it builds a tower that is either 3 blocks or 8 blocks.
8 blocks would happen in the case where the first 7 blocks are 4Blue and 3Red, so the 8th must be the green block.
so $E(8) = 4B,3R,1G = \binom{4}{4}*\binom{3}{3}*\binom{2}{1}= 2$
$E(7) = 4B,2G,1R = \binom{4}{4}*\binom{2}{2}*\binom{3}{1}= 3$
E(6) = let's reason the other way around. First $\binom{9}{6} = 84$. then the combinations of 6 blocks with only 2 colours is 4B,2R or 4B,2G or 3B,3R.
This makes:
$ \binom{4}{4}*\binom{3}{2} + \binom{4}{4}*\binom{2}{2}+ \binom{4}{3}*\binom{3}{3}= 8$.
So $E(6) = 84 - 8 = 76$
E(5) = 3B,1R,1G or 3R,1B,1G or 2B,2R,1G or 2B,2G,1R or 2R,2G,1B =
$E(5) = \binom{4}{3}*\binom{3}{1}*\binom{2}{1}+ \binom{3}{3}*\binom{4}{1}*\binom{2}{1}+ \binom{4}{2}*\binom{3}{2}*\binom{2}{1}+ \binom{4}{2}*\binom{2}{2}*\binom{3}{1}+ \binom{3}{2}*\binom{2}{2}*\binom{4}{1}= 98$
E(4) = 2B,1R,1G or 2R,1B,1G or 2G,1B,1R =
$\binom{4}{2}*\binom{3}{1}*\binom{2}{1}+ \binom{3}{2}*\binom{4}{1}*\binom{2}{1}+ \binom{2}{2}*\binom{4}{1}*\binom{3}{1}= 72$
$E(3) = \binom{4}{1}*\binom{3}{1}*\binom{2}{1}= 24$
total combinations $(T) = 24 + 72 + 98 + 76 + 3 + 2 = 275$.
The probability of stopping at 4 blocks is $E(4)/(T) = 72/275$
but this is a very long solution. Do you think that is correct? if so, is there any other method to make it shorter?
| We can easily solve using generating functions
For blocks of $4$ that have all three colors, the number of permutations is given by
coefficient of $x^4$ in $4!(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!})(x+\frac{x^2}{2!} + \frac{x^3}{3!})(x+\frac{x^2}{2!}) = 36$
while unrestricted permutations = $\dfrac{9!}{4!3!2!} = 1260$
Thus $Pr = \dfrac{36}{1260} = \dfrac{2}{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I solve $\frac{(1-x)(x+3)}{(x+1)(2-x)}\leq0$ and express solution in interval notation? The inequality I need to solve is
$$\frac{(1-x)(x+3)}{(x+1)(2-x)}\leq0$$
My attempt:
Case I: $x<2$
Then we have $(1-x)(x+3)\leq0$
$x\leq-3, x\geq1$
We reject $x\leq-3$ (not sure why exactly, I just looked at the graph on desmos and it shows that $x\geq1$
Case II: $x<-1$
Then we have $(1-x)(x+3)\geq0$
$x\geq-3, x\leq-1$
Reject $x\leq-1$ (again, not sure exactly why aside from the graph I saw on desmos)
So the final solution is $x\in[-3,-1) \cup [1,2)$
I know that my final solution is correct, however I am a little confused on why I reject certain values.
| The key insight here is that for a fraction to be negative you need to have the numerator and denominator have opposite signs. If you have $$\frac{A}{B}\le 0$$ then either you have $A\le 0$ and $B> 0$ or you have $A\ge 0$ and $B< 0$. This gives us two cases, with several sub-cases:
Case $1$:
$$(1-x)(x+3)\le 0\quad\text{and}\quad(x+1)(2-x)> 0$$
Here we need to remember that for a product to be negative the factors have opposite signs, and for a product to be positive then the factors have the same sign. We have some sub-cases, then:
Case $1a$:
\begin{align*}
1-x &\le 0 \implies x\ge 1\\
x+3 &\ge 0\implies x \ge -3\\
x + 1 &> 0\implies x> -1\\
2 - x &> 0\implies x< 2
\end{align*}
Looking at the overlap between these four inequalities we get $1\le x < 2$.
Case $1b$:
\begin{align*}
1-x &\le 0 \implies x\ge 1\\
x+3 &\ge 0\implies x \ge -3\\
x + 1 &< 0\implies x< -1\\
2 - x &< 0\implies x> 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Case $1c$:
\begin{align*}
1-x &\ge 0 \implies x\le 1\\
x+3 &\le 0\implies x \le -3\\
x + 1 &> 0\implies x> -1\\
2 - x &> 0\implies x< 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Case $1d$:
\begin{align*}
1-x &\ge 0 \implies x\le 1\\
x+3 &\le 0\implies x \le -3\\
x + 1 &< 0\implies x< -1\\
2 - x &< 0\implies x> 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Case $2$:
$$(1-x)(x+3)\ge 0\quad\text{and}\quad(x+1)(2-x)< 0$$
Case $2a$:
\begin{align*}
1-x &\ge 0 \implies x\le 1\\
x+3 &\ge 0\implies x \ge -3\\
x + 1 &> 0\implies x> -1\\
2 - x &< 0\implies x> 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Case $2b$:
\begin{align*}
1-x &\ge 0 \implies x\le 1\\
x+3 &\ge 0\implies x \ge -3\\
x + 1 &< 0\implies x< -1\\
2 - x &> 0\implies x< 2
\end{align*}
Looking at the overlap between these four inequalities we get $-3\le x < -1$.
Case $2c$:
\begin{align*}
1-x &\le 0 \implies x\ge 1\\
x+3 &\le 0\implies x \le -3\\
x + 1 &> 0\implies x> -1\\
2 - x &< 0\implies x> 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Case $2d$:
\begin{align*}
1-x &\le 0 \implies x\ge 1\\
x+3 &\le 0\implies x \le -3\\
x + 1 &< 0\implies x< -1\\
2 - x &> 0\implies x< 2
\end{align*}
These inequalities are inconsistent so we don't get any more solutions.
Overall, then, we get the solution $-3\le x < -1$ and $1\le x < 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$
The system says
$$x+y+z=0$$
$$xy +xz+yz=-1$$
$$xyz=-1$$
Find
$$x^8+y^8+z^8$$
With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$
trying with
$$(x + y + z)^3 =
x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 +
3 y z^2 + 6 x y z$$
taking advantage of the fact that there is an $xyz=-1$ in the equation, but I'm not getting anywhere, someone less myopic than me.how to solve it?
Thanks
Edit : Will there be any university way to solve this problem , they posed it to a high school friend and told him it was just manipulations of remarkable products.
His answers I understand to some extent but I don't think my friend understands all of it.
| Denoting with $e_k(x,y,z), k\geq 0$ the elementary symmetric polynomials
\begin{align*}
e_1(x,y,z)&=x+y+z=0\\
e_2(x,y,z)&=xy+xz+yz=-1\tag{1}\\
e_3(x,y,z)&=xyz=-1
\end{align*}
and with $p_k(x,y,z), k\geq 0$ the $k$-th power sum
\begin{align*}
p_k(x,y,z)=x^k+y^k+z^k\tag{2}
\end{align*}
we recall Newtons identities
admit a generating function representation of the power sums as
\begin{align*}
\color{blue}{\sum_{k=1}^\infty(-1)^{k-1}p_k\frac{t^k}{k}=\ln(1+e_1t+e_2t^2+e_3t^3+\cdots)}\tag{3}
\end{align*}
We obtain from (1) - (3)
\begin{align*}
\color{blue}{\left.x^8+y^8+z^8\right|_{{{e_1=0\ \ }\atop{e_2=-1}}\atop{e_3=-1}}}
&=(-8)[t^8]\ln\left(1-t^2-t^3\right)\\
&=(-8)[t^8]\ln\left(1-t^2(1+t)\right)\\
&=8[t^8]\sum_{j=1}^\infty\frac{t^{2j}(1+t)^j}{j}\tag{4.1}\\
&=8\sum_{j=1}^{4}\frac{1}{j}[t^{8-2j}](1+t)^j\tag{4.2}\\
&=8\sum_{j=1}^4\frac{1}{j}\binom{j}{8-2j}\tag{4.3}\\
&=8\left(\frac{1}{3}\binom{3}{2}+\frac{1}{4}\binom{4}{0}\right)\tag{4.4}\\
&\,\,\color{blue}{=10}
\end{align*}
Comment:
*
*In (4.1) we expand the logarithmic series.
*In (4.2) we apply the rule $[t^{p-q}]A(t)=[t^p]t^qA(t)$. We also set the upper limit of the series to $4$ since other terms do not contribute.
*In (4.3) we select the coefficient of $t^{8-2j}$.
*In (4.4) we use $\binom{p}{q}=0$ if $0<p<q$ are positive integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 2
} |
Finding binomial Coefficient using expansion If ${\left( {1 + x + {x^2}} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{2n - 1}}{x^{2n - 1}} + {a_{2n}}{x^{2n}}$, prove that $a_0=a_{2n},a_1=a_{2n-1},....a_n=a_{n+1}$
My approach is as follow ${\left( {1 + x + {x^2}} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{2n - 1}}{x^{2n - 1}} + {a_{2n}}{x^{2n}}$
${\left( {1 + x + {x^2}} \right)^n} = {\left( {1 + X} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{X^r}} = \sum\limits_{r = 0}^n {{}^n{C_r}{{\left( {x + {x^2}} \right)}^r}} = \sum\limits_{r = 0}^n {{}^n{C_r}\sum\limits_{g = 0}^r {{}^r{C_g}{x^{r - g}}.{x^{2g}}} } = \sum\limits_{r = 0}^n {{}^n{C_r}\sum\limits_{g = 0}^r {{}^r{C_g}{x^{r + g}}} } $
How do we proceed from here
| $$(1+x+x^2)^n=\sum_{k=0}^{2n} A_k x^k$$ Change $x$ to $1/x$ in this udentity, then
$$(1+x+x^2)^n=\sum_{k=0}^{2n} A_k x^{2n-k}=\sum_{k=0}^{2n} A_{2n-k} ~ x^k,$$ $k \to 2n-k$ used in above. Finally we get $A_{2n-k}=A_k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$? I was recently searching for interesting looking integrals. In my search, I came upon the following result:
$$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$$
and I wanted to try and prove it.
Inspired by this answer by Jack D'Aurizio, I took the Weierstrass product for $\cosh(x)$ to obtain
$$
\cosh\left(\frac{\pi x}{2} \right) = \prod_{n \ge 1}\left(1 + \frac{x^2}{(2n-1)^2} \right)
$$
And by logarithmically differentiating twice we get
$$
\frac{\pi^2}{4}\text{sech}^2\left(\frac{\pi x}{2} \right) = \sum_{n \ge 1} \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}
$$
Which means we get
\begin{align*}
\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx & =\frac{4}{\pi^2}\sum_{n\ge 1} \int_{0}^{\infty} \frac{(1-x^2)}{(1+x^2)^2}\left( \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}\right)\, dx
\end{align*}
However, after this, I couldn't figure out how to evaluate the resulting integral.
Does anyone know how I could continue this method? Or alternatively, does anyone know another way in which the result can be proven? Thank you very much!!
Edit:
Per jimjim's request, I'll add that I found this integral on the Wikipedia article for $\zeta(3)$. I believe the reference is to this text where the following formula is given
$$
(s-1) \zeta(s) = 2\pi \int_{\mathbb{R}}\frac{\left(\frac{1}{2} + xi \right)^{1-s}}{\left(e^{\pi x} +e^{-\pi x} \right)^2}\, dx
$$
which for the case of $s=3$ reduces to the surprisingly concise
$$
\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx = \frac{\zeta(3)}{\pi}
$$
And I presume that one can modify the previous equation to get to the original integral from the question, but it is not apparent to me how this may be done.
Edit 2:
Random Variable has kindly posted in the comments how to go from $\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx$ to $ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx$. Thank you very much!
| After playing around a bit more with the Weierstrass product method, I realized I could proceed to do the integration with partial fractions. Using that
$$
\int_{0}^{\infty} \frac{1}{(x^2+a^2)^n} \ dx = \frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2a^{2n-1}}, \quad a\ge0,\ n \in \mathbb{N}
$$
as shown in this answer, we get the following
$\require{cancel}$
\begin{align*}
I& \overset{\color{blue}{n\to n-1}}{=}\frac{4}{\pi^2} \sum_{n\ge \color{blue}{0}} \int_{0}^{\infty} \frac{(1-x^2)}{(1+x^2)^2}\left( \frac{4(2n\mathbin{\color{blue}{+}}1)^2}{\left(x^2 + (2n\mathbin{\color{blue}{+}}1)^2\right)^2} - \frac{2}{x^2 + (2n\mathbin{\color{blue}{+}}1)^2}\right)\, dx\\
& \overset{\color{purple}{a=2n+1}}{=}\frac{4}{\pi^2}\Bigg(\int_{0}^{\infty}\underbrace{\frac{8}{(x^2 +1)^4}-\frac{8}{(x^2 +1)^3} +\frac{2}{(x^2 +1)^2}}_{\color{red}{n=0}}\ dx +\sum_{n\ge \color{red}{1}} \Bigg[\frac{4(a^2 +1)}{(a^2 -1)^2}\int_{0}^{\infty} \frac{1}{(x^2 +1)^2}\, dx \\
& \qquad- \frac{2(a^4+6a^2 +1)}{(a^2 -1)^3}\int_{0}^{\infty} \frac{1}{x^2 +1}\, dx +\frac{2(a^4+6a^2 +1)}{(a^2 -1)^3}\int_{0}^{\infty} \frac{1}{x^2 +a^2}\, dx\\
&\qquad+\frac{4a^2(a^2 +1)}{(a^2 -1)^2}\int_{0}^{\infty} \frac{1}{(x^2 +a^2)^2}\, dx \Bigg]\Bigg)\\
& =\frac{4}{\pi^2} \Bigg( \frac{40\pi}{32} - \frac{24\pi}{16} +\frac{2\pi}{4}+\sum_{n\ge 1}\Bigg[\frac{4(a^2 +1)}{(a^2 -1)^2}\frac{\pi}{4} - \frac{2(a^4+6a^2 +1)}{(a^2 -1)^3}\frac{\pi}{2}+\frac{2(a^4+6a^2 +1)}{(a^2 -1)^3}\frac{\pi}{2 a }\\
&\qquad+\frac{4a^2(a^2 +1)}{(a^2 -1)^2}\frac{\pi}{4a^3}\Bigg]\Bigg)\\
& =\frac{4}{\pi} \left( \frac{1}{4}+\sum_{n\ge 1}\Bigg[\frac{a^5-a}{a(a^2 -1)^3} - \frac{a^5+6a^3 +a}{a(a^2 -1)^3}+\frac{a^4+6a^2 +1}{a(a^2 -1)^3} +\frac{a^4-1}{a(a^2 -1)^3}\Bigg]\right)\\
& =\frac{4}{\pi} \left( \frac{1}{4}+\sum_{n\ge 1}\frac{2\cancel{a}\cancel{(a-1)^3}}{\cancel{a}(a+1)^3\cancel{(a-1)^3}}\right)\\
& \overset{\color{purple}{a=2n+1}}{=}\frac{4}{\pi} \left( \frac{1}{4}+\frac{1}{4}\sum_{n\ge 1}\frac{1}{(n+1)^3}\right)\\
& =\frac{\zeta(3)}{\pi}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 1
} |
Find value of $x^{5} + y^{5} + z^{5}$ given the values of $x + y + z$, $x^{2} + y^{2} + z^{2}$ and $x^{3} + y^{3} + z^{3}$ If$$x+y+z=1$$
$$x^2+y^2+z^2=2$$
$$x^3+y^3+z^3=3$$
Then find the value of $$x^5+y^5+z^5$$
Is there any simple way to solve this problem ? I have tried all my tricks tried to multiply two equations , substitute $z=1-x-y$ , but things got messy nothing seems to work out .
| Another way.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$,where $v^2$ can be negative, and $xyz=w^3$.
Thus, $u=\frac{1}{3}$ and since
$$2=x^2+y^2+z^2=9u^2-6v^2,$$
we obtain $$v^2=-\frac{1}{6}.$$
Also, since $$3=x^3+y^3+z^3=27u^3-27uv^2+3w^3=1+\frac{3}{2}+3w^3,$$ we get $$w^3=\frac{1}{6}.$$
Id est, $$x^5+y^5+z^5=243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3=6.$$
I used the following identities, which easy to prove and which we always use in the $uvw$'s method.
About $uvw$ see here https://artofproblemsolving.com/community/c6h278791
$$x^2+y^2+z^2=9u^2-6v^2,$$
$$x^3+y^3+z^3=27u^3-27uv^2+3w^3$$ and
$$x^5+y^5+z^5=243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3.$$
There are also:
$$\sum_{cyc}(x^2y+x^2z)=9uv^2-3w^3,$$
$$\sum_{cyc}x^2y^2=9v^4-6uw^3$$ and more
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the minimum value of $a^8+b^8+c^8+2(a-1)(b-1)(c-1)$ Let $a,b,c$ be the lengths of the three sides of the triangle, $a+b+c=3$. Find the minimum value of $$a^8+b^8+c^8+2(a-1)(b-1)(c-1)$$
My attempts:
$\bullet$ The minimum value is $3$, equality holds iff $a=b=c=1$ so by AM-GM, we have: $$a^8+b^8+c^8\ge 8(a+b+c)-21=3$$
$\bullet$ We need to prove $$3+2(a-1)(b-1)(c-1)\ge3$$ or $$abc-(ab+bc+ca)+a+b+c-1\ge0$$
$\bullet$ Note that $ab+bc+ca\le \dfrac{(a+b+c)^2}{3}=3 $ so we need to prove: $$abc-3+3-1\ge0$$ or $$abc\ge1$$
But I have no idea from here, please help me
| The function $f(a,b,c)=a^8+b^8+c^8+2(a-1)(b-1)(c-1)$ has only one extrema minimum point $f(1,1,1)=3$ on the domain $a+b+c=3$. To prove that this one is the global minimum we need to check the value of $f$ at the boundary of the following domain:
*
*$0<a<\frac 32$
*$\frac32-a<b<\frac 32$
*$c=3-a-b$
and we have
*
*for $a\to0 \implies b,c\to\frac 32$
$$f\left(0, \frac 32, \frac 32\right)=2\left(\frac32\right)^8 -2\left(\frac12\right)^2>3$$
*
*for $a\to\frac32 \implies 0<b<\frac 32$ and $c=\frac 32-b$ we obtain
$$f\left(\frac32, b, \frac32-b\right)=\left(\frac32\right)^8+b^8+\left(\frac32-b\right)^8+2\left(\frac12\right)\left(b-1\right)\left(\frac12-b\right)>3$$
which by symmetry suffices, therefore the minimum value is attained at $(a,b,c)=(1,1,1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Non-probabilistic/non-combinatoric proof of $\sum_{k=n}^{2n} \frac{{k \choose n}}{2^k}=1$ The summation $$S_n=\sum_{k=n}^{2n} \frac{{k \choose n}}{2^k}=1~~~~(1)$$
has been proved using probabilistic/combinatoric method in MSE earlier:
Combinatorial or probabilistic proof of a binomial identity
Here we give an analytic proof for (1):
$$S_n=\sum_{k=n}^{2n} \frac{{k-1 \choose n-1}+{k-1 \choose n}}{2^k}~~~~~(2)$$
Let $k-1=p$, then
$$S_n=\sum_{p=n-1}^{2n-1} \frac{{p \choose n-1}+{p \choose n}}{2^{p+1}}~~~~(3)$$
Take the first term
$$\sum_{p=n-1}^{2n-1} \frac{{p \choose n-1}}{2^{p+1}}=\sum_{p=n-1}^{2n-2} \frac{{p \choose n-1}}{2^{p+1}}+\frac{{2n-1 \choose n-1}}{2^{2n-1}}=\frac{S_{n-1}}{2}+\frac{{2n-1 \choose n-1}}{2^{2n}}~~~~~(4)$$
Now take the second term
$$\sum_{p=n-1}^{2n-1} \frac{{p \choose n}}{2^{p+1}}=\sum_{p=n}^{2n} \frac{{p \choose n}}{2^{p+1}}+\frac{{n-1 \choose n}}{2^{n}}-\frac{{2n \choose n}}{2^{2n+1}}~~~~(5)$$
The second term in RHS vanishes and the third term is equal and opposite to the second term of the RHS of (4). Adding (4) and (5), we get
$$S_n=\frac{S_{n-1}}{2}+\frac{S_{n}}{2} \implies S_n=S_{n-1}$$So $S_n$ is a constant (independent of $n$) we can have $S_n=S_{n-1}=S_1=1$
The question is: What could be other analytic proofs of (1) without using the ideas of probability/combinatorics?
| Using algebra.
$$\sum_{k=n}^{2n} \binom{k}{n} x^k=(1-x)^{-(n+1)} x^n\Bigg[1-\binom{2 n+1}{n} ((1-x) x)^{n+1} \, _2F_1(1,2 n+2;n+2;x) \Bigg]$$
Developed as a series around $x=\frac 12$
$$\sum_{k=n}^{2n} \binom{k}{n} x^k=1+ \left(4 n+2-\frac{4\, \Gamma
\left(n+\frac{3}{2}\right)}{\sqrt{\pi }\, \Gamma
(n+1)}\right)\left(x-\frac{1}{2}\right)+O\left(\left(x-\frac{1}{2}\right)^2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to find a closed-form expression for this summation? Is it possible to find a closed-form expression for this summation?
$$\sum\limits_{k = 0}^n {\left( \left( {\matrix{
2n \cr
k \cr
} } \right)+ \left( {\matrix{
2n \cr
n+k-1 \cr
} } \right) \left( \frac{1}{3}\right)^\left(2n-2k+1 \right) \right)
\cdot\left( \frac{1}{3}\right)^k}$$
I can not find one that is not using hyper geometric functions, and splitting the sum did not help me, so there might be something I am missing. If anyone has advice that would be great.
| When you have this kind of summation, you cannot avoid at least gaussian hypergeometric functions and in most cases you will not obtain any closed form.
Relacing the $\frac 13$ by $x$ and splitting the sum, you have
$$\sum_{k=0}^n \binom{2 n}{k} x^k=(x+1)^{2 n}-\binom{2 n}{n+1} x^{n+1} \, _2F_1(1,1-n;n+2;-x)$$ The second term will be
$$\sum_{k=0}^n \binom{2 n}{n+1-k} x^{2 n-k}=\binom{2 n}{n+1} x^{2 n} \, _2F_1\left(1,-n-1;n;-\frac{1}{x}\right)-x^{n-1}$$ probably leading to what Wolfram Alpha gave for your specific case.
Now, for a given value of $n$, what you will obtain is a polynomial of degree $2n$ in $x$ with integer coefficients. The very first ones are
$$\left(
\begin{array}{cc}
n & \text{Polynomial} \\
1 & x^2+4 x+1 \\
2 & 4 x^4+6 x^3+10 x^2+4 x+1 \\
3 & 15 x^6+20 x^5+15 x^4+26 x^3+15 x^2+6 x+1 \\
4 & 56 x^8+70 x^7+56 x^6+28 x^5+78 x^4+56 x^3+28 x^2+8 x+1 \\
5 & 210 x^{10}+252 x^9+210 x^8+120 x^7+45 x^6+262 x^5+210 x^4+120 x^3+45 x^2+10
x+1
\end{array}
\right)$$
Very few patterns have been found in $OEIS$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Gambler's ruin recursive solution Suppose we have 2 players A and B, A starting with $a$ coins and B starting with $n-a$ coins. Assuming every game is independent, every time A wins, they win $2$ coins and every time they lose they lose 1 coin. What is the probability that A wins the game ?
I can solve this using a difference equation, but I was wondering if there is a way to write down the recurrence equation in a way that you can solve this without using difference equations. So far I could only come up with the following:
Considering $W_{a}$ the probability of A winning the game given they start with a coins.
Boundary equations are $W_{0} = 0$ and $W_{N} = 1$
$$W_{a} = pW_{a+2} + qW_{a-1}$$
$$(p+q)W_{a} = pW_{a+2} + qW_{a-1}$$
$$p(W_{a+2} - W_{a}) = q(W_{a} - W_{a-1})$$
$$W_{a+2} - W_{a} = \frac{q}{p}(W_{a} - W_{a-1})$$
If you extend this to a few terms, you can notice:
$$W_{3} - W_{1} = \frac{q}{p}(W_1 - W_0)$$
$$W_{4} - W_{2} = \frac{q}{p}(W_2 - W_1)$$
$$W_{5} - W_{3} = \frac{q}{p}(W_3 - W_2)$$
$$...$$
$$W_{a} - W_{a-2} = \frac{q}{p}(W_{a-2} - W_{a-3})$$
$$W_{a+1} - W_{a-1} = \frac{q}{p}(W_{a-1} - W_{a-2})$$
$$W_{a+2} - W_{a} = \frac{q}{p}(W_{a} - W_{a-1})$$
If you sum everything up, the terms cancel out and you're left with:
$$W_{a+2} + W_{a+1} - W_{2} - W_{1} = \frac{q}{p}W_{a}$$
Is there a way to find $W_{a}$ out of this or the difference equation is the only way to solve this ?
| (Note that we need $W_{n + 1} = 1$ as an additional boundary condition.)
Your idea is good; I’ll write it more compactly as follows:
$$0 = \sum_{i=1}^a (W_i - pW_{i+2} - qW_{i-1}) = pW_1 + pW_2 + qW_a - pW_{a+1} - pW_{a+2}.$$
We can generalize it by throwing in a $c^i$ coefficient, where $c$ is any root of $p - c^2 + c^3q = 0$ so that the sum still telescopes:
\begin{multline*}
0 = \sum_{i=1}^a c^i (W_i - pW_{i+2} - qW_{i-1}) \\
= \frac{p}{c}W_1 + pW_2 + qc^{a+1}W_a - pc^{a-1}W_{a+1} - pc^aW_{a+2}.
\end{multline*}
Knowing one root $c = 1$, we can factor $p - c^2 + c^3q = (1 - c)(p + cp - c^2q)$ to find the other two roots. This gives us three linear equations in five unknowns $W_1, W_2, W_a, W_{a+1}, W_{a+2}$. If we substitute $a = n - 1$, we can use our knowledge of $W_n = W_{n+1} = 1$ to solve for $W_1, W_2, W_{n-1}$. Then for general $a$, we can use our knowledge of $W_1, W_2$ to solve for $W_a, W_{a+1}, W_{a+2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
Solving differential equation $af'(x)^2-f''(x)=0$ I want to solve the differential equation $$af'(x)^2-f''(x)=0$$ where $a $ is constant and we have the boundary conditions $f(0)=0, f (1)=c $ for some positive $c$.
If $a=0$ we get $f (x)=cx$ but what if $a\ne 0$. Could someone show how to solve this differential equation?
| The differential equation can be written as
$$
\frac{f''(x)}{f'(x)^2} = a.
$$
Taking antiderivatives gives
$$
-\frac{1}{f'(x)} = ax + B,
$$
for some constant $B.$ Thus,
$$
f'(x) = -\frac{1}{ax+B}
$$
and taking another antiderivative gives
$$
f(x) = C - \frac{1}{a}\ln(ax+B)
$$
where $C$ is a constant.
Boundary condition $f(0)=0$ then gives
$$
0 = f(0) = C - \frac{1}{a}\ln(a\cdot 0+B) = C - \frac{1}{a}\ln B,
$$
i.e.
$$
C = \frac{1}{a}\ln B.
$$
Boundary condition $f(1)=c$ gives
$$
c = f(1) = C - \frac{1}{a}\ln(a\cdot 1+B) = \frac{1}{a}\ln B - \frac{1}{a}\ln (a+B) = \frac{1}{a}\ln\frac{B}{a+B},
$$
from which we can solve for $B$ and get
$$
B = \frac{a}{e^{-ac} - 1}.
$$
Thus we get the solution
$$
f(x) = \frac{1}{a}\ln \frac{a}{e^{-ac} - 1} - \frac{1}{a}\ln(ax+\frac{a}{e^{-ac} - 1}) \\
%= \frac{1}{a} \ln \frac{\frac{a}{e^{-ac} - 1}}{ax+\frac{a}{e^{-ac} - 1}}
= \frac{1}{a} \ln \frac{1}{x(e^{-ac} - 1)+1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
"solved" Confusion calculating $\mathop {\lim }\limits_{x\to 0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}}$ I can get the correct answer through one solution, but when I try the second method, it shows an obvious error, and I can't find where and why. Can someone know the reason, or can provide some useful suggestions? thanks, : )
Solution2
$$\begin{align*}
\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}}
& = \mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{({\rm{1 + }}\cos 2x{\rm{ - 1}})}^{\frac{1}{2}}}{{({\rm{1 + }}\cos 3x{\rm{ - 1}})}^{\frac{1}{3}}}}}{{{x^2}}}\\
& =\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x({\rm{1 + }}\frac{{\cos 2x{\rm{ - 1}}}}{{\rm{2}}})({\rm{1 + }}\frac{{\cos 3x{\rm{ - 1}}}}{{\rm{3}}})}}{{{x^2}}}\\
&=\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x}}{{{x^2}}}\\
&=\frac{{\rm{1}}}{{\rm{2}}}
\end{align*}$$
Maybe Solution1 is a bit informal, but all I want about it is just to talk about ideas
Solution1
$$\begin{align*}
\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}}
& = \mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x + \cos x(1 - {{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}})}}{{{x^2}}}\\
&=\frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - {{(\cos 2x)}^{\frac{1}{2}}}{\rm{ + }}{{(\cos 2x)}^{\frac{1}{2}}}(1 - {{(\cos 3x)}^{\frac{1}{3}}})}}{{{x^2}}}\\
&= \frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - {{(1 + \cos 2x - 1)}^{\frac{1}{2}}}{\rm{ + }}{{(\cos 2x)}^{\frac{1}{2}}}(1 - {{(1 + \cos 3x - 1)}^{\frac{1}{3}}})}}{{{x^2}}}\\
& = \frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - \cos 2x}}{{2{x^2}}}\mathop { + \lim }\limits_{x\to0} \frac{{1 - \cos 3x}}{{3{x^2}}}\\
&= \frac{1}{2} + 1 + \frac{3}{2} = 3
\end{align*}$$
I find my mistake is "forget considering the infinitesimal term when replacing"
thanks help for clear answer for @user
a wonderful and general answer for @CHAMSI
also, Parthib Ghosh's opinion is also useful
| When calculating limits, you can not just replace $ x $ in a part of the expression and leave it in the other part, this is a very common mistake.
I'll provide a solution to a more generalized limit, and won't use series expansions or L'hopital's rule. We must know that $ \frac{1-\cos{x}}{x^{2}}\underset{x\to 0}{\longrightarrow}\frac{1}{2} $, though.
Let $ n\in\mathbb{N}^{*}\left(=\mathbb{N}\setminus\left\lbrace 0\right\rbrace\right) $, we have :
\begin{aligned}\lim_{x\to 0}{\frac{1-\prod\limits_{k=1}^{n}{\cos^{1/k}{\left(kx\right)}}}{x^{2}}}&=\lim_{x\to 0}{\frac{\sum\limits_{p=1}^{n}{\left(\prod\limits_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}-\prod\limits_{k=1}^{p}{\cos^{1/k}{\left(kx\right)}}\right)}}{x^{2}}}\\ &=\lim_{x\to 0}{\sum_{p=1}^{n}{\frac{1-\cos^{1/p}{\left(px\right)}}{x^{2}}\prod_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}}\\&=\lim_{x\to 0}{\sum_{p=1}^{n}{\frac{\left(1-\cos^{1/p}{\left(px\right)}\right)\color{blue}{\times\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}}{x^{2}\color{blue}{\times\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}}\prod_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}}\\ &=\lim_{x\to 0}{\sum_{p=1}^{n}{\frac{1-\cos{\left(px\right)}}{x^{2}\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}\prod_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}}\\ &=\lim_{x\to 0}{\sum_{p=1}^{n}{\left(p^{2}\times\frac{1-\cos{\left(px\right)}}{\left(px\right)^{2}}\times\frac{\prod\limits_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}{\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}\right)}}\\&=\sum_{p=1}^{n}{\left(p^{2}\times\frac{1}{2}\times\frac{\prod\limits_{k=1}^{p-1}{1}}{\sum\limits_{j=0}^{p-1}{1}}\right)}\\ &=\frac{1}{2}\sum_{p=1}^{n}{p}\\ \lim_{x\to 0}{\frac{1-\prod\limits_{k=1}^{n}{\cos^{1/k}{\left(kx\right)}}}{x^{2}}}&=\frac{n\left(n+1\right)}{4}\end{aligned}
| {
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Inequality between integral of $(1-x)^k P(x)^2$ and integral of $(1-x)^{k+1} P'(x)^2$ Given a polynomial $P$ with real coefficients and an integer $k \ge 0$, define
$$I_k(P) = \int_0^1 \frac{(1-x)^k}{k!} P(x)^2 dx.$$
I would like to show that if $P(0) = 0$ and $P$ is not identically zero, then
$$I_k(P) \le \frac{4}{k+2} I_{k+1}(P').$$
The result is true for monomials $P(x) = x^r$. Indeed, we have
$$I_k(P) = \frac{(2r)!}{(k+2r+1)!}
\quad \text{and} \quad
I_{k+1}(P') = \frac{r^2(2r-2)!}{(k+2r)!},$$
so
$$\frac{I_{k-2}(P)}{I_{k-1}(P')} = \frac{2(2r-1)}{r(k+2r+1)} < \frac{4}{k+2}.$$
In terms of partial progress, I can show $I_k(P) \le 2 \sqrt{I_{k+1}(P) I_{k+1}(P')}$ and $I_k(P) \le \frac{1}{4k} I_k(P')$. Together, these give $I_k(P) \le \frac{1}{\sqrt{k+1}} I_{k+1}(P')$ (e.g. proving the result for $k \le 12$).
For the first, we may integrate by parts and then apply Cauchy-Schwarz to find
$$I_k(P) = 2 \int_0^1 \frac{(1-x)^{k+1}}{(k+1)!} P(x) P'(x) dx
\le 2 \left(\int_0^1 \frac{(1-x)^{k+1}}{(k+1)!} P(x)^2 dx\right)^{1/2}
\left(\int_0^1 \frac{(1-y)^{k+1}}{(k+1)!} P'(x)^2 dx\right)^{1/2}
= 2 \sqrt{I_{k+1}(P) I_{k+1}(P')}.$$
For the second, we may replace $P(x)$ with $\int_0^x P'(t) dt$ and then apply Cauchy-Schwarz to find
\begin{aligned}
I_k(P) &= \int_{0 \le t \le x \le 1} \frac{(1-x)^k}{k!} P(x) P'(t) \; dx \; dt \\
&\le \left(\int_{0 \le t \le x \le 1} \frac{(1-x)^{k+1}}{k!} P(x)^2 \; dx \; dt\right)^{1/2}
\left(\int_{0 \le t \le x \le 1} \frac{(1-x)^{k-1}}{k!} P'(t)^2 \; dx \; dt\right)^{1/2} \\
&= \left(\int_0^1 x(1-x) \frac{(1-x)^k}{k!} P(x)^2 \; dx\right)^{1/2}
\left(\int_0^1 \frac{(1-t)^k}{k \cdot k!} P'(t)^2 \; dt\right)^{1/2} \\
&\le \left(\frac{1}{4} I_k(P)\right)^{1/2} \left(\frac{1}{k} I_k(P')\right)^{1/2},
\end{aligned}
using in the last step that $x(1-x) \le \frac{1}{4}$.
| Using the inequality $I_k(P) \le 2 \sqrt{I_{k+1}(P) I_{k+1}(P')}$, it suffices to show $I_{k+1}(P) \le \left(\frac{2}{k+2}\right)^2 I_{k+1}(P')$.
Integrating by parts and applying Cauchy-Schwarz, we have
\begin{align}
I_{k+1}(P)
&= 2 \int_0^1 \frac{(1-x)^{k+2}}{(k+2)!} P(x) P'(x) dx \\
&\le 2 \left(\frac{k+3}{k+2}\right)^{1/2}
\left( \int_0^1 \frac{(1-x)^{k+3}}{(k+3)!} P(x)^2 dx \right)^{1/2}
\left( \int_0^1 \frac{(1-x)^{k+1}}{(k+1)!} P'(x)^2 dx \right)^{1/2} \\
&= 2 \left(\frac{k+3}{k+2}\right)^{1/2} I_{k+3}(P)^{1/2} I_{k+1}(P')^{1/2}.
\end{align}
Since $(1-x)^{k+3} \le (1-x)^{k+1}$ for $0 \le x \le 1$, we have $I_{k+3}(P) \le \frac{1}{(k+3)(k+2)}I_{k+1}(P)$, so altogether we have shown
\begin{align}
I_{k+1}(P)
&\le 2 \left(\frac{k+3}{k+2}\right)^{1/2} \left(\frac{1}{(k+3)(k+2)}I_{k+1}(P)\right)^{1/2} I_{k+1}(P')^{1/2} \\
&= \left(\frac{2}{k+2}\right) I_{k+1}(P)^{1/2} I_{k+1}(P')^{1/2}.
\end{align}
Squaring and dividing by $I_{k+1}(P)$ gives the desired $I_{k+1}(P) \le \left(\frac{2}{k+2}\right)^2 I_{k+1}(P')$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the function(s) $f$ satisfying the condition: $x^2f(x)+yf\left(y^2\right)=f(x+y)f\left(x^2-xy+y^2\right)$.
Find the function(s) $f$ satisfying the condition: $$x^2f(x)+yf\left(y^2\right)=f(x+y)f\left(x^2-xy+y^2\right)\text.$$
Of Course, I expect the function $f$ be $f(x)=x$ or $f\equiv0$.
My attempt:
\begin{align}
& P(0, 0): 0=f(0)^2 \implies f(0)=0 \text. \\
& P(x, 0): x^2f(x)=f(x)f\left(x^2\right) \text. & \tag{A} \label{A}\\
& P(0, y): yf\left(y^2\right)=f(y)f\left(y^2\right) \text. & \tag{B} \label{B}\\
& \text{If } f \equiv 0: \ \text{Solution.} \\
& \text{If } f \not\equiv 0: \\
\\
& \text{Let } f(s) \ne 0 \text. \\
& \text{ Putting } s \rightarrow x \text { in } \eqref{A} \text {: } s^2f(s)=f(s)f\left(s^2\right) \text. \\
& \therefore f\left(s^2\right)=s^2 \text. \\
\\
& \text{ Putting } s \rightarrow y \text { in } \eqref{B} \text {: } sf\left(s^2\right)=f(s)f\left(s^2\right) \text. \\
& \therefore s^3=s^2f(s) \text. \\
\\
& \text{Since }f(0)= 0 \text, \ s \ne 0 \text. \\
& \therefore f(s)=s \text. \\
\end{align}
Now, I have to show that $\text{there is only }s=0\text{ which is satisfying } f(s)=0,$ to make sure that $f(x)=x \text{ or } f \equiv 0$.
| You've correctly managed to show that if $ f ( x ) \ne 0 $ then $ f ( x ) = x $. As $ f \not \equiv 0 $, there is some $ x _ 0 \ne 0 $ with $ f ( x _ 0 ) = x _ 0 $. Consider an arbitrary $ y \ne 0 $. If $ f ( y ) = 0 $, by \eqref{B} you get $ f \left ( y ^ 2 \right ) = 0 $. Thus, $ P ( x _ 0 , y ) $ gives
$$ f ( x _ 0 + y ) f \left ( x _ 0 ^ 2 - x _ 0 y + y ^ 2 \right ) = x _ 0 ^ 3 \text . $$
As $ x _ 0 ^ 3 \ne 0 $, you must have $ f ( x _ 0 + y ) \ne 0 $ and $ f \left ( x _ 0 ^ 2 - x _ 0 y + y ^ 2 \right ) \ne 0 $, and therefore $ f ( x _ 0 + y ) = x _ 0 + y $ and $ f \left ( x _ 0 ^ 2 - x _ 0 y + y ^ 2 \right ) = x _ 0 ^ 2 - x _ 0 y + y ^ 2 $. This leads to $ x _ 0 ^ 3 + y ^ 3 = x _ 0 ^ 3 $, which is in contradiction with $ y \ne 0 $. Hence, $ f ( y ) = 0 $ is impossible, and you must have $ f ( y ) = y $. As you also have $ f ( 0 ) = 0 $, you get $ f ( x ) = x $ for all $ x $.
| {
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Induction inequality/ n in exponential I need to find out through induction for which $n \geq 0$ the following inequality holds
$$3n+2^{n} \leq 3^n$$
clearly it does not work for n=1 and n=2, so my induction hypothesis is that it works for $n \geq 3$,
however when performing the induction step, I am kind of stuck
because
$ n \rightarrow n+1$
$3 (n+1) + 2^{n+1} \leq 3^{n+1}$
$3(n+1) + 2^{n} \cdot 2 \leq 3^n \cdot 3$
I recreated the starting inequality within (overline)
$3n + 3 + 2^n + 2^n \leq (2+1) \cdot 3^n$
$2^n + 3 + \overline{3n + 2^n \leq 3^n} + 2\cdot 3^n$
I tried to show that what was added on the LHS is always smaller than on the RHS, therefor
$2n+3 \leq 2\cdot 3^n$
$2n + 3 \leq 3^n + 3^n$
so it is clear that this holds,but would the proof work that way? if yes, are there other, more elegant approaches?
| So I think the necessary steps in the proof are all there, it is just better to do it slightly differently. You initially write $3(n+1) + 2^{n+1} \leq 3^{n+1}$ and then manipulate it. I know that's what you are trying to show, but you never want to manipulate what you are trying to show. Its nice form to start with one side of the inequality, and make your way to the other side. You don't want to write both sides at the same time. I would write the induction step as follows:
We want to show $3(n+1) + 2^{n+1} \leq 3^{n+1}$. By the induction hypothesis we know $3^n \geq 3n + 2^n$ so we can say
\begin{align*}
3^{n+1}
&= 3\cdot 3^n\\
&\geq 3(3n + 2^n)\\
&= 3(3n) + 2^n\cdot 3
\end{align*}
Now since $n \geq 3$, we know $3n > n+1$ and of course $3 > 2$. So we make these substitutions and arrive at
\begin{align*}
&\geq 3(n+1) + 2^n\cdot 2\\
&= 3(n+1) + 2^{n+1}.
\end{align*}
Thus, we have shown the statement holds for $n+1$.
| {
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"source": "stackexchange",
"question_score": "2",
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Conditional expectation and expected value Given the joint distribution $f(x,y) = {x \choose y} (\frac{1}{2})^x \frac{x}{20}$ where $y=0,1,2 \cdots x$ and $x=2,3,4,5,6$.
Find $E(y\mid x)$ and $E(y)$
Wondering if I am missing something.
Using Binomial identity and direct definition of $f(x)$ (the limit was from the plot I made):
$f(x) = (\frac{1}{2})^x \frac{x}{20} \sum_{y=0}^x {x \choose y} =(\frac{1}{2})^x \frac{x}{20} 2^{x} = \frac{x}{20} $
Use $f(x)$ to compute $f(y\mid x)$, given that the joint distribution is given:
So that $f(y\mid x) = \frac{f(x,y)} {f(x)} = \frac{ {x \choose y} (\frac{1}{2})^x \frac{x}{20}}{ \frac{x}{20}} = {x \choose y} 2^{-x} $
$E(y \mid x) = \sum_{y=0}^{x} y f(y\mid x) = 2^{-x} \sum_{y=0}^{x} {x \choose y} y $
Then: Using the identity : $ \sum_{y=0}^{x} {x \choose y} y = x 2^{x-1} $
$E(y \mid x) = \sum_{y=0}^{x} y f(y\mid x) = 2^{-x} \sum_{y=0}^{x} {x \choose y} y = x (2^{-x} \cdot 2^{x-1}) = \frac{x}{2} $
Does this makes sense?
How does $E(y)$ follows, I guess $E(y) = E(E(y \mid x))$
$E(y) = E(E(y \mid x))$
$E(y) = E(E(y \mid x)) = \sum_{x=2}^6 \frac{x}{20} (2^{-x} \sum_{y=0}^{x} {x \choose y} y) = \frac{1}{40}\sum_{x=2}^6 x^2 $
| The easiest way to proceed is to observe that your joint pmf is
$$p(x,y)=p(y|x)p(x)=\binom{x}{y}\left(\frac{1}{2}\right)^x\times\frac{x}{20}$$
Thus $(Y|X=x)\sim Bin\left(x;\frac{1}{2}\right)$
And X is a discrete rv with support $x=\{2,3,4,5,6\}$ and pmf
$$p(X=x)=x/20$$
Thus
$$\mathbb{E}[Y|X=x]=x/2$$
And
$$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]= \mathbb{E}[X/2]= \frac{1}{2 }\mathbb{E}[X]
= \frac{1}{2 }\times\frac{9}{2}=\frac{9}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Strange/Unexpected behavior of an Infinite product Some friends and I were playing around with this continued fraction:
We noticed when writing it out for each next step, the end behavior went either to 1 (when there was an even number of terms) or to a linear dependence on x (when there were odd). This was expected - but fitting the linear dependence gave something we didn't expect.
We wrote the odd-term fraction as a product:
.
We expected that the tail behavior would go as $(x+2m+2)$, but insead we found it closer to $(x+m+1)$: https://www.desmos.com/calculator/7ye4ijgoef
Its clear both give identical limits for finite $m$ as $x \longrightarrow \infty$, but
is there an argument as to why the significantly more accurate behavior $(x+m+1)$ occurs over the expected behavior $(x+2m+2)$ for large but fininte $x$?
| Transform the product to a sum using logarithms
$$P_m=\prod_{n=0}^m \frac{x+2 n}{x+2 n+1} \implies \log(P_m)=\sum_{n=0}^m \log \left(\frac{x+2 n}{x+2 n+1}\right)$$ Ubsing Taylor for large values of $x$
$$\log \left(\frac{x+2 n}{x+2 n+1}\right)=-\frac 1 x+\frac{4 n+1}{2 x^2}-\frac{12 n^2+6 n+1}{3 x^3}+O\left(\frac{1}{x^4}\right)$$ Summing from $n=0$ to $n=m$
$$\log(P_m)=-\frac{m+1}{x}+\frac{(m+1) (2 m+1)}{2 x^2}-\frac{(m+1)^2 (4 m+1)}{3 x^3}+O\left(\frac{1}{x^4}\right)$$
$$P_m=e^{\log(P_m)}=1-\frac{m+1}{x}+\frac{(m+1) (3 m+2)}{2 x^2}-\frac{(m+1)^2 (5 m+2)}{2 x^3}+O\left(\frac{1}{x^4}\right)$$
$$(x+2m+2)P_m=(x+m+1)-\frac{(m+1) (m+2)}{2 x}+\frac{(m+1)^2 (m+2)}{2
x^2}+O\left(\frac{1}{x^3}\right)$$ which is even smaller that $(x+m+1)$.
Let us try with $x=10^3$ and $m=10$ the exact result is
$$\frac{4725919019374652598779904000}{4674801393661613629292257}=1010.934716$$
The truncated series gives
$$\frac{505467363}{500000}=1010.934726$$
Edit
Doing the same for a more general case, the first terms are
$$(x+am+b)P_m=\big[x +(a-1) m+(b-1)\big]-\frac{(m+1) ((2 a-3) m+2( b-1))}{2 x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Failure of L’Hospital’s rule to $\lim_{x\rightarrow\infty} x\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right)$? I rewrote the limit
$$\lim_{x\rightarrow\infty} x\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right) =\lim_{x\rightarrow\infty} \frac{\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right)}{\frac 1x},$$
noting that both the numerator and denominator tend to zero as $x\rightarrow \infty$ so that the conditions for L’Hospital’s rule are fulfilled. However, differentiating top and bottom then yields:
$$\lim_{x\rightarrow\infty} \frac{\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right)}{\frac 1x}=\lim_{x\rightarrow\infty} \frac{\frac{1}{1+\left(\frac{x+1}{x+2}\right)^2}}{-\frac 1{x^2}}=\lim_{x\rightarrow\infty} \frac{-x^2(x+2)^2}{(x+2)^2+(x+1)^2}=\infty,$$
which is incorrect. What went wrong?
| You forgot to use chain rule in the numerator:
$\frac{d}{dx}(arctan(\frac{x+1}{x+2})-\frac{\pi}{4}) = \dfrac{\frac{x+2-(x+1)}{(x+2)^2}}{1+(\frac{x+1}{x+2})^2} = \frac{1}{(x+2)^2+(x+1)^2}$
Now the limit exists.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$ in two ways gives different answers
Evaluate
$$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$$
I started by rationalising followed by dividing numerator and denominator by $x$.
$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$
=$\lim_{x\rightarrow-\infty}\frac{{(\sqrt{4x^2-x}+2x)}{(\sqrt{4x^2-x}-2x)}}{(\sqrt{4x^2-x}-2x)}$
=$\lim_{x\rightarrow-\infty}\frac{-x}{\sqrt{4x^2-x}-2x}$
=$\lim_{x\rightarrow-\infty}\frac{-1}{\sqrt{4-\frac{1}{x}}-2}$
and we get $\frac{-1}{√4-2}$ i.e.-1/0 form.
The answer however is $\frac{1}{4}$.
$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$
=$\lim_{x\rightarrow\infty}{\sqrt{4x^2+x}-2x}$
=$\lim_{x\rightarrow\infty}\frac{{(\sqrt{4x^2+x}-2x)}{(\sqrt{4x^2+x}+2x)}}{(\sqrt{4x^2+x}+2x)}$
=$\lim_{x\rightarrow\infty}\frac{x}{\sqrt{4x^2+x}+2x}$
=$\lim_{x\rightarrow\infty}\frac{1}{\sqrt{4+\frac{1}{x}}+2}$
=$\frac{1}{4}$
What is wrong in first method?
| The problem happens when you go from
$$\lim_{x\rightarrow-\infty}\frac{-x}{\sqrt{4x^2-x}-2x}$$
to
$$\lim_{x\rightarrow-\infty}\frac{-1}{\sqrt{4-\frac{1}{x}}-2}$$
The first line is actually not identical to the second line, and you can verify yourself by plugging in some value of $x$. For example, if $x=-20$, then
$$\frac{-x}{\sqrt{4x^2-x}-2x} = \frac{-(-20)}{\sqrt{4\cdot(-20)^2-(-20)}-2\cdot(-20)} = \frac{20}{\sqrt{1620}+40}\approx\frac{20}{40+40}=\frac14$$
while
$$\frac{-1}{\sqrt{4-\frac{1}{x}}-2} = \frac{-1}{\sqrt{4-\frac{1}{-20}}-2}=\frac{-1}{\sqrt{4+\frac{1}{20}}-2}\approx\frac{-1}{2+\epsilon-2}=\frac{-1}{\epsilon}$$ which is very negative, because $\epsilon$ is small.
So, you need to be very careful when going from one expression to the other. In particular, what you probably did was this:
$$\begin{align}
\frac{-x}{\sqrt{4x^2-x}-2x}&=\frac{\frac{-x}{x}}{\frac{\sqrt{4x^2-x}-2x}{x}}\\
&=\frac{-1}{\frac{\sqrt{4x^2-x}}{x}-\frac{2x}{x}}\\
&=\frac{-1}{\frac{\sqrt{4x^2-x}}{\sqrt{x^2}}-2}\\
&=\frac{-1}{\sqrt{\frac{4x^2-x}{x^2}}-2}\\
&=\frac{-1}{\sqrt{4-\frac1x}-2}
\end{align}$$
And you can see (again, just plug in $x=20$ for a sanity check) that only the first two lines of the above expression are correct.
The mistake comes when you replace $x$ with $\sqrt{x^2}$, which is an equality that does not hold for $x<0$. Instead, you must replace $x$ with $-\sqrt{x^2}$ to get
$$\frac{-1}{\frac{\sqrt{4x^2-x}}{-\sqrt{x^2}}-2} = \frac{-1}{-\sqrt{4-\frac1x}}-2$$
and then you get the correct result.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Evaluating $S_{N} = \sum_{i=0}^\infty \frac{i^N}{4^i}$ using recursion I am trying to obtain a formulae for a summation problem under section (d) given in a solutions manual for "Data Structures and Algorithm Analysis in C - Mark Allen Weiss", here's the screen shot
As the pdf is protected i could not download it.
Here's my attempt at it.
Let $S_{N} = \sum_{i=0}^\infty \frac{i^N}{4^i}$, then starting from $N = 4$ we have
$S_{0} = \frac{4}{3}, S_{1} = \frac{4}{9} , S_{2} = \sum_{i=0}^\infty \frac{2i + 1}{3*4^i}, S_{3} = \sum_{i=0}^\infty \frac{3i^2 + 3i + 1}{3*4^i},S_{4} = \sum_{i=0}^\infty \frac{4i^3 + 6i^2 + 4i + 1}{3*4^i}$.
Using recursion, we have
$S_{0} = \frac{4}{3}, S_{1} = \frac{1}{3}S_{0} , S_{2} = \frac{2S_{1} + S_{0}}{3} = \frac{5}{9}S_{0} ,
S_{3} = \frac{3S_{2} + 3S_{1} + S_{0}}{3} = \frac{11}{9}S_{0}, S_{4} = \frac{4S_{3} + 6S_{2} + 4S_{1} + S_{0}}{3} = \frac{95}{27}S_{0}$.
After cleaning up further we have
$$S_{0} = \frac{4}{3}, S_{1} = \frac{1}{3}S_{0} , S_{2} = \frac{5}{9}S_{0} ,
S_{3} = \frac{11}{9}S_{0}, S_{4} = \frac{95}{27}S_{0}$$
I dont see a pattern emerging to make a formulae.I may be doing something wrong, any help is greatly appreciated.
| The technique given in the question gives the recurrence:
$$3S_N=\sum_{i}\frac{(i+1)^N-i^N}{4^i}=\sum_{j=0}^{N-1} \binom Nj S_j$$
This follows from the binomial theorem: $$(i+1)^N-i^n=\sum_{j=0}^{N}\binom Nj i^j$$
But “solving” this recurrence is very painful. I don’t know how to solve it. It’s not even clear the book is asking you to solve the recurrence.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $\frac{p^2q^2(1-\epsilon^2\cos^2t)}{p^2\cos^2t+q^2\sin^2t}$, where $\epsilon=\sqrt{1-(q/p)^2}$ I am currently trying to show that the product of the distances from the focis of an ellipse to the tangent line at any point of the ellipse is a constant. While I thought that the computation is a straightforward plug-in and be done with it, this turned out to be harder than expected.
I know that the end result should be $q^2$, but I have no idea on how to manipulate the expression
$$\frac{p^2q^2(1 - \epsilon^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}$$ where $p > q > 0, \epsilon = \sqrt{1 - (q/p)^2}$.
Edit: I had mixed up $\sin$ and $\cos$ in the denominator, so instead of $p^2\cos^2(t) + q^2\sin^2(t)$ the expression should be $p^2\sin^2(t) + q^2\cos^2(t)$.
| Let us use that $\cos^2(t)+\sin^2(t)=1$ and that $\epsilon^2=1-(q/p)^2$:
$$\frac{p^2q^2(1 - \epsilon^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=\frac{p^2q^2(\cos^2(t)+ \sin^2(t) - (1-(q/p)^2)\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=$$
$$=\frac{p^2q^2(\cos^2(t)+\sin^2(t) - \cos^2(t)+(q/p)^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=\frac{p^2q^2(\sin^2(t)+(q/p)^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=$$
$$=\frac{1}{p^2}\frac{p^2q^2(p^2\sin^2(t)+q^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=q^2$$
Maybe you can think that it is hard come up with this solution since you have to notice that you have to use $\cos^2(t)+\sin^2(t)=1$.
I'll give you another solution, which is the standard method to show that a function is constant. The method consists on taking the derivative and see that this is equal to $0$ (since the only functions with derivative $0$ are the constant functions):
Let $f(t)=\frac{p^2q^2(1 - \epsilon^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}$, taking the derivative you will find that:
$$f'(t)=\frac{2p^2q^2(p^2\epsilon^2 - p^2+q^2)}{(p^2\sin^2(t) + q^2\cos^2(t))^2}$$
(As homework you can check that this is indeed the derivative). Now notice that:
$$p^2\epsilon^2 - p^2+q^2=p^2(1-(q/p)^2)-p^2+q^2=p^2-q^2-p^2+q^2=0$$
Hence $f'(t)=0$ and thus $f(t)=c$ with $c\in\mathbb{R}$ a constant. To find the value of $f$ you can plug any value for $t$, for example $f(0)=q^2$, and then $f(t)=q^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4294293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving Big O and Big Omega of a Polynomial I have a polynomial:
$f(n)=\frac{1}{4}n^2-24n-16$
I am supposed to show that it is $\Omega(n^2)$ and $O(n^2)$. Once I prove those, it should be easy to show that it is also $\Theta(n^2)$ using the theorems given, but I'm having difficulties with $\Omega$ and $O$. Mainly the subtraction part is throwing me off, as it has been hard finding a constant and $k$ value that works.
For showing $\Omega(n^2)$, I started with $f(n)=\frac{1}{4}n^2-24n-16$ and $g(n) = n^2$, and have been trying to show that $f(n) \geq C\cdot g(n)$, but the subtraction has made it hard for me to choose a value for $C$ and $k$. Can I just chose any value that makes it work?
Pretty similar for showing $O(n^2)$, I started with $f(n)=\frac{1}{4}n^2-24n-16$ and $g(n) = n^2$, and have been trying to show that $f(n) \leq C \cdot g(n)$, and again, I'm having troubles knowing where to go from here.
| For checking that $f(n)$ is $O(n^2)$ I think you have just made a simple mistake and it seems like you understand what to do: if $C = \frac{1}{4}$ then $f(n) \leq C g(n)$ becomes
$$
\frac{1}{4} n^2 - 24 n - 16 \leq \frac{1}{4} n^2
$$
which is automatically always true for $n \geq 0$.
They key point for showing that $f(n)$ is $\Omega(n^2)$ is that you only need to show that there exists $C$ so that $f(n) \leq C g(n)$ for all $n$ greater than some fixed $k$. Here is how we use this fact via basically a simplifying trick: observe that for all $n \geq 1$ we have
$
6 < 24 n
$.
This means that
$$
\frac{1}{4} n^2 - 24 n - 16
\geq \frac{1}{4} n^2 - 24 n - 24 n
= \frac{1}{4} n^2 - 48 n.
$$
(In effect this means that it's enough to solve the problem for $\frac{1}{4} n^2 - 48 n$ instead.)
Moreover also note that whenever $n > 4 \times 48$ we have $\frac{1}{16} n^2 > 48 n$. That means that
$$
\frac{1}{4} n^2 - 48 n \geq \frac{1}{4} n^2 - \frac{1}{16} n^2 = \frac{3}{16} n^2 \quad \text{whenever} \quad n > 4 \times 48.
$$
Putting this all together, if we set $C = \frac{3}{16}$ then we have just shown that
$$
f(n) = \frac{1}{4} n^2 - 24 n - 16
\geq \frac{3}{16} n^2 = \frac{3}{16} g(n)
$$
for all $n > k = 4 \times 48$, which is enough to establish that $f(n)$ is $\Omega(n^2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show this sequence always is rational number let $\{a_{n}\}$ such $a_{1}=-8$,and such
$$4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}}=3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$$
show that
$$a_{n}\in Q,\forall n\in N^{+}$$
I try let $a_{2}=x$,and for $n=1$, then we have
$$-8+5\sqrt[3]{x}=3\sqrt[3]{-49(x+1)}\Longrightarrow x=-1/8$$
and for $n=2$ I get $a_{3}=-\dfrac{389017}{4913}$ and so on
The first observation
i want use this well known identity: $$a^3+b^3+c^3=3abc ~~~~~~~~~~~~~~~~~~if~~~~~ a+b+c=0$$
let $a=4\sqrt[3]{a_{n}},b=5\sqrt[3]{a_{n+1}},c=-3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$,so
$$64a_{n}+125a_{n+1}-189(a_{n}+1)(a_{n+1}+1)=-180\sqrt[3]{7a_{n}a_{n+1}(a_{n}+1)(a_{n+1}+1)}$$
ADD it by 2021,11.6.PM.18:05
The second observation
and Now I have found this interesting:
if $x,y,p,q,a,b\ge 0$, Hölder's inequality :
$$(x^3+y^3)(p^3+q^3)(a^3+b^3)\ge (xpa+yqb)^3$$
$1+\sqrt[3]{e^{2a}}\sqrt[5]{e^{b}}\sqrt[15]{e^{2c}} \leq \sqrt[3]{(1+e^{a})^2}\sqrt[5]{1+e^{b}}\sqrt[15]{(1+e^{c})^2}$
the condition it's
$$(a_{n}+1)(1+a_{n+1})(4^3+5^3)= (4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}})^3$$
or
$$4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}}=3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$$
but this sequence $a_{n}<0$ or $a_{n}>0$,so this Hölder inequality seem can't hold
| By looking at the initial values, we hypothesize that
*
*$a_n = x_n^3, x_n \in \mathbb{Q}$
*$a_n + 1 = 7y_n^3, y_n \in \mathbb{Q}$
This suggests to study rational points on $7y^3 = x^3 + 1$.
We map $ a_i$ to the point $ ( \sqrt[3]{a_i}, \sqrt[3]{\frac{a_i + 1}{7} } )$.
We have the starting points $a_1: (-2, -1), a_2: (-1/2, 1/2)$.
Notice that we also have the rational points $ (-1, 0), (4/5, 3/5)$ on the curve. The latter was inspired by the coefficients in the question, and you will see how this is applicable in a moment.
We apply the usual tricks with group addition to determine the sequence.
Given a solution $(x,y)$, divide throughout by $x^3$ to get $ 7 (\frac{y}{x} ) ^3 = ( \frac{1}{x} ) ^3 + 1 $, so observe that $(\frac{1}{x}, \frac{y}{x} )$ is also a solution.
Given a solution $(x,y)$, we can construct the line that passes through $(x,y)$ and $(\frac{4}{5}, \frac{3}{5} )$ which intersects the cubic again at $(x', y')$.
Furthermore, if $(x,y)$ is a rational point, then so is $(x', y')$. (Prove this.)
We combine this with the previous observation and send $(x,y)$ (to $(x',y')$ and then ) to $(\frac{1}{x'} , \frac{y'}{x'} ) $.
Claim: $(x,y)$ and $(\frac{1}{x'} , \frac{y'}{x'} )$ satisfy $4x + \frac{5}{x'} = 21 y \frac{y'}{x'}$.
See Zhaohui's solution for a beautiful proof that if the cubic intersects a line at 3 points, then $x_1x_2x_3 + 1 = 7 y_1y_2y_3$.
Applying it to the 3 points on the line, we get $\frac{4}{5} x x' + 1 = 7 \times \frac{3}{5} y y'$, and hence the claim follows.
Corollary: This shows that we send $ a_n : (x_n, y_n) $ to $a_{n+1} : (x_{n+1}, y_{n+1}) $ via the above description.
Since the starting point $a_1$ is rational, hence all of these are rational points.
So $ x_n \in \mathbb{Q}$ and $ a_n \in \mathbb{Q}$.
Notes
*
*Someone who is more familiar with elliptic curves might be able to manipulate the equations better. (IE See Zhaohui's solution)
*This approaches illustrates the theory behind the question, and why the question isn't just "magic".
*If one knew the fact that Zhaohui showed, then this approach would have felt natural. My guess is that similar questions like this can be dealt with in a similar way, especially for quadratic (and cubic) terms. Previously, I mainly dealt with similar questions previously via heavy-handed induction + guessing what the sequence is (EG Zhaohui's formulas at the end). I'm excited to apply this to future questions.
*For $a_1$, the line is actually tangential at $a_1$. So the 3rd point of intersection is $a_1$ (which explain why $a_1, a_2$ are related via $ (\frac{1}{x}, \frac{y}{x} )$.
Uncompleted attempt at proof: The line that passes through $(x,y)$ and $(\frac{4}{5}, \frac{3}{5} )$ is $ (Y' - y) ( \frac{4}{5} - x ) = (X' - x) ( \frac{3}{5} - y ) $, or that $Y' = X'\frac{5y-3}{5x-4} + \frac{3x-4y}{5x-4} $.
Substituting this into $7Y^3 = X^3 + 1$, and applying vieta's formula to find the sum of roots, we get that $\frac{4}{5} + x + x' = \frac{3 (\frac{5y-3}{5x-4})^2 (\frac{3x-4y}{5x-4}) } {1 - (\frac{5y-3}{5x-4})^3 } $, so $ x' = -\frac{625 x^4 - 1000 x^3 - 625 x y^3 + 675 x y + 370 x + 1000 y^3 - 900 y^2 - 148}{5 (5 x - 5 y - 1) (25 x^2 + 25 x y - 55 x + 25 y^2 - 50 y + 37))}$.
I'm unwilling to continue this tedious calculations, but one can find $y'$ and then verify that the equation holds.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the sequence $(a_n)$ is Cauchy and find the limit. Let us define a sequence $(a_n)$ as follows:
$$a_1 = 1, a_2 = 2 \text{ and } a_{n} = \frac14 a_{n-2} + \frac34 a_{n-1}$$
Prove that the sequence $(a_n)$ is Cauchy and find the limit.
I have proved that the sequence $(a_n)$ is Cauchy. But unable to find the limit. I have observed that the sequence $(a_n)$ is decreasing for $n \ge 2$.
| Given that $a_1=1$ and $a_2=2$ such that $\displaystyle a_n=\frac{1}{4}a_{n-2}+\frac{3}{4}a_{n-1}$
for $n\geq3$
Now $\displaystyle a_{n}-a_{n-1}=-\frac{1}{4}\left(a_{n-1}-a_{n-2}\right)=\left(-\frac{1}{4}\right)^2\left(a_{n-2}-a_{n-3}\right)$
$\displaystyle \dots=\left(-\frac{1}{4}\right)^{n-2}(a_2-a_1)=\left(-\frac{1}{4}\right)^{n-2}$
So $\displaystyle \sum_{n=2}^k(a_{n}-a_{n-1})=\sum_{n=2}^k\left(-\frac{1}{4}\right)^{n-2}\Rightarrow a_k=1+\sum_{n=2}^k\left(-\frac{1}{4}\right)^{n-2}$
Now take limit we will get $\displaystyle \lim_{k\to\infty}a_k=1+\sum_{n=2}^{\infty}\left(-\frac{1}{4}\right)^{n-2}=1+\frac{4}{5}=\frac{9}{5}$
| {
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Solving an absolute value inequality with fractions I'm having a hard time figuring out this inequality: $\bigg|\dfrac{x-4}{x+5}\bigg| \le 4$
I use one of the absolute value properties, which results in: $-4 \leq \dfrac{x-4}{x+5} \leq 4$.
From there, I get $x \geq -\frac{16}{5}$ and $x \geq -8$, yet when I look at Wolfram alpha, the answer is $[-\frac{16}{5}, \infty)$ and $(-\infty, -8]$, which makes sense. What am I missing?
These are my steps to get $x \ge -8$:
\begin{align*}
&\frac{x - 4}{x + 5} \le 4 \\
&x - 4 \le 4(x + 5) \\
&x - 4 \le 4x + 20 \\
&x \le 4x + 24 \\
&-3x \le 24 \\
&x \ge -8.
\end{align*}
| From your steps, a couple of red flags jump out at me. First, as I predicted in my comment, it appears you have multiplied both sides by $x + 5$ without worrying whether $x + 5$ is positive or negative (it can't be zero, for obvious reasons!). Second, you are ignoring the other inequality $-4 \le \frac{x - 4}{x + 5}$, presumably because you thought that this simply resulted in $x \ge -\frac{16}{5}$, though I'm guessing that a similar error was committed there.
First, let me say that the best way to do this is to follow Macavity's suggestion. Start by squaring both sides. If you know that $a$ and $b$ are both non-negative, then $a \le b$ is equivalent to $a^2 \le b^2$ (Why? Because $x^2$ is a strictly increasing function on $[0, \infty)$, so it respects the order. Also, $\sqrt{x}$ is strictly increasing too.) Therefore,
$$\left|\dfrac{x-4}{x+5}\right| \le 4 \iff \left|\dfrac{x-4}{x+5}\right|^2 \le 4^2 \iff \frac{(x - 4)^2}{(x + 5)^2} \le 16.$$
The advantage now is that we can multiply both sides by $(x + 5)^2$, safe in the knowledge that, assuming $x \neq -5$ so that the expression is well-defined, we are multiplying both sides by a positive number, so we need not change the sign. This leads to a quadratic that can be factored (as Macavity does in the comment), and can be solved with cases from there (if the product is positive, then either both factors are positive, or both factors are negative).
However, one can proceed as you have, but we just need to be more cautious.
It's a good idea to consider $\frac{x - 4}{x + 5} \le 4$ and $\frac{x - 4}{x + 5} \ge -4$ separately, taking the intersections at the end. Let's consider the former first.
We wish to multiply both sides by $x + 5$. Let us first consider the case that $x + 5 > 0$. Then, under this assumption,
$$\frac{x - 4}{x + 5} \le 4 \iff x - 4 \le 4(x + 5) \iff x \ge -8,$$
as your working showed. But, this is all under the overarching assumption that $x > -5$. So, the only valid solutions in this case satisfy both $x > -5$ and $x \ge -8$, which is to say $x > -5$.
Next, consider the case $x + 5 < 0$. Then
$$\frac{x - 4}{x + 5} \le 4 \iff x - 4 \ge 4(x + 5) \iff x \le -8.$$
So, our valid solutions in this case satisfy both $x < -5$ and $x \le -8$, which is true precisely when $x \le -8$. Thus, the entire solution to $\frac{x - 4}{x + 5} \le 4$ is the union of these intervals, i.e.
$$(-\infty, -8] \cup (-5, \infty).$$
I'll leave it to you, but the next step would be to repeat this analysis (taking into account cases as we have done) for $-4 \le \frac{x - 4}{x + 5}$. You should get another union of intervals, as above. The full solution to $\left|\frac{x-4}{x+5}\right| \le 4$ will be the intersection of the two sets.
| {
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Expressing a solid in spherical coordinates I am trying to solve the following question:
The volume of the solid $E$ can be represented as
$$\int_{-3}^3 \int_0^{\sqrt{9-x^2}} 3 - \sqrt{x^2+y^2} dydx$$
Describe the solid in spherical coordinates.
I graphed the solid and it looks like this:
So clearly $0 \le \theta \le \pi$ and $0 \le \phi \le \frac{\pi}{2}.$ Also, for each $(\theta, \phi)$ the value of $\rho$ ranges from $0$ until the surface of the cone. The cone is $z = 3-\sqrt{x^2+y^2}$. Using the substitutions
\begin{align*}
x &= \rho \cos \theta \sin \phi\\
y &= \rho \sin \theta \sin \phi\\
z &= \rho \cos \phi
\end{align*}
we get
\begin{align*}
\rho \cos \phi &= 3 - \sqrt{\rho^2 \sin^2 \phi}\\
\rho &= \frac {3}{\cos \phi + \sin \phi}
\end{align*}
So my answer is
$$E_{\text{spherical}} = \{(\rho, \theta, \phi)| 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{2}, 0 \le \rho \le \frac {3}{\cos \phi + \sin \phi} \}$$
However, the correct answer is
$$E_{\text{spherical}} = \{(\rho, \theta, \phi)| 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{4}, 0 \le \rho \le 3 \csc\phi \}$$
Did I make any mistake?
Thanks!
| The solid is half of an inverted cone with vertex at $(0, 0, 3)$ and above $z = 0$.
Equation of the surface of the cone is $ ~\sqrt{x^2+y^2} = 3 - z$
That translates to $ \rho \sin\phi = 3 - \rho \cos\phi \implies \rho = \frac{3}{\cos\phi + \sin\phi}$
Please note that $0 \leq \phi \leq \pi/2$ as we are above $z = 0$.
Also given bounds of $x$ and $y$, you can see that the projection of the solid in xy-plane is in the first and the second quadrant. That leads to $0 \leq \theta \leq \pi$.
$E_{\text{spherical}} = \{(\rho, \theta, \phi)| 0 \le \rho \le \frac{3}{ \cos\phi + \sin\phi}, 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{2} \}$
But to evaluate the volume, it is easier to use $ ~ x = \rho \cos\theta\sin\phi, y = \rho \sin\theta \sin\phi, z = 3 - \rho \cos\phi$
That translates the cone to $ ~ \phi = \frac{\pi}{4}$
$\rho$ is bound by the plane $z = 3 - \rho \cos\phi = 0 \implies \rho = 3 \sec\phi$
So the integral to find volume in spherical coordinates should be,
$ \displaystyle \int_0^{\pi} \int_0^{\pi/4} \int_0^{3\sec\phi} ~ \rho^2 \sin\phi ~ d\rho ~ d\phi ~ d\theta$
| {
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Do I have the right bounds and function for this integral? Find the integral of function $f(x,y,z)=(x^2+y^2+z^2)^{3/2}$ inside the sphere $(z-2)^2+x^2+y^2=4$
My approach: by changing it to spherical coordinates, we have $0\le\rho\le2\cos\varphi$ $(0\le\theta\le2\pi,\ 0\le\varphi\le\frac\pi2)$, and the function becomes $f(\rho,\theta,\varphi)=ρ^3$, which, when multplied by the Jacobian, $ρ^2\sin\varphi$, becomes $ρ^5\sin\varphi$. In other words, we are integrating this function in the above $\rho,\theta,\varphi$ bounds. Is this correct？Why would Wolfram alpha's calculator give a much larger value when integrated in Cartesian coordinates?
| The region of integration is the sphere of radius $2$ centered at $(0,0,2)$
So in spherical coordinates, the change of variables is
$x = r \sin \theta \cos \phi$
$y = r \sin \theta \sin \phi$
$z = 2 + r \cos \theta$
The differential volume is $ r^2 \sin \theta dr d\theta d\phi $ and the limits of integration are $ r \in [0, 2], \theta \in [0, \pi] , \phi \in [0, 2 \pi] $
The integrand is $(x^2 + y^2 + z^2)^{\dfrac{3}{2}} = (4 + 4 r \cos \theta + r^2)^{\dfrac{3}{2} } $
Hence, the required integral is
$ I = \displaystyle \int_{0}^{2} \int_0^\pi \int_0^{2 \pi} (4 + 4 r \cos \theta + r^2)^{\dfrac{3}{2} } r^2 \sin \theta d \phi d\theta d r $
Since the integrand does not depend on $\phi$, this becomes
$ I = 2 \pi \displaystyle \int_{0}^{2} \int_0^\pi (4 + 4 r \cos \theta + r^2)^{\dfrac{3}{2} } r^2 \sin \theta d\theta d r $
| {
"language": "en",
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How to separate $ \prod_{k=1}^{\infty} \frac{(16k-15)^{\frac{1}{16k-15}}}{(16k-1)^{\frac{1}{16k-1}}} $ from Log-gamma function. Using the Fourier series expansion for $0 < z < 1$, one has:
$$ \ln (\Gamma(z)) = (\tfrac{1}{2}-z)(\gamma+\ln(2)) + (1-z) \ln(\pi)-\frac{1}{2}\ln( \sin(\pi z)) + \frac{1}{\pi} \sum_{n=1}^{\infty} \frac{\sin(2 \pi n z)}{n} \ln(n) .$$
Now taking the general form $$ \prod_{k=1}^{\infty} \frac{(\alpha k-(\alpha-1))^{\frac{1}{\alpha k-(\alpha -1)}}}{(\alpha k-1))^{\frac{1}{\alpha k-1}}} .$$
I have been able to isolate the general form from $\alpha = 2$ to $8$ but I have become stumped upon taking the next power of $2$ and attempting $16$.
For example, $$ \prod_{k=1}^{\infty} \frac{(4k-3)^{\frac{1}{4k-3}}}{(4k-1)^{\frac{1}{4k-1}}} = \left[\frac{ \Gamma(\frac{1}{4}) }{e^{\frac{\gamma}{4}} 2^{\frac{1}{2}} \pi^{\frac{3}{4}} }\right]^{\pi} ,
\\
\prod_{k=1}^{\infty} \frac{(8k-7)^{\frac{1}{8k-7}}}{(8k-1)^{\frac{1}{8k-1}}} = \left[\frac{\Gamma^{\sqrt{2}}(\frac{1}{8}) \: \Gamma^{1-\frac{1}{\sqrt{2}}} (\frac{1}{4}) \: \sin^{\frac{\sqrt{2}}{2}}(\frac{\pi}{8})}{e^{\frac{\gamma}{4}(1 + \sqrt{2})} \: 2^{\frac{1}{2}(1+\frac{1}{\sqrt{2}})} \: \pi^{\frac{1}{2}(\sqrt{2} + \frac{3}{2} )}}\right]^{\frac{\pi}{2} } .$$
If anyone has any tips or pointers, I would greatly appreciate it.
P.S. Just a heads up; In their product form they have extremely slow convergence.
| Let $$f(z)=\pi\left(\ln\Gamma(z)+\left(z-\frac{1}{2}\right)(\gamma+\ln 2)+(z-1)\ln\pi+\frac{1}{2}\ln\sin(\pi z)\right)=\sum_{k=1}^{\infty} \frac{\sin(2 \pi kz)}{k} \ln k$$
Then assuming $\alpha\in\mathbb{N}$ and $\alpha>2$,
$$\ln\prod_{k=1}^{\infty} \frac{(\alpha k-\alpha+1)^{\frac{1}{\alpha k-(\alpha -1)}}}{(\alpha k-1)^{\frac{1}{\alpha k-1}}}=\sum_{k=1}^\infty\left(\frac{\ln(\alpha k-\alpha+1)}{\alpha k-\alpha+1}-\frac{\ln(\alpha k-1)}{\alpha k-1}\right)=\sum_{k=1}^\infty s(k)\frac{\ln k}{k}$$
where
$$s(k)=\begin{cases}
1 & k\equiv 1\pmod{\alpha}\\
-1 & k\equiv -1\pmod{\alpha}\\
0 & \text{otherwise}
\end{cases}$$
We can express $s(k)$ as a sum of $\sin$ functions using the discrete fourier transform:
$$s(k)=\frac{1}{\alpha}\sum_{n=1}^{\alpha-1}s_n\sin\left(\frac{2\pi nk}{\alpha}\right)$$
where
$$s_n=\sum_{k=1}^{\alpha-1}s(k)\sin\left(\frac{2\pi nk}{\alpha}\right)=\sin\left(\frac{2\pi n}{\alpha}\right)-\sin\left(\frac{2\pi n(\alpha-1)}{\alpha}\right)=2\sin\left(\frac{2\pi n}{\alpha}\right)$$
Hence,
$$\sum_{k=1}^\infty s(k)\frac{\ln k}{k}=\frac{2}{\alpha}\sum_{n=1}^{\alpha-1}\sin\left(\frac{2\pi n}{\alpha}\right)\sum_{k=1}^\infty\sin\left(\frac{2\pi nk}{\alpha}\right)\frac{\ln k}{k}=\frac{2}{\alpha}\sum_{n=1}^{\alpha-1}\sin\left(\frac{2\pi n}{\alpha}\right)f\left(\frac{n}{\alpha}\right)$$
Constant terms in $f$ are annihilated in this formula, and $$\sin(2 x)\ln(\sin(x))=-\sin(2(\pi-x))\ln\sin(\pi-x),$$
so the $\ln\sin$ terms are annihilated as well to give
$$\frac{2\pi}{\alpha}\sum_{n=1}^{\alpha-1}\sin\left(\frac{2\pi n}{\alpha}\right)\left(\ln\Gamma\left(\frac{n}{\alpha}\right)+\left(\frac{n}{\alpha}\right)(\gamma+\ln 2\pi)\right)$$
Using
$$\sum_{n=1}^{\alpha-1}n\sin\left(\frac{2\pi n}{\alpha}\right)=-\frac{\alpha}{2}\cot\left(\frac{\pi}{\alpha}\right)$$
which comes from taking the imaginary part of $\sum_{n=0}^{\alpha-1} nz^n=\frac{2z}{(1-z)^2} +\frac{(\alpha-1)z}{1-z}$ with $z=\exp\left(\frac{2\pi i}{\alpha}\right)$, we get
$$\frac{2\pi}{\alpha}\sum_{n=1}^{\alpha-1}\sin\left(\frac{2\pi n}{\alpha}\right)\left(\frac{n}{\alpha}\right)(\gamma+\ln 2\pi)=-\frac{\pi}{\alpha}(\gamma+\ln 2\pi)\cot\left(\frac{\pi}{\alpha}\right)$$
so we can finally write
$$\prod_{k=1}^{\infty} \frac{(\alpha k-\alpha+1)^{\frac{1}{\alpha k-\alpha +1}}}{(\alpha k-1)^{\frac{1}{\alpha k-1}}}=\left(2\pi e^{\gamma}\right)^{-\frac{\pi}{\alpha}\cot\left(\frac{\pi}{\alpha}\right)}\prod_{n=1}^{\alpha-1}\Gamma\left(\frac{n}{\alpha}\right)^{\frac{2\pi}{\alpha}\sin\left(\frac{2\pi n}{\alpha}\right)}$$
| {
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Find the minimum value of the constant term.
Let $f(x)$ be a polynomial function with non-negative coefficients such that $f(1)=f’(1)=f’’(1)=f’’’(1)=1$. Find the minimum value of $f(0)$.
By Taylor’s formula, we can obtain
$$f(x)=1+(x-1)+\frac{1}{2!}(x-1)^2+\frac{1}{3!}(x-1)^3+\cdots+\frac{f^{(n)}(1)}{n!}(x-1)^n.$$
Hence $$f(0)=\frac{1}{2}-\frac{1}{6}+\sum_{k=4}^n\frac{f^{(k)}(1)}{k!}(-1)^k.$$
This will help?
| Motivation: First, consider $f(x) = a_0 + a_1 x + a_2x^2 + a_3 x^3 + a_4 x^4$.
Using $f(1) = f'(1) = f''(1) = f'''(1) = 1$,
we get $a_0 = \frac13 + a_4$.
Second, consider $f(x) = a_0 + a_1 x + a_2x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5$.
Using $f(1) = f'(1) = f''(1) = f'''(1) = 1$,
we get $a_0 = \frac13 + a_4 + 4a_5$.
Thus, we think we can eliminate $a_1, a_2, a_3$
from $f(1) = f'(1) = f''(1) = f'''(1) = 1$.
The operation $f(1) - f'(1) + \frac12 f''(1) - \frac16 f'''(1)$
can do this.
A solution without Taylor's theorem:
Let $f(x) = \sum_{k=0}^n a_k x^k$.
We have
\begin{align*}
f'(x) &= \sum_{k=1}^n k a_k x^{k - 1}, \\
f''(x) &= \sum_{k=2}^n k(k - 1) a_k x^{k - 2}, \\
f'''(x) &= \sum_{k=3}^n k(k - 1)(k - 2) a_k x^{k - 3}.
\end{align*}
Then we have
\begin{align*}
&f(x) - f'(x) + \frac12 f''(x) - \frac16 f'''(x)\\
=\,\, & a_0 + a_1(x - 1) + a_2(x - 1)^2 + a_3(x - 1)^3\\
&\quad + \sum_{k=4}^n a_k \left(x^k - k x^{k - 1} + \frac{k(k - 1)}{2}x^{k - 2} - \frac{k(k - 1)(k - 2)}{6}x^{k - 3}\right).
\end{align*}
Thus, we have
$$f(1) - f'(1) + \frac12 f''(1) - \frac16 f'''(1)
= a_0 - \sum_{k=4}^n \frac{(k - 1)(k - 2)(k - 3)}{6} a_k $$
which results in
$$a_0 = \frac13 + \sum_{k=4}^n \frac{(k - 1)(k - 2)(k - 3)}{6} a_k.$$
Thus, $f(0) = a_0 \ge \frac13$
with equality if $a_0 = \frac13$ and
$a_4 = a_5 = \cdots = a_n = 0$.
Using $f(x) = \frac13 + a_1 x + a_2 x^2 + a_3x^3$
and $f(1) = f'(1) = f''(1) = f'''(1) = 1$,
we have $a_1 = \frac12, a_2 = 0, a_3 = \frac16$.
Thus, $f(0) = \frac13$ is attained when $f(x) = \frac{1}{3} + \frac12 x + \frac16 x^3$.
| {
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Ellipse in the triangle Find the equation of an ellipse if its center is S(2,1) and the edges of a triangle PQR are tangent lines to this ellipse. P(0,0), Q(5,0), R(0,4).
My attempt: Let take a point on the line PQ. For example (m,0). Then we have an equation of a tangent line for this point: $(a_{11}m+a_1)x+(a_{12}m+a_2)y+(a_1m+a)=0$, where $a_{11}$ etc are coefficients of our ellipse: $a_{11}x^2+2a_{12}xy+a_{22}y^2+2a_1x+2a_2y+a=0$. Now if PQ: y=0, then $(a_{11}m+a_1)=0$, $a_{12}m+a_2=1$, $a_1m+a=0$.I've tried this method for other 2 lines PR and RQ and I got 11 equations (including equations of a center)! Is there a better solution to this problem?
| The triangle vertices are $P_1 (0,0), Q_1 (5, 0), R_1 (0, 4) $
The equation of the ellipse in matrix-vector form is
$ (r - C)^T Q (r - C) = 1 $
where $C = (2, 1) $ is the center , and $Q$ is a $2 \times 2$ symmetric matrix.
Drawing the triangle, we realize that the required ellipse is tangent to the $x$ axis, the $y$ axis, and the line $y = 5 - \frac{5}{4} x $.
Starting with the $x$ axis, we know that the gradient will be pointing in the $-j$ direction.
Now the gradient $g = 2 Q (r - C) $. If $r_1$ is the tangent point to the ellipse on the $x$ axis then we must have
$ Q(r_1 - C) = \alpha (-j) $
So that $(r_1 - C) = - \alpha Q^{-1} j $
Pluggin this into the ellipse equation gives us $\alpha = \dfrac{1}{\sqrt{ j^T Q^{-1} j }} $
Therefore, $ r_1 - C = - \dfrac{ Q^{-1} j }{\sqrt{ j^T Q^{-1} j } } $
Note that $r_1$ is a point on the $x$ axis, so its $y$ coordinate is zero, i.e. $j^T r_1 = 0 $. Hence
$ j^T (r_1 - C) = - j^T C = - C_y = - \sqrt{ j^T Q^{-1} j } $
Hence, $C_y = 1 = \sqrt{ j^T Q^{-1} j } = Q_{22}^{-1} $
Similar to the above reasoning, we find that $C_x = \sqrt{i^T Q^{-1} i} = Q_{11}^{-1} $
Hence, so far, $Q^{-1} =\begin{bmatrix} 4 && Q_{12}^{-1} \\ Q_{12}^{-1} && 1 \end{bmatrix} $
To find the last unknown, we use the the third condition, which is tangency with the line $y = 5 - \frac{5}{4} x $
The vector $QR = (-5, 4) $ so the unit normal vector to $QR$ is $n = \dfrac{1}{\sqrt{41}} (4, 5) $
Similar to the first tangency analysis with the $x$ axis, we have
$ r_3 - C = \gamma Q^{-1} n $
Plugging this into the ellipse equation results in $\gamma = \dfrac{1}{\sqrt{n^T Q^{-1} n }} $
Therefore, $ n^T (r_3 - C) = n^T (R_1 - C) = n^T (Q_1 - C) = \sqrt{ n^T Q^{-1} n } $
Now $ n^T (R_1 - C) = n^T ( (0,4) - (2, 1) ) = \dfrac{1}{\sqrt{41}} ((4, 5) \cdot (-2, 3) = \dfrac{7}{\sqrt{41}}$
And, $n^T Q^{-1} n = \dfrac{1}{41} ( 4 (4)^2 + 1 (5)^2 + 2 Q_{12}^{-1} (4)(5) )$
Hence $ 40 Q_{12}^{-1} = -40 $ From which $Q_{12}^{-1} = -1 $
Thus, $Q^{-1} = \begin{bmatrix} 4 && -1 \\ -1 && 1 \end{bmatrix} $
Inverting, we get $Q = \dfrac{1}{3} \begin{bmatrix} 1 && 1 \\ 1 && 4 \end{bmatrix} $
Plugging $Q$ in the ellipse equation and expanding gives
$ (x-2)^2 + 2 (x-2)(y-1) + 4 (y-1)^2 = 3 $
| {
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Find $a$ and $b$ such that $4(b^3-a^3)-3(b^4-a^4)=\frac{1}{2}$ and $b-a$ gets minimum. I was solving this problem that for $f(x)=12x^2-12x^3, 0<x<1$ that is probability density function we have $P(a<X<b)=\frac{1}{2}$. So we have
$$P(a<X<b)=\int_{a}^{b}f(x)dx=\frac{1}{2}$$
with calculation I reached to this: $4(b^3-a^3)-3(b^4-a^4)=\frac{1}{2}$. But I don't know how to find $a$ and $b$ such that $b-a$ becomes minimum.
| Let's rewrite our function in terms of $s = a + b$ and $d = b-a$. Then we have
$$
4(b^3 - a^3) + 3(b^4 - a^4) = \frac{d}{2}\left[d^2(2-3s) + 3(2-s)s^2\right] = \frac{1}{2}
$$
Now we can proceed by implicit differentiation. We can define $d(s)$ as satisfying the equation $d[d^2(2-3s) + 3(2-s)s^2] =1$. Differentiating with respect to $s$ gives
$$
\left[(2-3s)d^2+(2-s)s^2\right] d'(s)+ \left[d^2-s(4-3s)\right] = 0.
$$
When $d$ is minimized, $d'(s) = 0$, so we have at the minimum that $d^2 = s(4-3s)$. This can be used in the constraint equation to give a system of two equations in two unknowns:
$$
d^2 = s(4-3s)\\
2ds(3s^2 -6s + 4) = 1.
$$
Eliminating either variable then gives a nasty 8th order polynomial that can only be solved numerically, and doing so gives $d\approx 0.2955$. However, we can actually get a fairly good rough estimate of $d$ analytically. Since $d$ is less than $1$, $d^2$ is probably fairly small. $s$ probably isn't small, so we know from the first equation that $4 - 3s$ is small. That is, $s$ is a little less than $4/3$. Putting this into the second equation gives that $d$ should be a little more than $9/32 \approx 0.28$, which it is.
| {
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Write $x^k - y^k$ as the product of two factors. Write $x^k - y^k$ as the product of two factors.
$x^3 - y^3 = (x-y)(x^2+xy+y^2)$
$x^4 - y^4 = (x-y)(x^3+x^2y+xy^2+y^3)$
$x^5 - y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$
$x^6 - y^6 = (x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5)$
$x^k - y^k = (x-y)(x^{k-1}+\underbrace{x^{k-2}y + \cdots + xy^{k-2}}_{k-2} +y^{k-1})$
It's not exact what the middle terms should be, so I'm not sure how to write the expression for the second factor. I think I may need to use the parity of $k$ and some mention of mirroring, though mirroring sounds wrong.
| The sum you bracketed can be expressed as $$x^{k-2}y+\cdots +xy^{k-2} =\sum_{i=1}^{k-2}x^{k-2+1-i}y^i =\sum_{i=0}^{k-1}x^iy^{k-j-1} -x^{k-1}-y^{k-1}$$ For a formal proof: Notice that $$a^2-b^2 = a^2+ab-ab-b^2 = a(a+b)-b(a+b) = (a-b)(a+b)$$ so that our result holds for $n=2$.Now suppose that $$a^k -b^k = (a-b)(a^{k-1}+ba^{k-2}+b^2a^{k-3}+b^3a^{k-4} +\cdots +b^{k-3}a^2 +b^{k-2}a +b^{k-1})$$ for some $k\in \mathbf{N}$ with $k\geqslant 2.$ Then \begin{align*}
a^{k+1}-b^{k+1} & = aa^k-bb^k \\ &= aa^k-ba^k+ba^k-bb^k \\ &= a^k(a-b)+b(a^k-b^k) \\ &= a^k(a-b)+b\left((a-b)(a^{k-1}+ba^{k-2}+\cdots +b^{k-2}a +b^{k-1})\right) \hspace{8mm}\text{Inductive Hypothesis}\\ &= (a-b)(a^k +b(a^{k-1}+ba^{k-2}+\cdots +b^{k-2}a +b^{k-1}))\\ &= (a-b)(a^{k}+ba^{k-1}+b^2a^{k-2}+b^3a^{k-3} +\cdots +b^{k-2}a^2 +b^{k-1}a +b^{k}).
\end{align*}
Therefore, by the Principles of Mathematical Induction, we have that $$a^n -b^n = (a-b)(a^{n-1}+ba^{n-2}+b^2a^{n-3}+b^3a^{n-4} +\cdots +b^{n-3}a^2 +b^{n-2}a +b^{n-1})$$ for any $a$, $b\in \mathbf{R}$ and $n\in \mathbf{N}.\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{a}{b^2 + c}+\frac{b}{c^2 + a}+\frac{c}{a^2 + b} \ge \frac{3}{2}$ when $a+b+c=3$ I've tried solving Pham Kim Hung's famous inequality problem which he posted on aops in April 2007 for 2 days now. This is the problem: Show that $$\frac{a}{b^2 + c}+\frac{b}{c^2 + a}+\frac{c}{a^2 + b} \ge \frac{3}{2} \tag{*}$$
when $a, b, c > 0$ and $a+b+c=3$.
I've tried with many theorems and tricks (like AM-GM, C-S, etc.) but haven't really figured out a way.
The closest I could get was by solving a similar problem like this.
So, if you know the proof of (*) please share here. I'd be much thankful to you.
| $\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^4}{a^3b+a^3c^2}\geq\frac{(a^2+b^2+c^2)^2}{\frac{1}{3}(a^2+b^2+c^2)^2+\frac{1}{3}(a^2+b^2+c^2)^2}=\frac{3}{2}$
(For numerator we use Titu's Lemma and simplify denominator :) )
Thanks to arqady
For more information vist https://artofproblemsolving.com/community/c6h401290p2234611
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the general value of $\theta$ for the inverse trigonometric function $\text { Find the general value of } \theta, \text { when } 9 \sec ^{4} \theta=16$
My work-
Given $\sec ^{4} \theta=\frac{16}{9}$
or, $\sec ^{4} \theta = \frac{(4)^{2}}{(3)^{2}}$
$\implies \sec ^{2} \theta=\frac{4}{3} \Longrightarrow \theta=\sec ^{-1}\left(\sqrt{\frac{4}{3}}\right)$
hence, $\theta=2 n \pi \pm \frac{\pi}{6}$.
I am following Hobson's Trigonometry and it doesn't have the answer or the solution provided, neither any examples regarding inverse trigonometric functions. I got the feedback that my approach is incorrect but I cannot see how to fix it so any hints on where I am going mistaken ?
P.S. Hobson mentions that $x = \sec^{-1}(y)$ is a multi-valued function.
| $$
\begin{aligned}
9 \sec ^{4} \theta &=16 \\
\sec ^{4} \theta &=\frac{16}{9} \\
\sec \theta &=\pm \frac{2}{\sqrt{3}} \\
\cos \theta &=\pm \frac{\sqrt{3}}{2} \\
\theta &=n \pi \pm \frac{\pi}{6}=\frac{(6 n \pm 1) \pi}{6},
\end{aligned}
$$
where $n \in \mathbb{Z}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4318209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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