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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$ Please help me for prove this inequality:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$
| There is a basic inequality you should know:
$$(a_{1}+a_{2}+\cdots+a_{n}) \cdot \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots +\frac{1}{a_{n}}\right) \ge n^2 $$
that is elementarily solved by Cauchy-Schwarz in a single row proof
$$\left(\sqrt{a_{1}}\frac{1}{\sqrt{a_{1}}}+\cdots +\sqrt{a_{n}}\frac{1}{\sqrt{a_{n}}}\right)^2=n^2\le (a_{1}+\cdots+a_{n}) \cdot \left(\frac{1}{a_{1}}+\cdots +\frac{1}{a_{n}}\right) $$
Q.E.D.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$? Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?
| $$\tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}{\left(\frac{2+3}{1-2\cdot 3}\right)}=\tan^{-1}(-1)=n\pi-\frac \pi 4,$$ where $n$ is any integer.
Now the principal value of $\tan^{-1}(x)$ lies in $[-\frac \pi 2, \frac \pi 2]$ precisely in $(0, \frac \pi 2)$ if finite $x>0$.
So, the principal value of $\tan^{-1}(2)+\tan^{-1}(3)$ will lie in $(0, \pi) $.
So, the principal value of $\tan^{-1}(2)+\tan^{-1}(3)$ will be $\frac {3\pi} 4$.
Interestingly, the principal value of $\tan^{-1}(-1)$ is $-\frac {\pi} 4$.
But the general values of $\tan^{-1}(2)+\tan^{-1}(3)$ and $\tan^{-1}(-1)$ are same.
Alternatively, $$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}{\left(\frac{1+2+3-1\cdot 2\cdot 3}{1-1\cdot 2- 2\cdot 3 -3\cdot 1}\right)}=\tan^{-1}(0)=m\pi$$, where $m$ is any integer.
Now the principal value of $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)$ will lie in $(0 ,\frac {3\pi} 2)$ which is $\pi$.
The principal value of $\tan^{-1}(0)$ is $0\neq \pi$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it possible to take the absolute value of both sides of an equation? I have a problem that says: Suppose $3x^2+bx+7 > 0$ for every number $x$, Show that $|b|<2\sqrt21$.
Since the quadratic is greater than 0, I assume that there are no real solutions since
$y = 3x^2+bx+7$, and $3x^2+bx+7 > 0$, $y > 0$
since $y>0$ there are no x-intercepts. I would use the discriminant $b^2-4ac<0$.
I now have $b^2-4(3)(7)<0$
$b^2-84<0$
$b^2<84$
$b<\pm\sqrt{84}$
Now how do I change $b$ to $|b|$? Can I take the absolute value of both sides of the equation or is there a proper way to do this?
| From $b^2\lt 84$, you cannot conclude that $b\lt \pm\sqrt{84}$, whatever that may mean. It cannot mean that $b\lt \sqrt{84}$ or $b\lt -\sqrt{84}$, since $-100$ is less than each of $\sqrt{84}$ and $-\sqrt{84}$.
What you probably intend to say is that $b^2\lt 84$ iff $-\sqrt{84}\lt b\lt \sqrt{84}$. And we can rewrite this double inequality as $|b|\lt\sqrt{84}$.
More simply, note that for any $b$, we have $\sqrt{b^2}=|b|$. To check this is true, verify it holds when $b\ge 0$ and when $b\lt 0$.
So we can conclude directly from $b^2\lt 84$ that $|b|\lt \sqrt{84}$. And $\sqrt{84}=2\sqrt{21}$.
| {
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Sum and Product of Infinite Radicals How to compute $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}$$ & $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}$$?
I understand that $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}=(1^{1/2})(2^{1/4})(3^{1/8})\cdots$$
and
$$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}=(1+(2+(3+(\cdots))^{1/8})^{1/4})^{1/2} $$
How to Proceed further?
| About
$$ c=\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\ldots}}}} $$
we may notice that for any $n\geq 1$ the functions $f_n(x)=\sqrt{n+x}$ are contractions of $[\varphi,+\infty)$ with Lipschitz constant $\leq\frac{1}{2\sqrt{n+1}}$. Additionally, for any $n\geq 2$ we have
$$ \sqrt{n+\sqrt{n+1+\sqrt{n+2+\sqrt{n+3+\ldots}}}}<\sqrt{n+\sqrt{n^2+\sqrt{n^4+\sqrt{n^8+\ldots}}}}=\varphi\sqrt{n}$$
so
$$ \left|c-\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4}}}}\right|\leq\frac{\varphi}{2^4\sqrt{4!}} $$
and, in general, the "truncations" of the nested radical defining $c$ converge extremely fast to $c$:
$$ \left|c-\sqrt{1+\sqrt{2+\sqrt{\ldots+\sqrt{n}}}}\right|\leq\frac{\varphi}{2^n \sqrt{n!}}. $$
By considering $n=10$ we already get five correct figures: $c\approx 1.75793$.
| {
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Explain why 67 is prime based on the fact that order of 2 mod 67 is 66 Without using the fact that 67 is prime, show that the order of 2 mod 67 is 66. Explain why this result proves that 67 is prime
What I understand:
*
*The order of 2 in $\mathbb{Z}_{67}$(or mod $67$) $ = 66$ means that $66$ is the smallest power $2^x$ such that $2^x \equiv 1$ mod 67
*Lucas primality test states if we can find $a$ such that $a$ has order $n-1$ mod $n$ then $n$ is prime. Here the question states that $a = 2$ has order $67-1=66$
*This result proves $67$ is prime by Lucas test
Now the part I don't get is how can you show the order of $2$ mod $67$ is indeed $66$?
| $2^6\equiv64\pmod {67} \implies 2^6\equiv -3 \pmod {67}\implies {(2^6)}^{11}\equiv -3^{11}\pmod {67}$
Now, $3^4\equiv 14\pmod {67}\implies 3^8\equiv 196\pmod {67}\equiv{-5}\pmod {67}$
$\implies 3^{11}=3^8.3^3\equiv -135\pmod {67}$
Therefore, $2^{66}\pmod {67}\equiv-3^{11}\pmod {67}\equiv 135\pmod {67}\equiv 1\pmod {67}$
order of $2$ must divide $66=2*3*11$
Factors of $66$ are: $2,3,6,11,22,33,66$
Since, $2^2=4\not\equiv1\pmod {67}$, $2^3=8\not\equiv 1 \pmod {67}$,$2^6=64\equiv{-3}\pmod {67}$, $2^{11}=38\not\equiv 1\pmod {67}$, $2^{22}={(2^6)}^3.2^4\equiv{-30}\pmod {67}$ and
$2^{33}=2^{22}.2^{11}\equiv{-1}\pmod {67}$
Hence, order of $2\pmod {67}$ is $66$
| {
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For this non-linear differential equation: $ xy' = y + ax \sqrt{x^2+y^2} $, where $ x>0 $ and $ a>0$ is a constant This is a 3 part question (I seem to have found (a), but I am not confident about (b) and thus (c) as well):
(a) if y is a solution, show that v:= $yx^{-1}$ satisfies the differential equation:
$v'=a \sqrt{1+v^2}$
My solution is as follows:
$ v=\dfrac{y}{x}$
$v'=\dfrac{d}{dx}(\dfrac{y}{x})$
$ v'= \dfrac{y'}{x}-\dfrac{y}{x^2}$
Now substituting y', I have:
$v' = \dfrac{1}{x}(\dfrac{y}{x}+a\sqrt{x^2+y^2})-\dfrac{y}{x^2} $
Distributing the $\dfrac{1}{x}$, factoring out $x^2$ from the square root, and then substituting for v I get:
$v' = \dfrac{y}{x^2}+\dfrac{ax\sqrt{1+v^2})}{x}-\dfrac{y}{x^2} $
Cancelling gives:
$v'=a \sqrt{1+v^2}$
Does this sufficiently answer this question? (Note: This is a 2 point question)
(b) Use the differential equation in part (a) to get the general solution to:
$ xy' = y + ax \sqrt{x^2+y^2} $
This is a 3 point question and where I am having a little trouble, my strategy has been as follows:
Since $ v = \dfrac{y}{x}$, then:
$\int v'\,dv = a\int \sqrt{1+v^2}\,dv = \dfrac{y}{x}$
Substituting $v = sinhx$, (I am not sure if I can do this, or have used this correctly) I get:
$v= a\int coshx\, dx = asinhx+c=\dfrac{y}{x}$
Therefore, $y = axsinhx+c$
Now since $y'= \dfrac{y}{x}+a\sqrt{x^2 + y^2}$
substituting $y = axsinhx+c$, and pulling the constant out of the square root and simply adding it to the end, I have just observed and tried out that when $a=1$,
$y = (1)xsinhx+c$ seems to be a valid solution.
But I just guessed $a$, so I need a bit of guidance to give a solid answer to this.
(c) Find the particular solution of $ xy' = y + ax \sqrt{x^2+y^2} $ with $y(1)=0$
Perhaps trivially:
$0 = (1)sinh(1) + c$, gives $c=-sinh(1)$
Thus my final answer is: $y = (1)xsinhx-sinh(1)$
Any help/guidance on this would be greatly appreciated!
| After getting differential eqn $$\frac{dv}{dx}=a\sqrt{1+v^2}$$ $$\implies \int\frac{dv}{\sqrt{1+v^2}} =\int a dx$$ $$\implies \ln(v+\sqrt{1+v^2})=ax+c$$
Putting $v=\frac{y}{x}$ gives
$$\ln(\frac{y+\sqrt{x^2+y^2}}{x})=ax+c$$
Since $y(1)=0\implies \ln(1)=a+c\implies c=-a$ (as $\ln(1 )=0$)
Thus, particular solution is $$ \ln(\frac{y+\sqrt{x^2+y^2}}{x})=a(x-1)$$
EDIT:
If you want $y$ explicitly, then,
$$\frac{y+\sqrt{x^2+y^2}}{x}=e^{a(x-1)}\implies {y+\sqrt{x^2+y^2}}=xe^{a(x-1)}$$
$$\sqrt{x^2+y^2}=xe^{a(x-1)}-y$$
Squaring both sides,
$$x^2+y^2=x^2e^{2a(x-1)}+y^2-2xye^{a(x-1)}$$
$$\implies 2xye^{a(x-1)}=x^2(e^{2a(x-1)}-1)$$
$$\implies y=x(\frac{e^{a(x-1)}-e^{-a(x-1)}}{2})$$
But squaring usually changes the domain, so be careful.
| {
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Proof that $\sqrt[3]{3}$ is irrational Proof that $\sqrt[3]{3}$ (Just need a check)
Let $\sqrt[3]{3}= \frac{a}{b}$ both a,b are integers of course. $\Rightarrow 3=\frac{a^{3}}{b^{3}}$ $\Rightarrow$ $3b^{3}=a^{3}$ $\Rightarrow$ 3b=a $\Rightarrow$ $3b^{3}=27b^{3}$ and this is a contradiction because the cubing function is a 1-1 function. Does this work?
| Supposing $\sqrt[3]{3}$ is a rational solution of an equation, we can assume $x=\sqrt[3]{3}$ and $x^3=3$.
But $x^3=3 \Rightarrow x^3-3 =0$. The rational-root theorem tells us that for a polynomial $f(x) = a_nx^n + a_{n−1}x^{n−1} + \ldots + a_1x + a_0$ all of whose coefficients ($a_n$ through $a_0$) are integers, the real rational roots are always of the form $\frac{p}{q}$, where $p$ is an integer divisor of $a_0$ and $q$ is an integer divisor of $a_n$. In this equation the possible rational roots could be $3,-3,1,-1$ and there wasn't any $\sqrt[3]{3}$. So $\sqrt[3]{3}$ is an irrational number.
| {
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"timestamp": "2023-03-29T00:00:00",
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probability word problems with marbles Select 3 marbles at random from a jar with 6-blue, 4-green, and 3-red. What is probability that
A.All are green?
B.What is probability of selecting 2-blue and 1-red?
C.What is probability of selecting in exact order 1 blue, 1 green, and 1 red?
| There are $13$ balls, so $\dbinom{13}{3}$ ways to choose $3$ balls. All these ways are equally likely. There are $\dbinom{4}{3}$ ways to choose $3$ green. Divide.
B: Same denominator. There are $\dbinom{6}{2}$ ways to choose $2$ blue. For each of these ways, there are $\dbinom{3}{1}$ ways to choose $1$ red, for a total of $\dbinom{6}{2}\dbinom{3}{1}$. Now divide.
C: The probability that the first is blue is $\dfrac{6}{13}$. Given this has happened, the probability the next is green is $\dfrac{4}{12}$. Given these two things have happened, the probability the last is red is $\dfrac{3}{11}$. Multiply.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/204280",
"timestamp": "2023-03-29T00:00:00",
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Central Projection of the Sphere
Given the parameterization of the unit sphere $x^2+y^2+z^2=1$ as
$x = \displaystyle\frac{u}{\sqrt{1+u^2+v^2}} $
$y = \displaystyle\frac{v}{\sqrt{1+u^2+v^2}} $
$z = \displaystyle\frac{1}{\sqrt{1+u^2+v^2}} $
Find $ds^2=dx^2+dy^2+dz^2$ and using the metric computer the area of
the hemisphere $z\geq0$
I got:
$dx = \displaystyle\frac{v^2+1}{(u^2+v^2+1)^{3/2}}du-\displaystyle\frac{uv}{(u^2+v^2+1)^{3/2}}$dv
$dy = -\displaystyle\frac{uv}{(u^2+v^2+1)^{3/2}}du+\displaystyle\frac{u^2+1}{(u^2+v^2+1)^{3/2}}dv$
$dz = \displaystyle\frac{-u}{(u^2+v^2+1)^{3/2}}du-\displaystyle\frac{v}{(u^2+v^2+1)^{3/2}}dv$
And
$dx^2+dy^2+dz^2= \displaystyle\frac{(v^4+2v^2+1+u^2+u^2v^2)du^2+(u^4+2u^2+1+v^2+u^2v^2)dv^2+(-2uv-2u^3v-2uv^3)dudv}{(u^2+v^2+1)^3}$
But I can't see an obvious simplification
| Factor your coefficients:
$$
v^4 + 2 v^2 + 1 + u^2 + u^2 v^2 = (1+v^2)(1+u^2+v^2)
$$
$$
u^4 + 2 u^2 + 1 + v^2 + u^2 v^2 = (1+u^2) (1+u^2+v^2)
$$
$$
(-2 u v-2 u^2 v-2 u v^3) = - 2u v(1+u^2+v^2)
$$
Then cancel common factors of the numerator and the denominator.
| {
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"timestamp": "2023-03-29T00:00:00",
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How $v=(-\sin\frac{\alpha}{2}\sin\frac{\beta}{2},\cos\frac{\alpha}{2}\sin\frac{\beta}{2},\sin\frac{\alpha}{2}\cos\frac{\beta}{2})$ is derived from... In the book Quarternion and Rotation Sequences, I can't seem to work out how the final equation (colored in $\color{red}{red}$) is derived from the original equation (colored in $\color{blue}{blue}$).
I did check using Wikipedia's list of trigonometric identities as my references as well as the book errata but to no avail.
I copy/paste the text that is giving me problem literally below:
Thus our tracking transformation has axis of rotation given by
$$\color{blue}{v=\left(k,\frac{k\sin\alpha}{\cos\alpha-1},\frac{k\sin\beta}{\cos\beta-1}\right).}$$
Notice that in this computation we determine only the direction of the axis of rotation.
Should we wish to obtain a specific vector as the axis of rotation we may,
for instance, choose $k = -1$, to obtain
$$v=\left(-1,\frac{\sin\alpha}{1-\cos\alpha},\frac{\sin\beta}{1-cos\beta}\right).$$
We note that by using the trigonometric identity
$$1-\cos\alpha=2\sin^{2}\frac{\alpha}{2}$$
we may write the following expression for the axis of the rotation
$$\color{red}{v=\left(-\sin\frac{\alpha}{2}\sin\frac{\beta}{2},\cos\frac{\alpha}{2}\sin\frac{\beta}{2},\sin\frac{\alpha}{2}\cos\frac{\beta}{2}\right)}$$
Anyone have any idea?
| Facts:
*
*$1-\cos{2\theta}=2\sin^2\frac{\theta}{2}$.
*$\sin{\theta}=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$, which we can rearrange to $\displaystyle\frac{\sin{\theta}}{2\sin\frac{\theta}{2}}=\cos\frac{\theta}{2}$.
Steps:
So, starting with $$\color{blue}{\vec{v}=\left(k,\frac{k\sin\alpha}{\cos\alpha-1},\frac{k\sin\beta}{\cos\beta-1}\right)}.$$
Since it doesn't matter what $k$ we use, pick $k=-1$ and this becomes
$$\vec{v}=\left(-1,\frac{\sin{\alpha}}{1-\cos{\alpha}},\frac{\sin{\beta}}{1-\cos{\beta}}\right).$$
We can use fact 1 to replace the denominators of the second and third components, obtaining
$$\vec{v}=\left(-1,\frac{\sin{\alpha}}{2\sin^2\frac{\alpha}{2}},\frac{\sin{\beta}}{2\sin^2\frac{\beta}{2}}\right).$$
Now multiply through by $\sin\frac{\alpha}{2}\sin\frac{\beta}{2}$ and we get
$$\vec{v}=\left(-\sin\frac{\alpha}{2}\sin\frac{\beta}{2},\frac{\sin\alpha\sin\frac{\beta}{2}}{2\sin\frac{\alpha}{2}},\frac{\sin{\beta}\sin\frac{\alpha}{2}}{2\sin\frac{\beta}{2}}\right).$$
Finally, we use fact 2 on the second and third terms.
$$\color{red}{\vec{v}=\left(-\sin\frac{\alpha}{2}\sin\frac{\beta}{2},\cos\frac{\alpha}{2}\sin\frac{\beta}{2},\cos\frac{\beta}{2}\sin\frac{\alpha}{2}\right)}.$$
| {
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Proving $\sum\limits_{i=1}^{n}{\frac{i}{2^i}}=2-\frac{n+2}{2^n}$ by induction.
Theorem (Principle of mathematical induction):
Let $G\subseteq \mathbb{N}$, suppose that
a. $1\in G$
b. if $n\in \mathbb{N}$ and $\{1,...,n\}\subseteq G$, then $n+1\in G$
Then $G=\mathbb{N}$
Proof by Induction
Prove that
$$\sum_{i=1}^{n}{\frac{i}{2^i}}=2-\frac{n+2}{2^n},\quad \quad \forall n\in \mathbb{N} \tag{1}$$
$\color{darkred}{\mbox{Proof:}}$
let $P(n)$ be the statement $\sum_{i=1}^{n}{\frac{i}{2^i}}=2-\frac{n+2}{2^n}$
$$\sum_{i=1}^{1}{\frac{i}{2^i}}=\frac{1}{2}=2-\frac{1+2}{2^1}\tag{Basis step $P(1)$}$$
Thus $P(1)$ is true
We assume that $P(n)$ is true and prove that $P(n+1)$ is true
$$\sum_{i=1}^{n+1}{\frac{i}{2^i}}\stackrel{?}{=}2-\frac{(n+1)+2}{2^{n+1}} $$
We know that
$$
\begin{align*}
\sum_{i=1}^{n+1}{\frac{i}{2^i}}&=\color{darkred}{\underbrace{\sum_{i=1}^{n}{\frac{i}{2^i}}}_\text{P(n) is true}}+\frac{n+1}{2^{n+1}} \tag{LHS} \\
&=\color{darkred}{2-\frac{n+2}{2^n}} + \frac{n+1}{2^{n+1}}\\
&=\frac{2^{n+1}-n-2}{2^n} + \frac{n+1}{2^{n+1}}\\
&=\frac{2\cdot 2^{n+1}-2n-4+n+1}{2^{n+1}}\\
&=\frac{2\cdot 2^{n+1}-n-3}{2^{n+1}}\tag{2}\\
\end{align*}
$$
And
$$
\begin{align*}
2-\frac{(n+1)+2}{2^{n+1}}&=\frac{2\cdot 2^{n+1}-n-3}{2^{n+1}} \tag{RHS}
\end{align*}
$$
So
$$\sum_{i=1}^{n+1}{\frac{i}{2^i}}\stackrel{}{=}2-\frac{(n+1)+2}{2^{n+1}} \tag{3}$$
Hence, if $P(n)$ is true, then $P(n+1)$ is true.
Therefore $\sum\limits_{i=1}^{n}{\frac{i}{2^i}}=2-\frac{n+2}{2^n},\quad \quad \forall n\in \mathbb{N} $
$\hspace{8cm}$ ${\Large ▫}$
My Question
$(i)$ does my solution make sense (is it correct?)
$(ii)$ If so, is this an correct approach to proving formulas by induction?
Can someone give me some hints/tips?
| Absolutely correct, this is it.
| {
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Proving that a set is not an ordered field Problem:
Let $S=\left \{ 0,1,2 \right \}$. How can someone prove that there is unique way of defining addition and multiplication such that $S$ is a field if $0$ of the set $S$ has the meaning (for any element $a$ in $S$: $0+a=a$), and $1$ in $S$ has the meaning (for any element $a$ in $S$: $1.a=a$)? Also, can $S$ be an ordered field?
| The axioms for a ring already imply that $0\cdot a=0$ for all $a\in S$, so you have almost all of the multiplication table:
$$\begin{array}{c|cc}
\cdot&0&1&2\\ \hline
0&0&0&0\\
1&0&1&2\\
2&0&2
\end{array}$$
And $2$ has to have a multiplicative inverse, so we must have $2\cdot 2=1$.
For addition we automatically have this much:
$$\begin{array}{c|cc}
+&0&1&2\\ \hline
0&0&1&2\\
1&1&&\\
2&2&
\end{array}$$
Now $1$ must have an additive inverse, so either $1+1=0$, or $1+2=0$. Suppose that $1+1=0$; then what is $1+2$? If $1+2=0$, then $1=1+0=1+(1+2)=(1+1)+2=0+2=2$, which is absurd. If $1+2=1$, then $0=1+1=1+(1+2)=(1+1)+2=0+2=2$, which is also absurd. Thus, $1+2=2$. But we know that $2$ has an additive inverse $-2$, even if we don’t yet know what it is, so $1=1+\big(1+(-2)\big)=(1+2)+(-2)=2+(-2)=0$, which is also impossible. Thus, $1+1$ cannot be $0$. It also can’t be $1$ (why not?), so it must be $2$, and we have
$$\begin{array}{c|cc}
+&0&1&2\\ \hline
0&0&1&2\\
1&1&2&\\
2&2&
\end{array}$$
Clearly $2$ must be $-1$, giving us
$$\begin{array}{c|cc}
+&0&1&2\\ \hline
0&0&1&2\\
1&1&2&0\\
2&2&0
\end{array}$$
and it’s not hard to check that $2+2$ can now only be $1$.
Added: By the way, that last bit can be shortened to practically nothing if you recall that up to isomorphism there is only one group of order $3$, $\Bbb Z/3\Bbb Z$: that gives you the addition table right away.
How to see that $S$ cannot be an ordered field depends on how you’ve defined ordered field. If you’ve defined it in terms of a positive cone, note that $1+1=2=-1$, so $1$ can’t be in the positive cone: it’s closed under addition and never contains a non-zero element and its additive inverse. Similarly, $(-1)+(-1)=2+2=1$, so $-1$ can’t be in the positive cone, either. But this is also impossible: one of them has to be in it.
If you’ve defined it in terms of an order relation, you can get essentially the same contradictions. If $0<1$, then $1=0+1<1+1=2$, $2=1+1<2+1=0$, so by transitivity $0<0$; OOPS! A similar problem arises if $1<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/210260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving trigonometric equations of the form $a\sin x + b\cos x = c$ Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$.
How do you solve this equation without using the method that moves $b\cos x$ to the right side and squaring left and right sides of the equation?
And how does solving $\sqrt{3}\sin x + \cos x = 2$ equal to solving $\sin (x+ \frac{\pi}{6}) = 1$
| The idea is to use the identity $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$. You have $a\sin x+b\cos x$, so you’d like to find an angle $\beta$ such that $\cos\beta=a$ and $\sin\beta=b$, for then you could write
$$a\sin x+b\cos x=\cos\beta\sin x+\sin\beta\cos x=\sin(x+\beta)\;.$$
The problem is that $\sin\beta$ and $\cos\beta$ must be between $-1$ and $1$, and $a$ and $b$ may not be in that range. Moreover, we know that $\sin^2\beta+\cos^2\beta$ must equal $1$, and there’s certainly no guarantee that $a^2+b^2=1$.
The trick is to scale everything by $\sqrt{a^2+b^2}$. Let $A=\dfrac{a}{\sqrt{a^2+b^2}}$ and $B=\dfrac{b}{\sqrt{a^2+b^2}}$; clearly $A^2+B^2=1$, so there is a unique angle $\beta$ such that $\cos\beta=A$, $\sin\beta=B$, and $0\le\beta<2\pi$. Then
$$\begin{align*}
a\sin x+b\cos x&=\sqrt{a^2+b^2}(A\sin x+B\cos x)\\
&=\sqrt{a^2+b^2}(\cos\beta\sin x+\sin\beta\cos x)\\
&=\sqrt{a^2+b^2}\sin(x+\beta)\;.
\end{align*}$$
If you originally wanted to solve the equation $a\sin x+b\cos x=c$, you can now reduce it to $$\sqrt{a^2+b^2}\sin(x+\beta)=c\;,$$ or $$\sin(x+\beta)=\frac{c}{\sqrt{a^2+b^2}}\;,$$ where the new constants $\sqrt{a^2+b^2}$ and $\beta$ can be computed from the given constants $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/213545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 0
} |
Solve three simultaneous equations with three unknowns $0 = 9a + 3b + c$
$2 = 25a + 5b + c$
$6 = 49a + 7b + c$
These are my 3 equations that have 3 unknowns. How do I solve for these unknowns?
| (1) $9a+3b+c=0$
(2) $25a+5b+c=2$
(3) $49a+7b+c=6$
Using (2)-(1) we have:
(4) $16a+2b=2$
Using (3)-(2) we have:
(5) $24a+2b=4$
Using (5)-(4) we have:
(6) $8a=2$
It follows that $a=\frac{1}{4}$, $b=-1$ and $c=\frac{3}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/216593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Urgent - Find the equation of the lines tangent to a circle Question: 'Find the equation of the lines from point $P(0,6)$ tangent to the circle $x^2+y^2=4x+4$.
So what I did firstly is rewrite it to the form $(x-2)^2 + y^2 = 8$, and I saw that point $P$ is not on the circle. I learned that the equation of the line tangent to the circle $x^2+y^2=r^2$ from the point $P(a,b)$ is $xa+yb=r^2$
$ xa+yb=4x+4$
$x.o+y.6=2x+2.0 +4$ (This is the step I don't understand)
$6y=2x+4$
$y=\dfrac{1}{3}x + \dfrac{2}{3}$
So basically, my question is: Why did the $4x+4$ change into the $2x+4$?
| The equation of any line passing through $(0,6)$ can be written as $\frac{y-6}{x-0}=m$ where $m$ is the gradient, So, $y=mx+6$
If the $(h,k)$ be the point of contact, then $k=mh+6$ and $h^2+k^2-4h-4=0$
Replacing $k$ in the 2nd equation, $h^2+(mh+6)^2-4h-4=0$
or $(1+m^2)h^2+2h(6m-2)+32=0$, it is a quadratic equation in $h,$
For tangency, both the root should be same, so, $4(6m-2)^2=4\cdot(1+m^2)32$
$(6m-2)^2=(1+m^2)32, (3m-1)^2=8(1+m^2) \implies m^2-6m-7=0$
So, $m=7,-1$
If $m=7,\frac{y-6}{x-0}=7, 7x-y+6=0$
If $m=-1,\frac{y-6}{x-0}=-1, x+y-6=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/219298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$ Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$. Please show me the way you find it.
The answer in my textbook is $f(x)=\frac{1+x^2+x^4}{x\cdot \sqrt{1-x^2}}$
| Note that $(x+\frac1x)^2=x^2+2+\frac1{x^2}$.
Hence it looks like $f(x)=x^2-2$ is a good candidate.
Of course, $\left|x+\frac1x\right|\ge2$ implies that we cannot say anything about $f(x)$ if $|x|<2$.
But for $|x|\ge 2$, we can find a real number $t$ such that $t^2-xt+1=0$ (and hence $t+\frac1t=x$), namely $t=\frac{x\pm \sqrt{ x^2-4}}2$, and then see that indeed $f(x)=f(t+\frac1t)=t^2+\frac1{t^2}=x^2-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/220912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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How to relate $2\sin(3\pi/8)-2\sin(7\pi/8)$ and $\csc(3\pi/8)$? Trying to simplify $2\sin(3\pi/8)-2\sin(7\pi/8)$ down to $\csc(3\pi/8)$. The two expressions have equal decimal approximations but I'm literally at my wit's end trying to relate them based on trigonometric identities.
| $$\sin(7 \pi/8) = \sin(\pi/2 + 3 \pi/8) = \cos(3 \pi/8)$$
\begin{align}
\dfrac{\sin(3 \pi/8) - \sin(7 \pi/8)}{\csc(3 \pi/8)} & = \sin(3 \pi/8) (\sin(3 \pi/8) - \sin(7 \pi/8))\\
& = \sin(3 \pi/8) (\sin(3 \pi/8) - \cos(3 \pi/8))\\
& = \sin^2(3 \pi/8) - \sin(3 \pi/8) \cos(3 \pi/8)\\
& = \underbrace{\dfrac{1 - \cos(3 \pi/4)}{2}}_{\sin^2(\theta) = \frac{1 - \cos(2 \theta)}2} - \underbrace{\dfrac{\sin(3 \pi/4)}2}_{\sin(\theta) \cos(\theta) = \frac{\sin(2 \theta)}2}\\
& = \dfrac{1 + \dfrac1{\sqrt{2}} - \dfrac1{\sqrt{2}}}2\\
& = \dfrac12
\end{align}
which is what we want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/222267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all natural numbers with the property that... Find all natural numbers with the property that when the first digit is moved to the end, the resulting number is $3.5$ times the original number.
| Suppose that our original number is $k+1$ digits long. Let $a$ denote the first digit of the number. We represent the number as
$$10^ka + b$$
where $b < 10^k$ is the remaining portion of our number. The condition is thus expressed as
$$3.5(10^ka + b) = 10b + a$$
Clearing fractions and rearranging, we end up with
$$(7\cdot 10^k - 2)a = 13b$$
Clearly $13\nmid a$ and we must therefore have
$$7\cdot 10^k - 2 \equiv 0 \pmod{13} \implies 10^k \equiv 4 \pmod{13}$$
We find $k = 5$ as the smallest solution to the congruence. The order of $10$ modulo $13$ is $6$, so we have $k\equiv 5\pmod6$.
Getting back to our former equation, we note that
$$a \ge 2 \implies \frac{14}{13}10^k - 2 > 10^k$$
which will contradict the restrictions on $b$. We then have $a = 1$. Therefore the numbers are of the form
$$10^k + \frac{7\cdot 10^k - 2}{13} = \frac{2\cdot 10^{k+1} - 2}{13}$$
where $k \equiv 5 \pmod6$. The order of $10$ modulo $13$ is $6$ so the fraction $\frac{1}{13}$ repeats with period $6$. Denote the repeating block of $13^{-1}$ as $r = 076923$. Our number is of the form
$$2\times \frac{10^{6n}- 1}{13} = 2\times\underbrace{rr\cdots r}_{n \text{times}}$$
where the latter expression denotes the number formed by concatenating $r$ with itself $n$ times for some natural number $n$. Putting it all together, our number is of the form $153846$ appended with itself an arbitrary number of times,
$$\left\{153846,\ 153846153846,\ 153846153846153846,\ \cdots\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/223928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Positive integers and the number of their digits Let $a$, $b$, $c$ be positive integers and $s(a)$, $s(b)$, $s(c)$ denote the number of their digits (when the integers are written in decimal form) respectively. If,
$s(a)+s(b)=a\qquad$
$a + b + s(c) = c\qquad$
and
$4 + s(a) + s(b) + s(c) = b \qquad$
then what would be the possible values of $a$, $b$, $c$?
| If $a$ is large, then (1) says that $b\approx10^a$; then (3) says that $c\approx10^{10^a}$, and then (2) can't work out. Thus $a$ can't be large. Then the same arguments applied to (2) and (3) show that $b$ also can't be large, and then the same argument applied just to (2) shows that $c$ can't be large, either. Thus there are only finitely many solutions.
To get rough bounds on the magnitudes, note that $s(x)\le1+x/10$. Thus from (1)
$$s(b)=a-s(a)\ge a-\left(1+\frac a{10}\right)=\frac9{10}a-1\;.$$
and hence
$$
b\ge10(s(b)-1)\ge9a-20\;.
$$
Then substituting into (3) yields
$$
\begin{align}
s(c)
&=b-4-s(a)-s(b)
\\
&\ge b-4-\left(1+\frac a{10}\right)-\left(1+\frac b{10}\right)
\\
&
=\frac9{10}b-\frac1{10}a-6
\end{align}$$
and hence
$$
c\ge10(s(c)-1)\ge9b-a-70\;.
$$
But (2) yields
$$c=a+b+s(c)\le a+b+1+\frac c{10}$$
and thus
$$c\le\frac{10}9(a+b+1)\;.$$
Together, this is
$$\frac{10}9(a+b+1)\ge9b-a-70$$
or
$$
71b\le19a+640\;,
$$
so
$$
71(9a-20)\le19a+640\;,
$$
or
$$
a\le\frac{103}{31}\lt4\;.
$$
Since $a=1$ doesn't work in (1), that leaves $a=2$ or $a=3$. That implies $s(b)=1$ or $s(b)=2$, thus $b\le99$ and thus
$$c\le\frac{10}9(a+b+1)\lt\frac{10}9(4+99+1)=\frac{1040}9\lt1000\;.$$
Thus $s(c)\le3$, and then (3) yields $b\le10$, then (2) yields $c\le16$ and thus $s(c)\le2$, then (3) yields $b\le9$ and thus $s(b)=1$, which by (1) implies $a=2$. Now (3) becomes $b=6+s(c)$; that leaves only $b=7$ and $b=8$, and in both cases (2) yields $c\ge10$ and thus $s(c)=2$; then finally (3) yields $b=8$ and (2) yields $c=12$.
Thus the solution $a=2$, $b=8$, $c=12$ that Gerry found is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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The positive integer solutions for $2^a+3^b=5^c$ What are the positive integer solutions to the equation
$$2^a + 3^b = 5^c$$
Of course $(1,\space 1, \space 1)$ is a solution.
| If $a=0$, then it is clear there are no solutions. If $b=0$, then we need $2^a + 1 = 5^c$. It is easy to show in this case $a=2,c=1$ is the only solution by showing that we need $2^{a-2}|c$. When $c=0$ there are obviously no solutions.
Suppose $a=1$. Then $2 + 3^b = 5^c$ only has the solution of $b=1,c=1$. To show this check modulo $275$ to deduce $c=1$ and thus $b=1$ is forced.
Now, suppose $a \ge 3$. Then remark that by checking modulo $4$ we need $b$ to be even so let $b = 2b'$. So let's solve $2^a + 3^{2b'} = 5^c$. Checking modulo $8$ we get $c$ is even so let $c = 2c'$. Then: $$2^a = (5^{c'} - 3^{b'})(5^{c'}+3^{b'})$$
We get $5^{c'} - 3^{b'} = 2^m, 5^{c'}+3^{b'} = 2^n$ for some $m,n$.
But then $2 \cdot 5^{c'} = 2^m + 2^n$, forcing $m=1$. Thus we need $5^{c'} - 3^{b'} = 2$. But we already showed this only has the solution $b' = 1, c' = 1$. Thus it follows the only solution where $a \ge 2$ is with $a=4, b = 2, c= 2$.
Putting every together, we have proven the only solutions are $(2,0,1), (1,1,1), (4,2,2)$
EDIT: I realize I forgot to do the case of $a=2$. So we need to solve $4 + 3^{2b'} = 5^c$. Modulo $275$ happens to work again to force $c=1$ and thus we get no solutions when $b$ is nonzero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove inequality $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} ≤ 1.5$ Prove $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} ≥ 1.5$ with $a + b + c = 3 $ and $a,b,c > 0$
The correct question is $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} ≥ 1.5$ (I have proved it)
Can prove $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq 1.5$ now:
$\frac{a}{b^2 + 1} = a-\frac{ab^2}{b^2+1}\geq a-\frac{ab}{2}$
Do it the same for $\frac{b}{c^2 + 1}, \frac{c}{a^2 + 1}$, then sum...
$\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq a+b+c - \frac{ab+bc+ac}{2} = 3 - \frac{ab+bc+ac}{2}$.
In different side: $(a+b+c)^2 \geq 3(ab+bc+ca) => ab+bc+ac \leq3$.
So:
$\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq 3 - 1,5 = 1,5$
The following inequality is true already.
Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$. Prove that:
$$\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1}+\frac{63abc}{2(a^2+b^2+c^2+6(ab+ac+bc))}\leq3.$$
Also, there is the following very nice inequality.
Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$. Prove that:
$$\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1}+\frac{ab+ac+bc}{2}\geq3.$$
| The inequality does not hold! For $a = 2, b = c = 1/2$, we have
$$
\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq \frac{a}{b^2 + 1} = \frac 8 5 > \frac 3 2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What do I use to find the image and kernel of a given matrix? I had a couple of questions about a matrix problem. What I'm given is:
Consider a linear transformation $T: \mathbb R^5 \to \mathbb R^4$ defined by $T( \vec{x} )=A\vec{x}$, where
$$A = \left(\begin{array}{crc}
1 & 2 & 2 & -5 & 6\\
-1 & -2 & -1 & 1 & -1\\
4 & 8 & 5 & -8 & 9\\
3 & 6 & 1 & 5 & -7
\end{array}\right)$$
*
*Find $\mathrm{im}(T)$
*Find $\ker(T)$
My questions are:
What do they mean by the transformation?
What do I use to actually find the image and kernel, and how do I do that?
| After a long night of studying I finally figured out the answer to these. The previous answers on transformation were all good, but I have the outlined steps on how to find $\mathrm{im}(T)$ and $\ker(T)$.
$$A = \left(\begin{array}{crc}
1 & 2 & 2 & -5 & 6\\
-1 & -2 & -1 & 1 & -1\\
4 & 8 & 5 & -8 & 9\\
3 & 6 & 1 & 5 & -7
\end{array}\right)$$
(1) Find $\mathrm{im}(T)$
$\mathrm{im}(T)$ is the same thing as column space or $C(A)$. The first step to getting that is to take the Transpose of $A$.
$$
A^T = \left(\begin{array}{crc}
1 & -1 & 4 & 3 \\
2 & -2 & 8 & 6 \\
2 & -1 & 5 & 1 \\
-5 & 1 & -8 & 5 \\
6 & -1 & 9 & -7
\end{array}\right)$$
once that's done the next step is to reduce $A^T$ to Reduced Row Echelon Form
$$
\mathrm{rref}(A^T) = \left(\begin{array}{crc}
1 & 0 & 1 & -2 \\
0 & 1 & -3 & -5 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)$$
now on this step I honestly don't know the reasons behind it, but the thext thing you do is take the rows and that's your answer. so that:
$$\mathrm{im}(T)\ = \begin{align*}
\operatorname{span}\left\{\left(\begin{array}{crc}
1 \\
0 \\
1 \\
-2 \end{array}\right), \left(\begin{array}{crc}
0 \\
1 \\
-3 \\
-5 \end{array}\right)\right\}
\end{align*}$$
(2) Find $\ker(T)$
$\ker(T)$ ends up being the same as the null space of matrix, and we find it by first taking the Reduced Row Echelon Form of A
$$
\mathrm{rref}(A) = \left(\begin{array}{crc}
1 & 2 & 0 & 3 & -4\\
0 & 0 & 1 & -4 & 5\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{array}\right)$$
we then use that to solve for the values of $\mathbb R^5$ so that we get
$$\begin{align*}
\left(\begin{array}{crc}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \end{array}\right) = r\left(\begin{array}{crc}
-2 \\
1 \\
0 \\
0 \\
0 \end{array}\right) + s\left(\begin{array}{crc}
-3 \\
0 \\
4 \\
1 \\
0 \end{array}\right) + t\left(\begin{array}{crc}
4 \\
0 \\
-5 \\
0 \\
1 \end{array}\right)
\end{align*}$$
from that we arrange the vectors and get our answer the vectors and that gives us our answer
$$\begin{align*}
\ker(T) = \operatorname{span}\left\{\left(\begin{array}{crc}
-2 \\
1 \\
0 \\
0 \\
0 \end{array}\right), \left(\begin{array}{crc}
-3 \\
0 \\
4 \\
1 \\
0 \end{array}\right), \left(\begin{array}{crc}
4 \\
0 \\
-5 \\
0 \\
1 \end{array}\right)\right\}
\end{align*}$$
and that's that.
| {
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"url": "https://math.stackexchange.com/questions/236541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
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Express the volume of an n-sphere in terms of the volume of an n-1 dimensional ball It's an exercise from Munkres "Analysis on Manifolds" Chapter 5, "Integrating a scalar function over a manifold".
Due to the suggestion, I'm repeating the question here:
Express the volume of an n-sphere in terms of the volume of an n-1 dimensional ball.
| Express the volume of $S^n(a)$ in terms of the volume
$B^{n-1}(a)$. [\it Hint: \rm Follow the pattern of Example 2.]
\paragraph{\bf{sln}.}
We can write
\begin{eqnarray}
S^n(a) &=& \{ (x_1, \cdots x_{n+1}) \; , \; x_1^2 + \cdots x_{n+1}^2 = a^2 \} \\
&=& \{ (x_1, \cdots x_{n-1}) \; , \; x_1^2 + \cdots x_{n-1}^2 = a^2 \cos^2 \theta \}
\\
&& \times \{ (x_{n},x_{n+1}) \; , \; x_n^2 + x_{n+1}^2 = a^2 \sin^2 \theta \}
\end{eqnarray}
(this equality is easy to show and I will omit its proof)
Then we parametrized the sphere based on the single parameter $\theta$ which we integrate
between 0 and $\pi/2$. That is
\begin{equation}
v(S^n(a)) = \int_0^{\pi/2} v(S^{n-2}(a \cos \theta)) \, v(S^1(a \sin \theta)) J d \theta
\end{equation}
The product of the two volumes is taken because we are integrating over
cross product of independent spaces (for each fixed $\theta$). The
Jacobian $J=a$ comes from the transformation from rectuangular coordinates
to polar $(a, \theta)$ coordinates. To get this Jacobian requires
a good amount of work. It is obvious for 2D when we say that
$x_1^2 + x_2^2= a^2$ and then $x_1= a \cos \theta$ and
$x_2 = a \sin \theta$, then the Jacobian
\begin{equation}
J = \det \left ( \frac{\partial(x_1, x_2)}{\partial(a, \rho)} \right ) = a.
\end{equation}
\begin{eqnarray}
v(S^n(a)) &=& a \int_0^{\pi/2} v(S^{n-2}(a \cos \theta)) \;
2 \pi (a \sin \theta)) d \theta \\
&=& 2 \pi a \int_0^a v( S^{n-2}(\rho)) d \rho \\
&=& 2 a \pi v(B^{n-1}(a)).
\end{eqnarray}
with the substitution $\rho = a \cos \theta$,
$d \rho = - a\sin \theta d \theta$, and
recognizing that
\begin{equation}
v(B^{n-1}(a)) = \int_0^a v( S^{n-2}(\rho)) d \rho.
\end{equation}
This integral is easy to send{equation} as thinking that an onion is the union of all its
concentric shells.
To verify the result let us consider a few cases.
\begin{itemize}
\item For $n=2$
\begin{equation}
v(S^2(a)) = 4 \pi a^2, \
2 \pi B^1(a) = (2 \pi a) (2 \pi) (a) = 4 \pi a^2
\end{equation}
\item For $n=3$
\begin{equation}
v(S^3(a)) = 2 \pi^2 a^3, \\
2 \pi a B^2(a) = (2 \pi a) (\pi^2)a^2) = 2 \pi^2 a^3
\end{equation}
\item and for $n=4$
\begin{equation}
v(S^4(a)) = \frac{8}{3} \pi^2 a^4, \\
2 \pi a B^3(a) = (2 \pi a) (\frac{4}{3} \pi a^3) = \frac{8}{3} \pi^2 a^3
\end{equation}
\end{itemize}
So,
\begin{equation}
\frac{v(S^n(a))}{v(B^{n-1}(a)} = 2 \pi a.
\end{equation}
\end{document}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/236662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Positive integers less than $N$ not divisible by $4$ or $6$ How many positive integers less than $N$ are not divisible by $4$ or $6$ for some $N$?
| A naive and computational answer: look at congruence modulo 12.
*
*Empty (no natural numbers less than 1) = $0$
*${1}$ = $1$
*${1,2}$ = $2$
*${1, 2, 3}$ = $3$
*${1, 2, 3}$ = $3$
*${1, 2, 3, 5}$ = $4$
*${1, 2, 3, 5}$ = $4$
*${1, 2, 3, 5, 7}$ = $5$
*${1, 2, 3, 5, 7}$ = $5$
*${1, 2, 3, 5, 7, 9}$ = $6$
*${1, 2, 3, 5, 7, 9, 10}$ = $7$
*${1, 2, 3, 5, 7, 9, 10, 11}$ = $8$
From here, observe that any $N$ will be congruent to one of these guys, so you add $N \mod 12$ to 8 times the floor of $N/12$; ie $13 = 12 + 1$ so $n(13) = 8 \times 1 + 0 = 8$ and $n(155) = 8 \times 12 + n(11) = 96 + 7 = 103$.
And then see how you can improve my naive method to get a more "general" answer.
| {
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"answer_count": 3,
"answer_id": 2
} |
numbers' pattern It is known that
$$\begin{array}{ccc}1+2&=&3 \\ 4+5+6 &=& 7+8 \\
9+10+11+12 &=& 13+14+15 \\\
16+17+18+19+20 &=& 21+22+23+24 \\\
25+26+27+28+29+30 &=& 31+32+33+34+35 \\\ldots&=&\ldots
\end{array}$$
There is something similar for square numbers:
$$\begin{array}{ccc}3^2+4^2&=&5^2 \\ 10^2+11^2+12^2 &=& 13^2+14^2 \\ 21^2+22^2+23^2+24^2 &=& 25^2+26^2+27^2 \\ \ldots&=&\ldots
\end{array}$$
As such, I wonder if there are similar 'consecutive numbers' for cubic or higher powers.
Of course, we know that there is impossible for the following holds (by Fermat's last theorem):
$$k^3+(k+1)^3=(k+2)^3 $$
| We do have
$$ \eqalign{6^3 &= 3^3 + 4^3 + 5^3\cr
20^3 &= 11^3 + 12^3 + 13^3 + 14^3\cr
40^3 &= 3^3 + \ldots + 22^3\cr
70^3 &= 15^3 + \ldots + 34^3\cr
37^3 + 38^3 &= 5^3 + \ldots + 25^3\cr
30^3 + 31^3 + 32^3 &= 7^3 + \ldots + 24^3\cr
101^3 + 102^3 + 103^3 &= 61^3 + \ldots + 71^3\cr
15^3 + \ldots + 20^3 &= 11^3 + \ldots + 19^3\cr
681^3 + \ldots + 687^3 &= 566^3 + \ldots + 577^3\cr
\cr}$$
and many others.
EDIT: See also http://oeis.org/A062682
| {
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"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of the first $n$ terms of $\sum^n_{k=1}k^3$ The question:
Find the sum of the first $n$ terms of $$\sum^n_{k=1}k^3$$
[Hint: consider $(k+1)^4-k^4$]
[Answer: $\frac{1}{4}n^2(n+1)^2$]
My solution:
$$\begin{align}
\sum^n_{k=1}k^3&=1^3+2^3+3^3+4^3+\cdots+(n-1)^3+n^3\\
&=\frac{n}{2}[\text{first term} + \text{last term}]\\
&=\frac{n(1^3+n^3)}{2}
\end{align}$$
What am I doing wrong?
| Method 1
Using the binomial identity
$$
\sum_{k=m}^{n-j}\binom{n-k}{j}\binom{k}{m}=\binom{n+1}{j+m+1}\tag{1}
$$
with $j=0$ yields
$$
\sum_{k=0}^n\binom{k}{m}=\binom{n+1}{m+1}\tag{2}
$$
Using $(2)$ and the identity
$$
k^3=6\binom{k}{3}+6\binom{k}{2}+\binom{k}{1}\tag{3}
$$
we get that
$$
\begin{align}
\sum_{k=0}^nk^3
&=6\binom{n+1}{4}+6\binom{n+1}{3}+\binom{n+1}{2}\\
&=6\binom{n}{4}+12\binom{n}{3}+7\binom{n}{2}+\binom{n}{1}\\
&=\frac{n^2(n+1)^2}{4}\tag{4}
\end{align}
$$
Method 2
Using the Euler-Maclaurin Sum Formula, we get
$$
\sum_{k=0}^nk^3=\frac14n^4+\frac12n^3+\frac14n^2+C\tag{5}
$$
where we get $C=0$ by plugging in $n=1$.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Evaluate $\int_1^4\frac{dx}{x^2+x+1}$ I've to evaluate the integral $$\int_1^4 \frac{dx}{x^2+x+1}$$
but I can't find the answer. I checked with Wolfram Alpha but I still don't fully understand. Could you please explain the steps to me? I think I should use arctan in my answer.
| First note that
$$\dfrac1{x^2 + x + 1} = \dfrac1{(x+1/2)^2 + (\sqrt{3}/2)^2}$$
We now have that
\begin{align}
I = \int_1^4 \dfrac{dx}{x^2 + x + 1} = \int_1^4 \dfrac{dx}{(x+1/2)^2 + (\sqrt{3}/2)^2}
\end{align}
Now let $y = x+1/2$, then we get that
$$I = \int_{3/2}^{9/2} \dfrac{dy}{y^2 + (\sqrt{3}/2)^2}$$
Now let $y = \sqrt{3}/2 \tan(\theta)$. We then get that $dy = \sqrt{3}/2 \sec^2(\theta) d \theta$.
Hence,
$$I = \int_{\theta = \arctan(\sqrt{3})}^{\theta = \arctan(3 \sqrt{3})} \dfrac{\sqrt{3}/2 \sec^2(\theta) d \theta}{3/4 \sec^2(\theta)} = \dfrac2{\sqrt{3}} \left( \arctan(3 \sqrt{3}) - \arctan(\sqrt{3}) \right) = \dfrac2{\sqrt{3}} \left( \arctan(3 \sqrt{3}) - \dfrac{\pi}3 \right)$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
On Ceva's Theorem? The famous Ceva's Theorem on a triangle $\Delta \text{ABC}$
$$\frac{AJ}{JB} \cdot \frac{BI}{IC} \cdot \frac{CK}{EK} = 1$$
is usually proven using the property that the area of a triangle of a given height is proportional to its base.
Is there any other proof of this theorem (using a different property)?
EDIT: I would like if someone can use the proof of Menelaus' Theorem.
| The following proof is heavily influenced by my background in projective geometry. Use homogenous and choose the following affine basis, without loss of generality. You might also consider this as barycentric coordinates, the way Olivier Bégassat wrote in a comment.
\begin{align*}
A &= \begin{pmatrix}1\\0\\0\end{pmatrix} &
B &= \begin{pmatrix}0\\1\\0\end{pmatrix} &
C &= \begin{pmatrix}0\\0\\1\end{pmatrix}
\end{align*}
Based on these coordinates, you can specify the other three points as linear combinations of the corresponding triangle corners:
\begin{align*}
I &= \begin{pmatrix}0\\\lambda_I\\\mu_I\end{pmatrix} &
J &= \begin{pmatrix}\lambda_J\\\mu_J\\0\end{pmatrix} &
K &= \begin{pmatrix}\mu_K\\0\\\lambda_K\end{pmatrix}
\end{align*}
You can obtain the oriented length ratios from these parameters:
\begin{align*}
\frac{AJ}{JB} &= \frac{\mu_J}{\lambda_J} &
\frac{BI}{IC} &= \frac{\mu_I}{\lambda_I} &
\frac{CK}{KA} &= \frac{\mu_K}{\lambda_K}
\end{align*}
Now you can compute the connections of each of these points with the opposite triangle corner using a cross product:
\begin{align*}
A\times I &= \begin{pmatrix}0\\-\mu_I\\\lambda_I\end{pmatrix} &
B\times K &= \begin{pmatrix}\lambda_K\\0\\-\mu_K\end{pmatrix} &
C\times J &= \begin{pmatrix}-\mu_J\\\lambda_J\\0\end{pmatrix} \\
\end{align*}
To check whether these three lines are concurrent, you compute their determinant. If that determinant becomes zero, the lines go through the same point.
\begin{align*}
\begin{vmatrix}
0 & \lambda_K & -\mu_J \\
-\mu_I & 0 & \lambda_J \\
\lambda_I & -\mu_J & 0
\end{vmatrix} =
\lambda_I\lambda_J\lambda_K - \mu_I\mu_J\mu_K &= 0 \\
\lambda_I\lambda_J\lambda_K &= \mu_I\mu_J\mu_K \\
1 &= \frac{\lambda_I\lambda_J\lambda_K}{\mu_I\mu_J\mu_K} =
\frac{AJ}{JB}\cdot\frac{BI}{IC}\cdot\frac{CK}{KA}
\end{align*}
As the choice of coordinates was without loss of generality, the above equivalence (between the concurrence of the three lines and the product of oriented length ratios being one) will hold for any non-degenerate triangle $ABC$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Limits of functions with square root and sin Can someone help me calculate the following limits(without L'Hopital!) :
1) $\lim_{x\to 1 } \frac { \sqrt{x}-1}{\sqrt[3]{x}-1} $ . I have tried taking the logarithm of this limit, but without any success.
2) $\lim_{x\to\pi} \frac{\sin5x}{\sin3x}$ and the hint is : $\sin(\pi-x)=\sin x , \sin(2\pi-x)=-\sin x$.
Thanks everyone!
| For the first one, set $\sqrt[6]{x} = y$. This means $\sqrt{x} = y^3$ and $\sqrt[3]{x} = y^2$. We then get that
$$\lim_{x \to 1} \dfrac{\sqrt{x}-1}{\sqrt[3]{x}-1} = \lim_{y \to 1} \dfrac{y^3-1}{y^2-1} = \lim_{y \to 1} \dfrac{(y-1)(y^2+y+1)}{(y-1)(y+1)} = \lim_{y \to 1} \dfrac{(y^2+y+1)}{(y+1)} = \dfrac32$$
For the second one, as the hint suggests, let $y = \pi - x$. We then get that
\begin{align}
\lim_{x \to \pi} \dfrac{\sin(5x)}{\sin(3x)} & = \lim_{y \to 0} \dfrac{\sin(5(\pi-y))}{\sin(3(\pi-y))} = \underbrace{\lim_{y \to 0} \dfrac{\sin(5 \pi -5y)}{\sin(3\pi-3y)} = \lim_{y \to 0} \dfrac{\sin(5y)}{\sin(3y)}}_{(\star)}\\
& = \lim_{y \to 0} \dfrac{5 \times \dfrac{\sin(5y)}{5y}}{3 \times \dfrac{\sin(3y)}{3y}} = \dfrac{5 \times \lim_{y \to 0} \dfrac{\sin(5y)}{5y}}{3 \times \lim_{y \to 0} \dfrac{\sin(3y)}{3y}} = \dfrac53\\
\end{align}
where $(\star)$ follows from the fact that $\sin((2n+1)\pi-y) = \sin(y)$ and the last equality comes from the fact that $\displaystyle \lim_{\theta \to 0} \dfrac{\sin(\theta)}{\theta} = 1$.
EDIT
$$\sin((2n+1)\pi-y) = \overbrace{\sin(2n \pi + \pi -y) = \sin(\pi -y)}^{\text{Because $\sin$ is $2 \pi$ periodic.}}$$
Now $$\sin(\pi-y) = \sin(\pi) \cos(y) - \cos(\pi) \sin(y) = 0 - (-1) \sin(y) = \sin(y)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/244158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Value of $\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ I remember that a couple of years ago a friend showed me and some other people the following expression:
$$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$
As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:
What is the value of this limit and how do we prove it rigorously?
(Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)
I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.
(1) The sequence $a_n$ satisfies $a_n<1$ for all $n\ge 2$.
Note that $a_n<1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n<n^n$. I prove this second statement by induction. Observe that $1^2=1<4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n<n^n$ for some integer $n\ge 2$. Then
$$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}<(n-1)n^n+n^{n+1}<(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$
(2) The sequence $a_n$ is increasing for all $n\ge 2$.
We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)
(2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.
We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By Bernoulli's Inequality, we may find:
$$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$
A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.
We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:
$$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$
By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:
$$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}<\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$
(In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.
I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.
Thanks!
| We can write $a_n=\sum_{k=1}^{n-1}\left(1-\frac{k}{n}\right)^n$. The given limit can then be written in the form
$$
\lim_{n\to\infty}\sum_{k=1}^\infty\left[1-\frac{k}{n}\right]_+^n
$$
(where $[x]_+$ is the positive part, i.e., $[x]_+=x$ if $x\geq0$, $0$ otherwise).
I claim that for fixed $k$, $\left[1-\frac{k}{n}\right]_+^n$ is a non-decreasing function of $n$. This means we can pass the limit inside the sum. The limiting value of the $k$-th term is $e^{-k}$, so the value of the limit is
$$
\sum_{k=1}^\infty e^{-k}=\frac{1}{e-1}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 0
} |
Example of two dependent random variables that satisfy $E[f(X)f(Y)]=Ef(X)Ef(Y)$ for every $f$ Does anyone have an example of two dependent random variables, that satisfy this relation?
$E[f(X)f(Y)]=E[f(X)]E[f(Y)]$
for every function $f(t)$.
Thanks.
*edit: I still couldn't find an example. I think one should be of two identically distributed variables, since all the "moments" need to be independent: $Ex^iy^i=Ex^iEy^i$. That's plum hard...
| Here is a counterexample. Let $V$ be the set $\lbrace 1,2,3 \rbrace$. Consider random variables $X$ and $Y$ with values in $V$, whose joint distribution is defined by the following matrix :
$$
P=\left(
\begin{matrix}
\frac{1}{10} & 0 & \frac{1}{5} \\
\frac{1}{5} & \frac{1}{10} & 0 \\
\frac{1}{30} & \frac{7}{30} & \frac{2}{15}
\end{matrix}
\right)=
\left(
\begin{matrix}
\frac{3}{30} & 0 & \frac{6}{30} \\
\frac{6}{30} & \frac{3}{30} & 0 \\
\frac{1}{30} & \frac{7}{30} & \frac{4}{30}
\end{matrix}\right)
$$
Thus, for example, $P(X=1,Y=2)=0$ while $P(X=1)P(Y=2)=(\frac{1}{10} + \frac{1}{5})(\frac{1}{10} + \frac{7}{30}) >0$. So $X$ and $Y$ are not independent.
Let $f$ be an ARBITRARY (I emphasize this point because of a comment below) function defined on $X$ ; put $x=f(1),y=f(2),z=f(3)$. Then
$$
\begin{array}{rcl}
{\mathbf E}(f(X)) &=& \frac{3(x+y)+4z}{10} \\
{\mathbf E}(f(Y)) &=& \frac{x+y+z}{3} \\
{\mathbf E}(f(X)){\mathbf E}(f(Y)) &=& \frac{3(x+y)^2+7(x+y)z+4z^2}{30} \\
{\mathbf E}(f(X)f(Y)) &=& \frac{3x^2+6xy+3y^2+7xz+7yz+4z^2}{30} \\
\end{array}
$$
The last two are equal, qed.
| {
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"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 1
} |
Partial Differential with Integrating Factor I have the following:
$ \dfrac{dx}{dz} = \dfrac{x(x+y)}{(y-x)(2x+2y+z(x,y))} $
I've tried to solve through using an integrating factor, i.e. I rearranged to get:
$ \dfrac{dz(x,y)}{dx} - \dfrac{(y-x)}{x(x+y)}z(x,y) = \dfrac{2(y-x)}{x} $
and used IF $= \dfrac{(x+y)^2}{x} $. I do get an eventual (fairly messy, but ok) answer. I would appreciate any feedback on the validity of the method - in other words, because $z$ is a function of $x$ and $y$, is it 'acceptable' to rearrange it in this way and then use integrating factor.
Help/feedback very appreciated.
| It would be better in this context either to use partial derivative sign, or suppress $y$ as an argument and treat it purely as a parameter. To cross-check your solution a calculation by means of variating the constant of integration can be carried out.
First solve the homogeneous equation
$$\frac{dz}{z}=\frac{y-x}{x\left(x+y\right)}dx$$
$$\frac{dz}{z}=\frac{x+y-2x}{x\left(x+y\right)}dx$$
$$\frac{dz}{z}=\frac{dx}{x}-2\frac{d\left(x+y\right)}{x+y}$$
$$z=\frac{C(y)x}{\left(x+y\right)^{2}}\tag{1}$$
Now variate $C$ in $x$
$$C=C\left(x,y\right)$$
$$z_x=\frac{C_xx}{\left(x+y\right)^{2}}+C\left[\frac{1}{\left(x+y\right)^{2}}-\frac{2x}{\left(x+y\right)^{3}}\right]=\\\frac{C_x\left(x\right)x}{\left(x+y\right)^{2}}+C\frac{y-x}{\left(x+y\right)^{3}}$$
Inserting the results in the original equation:
$$
\frac{C_xx}{\left(x+y\right)^{2}}+C\frac{y-x}{\left(x+y\right)^{3}}-\frac{y-x}{x\left(x+y\right)}\frac{Cx}{\left(x+y\right)^{2}}=\frac{2\left(y-x\right)}{x}$$
$$\frac{C_xx}{\left(x+y\right)^{2}}=\frac{2\left(y-x\right)}{x}$$
Solving for $C_x$:
$$C_x=\frac{2\left(x+y\right)^{2}\left(y-x\right)}{x^{2}}=2\frac{\left(x^{2}+2xy+y^{2}\right)\left(y-x\right)}{x^{2}}=2\frac{x^{2}y+2xy^{2}+y^{3}-x^{3}-2x^{2}y-xy^{2}}{x^{2}}=2\frac{-x^{3}-yx^{2}+xy^{2}+y^{3}}{x^{2}}$$
$$C_x=2\left(-x-y+\frac{y^{2}}{x}+\frac{y^{3}}{x^{2}}\right)$$
Now integrate with respect to $x$:
$$C=2\left(-\frac{x^{2}}{2}-xy+y^{2}\ln x-\frac{y^{3}}{x}\right)+C_0(y)$$
Substitute in $(1)$ to get the final result.
| {
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"answer_id": 0
} |
Proof using Mod How can you prove that:
$$a^7\equiv a\:(\text{mod } 42)$$
I haven't been given any other information other than to use Fermat's theorem.
| There are two versions of Fermat's Theorem.
Version $1$: If $a$ is not divisible by $p$, then $a^{p-1}\equiv 1\pmod{p}$.
Version $2$: For any $a$, $a^p\equiv a\pmod p$.
The more common version is Version $1$. From it, you can easily deduce Version $2$. For if $a$ is not divisible by $p$, then $a^{p-1}\equiv 1\pmod{p}$. Multiplying by $a$, we get that $a^p\equiv a \pmod{p}$. On the other hand, if $a$ is divisible by $p$, we have that both $a^p$ and $a$ are congruent to $0$ modulo $p$, and therefore are congruent to each other.
We omit the proof that Version $2$ implies Version $1$. It is straightforward.
Using Version $2$, we find that for all $a$,
$$a^7 \equiv a\pmod{7}.\tag{$1$}$$
Now work modulo $2$. It is hardly necessary to use machinery. If $a$ is odd, then so is $a^7$, and if $a$ is even, then so is $a^7$. We conclude that for all $a$,
$$a^7 \equiv a\pmod{2}.\tag{$2$}$$
Now work modulo $3$. We can do a three-cases verification that $a^7\equiv a\pmod{3}$ for all $a$. Or else we can use Fermat's Theorem. If $a$ is divisible by $3$, then $a^7$ and $a$ are congruent to $0$ modulo $3$, so they are conruent to each other. If $a$ is not divisible by $3$, then $a^2\equiv 1\pmod{3}$, and therefore $a^6\equiv 1\pmod{3}$, and therefore $a^7\equiv a \pmod{3}$. we conclude that for all $a$
$$a^7 \equiv a\pmod{3}.\tag{$3$}$$
So $7$, $2$, and $3$ each divide $a^7-a$. Since they are pairwise relatively prime, it follows that their product $42$ divides $a^7-a$ for all $a$. This is what we wanted to prove.
| {
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Proof : $2^{n-1}\mid n!$ if and only if $n$ is a power of $2$. I want to prove that:
$2^{n-1}\mid n!$ if and only if $n$ is a power of $2$.
| write out n! and consider how many times each divides by 2 for example 16:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 = 16!
^ ^ ^ ^ ^ ^ ^ ^ divisible by 2
^ ^ ^ ^ divisible by 2^2
^ ^ divisible by 2^3
^ divisible by 2^4
so there are $16/2 = 8$ numbers there divisible by 2. $16/4 = 4$ numbers there divisible by 2^2. etc.
in total then, there are 16/2 + 16/2^2 + 16/2^3 + 16/2^4.
In general if 2^r is the highest power of 2 dividing n! then $r = \sum_{i=1} [n/2^i]$.
To show that $2^{n-1}\mid n!$ we thus want to prove $n-1 = \sum_{i=1} [n/2^i]$ iff $n = 2^k$. ($\Leftarrow$) is immediate from computation. For ($\Rightarrow$) Let $2^{k-1} < n < 2^k$ but then $r < n - 1$ by comparing each term of the summation for $n$ and the one for $2^{k}$.
| {
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Put into each cell one number 1 or one number -1 that the sums of number in each coloums each rows is 0. There is a 4 x 4 grid. In each cell of the grid, you are allowed to put either a 1 or a -1. The sum of the numbers of each column and each row must equal 0. How many such configurations of the grid are there?
| You can to start completing the first line of the square
\begin{array}{|c|c|c|c|}
\hline
.& . & .&. \\ \hline
.& .&. & .\\ \hline
. & . & .& .\\ \hline
.& . & .& .\\ \hline
\end{array}
And you can do this of $ 6 = \binom{4}{2} $ ways because you need to choose the 2 places between the 4 to put the numbers 1. For exemplo,
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 & -1 \\ \hline
. & .&.& .\\ \hline
. & .& .&.\\ \hline
.&.&. &.\\ \hline
\end{array}.
Then you can complete the fisrt column choosing one of the three places to put the number 1 . For exemplo,
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & & &\\ \hline
. & . & .&.\\ \hline
.& .& .&.\\ \hline
\end{array}.
Then, after you do this, your table must be
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & & &\\ \hline
-1 & & &\\ \hline
-1 & & &\\ \hline
\end{array}.
You can complete the column that contains the number 1. There are two signicant ways to do this. You can to put the 1 in the column that already countains the 1 or no. These cases are respectively of this forms
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & -1&1 &-1\\ \hline
-1 & & &\\ \hline
-1 & & &\\ \hline
\end{array}.
or for examplo
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & 1& -1&-1\\ \hline
-1 & & &\\ \hline
-1 & & &\\ \hline
\end{array}.
Notice that there are another way to do the second case, namely
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & -1& -1&1\\ \hline
-1 & & &\\ \hline
-1 & & &\\ \hline
\end{array}.
The case \begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & -1&1 &-1\\ \hline
-1 & & &\\ \hline
-1 & & &\\ \hline
\end{array}
is determined to be
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & -1&1 &-1\\ \hline
-1 &1 &-1 &1\\ \hline
-1 &1 &-1 &1\\ \hline
\end{array}.
And the second case
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 &- 1&-1 &1\\ \hline
-1 & & &\\ \hline
-1 & & &\\ \hline
\end{array}
and
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & 1& -1&-1\\ \hline
-1 & & &\\ \hline
-1 & & &\\ \hline
\end{array}.
also are to be determined by respectvely
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 &- 1&-1 &1\\ \hline
-1 & 1 & &\\ \hline
-1 & 1 & &\\ \hline
\end{array}
and
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & 1& -1&-1\\ \hline
-1 & & &1\\ \hline
-1 & & &1\\ \hline
\end{array}.
Then you need to choose the polace where you put the number 1 on the third and this determines the table. This can be done of 2 ways. Then the answer is
$$ 6 \cdot 3(1 + 2 \cdot 2) = 90 \quad \mbox{ways.} $$
| {
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Closed form for $\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$ How can I calculate the following sum involving binomial terms:
$$\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$$
Where the value of n can get very big (thus calculating the binomial coefficients is not feasible).
Is there a closed form for this summation?
| I'm even later to the party, but that's only because "absorption identity" kept yelling in my ear. :)
One application of the absorption identity gets one of the factors of $k+1$ out of the denominator:
$$\begin{align}
\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(k+1)^2} &= \frac{1}{n+1} \sum_{k=0}^n \binom{n+1}{k+1} \frac{(-1)^k}{k+1} \\
&= \frac{1}{n+1} \sum_{k=1}^{n+1} \binom{n+1}{k} \frac{(-1)^{k+1}}{k} \end{align}.$$
It would be nice to use the absorption identity again, but we need a $k+1$ in the denominator of the summand rather than a $k$. By using the basic binomial coefficient recursion formula, we can make that happen.
Let $\displaystyle f(n) = \sum_{k=1}^{n} \binom{n}{k} \frac{(-1)^{k+1}}{k}.$ Then looking at the difference of $f(n+1)$ and $f(n)$ gives us
$$\begin{align}
f(n+1) - f(n) &= \sum_{k=1}^{n+1} \binom{n+1}{k} \frac{(-1)^{k+1}}{k} - \sum_{k=1}^n \binom{n}{k} \frac{(-1)^{k+1}}{k} \\
&= \sum_{k=1}^{n+1} \binom{n}{k-1} \frac{(-1)^{k+1}}{k} \\
&= \sum_{k=0}^n \binom{n}{k} \frac{(-1)^{k}}{k+1} \\
&= \frac{1}{n+1}\sum_{k=0}^n \binom{n+1}{k+1} (-1)^{k} \:\:\:\: \text{ (absorption identity!)} \\
&= \frac{1}{n+1} \left(1 + \sum_{k=0}^{n+1} \binom{n+1}{k} (-1)^{k+1} \right)\\
&= \frac{1}{n+1},
\end{align}$$
where in the last step we used the fact that the alternating sum of the binomial coefficients is $0$.
Thus $$f(n+1) = \sum_{k=0}^n (f(k+1) - f(k)) = \sum_{k=0}^n \frac{1}{k+1} = H_{n+1}.$$
Therefore,
$$\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(k+1)^2} = \frac{H_{n+1}}{n+1}.$$
| {
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How to show this is relatively prime? Let $a_1= 2$, and for each
$y > 1$, define $a_{y+1} = a_y(a_y −1) +1$.
Prove that for all $x \ne y$, $a_x$ and $a_y$ are coprime.
| Hint:
$a_{y+1}=a_y(a_y-1)+1$
$\implies a_{y+1}-1=a_y(a_y-1) $
So, $ a_y-1=a_{y-1}(a_{y-1}-1)$ and $a_{y+1}-1=a_ya_{y-1}(a_{y-1}-1) $
$ a_{y-1}-1=a_{y-2}(a_{y-2}-1)$ and
so $a_{y+1}-1=a_ya_{y-1}a_{y-2}(a_{y-2}-1)=(a_1-1)\prod_{y\le x \le1}a_x=\prod_{y\le x \le1}a_x$ as $a_1=2$
As $(a_{y+1}-1,a_{y+1})=1, (\prod_{y\le x \le1}a_x,a_{y+1})=1\implies (a_x,a_{y+1})=1$ for $x\le y$
| {
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Evaluate $\lim\limits_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)$ Evaluate
$$
\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)
$$
The answer is $\frac{1}{2}$, have no idea how to arrive at that.
| Using $$\alpha-\beta=\frac{\alpha^2-\beta^2}{\alpha+\beta}$$
yields
$$\lim_{x \to \infty} \sqrt{\frac{x^3}{x-1}}-x=\lim_{x \to \infty} \frac{\frac{x^3}{x-1}-x^2}{\sqrt{\frac{x^3}{x-1}}+x}=\lim_{x \to \infty} \frac{x^3-x^3+x^2}{(x-1)(\sqrt{\frac{x^3}{x-1}}+x)}=
\lim_{x \to \infty} \frac{x^2}{\sqrt{x^4-x^3}+x^2-x}=\lim_{x \to \infty} \frac{x^2}{x^2(\sqrt{1-\frac1x}+1-\frac1x)}=\lim_{x \to \infty} \frac{1}{\sqrt{1-\frac1x}+1-\frac1x}=\frac12$$
| {
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How to integrate this $\int\frac{\mathrm{d}x}{{(4+x^2)}^{3/2}} $ without trigonometric substitution? I have been looking for a possible solution but they are with trigonometric integration..
I need a solution for this function without trigonometric integration
$$\int\frac{\mathrm{d}x}{{(4+x^2)}^{3/2}}$$
| Let $I=\int \frac{dx}{\sqrt{x^2+4}} \,;\, J=\int \frac{dx}{\sqrt{x^2+4}^3}$
We integrate $I$ by parts, we get:
$u=(x^2+4)^{-1/2}, dv=dx$
$du=-x(x^2+4)^{-3/2}, v=x$
Thus
$$I= \frac{x}{\sqrt{x^2+4}}+ \int \frac{x^2}{\sqrt{x^2+4}^3}= \frac{x}{\sqrt{x^2+4}}+ \int \frac{x^2+4-4}{\sqrt{x^2+4}^3}=$$
$$= \frac{x}{\sqrt{x^2+4}}+ I-4J=$$
Thus, canceling the $I$ we get
$$4J= \frac{x}{\sqrt{x^2+4}} +C$$
| {
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Using product and chain rule to find derivative. Find the derivative of
$$y =(1+x^2)^4 (2-x^3)^5$$
To solve this I used the product rule and the chain rule.
$$u = (1+x^2)^4$$
$$u' = 4 (1+x^2)^3(2x)$$
$$v= (2-x^3)^5$$
$$v' = 5(2-x^3)^4(3x^2)$$
$$uv'+vu'$$
$$((1+x^2)^4)(5(2-x^3)^4(3x^2)) + ((2-x^3)^5 )(4 (1+x^2)^3(2x))$$
The answer I got is: $$(15x^2)(1-x^2)^4(2-x^3)^4 + 8x(2-x^3)^5(1+x^2)^3$$.
Why is the answer $$8x(x^2 +1)^3(2-x^3)^5-15x^2(x^2)(X^2+1)^4(2-x^3)4$$?
How did the $15x^2$ become negative?
| Everything is correct in your answer, except for the chain rule for $v$.
The derivative of $2-x^3$ is $-3x^2$.
So $v'=5(2-x^3)^4(-3x^2)$ and this is why the $15x^2$ becomes negative.
| {
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Help Solving Trigonometry Equation I am having difficulties solving the following equation:
$$4\cos^2x=5-4\sin x$$
Hints on how to solve this equation would be helpful.
|
I am having difficulties solving the following equation:$${4\cos^2x=5-4\sin x}$$
First, substitute $4\cos^2(x)$ with $4\left(1 - \sin^2(x)\right) = 4 - 4\sin^2(x).$ We are left with $4 - 4\sin^2(x) = 5 - 4 \sin (x) .$ This can be rewritten as $-4\sin^2(x) + 4\sin(x) - 1 = 0.$
Observe that the equation can further be rewritten in the form $-4t^2 + 4t - 1 = 0$ where $t = \sin(x)$. Solve the quadratic equation for $t$ and then use $\sin(x) = t$ to solve for $x$.
| {
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What is the value of $N$? $N$ is the smallest positive integer such that for any integer $n > N$, the quantity $n^3 – 7n^2 + 11n – 5$ is positive. What is the value of $N$?
Note: $N$ is a single digit number.
| Start by trying to factorise the expression. One thing to note is that the sum of the coefficients is zero: $(+1) + (-7) + (+11) + (-5) = 0$, and so $n=1$ is a solution. We are able to factorise:
$$ n^3 - 7n^2 + 11n -5 \equiv (n-1)(\text{quadratic}) \, . $$
Assume the quadratic is of the form $a_2n^2+a_1n+a_0$, for some yet to be determined numbers $a_2$, $a_1$ and $a_0$. We may expand and compare coefficients. We see that $a_2 = 1,$ $a_1 - a_2 = -7$, $a_0-a_1 = 11$ and $-a_0 = -5.$ Solving these linear equations gives: $a_2 = 1,$ $a_1 = -6$ and $a_0 = 5$. Thus: $n^3-7n^2+11n + 5 \equiv (n-1)(n^2-6n+5).$ Again, we can factorise the quadratic term and we see that $n^2-6n+5 \equiv (n-1)(n-5)$. Thus:
$$n^3-7n^2+11n-5 \equiv (n-1)^2(n-5) \, . $$
The graph $y = x^3-7x^2+11x-5$ cuts the $x$-axis when $x=5$ and is tangent there when $x=1$. These are the critical values: $x=1$ and $x=5$. The question is: what happens when $x < 1$, when $1 < x < 5$ and when $x > 5.$ We can substitute values, say $x = 0$, $x = 2$ and $x = 6$.
$$
\begin{array}{ccc}
0^3 - 7\times 0^2 + 11\times 0 - 5 & = & -5 \, , \\
2^3 - 7\times 2^2 + 11\times 2 - 5 & = & -2 \, , \\
6^3 - 7\times 6^2 + 11\times 6 - 5 & = & 25 \, .
\end{array}
$$
It follows that $x^3-7x^2+11x-5 \le 0$ for all $x \le 5$ while $x > 0$ for all $x > 5$. Thus, for all integer $n > N = 5$ we have $n^3 - 7n^2 + 11n -5 > 0$.
| {
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Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Prove the following inequality: for
$a,b,c>0$
$$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$
What I tried is using substitution:
$p=a+b+c$
$q=ab+bc+ca$
$r=abc$
But I cannot reduce $a^2(b+c)(c+a)+b^2(a+b)(c+a)+c(a+b)(b+c) $ interms of $p,q,r$
| This is just Cauchy-Schwarz:
$$\left(\frac{a}{\sqrt{a+b}}\sqrt{a+b}+\frac{b}{\sqrt{b+c}}\sqrt{b+c}+\frac{c}{\sqrt{a+c}}\sqrt{a+c}\right)^2 \leq \left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a} \right)\big( (a+b)+(a+c)+(b+c)\big)$$
| {
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Show that the sum of the largest odd divisors of $n+1, n+2, \ldots, 2n$ (where $n$ is a natural number) is a perfect square? I've been given the solution but I don't understand it at all, could someone please explain?
| The first thing to understand is that every positive integer $N$ can be written uniquely in the form $2^ab$, where $a$ is a non-negative integer, and $b$ is a positive odd integer. For example, $12=2^2\cdot3$ ($a=2,b=3$), $11=2^0\cdot 11$ ($a=0,b=11$), and $8=2^3\cdot1$ ($a=3,b=1$). In this decomposition the number $b$ is called the odd part of $N$, so $3$ is the odd part of $12$, $11$ is the odd part of $11$, and $1$ is the odd part of $8$. Note that the odd part of $N$ is indeed the largest odd number that divides $N$.
Next, notice that if $N$ and $M$ have the same odd part, then one of $N$ and $M$ is a multiple of the other. Say that $N=2^ab$ and $M=2^cb$, where $b$ is the odd part of $N$ and $M$; then either $a\le c$, in which case $M$ is a multiple of $N$, or $c\le a$, in which case $N$ is a multiple of $N$.
Now look at the integers $n+1,n+2,\dots,2n$. The largest of these, $2n$, is too small to be a multiple of the smallest, $n+1$, so none of these integers can be a multiple of another. Thus, they must all have different odd parts. The odd part of any integer is at most as large as that integer, so the largest possible odd part of any of the integers $n+1,\dots,2n$ is $2n$ $-$ except that $2n$ isn’t odd, so the largest possible odd part here is actually only $2n-1$.
There are $n$ numbers in the set $\{n+1,n+2,\dots,2n\}$, and each has a different odd part that is at most $2n-1$. The first $n$ odd positive integers are $1,3,\dots,2n-1$. Thus, the odd parts of the integers $n+1,n+2,\dots,2n$ must be precisely these $n$ odd integers, $1,3,\dots,2n-1$. Thus,
$$\begin{align*}
\sum(\text{largest odd divisors of }n+1,n+2,\dots,2n)&=\sum(\text{odd parts of }n+1,n+2,\dots,2n)\\
&=1+3+\ldots+(2n-1)\;,
\end{align*}$$
the sum of the first $n$ positive odd integers.
The proof by induction that $$1+3+\ldots+(2n-1)=n^2$$ is quite straightforward and can be found in many places.
| {
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Origin of the constant $\phi$ in Binet's formula of the $n$-th term of the Fibonacci sequence I have read in a page that " To find the nth term of the Fibonacci series, we can use Binet's Formula"
F(n) = round( (Phi ^ n) / √5 ) provided n ≥ 0
where Phi=1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ... .
my question is,
*
*What is this Phi?
*How they generated this Phi?
| Using generating functions:
\begin{align}
\color{brown}{f(x) = \sum_0^\infty F_n x^n} &= F_0x^0 + F_1x^1+F_2 x^2 + F_3 x^3+ F_4 x^4+\cdots\\
xf(x) &= \quad\quad\quad\, F_0x^1+F_1x^2+F_2x^3+F_3x^4+\cdots\\
x^2f(x)&=\quad\quad\quad\quad\quad\quad\,\,F_0x^2+F_1x^3 + F_2x^4+\cdots
\end{align}
Since the value of $F_0=0$ and $F_1=F_2=1$, while the recursive formula is $F_{n+2}=F_{n+1}+F_{n}$,
\begin{align}
f(x) &= F_1x^1+F_2 x^2 + F_3 x^3+ F_4 x^4+\cdots\\
-\big[ xf(x) &= \quad\quad\quad\, F_1x^2+F_2x^3+F_3x^4+\cdots\big]\\
-\big[x^2f(x)&=\quad\quad\quad\quad\quad\quad\,\,F_1x^3 + F_2x^4+\cdots\big]\\
\quad\\
&\hline\\
(1-x-x^2)f(x)&=x\\
f(x)&=\frac{x}{1-x-x^2}
\end{align}
The denominator is a quadratic with roots $0=1-x-x^2$; $\large x=\frac{1\pm\sqrt{5}}{-2}$, or $-\phi$ and $-\psi:$
$$\color{red}{\varphi=\frac{1+\sqrt{5}}{2}} \text{ and } \psi=\frac{1-\sqrt{5}}{2}$$
Hence,
\begin{align}
(1-x-x^2)f(x)&=x\\
f(x)&=\frac{x}{1-x-x^2}\\
f(x)&=\frac{-x}{(x+\varphi)(x-\psi)}\\
f(x)&=\frac{A}{x+\varphi}+\frac{B}{x-\psi}\tag 1
\end{align}
Therefore,
$$-x=A(x-\psi)+B(x+\varphi)$$
If $x=-\varphi$, $A=-\frac{\varphi}{\sqrt{5}}$, and if $x=-\psi$, $B=\frac{\psi}{\sqrt{5}}.$ Going back to Eq. 1:
$$f(x)=\frac{1}{\sqrt{5}}\left(\color{blue}{\frac{\psi}{x+\psi}}-\color{red}{\frac{\varphi}{x+\varphi}}\right)\tag 2$$
Since $\varphi=-\frac{1}{\psi}$,
$$\color{blue}{\frac{\psi}{x+\psi}}= \frac{1}{1+\frac{x}{\psi}}
=\frac{1}{1-\varphi x}
=\sum_{n=0}^\infty \varphi^n x^n\tag{*}
$$
(*) using the formula for the geometric series, $\sum_0^\infty x^n= \frac{1}{1-x}.$ In the case of
$$\color{red}{\frac{\varphi}{x+\varphi}}=\sum_{n=0}^\infty \psi^n x^n$$
Going back to Eq.2,
$$f(x)=\frac{1}{\sqrt{5}}\left(\sum_{n=0}^\infty \varphi^n x^n-\sum_{n=0}^\infty \psi^n x^n\right)=\sum_0^\infty \frac{1}{\sqrt{5}}\left( \varphi^n - \psi^n \right)x^n=\color{brown}{\sum_0^\infty F_n x^n}$$
So we get Binet's formula:
$$\large F_n=\frac{1}{\sqrt{5}}(\varphi^n-\psi^n)\tag{Binet's formula}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/268056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
how many number like $119$ How many 3-digits number has this property like $119$:
$119$ divided by $2$ the remainder is $1$
119 divided by $3$ the remainderis $2 $
$119$ divided by $4$ the remainder is $3$
$119$ divided by $5$ the remainder is $4$
$119$ divided by $6$ the remainder is $5$
| Note that
$$x \equiv 1 \mod 2 \\ x \equiv 2 \mod 3 \\ x \equiv 3 \mod 4 \\ x \equiv 4 \mod 5 \\ x \equiv 5 \mod 6$$
is equivalent to
$$x \equiv -1 \mod 2 \\ x \equiv -1 \mod 3 \\ x \equiv -1 \mod 4 \\ x \equiv -1 \mod 5 \\ x \equiv -1 \mod 6$$
And by the Chinese Remainder Theorem, the solution to the latter is easily found to be $x \equiv -1 \mod{\mathrm{lcm}(2,3,4,5,6)}$, or $x \equiv -1 \mod 60$.
So just count the number of $3$-digit numbers $x$ satisfying $x \equiv -1 \mod 60$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/268619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Differential equation $xy^3y'=2y^4+x^4$ Solve the differential equation
$$xy^3y'=2y^4+x^4$$
| Another method is to realise that $y^3y'$ is, up to a constant factor, the derivative of $y^4$. So we obtain:
\begin{align*}
xy^3y' &= 2y^4 + x^4\\
x\frac{d}{dx}\left(\frac{y^4}{4}\right) - 2y^4 &= x^4\\
\frac{d}{dx}(y^4) - \frac{8}{x}y^4 &= 4x^3\\
\end{align*}
where, in the last step, we need $x \neq 0$. Now you can solve this by using an integrating factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/269086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
$ \displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$ $$ \ X_n=\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$$ Find $\displaystyle\lim_{n\to\infty} X_n$ using the squeeze theorem
I tried this approach:
$$
\frac{1}{\sqrt{n^3+1}}\le\frac{1}{\sqrt{n^3+1}}<\frac{n}{\sqrt{n^3+1}}
$$
$$
\frac{1}{\sqrt{n^3+1}}<\frac{2}{\sqrt{n^3+2}}<\frac{n}{\sqrt{n^3+1}}$$
$$\vdots$$
$$\frac{1}{\sqrt{n^3+1}}<\frac{n}{\sqrt{n^3+n}}<\frac{n}{\sqrt{n^3+1}}$$
Adding this inequalities:
$$\frac{n}{\sqrt{n^3+1}}\leq X_n<\frac{n^2}{\sqrt{n^3+1}}$$
And this doesn't help me much. How should i proced?
| Since
$$
X_n=\sum_{k=1}^n\frac{k}{\sqrt{n^3+k}}\ge\sum_{k=1}^n\frac{k}{\sqrt{n^3+n}}=\frac{n^2+n}{2\sqrt{n^3+n}},
$$
it follows that
$$
\lim_{n \to \infty}X_n \ge \lim_{n \to \infty}\frac{n^2+n}{2\sqrt{n^3+n}}=\infty,
$$
i.e. $\lim_{n \to \infty}X_n=\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/270216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Recurrence relation by substitution I have an exercise where I need to prove by using the substitution method the following
$$T(n) = 4T(n/3)+n = \Theta(n^{\log_3 4})$$
using as guess like the one below will fail, I cannot see why, though, even if I developed the substitution
$$T(n) ≤ cn^{\log_3 4}$$
finally, they ask me to show how to substract off a lower-order term to make a substitution proof work. I was thiking about using something like
$$T(n) ≤ cn^{\log_3 4}-dn$$
but again, I cannot see how to verify this recurrence.
What I did:
$$T(n) = 4T(n/3)+n$$
$$\qquad ≤ \frac{4c}{3}n^{\log_3 4} + n$$
and then from here, how to proceed and conclude that the first guess fails?
The complete exercise says:
Using the master method, you can show that the solution to the recurrence $T(n) = 4T(n/3)+n$ is $T(n)=\Theta(n^{log_3 4})$. Show that a substitution proof with the assumption $T(n) ≤ cn^{log_3 4}$ fails. Then show how to subtract off a lower-order term to make a substitution proof work.
| Let $n = 3^{m+1}$. Then we have
$$T \left(3^{m+1} \right) = 4T \left(3^{m} \right) + 3^{m+1}$$
Let $T\left(3^m\right) = g(m)$. We then get that
$$g(m+1) = 4g(m) + 3^{m+1} = 4(4g(m-1) + 3^m) + 3^{m+1} = 4^2 g(m-1) + 4\cdot 3^m + 3^{m+1}\\
= 4^2 (4g(m-2) + 3^{m-1}) + 4\cdot 3^m + 3^{m+1} = 4^3 g(m-3) + 4^2 \cdot 3^{m-1} + 4 \cdot 3^m + 3^{m+1}$$
Let $\lfloor m \rfloor = M$
Hence, by induction we get that
$$g(m+1) = 4^M g(m-M) + \sum_{k=0}^{M-1} 4^k \cdot 3^{m+1-k} = 4^M g(m-M) + 3^{m+1} \sum_{k=0}^{M-1} (4/3)^k\\
= 4^M g(m-M) + 3^{m+1} \dfrac{(4/3)^M-1}{4/3-1}= 4^M g(m-M) + 3^{m+2} ((4/3)^M-1)$$
Hence, we get that
$$g(m+1) = 4^M g(m-M) + 3^{m+2-M} \cdot 4^M - 3^{m+2}$$
To get the order, we can take $m$ to be an integer.
Hence, $m=M = \log_3(n)-1$ and we get that
$$g(m+1) = 4^{\log_3(n)-1} g(0) + 3^{2} \cdot 4^{\log_3(n)-1} - 3^{\log_3(n)+1}$$
Note that $4^{\log_3(n)} = n^{\log_3(4)}$. Hence, we get that
$$T(n) = \dfrac{g(0)}4 n^{\log_3(4)} + \dfrac94 \cdot n^{\log_3(4)} - 3n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/270591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find basis of im, ker and dim im, dim ker verification In my homework, I've to find basis of im and ker of linear transformation
$\varphi:\mathbb{R}^{3}\rightarrow\mathbb{R}^{2}, \varphi((x_{1},x_{2},x_{3}))=(2x_{1}+x_{2}-3x_{3},x_{1}+4x_{2}+2x_{3})$
My solution:
Kernel of $\varphi$ is described by a matrix representing set of equations
$\left[ \begin{matrix}2 & 1 & -3 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right] \sim\left[ \begin{matrix}0 & -7 & -7 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right] \sim\left[ \begin{matrix}0 & 1 & 1 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right]$
General solution : lin (2, -1, 1) - that's the ker $\varphi$ , dim ker $\varphi$ = 1
$M(\varphi)_{st}^{st}=\left[ \begin{matrix}2 & 1 & -3\\ 1 & 4 & 2 \end{matrix}\right]$
$(M(\varphi)_{st}^{st})^{T}=\left[ \begin{matrix}2 & 1\\ 1 & 4\\ -3 & 2 \end{matrix}\right] \sim\left[ \begin{matrix}1 & 4\\ 0 & 1\\ 0 & 0 \end{matrix}\right]$
that's im $\varphi$
dim im $\varphi$ = 2
Is idea of my solution correct?
| Kernel of $\varphi$ is described by a matrix representing set of equations
$\left[ \begin{matrix}2 & 1 & -3 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right] \sim\left[ \begin{matrix}0 & -7 & -7 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right] \sim\left[ \begin{matrix}0 & 1 & 1 & 0\\ 1 & 4 & 2 & 0 \end{matrix}\right]$
General solution : lin (2, -1, 1) - that's the ker $\varphi$ , dim ker $\varphi$ = 1
$M(\varphi)_{st}^{st}=\left[ \begin{matrix}2 & 1 & -3\\ 1 & 4 & 2 \end{matrix}\right]$
$(M(\varphi)_{st}^{st})^{T}=\left[ \begin{matrix}2 & 1\\ 1 & 4\\ -3 & 2 \end{matrix}\right] \sim\left[ \begin{matrix}1 & 4\\ 0 & 1\\ 0 & 0 \end{matrix}\right]$
that's im $\varphi$
dim im $\varphi$ = 2
Posted as answer. Thanks for verification!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/272762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Need help to solve the trigonometric equation $\tan\frac {1+x}{2} \tan \frac {1+x}{3}=-1$ How can I solve the equation $\tan\frac {1+x}{2} \tan \frac {1+x}{3}=-1$. Please give me some hint for that. Thank you.
| Assuming $\frac12,\frac13$ are measured in radian,
$$\tan\left(\frac{1+ x}2\right)=-\frac1{\tan\left(\frac{1+ x}3\right)}=-\cot\left(\frac{1+ x}3\right)=\tan\left(\frac\pi2+\frac{x+1}3\right)$$ as $\tan(\frac\pi2+C)=-\cot C$
So, $$\frac{1+ x}2=n\pi+\frac\pi2+\frac{x+1}3$$ where $n$ is any integer as $\tan A=\tan B\implies A=n\pi+B$.
Alternatively, we know $$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
If $\tan A\tan B=-1,\tan(A-B)=\infty\implies A-B=m\pi+\frac\pi2$ where $m$ is any integer
or, $$\tan A\tan B=-1\implies \frac{\sin A}{\cos A}=-\frac{\cos B}{\sin B}$$
$$\implies \cos A\cos B+\sin A\sin B=0\implies \cos(A-B)=0\implies A-B=(2r+1)\frac\pi2$$ where $r$ is any integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/273359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
I am trying to solve the inequality $\log_{\log{\sqrt{9-x^2}}} x^2 <0$ I am trying to solve the inequality
$$\log_{\log{\sqrt{9-x^2}}} x^2 <0.$$
I got $\mathrm{S.S}=(-\sqrt8 ,-1)\cup( 1,\sqrt8)$, but a friend got $\mathrm{S.S}=(-1,1)- \{0\}$.
Please, what is true?
| We shall consider two cases:
(1) Case ($\log \sqrt {9-x^2} >1$)
$$\log_{\log \sqrt{9-x^2}} x^2 <0 \Rightarrow $$
$$\left \{
\begin{array}{l}
0 \neq x^2 < 1 \\
\log \sqrt {9-x^2} >1 \\
\end{array}
\right. \Leftrightarrow
\left \{
\begin{array}{l}
-1 < x < 1 \quad \mathrm {and} \quad x\neq 0\\
\sqrt {9-x^2} > 10 \\
\end{array}
\right. \Leftrightarrow $$
$$ \Leftrightarrow
\left \{
\begin{array}{l}
-1 < x < 1 \quad \mathrm {and} \quad x\neq 0\\
-x^2 > 91 \\
\end{array}
\right. \Leftrightarrow
\left \{
\begin{array}{l}
-1 < x < 1 \quad \mathrm {and} \quad x\neq 0\\
x^2 < -91 \\
\end{array}
\right. \Leftrightarrow $$
$$ \Leftrightarrow S = \emptyset $$
(2) Case ($0 < \log \sqrt {9-x^2} < 1$)
$$\log_{\log \sqrt{9-x^2}} x^2 <0 \Rightarrow $$
$$\left \{
\begin{array}{l}
x^2 > 1 \\
0 < \log \sqrt {9-x^2} <1 \\
\end{array}
\right. \Leftrightarrow
\left \{
\begin{array}{l}
x< -1 \quad \mathrm{or} \quad x > 1\\
1< \sqrt {9-x^2} < 10 \\
\end{array}
\right. \Leftrightarrow $$
$$ \Leftrightarrow
\left \{
\begin{array}{l}
x< -1 \quad \mathrm {or} \quad x > 1\\
1< 9-x^2 < 100 \\
\end{array}
\right. \Leftrightarrow
\left \{
\begin{array}{l}
x< -1 \quad \mathrm {or} \quad x > 1\\
-8< -x^2 < 91 \\
\end{array}
\right. \Leftrightarrow $$
$$ \Leftrightarrow
\left \{
\begin{array}{l}
x< -1 \quad \mathrm {or} \quad x > 1\\
8>\ x^2 > -91 \\
\end{array}
\right. \Leftrightarrow
\left \{
\begin{array}{l}
x< -1 \quad \mathrm {or} \quad x > 1\\
-\sqrt{8}< x < \sqrt{8} \\
\end{array}
\right. \Leftrightarrow $$
$$ \Leftrightarrow S = (-\sqrt{8},-1) \cup (1,\sqrt{8}) $$
Therefore you are right and not your friend.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding polynomial function's zero values not native English speaker so I may get some terms wrong and so on.
On to the question:
I have as an assignment to find a polynomial function $f(x)$ with the coefficients $a$, $b$ and $c$ (which are all integers) which has one root at $x = \sqrt{a} + \sqrt{b} + \sqrt{c}$.
I've done this with $f(x) = 0$ where $x = \sqrt{a} + \sqrt{b}$ through an iterative method which looks like this (forgive me for my pseudo code):
*
*var x = sqrt(a) + sqrt(b)
*Set x to x multiplied by the conjugate of x
*Repeat from step 2 until all square roots are gone
The full calculation looks like this (the exponent signs disappeared, sorry about that):
$(x-(\sqrt{a}+\sqrt{b}))(x+(\sqrt{a}+\sqrt{b}))
= x^2-(\sqrt{a} +\sqrt{b})^2 \\
= x^2-(a+2\sqrt{a}\sqrt{b} +b) \\
= x^2-a-b-2\sqrt{a}\sqrt{b} $
$((x^2-a-b)-(2\sqrt{a}\sqrt{b} ))(( x^2-a-b)+(2\sqrt{a}\sqrt{b} ))
= x^4-2ax^2-2bx^2+2ab+a^2+b^2-4ab \\
= x^4-2ax^2-2bx^2-2ab+a^2+b^2 \\
= x^4-2x^2(a+b)-2ab+a^2+b^2$
$p(x) = x^4-2x^2(a+b)-2ab+a^2+b^2$
And then I used the factor theorem to calculate the remaining roots, which gave the following results:
$x = (\sqrt{a} +\sqrt{b} )$, $x = -(\sqrt{a} +\sqrt{b} )$, $x = (\sqrt{a} -\sqrt{b} )$ and $x = -(\sqrt{a} - \sqrt{b})$.
When I try the same method on $x = \sqrt{a} + \sqrt{b} + \sqrt{c}$, the calculations just become absurd. Any kind of help would be enormously appreciated!
| You certainly have the right idea. Following Gerry Myerson's suggestion above, we have
\begin{align}
x&=\sqrt{a}+\sqrt{b}+\sqrt{c}\\
x-\sqrt{a}&=\sqrt{b}+\sqrt{c}\\
(x-\sqrt{a})^2=x^2-2x\sqrt{a}+a&=b+2\sqrt{bc}+c=(\sqrt{b}+\sqrt{c})^2\\
x^2+a-b-c&=2x\sqrt{a}+2\sqrt{bc}\\
\left(x^2+a-b-c\right)^2&=4x^2a+8x\sqrt{abc}+4bc\\
\left(x^2-a+b+c\right)^2-4x^2a-4bc&=8x\sqrt{abc}\\
\left(\left(x^2-a+b+c\right)^2-4x^2a-4bc\right)^2&=64abc\cdot x^2.
\end{align}
Expanding this will yield a polynomial with integer coefficients (assuming $a$, $b$, and $c$ are integers) which has a zero at $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/276111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is my limit evaluation correct? I'm studying for my calculus exam and I have the following limit:
$$\lim\limits_{n \to \infty} \left ( \frac{1}{\sqrt{n^3 +3}}+\frac{1}{\sqrt{n^3 +6}}+ \cdots +\frac{1}{\sqrt{n^3 +3n}} \right )$$
My solution is:
$$\begin{align*}
&\lim\limits_{n\ \to \infty} \left ( \frac{1}{\sqrt{n^3 +3}}+\frac{1}{\sqrt{n^3 +6}}+ \cdots +\frac{1}{\sqrt{n^3 +3n}} \right )\\&= \lim_{n \to \infty}\left ( \frac{1}{\sqrt{n^3 +3}} \right ) + \lim_{n \to \infty}\left ( \frac{1}{\sqrt{n^3 + 6}} \right ) + \cdots + \lim_{n \to \infty} \left ( \frac{1}{\sqrt{n^3 +3n}} \right )\\
&=0 + 0 + \cdots + 0\\
&= 0
\end{align*}$$
It turned out to be suspiciously easy to solve.
Is this correct? If it isn't, what is wrong and how can I solve it correctly?
| The answer is correct, the reasoning is not. First we solve the problem.
The sum has $n$ terms. Each term is $\lt \frac{1}{n^{3/2}}$. It follows that if $S_n$ is our sum, then
$$0\lt S_n \lt \frac{n}{n^{3/2}}=\frac{1}{n^{1/2}}.$$
Now let $n\to\infty$. Note that $\frac{1}{n^{1/2}}\to 0$, so by Squeezing $S_n\to 0$.
Remark: It is true that each term in the sum $S_n$ approaches $0$. However, the number of terms in the sum increases. Note for example that the sum
$$T_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
does not approach $0$, for clearly the sum is always $\ge \frac{1}{2}$. Yet if we applied the "$0+0+\cdots+0$" reasoning, we would conclude incorrectly that $\lim_{n\to\infty} T_n =0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/276178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to show that $\frac{x^2}{x-1}$ simplifies to $x + \frac{1}{x-1} +1$ How does $\frac{x^2}{(x-1)}$ simplify to $x + \frac{1}{x-1} +1$?
The second expression would be much easier to work with, but I cant figure out how to get there.
Thanks
| A general $^1$ method is to perform the polynomial long division algorithm in the following form or another equivalent one:
to get
$$
x^{2}=(x-1)(x+1)+1.
$$
Divide both sides by $x-1$:
$$\begin{eqnarray*}
\frac{x^{2}}{x-1} &=&\frac{(x-1)(x+1)+1}{x-1}=\frac{(x-1)(x+1)}{x-1}+\frac{1}{x-1} \\
&=&\frac{x+1}{1}+\frac{1}{x-1}=x+1+\frac{1}{x-1}=x+\frac{1}{x-1}+1
\end{eqnarray*}
$$
--
$^1$ The very same method can be applied to find (or show) that
$$\frac{x^{4}-2x^{3}-6x^{2}+12x+15}{x^{3}+x^{2}-4x-4}=x-3+\frac{
x^{2}+4x+3}{x^{3}+x^{2}-4x-4};$$
In this case the polynomial long division algorithm corresponds to the following computation (or another equivalent one):
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/278481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
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Some weird equations In our theoreticall class professor stated that from this equation $(C = constant)$
$$
x^2 + 4Cx - 2Cy = 0
$$
we can first get:
$$
x = \frac{-4C + \sqrt{16 C^2 - 4(-2Cy)}}{2}
$$
and than this one:
$$
x = 2C \left[\sqrt{1 + \frac{y}{2C}} -1\right]
$$
How is this even possible?
| This comes from the quadratic formula. See here for more details.
In your case, $$x^2 + 4Cx -2Cy = x^2 + 4Cx +(2C)^2 - (2C)^2-2Cy = (x+2C)^2 - (2C)^2 \left(1 + \dfrac{y}{2C} \right) = 0$$
Hence, we have that
$$(x+2C)^2 = (2C)^2 \left(1 + \dfrac{y}{2C} \right) \implies x + 2C = \pm 2C \sqrt{\left(1 + \dfrac{y}{2C} \right)}$$
Hence, $$x = -2C \pm 2C \sqrt{\left(1 + \dfrac{y}{2C} \right)} = 2C \left( \pm \sqrt{\left(1 + \dfrac{y}{2C} \right)} - 1\right)$$
If you have a constraint that $x$ has the same sign as $C$ (for instance if $x,C > 0$), then $$x = 2C \left(\sqrt{\left(1 + \dfrac{y}{2C} \right)} - 1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/278780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Isolate middle value from $3\times 3$ matrix Not trained in math the solution to this problem is not immediately apparent, plus I am working on a larger problem which I'd rather get to.
I'm trying to isolate the middle value from a $3\times 3$ matrix
Suppose my matrix is
$$\begin{pmatrix}
\\ 5 & 7 & 3
\\ 4 & 13 & 9
\\ 9 & 9 & 1
\end{pmatrix}$$
I'd like to produce a matrix using linear algebra methods to get
$$\begin{pmatrix}
\\ 0 & 0 & 0
\\ 0 & 13 & 0
\\ 0 & 0 & 0
\end{pmatrix}$$
I do not want a simple answer like subtract
$$\begin{pmatrix}
\\ 5 & 7 & 3
\\ 4 & 0 & 9
\\ 9 & 9 & 1
\end{pmatrix}$$
Additionally, what is the terminology for this kind of derived matrix if there is one?
$$\begin{pmatrix}
\\ 0 & 0 & 0
\\ 0 & 13 & 0
\\ 0 & 0 & 0
\end{pmatrix}$$
| $$\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}\begin{pmatrix}5&7&3\\4&13&9\\9&9&1\end{pmatrix}\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix} = \begin{pmatrix}0&0&0\\0&13&0\\0&0&0\end{pmatrix} $$
The first matrix isolates the middle row and the last isolates the middle column.
Edit: These types of matrices are sometimes called $E_{ij}$, where $i$ is the row and $j$ is the column. So these would be $E_{22}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Studying $ u_n = \int_0^1 (\arctan x)^n \mathrm dx$ I would like to find an equivalent of:
$$ u_n = \int_0^1 (\arctan x)^n \mathrm dx$$
which might be: $$ u_n \sim \frac{\pi}{2n} \left(\frac{\pi}{4} \right)^n$$
$$ 0\le u_n\le \left( \frac{\pi}{4} \right)^n$$
So $$ u_n \rightarrow 0$$
In order to get rid of $\arctan x$ I used the substitution
$$x=\tan \left(\frac{\pi t}{4n} \right) $$
which gives:
$$ u_n= \left(\frac{\pi}{4n} \right)^{n+1} \int_0^n t^n\left(1+\tan\left(\frac{\pi t}{4n} \right)^2 \right) \mathrm dt$$
But this integral is not easier to study!
Or: $$ t=(\arctan x)^n $$
$$ u_n = \frac{1}{n} \int_0^{(\pi/4)^n} t^{1/n}(1+\tan( t^{1/n})^2 ) \mathrm dt $$
How could I deal with this one?
| $$
\begin{align}
&\int_0^1\arctan^n(x)\,\mathrm{d}x\\
&=\int_0^{\pi/4}x^n\sec^2(x)\,\mathrm{d}x\\
&=\int_0^{\pi/4}\left(\frac\pi4-x\right)^n\sec^2\left(\frac\pi4-x\right)\,\mathrm{d}x\\
&=\frac{(\pi/4)^{n+1}}{n}\int_0^ne^{-x}\color{#C00000}{e^x\left(1-\frac{x}{n}\right)^n}\color{#00A000}{\sec^2\left(\frac\pi4-\frac{\pi x}{4n}\right)}\,\mathrm{d}x\\
&=\frac{(\pi/4)^{n+1}}{n}\int_0^ne^{-x}\color{#C00000}{\small\left(1-\frac{x^2}{2n}-\frac{8x^3-3x^4}{24n^2}+O\left(\frac1{n^3}\right)\right)}\color{#00A000}{\small\left(2-\frac{\pi x}{n}+\frac{\pi^2x^2}{2n^2}+O\left(\frac1{n^3}\right)\right)}\,\mathrm{d}x\\
&=\frac{(\pi/4)^{n+1}}{n}{\small\left(2-\frac{\pi+2}{n}+\frac{2+3\pi+\pi^2}{n^2}+O\left(\frac1{n^3}\right)\right)}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/281017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Finding $\frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3}+\cdots+\frac{1}{S_{2013}}$ Assume $S_1=1 ,S_2=1+2,S=1+2+3+,\ldots,S_n=1+2+3+\cdots+n$
How to find :
$$\frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3}+\cdots+\frac{1}{S_{2013}}$$
| $$\sum_{n=1}^{2013}\frac{1}{n(n+1)/2}=\sum_{n=1}^{2013}\frac{2}{n(n+1)}=2\sum_{n=1}^{2013}\frac{1}{n(n+1)}=2\sum_{n=1}^{2013}\left(\frac{1}{n}-\frac{1}{n+1}\right)=$$
$$2\sum_{n=1}^{2013}\frac{1}{n}-2\sum_{n=1}^{2013}\frac{1}{n+1}=2\sum_{n=1}^{2013}\frac{1}{n}-2\sum_{n=2}^{2014}\frac{1}{n}=2\sum_{n=1}^{2013}\frac{1}{n}-2\left(\sum_{n=1}^{2013}\frac{1}{n}-1+\frac{1}{2014}\right)=$$
$$=2-\frac{2}{2014}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/281318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a,b$ and $c$ be real numbers.evaluate the following determinant: |$b^2c^2 ,bc, b+c;c^2a^2,ca,c+a;a^2b^2,ab,a+b$| Let $a,b$ and $c$ be real numbers. Evaluate the following determinant:
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}$$
after long calculation I get that the answer will be $0$. Is there any short processs? Please help someone thank you.
| We can assume $\,abc\neq 0\,$ , otherwise the determinant is zero at once (why? For example, suppose $\,b=0\,$ . Then either $\,ac=0\,$ and we get a row of zeros, or the 3rd row is a multiple of the 1st one...and likewise if $\,a=0\vee c=0\,$ ):
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}\stackrel{aR_1\,,\,bR_2\,,\,cR_3}{\longrightarrow}\begin{vmatrix}ab^2c^2 &abc& ab+ac\cr a^2bc^2&abc&ab+bc\cr a^2b^2c&abc&ac+bc\cr\end{vmatrix}\stackrel{R_2-R_1\,,\,R_3-R_1}\longrightarrow$$
$$\begin{vmatrix}ab^2c^2 &abc& ab+ac\cr abc^2(a-b)&0&-c(a-b)\cr ab^2c(a-c)&0&-b(a-c)\cr\end{vmatrix}\stackrel{\text{develop by 2nd column}}\longrightarrow$$
$$-abc\begin{vmatrix}abc^2(a-b)&-c(a-b)\\ab^2c(a-c)&-b(a-c)\end{vmatrix}$$
And now factor out $\,c(a-b)\,\,,\,\,b(a-c)\,$ from first and second row resp. , and get
$$-ab^2c^2(a-b)(a-c)\begin{vmatrix}abc&-1\\abc&-1\end{vmatrix}=0$$
Finally, pay attention to the fact that the original determinant is zero (or not) without any regard to the above operations
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Derivative for log I have the following problem:
$$
\log \bigg( \frac{x+3}{4-x} \bigg)
$$
I need to graph the following function so I will need a starting point, roots, zeros, stationary points, inflection points and local minimum and maximum and I need to know where the function grows and declines.
I calculated roots zeros $ x + 3= 0$, $x=-3$ and roots $ 4-x=0$,$x=-4 $.
Now I sort of know how to graph the function from here but how do I get the stationary points do I have to find the derivative of $\log \left( \frac{x+3}{4-x} \right)$ or just $ \left( \frac{x+3}{4-x} \right)$.
I don't fully understand how to find the derivative of $\log$. Can i use the $\log(x)' = \frac{1}{x} $ rule here to get $ \frac{1}{\frac{x+3}{4-x}} $ and then find stationary points here ?
| For the stationary points you need to find $\frac{d}{dx} \left(\log \bigg( \frac{x+3}{4-x} \bigg)\right)$. You need to use the chain rule here $\frac{d}{dx}\log(f(x))=\frac{1}{f(x)}\cdot f'(x)$ which would give:
$$\frac{d}{dx} \left(\log \bigg( \frac{x+3}{4-x} \bigg)\right)= \frac{1}{\frac{x+3}{4-x}} \cdot \frac{d}{dx}\left(\frac{x+3}{4-x}\right)=\frac{4-x}{x+3}\left( \frac{(4-x)+(x+3)}{(4-x)^2}\right)=\frac{7}{(x+3)(4-x)}$$
Hence, the function has no stationary points. Also, $ \log \bigg( \frac{x+3}{4-x} \bigg)$ has zeros when $\frac{x+3}{4-x}=1$ because $\log(1)=0$. So for the zeroes you need to solve $x+3=4-x$ giving $x=\frac{1}{2}$. It will also have asymptotes where the derivative goes to infinity i.e. at $x=-3$ and $x=4$, the first going to $-\infty$ and the second to $+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/284193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the solution of the differential equation
Find the solution of the differential equation
$$\frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)}, y(0)=1$$
Trial: $$\begin{align} \frac{dy}{dx}=-\frac{x(x^2+y^2-10)}{y(x^2+y^2+5)} \\ \implies \frac{dy}{dx}=-\frac{1+(y/x)^2-10/x^2}{(y/x)(1+(y/x)^2+5/x^2)} \\ \implies v+x\frac{dv}{dx}=-\frac{1+v^2-10/x^2}{v(1+v^2+5/x^2)} \end{align}$$ I can't seperate $x$ and $v$.
| Your differential equation can be written as
$$ (x^{3}+x y^{2}-10 x)dx+(x^{2}y+y^{3}+5y)dy=0$$
So it is of the form $M(x,y)dx+N(x,y)dy=0$
Now, $$M_{y}=2xy$$ and $$N_{x}=2xy.$$
$\therefore$ The given differential equation is exact.
So its general solution is
$$\int M dx+\int (\text{Terms in $N$ which does not contain $x$})dy=C$$
$$\Rightarrow \int (x^{3}+x y^{2}-10 x)dx+\int 0 dy=C$$
$$\Rightarrow \frac{x^{4}}{4}+\frac{x^{2}y^{2}}{2}-5x^{2}=c$$
Now $y(0)=1$, we have $C=0$.
$\therefore$ The solution of given differential equation is
$\displaystyle \frac{x^{4}}{4}+\frac{x^{2}y^{2}}{2}-5x^{2}=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$? Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$ ?
$f(x)=x^4+x^3+x^2+x+1$
$f(x^5)=x^{20}+x^{15}+x^{10}+x^5+1$
| Rewrite the expression as
$$(x^{20}-1) +(x^{15}-1)+(x^{10}-1)+(x^5-1)+1+4.$$
The expression $x^4+x^3+x^2+x+1$ divides the first four terms, since $x^5-1$ does. So the remainder is $5$.
The idea obviously generalizes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to show $AB^{-1}A=A$
Let $$A^{n \times n}=\begin{pmatrix} a & b &b & \dots & b \\ b & a &b & \dots & b \\ b & b & a & \dots & b \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \dots &a\end{pmatrix}$$
where $a \neq b$ and $a + (n - 1)b = 0$.
Suppose $B=A+\frac{11'}{n}$ , where $1=(1,1,\dots,1)'$ is an $n \times 1$ vector. Show that $AB^{-1}A=A$
Trial: Here $B^{-1}=\frac{1}{\alpha-\beta}I_n-\frac{\beta~11'}{(\alpha-\beta)(\alpha+(n-1)\beta)}$ Where $(\alpha-\beta)=(a+\frac{1}{n}-b-\frac{1}{n})=(a-b)$ and $\alpha+(n-1)\beta=1$ So, $B^{-1}=\frac{1}{a-b}[I_n-(b+\frac{1}{n})11']$. Then pre and post multipling $A$ I can't reach to the desire result . Please help.
| Writing replacing $a$ with $(1-n)b$ and $\mathbf{1}\mathbf{1}^T$ with $E$, we have
$$A = b\left( -n I + E \right) \qquad B = - n b I + \frac{1+nb}{n} E$$
Note that $E^2=nE$, so that
$$A^2 = b^2 \left(n^2I-2nE+E^2\right) = -nbA \qquad \text{and}\qquad AE=0$$
By the ever-popular Sherman-Morrison identity,
$$B^{-1} = -\frac{1}{nb} I + k E$$
for some $k$. Whence,
$$AB^{-1}A = -\frac{1}{nb}A^2 + 0= A$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/287917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
Counting Card hands with various restrictions I would like to know if my solutions are correct for the following three combinatorial card questions. In each question, assume we have a standard deck of cards (13 ranks, and 4 suits).
*
*How many ways are there to select $7$ cards so that we have at least $2$ distinct suits?
${\bf Solution:}$ Count the total number of ways to select $7$ cards from $52$ and subtract the number of ways to select only one suit. This is:
\begin{align}
{52 \choose 7} - {13 \choose 7}
\end{align}
*
*Count the number of ways to select $7$ cards so that no $2$ cards have the same rank?
${\bf Solution:}$ By definition of ${n\choose k}$ this is just $13 \choose 7$
*
*How many ways are there to select $7$ cards so that two distinct ranks occur twice, and three distinct ranks occur once?
${\bf Solution:}$ First we choose $5$ distinct ranks which is ${13 \choose 5}$. Then choose the $2$ suits for one rank, $2$ suits for the other rank, and then three remaining suits. This is: \begin{align} {13 \choose 5}\cdot{4 \choose 2}\cdot{4 \choose 2}\cdot4\cdot4\cdot4
\end{align}
| For the first question, there are $4\dbinom{13}{7}$ "bad" (one suit) hands, so the answer needs to be changed somewhat.
For the second question, there are $\dbinom{13}{7}$ ways to select the ranks that will be represented. For each rank, there are $4$ ways to choose the actual card, for a total of $\dbinom{13}{7}4^7$.
The last question is trickier. There are $\dbinom{13}{2}$ ways to choose which two ranks we will have $2$ cards in, and for each choice there are $\dbinom{11}{3}$ ways to choose the ranks we will have singles of.
For each "double" rank selected, there are $\dbinom{4}{2}$ ways to choose the actual cards. For each "single" ranks selected, there are $\dbinom{4}{1}$ (that is, $4$) ways to choose the actual card, for a total of
$$\binom{13}{2}\binom{11}{3}\binom{4}{2}^2\binom{4}{1}^3.$$
Your calculation for the last question almost works. If, as you did, we first select $5$ ranks, we must then choose from these $5$ the $2$ that we will have doubletons in. If we replace your $\dbinom{13}{5}$ by $\dbinom{13}{5}\dbinom{5}{2}$, we get an expression which is equal to my $\dbinom{13}{2}\dbinom{11}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A functional equation problem on $\mathbb{Q}^{+}$: $f(x)+f\left(\frac{1}{x}\right)=1$ and $f(2x)=2f\bigl(f(x)\bigr)$
Let $f$ be a function which maps $\mathbb{Q}^{+}$ to $\mathbb{Q}^{+}$ and satisfies
$$
\begin{cases}
f(x)+f\left(\frac{1}{x}\right)=1\\
f(2x)=2f\bigl(f(x)\bigr)
\end{cases}
$$
Show that $f\left(\frac{2012}{2013}\right)=\frac{2012}{4025}$.
I tried to prove that $f(x)=\frac{x}{x+1}$(which satisfies all these comditions), but failed.
So I tried induction:
Surely, $f(1)=\frac{1}{2}$ by $f(1)+f\left(\frac{1}{1}\right)=1$.
By $f(2)=2f\bigl(f(1)\bigr)=2f\left(\frac{1}{2}\right)$, we know $f\left(\frac{1}{2}\right)=\frac{1}{3}$ and $f(2)=\frac{2}{3}$.
Note that $f(1)=2f\Bigl(f\left(\frac{1}{2}\right)\Bigr)$, we have $f\left(\frac{1}{3}\right)=\frac{1}{4}$.
By $f(4)=2f\bigl(f(2)\bigr)=2f\left(\frac{2}{3}\right)$ as well as $f\left(\frac{2}{3}\right)=2f\left(\frac{1}{4}\right)$, we know $f\left(\frac{1}{4}\right)=\frac{1}{5}$, $f\left(\frac{2}{3}\right)=\frac{2}{5}$ and $f(4)=\frac{4}{5}.$
Unfortunately, these procedures seems to have no similarity, and the larger the denominator, the more complex the procedure is. Is there any hints or solutions? Thanks for attention!
| Use induction on $q$. There are gaps in the proof which I cannot fill yet.
First establish the hypothesis for $q=1$.
We already know that $f(1)=1$ and $f(2)=2/3$. Let's assume that $f(x)=\frac{x}{x+1}$ for all integer $x<=2n$. Then we have: $
f(\frac{1}{2n})=\frac{1}{2n+1}; $ hence $2f(f(\frac{1}{2n}))=2f(\frac{1}{2n+1})$ or $f(\frac{1}{n}) = 2f(\frac{1}{2n+1})$ Hence we've proven it for $x=2n+1$.
I cannot yet establish the case $x=2n+2$ Once this is done we'll have established the case $q=1$.
Next establish the hypothesis for $q=2$. The first few cases are calculated by hand. Assume the hypothesis $f(p/2)=p/(p+2)$ holds for $p<=2n$ and we'll prove it for $p=2n+1$. Use that because of the induction we have: $\frac{2}{2n+1}=f(\frac{2}{2n-1})$. This may be enough to prove this step.
Now for the main induction step we will try to prove that $f(p/q)=p/(p+q)$ Note that: $p/(q+1)=f(p/(q+1-p))$ due to the induction. Apply $f$ to both sides, multiply by 2 and use the second functional equation to calculate $f(p/(q+1))$. That will prove the induction step.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The figure below shows the graph of a function $f(x)$. How many solutions does the equation $f(f(x)) = 15$ have? From the graph we can find that, $f(-5)=f(-1)=f(9)=0$
| We are to find the solutions for $f(f(x)) = 15$.
From the graph, $f(4) = 15$ and $f(12) = 15$.
The required solutions will be those values of $x$ for which $f(x) = 4$ and $f(x) = 12$.
From the graph, the value of function $f(x) = 4$ at four different values of $x$, namely $x = –8, 1, 7.5, 10$.
The value of the function $f(x) = 12$ at three different points, namely $x = 3.5, 5.25, 11.5$.
Hence, the given equation has 7 solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/292119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing the value of $\operatorname{Li}_{3}\left(\frac{1}{2} \right) $
How to prove the following identity
$$ \operatorname{Li}_{3}\left(\frac{1}{2} \right) = \sum_{n=1}^{\infty}\frac{1}{2^n n^3}= \frac{1}{24} \left( 21\zeta(3)+4\ln^3 (2)-2\pi^2 \ln2\right)\,?$$
Where $\operatorname{Li}_3 (x)$ is the trilogarithm also the result from above can be found here in $(2)$.
In particular for $x=\frac12$ we have:$$\operatorname{Li}_3\left(\frac12\right)=\sum_{n=1}^\infty \frac{1}{n^3}\frac{1}{2^n}=\frac14 \sum_{n=1}^\infty \frac{1}{2^{n-1}} \int_0^\frac12 x^{n-1}\ln^2 xdx=\frac14 \int_0^1 \frac{\ln^2 x}{1-\frac{x}{2}}dx$$
| Here is a partial solution:
Substitute $x=2\sin^2 \theta$ to obtain:
$$\int_0^{\frac{\pi}{4}} \left(\ln^2 2 + 4 \ln 2 \ln (\sin \theta) + 4 \ln^2 (\sin \theta)\right) \tan \theta \, \text{d}\theta$$
The $\ln^2 2$ part is trivial and evaluates to $\frac{\ln^3 2}{2}$.
The $\ln 2 \ln (\sin \theta)$ part is relatively straightforward.
Let $\displaystyle I = 4\ln 2 \int_0^\frac{\pi}{4} \ln (\sin \theta) \tan{\theta} \, \text{d}\theta$.
Integrate by parts to obtain $\displaystyle I = -\ln^3 2 + 4\ln 2 \int_0^\frac{\pi}{4} \ln (\cos \theta) \cot{\theta} \, \text{d}\theta$
Substitute $\theta \mapsto \frac{\pi}{2} - \theta$ to obtain $\displaystyle I = -\ln^3 2 + 4\ln 2 \int_\frac{\pi}{4}^\frac{\pi}{2} \ln (\sin \theta) \tan{\theta} \, \text{d}\theta$
which can then be averaged to obtain $\displaystyle I = -\frac{\ln^3 2}{2} + 2\ln 2 \int_0^\frac{\pi}{2} \ln (\sin \theta) \tan{\theta} \, \text{d}\theta$
Then use the substitution $u = \cos^2 \theta$ or $u = \sin^2 \theta$ to obtain the classic dilogarithm integral for $\zeta(2)$.
This simplifies down to $\displaystyle I = -\frac{\ln^3 2}{2} - \frac{\pi^2 \ln 2}{12}$
It then remains to show that $\displaystyle 4 \int_0^\frac{\pi}{4} \ln^2 (\sin \theta) \tan \theta = \frac{\ln^3 2}{6} + \frac{7}{8} \zeta(3)$.
I will continue attempting this integral and come back when/if I progress.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factoring $3x^2 - 10x + 5$ How can $3x^2 - 10x + 5$ be factored? FOIL seemingly doesn't work (15 has no factors that sum to -10).
| You can always use the general formula for a quadratic equation, if we have $ax^2+bx+c$ then let $$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ and $$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$ then we can write $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ so in particular if we have $3x^2-10x+5$ then $$x_1=\frac{10+\sqrt{40}}{6}$$ and $$x_2=\frac{10-\sqrt{40}}{6}$$ which implies that $$3x^2-10x+5=3(x-\frac{10+\sqrt{40}}{6})(x-\frac{10-\sqrt{40}}{6}).$$
| {
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"url": "https://math.stackexchange.com/questions/297667",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defined a function $f(x)$ such that
$$
f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}}
$$
We can see that
$$\begin{align}
f(0) &= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\
&= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
\end{align}$$
And we begin solving by
$$\begin{align}
f(x) &= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\
f(x)^2 &= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\
&= 2^x + f(x + 1) \\
f(x + 1) &= f(x)^2 - 2^x
\end{align}$$
At this point I find myself stuck, as I have little experience with recurrence relations.
How would this recurrence relation be solved? Would the method extend easily to
$$\begin{align}
f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\
f_n(x)^2 &= n^x + f_n(x + 1)~\text ?
\end{align}$$
| I'd rather try this:
If
$$x=\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}},$$
then
$$\begin{align}ux&=u\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+u^2\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+u^4\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+u^8\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+\sqrt{8u^{16}+\ldots}}}}\\
\end{align}$$
Thus if $u=\frac1{\sqrt 2}$ then
$$ \frac x{\sqrt2}=\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\ldots}}}}$$
The right hand side $y$ has the property that $y^2-\frac12=y$, i.e. can be found by solving a quadratic.
| {
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"url": "https://math.stackexchange.com/questions/300299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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inverse of function Thanks for the help! Here is the solution..
i had a problem:
$$f(x)=\frac{(\sqrt x+8)}{(5-\sqrt x)}$$
i had to find the inverse, so lets begin...
1) first i write in terms of $y$
$$y=\frac{(\sqrt x+8)}{(5-\sqrt x)}$$
2) now try to get $x$ by itself
$$(5-\sqrt x)y=\sqrt x+8$$
3)Distribute y along the left hand side both sides
$$(5y-y\sqrt x)=\sqrt x+8$$
4)subtract 8 and from both sides and add $$ y\sqrt x $$ to both sides
$$ 5y-8=y\sqrt{x} + \sqrt x$$
5) factor right hand side
$$ 5y-8=\sqrt{x} (y + 1)$$
6)divide out $$ (y+1) $$ from right side
$$ ((5y-8)/(y+1))=\sqrt x $$
7) square both sides leaving
$$ x=((5y-8)/(y+1))^2$$
8) so the solution is $$f^-1(x)=((5x-8)/(x+1))^2$$
| I have a problem:
$$f(x)= \frac{\sqrt{x}+8}{5-\sqrt{x}}$$
I have to find the inverse but my calculations are off. I have listed what I did below, could someone tell me where I went wrong? Thank You in advance.
1) Write in terms of y
$$y = \frac{\sqrt{x}+8}{5-\sqrt{x}}$$
2) Now try to get $x$ by itself
$$ y(5-\sqrt{x}) = \sqrt{x}+8 $$
3) Multiply the $y$ through the expression
$$ 5y-y\sqrt{x} = \sqrt{x}+8 $$
5) Subtract 8 from both sides
$$ 5y-y\sqrt{x} -8= \sqrt{x} $$
5) Add $y\sqrt{x}$ to both sides
$$ 5y - 8= \sqrt{x} +y\sqrt{x}$$
6) Factor out $\sqrt{x}$
$$ 5y - 8= \sqrt{x}(1+y)$$
7) Divide by $(1+y)$
$$ \frac{5y - 8}{(1+y)}= \sqrt{x}$$
8) Square both sides
$$ \frac{(5y - 8)^2}{(1+y)^2}= x$$
9) Replace $x$ with $y$
$$ f^{-1} = \frac{(5x - 8)^2}{(1+x)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$ for $x$ Is there any smart way to solve the equation: $$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$
Use Maple I can find $x \in \{1;ab+bc+ca\}$
| $$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$
$$\Leftrightarrow (b-c)(1+a^2)+\frac{(x+a^2)(c-a)(1+b^2)}{x+b^2}+\frac{(x+a^2)(a-b)(1+c^2)}{x+c^2}=0$$
$$\Leftrightarrow (b-c)(1+a^2)(x+b^2)+(x+a^2)(c-a)(1+b^2)+\frac{(x+a^2)(x+b^2)(a-b)(1+c^2)}{x+c^2}=0$$
$$\Leftrightarrow (b-c)(1+a^2)(x+b^2)(x+c^2)+(x+c^2)(x+a^2)(c-a)(1+b^2)+(x+a^2)(x+b^2)(a-b)(1+c^2)=0$$
$$\Leftrightarrow (b-c)(1+a^2)(x^2+x(b^2+c^2)+(bc)^4)+(x^2+x(a^2+c^2)+(ac)^4)(c-a)(1+b^2)+(x^2+x(b^2+a^2)+(ba)^4)(a-b)(1+c^2)=0$$
$$\Leftrightarrow (b-c)(1+a^2)(x^2)+x(b-c)(1+a^2)(b^2+c^2)+(bc)^4(b-c)(1+a^2)+(x^2)(c-a)(1+b^2)+x(c-a)(1+b^2)(a^2+c^2)+(ac)^4(c-a)(1+b^2)+(x^2)(a-b)(1+c^2)+x(b^2+a^2)(a-b)(1+c^2)+(ba)^4(a-b)(1+c^2)=0$$
$$\Leftrightarrow [(a-b)(1+c^2)+(b-c)(1+a^2)+(c-a)(1+b^2)](x^2)+x[(b^2+a^2)(a-b)(1+c^2)+(b-c)(1+a^2)(b^2+c^2)+(b^2+a^2)(a-b)(1+c^2)]+(bc)^4(b-c)(1+a^2)+(ac)^4(c-a)(1+b^2)(ba)^4(a-b)(1+c^2)=0$$
$$\Leftrightarrow x^2+\frac{(b^2+a^2)(a-b)(1+c^2)+(b-c)(1+a^2)(b^2+c^2)+(b^2+a^2)(a-b)(1+c^2)}{(a-b)(1+c^2)+(b-c)(1+a^2)+(c-a)(1+b^2)}x+\frac{(bc)^4(b-c)(1+a^2)+(ac)^4(c-a)(1+b^2)(ba)^4(a-b)(1+c^2)}{(a-b)(1+c^2)+(b-c)(1+a^2)+(c-a)(1+b^2)}=0$$
You can now simplify and solve the quadratic.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find $\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}.$ Find
$$\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}\;.$$
| $$
\frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = \frac{1}{\log n}\sum_{m=1}^n\left(2\sum_{i=1}^m \frac{m+i}{m^3+i^3} - \frac{1}{m^2}\right)
$$
Since
$$
\lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n \frac{1}{m^2} = 0,
$$
we conclude
$$
\lim_{n\to\infty} \frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = 2 \lim_{n\to\infty}\frac{1}{\log n}\sum_{m=1}^n\left(\sum_{i=1}^m \frac{m+i}{m^3+i^3}\right).
$$
Define
$$
a_m = m \cdot \sum_{i=1}^m \frac{m+i}{m^3+i^3},
$$
so
$$
(1)\lim_{n\to\infty} \frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = \lim_{n\to\infty} 2 \frac{1}{\log n} \sum_{m=1}^n \frac{a_m}{m}.
$$
We can approximate $a_m$ using a Riemann integral, giving
$$
(2) \lim_{m\to\infty} a_m = \int_0^1 \frac{1+x}{1+x^3} = \frac{2\pi}{3\sqrt{3}}.
$$
(I forget how to do that last integral. I used Mathematica.) Recall that
$$
\lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n \frac{1}{m} = 1.
$$
For any sequence $b_m\to 0$,
$$
(3) \lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n b_m \frac{1}{m} = 0.
$$
To see this, suppose $|b_m|<\epsilon$ for $m>N$. Then
$$
\left|\frac{1}{\log n}\sum_{m=1}^n b_m \frac{1}{m}\right| <= \left|\frac{1}{\log n}\sum_{m=1}^N b_m \frac{1}{m}\right| + \left|\frac{1}{\log n}\sum_{m>N}^n b_m \frac{1}{m}\right| \\
<= \left|\frac{1}{\log n}\sum_{m=1}^N b_m \frac{1}{m}\right| + \frac{\epsilon}{\log n}\sum_{m>N}^n \frac{1}{m}.
$$
The limit of the right hand side of this inequality is $\epsilon$. Since $\epsilon$ can be chosen arbitrarily small, (3) follows. As an immediate corollary, if $b_m\to L$, then
$$
(4) \lim_{n\to\infty} \frac{1}{\log n}\sum_{m=1}^n b_m \frac{1}{m} = L.
$$
It follows from (1), (2), and (4) that
$$
\lim_{n\to\infty} \frac{1}{\log n}\sum_{j,k=1}^n \frac{j+k}{j^3 +k^3} = \frac{4\pi}{3\sqrt{3}}.
$$
| {
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"url": "https://math.stackexchange.com/questions/306831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How many zeros are there at $1000!$ in the base $24$ I know $1000!$ has $\frac{1000}{5}+\frac{1000}{25}+\frac{1000}{125}+\frac{1000}{625}=249$ terminal zeros in decimal notation, but what if we write $1000!$ in base $24$, how many terminal zeros would it have? Is there a way to compute it?
Regards
| The power of $2$ in $1000!$ is
$$a=\left\lfloor \frac{1000}{2}\right\rfloor+\left\lfloor\frac{1000}{4}\right\rfloor+\left\lfloor\frac{1000}{8}\right\rfloor+\left\lfloor\frac{1000}{16}\right\rfloor+\left\lfloor\frac{1000}{32}\right\rfloor+\left\lfloor\frac{1000}{64}\right\rfloor+\left\lfloor\frac{1000}{128}\right\rfloor+\left\lfloor\frac{1000}{256}\right\rfloor+\left\lfloor \frac{1000}{512}\right\rfloor$$
The power of $3$ in $1000!$ is
$$b=\left\lfloor\frac{1000}{3}\right\rfloor+\left\lfloor\frac{1000}{9}\right\rfloor+\left\lfloor\frac{1000}{27}\right\rfloor+\left\lfloor\frac{1000}{81}\right\rfloor+\left\lfloor\frac{1000}{243}\right\rfloor+\left\lfloor\frac{1000}{729}\right\rfloor$$
Thus the number of zeroes is $\min \{ \lfloor \frac{a}{3} \rfloor, b \}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Help with Euler Substitution Let $a < 0$. Find the following indefinite integral by using the third Euler substitution.
$$\int \frac{dx}{(x^2 + a^2) \sqrt{x^2 - a^2}}$$
Where the third Euler substitution is defined by:
Given an integral of the form $\int R(x, \sqrt{ax^2 + bx + c})dx,$ $a \neq 0$.
If the quadratic polynomial under the radical has 2 distinct real roots $\alpha_1$ and $\alpha_2$ i.e.,
$$ax^2 + bx + c = a(x-\alpha_1)(x - \alpha_2),$$
then set
$$\sqrt{ax^2 + bx + c} = t(x - \alpha_{1 \text{ or } 2}).$$
Hint: $t^4 + 1 = (t^2 - \sqrt 2t + 1)(t^2 + \sqrt 2t + 1)$.
I tried beginning by setting $$\sqrt{x^2 - a^2} = t(x-a)$$ and attempted to solve for $x$ in order to find some function to define $dx$ in terms of $dt$ but I'm having some trouble. I'm not quite sure how to apply the hint here.
Thanks in advance.
| I went with $\sqrt{x^2-a^2}=t (x+a)$, although I doubt it matters. In any case, there's just a bunch of algebra to work through:
$$t=\sqrt{\frac{x-a}{x+a}} \implies x = a \frac{1+t^2}{1-t^2}$$
$$dx = \frac{4 a t}{(1-t^2)^2} dt$$
$$x^2+a^2 = 2 a^2 \frac{1+t^4}{(1-t^2)^2}$$
$$x^2-a^2 = a^2 \frac{4 t^2}{(1-t^2)^2}$$
so that
$$\begin{align} \int \frac{dx}{(x^2+a^2) \sqrt{x^2-a^2}} &= \frac{2a}{a^3} \int dt \frac{t}{(1-t^2)^2} \frac{(1-t^2)^2}{1+t^4} \frac{1-t^2}{2 t}\\ &= \frac{1}{a^2} \int dt \frac{1-t^2}{1+t^4} \end{align}$$
Now put that hint to good use. Actually, I'll give you a better one:
$$\frac{1-t^2}{1+t^4} = \frac{1}{2 \sqrt{2}} \left ( \frac{2 t + \sqrt{2}}{t^2+\sqrt{2} t+1} - \frac{2 t - \sqrt{2}}{t^2-\sqrt{2} t+1} \right )$$
Note that the fractions are of the form $h'(t)/h(t)$ for some function $h$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Implicit differentiation 2 questions? Hello everyone I have two questions on implicit differentiation.
My first one is express $\frac{dy}{dx}$ in terms of $x$ and $y$ if $x^2-4xy^3+8x^2y=20$
what I did is this
$2x-4x(3y^2)(\frac{dy}{dx})+4y^3+8x^2\frac{dy}{dx}+16xy=0$
$-4x(3y^2\frac{dy}{dx})+8x^2\frac{dy}{dx}=-2x-4y^3-16$
Finally I got $\frac{dy}{dx}=\frac{2x-4y^3-16}{-4(x)(3y^2)+8x^2}$
My second question is
Find an equation for the tangent line to the graph of $2x^2-5y^2+xy=5$ at the point $(2,1)$
Anyway I simplified the equation to
$\frac{dy}{dx}=\frac{-4x-1}{-10y+x}$ plugging in x and y I got $\frac{9}{8}$
so my line is $y-1=\frac{9}{8}(x-2)$ but I am unsure if if I did this correctly.
| There are slight computation errors and omission errors; Other than that you are doing fine. If you experience such errors, it is better to write every step.
In the first one,
when you differentiate both sides you get $2x-4(x3y^2 dy/dx +y^3)+8(2xy+x^2 dy/dx)=0$. Thus $(-4x3y^2+8x^2)dy/dx=-2x+4y^3-16xy$ so that $dy/dx=\frac{-2x+4y^3-16xy}{-12xy^2+8x^2}=\frac{-x+2y^3-8xy}{-6xy^2+4x^2}$
In the second one, a similar error is there.
Differentiating gives $4x-10y dy/dx+(x dy/dx +y)=0$ and thus $(-10y+x)dy/dx=-4x+y$ so that $dy/dx=\frac{-4x-y}{-10y+x}$. The value you get after substitution is accidentally(?) same here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does this sum go to $0$? http://www.math.chalmers.se/Math/Grundutb/CTH/tma401/0304/handinsolutions.pdf
In problem (2), at the very end it says
$$\left(\sum_{k = n+1}^{\infty} \frac{1}{k^2}\right)^{1/2} \to 0$$
I don't see how that is accomplished. I understand the sequence might, but how does the sum $$\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$?
Is one allowed to do the following?
$$\lim_{n \to \infty}\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$
$$=\left (\lim_{n \to \infty}\frac{1}{(n+1)^2}+ \lim_{n \to \infty}\frac{1}{(n+2)^2} + \lim_{n \to \infty}\frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$
$$(0 + 0 + 0 + \dots)^{1/2} = 0$$
| A related problem. Note that,
$$ \sum_{k=n+1}^{\infty} \frac{1}{k^2}= \sum_{k=1}^{\infty} \frac{1}{(k+n)^2} \leq \sum_{k=1}^{\infty} \frac{1}{k^2} <\infty, $$
which implies that the series
$$ \sum_{k=1}^{\infty} \frac{1}{(k+n)^2} $$
converges uniformly. So, we have
$$ \lim_{n\to \infty} \sqrt{ \sum_{k=n+1}^{\infty}\frac{1}{k^2} } = \sqrt{\lim_{n\to \infty} \sum_{k=n+1}^{\infty}\frac{1}{k^2} } = \sqrt{\lim_{n\to \infty} \sum_{k=1}^{\infty}\frac{1}{(k+n)^2} } = \sqrt{ \sum_{k=1}^{\infty}\lim_{n\to \infty}\frac{1}{(k+n)^2} }=0. $$
Notice that, the function $\sqrt{x}$ is a continuous function for $x>0$, which justifies changing the operation $\lim f(a_n) = f( \lim a_n )$.
| {
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"url": "https://math.stackexchange.com/questions/311528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maclaurin series for $\frac{x}{e^x-1}$ Maclaurin series for
$$\frac{x}{e^x-1}$$
The answer is
$$1-\frac x2 + \frac {x^2}{12} - \frac {x^4}{720} + \cdots$$
How can i get that answer?
| An other approach would be to take the inverted first:
$$ \frac{e^x-1}{x} = 1 + \frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24} + \frac{x^4}{120} \cdots $$
and then you go:
$$\frac{x}{e^x-1} = (\frac{e^x-1}{x})^{-1} = [1 + (\frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24} +\frac{x^4}{120})]^{-1} $$
and then expand this as in the binomial expansion $(1+u)^{-1}$ and since here $ u <1$ the previous gets expanded to:
$$ 1 - (\frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24} +\frac{x^4}{120}) +(\frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24})^2 - (\frac{x}{2} + \frac{x^2}{6})^3 + (\frac{x}{2})^4 \tag1 $$ Notice that i only take terms accordingly until the desired $x^4$ terms neglecting the other higher order terms. Now keep expanding $(1)$ while still not taking into account calculations that will result in terms higher than $x^4$ (to avoid unnecessary trouble) and it won't be long until you reach the correct result.
| {
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"url": "https://math.stackexchange.com/questions/311817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of inequality involving logarithms How could we show that
$$\left|\log\left( \left({1 + \frac{1}{n}}\right)^{n + \frac{1}{2}}\cdot \frac{1}{e}\right)\right| \leq \left|\log\left( \left({1 - \frac{1}{n}}\right)^{n - \frac{1}{2}}\cdot \frac{1}{e}\right)\right| ,\; \forall n \text{ sufficiently large?} $$
I already calculated in wolfram the limit of the quotient of the logs when $n \rightarrow \infty$. And it is zero. However, I can't prove it.
| The left hand side is
$$
\begin{align}
&\left|\,\left(n+\frac12\right)\log\left(1+\frac1n\right)-1\,\right|\\
&=\left(n+\frac12\right)\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}-\frac1{4n^4}+\dots\right)-1\\
&=\frac1{3\cdot4n^2}-\frac2{4\cdot6n^3}+\frac3{5\cdot8n^4}-\frac4{6\cdot10n^5}+\dots+\frac{(-1)^k(k-1)}{2k(k+1)n^k}+\dots\tag{1}
\end{align}
$$
The right hand side is
$$
\begin{align}
&\left|\,\left(n-\frac12\right)\log\left(1-\frac1n\right)-1\,\right|\\
&=\left(n-\frac12\right)\left(\frac1n+\frac1{2n^2}+\frac1{3n^3}+\frac1{4n^4}+\dots\right)+1\\
&=2+\frac1{3\cdot4n^2}+\frac2{4\cdot6n^3}+\frac3{5\cdot8n^4}+\frac4{6\cdot10n^5}+\dots+\frac{(k-1)}{2k(k+1)n^k}+\dots\tag{2}
\end{align}
$$
Thus, the left hand side tends to $0$ and the right hand side tends to $2$.
However, assuming that the inequality is actually
$$
\left|\,\log\left(\left(1+\frac1n\right)^{n+\frac12}\cdot\frac1e\right)\,\right| \le\left|\,\log\left(\left(1-\frac1n\right)^{n-\frac12}\cdot e\right)\,\right|\tag{3}
$$
The right hand side is
$$
\begin{align}
&\left|\,\left(n-\frac12\right)\log\left(1-\frac1n\right)+1\,\right|\\
&=\left(n-\frac12\right)\left(\frac1n+\frac1{2n^2}+\frac1{3n^3}+\frac1{4n^4}+\dots\right)-1\\
&=\frac1{3\cdot4n^2}+\frac2{4\cdot6n^3}+\frac3{5\cdot8n^4}+\frac4{6\cdot10n^5}+\dots+\frac{(k-1)}{2k(k+1)n^k}+\dots\tag{4}
\end{align}
$$
Thus, the left hand side is still smaller than the right hand side, but the difference is much smaller.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_0^1 x(1-x)\log(x(1-x))dx=?$ I would like to compute the following integral. The answer must be $-\frac{5}{18}$ but I do not know how to evaluate.
$$\int_0^1 x(1-x)\log(x(1-x))dx$$
Thank you for your help in advance.
| Hint:
$$\int_0^1 x^k \log{x} = -\frac{1}{(k+1)^2}$$
Taylor expand the $\log(1-x)$ piece about $x=0$.
$$\begin{align}\int_0^1 x(1-x)\log(x(1-x))dx &= \int_0^1 x(1-x)\log{x} + \int_0^1 x(1-x)\log{(1-x)}\\&= -\frac{1}{4} + \frac{1}{9} - \sum_{k=1}^{\infty} \frac{1}{k}\int_0^1 dx \: x(1-x) x^k \\ &= -\frac{5}{36} - \sum_{k=1}^{\infty} \frac{1}{k} \left (\frac{1}{k+2}-\frac{1}{k+3} \right ) \\ &= -\frac{5}{36} - \frac{1}{2} \sum_{k=1}^{\infty} \left (\frac{1}{k}-\frac{1}{k+2} \right ) + \frac{1}{3} \sum_{k=1}^{\infty} \left (\frac{1}{k}-\frac{1}{k+3} \right ) \\ &= -\frac{5}{36} - \frac{1}{2} \left ( 1 + \frac{1}{2} \right )+ \frac{1}{3} \left ( 1 + \frac{1}{2}+ \frac{1}{3} \right )\\ &= -\frac{5}{36} - \frac{3}{4}+ \frac{11}{18}\\\therefore \int_0^1 x(1-x)\log(x(1-x))dx &= -\frac{5}{18} \end{align}$$
| {
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Prove that $ \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.
Given $a,b,c>0$, prove that $\displaystyle \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.
I expanded the LHS, and realized I have to prove $\displaystyle\frac a b +\frac a c +\frac b c +\frac b a +\frac c a +\frac c b \geq \frac{2(a+b+c)}{\sqrt[3]{abc}}$, but I don't know how. Please help. Thank you.
| Hint: $$\frac{a}{b} + \frac{a}{c} + 1 \ge 3 \frac{a}{\sqrt[3]{abc}}$$ by AM-GM.
| {
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What is the limit of this series? assume $$x_n=\frac{n+1}{2^{n+1}}\sum_{k=1}^n\frac{2^k}{k} ,n=1,2,.....$$
how compute $\lim_{n\to +\infty}x_n$?
Thanks for any hint
| Since the answer is straightforward if we exploit the Cesaro-Stolz theorem, here I want to present a solution that does not use this theorem.
Notice first that whenever $1<r<2$, we have
$$ \lim_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{r^k}{k}}{\frac{2^{n+1}}{n+1}} ＝ 0.$$
Indeed, the numerator is dominated by $n r^n$, thus we have
$$ 0 \leq \frac{\sum_{k＝1}^{n} \frac{r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \frac{n(n+1)}{2} \left(\frac{r}{2}\right)^n \xrightarrow[]{n\to\infty} 0.$$
This shows that, the limit in question exists with value $\ell$ if and only if
$$ \lim_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} ＝ \ell.$$
But note that
$$ \sum_{k＝1}^{n} \frac{2^k - r^k}{k} ＝ \sum_{k＝1}^{n} \int_r^2 x^{k-1} \; dx ＝ \int_r^2 \frac{x^{n} - 1}{x-1} \; dx. $$
This in particular shows that
$$ \sum_{k＝1}^{n} \frac{2^k - r^k}{k} \leq \int_r^2 \frac{x^{n}-1}{r-1} \; dx \leq \frac{2^{n+1}-r^{n+1}-(n+1)(2-r)}{(r-1)(n+1)} $$
and
$$ \sum_{k＝1}^{n} \frac{2^k - r^k}{k} \geq \int_r^2 (x^{n}-1) \; dx \geq \frac{2^{n+1}-r^{n+1}-(n+1)(2-r)}{n+1} $$
These equalities show that
$$ 1\leq \liminf_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \limsup_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \frac{1}{r-1}. $$
Now since $r$ was arbitrary, taking $r\to2^{-}$ yields the limit $\ell ＝ 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/315122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the Taylor series of $\frac{1}{x+1} $ at $x=2$ This is what I did:
$\begin{align*}
f(x)&=&(x+1)^{-1}\\
f'(x)&=&-(x+1)^{-2}\\
f''(x)&=&2(x+1)^{-3}\\
f'''(x)&=&-6(x+1)^{-4}\\
f''''(x)&=&24(x+1)^{-5}\\
&\vdots\\
f^{(n)}(x)&=&(-1)^nn!(x+1)^{-(n+1)}
\end{align*}$
Then I substituted $x=2$:
$f^{(n)}(2)=(-1)^nn!(2+1)^{-(n+1)}=(-1)^nn!(3)^{-(n+1)} $
Then I put it all together.
\begin{align*}
\frac{1}{x+1}&= &\sum_{n=0}^{\infty}\frac{f^{(n)}(2)}{n!}(x-2)\\
&=&\sum_{n=0}^{\infty}\frac{(-1)^nn!(3)^{-(n+1)}}{n!}(x-2)\\
&=&\sum_{n=0}^{\infty}(-1)^n(3)^{-(n+1)}(x-2)
\end{align*}
I am not sure if this is correct.
| This is the approach mentioned by Andre:
$$\frac{1}{x+1}=\frac{1}{3+(x-2)}=\frac{1}{3}\frac{1}{1+\frac{x-2}{3}}=\frac{1}{3}\left(1-\frac{x-2}{3}+\left(\frac{x-2}{3}\right)^2-\ldots\right)=$$
$$=\frac{1}{3}\sum_{n=0}^\infty(-1)^n\left(\frac{x-2}{3}\right)^n$$
The above being true whenever
$$\left|\frac{x-2}{3}\right|<1\Longleftrightarrow |x-2|<3\Longleftrightarrow -1<x<5$$
which is a reasonable assumption as we want $\,x\,$ to be "close" to $\,2\,$ ("close" may be in the eyes of the beholder...)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/316631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $x^{11}+x^8+5\equiv 0\pmod{49}$ Solve $x^{11}+x^8+5\pmod{49}$
My work
$f(x)=x^{11}+x^8+5$
consider the polynomial congruence $f(x) \equiv 0 \pmod {49}$
Prime factorization of $49 = 7^2$
we have $f(x) \equiv 0 \mod 7^2$
Test the value $x\equiv0,1,2,3,4,5,6$ for $x^{11}+x^8+5 \equiv 0\pmod 7$
It works for $x\equiv1$, $x\equiv1\pmod7$ is a solution.
We proceed to lift the solution $\mod 7^2$
$f'(x) = 11x^{10} + 8x^7 +5$, we have $f(1) = 7$ , $f'(x) = 24$
Since $7$ can not divide $f'(1)$, we need to solve $24t \equiv 0 \mod 7$
we get $t \equiv 0 \pmod 5$
and then ..................... ?????
| If $x = 1 + 7 t$ for an integer $t$, then $x^{11} \equiv 1 + 11 (7 t) \mod 49$,
$x^8 \equiv 1 + 8 (7 t) \mod 49$, and so $f(x) \equiv 7 + 19 (7t) \mod 49$.
Thus you want $1 + 19 t \equiv 0 \mod 7$, ...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Solve $a^3-5a+7=3^b$ over the positive integer Solve $a^3-5a+7=3^b$ over the positive integer
I don't know how to solve such equation, please help me. Thanks
| When $b=1$, we already know the result.
I can prove when $3\mid b$, there is no solution.
You may see
if we try $a = 3^{b/3}$, then
$a^3-5a+7 = 3^b-5\cdot3^{b/3}+7<3^b$, if $b\ge3$.
If we try $a = 3^{b/3}+1$, then
$a^3-5a+7 = 3^b+3\cdot 3^{2b/3}-2\cdot 3^{b/3}+3>3^b$, if $b\ge 3.$
Thus $3^{b/3}<a<3^{b/3}+1$, there is no $a$ satisfying this, when $3\mid b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find point closest to the given point I need to find the closest point of the graph $y=\sqrt x$ to the point $(4,0)$.
This is how I understand it:
I assume that triangle with the sides $a$, $x$ and $x/2$ and the largest triangle are similar triangles. From that I conclude that the smaller triangle's one side is $x/2$.
The closest point on the graph from $(4,0)$ must be in the intersection of the lines $a$, $b$ and $c$. $$a = \sqrt{x^2+(x/2)^2},\quad b = \sqrt{x^2 + (2-x/2)^2},\quad c = \sqrt{(4-x)^2+(x/2)^2}$$ by the Pythagorean theorem.
I have to minimize the sum $a+b+c$ by diferentiating and then solving for $x$, and then subtracting $x$ from 4 would get the abscissa of the wanted point.
Solving with wolframalpha $x \approx 0.57$, so the abscissa is $4-0.57=3.43$, ordinate is $0.57/2=0.285$.
Is this correct?
| The distance from a point on the graph $y=\sqrt{x}$ to the point $(4,0)$ is $$\sqrt{(x-4)^2+(\sqrt{x}-0)^2}.$$ This will be minimized precisely when $$(x-4)^2+(\sqrt{x}-0)^2=x^2-8x+16+x=x^2-7x+16$$ is minimized, since $\sqrt{\cdot}$ is an increasing function. Do you know how to find $x\geq 0$ that minimizes $x^2-7x+16$?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Understanding a particular evaluation of $\prod\limits_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right)$ I'm having a hard time understanding the following evaluation of the infinite product $$ \prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right).$$
In particular, I don't understand how you go from line 2 to line 3.
Here $\omega = -\frac{1}{2}+ i \frac{\sqrt{3}}{2}$, which is a primitive third root of unity.
$$\begin{align}\prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right) &=\prod_{n=2}^{\infty}\frac{(n-1)(n^2+n+1)}{n^3} \\ &=\lim_{m\to\infty}\frac{1}{m}\prod_{n=2}^m\frac{(n-\omega)(n-\omega^2)}{n^2} \\ &=\lim_{m\to\infty}\frac{\Gamma(m+1-\omega)\Gamma(m+1-\omega^2)}{m(m!)^2\Gamma(-\omega)\Gamma(-\omega^2)(1-\omega)(1-\omega^2)(-\omega)(-\omega^2)} \\ &=\frac{1}{3\Gamma(-\omega)\Gamma(-\omega^2)} \\ &=\frac{\sin{\pi(-\omega)}}{3\pi} \\ &=\frac{\cosh (\frac{\sqrt{3}\pi}{2})}{3\pi } \end{align}$$
EDIT: I think I'm starting to make sense out of this by writing out the terms.
$$ \begin{align} \prod_{n=2}^{m} \frac{(n- \omega)(n-\omega^{2})}{n^{2}} &= \frac{(2-\omega)(3-\omega) \cdots (m- \omega)(2-\omega^{2})(3-\omega^{2})\cdots (m- \omega^{2})}{2^{2} \cdot 3^{2} \cdot \cdots \cdot m^{2}} \\ &= \frac{1}{(m!)^{2}} \frac{\Gamma(m+1-\omega)}{(1-\omega)(-\omega)\Gamma(-\omega)} \frac{\Gamma(m+1-\omega^{2})}{(1-\omega^{2})(-\omega^{2})\Gamma(-\omega^{2})} \end{align}$$
Now I have to figure out how to take the limit as $m \to \infty$.
| Here's my understanding of that evaluation.
$$ \begin{align} \prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right) &= \prod_{n=2}^{\infty}\frac{(n-1)(n^2+n+1)}{n^3} \\ & = \prod_{n=2}^{\infty} \frac{(n-1)(n- \omega)(n-\omega^{2})}{n^{3}} \\ &= \lim_{m \to \infty} \prod_{n=2}^{m} \frac{n-1}{n} \frac{(n- \omega)(n-\omega^{2})}{n^{2}} \\ &= \lim_{m \to \infty} \frac{1}{m}\frac{(2-\omega)(3-\omega) \cdots (m- \omega) (2-\omega^{2})(3-\omega^{2})\cdots (m- \omega^{2})}{2^{2} \cdot 3^{2} \cdot \cdots \cdot m^{2}} \\ &= \lim_{m \to \infty} \frac{1}{m} \frac{1}{(m!)^{2}} \frac{\Gamma(m+1-\omega)}{(1-\omega)(-\omega)\Gamma(-\omega)} \frac{\Gamma(m+1-\omega^{2})}{(1-\omega^{2})(-\omega^{2})\Gamma(-\omega^{2})} \\ & \stackrel{(1)}= \frac{1}{3\Gamma(-\omega) \Gamma(-\omega^{2})}\lim_{m \to \infty} \frac{\Gamma(m+1- \omega) \Gamma(m+1-\omega^{2})}{m \Gamma(m+1) \Gamma(m+1) } \\ & \stackrel{(2)}= \frac{1}{3\Gamma(-\omega) \Gamma(-\omega^{2})}\lim_{m \to \infty} \frac{\Gamma(m+1- \omega) \Gamma(m+1-\omega^{2})}{m^{-\omega}m^{-\omega^{2}} \Gamma(m+1) \Gamma(m+1) } \\ &= \frac{1}{3\Gamma(-\omega) \Gamma(-\omega^{2})}\lim_{m \to \infty} \frac{(m-\omega)(m-\omega^{2})}{m^{2}}\frac{\Gamma(m- \omega)}{\Gamma(m) m^{-\omega} } \frac{\Gamma(m-\omega^{2})}{ \Gamma(m)m^{-\omega^{2}} } \\ & \stackrel{(3)}= \frac{1}{3\Gamma(-\omega) \Gamma(-\omega^{2})} (1)(1)(1) \\ & = \frac{1}{3 \Gamma \left(\frac{1}{2} - i\frac{\sqrt{3}}{2} \right) \Gamma \left(\frac{1}{2} + i\frac{\sqrt{3}}{2} \right)} \\& = \frac{1}{3 \Gamma \left(1- \frac{1+ i \sqrt{3}}{2} \right) \Gamma \left(\frac{1}{2} + i\frac{\sqrt{3}}{2} \right)} \\ & \stackrel{(4)}= \frac{\sin \left(\pi \, \frac{1+i \sqrt{3}}{2} \right)}{3 \pi} \\&= \frac{\sin \left( \frac{\pi}{2}\right) \cos \left(\frac{i \sqrt{3} \pi}{2} \right) + \cos \left( \frac{\pi}{2} \right) \sin \left(\frac{i \sqrt{3} \pi }{2}\right)}{3 \pi} \\ &= \frac{\cosh \left( \frac{\sqrt{3} \pi}{2} \right)}{3 \pi} \end{align}$$
$(1)$ $(1-\omega)(1-\omega^{2}) = 3$ and $(-\omega)(-\omega^{2})=1 $
$(2)$ $-\omega - \omega^{2} =1$
$(3)$ $\lim_{n \to \infty} \frac{\Gamma(n+ \alpha)}{\Gamma(n)n^{\alpha}}=1 $
$(4)$ Reflection formula for the gamma function
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 1,
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For this matrix $A$, what is $A^n$? $$A = \begin{pmatrix}0 & a & b \\ 0& 0 & c \\ 0& 0 &0\end{pmatrix}$$
What is $A^n$ (for $n\geq 1)$?
| This matrix is a nilpotent one, so we know that $A^3=0$. Now we calculate
$$A^2=A\cdot A = \begin{pmatrix} 0 & a & b \\ 0 & 0& c \\0 & 0 & 0 \end{pmatrix}\cdot
\begin{pmatrix} 0 & a & b \\ 0 & 0& c \\0 & 0 & 0 \end{pmatrix}=
\begin{pmatrix} 0 & 0 & ac \\ 0 & 0& 0 \\0 & 0 & 0 \end{pmatrix}$$
As $A^3=0$ $A^n=0$ for $n>2$
Instead of using that every strict upper triagle matrix is nilpotent we could calculate $A^3$
$$A^3 = A \cdot A^2= \begin{pmatrix} 0 & a & b \\ 0 & 0& c \\0 & 0 & 0 \end{pmatrix} \cdot
\begin{pmatrix} 0 & 0 & ac \\ 0 & 0& 0 \\0 & 0 & 0 \end{pmatrix}=
\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0& 0 \\0 & 0 & 0 \end{pmatrix} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Algorithm for finding the square root of a polynomial... I'm going through Wallace Clarke Boyden's A First Book in Algebra, and there's a section on finding the square root of a perfect square polynomial, eg. $4x^2-12xy+9y^2=(2x-3y)^2$. He describes an algorithm for finding the square root of such a polynomial when it's not immediately apparent, but despite my best efforts, I find the language indecipherable. Can anyone clarify the process he's describing? The example I'm currently wrestling with is $x^6-2x^5+5x^4-6x^3+6x^2-4x+1$.
It's a lot of language to parse, but if anyone wants to take a stab at it, here's the original text:
To find the square root of a polynomial, arrange the terms with reference to the powers of some number; take the square root of the first term of the polynomial for the first term of the root, and subtract its square from the polynomial; divide the first term of the remainder by twice the root found for the next term of the root, and add the quotient to the trial divisor; multiply the complete divisor by the second term of the root, and subtract the product from the remainder. If there is still a remainder, consider the root already found as one term, and proceed as before.
I did some hunting online but didn't turn up anything useful. Is it possible this is an outdated method that's been abandoned for something cleaner?
| It's just saying start with the highest degree and work down. This may not be the fastest but will work or tell you that the thing is not really a square. So, begin with $x^3,$ since the square must be $x^6$ and we get one free choice, $\pm x^3.$ Next, $(x^3 + A x^2)^2 = x^6 + 2 A x^5 + \mbox{more}.$ So $2A = -2, A = -1.$
Alright, $(x^3 -x^2 + B x)^2 = x^6 - 2 x^5 + (2B+1)x^4 + \mbox{more}.$ So $2B+1 = 5$ and $B=2.$
Finally $(x^3 - x^2 + 2 x + C)^2 = x^6 - 2 x^5 + 5 x^4 +(2C -4)x^3 + \mbox{more}.$ So $2C-4 = -6$ and $C=-1.$
Then check
$$ (x^3 - x^2 + 2 x -1)^2 = x^6 - 2 x^5 +5 x^4 - 6 x^3 + 6 x^2 - 4 x + 1. $$
So it worked.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help determining the parameterized solution to a system of linear equation. I am quite new to this area so please bear with me if I am overlooking something glaringly obvious here :)
I am trying to solve the following equation system:
$$\begin{array}{lcl} x + y + z & = & 30 \\ x + 2y & = & 25 \\ 2x + 3y +x& = & 55 \end{array}$$
I am using the matrix notation and write the equation system as (i do not know how to typeset the divisor line between the equality part, sorry about that):
$$\begin{pmatrix} 1 & 1 & 1 & 30 \\ 1 & 2 & 0 & 25\\2 & 3 & 1 & 55\end{pmatrix}$$
This matrix can be written as this: (by multiplying row $2$ with $-1$ and adding it to row $1$, and multiplying the new row $1$ with $-1$ and adding it to row $3$, and finally by multiplying row $2$ with $-1$ and adding it to row $3$).
$$\begin{pmatrix} 1 & 1 & 1 & 30 \\ 1 & 2 & 0 & 25\\2 & 3 & 1 & 55\end{pmatrix}\longleftrightarrow\begin{pmatrix} 0 & -1 & 1 & 5 \\ 1 & 2 & 0 & 25\\2 & 3 & 1 & 55\end{pmatrix} \longleftrightarrow \begin{pmatrix} 0 & -1 & 1 & 5 \\ 1 & 2 & 0 & 25\\2 & 4 & 0 & 50\end{pmatrix} \longleftrightarrow \begin{pmatrix} 0 & -1 & 1 & 5 \\ 1 & 2 & 0 & 25\\0 & 0 & 0 & 0\end{pmatrix}$$
This can be written as:
$$\begin{array}{lcl} -y + z & = & 5 \\ x + 2y & = & 25 \\ 0& = & 0 \end{array}$$
according to my textbook $z$ is a parameter in this equation system. But from the development above, could not $y$ just as well be the parameter?
By solving the equation system differently you could get the final result as:
$$\begin{pmatrix} 1 & 0 & 2 & 35 \\0 & 1 & -1 & -5\\0 & 0 & 0 & 0\end{pmatrix}\longleftrightarrow \begin{array}{lcl} x + 2z & = & 35 \\ y + -z & = & -5 \\ 0& = & 0 \end{array}$$
From this development is is much more clear that $z$ should be the parameter since $x, y$ can be expressed as linear functions of $z$?
I would please like to get some help with how to think about this.
Thank you kindly!
|
From this development is is much more clear that z should be the parameter since x,y can be expressed as linear functions of z?
Precisely. $z$ is a "free" variable (as you can tell by the last row of zeros in both approaches) on which the precise values of $x, y$ depend. Once a value for $z$ is chosen (among infinitely many possible values for $z$, then we can determine $x, y$.
So we can write the solutions as follows:
\begin{pmatrix}
35 - 2z \\
z - 5 \\
z
\end{pmatrix}
You'll often see the following: Given any assignment $z = \alpha$, then the solution can be expressed:
\begin{pmatrix}
35 - 2\alpha \\
\alpha - 5 \\
\alpha
\end{pmatrix}
I used the latter solution to the system of equations as it is in the preferred reduced row echelon form. But the first approach will yield the same parametrization, which you can see when you write $y$ as a function of $z$, since then $x$ is also a function of $z$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Intersection points of a Triangle and a Circle How can I find all intersection points of the following circle and triangle?
Triangle
$$A:=\begin{pmatrix}22\\-1.5\\1 \end{pmatrix} B:=\begin{pmatrix}27\\-2.25\\4 \end{pmatrix} C:=\begin{pmatrix}25.2\\-2\\4.7 \end{pmatrix}$$
Circle
$$\frac{9}{16}=(x-25)^2 + (y+2)^2 + (z-3)^2$$
What I did so far was to determine the line equations of the triangle (a, b and c):
$a : \overrightarrow {OX} = \begin{pmatrix}27\\-2.25\\4 \end{pmatrix}+ \lambda_1*\begin{pmatrix}-1.8\\0.25\\0.7 \end{pmatrix} $
$b : \overrightarrow {OX} = \begin{pmatrix}22\\-1.5\\1 \end{pmatrix}+ \lambda_2*\begin{pmatrix}3.2\\-0.5\\3.7 \end{pmatrix} $
$c : \overrightarrow {OX} = \begin{pmatrix}22\\-1.5\\1 \end{pmatrix}+ \lambda_3*\begin{pmatrix}5\\-0.75\\3 \end{pmatrix} $
But I am not sure what I have to do next...
| If $\,\;\;\vec a+t\vec b\,,\;t\in\Bbb R\;\;$ is a line in $\,\Bbb R^3\,$, then each of its points can be expressed as
$$\begin{pmatrix}\alpha_1:=a_1+tb_1\\\alpha_2:=a_2+tb_2\\\alpha_3:=a_3+tb_3\end{pmatrix}\;,\;\;a_i,b_i,t\in\Bbb R$$
then a point as above belongs to the circle
$$(x-25)^2+(y+2)^2+(z-3)^2=9\iff (\alpha_1-25)^2+(\alpha_2+2)^2+(\alpha_3-3)^2=9$$
so you have to substitute above, get a quadratic in $\,t\,$ and solve...of course, at most two different points for each straight line.
| {
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Need help proving this integration If $a>b>0$, prove that :
$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \frac{2\pi}{b^2} \left(a-\sqrt{a^2-b^2} \right) $$
| You can solve it with direct integration some subsitutions gives you the antiderivative:
$$\frac{-2 \sqrt{b^2-a^2} \tanh ^{-1}\left(\frac{(a-b) \tan \left(\frac{\theta }{2}\right)}{\sqrt{b^2-a^2}}\right)+a \theta -b \sin (\theta )}{b^2}$$
use a substituion like
$t=\tan(\theta)$. (and a lot of the trigonometric identies.)
We use the identies
\begin{align*}
\cos(\arctan(t))&=\frac{1}{\sqrt{1+t^2}}\\
\sin(\arctan(t))&=\frac{t}{\sqrt{1+t^2}}
\end{align*}
So we have
$$\int \frac{\sin^2(\theta)}{a+b\cos(\theta)}\, \mathrm{d}\theta= \int
\frac{t^2}{a + b \frac{1}{\sqrt{1+t^2}}} \, \mathrm{d} t$$
With residue theorem it should work like this :
$$\int_0^{2\pi} \frac{\sin^2(\theta)}{a+ b \cos(\theta)}\, \mathrm{d}\theta=
-\int_0^{2\pi} \frac{1}{4}\cdot \frac{(\exp(i \theta)- \exp(-i \theta))^2}{a+\frac{b}{2} (\exp(i\theta) +\exp(-i\theta)} \, \mathrm{d}\theta$$
Chosing the way $\gamma(\theta) = e^{i\theta}$we see that it equal
$$-\frac{1}{4i} \int_\gamma \frac{(z-z^{-1})^2}{a+ \frac{b}{2} (z+z^{-1})} \cdot \frac{1}{z} \, \mathrm{d} z$$
Expanding gives us
$$-\frac{1}{4i} \int_\gamma \frac{z^2 -2 + \frac{1}{z^2}}{a+ \frac{b}{2} (z+\frac{1}{z})} \cdot \frac{1}{z} \, \mathrm{d}z=-\frac{1}{4i} \int_\gamma \frac{z^2 -2 + \frac{1}{z^2}}{za+ \frac{b}{2} (z^2+1)} \, \mathrm{d}z$$
| {
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"source": "stackexchange",
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Solve $x^{10} + 4x^3 +3x + 4 \equiv 0 \pmod {(4\cdot3)}$ Solve $x^{10} + 4x^3 +3x + 4 \equiv 0 \pmod {(4\cdot3)}$
Work:
Let $P(x) = x^{10} +4x^3+3x+4 \equiv 0 \pmod 2$ and $12=4\cdot3=2^2\cdot3$
We have $P(x) \equiv0 \pmod {2^2}$ and $P(x)\equiv0\pmod3$
For $\pmod {2^2}$,
$x^{10} + 4x^3 +3x + 4 \equiv 0 \pmod 2$
Try $[0],[1]$,
$[0]$ works since $P(0) =4 \equiv0\pmod2$, say $a_1 =0$
$[0]$ lifts to a unique solution to a class that $[a_2] \in \mathbb{Z}_{2^2}$ so that $P(a_2) \equiv 0 \pmod {2^2}$
Solving $P'(a_1)\cdot x =3\cdot{x}\equiv1\pmod2$, $\overline{P'(a_1)}=x=1$
any representative in class of
$a_2 = a_1 -\overline{P'(a_1)}\cdot {P'(a_1)}=3-1\cdot3=0\equiv{0}\pmod{2^2}$
$0\pmod{2^2}$ the only solution to $P(x)\equiv{0}\pmod{2^2}$
For $\pmod3$,
$x^{10} + 4x^3 +3x + 4 \equiv 0 \pmod 3$
Try $[0],[1],[2]$, $[1]$ works. Say $a_2=1$
$P(1)$ is a solution to $\pmod3$
and how do I combine 2 cases??
| If you want $n \equiv 1 \pmod 3$ and $n \equiv 0 \pmod 4$ you can just try the multiples of $4$ until you find one that works. In this case it is $4$. So $P(4)\equiv 0 \pmod {12}$. Here is a check from Alpha
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/328090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Representing Functions as Power Series Rewrite $$f(x)=(1+x)/(1-x)^2$$ as a power series.
Work thus far:
I separated it into two parts:
$$1/(1-x)^2 + x/(1-x)^2$$
I realize that the first expression is the derivative of $1/(1-x)$ and come up with this sum of series:
$$\sum_{n=0}^\infty x^n$$
Since a derivative was involved, we must derive the series:
$$\displaystyle\sum_{n=0}^\infty nx^{n-1} + x \displaystyle\sum_{n=0}^\infty nx^{n-1}$$
$$=\displaystyle\sum_{n=0}^\infty nx^{n-1} + \displaystyle\sum_{n=0}^\infty nx^n$$
These series need to be added, but how?
The final answer is $\displaystyle\sum_{n=0}^\infty (2n+1)x^n$
| $$
\frac{1}{1-x}=1+x+x^2+x^3+\ldots
$$
$$
\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+\ldots
$$
$$
\frac{1+x}{(1-x)^2}=1+2x+3x^2+\ldots +x+2x^2+ 3x^3+\ldots=1+3x+5x^2+7x^3+\ldots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/329070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to compute $\int_a^b (b-x)^{\frac{n-1}{2}}(x-a)^{-1/2}dx$? I wonder how to compute the following integral (for any natural number $n\geq 0$):
$$\int_a^b (b-x)^{\frac{n-1}{2}}(x-a)^{-1/2}dx.$$
Does anyone know the final answer and how to get there? Is there a trick?
Thank you very much!
| $$I=\int_a^b\frac{\sqrt{b-x}^{\,n}}{\sqrt{(b-x)(x-a)}}\,dx$$
$$x=a\cos^2 t+b\sin^2 t:$$
$$\begin{align*}I&=\int_0^{\pi/2}\frac{2(b-a)\cos t\sin t\cdot\sqrt{(b-a)\cos^{2}t}^{\,n}}{\sqrt{(b-a)^2\cos^2 t\sin^2 t}}\,dt\\[7pt]&=2\sqrt{b-a}^{\,n}\int_0^{\pi/2}\cos^{n}t\,dt\\[7pt]&= \left\{
\begin{array}{l l}
\pi\sqrt{b-a}^{\,n}\frac{1\cdot 3\cdots (n-1)}{2\cdot 4\cdots n}\quad(n\;\text{even})\\[7pt]
2\sqrt{b-a}^{\,n}\frac{2\cdot 4\cdots (n-1)}{1\cdot 3\cdots n}\quad(n\;\text{odd})
\end{array} \right.\end{align*}$$
$$\star$$
The latter integral is well known, and can easily be evaluated by parts:
$$\begin{align*}I_n=\int_0^{\pi/2}\cos^nt\,dt\Rightarrow I_n&=(n-1)\int_0^{\pi/2}\sin^2 t\cos^{n-2}t\,dt\\[7pt]&=(n-1)I_{n-2}-(n-1)I_{n}\\[7pt]&\qquad\Rightarrow I_n=\frac{n-1}{n}I_{n-2}\\[7pt]&\qquad\qquad\Rightarrow\left\{
\begin{array}{l l}
I_n=\frac{1\cdot 3\cdots (n-1)}{2\cdot 4\cdots n}I_0\;(n\;\text{even})\\[7pt]
I_n=\frac{2\cdot 4\cdots (n-1)}{1\cdot 3\cdots n}I_1\;(n\;\text{odd})\end{array}\right. \\[7pt]&\qquad\qquad\qquad\big(I_0=\dfrac{\pi}{2}\,\;\;I_1=1\big)\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/329561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Derive the polynomial Given that the solutions to a cubic equation using Cardano's method are $$x_1, x_2, x_3=\sqrt[3]{-\frac{260}{9}i\sqrt{3}-21}+\sqrt[3]{\frac{260}{9}i\sqrt{3}-21}-3$$ derive the cubic polynomial and its factors using algebraic methods only, i.e. without using trigonometric functions.
My initial thought is I need to find the cube roots of the nested radicals which I think could be done with DeMoivre’s formula. But since that is not permitted I am unsure of where to start to get the cube roots.
Thanks for your help
| A way to get to the cube roots of the nested radicals in the question, thus getting to the three separate roots "buried" in the Cardano formula above, without using DeMoivre's formula to find the cube roots of the nested radicals is, first set
$x=-21$ and $y=-\frac{260}{9}i\sqrt{3}$ and (without going into how the equation below was derived) then find any real roots of the following equation
$$\frac{(-64a^9+(48x)a^6+((15(x)^2)-3(3y)^2)a^3+(x)^3)}{-64} = 0$$
This expands to
$$a^9 +\frac{63}{4}a^6 - \frac{74215}{64}a^3 +\frac{9261}{64} = 0$$
and as luck would have it has three rational roots $a_1=3; $ $a_2=\frac{1}{2}$; $a_3=-\frac{7}{2}$.
Next solve the equations
$$a_1^3+3a_1b_1^2 = -21$$
$$a_2^3+3a_2b_2^2 = -21$$
$$a_3^3+3a_3b_3^2 = -21$$
(where $-21$ is the value set as $x$) for $b_1$, $b_2$ and $b_3$, or $$b_1=\pm\frac{4i\sqrt{3}}{3}$$ $$b_2=\pm\frac{13i\sqrt{3}}{6}$$ $$b_3=\pm\frac{5i\sqrt{3}}{6}$$ So the three cube roots of $$\sqrt[3]{-\frac{260}{9}i\sqrt{3}-21}$$ are $a$+$b$ or
$$a_1+b_1=-\frac{4i\sqrt{3}}{3}+3$$
$$a_2+b_2=\frac{13i\sqrt{3}}{6}+\frac{1}{2}$$
$$a_3+b_3=-\frac{5i\sqrt{3}}{6}-\frac{7}{2}$$
The same method would be used to derive the cube roots of $$\sqrt[3]{\frac{260}{9}i\sqrt{3}-21}$$
Summarize all the calculations including the cube roots of of $$\sqrt[3]{\frac{260}{9}i\sqrt{3}-21}$$, (which we did not calculate here), to get the following roots of the equation
$$x_1=\left(-\frac{4i\sqrt{3}}{3}+3\right) +\left(\frac{4i\sqrt{3}}{3}+3\right) -3 = 3$$
$$x_2=\left(\frac{13i\sqrt{3}}{6}+\frac{1}{2}\right)+\left(-\frac{13i\sqrt{3}}{6}+\frac{1}{2}\right) -3 = -2$$
$$x_3=\left(-\frac{5i\sqrt{3}}{6}-\frac{7}{2}\right)+\left(\frac{5i\sqrt{3}}{6}-\frac{7}{2}\right)-3=-10$$ Multiplying the factors $$(x-3)(x+2)(x+10)=0$$ equals
$$x^3+9x^2-16x-60=0$$ the polynomial the question is seeking to derive.
Indeed, Cardano's formula for this cubic equation is the same as that presented in the original question and the three cube roots of the nested radicals in the equation are in the solutions found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/331601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve this system of equations? How to solve this system of equations?
$$\begin{cases}
1+\sqrt{2 x+y+1}=4 (2
x+y)^2+\sqrt{6 x+3 y},\\
(x+1) \sqrt{2 x^2-x+4}+8 x^2+4
x y=4.
\end{cases}$$
| Hint
Define $U=2x+y$ in first equation
You will get
$1+\sqrt{U+1}=4 U^2+\sqrt{3U}$ solve $U$
here you need to solve $U$
$$(x+1) \sqrt{2 x^2-x+4}+8 x^2+4xy=4$$
$$(x+1) \sqrt{2 x^2-x+4}+4x(2x+y)=4$$
Then put U in second equation and find $x$
$$(x+1) \sqrt{2 x^2-x+4}+4xU=4$$
after finding $x$ , you can get $y$ from $U=2x+y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/332059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find $\lim_{x\to0^+}\frac{1 - \frac{2}{\pi} \arcsin(_{2}F_{1}(1/2, 1/2, 3/2, x^4))}{x^2}$? Does the limit exist?
$$\lim_{x\to0^+}\frac{1 - \displaystyle \frac{2}{\pi} \arcsin(_{2}F_{1}(1/2, 1/2, 3/2, x^4))}{x^2}$$
| We may use a classical hypergeometric identity (equation $(14)$ from Cook's 'Notes on hypergeometric functions') :
$$_{2}F_{1}\left(\frac 12, \frac 12; \frac 32; z^2\right)=\frac {\arcsin(z)}{z}$$
to rewrite your limit as :
$$\lim_{x\to0^+}\frac{1 - \displaystyle \frac{2}{\pi} \arcsin\left(\frac {\arcsin\left(x^2\right)}{x^2}\right)}{x^2}$$
and use the Taylor expansion : $\,\displaystyle\frac {\arcsin\left(x^2\right)}{x^2}=1+\frac{x^4}6+O\bigl(x^8\bigr)$
Note that you could obtain this expansion directly from the hypergeometric expansion :
$$_2F_1\left(\begin{array}{c} \frac 12, & \frac 12\ \\\qquad \frac 32 \end{array};\ x^4\right)=\sum_{j=0}^{\infty}\frac{\bigl(\frac 12\bigr)_j\bigl(\frac 12\bigr)_j}{\bigl(\frac 32\bigr)_j}\frac {x^{4j}}{j!}$$
with $(a)_0:=1,\ (a)_j:=a(a+1)(a+2)\cdots(a+j-1)\,$ for $j>0$ .
Anyway you need only to compute the limit :
$$\lim_{x\to0^+}\frac{1 - \displaystyle \frac{2}{\pi} \arcsin\left(1+\frac{x^4}6+O\bigl(x^8\bigr)\right)}{x^2}$$
The numerator becomes $0$ as $x\to 0^+$ and you may apply for example l'Hôpital's rule to get :
$$L=\lim_{x\to0^+}\frac {-\frac 2{\pi} \frac{4x^3+O\bigl(x^7\bigr)}{6\sqrt{1-\left(1+\frac{x^4}6+O\bigl(x^8\bigr)\right)^2}}}{2x}=\frac {-2}{\sqrt{-3}\pi}$$
The choice of $\sqrt{3}\;i$ or $-\sqrt{3}\;i$ for the root of the negative value $-3$ remains to you (this imaginary value comes from evaluating $\arcsin$ for a value greater than $1$, this could be too a sign missing in your problem...)
Hoping this helped,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/332141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplify $ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $ Please help me find the sum
$$
\frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}}
$$
| $$
\left(\left(\left(\left(\left(\frac{1}{x-y}+\frac{1}{x+y}\right)+\frac{2x}{x^2+y^2}\right)+\frac{4x^3}{x^4+y^4}\right)+\frac{8x^7}{x^8+y^8}\right)+\frac{16x^{15}}{x^{16}+y^{16}}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/332191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Calculating $\lim_{x \rightarrow 1}(\frac{23}{1-x^{23}} - \frac{31}{1-x^{31}})$ How to calculate following limit?
$$\lim_{x \rightarrow 1}\left(\frac{23}{1-x^{23}} - \frac{31}{1-x^{31}}\right)$$
| Generalized form
$$\lim_{x \rightarrow 1}\left(\frac{m}{1-x^{m}} - \frac{n}{1-x^{n}}\right) = \frac{n-m}{2}$$
$Proof:$
$Let$ $L = \lim_{x \rightarrow 1}\left(\frac{m}{1-x^{m}} - \frac{n}{1-x^{n}}\right)$
$Let x = \frac{1}{y}$
$if x\rightarrow 1$ $then$ $y\rightarrow 1$
$L =\lim_{x \rightarrow 1}\left(\frac{m}{1-\frac{1}{y^{m}}} - \frac{n}{1-\frac{1}{y^{n}}}\right) $
$L =\lim_{x \rightarrow 1}\left(\frac{my^{m}}{y^{m}-1} - \frac{ny^{n}}{y^{n}-1}\right)$
$L =\lim_{x \rightarrow 1}\left(\frac{my^{m}-m+m}{y^{m}-1} - \frac{ny^{n}-n+n}{y^{n}-1}\right)$
$L =\lim_{x \rightarrow 1}\left(\frac{(-m)(1-y^{m})+m}{y^{m}-1} - \frac{(-n)(1-y^{n})+n}{1-y^{n}}\right)$
$L =\lim_{x \rightarrow 1}\left((-m)+\frac{m}{y^{m}-1} - (-n)-\frac{n}{y^{n}-1}\right)$
$L =\lim_{x \rightarrow 1}\left((n-m)+\frac{m}{y^{m}-1} - \frac{n}{y^{n}-1}\right)$
$L =\lim_{x \rightarrow 1}(n-m)+\lim_{x \rightarrow 1}\left(-\frac{m}{1-y^{m}} + \frac{n}{1-y^{n}}\right)$
$L =(n-m)-\lim_{x \rightarrow 1}\left(\frac{m}{1-y^{m}} - \frac{n}{1-y^{n}}\right)$
$L =(n-m)-L$
$2L =(n-m)$
$L =\frac{(n-m)}{2}$
For above question m = 31, n = 23
hence $L= \frac{(23-31)}{2} = (-4)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/335423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
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