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prove $\int_0^\infty \frac{\log^2(x)}{x^2+1}\mathrm dx=\frac{\pi^3}{8}$ with real methods Context: I looked up "complex residue" on google images, and saw this integral. I, being unfamiliar with the use of contour integration, decided to try proving the result without complex analysis. Seeing as I was stuck, I decided to ask you for help.
I am attempting to prove that
$$J=\int_0^\infty\frac{\log^2(x)}{x^2+1}\mathrm dx=\frac{\pi^3}8$$
With real methods because I do not know complex analysis.
I have started with the substitution $x=\tan u$:
$$J=\int_0^{\pi/2}\log^2(\tan x)\mathrm dx$$
$$J=\int_0^{\pi/2}\log^2(\cos x)\mathrm dx-2\int_{0}^{\pi/2}\log(\cos x)\log(\sin x)\mathrm dx+\int_0^{\pi/2}\log^2(\sin x)\mathrm dx$$
But frankly, this is basically worse.
Could I have some help? Thanks.
Update: Wait I think I actually found a viable method
$$F(\alpha)=\int_0^\infty \frac{x^{\alpha}}{x^2+1}\mathrm dx$$
As I have shown in other posts of mine,
$$\int_0^\infty\frac{x^{2b-1}}{(1+x^2)^{a+b}}\mathrm dx=\frac12\mathrm{B}(a,b)=\frac{\Gamma(a)\Gamma(b)}{2\Gamma(a+b)}$$
so
$$F(\alpha)=\frac12\Gamma\left(\frac{1+\alpha}2\right)\Gamma\left(\frac{1-\alpha}2\right)$$
And from
$$\Gamma(1-s)\Gamma(s)=\frac\pi{\sin \pi s}$$
we see that
$$F(\alpha)=\frac\pi{2\cos\frac{\pi \alpha}{2}}$$
So $$J=F''(0)=\frac{\pi^3}8$$
Okay while I have just found a proof, I would like to see which ways you did it.
| \begin{align}
J&=\int_0^\infty\frac{\ln^2 x}{x^2+1}\mathrm dx\\
K&=\int_0^\infty\frac{\ln(x)}{x^2+1}\mathrm dx\\
K^2&=\int_0^\infty\frac{\ln x}{x^2+1}\mathrm dx\int_0^\infty\frac{\ln y}{y^2+1}\mathrm dy\\
&=\int_0^\infty \int_0^\infty\frac{\ln x\ln y}{(x^2+1)(y^2+1)}\mathrm dx\mathrm dy\\
&=\frac{1}{2}\int_0^\infty \int_0^\infty\frac{\ln(xy)^2-\ln^2 x-\ln^2 y}{(x^2+1)(y^2+1)}\mathrm dx\mathrm dy\\
&=\frac{1}{2}\int_0^\infty \int_0^\infty\frac{\ln(xy)^2}{(x^2+1)(y^2+1)}\mathrm dx\mathrm dy-\int_0^\infty \int_0^\infty\frac{\ln^2 x}{(x^2+1)(y^2+1)}\mathrm dx\mathrm dy\\
&=\frac{1}{2}\int_0^\infty \frac{1}{y^2+1}\left(\int_0^\infty\frac{\ln(xy)^2}{x^2+1}\mathrm dx\right)\mathrm dy-\frac{\pi}{2}J
\end{align}
Perform the change of variable $u=xy$ ($x$ is the variable)
\begin{align}K^2&=\frac{1}{2}\int_0^\infty \int_0^\infty\frac{y\ln^2 u}{(y^2+1)(y^2+u^2)}\mathrm du\mathrm dy-\frac{\pi}{2}J\\
&=\frac{1}{4}\int_0^\infty\frac{\ln^2 u}{1-u^2}\left[\ln\left(\frac{u^2+y^2}{1+y^2}\right)\right]_{y=0}^{y=\infty}\mathrm du-\frac{\pi}{2}J\\
&=-\frac{1}{2}\int_0^\infty\frac{\ln^3 u}{1-u^2}\mathrm du\\
&=-\frac{1}{2}\left(\int_0^1\frac{\ln^3 u}{1-u^2}\mathrm du+\int_1^\infty\frac{\ln^3 u}{1-u^2}\mathrm du\right)-\frac{\pi}{2}J\\
\end{align}
In the second integral perform the change of variable $v=\dfrac{1}{u}$
\begin{align}K^2&=-\int_0^1\frac{\ln^3 u}{1-u^2}\mathrm du-\frac{\pi}{2}J\\
&=\frac{1}{2}\int_0^1\frac{2u\ln^3 u}{1-u^2}\mathrm du-\int_0^1\frac{\ln^3 u}{1-u}\mathrm du-\frac{\pi}{2}J
\\\end{align}
In the first integral perform the change of variable $v=u^2$,
\begin{align}K^2&=\frac{1}{16}\int_0^1\frac{\ln^3 u}{1-u}\mathrm du-\int_0^1\frac{\ln^3 u}{1-u}\mathrm du-\frac{\pi}{2}J\\
&=-\frac{15}{16}\int_0^1\frac{\ln^3 u}{1-u}\mathrm du-\frac{\pi}{2}J\\
&=-\frac{15}{16}\times -6\zeta(4)-\frac{\pi}{2}J\\
&=\frac{45}{8}\zeta(4)-\frac{\pi}{2}J\\
\end{align}
Moreover, $K=0$ (perform the change of variable $v=\frac{1}{x}$)
Therefore,
\begin{align}J&=\frac{45}{4\pi}\zeta(4)\\
&=\frac{45}{4\pi}\times \frac{\pi^4}{90}\\
&=\boxed{\frac{\pi^3}{8}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Integral $\int\frac{1}{1+x^3}dx$
Calculate$$\int\frac{1}{1+x^3}dx$$
After calculating the partial fractions I got:
$$\frac{1}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2-x}{x^2-x+1}dx=\frac{1}{3}\ln(x+1)+\frac{1}{3}\int\frac{2-x}{x^2-x+1}dx$$
I have no idea on how to proceed. Am I missing a substitution or something?
| To integrate $\frac{2- x}{x^2- x+ 1}$, complete the square in the denominator: $\int \frac{2- x}{x^2- x+ 1}dx= \int\frac{2- x}{x^2- x+ \frac{1}{4}-\frac{1}{4}+ 1}dx= \int\frac{2-x}{\left(x- \frac{1}{2}\right)+ \frac{3}{4}}dx$.
Now let $u= x- \frac{1}{2}$ so that $du= dx$ and $x= u+ \frac{1}{2}$ and $2- x= \frac{3}{2}- u$ and $dx= -dy.
$\int \frac{2- x}{x^2- x+ 1}dx= $$-\int \frac{\frac{3}{2}- u}{u^2+ \frac{3}{4}}du= $$-\frac{3}{2}\int\frac{du}{u^2+ \frac{3}{4}}+ \int\frac{u du}{u^2+ \frac{3}{4}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$ solve $$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$$
my attempt
$$t = \frac{1}{2}\tan(u)$$
$$dt = \frac{1}{2}\sec^2(u)du\\$$
$$\begin{align}
\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt&=\int_{0}^{1} \frac{\tan^2(u)}{4} \sqrt{1+\tan^2(u)}\frac{1}{2}\sec^2(u)du\\
&=\frac{1}{8}\int_{0}^{1} \tan^2(u)\sec^{3}(u)du\\
&= \frac{1}{8} \int_{0}^{1} (\sec^2(u) - 1)(\sec^{3}(u))du\\
&=\frac{1}{8}\int_{0}^{1} \sec^5(u)du - \frac{1}{8} \int_{0}^{1}\sec^3(u)du
\end{align}$$
what now?
| To compute the integral in question first consider that $$\cosh^2(x)-\sinh^2(x)=1,$$ so $$\cosh^2(x)=1+\sinh^2(x).$$
Making the substitution $t=\frac{\sinh(x)}{2},$ then $4t^2=\sinh^2(x),$ and we get $dt=\frac{\cosh(x)}{2}dx$
$$\int t^2 \sqrt{(1+4t^2)}dt=\frac{1}{8}\int{\sinh^2(x)}\sqrt{1+\sinh^2(x)}\cosh(x)dx$$
$$=\frac{1}{8}\int{\sinh^2(x)}\cosh^2(x)dx=\frac{1}{8}\left(\frac{1}{32}\sinh(4x)-4x\right) +C.$$
I leave the change of parameter for $t=0$ and $t=1$ for you to finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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} |
Prove $\sqrt{b} - \sqrt{a} < \sqrt{b-a}$
Prove that if $0 < a < b$ then
$$\sqrt{b} - \sqrt{a} < \sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2\sqrt{ab} < b-a$
subtract $b$ from both sides $a-2\sqrt{ab} < -a$
add $a$ to both sides $2a-2\sqrt{ab} < 0$
than add $2 \sqrt{ab}$ to both sides and get $2a < 2\sqrt{ab}$
divide by $2$ and get $a < \sqrt{ab}$
Thus we know $\sqrt{ab}$ is bigger than $a$ so that $\sqrt{a \cdot a} < \sqrt{ab}$ which means $\sqrt{a} < \sqrt{b}$.
Therefore together with the given $a < b$, we have $\sqrt{b} - \sqrt{a} < \sqrt{b - a}$
I'm confused as to where I went wrong and how to fix this.
| The reasoning of your argument is basically correct but you can't argue from the conclusion to be the beginning without clarifying why.
(particularly you need to state what is known and why from what is being speculated and what required conditions are necessary for the speculation. You are stating things that you don't know are true as though they are true, then concluding things that would make them true as though those are results of them being true and finally stating a statement that is neither the result you are trying to prove nor your hypothesis without saying why it is relevant. [Your intent is that it is easily verified by the premise, and once verified we have already shown acceptance of it is sufficient for us to accept the conclusion.... Now if you are having trouble following the sequence of that argument... well, that's the exact same problem a reader of your proof will have.])
Use what you have done as a "rough draft" and write your argument in the correct "forward" direction.
$0 < a < b$ means that $a\cdot a < a\cdot b < b\cdot b$ and so $a^2 < ab < b^2 $ and $a = \sqrt a^2 < \sqrt {ab} < \sqrt b^2 = b$.
So $a < \sqrt {ab}$
$2a < 2\sqrt {ab}$
$ - 2\sqrt{ab} +2a< 0$
$ -2\sqrt{ab} +a< -a$
$b - 2\sqrt{ab} + a < b - a$
$(\sqrt b -\sqrt a)^2 < b-a$
$|\sqrt b - \sqrt a| < \sqrt{b-a}$.
But as $b > a > 0$ then $\sqrt b > \sqrt a$ and $|\sqrt b - \sqrt a| = \sqrt b - \sqrt a$.
So
$\sqrt b - \sqrt a < \sqrt{b-a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
maximum and minimum of $x^2+y^4$ for real $x,y$ If $y^2(y^2-6)+x^2-8x+24=0$ then maximum and minimum value of $x^2+y^4$ is
what i try
$y^4-6y^2+9+x^2-8x+16=1$
$(x-4)^2+(y^2-3)^2=1\cdots (1)$
How i find maximum and minimum of $x^2+y^4$ from $(1)$ relation
help me to solve it please
| By C-S $$0=y^2(y^2-6)+x^2-8x+24=$$
$$=x^2+y^4-8x-6y^2+24\geq$$
$$\geq x^2+y^4-\sqrt{(8^2+6^2)(x^2+y^4)}+24=$$
$$=x^2+y^4-10\sqrt{x^2+y^4}+25-1=\left(\sqrt{x^2+y^4}-5\right)^2-1,$$
which gives
$$4\leq\sqrt{x^2+y^4}\leq6$$ and
$$16\leq x^2+y^4\leq36.$$
The equality occurs for $(|x|,y^2)||(8,6),$ which says that $36$ is a maximal value and $16$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
equating coefficients in algebraic expansion If $\displaystyle \bigg(\frac{1+x}{1-x}\bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+\cdots +\infty,$ then value of
$(1)\; \displaystyle \frac{3b_{3}-b_{1}}{b_{2}}$
$(2)\; \displaystyle \frac{2b_{4}-b_{2}}{b_{3}}$
$(3)\; \displaystyle \frac{3b_{6}-2b_{4}}{b_{5}}$
$(4)\; \displaystyle \frac{5b_{10}-4b_{8}}{b_{9}}$
Answers given in $n,2n,3n,4n$ formats
What I tried:
$\displaystyle (1+x)^n(1-x)^{-n}=1+b_{1}x+b_{2}x^2+b_{3}x^3+\cdots \cdots \infty$
$\displaystyle \bigg[1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}\cdots \bigg]\bigg[1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+\cdots \bigg]=1+b_{1}x+b_{2}x^2+b_{3}x^3+\cdots $
$\displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$
How do I find other coefficients? Help me, please.
| Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^j]\left(\frac{1+x}{1-x}\right)^n}
&=[x^j](1-x)^{-n}(1+x)^n\\
&=[x^j]\sum_{k=0}^\infty\binom{-n}{k}(-x)^k(1+x)^n\tag{1}\\
&=[x^j]\sum_{k=0}^\infty\binom{n+k-1}{k}x^k(1+x)^n\tag{2}\\
&=\sum_{k=0}^j\binom{n+k-1}{k}[x^{j-k}](1+x)^n\tag{3}\\
&\,\,\color{blue}{=\sum_{k=0}^j\binom{n+k-1}{k}\binom{n}{j-k}}\tag{4}
\end{align*}
Comment:
*
*In (1) we apply the binomial series expansion
*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$
*In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.
*In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]\sum_{t=0}^n\binom{n}{t}x^t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can these bounds in terms of the abundancy index and deficiency functions be improved for deficient-perfect numbers? Let
$$\sigma(x) = \sum_{e \mid x}{e}$$
denote the sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$, and the deficiency of $x$ by $D(x)=2x-\sigma(x)$. A positive integer $N$ is said to be deficient-perfect if $D(N) \mid N$.
Here is my question:
Can these bounds in terms of the abundancy index and deficiency functions be improved for deficient-perfect numbers $N > 1$?
$$\frac{2N}{N + D(N)} < I(N) < \frac{2N + D(N)}{N + D(N)}$$
(Note that the inequality
$$\frac{2N}{N + D(N)} < I(N) < \frac{2N + D(N)}{N + D(N)}$$
is true if and only if $N$ is deficient.)
References
A Criterion for Deficient Numbers Using the Abundancy Index and Deficiency Functions, Journal for Algebra and Number Theory Academia, Volume 8, Issue 1, February 2018, pages 1-9
| ILLUSTRATING VIA A TOY EXAMPLE
Let $M$ be an odd perfect number given in the so-called Eulerian form
$$M = p^k m^2$$
(i.e. $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$).
It is known that the non-Euler part $m^2$ is deficient-perfect if and only if the Descartes-Frenicle-Sorli conjecture that $k=1$ holds. (See this paper for a proof of this fact.)
So, suppose that $k=1$. Then $m^2$ is deficient-perfect.
In particular, $m^2$ is deficient, so that the criterion in this paper applies.
We have
$$\frac{2m^2}{m^2 + D(m^2)} < I(m^2) < \frac{2m^2 + D(m^2)}{m^2 + D(m^2)}.$$
Under the hypothesis that $k=1$, $m^2$ is deficient-perfect, with deficiency
$$D(m^2) = \frac{m^2}{(p+1)/2}.$$
We also have
$$I(m^2) = \frac{2}{I(p)} = \frac{2p}{p+1}.$$
Putting these all together, we have
$$\frac{m^2}{D(m^2)} = \frac{p+1}{2}$$
$$\frac{2p}{p+1} = I(m^2) > \frac{2m^2}{m^2 + D(m^2)} = \frac{2\bigg(\frac{m^2}{D(m^2)}\bigg)}{\frac{m^2}{D(m^2)} + 1} = \frac{2\bigg(\frac{p+1}{2}\bigg)}{\bigg(\frac{p+1}{2}\bigg) + 1} = \frac{p+1}{\frac{p+3}{2}} = \frac{2(p+1)}{p+3}$$
which implies that
$$p^2 + 3p = p(p+3) > (p+1)^2 = p^2 + 2p + 1$$
$$p > 1$$
(This last inequality is trivial as $p$ is prime with $p \equiv 1 \pmod 4$ implies that $p \geq 5$.)
$$\frac{2p}{p+1} = I(m^2) < \frac{2m^2 + D(m^2)}{m^2 + D(m^2)} = \frac{2\bigg(\frac{m^2}{D(m^2)}\bigg) + 1}{\frac{m^2}{D(m^2)} + 1} = \frac{2\bigg(\frac{p+1}{2}\bigg) + 1}{\bigg(\frac{p+1}{2}\bigg) + 1} = \frac{p+2}{\frac{p+3}{2}} = \frac{2(p+2)}{p+3}$$
which implies that
$$p^2 + 3p = p(p+3) < (p+1)(p+2) = p^2 + 3p + 2$$
$$0 < 2.$$
This example illustrates my interest in improvements to the bounds in terms of the abundancy index and deficiency functions of $N$, when $N > 1$ is deficient-perfect.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does $d_1$ equal $\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$? I'm going over the proof of the midpoint formula and the solution in my textbook solves its first distance as follows
$$d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}$$
$$=\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$$
How does $d_1$ equal $\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$ ?
I tried to do it on paper but end up with $d_1$ = $\sqrt{\frac{1}{2}\left(x_2-x_1\right)^2 + \frac{1}{2}\left(y_2-y_1\right)^2} = \sqrt{\frac{1}{2}\left(\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2\right)}$
I'm sure you cant just factor out a multiple right?
| You forgot that the $\frac{1}{2}$ had to be squared when you were factoring it out:
$$
d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}=
\sqrt{\left(\frac{x_1+x_2}{2}-\frac{2x_1}{2}\right)^2 + \left(\frac{y_1+y_2}{2}-\frac{2y_1}{2}\right)^2}=\\
\sqrt{\left(\frac{x_1+x_2-2x_1}{2}\right)^2 + \left(\frac{y_1+y_2-2y_1}{2}\right)^2}=\\
\sqrt{\frac{\left(x_2-x_1\right)^2}{2^2} + \frac{\left(y_2-y_1\right)^2}{2^2}}=
\sqrt{\frac{1}{4}\left(x_2-x_1\right)^2 + \frac{1}{4}\left(y_2-y_1\right)^2}=\\
\sqrt{\frac{1}{4}}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}=
\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can I find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$ efficiently with combinatorics? To find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$,
I used factorization on $(1+x+\frac{x^2}{2})$ to obtain $\frac{((x+(1+i))(x+(1-i)))}{2}$, then simplified the question to finding the coefficient of $x^6$ in $(x+(1+i))^{10}(x+(1-i))^{10}$, then dividing by $2^{10}$.
Then, we find that the coefficient of $x^6$ would be:
$$\sum_{i=0}^{6} \binom{10}{6-i} \binom{10}{i} (1-i)^{10-(6-i)} (1+i)^{10-i}$$
with the knowledge that $(1-i)(1+i)=2$, I simplified to
$$\binom{10}{6}\binom{10}{0}2^4((1-i)^6+(1+i)^6)+\binom{10}{5}\binom{10}{1}2^5((1-i)^4+(1+i)^4)+\binom{10}{4}\binom{10}{2}2^6((1-i)^2+(1+i)^2)+\binom{10}{3}\binom{10}{3}2^7$$
Note: the formula $(1+i)^x+(1-i)^x$ gives:
$ 2(2^{\frac{x}{2}}) \cos(\frac{x\pi}{4})$
After simplifying and reapplying the division by $2^{10}$, I get $(\frac{0}{1024}) + (-8)(2520)(\frac{32}{1024}) + (\frac{0}{1024}) + (120)(120)(2^7)$, which gives $0-630+0+1800,$ which is 1170, and I checked this over with an expression expansion calculator.
If the original equation was $(1+x+x^2)^{10}$, I would have used binomials to find the answer, however, the $x^2$ was replaced by $\frac{x^2}{2}$.
My question is whether anyone has a combinatorics solution to this question, rather than just algebra. It would be nice if the solution did not require complex numbers.
| You are looking for weak compositions of $6$ with $10$ parts of at most $2$. The nonzero entries can be $(2,2,2), (2,2,1,1), (2,1,1,1,1)$ or $(1,1,1,1,1,1)$ You can fill those out to $10$ with zeros. Now sum up multinomial coefficients and you are there.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = n^2$, then find the value of $\frac{m^2 - n^2}{n^2}$ I am a beginner at trigonometry, I want to know the answer to this question.
If $m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = n^2$, then find the value of $\frac{m^2 - n^2}{n^2}$.
These are the steps I have tried and got stuck in the middle.
$$m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{7\pi}{15}}\cos{\frac{\pi}{15}} = n^2 $$
$$m^2(\frac{1}{4})(2\cos{\frac{2\pi}{15}}\cos{\frac{\pi}{15}})(2\cos{\frac{4\pi}{15}}\cos{\frac{7\pi}{15}}) = n^2$$
$$m^2(\frac{1}{4})(\cos{\frac{3\pi}{15}} + \cos{\frac{\pi}{15}})(\cos{\frac{11\pi}{15}} + \cos{\frac{3\pi}{15}}) = n^2$$
Fro here onwards I could not continue. These steps may be wrong so please check if my method of solving is correct and help me solve the question. Thanks:)
| You can use Morrie's law:
*
*$2^n\prod_{k=0}^{n-1}\cos(2^k a) = \frac{\sin(2^na)}{\sin a}$
To do so, note that
*
*$\cos\left(\frac{14}{15}\pi \right) = -\cos\left(\frac{1}{15}\pi\right)$ and
*$\sin\left(\pi + \frac{1}{15}\pi \right) = -\sin\left(\frac{1}{15}\pi \right) $
It follows
$$\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = -2^4\prod_{k=0}^{3}\cos\left(2^k \frac{1}{15}\pi\right)$$ $$ = \frac{\sin\left(2^4 \frac{1}{15}\pi\right)}{\sin \frac{1}{15}\pi}= \frac{\sin\left(\pi +\frac{1}{15}\pi\right)}{\sin \frac{1}{15}\pi} = -1$$
Hence
$$\boxed{p :=\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = \frac{1}{16}}$$
Now, remember that
*
*$m^2p = n^2 \Rightarrow \color{blue}{\frac{m^2-n^2}{n^2}}= \frac{m^2}{n^2}-1 = \frac{1}{p}-1 \color{blue}{= 15}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ .
I think that must I use from $\dfrac{a^2}{2}+\dfrac{b^2}{2} \geq ab$ but no result please help me .!
| A slightly different approach using $\frac{a^2}2+\frac{b^2}2\ge ab$ :
$$\begin{align*}
\frac{a^2}{2}+\frac{b^3}{3}+\frac{c^6}{6}+\frac{abc}3&=\frac13 a^2+\frac13 abc+\left(\frac16 a^2+ \frac16 b^3\right)+\left(\frac16 b^3+\frac16 c^6\right)\\&\ge\frac13a^2+\frac13abc+\frac13ab^{3/2}+\frac13b^{3/2}c^3 \\
&\ge\frac23a^{3/2}b^{1/2}c^{1/2}+\frac23a^{1/2}b^{3/2}c^{3/2}\\
&\ge \frac43abc.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Characters of $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ From the Cayley table:
\begin{align*}
\begin{array}{c | c c c c }
& (0,0) & (0,1) & (1,0) & (1,1)\\
\hline
(0,0) & (0,0) & (0,1) & (1,0) & (1,1)\\
(0,1) & (0,1) & (0,0) & (1,1) & (1,0)\\
(1,0) & (1,0) & (1,1) & (0,0) & (0,1)\\
(1,1) & (1,1) & (1,0) & (0,1) & (0,0)\\
\end{array}
\end{align*}
How would I construct the characters of this group, $G =\mathbb{Z}_2 \oplus \mathbb{Z}_2$?
EDIT: Since $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ is abelian, all characters are one-dimensional so they take on values $\pm 1$. So we have the same character table as the Klein-4 group:
\begin{align*}
\begin{array}{c | c c c c }
& (0,0) & (0,1) & (1,0) & (1,1)\\
\hline
\chi_{(0,0)} & 1 & 1 & 1 & 1\\
\chi_{(0,1)} & 1 & 1 & -1 & -1\\
\chi_{(1,0)} & 1 & -1 & 1 & -1\\
\chi_{(1,1)} & 1 & -1 & -1 & 1\\
\end{array}
\end{align*}
If this is correct, I can put it as a solution rather than an edit however feel free to critique my attempt.
| Since $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ is abelian, all characters are one-dimensional so they take on values $\pm 1$. So we have the same character table as the Klein-4 group:
\begin{align*}
\begin{array}{c | c c c c }
& (0,0) & (0,1) & (1,0) & (1,1)\\
\hline
\chi_{(0,0)} & 1 & 1 & 1 & 1\\
\chi_{(0,1)} & 1 & 1 & -1 & -1\\
\chi_{(1,0)} & 1 & -1 & 1 & -1\\
\chi_{(1,1)} & 1 & -1 & -1 & 1\\
\end{array}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simultaneous congruences $3x \equiv 2 \pmod{5}$, $3x \equiv 4 \pmod{7}$, $3x \equiv 6 \pmod{11}$ I am stuck in a simultaneous linear congruence problem:
\begin{cases}
3x \equiv 2 \pmod{5} \\[4px]
3x \equiv 4 \pmod{7} \\[4px]
3x \equiv 6 \pmod{11}
\end{cases}
Using the Chinese remainder theorem,
I started with the 'highest' divisor: $11$. Since $(3, 11) =1$ there is a unique solution. $x= 6 \cdot 3 ^{\phi (11) -1} \equiv 6 \cdot 3^4\pmod{11}$ But to be honest, I have no clue how to continue.
Perhaps, cancel out the last equation to: $x \equiv 2 \pmod{11}$?
| This is an old-fashioned solution.
$$3x \equiv 2 \mod 5 \qquad 3x \equiv 4 \mod 7 \qquad 3x \equiv 6 \mod{11}$$
\begin{array}{|r|rrr|}
\hline
385 & 5 & 7 & 11 \\
\hline
77 & 2 & 0 & 0 \\
55 & 0 & 6 & 0 \\
35 & 0 & 0 & 2 \\
\hline
231 & 1 & 0 & 0 \\
-55 & 0 & 1 & 0 \\
-175 & 0 & 0 & 1 \\
\hline
\end{array}
$$3x \equiv 2(231) -4(55) -6(175) \equiv -808 + 385^{\#} \equiv -423 \pmod {385}$$
$$x \equiv -141 \equiv 244 \pmod {385}$$
($\#:$ Adding $385$ makes the answer, $-423$, a multiple of $3$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $f$ in the vector space of cubic spline functions? Let $S_{X,3}$ be the vector space of cubic spline functions on $[-1,1]$ in respect to the points $$X=\left \{x_0=-1, x_1=-\frac{1}{2}, x_2=0, x_3=\frac{1}{2}, x_4\right \}$$ I want to check if the function $$f(x)=\left ||x|^3-\left |x+\frac{1}{3}\right |^3\right |$$ is in $S_{X,3}$.
We have that \begin{align*}f(x)&=\left ||x|^3-\left |x+\frac{1}{3}\right |^3\right |\\ & =\begin{cases}|x|^3-\left |x+\frac{1}{3}\right |^3 , & |x|^3-\left |x+\frac{1}{3}\right |^3\geq 0 \\-|x|^3+\left |x+\frac{1}{3}\right |^3 , & |x|^3-\left |x+\frac{1}{3}\right |^3<0\end{cases} \\ & =\begin{cases}|x|^3-\left |x+\frac{1}{3}\right |^3 , & |x|^3\geq \left |x+\frac{1}{3}\right |^3 \\-|x|^3+\left |x+\frac{1}{3}\right |^3 , & |x|^3<\left |x+\frac{1}{3}\right |^3\end{cases} \\ & = \begin{cases}|x|^3-\left |x+\frac{1}{3}\right |^3 , & |x|\geq \left |x+\frac{1}{3}\right | \\-|x|^3+\left |x+\frac{1}{3}\right |^3 , & |x|<\left |x+\frac{1}{3}\right |\end{cases}\end{align*}
The function is piecewise a polynomial of degree smaller or equal to $3$, right?
Now we have to check if $f$ is continuous on $[-1,1]$.
How could we continue to get the definition of $f$ ?
| For $x\in(-1/3,0)$ we have
$$f(x)=\left|x^3+\left(x+\frac{1}{3}\right)^3\right|$$
which has strictly positive slope, because $x\mapsto x^3$ is monotonous with zero derivative only at $x=0$.
On $(-1/3,0)$ the first term goes from negative to zero, and the second term goes from zero to positive.
Hence the sum goes from negative to positive, and somewhere within $(-1/3,0)$ the sum is zero with positive slope. Taking the absolute value results in non-differntiability at such a point, so it is not a spline, because you have no knots in the interval $(-1/3,0)$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Symmetric matrix factorization of a $2 \times 2$ symmetric matrices Let S be a symmetric matrix such that
$$S= \begin{bmatrix}
A & B\\
B & C
\end{bmatrix}, \text{where A, B, C $\in$ $\mathbb{Z_P},$}$$
where p is prime.
I am tasked to get the solution(s) of the symmetric matrix factorization of $2 \times 2$ symmetric matrices of the forms,
$$S = X^2(\mod p)$$
and
$$S = XY(\mod p),$$
where $X$ $\neq$ $Y$ but are both symmetric.
For example, under $\mathbb{Z_3}$,
$$S = \begin{bmatrix}
2 & 2\\
2 & 1
\end{bmatrix} = \begin{bmatrix}
1 & 2\\
2 & 0
\end{bmatrix} \begin{bmatrix}
1 & 2\\
2 & 0
\end{bmatrix}(\mod 3) = X^2$$
and
$$S = \begin{bmatrix}
1 & 2\\
2 & 2
\end{bmatrix} = \begin{bmatrix}
2 & 0\\
0 & 1
\end{bmatrix} \begin{bmatrix}
2 & 1\\
2 & 2
\end{bmatrix}(\mod 3) = XY$$
I know that these pairs are not the only possible factors of the given matrix S. I've also done programming the matrix multiplication which is the inverse of matrix factorization because I have no idea how to start with matrix factorization.
Any suggestions will be much appreciated. Thanks in advance!
| First, if $S$ is diagonal then this is fairly straightforward. We distinguish three cases:
*
*If $S=0$ then there are $(p-1)^2(p+1)$ pairs $(X,Y)$ of nonzero matrices such that $S\equiv XY\pmod{p}$, and $2p^4-1$ such pairs $(X,Y)$ with either $X=0$ or $Y=0$.
*If $B=0$ and $A=C\neq0$ then $S=AI$, and hence the solutions to
$$S\equiv XY\pmod{p},$$
are precisely the pairs $(X,Y)$ where $X$ is invertible and $Y=AX^{-1}$. In particular there are $(p-1)^2p(p+1)$ solutions $(X,Y)$.
*If $B=0$ and $A\neq C$ then both $X$ and $Y$ must be diagonal and so
$$A\equiv x_{11}y_{11}\pmod{p}\qquad\text{ and }\qquad B\equiv x_{22}y_{22}\pmod{p}.$$
There are $(p-1)^2$ solutions if $A\neq0$ and $C\neq0$, and $(p-1)(2p-1)$ solutions if $AC=0$.
From here one we assume that $B\neq0$. If $A=C=0$ then for the matrix $R:=\binom{0\ 1}{1\ 0}$ we have
$$SR=\begin{pmatrix}
0&B\\B&0
\end{pmatrix}
\begin{pmatrix}
0&1\\1&0
\end{pmatrix}
=\begin{pmatrix}
B&0\\0&B
\end{pmatrix},$$
so $SR$ is diagonal. The above tells us the solutions $(X,Y)$ for $SR$, and hence because $R^2=I$ we have
$$S\equiv XYR\pmod{p},$$
so the solutions for $S$ are the pairs $(X,YR)$ where $SR\equiv XY\pmod{p}$.
If $A$ and $C$ are not both zero then wlog $A\neq0$. For the matrix $T:=\binom{1\ \hphantom{-}B}{0\ -A}$ we have
$$T_A^{\intercal}ST_A=\begin{pmatrix}
1&0\\B&-A
\end{pmatrix}\begin{pmatrix}
A&B\\B&C
\end{pmatrix}
\begin{pmatrix}
1&B\\0&-A
\end{pmatrix}
=\begin{pmatrix}
A&0\\0&A|S|
\end{pmatrix},$$
where $|S|=\det S$, so $T_A^{\intercal}ST_A$ is diagonal. Again the argument above tells us the solutions for $T_A^{\intercal}ST_A$. Note that $T_A$ is invertible because $A\neq0$, and so
$$S\equiv T_A^{-\intercal}XYT_A^{-1}\pmod{p},$$
and hence the solutions for $S$ are the pairs $(T_A^{-\intercal}X,YT_A^{-1})$ where $T_A^{\intercal}ST_A\equiv XY\pmod{p}$.
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $a,b$ at $f(x)=\frac{x^2+x-12}{x^2-ax+b}$ An High school question:
Given :$$f(x)=\frac{x^2+x-12}{x^2-ax+b}$$ it's given that $x=3$ is a vertical asymptote find $a$ and $b$.
I tried:
Since $x=3$ is a vertical asymptote then $3^2-3a+b=0$, but now what
| Note that $x^2+x-12$ also has $3$ as a root, so for an asymptote to exist we require $3$ to be a repeated root. Hence $x^2-ax+b=(x-3)^2\implies a=6,\,b=9$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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} |
Proving that two determinants are equal without expanding them So I need to prove that
$$
\begin{vmatrix}
\sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\
\sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\
\sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\
\end{vmatrix}
$$
$$
= \begin{vmatrix}
\sin(\alpha) & \cos(\alpha) & \sin(\alpha + \delta) \\
\sin(\beta) & \cos(\beta) & \sin(\beta + \delta) \\
\sin(\gamma) & \cos(\gamma) & \sin(\gamma + \delta) \\
\end{vmatrix}
$$
Now,
$$
\begin{vmatrix}
\sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\
\sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\
\sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\
\end{vmatrix} = \begin{vmatrix}
\sin^2(\alpha) & \cos^2(\alpha) - \sin^2(\alpha) & \cos^2(\alpha) \\
\sin^2(\beta) & \cos^2(\beta) - \sin^2(\beta) & \cos^2(\beta) \\
\sin^2(\gamma) & \cos^2(\gamma) - \sin^2(\gamma) & \cos^2(\gamma) \\
\end{vmatrix}
$$
Adding column $1$ to column $2$ then makes column $2$ and column $3$ equal and hence the first determinant $ = 0$.
I'm stuck in trying to prove that the second determinant is also zero, so I need help in that.
| HINT: Relate the third column to the first two using the trigonometric identity
$$\sin(\theta+\delta)=\sin\theta\cos\delta+\cos\theta\sin\delta.$$
| {
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"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $x = a \cos t^3 , y = b \sin t^3$ then what is $d^3y/dx^3$? If $ x = a \cos t^3 $, $ y = b \sin t^3 $, then what is $ \frac{d^3y}{dx^3} $?
I tried doing this problem by dividing $ \frac{d^3y}{dt^3} $ by $ \frac{d^3x}{dt^3} $ and got $ \frac{b}{a} $.
However my book says the third derivative doesn't exist. Why is this so?
| We can in fact obviate the parameterisation altogether. Since $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $\frac{x}{a^2}+\frac{y}{b^2}\frac{dy}{dx}=0$ so $\frac{dy}{dx}=-\frac{b^2x}{a^2y}$ and $$\frac{d^2y}{dx^2}=-\frac{b^2}{a^2}\left(\frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}\right)=-\frac{b^4x^2}{a^4y^3}-\frac{b^2}{a^2y},\\\frac{d^3y}{dx^3}=-\frac{2b^4x}{a^4y^3}+\frac{3b^4x^2}{a^4y^4}\frac{dy}{dx}+\frac{b^2}{a^2y^2}\frac{dy}{dx}=-\frac{3b^4x}{a^4y^3}-\frac{3b^6x^3}{a^6y^5}=-\frac{3b^4x(a^2y^2+b^2x^2)}{a^6y^5}.$$We can simplify this further with $a^2y^2+b^2x^2=(ab)^2$ to $-\frac{3b^6x}{a^4y^5}$, as I mentioned in a comment on another answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Understanding a proof from the APMO 1998 on inequalities. I was having trouble with proving the following inequality.The question was from the book Secrets to Inequalities by Pham Kim Hung.
$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{xyz} }$
I know that the AM-GM inequality must be applied but I am having a hard time proving the the above inequality. In the book the author was able to show that by applying the trick shown below and then applying AM-GM we can prove the inequality holds.
$3\left (\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \right ) = \left ( \frac{2x}{y} +\frac{y}{z}\right ) + \left ( \frac{2y}{z}+\frac{z}{x} \right ) + \left ( \frac{2z}{x}+\frac{x}{y} \right ) \geq \frac{3x}{\sqrt[3]{xyz}} + \frac{3y}{\sqrt[3]{xyz}} + \frac{3z}{\sqrt[3]{xyz}}$
The problem I had was after applying the AM-GM to the LHS I could not prove the RHS.
| It is by AM-GM: $$\frac{x}{y}+\frac{x}{y}+\frac{y}{z}\geq 3\sqrt[3]{\frac{x^2y}{y^2z}}=3\sqrt[3]{\frac{x^3}{xyz}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116317",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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How to find the jacobian of the following? I am stuck with the following problem that says :
If $u_r=\frac{x_r}{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}$ where $r=1,2,3,\cdot \cdot \cdot ,n$, then prove that the jacobian of $u_1,u_2,\cdot \cdot, u_n$ with respect to $x_1,x_2,\cdot \cdot, x_n$ is $(1-x_1^2-x_2^2-\cdot \cdot \cdot -x_n^2)^{-\frac12}$
My try: Now, I have to calculate the value of
\begin{vmatrix}
\frac{\delta u_1}{\delta x_1} & \frac{\delta u_1}{\delta x_2} & \frac{\delta u_1}{\delta x_3} & \cdots & \frac{\delta u_1}{\delta x_n} \\
\frac{\delta u_2}{\delta x_1} & \frac{\delta u_2}{\delta x_2} & \frac{\delta u_2}{\delta x_3} & \cdots & \frac{\delta u_2}{\delta x_n} \\
\vdots & \vdots & \vdots & \ddots &\vdots \\
\frac{\delta u_n}{\delta x_1} & \frac{\delta u_n}{\delta x_2} & \frac{\delta u_n}{\delta x_3} & \cdots & \frac{\delta u_n}{\delta x_n} \\
\end{vmatrix}
Now,the value of $\frac{\delta u_1}{\delta x_1}=\frac{1-2x_1^2}{\{1-x_1^2\}^\frac32}$ , $\frac{\delta u_1}{\delta x_2}=\cdot \cdot =\frac{\delta u_1}{\delta x_n}=0$..
So, things are getting complicated. Can someone show me the right direction?
| The partial derivatives seems easy to find due to symmetry:-
Applying chain rule we get $$ u_{r,i}=\frac{x_ix_r}{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}$$ if i
!= r
$$u_{i,i}=\frac{1}{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}-\frac {x_i^2}{{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}}$$
Then finding the determinant is the task ......
here $u_{i,r} $ denotes partial derivative.
| {
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"source": "stackexchange",
"question_score": "1",
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To determine a constant in an ODE
Let $w(r)$ be a function of $r$, we have the following ODE:
$$r^{n-1}w'+\frac{1}{2}r^nw=a$$
for a constant $a$.
Assume the equation holds for all positive integer $n$. The book claims that if assuming $\lim\limits_{r\to \infty}w= 0$ and $\lim\limits_{r\to \infty}w'= 0$, we have $a=0$. I cannot see why this holds. Any help is appreciated.
| Assume $a=1$ and $n=3$, then you have
\begin{align}
w'+\frac{r}{2}w= \frac{1}{r^2}.
\end{align}
Let us assume $w(1) = 1$, then we see that
\begin{align}
w(r) = \exp\left( \frac{1-r^2}{4}\right)+\exp\left(\frac{-r^2}{4} \right) \int^r_1 t^{-2} \exp\left(\frac{t^2}{4} \right)\ dt
\end{align}
and
\begin{align}
w'(r) = -\frac{r}{2}\exp\left( \frac{1-r^2}{4}\right)-\frac{r}{2}\exp\left( \frac{-r^2}{4}\right)\int^r_1 t^{-2} \exp\left(\frac{t^2}{4} \right)\ dt + \frac{1}{r^2}.
\end{align}
Observe
\begin{align}
\lim_{r\rightarrow \infty}w(r) =&\ \lim_{r\rightarrow \infty}\exp\left(\frac{-r^2}{4} \right) \int^r_1 t^{-2} \exp\left(\frac{t^2}{4} \right)\ dt\\
=&\ \lim_{r\rightarrow \infty} \frac{r^{-2}\exp\left( \frac{r^2}{4}\right)}{\frac{r}{2}\exp\left( \frac{r^2}{4}\right)} = \lim_{r\rightarrow \infty}\frac{2}{r^3}=0
\end{align}
and
\begin{align}
\lim_{r\rightarrow \infty} w'(r) = -\frac{1}{2}\lim_{r\rightarrow \infty}r\exp\left( \frac{-r^2}{4}\right)\int^r_1 t^{-2} \exp\left(\frac{t^2}{4} \right)\ dt =0.
\end{align}
Clearly $a=1\neq 0$. So I am not sure what your textbook is saying. Maybe limit as $r\rightarrow 0$ not $r\rightarrow \infty$?
| {
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"source": "stackexchange",
"question_score": "1",
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Is my integral in fully reduced form?
I have to integrate this:
$$\int_0^1 \frac{x-4}{x^2-5x+6}\,dx$$
Now
$$\int_0^1 \frac{x-4}{(x-3)(x-2)}\, dx$$
and by using partial fractions we get
$$\frac{x-4}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$$
$$x-4 = A(x-2) + B(x-3)$$
$$= Ax - 2A + Bx - 3B$$
$$x-4 = (A+B)x - 2A - 3B$$
so
$$A+B=1$$ or $$2X+2B = 2$$
$-B = -2$ and $B = 2$ and $A = -1$
Then
$$\int_0^1 \frac{x-4}{x^2-5x+6} = \int_0^1 \frac{-1}{x-3}dx + \int_0^1 \frac{2}{x-2} dx$$
usub using $u = x-3$ and $du = dx$
so $$ -1 \ln | x-3 | \rbrack_0^1 + 2 \ln |x-2| \rbrack_0^1$$
$$-1 ( \ln 2 - \ln 3) + 2 (\ln 1 - \ln 2)$$
$$-\ln 2 + \ln 3 + 2\ln 1 - 2\ln 2 = -3 * \ln 2 + \ln 3 + 2\ln 1$$
Is there anyway to simplify this further? Wolfram has the answer at $-\ln(8/3)$ and I'm not sure how to simplify to here? How does my work look?
| First, $\log 1= 0$. And there is a mistake when you compute the definite integral:
$$
-\log 2 +\log 3 -2\log2 = -3\log 2 +\log 3 = -\log 8+\log3 = -\log\frac 83.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Amidst $7$ prime numbers, difference of the largest and the smallest prime number is $d$. What is the highest possible value of $d$?
Let $a, b, c, b+c-a, c+a-b, a+b-c$ and $a+b+c$ be $7$ distinct prime numbers. Among $a+b, b+c$ and $c+a$, only one of the three numbers is equal to $800$. If the difference of the largest prime number and the smallest prime number among the $7$ distinct prime numbers is described as $d$, then what is the highest possible value of $d$?
SOURCE: Bangladesh Math Olympiad
I didn't understand the pattern of rest prime numbers except $a, b$ and $c$ and their construction with the variable $a, b$ and $c$. Moreover, I couldn't catch out the probable number of the three numbers which were constructed by $a+b, b+c$ and $c+a$ whose real value is equal to $800$.
How can I get that $7$ prime numbers by applying any method or with that condition?
So, I really need some help and any reference to the post will be very helpful for my reaching to the conclusion. Thanks in advance.
| Say $a+b=800$.
If $c\equiv 0\pmod 3$ then $c=3$ and then $a+b+c=803$ which is not a prime
If $c\equiv 1\pmod 3$ then $800+c\equiv 0\pmod 3$ so $800+c =3$ impossibile.
If $c\equiv -1\pmod 3$ then $800-c\equiv 0\pmod 3$ so $800-c =3\implies c=797$ so $a+b+c =1597$ and thus $\boxed{d= 1597-3 = 1594}$
Such a prime numbers exsist:
$a=787, b=13, c= 797,a+b+c=1597$
$a+b-c =3,\;\; a+c-b= 1571$ and $b+c-a = 23$
| {
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"timestamp": "2023-03-29T00:00:00",
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Disprove or prove using delta-epsilon definition of limit that $\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$ I want to prove if the following limit exists, using epsilon-delta definition, or prove it doesn't exist:$$\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$$
My attempt:
First I proved some directional limits, like for $y=mx$ , and $y=ax^n$, and for all of them I got 0. So I conjectured that this limit exists and it's 0. Then I have to prove:
$$\forall \delta \gt 0 : \exists \epsilon \gt 0 : \|(x,y)\| \lt \epsilon \rightarrow \left| \frac{x^3-y^3}{x^2-y^2}\right| \lt \delta$$
First I noted that $\frac{x^3-y^3}{x^2-y^2} = \frac{(x^2+xy+y^2)(x-y)}{(x+y)(x-y)} = \frac{x^2+xy+y^2}{x+y}$.
Then I did $\left|\frac{x^2+xy+y^2}{x+y}\right| \leq \left|\frac{x(x+y)}{x+y}\right|+\frac{y^2}{\vert x+y\vert} = \vert x \vert + \frac{y^2}{\vert x+y\vert}$
Using $\|(x,y)\| = \vert x\vert + \vert y\vert$ and assuming $\|(x,y)\| \lt \epsilon$
$\vert x\vert + \vert y\vert = \vert x \vert + \frac{ y^2}{\vert y \vert} \geq \vert x \vert+\frac{y^2}{\vert y\vert+\vert x\vert}$
but I can't continue from that since $\vert x + y \vert \leq \vert x \vert + \vert y \vert$
.
I don't know what else to try.
| Use polar coordinates $x=r\cos\theta$, $y=r\sin\theta$. Because of the domain we have $\cos\theta\ne\pm\sin\theta$. Substituting into your formula gives
$$\frac{x^3-y^3}{x^2-y^2}=r\,\frac{\cos^3\theta-\sin^3\theta}{\cos^2\theta-\sin^2\theta}\ .$$
Now it is not hard to prove that
$$\frac{\cos^3\theta-\sin^3\theta}{\cos^2\theta-\sin^2\theta}\to\infty\quad
\hbox{as}\quad \theta\to\Bigl(\frac{3\pi}{4}\Bigr)^{\textstyle-}\ .$$
So consider the path given by
$$r=\frac{\cos^2\theta-\sin^2\theta}{\cos^3\theta-\sin^3\theta}\ ,\qquad
\theta\to\Bigl(\frac{3\pi}{4}\Bigr)^{\textstyle-}\ .$$
This will approach the origin, but for every point on this path you have
$$\frac{x^3-y^3}{x^2-y^2}=1\ .$$
So the limit does not exist.
| {
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Finding out the remainder of $\frac{11^\text{10}-1}{100}$ using modulus
If $11^\text{10}-1$ is divided by $100$, then solve for '$x$' of the below term $$11^\text{10}-1 = x \pmod{100}$$
Whatever I tried:
$11^\text{2} \equiv 21 \pmod{100}$.....(1)
$(11^\text{2})^\text{2} \equiv (21)^\text{2} \pmod{100}$
$11^\text{4} \equiv 441 \pmod{100}$
$11^\text{4} \equiv 41 \pmod{100}$
$(11^\text{4})^\text{2} \equiv (41)^\text{2} \pmod{100}$
$11^\text{8} \equiv 1681 \pmod{100}$
$11^\text{8} \equiv 81 \pmod{100}$
$11^\text{8} × 11^\text{2} \equiv (81×21) \pmod{100}$ ......{from (1)}
$11^\text{10} \equiv 1701 \pmod{100} \implies 11^\text{10} \equiv 1 \pmod{100}$
Hence, $11^\text{10} -1 \equiv (1-1) \pmod{100} \implies 11^\text{10} - 1 \equiv 0 \pmod{100}$ and thus we get the value of $x$ and it is $x = 0$ and $11^\text{10}-1$ is divisible by $100$.
But this approach take a long time for any competitive exam or any math contest without using calculator. Any easier process on how to determine the remainder of the above problem quickly? That will be very much helpful for me. Thanks in advance.
| $$11^{10}=(10+1)^{10}=10^{10}+10×10^9+\frac {(10×9)}{2}×10^9+\cdots+(10×10)+1$$(using binomial expansion ). Now note that every term except last one is a multiple of $100$.
| {
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Prove that if $x$, $y$, and $z$ are real numbers such that $x^2(y-z)+y^2(z-x)+z^2(x-y)=0,$ then at least two of them are equal
Prove that if $x$, $y$, and $z$ are real numbers such that $x^2(y-z)+y^2(z-x)+z^2(x-y)=0,$ then at least two of them are equal
This question was asked in a past international math exam. My idea to solve it was to factorize it and then end up showing that the polynomial that was formed would directly show that at least two of the numbers $x$, $y$, $z$ are equal. However, it was to no avail.
I researched for theorems, that may help me in my quest to solving this problem, but none of them seemed applicable to this problem. Can someone please show me, a method of either finishing my thought, or if you believe that I was on the completely wrong trail of thought, can you please show me another method of solving it?
| Factoring is hard and I'm not good at it.
But if:
$x^2(y-z)+y^2(z-x) + z^2(x-y) = 0$
$x^2(y-z)+y^2(z-x)= z^2(y-x)$
If $y =x$ then $y=x$. But if $y\ne x$ then $y-x \ne 0$.
$\frac {x^2(y-z) + y^2(z-x)}{y-x} = z^2$
So $\frac {xy*y - x^2z + y^2z - xy*y}{y-x} =$
$ -xy + \frac{y^2z -x^2z}{y-x} = $
$-xy + z\frac{(y-x)(y+x)}{y-x} =$
$-xy + z(y+x)=$
$-xy + zy + xz = z^2$
And so
$xz -xy = z^2 - zy$
$x(z-y) = z(z-y)$.
If $z = y$ then $z=y$ but if $z \ne y$ the $z - y\ne 0$ and
$x = z$.
So either $x=y$ or if not then either $z=y$ or if not $x = z$.
....
Which actually helps me figure out how to factor. So we could have done the following instead:
We can factor $x-y$ (or equivalently $y-x$) from $x^2(y-z) + y^2(z-x)$ via
$x^2(y-z) + y^2(z-x) =$
$xy*x - xy*y - (z*x^2 - z*y^2) = $
$(x-y)(xy - z(x+y))$
So $x^2(y-z) + y^2(z-x) + z^2(x-y) = $
$(x-y) (xy -z(x+y) + z^2)$
And we can factor $y-z$ from $xy - z(x+y) + z^2$ via
$xy - z(x+y) + z^2 = $
$xy - zx + z^2 - zy =$
$x(y-z) + z(z-y) =$
$(y-z)(x-z)$
And thus:
$x^2(y-z) + y^2(z-x) + z^2(x-y)= (x-y)(y-z)(x-z) = 0$
.... which means one of $x-y$ or $y-z$ or $x - z=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove/show this actually defines a homomoprhism
We define the homomorphism $f: \text{SL}_2(\mathbb Z / 2 \mathbb Z) \to \text{SL}_2(\mathbb Z / 2 \mathbb Z)$ that maps the generators to:
$ \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
$ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \to \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
How do we know this is a homorphism? In a previous exercise I explored that these two elements generate the group, the first has order 3, the second has order 2. Essentially this maps any power of the first matrix to the identity, and any power of the second to a power of the matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ is it as simple as:
$$f\left(\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^{m \bmod 3}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}^{n \bmod 2}\right)=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}^{n \bmod 2}$$
But since the matrices don't commute I find it hard to prove this is a homomorphism ($f(AB)=f(A) f(B))$.
Essentially I want to prove that:
$f$ defined by:
$$\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^2= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^3 =\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \to \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$
is a homomorphism.
| Note that$$\mathrm{SL}_2(\mathbb{Z}_2)=\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}1&1\\1&0\end{bmatrix},\begin{bmatrix}1&0\\1&1\end{bmatrix},\begin{bmatrix}1&1\\0&1\end{bmatrix},\begin{bmatrix}0&1\\1&1\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix}\right\}.$$Here, the first element is $\operatorname{id}_{\mathrm{SL}_2(\mathbb{Z}_2)}$, the next three have order $2$ and the last two ones have order $3$. So, $\mathrm{SL}_2(\mathbb{Z}_2)\simeq S_3$. And your question basically becomes: is there an endomorphism of $S_3$ whose kernel contains an element of order $3$ and such that an element $g$ of order $2$ is mapped into itself? Yes, there is. First you consider the projection $S_3\longrightarrow S_3/A_3\simeq S_2$ and then you composite it with the homomorphism from $S_2$ into $S_3$ which maps the non-identity element into $g$. Then you will get the map\begin{align}\operatorname{id}&\mapsto\operatorname{id}\\(1\ \ 2)&\mapsto(1\ \ 2)\\(1\ \ 3)&\mapsto(1\ \ 2)\\(2\ \ 3)&\mapsto(1\ \ 2)\\(1\ \ 2\ \ 3)&\mapsto\operatorname{id}\\(1\ \ 3\ \ 2)&\mapsto\operatorname{id}.\end{align}You can now translate this back into $\textrm{SL}_2(\mathbb{Z}_2)$.
| {
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For matrices, under what conditions can we write $AB = BC$? If $A$ and $B$ are real matrices, when does there exist a matrix $C$ such that $AB = BC$? I understand that there is the case where $A$ and $B$ commute so $AB =BA$, but is there a more general rule (ie. necessary and sufficient conditions)?
Thanks!
| One could interpret the question as: given a matrix $B \in M^{m\times n}$, describe the subspace of $A \in M^{m \times m}$ for which there exists $C \in M^{n \times n}$ such that $AB = BC$.
If $B$ has rank $r$ and $P \in \text{GL}_n$ and $Q \in \text{GL}_m$ are such that $B = P \begin{pmatrix}I_r & 0_{r \times n-r} \\ 0_{m-r \times r} & 0_{m-r \times n-r} \end{pmatrix} Q$, then this is equivalent to
$$P^{-1}AP \begin{pmatrix}I_r & 0_{r \times n-r} \\ 0_{m-r \times r} & 0_{m-r \times n-r} \end{pmatrix} = \begin{pmatrix}I_r & 0_{r \times n-r} \\ 0_{m-r \times r} & 0_{m-r \times n-r} \end{pmatrix} QCQ^{-1} \tag{1}$$
Writing $PAP^{-1} = \begin{pmatrix}X_{r \times r} & X_{r \times m-r} \\ X_{m-r \times r} & X_{m-r \times m-r} \end{pmatrix}$ and $QCQ^{-1} = \begin{pmatrix}Z_{r \times r} & Z_{r \times n-r} \\ Z_{n-r \times r} & Z_{n-r \times n-r} \end{pmatrix}$, $(1)$ becomes
$$\begin{pmatrix}X_{r \times r} & 0_{r \times n-r} \\ X_{m-r \times r} & 0_{m-r \times n-r} \end{pmatrix} = \begin{pmatrix}Z_{r \times r} & Z_{r \times n-r} \\ 0_{m-r \times r} & 0_{m-r \times n-r} \end{pmatrix}$$
so that $X_{m-r \times r} = 0$, and this is the only restriction.
Conclusion: The said subspace of $A \in M^{m \times m}$ is
$$P^{-1}\begin{pmatrix}*_{r \times r} & *_{r \times m-r} \\ 0_{m-r \times r} & *_{m-r \times m-r} \end{pmatrix}P$$
It has dimension $m^2-mr+r^2$.
| {
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Solve for $x$ : $\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$? I want to solve the following equation for $x$ :
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$$
My approach:
Let the given eq.:
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9 \tag {i}$$
On rearranging, we get:
$$\sqrt{x-6} \, + \, \sqrt{x+6} = 9 \, - \, \sqrt{x-1} $$
On Squaring both sides, we get:
$$(x-6) \, + \, (x+6) + 2 \,\,. \sqrt{(x^2-36)} = 81 + (x-1)\, - 18.\, \sqrt{x-1}$$
$$\implies 2x + 2 \,\,. \sqrt{(x^2-36)}= 80 + x - 18.\, \sqrt{x-1}$$
$$\implies x + 2 \,\,. \sqrt{(x^2-36)}= 80 - 18.\, \sqrt{x-1} \tag{ii}$$
Again we are getting equation in radical form.
But, in Wolfram app, I am getting its answer as $x=10$, see it in WolframAlpha.
So, how to solve this equation? Please help...
| You're fine so far. Now from $(ii)$, you have
$$18 \sqrt{x-1} + 2 \sqrt {x^2-36} = 78-x.$$
You'll end up needing to square both sides a couple of more times but you'll be able to clear the radicals. Then you need to confirm that none of your potential solutions are spurious.
| {
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$ \lim_{(x,y)\rightarrow (0,0)} \frac{x^{5}y^{3}}{x^{6}+y^{4}}. $ Does it exist or not? I have this limit:
$$ \lim_{(x,y)\rightarrow (0,0)} \frac{x^{5}y^{3}}{x^{6}+y^{4}}. $$
I think that this limit does not exist (and wolfram|alpha agrees with me). But I can't find a way to prove it. I chose some paths and the limit was always equal to zero. I tried polar coordinates but I couldn't get an answer. Any tips?
(I also have this one: $$ \lim_{(x,y)\rightarrow (0,0)} x^{y^{2}} $$
Can I get 2 different paths to show that the limit does not exist?
$$x=0: \lim_{(x,y)\rightarrow (0,0)} 0^{y^{2}} = \lim_{(x,y)\rightarrow (0,0)} 0 = 0, $$ $$y=0: \lim_{(x,y)\rightarrow (0,0)} x^{0^{2}} = \lim_{(x,y)\rightarrow (0,0)} 1= 1. $$ Am I right? )
| The first limit actually exists and is zero. You can show that by definition.
Namely,
$$\left(\frac{x^5y^3}{x^6+y^4}\right)^2 = \frac{|x^5 y^3|}{x^6+y^4} \cdot \frac{|x^5y^3|}{x^6+y^4}$$
Because
$$x^6+y^4\ge x^6 \implies \frac{1}{x^6+y^4} \le \frac{1}{x^6}$$
$$x^6+y^4\ge y^4 \implies \frac{1}{x^6+y^4} \le \frac{1}{y^4}$$
assuming $x$ and $y$ are both non-zero, we have that
$$\left(\frac{x^5y^3}{x^6+y^4}\right)^2 \le \frac{|x^5 y^3|}{x^6} \cdot \frac{|x^5y^3|}{y^4} = x^4y^2$$
Taking the square root of both sides we get
$$\left|\frac{x^5y^3}{x^6+y^4}\right| \le x^2|y|$$
You can see that this inequality also holds when one of $x$, $y$ is zero (both cannot be zero at the same time). For this to be smaller than $\epsilon$ it suffices to take $\delta=\sqrt[3] \epsilon$.
The second limit does not exist and your proof is correct.
| {
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how to find the point obtained by reflecting over a line? Given the point $A=(-2,6)$ and the line $y=2x$,what are the coordinates of the point B obtained by reflecting A over the line $y=2x$ ?
Can someone teach me how to solve this question please?
| Note that AB has a gradient of $-\frac{1}{2}$, and that AB, which is $y=-\frac{1}{2}x+c$ intersects $y=2x$ at the midpoint of AB.
Sub $(-2,6)$ into $y=-\frac{1}{2}x+c$, and solving, $c=5$.
Equate the two equations.
$-\frac{1}{2}x+5=2x$
Coordinates of midpoint of AB: $x=2$, $y=4$
Coordinates of B: $x=2+2+2=6$, $y=4-(6-4)=2$
| {
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Find $\max\{y-x\}$
If $x+y+z=3, $ and $x^2+y^2+z^2=9$ , find $\max\{y-x\}$.
I tried to do this geometrically, $x+y+z=3$ is a plane in $\Bbb{R}^3$ and $x^2+y^2+z^2=9$ is a ball with radius 3 and center of origin . So the candidate points for $y-x$ are on the intersection of the plane and the ball. But now I am confused how to choose to make $y-x$ maximized.
| Our conditions give $$x^2+y^2+z^2=(x+y+z)^2$$ or
$$xy+xz+yz=0.$$
Now, let $y-x=t$.
Thus, $y=x+t$, $z=3-x-y=3-2x-t$ and we obtain that the equation
$$x(x+t)+(3-2x-t)(x+x+t)=0$$
has real roots $x$, which says that $\Delta\geq0.$
We obtain:
$$3x^2+3(t-2)x+t^2-3t=0,$$ which gives
$$9(t-2)^2-12(t^2-3t)\geq0$$ or
$$-2\sqrt3\leq t\leq2\sqrt3.$$
| {
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Proving the inequality $\angle A+\angle COP < 90^\circ$ in $\triangle ABC$
In an acute angled $\triangle ABC$, $AP \perp BC$and $O$ is its circumcenter. If $\angle C \ge \angle B + 30^\circ$, then prove that $$\angle A + \angle COP < 90^\circ$$
My Attempt:
Extending the line $AP$ to the circumferential point $D$, I connected $O,D$ and $C,D$ and got two lines $OD$ and $CD$ as such below
Given that, $\angle C \ge \angle B + 30^\circ$ and so from that I got
$\angle C +\angle B+ \angle A \ge \angle B + \angle B +\angle A + 30^\circ \implies 180^\circ \ge 2\angle B + \angle A + 30^\circ \implies 150^\circ - \angle A \ge 2\angle B$
$2\angle B \le 150^\circ - \angle A$.....(1)
In right angled $\triangle APC, \angle APC = 90^\circ$
So, $\angle PAC = 90^\circ - \angle C$
After that, we know that $\angle COD = 2\angle DAC$
$\angle COD = 2(90^\circ - \angle C) \implies 180^\circ - \angle COD = 2\angle C$
$180^\circ - \angle COD \ge 2\angle B + 60^\circ \implies 180^\circ - 60^\circ - \angle COD \ge 2\angle B$
$120^\circ - \angle COD \ge 2\angle B$.....(2)
Notice that, both ($150^\circ - \angle A$) and ($120^\circ - \angle COD$) are greater than or equal to $2\angle B$.
So, how could I show the relation between ($150^\circ - \angle A$) and ($120^\circ - \angle COD$). Or, any other way to prove for the desired inequality? Thanks in advance.
| Well, afterall this is an IMO 2001 geometry problem and you can find two solutions, along with a solutions of the other problems from that year, here:
https://sms.math.nus.edu.sg/Simo/IMO_Problems/01.pdf
| {
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Finding integer solutions out of a a | b Determine all positive integer values of (n) such that
$$ { n \choose 0 } + { n \choose 1 } + { n \choose 2 } + { n \choose 3 } \ \bigg| \ 2 ^ { 2008 } $$
What is the sum of all these values?
CURRENT PROGRESS:
I was able to find out that this is equivalent to $(n+1)(n^2 - n + 6) \,| \,3\times (2)^{2009}$ and opened in 2 cases, $n+1 = 2^a$ and $n+1 = 3\times 2^a$, trying to solve like: $n = 2^a - 1$, then $ n^2 - n + 6 = 2^{2a} - 3\times2^a + 8$, doing the same to the 2nd case but couldn't find solutions. Something that should be useful is that $ n^2 - n + 6 = 2^{2a} - 3\times2^a + 8 | \space\space 3\times2^{2019-a}$, it has a factor 3 in it.
| Set $n+1=m$
$n^2-n+6=(m-1)^2-(m-1)+6=m^2-3m+8$
$g=(m, m^2-3m+8)=(m,8)$
Check all the four possible cases
For example,
If $g=4$
$\dfrac{n+1}4\cdot\dfrac{ n^2-n+6}4$ will divide $3\cdot2^{2008-4}$
where the two factors are relatively prime
i.e. one of the factors must be odd $1$ or $3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int \frac{\sqrt{x^2-1}}{x^4}dx$ I am trying to integrate $\int \frac{\sqrt{x^2-1}}{x^4}dx$ via trig substitution. I decided to substitute $x = \sec\theta$ into the square root and $dx = \sec\theta \tan\theta\,d\theta$.
$$\int \frac{\sqrt{\sec^2 \theta-1^2}}{\sec^4\theta} \,dx
= \int \frac{\sqrt{\tan^2\theta + 1 - 1}}{\sec^4\theta}\,dx
= \int \frac{\tan\theta}{\sec^4\theta} \sec\theta \tan\theta\,d\theta = \int \dfrac{\tan^2\theta}{\sec^3\theta}\,d\theta$$
Here is where I am currently stuck. I attempted substitution with $u = \sec\theta, du = \sec x \tan x dx$ but that didn't seem to work out. I wasn't able to get an integration by parts strategy working either.
I think the answer lies in some sort of trigonometry regarding $\int \frac{\tan^2\theta}{\sec^3\theta}\,d\theta$ that I am overlooking to further simplify the problem, but no idea what it is
| Alternative solution (without trigonometry). Note that by integration by parts
$$\begin{align}
\int \frac{\sqrt{x^2-1}}{x^4}dx&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{3}\int \frac{1}{x^2\sqrt{x^2-1}}dx\\
&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{6}\int \frac{D(1-1/x^2)}{\sqrt{1-1/x^2}}dx\\
&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{\sqrt{x^2-1}}{3x}+c=\frac{(x^2-1)^{3/2}}{3x^3}+c.
\end{align}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Evaluate $I=\int_{0}^{\infty}\frac{(x^2+x+1)dx}{(x^3+x+1)\sqrt{x}}$ Evaluate $$I=\int_{0}^{\infty}\frac{(x^2+x+1)dx}{(x^3+x+1)\sqrt{x}}$$
My try:
Letting $\sqrt{x}=t$ we get
$$I=2\int_{0}^{\infty} \frac{t^4+t^2+1}{(t^6+t^2+1)}\,dt$$
Now
$$t^4+t^2+1=(t^2+1)^2-(t^2+1)+1$$
$$t^6+t^2+1=(t^2+1)^3-2(t^2+1)^2+(t^2+1)+1$$
Any clue then?
| According to Wolfram Alpha, the answer is
$$\frac{\pi}{\sqrt{186}}\left(\sqrt{40-14\sqrt[3]{\frac{2}{47-3\sqrt{93}}}-\sqrt[3]{4\left(47-3\sqrt{93}\right)}}+\sqrt{80+14\sqrt[3]{\frac{2}{47-3\sqrt{93}}}+\sqrt[3]{4\left(47-3\sqrt{93}\right)}+36\sqrt{\frac{186}{40-14\sqrt[3]{\frac{2}{47-3\sqrt{93}}}-\sqrt[3]{4\left(47-3\sqrt{93}\right)}}}}\right)$$
which is approximately $4.34952$. Alternatively, it is simply the positive solution of
$$\frac{31}{4\pi^4}x^4 - \frac{10}{\pi^2}x^2 - \frac{6}{\pi}x - 1 = 0 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find the coefficient of $x^{13}$ in the convolution of two generating functions
Four thiefs have stolen a collection of 13 identical diamonds. After the
theft, they decided how to distribute them.
3 of them have special requests:
*
*One of them doesn't want more than 2 diamonds ($\leq2$).
*The other one only wants a number of diamonds that's a multiple of 3.
*And the other one wants an odd number of diamonds greater or equal than 3.
Find in how many ways they can distribute the diamonds.
My first thought was to use generating functions to find the coefficient of $x^{13}$, for this problem it would be:
$f(x)=(1+x+x^2+x^3+...)(1+x+x^2)(1+x^3+x^6+...)(x^3+x^5+x^7+...)=\frac{1}{1-x} \frac{1-x^3}{1-x} \frac{1}{1-x^3}\frac{x^3}{1-x^3}=\frac{1}{(1-x)^{2}}\frac{x^3}{(1-x^2)}$
and that would be equivalent to finding the coefficient of $x^{10}$ in $\frac{1}{(1-x)^{2}}\frac{1}{(1-x^2)}$.
I know that I could use the binomial theorem, but the solution I have says I should be using convolution of these two generating functions, but I have no idea how to use it to find the coefficient of $x^{10}$.
| It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain
\begin{align*}
\color{blue}{[x^{13}]\frac{x^3}{(1-x)^2(1-x^2)}}
&=[x^{10}]\frac{1}{(1-x)^2(1-x^2)}\tag{1}\\
&=[x^{10}]\sum_{j=0}^5 x^{2j}\sum_{k=0}^\infty \binom{-2}{k}(-x)^k\tag{2}\\
&=\sum_{j=0}^5[x^{10-2j}]\sum_{k=0}^\infty\binom{k+1}{k}x^k\tag{3}\\
&=\sum_{j=0}^5[x^{2j}]\sum_{k=0}^\infty(k+1)x^k\tag{4}\\
&=\sum_{j=0}^5(2j+1)\tag{5}\\
&=2\left(\frac{5\cdot6}{2}\right)+6\\
&\,\,\color{blue}{=36}
\end{align*}
Comment:
*
*In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
*In (2) we do a geometric and a binomial series expansion.
*In (3) we select the coefficient of $x^{2j}$ and restrict the upper limit of the outer sum to $5$ since other values do not contribute to $[x^{10}]$. We also use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (4) we change the order of summation $j\to 5-j$ and use $\binom{k+1}{k}=\binom{k+1}{1}=k+1$.
*In (5) we select the coefficient of $x^{2j}$.
*In (6) we apply the formula $\sum_{j=0}^n j=\frac{n(n+1)}{2}$.
| {
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"url": "https://math.stackexchange.com/questions/3142542",
"timestamp": "2023-03-29T00:00:00",
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How many permutations are there of M, M, A, A, A, T, T, E, I, K, so that no two consecutive letters are the same? How many permutations are there of
$$ M, M, A, A, A, T, T, E, I, K $$
so that there are no two consecutive letters are the same?
I would use the Inclusion-exclusion principle where
$$ A_{i} = \{ \text{on} \ i-\text{th} \text{ and }(i+1)-\text{th} \text{ position, there are two same consecutive letters} \}. $$
So my answer would be
$$ \frac{10!}{2!3!2!} - 9 \cdot \left(\frac{9!}{3!2!} + \frac{9!}{3!2!} +\frac{9!}{2!2!}\right)+ 8 \cdot \left(\frac{8!}{2!2!}\right) +8 \cdot 7 \cdot \left(\frac{8!}{3!} + \frac{8!}{2!} +\frac{8!}{2!}\right) - 7 \cdot 6 \cdot 5 \cdot 7!$$
| As you observed, there are $10$ letters, of which $3$ are $A$s, $2$ are $M$s, $2$ are $T$s, $1$ is an $E$, $1$ is an $I$, and $1$ is a $K$.
If there were no restrictions, we would choose three of the ten positions for the $A$s, two of the remaining seven positions for the $M$s, two of the remaining five positions for the $T$s, and arrange the $E$, $I$, $K$ in the remaining three positions in
$$\binom{10}{3}\binom{7}{2}\binom{5}{2}3! = \frac{10!}{3!7!} \cdot \frac{7!}{2!5!} \cdot \frac{5!}{2!3!} \cdot 3! = \frac{10!}{3!2!2!}$$
in agreement with your answer.
From these, we must subtract those arrangements in which or more pairs of identical letters are adjacent.
Arrangements with a pair of adjacent identical letters: We have to consider cases, depending on whether the identical letters are $A$s, $M$s, or $T$s.
A pair of $A$s are adjacent: We have nine objects to arrange: $AA, A, M, M, T, T, E, I, K$. Choose two of the nine positions for the $M$s, two of the remaining seven positions for the $T$s, and then arrange the five distinct objects $AA$, $A$, $E$, $I$, $K$ in the remaining five positions, which can be done in
$$\binom{9}{2}\binom{7}{2}5!$$
ways.
A pair of $M$s are adjacent: We have nine objects to arrange: $A, A, A, MM, T, T, E, I, K$. Choose three of the nine positions for the $A$s, two of the remaining six positions for the $T$s, and arrange the four distinct objects $MM, E, I, K$ in the remaining four positions, which can be done in
$$\binom{9}{3}\binom{6}{2}4!$$
ways.
A pair of $T$s are adjacent: By symmetry, there are
$$\binom{9}{3}\binom{6}{2}4!$$
such arrangements.
Arrangements with two pairs of adjacent identical letters: This can occur in two ways. Either the pairs are disjoint or they overlap.
Two pairs of $A$s are adjacent: This can only occur if the three $A$s are consecutive. Thus, we have eight objects to arrange: $AAA, M, M, T, T, E, I, K$. Choose two of the eight positions for the $M$s, two of the remaining six positions for the $T$s, and arrange the four distinct objects $AAA, E, I, K$ in the remaining four positions, which can be done in
$$\binom{8}{2}\binom{6}{2}4!$$
ways.
A pair of $A$s are adjacent and a pair of $M$s are adjacent: We have eight objects to arrange: $AA, A, MM, T, T, E, I, K$. Choose two of the eight positions for the $T$s and arrange the six distinct objects $AA, A, MM, E, I, K$ in the remaining six positions in
$$\binom{8}{2}6!$$
ways.
A pair of $A$s are adjacent and a pair of $T$s are adjacent: By symmetry, there are
$$\binom{8}{2}6!$$
such arrangements.
A pair of $M$s are adjacent and a pair of $T$s are adjacent: We have eight objects to arrange: $A, A, A, MM, TT, E, I, K$. Choose three of the eight positions for the $A$s and then arrange the remaining five distinct objects $MM, TT, E, I, K$ in the remaining five positions in
$$\binom{8}{3}5!$$
ways.
Arrangements with three pairs of adjacent identical letters: We again consider cases.
Two pairs of $A$s are adjacent and a pair of $M$s are adjacent: We have seven objects to arrange, $AAA, MM, T, T, E, I, K$. Choose two of the seven positions for the $T$s and arrange the five distinct objects $AAA, MM, E, I, K$ in the remaining five positions in
$$\binom{7}{2}5!$$
ways.
Two pairs of $A$s are adjacent and a pair of $T$s are adjacent: By symmetry, there are
$$\binom{7}{2}5!$$
such arrangements.
A pair of $A$s are adjacent, a pair of $M$s are adjacent, and a pair of $T$s are adjacent: We have seven objects to arrange: $AA, A, MM, TT, E, I, K$. Since all the objects are distinct, there are
$$7!$$
such arrangements.
Arrangements containing four pairs of adjacent identical letters: We have six objects to arrange: $AAA, MM, TT, E, I, K$. Since all the objects are distinct, there are
$$6!$$
such arrangements.
By the Inclusion-Exclusion Principle, there are
$$\binom{10}{3}\binom{7}{2}\binom{5}{2}3! - \binom{9}{2}\binom{7}{2}5! - \binom{9}{3}\binom{6}{2}4! - \binom{9}{3}\binom{6}{2}4! + \binom{8}{2}\binom{6}{2}4! + \binom{8}{2}6! + \binom{8}{2}6! + \binom{8}{3}5! - \binom{7}{2}5! - \binom{7}{2}5! - 7! + 6!=47760$$
arrangements in which no two adjacent letters are identical.
| {
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"url": "https://math.stackexchange.com/questions/3143471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $T_n$ an unbiased estimator of $\theta$? Prove your answer.
Let $(x_1, x_2, \dots, x_n)$ be an observed sample from a distribution with probability density function given by $$f(x) =\begin{cases}\displaystyle\frac{1}{5\theta+8}&\text{if }0 \leq x \leq 5 \theta + 8\\0&\text{otherwise. }\end{cases}$$
Where $\theta \in \mathbb R^+$ is an unknown parameter. It can be shown that $T_n = \frac{X_{(n)}-8}{5}$ is an $MLE$ of $\theta$, where $X_{(n)}$ is the $n^{th}$ order statistic. $\big($i.e. max$\{X_1, X_2, \dots, X_n\}$$\big)$.
(a) Is $T_n$ an unbiased estimator of $\theta$? Prove your answer.
(b) Is $T_n$ asymptotically unbiased for $\theta$? Give reasons for your answer.
(c) Is $T_n$ consistent for $\theta$ in probability? Prove your answer.
ATTEMPT
$(a)$
$$F_{X}(x) = \int_{0}^{x} f(t)dt = \int_{0}^{x} \frac{1}{5 \theta +8} dt = \frac{1}{5\theta+8}t|_{0}^{x} = \frac{1}{5\theta+8}x$$
$$F_{X_{(n)}}(x) = \left(\frac{x}{5\theta+8}\right)^n$$
$$f_{X_{(n)}}(x) = \frac{dF_{X_{(n)}}(x)}{dx} = \frac{n}{(5 \theta + 8)^n}x^{n-1}, 0 \leq x \leq 5 \theta + 8$$
$$\mathsf{E}T_n = \mathsf{E}\left[\left(\frac{X_{(n)} - 8}{5}\right)\right] = 1/5(\mathsf{E}(X_{(n)}) - 8)$$
$$\begin{align}\mathsf{E}(X_{(n)}) = \int_{0}^{5 \theta + 8} x f_{X_{(n)}}(x)dx &= \int_{0}^{5 \theta + 8}\frac{n}{(5 \theta + 8)^n}x^{n}dx \\ &= \frac{n}{(5 \theta + 8)^n(n+1)}x^{n+1}\bigg|_{0}^{5 \theta+8} = \frac{n(5 \theta + 8)^{n+1}}{(5 \theta + 8)^n(n+1)} = \frac{(5 \theta + 8)n}{n+1}\end{align}$$
$$\mathsf{E}T_n = 1/5(\mathsf{E}(X_{(n)}) - 8) = 1/5 \left (\frac{(5 \theta + 8)n}{n+1} - 8 \right) \neq \theta,$$ so $T_n$ is not an unbiased estimator of $\theta.$
$(b)$ Yes, because $\lim\limits_{n\to\infty} \mathsf{E}T_n = \lim\limits_{n\to\infty} 1/5 \left (\dfrac{(5 \theta + 8)n}{n+1} - 8 \right) = \theta$
$(c)$ $$\begin{align}\mathsf{E}(X^2_{(n)}) = \int_{0}^{5 \theta + 8} x^2 f_{X_{(n)}}(x)dx &= \int_{0}^{5 \theta + 8}\frac{n}{(5 \theta + 8)^n}x^{n+1}dx \\ & = \frac{n}{(5 \theta + 8)^n(n+2)}x^{n+2}\bigg|_{0}^{5 \theta+8} = \frac{n(5 \theta + 8)^{n+2}}{(5 \theta + 8)^n(n+2)}= \frac{(5 \theta + 8)^2n}{n+2}\end{align}$$
$$\begin{align}\lim\limits_{n\to\infty} \mathsf{Var}(X_{(n)}) = \lim\limits_{n\to\infty} \left(\mathsf{E}X^2_{(n)} - (\mathsf{E}X_{(n)})^2\right) = \lim\limits_{n\to\infty} \left(\frac{(5 \theta + 8)^2n}{n+2} - \left(\frac{(5 \theta + 8)n}{n+1}\right)^2\right) = 0\end{align}$$
$T_n$ is consistent for $\theta$ in probability because $(1)$ $T_n$ is asymptotically unbiased for $\theta$ and $(2)$ $\lim\limits_{n\to\infty} \mathsf{Var}(X_{(n)}) = 0$
Is this correct?
| The computation in (a) is not strictly necessary, because $$\Pr[T_n > \theta] = 0,$$ whereas $\Pr[T_n < \theta] > 0$; therefore, $\operatorname{E}[T_n] < \theta$ hence is trivially biased.
However, once the computation is performed, (b) immediately follows, and (c) is also correct.
| {
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Equilateral triangle with vertices on 3 concentric circles Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of the equilateral triangle?
My idea is to set a point at the middle of the triangle, then use the distance of it to the vertices given that the three concentric circles are set as $$x^2 + y^2 = 1$$$$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 9$$ I will manipulate the formula afterwards,,,
| While the posted geometric solutions are much simpler, it is possible to do this with algebra and coordinate geometry.
Centering the circles at the origin, we get the equations that you provided: $$x^2+y^2=1$$
$$x^2+y^2=4$$
$$x^2+y^2=9$$
Let's choose an arbitrary point on the smallest circle, say $(0, 1)$ for simplicity. Let $l$ be the length of each side of the equilateral triangle. So the vertices on the other two circles must be a distance of $l$ from our chosen point $(0, 1)$. Equivalently, the two vertices must be on the circle with radius $l$ centered at $(0, 1)$ We can set up an equation to represent this:
$$x^2+(y-1)^2=l^2$$
Red is the circle of radius 1, Blue is the circle of radius 2, Green is the circle of radius 3, Dotted Black is the circle centered at $(0, 1)$ with radius $l$.
Finding the intersection of this circle with the other two circles, we get the following two equations to represent the vertices:
$$x^2+y^2-4=x^2+(y-1)^2-l^2$$
$$x^2+y^2-9=x^2+(y-1)^2-l^2$$
Solving the equations for $y$, we get the following. $y_1$ is the y-coordinate of the vertex on the circle of radius 2, and $y_2$ is the y-coordinate of the vertex on the circle of radius 3:
$$y_1=\frac{5-l^2}{2}$$
$$y_2=\frac{10-l^2}{2}$$
We can plug this into their respective equations to find the x-coordinates:
$$x_1=\sqrt{4-\left(\frac{5-l^2}{2}\right)^2}$$
$$x_2=\sqrt{9-\left(\frac{10-l^2}{2}\right)^2}$$
These coordinates are a distance of $l$ from the point on the smallest circle. It now remains to make these two points a distance of $l$ from each other:
$$l=\sqrt{\left(\sqrt{9-\left(\frac{10-l^2}{2}\right)^2}-\sqrt{4-\left(\frac{5-l^2}{2}\right)^2}\right)^2+\left(\frac{10-l^2}{2}-\frac{5-l^2}{2}\right)^2}$$
Solving this equation for $l$ yields the answer of $l=\sqrt{7}$
| {
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"url": "https://math.stackexchange.com/questions/3149465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\sqrt 7 + 2 \sqrt 5 < 2 + \sqrt{35}$ How can I prove that $\sqrt 7 + 2 \sqrt 5 < 2 + \sqrt{35}$?
I can't convert the right side to get the $\sqrt 5$.
| Since $2^2 <7$. we get
$2=\sqrt{2^2} <√7$ since $√$ is strictly increasing.
Also since $1 < 5$ we get $1 <√5.$
Finally :
$2(√5-1) < √7(√5-1)$; (Why?)
$√7 +2√5 < 2 +\sqrt{35}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence behavior of a rational iterative procedure of the form $\frac{1}{2}(z+1/z)$. I am trying to prove the following result.
First, we have an auxiliary sequence satisfying
$E_{n+1}=\dfrac{E_n^2}{(1+\sqrt{1-E_n^2})^2}$ with $E_0<1$.
It is called the Landen transformation.
Let $S_0=1+r\exp(i\theta)$ with $r\le F_0$, and $F_0=E_0$.
The interested iteration relation writes
$$S_{n+1}=\frac{\sqrt{1-E_n^2}}{1+\sqrt{1-E_n^2}}(\frac{S_n}{\sqrt{1-E_n^2}}+\frac{\sqrt{1-E_n^2}}{S_n})$$
I intend to find out how the $S_{n}$ converges to 1.
Since $E_n$ satisfies the Landen transformation, it decreases to zero rapidly.
Thus, the iteration is approximately to be $\frac{1}{2}(z+1/z)$ and in this case, the problem can be handled if $|S_n-1|<0.1$ for some $n$.
However, I have no idea how to analysis the first several terms when $E_n$ and $F_n$ is near $1$.
Notice that if $S_n=1+E_n$, then $S_{n+1}=1+E_{n+1}$. Thus，we set
$$z_n=\frac{S_n}{1+E_n}$$
From the iteration relation
\begin{equation*}
S_{n+1}=\frac{\sqrt{1-E_n^2}}{1+\sqrt{1-E_n^2}}(\frac{S_n}{\sqrt{1-E_n^2}}+\frac{\sqrt{1-E_n^2}}{S_n}),
\end{equation*}
we deduce that
\begin{align*}
\frac{S_{n+1}}{1+E_{n+1}}&=\frac{\frac{\sqrt{1-E_n^2}}{1+\sqrt{1-E_n^2}}}{1+E_{n+1}}\left(
\frac{S_n}{1+E_n}\sqrt{\frac{1+E_n}{1-E_n}}+\frac{\sqrt{\frac{1-E_n}{1+E_n}}}{\frac{S_n}{1+E_n}}
\right)\\
&=\frac{\sqrt{1-E_n^2}}{2}\left(
\frac{S_n}{1+E_n}\sqrt{\frac{1+E_n}{1-E_n}}+\frac{\sqrt{\frac{1-E_n}{1+E_n}}}{\frac{S_n}{1+E_n}}
\right),
\end{align*}
which can be reformulated as
\begin{equation*}
z_{n+1}=\frac{1+E_n}{2}
\left[
z_n+\frac{{\frac{1-E_n}{1+E_n}}}{z_n}\right].
\end{equation*}
Then, the problem is to show how $z_n\to 1$.
Next, we map $z_n$ into the unit disk by setting
\begin{equation*}
w_n=\frac{z_n-1}{z_n+1}.
\end{equation*}
Then the inverse transform writes
\begin{equation*}
z_n=\frac{1+w_n}{1-w_n}
\end{equation*}
Then we have
\begin{align*}
w_{n+1}&=\frac{z_{n+1}-1}{z_{n+1}+1}=\frac{
\frac{1}{2}\left[(1+E_n)z_n+\frac{1-E_n}{z_n}\right]-1}{
\frac{1}{2}\left[(1+E_n)z_n+\frac{1-E_n}{z_n}\right]+1}\\
&=\frac{
\left[(1+E_n)z_n^2+1-E_n\right]-2z_n}{
\left[(1+E_n)z_n^2+1-E_n\right]+2z_n}\\
&=\frac{(z_n-1)(z_n-1+E_n(z_n+1))}{(z_n+1)(z_n+1+E_n(z_n-1))}\\
&=w_n\frac{w_n+E_n}{1+E_nw_n}.
\end{align*}
Next, we prove that how $w_n\to 0$.
For the initial data, we have
\begin{equation*}
z_0=\frac{S_0}{1+E_0},\quad w_0=\frac{z_0-1}{z_0+1}.
\end{equation*}
We may take the assumption that $\Re(S_0)>1-E_0$.
Thanks in advance.
| Suppose that
$|w_n|<F_n$, then we set
\begin{equation*}
w_n=r\exp(i \theta), \quad r\le F_n.
\end{equation*}
Thus,
\begin{align*}
w_{n+1}=w_n\frac{w_n+E_n}{1+E_nw_n}=
r\exp(i \theta)\frac{r\exp(i \theta)+E_n}{1+E_nr\exp(i \theta)}
\end{align*}
Then
\begin{align*}
|w_{n+1}|&=r\left|\frac{r\exp(i \theta)+E_n}{1+E_nr\exp(i \theta)} \right|=r
\sqrt{\frac{E_n^2+r^2+2E_nr\cos\theta}{1+E_n^2r^2+2E_nr\cos\theta}}\\
&\le \frac{r(E_n+r)}{(1+E_nr)}
\le \frac{F_n(E_n+F_n)}{(1+E_nF_n)}.
\end{align*}
The inequality maight be coarse.
Moreover,
$$\frac{E_n+F_n}{1+E_nF_n}<1.$$
and the fractional tends to zero.
Then we conclude that $F_n\to 0$.
However, if we know for which $n$, $F_n\le \epsilon$, the result will be better. But I cannot accomplish this task.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Area between curves. I have to calculate area bounded by curves :
$(x^3+y^3)^2=x^2+y^2 $ for $ x,y \ge 0 $.
I tried to use polar coordinates, but I have :
$r^4(\cos^6\alpha +2\sin^3\alpha\cos^3\alpha + \sin^6\alpha)=1$
| By your work: $$r=\frac{1}{\sqrt{\sin^3\alpha+\cos^3\alpha}}.$$ Since it's symmetric in respect to $y=x$, we obtain that the needed area it's
$$2\int\limits_0^{\frac{\pi}{4}}d\alpha\int\limits_0^{\frac{1}{\sqrt{\sin^3\alpha+\cos^3\alpha}}}rdr=\int_0^{\frac{\pi}{4}}\frac{1}{\sin^3\alpha+\cos^3\alpha}d\alpha.$$
Can you end it now?
| {
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Prove without induction that $2×7^n+3×5^n-5$ is divisible by $24$. I proved this by induction. But I want to show it using modular arithmetic. I tried for sometime as follows
$$2×7^n-2+3×5^n-3\\
2(7^n-1)+3(5^n-1)\\
2×6a+3×4b\\
12(a+b)$$
In this way I just proved that it is divisible by 12 but it is not enough. Am I missing something or it will solved by some other method.
| Yes, it can be done by another method. Note that $7^2=2\times24+1$ and that $5^2=24+1$ and that therefore$$7^n\equiv\begin{cases}7\pmod{24}&\text{ if $n$ is odd}\\1\pmod{24}&\text{ otherwise}\end{cases}$$and$$5^n\equiv\begin{cases}5\pmod{24}&\text{ if $n$ is odd}\\1\pmod{24}&\text{ otherwise.}\end{cases}$$So:
*
*if $n$ is odd, then $2\times7^n+3\times5^n-5\equiv2\times7+3\times5-5=24\equiv0\pmod{24}$;
*otherwise, $2\times7^n+3\times5^n-5\equiv2\times1+3\times1-5\equiv0\pmod{24}$.
| {
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About the fact that $\frac{a^2}{b} + \frac{b^2}{a} + 7(a + b) \ge 8\sqrt{2(a^2 + b^2)}$ There's a math competition I participated yesterday (19/3/2019).
In these kinds of competitions, there will always be at least one problem about inequalities.
Now this year's problem about inequality is very easy. I am more interested in last year's problem, which goes by the following:
If $a$ and $b$ are positives then prove that $$ \frac{a^2}{b} + \frac{b^2}{a} + 7(a + b) \ge 8\sqrt{2(a^2 + b^2)}$$
Now I know what you are thinking. It's a simple problem.
By the Cauchy - Schwarz inequality, we have that $\dfrac{a^2}{b} + \dfrac{b^2}{a} \ge \dfrac{(a + b)^2}{a + b} = a + b$. $$\implies \dfrac{a^2}{b} + \dfrac{b^2}{a} + 7(a + b) \ge (a + b) + 7(a + b) = 8(a + b) \le 8\sqrt{2(a^2 + b^2)}$$
And you have fallen into the traps of the people who created the test. Almost everyone in last year's competition did too.
But someone came up with an elegant solution to the problem. He was also the one winning the contest.
You should have 15 minutes to solve the problem. That's what I also did yesterday.
| Squaring the given inequality and factorizing we get
$${\frac { \left( {a}^{2}+18\,ba+{b}^{2} \right) \left( a-b \right) ^{4
}}{{b}^{2}{a}^{2}}}
\geq 0$$ which is true.
| {
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"timestamp": "2023-03-29T00:00:00",
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GCD of cubic polynomials I would appreciate some help finding $GCD(a^3-3ab^2, b^3-3ba^2)$; $a,b \in \mathbb{Z}$. So far I've got here: if $GCD(a,b)=d$ then $\exists \alpha, \beta$ so that $GCD(\alpha, \beta)=1$ and $\alpha d=a$, $\beta d=b$.
Therefore we know that $GCD(a^3-3ab^2, b^3-3ba^2)=d^3 GCD(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2)$. However I don't know to figure out $GCD(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2)$ given that $GCD(\alpha,\beta)=1$.
| I'll write $\alpha=m,\beta=n$ for the ease of typing
If $d(\ge1)$ divides $m^3-3mn^2,n^3-3m^2n$
$d$ will divide $n(m^3-3mn^2)+3m(n^3-3m^2n)=-8m^3n$
and $3n(m^3-3mn^2)+m(n^3-3m^2n)=-8mn^3$
Consequently, $d$ must divide $(-8m^3n,-8mn^3)=8mn(m^2,n^2)=8mn$
So, $d$ will divide $8$
As $(m,n)=1,$ both $m,n$ cannot be even
If $m$ is even, $n^3-3m^2n$ will be odd $\implies d=1$
If both $m,n$ are odd,. $m^2,n^2\equiv1\pmod8$
$m(m^2-3n^2)\equiv m(1-3)\equiv-2m\pmod8$
Similarly, we can establish the highest power of $2$ in $m^3-3mn^2$ will be $2$
$\implies d=2$ if $m,n$ are odd
| {
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"timestamp": "2023-03-29T00:00:00",
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Quotient rule and implicit differentiation Find $\frac{dy}{dx}$ for $x^2=\frac{x-y}{x+y}$.
I have solved this in two ways.
First, I multiplicated the whole equation by $x+y$ and then I calculated the implicit derivative. I got the following solution:
$\frac{1-3x^2-2xy}{x^2+1}$
So far so good. When I calculated the implicit derivative of the original expression using the quotient rule though, I got a different solution, i.e.:
$-\frac{x(y+x)^2-y}{x}$
I have tried using Wolfram and I got the same results.
Can anyone explain to me why I get different solutions ?
| The two solutions that you found are equal. So where is no contradiction.
Note that solving $\quad x^2=\frac{x-y}{x+y}\quad$ for $y$ gives $\quad y=x\frac{1-x^2}{1+x^2}$
Your first solution :
$$\frac{1-3x^2-2xy}{x^2+1} =\frac{1-3x^2-2x\left(x\frac{1-x^2}{1+x^2} \right)}{x^2+1} = \frac{1-4x^2-x^4}{(1+x^2)^2}$$
Your second solution :
$$-(x+y)^2+\frac{y}{x}=-\left(x+x\frac{1-x^2}{1+x^2}\right)^2+\frac{1}{x}\left(x\frac{1-x^2}{1+x^2}\right)=\frac{1-4x^2-x^4}{(1+x^2)^2}$$
Thus
$$\frac{1-3x^2-2xy}{x^2+1}=-(x+y)^2+\frac{y}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157869",
"timestamp": "2023-03-29T00:00:00",
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Showing $\int_{a}^{b} \left\lfloor x \right\rfloor dx + \int_{a}^{b} \left\lfloor -x \right\rfloor dx=a-b$ I want to show
$$\int_{a}^{b} \left\lfloor x \right\rfloor dx + \int_{a}^{b} \left\lfloor -x \right\rfloor dx=a-b$$
I know that
\begin{equation}
\left\lfloor -x \right\rfloor = \begin{cases} -\left\lfloor x \right\rfloor & \text{if } x \in \mathbb{Z} \\ -\left\lfloor x \right\rfloor-1 & \text{if } x \notin \mathbb{Z}. \end{cases}
\end{equation}
In this case do I use $-\left\lfloor x \right\rfloor$ or $-\left\lfloor x \right\rfloor-1$? I think I am confused about some definitions, one of the solutions said $\left\lfloor x \right\rfloor$ is constant on the open subintervals of the partition
$$P=\left(a, \left\lfloor a \right\rfloor+1 \cdots \left\lfloor a \right\rfloor + \left\lfloor b-a \right\rfloor, b\right)$$
and since there are no integers in the open subintervals of P, then we would use $-\left\lfloor x \right\rfloor - 1$.. I don't think I quite understand this point here. I know I can solve it and say
\begin{align*}
\int_{a}^{b} \left\lfloor x \right\rfloor dx + \int_{a}^{b} \left\lfloor -x \right\rfloor dx= \int_{a}^{b} \left\lfloor x \right\rfloor dx + \int_{a}^{b} -\left\lfloor x \right\rfloor -1 \; dx= a-b
\end{align*}
but I don't understand why.
| What happens at the integers does not matter since they have content 0. Suppose
$$n < x < n + 1.$$
then $\lfloor x \rfloor = n$, and $-n > x > -(n+1)$, so $\lfloor -x \rfloor = -(n+1)$. Adding gives $\lfloor x\rfloor + \lfloor -x \rfloor = -1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to integrate $\frac{1}{\sqrt{x^2+x+1}}$
How to integrate $$\frac{1}{\sqrt{x^2+x+1}}$$
I tried to solve this integral as follows
$\displaystyle
\int \frac{1}{\sqrt{x^2+x+1}} \ dx=
\int \frac{1}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}} \ dx=
\int \frac{1}{\sqrt{(\frac{2x+1}{2})^2+\frac{3}{4}}} \ dx=
\int \frac{1}{\sqrt{\frac{3}{4}((\frac{2x+1}{\sqrt{3}})^2+1)}} \ dx=
\int \frac{1}{\sqrt{(\frac{2x+1}{\sqrt{3}})^2+1}}\frac{2}{\sqrt{3}} \ dx$
Substitution $t=\frac{2x+1}{\sqrt{3}} ;dt=\frac{2}{\sqrt{3}} \ dx$
$\displaystyle\int \frac{1}{\sqrt{(\frac{2x+1}{\sqrt{3}})^2+1}}\frac{2}{\sqrt{3}} \ dx=
\int \frac{1}{\sqrt{t^2+1}} \ dt$
Substitution $\sqrt{u-1}= t;\frac{1}{2\sqrt{u-1}} \ du= dt$
$\displaystyle
\int \frac{1}{\sqrt{t^2+1}} \ dt=
\int \frac{1}{2 \sqrt{u(u-1)}} \ du=
\frac{1}{2}\int \frac{1}{\sqrt{u^2-u}} \ du=
\frac{1}{2}\int \frac{1}{\sqrt{(u-\frac{1}{2})^2-\frac{1}{4}}} \ du=
\int \frac{1}{\sqrt{(2u-1)^2-1}} \ du$
Substitution $g= 2u-1;dg= 2 \ du$
$\displaystyle
\int \frac{1}{\sqrt{(2u-1)^2-1}} \ du=
\frac{1}{2}\int \frac{2}{\sqrt{(2u-1)^2-1}} \ du=
\frac{1}{2}\int \frac{1}{\sqrt{g^2-1}} \ dg=
\frac{1}{2}\arcsin g +C=\frac{1}{2}\arcsin (2u-1) +C= \frac{1}{2}\arcsin (2(t^2+1)-1) +C=\frac{1}{2}\arcsin (2((\frac{2x+1}{\sqrt{3}})^2+1)-1) +C$
However when I tried to graph it using desmos there was no result, and when i used https://www.integral-calculator.com/ on thi problem it got the result $$\ln\left(\left|2\left(\sqrt{x^2+x+1}+x\right)+1\right|\right)$$
where have I made a mistake?
| You have got mixed up with the $\arcsin$. Note that $$\int\frac{1}{\sqrt{\color{red}{1-g^2}}}\, dg = \color{red}{\arcsin g},$$
but
$$\int\frac{1}{\sqrt{\color{blue}{g^2-1}}}\, dg = \color{blue}{\ln\left(\left|g+\sqrt{g^2 -1}\right|\right)}.$$
You can verify this by differentiating the right-hand side, or try to derive this by using a trig. substitution like $g = \sec \theta$. See the bottom two answers here for approaches that don't involve hyperbolic functions.
| {
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"timestamp": "2023-03-29T00:00:00",
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For which $a>0$ series is convergent?
For which $a>0$ series $$\sum { \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right) \right)^a } $$ $(n \in \mathbb N)$ is convergent?
My try:From Taylor theorem I know that:$$a_{n}={ \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right) \right)^a } = (\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a}$$
Then I have:
$$(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a} \le (\frac{1}{n^{4}}+o(\frac{1}{n^{8}}))^{a}$$At this point, my problem is that if I had:
$$(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a} \le (\frac{1}{n^{4}})^{a}$$I could say that $0 \le a_{n} \le (\frac{1}{n^{4}})^{a}$ so for $a>\frac{1}{4}$ this series is convergent.However in this task I have also $o(\frac{1}{n^{8}}))^{a}$ and I don't know what I can do with it to finish my sollution.Can you help me?
| It is much simpler to use equivalents, which can be found with Taylor's formula:
*
*$2(1-\cos x)=x^2-\dfrac{x^4}{12}+o(x^4)$,
*$\sin(\sin x)=\sin\Bigl(x-\dfrac{x^3}6+o(x^3)\Bigr)=\Bigl(x-\dfrac{x^3}6\Bigr)-\frac16\Bigl(x-\dfrac{x^3}6\Bigr)^{\!3}+o(x^3)=x-\dfrac{x^3}3+o(x^3)$, so
$$x\sin(\sin x)=x^2-\frac{x^4}3+o(x^4).$$
Now, replacing $x$ with $\frac 1n$, we obtain
$$2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)=\frac1{n^2}-\frac1{12n^4}-\frac1{n^2}+\frac1{3n^4}+o\biggl(\frac1{n^4}\biggr)=\frac1{4n^4}+o\biggl(\frac1{n^4}\biggr)$$
so an asymptotic equivalent for the general term of the series is
$$\left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left(\sin\frac{1}{n}\right) \right)^a\sim_\infty\frac1{4^a n^{4a}}.$$
Knowing that series (with positive terms) which have asymptotically equivalent general terms both converge or both diverge, can you conclude?
| {
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"timestamp": "2023-03-29T00:00:00",
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induction proof: number of compositions of a positive number $n$ is $2^{n-1}$ So my problem was to find the formula for the number of composition for a positive number $n$. Then prove its validity. I found the formula which was $2^{n-1}$. Having trouble with the induction.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
When $n=5$ $$ 1+2+4+(2^{5-2}+1)=2^{5-1}$$
$$ 1+2+4+(2^{3}+1)=2^{4}$$
$$ 1+2+4+(8+1)=16$$
$$ 16=16$$
Assume that $n=k$
$$2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}$$
Now show for $k+1$
$$2^0+2^1+2^2+...+(2^{k-1}+1)=2^{k}$$
Don't know what to do after. Help is appreciated.
$$2^0+2^1+2^2+...+(2^k)+(2^{k-1}+1)=2^{k}$$
| What you know: $\color{blue}{2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}}$
What you want to show: $\color{green}{2^0+2^1+2^2+...+(2^{k-1}+1)=2^{k}}$
Use what you know:
$\color{green}{2^0+2^1+2^2+...+2^{k-2} + (2^{k-1}+1)} =$
$\color{blue}{2^0+2^1+2^2+...+2^{k-2}} + (\color{green}{2^{k-1}} + \color{blue}1) =$
$\color{blue}{2^0+2^1+2^2+...+(2^{k-2}+1)}+\color{green}{2^{k-1}}=$
$\color{blue}{2^{k-1}} + \color{green}{2^{k-1}}=$
$\color{green}{2\times 2^{k-1}} =$
$\color{green}{2^{(k-1) + 1}} =$
$\color{green}{2^k}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integers $n$ satsifying $\frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}}$
If $\displaystyle \frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}},n\in \mathbb{Z}$, then number of $n$ satisfies given equation ,is
What I tried:
Let $\displaystyle \frac{\pi}{n}=x$ and equation is $\sin 5x=\sin 3x$
$\displaystyle \sin (5x)-\sin (3x)=2\sin (4x)\cos (x)=0$
$\displaystyle 4x= m\pi$ and $\displaystyle x= 2m\pi\pm \frac{\pi}{2}$
$\displaystyle \frac{4\pi}{n}=m\pi\Rightarrow n=\frac{4}{m}\in \mathbb{Z}$ for $m=\pm 1,\pm 2\pm 3,\pm 4$
put into $\displaystyle x=2m\pi\pm \frac{\pi}{2}$
How do i solve my sum in some easy way Help me please
| No need to go with sum-to-product formulas. From $\sin5x=\sin3x$ you get either
$$
5x=3x+2k\pi
$$
or
$$
5x=\pi-3x+2k\pi
$$
The former yields $x=k\pi$, so $1=kn$, whence $n=\pm1$.
The latter yields
$$
\frac{\pi}{n}=\frac{1}{8}(2k+1)\pi
$$
that is, $(2k+1)n=8$. Hence $n=\pm8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164257",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Does it follow that a+b+c=x+y+z=m+n+p If
$$abc=xyz=mnp$$
$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2=a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=m^4+n^4+p^4-2m^2n^2-2n^2p^2-2p^2m^2$
$$\frac{x}{m}=\frac{n}{b}=\frac{c}{z}$$
Then $a+b+c=x+y+z=m+n+p$?
$a,b,c$ positive and are lenghts of triangles,
$x,y,z$ same, $m,n,p$ same.
I edited it.
| Hint: Note that
$$
a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a + b + c)(a + b - c)(a - b + c)(a - b - c).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) \ge 3/2$ Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c \ge \frac{3}{2}\cdot [(a+b)(a+c)(b+c)]^{\frac{1}{3}}$$
Hence or otherwise prove:
$$\color{blue}{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}}$$
| To minimize $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ , we need to have
$$
\begin{align}
0
&=\delta\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\\
&=\left(\frac1{b+c}-\frac{b}{(c+a)^2}-\frac{c}{(a+b)^2}\right)\delta a\\
&+\left(\frac1{c+a}-\frac{c}{(a+b)^2}-\frac{a}{(b+c)^2}\right)\delta b\\
&+\left(\frac1{a+b}-\frac{a}{(b+c)^2}-\frac{b}{(c+a)^2}\right)\delta c\tag1
\end{align}
$$
for all $\delta a,\delta b,\delta c$. That means
$$
\begin{align}
\frac1{a+b}&=\frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}\tag2\\
\frac1{b+c}&=\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\tag3\\
\frac1{c+a}&=\frac{c}{(a+b)^2}+\frac{a}{(b+c)^2}\tag4
\end{align}
$$
Subtract $(4)$ from the sum of $(2)$ and $(3)$:
$$
\frac1{b+c}+\frac1{a+b}-\frac1{c+a}=\frac{2b}{(c+a)^2}\tag5
$$
Add $\frac2{c+a}$ and divide by $2(a+b+c)$:
$$
\frac1{2(a+b+c)}\left(\frac1{b+c}+\frac1{a+b}+\frac1{c+a}\right)=\frac1{(c+a)^2}\tag6
$$
By symmetry,
$$
\frac1{(a+b)^2}=\frac1{(b+c)^2}=\frac1{(c+a)^2}\tag7
$$
from which we get $a=b=c$. Thus, we get
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac32\tag8
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Let $y = \sqrt{x + 3} \implies 3 = y^2 - x$.
$$\large \begin{align}
&13x + 2(3x + 2)\sqrt{x + 3} +42\\
= &14(x + 3) + (6x + 4)y - x\\
= &14y^2 + [6(x + 3) - 14]y - x\\
= &14y(y - 1) - (y^2 - x - 9)y^3 - x\\
= &14y(y - 1) + x(y^3 - y) - y^3(y^2 - 1) + 8y^3\\
= &14y(y - 1) + (xy^2 + xy + x)(y - 1) - (y^4 + y^3)(y - 1) + 8y^3\\
= &(-y^4 + y^3 + xy^2 + xy + x + 14y)(y - 1) + 8y^3\\
\end{align}$$
And I'm stuck here.
| I have no idea where you are trying to go with your approach, but I would suggest squaring both sides of
$$13x+42=-2(3x+2)\sqrt{x+3}.$$
This yields a cubic equation in $x$ which can be solved by standard methods; it turns out that all roots are rational so the rational root theorem yields them all. It remains to check whether these three roots are in fact solutions to the original equation.
| {
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Show that the equation $x^5+x^4=1$ has a unique solution. Show that the equation $x^5 + x^4 = 1 $ has a unique solution.
My Attempt:
Let $f(x)=x^5 + x^4 -1 $
Since $f(0)=-1$ and $f(1)=1$, by the continuity of the function $f(x)$ has at least one solution in $(0,1)$.
| Let $f(x) = x^5+x^4-1$.
Then we have $f'(x)=5x^4+4x^3=x^3(5x+4)$
There are two turning points, one is at the location when $x=-\frac45$ and the corresponding $f$ value is $-\frac{4^5}{5^5}+\frac{4^4}{5^4}-1 =\frac{-4^5+5(4^4)}{5^5}-1=\frac{4^4-5^5}{5^5}<0$ and the other is at the location when $x$ takes value $0$ with the corresponding $f$ value $-1$.
By observing the sign of the gradient, the graph increases on the interval $(-\infty, -\frac45)$ then reduces on $(-\frac45, 0)$ then increases on $(0,\infty)$. There is a local maximum at $x=-\frac45$ and a local minimum at $x=0$. Hence all $f$ values on $(-\infty, 0)$ is upper bounded by $f(-\frac45)<0$. there is no negative root.
The graph then increases to positive infinity, hence there is exactly one root which is positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding null space and image Let $T(f(x)) = g(x) = \int_{-1}^{1}f(t)(t-x)^2dt$ be a linear transformation from $V$ to $V$ where $V$ is the vector space of continuous functions on $[-1 , 1]$ . Find $\text{Nul}(T)$ and $\text{Im}(T)$ . I really have no idea about the solution . Can $x$ be considered constant in the integral ? And then how to evaluate integral ?
| As other have pointed out the image of $T$ is a subspace of the $2^{nd}$ degree polynomials
To find the image let us only look at how $T$ operates on polynomials
$$ f(t) = \sum_{k=1}^n a_kt^k$$
\begin{align*}
\int_{-1}^1 f(t) (t-x)^2 dt &= \int_{-1}^1f(t)(t^2 + x^2 -2tx) \;dt \\
&= \int_{-1}^1 \sum_{k = 1}^n a_k t^{k+2} \;dt + x^2 \int_{-1}^1 \sum_{k = 1}^n a_k t^k \; dt -2x \int_{-1}^1 \sum_{k = 1}^n a_k t^{k+1} \; dt \\
&= \sum_{k =1}^n \frac{a_k}{k+3} (1 - (-1)^{k+3}) + x^2 \sum_{k = 1}^n \frac{a_k}{k+1} (1 - (-1)^{k+1}) -2x \sum_{k = 1}^n \frac{a_k}{k+2} (1-(-1)^{k+2}) \\
&= \sum_{k =1 \\k \text{ even}}^n \frac{2a_k}{k+3} + x^2 \sum_{k = 1 \\ k \text{ even}}^n \frac{2a_k}{k+1} +x \sum_{k = 1 \\ k \text{ odd}}^n \frac{-4a_k}{k+2} \\
&= \Big( \sum_{k = 1 \\ k \text{ even}}^n \frac{2a_k}{k+1} \Big) x^2 + \Big( \sum_{k = 1 \\ k \text{ odd}}^n \frac{-4a_k}{k+2}\Big) x + \sum_{k =1 \\k \text{ even}}^n \frac{2a_k}{k+3}
\end{align*}
Notice that this is actually a polynomial in $x$. So $T(f)(x) = ax^2 +bx +c$ is actually a linear system of equations in $a_1,a_2,...a_n$
Also notice the equations are linearly independent for $n$ greater than $4$.
If $n = 4$ and the variables $a_1 = x, a_2 = y, a_3 = z, a_4 =t$ we have the following system
Let $a,b,c \in \mathbb{R}$
\begin{align*}
\frac{2}{3} y + \frac{2}{5} t &= a \\
\frac{-4}{3} x + \frac{-4}{5} z&= b \\
\frac{2}{5} y + \frac{2}{7} t &= c
\end{align*}
By solving the two equations in $y$ and $t$ first we get a unique solution for $y$ and $t$. Then we can always chose $x$ and $z$ such that the last equation is satisfied.
Therefore
$$ Im(T) = \{f(x) \in \mathbb{R}[X] : deg(f)\leq 2\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3173719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What is the easiest way to get: $2+ \sqrt{-121} = (2+ \sqrt{-1})^3$ I was reading the book Seventeen equations have changed the world.
At some point, while the book was talking about complex numbers, I see this equation:
$2+ \sqrt{-121} = (2+ \sqrt{-1})^3$
Even if it's easy to proof the truth of this equivalence (it is enough to develop the two members),
I can't find an easy/good/fast way to obtain straight the identity.
Can you help me? Does there exist a mathematical property that I'm missing?
| Not sure how you approached it before.
If you want to utlise the binomial expansion
$$
(x+y)^n = \sum_{k=0}^n\left(\matrix{n\\k}\right)x^ky^{n-k}
$$
then we have
$$
\begin{align}
(x+y)^3 &= \left(\matrix{3\\0}\right)2^3 +\left(\matrix{3\\1}\right)2^2\sqrt{-1} + \left(\matrix{3\\2}\right)2^1(\sqrt{-1})^2 + \left(\matrix{3\\3}\right)(\sqrt{-1})^3 \\
&= 2^3 + 3\cdot 2^2\sqrt{-1} + 3\cdot 2\cdot-1 + \sqrt{-1}\cdot(\sqrt{-1})^2\\
&=8+12\sqrt{-1} -6 -\sqrt{-1} = 2 +11\sqrt{-1} = 2 + \sqrt{11^2\cdot - 1} \\
&=2+\sqrt{-121}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Given $\cos 3\theta = 4(\cos\theta)^3 - 3\cos\theta$, solve $4x^3 - 9x - 1 = 0$, correct to 3 decimal places. I am trying to solve the following problem:
Given $\cos 3\theta = 4(\cos\theta)^3 - 3\cos\theta$, solve the equation $4x^3 - 9x - 1 = 0$, correct to 3 decimal places.
Assuming that $x = \cos 3\theta$, you can substitute it into the equation so that you get:
$4(\cos\theta)^3 - 9(\cos\theta) = 1$
The problem with the above is that there is a $9$ in front of the $\cos\theta$. If there was a $3$ instead, I could have replaced the LHS of the equation with $\cos 3\theta$, and then solved for $\theta$. How can I solve this equation? Any insights are appreciated.
| As you received as answers shows that
$$x=\sqrt{3} \cos \left(\frac{1}{3} \cos ^{-1}\left(\frac{1}{3 \sqrt{3}}\right)\right)$$ what you want to evaluate to a given accuracy assuming that your calculator does not have trigonoetric functions at all.
Let us write
$$x=\sqrt{3} \cos \left(\frac{1}{3} \cos ^{-1}(\epsilon )\right)\qquad \text{where} \qquad \epsilon=\frac{1}{3 \sqrt{3}}$$ and use composition of Taylor series
$$\cos ^{-1}(\epsilon )=\frac{\pi }{2}-\epsilon -\frac{\epsilon ^3}{6}-\frac{3 \epsilon
^5}{40}+O\left(\epsilon ^7\right)$$
$$\frac{1}{3} \cos ^{-1}(\epsilon )=\frac{\pi }{6}-\frac{\epsilon }{3}-\frac{\epsilon ^3}{18}-\frac{\epsilon
^5}{40}+O\left(\epsilon ^7\right)$$
$$\cos \left(\frac{1}{3} \cos ^{-1}(\epsilon )\right)=\frac{\sqrt{3}}{2}+\frac{\epsilon }{6}-\frac{\epsilon ^2}{12 \sqrt{3}}+\frac{2
\epsilon ^3}{81}-\frac{35 \epsilon ^4}{1296 \sqrt{3}}+O\left(\epsilon ^5\right)$$
$$x=\frac{3}{2}+\frac{\epsilon }{2 \sqrt{3}}-\frac{\epsilon ^2}{12}+\frac{2 \epsilon
^3}{27 \sqrt{3}}-\frac{35 \epsilon ^4}{1296}+O\left(\epsilon ^5\right)$$
Now, making $\epsilon=\frac{1}{3 \sqrt{3}}$, computing each term separately, we have
$$x=\frac{3}{2}+\frac{1}{18}-\frac{1}{324}+\frac{2}{6561}$$ and you see that the last term is already smaller than what is required (remember that we face an alternating series). So, with thiese numbers
$$x=\frac{1467001}{944784}\approx 1.55274$$ while the exact value would be ... the same.
If you forget the last term, $x=\frac{503}{324}\approx 1.55247$ is more than correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Roots of a Complex Number expression If $z_1$, $z_2$, $z_3$, $z_4$ are roots of the equation $z^4+z^3+z^2+z+1=0$,
then what is the least value of $\lfloor mod(z_1 + z_2)\rfloor + 1$?
($\lfloor.\rfloor$ denotes Greatest Integer Function)
| $$z^4+z^3+z^2+z+1=\frac{z^5-1}{z-1}=0$$
$$\therefore z^5=1\text{ and } z\ne1$$
These solutions can be plotted on an Argand diagram which more easily shows that the solution is given by the two complex conjugates with the smallest magnitude real parts - $e^{\frac{2i\pi}5}$ and $e^{\frac{-2i\pi}5}$. Adding these values gives
$$e^{\frac{2i\pi}5}+e^{-\frac{2i\pi}5}=\cos{\left({\frac{2\pi}5}\right)}+i\sin{\left({\frac{2\pi}5}\right)}+\cos{\left({\frac{2\pi}5}\right)}-i\sin{\left({\frac{2\pi}5}\right)}=2\cos{\left({\frac{2\pi}5}\right)}$$
So the smallest value is given by
$$\lfloor z_1+z_2\rfloor+1=\left\lfloor 2\cos{\left({\frac{2\pi}5}\right)}\right\rfloor+1=\lfloor 0.618...\rfloor+1=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3178747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the last three digits of $p$ if the equations $x^6 + px^3 + q = 0$ and $x^2 + 5x - 10^{2013} = 0$ have common roots.
Find the last three digits of $p$ if the equations $x^6 + px^3 + q = 0$ and
$x^2 + 5x - 10^{2013} = 0$ have common roots.
Let $a,b $ be the solutions of second equation, then by Vieta we have $a+b =-5$ and $ab=-10^{2013}$
Since $ a^6+a^3p+q=0$ and $b^6+b^3p+q=0$ we have $ a^6-b^6+p(a^3-b^3)=0$ so $ -p = a^3+b^3 = (a+b)(a^2-ab+b^2)$
So $ p = 5((a+b)^2-3ab) = 5(25+3\cdot 10^{2013})$ and thus last three digits are $125$.
| Clearly $5|x$; when $x_1+x_2=5$, that means $x_1$ is $10^u+5$ and $x_2=10^v$. Hence $x_1^3=10 k_1+125$ and $x_2^3=10^{3v}$. Now in equation $x^6+px^3+q$ we have $p=x_1^3+x_2^3$ if $x_1$ and $x_2$ are common roots. The result is that $p=10^t+125$, that is the last three digits of $p$ is $125$.Generally $p=10^t+5^n$ in equations like $x^{2n}+px^n+q=0$ if they have common roots withe equation $x^2+5x-10^{2013}=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$
Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$
I know that I should use Taylor's theorem and create power series. However I don't have idea how I can find $a_{n}$ such that $f(x)=\sum_{n=1}^{+\infty} a_{n}x^{n}$. I know only that $f(x)=\frac{x}{(x-1)(x^{2}+2)}$ but don't know how I can continue it because $f(x)=\frac{A}{(x^2+2)}+\frac{B}{(x-1)}$ and I get that $A=1, B=0, A=2B$ so it is conflict.Have you some tips how I can write this function and finish this task? Updated: Thanks to @gt6989b I know that $f(x)=\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}+\frac{1}{3} \cdot \frac{1}{1-x}=\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}+\frac{1}{3} \cdot (1+x+x^{2}+...)$ but I still have a problem with $\frac{\frac{-1}{3}x+\frac{2}{3}}{x^{2}+2}$
| By way of a contrast, here's an intuitve method that lets you see the bigger picture rather than getting bogged down in technical details,
$$f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$$
$$=\frac{1}{1-x}\times \frac{x}{2+x^2}$$
$$=\frac{1}{1-x}\times \frac{x}{2(1+(\frac{x}{\sqrt{2}})^2}$$
$$=\frac{1}{1-x}\times \frac{1}{\sqrt{2}}\times\frac{\frac{x}{\sqrt{2}}}{(1+(\frac{x}{\sqrt{2}})^2}$$
Now, standard result,
$$\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8-x^{10}+\dots$$
$$\frac{x}{1+x^2}=x-x^3+x^5-x^7+x^9-x^{11}+\dots$$
$$\frac{\frac{x}{\sqrt{2}}}{(1+(\frac{x}{\sqrt{2}})^2}=\frac{x}{\sqrt{2}}-\big(\frac{x}{\sqrt{2}}\big)^3+\big(\frac{x}{\sqrt{2}}\big)^5-\big(\frac{x}{\sqrt{2}}\big)^7+\big(\frac{x}{\sqrt{2}}\big)^9-\big(\frac{x}{\sqrt{2}}\big)^{11}+\dots$$
$$\frac{1}{\sqrt{2}}\times\frac{\frac{x}{\sqrt{2}}}{(1+(\frac{x}{\sqrt{2}})^2}=\frac{x}{2}-\frac{x^3}{4}+\frac{x^5}{8}-\frac{x^7}{16}+\dots$$
Another standard result; multiplying by $\frac{1}{1-x}$ forms the partial sums,
$$\frac{x}{x^{3}-x^{2}+2x-2}=\frac{x}{2}+\frac{x^3}{4}+\frac{3x^5}{8}+\frac{5x^7}{16}+\frac{11x^9}{32}+\dots$$
and I'll leave it to the reader to fight through to the end of the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3184465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\tan(\sin(x))=x-\frac{x^{3}}{6}+o(x^{3})$
Prove that $\tan(\sin(x))=x-\frac{x^{3}}{6}+o(x^{3})$
I know that when I calculate derivative then I get true answer. However I want to know why the way that I will present soon does not work. My try:$$\tan(x)=x+r_{1}(x), \quad r_{1}(x)=o(x)$$ $$\tan(\sin(x))=\sin(x)+r_{1}(\sin x),\quad r_{1}(\sin x)=o(\sin x)$$ $$\tan(\sin(x))=x-\frac{x^{3}}{6}+r_{2}(x)+r_{1}(x-\frac{x^{3}}{6}+r_{2}(x)),\quad r_{2}(x)=o(x^{3})$$ In this moment I should find: $$o(?)=r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))$$I suspect that $o(?)=o(x^{3})$ so I write: $$\frac{r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))}{(x-\frac{x^{3}}{6}+r_{2}(x))}\cdot \frac{(x-\frac{x^{3}}{6}+r_{2}(x))}{x^{3}}$$ But in this moment I have $0 \cdot (+\infty)$. Moreover I see that $$r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))=o(x)$$ because $$\frac{r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))}{(x-\frac{x^{3}}{6}+r_{2}(x))}\cdot \frac{(x-\frac{x^{3}}{6}+r_{2}(x))}{x} \rightarrow 0$$Can someone tell me where I'm making a mistake, that I'm getting $ o (x) $ instead of $ o (x ^ {3}) $?
| Your algebraic manipulation is correct, but your desired result is wrong (as mentioned in another answer). The problem with your approach is that although it is correct it does not lead you to anything useful. The equation $$\tan(\sin x) =x-\frac{x^3}{6}+o(x^3)+o(\sin x) \tag{1}$$ although correct is much less useful compared to the equation $$\tan(\sin x) =x+\frac{x^3}{6}+o(x^3)\tag{2}$$ precisely because equation $(1)$ contains two errors terms of different orders. The typical mechanism in these scenarios is use to expansions upto desired order and then perform algebraic manipulation.
So here it goes
\begin{align*}
\tan\sin x &= \sin x+\frac{\sin^3x}{3}+o(\sin^3x)\\
&=x-\frac{x^3}{6}+o(x^3)+\frac{\sin^3x}{3}+o(x^3)\tag{3}\\
&=x-\frac{x^3}{6}+\frac{1}{3}\left(x-\frac{x^3}{6} +o(x^3)\right)^3+o(x^3)\\
&=x+\frac{x^3}{6}+o(x^3)
\end{align*}
You should observe the equality $o(x^3)=o(\sin^3x)$ used in $(3)$ which comes as a result of $\sin x=x+o(x) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What causes the extraneous root when intersecting parabola $y^2=4ax$ with circle $x^2+y^2=9a^2/4$? If I solve the parabola: $y^2=4ax$ and the circle: $x^2+y^2=\frac{9a^2}{4}$ I get a quadratic in $x$; ie:
$$4x^2+16ax-9a^2=0$$
which has roots $x=\frac{a}{2},-\frac{9a}{2}$. But if we see the graphs of these two, both the intersections occur at same $x$-coordinate so why does this extra root $x=-\frac{9a}{2}$ come?
So my question is that if the circle was unknown say $x^2+y^2=b^2$ then I would have put the discriminant of the quadratic in $x$ as $0$ but that would be wrong so how do I judge that an extraneous root will be present and discriminant won't be zero?
| The full solution is this $$\begin{cases}x^2+y^2=\frac94a^2\\ y^2=4ax\end{cases}\iff \begin{cases}x^2+4ax-\frac94a^2=0\\ y^2=4ax\end{cases}\iff\begin{cases}x=\frac a2\\ y^2=2a^2\end{cases}\lor\begin{cases}x=-\frac 92a\\ y^2=-18a^2\end{cases}$$
Now the system must be "discussed" in terms of the parameter, in the sense that it is equivalent to the following: $$\begin{cases}x=\frac a2\\ y^2-2a^2=0\end{cases}\lor\begin{cases}x=-\frac 92a\\ y^2+18a^2=0\end{cases}\iff\begin{cases}x=\frac a2\\ y=\sqrt2a\end{cases}\lor\begin{cases}x=\frac a2\\ y=-\sqrt2a\end{cases}\lor\begin{cases}x=0\\ y=0\\ a= 0\end{cases}$$
And you decide whether or not to accept the degenerate case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Minimizing lengths of cevians in an isosceles right triangle Consider isosceles right triangle $ABC$ with $BC$ as the hypotenuse and $AB=AC=6$. $E$ is on $BC$ and $F$ is on $AB$ such that $AE+EF+FC$ is minimized. Compute $EF$.
My thought process:
I reflected triangle $ABC$ across $BC$ to get a square $ABA'C$. Then, I messed around with the placement of $F$ on $AB$, to see what results I could produce, with $E$ always being the midpoint. May someone help me on this?
|
Let $|AB|=|AC|=6=b$.
It is known that the minimal path would be
the path of the light ray from the point $C$
to point $A$, reflected at points $F$ and $E$.
Consider points $C_1,A_1$ and $E_1$
as reflection of the points $C,A$ and $E$
with respect to $AB$.
Let
$\angle A_1CC_1=\angle CFG=\angle GFE=\phi$,
$\angle FEH=\angle HEA=\angle AE_1H=\angle HE_1C=\theta$.
Then
\begin{align}
|EF|&=|E_1F|,\quad |EA|=|E_1A|=|E_1A_1|
,\\
|AE|+|EF|+|FC|&=
|A_1E_1|+|E_1F|+|FC|
\\
&=|CA_1|
=\sqrt{(2b)^2+b^2}=\sqrt5\,b
.
\end{align}
\begin{align}
\triangle BEF:\quad
\angle BEF=90^\circ-\theta
,\quad
\angle EFB=90^\circ-\phi
\quad\Rightarrow\quad
\theta=45^\circ-\phi
,
\end{align}
\begin{align}
\triangle A_1CC_1:\quad
\phi&=\arctan\tfrac12
=\arcsin\tfrac{\sqrt5}5
=\arccos\tfrac{2\sqrt5}5
,\\
\theta&=45^\circ-\arcsin\tfrac{\sqrt5}5
=45^\circ-\arccos\tfrac{2\sqrt5}5
.
\end{align}
\begin{align}
\triangle BEF:\quad
\frac{|EF|}{\sin45^\circ}
&=
\frac{|BF|}{\sin(90^\circ-\theta)}
=
\frac{\tfrac12|AB|}{\cos\theta}
,\\
|EF|&=\frac{b\sqrt2}{4\cos(45^\circ-\arctan\tfrac12)}
\\
&=
\tfrac{\sqrt2}4\,
\frac{b}{
\cos 45^\circ \cos(\arccos\tfrac{2\sqrt5}5)
+
\sin 45^\circ \sin(\arcsin\tfrac{\sqrt5}5)
}
\\
&=
\tfrac{\sqrt2}4\,
\frac{b}{
\tfrac{\sqrt2}2
\tfrac{2\sqrt5}5
+
\tfrac{\sqrt2}2
\tfrac{\sqrt5}5
}=b\,\tfrac{\sqrt5}6
,
\end{align}
and since $b=6$,
the answer is $|EF|=\sqrt5$.
| {
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"url": "https://math.stackexchange.com/questions/3189411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\lim_{n\to\infty}\left(\frac{\sin(2\sqrt 1)}{n\sqrt 1\cos\sqrt 1} +\cdots+\frac{\sin(2\sqrt n)}{n\sqrt n\cos\sqrt n}\right)$ Using the trigonometric identity of $\sin 2\alpha = 2\sin \alpha \cos \alpha$, I rewrote the expression to:
$$\lim_{n\to\infty}\left(\frac{\sin(2\sqrt 1)}{n\sqrt 1\cos\sqrt 1} + \cdots+\frac{\sin(2\sqrt n)}{n\sqrt n\cos\sqrt n}\right) = \lim_{n\to\infty}\left(\frac{2\sin(\sqrt 1)}{n\sqrt 1} +\cdots+\frac{2\sin(\sqrt n)}{n\sqrt n}\right)$$
I then tried using the squeeze theorem to get the limit, but I can't get an upper sequence that converges to zero, and the best I could get is a sequence that converges to two:
$$\frac{2\sin(\sqrt 1)}{n\sqrt 1} + \cdots+\frac{2\sin(\sqrt n)}{n\sqrt n} \leq \frac{2}{n\sqrt 1} + \cdots+\frac{2}{n\sqrt n} \leq \frac{2}{n}\cdot n \longrightarrow 2$$
What am I missing?
| The inequality
$$\sum_{k=1}^{n}{\frac{1}{\sqrt k}} \le 1+\int_{1}^{n}{\frac{1}{\sqrt x}}\,\mathrm{d}x \leq 2\sqrt{n},$$
yields
$$\left|\frac{2\sin(\sqrt 1)}{n\sqrt 1} + ...+\frac{2\sin(\sqrt n)}{n\sqrt n}\right| \leq 4\frac{1}{\sqrt n} \to 0.$$
Note that $\sin(\cdot)$ can change signs, so absolute value is needed in the estimate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Complex number cannot arive at $\frac{9}{2}-\frac{9}{2}i$ with problem $\frac{4+i}{i}+\frac{3-4i}{1-i}$ I am asked to evaluate: $\frac{4+i}{i}+\frac{3-4i}{1-i}$
The provided solution is: $\frac{9}{2}-\frac{9}{2}i$
I arrived at a divide by zero error which must be incorrect. My working:
$\frac{4+i}{i}$, complex conjugate is $-i$ so:
$\frac{-i(4+i)}{-i*i}$
= $\frac{-4i+i^2}{i^2}$
= $\frac{-4i--1}{-1}$
= $-4i+1$
Then the next part:
$\frac{3-4i}{1-i}$ complex conjugate is $1+i$ so:
$\frac{(1+i)(3-4i)}{(1+i)(1-i)}$
= $\frac{3-4i+3i-4i^2}{1-i^2}$
= $\frac{7-i}{0}$ # 1 + -1 = 0
How can I arrive at $\frac{9}{2}-\frac{9}{2}i$?
| You have made a mistake in the second last step of your simplification.
$$1 - i^2 = 1 - (-1)=1 + 1 = 2$$
Then,
$$-4i + 1 + \frac{7 - i}{2}
= 1 + \frac{7}{2} + i(-4 - \frac{1}{2})
= \frac{9}{2} - \frac{9}{2}i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How can I find the P matrix in $P^{-1}AP $= D? I have the following exercise:
Let a matrix A = $\begin{pmatrix}
4 & 0 & 1\\
2 & 2 & 1\\
0 & 4 & 1
\end{pmatrix}$
1) Determine its eigenvalues and their multiplicity.
2) Give a basis of the eigen spaces associated with each of the
distinct eigenvalues of A.
3) find P and D such that $P^{-1}AP$ = D is diagonal.
I've done 1) and 2) and found the basis
B ={ $\begin{pmatrix}
-1 \\
-1 \\
4
\end{pmatrix}$ , $\begin{pmatrix}
-2 \\
1 \\
4
\end{pmatrix}$ , $\begin{pmatrix}
-1 \\
1 \\
1
\end{pmatrix}$ }
Now, in my notes they say that the matrix P is constituted of the vectors of the basis found.
So, P = $\begin{pmatrix}
-1 & -2 & -1\\
-1 & 1 & 1\\
4 & 4 & 1
\end{pmatrix}$
Also, because we have a 3x3 matrix and we've found three distinct eigen values, the matrix D exists and as such we put the eigen values on the diagonal line of the matrix. ($E_(\lambda_1) = 0$, $E_(\lambda_2) = 2$, $E_(\lambda_3) = 5$ )
D = $\begin{pmatrix}
0 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 5
\end{pmatrix}$
Finally to verify that this is all true I compute $AP = PD$. However I didn't find the same result on the left side and the right side of the equation. Where did I go wrong?
$AP \ne PD$
$\begin{pmatrix}
0 & -4 &-3\\
0 & 2 & 1 \\
0 & 8 & 5 \\
\end{pmatrix} \ne \begin{pmatrix}
0 & -4 &-5 \\
0 & 2 & 5 \\
0 & 8 & 5 \\
\end{pmatrix} $
| A quick look at your calculations, I think your last eigen vector is wrong. The last eigen vector is (1, 1, 1) and not (-1, 1, 1). I think that would solve your problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a remainder when $(x^5+1)^{100} + (x^5-1)^{100}$ is divided by $x^4+x^2+1$ The question is:
Find a remainder when $f(x)=(x^5+1)^{100} + (x^5-1)^{100}$ is divided by $x^4+x^2+1$
I first began by decomposing $$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$$
and using $$x^3-1=(x-1)(x^2+x+1)\\x^3+1=(x+1)(x^2-x+1)$$
and could get remainders when divided by each factors.
\begin{align}
(x^5+1)^{100}&\equiv(x^2+1)^{100}&(\mathrm{mod}\,x^3-1)\\
&\equiv(-x)^{100}&(\mathrm{mod}\,x^2+x+1)\\
&=x^{100}=x\cdot(x^3)^{33}\\
&\equiv x&(\mathrm{mod}\,x^3-1)\\\\
(x^5-1)^{100}&\equiv(x^2-1)^{100}&(\mathrm{mod}\,x^3-1)\\
&=(x+1)^{100}(x-1)^{100}\\
&=(x^2+x+1+x)^{50}\cdot(x^2+x+1-3x)^{50}\\
&\equiv x^{50}\cdot(-3x)^{50}&(\mathrm{mod}\,x^2+x+1)\\
&=3^{50}(x^2)^{50}=3^{50}x\cdot(x^3)^{33}\\
&\equiv 3^{50}x&(\mathrm{mod}\,x^3-1)\\
\end{align}
Therefore
$$f(x)\equiv (3^{50}+1)x\quad(\mathrm{mod}\,x^2+x+1)\\$$
Similarly,
$$f(x)\equiv -(3^{50}+1)x\quad(\mathrm{mod}\,x^2-x+1)\\$$
So
\begin{align}
f(x) &= (x²+x+1)p(x) + (3^{50}+1)x\\
f(x) &= (x²-x+1)q(x) - (3^{50}+1)x
\end{align}
As $f(x)$ is an even function, I could derive that $$q(x)=p(-x)$$
And then, my plan was to find a remainder when $p(x)$ is devided by $x²-x+1$ and so on
\begin{align}
p(x)&=(x²-x+1)A(x)+Cx+D\\
q(x)&=(x²+x+1)B(x)+Ex+F\\
\end{align}
So that I could conclude that $$f(x)\equiv (x^2+x+1)(Cx+D)+(3^{50}+1)x\quad(\mathrm{mod}\,x^4+x^2+1)\\$$
Using $p(x)=q(-x)$, I could derive that $A(x)=B(-x)$, $C=-E$, and $D=F$. And further using $f(x)=f(-x)$, I could do
\begin{align}
p(x)&=e(x)+o(x)\quad(e(x)=e(-x)\,,o(x)=-o(-x))\\
\rightarrow&(x²+x+1)(e(x)+o(x)) + (3^{50}+1)x = (x²-x+1)(e(x)-o(x)) - (3^{50}+1)x\\
\rightarrow&2xe(x)+(x²+1)o(x) + 2(3^{50}+1)x = 0\\
p(x)&=\frac{(-x²+2x-1)o(x)}{2x} - (3^{50}+1) = \frac{(-x²+x-1 + x)o(x)}{2x} - (3^{50}+1)\\
&= \frac{(-x²+x-1)o(x)}{2x} + \frac{o(x)}2 - (3^{50}+1)
\end{align}
And that was it. I couldn't proceed further.
|
Therefore, $ $ with $\,n = 3^{50}\!+1$
$$\begin{align} &f(x)\,\equiv\ \ \ n\,\color{#c00}x\quad(\mathrm{mod}\ \ x^2\!+\!x\!+\!1)\\[.2em]
&f(x)\,\equiv -n\,\color{#0a0}x\quad(\mathrm{mod}\ \ x^2\!-\!x\!+\!1)\end{align}\qquad\qquad$$
Hint: make the following substitutions $$\begin{align}\color{#c00}x &\equiv -(x^2\!+\!1)\!\pmod{x^2\!+\!x\!+\!1}\\ \color{#0a0}x&\equiv \ \ \ \ \ x^2\!+\!1\,\pmod{x^2\!-\!x\!+\!1}
\end{align}$$
which yields that $\, f\equiv -n(x^2\!+\!1)\,$ mod both, so also mod their lcm = product. QED $ $ Good work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Show that $c^2 + a^2d=abc$ for a monic quartic polynomial I apologize in advance for asking a homework question, but I have genuinely no idea on how to approach part b). The question is as follows:
Consider the polynomial equation $\rm{P}(x)=x^4 + ax^3 + bx^2 + cx+d$, with a root $ki$. In addition, $a,b,c,d\in Z$ and $k\in R,\; k\neq 0$
(a) Show that $c=k^2a$
(b) Show that $c^2 + a^2d=abc$
Part (a) is quite trivial and can be done by knowing that $ki$ and $\bar{ki}$ are roots since all the coefficients are real. However I am unsure of how to do part (b), do I square the expression in part (a). I tried working backwards from the solution to a true statement here is what I had:
$$
\begin{align}
c^2 + a^2d &=abc \\
k^4a^2+a^2d &=abc \\
a^2\left(k^4 +d\right) &=abc \\
a\left(k^4 +d\right) &=bc \tag{assuming $a\neq 0$, though this may not be true} \\
\dots &\dots \\
k^4 &=\frac{bc-ad}{a}
\end{align}
$$
This gets me nowhere
| We know that $ki$ is a solution to $P(x)=0$ so in particular,
$$\begin{split}
(ki)^4+a(ki)^3+b(ki)^2+c(ki)+d&=0 \\
\Leftrightarrow k^4-ak^3i-bk^2+cki+d&=0 \\
\Leftrightarrow k^4-bk^2+d+ki(c-ak^2)&=0
\end{split}$$
Using part (a), this gives $k^4-bk^2+d=0$. Now add $bk^2$ on both sides and multiply through with $a^2$ to get $$a^2k^4+a^2d=a^2bk^2=abak^2.$$ Due to part (a) we can substitute $c^2$ for $a^2k^4$ as well as $c$ for $ak^2$ and obtain
$$c^2+a^2d=abc.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Given three non-negatve numbers $a,b,c$. Prove that $1+a^{2}+b^{2}+c^{2}+4abc\geqq a+b+c+ab+bc+ca$ .
Given three non-negatve numbers $a, b, c$. Prove that:
$$1+ a^{2}+ b^{2}+ c^{2}+ 4abc\geqq a+ b+ c+ ab+ bc+ ca$$
Let $t= a+ b+ c$, we have to prove
$$\left(\!\frac{1}{t^{3}}- \frac{1}{t^{2}}+ \frac{1}{t}\!\right)\sum a^{3}+ \left(\!\frac{3}{t^{3}}- \frac{3}{t^{2}}\!\right)\left (\!\sum a^{2}b+ \sum a^{2}c\!\right )+ \left(\!\frac{6}{t^{3}}- \frac{6}{t^{2}}- \frac{3}{t}+ 4\!\right)abc\geqq 0$$
If $0< t< 1$ so
$${\rm LHS}\geqq \left(\frac{3}{t}+ 1\right)\left(\frac{3}{t}- 2\right)^{2}abc\geqq 0$$
If $1< t$ so
$${\rm LHS}= \left(\!\frac{3}{t^{2}}- \frac{3}{t^{3}}\!\right)(\!{\rm Schur.3}\!)+ \frac{1}{t}\left(\!\frac{2}{t}- 1\!\right)^{2}(\!{\rm a.m.}- {\rm g.m.}\!)+ \left(\!\frac{3}{t}+ 1\!\right)+ \left(\!\frac{3}{t}- 2\!\right)^{2}abc\geqq 0$$
(Can you find the way without deviding two cases as above?)
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$1+9u^2-6v^2+4w^3\geq3u+3v^2,$$ which is a linear inequality of $w^3$.
Id est, it's enough to prove our inequality for the extreme value of $w^3$, which happens in the following cases.
*
*$w^3=0$.
Let $c=0$.
Thus, we need to prove that
$$1+a^2+b^2\geq a+b+ab$$ or
$$(a-1)^2+(b-1)^2+(a-b)^2\geq0,$$ which is obvious;
*Two variables are equal.
Let $b=a$.
Thus, we need to prove that:
$$(1+4c)a^2-2(1+c)a+c^2-c+1\geq0$$ for which it's enough to prove that
$$(1+c)^2-(1+4c)(c^2-c+1)\leq0$$ or
$$c(2c-1)^2\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving polynominals equations (relationship of roots)
The roots of $x^3-4x^2+x+6$ are $\alpha$, $\beta$, and $\omega$.
Find (evaluate):
$$\frac{\alpha+\beta}{\omega}+\frac{\alpha+\omega}{\beta}+\frac{\beta+\omega}{\alpha}$$
So far I have found:
$$\alpha+\beta+\omega=\frac{-b}{a} = 4 \\
\alpha\beta+\beta\omega+\alpha\omega=\frac{c}{a} = 1 \\
\alpha×\beta×\omega=\frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$\frac{\alpha^2\beta+\alpha\beta^2+\alpha^2\omega+\alpha\omega^2+\beta^2\omega+\beta\omega^2}{\alpha\beta\omega}$$
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-\dfrac{11}{3}$
| $$\frac{\alpha + \beta}{\omega} + \frac{\beta + \omega}{\alpha} + \frac{\alpha + \omega}{\beta}$$
$$= \frac{\alpha + \beta + \omega - \omega}{\omega} + \frac{\beta + \omega + \alpha - \alpha}{\alpha} + \frac{\alpha + \omega + \beta - \beta}{\beta}$$
$$ = (\alpha + \beta + \omega) \left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\omega}\right) - 3$$
$$ = (\alpha + \beta + \omega) \left(\frac{\beta\omega}{\alpha\beta\omega} + \frac{\alpha\omega}{\alpha\beta\omega} + \frac{\alpha\beta}{\alpha\beta\omega}\right) - 3$$
$$ = \frac{\alpha + \beta + \omega}{\alpha\beta\omega}(\beta\omega + \alpha\omega + \alpha\beta) - 3$$
I think you should be able to take it from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Given $(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$ Find $\frac{a_7}{a_{13}}$ Given $$(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$$ Find Value of $\frac{a_7}{a_{13}}$
My try:
I assumed $A=2x^2$,$B=3x$ and $C=4$
Then we have the following cases to collect coefficient of $x^7$:
Case $1.$ $A^3 \times B^1 \times C^6$
Case $2.$ $A^2 \times B^3 \times C^5$
Case $3.$ $A^1 \times B^5 \times C^4$
case $4.$ $A^0 \times B^7 \times C^3$
Using multinomial theorem we get Coefficient of $x^7$ as:
$$a_7=10! \times \left(\frac{2^33^14^6}{3!6!}+\frac{2^13^54^4}{1!5!4!}+\frac{2^23^34^5}{2!3!5!}+\frac{2^03^74^3}{3!7!}\right)$$
Like wise we need to find $a_{13}$
But is there any better way?
| This approach is from the idea of Pascal's triangle. We write
$$
(2x^2+3x+4)^n = \sum_r a_{n,r} x^r
$$
Then
$$
(2x^2+3x+4)^{n+1}=(\sum_r a_{n,r} x^r )(2x^2+3x+4) = \sum_r (2a_{n,r-2}+3a_{n,r-1}+4a_{n,r})x^r.$$
Thus, by comparing coefficients, we obtain a recurrence relation
$$
a_{n+1,r}=2a_{n,r-2}+3a_{n,r-1}+4a_{n,r}
$$
with an initial condition $a_{0,0}=1$, $a_{1,2}=2$, $a_{1,1}=3$, and $a_{1,0}=4$.
I am not sure if this is a better way. But, it will be more convenient than computing factorials many times. Especially for small numbers, this method will give answers quickly.
However, for large numbers, the number of recursion steps will increase enormously with $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$
If $$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}= \frac{3\sqrt{3}}{4}$$ for $a,b,c,d>0$
Then
Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$ is
Try: using A.M G.M Inequality
$$1+a^2\geq 2a\;, 1+b^2\geq 2b\;,1+c^2\geq 2c\; 1+d^2\geq 2d$$
$$\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}\geq 4\sqrt{abcd}$$
$$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}\leq \frac{a+b+c+d}{4\sqrt{abcd}}$$
I have edited it, please have a look
Could some help me to solve it , Thanks
| With thanks to Michael for pointing this beautiful topic.
For proving
$$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2,$$
I have used the following result discovered by me:
$$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2+$$
$$+9\sum_{1\leq i<j\leq4}{(xy-1)^2}+3\sum_{1\leq i<j<k\leq4}{(xyz-1)^2}+(xyzt-1)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Minimizing $\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2$ While solving a problem I came across this task, minimizing
\begin{align}
\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2.
\end{align}
One can easily do it with calculus to show that the minimum value is $12.5$.
I tried to do it using trigonometric identities and fundamental inequalities (like AM-GM, Cauchy-Schwarz, etc.) but failed. Can someone help me to do it using trig identities and inequalities?
| Knowing the answer it's not that difficult to get it another way. Let $y = \cos 2x$. We have $\sin^2 x = \frac{1-y}{2}$, $\cos^2 x =\frac{1+y}{2}$. Then
\begin{align} \left(\sin^2 x + \frac{1}{\sin^2 x}\right)^2 + \left(\cos^2 x + \frac{1}{\cos^2 x}\right)^2 &= \left(\frac{1-y}{2} + \frac{2}{1-y}\right)^2 + \left(\frac{1+y}{2} + \frac{2}{1+y}\right)^2 = \\
&= \frac{y^6+7y^4-y^2+25}{2(1-y^2)^2} = \\
&= \frac{25}{2} + \frac{y^2(y^4-18y^2+49)}{2(1-y^2)^2} = \\
&= \frac{25}{2} + y^2\Big(\frac{1}{2} + \frac{8}{1-y^2} + \frac{16}{(1-y^2)^2}\Big) \end{align}
Since $y^2 \le 1$, the expression in the brackets is strictly positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find the limit without using Lhopital this question came out on my analysis exam: Evaluate
$$\lim_{x\to 0}\left(\frac{5^{x^2}+7^{x^2}}{5^x+7^x}\right)^{\frac{1}{x}} $$
I did it using L'hopital rule but is there another way to do this?
| We call your limit A.
By using the exp, you get :
$$A=\lim_{x \rightarrow 0} \exp(\frac{1}{x}\ln\frac{e^{x^2\ln{5}}+e^{x^2\ln{7}}}{e^{x\ln{5}}+e^{x\ln{7}}})$$
Which gives you the following limit, as PierreCarre mentionned.
$$\lim_{x \rightarrow 0} \frac{1}{5}\Big(\frac{1+a^{x^2}}{1+a^x}\Big)^{\frac{1}{x}}$$
$$=\frac{1}{5}\lim_{x \rightarrow 0 } \exp\Big(\frac{1}{x}\ln\big(\frac{1+a^{x^2}}{1+a^x}\big)\Big)$$
By using this little tip :
$$\frac{1+a^{x^2}}{1+a^x}=\frac{1+a^{x^2}-a^x+a^x}{1+a^x}=1+\frac{a^{x^2}-a^x}{1+a^x}$$
You get :
$$\ln\Big(\frac{1+a^{x^2}}{1+a^x}\Big)\sim_0 \Big( \frac{a^{x^2}-a^x}{1+a^x}\Big) $$
As $\frac{a^{x^2}-a^x}{1+a^x}$ tends to $0$.
Then,
$$A=\frac{1}{5}\lim_{x \rightarrow 0}\exp\Big(\frac{1}{x}.\frac{a^{x^2}-a^x}{1+a^x}\Big)$$
By using $a^{x^2}=e^{x^2\ln a}=1+x^2 \ln{a} +o(x^2\ln{a})$ and $a^{x}=e^{x\ln a}=1+x \ln{a} +o(x\ln{a})$
$$\frac{a^{x^2}-a^x}{x+xa^x}=\frac{1+x^2 \ln{a} +\ln a.o(x^2)-(1+x \ln{a} +\ln a.o(x))}{x(1+1+x \ln{a} +\ln a.o(x))}=B(x)$$
$$B(x)=\frac{x \ln{a} +\ln a.o(x)-(\ln{a} +\ln a.o(1))}{(1+1+x \ln{a} +\ln a.o(x))}$$
And $$\lim_{x \rightarrow 0 }B(x)=-\frac{\ln a}{2}$$
So we get :
$$A=\frac{1}{5}\exp{\frac{-\ln a}{2}}\qquad a =\frac{7}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
Here's my working:
Start with $$4+\frac{1}{x}-\frac{1}{x^2}=0$$
Rearranging into standard form:
$$-\frac{1}{x^2}+\frac{1}{x}+4=0$$
Multiply by $-1$ to get a positive leading coefficient $a$:
$$\frac{1}{x^2}-\frac{1}{x}-4=0$$
I'm not sure how to determine my inputs $a,b$ and $c$ with these fractions but I guess $a=\dfrac{1}{x^2}$, $b=\dfrac{1}{x}$ and $c=-4$.
Plugging into quadratic function:
$$x = \frac{-\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
I find this challenging due to the coefficients $a$ and $b$ being fractions.
How can I apply the quadratic formula to $4+\dfrac{1}{x}-\dfrac{1}{x^2}=0$ to arrive at $\dfrac{-1\pm\sqrt{17}}{8}$?
| Note that $$4+\frac1x-\frac1{x^2}=4+\frac1x-\left(\frac1x\right)^2,$$ so making the substitution $y=\frac1x,$ we obtain the quadratic $$-y^2+y+4=0,$$ which should be more familiar. Solve for $y,$ and since neither solution for $y$ should be equal to $0,$ use $x=\frac1y$ to solve for $x.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$? I need to remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$, and find out what $f(7)$ equals. I am not sure what I've done wrong, but I'm getting 33, which the website I'm using tells me is wrong.
Please help me figure out what I'm doing wrong here.
My Steps:
$$\frac{x^2-11x+28}{x-7} \cdot \frac{x+7}{x+7}$$
$$=\frac{x^3+7x^2-11x^2-77x+28x+196}{x^2-49}$$
$$=\frac{x^3-4x^2-49x+196}{x^2-49}$$
$$=\frac{x^2(x-4)-49(x+4)}{x^2-49}$$
$$=(x-4)(x+4)$$
plug in $7$ for $x$.
$$(7-4)(7+4)=7^2-16=33$$
| You probably want to remove it at $7$ and not at $f(7)$. Write $$ f(x) = {(x-7)(x-4)\over x-7 } = x-4$$
So $$\lim_{x\to 7} f(x) =3$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Limit of $n(n\pi + \frac{\pi}{2} - x_n)$, where $x_n$ is the solution of $x = \tan(x)$ in the interval $(n \pi, n\pi + \frac{\pi}{2})$ Let $x_n$ denote the solution to $x = \tan(x)$ in the interval $(n \pi,n \pi +\frac{\pi}{2}).$ Find
$$ \lim_{n \to \infty}n \left(n\pi + \frac{\pi}{2} - x_n\right)$$
| $x_n\in(n\pi,n\pi+\frac{\pi}{2})\implies n\pi+\frac{\pi}{2}-x_n\in(0,\frac{\pi}{2})$ and $\frac{1}{n\pi+\frac{\pi}{2}}<\frac{1}{x_n}<\frac{1}{n\pi}$.
\begin{align*}
\tan\left(n\pi+\frac{\pi}{2}-x_n\right)
&=\tan\left(\frac{\pi}{2} - x_n\right)\\
&=\frac{1}{\tan x_n}\\
&=\frac{1}{x_n}.
\end{align*}
So
$$n\pi+\frac{\pi}{2}-x_n=\arctan\frac{1}{x_n}\sim\frac{1}{x_n}\quad (n\to\infty).$$
$$\lim_{n\to\infty}n\left(n\pi+\frac{\pi}{2}-x_n\right)
=\lim_{n\to\infty}n\arctan\frac{1}{x_n}
=\lim_{n\to\infty}\frac{n}{x_n}=\frac{1}{\pi}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve this D.E $y''(\frac{x}{2})+y'(\frac{x}{2})+y(x)=x$ I know how to slove $y''(x)+y'(x)+y(x)=x$
But I couldn't solve this
$$y''(\frac{x}{2})+y'(\frac{x}{2})+y(x)=x$$
any hint to help me?
Thanls
| Here is an incomplete attempt that I might be able to rectify later.
Write the equation as:
$$y''(x)+y'(x)+y(2x) = 2x$$ and let $y(x) = f(x)+x-1.$ Then we get an equation for $f:$
$$f''(x)+f'(x)+f(2x) = 0.$$
Now assume $x\neq 0$ and By Taylor's theorem:
$$f(2x) = f(x)+xf'(x)+\frac{x^2}{2}f''(x)+x^2h(2x)$$
with $h(t)\to 0$ as $t\to 0.$ Substituting this in the differential equation:
$$\dfrac{f''(x)}{x}+\dfrac{2+2x}{x(2+x^2)}f'(x)+\dfrac{2}{x(2+x^2)}f(x) = -h(2x)\cdot\dfrac{2x}{2+x^2}.$$
Now again by Taylor's, we have $f(x) = f(0)+xf'(0)+o(x^2)$ and $f'(x) = f'(0)+xf''(0)+o(x^2).$ Plug them in and we get:
$$\dfrac{f''(x)}{x}+\dfrac{2+2x}{2+x^2}\left(\dfrac{f'(0)}{x}+f''(0)+o(x)\right)+\dfrac{2}{2+x^2}\left(\dfrac{f(0)}{x}+f'(0)+o(x)\right) = -h(2x)\cdot\dfrac{2x}{2+x^2}.$$
Now, we take the limit $x\to 0$ from both sides and that leaves us with:
$$\lim\limits_{x\to 0}\dfrac{1}{x}\left(f''(x)+\dfrac{2+2x}{2+x^2}f'(0)+ \dfrac{2}{2+x^2}f(0)\right) = 0.$$
The plan of attack is to prove $f''(x) = 0$ and this leaves us only linear functions as possible answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Indefinite integral through factorization So I have
$$\int \frac{1}{(x^2+1)^2}dx$$
And the professor does some magic
I'm confused. what's with the derivative?
I solved the integral via substitution but I'm curious how this works, so I can learn it
thanks!
| It is possible to shorten this magic as follows
\begin{eqnarray*} \int \frac{1}{(x^2+1)^2}dx
& = & \int \frac{1+x^2 - x^2}{(x^2+1)^2}dx \\
& = & \int \frac{1}{x^2+1}dx - \frac{1}{2}\int x\frac{2x}{(x^2+1)^2}dx \\
& = & \arctan x - \frac{1}{2}\left(- \frac{x}{x^2+1} + \int \frac{1}{x^2+1}dx\right) \\
& = & \arctan x + \frac{1}{2}\frac{x}{x^2+1} - \frac{1}{2}\arctan x \;(+C) \\
& = & \frac{1}{2}\left(\arctan x + \frac{x}{x^2+1}\right) \;(+C) \\
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3218154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Number of possible integer values of $x$ for which $\frac{x^3+2x^2+9}{x^2+4x+5}$ is integer How many integer numbers, $x$, verify that the following
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5}
\end{equation*}
is an integer?
I managed to do:
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5} = x-2 + \frac{3x+19}{x^2+4x+5}
\end{equation*}
but I cannot go forward.
| Here is another method that works well when Diophantine quadratics $ax^2+bx+c=0$ are involved, it consists in saying that since a solution should exists then $\exists \delta\in\mathbb Z\mid b^2-4ac=\delta^2$.
In this problem we'd like $\quad\dfrac{3x+19}{x^2+4x+5}=n\quad$ to be an integer.
Thus applying the method to $$nx^2+(4n-3)x+(5n-19)=0$$
gives $\quad(4n-3)^2-4n(5n-19)=p^2\iff 4n^2-52n+(p^2-9)=0$
Which by itself should have solutions in $n$ : $\quad(-52)^2-4(4)(p^2-9)=q^2\iff 2848-16p^2=q^2$
We can divide by $16$ (setting $q=4r$) to get $$p^2+r^2=178$$
This has solutions $(\pm 3,\pm 13)$.
*
*case 1 : $p^2=3^2=9$
Leads to $4n^2-52n=0\iff 4n(n-13)=0\iff n=0,\ 13$
$n=0\implies 3x+19=0$ impossible
$n=13\implies 13x^2+49x+46=(x+2)(13x+23)=0$ and we verify that $x=-2$ is solution.
*
*case 2: $p^2=13^2=169$
Leads to $4n^2-52n+160=4(n-5)(n-8)=0\iff n=5,\ 8$
$n=5\implies 5x^2+17x+6=(x+3)(5x+2)=0$ and we verify that $x=-3$ is solution.
$n=8\implies 8x^2+29x+21=(x+1)(8x+21)=0$ and we verify that $x=-1$ is solution.
Finally we have found all solutions $x\in\{-3,-2,-1\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Count the number of permutations of certain cycles type Suppose we have $n$ elements, assume there is a permutations over $k$ elements among the $n$ elements so $n-k$ are fixed. Let that the permutation over the k elements is represented by permutation cycles so the length of all permutation cycles $=k$.
As an example: Suppose we have the following permutation
$$ x = \left( {\begin{array}{c}
x_1 = \left( {\begin{array}{c}
1 \\
2 \\
\end{array} } \right) \\
x_2 = \left( {\begin{array}{c}
3 \\
4 \\
5 \\
\end{array} } \right) \\
x_3 = \left( {\begin{array}{c}
6 \\
7 \\
\end{array} } \right) \\
8 \\
9 \\
\vdots \\
15 \\
\end{array} } \right)$$
My question: What is the number of permutations we can construct from the $n$ elements where each permutation should consists of the same cycles type?
Addition: I know that the number of $k-$cycles in the symmetric group $S_n$ is $\binom{n}{k}(k-1)!$ but I don't know what to do for the constraint asking that each permutation cycle has the same length in all permutations!
| Let's take your example of a cycle structure corresponding to the partition $3^1 + 2^2 + 1^8 \vdash 15$. I think the easiest way to handle the $a_i$ cycles of a given length $i$ is to select the lot (e.g. for the two two-cycles select four elements) and then repeatedly force the lower unselected element to be in the next $i$-cycle.
So we initially have 15 elements. We select three for the $3$-cycle in $\binom{15}{3}$ ways, leaving 12. Now we select four for the two $2$-cycles in $\binom{12}{4}$ ways, assign the lowest to one cycle along with one of the $\binom{3}{1}$ remaining ones; then assign the lowest to the next cycle along with the $\binom{1}{1}$ remaining one. Overall we get $$\left[\binom{15}{3} \binom{2}{2} \right] \left[ \binom{12}{4} \binom{3}{1} \binom{1}{1} \right] \left[ \binom{8}{8} \binom{7}{0} \binom{6}{0} \binom{5}{0} \binom{4}{0} \binom{3}{0} \binom{2}{0} \binom{1}{0} \binom{0}{0}\right]$$
In general, if we have $k$ items remaining to assign and $a_i$ cycles of length $i$ the term is $$\binom{k}{a_i i} \prod_{j=0}^{a_i - 1} \binom{(a_i - j) i - 1}{i - 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The odds in rolling two dice together. Two dice are rolled. I want to see the odds of the following:
$1.$ A sum of $5$.
$2.$ A sum of $8$ or $10$.
$3.$ A sum less than $6$.
$4.$ Not a sum of $7$.
Solution: $1.$ A sum of $5$. $5 = 1+4,2+3,3+2,4+1$. So the odds is $4/36 = 1/9.$
$2.$ A sum of $8$ or $10$. We can express $8 = 2+6,3+5,4+4,5+3,6+2$ and $10 = 4+6,5+5,6+4$. So the odds is $8/36 = 2/9$.
$3.$ A sum less than $6$. We can express $2=1+1$, $3=1+2,2+1$, $4=1+3,2+2,3+1$ and $5 = 1+4,2+3,3+2,4+1$. So the odds is $10/36 = 5/18$.
$4.$ Not a sum of $7$. We can express $7 = 1+6,2+5,3+4,4+3,5+2,6+1$. So the odds is $\frac{36-6}{36} = \frac{30}{36} = \frac56$
Is the solution correct?
| As stated in the comments, you have correctly calculated the probabilities.
When two dice are rolled, there are $6 \cdot 6 = 36$ possible sums. Of these, four outcomes yield a sum of $5$, while the remaining $36 - 4 = 32$ do not. Assuming each of the $36$ possible sums are equally likely to occur, the odds for a sum of $5$ being obtained is the ratio of the number of favorable events to the number of unfavorable events, which in this case is $4 : 32 = 1 : 8$. The odds against a sum of $5$ being obtained is the ratio of the number of unfavorable events to the number of favorable events, which in this case, is $8 : 1$.
Using the same reasoning, the odds for a sum of $8$ or $10$ is $8 : 28 = 2 : 7$ and the odds against the same outcome is $7 : 2$. The odds for a sum less than $6$ is $10 : 26 = 5 : 13$ and the odds against the same outcome is $13 : 5$. The odds for a sum other than $7$ is $30 : 6 = 5 : 1$ and the odds against the same outcome is $1 : 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan C $ and $ \tan B $
Settings discriminant greater than equal to zero.
$ { (p-1)}^2 - 4p \ge 0 $
This gives $ p \le 3 - 2\sqrt2 $. Or $ p \ge 3 + 2\sqrt2 $
solving both equation
$ A + B + C = \pi $
$ C + B + \frac{\pi}{4} = \pi $
$ C + B = \frac{3\pi}{4} $
Using this to solve both the equation give $ p \in $ real
I found this on Quora.
https://www.quora.com/Let-A-B-C-be-three-angles-such-that-A-frac-pi-4-and-tan-B-tan-C-p-What-are-all-the-possible-value-of-p-such-that-A-B-C-are-the-angles-of-the-triangle
the right method
$ 0 \lt B , C \lt \frac{3\pi}{4} $
Converting tan into sin and cos gives
$ \dfrac {\sin B \sin C}{\cos B \cos C} = p $
Now using componendo and dividendo
$ \frac{\cos (B-C) }{- \cos(B+C) } = \frac{p+1}{p-1} $
We know $ \cos (B+C) = 1/\sqrt2 $
We know the range of $B$ and $C$ $(0, 3π/4)$
Thus the range of $B - C$. $(0, 3π/4 )$
Thus range of $\cos(B+C)$ is $ \frac{ -1}{\sqrt2} $ to $1$
Thus using this to find range gives
$ P \lt 0 $ or $ p \ge 3+ 2\sqrt2 $
| Let $C\leq B$ and $B+C=\frac{3\pi}{4}$.
i) When $\frac{3\pi}{8}\leq B<\frac{\pi}{2}$, then
$\frac{\pi}{4}<C\leq \frac{3\pi}{8}$.
Define $$f=\tan\ B\tan\ ( \frac{3\pi}{4}-B)$$ The range of $f$ is
$\{ \tan^2\frac{3\pi}{8} \leq t<\infty\}$, by (1) continuity, (2)
considering $B\rightarrow \frac{\pi}{2},\ B\rightarrow
\frac{3\pi}{8}$ and (3)
\begin{align*} f' &=\frac{-\cos\ (2B-\frac{\pi}{4} ) }{\sqrt{2}\cos^2 B\sin^2
(B-\frac{\pi}{4}) } >0
\end{align*} when $\frac{3\pi}{8}<B$
ii) When $\frac{\pi}{2} < B<\frac{3\pi}{4}$, then $
-\infty<\tan\ B<T<0$ for some $T$. And $0<\tan\ C<1$ so that range
of $f$ contains $\{-\infty < t<0\}$.
Hence note that these $B,\ C$ can be angles in a triangle so that $
\{ \tan^2\frac{3\pi}{8} \leq t<\infty\ {\rm or}\ -\infty<t<0\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
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Compute polynomial $p(x)$ if $x^5=1,\, x\neq 1$ [reducing mod $\textit{simpler}$ multiples] The following question was asked on a high school test, where the students were given a few minutes per question, at most:
Given that,
$$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$
and,
$$Q(x)=x^4+x^3+x^2+x+1$$
what is the remainder of $P(x)$ divided by $Q(x)$?
The given answer was:
Let $Q(x)=0$. Multiplying both sides by $x-1$:
$$(x-1)(x^4+x^3+x^2+x+1)=0 \implies x^5 - 1=0 \implies x^5 = 1$$
Substituting $x^5=1$ in $P(x)$ gives $x^4+x^3+x^2+x+1$. Thus,
$$P(x)\equiv\mathbf0\pmod{Q(x)}$$
Obviously, a student is required to come up with a “trick” rather than doing brute force polynomial division. How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem?
| This may be accessible to a high school student:
$x^{104}+x^{93}+x^{82}+x^{71}+1$
$ = (x^{104}-x^4)+(x^{93}-x^3)+(x^{82}-x^2)+(x^{71}-x)+(x^4+x^3+x^2+x+1)$
$=x^4(x^{100}-1)+x^3(x^{90}-1)+x^2(x^{80}-1)+ x(x^{70}-1)+(x^4+x^3+x^2+x+1)$
We know that $(x^n-1)|(x^{mn}-1), m,n \in \mathbb{N}$ so $x^5-1$ divides $x^{100}-1, x^{90}-1$ etc.
In turn $x^5-1$ is divisible by $(x^4+x^3+x^2+x+1)$ which concludes the proof
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "63",
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Evaluating $\int_0^1 \frac{3x}{\sqrt{4-3x}} dx$
So this is the integral I must evaluate:
$$\int_0^1 \frac{3x}{\sqrt{4-3x}} dx$$
I have this already evaluated but I don't understand one of the steps in its transformation.
I understand how integrals are evaluated, but I don't understand some of the steps when it is being broken down and integrated.
The steps are as follows:
$$ - \left( \frac {-3x} {\sqrt{4-3x}}\right)$$
$$ - \left( \frac {4-3x-4}{\sqrt{4-3x}}\right) $$
$$ - \sqrt {4-3x} + \frac {4}{\sqrt{4-3x}}$$
$\mathbf {Question1} $ In these three steps the first thing I don't understand how it got broken down into two terms in the third step. If I add together the terms of the third step I get back the original one but I don't understnad how the author reached this point in the first place, like how can I know how to break it down into which and which terms?
After this it is put back into the original equation:
$$ \int_0^1 \frac {3x}{\sqrt{4-3x}} dx = - \int_0^1 (4-3x)^\frac {1}{2} dx + 4 \int_0^1 (4-3x)^\frac{-1}{2} dx $$
$$ = \frac {1}{3} \int_0^1 (4-3x)^\frac {1}{2} (-3 dx) - \frac{4}{3} \int_0^1 (4-3x)^\frac{-1}{2} (-3dx) $$
After this it is integrated as usual with $-3dx$ term disappearing in both and $(4-3x)^\frac{1}{2}$ and $(4-3x)^\frac{-1}{2}$ get integrated with n+1 formula.
$\mathbf{Question 2}$ Why was $-3$ multiplied and divided in second step. The dx doesn't change to du so it clearly isn't substitution. So what exactly is happening here?
| Addressing question 1:
\begin{align*}
& - \left( \frac {4-3x-4}{\sqrt{4-3x}}\right) \\
= & - \left( \frac {4-3x}{\sqrt{4-3x}} - \frac{4}{\sqrt{4 - 3x}}\right) \\
= & - \frac {(\sqrt{4-3x})^2}{\sqrt{4-3x}} - \left(- \frac{4}{\sqrt{4 - 3x}}\right) \\
= & - \sqrt {4-3x} + \frac {4}{\sqrt{4-3x}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
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Evaluate $(\sum_{k=1}^{7} \tan^2(\frac{k\pi}{16})) - \tan^2(\frac{4\pi}{16})$ The question is:
Evaluate
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$
The given answer:$34$
What I've tried:
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+\tan^2\left(\frac{3\pi}{16}\right)+\tan^2\left(\frac{5\pi}{16}\right)+\tan^2\left(\frac{6\pi}{16}\right)+\tan^2\left(\frac{7\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+...+\tan^2\left(\frac{\pi}{2}-\frac{2\pi}{16}\right)+\tan^2\left(\frac{\pi}{2}-\frac{\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\cot^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+\cot^2\left(\frac{2\pi}{16}\right)+\tan^2\left(\frac{3\pi}{16}\right)+\cot^2\left(\frac{3\pi}{16}\right)$$
How do I proceed from here?
|
The summation is 35 and subtract -1 because tan pi/4 is 1
Finally the result is 34
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$ if $3\leq aI don't really know if I should use brute force or some kind of theorem, it comes on a calculus past exam and it says:
suppose: $3≤a<b≤8$
prove that $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$
| Note: This answer was finished after farruhota's post and has a similar argument.
$\quad \frac{b-a}{6} \le \sqrt{1+b} - \sqrt{1+a} \; \text{ iff }$
$\quad\quad\quad 1 \le 6 \frac{\sqrt{1+b} - \sqrt{1+a}}{b-a} = \frac{6}{ \sqrt{1+b} + \sqrt{1+a}}$
Ok, the denominator, $\sqrt{1+b} + \sqrt{1+a}$, of the RHS of the last inequality is always less than
$$ \sqrt{1+8} + \sqrt{1+8} = 3 + 3$$
and $1 \le \frac{6}{3 + 3}$.
To show the second inequality holds:
$\quad \frac{b-a}{4} \ge \sqrt{1+b} - \sqrt{1+a} \; \text{ iff }$
$\quad\quad\quad 1 \ge 4 \frac{\sqrt{1+b} - \sqrt{1+a}}{b-a} = \frac{4}{ \sqrt{1+b} + \sqrt{1+a}}$
Ok, the denominator, $\sqrt{1+b} + \sqrt{1+a}$, of the RHS of the last inequality is always greater than than
$$ \sqrt{1+3} + \sqrt{1+3} = 2 + 2$$
and $1 \ge \frac{4}{2 + 2}$.
We've shown that the OP's (actually strict) inequalities hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fourier Series Representation for piecewise function I've been posed the following question:
$$
f(x)=
\begin{cases}
1-x^2, & 0 \leqslant|x|<1,\\
0, & 1\leqslant|x|<2\\
\end{cases}
$$
I'm trying to find the Fourier series representation but I am having trouble with $ a_n$ and $b_n$ coefficients.
For reference, I obtained $a_0 = \frac{4}{3}$ and I'm pretty sure this is correct.
For $a_n$, I have tried solving the following
$$\frac{1}{2} \int_{-2}^{2} (1-x^2)\cos \left(\frac{\pi m x}{2} \right) dx$$
to obtain
$$\frac{16}{\pi^2 m^2} \cos(2 \pi m) \cos \left(\frac{\pi n x}{2} \right)$$
but am fairly certain I did not obtain the correct answer.
For $b_n$, the function $1-x^2$ is even and hence $b_n = 0$. However, I'm not sure if this is right either.
Any help will be much appreciated!
| As you said $a_0=\dfrac43$ and $b_n=0$. Also integrating by parts shows
$$a_n=\dfrac{1}{2}\int_{-2}^{2} (1-x^2)\cos\dfrac{m\pi x}{2}\ dx
= \dfrac{1}{2}
\left[
(1-x^2)\left(\dfrac{2}{m\pi}\right)\sin\dfrac{m\pi x}{2}
-2x\left(\dfrac{2}{m\pi}\right)^2\cos\dfrac{m\pi x}{2}
+2\left(\dfrac{2}{m\pi}\right)^3\sin\dfrac{m\pi x}{2}
\right]_{-1}^{1}
= \dfrac{1}{2}
\left[
-2\left(\dfrac{2}{m\pi}\right)^2\cos\dfrac{m\pi}{2}
-2\left(\dfrac{2}{m\pi}\right)^2\cos\dfrac{m\pi}{2}
+2\left(\dfrac{2}{m\pi}\right)^3\sin\dfrac{m\pi}{2}
+2\left(\dfrac{2}{m\pi}\right)^3\sin\dfrac{m\pi}{2}
\right]
= \color{blue}{
-2\left(\dfrac{2}{m\pi}\right)^2\cos\dfrac{m\pi}{2}
+2\left(\dfrac{2}{m\pi}\right)^3\sin\dfrac{m\pi}{2}
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve for $x, y \in \mathbb R$: $x^2+y^2=2x^2y^2$ and $(x+y)(1+xy)=4x^2y^2$
Solve the following system of equations. $$\large \left\{ \begin{aligned} x^2 + y^2 &= 2x^2y^2\\ (x + y)(1 + xy) &= 4x^2y^2 \end{aligned} \right.$$
From the system of equations, we have that
$$\left\{ \begin{align*} (x + y)^2 \le 2(x^2 + y^2) = 4x^2y^2\\ 4x^2y^2 = (x + y)(1 + xy) \le \frac{[(x + 1)(y + 1)]^2}{4} \le \frac{(x + y + 2)^4}{4^3} \end{align*} \right.$$
$$\implies (x + y)^2 \le \frac{(x + y + 2)^4}{4^3} \implies |x + y| \le \left|\frac{x + y + 2}{2^3}\right|$$
I don't know what to do next. Please help me solve this problem.
|
My simple solution is $(\!x^{2}+ y^{2}- 2x^{2}y^{2}\!)+ \left (\!(\!x+ y\!)(\!1+ x^{2}y^{2}\!)\!\right )= (\!x+ y- 2xy\!)(\!3xy+ x+ y+ 1\!)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
prove $\ln(1+x^2)\arctan x=-2\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{2n+1}x^{2n+1}$ I was able to prove the above identity using 1) Cauchy Product of Power series and 2) integration but the point of posting it here is to use it as a reference in our solutions.
other approaches would be appreciated.
| knowing that fact that
$$2\sum_{n=1}^\infty f(2n)=\sum_{n=1}^\infty f(n)(1+(-1)^n)$$
then
\begin{align}
2\sum_{n=1}^\infty (-1)^nx^{2n}H_{2n}&=2\sum_{n=1}^\infty (i)^{2n}x^{2n}H_{2n}\\
&=\sum_{n=1}^\infty (ix)^nH_{n}+\sum_{n=1}^\infty (-ix)^nH_{n}\\
&=-\frac{\ln(1-ix)}{1-ix}-\frac{\ln(1+ix)}{1+ix}\\
&=-\frac{\ln(1-ix)+\ln(1+ix)+ix(\ln(1-ix)-\ln(1+ix))}{1+x^2}\\
&=-\frac{\ln(1+x^2)+ix(-2i\arctan x)}{1+x^2}\\
&=-\frac{\ln(1+x^2)}{1+x^2}-\frac{2x\arctan x}{1+x^2}
\end{align}
integrate both sides from $x=0$ to $z$
\begin{align}
2\sum_{n=1}^\infty (-1)^nH_{2n}\int_0^zx^{2n}\ dx&=2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}z^{2n+1}\\
&=-\int_0^z\left(\frac{\ln(1+x^2)}{1+x^2}+\frac{2x\arctan x}{1+x^2}\right)\ dx\\
&=-\int_0^zd(\ln(1+x^2)\arctan x)\\
&=-\ln(1+z^2)\arctan z
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Use real integral to calculate $\int_R \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$ Problem :
Evaluate $$\int_{-\infty}^{\infty} \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$$
Use only real integral.
What I did :
$$\int_{-\infty}^{\infty} \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$$
$$=2\int_{0}^{\infty} \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$$
$$=2\int_{0}^{\infty} \left( \frac{2\cos(\pi x)}{x^2+2} - \frac{\cos (\pi x)}{x^2 + 1}\right) dx$$
Any easy way to calculate this? or idea like differentiate under the integral sign?
| Just A Bit of a Generalization
We may actually evaluate the integral
$$J(a,b;t)=2\int_0^\infty \frac{x^2\cos(tx)dx}{(x^2+a^2)(x^2+b^2)}$$
for some $a, b>0$, $a\ne b$. We see that
$$J(a,b;t)=\frac{2a^2}{a^2-b^2}f(a;t)+\frac{2b^2}{b^2-a^2}f(b;t)$$
where $$f(q;t)=\int_0^\infty \frac{\cos(tx)}{x^2+q^2}dx\, .$$
We take the Laplace transform of the remaining integral:
$$\begin{align}
\mathcal{L}\{f\}(s)&=\int_0^\infty e^{-st}\int_0^\infty \frac{\cos(tx)}{x^2+q^2}dxdt\\
&=\int_0^\infty \frac{1}{x^2+q^2}\int_0^\infty \cos(tx)e^{-st}dtdx\\
&=\int_0^\infty \frac{1}{x^2+q^2}\text{Re}\int_0^\infty e^{-(s-ix)t}dtdx\\
&=\int_0^\infty \frac{1}{x^2+q^2}\text{Re}\left[\frac1{s-ix}\right]dx\\
&=s\int_0^\infty \frac{dx}{(x^2+q^2)(x^2+s^2)}\\
&=\frac{s}{s^2-q^2}\left[\int_0^\infty \frac{dx}{x^2+q^2}-\int_0^\infty\frac{dx}{x^2+s^2}\right]\\
&=\frac{\pi}{2}\frac{s}{s^2-q^2}\left[\frac{1}{q}-\frac{1}{s}\right]\\
&=\frac{\pi}{2}\left[\frac{1}{q}\frac{s}{s^2-q^2}-\frac{1}{s^2-q^2}\right]\, .
\end{align}$$
Then we define
$$C(z;s)=\int_0^\infty e^{-st}\cosh(zt)dt=\mathcal{L}\{\cosh(zt)\}(s)$$
$$S(z;s)=\int_0^\infty e^{-st}\sinh(zt)dt=\mathcal{L}\{\sinh(zt)\}(s)$$
So that
$$C(z;s)+S(z;s)=\int_0^\infty e^{-(s-z)t}dt=\frac{s+z}{s^2-z^2}$$
and
$$C(z;s)-S(z;s)=\int_0^\infty e^{-(s+z)t}dt=\frac{s-z}{s^2-z^2}\ .$$
Hence
$$C(z;s)=\frac{s}{s^2-z^2}$$
$$S(z;s)=\frac{z}{s^2-z^2}$$
And
$$\begin{align}
\mathcal{L}\{f\}(s)&=\frac{\pi}{2q}\left[\mathcal{L}\{\cosh(qt)\}(s)-\mathcal{L}\{\sinh(qt)\}(s)\right]\\
&=\frac{\pi}{2q}\mathcal{L}\{e^{-qt}\}(s)\, .
\end{align}$$
Therefore
$$f(q;t)=\frac{\pi}{2q}e^{-qt}$$
and
$$J(a,b;t)=\frac{\pi}{a^2-b^2}\left[ae^{-at}-be^{-bt}\right]$$
and your integral is given by
$$J(\sqrt{2},1;\pi)=\int_{-\infty}^\infty \frac{x^2\cos(\pi x)dx}{(x^2+2)(x^2+1)}=\pi\sqrt{2}e^{-\pi\sqrt{2}}-\pi e^{-\pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Calculate $\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}dx$
Calculate $$\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}dx$$
I tried to do this task in several ways, but none of them proved to be effective. For example:
$$\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}dx=\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2(1-(\sin x)^2)}{1-(\sin x)^4}dx=\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2}{1+(\sin x)^2}dx=\int ^{4\pi} _{-4\pi} \frac{1}{1+\frac{1}{(\sin x)^2}}dx$$
However I don't know what I can do the next to finish this task. When I use $u=(\sin x)^2 $ I have $du=\cos x dx$ so I can't use it.Have you got some intelligent way to do this task?
| Hint:
$$ \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}= 1-\frac{1-\sin^2 x}{1-(\sin x)^4}=1-\frac{1}{1+\sin^2x}=1-\frac{\sec^2x}{1+2\tan^2x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Difference in my and wolfram's integration. Calculate
$$\int \frac{\sin ^3(x)+1}{\cos (x)+1} \, dx$$
Let $$u = \tan(x/2)$$
$\int \frac{\sin ^3(x)+1}{\cos (x)+1} \, dx = \int \frac{2\left(\frac{8u^3}{(u^2+1)^3}+1 \right)}{(u^2+1)\left( \frac{1-u^2}{u^2+1} + 1 \right)} \, du = \int \frac{8u^3}{(u^2+1)^3 } + 1 \, du$
$$ s := u^2 \wedge ds = 2u \, du $$
$ u + \int \frac{8su}{(s+1)^3 } (2u)^{-1} \, ds = u + 4\int \frac{s}{(s+1)^3 } \, ds = $
$ u + 4\int \frac{1}{(s+1)^2 } - \frac{1}{(s+1)^3 } \, ds = u + 4\left(\frac{-1}{s+1} +\frac{1}{2(s+1)^2} \right) + C =$
$ u + 4\left(\frac{-1}{u^2+1} +\frac{1}{2(u^2+1)^2} \right) + C = $
$ \tan(x/2) + 4\left(\frac{-1}{(\tan(x/2))^2+1} +\frac{1}{2((\tan(x/2))^2+1)^2} \right) + C := F(x)$
$$F(\pi/2) - F(-\pi/2) = 2 $$
Wolfram tells that result ($=2$) is okay, but my integral is different from wolfram result:
$$ \frac{1}{8} \sec \left(\frac{x}{2}\right) \left(8 \sin \left(\frac{x}{2}\right)-4 \cos \left(\frac{x}{2}\right)-3 \cos \left(\frac{3 x}{2}\right)+\cos \left(\frac{5 x}{2}\right)\right) $$
Where I failed?
| There's no error in your solution. Both results (yours and Wolfram's) are equal up to a constant (which for an indefinite integral is arbitrary). You can check that using the equalities
$$ \cos (5\alpha) = \cos^5 \alpha - 10\cos^3\alpha \sin\alpha + 5\cos\alpha \sin^4\alpha = 16\cos^5\alpha -20\cos^3\alpha + 5\cos\alpha$$
$$ \cos (3\alpha) = \cos^3\alpha - 3\cos\alpha\sin^2\alpha = 4\cos^3\alpha -3\cos\alpha$$
$$ \frac{1}{\tan^2\alpha +1} = \cos^2\alpha$$
$$ \sec\alpha = \frac{1}{\cos\alpha}$$
Just write both results using only $\cos(\frac{x}{2})$ and compare.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$. This question is taken from book: Exercises and Problems in Calculus, by
John M. Erdman, available online, from chapter 1.1, question $4$.
Request help, as not clear if my approach is correct.
(4) Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$.
Have two conditions, leading to three intervals;
(i) $x \lt \sqrt{-2}$
(ii) $\sqrt{-2} \le x \lt \sqrt{11}$
(iii) $x \ge \sqrt{11}$
(i) no soln.
(ii) $2x^2 = 9 \implies x = \pm\sqrt{\frac 92}$
(iii) no soln.
Verifying:
First, the two solutions must satisfy that they lie in the given interval,
this means $\sqrt{-2} \le x \lt \sqrt{11}\implies \sqrt{2}i \le x \lt \sqrt{11}$.
Am not clear, as the lower bound is not a real one. So, given the two real values of $x = \pm\sqrt{\frac 92}$ need only check with the upper bound.
For further verification, substitute in values of $x$, for interval (ii):
a) For $x = \sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
b) For $x = -\sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
| Your solution is not correct. But contains good ideas.
You write $\sqrt{2}i\leq x\leq \sqrt{11}$.
You can not do that, because you can not order complex numbers! So that is a mistake.
Also it is not clear, how you get your three conditions, which are also the wrong intervals.
It is $x^2+2>0$ for all $x\in\mathbb{R}$.
So $|x^2+2|=x^2+2$.
Now: $|x^2-11|=\begin{cases} -x^2+11, \sqrt{11}\geq x\geq-\sqrt{11}\\ x^2-11, x\in\mathbb{R}\setminus (-\sqrt{11},\sqrt{11})\end{cases}$
Now seperate the cases and check, if
$x^2+2=-x^2+11$ has solutions in $(-\sqrt{11},\sqrt{11})$
or
$x^2+2=x^2-11$ has solutions in $\mathbb{R}\setminus (-\sqrt{11},\sqrt{11})$
For the first equation, you get $2x^2=9\Leftrightarrow x_{1,2}=\sqrt{\frac92}\in (-\sqrt{11},\sqrt{11})$
The second equation has obviously no solutions, because it is equivalent to $2=-11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Given $f(x)=ax^3-ax^2+bx+4$ Find the Value of $a+b$ Let $f(x)=ax^3-ax^2+bx+4$. If $f(x)$ divided by $x^2+1$ then the remainder is $0$. If $f(x)$ divided by $x-4$ then the remainder is $51$. What is the value of $a+b$?
From the problem I know that $f(4)=51$.
Using long division, I found that remainder of $\frac{ax^3-ax^2+bx+4}{x^2+1}$ is $a+b+x(b-a)$.
Then
$$a+b+x(b-a)=0$$
I can't proceed any further so I'm guessing the other factor of $f(x)$ is $ax+4$.
Then
$$f(x)=(ax+4)(x^2+1)=ax^3+4x^2+ax+4=ax^3-ax^2+bx+4$$
I found that $a=-4$ and $b=a=-4$. Then $f(x)=-4x^3+4x^2-4x+4$. But I doesn't satisfy $f(4)=51$
| $f(x)=ax^3-ax^2+bx+4$
Since $f(4) = 51$,
$51
=a(64-16)+4b+4
=48a+4b+4
$
so
$12a+b
=47/4
$.
Since $f(i) = 0$,
$0
=a(-i+1)+ib+4
=i(b-a)+a+4
$
so
$ a+4 = 0,
a=-4,
b-a = 0,
b=a=-4
$.
Therefore
$f(x) = -4x^3+4x^2-4x+4
$.
But this does not satisfy
$f(4) = 51$.
Therefore the problem is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How to solve $x^2+y^2 \equiv 8\pmod 9$? How to solve $x^2+y^2 \equiv 8 \pmod 9$?
I know the Chinese Remainder Theorem, but how do I apply it here?
| If we consider each possible residue of $x^2\mod9$, we see that they are
$0^2\equiv 0$
$1^2\equiv 1$
$2^2\equiv 4$
$3^2\equiv 0$
$4^2\equiv 7$
$5^2\equiv 7$
$6^2\equiv 0$
$7^2\equiv 4$
$8^2\equiv 1$
Obviously (in a minor abuse of notation) the solutions have to be $x\in\{4,5\},y\in\{1,8\}$ (or vice versa) (where numbers represent their residue classes) or $x,y\in\{2,7\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\sum_{k=1}^{n} \frac{1}{k(k+1)}$ I have just started learning sums. I need to evaluate the following sum:
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k(k+1)}$$
$$a_{1} = \frac{1}{2}, a_{2} = \frac{1}{6}, a_{3} = \frac{1}{12}, a_{4} = \frac{1}{20}, a_{5} = \frac{1}{30}, a_{6} = \frac{1}{42}, ...$$
I tried splitting it to partial fractions:
$$\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}$$
$$1 = A(k+1) + Bk$$
$$1 = Ak + A + Bk$$
$$\begin{cases} A = 1 \\ 0 = A + B \quad \Rightarrow B = -1\end{cases}$$
$$\Longrightarrow S_{n} = \sum_{k=1}^{n} \Big( \frac{1}{k} - \frac{1}{k+1} \Big)$$
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k+1}$$
Wolframalpha gives answer $S_{n} = \frac{n}{n+1}$
Help's appreciated. P.S: Are the tags okay?
| $\require{cancel}$Note that$$S_n=1-\cancel{\frac12}+\cancel{\frac12}-\cdots-\cancel{\frac1n}+\cancel{\frac1n}-\frac1{n+1}=1-\frac1{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to prove the equality $|\sqrt{x}-\sqrt{a}| = \frac{|x-a|}{\sqrt{x}+\sqrt{a}}$ How to prove this equality?
$|\sqrt{x}-\sqrt{a}| = \frac{|x-a|}{\sqrt{x}+\sqrt{a}}$
related to this post
| Assuming both $a$ and $x$ are positive. Two cases (1) $x \gt a$ and (2) $x\le a$.
Case(1) becomes $\sqrt{x}-\sqrt{a}=\frac{x-a}{\sqrt{x}+\sqrt{a}}$. Clear the denominator and get $x-a=x-a$. Since $x\gt a$, $x-a=|x-a|$
Case(2) becomes $\sqrt{a}-\sqrt{x}=\frac{a-x}{\sqrt{x}+\sqrt{a}}$. Clear the denominator and get $a-x-a-x$. Since $x\le a$, $a-x=|a-x|=|x-a|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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