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$\int_0^{2\pi}\frac{\sin^2x}{5+4\cos x}\ dx$ Using complex integration Here is the integral I am trying to evaluate:
$$\int_0^{2\pi}\dfrac{\sin^2x}{5+4\cos x}\ dx$$
I am getting the final answer as $-{\pi}/8$ but the correct answer is ${\pi}/4$
Here are the steps I did:
$$\begin{align}
\dfrac{\sin^2x}{5+4\cos x} dx &= \\\\
&=\dfrac{\left(\dfrac{z-1/z}{2i}\right)^2}{5+4\dfrac{z+1/z}{2}} \dfrac{dz}{iz} \\\\
&=\dfrac{-1}{4i}\dfrac{{z^2+1/z^2-2}}{5z+2(z^2+1)} dz \\\\
&=\dfrac{-1}{4i}\dfrac{{z^4+1-2z^2}}{{z^2}(5z+2({z^2+1}))} dz \\\\
&=\dfrac{-1}{4i}\dfrac{{z^4+1-2z^2}}{{z^2}(2z+1)(z+2)} dz
\end{align}$$
Clearly, singularities within $|z|=1$ are $0$, and $\frac{-1}{2}$. After applying Residue theorem, I am getting $\frac{-\pi}{8}$
Are my steps correct?
| So far the steps you showed are correct except for a little typo I corrected.
But you didn't show the calculation of the residues.
With $f(z) =\frac{-1}{4i}\frac{{z^4+1-2z^2}}{{z^2}(2z+1)(z+2)}$ I get
$$\operatorname{Res}_{z=0}f(z)= -\frac{5}{16}i \mbox{ and }\operatorname{Res}_{z=-\frac 12}f(z)= \frac{3}{16}i$$
Hence,
$$\int_0^{2\pi}\frac{\sin^2x}{5+4\cos x}\ dx = 2\pi i\left( -\frac{5}{16}i + \frac{3}{16}i\right) =\frac{\pi}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Does $\prod_{m=1}^\infty \frac{1}{m^2}$ have a closed form? We know that $$\sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6},$$ but what about the product of the reciprocal of the squares:
$$\prod_{m=1}^\infty \frac{1}{m^2}?$$
Do we use a different product representation to compute this? Maybe the cosine product
$$\cos{x} = \prod_{m=1}^\infty \left(1-\frac{x^2}{\pi^2\left(m-\frac{1}{2}\right)^2}\right).$$
| For a sum to converge, it is necessary that the terms converge to $0$ (the neutral element for the sum). $\frac{1}{n^2}$ satisfies this.
For a product to converge to a nonzero value, it is necessary that the factors converge to $1$ (the neutral element for the product). $\frac{1}{n^2}$ does not satisfies this, so the product diverges. In this case, the product is $0$.
But because $\frac{1}{n^2} \to 0$, we do of course have that
$$1+\frac{1}{n^2} \to 1$$
and
$$1-\frac{1}{n^2} \to 1$$
So perhaps more interesting product analogies of the sum you mention are
$$\begin{aligned}
\prod_{n=1}^\infty 1+\frac{1}{n^2} &=\frac{\sinh(\pi)}{\pi} = \frac{-e^{-\pi}}{2\pi} + \frac{e^\pi}{2\pi}\\
\prod_{n=2}^\infty 1-\frac{1}{n^2} &= \frac{1}{2}
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why do we use this complicated algorithm for finding values of trigonometric functions? My Mathematics Textbook covers the topic of Values of Trigonometric Functions at Allied Angles using some general formulae first and then goes on to the topic of finding the values of trigonometric functions at allied angles using an algorithm.
The cases discussed in finding the values using some general formulae are :
* At $(-x)$
* At $\Big (\dfrac{\pi}{2} \pm x \Big )$
* At $(\pi \pm x)$
* At $\Big ( \dfrac {3\pi}{2} \pm x \Big )$
* At $(2\pi \pm x)$ which can also be written as $(\pm $ $x)$
The algorithm is as follows :
* Let the angle be $x$
* If $x<0$ and $x = (-a)$, continue with the further steps as $a$ in place of $x$ and when the final result arrives : if $f$ is an even function, then $f(x) = f(-x)$, so $f(x) = f(a)$ and if $f$ is an odd function, $f(-x) = -f(x)$, so $f(x) = -f(-x) = -f(a)$
* Express $x$ (or $a$) in the form of $\dfrac {n\pi}{2} \pm \alpha$, where $0<\alpha<\dfrac{\pi}{2}$ or $\alpha \in \Big (0, \dfrac {\pi}{2} \Big )$
* If $n$ is odd, then $\sin x = \pm \cos \alpha$, $\cos x = \pm \sin \alpha$, $\tan x = \pm \cot \alpha$, $\cot x = \pm \tan \alpha$, $\sec x = \pm \csc \alpha$ and $\csc x = \pm \sec \alpha$
* If $n$ is even, then $\sin x = \pm \sin \alpha$, $\cos x = \pm \cos \alpha$, $\tan x = \pm \tan \alpha$, $\cot x = \pm \cot \alpha$, $\sec x = \pm \sec \alpha$ and $\csc x = \pm \csc \alpha$
* Determine the quadrant that $x$ lies in and then decide the sign of the value
Let's take an example : Find the value of $\sin \dfrac{7\pi}{4}$.
One method to do this will be using the first method.
$\sin \dfrac{7\pi}{4} = \sin \Big (2\pi - \dfrac{\pi}{4} \Big )$
We know that $\sin (2\pi-x)=(-\sin x)$. So, $\sin \Big (2\pi - \dfrac{\pi}{4} \Big ) = \Big ( -\sin \dfrac {\pi}{4} \Big ) = -\dfrac {1}{\sqrt{2}}$
Another method would be to use the algorithm
$\dfrac{7\pi}{4}=\dfrac{3\pi}{2}+\dfrac{\pi}{4}$, so $\dfrac {3\pi}{2} < \dfrac{7\pi}{4} < 2\pi$ and $\dfrac{7\pi}{4}$ lies in the $IV$ quadrant, which means that $\sin \dfrac{7\pi}{4} < 0$
Now, $\dfrac{7\pi}{4} = \dfrac {3.\pi}{2} + \dfrac{\pi}{4}$. $3$ is odd, so $\sin \dfrac{7\pi}{4} = -\sin \dfrac {\pi}{4} = -\dfrac{1}{\sqrt{2}}$
Now, this algorithm seems like something extremely complex for solving simple questions like these. So, why do we use this algorithm when we can just use the simple formulae that help us find the values of trigonometric functions at allied angles? Are there some advantageous applications of this algorithm?
I feel like it's just a generalization for all the cases that appear in the case of allied angles, just like the lens formula is a generalization of all the cases of image formation through lenses.
Thanks!
EDIT : Also, when should I use which method?
| What, precisely, are these "simple formulae" that you are referring to?
The algorithm here essentially breaks down all the periodicity, symmetry, and reflection-type changes you can do to the angle (some change the value in simple ways).
The point is to normalize the angle you're working with, ultimately, to the first octant $0\leq\theta\leq\frac{\pi}{2}$, so that you can use one table of special values.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\lim_{x\to a} \frac{1}{(a^2-x^2)^2}(\frac{a^2+x^2}{ax}-2\sin( \frac{a\pi}{2})\sin(\frac{x\pi}{2}))$
Without using the L'Hopital Rule, evaluate the following limit.
$$\lim_{x\to a} \frac{1}{(a^2-x^2)^2}\left(\dfrac{a^2+x^2}{ax}-2\sin\left( \frac{a\pi}{2}\right)\sin\left(\frac{x\pi}{2}\right) \right)$$
Where $a$ is an odd integer.
Answer: $\dfrac{\pi^2a^2+4}{16a^4}$
My attempt:
I considered the substitution $x=\lim_\limits{h\to 0}a\cos h$ and got the limit to be:
$$\lim_\limits{h\to 0}\frac{1}{a^4h^4}\left(\frac{1+\cos^2 h}{\cos h}-2\cos^2(ah^2\pi)\right)$$
Aaaand I'm stuck. Please nudge me in the right direction. Profuse thanks.
| Let, $x-a=z$.
So, the limit changes to
$\lim \limits_{z \to 0} \frac{\frac{a^2+x^2}{ax}-2\text{sin}(\frac{aπ}{2})\text{sin}(\frac{xπ}{2})}{(a^2-x^2)^2}$
$=\lim \limits_{z \to 0} \frac{\frac{(x-a)^2}{ax}+2-2\text{sin}(\frac{aπ}{2})\text{sin}(\frac{xπ}{2})}{(z(x+a))^2} $
$=\frac{1}{4a^2}(\lim \limits_{z \to 0} \frac{\frac{z^2}{a^2}+1-\text{cos}(\frac{z\pi}{2})+1+\text{cos}(aπ+\frac{z\pi}{2})}{z^2})$
(Because $x+a=2a+z$)
$=\frac{1}{4a^2}(\lim \limits_{z \to 0} \frac{1}{a^2}+\frac{1-\text{cos}(\frac{zπ}{2})+1-\text{cos}(\frac{zπ}{2})}{z^2})$
(as $a$ is odd, $\text{cos}(aπ+\frac{zπ}{2}))=-\text{cos}(\frac{zπ}{2})$)
$=\frac{1}{4a^2}(\lim \limits_{z \to 0} \frac{1}{a^2}+\frac{4\text{sin}^2(\frac{zπ}{4})}{z^2})$
(For odd $a$, $\text{cos}(\frac{aπ}{2})=0$)
$=\frac{1}{4a^2}(\lim \limits_{z \to 0} \frac{1}{a^2}+\frac{π^2\text{sin}^2(\frac{zπ}{4})}{4(\frac{zπ}{4})^2})$
$=\frac{(πa)^2+4}{16a^4}$ .... proved .
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^{\infty} \frac{ dx }{(x^4+c)(x^2+1) } $ Consider $$\int_0^{\infty} \dfrac{ dx }{(x^4+c)(x^2+1) } $$
where $c \in \mathbb{R}$. Now, it is possible to evaluate this using complex analysis, but I was wondering if there is way to compute this integral without complex analysis.
I was thinking on partial fractions, but there is no easy form to decompose it too.
Here is a way I was thinking: Write $1 = \dfrac{c+x^4 - x^4 }{c}$ to obtain
$$ \frac{1}{c} \int_0^{\infty} \frac{dx}{x^2+1} - \dfrac{1}{c} \int_0^{\infty} \dfrac{x^4}{(x^4+c)(x^2+1)} = \dfrac{ \pi }{2c} - \dfrac{1}{c} \int_0^{\infty} \dfrac{x^4}{(x^4+c)(x^2+1)}$$
I dont know if this is "progress", what do you think?
| Assume $c>0$ for convergence.
$$I = \int_0^{\infty} \dfrac{ dx }{(x^4+c)(x^2+1) }
= \frac1{c+1}\int_0^{\infty}\left( \frac{ 1}{x^2+1}- \frac{ x^2-1}{x^4+c }\right) dx \\
= \frac{\pi}{2(c+1)} - \frac{1}{c+1}\int_0^{\infty}\frac{ x^2-1}{x^4+c } dx$$
where
\begin{align}
\int_0^{\infty}\frac{ x^2-1}{x^4+c } dx
& =\frac{\sqrt c+1}{2\sqrt c} \int_0^{\infty}\frac{ 1-\frac{\sqrt c}{x^2}}{x^2+\frac c{x^2} } dx
+ \frac{\sqrt c-1}{2\sqrt c} \int_0^{\infty}\frac{ 1+\frac{\sqrt c}{x^2}}{x^2+\frac c{x^2} } dx \\
& =\frac{\sqrt c+1}{2\sqrt c} \int_0^{\infty}\frac{d(x+\frac{\sqrt c}{x})}{(x+\frac {\sqrt c}{x} )^2-2\sqrt c}
+ \frac{\sqrt c-1}{2\sqrt c} \int_0^{\infty}\frac{d(x-\frac{\sqrt c}{x})}{(x-\frac {\sqrt c}{x} )^2+2\sqrt c} \\
& =0+ \frac{(\sqrt c-1)\pi}{(2\sqrt c)^{3/2}}
\end{align}
Thus,
$$I = \frac{\pi}{2(c+1)} - \frac{1}{c+1}\frac{(\sqrt c-1)\pi}{(2\sqrt c)^{3/2}}
=\frac{\pi}{2(c+1)} \left( 1- \frac{\sqrt c-1}{\sqrt2 c^{3/4}} \right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $AD\cdot BD \cdot CD \leq \dfrac{32}{27}$ where $ABC$ is a triangle of circumradius 1 and $D\in (BC)$.
Let triangle $ABC$ of circumradius $1$ and $D$ a point on side $(BC)$.
Prove that $$AD\cdot BD\cdot CD\leq \dfrac{32}{27}.$$
My idea. By letting $\alpha = \dfrac{BD}{BC}$ (of course $0<\alpha <1$) we get $BD=BC\cdot \alpha
, \enspace CD=BC\cdot(1-\alpha)\tag{1}$ and also
$$\overrightarrow{AD}=(1-\alpha)\cdot\overrightarrow{AB}+\alpha\cdot \overrightarrow{AC}.$$
By squaring this relation we have that $$AD^2=AB^2(1-\alpha)+AC^2\alpha+BC^2(\alpha^2-\alpha). \tag{$2$}$$
By law of sines we also have $AB=2\sin C$, $AC=2\sin B$ and $BC=2\sin A$.
Now combining with $(1)$ and $(2)$ we may rewrite the desired inequality as follows:
$$((1-\alpha)\sin^2C+\alpha\sin^2B+(\alpha^2-\alpha)\sin^2A)\cdot\alpha^2(1-\alpha)^2\sin^4A\leq \dfrac{2^4}{27^2}.$$
This is where I got stuck. Maybe we could also use the fact that $\sin A=\sin (\pi -B-C)=-\sin(B+C)=-(\sin B\cos C+\sin C\cos B)$ to get rid of $\sin A$?
Thank you in advance!
|
It is a really bad practice to ask a geometry problem without a picture.
Let $OD = x.$ Then $BD\cdot CD = 1 - x^2.$ If $\angle AOD = \alpha,$ then:
$$AD^2 = 1+x^2-2x\cos\alpha.$$
So you need to prove:
$$(1-x^2)\sqrt{1+x^2-2x\cos\alpha}\leq\dfrac{32}{27}.$$
But this is just AM-GM:
$$(1-x^2)(1+x^2-2x\cos\alpha)^{\frac 12}\leq(1-x^2)(1+x)=4(1-x)\cdot\dfrac{1+x}{2}\cdot\dfrac{1+x}{2}\leq 4\cdot \left(\dfrac{2}{3}\right)^3 = \dfrac{32}{27}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$x-\sin(x) \geq \dfrac{x^3}{(x+\pi)^2}$ Let $x \geq 0.$ I need to prove that $x-\sin(x)\geq\dfrac{x^3}{(\pi+x)^2}.$
I tried the derivative, of $f(x)=x-\sin(x)-\dfrac{x^3}{(\pi+x)^2}$ which is $1-\cos(x)-\dfrac{x^2(x+3\pi)}{(\pi+x)^3},$ but it has a complicated formula.
Any ideas, hints?
Edit: sorry, there was a mistake in the derivative, I corrected it.
| Maybe an easier way to do it, though conceptually similar with @Integrand
By multiplying out and simplifying the terms, while noting that $x(x+\pi)^2-x^3=\pi x(x+2\pi)=\pi(x+\pi)^2-\pi^3$, the inequality is equivalent to:
$(\pi-\sin x)(x+\pi)^2 \ge \pi^3, x \ge 0$
But now $\pi-\sin x \ge \pi -1$ and we see that $(\pi+1)^2(\pi-1)=(\pi^2-1)(\pi+1)=\pi^3+\pi^2-\pi-1>\pi^3$, so the inequality is automatically true for $x \ge 1$
Then taking derivatives for $f(x)=(\pi-\sin x)(x+\pi)^2$ we have $f'(x)=2(x+\pi)(\pi-\sin x)-(x+\pi)^2 \cos x$ and for $0 \le x \le 1$, $2(\pi-\sin x)-(x+\pi) \cos x \ge 2\pi-2 -\pi-1= \pi -3 >0$ so $f$ is increasing there, hence $f(x) \ge f(0)=\pi^3$ and the inequality holds in this interval too. so done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a+b=2 \text{ and } ab = -1.$ Determine $a^{10} + b^{10}$
Let $a,b \in \mathbb{R}$ for which $$a+b=2 \text{ and } ab = -1.$$ Determine $a^{10}+b^{10}.$
I tried to approach this the following way:
From $ab=-1$ we get that $a = -\frac{1}{b}$ and substituting this to $a+b=2$ yields in $b^2-2b-1=0$. This quadratic has the solutions $b= 1 \pm \sqrt{2}$
Subsituting now $b$ we have that $a= - \frac{1}{1\pm \sqrt{2}}$
I have the feeling that I'm not going in the right direction here. Is there something I'm not seeing since the quadratic formula seems to yield a bit too much work?
| Another way: Following your tack, if you choose $b = 1-\sqrt{2}$ then $a=-1/(1-\sqrt{2}) = 1+\sqrt{2}$ after you rationalize the denominator. So $a$ and $b$ are conjugates.
If you expand $(1+\sqrt{2})^{10} + (1-\sqrt{2})^{10}$ with binomial theorem, all the odd terms will cancel out. The 10th row of Pascal's triangle is:
1 10 45 120 210 252 210 120 45 10 1
Eliminate the odd terms and double the even and you have
$$a^{10}+b^{10} = 2\left(1 + 45\cdot 2+210\cdot 4 + 210\cdot 8 + 45\cdot 16 + 32\right) $$
$$=6726.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$ Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$
So, using AM-GM, or just pop out squares under square roots we can show:
$$\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}),$$
i.e. we need next to show that $(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge (a + b + c)$, but i don't know how to do it.
Any help appreciated
| The problem has appeared here various times and maybe elsewhere too for sure. The key to this is to observe the relationship between: $(a-b)^2 \ge 0$ and $a^2+ab+b^2 \ge \dfrac{3(a+b)^2}{4}$. They are equivalent. Using the latter but taking the square root first $3$ times for $(a,b), (b,c),(c,a)$ and add up, yielding the result.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a_1a_2 = 1, a_2a_3 = 2, a_3a_4 = 3 \cdots$ and $ \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$, find $|a_1|$
If $a_1a_2 = 1, a_2a_3 = 2, a_3a_4 = 3 \cdots$ and $\displaystyle \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$. Find $|a_1|$
I could conclude that $\displaystyle \lim_{n \to \infty}a_n$ must be $\infty$. But couldn't get any further. Any help is welcome.
| Note that
$$
\frac{a_3}{a_1} = \frac{a_2a_3}{a_1a_2} = \frac21,\quad
\frac{a_5}{a_3} = \frac{a_4a_5}{a_3a_4} = \frac43,\quad
\frac{a_7}{a_5} = \frac{a_6a_7}{a_5a_6} = \frac65,\quad\dots
$$
and therefore
$$
\frac{a_{2k+1}}{a_1} = \frac{a_3}{a_1} \frac{a_5}{a_3}\cdots \frac{a_{2k+1}}{a_{2k-1}} = \frac21 \frac43 \cdots \frac{2k}{2k-1} = \frac{4^k(k!)^2}{(2k)!}.
$$
Similarly,
$$
\frac{a_{2k+2}}{a_2} = \frac{a_4}{a_2} \frac{a_6}{a_4} \cdots \frac{a_{2k+2}}{a_{2k}} = \frac32 \frac54 \cdots \frac{2k+1}{2k} = \frac{(2k+1)!}{4^k(k!)^2}.
$$
Therefore
$$
1 = \lim_{k\to\infty} \frac{a_{2k+1}}{a_{2k+2}} = \lim_{k\to\infty} \frac{4^k(k!)^2 a_1/(2k)!}{(2k+1)!a_2/4^k(k!)^2} = \lim_{k\to\infty} \frac{a_1}{a_2} \frac{16^k(k!)^4}{(2k)!(2k+1)!} = \frac{a_1}{a_2} \frac\pi2
$$
(using Stirling's formula),
which forces $a_1/a_2 = 2/\pi$; together with $a_1a_2=1$ this yields $a_1=\pm\sqrt{2/\pi}$. (One can reality check $\lim_{k\to\infty} \frac{a_{2k}}{a_{2k+1}}$ as well if desired.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int \cos^2(x)\tan^3(x) dx$ using trigonometric substitution How would I integrate to evaluate $\int \cos^2(x)\tan^3(x) dx$ using trigonometric substitution?
I made an attempt by making substitutions such as $$\cos^2(x)=1-\sin^2(x)$$
$$\tan(x) = \frac{\sin(x)}{\cos(x)}$$ and $$\tan^2(x)=1+\sec^2(x)$$
But I couldn't find a way to make it look like an integral I could solve using a $u$ substitution or identity.
Could I get some help on this one?
| Just for fun:
$$\int \cos^2 x \tan^3x \ \mathrm{d} x$$
$$=\int \frac{\tan^3 x}{\sec^2 x} \ \mathrm{d} x$$
Now let $u = \tan^2 x, \mathrm du = 2 \tan x \sec^2 x \ \mathrm dx$:
$$=\int \frac{u \tan x }{\sec^2 x} \cdot \frac{\mathrm{d} u}{2 \tan x \sec^2 x}$$
$$=\frac{1}{2} \int \frac{u}{\sec^4 x} \mathrm d u$$
$$=\frac{1}{2} \int \frac{u}{(1+u)^2} \ \mathrm d u$$
$$=\frac{1}{2} \int \frac{1+u}{(1+u)^2} -\frac{1}{(1+u)^2} \mathrm d u$$
$$=\frac{1}{2} \left( \ln(\tan^2 x + 1) + \frac{1}{1+\tan^2 x} \right) +C$$
$$=\frac{1}{2} \left( \ln | \sec^2 x| + \cos^2 x \right) + C$$
where we have used the identity $1 + \tan^2 x = \sec^2 x$ twice.
Further simplifying gives the accepted answer:
$$=\frac{1}{2} \left( \ln | \cos^{-2} x| + \cos^2 x \right) + C$$
$$=\frac{1}{2} \left(-2 \ln | \cos x | + \cos^2 x \right) + C$$
$$=\frac{\cos^2 x}{2} - \ln | \cos x | + C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Surface integral of piecewise volume boundary? How should I go about solving the surface integral
$$\iint_{\partial V} 2(x^2+y^2) \, dS,$$ where $V$ is the region bounded by the paraboloid $$z=\frac12-x^2-y^2$$ and the cone $$z^2=x^2+y^2?$$
I have done surface integrals before, but I am unsure how to proceed with finding the boundary of the solid $V.$
| I assume the surface in question lies above the plane $z = 0;$ however, if not, you can easily adapt this solution to the case that the surface contains both the upper and lower nappe of the cone.
Graphing the two surfaces shows that the surface for which we would like to compute the surface area resembles an ice cream cone: it consists of the cap $\mathcal C$ of the paraboloid $z = \frac 1 2 - x^2 - y^2$ and part of the upper nappe $\mathcal N$ of the cone $z^2 = x^2 + y^2.$ Ultimately, we will seek to compute $$\iint_{\partial V} 2(x^2 + y^2) \, dS = \iint_\mathcal C 2(x^2 + y^2) \, dS + \iint_\mathcal N 2(x^2 + y^2) \, dS.$$
Observe that $\mathcal C$ and $\mathcal N$ intersect if and only if $z^2 = x^2 + y^2$ and $z = \frac 1 2 - x^2 - y^2$ if and only if $z = \frac 1 2 - z^2$ if and only if $z^2 + z - \frac 1 2 = 0$ if and only if $z = \frac{\sqrt 3 - 1}{2}.$ Consequently, we have that $$\mathcal C = \biggl \{(x, y, z) \,|\, z = \frac 1 2 - x^2 - y^2 \text{ and } \frac{\sqrt 3 - 1}{2} \leq z \leq \frac 1 2 \biggr \} \text{ and}$$ $$\mathcal N = \biggl \{(x, y, z) \,|\, z^2 = x^2 + y^2 \text{ and } 0 \leq z \leq \frac{\sqrt 3 - 1}{2} \biggr \}. \phantom{\text{ and butt }}$$ Geometrically, the cap is a "deformation" of disk in the $xy$-plane, so we may parametrize $\mathcal C$ by polar coordinates $F(r, \theta) = \bigl \langle r \cos \theta, r \sin \theta, \frac 1 2 - r^2 \bigr \rangle$ for $0 \leq r \leq \sqrt{1 - \frac{\sqrt 3}{2}}$ and $0 \leq \theta \leq 2 \pi.$
Likewise, the upper nappe $\mathcal N$ of the cone can be parametrized most easily by polar coordinates $G(r, \theta) = \langle r \cos \theta, r \sin \theta, r \rangle$ for $0 \leq r \leq \frac{\sqrt 3 - 1}{2}$ and $0 \leq \theta \leq 2 \pi.$
Can you finish the solution from here? Use the definition of the surface integrals $$\iint_\mathcal C 2(x^2 + y^2) \, dS = \int_0^{2 \pi} \int_0^{\sqrt{1 - \sqrt 3 /2}} 2r^2 ||F_r(r, \theta) \times F_\theta(r, \theta)|| \cdot r \, dr \, d \theta \text{ and}$$ $$\iint_\mathcal N 2(x^2 + y^2) \, dS = \int_0^{2 \pi} \int_0^{(\sqrt 3 - 1)/2} 2r^2 ||G_r(r, \theta) \times G_\theta(r, \theta)|| \cdot r \, dr \, d \theta. \phantom{\text{ and }}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do you find appropriate trig substitution for $\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$? I want to solve the below:
$$\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$$
I know for trig substitution, if I have something in the form of $\sqrt{x^2-a^2}$, I can use $x = a\sec{u}$; it just so happens my integral has a numerator in this form: $\sqrt{16x^2 - 3^2}$ so I know to use $x = 3\sec u$:
$$
\begin{align}
& \int \frac{\sqrt{16x^2 - 9}}{x} \, dx \\
= {} & \int \frac{\sqrt{16x^2 - 3^2}}{x} \, dx \\
= {} & \int \frac{\sqrt{16(3\sec u)^2 - 3^2}}{3\sec u} 3\sec u\tan u \, du \\
= {} & \int \frac{(\sqrt{16(3\sec u)^2 - 3^2)}(3\sec u\tan u)}{3\sec u} \, du \\
= {} & \int \sqrt{(16(3\sec u)^2 - 3^2)}(\tan u) \, du
\end{align}
$$
This doesn't seem to make it easy. However, using a calculator online, it suggests I instead use $x = \dfrac{3}{4}\sec{u}$ which simplifies the integral to a crisp $\int 3\tan^2 u \, du$.
My question is, how did the calculator get $a = \dfrac{3}{4}$ and is there a way to determine an ideal trig substitution for a given function?
| $$
16x^2 - 9 = 9\left( \left( \tfrac{4x}{3} \right)^2 - 1 \right) = 9(\sec^2\theta - 1) = 9\tan^2\theta.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 2
} |
Sum of complex roots' fractions According to this:
If $\omega^7 =1$ and $\omega \neq 1$ then find value of
$\displaystyle\frac{1}{(\omega+1)^2} +
\frac{1}{(\omega^2+1)^2} +
\frac{1}{(\omega^3+1)^2} +
... + \frac{1}{(\omega^6+1)^2}=?$
First I try like
$\displaystyle\frac{1}{\omega+1} +
\frac{1}{\omega^2+1} +
\frac{1}{\omega^3+1} +
... + \frac{1}{\omega^6+1} = 3
$
I have done distribution them and finally got the solution $\dfrac{5}{3}$
However, this is, without a doubt, a time-consuming way.
Can someone please suggest easier way to solve this one.
| Let $\dfrac1{w+1}=x\implies w=\dfrac{1-x}x$
$$\implies\left(\dfrac{1-x}x\right)^7=1$$
As $x\ne0,$ $$x^6-3x^5+5x^4-\cdots=0$$
We need $$\sum_{r=1}^6x_r^2=\left(\sum_{r=1}^6x_r\right)^2-2\sum_{1\le i< j\le6}x_ix_j=\left(\dfrac31\right)^2-2\cdot\dfrac51$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Proving $\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)} $ For $a,b,c \in (0,\infty).$ Prove$:$
$$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)} $$
My proof by SOS$:$
$$ \left( {a}^{2}+bc \right) \left( ac+{b}^{2} \right) \left( ab+{c}^{ 2} \right) \left( a+b \right) \left( b+c \right) \left( c+a \right)\, \cdot \,(\text{LHS}-\text{RHS})$$
$$=\frac{5}{4} abc \sum\limits_{cyc} c^2 (a+b-2c)^2 (a-b)^2 +\frac{1}{4} \sum\limits_{cyc} {c}^{3} \left( 4\,{a}^{2}+3\,ab+4\,{b}^{2} \right) \left( a-b
\right) ^{4}$$
However$,$ it's hard to find this SOS's form without computer.
So I am looking for alternative solution without $uvw.$ Thanks very much!
| $$LHS-RHS=\frac{4abc(b-c)^{2}(c-a)^{2}(a-b)^{2}}{(b+c)(c+a)(a+b)(a^{2}+bc)(b^{2}+ca)(c^{2}+ab)}+\frac{(a^{2}+b^{2}+c^{2}-bc-ca-ab)(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}-a^{2}bc-b^{2}ca-c^{2}ab)}{(a^{2}+bc)(b^{2}+ca)(c^{2}+ab)}\geq 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3708222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find value of $\cot(\theta-\alpha)$ Let $\theta$ and $\alpha$ be the solutions to $$\dfrac{\sin^2\dfrac{x}{2}}{1-\cot\dfrac{x}{2}}+\dfrac{\cos^2\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}=\dfrac{3\cos2x+6}{10}$$
find the value of $\cot(\theta-\alpha)$, where $\dfrac{\pi}{2}<\theta<\alpha<\dfrac{3\pi}{2}$
My attempt :
Let $A=\dfrac{x}{2}$, $$\dfrac{\sin^2A}{1-\cot A}+\dfrac{\cos^2A}{1-\tan A}=\dfrac{3\cos 4A+6}{10}$$
$$\dfrac{\sin A\cos A-\sin^4A-\cos^4A}{2\sin A\cos A-\cos^2A-\sin^2A}=\dfrac{12\cos^2A-3}{10}$$
$$\dfrac{\sin A\cos A+\cos^2A-\sin^2A}{2\sin A\cos A-1}=\dfrac{12\cos^3A-3}{10}$$
Are there any better ways for approaching this problem? or I've come the right way.
| Hint:
For $\sin \dfrac x2=s,\cos\dfrac x2=c$ and $c\ne s$
$$\dfrac{s^2}{1-\dfrac cs}+\dfrac{c^2}{1-\dfrac sc}=\dfrac{s^3-c^3}{s-c}=\cdots=1+\dfrac{\sin x}2$$
Now use $\cos2x=1-2\sin^2x$ to form a Quadratic Equation in $\sin x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\iint\frac{dxdy}{(1+x^2+y^2)^2}$ over a triangle
Calculate
$$\iint\frac{dxdy}{(1+x^2+y^2)^2}$$ over the triangle $(0,0)$, $(2,0)$, $(1,\sqrt{3})$.
So I tried changing to polar coordinates and I know that the angle is between $0$ and $\frac{\pi}{3}$ but I couldn't figure how to set the radius because it depends on the angle.
| Here is an alternative to Robert's nice way to find how $r(\theta)$ depends on $\theta$.
Let $A(0,0)$, $B(2,0)$ and $C(1,\sqrt{3})$ be the three vertices of the triangle. Imagin, or see the picture below, a ray starting from $A$ and intersects with the side $CB$ at $D$. Suppose the angle $\angle DAB=\theta$. You want to find the length of $AD$ in terms of $\theta$. You can apply the law of sines here to the triangle $ABD$:
$$
\frac{\sin(\pi-\theta-\pi/3)}{2}=\frac{\sin (\pi/3)}{f(\theta)},
$$
Since $\sin(\pi-a)=\sin(a)$, and $\sin(\pi/2) = \sqrt{3}/2$, it follows that
$$
f(\theta) = \frac{\sqrt{3}}{\sin(\theta+\pi/3)}\;.
$$
Let us continue the calculations that are done in Robert's answer:
$$\iint_T\frac{dxdy}{(1+x^2+y^2)^2}
=\int_{0}^{\pi/3}\left(\int_{0}^{f(\theta)}\frac{\rho d\rho}{(1+\rho^2)^2}\right)\;d\theta
=-\frac{1}{2}\int_{0}^{\pi/3}
\left[
\frac{1}{1+\rho^2 }
\right]_{\rho=0}^{\rho=f(\theta)}\,d\theta=:\frac12 I\;.
$$
where
$$
I=-\int_{0}^{\pi/3}
\frac{1}{1+f^2(\theta) }-1\,d\theta
=\int_{0}^{\pi/3}
\frac{f^2(\theta)}{1+f^2(\theta) }\,d\theta
=\int_{0}^{\pi/3}\frac{3}{3+\sin^2(\theta+\pi/3)}\;d\theta=:3J\;.
$$
Up to this point, you can go directly to the general method of Weierstrass substitution. But in this specific case, some trig substitutions makes the integral easier.
Observe that $\cos(\pi/2-a)=\sin(a)$. So
$$
\begin{align}
J &= \int_{0}^{\pi / 3} \frac{1}{\cos ^{2}\left(\frac{\pi}{6}-x\right)+3} dx
= \int_{-\pi / 6}^{\pi / 6} \frac{1}{\cos ^{2}(u)+3} du
= \int_{-\pi / 6}^{\pi / 6} \frac{\sec ^{2}(u)}{3 \sec ^{2}(u)+1} du\\
&= \int_{-\pi / 6}^{\pi / 6} \frac{\sec ^{2}(u)}{3 \tan ^{2}(u)+4} du
\quad (\sec^2u = \tan^2u+1)\\
&= \int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{1}{3 s^{2}+4} ds
\quad (d(\tan u)=\sec^2u\;du)\\
&= \frac{1}{4} \int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{1}{\frac{3 s^{2}}{4}+1} ds
=\frac{\sqrt{3}}{6} \int_{-1 / 2}^{1 / 2} \frac{1}{p^{2}+1} d p\\
&=\frac{\sqrt{3}}{3} \tan ^{-1}\left(\frac{1}{2}\right)
= \frac{\sqrt{3}}{3} \cot ^{-1}(2)\;.
\end{align}
$$
So the result is
$$
\frac32J = \frac{\sqrt{3}}{2} \cot ^{-1}(2)\;.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solve the following equation in integers $x,y:$ $x^2+6xy+8y^2+3x+6y=2.$ Question: Solve the following equation in integers $x,y:$ $$x^2+6xy+8y^2+3x+6y=2.$$
Solution: For some $x,y\in\mathbb{Z}$ $$x^2+6xy+8y^2+3x+6y=2\\\iff x^2+2xy+4xy+8y^2+3x+6y=2\\\iff x(x+2y)+4y(x+2y)+3(x+2y)=2\\\iff(x+4y+3)(x+2y)=2.$$
Now if $(x+4y+3)(x+2y)=2$, then either $$\begin{cases} x+4y+3=1\\ x+2y=2\end{cases}\text{ or }\begin{cases} x+4y+3=2\\ x+2y=1\end{cases}\text{ or }\begin{cases} x+4y+3=-1\\ x+2y=-2\end{cases}\text{ or }\begin{cases} x+4y+3=-2\\ x+2y=-1\end{cases}.$$
We have $$\begin{cases} x+4y+3=1\\ x+2y=2\end{cases}\iff (x,y)=(6,-2), \\\begin{cases} x+4y+3=2\\ x+2y=1\end{cases}\iff (x,y)=(3,-1), \\\begin{cases} x+4y+3=-1\\ x+2y=-2\end{cases}\iff (x,y)=(0,-1),\\\begin{cases} x+4y+3=-2\\ x+2y=-1\end{cases}\iff (x,y)=(3,-2).$$
Now since, all the four pairs $(6,-2),(3,-1),(0,-1),(3,-2)$ satisfies the integer equation $(x+4y+3)(x+2y)=2$, thus we can conclude that $(x+4y+3)(x+2y)=2\iff (x,y)=(6,-2),(3,-1),(0,-1),(3,-2).$
Hence, we can conclude that the integer equation $x^2+6xy+8y^2+3x+6x=2$ is satisfied if and only if $(x,y)=(6,-2),(3,-1),(0,-1),(3,-2)$, and we are done.
Is the solution correct and rigorous enough? And, I am always confused while solving equations regarding the usage of the if and only if arguments, which I feel is very necessary in order to have a complete and rigorous solution, but I rarely find it's usage in any book while solving equations of any kind. So, is it necessary? Also, is there a better solution than this?
| Another solution is to let $k=2y$ and get
$$x^2+3xk+2k^2+3x+3k=2$$
$$\iff (x^2+2xk+k^2)+(k^2+3xk)+3(x+k)=2\iff (x+k)^2+(x+k)(k+3)=2$$
So now we can let $m=x+k$ and get
$$m(m+k+3)=2 \implies (m,k) \in \{(1,-2),(-1,-4),(2,-4),(-2,-2)\}$$
Noting that $k=\frac{2}{m}-m-3$ in our calculations. This gives (after dividing the solutions of $k$ by $2$ and putting $x=m-k$)
$$(x,y) \in \{(3,-1),(3,-2),(6,-2),(0,-1)\}$$
This is much simpler.
Or else, I guess we can stop at the point where
$$(x+4y+3)(x+2y)=2$$
and let $a$ be an integer such that $a=x+2y$, this simplifies a lot:
$$a(a+2y+3)=2$$
And we note that $a \in \{1,-1,2,-2\}$ and $y=\frac{1}{a}-\frac{a}{2}-\frac{3}{2}$
$$\implies (a,y) \in \{(1,-1),(-1,-2),(2,-2),(-2,-1)\}$$
$$x=a-2y\implies (x,y) \in \{(3,-1),(3,-2),(6,-2),(0,-1)\}$$
That makes stuff a lot easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$
My try:
we have $$\frac{13}{32}=\frac{2^2+3^2}{2^5}=\frac{1}{8}\left(1+(1.5)^2)\right)$$
Let $x=1.5$
Now consider the function $$f(x)=\frac{1+x^2}{8}-\ln x$$
$$f'(x)=\frac{x}{4}-\frac{1}{x}$$ So $f$ is Decreasing in $(0,2)$
any help here?
| You can use the series
$$\ln(3/2) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2^nn} = \frac{1}{2} - \frac{1}{8} + \frac{1}{24} - \frac{1}{64} + \dotsc$$
Clearly the partial sums fluctuate closer and closer to $\ln(3/2)$, and each new term has a strictly smaller magnitude than the previous one. Thus, since
$$\sum_{n=1}^{6}\frac{(-1)^{n+1}}{2^nn} = \frac{259}{640} < \frac{909}{2240} = \sum_{n=1}^{7}\frac{(-1)^{n+1}}{2^nn} < \frac{13}{32},$$ it follows that $\ln(3/2) < \dfrac{13}{32}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 5
} |
How to compute the integral $\int_0^1 \frac{x\ln x}{\ln (1-x)}dx$?
How to compute the integral $\int_0^1 \frac{x\ln x}{\ln (1-x)}dx$?
We can write
\begin{align*}
\int_0^1 \frac{x\ln x}{\ln (1-x)}dx=\int_0^1\frac{(1-u)\ln (1-u)}{\ln u}du
=-\int_0^1\frac{1-u}{\ln u}\sum_{n=1}^\infty \frac{u^n}{n}du
\end{align*}
Seeing
\begin{align*}
\int_0^1\frac{u^{a-1}-u^{b-1}}{\ln u}du=\ln\frac{a}{b},\forall a,b>0,
\end{align*}
we get
\begin{align*}
\int_0^1 \frac{x\ln x}{\ln (1-x)}dx=\sum_{n=1}^\infty \frac{1}{n}\ln \frac{n+2}{n+1}.
\end{align*}
I have tried hard to compute the series above ,but without any progress.
| One thig we could do is to expand as series
$$\frac{x}{\log (1-x)}=-1+\frac{x}{2}+\frac{x^2}{12}+\frac{x^3}{24}+\frac{19 x^4}{720}+\frac{3
x^5}{160}+\frac{863 x^6}{60480}+\frac{275 x^7}{24192}+\frac{33953
x^8}{3628800}+O\left(x^9\right)$$ where the coefficients correspond to the absolute value of Gregory coefficients, that is to say
$$\frac{x \log (x)}{\log (1-x)}=-\log(x)+\sum_{n=1}^\infty |G_n| x^n \log(x)$$ Integrating termwise
$$\int_0^1\frac{x \log (x)}{\log (1-x)}\,dx=1-\sum_{n=1}^\infty |G_n|\frac{H_{n+1}}{n+1}$$ which generates the sequence
$$\left\{1,\frac{7}{8},\frac{187}{216},\frac{2983}{3456},\frac{372419}{432000},\frac{
186097}{216000},\frac{1994053}{2315250},\frac{2721985571}{3161088000},\frac{1984
061289739}{2304433152000},\cdots\right\}$$ which seems to converge quite fast.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does the polynomial $r^3+s^3+t^3-3rst-1$ have rational roots? I am looking at natural numbers $n$ of the form
$$n=a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc) = \det \left(\begin{array}{rrr}a & b & c \\c & a & b \\b & c & a\end{array}\right) $$
with $a,b,c \in \mathbb{Q}$.
Suppose we can find rational numbers $r,s,t$ such that:
$$1 = r^3+s^3+t^3-3rst = (r+s+t)(r^2+s^2+t^2-rs-rt-st) = \det\left(\begin{array}{rrr} r & s & t \\ t & r & s \\ s & t & r \end{array}\right) $$
Then we would have a new "solution" for $n$:
$$n = n\cdot 1 = (a^3+b^3+c^3-3abc)(r^3+s^3+t^3-3rst) = u^3+v^3+w^3-3uvw$$
So the question is if the group:
$$G := \{ (r,s,t) | r,s,t \in \mathbb{Q} , \det(r,s,t) = 1 \}$$
has some interesting points?
Thanks for your help!
| I hope, that this it what you are looking for:
$\gamma:=e^{i\frac{2\pi}{3}}$
$A^3+B^3+C^3-3ABC=D^3 \enspace$ with $\enspace D:=(a+b+c)(a+\gamma b+\gamma^2 c)(a+\gamma^2 b+\gamma c)$
and therefore $\enspace D= a^3+b^3+c^3-3abc\,$ .
So, you'll get $\,A,B,C\,$ expressed by $\,a,b,c\,$ which means, that with rational numbers $\,a,b,c\,$ the numbers $\,A,B,C\,$ and $\,\frac{A}{D},\frac{B}{D},\frac{C}{D}\,$ are also rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the sum $\sum_{n=1}^{\infty} \frac{3^n}{5^n-2^{2n}}$. Can someone help me with this sum $\sum_{n=1}^{\infty} \frac{3^n}{5^n-2^{2n}}$. I can't find $S_n$
| You can write $$\frac{3^n}{5^n-2^n} = \frac{(3/5)^n}{1 - (2/5)^n} = \sum_{k=0}^\infty (3/5)^n (2/5)^{kn}$$
so your sum is
$$ \eqalign{ \sum_{n=1}^\infty \sum_{k=0}^\infty (3/5)^n (2/5)^{kn}
&= \sum_{k=0}^\infty \sum_{n=1}^\infty (3/5)^n (2/5)^{kn}\cr
&= \sum_{k=0}^\infty \frac{3 \cdot 2^k}{5^{1+k}-3 \cdot 2^k}}$$
But that's no closer to a "closed form" than the original.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Is $\int_0^1\int_0^1 \frac{Li_3(xy)}{(xy)^2}dxdy$ equal to $3\zeta(2)+\zeta(3)-6$ or to infinity? I am comparing two commercial Computer Algebra Sytems on the integral $\int_0^1\int_0^1 \frac{Li_3(xy)}{(xy)^2}dxdy$. One says it equals $3\zeta(2)+\zeta(3)-6=0.1368...$, the other that it does not converge.
which one is correct?
Integrating on the square $[0+\epsilon;1-\epsilon]\times[0+\epsilon;1-\epsilon]$ numerically for smaller and smaller $\epsilon$ gives larger and larger values, which hints at non convergence, but there may well be a cancellation of singularities in the limit.
Any productive comment or answer would be appreciated!
| Thanks to metamorphy and Angina Seng, I can answer completely the question:
*
*the integral diverges, since $\int_0^1\int_0^1 \frac{Li_3(xy)}{(xy)^2}dxdy = \int_0^1\int_0^1 \frac{Li_3(xy)-xy}{(xy)^2}dxdy +\int_0^1\int_0^1 \frac{xy}{(xy)^2}dxdy$, where:
a) the second integral on the rhs diverges
b) the first one on the rhs converges, since $Li_3(xy)=\frac{xy}{1^3} + \frac{(xy)^2}{2^3}+...$ so by canceling the term $xy$ and then simplifying numerator and denominator by $(xy)^2$ the integrand becomes a continuous function on a bounded domain of $\mathbb{R}^2$.
*in fact, we can prove that $\int_0^1\int_0^1 \frac{Li_3(xy)-xy}{(xy)^2}dxdy=3\zeta(2)+\zeta(3)-6$. Indeed, performing explicitely the double integration of the series $\frac{Li_3(xy)-xy}{(xy)^2}$ by $x$ then $y$ we get successively $\int_0^1[\frac{1}{2^3}x+\frac{x^2}{2}\frac{y}{3^3}+\frac{x^3}{3}\frac{y^2}{4^3}+...]_0^1 dy =\int_0^1 (\frac{1}{2^3}+\frac{1}{2}\frac{y}{3^3}+\frac{1}{3}\frac{y^2}{4^3}+...)dy $
$=\frac{1}{2^3}+\frac{1}{2^2}\frac{1}{3^3}+\frac{1}{3^2}\frac{1}{4^3}+... = \frac{1}{2^3}+\sum_{n=2}^{\infty}\frac{1}{n^2(n+1)^3} (*)$.
Now, we have the partial fraction decomposition $\frac{1}{n^2(n+1)^3}=\frac{1}{n^2}+\frac{3}{n+1}+\frac{2}{(n+1)^2}+\frac{1}{(n+1)^3}-\frac{3}{n}$. So the sum from $n=2$ to infinity, taking into account the telescoping cancelations of the terms of $\frac{3}{n+1}$ and $-\frac{3}{n}$, becomes $[\zeta(2)-\frac{1}{1}]+2[\zeta(2)-\frac{1}{1}-\frac{1}{2^2}]+[\zeta(3)-\frac{1}{1}-\frac{1}{2^3}]+[-\frac{3}{2}]$. So when we add the $\frac{1}{2^3}$ from $(*)$ we get:
$\int_0^1\int_0^1 \frac{Li_3(xy)-xy}{(xy)^2}dxdy=3\zeta(2)+\zeta(3)-6$.
And so the first software was somehow computing this (it is not a typo in my code, I've checked by computing it too).
| {
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"url": "https://math.stackexchange.com/questions/3721608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculating $\cos\frac\pi4$ from the half-angle formula gives $\sqrt{\frac12}$ instead of $\frac{\sqrt{2}}2$. What went wrong? I am using the formula $$\cos\left(\frac x2\right)=\sqrt{\frac{1+\cos(x)}2}$$ to find $\cos\left(\frac{pi}{4}\right)$ but it does not give me the correct result.
$$
\cos\left(\frac{\pi}{4}\right)
= \sqrt{\frac{1+\cos\left(\frac{\pi}{2}\right)}2}
= \sqrt{\frac{1+0}2}
= \sqrt{\frac12}
$$
This contradicts $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $. What did I do wrong?
| $\sqrt {\frac 12} = \frac{\sqrt 1} {\sqrt 2} = \frac 1 {\sqrt 2} = \frac {\sqrt 2} {\sqrt 2 \cdot \sqrt 2} = \frac {\sqrt 2} {2} $
This process is called 'rationalising the denominator' and you should try to do this when the denominator is an algebraic irrational number.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration of $\frac{1}{u^4 + (4\zeta^2-2)u^2 + 1}$ I am trying to compute
$$I(\zeta) = \int_{-\infty}^{\infty} \frac{1}{u^{4} + \left(4 \zeta^{2} - 2\right)u^{2} + 1}\, du$$
for positive real $\zeta$. Can anyone help?
I'm way out of practice for integrals except for simple stuff like $\int 1/(1+u^2)\, du = \tan^{-1} u + C$.
Sympy fails on the definite integral and gives me this weird RootSum expression for the indefinite integral:
$$\operatorname{RootSum} {\left(t^{4} \left(4096 \zeta^{8} - 8192 \zeta^{6} + 4096 \zeta^{4}\right) + t^{2} \left(256 \zeta^{6} - 384 \zeta^{4} + 128 \zeta^{2}\right) + 1, \left( t \mapsto t \log{\left (- 512 t^{3} \zeta^{6} + 768 t^{3} \zeta^{4} - 256 t^{3} \zeta^{2} - 32 t \zeta^{4} + 32 t \zeta^{2} - 4 t + u \right )} \right)\right)}$$
Wolfram Alpha gives me the following for the indefinite integral :
$$\begin{align}
& \frac{\frac{1}{a_1}\tan^{-1} \frac{u}{a_1} - \frac{1}{a_2}\tan^{-1} \frac{u}{a_2}}{4\zeta\sqrt{\zeta^2-1}} + C \\
\\
a_1 &= \sqrt{2\zeta^2-2\zeta\sqrt{\zeta^2-1}-1} = \sqrt{b-c}\\
a_2 &= \sqrt{2\zeta^2+2\zeta\sqrt{\zeta^2-1}-1} = \sqrt{b+c}\\
\end{align}$$
(with $b=2\zeta^2-1$ and $c=2\zeta\sqrt{\zeta^2-1}$) but I'm a bit lost how it got there, and then I'm not exactly sure what to do if $\zeta \le 1$ (is the formula still valid?!)
edit: OK, partial fraction expansion is sloooowwwwly coming back to me. It looks like $a_1a_2 = 1$ and $a_1{}^2 + a_2{}^2 = 4\zeta^2-2$, so I guess they used the expansion
$$
\frac{1}{u^{4} + \left(4 \zeta^{2} - 2\right)u^{2} + 1} = \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left(\frac{1}{u^2+a_1{}^2} - \frac{1}{u^2+a_2{}^2}\right)
$$
| Partial fraction expansion gives
$$
\frac{1}{u^{4} + \left(4 \zeta^{2} - 2\right)u^{2} + 1} = \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left(\frac{1}{u^2+a_1{}^2} - \frac{1}{u^2+a_2{}^2}\right)
$$
and since
$$\int \frac{du}{u^2+a^2}\, = \frac{1}{a}\int \frac{a\, du}{u^2+a^2}\,= \frac{1}{a}\int \frac{a^2\, dx}{a^2x^2+a^2}\, = \frac{1}{a}\tan^{-1}x+C = \frac{1}{a}\tan^{-1}\frac{u}{a}+C$$
(with $u = ax$) then I get
$$\begin{align}
I(\zeta) &= \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left[\frac{1}{a_1}\tan^{-1}\frac{u}{a_1} - \frac{1}{a_2}\tan^{-1}\frac{u}{a_2}\right]_{-\infty}^{\infty} \\
&= \frac{\pi}{4\zeta\sqrt{\zeta^2-1}}\left[\frac{1}{a_1} - \frac{1}{a_2}\right] \\
&= \frac{\pi(a_2 - a_1)}{4\zeta\sqrt{\zeta^2-1}} \end{align}$$
I can simplify further since $a_1= \sqrt{b-c}$ and $a_2=\sqrt{b+c}$ with $b=2\zeta^2-1$ and $c=2\zeta\sqrt{\zeta^2-1}$:
$\sqrt{b+c}-\sqrt{b-c} = \frac{(b+c) - (b-c)}{\sqrt{b+c}+\sqrt{b-c}} = \frac{2c}{\sqrt{b+c}+\sqrt{b-c}}$
so we have
$$\begin{align}
I(\zeta) &= \frac{\pi}{4\zeta\sqrt{\zeta^2-1}}\cdot\frac{2c}{\sqrt{b+c}+\sqrt{b-c}} \\
&= \frac{\pi}{\sqrt{b+c}+\sqrt{b-c}} \\
&= \frac{\pi}{\sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}}. \\
\end{align}$$
For positive real $\zeta$ the denominator can be simplified to $2\zeta$ (see https://math.stackexchange.com/a/3724215/120)
so we have $I(\zeta) = \frac{\pi}{2\zeta}.$
| {
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Find all pairs of positive integers $(m, n)$ for which $X^m + X + 1$ divides $X^n + 1$ and pairs $(m,n)$ for which $X^m +X −1$ divides $X^n +1$?
The special case I can think of is when $n=3$, and $m=2$ for the first part. But I don't know if other cases exist. Any help would be appreciated. Thanks!
| Let $r$ be a root of $X^m+X+1.$ The roots of $X^n+1$ lie on the unit circle, so $r$ does as well. Since $\Im(r+r^m) = \Im(-1) = 0$ and $r^m, r$ lie on the unit circle, $r, r^m$ are conjugates, so $r = r^{-m} \Rightarrow r^{m+1} = 1.$ But we also have $r^{m+1} = -r^2 - r,$ so $r^2+r+1 = 0 \Rightarrow r = w, w^2$ where $w \ne 1$ is a $3$rd root of unity. By differentiation, we see $X^m+X+1$ has no double roots, hence at most $2$ roots, which means $m \le 2.$ Furthermore, $w$ is a root iff $\overline{w} = w^2$ is a root, so $m=2.$ But $w^n + 1 \in \{2, w+1, w^2+1\},$ none of which are zero, so no value of $m$ works.
The 2nd problem is similar. We get $r^{m+1} = 1 \Rightarrow 1 = r-r^2 \Rightarrow r^2-r+1 = 0 \Rightarrow r=w, w^5$ where $w$ is a primitive $6$th root of unity. By differentiation, there are no double roots, so $m \le 2.$ Again, $w$ is a root iff $\overline{w} = w^5$ is a root, so $m=2.$ But then $w^2 + w - 1 = w^2 - w + 1 = 0 \Rightarrow w^2=0,$ contradiction, so there are no solutions again.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)=\frac{-3}{2}$,show that $\cos\alpha+\cos\beta+\cos\gamma=\sin\alpha+\sin\beta+\sin\gamma=0$ I think that I've done a major part of the problem but I'm stuck at a point.
Here's what I've done :
It's given to us that
$$\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha) = \dfrac{-3}{2}$$
Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$, we obtain :
$$\cos\alpha\cos\beta + \sin\alpha\sin\beta + \cos\beta\cos\gamma + \sin\beta\sin\gamma + \cos\gamma\cos\alpha + \sin\gamma\sin\alpha = \dfrac{-3}{2}$$
Multiplying both sides by $2$, we obtain :
$$2\cos\alpha\cos\beta + 2\cos\beta\cos\gamma + + 2\cos\gamma\cos\alpha + 2\sin\alpha\sin\beta + 2\sin\beta\sin\gamma + 2\sin\gamma\sin\alpha = -3$$
Adding $\sin^2\alpha+\sin^2\beta+\sin^2\gamma+\cos^2\alpha+\cos^2\beta+\cos^2\gamma$ to both sides, we obtain :
$$\text{LHS : } (\cos^2\alpha + \cos^2\beta + \cos^2\gamma + 2\cos\alpha\cos\beta + 2\cos\beta\cos\gamma + 2\cos\gamma\cos\alpha)$$
$$ + (\sin^2\alpha + \sin^2\beta + \sin^2\gamma + 2\sin\alpha\sin\beta + 2\sin\beta\sin\gamma + 2\sin\gamma\sin\alpha)$$
$$\text{RHS : } -3 + (\cos^2\alpha + \sin^2\alpha) + (\cos^2\beta + \sin^2\beta) + (\cos^2\gamma + \sin^2\gamma)$$
On simplifying,
$$\text {LHS : } (\cos\alpha + \cos\beta + \cos\gamma)^2 + (\sin\alpha + \sin\beta + \sin\gamma)^2$$
$$\text{RHS : } -3+1+1+1 = -3+3 = 0$$
So, we obtain :
$$(\cos\alpha + \cos\beta + \cos\gamma)^2 + (\sin\alpha + \sin\beta + \sin\gamma)^2 = 0$$
$$\implies (\cos\alpha + \cos\beta + \cos\gamma)^2 = -(\sin\alpha + \sin\beta + \sin\gamma)^2$$
Now, square rooting both sides would involve $\iota$ i.e. $\sqrt{-1}$ but I haven't learnt about complex numbers yet and I think that the solution can be continued without using complex numbers but I don't know how.
Any help would be appreciated.
Thanks!
| If the sum of squares of two real numbers is zero, it implies that both numbers are zero. If you want you can simply prove this using Reductio-Ad-Absurdum.
| {
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Proof of the following approximation for $|x| \leqslant 1$ we have $1+x \leqslant e^x \leqslant1+x+x^2$ I was going through the text Introduction to Algorithms by Cormen et. Al. where I came across the following approximation...
for $|x| \leqslant 1$ we have $1+x \leqslant e^x \leqslant 1+x+x^2$
Now I could prove the left side of the inequality and it holds for all real $x$. I proved using Mean Value Theorem( Taylor Series).
Let $f(x)= e^x$
So by Mean Value Theorem we have,
$$f(x) = f(0) + x.f'(\vartheta x) = e^0 + x.e^{\vartheta x} , 0< \vartheta <1 $$
$$ 0< \vartheta < 1 $$
Case 1:
(if $x \geqslant 0$)
$$\begin{aligned}\implies &0 \leqslant \vartheta .x \leqslant x\\\implies& e^0 \leqslant e^{\vartheta .x} \leqslant e^x\\\implies &x.1 \leqslant x.e^{\vartheta .x} \leqslant x.e^x\\\implies& 1+x \leqslant 1+x.e^{\vartheta .x} \leqslant 1+x.e^x\end{aligned}$$
$$\implies 1+x \leqslant f(x) \leqslant 1+x.e^x \ldots\tag 1$$
Case 2:
(if $x \lt 0$)
$$\begin{aligned}\implies &0 \gt \vartheta .x \gt x\\\implies& e^0 \gt e^{\vartheta .x} \gt e^x\\\implies& x.1 \lt x.e^{\vartheta .x} \lt x.e^x\\\implies& 1+x \lt 1+x.e^{\vartheta .x} \lt 1+x.e^x\end{aligned} $$
$$\implies 1+x \lt f(x) \lt 1+x.e^x \ldots\tag 2$$
So from cases $(1)$ and $(2)$ we have that $e^x \geqslant 1+x\ \forall x\in\Bbb R$ and hence for $|x| \leqslant 1$ .
Now my attempt to prove the right inequality.
Here again I attempt using Taylor Series,
$$f(x) = f(0) + x.f'(0) + \frac{x^2}{2!}.f''(\theta .x) = e^0 + x.e^0+ \frac{x^2}{2!}.e^{\vartheta .x} , 0< \vartheta <1 $$
$$\implies f(x) = 1 + x + \frac{x^2}2.e^{\vartheta .x} $$
(if $0\leqslant x \leqslant 1$)
$$\begin{aligned}\implies& 0 \leqslant \theta .x \leqslant x\\\implies& e^0 \leqslant e^{\vartheta .x} \leqslant e^x\\\implies& \frac{x^2}2.1 \leqslant \frac{x^2}2.e^{\vartheta .x} \leqslant \frac{x^2}{2}.e^x\\\implies& 1+ x +\frac{x^2}{2} \leqslant 1+x+\frac{x^2}{2}.e^{\vartheta .x} \leqslant 1+x+\frac{x^2}2.e^x\end{aligned}$$
$$\implies 1+ x +\frac{x^2}2\leqslant f(x) \leqslant 1+x+\frac{x^2}2.e^x\ldots\tag 3$$
Now,
$$\begin{aligned}0\leqslant x \leqslant 1 \implies& 1 \leqslant e^x \leqslant e \implies \frac12 \leqslant \frac{e^x}2 \leqslant \frac{e}2(\approx 1.4)\\\implies &1+x+\frac12.x^2 \leqslant 1+x+\frac{e^x}2.x^2 \leqslant 1+x+\frac{e}2x^2 \lt 1+x+ 1.4 x^2\end{aligned}$$
How to show a more tighter bound for f(x)? I don't get it. Maybe I made some mistake.
(if $-1\leqslant x \lt 0$)
$$\begin{aligned}\implies 0 \gt \vartheta .x \gt x\\\implies& e^0 \gt e^{\vartheta .x} \gt e^x\\\implies&\frac{x^2}{2}.1 \gt \frac{x^2}2.e^{\vartheta .x} \gt \frac{x^2}2.e^x\\\implies &1+ x +\frac{x^2}{2} \gt 1+x+\frac{x^2}2.e^{\vartheta .x} \gt 1+x+\frac{x^2}2.e^x\\\implies &1+ x +\frac{x^2}{2} \gt f(x) \gt 1+x+\frac{x^2}2.e^x\end{aligned}$$
$$\implies 1+x+\frac{x^2}2.e^x \lt f(x) \lt 1+ x +\frac{x^2}2 \lt 1+ x +x^2 \ldots\tag 4$$
| For $0\leq x\leq 1$, we have
\begin{align*}
e^x - 1 - x - x^2 & = - \frac{{x^2 }}{2} + \sum\limits_{n = 3}^\infty {\frac{{x^n }}{{n!}}} \le - \frac{{x^2 }}{2} + \sum\limits_{n = 3}^\infty {\frac{{x^2 }}{{n!}}} \\ & = - \frac{{x^2 }}{2} + x^2 \left( {e - 1 - 1 - \frac{1}{2}} \right) = x^2 (e - 3) \leq 0.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove: $\sqrt{\frac{bc}{a(3b+a)}} + \sqrt{\frac{ac}{b(3c+b)}} + \sqrt{\frac{ab}{c(3a+c)}} \ge \frac{3}{2}$. Prove: $\sqrt{\dfrac{bc}{a(3b+a)}} + \sqrt{\dfrac{ac}{b(3c+b)}} + \sqrt{\dfrac{ab}{c(3a+c)}} \ge \dfrac{3}{2}$ with $a, b, c$ are positive real numbers.
Let $a \le b \le c$:
\begin{align*}
\sqrt{\dfrac{bc}{a(3b+a)}} = \sqrt{\dfrac{b}{a}}\sqrt{\dfrac{c}{3b+a}} &\ge \sqrt{\dfrac{b}{a}}\sqrt{\dfrac{c}{3c+c}} = \dfrac{1}{2} \sqrt{\dfrac{b}{a}} && (1)
\\
\sqrt{\dfrac{ac}{b(3c+b)}} = \sqrt{\dfrac{a}{b}}\sqrt{\dfrac{c}{3c+b}} &\ge \sqrt{\dfrac{a}{b}}\sqrt{\dfrac{c}{3c+c}} = \dfrac{1}{2}\sqrt{\dfrac{a}{b}} && (2)
\\
\sqrt{\dfrac{ab}{c(3a+c)}} \ge \sqrt{\dfrac{ab}{c(3c+c)}} &\ge \dfrac{1}{2} \dfrac{\sqrt{ab}}{c} && (3)
\end{align*}
With $(1)+(2)+(3)$, we have:
$$
\sqrt{\dfrac{bc}{a(3b+a)}} + \sqrt{\dfrac{ac}{b(3c+b)}} + \sqrt{\dfrac{ab}{c(3a+c)}} \ge \dfrac{1}{2}\left(\sqrt{\dfrac{b}{a}} + \sqrt{\dfrac{a}{b}} + \dfrac{\sqrt{ab}}{c} \right).
$$
And I can not find the method to finish this problem. I am trying to think another method.
| From an old Vasile Cirtoaje inequality (see solution here, or here)
If $x,\,y,\,z$ are positive numbers, such that $xyz=1,$ then $$\sqrt{\frac{x}{y+3}}+\sqrt{\frac{y}{z+3}}+\sqrt{\frac{z}{x+3}}\geqslant\frac{3}{2}. \quad (1)$$
Beause $abc=1,$ we can put $x=\frac ab,\,y = \frac bc, \, z=\frac ca.$ Now, the inequality $(1)$ become
$$\sqrt{\frac{bc}{a(a+3b)}} + \sqrt{\frac{ac}{b(b+3c)}} + \sqrt{\frac{ab}{c(c+3a)}} \geqslant \frac{3}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(x) = x^4 - x^2 + 1$, find the values of $x$ such that $f(f(f(x))) \le x^8$
If $$f(x) = x^4 - x^2 + 1$$ find the values of $x$ such that $f(f(f(x))) \le x^8$
I noticed that $f(\pm 1) = 1 \implies \underbrace{f(f(f(... f(\pm 1)..)))}_{\text{n times}} = 1$. thus, $f(f(f(x))) = x^8$ at $x = \pm 1$. Fortunately, these were the only two solutions to this problem. However, this is hacky at best and incomplete at worst.
Can somebody provide a rigorous proof for this problem?
| $f(x)$=$x^2(x^2+1/x^2-1)$ ,
Now notice that
$(x^2+1/x^2-1)\ge1$ for all real x . (Apply $AM \ge GM$ for $x^2$ and $1\over x^2$)
That is $f(x)\ge x^2$ , which implies that $f(f(f(x))) \ge x^8$
But in the question it is given that $f(f(f(x))) \le x^8$
Therefore , $f(f(f(x))) = x^8$ , which will happen when $(x^2+1/x^2-1)=1$ , hence $x = \pm 1$ are the only solutions.
| {
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"url": "https://math.stackexchange.com/questions/3729878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that $\sum\limits_{j=0}^\infty a_j=\sum\limits_{k=0}^\infty b_k$ Show when $\sum\limits_{j=0}^\infty a_j$ is absolute convergent, than $\sum\limits_{k=0}^\infty b_k$ with $b_k:=(a_0+2a_1+...+2^ka_k)/2^{k+1}$ is also absolute convergent,
and even $\sum\limits_{j=0}^\infty a_j=\sum\limits_{k=0}^\infty b_k$
My attempt:
$\sum\limits_{k=0}^\infty b_k=\frac{a_0}{2}+\left(\frac{a_0}{2^2}+\frac{2a_1}{2^2}\right)+\left(\frac{a_0}{2^3}+\frac{a_1}{2^3}+\frac{2^2a_2}{2^3}\right)+...=\sum\limits_{k=0}^\infty \left (\sum\limits_{i=0}^k \frac{1}{2^{k+1}}2^ia_i\right)$
$$\lim\limits_{k\rightarrow\infty}\left |\sqrt[k]{\sum\limits_{i=0}^{k} \left ( \frac{1}{2^{k+1}}2^ia_i\right)}\right|=0<1$$
Because $\lim\limits_{k\rightarrow\infty}\left |\sum\limits_{i=0}^{k} \left ( \frac{1}{2^{k+1}}2^ia_i\right)\right|=\left |\sum\limits_{i=0}^{\infty} \left ( 0*2^ia_i\right)\right|=0$
$\sum\limits_{k=0}^\infty \left (\sum\limits_{i=0}^k \frac{1}{2^{k+1}}2^ia_i\right)=\sum\limits_{k=0}^\infty b_k$ is absolute convergent
Since $\sum\limits_{k=0}^\infty b_k$ is absolute convergent, we can rearrange its pieces:
$$\sum\limits_{k=0}^\infty b_k=\frac{a_0}{2}+\left(\frac{a_0}{2^2}+\frac{2a_1}{2^2}\right)+\left(\frac{a_0}{2^3}+\frac{2a_1}{2^3}+\frac{2^2a_2}{2^3}\right)+...=a_0\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...\right)+a_1\left(\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+...\right)+...$$
$$=a_0\left(\sum\limits_{k=1}^\infty \left(\frac{1}{2}\right)^k\right)+2a_1\left(\sum\limits_{k=2}^\infty \left(\frac{1}{2}\right)^k\right)+2^2a_2\left(\sum\limits_{k=3}^\infty \left(\frac{1}{2}\right)^k\right)+...$$
$=a_0\left(2-1\right)+2a_1\left(2-1-\frac{1}{2}\right)+2^2a_2\left(2-1-\frac{1}{2}-\frac{1}{4}\right)+...=a_0+\frac{1}{2}2a_1+\frac{1}{4}2^2a_2+...=\sum\limits_{k=0}^\infty a_k$
$\Box$
Hello could someone look over my work and give some feedback :)? Is my solution correct, and if not, what could I improve?
| When you want to prove the absolute convergence of the series $\sum_k b_k$ you should prove that $\sum_k |b_k|$ is finite. When you only have non-negative terms, you can rearrange them in any order. Here
$$
\begin{aligned}
\sum\limits_{k=0}^\infty |b_k| &= \sum\limits_{k=0}^\infty \left |\sum\limits_{i=0}^k \frac{1}{2^{k+1}}2^ia_i\right|
\leq \sum\limits_{k=0}^\infty \sum\limits_{i=0}^k \frac{1}{2^{k+1}}2^i |a_i|
= \sum_{0 \leq i \leq k} \frac{1}{2^{k+1}}2^i |a_i| \\
&=\sum\limits_{i=0}^\infty 2^i |a_i| \underbrace{\left (\sum\limits_{k=i}^\infty \frac{1}{2^{k+1}}\right)}_{=\frac{1}{2^j}}
= \sum\limits_{i=0}^\infty |a_i| < \infty
\end{aligned}
$$
so the series $\sum_k b_k$ is absolutely convergent. Once you have proved this, you are guaranteed that you can rearrange the terms with $\sum\limits_{k=0}^\infty b_k$ so the same computations works, but you have an equality instead of an inequality in the second step so you get that $\sum_k b_k = \sum_k a_k$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show the $\arcsin$ identity: $ \arcsin(1 - 2x) + 2\arcsin(\sqrt{x}) = \pi / 2$ Can somebody find an elementary proof of the following identity:
$$
\arcsin ( 1 - 2x) + 2 \arcsin(\sqrt{x}) = \frac\pi2
$$
I noticed it while solving the following integral:
$$ I = \int \frac{\sqrt{x}}{\sqrt{1 - x}} = -2 \sqrt{1 - x}\sqrt{x} + \int \frac{\sqrt{1 - x}}{\sqrt{x}}
$$
where the first equality follows after applying an integration by parts with $f = \sqrt{x}$ and $\mathrm{d} g = 1/\sqrt{1 - x}$. For the sake of simplicity we omit $\mathrm{d}x$ in each integral. Adding the integral to itself:
\begin{align} 2I = \int \frac{\sqrt{x}}{\sqrt{1 - x}} &= -2 \sqrt{1 - x}\sqrt{x} + \int \left(\frac{\sqrt{1 - x}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{1- x}}\right) \\&= -2 \sqrt{1-x}\sqrt{x} + \int\frac{1}{\sqrt{x}\sqrt{1-x}}\end{align}
The last integral on the RHS evaluates to $\arcsin(1 - 2x)$, so $$I = - \sqrt{1-x}\sqrt{x} + \frac{1}{2}\arcsin(1 - 2x) + C.$$
On the other hand, the integral can also be evaluated by applying a $u$-sub with $u = \sqrt{x}$. We find that:
$$
I = 2 \int\frac{u^2}{\sqrt{1 - u^2}} =2 \left(-u \sqrt{1 - u^2} + \int\sqrt{1 - u^2}\right) = - u\sqrt{1 - u^2} + \arcsin u + C_2
$$
So then it follows that $I$ is also equal to $-\sqrt{x}\sqrt{1-x} + \arcsin{\sqrt{x}} + C_2$. Equate the results of the two methods and plug in a random point to find $C - C_2$ and the subsequent identity.
| Let $x=u^2$ with $0\le u\le1$ and rewrite the equation $\arcsin(1-2x)+2\arcsin(\sqrt x)=\pi/2$ as
$$\arcsin(1-2u^2)={\pi\over2}-2\arcsin u$$
Think of each side as an angle, $\theta_L$ and $\theta_R$. Clearly $-\pi/2\le\theta_L\lt\pi/2$, that being the range of the arcsine function. Because $u$ is nonnegative, we have $0\le2\arcsin u\le\pi$, and hence $-\pi/2\le\theta_R\le\pi/2$ as well. Thus to prove that $\theta_L=\theta_R$, it suffices to show that $\sin\theta_L=\sin\theta_R$. Obviously
$$\sin\theta_L=\sin(\arcsin(1-2u^2))=1-2u^2$$
while standard trig identities take care of $\theta_R$:
$$\sin\theta_R=\sin\left({\pi\over2}-2\arcsin u \right)=\cos(2\arcsin u)=1-2\sin^2(\arcsin u)=1-2u^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
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} |
Evaluate the following limit: $\lim\limits_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})$ How can I evaluate following limit $$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})=?$$
My first try:
$$\lim_{x\to\infty}(12x^2-2)\to \infty$$
$$\lim_{x\to\infty}(6x\sqrt{3x^2-2})\to \infty$$so
$$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})=\infty-\infty=0.$$
My answer $0$ is correct, but I don't know whether my method is correct.
My second try:
I substituted $3x^2=2\sec^2\theta$
So limit becomes
$$\lim_{x\to\pi/2}(8\sec^2\theta-2-4 \sqrt{3}\sec\theta\tan\theta)$$
I got stuck. I also can't see application of L'Hospital rule here. Can someone please help me solve this limit? Thanks
| Your method is not correct. It's not true that $\infty - \infty =0$. You may proceed by rationalizing the given function as follows:
$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})=\lim_{x\to\infty} \frac{(12x^2-2-6x\sqrt{3x^2-2})( (12x^2-2+6x\sqrt{3x^2-2}) ) } { 12x^2-2+6x\sqrt{3x^2-2} } $
$=\lim_{x\to \infty} \frac{(12x^2-2)^2-36x^2(3x^2-2)} {(12x^2-2+6x\sqrt{3x^2-2})} =\lim_{x\to \infty} \frac{36x^4+24x^2+4}{ (12x^2-2+6x\sqrt{3x^2-2}) }=\lim_{x\to\infty} \frac{36+24x^{-2}+4x^{-4}} {12x^{-2}-2x^{-4}+6\sqrt{3x^{-4}-2x^{-6}} }$, which is a $\frac{\text{36}} {0}$ form and hence the limit is equal to $\infty$ on set of extended real nos. (i.e. on the domain $\mathbb R\cup \{-\infty, \infty\}) $ but the limit does not exist on $\mathbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$ with Fubini's theorem I have to solve the following double integral
$$\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$$
with $A= \left[0,+\infty\right[ \times [0,1].$ So far I've tried to solve it integrating w.r.t. $y$ first.
$$\iint_0^1\frac{dy\,dx}{(1+x^2)(1+x^2 y^2)} = \int_0^\infty\frac{1}{1+x^2}\int_0^1\frac{dy}{1+x^2 y^2} \, dx. $$
I've solved the internal integral by substitution, remembering that $\int\frac{du}{1+u^2}=\arctan u$
Substitution:
$$x^2 y^2= u^2 \to y=\frac{1}{x}u \to dy=\frac{1}{x}du.$$
$$y=0 \to u=0, \qquad y=1 →u=x$$
So:
\begin{align}
& \int_0^∞\frac{1}{1+x^2} \left( \int_0^x\frac{1}{x}\frac{1}{1+u^2}\,du \right) \, dx \\[8pt]
= {} & \int_0^\infty\frac{1}{x}\frac{1}{1+x^2}[\arctan u] \, dx \\[8pt]
= {} & \int_0^\infty\frac{\arctan x}{x(1+x^2)} \, dx.
\end{align}
Now I have to solve this last integral with the Fubini's theorem but I don't know how to do it.
| \begin{align} & \iint\limits_{[0,+\infty) \times [0,1]} \frac{d(x,y)}{(1+x^2)(1+x^2 y^2)} \\[8pt] = {} & \int_0^1 \left( \int_0^\infty \frac{dx}{(1+x^2)(1+x^2y^2)} \right) \, dy \end{align}
So we need partial fractions:
\begin{align}
& \frac 1 {(1+x^2)(1+x^2 y^2)} = \frac {Ax+B} {1+x^2} + \frac{Cx+D}{1+x^2y^2} \\[10pt]
1 & = (Ax+B)(1+x^2y^2) + (Cx+D)(1+x^2) \\[8pt]
& = (Ay^2+C)x^3 + (By^2+D)x^2 + (A+C)x + (B+D) \\[8pt]
\text{So } & Ay^2 + C=0 \\
& By^2+D=0 \\
& A+C=0 \\
& B+D=1 \\[8pt]
\text{and so } & A=0, \quad C=0, \quad B=1/(1-y^2), \quad D= y^2/(y^2-1). \\[8pt]
& \frac 1 {(1+x^2)(1+x^2 y^2)} = \frac {1/(1-y^2)} {1+x^2} + \frac{y^2/(y^2-1)}{1+x^2y^2}
\end{align}
And so the first integral:
\begin{align}
& \int_0^\infty \frac{dx}{(1+x^2)(1+x^2y^2)} \\[8pt]
= {} & \int_0^\infty \left( \frac{1/(1-y^2)} {1+x^2} + \frac{y^2/(y^2-1)}{1+x^2y^2} \right) \, dx \\[8pt]
= {} & \frac \pi {2(1+y)}
\end{align}
and so on.
Corollary: In view of one of my comments under the question, we conclude that $$ \int_0^{\pi/2} \frac{w}{\tan w} \, dw = \frac \pi 2 \log 2. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\sum\limits_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$? I am trying to prove this. I used the telescoping method, but the problem is I need the first fraction to be $\frac{1}{4(n+1)}$ and I also tried to relate it to alternating Harmonic series which didn't work. Any hint would be greatly appreciated.$$\sum_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$$
| Another approach is to use the Anti-difference concept.
Since the digamma function is defined as
$$
\psi \left( z \right) = {d \over {dz}}\ln \Gamma \left( z \right)
$$
its functional equation is
$$
\Delta \psi \left( z \right) = \psi \left( {z + 1} \right) - \psi \left( z \right) = {d \over {dz}}\ln \left( {z\,\Gamma \left( z \right)} \right) - {d \over {dz}}\ln \left( {\Gamma \left( z \right)} \right)
= {d \over {dz}}\ln z = {1 \over z}
$$
It follows that
$$
\sum\limits_{n = 1}^N {{1 \over {n + a}}} = \sum\limits_{n = 1}^N {\psi \left( {n + a + 1} \right) - \psi \left( {n + a} \right)} = \psi \left( {N + a + 1} \right) - \psi \left( {1 + a} \right)
$$
and therefore
$$
\sum\limits_{n = 1}^N {{1 \over {4n + 1}} - {1 \over {4n}}} = {1 \over 4}\sum\limits_{n = 1}^N {{1 \over {n + 1/4}} - {1 \over n}}
= {1 \over 4}\left( {\psi \left( {N + 5/4} \right) - \psi \left( {5/4} \right) - \psi \left( {N + 1} \right) + \psi \left( 1 \right)} \right)
$$
Since $\psi(z)$ is holomorphic for $0<\Re(z)$, then
$$
\psi \left( {N + 5/4} \right) - \psi \left( {N + 1} \right) = \psi ^{\,\left( 1 \right)} \left( {N + 1} \right){1 \over 4} + {{\psi ^{\,\left( 2 \right)} \left( {N + 1} \right)} \over {2!}}\left( {{1 \over 4}} \right)^2 + \cdots
$$
and since
$$
\mathop {\lim }\limits_{N \to \infty } \psi ^{\,\left( k \right)} \left( {N + 1} \right) = 0\quad \left| {\;1 \le k} \right.
$$
Therefore
$$
\mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 1}^N {{1 \over {4n + 1}} - {1 \over {4n}}} = {1 \over 4}\left( {\psi \left( 1 \right) - \psi \left( {5/4} \right)} \right) = {3 \over 4}\ln 2 + {\pi \over 8} - 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
$ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?
| If you want to go through L'Hospital you may separate $\frac{1}{\cos x}$ the limit of which is $1$ for $x \to 0$:
$$\frac{\tan x - \sin x}{x^3} = \frac 1{\cos x}\cdot \frac{\sin x - \sin x \cos x}{x^3}$$
So, you only need to calculate the limit for $x \to 0$ of
$$\frac{\sin x - \sin x \cos x}{x^3}= \frac{\sin x - \frac 12\sin 2x }{x^3}$$ $$\stackrel{3\times L'Hosp.}{\sim}\frac{-\cos x+4\cos 2x}{6}\stackrel{x \to 0}{\longrightarrow} \frac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
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Factoring $ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a$
Factor $$ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a$$
I used the factor theorem to get factors
$$f(a, b, c)=(a-b)(b-c)(a-c)\;g (a, b, c)$$
for some polynomial $g (a, b, c)$.
How can I continue using this method?
(sorry for the previously messed up question I'm new to this website and didn't fully understand the guidelines).
| $$ab^3-a^3b+bc^3-b^3c+ac^3-a^3c$$ $$=ab(a+b)(b-a)+bc (b+c)(c-b)+ac (a+c)(a-c)$$ $$=ab( (a+b+c)-c)(b-a)+bc ((a+b+c)-a)(c-b)+ac ((a+b+c)-b)(a-c)$$ $$=(a+b+c) [ab (b-a)+bc (c-b)+ac (a-c)]-abc (b-a+c-b+a-c)$$ $$=(a+b+c) [ab (b-a)+bc (c-b)+ac (a-c)]+0$$$$=(a+b+c)[(a-b)(b-c)(c-a)]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Remainder when divided by $7$ What would be the remainder when
$12^1 + 12^2 + 12^3 +\cdots + 12^{100}$ is divided by $7$ ?
I tried cyclic approach (pattern method), but I couldn't solve this particular question.
| We have
$$1+\sum_{i=1}^{100}12^i=\frac{12^{101}-1}{12-1}=\frac{12^{101}-1}{11}$$
Since $\gcd(7,11)=1$, we need only find the remainer of $12^{101}-1$ with $7$.
We have $12^{101}\equiv 5^{101}\bmod 7$ and $5^{101}=25^{50}\cdot5\equiv 5\mod7$
Thus $12^{101}-1\equiv 4\mod7$
and as we explained this gives $\frac{12^{101}-1}{11}\equiv4\mod7$
So $\sum_{i=1}^{100}12^i\equiv 3\mod7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Showing that the Diophantine equation $m(m-1)(m-2)(m-3) = 24(n^2 + 9)$ has no solutions
Consider the Diophantine equation $$m(m-1)(m-2)(m-3) = 24(n^2 + 9)\,.$$ Prove that there are no integer solutions.
One way to show this has no integer solutions is by considering modulo $7$ (easy to verify with it).
I am curious whether there is a slightly less $``$random$``$ way to solve this problem such as using the fact that if $p\equiv 3 \pmod 4$ divides $x^2 + y^2$, then $p$ must divide both $x$ and $y$. This looks convenient since the left-hand side has a multiplier which is $\equiv 3 \pmod 4$ (and hence such a $p$ surely exists) and we will be done provided we can take $p\neq 3$ (since the only prime $p\equiv 3 \pmod 4$ which divides $y=3$ is $3$ itself). Any idea if this method could work?
I am of course also open to see other ideas. Any help appreciated!
| Here is one approach that is more motivated. Let $k$ be the integer in between $m-1$ and $m-2.$ So $$k=\frac{(m-1)+(m-2)}{2}=m-\frac{3}{2}.$$ Then the left side of the equation is
\begin{align*}
m(m-1)(m-2)(m-3) &= \left(k+\frac{3}{2}\right)\left(k+\frac{1}{2}\right)\left(k-\frac{1}{2}\right)\left(k-\frac{3}{2}\right)\\
&=\left(k^2-\frac{9}{4}\right)\left(k^2-\frac{1}{4}\right)\\
&= k^4 -\frac{10}{4}k^2+\frac{9}{16}\\
&= k^4 -\frac{5}{2}k^2+\frac{25}{16}-\frac{25}{16}+\frac{9}{16}\\
&= \left(k^2 -\frac{5}{4}\right)^2 - 1\\
&= \left(\left(m-\frac{3}{2}\right)^2-\frac{5}{4}\right)^2-1\\
&= (m^2-3m+1)^2 -1.
\end{align*}
So the equation is $$(m^2-3m+1)^2 - 24n^2=24\cdot 9 +1 =7\cdot 31.$$ This is a good enough reason to try $\mod 7.$ Then $$(m^2-3m+1)^2\equiv 3n^2 \pmod{7}.$$ The quadratic residues modulo $7$ are $0,1,2,4.$ The only two residues where one is three times the other are $0$ and $0.$ So $m^2-3m+1$ and $n$ are both divisible by $7.$ In the first case $$m^2-3m+1\equiv 0\pmod 7.$$ Equivalently, $$(m+2)^2\equiv 3\pmod{7}.$$ But none of the quadratic residues are $3$ so that's a contradiction to the existence of a solution $(m,n).$
| {
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"url": "https://math.stackexchange.com/questions/3741381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to solve $\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}$?
How can i evaluate the following integral $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=?$$
This is taken from a definite integral where $x$ varies from $0$ to $1$.
My attempt:
multiplied by conjugate
$$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}$$
$$=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{1+x-1+x}$$
$$=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{2x}$$
*
*if i use $x=\sin^2\theta$
$$\int \frac{(\sqrt{1+\sin^2\theta}+\cos\theta)}{2\sin^2\theta}\sin2\theta\ d\theta=\int (\sqrt{1+\sin^2\theta}+\cos\theta)\cot\theta d\theta$$
*if i use $x=\tan^2\theta$
$$\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{2\tan^2\theta}2\tan\theta\sec^2\theta d\theta\ d\theta=\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{\sin\theta\cos\theta} d\theta$$
Should I use substitution $x=\sin^2\theta$ or $x=\tan^2\theta$?. I can't decide which substitution will work further. Please help me solve this integration.
Thanks
| $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{dx}{\sqrt{(\sqrt{1+x}-\sqrt{1-x})^2}}$$
$$=\int \frac{dx}{\sqrt{2-2\sqrt{1-x^2}}}$$
Let $x=\sin\theta\implies dx=\cos\theta d\theta$
$$=\int \frac{\cos\theta d\theta}{\sqrt{2-2\cos\theta}}$$
$$=\int \frac{\cos\theta d\theta}{\sqrt{4\sin^2\frac{\theta}{2}}}\quad \quad \left(\because \cos\theta=1-2\sin^2\frac{\theta}{2}\right)$$
$$=\int \frac{\left(1-2\sin^2\frac{\theta}{2}\right)d\theta}{2\sin\frac{\theta}{2}}$$
$$=\int \left(\frac12\csc\frac{\theta}{2}-\sin\frac{\theta}{2}\right)\ d\theta$$
$$=\ln \left|\tan\frac{\theta}{4}\right|+2\cos\frac{\theta}{2}+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove that these two functions do not intersect? I have this function $\;g(x)-x\;$ on the domain $\;0<x<\frac{1}{3}\;$, where
$g(x)=\frac{\left(9 x^2+1\right) \cosh ^{-1}\left(\frac{36 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi x)}{\left(9 x^2+1\right)^2}\right) \sqrt{81 x^4+54 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi x)+1}}{2 \sqrt{2} \pi \cosh (\pi x)}+\frac{18 \sqrt{2} x^2 \left(9 \pi x^3+4 \tanh (\pi x)\right)}{2 \sqrt{2} \pi }.$
I need to prove that $\;g(x)\;$ does not intersect $\;x\;$ over the mentioned domain. Calculating the first derivative does not work since it makes the problem more complicated. The plot of the functions is attached.
Any hint or suggestion is welcome.
| Partial answer
Denote
$$g_1(x)=\frac{\left(9 x^2+1\right) \cosh ^{-1}\left(\frac{36 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi x)}{\left(9 x^2+1\right)^2}\right) \sqrt{81 x^4+54 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi x)+1}}{2 \sqrt{2} \pi \cosh (\pi x)}$$ and
$$g_2(x) = \frac{18 \sqrt{2} x^2 \left(9 \pi x^3+4 \tanh (\pi x)\right)}{2 \sqrt{2} \pi }$$
$g_1, g_2$ are positive maps and we have $g = g_1+g_2$.
$g_2(x)/x$ is increasing on $(0, 1/3)$ as a product of positive increasing maps. Moreover $g_2(0.06) / 0.06 >1.0$. Hence we have the partial result that $g(x) - x$ is positive on $(0.06, 1/3)$. Remains to prove the result on $(0, 0.06]$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Olympiad inequality proof issue Prove that $(a^2+b^2)^2\geq(a+b+c)(a+b-c)(b+c-a)(c+a-b)\ \forall \ a,b,c\in\mathbb{R^+} $.
I, forgetting to consider whether $a_1$ and $a_2$ are strictly non-negative (don't think they are), found a proof (almost) using the AM-GM inequality with $a_{1}=(a+b+c)(a+b-c)=(a+b)^2-c^2$ and $a_2=(b+c-a)(c+a-b)=c^2-(a-b)^2$, leading to (after manipulation):
$4a^2b^2 \geq (c^2-(a-b)^2)((a+b)^2-c^2)=(a+b+c)(a+b-c)(b+c-a)(c+a-b)$
$4a^2b^2\leq(a^2+b^2)^2$, clearly, so initial inequality proven.
However, I just realised AM-GM only holds for non-negative reals, and my $a_1$ and $a_2$ are both non-negative only if $b-a\leq c\leq b+a$.
Is there another way to prove this using AM-GM, or an extension of the same idea that covers the cases where $c\gt b+a \gt b-a$ or $b+a \gt b-a \gt c$? Also, why does the proof so cleanly yield the result?
| We are to prove (continuing from your work)
$$ \left(c^2-(a+b)^2\right)\left(c^2-(a-b)^2\right)+4a^2b^2\geqslant 0\iff\left(c^2-(a^2+b^2)\right)^2\geq 0$$
which is trivially true.
Your proof works if $b+c\geqslant a$, $c+a\geqslant b$ and $a+b\geqslant c$. Otherwise, if exactly one of the three terms containing a minus sign on the right-hand side of the inequality is negative, then the inequality is trivially true. Other cases force at least one of the variables to be negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a_{n} = \sqrt{n^{2}+n} - n$, for $n\in\textbf{N}$. Is the sequence $(a_{n})_{n=1}^{\infty}$ monotonic? Let $a_{n} = \sqrt{n^{2}+n} - n$, for $n\in\textbf{N}$. Show that $a_{n}$ converges as $n\to\infty$. What is the limit? Is the sequence $(a_{n})_{n=1}^{\infty}$ monotonic?
MY ATTEMPT
The answer to the first question is $a_{n}\to 1/2$ as $n\to\infty$. Indeed,
\begin{align*}
\lim_{n\to\infty}\sqrt{n^{2}+n} - n = \lim_{n\to\infty}\frac{n}{\sqrt{n^{2} + n} + n} = \lim_{n\to\infty}\frac{1}{\sqrt{1 + 1/n} + 1} = \frac{1}{2}
\end{align*}
To test for monotonicity, we can try to study the behavior of the quotient:
\begin{align*}
\frac{a_{n+1}}{a_{n}} & = \frac{\sqrt{(n+1)^{2} + n + 1} - n - 1}{\sqrt{n^{2} + n} - n}\\\\
& = \frac{(n+1)^{2} + n + 1 - (n+1)^{2}}{n^{2} + n - n^{2}}\times\frac{\sqrt{n^{2} + n} + n}{\sqrt{(n+1)^{2} + n + 1} + n + 1}\\\\
& = \frac{n+1}{n}\times\frac{\sqrt{n^{2} + n} + n}{\sqrt{(n+1)^{2} + n + 1} + n + 1}
\end{align*}
But then I get stuck, because the first factor is greater than one and the second is smaller than one.
Can someone please finish my attempt or provide an alternative approach?
| Let $f(x)=\sqrt{x^2+x}-x$.
By AM-GM we have $\sqrt{x(x+1)}< \dfrac{x+(x+1)}{2}\ $ thus $\ f(x)< \dfrac 12\ $ and $\ a_n< \dfrac 12$
This also means $A=2x+1-2\sqrt{x(x+1)}>0$
Since $f'(x)=\dfrac{A}{2\sqrt{x(x+1)}}$ then $f$ is $\nearrow$ and so is $a_n$
| {
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Closed-form solution for the determinant of a Vandermonde-like matrix I'm trying to find a closed-form solution $\forall$ odd integer $n\ge 3$ for the determinant of a matrix with some structure on it. After some manipulation, I've reduced it to the following matrix:
$\small\begin{bmatrix}\boldsymbol{t_{1}^{n}-t_{a}^{n}} & \boldsymbol{t_{2}^{n}-t_{a}^{n}} & \boldsymbol{\cdots} & \boldsymbol{t_{a-1}^{n}-t_{a}^{n}} & nt_{1}^{n-1} & \cdots & nt_{a-1}^{n-1} & nt_{a}^{n-1}\\
\boldsymbol{t_{1}^{n-1}-t_{a}^{n-1}} & \boldsymbol{t_{2}^{n-1}-t_{a}^{n-1}} & \boldsymbol{\cdots} & \boldsymbol{t_{a-1}^{n-1}-t_{a}^{n-1}} & (n-1)t_{1}^{n-2} & \cdots & (n-1)t_{a-1}^{n-2} & (n-1)t_{a}^{n-2}\\
\boldsymbol{\vdots} & \boldsymbol{\vdots} & \boldsymbol{\ddots} & \boldsymbol{\vdots} & \vdots & \ddots & \vdots & \vdots\\
\boldsymbol{t_{1}^{2}-t_{a}^{2}} & \boldsymbol{t_{2}^{2}-t_{a}^{2}} & \boldsymbol{\cdots} & \boldsymbol{t_{a-1}^{2}-t_{a}^{2}} & 2t_{1} & \cdots & 2t_{a-1} & 2t_{a}\\
\boldsymbol{t_{1}-t_{a}} & \boldsymbol{t_{2}-t_{a}} & \boldsymbol{\cdots} & \boldsymbol{t_{a-1}-t_{a}} & 1 & \cdots & 1 & 1
\end{bmatrix}_{n\times n}$
where $a:=\frac{n+1}{2}$, the bold block is $n\times(\frac{n+1}{2}-1)$, and the non-bold block is $n \times \frac{n+1}{2}$.
Although it has some similarities with the Vandermonde Matrix or some generalizations, it's not the same. Using some values of n, its determinant looks pretty simple, which leads me to think that there should be a closed-form solution:
$n=3$:
$$
det\left(
\left[\begin{array}{ccc} {t_{1}}^3-{t_{2}}^3 & 3\,{t_{1}}^2 & 3\,{t_{2}}^2\\ {t_{1}}^2-{t_{2}}^2 & 2\,t_{1} & 2\,t_{2}\\ t_{1}-t_{2} & 1 & 1 \end{array}\right]
\right)=
-{\left(t_{1}-t_{2}\right)}^4
$$
$n=5$:
$$
det\left(
\left[\begin{array}{ccccc} {t_{1}}^5-{t_{3}}^5 & {t_{2}}^5-{t_{3}}^5 & 5\,{t_{1}}^4 & 5\,{t_{2}}^4 & 5\,{t_{3}}^4\\ {t_{1}}^4-{t_{3}}^4 & {t_{2}}^4-{t_{3}}^4 & 4\,{t_{1}}^3 & 4\,{t_{2}}^3 & 4\,{t_{3}}^3\\ {t_{1}}^3-{t_{3}}^3 & {t_{2}}^3-{t_{3}}^3 & 3\,{t_{1}}^2 & 3\,{t_{2}}^2 & 3\,{t_{3}}^2\\ {t_{1}}^2-{t_{3}}^2 & {t_{2}}^2-{t_{3}}^2 & 2\,t_{1} & 2\,t_{2} & 2\,t_{3}\\ t_{1}-t_{3} & t_{2}-t_{3} & 1 & 1 & 1 \end{array}\right]
\right)=
-{\left(t_{1}-t_{2}\right)}^4\,{\left(t_{1}-t_{3}\right)}^4\,{\left(t_{2}-t_{3}\right)}^4
$$
I was wondering if there is a known closed-form solution for this determinant, or if it could be found using the determinant of a generalized Vandermonde matrix
Thanks!
| These seem to be a direct consequence of Schendel's 1891 theorem about "confluent Vandermonde matrices", as explained in (say) "On a Recursion Formula Related to Confluent Vandermonde",
Shui-Hung Hou and Edwin Hou, The American Mathematical Monthly,
Vol. 122, No. 8 (October 2015), pp. 766-772, or here or here. An example of a confluent Vandermonde determinant is
$$
\begin{vmatrix}x^3&3x^2&y^3&3y^2\\x^2&2x&y^2&2y\\x&1&y&1\\1&0&1&0\end{vmatrix}$$ where certain columns of an ordinary Vandermonde determinant have been replaced by derivatives of others.
Schendel's formula is that this determinant is equal to $(x-y)^4$. If you subtract the third column from the first, and then expand by minors, you end up with the identities
$$
\begin{vmatrix}x^3&3x^2&y^3&3y^2\\x^2&2x&y^2&2y\\x&1&y&1\\1&0&1&0\end{vmatrix}
=
\begin{vmatrix}x^3-y^3&3x^2&y^3&3y^2\\x^2-y^2&2x&y^2&2y\\x-y&1&y&1\\0&0&1&0\end{vmatrix}
=
-\begin{vmatrix}x^3-y^3&3x^2&3y^2\\x^2-y^2&2x&2y\\x-y&1&1\end{vmatrix}.
$$
This is the OPs $n=3$ example.
Schendel's general formula for confluent Vandermonde determinants is $\prod_{i<j}(x_i-x_j)^{n_in_j}$, where the value $x_i$ is used to form $n_i$ columns consisting of the first $n_i$ derivatives (the $0$-th up through the $n_i-1$-th derivative) of the usual Vandermonde column $(1,x_i, x_i^2,\ldots)'$. In the OP's case, all of the $n_i=2$. The ordinary Vdm formula has all $n_i=1$.
| {
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Simplifying $\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$
Simplify $$\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}\,.$$
I tried very hard but I am not being able to solve it easily I opened up everything and multiplied all of it and got the answer -2. But it took me 1 hour and I also made many silly mistakes. Is there a quicker way than brute force?
| Here we use $$ab(a-b)+bc(b-c)+ca(c-a)=a^2(b-c)+b^2(c-a)+c^2(a-b)=-(a-b)(b-c)(c-a)$$
In short we cab write $$\sum ab(a-b)=\sum a^2(b-c)=-(a-b)(b-c)(c-a)~~~~(1)$$
Then $$F=\frac{a^2+b^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-a)(b-c)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$$
$$\implies F=\frac{(a^2+b^2-a^2)(c-b)+(c^2+a^2-b^2)(a-c)+(a^2+b^2-c^2)(b-a)}{(a-b)(b-c)(c-a)}$$
upon simplification in the numerator six terms: $(\pm a^3\pm b^3 \pm c^3)$ will cancel each other, we will get $6+6=12$ terms in the numerator as
$$F=\frac{-\sum a^2(b-c)-\sum ab(a-b)}{(a-b)(b-c)(c-a)}= 2,$$
on using (1).
| {
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"answer_id": 2
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How can I evaluate $\int _0^1\frac{\tan ^{-1}\left(3\sqrt{\frac{a}{4-a}}\right)}{\sqrt{a}\sqrt{4-a}}\:\mathrm{d}a$ I have the integral:
$$\int _0^1\frac{\tan ^{-1}\left(3\sqrt{\frac{a}{4-a}}\right)}{\sqrt{a}\sqrt{4-a}}\:\mathrm{d}a.$$
If I use $u=3\sqrt{\frac{a}{4-a}}$, I get
$$6\int _0^{\sqrt{3}}\:\frac{\tan ^{-1}\left(u\right)}{u^2+9}\mathrm{d}u$$
But I cant get past this. I tried integration by parts but nothing seems to come out from that.
| Using $$ \tan^{-1}(x) = \frac{i}{2} \, \ln\left( \frac{1 - i \, x}{1 + i \, x} \right) $$ then
\begin{align}
\int \frac{\tan^{-1}(x)}{x^2 + a^2} \, dx &= \frac{i}{2} \, \left( \int \frac{\ln(1 - i \, x)}{x^2 + a^2} \, dx - \int \frac{\ln(1 + i \, x)}{x^2 + a^2} \, dx \right).
\end{align}
Using
\begin{align}
\int \frac{\ln(1 + i \, x)}{x^2 + a^2} \, dx &= \frac{i}{2 \, a} \, \left[ - \text{Li}_{2}\left(\frac{1+ i \, x}{1 - a}\right) + \text{Li}_{2}\left( \frac{1 + i \, x}{1 + a}\right) + \ln(1 + i \, x) \, \ln\left( \frac{a-1}{a+1} \cdot \frac{a - i \, x}{a + i \, x}\right) \right] \\
\int \frac{\ln(1 - i \, x)}{x^2 + a^2} \, dx
&= \frac{i}{2 \, a} \, \left[ - \text{Li}_{2}\left(\frac{1- i \, x}{1 + a}\right) + \text{Li}_{2}\left( \frac{1 - i \, x}{1 - a}\right) + \ln(1 - i \, x) \, \ln\left( \frac{a+1}{a-1} \cdot \frac{a - i \, x}{a + i \, x}\right) \right]
\end{align}
then the result can be found in closed form with:
\begin{align}
\int \frac{\tan^{-1}(x)}{x^2 + a^2} \, dx &= \frac{-1}{4 \, a} \, \left[ - \text{Li}_{2}\left(\frac{1- i \, x}{1 + a}\right) - \text{Li}_{2}\left(\frac{1 + i \, x}{1 + a}\right) + \text{Li}_{2}\left(\frac{1 - i \, x}{1 - a}\right) + \text{Li}_{2}\left(\frac{1 + i \, x}{1 - a}\right) \\+ \ln(1 + x^2) \, \ln\left(\frac{a+1}{a-1}\right) - 4 \, \tan^{-1}(x) \, \tan^{-1}\left(\frac{x}{a}\right) \right]
\end{align}
The integral with the desired limits is:
$$ \int_{0}^{\sqrt{3}} \frac{\tan^{-1}(x)}{x^2 + 3^2} \, dx = \frac{1}{6} \, \left( \frac{\zeta(2)}{3} - \frac{\ln(2) \, \ln(3)}{2} - \frac{\text{Li}_{2}(-3)}{4} \right). $$
Also:
$$ \int_{0}^{1} \tan^{-1}\left( 3 \, \sqrt{\frac{x}{4-x}} \right) \, \frac{dx}{\sqrt{x \, (4-x)}} = \frac{\zeta(2)}{3} - \frac{\ln(2) \, \ln(3)}{2} - \frac{\text{Li}_{2}(-3)}{4}. $$
| {
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} |
Finding $\sum_{r=1}^{\infty}\left(\frac{2r+5}{r^2+r}\right)\left(\frac{3}{5}\right)^{r+1}$
$\text{Find the value of}$ $$\lim_{n\to \infty}\left(\sum_{r=1}^{n}\left(\frac{2r+5}{r^2+r}\right)\left(\frac{3}{5}\right)^{r+1}\right)$$
$\text{Answer}: \frac{9}{5}$
Firstly I split "linear-upon-quadratic" term:
$$\frac{2r+5}{r^2+r}=\frac{5}{r}-\frac{3}{r+1}
\\ =\frac{2}{r}+3\left(\frac {1}{r} -\frac {1}{r+1}\right)$$
If it hadn't been for the $\left(\frac 35\right)^{r+1}$ term, the above step would have been very useful - splitting the single summation into two and summing individually.
Unfortunately, that's not the case. I'm unable to proceed further, though my gut says telescoping is the way.
Thanks in advance.
| Your idea was right:$$\sum_{r=1}^{\infty}\left(\frac{5}{r}-\frac{3}{r+1}\right)\left(\frac{3}{5}\right)^{r+1}$$
$$=\sum_{r=1}^{\infty}\left(\frac{5}{r}\right)\left(\frac{3}{5}\right)^{r+1}-\sum_{r=1}^{\infty}\left(\frac{3}{r+1}\right)\left(\frac{3}{5}\right)^{r+1}$$
$$=3\sum_{r=1}^{\infty}\left(\frac{1}{r}\right)\left(\frac{3}{5}\right)^{r}-3\sum_{r=2}^{\infty}\left(\frac{1}{r}\right)\left(\frac{3}{5}\right)^{r}$$
$$
=3\cdot \underbrace{\frac{3}{5}}_{r=1}=\frac{9}{5}
$$
| {
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} |
How to prove that $f(x)-x f'(x)\neq 0$? I have this function:
$f(x)=\cosh ^{-1}\left(\frac{4 a^2 x^2+\left(a^2 x^2-1\right)^2 \cosh (2 \pi x)}{\left(a^2 x^2+1\right)^2}\right),$
where $0<x<\frac{1}{a}$ and $a$ is a positive real number. I want to prove that $\;f(x)-x f'(x)>0$, or at least to prove that $f(x)-x f'(x)\neq 0$. Is there any way to prove this?
| I hope and wish that you will receive simpler answers.
If you compose Taylor series around $x=0$ (I skip the intermediate steps)
$$A=\frac{4 a^2 x^2+\left(a^2 x^2-1\right)^2 \cosh (2 \pi x)}{\left(a^2
x^2+1\right)^2}=1+2 \pi ^2 x^2+\left(\frac{2 \pi ^4}{3}-8 \pi ^2 a^2\right) x^4+O\left(x^6\right)$$
$$f(x)=\cosh ^{-1}(A)=2 \pi x-4 \pi a^2 x^3+O\left(x^5\right)$$
$$f(x)-x f'(x)=8 \pi a^2 x^3+O\left(x^5\right)$$
Edit
Doing the same around $x=\frac 1a$, we have
$$A=1+2 a^2 \left(x-\frac{1}{a}\right)^2 \sinh ^2\left(\frac{\pi }{a}\right)+O\left(\left(x-\frac{1}{a}\right)^3\right)$$
$$f(x)=\cosh ^{-1}(A)=-2 \left(x-\frac{1}{a}\right) \left(a \sinh \left(\frac{\pi
}{a}\right)\right)+O\left(\left(x-\frac{1}{a}\right)^3\right)$$
$$f(x)-x f'(x)=2 \sinh \left(\frac{\pi }{a}\right)+O\left(\left(x-\frac{1}{a}\right)^2\right)$$
| {
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How can i solve $\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$ How can i solve following $$\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$$
My work:
I substituted $x=\tan\theta$, $dx=\sec^2\theta d\theta $
integral becomes
$\int \dfrac{\tan^3\theta+2\tan \theta-7}{\sqrt{\tan^2\theta+1}}\ \sec^2\theta d\theta$
$\int \dfrac{\tan\theta(\tan^2\theta+1)+\tan \theta-7}{\sec\theta}\sec^2\theta d\theta$
$\int (\tan\theta(\sec^2\theta)+\tan \theta-7)\sec\theta d\theta$
$\int \tan\theta\sec^3\theta\ d\theta+\int \sec\theta \tan \theta\ d\theta-7\int \sec\theta d\theta$
$\int \tan\theta\sec^3\theta+\sec\theta -7\ln|\sec\theta+\tan\theta|+C$
I got stuck here in solving first part of above integral. I can't see the way to solve it. please help me solve it by substitution or other method. thanks
| $$\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx$$
$$=\int \frac{x(x^2+1)+x-7}{\sqrt{x^2+1}}\ dx$$
$$=\int x\sqrt{x^2+1}\ dx+\int \frac{x}{\sqrt{x^2+1}}\ dx-\int \frac{7}{\sqrt{x^2+1}}\ dx$$
$$=\frac12\int \sqrt{x^2+1}\ d(x^2+1)+\frac12\int \frac{d(x^2+1)}{\sqrt{x^2+1}}-7\int \frac{dx}{\sqrt{x^2+1}}$$
$$=\frac13(x^2+1)^{3/2}+\sqrt{x^2+1}-7\sinh^{-1}(x)+C$$
| {
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Find the approximate sum of the series $\sum_{n=0}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}$ Prove that the series converges and find the approximate sum. $$\sum_{n=0}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}$$
To prove that converges what I did is to use the Cauchy's convergence test.
$$\lim_{x\to \infty} \sqrt[n]{a_n} =\lim_{x\to \infty} \frac{1}{\sqrt[n]{3^{n}\sqrt{n+1}}} =\lim_{x\to \infty} \frac{1}{3\sqrt[2n]{n+1}} = \frac{1}{3\sqrt{\lim_{x\to \infty}\sqrt[n]{n+1}}} = \frac{1}{3\sqrt{\lim_{x\to \infty}\sqrt[n]{n(1+\frac{1}{n})}}} = \frac{1}{3\sqrt{\lim_{x\to \infty}\sqrt[n]{n}\sqrt[n]{(1+\frac{1}{n})}}} = \frac{1}{3} \lt 1$$
Since $\lim_{x\to \infty} \sqrt[n]{n} = 1$ and $\lim_{x\to \infty} \sqrt[n]{(1+\frac{1}{n})} = 1$
Then, the series converges. Now, I'm not sure how to find the approximate sum of the series since it's not an alternating series. Is there a unique method?
Thanks in advance.
| Another way to check convergence is the comparison test
$$\sum_{n=0}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}<\sum_{n=0}^{\infty} \frac{1}{3^n}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$$
This also provides a way to approximate the sum to as many places as required. Lets say you want to approximate the sum to $\epsilon$ accuracy. Then it is sufficient to find $N$ such that
$$ \sum_{n=N}^{\infty} \frac{1}{3^n}\leq \epsilon$$
as
$$\sum_{n=N}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}<\sum_{n=N}^{\infty} \frac{1}{3^n}\leq \epsilon$$
Of course, this geometric series is easily found to be
$$\sum_{n=N}^{\infty} \frac{1}{3^n}=\frac{\frac{1}{3^N}}{1-\frac{1}{3}}=\frac{2}{3^{N-1}}$$
Solving, we get
$$N\geq \left\lceil\frac{\ln(6/\epsilon)}{\ln(3)}\right\rceil$$
(as $N$ is an integer). We conclude that if $N$ is given by the equation above, then
$$\left|\sum_{n=0}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}-\sum_{n=0}^{N-1} \frac{1}{(3^n)\sqrt{n+1}}\right|=\sum_{n=N}^{\infty} \frac{1}{(3^n)\sqrt{n+1}}<\epsilon$$
To find your approximation, simply calculate
$$\sum_{n=0}^{N-1} \frac{1}{(3^n)\sqrt{n+1}}$$
manually.
| {
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Sums of powers of cosines and sines shifted by $2\pi/3$ I have stumbled across these two identities
$$
\begin{split}
\cos^2(x)+\cos^2(x+2\pi/3)+\cos^2(x+4\pi/3) &= 3/2,\\
\cos^4(x)+\cos^4(x+2\pi/3)+\cos^4(x+4\pi/3) &= 9/8.
\end{split}
$$
There is also the more intricate
$$
\begin{split}
\cos^2(x)\sin^2(x)+\cos^2(x+2\pi/3)\sin^2(x+2\pi/3)+\cos^2(x+4\pi/3)\sin^2(x+4\pi/3) &= 3/8,\\
\cos^4(x)\sin^4(x)+\cos^4(x+2\pi/3)\sin^4(x+2\pi/3)+\cos^4(x+4\pi/3)\sin^4(x+4\pi/3) &= 9/128,
\end{split}
$$
and of course the most elementary
$$
\cos(x)+\cos(x+2\pi/3)+\cos(x+4\pi/3)=0.
$$
The last identity admits a rather intuitive interpretation in terms of unitary complex numbers centered about the origin. My questions are:
*
*Do the other identities admit similar more or less intuitive interpretations as well?
*Do such identities have names?
*Not all powers and combinations produce a constant; What is the general form of the expressions that do?
Context: The first two identities came up while calculating the elastic response of a two-dimensional truss (a planar lattice of nodes connected with springs) that is invariant by rotations of order 3, in which case $x$ describes the orientation of the truss. We know that such trusses must exhibit an isotropic response and that justifies, in a rather convoluted manner, that these expressions must be constants. The other expressions I found by trial and error. I am looking for a satisfying, non-brute-force, non-too-group-theoretic, explanation.
| This is an answer to question 3.
Let
$$f_n(x):=\cos^n(x)+\cos^n\bigg(x+\frac{2\pi}{3}\bigg)+\cos^n\bigg(x+\frac{4\pi}3\bigg)$$
$$\small g_n(x):=\cos^n(x)\sin^n(x)+\cos^n\bigg(x+\frac{2\pi}3\bigg)\sin^n\bigg(x+\frac{2\pi}3\bigg)+\cos^n\bigg(x+\frac{4\pi}3\bigg)\sin^n\bigg(x+\frac{4\pi}3\bigg) $$
where $n$ is a positive integer.
This answer proves the following two claims :
Claim 1 : $f_n(x)$ is a constant function if and only if $n=1,2,4$.
Claim 2 : $g_n(x)$ is a constant function if and only if $n=1,2,4$.
Claim 1 : $f_n(x)$ is a constant function if and only if $n=1,2,4$
Proof :
You already noticed that $f_1(n),f_2(n)$ and $f_4(n)$ are constant functions.
Now, let us prove that if $f_n(x)$ is a constant function, then $n=1,2,4$ as follows :
$$\begin{align}&\text{$f_n(x)$ is a constant function}
\\\\&\implies f_n(0)=f_n\bigg(\frac{\pi}{6}\bigg)
\\\\&\implies 1+\bigg(-\frac 12\bigg)^n+\bigg(-\frac 12\bigg)^n=\bigg(\frac{\sqrt 3}{2}\bigg)^n+\bigg(-\frac{\sqrt 3}{2}\bigg)^n+0
\\\\&\implies 2^n+2(-1)^n-(\sqrt 3)^n-(-\sqrt 3)^n=0
\\\\&\implies \begin{cases}2^n-2=0&\text{if $n$ is odd}\\\\2(\sqrt 3)^{n-1}\bigg(\bigg(\frac{2}{\sqrt 3}\bigg)^{n-1}-\sqrt 3\bigg)+2=0&\text{if $n$ is even}\end{cases}
\\\\&\implies n=1,2,4\end{align}$$
since for odd $n$ , we have $2^n-2=0\implies n=1$, and for even $n$, letting $h(n):=2(\sqrt 3)^{n-1}\bigg(\bigg(\frac{2}{\sqrt 3}\bigg)^{n-1}-\sqrt 3\bigg)+2$, we see that $h(2)=h(4)=0$ and that $h(n)$ is increasing for $n\ge 6$ with $h(6)=12$.
Claim 2 : $g_n(x)$ is a constant function if and only if $n=1,2,4$.
Proof :
You already noticed that $g_2(n)$ and $g_4(n)$ are constant functions. We have $g_1(n)=0$.
Now, let us prove that if $g_n(x)$ is a constant function, then $n=1,2,4$ as follows :
$$\small\begin{align}&\text{$g_n(x)$ is a constant function}
\\\\&\implies g_n(0)=g_n\bigg(\frac{\pi}{4}\bigg)
\\\\&\implies 0+\bigg(-\frac 12\bigg)^n\bigg(\frac{\sqrt 3}{2}\bigg)^n+\bigg(-\frac 12\bigg)^n\bigg(\frac{-\sqrt 3}{2}\bigg)^n\\&\qquad\qquad =\bigg(\frac{1}{\sqrt 2}\bigg)^n\bigg(\frac{1}{\sqrt 2}\bigg)^n+\bigg(-\frac{1+\sqrt 3}{2\sqrt 2}\bigg)^n\bigg(\frac{\sqrt 3-1}{2\sqrt 2}\bigg)^n+\bigg(\frac{\sqrt 3-1}{2\sqrt 2}\bigg)^n\bigg(-\frac{1+\sqrt 3}{2\sqrt 2}\bigg)^n
\\\\&\implies 2^n+2(-1)^n-(\sqrt 3)^n-(-\sqrt 3)^n=0
\\\\&\implies n=1,2,4\end{align}$$
where the last step is the same as that of the proof for claim 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Evaluating limit of the function at $\frac{\pi}{2}$ I'm trying to solve this
$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}}{(x-\frac{\pi}{2})^3}$$
I have tried using the L'Hôpital's rule
But I'm stuck at
$$\lim_{x \to \frac{\pi}{2}} \frac{-\sin{x}}{3(x-\frac{\pi}{2})^2}$$
Since the above equation is not in the $\frac{0}{0}$ , $\frac{\infty}{\infty}$
or $\frac{anything}{\infty}$ form
Then I tried expanding the $\cos{x}$ as taylor series at $x=\frac{\pi}{2}$. Which on simplifying I am left with
$$ \lim_{x \to \frac{\pi}{2}} \frac{-1}{(x - \frac{\pi}{2})^2} + \frac{1}{6} - \frac{1}{120} (x - \frac{\pi}{2})^2 + \frac{(x - \frac{\pi}{2})^4}{5040} - ... $$
and I'm stuck again. How do I proceed ahead?
| welcome to MSE
AS a hint
$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}}{(x-\frac{\pi}{2})^3}=\lim_{x \to \frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)}{(x-\frac{\pi}{2})^3}$$now take $x-\frac{\pi}{2}=a $ when $x$ tends to $\frac{\pi}{2}$ ,a tends to zero
$$\lim_{x \to \frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)}{(x-\frac{\pi}{2})^3}=\\
\lim_{a\to 0} \frac{\sin(-a)}{(a)^3}\\=
\lim_{a\to 0} \frac{-\sin(a)}{(a)^3}\\=
\lim_{a\to 0} \frac{-1}{(a)^2}\to -\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the error terms in the asymptotic expansion of a binomial coefficient are uniformly bounded I'm interested in the asymptotic behavior of
$$ a_{n,k} :=\frac{1}{2^n} \binom{n}{\frac{n}{2} +k},$$
where $k$ is within a constant times $\sqrt{n}$, say
$$ |k| < C \sqrt{n }.$$
Using Stirling's approximation, I found
$$ a_{n,k} = \sqrt{\frac{2}{\pi n}} e^{-\frac{2k^2}{n}}\left(1 + O\left( \frac{1}{\sqrt{n}} \right)\right).$$
In each $a_{n,k}$, let us call the relative error term $b_{n,k}$, so
$$ a_{n,k} = \sqrt{\frac{2}{\pi n}} e^{-\frac{2k^2}{n}}\left(1 + b_{n,k}\right).$$
My Question: How do I show that the $\{b_{n,k}\}$ are all bounded by the same constant divided by $\sqrt{n}$.
Clarification: I'd like to claim: there's a constant $K>0$, and an index $n_0\in \mathbb{N}$, such that for all $n>n_0$, and all $k$ satisfying $|k|< C\sqrt{n}$,
$$ |b_{n,k}| < \frac{K}{\sqrt{n}} .$$
| The desired uniform bound follows from essentially the same computation that you might have used to obtain the formula for $a_{n,k}$.
Indeed, the Stirling's approximation tells that
\begin{align}
n! = \sqrt{2\pi} \, n^{n+\frac{1}{2}} e^{-n} \left( 1 + \mathcal{O}\left(n^{-1}\right) \right),
\end{align}
where the implicit bound is absolute. Next, fix $C > 0$. Then uniformly in $|k| \leq C\sqrt{n}$, we have
\begin{align}
\left(\tfrac{n}{2}\pm k\right)!\
&= \sqrt{2\pi} \, \left(\tfrac{n}{2}\pm k\right)^{\frac{n}{2}\pm k+\frac{1}{2}} e^{-(\frac{n}{2}\pm k)} \left( 1 + \mathcal{O}\left((\tfrac{n}{2}\pm k\right)^{-1}) \right) \\
&= \sqrt{2\pi} \, \left(\tfrac{n}{2}\pm k\right)^{\frac{n}{2}\pm k+\frac{1}{2}} e^{-(\frac{n}{2}\pm k)} \left( 1 + \mathcal{O}_C\left(n^{-1}\right) \right),
\end{align}
where the implicit bound for $\mathcal{O}_C\left(n^{-1}\right)$ depends only on $C$. Plugging this to $a_{n,k}$,
$$ a_{n,k} = \sqrt{\frac{2}{\pi n}} \left(1+\frac{2k}{n}\right)^{-(\frac{n}{2} + k+\frac{1}{2})} \left(1 - \frac{2k}{n}\right)^{-(\frac{n}{2}- k+\frac{1}{2})} \left( 1 + \mathcal{O}_C(n^{-1}) \right). $$
Now, by the Taylor's theorem, we have
$$ \log\left(1 \pm \frac{2k}{n} \right) = \pm\frac{2k}{n} - \frac{2k^2}{n^2} + \mathcal{O}_C(n^{-3/2}). $$
Plugging this back shows that
$$ a_{n,k} = \sqrt{\frac{2}{\pi n}} e^{-\frac{2k^2}{n}} \left( 1 + \mathcal{O}_C(n^{-1/2}) \right) $$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Simplify $\sqrt{8-\sqrt{63}}$ I simplified the expression into $$\sqrt{8-3\cdot \sqrt{7}}$$ but my tutor said it wasn't the answer he was looking for. Can someone help me?
| Note that $63=9 \times 7$ and $8=\frac{1}{2}(9+7)$. Therefore,
$$ 9+7-2\sqrt{9 \times 7} = (\sqrt{9}-\sqrt{7})^2, $$
so that
$$ 8-\sqrt{63} = \frac{1}{2}(16-2\sqrt{63}) = \frac{1}{2}(3-\sqrt{7})^2 $$
and
$$ \sqrt{8-\sqrt{63}} = \frac{3-\sqrt{7}}{\sqrt{2}} = \frac{3\sqrt{2}-\sqrt{14}}{2}. \quad \blacksquare $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integrate: $\int \frac{x}{\left(x^2-4x-13\right)^2}dx$.
Integrate:
$$\int \frac{x}{\left(x^2-4x-13\right)^2}dx$$
Here's my attempt:
I first completed the squares for the denominator:
$$\left(x^2-4x-13\right)^2=(x-2)^2-17 \implies \int \frac{x}{\left(\left(x-2\right)^2-17\right)^2}dx$$
I then used $u$-subsituition:
$$u=x-2 \implies \int \frac{u+2}{\left(u^2-17\right)^2}du = \int \frac{u}{\left(u^2-17\right)^2}du+\int \frac{2}{\left(u^2-17\right)^2}du$$
The first part of the new integral is quite simple:
$$\int \frac{u}{\left(u^2-17\right)^2}du=\frac{-1}{2(u^2-17)}$$
Then I did the second part:
$$\int \frac{2}{\left(u^2-17\right)^2}du = -\frac{1}{2\left(u^2-17\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|u+\sqrt{17}\right|-\frac{1}{68\left(u+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|u-\sqrt{17}\right|-\frac{1}{68\left(u-\sqrt{17}\right)}\right) = -\frac{1}{2\left(\left(x-2\right)^2-17\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|x-2+\sqrt{17}\right|-\frac{1}{68\left(x-2+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|x-2-\sqrt{17}\right|-\frac{1}{68\left(x-2-\sqrt{17}\right)}\right) = -\frac{1}{2\left(x^2-4x-13\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|x-2+\sqrt{17}\right|-\frac{1}{68\left(x-2+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|x-2-\sqrt{17}\right|-\frac{1}{68\left(x-2-\sqrt{17}\right)}\right) + C, C \in \mathbb{R}$$
Is this working out correct? I'm not really sure how WolframAlpha works, so I didn't check it on there.
| Express the second integral as
\begin{align}
\int \frac{2}{\left(u^2-17\right)^2}du
&= -\frac2{17}\int \frac{du}{u^2-17} +\frac2{17}\int \frac{u^2du}{(u^2-17)^2}
\end{align}
with
\begin{align}
\int \frac{u^2du}{(u^2-17)^2}
= -\frac12\int u d\left(\frac1{u^2-17} \right)
= -\frac u{2(u^2-17)}+\frac12\int \frac {ud u}{u^2-17}
\end{align}
Then, combining with the first integral to get
\begin{align}
\int \frac{(u+2)du}{\left(u^2-17\right)^2}=
\frac {-u}{17(u^2-17)} -\frac2{17}\int \frac{du}{u^2-17}
& +\frac1{17}\int \frac {ud u}{u^2-17} + \int \frac{udu }{\left(u^2-17\right)^2}
\end{align}
where each piece can be integrated steadily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Sum of Continued Fractions Let $x$ be a positive integer.
Consider the following sum (maybe there is a better notation with continued fractions, but I am not aware of it):
$\frac{1}{x} + \frac{1}{x- \frac{1}{x}} + \frac{1}{x-\frac{1}{x}-\frac{1}{x-\frac{1}{x}}} +\frac{1}{x-\frac{1}{x}-\frac{1}{x-\frac{1}{x}} - \frac{1}{x- \frac{1}{x} - \frac{1}{x-\frac{1}{x}}}} + \dots $
My question is: How many summands do I need such that the sum is at least $x$? $x^2$ is a trivial upper bound here, because every summand is $\geq \frac{1}{x}$. Is there a way to get a better bound here?
| If $\{a_i\}$ is the sequence of terms, and $s_n$ is the sum of first $n$ terms, then we get $$a_{n+1} = \frac{1}{x-s_{n}}$$ for $n\geq1$ and $a_1= \frac{1}{x}$. Rearranging this, $s_n = x-\frac{1}{a_{n+1}}$. Each term is $\geq 1/x$, so their reciprocals are $\leq x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if the roots of $x^3+ax^2+bx+c=0$ form an arithmetic sequence, then $2a^3+27c=9ab$
Prove that if the roots of $x^3+ax^2+bx+c=0$ form an arithmetic sequence, then $$2a^3+27c=9ab$$
So far, I let the roots of $x^3+ax^2+bx+c=0$ be $r_1, r_2,$ and $r_3$. $r_1=r_2-d$ and $r_3=r_2+d$ because they form an arithmetic sequence with $d$ being the difference. the sum of the roots is $-a$. So, $r_2=-a/3$. We can let the product of the roots be $-c$. So, $(r_2-d)(r_2)(r_2+d)=-c$. Plugging in $r_2=-a/3$ we get $(-a/3-d)(-a/3)(-a/3+d)$. How do I continue with this method?
EDIT: I used hamam_abdallah's hint to get $\frac{-a^3}{27} + \frac{ad^2}{3} = -c$ what do i do after applying vieta's formulas?
| Let's do this thing
$(r+d)^3 +a(r+d)^2 +b(r+d) + c =(r-d)^2 + a(r-d)^2 + b(r-d) + c = r^3 + ar^2 +br + c = 0$ (read the hidden for the insight as to how I thought to do the next step... or just go on to the next step)
Insight: Now the standard trick is to note that the coefficients of the even powers of $d$ in $(r+d)^3 +a(r+d)^2 +b(r+d) + c$ and in $(r-d)^2 + a(r-d)^2 + b(r-d) + c $ are the same and the coefficients of the odd powers of $d$ are equal be opposite signs. Yet they both sum to the same value. So the sum of the coefficients of the odd powers add to $0$. That is:
so $\frac {[(r+d)^3 +a(r+d)^2 +b(r+d) + c]-[(r-d)^2 + a(r-d)^2 + b(r-d) + c]}2=0$ so
$3r^2d + d^3 + 2ard + bd = 0$
Insight: Now we have $(r+d)^3 +a(r+d)^2 +b(r+d) + c = r^3 + ar^2 + br +c$ and this is a little less of a standard trick but that means the sum of the non-negative powers of $d$ add to zero. And as we know the sum of the odd powers add to zero, that means the sum of the even powers add to zero as well. That is:
And $ [(r+d)^3 +a(r+d)^2 +b(r+d) + c]-[3r^2d + d^3 + 2ard + bd]-[r^3 + ar^2 +br + c]=0$ so
$3rd^2 + ad^2= 0$
If $d$ is the incremental of an arithmetical sequence then $d$, presumably does not equal $0$. Otherwise the "sequence" is constant. Which.... technically is an arithmetic sequence let's put a pin in that...
and if we assume $d\ne 0$
so $r=-\frac a3$.... ooookay.... I wasn't expecting that....
$-\frac {a^3}{27} + a\frac {a^2}9 -b\frac a3 + c = \frac {2a^3}{27}-\frac {ab}3 + c =0$ so
$2a^3 +27c = 9ab$.
.....
And if $d = 0$ then $r+d = r=r-d$ and $r$ is a triple root.
So $x^3 + ax^2 + bx + c = (x-r)^3$
So $a = -3r$ and $b=3r^2$ and $c= -r^3$. So $2a^3 + 27c = -2*27r^3 - 27r^3 = -81r^3$. And $9ab= -9*3r*3r^2 =-81r^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find coordinates of a point Q on the graph $\sin (x) + \cos (y) = 0.5$ given that the gradient of its tangent is perpendicular to point P. Note:
Point $P$ is on the $y$-axis and above the $x$-axis
$\frac{-\pi}{6}\le x \le\frac{7\pi}{6}$
$\frac{-2\pi}{3}\le y\le\frac{2\pi}{3}$
What I have done so far:
Solving for $P$:
$$x = 0
\\ \sin (0) + \cos (y) = 0.5
\\ 0 + \cos (y) = 0.5
\\ y= \pm\frac{\pi}{3} $$
For $P$, $y \gt 0$
$\therefore y = \frac{\pi}{3}$
Solving for $\frac{dy}{dx}$:
$$\sin(x) + \cos(y) = 0.5
\\ \cos(x) - \sin(y)\frac{dy}{dx} = 0$$
$\therefore \frac{dy}{dx} = \frac{\cos(x)}{\sin (y)}$
Derivative at $P$:
$$\frac{dy}{dx} = \frac{\cos(x)}{\sin (y)}
= \frac{\cos(0)}{\sin(\frac{\pi}{3})}
= \frac{2}{\sqrt3}$$
As for the gradient of the tangent line at $Q$ is perpendicular to that at $P$:
$\frac{dy}{dx} = \frac{-\sqrt3}{2}$
How do I solve for the coordinates of $Q$ after this?
| Note that the curve (one of an infinite number of such "loops" satisfying the given equation) is symmetrical about the point $ \ \left(\frac{\pi}{2} \ , \ 0 \right) \ ; $ this is a consequence of the trigonometric function properties $ \ \cos (-y) \ = \ \cos y \ $ and $ \ \sin \left(\frac{\pi}{2} + \xi \right) \ = \ \sin \left(\frac{\pi}{2} - \xi \right) \ \ . $ If we use this latter symmetry to define a coordinate $ \ x \ = \ \frac{\pi}{2} + \xi \ \Rightarrow \ \xi \ = \ x - \frac{\pi}{2} \ \ , $ then the curve equation becomes $ \ \sin \left(\frac{\pi}{2} + \xi \right) + \cos y \ = \ 0.5 \ $ over the intervals $ \ \frac{-2\pi}{3} \ \le \ \xi \ , \ y \ \le \ \frac{2\pi}{3} \ \ . $ The slope of the tangent line at $ \ P(x,y) \ $ is then
$$ \frac{dy}{dx}|_P \ \ = \ \ \frac{dy}{d \xi}|_P \ \ = \ \ \frac{\cos\left(\frac{\pi}{2} + \xi_P \right)}{\sin (y_P)} $$
and the slope of the normal line (which is equal to the slope of the tangent line at $ \ Q \ $ ) is
$$ \frac{dy}{dx}|_Q \ \ = \ \ -\frac{\sin (y_P)}{\cos\left(\frac{\pi}{2} + \xi_P \right)} \ \ = \ \ \frac{\cos\left(\frac{\pi}{2} + \xi_Q \right)}{\sin (y_Q)} \ \ , $$
The "angle-addition" formula for cosine yields $ \ \cos\left(\frac{\pi}{2} + \xi_Q \right) \ = \ \cos\left(\frac{\pi}{2} \right)·\cos \xi_Q \ - \ \sin\left(\frac{\pi}{2}\right) ·\sin \xi_Q $ $ = \ -\sin \xi_Q \ \ , $ and similarly, $ \ \cos\left(\frac{\pi}{2} + y_Q \right) \ = \ -\sin y_Q \ \ . $ Thus, we may write
$$ \frac{dy}{dx}|_Q \ \ = \ \ \frac{-\sin \xi_Q}{-\cos\left(\frac{\pi}{2} + y_Q \right)} \ \ = \ \ \frac{\sin \xi_Q}{\cos\left(\frac{\pi}{2} + y_Q \right)} \ \ = \ \ -\frac{\sin (y_P)}{\cos\left(\frac{\pi}{2} + \xi_P \right)} \ \ , $$
which now suggests how to make correspondences between coordinates.
You identified $ \ P \ $ as one of the $ \ y-$intercepts $ \ (x_P \ = \ 0 \Rightarrow \ \xi_P \ = \ -\frac{\pi}{2} ) \ $ of the curve, which has $ \ \cos(y_P) \ = \ 0.5 \ \Rightarrow \ y_P \ = \ \frac{\pi}{3} \ \ . $ We may then take
$$ \sin(y_P) \ \ = \ \ \frac{\sqrt3}{2} \ \ = \ \ \sin(\xi_Q) \ \ \ , \ \ \ \cos\left(\frac{\pi}{2} + \xi_P \right) \ \ = \ \ 1 \ \ = \ \ -\cos\left(\frac{\pi}{2} + y_Q \right) \ \ , $$
or
$$ \sin(y_P) \ \ = \ \ \frac{\sqrt3}{2} \ \ = \ \ -\sin(\xi_Q) \ \ \ , \ \ \ \cos\left(\frac{\pi}{2} + \xi_P \right) \ \ = \ \ 1 \ \ = \ \ \cos\left(\frac{\pi}{2} + y_Q \right) \ \ . $$
The first pair of equations gives us $ \ \sin(\xi_Q) \ = \ \frac{\sqrt3}{2} \ \Rightarrow \ \xi_Q \ = \ \frac{\pi}{3} \ , \ \frac{2\pi}{3} \ \Rightarrow \ x_Q \ = \ \frac{5\pi}{6} \ , \ \frac{7\pi}{6} \ $ and $ \ \cos\left(\frac{\pi}{2} + y_Q \right) \ = \ -1 \ \Rightarrow \ \frac{\pi}{2} + y_Q \ = \ \pi \ \Rightarrow \ y_Q \ = \ \frac{\pi}{2} \ \ ; $ of these two points, $ \ \sin \left( \frac{5\pi}{6} \right) \ + \ \cos \left( \frac{\pi}{2} \right) \ = \ 0.5 \ $ satisfies the curve equation. On the other hand, the second pair produces $ \ \sin(\xi_Q) \ = \ -\frac{\sqrt3}{2} $ $ \Rightarrow \ \xi_Q \ = \ -\frac{\pi}{3} \ , \ -\frac{2\pi}{3} \ \Rightarrow \ x_Q \ = \ \frac{\pi}{6} \ , \ -\frac{\pi}{6} \ $ and $ \ \cos\left(\frac{\pi}{2} + y_Q \right) \ = \ 1 \ \Rightarrow \ \frac{\pi}{2} + y_Q \ = \ 0 \ \Rightarrow \ y_Q \ = \ -\frac{\pi}{2} \ \ , $ for which only $ \ \sin \left( \frac{\pi}{6} \right) \ + \ \cos \left( -\frac{\pi}{2} \right) \ = \ 0.5 \ $ works in the equation.
This tells us that the location of point $ \ Q \ $ is $ \ \left(\frac{5\pi}{6} \ , \ \frac{\pi}{2} \right) \ \ . $ We have also found, as we would expect from the symmetry of this loop about $ \ \left(\frac{\pi}{2} \ , \ 0 \right) \ , $ that there is a second point $ \ Q' \ \left(\frac{\pi}{6} \ , \ \frac{-\pi}{2} \right) \ $ at which the tangent line is perpendicular to the tangent line at $ \ P \ \ . $
By a related argument, we can also show that there is a point $ \ P' \ \left( \pi \ , \ \frac{-\pi}{3} \right) \ $ with a tangent line parallel to the one at $ \ P \ \ . $ Naturally, because the full "curve" described by the equation covers the plane with "loops" with periodicity $ \ 2 \pi \ $ in the $ \ x-$ and $ \ y-$directions, there are "families" of points $ \ \mathcal{P} \ \left(2 m \pi \ , \ \frac{\pi}{3} + 2 n \pi \right) \ \ $ and $ \ \mathcal{Q} \ \left(\frac{5\pi}{6} + 2 m \pi \ , \ \frac{\pi}{2} + 2 n \pi \right) \ \ , \ m \ \ \text{and} \ \ n \ $ being integers, which have the prescribed relationship.
| {
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"url": "https://math.stackexchange.com/questions/3761928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Definite integral of $\int_{-2}^{2} \frac{5}{(x^2+4)^2}\,dx$ using the substitution of $x=2\tanθ$. Can someone help with the integral $\int_{-2}^{2} \frac{5}{(x^2+4)^2}\,dx$?
I'm supposed to find the definite integral for this using the substitution $x=2\tanθ$.
This is what I've done so far:
$$\longrightarrow \frac{dx}{dθ}=2\sec^2θ$$
$$\longrightarrow x=2 \rightarrow θ=\frac{\pi}{4}$$
$$\longrightarrow x=-2 \rightarrow θ=-\frac{\pi}{4}$$
$$\therefore \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{10\sec^2θ}{(4\tan^2θ+4)^2}\,dθ$$
Using; $$t=\tan^2θ+1,$$
$$=\int_{2}^{2} \frac{10}{32t^2\sqrt{t-1}}\,dt$$
After this, I don't know how to finish. Does anyone know how to finish?
*Note: The first substitution is the one the exercise is telling me to use, the other is one I used myself.
| Alternate easier method: use reduction formula:
$\color{blue}{\int \frac{dt}{(t^2+a^2)^n}=\frac{t}{2(n-1)a^2(t^2+a^2)^{n-1}}+\frac{2n-3}{2(n-1)a^2}\int \frac{dt}{(t^2+a^2)^{n-1}}}$ as follows
$$\int_{-2}^2 \dfrac{5}{(x^2+4)^2}dx$$$$=10\int_{0}^2 \dfrac{dx}{(x^2+2^2)^2}$$
$$=10\left[\frac{x}{2(2-1) 2^2(x^2+2^2)}+\frac{2\cdot 2-3}{2(2-1)2^2}\int \frac{dx}{x^2+2^2}\right]_0^2$$
$$=10\left[\frac{x}{8(x^2+4)}+\frac{1}{8}\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)\right]_0^2$$
$$=10\left[\frac{2}{64}+\frac{1}{16}\frac{\pi}{4}-0\right]$$
$$=\bbox[15px,#ffd,border:1px solid green]{\frac{5(\pi+2)}{32}}$$
| {
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"url": "https://math.stackexchange.com/questions/3762400",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Proof that $\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0$ How should I prove
$$\forall n\in\mathbb{N}:\, \frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0?$$
My attempt:
$$\begin{align}\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-n\pi i-\frac{\pi i}{2}+\frac{\pi i}{4}}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-i\left(n\pi +\frac{\pi}{4}\right)}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1}{\sqrt{2}}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}+\frac{1}{\cos\left(n\pi +\frac{\pi}{4}\right)+i\sin\left(n\pi +\frac{\pi}{4}\right)}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{\sqrt{2}}\frac{1}{\frac{1}{\sqrt{2}}(\cos n\pi -\sin n\pi )+\frac{i}{\sqrt{2}}(\cos n\pi +\sin n\pi)}-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{\cos n\pi +i\cos n\pi}-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{i^{2n}+i^{2n+1}}-i^{n^2}\end{align}$$
How should I proceed?
| To proceed:
$$\begin{align}
\frac{1+i}{2}+\frac{1}{i^{2n}+i^{2n+1}}-i^{n^2}&=\frac{1+i}{2}+\frac{1}{i^{2n}(1+i)}-i^{n^2}\\
&=\frac{1+i}{2}+\frac{(-1)^n}{1+i}\frac{1-i}{1-i}-i^{n^2}\\
&=\frac{1+i}{2}+\frac{(-1)^n(1-i)}{2}-i^{n^2}\\
&=\frac{1+(-1)^n}{2}+i\frac{1-(-1)^n}{2}-i^{n^2}
\end{align}$$
For $n$ even, $4\mid n^2$ and so $i^{n^2}=1$, and for $n$ odd, $4\mid n^2-1$ and so $i^{n^2}=i\cdot i^{n^2-1}=i$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a, b, c\in\mathbb R^+,$ then prove that: $\sum_{cyc} \frac1{c-b}\left(\frac1{\sqrt{a+2b}}-\frac1{\sqrt{a+2c}}\right)\ge\frac3{\sqrt{(a+b+c)^3}}.$ We have:$$\sum_{cyc}\frac1{c-b}\left(\frac1{\sqrt{a+2b}}-\frac1{\sqrt{a+2c}}\right)\\\\=\sum_{cyc}\frac{\sqrt{a+2c}-\sqrt{a+2b}}{(c-b)\sqrt{a+2b}\sqrt{a+2c}}\\\\=\sum_{cyc}\frac{2(c-b)}{(c-b)\sqrt{a+2b}\sqrt{a+2c}\left(\sqrt{a+2c}+\sqrt{a+2b}\right)}\\\\=\underbrace{\sum_{cyc}\frac2{(a+2c)\sqrt{a+2b}+(a+2b)\sqrt{a+2c}}}_{=E\text{ (say)}}.$$ So we need to show that: $E\ge\frac3{\sqrt{(a+b+c)^3}}.$
I tried AM $\ge$ HM and obtained: $$E\ge\frac{18}{\sum_{cyc}\left\{(a+2c)\sqrt{a+2b}+(a+2b)\sqrt{a+2c}\right\}}.$$ But now I'm confused. How to tackle it further!?
Please suggest me what to do next.. Thanks in advance.
| Now, by AM-GM and Jensen we obtain:
$$\sum_{cyc}\frac{2}{\sqrt{(a+2b)(a+2c)}(\sqrt{a+2b}+\sqrt{a+2c})}\geq$$
$$\geq\sum_{cyc}\frac{2}{\frac{a+2b+a+2c}{2}\cdot\sqrt{2(a+2b+a+2c)}}=\frac{3}{\sqrt{(a+b+c)^3}}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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plot of $\sin(x) + \sin(y)= \cos(x) + \cos(y)$ I was playing arround with implicit plots of the form $f(x,y) = g(x,y)$, and I noticed that if you plot in the plane the following equation: $\sin(x) + \sin(y)= \cos(x) + \cos(y)$ you get the following graph:
My question is why does this trigonometric functions give us this squares spanning the entire plane?
| Using Prosthaphaeresis Formulas
$$2\sin\dfrac{x+y}2\cos\dfrac{x-y}2=\cos\dfrac{x+y}2\cos\dfrac{x-y}2$$
If $\cos\dfrac{x-y}2=0\implies\dfrac{x-y}2=(2n+1)\dfrac\pi2, x-y=(2n+1)\pi$
else $\sin\dfrac{x+y}2=\cos\dfrac{x+y}2\iff\tan\dfrac{x+y}2=1\implies\dfrac{x+y}2=m\pi+\dfrac\pi4\iff x+y=\dfrac{(4m+1)\pi}2$
So we are getting continuous perpendicular & equidistant straight lines.
In the first case, the distance between two consecutive lines is $$\dfrac{2(m+1)+1-(2m+1)}{\sqrt2}\cdot\pi$$
and in the second, $$\dfrac{2\pi}{\sqrt2}$$
So, we get infinite number of squares with each side $=\sqrt2\pi$
| {
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"timestamp": "2023-03-29T00:00:00",
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On Resolving a Fuss on Squares and Fractions over a few Inequalities Firstly, only AM-GM and C-S are to be sought.
$1.$Let $a, b, c, d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Show that
$${a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge 4$$
In my textbook, this problem is credited to Michael Rozenberg, but I couldn't find a solution to it by myself or on this site, so decided to ask.
I tried my best with fallacy-
$$4(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2 \Rightarrow 4 \ge a+b+c+d$$ with the constraint given. Proceeding-
$$(a+b+c+d)\left({a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a}\right) \ge (a+b+c+d)^2 $$
$$\Rightarrow {a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge a+b+c+d$$
That means, I need $a+b+c+d\ge 4$ to complete the proof but instead got $4 \ge a+b+c+d$ !
$2.$Let $a, b, c$ be positive real numbers such that $abc = 1$.
Show that
$${1\over b(a+b)}+{1\over c(b+c)}+{1\over a(c+a)}\ge \frac{3}{2}.$$
$3.$If a, b, c and d are positive real numbers such that $a + b + c + d = 4$. Prove that
$$ {a \over 1+b^2c}+{b \over 1+c^2d}+{c \over 1+d^2a}+{d \over 1+a^2b} \ge 2. $$
The #2 is credited to the Zhautykov Olympiad 2008 and there is no need for revealing my attempts as I've no idea what to do, Lastly,
This is a doubt, not an question that definitely has an answer, but if the doubt comes out to have an answer, nevertheless, this is a problem.
$4.$ Prove without Induction:
$$\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2} + ...+\sqrt{a_n^2+b_n^2} \ge \sqrt{(a_1+a_2+...+a_n)^2+(b_1+b_2+...+b_n)^2}$$
| The first problem.
By Holder $$\left(\sum_{cyc}\frac{a^2}{b}\right)^2\sum_{cyc}a^2b^2\geq\left(\sum_{cyc}a^2\right)^3.$$
Thus, it's enough to prove that $$\sum_{cyc}a^2b^2\leq4,$$ which is true by AM-GM:
$$\sum_{cyc}a^2b^2=(a^2+c^2)(b^2+d^2)\leq\left(\frac{a^2+c^2+b^2+d^2}{2}\right)^2=4.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove that $\binom{k+1}{1} S_k(n)+\binom{k+1}{2} S_{k-1}(n)+.........+\binom{k+1}{k+1} S_0(n)=(n+1)^{k+1}-1$ For $n,k \in N$,we define
$S_k(n)=1^k +2^k+... +n^k$
if $(1+x)^p=1+\binom{p}{1}x+\binom{p}{2}x^2+.... +\binom{p}{p}x^p,p\in N$,then we have to prove that
$\binom{k+1}{1} S_k(n)+\binom{k+1}{2} S_{k-1}(n)+...+\binom{k+1}{k+1} S_0(n)=(n+1)^{k+1}-1$
this looks similar to $\binom{k+1}{1}n+\binom{k+1}{2}n^2+...+\binom{k+1}{k+1}n^{k+1}= (1+n)^{k+1}-\binom{k+1}{0}=(1+n)^{k+1}-1$
what next?
| By binomial theorem:
$$(1+x)^{m+1}-x^m=1+{n \choose 1}(x)+{n \choose 2} (x^2)+....+{m+1 \choose m} (x^m)$$
putting $x=1,2,3,,..n$ and adding them, we get
$$(n+1)^{n+1}-1=(1+1+1...+1)+{m+1 \choose 1}(1+2+3+...+n)+{m+1 \choose 2}(1^2+2^2+3^2++...+n^2)+{m+1 \choose 3} (1^3+2^3+3^3+4^3+...+n^3)+.....+{m+1 \choose m}(1^m+2^m+3^m+4^m+...+n^m)$$
$$\implies \sum_{k=0}^{m} S_k {m+1 \choose k}=(n+1)^{m+1}-1,~~ S_k=1^k+2^k+3^k+4^k+...n^k$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Challenging Problem: $\int_{-\infty} ^{\infty} \frac{x \sin{3x} }{x^4 +1}dx =\pi^a e^{\frac{-b}{\sqrt{c}}}\sin \big({\frac {d}{\sqrt{e}}}\big)$ $$ \int_{-\infty} ^{\infty} \frac{x \sin{3x} }{x^4 +1}dx =\pi^a e^{\frac{-b}{\sqrt{c}}}\sin \big({\frac {d}{\sqrt{e}}}\big) $$ where $a,b,c,d,e$ are positive integers and $c$ and $e$ are square free numbers. Find $a+b+c+d+e$
My Attempt
$$ \int_{-\infty} ^{\infty} \frac{x \sin{3x} }{x^4 +1}dx =
\int_{-\infty} ^{\infty} \frac{16x \sin{3x} }{(2x-\sqrt{2}i-\sqrt{2})(2x-\sqrt{2}i+\sqrt{2})(2x+\sqrt{2}i-\sqrt{2})(2x+\sqrt{2}i+\sqrt{2})}dx =$$
by partial fractions
$$=\int -\frac {i\sin(3x)}{2 \big(2x+\sqrt{2}+\sqrt{-2}\big) } +\frac {i\sin(3x)}{2 \big(2x+\sqrt{2}-\sqrt{-2}\big) }+\frac {i\sin(3x)}{2 \big(2x-\sqrt{2}+\sqrt{-2}\big) }-\frac {i\sin(3x)}{2 \big(2x-\sqrt{2}-\sqrt{-2}\big) }dx=\frac{i}{2}\int \frac {\sin(3x)}{ \big(2x+\sqrt{2}+\sqrt{-2}\big) }dx +\frac{i}{2}\int\frac {\sin(3x)}{ \big(2x+\sqrt{2}-\sqrt{-2}\big) }dx+\frac{i}{2}\int\frac {\sin(3x)}{ \big(2x-\sqrt{2}+\sqrt{-2}\big) }dx-\frac{i}{2}\int\frac {\sin(3x)}{ \big(2x-\sqrt{2}-\sqrt{-2}\big) }dx$$
Solving
$$\int \frac {\sin(3x)}{ \big(2x+\sqrt{2}+\sqrt{-2}\big) }dx$$ Substitute $u=2x+\sqrt{2}+\sqrt{-2}\longrightarrow \frac{du}{dx}=2$
$$\Rightarrow \int \frac {\sin(3x)}{ \big(2x+\sqrt{2}+\sqrt{-2}\big) }=\frac{1}{2} \int \frac {\sin(\frac{3u}{2}-\frac{3i}{\sqrt{2}}-\frac{3}{\sqrt{2}})}{ u }du$$
applying the addition formula
$$=\int\frac {\cos \big(\frac{3i}{\sqrt{2}}+\frac{3}{\sqrt{2}}\big) \sin \big(\frac{3u}{2}\big) -\sin \big(\frac{3i}{\sqrt{2}}+\frac{3}{\sqrt{2}}\big) \cos \big(\frac{3u}{2}\big) }{ u }du$$ I understand the integral will not have an antidervative with sin(x) and cos(x) over x but, how do you proceed from here?
| Without contour integration (the simplest, for sure)
This is just for your curiosity : you ended with a series of antiderivatives
$$I_k=\int \frac{\sin(nx) }{x+k}dx$$ where $k$ is a complex number. Let $y=x+k$ to make
$$I_k=\int \frac{\sin (n (y-k))}{y}\,dy=\cos (k n)\int\frac{ \sin (n y)}{y}dy-\sin (k n)\int\frac{ \cos (n y)}{y} dy$$ Make $y=\frac t n$
$$I_k=\cos (k n)\int\frac{ \sin (t)}{t}dt-\sin (k n)\int\frac{ \cos (t)}{t} dt$$
$$I_k=\cos (k n)\,\text{Si}(t)-\sin (k n)\,\text{Ci}(t)$$ where appear the sine and cosine integrals (the are non-elementary functions). Now, back to $x$, let $k=a+ib$ and using trigonometric properties and bounds you should end with
$$J=\int_{-\infty}^{+\infty} \frac{\sin(nx) }{x+a+ib}dx=\pi e^{n (b-i a)} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of a diagonal matrix and a symmetric positive semidefinite matrix Let the following symmetric matrix
$A = \left[ \matrix{a_{12}+a_{13} & -a_{12} & -a_{13} \\
-a_{12} & a_{12}+a_{23} & -a_{23} \\
-a_{13} & -a_{23} & a_{13}+a_{23} } \right]$,
where each $a_{ij}$ is a nonnegative real number, and hence, $A$ is a symmetric positive semidefinite matrix, because:
$
x^T A x = (a_{12}+a_{13}) x_{1}^2 +
(a_{12}+a_{23}) x_{2}^2 +
(a_{13}+a_{23}) x_{3}^2
- 2( a_{12} x_1 x_2 + a_{13} x_1 x_3 + a_{23} x_2 x_3 ) \\
= a_{12}(x_1 - x_2)^2 + a_{13}(x_1 - x_3)^2 + a_{23}(x_2 - x_3)^2 \geq 0 $
On the other hand:
Let $B = A + D =
\left[ \matrix{a_{12}+a_{13} & -a_{12} & -a_{13} \\
-a_{12} & a_{12}+a_{23} & -a_{23} \\
-a_{13} & -a_{23} & a_{13}+a_{23} } \right] +
\left[ \matrix{d_{11} & & \\
& d_{22} & \\
& & d_{33} } \right]$,
where $d_{ii} \in \mathbb{R}$, that is, $D$ is a diagonal matrix with positive and negative real numbers.
What are the limits of $d_{ii}$ so that B is still a positive semidefinite matrix?
| By Sylvester's criterion, $A+D$ is positive semidefinite if and only if all principal minors of $A+D$ are nonnegative. In other words, $A+D$ is PSD iff
\begin{aligned}
&d_1+a_{12}+a_{13}\ge0,\\
&d_2+a_{12}+a_{23}\ge0,\\
&d_3+a_{13}+a_{23}\ge0,\\
&(d_1+a_{12}+a_{13})(d_2+a_{12}+a_{23})\ge a_{12}^2,\\
&(d_1+a_{12}+a_{13})(d_3+a_{13}+a_{23})\ge a_{13}^2,\\
&(d_2+a_{12}+a_{23})(d_3+a_{13}+a_{23})\ge a_{23}^2\\
\end{aligned}
and
\begin{aligned}
&(d_1+a_{12}+a_{13})(d_2+a_{12}+a_{23})(d_3+a_{13}+a_{23})\\
-&a_{23}^2(d_1+a_{12}+a_{13})
-a_{13}^2(d_2+a_{12}+a_{23})
-a_{12}^2(d_3+a_{13}+a_{23})
-2a_{12}a_{13}a_{23}
\ge0.
\end{aligned}
The region bounded by these surfaces is typically not a translated octant of $\mathbb R^3_+$. Therefore, we cannot find a vector $(d_1',d_2',d_3')$ such that $A+D$ is positive semidefinite if and only if $(d_1,d_2,d_3)\ge(d_1',d_2',d_3')$.
To illustrate, consider the analogous case where $A$ is $2\times2$. If $A$ is nonzero, by scaling $A$, we may assume that
$$
A=\pmatrix{1&-1\\ -1&1}.
$$
In this case, if $D=\operatorname{diag}(d_1,d_2)$, then $A+D$ is positive semidefinite if and only if $d_1+1\ge0,\ d_2+1\ge0$ and $(d_1+1)(d_2+1)\ge1$. That is, the feasible region of $(d_1,d_2)$ is the region above the upper right branch of the hyperbola $(d_1+1)(d_2+1)\ge1$. The hyperbola simply hasn't any "bottom left corner" $(d_1',d_2')$ that is entrywise less than or equal to all $(d_1,d_2)$ in the feasible region.
| {
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Finding a polynomial $f(x)$ of degree 5 such that $f(x)$ is divisible by $x^3$ and $f(x)+2$ is divisible by $(x+1)^3.$ There is some polynomial $f(x)$ of degree $5$ such that both of these properties hold:
$f(x)$ is divisible by $x^3$.
$f(x)+2$ is divisible by $(x+1)^3.$
Find that polynomial.
I know that because $f(x)$ is divisible by $x^3$ our polynomial is in the form of $ax^5+bx^4+cx^3.$ However, I'm not very sure how our second condition comes into use. Any help?
| If $f(x)+2=ax^5+bx^4+cx^3+2=(x^3+3x^2+3x+1)(dx^2+ex+f)$,
then multiplying out and equating coefficients yields
$a=d, b=e+3d, c=f+3e+3d, 0=d+3e+3f, 0=e+3f,$ and $ 2=f$.
This system is easy to solve for $a, b, c, d, e, $ and $f, $
and then the answer is $f(x)=ax^5+bx^4+cx^3$.
| {
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Prove $\cos^2(\theta)+\sin^2(\theta) = 1$ $$\cos^2(\theta) + \sin^2(\theta) = 1$$
I solved this by using right triangle,
$$\sin(\theta) = \frac{a}{c}, \quad \cos(\theta) = \frac{b}{c}$$
$$\cos^2(\theta) + \sin^2(\theta) = 1$$
$$\Bigl(\frac{b}{c}\Bigr)^2 + \Bigl(\frac{a}{c}\Bigr)^2 = 1 $$
$$\frac {a^2 + b^2} {c^2} = 1 $$
now using Pythagorean identity: $a^2 + b^2 = c^2$
$$\frac {c^2} {c^2} = 1, \quad 1 = 1 $$
| My start, for my students is the real circunference $\mathcal C$ of center $O(0,0)$ and radius $r$, $P\equiv(x,y)\in \mathcal C$ $$\mathcal C\colon \quad x^2+y^2=r^2$$ $r>0$ and $x=r\cos\theta, y=r\sin\theta$, $\theta\in[0,2\pi[$, that it gives: $$r^2\cos^2\theta+r^2\sin^2\theta=r^2 \iff r^2(\cos^2\theta+\sin^2\theta)=r^2$$
Hence dividing by $r^2>0$ you have:
$$\cos^2\theta+\sin^2\theta=1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the position vector for the point of intersection of a line and the perpendicular line through a point C
How could I find the exact coordinates of the point N for example which is the point of intersection of the line $L=\begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix} +t\begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}$ and a perpendicular line that passes through a point C(1,2,3) not on the line.
What I have done so far is find the value of point N in terms of $t$ to find the vector $\overrightarrow{CN}$:
$$\longrightarrow N\begin{pmatrix} -3t \\ -2t-1 \\ -3t+2 \end{pmatrix}$$
$$\longrightarrow \overrightarrow{CN}=\begin{pmatrix} -3t \\ -2t-1 \\ -3t+2 \end{pmatrix}-\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$
$$=\begin{pmatrix} -3t-1 \\ -2t-3 \\ -3t-1 \end{pmatrix}$$
After this I made the scalar product eual zero:
$$\overrightarrow{CN} \cdot \begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}=0$$
With this I find that, $$t=\frac{-6}{11}$$
But my textbook says that $$t=\frac{5}{11}$$
Can anyone confirm which answer for $t$ is right and if it is $t=\frac{5}{11}$ then what did I do wrong?
| Your derivation seems correct indeed, by a direct check, for $t=-\frac 6{11}$ we obtain
$$N=\begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix} -\frac 6{11}\begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}=\begin{pmatrix} \frac {18}{11} \\ \frac {1}{11} \\ \frac {40}{11} \end{pmatrix}$$
then
$$\overrightarrow{CN}=\begin{pmatrix} \frac {7}{11} \\ -\frac {21}{11} \\ \frac {7}{11} \end{pmatrix}$$
and
$$\begin{pmatrix} \frac {7}{11} \\ -\frac {21}{11} \\ \frac {7}{11} \end{pmatrix} \cdot \begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}= -\frac {21}{11}+\frac {42}{11}-\frac {21}{11}=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$ I need help with the following question:
Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$
My solution: First I know that $143=11\cdot 13$ then because $\gcd (11,13) = 1$ then $3x^2+18x+95\equiv 0\pmod {143}$ if, and only if $$3x^2+18x+95\equiv 3x^2+7x+7\equiv 0\pmod {11} \\ 3x^2+18x+95 \equiv 3x^2+5x+4\equiv 0\pmod {13}$$
I don't know how to solve those equations and I don't know how to combine it to the big solution for the real question (I know about the CRT, but I didn't realy understood how to use it, I'd love help with this).
thanks in advance
| You are allowed to use the quadraditic formula.
$3x^2 + 18x + 95 \equiv 0 \pmod {143}$ means
$x \equiv \frac {18\pm \sqrt{18^2 -4*95*3}}{6} \pmod {143}$.
Now use Chinese remainder theorem on $143 = 13*11$
$\frac {18\pm \sqrt {18^2 -4*95*3}}{6} \pmod {11}\equiv$
$\frac {7\pm {7^2 - 4*7*3}}{6}*12\pmod {11}\equiv$
$2(7\pm \sqrt{49 -12*7}) \equiv 14\pm 2\sqrt {5-7}\equiv$
$3 \pm 2\sqrt {-2}\equiv 3\pm 2\sqrt{9,64} \equiv 3\pm 6,16\equiv 9,8\pmod{11}$
And $\frac {18\pm \sqrt {18^2 -4*95*3}}{6} \pmod {13}\equiv$
$\frac {5\pm \sqrt {5^2 - 12*4}}{6}\pmod {13}\equiv$
$-12(\frac {5\pm \sqrt {25 + 4}}{6}\equiv -2(5\pm \sqrt{29})\equiv$
$-10 \pm2\sqrt{3}\equiv 3 \pm 2\sqrt{16,81}\equiv 3\pm 8,18\equiv 11,8$
So you have four solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
understand the proof of $\frac{2 n}{3} \sqrt{n}<\sum_{k=1}^{n} \sqrt{k}<\frac{4 n+3}{6} \sqrt{n}$
If $n \in \mathbb{N}^*$, prove that \begin{align*}\frac{2 n}{3} \sqrt{n}<\sum_{k=1}^{n} \sqrt{k}<\frac{4 n+3}{6} \sqrt{n}.\end{align*}
I am having trouble understanding the following proof of this problem:
Let
$a_{n}=(n+\lambda) \sqrt{n}, n \in \mathbb{N}^{*}$,
then
\begin{align}
\frac{\sqrt{n}}{a_{n}-a_{n-1}} &=\frac{\sqrt{n}}{(n+\lambda) \sqrt{n}-(n-1+\lambda) \cdot \sqrt{n-1}} \\
&=\frac{\sqrt{n}}{(\sqrt{n}-\sqrt{n-1})(2 n-1+\sqrt{n(n-1)})+\lambda(\sqrt{n}-\sqrt{n-1})} \\
&=\frac{n+\sqrt{n(n-1)}}{2 n+\sqrt{n(n-1)}-1+\lambda} \\
&=\frac{2}{3}+\frac{\sqrt{n(n-1)}-n+2(1-\lambda)}{6 n+3 \sqrt{n(n-1)}-3(1-\lambda)} \tag{1}
\end{align}
for $n \in \mathbb{N}^*$, we have
$$\frac{1}{2}<n-\sqrt{n(n-1)} \leqslant 1 .$$
Substitute $\lambda=\frac{1}{2}, \frac{3}{4}$ into the formula (1), respectively, then we obtain that
$$ \tag{2}
\frac{2}{3}\left(n+\frac{1}{2}\right) \sqrt{n} \leqslant \sum_{k=1}^{n} \sqrt{k}<\frac{2}{3}\left(n+\frac{3}{4}\right) \sqrt{n}.
$$
Can anyone explain to me how the inequality (2) was obtained. Any hlep would be appreciated.
| If $\lambda = \frac 12,$ by using the inequality they provided :
$$\dfrac{\sqrt{n}}{a_n-a_{n-1}}>\frac 23\iff \sqrt{n} >\frac 23\left(a_n-a_{n-1}\right ).$$
Sum this up for and then you will get one side by telescoping. You can do a similar thing for $\lambda=\frac 34$ as well.
| {
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"question_score": "1",
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} |
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
What I Tried :- Since $x^2 + y^2 = 2x - 2y + 2$ , we have $2x - 2y + 2 + \sqrt{32}$
=> $2(x - y + 1 + 2√2)$ . From this step I am not sure how to move forward .
Also I tried to express $x^2 + y^2 + \sqrt{32} \leq S$ , so that in that way we can say that $x^2 + y^2 + \sqrt{32}$ is maximum at $S$ , but I couldn't do it .
Can anyone help me ? Some hints or suggestions to this problem will be appreciated !!
| Using the Lagrange multiplier method, we make the expression $F(x,y,\lambda)=x^2+y^2+\sqrt{32}+\lambda(x^2+y^2-2x+2y-2)$ and then calculate partial derivatives on $x$, $y$ and $\lambda$, require them to be $0$, then solve for $x$ and $y$ (and $\lambda$):
$$0=\frac{\partial F}{\partial x}=2x+2\lambda x-2\lambda$$
$$0=\frac{\partial F}{\partial y}=2y+2\lambda y+2\lambda$$
$$0=\frac{\partial F}{\partial\lambda}=x^2+y^2-2x+2y-2$$
So, $x=\frac{\lambda}{1+\lambda}$ and $y=-\frac{\lambda}{1+\lambda}$ and so $y=-x$, which you can in turn substitute in the third equation. You will end up with two solutions: $(1+\sqrt 2,-1-\sqrt 2)$ and $(1-\sqrt 2,-1+\sqrt 2)$, our of which the first one gives the larger value of $x^2+y^2+\sqrt{32}$ - I will leave it to you to finish off the calculation.
| {
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"url": "https://math.stackexchange.com/questions/3775353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Finding the local and absolute values of a polynomial on a closed finite interval Problem:
Find the local and absolute extreme values of the $y$ on the interval $[0,5]$.
$$ y = x^3 - 9x^2 + 24x - 2 $$
Answer:
To see a plot of this function, click on the link below.
enter link description here
Based upon the plot, I claim that $x = 0$ is a local minimum and $x = 5$ is a local maximum. Now, I am going to look
for another local minimum or another local maximum.
\begin{align*}
y' &= 3x^2 - 18x + 24 \\
y' &= 0 \\
3x^2 - 18x + 24 &= 0 \\
x^2 - 6x + 8 &= 0 \\
(x-4)(x-2) &= 0 \\
\end{align*}
Hence, I claim that $x = 2$ is a local maximum and $x = 4$ is a local maximum. Now, I find $y(2)$.
\begin{align*}
y(2) &= 2^3 - 9(4) + 24(2) - 2 \\
y(2) &= 8 - 36 + 48 - 2 \\
y(2) &= 18 \\
y(5) &= 5^3 - 9(25) + 24(5) - 2 \\
y(5) &= 125 - 225 + 24(5) - 2 \\
y(5) &= 18
\end{align*}
Hence the absolute extreme values of the function on the interval is $x = 2$ and $x = 5$.
Do I have this right?
| For $x\in(0,5)$ we have
$$y' = 3x^2 - 18x + 24=0 \implies x_1=2\:,x_2=4 \implies f(x_1)=18\:, f(x_2)=14$$
and since
$$y'' = 6x - 18\implies f''(x_1)<0\:, f''(x_2)>0$$
then on the boundary
$$x_0=0\:,x_3=5 \implies f(x_0)=-2\:, f(x_3)=18$$
Therefore
*
*$x_1$ is a point of local maximum
*$x_2$ is a point of local minimum
*$x_0$ is a point of absolute minimum $(-2)$
*$x_1$ and $x_3$ are point of absolute maximum $(18)$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequality with a High Degree Constraint This question-
Suppose that $x, y, z$ are positive real numbers and $x^5 + y^5 + z^5 = 3$. Prove that $$ {x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \ge 3 $$
The inequality has a high degree constraint which can convert a $5$-degree polynomial to a $0$-degree term and makes it difficult.
On trying C-S to manage-
$$ \left({x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3}\right)\left(x^5 + y^5 + z^5\right) \ge 9 \Rightarrow
\left(x^2y+y^2z+z^2x\right)^2\geq9 \Rightarrow x^2y+y^2z+z^2x\geq3 $$Still gives a third degree inequality and not a useful fifth degree.
How can I do it and solve the problem?
| We'll prove that $$x^{6.5}+y^{6.5}+y^{6.5}\geq x^6+y^6+z^6,$$ for which it's enough to prove that:
$$\sum_{cyc}\left(10x^{6.5}-10x^6-x^5+1\right)\geq0$$ or
$$5\sum_{cyc}\left(2x^{1.5}-3x+1\right)x^{5}+\sum_{cyc}\left(5x^6-6x^5+1\right)\geq0,$$ which is true by AM-GM.
Now, by C-S and Vasc we obtain:
$$\sum_{cyc}\frac{x^4}{y^3}=\sum_{cyc}\frac{x^{13}}{x^9y^3}\geq\frac{\left(\sum\limits_{cyc}x^{6.5}\right)^2}{\sum\limits_{cyc}x^9y^3}\geq\frac{\left(\sum\limits_{cyc}x^{6}\right)^2}{\sum\limits_{cyc}x^9y^3}\geq3.$$
The following inequality is also true.
Let $x$, $y$ and $z$ be positive numbers such that $x^{34}+y^{34}+z^{34}=3.$ Prove that:
$$\frac{x^4}{y^3}+\frac{y^4}{z^3}+\frac{z^4}{x^3}\geq3$$
But it's not for contest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776396",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Show $\binom{n}{[n/2]} = \binom{n-1}{[(n-1)/2]} + \sum_{i=0}^{[n/2] - 1} \frac{1}{i+1} \binom{2i}{i} \binom{n-2i-2}{[n/2]- i - 1}$ The identity
$$\binom{n}{\lfloor n/2\rfloor} = \binom{n-1}{\lfloor (n-1)/2\rfloor} + \sum_{i=0}^{\lfloor n/2\rfloor - 1} \frac{1}{i+1} \binom{2i}{i} \binom{n-2i-2}{\lfloor n/2\rfloor - i - 1}$$
came up in a certain combinatorial calculation. Here $n \in \mathbb{Z}_{\geq 0}$. Surely this must be listed in some reference work, though I didn't see it when looking through various sources (Wikpedia, Gould's combinatorial identities, MathWorld, etc.).
It looks to me like a discrete fundamental theorem of calculus identity. It may also be related to the Gamma function at half-integers, e.g. Mathematica simplifies the sum when $n=2k$ to
$$2^{2k-1} \frac{\Gamma\left(\frac{1}{2} + k\right)}{k!}.$$
| In trying to evaluate
$$\sum_{q=0}^{\lfloor n/2 \rfloor -1}
\frac{1}{q+1} {2q\choose q}
{n-2q-2\choose \lfloor n/2 \rfloor - q - 1}$$
we get for $n=2m$
$$\sum_{q=0}^{m -1}
\frac{1}{q+1} {2q\choose q}
{2m-2q-2\choose m - q - 1}
\\ = [z^{m-1}] (1+z)^{2m-2} \sum_{q=0}^{m -1}
\frac{1}{q+1} {2q\choose q}
z^q (1+z)^{-2q}.$$
We will use formal power series throughout.
Here the coefficient extractor enforces the range and we recognize the
Catalan number OGF
$$C(w) = \frac{1-\sqrt{1-4w}}{2w}$$
so that we obtain
$$ [z^{m-1}] (1+z)^{2m-2}\sum_{q\ge 0}
\frac{1}{q+1} {2q\choose q}
z^q (1+z)^{-2q}
\\ = [z^{m-1}] (1+z)^{2m-2}
\frac{1-\sqrt{1-4z/(1+z)^2}}{2z/(1+z)^2}
\\ = [z^{m-1}] (1+z)^{2m-1}
\frac{1+z-\sqrt{(1+z)^2-4z}}{2z}
\\ = [z^{m-1}] (1+z)^{2m-1}
\frac{1+z-(1-z)}{2z}
= [z^{m-1}] (1+z)^{2m-1} = {2m-1\choose m-1}.$$
Similarly for $n=2m+1$ we get
$$\sum_{q=0}^{m -1}
\frac{1}{q+1} {2q\choose q}
{2m+1-2q-2\choose m - q - 1} = {2m\choose m-1}.$$
Joining these two we get the closed form
$$\bbox[5px,border:2px solid #00A000]{
{n-1\choose \lfloor n/2 \rfloor - 1}.}$$
We still have to verify two cases, from the given formula for the sum
which is
$${n\choose \lfloor n/2\rfloor}-{n-1\choose \lfloor (n-1)/2\rfloor}$$
first for $n=2m$
$${2m\choose m} - {2m-1\choose m-1} = {2m-1\choose m-1}$$
which is $\frac{2m}{m} {2m-1\choose m-1} = 2 {2m-1\choose m-1}$
and holds by inspection and second for $n=2m+1$
$${2m+1\choose m} - {2m\choose m} = {2m\choose m-1}$$
which also holds by inspection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782516",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is there a simpler form for $\sqrt{(-1 +\sqrt{1+8x^2} + 4x^2)/8} \cdot (3 + \sqrt{1+8x^2})/4|x|$ I'm trying to find a simpler form for $y=\sqrt{(-1 +\sqrt{1+8x^2} + 4x^2)/8} \cdot (3 + \sqrt{1+8x^2})/4|x|$
This is a "V"-shaped curve that:
*
*goes through $x=0, y=1$
*asymptotically approaches $y=\frac{|x|+\sqrt{2}}{2} + \frac{1}{8x+\sqrt{8}}$ as $|x|\to \infty$
but beyond that, I'm stuck.
Any suggestions?
| Let $x$ be a positive quantity, so we have;
$$f(x)=\left(\frac{3+\sqrt{8x^{2}+1}}{8x}\right)\sqrt{\frac{4x^{2}-1+\sqrt{8x^{2}+1}}{2}}$$
Another thing to consider is that for $T_{n-1}=n(n-1)/2$ or the $n$-th triangular number, we have
$$\sqrt{8T_{n-1}+1}=2n-1$$
Now keeping this in mind, If we let $x^2=T_{n-1}=n(n-1)/2$,
$$\begin{align}
f(x) & = \left(\frac{3+\sqrt{8T_{n-1}+1}}{8\sqrt{T_{n-1}}}\right)\sqrt{\frac{4T_{n-1}-1+\sqrt{8T_{n-1}+1}}{2}} \\
& = \left(\frac{3+\left(2n-1\right)}{8\sqrt{T_{n-1}}}\right)\sqrt{\frac{4T_{n-1}-1+\left(2n-1\right)}{2}} \\
& = \left(\frac{1+n}{4\sqrt{T_{n-1}}}\right)\sqrt{2T_{n-1}+\left(n-1\right)} \\
& = \left(\frac{1+n}{2\sqrt{2}\sqrt{n\left(n-1\right)}}\right)\sqrt{n\left(n-1\right)+\left(n-1\right)} \\
& = \left(\frac{1+n}{2\sqrt{2}\sqrt{n\left(n-1\right)}}\right)\sqrt{n\left(n-1\right)+\left(n-1\right)} \\
& = \left(\frac{1+n}{2\sqrt{2}\sqrt{n\left(n-1\right)}}\right)\sqrt{\left(n-1\right)\left(n+1\right)} \\
& = \frac{1}{2}\sqrt{\frac{\left(n+1\right)^{3}}{2n}}\tag{1}
\end{align}$$
Now what we have to do is substitute the value of $n$ in $x$. We have defined $n$ as;
$$\frac{n(n-1)}{2}=x^2$$
$$\Rightarrow n^{2}-n-2x^{2}=0$$
Evaluate this as a quadratic equation and find $n$ using the quadratic formula:
$$n=\frac{1}{2}\left(1+\sqrt{8x^{2}+1}\right)$$
Substitute this in $(1)$ and we finally get;
$$\small{\left(\frac{3+\sqrt{1+8x^{2}}}{8x}\right)\sqrt{\frac{4x^{2}-1+\sqrt{1+8x^{2}}}{2}}=\frac{1}{4\sqrt{2}}\cdot\sqrt{\frac{\left(3+\sqrt{8x^{2}+1}\right)^{3}}{1+\sqrt{8x^{2}+1}}}}$$
EDIT: I just realized, in your question there is a near miss, we can denest:
$$\sqrt{4x^{2}\color{red}{+}1+\sqrt{8x^{2}+1}}=\frac{1+\sqrt{8x^{2}+1}}{\sqrt{2}}$$
| {
"language": "en",
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Proving $\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}\geq \sum_{cyc} a\sqrt{2a^2+bc}$ for non-negative $a$, $b$, $c$ I was trying this question with factorization and other similar methods,
Let $a, b, c \geq 0$. Prove that
$$\begin{array}{c}
\sqrt{a^4+a^2b^2+b^4}+\sqrt{b^4+b^2c^2+c^4}+\sqrt{c^4+c^2a^2+a^4} \\[4pt]
\geq a\sqrt{2a^2+bc}+b\sqrt{2b^2+ca}+c\sqrt{2c^2+ab}
\end{array}$$
This is one of Hoojoo-Lee's Inequality. This seems very intuitive at first as if we square each term, $$ 2\sum_{cyc}{a^4}+\sum_{cyc}{a^2b^2} \geq 2\sum_{cyc}{a^4}+\sum_{cyc}{a^2bc} \Rightarrow \sum_{cyc}{a^2b^2} \geq \sum_{cyc}{a^2bc} $$ which is quite clear. I noticed it but can not exploit it. May be taking the square on each side could help? But I couldn't find a solution.
Please help!
| By the Minkowski's inequality, we have
\begin{align}
\mathrm{LHS} &= \sum_{\mathrm{cyc}}\sqrt{(a^2 + b^2/2)^2 + (b^2\sqrt{3}/2)^2}\\
&\ge \sqrt{\left(\sum_{\mathrm{cyc}} (a^2 + b^2/2)\right)^2 +
\left(\sum_{\mathrm{cyc}} b^2\sqrt{3}/2\right)^2 }\\
&= \sqrt{3(a^2+b^2+c^2)^2}.
\end{align}
Since $x\mapsto \sqrt{x}$ is concave, we have
\begin{align}
\mathrm{RHS} &= (a+b+c)\sum_{\mathrm{cyc}} \frac{a}{a+b+c}\sqrt{2a^2+bc}\\
&\le (a+b+c)\sqrt{\sum_{\mathrm{cyc}} \frac{a}{a+b+c}(2a^2+bc) }\\
&= \sqrt{a+b+c}\sqrt{2(a^3+b^3+c^3) + 3abc}.
\end{align}
Thus, it suffices to prove that
$$3(a^2+b^2+c^2)^2 \ge (a+b+c)[2(a^3+b^3+c^3) + 3abc]. \tag{1}$$
Let $p=a+b+c, q=ab+bc+ca, r=abc$. (1) is written as
$$p^4-6p^2q-9pr+12q^2 \ge 0.$$
Since $q^2 \ge 3pr$, it suffices to prove that
$$p^4-6p^2q-3q^2+12q^2 \ge 0$$
that is
$$(p^2-3q)^2\ge 0.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Convergence of $\sum_{n=1}^\infty 2^n\sin\frac{1}{3^nz}$ The problem is:
prove $\sum_{n=1}^\infty 2^n\sin\frac{1}{3^nz}$ converges absolutely for all $z\neq 0$, but does not converge uniformly near $z=0$.
Proof:
for all $z\neq 0$
$$\left|2^n\sin\frac{1}{3^nz}-2^n\frac{1}{3^nz}\right|
= \left|
2^n\left(
-\frac{(\frac{1}{3^nz})^3}{3!}+\frac{(\frac{1}{3^nz})^5}{5!}\dots
\right)
\right|, \ \ \ \ (1)$$
$\exists~ N$ such that when n>N, $\frac{1/z}{3^n}<1$, so that (1) is less than
$$\left|
2^n\left(
\frac{|\frac{1}{3^nz}|^3}{3!}+\frac{|\frac{1}{3^nz}|^5}{5!}\dots
\right)
\right|
<\left|
\frac{2^n}{3^nz}\left(
\frac{|\frac{1}{3^nz}|^2}{1-|\frac{1}{3^nz}|^2}
\right)
\right|
<\frac{2^n}{3^n}\left|
\frac{1}{z}\left(
\frac{|\frac{1}{3^nz}|^2}{1-|\frac{1}{3^nz}|^2}
\right)
\right|,
$$
$\forall~ \epsilon, \exists~ N_1>-\log(\epsilon^{1/2} z^{3/2})$, such that when $n>N_2=\max\{N, N_1\}$, $|\frac{1}{z}||\frac{1}{3^nz}|^2<\epsilon$, and so (1) is less $\frac{2^n}{3^n}\epsilon$.
Therefore, we have $N_2(\epsilon)$ satisfying that $\forall~ p,$
$$\left|\sum_{n=N_2}^{N_2+p} 2^n\sin\frac{1}{3^nz}-\sum_{n=N_2}^{N_2+p} 2^n\frac{1}{3^nz}\right|
<\sum_{n=N_2}^{N_2+p} \left|2^n\sin\frac{1}{3^nz}-2^n\frac{1}{3^nz}\right|\ \ \ \ (2)\\
<\sum_{n=N_2}^{N_2+p}\frac{2^n}{3^n}\epsilon
\leq 2\epsilon,
$$
and so $\sum_{n=1}^\infty 2^n\sin\frac{1}{3^nz}$ converges absolutely. (A step seems to be missing. One should, instead of $\sum_{n=N_2}^{N_2+p}\frac{2^n}{3^n}$, use something like 2/z (plus a constant), which is the limit of the former.)$\blacksquare$
A possibly trivial question is whether it is proper to prove this way: given n sufficiently large, $u_n<f(n)\epsilon$ (different from that $u_n<\epsilon$, or that $u_n/f(n)<\epsilon$ and so $u_n<f(n)\epsilon$).
There are other questions that I may post somewhere else.
The following is a proof that the series doesn't converge uniformly. It's unnecessarily for answering my questions, but I put it here for completeness of proof.
Proof:
However large $N_2$, we can find $n_0$>$N_2$ and $z=\frac{2}{\pi 3^{n_0}}$ such that $2^{n_0}\sin\frac{1}{3^{n_0}z}=2^{n_0}$, and so (say the limit function is $f(z)=\frac{2}{z}+C$, where $C$ is a constant; I suddenly realize C seems also to be a function of z, that could cause some issues),
$$\left|\sum_{n=1}^{\infty} 2^n\sin\frac{1}{3^nz}-f(z)\right|
>\left|\sum_{n=N_2}^{N_2+p} 2^n\sin\frac{1}{3^nz}-\sum_{n=N_2}^{N_2+p} 2^n\frac{1}{3^nz}-C\right|>|2^{n_0}-\frac{2}{z}-C|>\epsilon.$$
| Further to my comment, we can use that $\sin(x)\to x$ for $x\to0$ and look at the ratio of terms:
$$\lim_{n\to\infty}\frac{u_{n+1}}{u_n}=\lim_{n\to\infty}\frac{2^{n+1}\times z\times3^n}{2^n\times z\times 3^{n+1}}=\frac 23<1$$
As you mentioned the problem is as $z\to0$, $\sin\left(\frac 1{3^nz}\right)\not\to0$
| {
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"source": "stackexchange",
"question_score": "3",
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An uncertainty for $a^3+b^3+c^3-3abc$ I have a doubt regarding $a^3+b^3+c^3-3abc$.
For factoring, it is easy that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$ as $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
But on using AM-GM inequality, we see-
$$ {a^3+b^3+c^3\over 3}\geq abc \Rightarrow a^3+b^3+c^3\ge3abc $$
AM-GM inequality ensures that equality holds if and only if all variables are equal.
So, equality holds if and only if $a=b=c$, which is trivial.
But we see by factoring that equality holds also if $a+b+c=0$. As the inequality is nothing more than a mere AM-GM, so equality should hold where the inequality ensures us it holds. But it also holds for $a+b+c=0$.
How is it possible and if it is, how can I find such equality cases?
| AM-GM only work for non-negative nunber. Where the equality cases hold when all of variables are equal.
On the other hand, if $a,b,c\geqslant 0 $ then $a+b+c \geqslant 0$ if and only if $a=b=c=0$ (if one of them are greater $0$ then $a+b+c>0$).
So the equality also follow by AM-GM.
I hope my answer is useful for you. Thanks.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int\frac{\mathrm{d}x}{{(x^4+2x+10)}^4}$ I am having trouble with this integral $\displaystyle \int \frac{\mathrm{d}x}{{(x^4+2x+10)}^4}$
I know this can be solved using the "ostrogradsky" method. But that is too lengthy.
i tried using integration by parts by multiplying and dividing by $4x^3+2$ but nothing popped up .
Could anyone put me on the right track.
| In the same spirit as @lab bhattacharjee, because of the fourth power in denominatour, assume that $$\int\frac{dx}{{(x^4+2x+10)}^4}=\frac{P_n(x)}{{(x^4+2x+10)}^3}$$ Differentiate both sides and remove the denominator to get
$$0=-1+\left(x^4+2 x+10\right) P_n'(x)-6 \left(2 x^3+1\right) P_n(x)\tag 1$$ For me, this differential equation does not admit a polynomial solution because of the $1$.
Trying with
$$P_6(x)=a+b x+c x^2+d x^3+e x^4+f x^5+g x^6$$ and expanding $(1)$, we then have
$$(-6 a+10 b-1)+x (20 c-4 b)+x^2 (30 d-2 c)+x^3 (40 e-12 a)+x^4 (-11 b+2 e+50
f)+x^5 (-10 c+4 f+60 g)+x^6 (6 g-9 d)-8 e x^7-7 f x^8-6 g
x^9$$ Starting from the end, the last terms give $e=f=g=0$; since $g=0$ then $d=0$; then $c=0$; then $b=0$ and finally $a=0$ and what remains if the $-1$ !
So, and I am ready to bet that there is a typo in the problem, the solution could be to consider
$$\frac{1}{{(x^4+2x+10)}^4}=\frac 1 {(x-a)^4(x-b)^4(x-c)^4(x-d)^4}$$ where $a,b,c,d$ are the complex roots of $x^4+2x+10=0$ and use partial fraction decomposition (this would be a pure nightmare) to end with a bunch of integrals looking like
$$I=\int \frac {dx} {(x-k)^n} \qquad \text{with} \qquad n=1,2,3,4\qquad \text{and} \qquad k=\text{complex number}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why If $x^2=-\frac{1}{3}$ then $x=\pm\frac{i\sqrt{3}}{3}$ If $x^2=-\frac{1}{3}$ then $x=\pm\frac{i\sqrt{3}}{3}$ according to my text book.
I understand that $x=\pm\sqrt{-\frac{1}{3}}$ but do not fully grasp how to express this as $\pm\frac{i\sqrt{3}}{3}$
I understand that $i$ is $\sqrt{-1}$
As far as I can get with my understanding starting from:
$$x=\pm\sqrt{-\frac{1}{3}}$$
$$x=\pm\frac{\sqrt{-1}}{\sqrt{-3}}$$
$$x=\pm\frac{i}{\sqrt{3}i}$$
$$x=\pm\sqrt{3}$$
Where did I go wrong and how can I arrive at $x=\pm\frac{i\sqrt{3}}{3}$
[Edit]
From the comments I now know that $-\frac{1}{3}$ = $\frac{-1}{3}$
Taking that back into my working I still arrive at a different answer:
$$x=\pm\sqrt{-\frac{1}{3}}$$
$$x=\pm\sqrt{\frac{-1}{3}}$$
$$x=\pm\frac{\sqrt{-1}}{3}$$
$$x=\pm\frac{i}{3}$$
How can I get to $x=\pm\frac{i\sqrt{3}}{3}$
| After your edit; compare your simplification with
\begin{align}
x&=\pm\sqrt{-\frac{1}{3}}\\
x&=\pm\frac{\sqrt{-1}}{\sqrt{3}}\\
x&=\pm\frac{i}{\sqrt{3}}\\
x&=\pm i \frac{\sqrt{3}}{{3}}
\end{align}
The last step is achieved since
\begin{align}
\sqrt{3} \sqrt{3} = 3
\end{align}
Thus
$$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How does $x^2-8x+17=0$ have nonreal solutions?
The solutions of $x^2-8x+17=0$ are $4 + i$ and $4 - i$.
Well, I calculated and the results are different.
$$\begin{align}
x^2-8x+17 &= 0 \\
x^2-8x &=17 \\
x(x-8) &= 17
\end{align}$$
So the roots are $x=17$ or $x=17+8=25$.
Why $i$ comes from the problem? Could you please explain about it?
| Also note:
If the solutions are $a, b$ then $(x-a)(x-b) = 0$ and $x^2 -(a+b)x + ab = 0$ and as $x^2 - 8x +17 =0$ we would have $a+b = 8$ and $ab = 17$.
Option 1: $4 + i$ and $4- i$. Then $(4+i) + (4-i) = (4+4) + (i-i) = 8$. So for so good. $(4-i)(4+i) = 16 + 4i - 4i -i^2 = 16 -i^2 = 16 -(-1) =16+1 = 17$. That works.
Option 2: $-8, 17$. Then $-8 + 17 =9 \ne 8$ so no good. ANd $(-8)\cdot 17=-onehundredsomethingbig \ne 17$. Really no good.
Option 3: $4$ and $-4$. Then $4+ (-4) = 0\ne 8$. No good. And $4\times (-4) = -16 \ne 17$. No good.
Option 4: $\sqrt{17},-\sqrt{17}$. Then $\sqrt{17}+(-\sqrt{17}) = 0\ne 8$. And $\sqrt{17}\cdot (-\sqrt{17}) = -17\ne 17$. So no good.
| {
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"timestamp": "2023-03-29T00:00:00",
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Where is the error in this "proof" that 3=0? I saw this video (link at bottom), with a supposed "proof" that $3=0$. It goes as follows:
Let $x$ be a solution of $$x^2+x+1=0 \tag1$$
Since $x\neq0$, we can divide both sides by $x$:
$$\frac{x^2+x+1}{x}=\frac0x\implies x+1+\frac1x=0 \tag2$$
From $(1)$,
$$x^2+x+1=0\implies x+1=-x^2$$
Substitute $x+1=-x^2$ into $(2)$
$$\begin{align*}
-x^2+\frac1x&=0 \tag3\\
\frac1x&=x^2\\
1&=x^3\implies x=1 \tag4
\end{align*}$$
Substitute $x=1$ into $(1)$
$$\begin{align*}
1^2+1+1&=0\\
3&=0
\end{align*}$$
The explanation given in the video is
Substituting $x+1=-x^2$ into $(2)$ creates the extraneous solution $x=1$ which is not a solution to the original equation $(1)$, $x^2+x+1=0$.
Equations $(1)$ and $(2)$ have solutions $\frac{-1\pm i\sqrt3}{2}$, but after the substitution, equation $(3)$ has these two solutions and $1$.
Basically, it is saying that the issue is substituting $x+1=-x^2$, but I'm not sure if this is actually the problem. How can a substitution cause an issue if everything before the substitution is correct?
After reading the comments, I realised many of them say that the real issue is $(4)$, because $1=x^3$ could also mean that $x=\frac{-1\pm i\sqrt3}{2}$. Not considering these solutions is the issue with the "proof". One also needs to check these solutions before making conclusions, and "pick" whichever one is correct.
So, my question is, what is the issue with the above "proof" that $3=0$?
Video: "Prove" 3 = 0. Can You Spot The Mistake? https://www.youtube.com/watch?v=SGUZ-8u1OxM.
| The problem is $x^3=1$ does not imply that $x=1$. The equation $x^3-1=0$ has three possible roots and the root $x=1$ is an additionally generated root.
| {
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Is there an efficient way of showing $\int_{-1}^{1} \ln\left(\frac{2(1+\sqrt{1-x^2})}{1+x^2}\right)dx = 2$? Hi I recently came across the following integral:
$$
\int_{-1}^{1} \ln\left(\frac{2(1+\sqrt{1-x^2})}{1+x^2}\right)dx
$$
When an integral calculator finds the antiderivative of the equation (if it even can) it comes out as this crazy formula.
Anyways the definite integral happens to be equal to 2 and I was wondering if there might be an elegant way of showing that this is the case.
Thank you.
| Integrate by parts
\begin{align}
\int_{-1}^{1} \ln \frac{2(1+\sqrt{1-x^2})}{1+x^2}dx=&
- \int_{-1}^1 x \>d\left( \ln \frac{2(1+\sqrt{1-x^2})}{1+x^2} \right)dx\\
=& \int_{-1}^1 \left(\frac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}}
+\frac{2x^2}{1+x^2} \right)dx\\
=&\int_{-1}^1 \left( 1+ \frac1{\sqrt{1-x^2}}-\frac2{1+x^2} \right)dx\\
= &(x+\sin^{-1}x-2\tan^{-1}x)| _{-1}^1=2
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Use factor theorem to find all zeros of polynomial $2x^3+3x^2+x+6$ with one known factor $x+2$ I am to find the factors of $2x^3+3x^2+x+6$ where I am told that$x+2$ one of the factors.
Using synthetic division to divide $2x^3+3x^2+x+6$ by $x+2$ I confirm there is no remainder so it is a zero and the new quotient is $2x^2-x+3$
So I have: $(x+2)(2x^2-x+3)$
I would now like to factor $(2x^2-x+3)$ but am having a hard time. Since my leading coefficient is not 1, I know that in order to factor by grouping I must find two numbers whose sum is -1 and whose product is 6 (leading coefficient 2 * constant term 3).
I cannot find any so do not know how to proceed with factoring $(2x^2-x+3)$.
I considered:
-1 & 6: product = -6, sum 5
1 & -6: product = -6, sum -5
2 & -3: product = -6, sum -1 # close
-2 & 3: product = -6, sum 1 # also close
-2 & -3: product = 6, sum 5
How can I factor $(2x^2-x+3)$?
| $2x^2-x+3=2(x^2-\frac{1}{2}x+\frac{3}{2})$ where the discriminant is $\Delta=(\frac{-1}{2})^2-4\times 1\times \frac{3}{2}=\frac{1}{4}-6<0$. Therefore $x^2-\frac{1}{2}x+\frac{3}{2}$ has no real roots and so it is an irreducible polynomial over $\mathbb{R}$.
| {
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Is $f(x,y)=\frac{xy^3}{x^2+y^6}$ differentiable at $(0,0)$? Is the following function differentiable at $(0,0)$?
$$ \
f(x,y) =
\begin{cases}
\frac{xy^3}{x^2+y^6} & \text{if } (x,y) \ne (0,0), \\
0 & \text{if } (x,y) = (0,0).
\end{cases}
$$
I found that both of the partial derivatives are $0$, and then tried to calculate the following limit:
$$\lim_{(x,y) \to (0,0)} \frac{\frac{xy^3}{x^2+y^6}}{\sqrt{x^2+y^2}} = \lim_{(x,y) \to (0,0)} \frac{xy^3}{(x^2+y^6) \sqrt{x^2+y^2}}$$
And then I got stuck. I tried the squeeze theorem, but I still couldn't calculate it.
How can I calculate this limit?
| Recall that continuity is a necessary condition for differentiability since differentiability implies continuity and by $y^3=v \to 0$ using polar coordinates we have
$$\frac{xy^3}{x^2+y^6}=\frac{xv}{x^2+v^2}=\cos\theta\sin \theta$$
| {
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How's my proof, are there any mistakes I have made? Compute $\int \sin(x) \left( \frac{1}{\cos(x) + \sin(x)} + \frac{1}{\cos(x) - \sin(x)} \right)\,dx. $
First, simplify the two fractions into one using a common denominator
\begin{align*}
\int\sin(x)& \left(\frac1{\cos(x)+\sin(x)}+\frac1{\cos(x)-\sin(x)}\right)dx\\
&= \int\sin(x)\left(\frac{\cos(x)-\sin(x)}{\cos^2(x)-\sin^2(x)}
+\frac{\cos(x)+\sin(x)}{\cos^2(x)-\sin^2(x)}\right)dx \\
&= \int\frac{2\sin(x)\cos(x)}{\cos(2x)}dx \\
&= \int\frac{\sin(2x)}{\cos(2x)}dx \\
\end{align*}
Let $u = 2x$. Taking the derivative, $\frac{du}{dx} = 2,$ or $dx = \frac{du}{2}.$
Plugging in $u$ for $2x,$ and $\frac{du}{2}$ for $dx,$ we get $\frac{1}{2}\int \frac{\sin(u)}{\cos(u)}\,du.$
Taking the integral of $\tan(u)$, as solved in problem 10(assume this is true), we get $-\log|\cos(u)|+C.$
Therefore, $\frac{1}{2}\int \frac{\sin(u)}{\cos(u)}\,du = -\frac{1}{2}\log|\cos(u)|+C.$
Finally, subbing $2x$ for $u$, we get $\boxed{-\frac{1}{2}\log|\cos(2x)|+C}$ where $C$ is a constant
| Your solution is correct.
Alternatively you could've used the substitution $u=\cos(2x)$.
| {
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Evaluating $\int \ln(2x+3) \mathrm{d}x$ Evaluate $$\int \ln(2x+3)\mathrm{d}x$$
Set $r = 2x+3 \rightarrow \mathrm{d}r = 2\mathrm{d}x$
So integral becomes $\displaystyle \frac{1}{2}\int \ln(r)\mathrm{d}r$
Set $u=\ln(r)$, $\mathrm{d}v=\mathrm{d}r$, so $\mathrm{d}u = \frac{1}{r}\mathrm{d}r$ and $v=r$
$\implies \displaystyle \frac{1}{2}\int \ln(r)\mathrm{d}r=\frac{1}{2}\int u\, \mathrm{d}v=\frac{1}{2}(uv-\int v\, \mathrm{d}u) = \frac{1}{2}\left(\ln(r)r-\int \mathrm{d}r\right) = \ln(2x+3)\left(x+\frac{3}{2}\right)-\left(x+\frac{3}{2}\right)+C$
But the correct answer is $\ln(2x+3)\left(x+\frac{3}{2}\right)-x+C$
Can somebody show me where my mistake is and also a better way to do the problem? Thanks!
| Your solution is correct as a constant plus another constant can be represented by a different constant, so $-\frac{3}{2}+C=C_1$.
As an alternative, you could integrate by parts and let $u=\ln(2x+3)$ and $dv=dx$. Then $du=\frac{2}{2x+3}$ and we can take $v=x+\frac{3}{2}$. It follows that
\begin{align}\int \ln(2x+3)\,dx&=\ln(2x+3)\left(x+\frac{3}{2}\right)-\int \left(x+\frac{3}{2}\right)\frac{2}{2x+3}\,dx\\&=
\ln(2x+3)\left(x+\frac{3}{2}\right)-\int \,dx\\&=
\ln(2x+3)\left(x+\frac{3}{2}\right)-x+C,
\end{align}
as expected!
| {
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How would you calculate a derivative of $ f(x)= \frac{\sqrt{x+1}}{2-x}$ by the limit definition? I have a function defined as follows:
$$
f(x)= \frac{\sqrt{x+1}}{2-x}
$$
I tried to calculate the derivative using the limit definition using four methods, but I was unsuccessful in any. Could someone help me calculate it and explain the method?
$$
1) \lim_{h\to 0} =\frac{\frac{\sqrt{(x+h)+1}}{2-(x+h)}-\frac{\sqrt{x+1}}{2-x}}h
$$
$$
2)\lim_{z\to x} =\frac{\frac{\sqrt{z+1}}{2-z}-\frac{\sqrt{x+1}}{2-x}}{z-x}
$$
$$
3)\;f(x)= \frac{\sqrt{x+1}}{2-x}; u=\sqrt{x+1}
$$
$$
\lim_{h\to 0} =\frac{\frac{u+h}{3-(u+h)^2}-\frac{u}{3-u^2}}h
$$
$$
4)\;f(x)= \frac{\sqrt{x+1}}{2-x}; u={x+1};
$$
$$
\lim_{h\to 0} =\frac{\frac{\sqrt{u+h}}{3-(u+h)}-\frac {\sqrt{u}}{3-u}}h
$$
| Taking the first definition :
$$\lim_{h\to 0} \frac{\frac{\sqrt{(x+h)+1}}{2-(x+h)}-\frac{\sqrt{x+1}}{2-x}}h = \lim_{h\to 0} \frac{\sqrt{(x+h)+1}(2-x)-\sqrt{x+1}(2-(x+h))}{(2-(x+h))(2-x)h}$$$$=
\lim_{h\to 0} \frac{(x+h+1)(2-x)^2-(x+1)(2-x-h)^2}{(2-x-h)(2-x)h( \sqrt{(x+h)+1}(2-x)+\sqrt{x+1}(2-(x+h)) )} $$
$$= \lim_{h\to 0} \frac{h^2(-x-1)+h(-x^2-2x+8)}{(2-x-h)(2-x)h( \sqrt{(x+h)+1}(2-x)+\sqrt{x+1}(2-(x+h)) )} $$ $$= \frac{-x^2-2x+8}{(2-x)^2\cdot(2\sqrt{x+1}(2-x))}=-\frac{x+4}{2(2-x)^2\sqrt{x+1}}
$$
Too heavy calculations though, don't use the definition for such derivatives.
| {
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Solving a recurrence relation when $n$ is a power of $7$ I'm working through some recurrence relation examples and am struggling with this question that assumes inputs of $n$ are powers of $7$.
Essentially we have
$T(n) = T({\lfloor}n/7{\rfloor}) + \log_{3}(n)$
$T(1) = 0$
Using the bottom up method I have found the following
$T(7) = T(1) + \log_{3}(7) = \log_{3}(7)$
$T(49) = T(7) + \log_{3}(49) = \log_{3}(7) + \log_{3}(49)$
$T(343) = T(49) + \log_{3}(343) = \log_{3}(7) + \log_{3}(49)
+ log_{3}(343)$
I'm struggling to turn this into an equation though. I can see that essentially we have
$T(n) = \log_{3}(7^1) + \log_{3}(7^2) + ... + \log_{3}(7^k)$ where $7^k = n$.
How would I transform that logic into an equation?
| It seems more helpful to consider the function
$f(0) = 1$
$f(n + 1) = f(n) + \log_3(7^{n + 1}) = f(n) + (n + 1) \log_3(7)$
Here, we have $f(n) = T(7^n)$.
Then it is immediate that $f(n) = 1 + \sum\limits_{i = 1}^n n \log_3(7) = 1 + \log_3(7) \sum\limits_{i= 1}^n n$. Using the identity $\sum\limits_{i = 1}^n n = \frac{n (n + 1)}{2}$, we have
$f(n) = 1 + \frac{n(n+1) \log_3(7)}{2}$
hence
$T(n) = 1 + \frac{\log_7(n) (\log_7(n) + 1) \log_3(7)}{2} = 1 + \frac{\log_3(n) (\log_7(n) + 1)}{2}$
whenever $n$ is a power of 7.
| {
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Polynomial Question sum of powers Let $f(x)$ be a quadratic polynomial such that $f(-4) = -22,$ $f(-1)=2$, and $f(2)=-1.$ Let $g(x) = f(x)^{16}.$ Find the sum of the coefficients of the terms in $g(x)$ with even exponents. (For example, the sum of the coefficients of the terms in $-7x^3 + 4x^2 + 10x - 5$ with even exponents is $(4) + (-5) = -1.$)
I have determined the quadratic expression, which is $$f(x)=-\frac{3x^2}{2}+\frac{x}{2}+4.$$but I don't know how to proceed.
EDIT: What I am learning is related to odd and even polynomials. Is there a way to solve this problem by reducing $g(x)$ to an even polynomial?
| First of all, we need to determine what quadratic $f(x)$ is. We can write some equations in the form of a general quadratic($ax^2+bx+c$). We plug in the values stated in the question into our expression to get the following system of linear equations:
*
*$16a-4b+c=-22.$
*$a-b+c=2.$
*$4a+2b+c=-1.$
Now, we can solve this set of equations and find out what the coefficients of our expression is. We find that:
$$a=-\frac{3}{2}, b=\frac{1}{2}, c=4.$$So:
$$f(x)=-\frac{3x^2}{2}+\frac{x}{2}+4.$$
Now that we have our quadratic expression, we can look at $g(x)$. Since we only want the sum of the even exponents, we can try the values $f(1)$ and $f(-1)$. The first value works in all cases, while the negative value will resulting in the sum of the odd exponents subtrated from the even exponents.
We substitute $x=1$ into our quadratic to get $f(1)=3$, so:
$$(f(1))^{16}=3^{16}.$$
Now, we substitute $x=-1$, to get $f(-1)=2$, so:
$$(f(-1))^{16}=2^{16}.$$
Now that we have our two equations, we know that $f(1)-f(-1)=3^{16}-2^{16}$, which is also two times the sum of the coefficients of the odd terms. So the sum of the coefficients of the odd terms is:
$$\frac{3^{16}-2^{16}}{2}.$$
We subtract this from $f(1)$ and simplify to get the sum of the even exponents.
$$3^{16}-\frac{3^{16}-2^{16}}{2}=\frac{3^{16}+2^{16}}{2}.$$
Thus, the sum of the coefficients of the even exponents in $g(x)$ is $\boxed{\frac{3^{16}+2^{16}}{2}}$*.
*I used a calculator to find the result in numerical form: $\boxed{21556128.5}$.
| {
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Proving $3(1−a+a^2)(1−b+b^2)(1−c+c^2)≥1+abc+a^2b^2c^2$ My task was to prove the question above over real variables.
I thought that this minor inequality should help-
$$ 3(1 − a + a^2)(1 − b + b^2) ≥ 2(1 − ab + a^2 b^2). $$
which is true.
By this inequality, the original inequality is converted to-
$$ (1 - ab)^2 (1 - c)^2 + (ab-c)^2 + abc \geq 0 $$
This proves the inequality for $abc\geq 0$.
I want to prove this Inequality for $abc\lt0$. But I couldn't find a solution for $abc\lt0$.
Any extensions for $abc\lt0$ are thankfully accepted.
| Two SOS solutions with the help of computer
*
*According to Vasc's solution in @Michael Rozenberg's answer, we have a simple SOS expression:
\begin{align}
&3(a^2-a+1)(b^2-b+1)(c^2-c+1) - (1 + abc + a^2b^2c^2)\\
=\ & \frac{1}{8}(abc-3c+2)^2 + \frac{3}{8}(abc-2ab+c)^2 + \frac{3}{8}(a-1)^2(b-1)^2(2c-1)^2\\
&\quad + \frac{9}{8}(a-1)^2(b-1)^2 + \frac{3}{8}(a-b)^2(2c-1)^2 + \frac{9}{8}(a-b)^2.
\end{align}
*Without using Vasc's solution, I can obtain a complicated SOS expression
$$ 3(a^2-a+1)(b^2-b+1)(c^2-c+1) - (1 + abc + a^2b^2c^2) = \frac{1}{2}z^\mathsf{T}Qz$$
where $z = [1, a, b, c, ab, ca, bc, abc]^\mathsf{T}$ and
$$Q = \left(\begin{array}{rrrrrrrr} 4 & -3 & -3 & -3 & 2 & 2 & 2 & -1\\ -3 & 6 & 1 & 1 & -3 & -3 & -1 & 2\\ -3 & 1 & 6 & 1 & -3 & -1 & -3 & 2\\ -3 & 1 & 1 & 6 & -1 & -3 & -3 & 2\\ 2 & -3 & -3 & -1 & 6 & 1 & 1 & -3\\ 2 & -3 & -1 & -3 & 1 & 6 & 1 & -3\\ 2 & -1 & -3 & -3 & 1 & 1 & 6 & -3\\ -1 & 2 & 2 & 2 & -3 & -3 & -3 & 4 \end{array}\right).$$
Remarks: $Q$ is positive semidefinite.
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Find positive integers $a$ which there exists an integer $b, 0 \leq b \leq 2002$, such both quadratics $x^2+ax+b$ and $x^2+ax+b+1$ have integer roots. Compute the number of positive integers $a$ for which there exists an integer $b, 0 \leq b \leq 2002$, such that both of the quadratics $x^2+ax+b$ and $x^2+ax+b+1$ have integer roots.
First the conditions that satisfy integer roots: discriminant being $\geq 0$
$a^2-4b \geq 0$ and $a^2-4(b+1) \geq 0$
$a^2 \geq 4b$ and $a^2\geq 4(b+1)$
$a^2 \geq 4b$ and $a^2\geq 4(b+1)$
In order to satisfy both quadratics we consider $a^2\geq 4(b+1)$ only
$\therefore 0 \leq a\leq 2\sqrt{b+1}$
If $b=0$
$0 \leq a\leq 2$
If $b=2002$
$0 \leq a\leq 2\sqrt{2002+1}$
$\therefore 2\leq a \leq 2\sqrt{2002+1} \approx89.5$ 3sf
$(89.5-2)+1=88.5=88$
$\therefore$ the number of positive integers $a:88$
No idea if it's correct or not can anyone please chime in, greatly appreciated.
| For them to be integers both $a^2-4b$ and $a^2-4(b+1)$ have to be perfect squares. Let them be $k^2$ and $l^2$ respectively. Then $k^2-4=l^2 \implies k^2-l^2=4 \implies (k-l)(k+l)=4$
Since $a$ and $b$ both are integers, then $k$ and $l$ are also trivially integers. These two integers multiply to give $4$. hence they must be factors of $4$. Match them with all the possible factor pairs of $4$ and you will get your $k$ and $l$ values. Since there are only squares of $k$ and $l$ here, we do not need to worry about the negative values of $k$ and $l$. After doing all that you will get that $k=2$ and $l=0$ is the only solution. $a^2-4b=4 \implies a^2= 4(b+1)$. Using the bounds given, find the values of $a$. I leave the rest to you.
You can just count the number of cases yourself. Just remove all the odd squares from $1$ to $89$ and you got your answer. (The answer is $44$ values and they are $a^2=4k^2$ where $k$ is an integer from $1$ to $44$).
| {
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"url": "https://math.stackexchange.com/questions/3799829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $(a^2 + 1)(b ^2 + 1)(c ^2 + 1) ≥ 2(ab + bc + ca)$ where $a,b,c$ are real numbers. The inequality above seems very compelling for the pqr-method.
So this was my attempt-
$$ LHS = (a^2 + 1)(b^2 + 1)(c^2 + 1) = 1 + a^2 + b^2 + c^2 + a^2b^2 + b^2c^2 + c^2a^2 + a^2b^2c^2 $$
Now substituting $p = a+b+c$ , $q = ab+bc+ca$ and $r = abc$.
$$ LHS = 1 + p^2 - 2q + q^2 - 2pr + r^2 \geq 2q \Rightarrow 1 + p^2 + q^2 + r^2 \geq 4q + 2pr $$
It's quite well-known that $p^2\geq 3q$ and $q^2\geq 3pr$. So,
$$ 1 + 3q + 3pr + r^2 \geq 4q + 2pr \Rightarrow 1 + pr + r^2 \geq q $$
But I don't know how to prove it. It can also be seen that $a\ge b\ge c$, but I can't exploit symmetry.
Any help is thankfully welcome.
| If you write LHS - RHS as a quadratic in $a$, then the discriminant is:
$$D = 4(b+c)^2 - 4((b^2+1)(c^2+1)-2bc)(b^2+1)(c^2+1).$$
But $(b^2+1)(1+c^2)\geq (b+c)^2$ by C-S and $(b^2+1)(c^2+1)-2bc = b^2c^2+(b-c)^2+1\geq 1,$
so the discriminant is non-positive.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When its possible to write a number $a+b\sqrt{c}$ as the square of a sum? I was teaching my high school students how to find the roots of 2nd degree equations like the following:
$\frac{x^2}{2}+\sqrt{3}x-\sqrt{2}=0$
In this case, using the formula we get:
$x=-\sqrt{3}\pm\sqrt{3-2\sqrt{2}}$
To simplify I writted $3-2\sqrt{2}$ as $(\sqrt{2}-1)^2$, so I get:
$x=-\sqrt{3}\pm(\sqrt{2}-1)$
In this case was simple to write $3-2\sqrt{2}$ as $(\sqrt{2}-1)^2$, there are some harder cases to find this factorization.
My question is when it's possible to write $a+b\sqrt{c}$ as $(m+n)^2$,
and how to find these numbers $m$ and $n$?
| Suppose $a+b\sqrt{c} = (m+n\sqrt{c})^2$, where $a$, $b$, $c$, $m$, and $n$ are integers and further that $c > 0$ (otherwise we are not talking about the real-valued square root or we are talking about $a+0 = a$). Then \begin{align*}
a+b\sqrt{c} &= (m+n\sqrt{c})^2 \\
&= m^2 + cn^2 + 2 m n \sqrt{c} \\
\end{align*}
This forces $b$ even. So, not possible if $b$ is odd. Then for each (partial) factorization $b = 2 x y$, where $x$ and $y$ are integers, can be $\pm 1$, and need not be prime, we have a solution if $x^2 + cy^2 = a$ or $y^2 + cx^2 = a$. In the former case, take $m = x$ and $n = y$. In the latter, take $m = y$ and $n = x$.
Applied to your example... We attempt to simplify $\sqrt{3+2\sqrt{2}}$, so $a = 3$, $b = c = 2$. Since the $2$ coefficient of the radical is even, there may be a solution. The (partial) factorizations of $2$ of the suitable form are $2 \cdot 1 \cdot 1$ and $2 \cdot -1 \cdot -1$. In the first case, $1^2 + 1^2 \cdot c = 3 = a$, as desired and we obtain $3 + 2\sqrt{2} = (1+1\sqrt{2})^2$. (The latter (partial) factorization gives $3+2\sqrt{2} = (-1-\sqrt{2})^2$, which is also true.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find last two digits of $302^{46}$ I need to find the last two digits of $302^{46}$ without resorting to Euler's theorem or Chinese remainder theorem (they have not been introduce so far in the course; I can user Fermat's little theorem though). This is what I tried:
We have to work $\pmod{100}$ and it is easy to see that:
$302 = 2 \pmod{100}$
So I can write
$302^{46} = 2^{46} \pmod{100}$
I'm stuck here I don't know know to further reduce $2^{46}$.
| $2^{46} \equiv 4^{23} \equiv 4 \cdot 4^{22} \equiv 4 \cdot 16^{11} \equiv 4 \cdot 16 \cdot 16^{10} \equiv 4 \cdot 16 \cdot 56^{5} \equiv$
$\quad 4 \cdot 16 \cdot 56 \cdot 56^{4} \equiv 4 \cdot 16 \cdot 56 \cdot 36^{2} \equiv$
$\quad 4 \cdot 16 \cdot 56 \cdot 96 \equiv^\text{algorithm complete / now applying discretionary techniques}$
$\quad 4 \cdot 16 \cdot 56 \cdot (-4) \equiv (-1) \cdot 4\cdot 4 \cdot 16 \cdot 56 \equiv (-1) \cdot 16 \cdot 16 \cdot 56 \equiv$
$\quad (-1) \cdot 56 \cdot 56 \equiv -36 \equiv 64 \pmod{100}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 7
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What is the fastest way of calculating whether vectors are linearly independent?
Is the following family $(u_1, u_2, u_3)$ linearly independent?
$$ u_1 := \begin{pmatrix} 1 \\ 3 \\ 5 \\ -1 \end{pmatrix} ,
u_2 := \begin{pmatrix} 1 \\ -1 \\ -3 \\ 3 \end{pmatrix},
u_3 := \begin{pmatrix} 3 \\ 2 \\ 1 \\ 4 \end{pmatrix}$$
So, do I have to form a matrix $U$ with columns $u_1$, $u_2$ and $u_3$ and use gaussian elimination to see whether in every column is a pivot or is there a faster way? The problem is meant to be solved fast, that is why I am asking. I have the feeling that I am missing some important information.
| With Gaussian elimination, it is not very long to prove the rank is $2$:
\begin{align}
\begin{bmatrix}
1&1&3\\ 3&-1&2\\ 5&-3&1\\-1&3&4
\end{bmatrix} \rightsquigarrow
\begin{bmatrix}
1&3&1\\-1&2&3\\-3&1&5\\3&4&-1
\end{bmatrix} \rightsquigarrow
\begin{bmatrix}
1&3&1\\ 0& 5 & 4\\-3&1&5\\ 0 & 5& 4
\end{bmatrix} \rightsquigarrow
\begin{bmatrix}
1&3&1\\ 0& 5 & 4\\ 0&10&8\\ 0 & 0& 0
\end{bmatrix} \rightsquigarrow
\begin{bmatrix}
1&3&1\\ 0& 5 & 4\\ 0&0&0 \\ 0 & 0& 0
\end{bmatrix}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$ (without L'Hopital)? I am trying to evaluate the following limit:
$$ \lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$$
Approach #1
$ \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \\ \sqrt{x+2} + \frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}}\cdot\frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{\sqrt{x^2-2x}-\sqrt{2x-4}} = \\ \sqrt{x+2} + \frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{x-2} $
But I still end up at the indefinite form $\frac12 + \frac{0}{\infty}$
My Approach #2
$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot\frac{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}= \frac1{\sqrt{x-2}}\frac{x^2-x+2\sqrt{2x}-2}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}$
Which also seems to be a dead end.
Any ideas on how to evaluate this?
| For $y=x-2$, you have $$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}=\frac{\sqrt{(y+2)^2-4}+\sqrt{y+2}-\sqrt{2}}{\sqrt{y}} = \frac{\sqrt{y^2+4y}+\sqrt{y+2}-\sqrt{2}}{\sqrt{y}}$$
When $x$ tends to $2$, $y$ tends to $0$, so you have
$$\frac{\sqrt{y^2+4y}+\sqrt{y+2}-\sqrt{2}}{\sqrt{y}} = \frac{2\sqrt{y}\sqrt{1 + \frac{y}{4}}+\sqrt{2}\sqrt{1 + \frac{y}{2}}-\sqrt{2}}{\sqrt{y}}$$
$$ = \frac{2\sqrt{y}(1+ \frac{y}{8} + o \left( y\right))+\sqrt{2}(1 + \frac{y}{4} + o(y))-\sqrt{2}}{\sqrt{y}}$$
$$= 2 \left(1+ \frac{\sqrt{y}}{8} + o \left( \sqrt{y}\right)\right)+\sqrt{2}\left( \frac{\sqrt{y}}{4} + o(\sqrt{y})\right)$$
$$=2 + \sqrt{y} \left( \frac{1+\sqrt{2}}{4} \right) + o \left( \sqrt{y}\right)$$
Therefore the limit when $y$ tends to $0$ is equal to $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806082",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt{8}$ is irrational in different method I tried to prove that $\sqrt{8}$ is irrational. I said let $\sqrt{8}$ be rational then $\sqrt{8}$ = $a/b$ where $a$ and $b$ are relatively prime.
Then $2\sqrt{2}=a/b$ , and $\sqrt{2} =a/(2b)$. it is obvious that $RHS$ is rational and $LHS$ is irrational (assumed that $\sqrt{2}$ is proved). So there is a contradiction and proof done.
My question is that is there other ways to prove that $\sqrt{8}$ is irrational?
| let $\sqrt{8}$ is equal to $\frac {a}{b}$ where $(a,b)=1$
Then $8 = \frac {(a^2)}{(b^2)}$ , so $8(b^2)= a^2$
Because of the fact that $(a,b)$ are relatively prime $b$ cannot divide $a$.Hence $8|a^2$.
Then ,let's say $a=2k$ where $k$ is a positive integer.
Then $8(b^2)=(2k)^2$. Then,by simplification $2(b^2)=k^2$ .It is obvious that $k$ is an even integer, so k=2m where m is positive integer.
so $2(b^2)=(k^2)=(2m)^2$.then $b=8m b^2 = 2(m^2)$.As a result $2|b$ .
we concluded that a and b have common divisor which is $2$. it is a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limit of $(x+1+\sqrt{(x+1)^2 +1})$ as $x\to-\infty$ This seemed easy at first, but then I confronted my result with the one I got from WolframAlpha and they are different:
$$
\lim_{x\to-\infty}(x+1+\sqrt{(x+1)^2 +1})
$$
WolframAlpha says the limit is equal to 0, though for me it seems to be negative infinity. I tried some simple algebraic manipulation, namely multiplying each factor by 1:
$$
\frac{x\sqrt{(x+1)^2+1}}{\sqrt{(x+1)^2+1}}+\frac{\sqrt{(x+1)^2+1}}{\sqrt{(x+1)^2+1}}+[(x+1)^2+1]*\frac{1}{\sqrt{(x+1)^2+1}}
$$
The last part is 0, because the fraction goes to 0:
$$
\lim_{x\to-\infty}([(x+1)^2+1]*\frac{1}{\sqrt{(x+1)^2+1}})=0
$$
The middle part is 1 and the first part should be negative infinity, as the only factor left standing is x and it goes to negative infinity. What am I missing?
| Your mistake was in writing $\lim_{x\to -\infty}\frac{(x+1)^2+1}{\sqrt{(x+1)^2+1}}=0$. The expression $\frac{(x+1)^2+1}{\sqrt{(x+1)^2+1}}$ equals $\sqrt{(x+1)^2+1}$, which tends to $\infty$ as $x\rightarrow-\infty$.
You can attack this problem using other techniques. One approach is to multiply and divide by the conjugate of $x+1+\sqrt{(x+1)^2+1}$, namely $x+1-\sqrt{(x+1)^2+1}$, then use the difference of squares formula to simplify the result. Alternatively, if you are comfortable with hyperbolic functions and the exponential function $e^x$, you could write
$$x+1+\sqrt{(x+1)^2+1}=e^{\sinh ^{-1}(x+1)}$$
and use the fact that $\lim_{x\to -\infty}\sinh ^{-1}(x)=-\infty$ (this limit can be proven without using the logarithmic form of $\sinh ^{-1}x$). Since $\lim_{x\to -\infty}e^x=0$, we have that
$$\lim_{x\to -\infty}e^{\sinh ^{-1}(x+1)}=0$$
Thus,
$$\lim_{x\to -\infty}\left(x+1+\sqrt{(x+1)^2+1}\right)=0$$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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