Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Could someone explain me this "ownership" of the arctangent someone could explain to me this:
$$\int { \arctan { \left( \frac { 1 }{ { u }^{ 2 } } \right) } } \,du=\int { \frac { \pi }{ 2 } } -\arctan { \left( { u }^{ 2 } \right) } \, du$$
| $$\boxed{\tan^{-1} \dfrac 1 x = \cot^{-1}x = \dfrac {\pi} 2 - \tan^{-1} x}$$
Proof of first equality: $y = \tan^{-1} \dfrac 1 x \Rightarrow \tan y = \dfrac 1 x \Rightarrow \cot y = x \Rightarrow y = \cot^{-1} x$.
Proof of second equality: $\cot \left(\dfrac {\pi} 2 - \theta\right) = \tan \theta \Rightarrow \cot \left(\dfrac {\pi} 2 - \tan^{-1} x \right) = \tan \tan^{-1}x = x \Rightarrow \dfrac {\pi} 2 - \tan^{-1} x = \cot^{-1} x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/848565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to find length of the sides of a triangle given the ratio of the sines of the angles? Consider $\triangle ABC$.
Let $\dfrac{\sin A}{\sin B} = \dfrac56$ and
$\dfrac{\sin B}{\sin C} = \dfrac45$.
Find $\dfrac{\vert AC\vert\cdot \vert AB\vert}{\vert BC\vert}$.
If there is no definite answer, express in terms of one of the sides.
| From the Law of Sinus, we know that:
$\frac{\text{sin}(A)}{\text{sin}(B)}=\frac{|BC|}{|AC|}=\frac{5}{6}$
$\frac{\text{sin}(B)}{\text{sin}(C)}=\frac{|AC|}{|AB|}=\frac{4}{5}$
$\frac{\text{sin}(A)}{\text{sin}(C)}=\frac{|BC|}{|AB|}=\frac{2}{3}$
From this we get:
$6|BC|-5|AC|=0$
$5|AC|-4|AB|=0$
$3|BC|-2|AB|=0$
We can solve these equations with linear algebra:
$$
\left[
\matrix
{
6& -5& 0 & 0 \\
0& 5& -4 & 0 \\
3& 0& -2 & 0
}
\right]\rightarrow\left[
\matrix
{
1& -\frac{5}{6}& 0 & 0 \\
0& 1& -\frac{4}{5} & 0 \\
0& 0& 0 & 0
}
\right]
$$
Therefore, if t=|BC|, the solutions is given by:
$\left[
\matrix
{
|BC| \\
|AC| \\
|AB|
}
\right]=t\left[
\matrix
{
1 \\
\frac{5}{6} \\
\frac{2}{3}
}
\right]$
And the desired product is:
$\frac{5}{9t}= \frac{5}{9|BC|}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove trigonometric identity: $\sin(A-D)\sin(B-D)\sin(C-D)=[\sin(D)]^3$ If $ A+B+C= 180^\circ$ and $\cot(D) = \cot(A)+ \cot(B) + \cot(C)$
then show that
$$\sin(A-D)\sin(B-D)\sin(C-D)=[\sin(D)]^3$$
I couldn't do much about the second given statement!!
| \begin{align}
&\frac{\sin (A-D) \sin (B-D) \sin (C-D)}{\sin^3D} \\
&= [\sin (A) \cot (D)-\cos (A)]\cdot[\sin (B) \cot (D)-\cos (B)]\cdot[\sin (C) \cot (D)-\cos (C)]
\end{align}
Substitute $\cot(D) = \cot(A)+ \cot(B) + \cot(C)$ into above equation. (The algebraic detail is omitted as it's very tedious and not at all illuminating.) We can get
\begin{align}
&\frac{\sin (A-D) \sin (B-D) \sin (C-D)}{\sin^3D}
\\&=\frac{ \sin (A+B) \sin (A+C) \sin (B+C)}{\sin (A)\sin (B)\sin (C)}
\end{align}
Since $ A+B+C= 180^\circ$, thus $\sin(A+B)=\sin(180^\circ-C)=\sin C$, $\sin(B+C)=\sin(180^\circ-A)=\sin A$ and $\sin(A+C)=\sin(180^\circ-B)=\sin B$
Finally, we can get
\begin{align}
&\frac{\sin (A-D) \sin (B-D) \sin (C-D)}{\sin^3D} =1
\end{align}
QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/854171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determining Laurent Series expansion and residues Determining Laurent Series expansion and residues of $f(z)=\frac{z}{(z+1)(z+2)}$ around $z = -2$. What is the validity of the expanded region? What is $res(f, -2)$??
| A related technique. Here is how you advance.
$$ f(z)=\frac{z}{(z+1)(z+2)} = \frac{(z+2)-2}{(z+1)(z+2)}$$
$$= \frac{1}{(z+1)} - \frac{2}{(z+2)( (z+2) - 1 ) } $$
$$ = \frac{1}{((z+2)-1)} - \frac{2}{(z+2)( (z+2) - 1 ) } $$
$$ = \frac{1}{(z+2)(1-1/(z+2))} - \frac{2}{(z+2)^2( 1-1/(z+2) ) } $$
$$ = \frac{1}{z+2}(1-1/(z+2))^{-1} - \frac{1}{(z+2)^2}(1-1/(z+2))^{-1}. $$
Now you need to expand $(1-1/(z+2))^{-1}$ using the geometric series. For residues see here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/855938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Ratio of 2 Gammas, approximation with power Find all value of $\alpha$ such that $\lim\limits_{x\rightarrow +\infty}\left(\frac{\Gamma(x+\alpha)}{\Gamma(x)}-x^{\alpha}\right)=0$. (note: $\alpha$ is a constant with respect to $x$)
By restricting to just the integer case ($\Gamma$ reduce to factorial) it is suggestive that $\alpha\leq 1$ is the answer. I am having trouble upgrading this to the full $\Gamma$, as I don't really know much about this function.
My attempt:
EDIT: now I kinda solved it, but I'm looking for alternative method that are simpler and more accessible. My current solution is as follow:
Case $\alpha<0$: using Stirling's approximation (such as shown here: Quotient of gamma functions?) to show $\lim\limits_{x\rightarrow +\infty}\frac{\Gamma(x+\alpha)}{\Gamma(x)}=0$. Easily show that $\lim\limits_{x\rightarrow +\infty}x^{\alpha}=0$.
Case $\alpha=0$: direct calculation.
Case $0<\alpha<1$: apply Gautschi's inequality found here: How do you prove Gautschi's inequality for the gamma function?. After some substitution we can write it as $(x+\alpha-1)^{\alpha}<\frac{\Gamma(x+\alpha)}{\Gamma(x)}<(x+\alpha)^{\alpha}$ with $0<\alpha<1$. Then show that $\lim\limits_{x\rightarrow\infty}((x+\alpha)^{\alpha}-(x+\alpha-1)^{\alpha})$ using mean value theorem: $(x+\alpha)^{\alpha}-(x+\alpha-1)^{\alpha}$ is the derivative of $x^{\alpha}$ at some point in $(x+\alpha-1,x+\alpha)$, which go to $0$. Then squeeze in.
Case $\alpha=1$: use the functional equation $\Gamma(x+1)=x\Gamma(x)$.
Case $\alpha>1$: the case $\alpha$ is integer is trivial: simply restrict domain to integer $x$ and rewrite as factorial. Otherwise first rewrite $\frac{\Gamma(x+\alpha)}{\Gamma(x)}=(x+\beta)(x+\beta+1)\ldots(x+\alpha-1)\frac{\Gamma(x+\beta)}{\Gamma(x)}$ where $0<\beta=\alpha-[\alpha]<1$ (the bracket is floor). Then use Gautschi's inequality to get $(x+\beta)(x+\beta+1)\ldots(x+\alpha-1)\frac{\Gamma(x+\beta)}{\Gamma(x)}>(x+\beta)(x+\beta+1)\ldots(x+\alpha-1)(x+\beta-1)^{\beta}$. Now we can show $\lim\limits_{x\rightarrow +\infty}x^{\alpha}-x^{[\alpha]}(x+\beta-1)^{\beta}=\lim\limits_{x\rightarrow +\infty}\frac{x^{\beta}-(x+\beta-1)^{\beta}}{x^{[\alpha]}}=0$ using L'Hospital. Hence the problem reduced to showing $(x+\beta)(x+\beta+1)\ldots(x+\alpha-1)(x+\beta-1)^{\beta}-x^{[\alpha]}(x+\beta-1)^{\beta}$ go to infinity. Since $(x+\beta-1)^{\beta}$ go to infinity, we just need to show $(x+\beta)(x+\beta+1)\ldots(x+\alpha-1)-x^{[\alpha]}$ stay away from $0$, which is simply just expanding out the term. This show that none of the $\alpha>1$ work.
However, this proof is awfully complicated, and relying on more obscure results. And it does not intuitively showed why the cut off point is at $\alpha=1$. I am looking for a better proof.
Thank you for your help.
| Let $\alpha \in \mathbb R$ be fixed. We'll assume that $\alpha \neq 0$ and $\alpha \neq 1$ --- in these cases the behavior of $\frac{\Gamma(x+\alpha)}{\Gamma(x)} - x^\alpha$ is trivial anyway.
Stirling's formula for the gamma function tells us that
$$
\Gamma(x) = \sqrt{\frac{2\pi}{x}} \left(\frac{x}{e}\right)^x \left[1 + \frac{1}{12x} + O\left(\frac{1}{x^2}\right)\right] \tag{1}
$$
as $x \to +\infty$. If $x \to +\infty$ then $x+\alpha \to +\infty$, so we also have the formula
$$
\Gamma(x+\alpha) = \sqrt{\frac{2\pi}{x+\alpha}} \left(\frac{x+\alpha}{e}\right)^{x+\alpha} \left[1 + \frac{1}{12(x+\alpha)} + O\left(\frac{1}{(x+\alpha)^2}\right)\right], \tag{2}
$$
valid as $x \to +\infty$. One can check that
$$
1 + \frac{1}{12(x+\alpha)} + O\left(\frac{1}{(x+\alpha)^2}\right) = 1 + \frac{1}{12x} + O\left(\frac{1}{x^2}\right)
$$
as $x \to +\infty$. This, together with a few algebraic manipulations of $(2)$, yields
$$
\Gamma(x+\alpha) = x^\alpha \left(1+\frac{\alpha}{x}\right)^{\alpha-1/2} \frac{\left(1+\frac{\alpha}{x}\right)^x}{e^\alpha} \sqrt{\frac{2\pi}{x}} \left(\frac{x}{e}\right)^x \left[1 + \frac{1}{12x} + O\left(\frac{1}{x^2}\right)\right] \tag{3}
$$
as $x \to +\infty$. By dividing $(3)$ by $(1)$ we obtain
$$
\begin{align}
\frac{\Gamma(x+\alpha)}{\Gamma(x)} &= x^\alpha \left(1+\frac{\alpha}{x}\right)^{\alpha-1/2} \frac{\left(1+\frac{\alpha}{x}\right)^x}{e^\alpha} \frac{1 + \frac{1}{12x} + O\left(\frac{1}{x^2}\right)}{1 + \frac{1}{12x} + O\left(\frac{1}{x^2}\right)} \\
&= x^\alpha \left(1+\frac{\alpha}{x}\right)^{\alpha-1/2} \frac{\left(1+\frac{\alpha}{x}\right)^x}{e^\alpha} \left[1 + O\left(\frac{1}{x^2}\right)\right] \tag{4}
\end{align}
$$
as $x \to +\infty$.
By the binomial theorem we have
$$
\left(1+\frac{\alpha}{x}\right)^{\alpha-1/2} = 1 + \frac{(\alpha-1/2)\alpha}{x} + O\left(\frac{1}{x^2}\right) \tag{5}
$$
as $x \to +\infty$.
To approximate the factor involving the exponential function we proceed like
$$
\begin{align}
\left(1+\frac{\alpha}{x}\right)^x &= \exp\left\{x \log\left(1+\frac{\alpha}{x}\right)\right\} \\
&= \exp\left\{x \left[\frac{\alpha}{x} - \frac{\alpha^2}{2x^2} + O\left(\frac{1}{x^3}\right)\right]\right\} \\
&= \exp\left\{\alpha - \frac{\alpha^2}{2x} + O\left(\frac{1}{x^2}\right)\right\} \\
&= e^\alpha \exp\left\{- \frac{\alpha^2}{2x} + O\left(\frac{1}{x^2}\right)\right\} \\
&= e^\alpha \left[1 - \frac{\alpha^2}{2x} + O\left(\frac{1}{x^2}\right)\right], \tag{6}
\end{align}
$$
where the steps are valid as $x \to +\infty$.
Plugging $(5)$ and $(6)$ into $(4)$ yields
$$
\begin{align}
\frac{\Gamma(x+\alpha)}{\Gamma(x)} &= x^\alpha \left[1 + \frac{(\alpha-1/2)\alpha}{x} + O\left(\frac{1}{x^2}\right)\right] \left[1 - \frac{\alpha^2}{2x} + O\left(\frac{1}{x^2}\right)\right] \left[1 + O\left(\frac{1}{x^2}\right)\right] \\
&= x^\alpha \left[1 + \frac{\alpha^2-\alpha}{2x} + O\left(\frac{1}{x^2}\right)\right] \\
&= x^\alpha + \frac{\alpha^2-\alpha}{2} x^{\alpha-1} + O(x^{\alpha-2})
\end{align}
$$
as $x \to +\infty$. Thus
$$
\frac{\Gamma(x+\alpha)}{\Gamma(x)} - x^\alpha \sim \frac{\alpha^2-\alpha}{2} x^{\alpha-1}
$$
as $x \to +\infty$, and the right-hand side tends to zero if and only if $\alpha - 1 < 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/856035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Infinite limit terms under root Suggest me a hint to solve:$$\psi=\lim_{x\to0}{\frac{\sqrt{1-\cos x+\sqrt{1-\cos x+\sqrt{1-\cos x+\cdots}}}-1}{x^2}}$$
My try,
| Hint
If you use Taylor series, you can start from lab bhattacharjee's answer
$$y=\frac{1\pm\sqrt{5-4\cos x}}2$$ and use the fact that around $x=0$, $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^5\right)$. So,$$5-4\cos (x)=1+2 x^2-\frac{x^4}{6}+O\left(x^5\right)$$ Now, use the fact that $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ and replace $y$ by $(2 x^2-\frac{x^4}{6})$. So,$$\sqrt{5-4\cos (x)}=1+x^2-\frac{7 x^4}{12}+O\left(x^5\right)$$
I am sure that you can take from here and easily find lab bhattacharjee's results.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/856791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
finding $ \int_{C(0,2)^+} \frac{z^3}{z^3+z^2-z-1} $ I want to find
$$
I \ = \ \int_{C(0,2)^+} \frac{z^3}{z^3+z^2-z-1}
$$
First of all, I know that $z^3+z^2-z-1 = (z+1)^2(z-1)$. I split up the integral as a sum of residues:
$$
I \ = \
2\pi i \cdot Res_{z=-1}\frac{z^3}{(z+1)^2\cdot(z-1)}
\ + 2 \pi i \cdot Res_{z=1}\frac{z^3}{(z+1)^2\cdot(z-1)}
$$
The rightmost part becomes $2 \pi \cdot 1^3/(1+1)^2 \ = \ \pi i /2$. The same trick can't be applied for the the other part. There was another lemma that was useful though:
$$
Res_{z=1}f(x) \ = \ \frac{1}{(2-1)!} \cdot
\lim_{z \rightarrow -1}\left( (z+1)^2 \cdot \frac{z^3}{(z+1)^2(z-1)}\right)
\ = \ \frac{1}{(1+1)^2} \ = \ \frac{1}{4}
$$
Now I should multiply this by $2 \pi i$, which gives me $\pi \cdot i /2$.
Adding gives us $\pi i $.
Could you please check what I did and tell me if I this is the right way to solve this?
| You misread the other lemma, you have
$$\operatorname{Res}\left(\frac{f(z)}{(z+1)^2}; -1\right) = \frac{1}{(2-1)!}\lim_{z\to -1}\left(\frac{d}{dz}\right)^{2-1}\left((z+1)^2\frac{f(z)}{(z+1)^2}\right) = f'(-1);$$
you forgot to take the appropriate derivative.
Here $f(z) = \frac{z^3}{z-1}$, so
$$f'(z) = \frac{(z-1)3z^2 - z^3}{(z-1)^2} = \frac{2z^3 - 3z^2}{(z-1)^2}$$
and $f'(-1) = -\frac{5}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/857723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Trigonometry - Finding value of expression Question: Given $\sin x + \sin^2 x = 1$ then find the value of $$\cos^{12} x + 3\cos^{10} x + 3\cos^8 x + 6\cos^6 x + 2\cos^4 x + \cos^2 x -2$$
I have no idea where to start on this question. Please help me!
| HINT:
$$\sin x=1-\sin^2x=\cos^2x$$
Squaring we get $$\sin^2x=(\cos^2x)^2\iff 1-c^2=c^4\iff c^4+c^2-1=0$$ where $c=\cos x$
Now establish that $c^4+c^2-1$ is a factor of $$c^{12} + 3c^{10} + 3c^8 + 6c^6 + 2c^4 + c^2 -2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/859029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$
Calculate the following integral:
\begin{equation}
\int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx
\end{equation}
I am having trouble to calculate the integral. I tried to use by parts method but it didn't help. Wolfram Alpha gives me $0$ as the answer but I don't know how to get it. I also tried to search the similar question here and I got this:
$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx,$$
yet it didn't help much. Besides, I don't understand the answers there. Could anyone here please help me to calculate the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
| Similar problem (two answers)
$$
I = \int_1^{\sqrt{2}}\frac{1}{x}\log\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)\mathrm{d}x
$$
Integrate by parts
$$
I = \left.\log(x)\log\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)\right|_1^{\sqrt{2}}- \int_1^{\sqrt{2}}\log(x)\left(\frac{x^6-4 x^5+8 x^4-10 x^2+8 x-4}{x \left(x^2-2 x+2\right) \left(x^4-2 x^2+2\right)}\right)\mathrm{d}x
$$
Partial fractions
$$I = \frac{1}{2} \log (2) \log \left(\frac{2}{4 \sqrt{2}-4}\right) - \int_1^{\sqrt{2}}\log(x)\left(-\frac{2 (x-1)}{x^2-2 x+2}+\frac{4 \left(x^3-x\right)}{x^4-2 x^2+2}-\frac{1}{x}\right)\mathrm{d}x$$
It should be easy enough to show that $\int_1^\sqrt{2} \log(x)/x \,\mathrm{d}x = \left.\tfrac12 \log(x)^2\right|_1^\sqrt{2} = \tfrac18\log(2)^2$.
We are left with (after simplification)
$$I = -\frac{1}{8} \log (2) \left(\log (8)-4 \log \left(1+\sqrt{2}\right)\right) - \int_1^{\sqrt{2}}\log(x)\left(-\frac{2 (x-1)}{x^2-2 x+2}+\frac{4\! \left(x^3-x\right)}{x^4-2 x^2+2}\right)\mathrm{d}x$$
I will now show how to finish the evaluation$\def\I{\mathcal{I}}$
$$I_1 = \int_1^\sqrt{2} \frac{2(x-1)\log(x)}{x^2-2x+2} \mathrm{d}x$$
$$I_2 = \int_1^\sqrt{2} \frac{4x(x^2-1)\log(x)}{x^4-2x^2+2} \mathrm{d}x$$
The technique employ involves a partial fractions into the complex numbers. The first expression has a quadratic in the denominator and the second has a quadratic in $x^2$ in the denominator. It should be clear that it is sufficient to solve the equation in terms of
$$\I_1(a) = \int \frac{\log(x)}{x-a} \mathrm{d}x$$
$$\I_2(b) = \int \frac{x\log(x)}{x^2-b^2} \mathrm{d}x$$
The technique used will be geometric series expansions.
$$\I_1(a) = \int \frac{\log(x)}{1-x/a} \mathrm{d}x = -\frac1a \sum_{n=0}^{\infty}\int\log(x)\frac{x^n}{a^n} \mathrm{d}x$$
Integrate by parts
$$\I_1(a) = -\sum_{n=0}^\infty \frac{a^{-(n+1)} x^{n+1}}{(n+1)^2}-\sum_{n=0}^\infty\frac{a^{-(n+1)} x^{n+1} \log (x)}{(n+1)}$$
The first sum is a series expansion of the polylogarithm function of order $2$. The second is also a series for the polylogarithm but it is of order $1$ so it is easily expressible in terms of the logarithm. In all,
$$\I_1(a) = \operatorname{Li}_2\!\left(\frac{x}{a}\right) + \log (x) \log \!\left(1-\frac{x}{a}\right)$$
A similar procedure shows that
$$\I_2(b) = \frac{1}{4}\! \left(\operatorname{Li}_2\left(\frac{x^2}{b^2}\right)+2\log (x) \log \left(1-\frac{x^2}{b^2}\right)\right)$$
Without adding unnecessary details, we need to find
$$2\I_2(\sqrt{1+i}) + 2\I_2(\sqrt{1-i}) -\I_1(1+i) - \I_1(1-i)$$
and then evaluate at $x = 1$ and $x = \sqrt{2}$. A large part of the evaluation goes to zero at $x = 1$ but we are not so fortunate at $x = \sqrt{2}$. What remains is a product of logarithms and a sum of several dilogarithms. These dilogarithms are very easy to evaluate because they all fit under certain dilogarithm identities which can be found here.
For the time being, I will not explicitly evaluate the final sum but that may come in the near future. With the dilogaithm identities, evaluation is more of a nuisance than anything. I will assure you that you do come across the value which zeros out the integral. Feel free to ask if you have any questions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/870751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
$\ \sqrt{x+39}-\sqrt{x+7}=4 $ So I tried to solve this problem for x
$\ \sqrt{x+39}-\sqrt{x+7}=4 $
I multiplied both sides ($\ \sqrt{m}\cdot\sqrt{n}=\sqrt{mn} $)
$\ (\sqrt{x+39}-\sqrt{x+7})^2=16 $
$\ (x+39)-2(x^2+46x+273)-(x+7) $
$\ 0x+32+(-2x^2-92x-546) $
$\ -2x^2-92x-514 $
divide the 2 out
$\ x^2+46x+257=-8 $
$\ x^2+46x+265=0 $
Use the quadratic formula (or scientific calculator) and the answers are -6.752 and -39.248. I know the answer is exactly -3. What went wrong?
| $$
\sqrt{x + 39} - \sqrt{x + 7} = 4 \quad \Rightarrow \quad (\sqrt{x + 39})^2 = (\sqrt{x + 7} + 4)^2 \quad \Rightarrow
$$
$$
x + 39 = x + 7 + 8\sqrt{x + 7} + 16 \quad \Rightarrow \quad 8\sqrt{x + 7} = 16 \quad \Rightarrow \quad x = -3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to tell if a log series converges? I have the following series.
$$(-1)^n \times \ln\Bigg(\frac{8n+5}{7n+3}\Bigg)$$
I tried the root, ratio and integral tests, but am doing something wrong because I am unable to tell if this series converges.
| Considering a large value of $n$, rewrite $$ \ln\Bigg(\frac{8n+5}{7n+3}\Bigg)=\ln(8n+5)-\ln(7n+3)=\ln(8n)+\ln\Big(1+\frac{5}{8n}\Big)-\ln(7n)-\ln\Big(1+\frac{3}{7n}\Big)$$ $$ \ln\Bigg(\frac{8n+5}{7n+3}\Bigg)=\ln\Big(\frac{8}{7}\Big)+\ln\Big(1+\frac{5}{8n}\Big)-\ln\Big(1+\frac{3}{7n}\Big)$$ Now, consider, for small values of $y$, the Taylor series $$\ln(1+y) \simeq \frac{y}{1}-\frac{y^2}{2}+\frac{y^3}{3}$$ replace $y=\frac{5}{8n}$ in the first logarithm and $y=\frac{3}{7n}$ in the second logarithm. You will end with $$ \ln\Bigg(\frac{8n+5}{7n+3}\Bigg)= \log \left(\frac{8}{7}\right)+\frac{11}{56 n}-\frac{649}{6272
n^2}+\frac{29051}{526848 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$
Applying the same approach to the most general case, you could find that $$ \ln\Bigg(\frac{an+b}{cn+d}\Bigg)= \log \left(\frac{a}{c}\right)+\frac{\frac{b}{a}-\frac{d}{c}}{n}+\frac{\frac{d^2}{2
c^2}-\frac{b^2}{2 a^2}}{n^2}+\frac{\frac{b^3}{3 a^3}-\frac{d^3}{3
c^3}}{n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$
| {
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How to find the maximum value of $12\sin x -9\sin^2x$ How to find the maximum value of $12\sin x -9\sin^2x$ ;
My approach :
This can be written as $-[(3\sin x -2)^2-4]$.
It means that the function will be maximum when $(3\sin x-2)^2 <4$ due to negative sign outside bracket.
But I am not getting how to proceed from here, please suggest. Thanks.
| Note
$$ 12\sin x-9\sin^2x=3\sin x(4-3\sin x)\le \left(\frac{3\sin x+(4-3\sin x)}{2}\right)^2=4 $$
and the equal sign holds if and only if $3\sin x=4-3\sin x$ or $\sin x=\frac{2}{3}$.
Thus the max is 4 when $\sin x=\frac{2}{3}$.
| {
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Solving $\frac{d f(x)}{dx} + f(x-1) = x^2$ Given following differential equation:
$$\frac{d f(x)}{dx} + f(x-1) = x^2$$
where $ f(x)=0 $ for $x \leq 0 $. How do I find the solution for $ x \geq 0 $ ?
I understand that for $ 0 \leq x \leq 1 $ the solution is $ f(x)= \frac{x^3}{3} $.
| On $[0,1]$, the differential equation reduces to $\frac{d}{dx}f(x)=x^2$, which can be solved: $f(x)=\frac13x^3+C$. If the solution is continuous, then $C=0$.
Then on $[1,2]$, the differential equation reduces to $\frac{d}{dx}f(x)=x^2-\frac13(x-1)^3$. So $f(x)=\frac13x^3-\frac1{12}(x-1)^4+C$. To be continuous at $1$, $C$ must be $0$.
Now on $[2,3]$, the differential equation reduces to $\frac{d}{dx}f(x)=x^2-\frac13(x-1)^3+\frac1{12}(x-2)^4$. And repeat. (Yielding $f(x)=\frac13x^3-\frac1{12}(x-1)^4+\frac1{60}(x-2)^5$.)
It would appear that on $[n,n+1]$, that $f(x)=\sum_{k=0}^{n}(-1)^n\frac{2}{(n+3)!}(x-k)^{k+3}$. Induction could formally prove this.
| {
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Solution of the Legendre's ODE using Frobenius Method This is the Legendre's differential equation given in my book:
$(1-x)^{2}\ddot{y}-2x\dot{y}+k(k+1)y=0$
I solved this equation by taking:
$y=x^{c}\{a_{0}+a_{1}x+a_{2}x^{2}+.....+a_{r}x^{r}+.....\}$
Therefore, each term in the equation becomes (I deliberately didn't use the sum notation to see it clearly):
$k(k+1)y=k(k+1)a_{0}x^{c}+k(k+1)a_{1}x^{c+1}+k(k+1)a_{2}x^{c+2}+.....+k(k+1)a_{r}x^{c+r}+.....$
$-2x\dot{y}=-2ca_{0}x^{c}-2(c+1)a_{1}x^{c+1}-2(c+2)a_{2}x^{c+2}-.....-2(c+r)a_{r}x^{c+r}-.....$
$\ddot{y}=(c-1)ca_{0}x^{c-2}+c(c+1)a_{1}x^{c-1}+(c+1)(c+2)a_{2}x^{c}+.....+(c+r-1)(c+r)a_{r}x^{c+r-2}+.....$
$-x^{2}\ddot{y}=-(c-1)ca_{0}x^{c}-c(c+1)a_{1}x^{c+1}-(c+1)(c+2)a_{2}x^{c+2}-.....-(c+r-1)(c+r)a_{r}x^{c+r}-.....$
The indicial equation is $(c-1)ca_{0}x^{c-2}=0$. Therefore, there are two solutions for $c=0$ and $c=1$. Since,
$(c+r+2)(c+r+1)a_{r+2}+k(k+1)a_{r}-2(c+r)a_{r}-(c+r)(c+r-1)a_{r}=0$
general recurrence equation is as follows:
$a_{r+2}=\frac{[(c+r-k)(c+r+k+1)]a_{r}}{(c+r+2)(c+r+1)}$
There is no problem so far. But my book gives the following answer for $c=0$:
$y=a_{0}\left\{ 1-\frac{k(k+1)}{2}x^{2}+\frac{k(k-2)(k+1)(k+3)}{4!}x^{4}-.....\right\}$
This is only possible for $a_1=0$. But I can't see that. I will be very glad if anyone can show me why $a_1$ should be equal to zero when $c=0$.
Answer:
I solved both for $c=0$ and $c=1$ as @Semiclassical suggested. For $c=0$, $a_1$ is indeterminate because of the term: $c(c+1)a_1x^{c-1}$ (It can be any value). Therefore, the solution is:
$u=a_{0}\left\{ 1-\frac{k(k+1)}{2!}x^{2}+\frac{k(k-2)(k+1)(k+3)}{4!}x^{4}-.....\right\} +a_{1}\left\{ x-\frac{(k-1)(k+2)}{3!}x^{3}+\frac{(k-3)(k-1)(k+2)(k+4)}{5!}x^{5}-.....\right\}$
For $c=1$, $a_1=0$ because of the term: $c(c+1)a_1x^{c-1}$, thus the solution is:
$w=a_{0}\left\{ x-\frac{(k-1)(k+2)}{3!}x^{3}+\frac{(k-3)(k-1)(k+2)(k+4)}{5!}x^{5}-.....\right\}$
As it is seen this is not a distinct solution. It is already available in the solution for $c=0$. Therefore, the solution for $c=0$ in my book is incomplete. Actually, it should give the complete solution as above. I learned that this happens when indicial roots differ by an integer value.
| Hint: Write out the solution for $c=1$ (which necessarily has no constant term and leading term $a_0 x$) and compare with what you get for your solution at $c=0$ if you don't assume $a_1=0$.
| {
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Integrating 1/x The standard definition of integrating $\frac{1}{x}$ is:
$$
\int \frac{dx}{ax + b} = \frac {1}{a} \ln |ax + b| + K
$$
Now, if I'm understanding the "constant factor rule", that is:
$$
\int k \frac{dy}{dx} dx = k\int \frac{dy}{dx} dx
$$
Then what if we construct a new fraction:
$$
\frac{1}{ax + b} = \frac{1}{a(x + \frac{b}{a})}
$$
Then we integrate, using the "constant factor rule", such that:
$$
\int \frac{dx}{a(x + \frac{b}{a})} = \frac {1}{a} \int \frac{dx}{(x + \frac{b}{a})} = \frac{1}{a} \ln |x+\frac{b}{a}| + K
$$
Given that
$$
\frac{1}{a(x + \frac{b}{a})} =\frac{1}{ax + b}
$$
Why is it that their integrals are not equal?
$$
\frac{1}{a} \ln |x+\frac{b}{a}| + K = \int \frac{dx}{a(x + \frac{b}{a})} \neq \int \frac{dx}{ax + b} = \frac {1}{a} \ln |ax + b| + K
$$
I'm pretty sure I'm missing something super obvious here, I just don't know what.
| But they are equal:
$$\ln|ax + b| = \ln\Big(|a| |x + \frac b a|\Big) = \ln |a| + \ln |x + \frac b a|$$
Now just roll $\ln |a|$ into the arbitrary constant.
| {
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Finding $\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $ Help me to simplify:$$\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $$
I got a hunch that it will depend on whether $n$ is a multiple of $6$ and equals to $\frac{2^n+2}{3}$ when $n$ is a multiple of $6$.
| Here's a method that doesn't (explicitly) involve complex numbers.
Let $A_n=\sum_k \binom{n}{3k}$, $B_n=\sum_k \binom{n}{3k+1}$, $C_n=\sum_k \binom{n}{3k+2}$.
Then $A_n+B_n+C_n=2^n$ by the binomial theorem. Moreover, Pascal's identity implies the following:
$$
\begin{eqnarray}
A_n&=&A_{n-1} + C_{n-1} =2^{n-1}-B_{n-1} \\
B_n&=&B_{n-1} + A_{n-1} = 2^{n-1}-C_{n-1} \\
C_n&=&C_{n-1} + B_{n-1} = 2^{n-1}-A_{n-1}
\end{eqnarray}
$$
Putting all these together,
$$A_n = 2^{n-1}-(2^{n-2} - C_{n-2})=2^{n-2} + C_{n-2} = 2^{n-2} + 2^{n-3} - A_{n-3} = 3(2^{n-3}) - A_{n-3} \, .$$
Also, $A_0=A_1=A_2=1$.
It follows that $A_{3n}=3(2^{n-3}-2^{n-6}+\dots \pm 2^0) \mp 1$, with similar expressions for $A_{3n+1}$ and $A_{3n+2}$. Obviously, you could use a geometric series formula to further simplify this.
| {
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Proof by induction $\sum_{k=1}^{n}$ $k \binom{n}{k}$ $= n\cdot 2^{n-1}$ for each natural number $n$ Prove by induction that $\sum_{k=1}^{n}$ $k \binom{n}{k}$ $= n\cdot 2^{n-1}$ for each natural number $n$.
| Observe that: $\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$. Thus: $\displaystyle \sum_{k=1}^{n+1} k\binom{n+1}{k} = \displaystyle \sum_{k=1}^{n+1} k\binom{n}{k} + \displaystyle \sum_{k=1}^{n+1} k\binom{n}{k-1} = \displaystyle \sum_{k=1}^n k\binom{n}{k} + \displaystyle \sum_{k=0}^n (k+1)\binom{n}{k} = 2\displaystyle \sum_{k=1}^n k\binom{n}{k} + \displaystyle \sum_{k=0}^n \binom{n}{k} = 2\cdot n2^{n-1} + 2^n = (n+1)2^n$
| {
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Which one is correct for $\sqrt{-16} \times \sqrt{-1}$? $4$ or $-4$? As we can find in order to evaluate $\sqrt{-16} \times \sqrt{-1}$, we can do it in two ways.
FIRST
\begin{align*}
\sqrt{-16} \times \sqrt{-1} &= \sqrt{(-16) \times (-1)}\\
&= \sqrt{16}\\
&=4
\end{align*}
SECOND
\begin{align*}
\sqrt{-16} \times \sqrt{-1} &= \sqrt{16i^2} \times \sqrt{i^2}\\
&= 4i \times i\\
&=4i^2\\
&=-4
\end{align*}
Incidentally if the above is input in complex mode of Casio scientific calculator, the result comes out as $-4$.
Which of the above solutions is correct?
| The answer to each sqrt after removing i is +-, so for a simple explanation you have either (+)(+) or (-)(+) or (-)*(-), depending on the domain or context of the particular problem.
| {
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Easy algebra question: show equality? Let $c:=(u+d)/2$ and $-1<d<0<u$.
Show:
Can you show the steps to arrive from lhs to rhs?
| Expanding the left hand side:
$$\frac{1}{2}(1+u+2c+2uc)+\frac{1}{2}(1+d+dc+c)$$
$$\frac{1}{2}(2+u+c+ 2c + uc + uc +dc +d)$$
$$\frac{1}{2}(c+uc) + \frac{1}{2}(2+u+2c+uc+ dc+d)$$
Taking into account that $2c-d=u$:
$$\frac{1}{2}(c+uc) + \frac{1}{2}(2+4c-d+(2c-d)c+dc+d)$$
$$\frac{1}{2}(c+uc)+\frac{1}{2}(2+4c+2c^2)$$
$$\frac{1}{2}(c+uc)+(c+1)^2$$
| {
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$\nabla \cdot (\mathbf{B}\mathbf{B} - \frac{1}{2}B^2 \tilde{1})=(\nabla \cdot \mathbf{B})\mathbf{B} - \mathbf{B} \times (\nabla \times \mathbf{B})$ Does someone know how to show this identity?
$\nabla \cdot (\mathbf{B}\mathbf{B} - \frac{1}{2}B^2 \tilde{1})=(\nabla \cdot \mathbf{B})\mathbf{B} - \mathbf{B} \times (\nabla \times \mathbf{B})$
| As an alternative to Semiclassical, the problem can be tackled entirely using the machinery of clifford algebra. Let $a$ be an arbitrary vector, and let $\underline B(a) = \overline B(a) = (B \cdot a) B$. Then we have
$$\nabla \cdot [\underline B(a) - \frac{1}{2} B^2 a] = a \cdot [\overline B(\nabla) - \frac{1}{2} \nabla B^2] = a \cdot [B (\nabla \cdot B) + (B \cdot \nabla) B - \frac{1}{2} \nabla B^2]$$
Expansion of the $\nabla B^2$ term requires a common identity:
$$\nabla (A \cdot B) = (\nabla \wedge B) \cdot A + (\nabla \wedge A) \cdot B + (A \cdot \nabla) B + (B \cdot \nabla) A$$
You can see that, once this is proved, for $A = B$, the required expansion falls out immediately. The hard part, then, is merely the proof of this identity:
This is where clifford algebra becomes useful. We can write
$$\nabla (A \cdot B) = \nabla (AB) - \nabla \cdot (A \wedge B)$$
where $AB$ is understood as using the geometric product. The resulting expansion gives
$$\nabla (AB) = (\nabla \cdot A) B + (\nabla \wedge A) \cdot B + (A \cdot \nabla) B - (A \wedge \nabla) \cdot B$$
and
$$\nabla \cdot (A \wedge B) = (A \cdot \nabla) B + (\nabla \cdot A) B - (B \cdot \nabla) A - (\nabla \cdot B) A$$
We can see some terms canceling already. We need a slightly better expansion of terms like $(A \wedge \nabla) \cdot B$, though. See that
$$\langle A (\nabla B )\rangle_1 = (A \cdot \nabla) B + (A \wedge \nabla) \cdot B =A (\nabla \cdot B) + A \cdot (\nabla \wedge B) $$
Putting this into the $\nabla(AB)$ expansion gives
$$\nabla(AB) = (\nabla \cdot A) B + (\nabla \wedge A) \cdot B + 2 (A \cdot \nabla) B - A (\nabla \cdot B) - A \cdot (\nabla \wedge B)$$
This, combined with the previous expansion for $\nabla \cdot (A \wedge B)$, immediately gives the desired result. In case it's hard to keep track of all the terms, though, the following should be better formatted to convince you of the result:
$$\begin{alignat*}{6}
\nabla(AB) &= (\nabla \cdot A) B &- (\nabla \cdot B) A &+ 2(A \cdot \nabla) B &+ 0 &+ (\nabla \wedge A) \cdot B &+ (\nabla \wedge B) \cdot A \\
\nabla \cdot (A \wedge B) &= (\nabla \cdot A) B &- (\nabla \cdot B) A &+ (A \cdot \nabla) B &- (B \cdot \nabla) A &
\end{alignat*}$$
The resulting identify for $\nabla (A \cdot B)$ is then proven, and the special case needed for $A = B$ can be taken to prove the result. All of this was proven, without resorting to index notation, thanks to the power of clifford algebra and its associated calculus.
| {
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Maximum value of $f(x) = \left|\sqrt{\sin^2 x+2a^2} - \sqrt{2a^2-1-\cos^2 x}\right|\;\;,$ Where $a,x\in \mathbb{R}$ Calculation of Maximum value of $\displaystyle f(x) = \left|\sqrt{\sin^2 x+2a^2} - \sqrt{2a^2-1-\cos^2 x}\right|\;\;,$ Where $a,x\in \mathbb{R}$
$\bf{My\; Try::}$ We Can Write It as $f(x) = \left|\sqrt{2a^2-\cos^2 x+1}-\sqrt{2a^2-\cos^2 x-1}\right|$
Now Let $\left(2a^2-\cos^2 x\right) = b\in \mathbb{R}\;,$ Then Expression is $f(b) = \left|\sqrt{b+1}-\sqrt{b-1}\right|\;,$ where $b\in \mathbb{R}.$
Now How Can I find max. value of $f(b)$
Help me
Thanks
| Hint
the maximum value of $f(b) = \left|\sqrt{b+1}-\sqrt{b-1}\right|\;,$ is at $b=1$ which is $\sqrt{2}$ because it is decreasing.
| {
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If the numerator of a fraction is increased by $2$ and the denominator by $1$, it becomes $5/8$.... If the numerator of a fraction is increased by $2$ and the denominator by $1$, it becomes $\displaystyle \frac{5}{8}$ and if the numerator and the denominator of the same fraction are each increased by $1$, the fraction becomes equal to $\displaystyle \frac{1}{2}$. Find the fraction.
I tried,
Let the numerator of the fraction be $x$,
Let the denominator of the fraction be $y$
Therefore, Original fraction is $\displaystyle \frac{x}{y}$.
$$
\left\{
\begin{eqnarray*}
\frac{x+2}{y+1}=\frac{5}{8} \\[2mm]
\frac{x+1}{y+1}=\frac{1}{2} \\
\end{eqnarray*}
\right.
$$
What should I do now?
| That's a fine start. Now you've got two equations in two unknowns. $$\frac{x+2}{y+1} = \frac 58\tag{1}$$
$$\frac{x+1}{y+1} = \frac 12 \tag{2}$$
Now, we have to assume that $y+1 \neq 0$, given the information that's impossible, so we can take $(1)$ for example, and "cross multiply" to get: $$ 8(x+2) = 5(y+1) \iff \color{blue}{y + 1} = \frac{8(x+2)}{5}\tag{1}$$
From the second equation, we have
$$2(x+1) = \color{blue}{y+1}\tag{2}$$
That means $\color{blue}{y+1 = y+1}$, so that $$\frac {8(x+2)}5 = 2(x +1)\tag{3}$$
Now solve for $x$: $$\frac{8(x + 2)}5 = 2(x+1)\iff 5(x+1) = 4(x+2) \iff 5x+5 = 4x + 8 \iff x = 3$$ Then use $x=3$ to solve for $y$, using the fact that $y + 1 = 2(x + 1) \iff y = 2x + 1\implies y = 2(3) + 1 = 7$.
So the original fraction $$\frac xy = \frac 37$$
| {
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Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$? Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$?
I am learning trigonometric identities one identity I have to proof is the next:
$ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$
so I tried to resolve the identity for the left:
$ 1 + \sin^2 \alpha+ \cos^2\alpha - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha $
$= 1 + 1 - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha $
$= 2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)$
And I got stuck, I did not know what to do, so I went to see the problem's answer and I was going fine, the part I was not able to resolve is the next one:
$ 2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)$
$= 2 (1 + \cos\alpha - \sin\alpha(1+\cos\alpha))$
$= 2 (1 - \sin \alpha)(1+ \cos \alpha)$
So the question is how did the teacher do the last three steps? I cant figure it out.
| You calculated that the left side is equal to $$2(1-\sin\alpha + \cos\alpha -\sin\alpha\cos\alpha).$$
Now, try to prove that the left side is equal to that as well. Try expanding $$(1-\sin\alpha)(1+\cos\alpha)$$
| {
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reduction of $6xy + 8 y^2 -12x-26y + 11 = 0$ to canonical one of a second-order curve I have this polynomial
$$
6xy + 8 y^2 -12x-26y + 11 = 0
$$
and I need to reduce it to a canonical equation of a second-order curve. The correct answer from the textbook is that it is a hyperbola
$$
\frac{X^2}{1} - \frac{Y^2}{9} = 1
$$
and the coordinate system must be rotated to an angle equal to
$$
\arctan{3}
$$
and the origin of the new coordinate system must be moved to the point $O'(-1,2)$. I checked this in Mathematica® program, so I think it is safe to assume that there are no errors in the textbook.
And now about the problem itself. I calculated the angle of rotation according to the formula in my textbook. It is (the expression above is the example of a polynom to explain indexes like $a,b,c$ etc.)
$$
ax^2 + 2bxy + cy^2 + 2dx + 2ey +g = 0 \\
\cot{2\alpha} = \frac{a-c}{2b}
$$
I solved this equation and got the correct answer: $\alpha = \arctan{3}$
Then I substituted $x$ and $y$ variables with their values according to the formulae:
$$
x = \cos{\alpha}x' - \sin{\alpha}y'\\
y = \sin{\alpha}x' + \cos{\alpha}y'
$$
Let me show this incrementally for convenience
$$
6xy + 8 y^2 =\\
6(\cos{\alpha}x' - \sin{\alpha}y')(\sin{\alpha}x' + \cos{\alpha}y') + 8(\sin{\alpha}x' + \cos{\alpha}y')^2 =\\
6(\sin{\alpha}\cos{\alpha} * x'^2 +\cos^2{\alpha} * x'y' -\sin^2{\alpha} * x'y' - \sin{\alpha}\cos{\alpha} * y'^2) + 8(\sin^2{\alpha} * x'^2 + 2\sin{\alpha}\cos{\alpha} * x'y' + \cos^2{\alpha} * y'^2) =\\
6\sin{\alpha}\cos{\alpha} * x'^2 + 8\sin^2{\alpha} * x'^2 - 6\sin{\alpha}\cos{\alpha} * y'^2 + 8\cos{\alpha}^2 * y'^2 + 6\cos{\alpha}^2 * x'y' -6\sin^2{\alpha} * x'y' + 16\sin{\alpha}\cos{\alpha} * x'y' = \\
(6\sin{\alpha}\cos{\alpha} + 8\sin^2{\alpha})x'^2 + (- 6\sin{\alpha}\cos{\alpha} + 8\cos^2{\alpha})y'^2 + (6\cos{\alpha}^2 -6\sin^2{\alpha} + 16\sin{\alpha}\cos{\alpha})x'y'
$$
the factor at $x'y'$ evaluates to 0 given $\alpha = \arctan{3}$, I checked this in Mathematica, so from the part above only this remains:
$$
(6\sin{\alpha}\cos{\alpha} + 8\sin^2{\alpha})x'^2 + (- 6\sin{\alpha}\cos{\alpha} + 8\cos^2{\alpha})y'^2
$$
And the remaining part is:
$$
-12x-26y + 11 = \\
-12(\cos{\alpha}*x' - \sin{\alpha}*y') - 26(\sin{\alpha}*x' + \cos{\alpha}*y') + 11 = \\
-12\cos{\alpha}*x' + 12\sin{\alpha}*y' -26\sin{\alpha}*x' - 26\cos{\alpha}*y' + 11 = \\
(-12\cos{\alpha} - 26\sin{\alpha})*x' + (12\sin{\alpha} - 26\cos{\alpha})*y' + 11 = \\
2((-6\cos{\alpha} - 13\sin{\alpha}))*x' + 2(6\sin{\alpha} - 13\cos{\alpha})*y' + 11
$$
So now this curve has the next equation in the rotated coordinate system (let me use $x$ and $y$ instead of $x'$ and $y'$):
$$
\alpha = \arctan{3}\\
a = 6\sin{\alpha}\cos{\alpha} + 8\sin^2{\alpha} \\
c = - 6\sin{\alpha}\cos{\alpha} + 8\cos^2{\alpha} \\
d = -6\cos{\alpha} - 13\sin{\alpha} \\
e = 6\sin{\alpha} - 13\cos{\alpha}\\
g = 11\\
ax^2 + cy^2 + 2dx + 2ey + g = 0
$$
Then it is stated in my textbook that one can further reduce this equation by moving the origin of a coordinate system to the point $O(-\frac{d}{a}, -\frac{e}{c})$
And this is the place where I can't get the correct answer (which is $O(-1,2)$), cause this $-\frac{d}{a}$ gives me 1.58114 instead of -1 and this $-\frac{e}{c}$ gives 1.58114 instead of 2.
I've checked this in Mathematica.
Could anyone explain me what where I'm wrong, please?
Thank you in advance.
| I always try to complete the squares:
$$6xy+8y^2 −12x−26y+11=0$$
$$\left[ 8y^2+2.2 \sqrt{2}y\left(\frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}}\right)\right]-12x+11=0 $$
$$\left[ 8y^2+2.2 \sqrt{2}y\left(\frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}}\right)+\left(\frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}}\right)^2\right]-\left(\frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}}\right)^2-12x+11=0 $$
$$\left[2\sqrt{2}y + \frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}} \right]^2-\frac{1}{8}(9x^2+18x+81)=0 $$
$$\left[2\sqrt{2}y + \frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}} \right]^2-\frac{1}{8}(3x+3)^2=9 $$
$$\frac{1}{8}(8y+3x-13)^2-\frac{1}{8}(3x+3)^2=9 $$
$$(8y+3x-13)^2-(3x+3)^2=72$$
$$(8y+3x-13)^2-\frac{(x+1)^2}{9}=72$$
$$X^2-\frac{Y^2}{9}=72$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/881437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Need help getting started with a complicated limits problem So I've got this thing
$\lim_{x \rightarrow 3} \frac{\sqrt[3]{x^2+6x}-3}{x-3}$
I tried rationalizing but then it ended up in a huge mess and I didn't get the correct answer. I don't know if there's something else to do before resorting to rationalization. I just need something to get started.
Aside from that I know that something on the numerator will cancel out because if I evaluate the limit it ends up as $\frac{0}{0}$
| Hint
$$\lim_{x \to 3} \frac{\sqrt[3]{x^2+6x}-3}{x-3}=\lim_{x\to 3} \frac{\left(\sqrt[3]{x^2+6x}-3\right)\left(\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9\right)}{\left(x-3\right)\left(\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9\right)}$$
$$=\lim_{x\to 3} \frac{x^2+6x-27}{\left(x-3\right)\left(\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9\right)}=\lim_{x\to 3} \frac{x+9}{\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Factoring $x^4+x^2+2x+6$ We have to factor $x^4+x^2+2x+6$.Factoring through factor theorem is not helpful in this question as the question does not follow the integral root theorem i.e. the root of this expression is not any of the factors of 6. I am totally confused.Please help in factoring it!
| $$x^4+x^2+2x+6=x^4+2ax^2+a^2-\left( (2a-1)x^2-2x+a^2-6 \right)$$
Now, $\left( (2a-1)x^2-2x+a^2-6 \right)$ is a constant times perfect square if and only if $\Delta =0$ if and only if
$$4-4(2a-1)(a^2-6)=0$$
if and only if
$$(2a-1)(a^2-6)=1$$
Factoring and using the rational root test you will find that $a=\frac{5}{2}$ is (the only) rational root.
With this choice of $a$, your polynomial becomes a difference of squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
prove that : $\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3$
For $a^2+b^2+c^2 =3$, with $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3.$$
Can any one help me with this problem?
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$\sum_{cyc}\frac{a^2+b^2}{a+b}\geq\sqrt{3(a^2+b^2+c^2)}$$ or
$$\frac{\sum\limits_{cyc}(a^2+b^2)(c^2+ab+ac+bc)}{\prod\limits_{cyc}(a+b)}\geq3\sqrt{3u^2-2v^2}$$ or
$$\frac{\sum\limits_{cyc}(a^2+b^2)(c^2+3v^2)}{9uv^2-w^3}\geq3\sqrt{3u^2-2v^2}$$ or
$$\frac{2(9v^4-6uw^3)+6v^2(9u^2-6v^2)}{9uv^2-w^3}\geq3\sqrt{3u^2-2v^2}$$ or
$$2(9u^2v^2-3v^4-2uw^3)\geq(9uv^2-w^3)\sqrt{3u^2-2v^2}$$ or
$$ 2(9u^2v^2-3v^4)-9uv^2\sqrt{3u^2-2v^2}\geq(4u-\sqrt{3u^2-2v^2})w^3$$
and since $4u-\sqrt{3u^2-2v^2}>0$ it's enough to prove our inequality
for a maximal value of $w^3$, which happens for equality case of two variables.
Since $\sum\limits_{cyc}\frac{a^2+b^2}{a+b}\geq\sqrt{3(a^2+b^2+c^2)}$ is homogeneous, we can assume $b=c=1$, which gives
$$\frac{2(a^2+1)}{a+1}+1\geq\sqrt{3(a^2+2)}$$ or
$$(a-1)^2(a^2+3)\geq0.$$
Done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to prove $~\sqrt{3}\sqrt{4a^{3}-1}~$ isn't an integer? I'm trying solve: $~a^3 + b^3 = c^3~$ has no nonzero integer solutions.
If $~(c−b)=1~$ then $~c^3-b^3=3c^2-3c+1=a^3,~$
from Wolframalpha get:
$$
c = \dfrac{3- \sqrt{3}\sqrt{4a^{3}-1}}{6} \\
c = \dfrac{\sqrt{3}\sqrt{4a^{3}-1}+3}{6}
$$
How to prove $~\sqrt{3}\sqrt{4a^{3}-1}~$ isn't an integer? (eidt: while $~a,\ b,\ c ~$ are nonzero integers)
| Multiplying with $\sqrt{3}$ only returns an integer when multiplying with certain multiples of three.
Assuming $~4a^3 \equiv 1 (3) \equiv 4(3)$ we check for divisibility by three:
$$4a^3 \equiv4(3) \iff \exists k \in \mathbb N:4a^3-4=3k$$
$$3|4(a^3-1) \iff 3|a^3-1 \iff \Big( 3|a^{3/2}-1 ~~~~\vee~~~~ 3|a^{3/2}+1 \Big)$$
That is not true for $\forall a\not = 1$ and thus your product cannot be an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/884565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How to prove $(c - b) ^ 2 + 3cb = x^3$ has no nonzero integer solutions? I'm trying solve:
$a^3 + b^3 = c^3$ has no nonzero integer solutions.
Only one problem left: because
$c^3 - b ^ 3 = (c - b)((c - b) ^ 2 + 3cb) = a ^ 3,\quad (1)$
if $~c-b~$ is a cubic number, divide both side of $~(1)~$ by $~(c - b)~$ get
$(c - b) ^ 2 + 3cb = x^3.\quad (2)$
How to prove equation $~(2)~$ has no nonzero integer solutions for $~c,\ b,\ x$?
Edit: if $~c,\ b,\ x~$ are integers, we can assume they are coprime.
| Fermat's theorem cannot be proved, if you do not know how to solve Diophantine equations.
This approach cannot be used. This equation has a solution. $$(c-b)^2+3cb=x^3$$
The solutions have the form:
$$x=3(3p^2+s^2)^2$$
$$c=(3p^2+s^2)(36p^2s^2+36ps(3p^2-s^2)-3(3p^2-s^2)^2)$$
$$b=(3p^2+s^2)(36p^2s^2-36ps(3p^2-s^2)-3(3p^2-s^2)^2)$$
$p,s$ - any integer.
You can write another solution. This is equivalent to the equation:
$$x^2+xy+y^2=z^3$$
The solutions have the form:
$$x=s^3+3ps^2-p^3$$
$$y=p^3+3sp^2-s^3$$
$$z=p^2+ps+s^2$$
$$...$$
$$x=s(p^2+ps+s^2)$$
$$y=p(p^2+ps+s^2)$$
$$z=p^2+ps+s^2$$
$$...$$
Will make a replacement.
$$b=3p^2+6ps+2s^2$$
$$t=6s^2+6ps$$
$$q=3p^2+6ps+4s^2$$
The solution has the form:
$$x=q(3b^2-6bt-t^2)$$
$$y=q(3b^2+6bt-t^2)$$
$$z=3q^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Equality of two iterated square roots Solve for $x$:
$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}}=\sqrt{x\sqrt{x\sqrt{x\sqrt{x\dots}}}}$
My attempt:
The L.H.S is equal to $\dfrac{1+\sqrt{4x+1}}{2}$ and R.H.S equals $x^2$
Equating both sides:
$\implies 4x+1=(2x^2-1)^2$
$\implies 4x+1=4x^4-4x^2+1$
$\implies 4x^4-4x^2-4x=0$
$\implies x(x^3-x-1)=0$
Disregarding the complex roots,
$\implies x=0$ 0r $\dfrac{1}{3}\sqrt[3]{\dfrac{27-3\sqrt{69}}{2}}+\dfrac{\sqrt[3]{\dfrac{9+\sqrt{69}}{2}}}{3^{2/3}}$
Is my solution correct? By the way I would like to see other methods to solve it. Thanks!
| Note that RHS is wrong:
$$y=\sqrt{x\sqrt {x\cdots}}\implies y^2=x\sqrt{x\sqrt {x\cdots}}=xy\implies y=0 \text{ or }x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/888130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ I'm looking for an asymptotic equivalent of
$$\sum_{0 < k \le n} \frac{2^k}{k}$$
as $n \to \infty$. A plausible candidate seems to be $\frac{2^{n+1}}{n+1}$ (WolframAlpha plot, and the intuitive similarity with $\sum_{k \le n} 2^k = 2^{n+1}$ is also appealing), but my usual tricks seem powerless here.
I've tried:
*
*Interpreting $x^k/k$ as the primitive of $x^{k-1}$ and setting $x=2$
*Replacing $2^k/k$ with $\int_{x=0}^2 x^{k-1}$
*Reordering the sum's terms to expose log-resembling sub-sums like $\sum_k 1/k$
*Finding lower and upper bounds asymptotically equivalent to $\frac{2^{n+1}}{n+1}$ -- the lower bound is easy ($\sum\limits_{k \le n} \frac{2^k}{k} \ge \sum\limits_{k \le n} \frac{2^k}{n+1} \ge \frac{\sum\limits_{k \le n} 2^k}{n+1} \ge \frac{2^{n+1}}{n+1}$), but the upper bound seems trickier (I couldn't think of a sequence $\varepsilon_k \in o(2^k/k)$ that would make it easy to estimate $\sum_{0 < k \le n} 2^k/k - \varepsilon_k$)
*... and a few others, to no avail
Any hints?
| An asymptotic expansion for large $n$ may also be derived as follows:
\begin{align*}
\sum\limits_{k = 1}^n {\frac{{2^k }}{k}} & = \frac{{2^n }}{n}\sum\limits_{k = 0}^{n - 1} {\frac{{2^{ - k} }}{{1 - k/n}}} = \frac{{2^n }}{n}\sum\limits_{k = 0}^{n - 1} {2^{ - k} \sum\limits_{j = 0}^\infty {\frac{{k^j }}{{n^j }}} } \sim \frac{{2^n }}{n}\sum\limits_{j = 0}^\infty {\frac{1}{{n^j }}\sum\limits_{k = 0}^\infty {k^j 2^{ - k} } } \\ & = \frac{{2^{n + 1} }}{n}\bigg( 1 + \sum\limits_{j = 1}^\infty {2^{j} A_j \!\left( {\tfrac{1}{2}} \right)\frac{1}{{n^j }}} \bigg)= \frac{{2^{n + 1} }}{n}\bigg( 1 + \sum\limits_{j = 1}^\infty {\frac{A_j (2)}{{n^j }}} \bigg).
\end{align*}
Here $A_j$ denotes the $j$th Eulerian polynomial. In terms of the Stirling numbers of the second kind
$$
A_j (2) = \sum\limits_{k = 1}^j k! \,S(j,k) .
$$
Thus,
$$\tag{1}
\sum\limits_{k = 1}^n {\frac{{2^k }}{k}} \sim \frac{{2^{n + 1} }}{n}\left( {1 + \frac{1}{n} + \frac{3}{{n^2 }} + \frac{{13}}{{n^3 }} + \frac{{75}}{{n^4 }} + \frac{{541}}{{n^5 }} + \ldots } \right)
$$
as $n\to +\infty$. The numbers $1,3,13,75,541,\ldots$ are the ordered Bell numbers. It follows from their asymptotic behaviour that the best approximation coming from $(1)$ is achieved by taking only $
\left\lfloor n\log 2 \right\rfloor+1$ terms in the asymptotic series.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 4
} |
How to simplify this Boolean expression F=(A+B+C)(A+B+C')(A+B'+C')
I used sop method and I am left with A+BC', so the above expression should leave me with (A+B)(A+C'). Iam not able to get to this answer. Help is appreciated.
| $\begin{align}
F & =(A+B+C)(A+B+C')(A+B'+C')
\\ & = (A+B+C)\Bigl((A+B+C')(A+B+C')\Bigr)(A+B'+C')
\\ & = \Bigl((A+B+C)(A+B+C')\Bigr)\Bigl((A+B+C')(A+B'+C')\Bigr)
\\ & = (A+B)(A+C')
\\ & = A+AB+AC'+BC'
\\ & = A + BC'
\end{align}$
| {
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"url": "https://math.stackexchange.com/questions/889925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence of infinite $ \sum (\frac{n}{n+1})^{n^2} $? I have being trying to solve this convergence but with no success. Using the ratio test I have reached here:
$$
a_n = \left( \frac{n}{n+1} \right)^{n^2}
$$
And
$$
\frac{1}{a_n} = \left( \frac{n+1}{n} \right)^{n^2}
$$
Also
$$
a_{n+1} = \left(\frac{n+1}{n+2}\right)^{{(n+1)}^2} = \left(\frac{n+1}{n+2}\right)^{n^2+2n+1} = \\ a_{n+1} = \left(\frac{n+1}{n+2}\right)^{n^2} \cdot \left(\frac{n+1}{n+2}\right)^{2n} \cdot \left(\frac{n+1}{n+2}\right)
$$
Therefore
$$
\lim \frac{a_{n+1}}{a_n} = \lim \frac{1}{a_n} \cdot a_{n+1} = \\
\lim \left[ \left( \frac{n+1}{n} \right)^{n^2} \cdot \left(\frac{n+1}{n+2}\right)^{n^2} \cdot \left(\frac{n+1}{n+2}\right)^{2n} \cdot \left(\frac{n+1}{n+2}\right) \right]
$$
And I dont know how to go any further, I have tried many possibilities none of them reaching any result.
| The Bernoulli inequality proves that the sequence $\{a_n\}_{n\in\mathbb{N}^+}$ defined by:
$$ a_n = \left(1+\frac{a}{n}\right)^n $$
is an increasing sequence converging towards $e^a$. By taking $a=-1$, this gives that your series is positive and bounded by:
$$\sum_{n\geq 1}e^{1-n} = \frac{e^2}{e-1}.$$
| {
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Matrix Multiplication Understanding I have to multiply two 3x3 matrices. I don't understand the answer.
$$A=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$$
$$B=\begin{pmatrix}1&0&-1\\ 0&1&-1\\ 0&0&1 \end{pmatrix}$$
The answer given online is $$AB=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$$
I don't understand how this is so. I have said that:
$$AB=\begin{pmatrix}(1)(1)+(0)(0)+(1)(0) & (1)(0)+(0)(1)+(1)(0) & (1)(-1)+(0)(-1)+(1)(1)\\(0)(1)+(1)(0)+(0)(0) & (0)(0)+(1)(1)+(1)(0) & (0)(-1)+(1)(-1)+(1)(1) \\ (0)(1)+(0)(0)+(1)(0) & (0)(0)+(0)(1)+(1)(0) & (0)(-1)+(0)(-1)+(1)(1) \end{pmatrix}$$
$$AB=\begin{pmatrix}1 & 0 & (1)(-1)+(0)(-1)+(1)(1)\\0 & 1 & (0)(-1)+(1)(-1)+(1)(1) \\ 0 & 0 & 1 \end{pmatrix}$$
$$AB=\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
Can anyone tell me where I am going wrong?
| You have the right answer, but in your solution process there is one little mistake.
$$AB = \begin{pmatrix}
(1)(1)+(0)(0)+(1)(0) & (1)(0)+(0)(1)+(1)(0) & (1)(-1)+(0)(-1)+(1)(1) \\
(0)(1)+(1)(0)+\color{red}{(0)}(0) & (0)(0)+(1)(1)+(1)(0) & (0)(-1)+(1)(-1)+(1)(1) \\
(0)(1)+(0)(0)+(1)(0) & (0)(0)+(0)(1)+(1)(0) & (0)(-1)+(0)(-1)+(1)(1)
\end{pmatrix}$$
should be:
$$AB = \begin{pmatrix}
(1)(1)+(0)(0)+(1)(0) & (1)(0)+(0)(1)+(1)(0) & (1)(-1)+(0)(-1)+(1)(1) \\
(0)(1)+(1)(0)+\color{red}{(1)}(0) & (0)(0)+(1)(1)+(1)(0) & (0)(-1)+(1)(-1)+(1)(1) \\
(0)(1)+(0)(0)+(1)(0) & (0)(0)+(0)(1)+(1)(0) & (0)(-1)+(0)(-1)+(1)(1)
\end{pmatrix}$$
Luckily, this doesn't seem to affect the final answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of elements of order $2$ in $S_n$
How many elements of order $2$ are there in $S_n$?
Using combinatorics I arrived at this:
For $n$ even ($n=2k$) there are ${n\choose2}+{n\choose 2}{n-2\choose 2}\dfrac{1}{2!}+{n\choose 2} {n-2\choose 2}{n-4\choose 2}\dfrac{1}{3!}+\cdots+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\cdots{2\choose 2}\dfrac{1}{k!}$.
For $n$ odd ($n=2k+1$) there are ${n\choose 2}+{n\choose 2}{n-2\choose 2}\dfrac{1}{2!}+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\dfrac{1}{3!}+\cdots+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\cdots{3\choose 2}\dfrac{1}{k!}$
But how do I find the sums?
Seems like I have to use induction. But not quite upto there.
Thanks for the help!!
| Let $n$ be a integer and $J = [\frac{n}{2}]$ and let $T_n^1 = \frac{n(n-1)}{2}$. Then the number of $j$-transpositions ($1< j\leq J$) in $S_n$ is $T_n^j=(n-1)T_{n-j} $. Therefore,
the number of possible transpositions is $$\sum_{j=1}^J T_n^j = \sum_{j=1}^J (n-1)T_{n-j}.$$ For example, if $n=10$ then $\sum_{j=1}^5 T_{10}^j = 45+ 9\big(28+21+15+10 \big)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $(1-a)(1-b)(1-c)(1-d)\geq abcd$ if $a^2+b^2+c^2+d^2=1$ Let $a,b,c,d\geq0$, $a^2+b^2+c^2+d^2=1$
Prove
$\displaystyle (1-a)(1-b)(1-c)(1-d)\geq abcd$
I mutiplied both with $\displaystyle (1+a)(1+b)(1+c)(1+d)$ to use $1-a^2=b^2+c^2+d^2$ and try using the cauchy-schwarz and holder but it is doesn't work.
| Observe that $f(x)=\log\left(\frac1x-1\right)$ is concave for $0<x<1$ (its derivative is $\frac1{x(1-x)}$).
Taking $\log$, the inequality is equivalent to
$$\sum\log\left(\frac1a-1\right)\geq0.$$
By Jensen's inequality,
$$\sum\log\left(\frac1a-1\right)\geq4\log\left(\frac4{a+b+c+d}-1\right).$$
By Cauchy-Schwarz/Power Mean/Hölder, $a+b+c+d\leq2\sqrt{a^2+b^2+c^2+d^2}=2$, hence
$$LHS\geq4\log\left(\frac42-1\right)=0,$$
as was to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/895924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Radical Inequality $\sqrt{2x-1}$ + $\sqrt{3x-2}$ > $\sqrt{4x-3}$ + $\sqrt{5x-4}$
I have attempted to solve this by squaring each side, resulting in
$5x + 2\sqrt{2x-1}\sqrt{3x-2} - 3 > 9x + 2\sqrt{(4x-3)(5x-4)} - 7 $
$4 + 2\sqrt{2x-1}\sqrt{3x-2} > 4x + 2\sqrt{(4x-3)(5x-4)}$
$1 + 1/2\sqrt{2x-1}\sqrt{3x-2} > x + 1/2\sqrt{(4x-3)(5x-4)}$
After this my thought was to square again, but I don't think that'd help too much - and I don't even know if this is correct anymore.
| Note that
$$ x > y \Leftrightarrow \sqrt{x} > \sqrt{y}.
$$
So if $x \geq 1$,
$$
4x-3 = 2x-1+2(x-1) \geq 2x-1
$$
and thus
$\sqrt{4x-3} \geq \sqrt{2x-1}.
$
Similarly $\sqrt{5x-4} \geq \sqrt{3x-2}$. Just put those together. If $x < 1$, inequalities are reversed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/897029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Packing of rectangular blocks There are 3 cubes side length of each a, b and c. There are also 3 identical rectangular blocks with sides measuring (a + b ), (b + c) and (c + a ).
Can they be assembled together to form a cube of side (a + b + c ) ? ...
by virtue of the identity (a+b+c)^3 = a^3 + b ^3 + c^3 + 3( a + b )(b +c)( c+a).
If so how, if not, why not?
| No.
The point is that if it could be done, faces of the whole cube would be composed of the faces of original blocks. Every block can contribute maximum of half of it's faces for the cubes faces (no two opposite faces can lie on the cubes faces), so the total area of the faces of the cube is at most the half of the sum of the areas of faces of the blocks.
The former is
$$
6(a+b+c)^2 = 6a^2+6b^2+6c^2+12ab+12bc+12ca
$$
and the latter
$$
3a^2+3b^2+3c^2+3((a+b)(b+c)+(b+c)(c+a)+(c+a)(a+b)) \\
= 6a^2+6b^2+6c^2+9ab+9bc+9ca.
$$
So, basically, the bricks don't have enough area for the cube.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/901226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to find $\lim_{x\to 0}\frac{\tan 3x}{\tan 5x}$? I am asked to find the following limit:
$$ \lim_{x \to 0} \frac{\tan 3x}{\tan 5x}$$
My problem is in simplifying the function. I followed two different approaches to solve the problem. But both seems incorrect.
Apprach 1)
Since $\tan \theta = \frac{sin \theta}{cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$, we have:
$$\frac{\tan 3x}{\tan 5x} = \frac{\sin 3x \times \cos 5x}{\sin 5x \times \cos 3x}$$
This approach does not work well, because I cannot simplify more.
Approach 2)
Since $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \times \tan y)}$, we have:
$$\frac{\tan 3x}{\tan 5x} = \frac{\tan 3x}{\tan 3x + \tan 2x} = \frac{\tan 3x}{\frac{\tan 2 + \tan3}{1 - \tan 2 \times \tan 3}}$$
What am I doing wrong?
| Note that
$$\lim_{x \rightarrow 0} \frac{\sin(3x)}{\sin(5x)} =
\lim_{x \rightarrow 0} \frac{3\cos(3x)}{5\cos(5x)} = \frac{3}{5}$$
by L'Hopital's rule. You can use this fact in approach 1 since $\cos(3x) \rightarrow 1$ and $\cos(5x) \rightarrow 1$.
Details:
$$\begin{align}
\lim_{x \rightarrow 0} \frac{\sin(3x) \cos(5x)}{\sin(5x)\cos(3x)}
&= \left(\lim_{x \rightarrow 0} \frac{\sin(3x)}{\sin(5x)}\right)
\left(\lim_{x \rightarrow 0} \frac{\cos(5x)}{\cos(3x)}\right) \\
&= \left(\frac{3}{5}\right) \left(\frac{1}{1}\right) \\
&= \frac{3}{5} \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/905745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Differentiate with product rule Question: differentiate $x(x^2 +3x)^3$
I have gotten to the point where i've used the product rule and i've gotten
$$(x^2 + 3x)^3 + x\cdot(3x+9)(x^2 + 3x)^2$$
but now that it comes to the simplifying im completely at a loss, any help would greatly appreciated
| Use the $\color{red}{\text{product rule}}$ and the $\color{blue}{\text{chain rule}}$
$$\begin{array}{rcl}(x(x^2+3x)^3)' &\color{red}{=}& x'\cdot (x^2+3x)^3+((x^2+3x)^3)'\cdot x \\ &\color{blue}{=}& (x^2+3x)^3+x(3(x^2+3x)^2\cdot (x^2+3x)') \\ &=& (x^2+3x)^3+x(3(x^2+3x)^2(2x+3))\\ &=& (x^2+3x)^3+3x(x^2+3x)^2(2x+3)\\ &=&x^3 (x+3)^2 (7 x+12) \end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/905909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Make $n$ cents with $1$-cent, $2$-cent, and $3$-cent coins I encountered the following problem in Herber Wilf's book Generatingfunctionology:
Prove that, in country that has $1$-cent, $2$-cent, and $3$-cent coins
only, the number of ways of changing $n$ cents is exactly the integer
nearest to $ \frac{(n+3)^2}{12} $.
Of course, the solution provided was one that makes use of a generating function. However, it is unbelievably tedious. Is there a way to do this without making use of a generation function?
For reference, here is the solution given in the book.
If $f(n)$ is that number, then
$$ \begin {gather*}
\sum_{n\ge 0} f(n) x^n =
\frac {1}{(1-x)(1-x^2)(1-x^3)}\qquad\qquad\qquad\qquad \\
= \frac{1}{6(1-x)^3} + \frac{1 }{4(1-x)^2} +
\frac {17}{72(1-x)} + \frac {1}{8(1+x)} + \frac {1}{9(1-\omega x)} +
\frac {1}{9(1-\overline{\omega}x)}.
\end {gather*} $$
If we expand each of
the fractions on the right, we find the formula $$ f(n) = \frac {1}{6}
\dbinom {n+2}{2} + \frac {1}{4} (n+1) + \frac {17}{72} + \frac
{(-1)^n}{8} + \frac {2}{9} \cos \left( \frac {2n \pi}{3} \right), $$
which can be rewritten as
$$ f(n) = \frac {(n+3)^2}{12} + \frac {-7 + 9 (-1)^n
+ 16 \cos \left( \frac {2n\pi}{3} \right)}{72}. $$
The second
fraction cannot exceed $ \frac {32}{72} < \frac {1}{2} $ in absolute
value, so $f(n)$ is the unique integer who distance from
$ \frac {(n+3)^2}{12} $ is less than $\frac{1}{2}$, as required. $\Box$
| It is possible to shorten a bit the argument by noticing that
$$\operatorname{Res}\left(\frac{1}{(1-z)(1-z^2)(1-z^3)},z=1\right)=-\frac{17}{72},$$
hence the contribute given by the simple poles $-1,\omega,\omega^2$ lying on the unit circle cannot exceed $\frac{17}{72}$, and is exactly equal to $\frac{17}{72}$ when $6\mid n$. The pole in $z=1$ is a triple pole, hence $f(n)$ grows like a second-degree polynomial. By interpolating the values of $f(n)$ on $n=0,n=6,n=12$ we can derive our formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $ \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\;,$ Then the Remainder when $x$ is Divided by $11$
If $\displaystyle \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\;,$ Then the Remainder when $x$ is Divided by $11$.
$\bf{My\; Try::}$ Given $\displaystyle \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\Rightarrow 13!\left(1+\frac{1}{2}+\frac{1}{3}+................+\frac{1}{13}\right) = x$
So $\displaystyle x = \left(13!+\frac{13!}{2}+\frac{13!}{3}+.................+12!\right)$
Now How Can I solve after that
Help me
Thanks
| Each term in your expression (each $\frac{13!}{k}$) is an integer, and except for the $k=11$ case is expressed as a rational fraction where the numerator contains a factor of 11 but the denominator does not. Hence, each term except $k=11$ is 0 mod 11. So the answer is going to be
$$
\frac{13!}{11} \mod 11 = (13 \cdot 12 \cdot \prod_{n=1}^{10}n) \mod 11
= (2 \cdot 1 \cdot (-1)) \mod 11 = -2 \equiv 9 \mod 11
$$
Here I used the fact (Fermat) that for $p$ prime, $p! \equiv -1 \mod p$
So the remainder will be 9.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of $\arcsin$ of a rational function, using integration by parts I'm a class 12 student and this a question from my textbook:
$$I=\int{\arcsin{2x\over 1+x^2}}\mathrm{d}x$$
I did it using integration by parts like this:
$$I=\arcsin{\left(2x\over 1+x^2\right)}\cdot\int1\cdot\mathrm{d}x-\int(\frac{\mathrm{d}}{\mathrm{d}x}\arcsin{\left(2x\over1+x^2\right)}\int1\cdot\mathrm{d}x)\mathrm{d}x+c$$
Now,
$$\frac{\mathrm{d}}{\mathrm{d}x}\arcsin{\left(2x\over1+x^2\right)}\
= {1\over\sqrt{1-\left(2x\over1+x^2\right)^2}}
{ 2(1+x^2) - (2x)(2x)\over(1+x^2)^2}\
={1+x^2(2(1-x^2)\over(1-x^2)^2(1+x^2)^2}\
={2\over1-x^4}$$
So
$$I=\arcsin{\left(2x\over 1+x^2\right)}(x)-\int{2\over1-x^4}{x}\mathrm{d}x + c$$
let $I_1=\int{(2x)\over(1-x^4)}\mathrm{d}x$
$$I_1=\int{(2x)\over(1-x^2)(1+x^2)}\mathrm{d}x$$
Let $x^2=t$
So $2x\mathrm{d}x = \mathrm{d}t$ and
$$I_1=\int{\mathrm{d}t\over(1-t)(1+t)}\
=$\int{\mathrm{d}t\over1-t^2}\
={1\over2}\log{|1+t|\over|1-t|}+c_2\
={1\over2}\log{1+x^2\over1-x^2}+c_2$$
From all of this, we conclude
$$I=x \arcsin{\left(2x\over 1+x^2\right)}-{1\over2}\log{|1+x^2|\over|1-x^2|}+c$$
But the answer given in the book is :
$$(2x)\arctan x-{\log(1+x^2)} + c$$
I know they have done this using
$$\arcsin{\left(2x\over 1+x^2\right)}=2\arctan x$$
And then applied integration by parts, but I'd very much like to know where I went wrong.
Any help appreciated.
| $$\dfrac{\mathrm{d}}{\mathrm{d}x} \arcsin{\frac{2x}{1+x^2}}= {\dfrac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}} \cdot { 2(1+x^2) - (2x)(2x)\over(1+x^2)^2}=$$
$$=\color{red}{ \dfrac{2(1+x^2)(1-x^2)}{|1-x^2|(1+x^2)^2}} =$$
1)$\color{red}{\dfrac{2}{1+x^2}}, x\in (-1,1)$
2)$\color{red}{-\dfrac{2}{1+x^2}}, x\in (-\infty,-1)\cup(1, \infty).$
Then integrating by parts, as you started, expressing full that the three intervals...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the sum of this series May I know how I should go about finding the sum of this series?
$\displaystyle\sum_{n=1}^\infty$ $\dfrac{n}{2^{n-1}}$
I am really stuck. Thanks!
| Here is a way without differentiation.
Let
$$S=\sum_{n=1}^\infty \frac{n}{2^{n-1}}=\frac{1}{1}+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\dotsb$$
Now, let's see if we can get $\displaystyle\sum_{n=1}^\infty \frac{n-1}{2^{n-1}}$ in terms of $S$ in two different ways.
First way:
\begin{align}
\sum_{n=1}^\infty \frac{n-1}{2^{n-1}}&=\sum_{n=1}^\infty \frac{n}{2^{n-1}}-\sum_{n=1}^\infty \frac{1}{2^{n-1}} \\
&=S-\left(1+\frac{1}{2}+\frac{1}{4}+\dotsb\right) \\
&=S-\frac{1}{1-\frac{1}{2}}=S-2
\end{align}
Second way:
\begin{align}
\sum_{n=1}^\infty \frac{n-1}{2^{n-1}}&=\frac{1}{2}\sum_{n=1}^\infty \frac{n-1}{2^{n-2}} \\
&=\frac{1}{2}\left(0+\frac{1}{1}+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\dotsb\right) \\
&=\frac{1}{2}S
\end{align}
Since these must be equal, we have:
$$S-2=\frac{1}{2}S$$
Solving, we get $S=4$, which is the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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summing the powers of a complex number Let $z=e^{\frac{2\pi i}{5}}$, then $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=?$
I am kind of confused since by drawing a graph, $1+z+z^2+z^3+z^4$ should be zero, but using computational softwares the result is different, and hence I do not know how to solve this problem.
Thanks for helping!
| According to wolframalpha you are right about $1+z+z^2+z^3+z^4 = 0$
For the sum :
$1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9$
$(1+z+z^2+z^3+z^4)+4z^4+4z^5+4z^6+4z^7+4z^8+5z^9$
$0+4z^4(1+z+z^2+z^3+z^4)+5z^5*z^4$
$4z^4(0)+5*1*z^4$
$5*z^4$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\int_0^\pi \sin(x)\,dx$ explicitly A book asks me to prove that:
$$\int_0^{\pi}\sin(x)\,dx = 2$$
Using the identity:
$$\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{2\pi}{n}\right) + \cdots + \sin\left(\frac{n\pi}{n}\right) = \frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}$$
And the famous $\lim_{x\to0}\frac{\sin(x)}{x} = 1$
What I tried:
Using the Right Riemann Sum Method:
$$\int_a^{b}\sin(x)\,dx \approx \Delta x\left[f(a + \Delta x) + f(a + 2\,\Delta x) + \cdots + f(b)\right]$$
By taking $\Delta x = \frac{\pi}{n}$, $a = 0$ and $b = \pi$ we have:
$$\int_0^{\pi}\sin(x)\,dx \approx \Delta x\left[f(\Delta x) + f(2\,\Delta x) + \cdots + f\left(\frac{n\pi}{n}\right)\right] = \frac{\pi}{n}\left[\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{2\pi}{n}\right) + \cdots + \sin\left(\frac{n\pi}{n}\right)\right] = \frac{\pi}{n}\left[\frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}\right]$$
So
$$\int_0^\pi \sin(x)\,dx = \lim_{n\to\infty} \frac{\pi}{n}\frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}$$
I can't see, however, how to prove this limit to be $=2$
| Using $$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$
with $$\lim_{n\to\infty}\frac{\pi/(2n)}{\sin(\pi/(2n))}=1,$$
We have $$\frac{\pi/(2n)}{\sin(\pi/(2n))}\cdot (-2)\sin\left\{\frac{1+(2/n)}{2}\pi\right\}\sin\left(-\frac{\pi}{2}\right)\to 1\cdot (-2)\cdot 1\cdot (-1)=2\ (n\to \infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/913501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Find the side of an equilateral triangle given only the distance of an arbitrary point to its vertices Triangle $ABC$ is an equilateral triangle and $P$ is an arbitrary point inside it. The distance from $P$ to $A$ is $4$ and the distance from $P$ to $B$ is $6$ and the distance from $P$ to $C$ is $5$. How to find the side of an equilateral triangle from this information?
| Let $P=0\in{\mathbb C}$, and let $M={d\over\sqrt{3}}>0$ be the center of the triangle. When $s$ is the side length of the triangle then there is an argument $\phi$ such that the three vertices are given by
$$A={1\over\sqrt{3}}(d+se^{i\phi}),\quad B={1\over\sqrt{3}}(d+s\omega e^{i\phi}),\quad C={1\over\sqrt{3}}(d+s\bar\omega e^{i\phi})\ ,$$
where $\omega=e^{2\pi i/3}$ is a third root of unity. When $a$, $b$, $c$ are the given distances we therefore have
$$\eqalign{
a^2&={1\over3}(d^2 + ds(e^{i\phi}+e^{-i\phi})+s^2) \cr
b^2&={1\over3}(d^2 + ds(\omega e^{i\phi}+\bar\omega e^{-i\phi})+s^2) \cr
c^2&={1\over3}(d^2 + ds(\bar\omega e^{i\phi}+\omega e^{-i\phi})+s^2) \cr}\tag{1}$$
Adding these three equations gives
$$a^2+b^2+c^2=d^2+s^2\ .\tag{2}$$
Multiplying the equations $(1)$ with $1$, $\omega$, $\bar\omega$ (resp. with $1$, $\bar\omega$, $\omega$), and adding we obtain
$$\eqalign{
a^2+\omega b^2+\bar\omega c^2&=dse^{-i\phi}\cr
a^2+\bar\omega b^2+\omega c^2&=dse^{i\phi}\ .\cr}\tag{3}$$
In order to eliminate $\phi$ we now multiply the two equations $(3)$. Using $\omega+\bar\omega=-1$ we get
$$a^4+b^4+c^4-(a^2b^2+b^2c^2+c^2a^2)=d^2s^2\ .\tag{4}$$
Writing
$$a^2+b^2+c^2=:\sigma_1,\qquad a^2b^2+b^2c^2+c^2a^2=:\sigma_2$$
we can condense $(2)$ and $(4)$ to
$$d^2+s^2=\sigma_1,\qquad d^2s^2=\sigma_1^2-3\sigma_2\ .$$
This leads to
$$s^4-\sigma_1s^2+\sigma_1^2-3\sigma_2=0\ ,$$
from which $s$ can be determined.
For the given data we obtain $\sigma_1=77$, $\sigma_2=1876$, and this leads to
$s^4-77s^2+301=0$, or $s_1=8.536$, $s_2=2.0324$. (It seems that $s_2$ corresponds to a solution where $P$ is outside the triangle.)
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Evaluation of Integral $ \int\frac{1}{(2+3\sin x)^2}dx$ Evaluation of Integral $\displaystyle \int\frac{1}{(2+3\sin x)^2}dx$
$\bf{My\; Try::}$ Using Integration by parts,
Let $$\displaystyle I=\int\frac{1}{(2+3\sin x)^2}dx = \int \frac{1}{\cos x}\cdot \frac{\cos x}{(2+3\sin x)^2}dx$$
$$\displaystyle I = -\frac{1}{3 \cos x}\cdot \frac{1}{(2+3\sin x)}+\frac{1}{3}\int \frac{\sin x}{\cos^2 x}\cdot \frac{1}{(2+3\sin x)}dx$$
Now Let $$\displaystyle J=\int \cdot \frac{\sin x}{(2+3\sin x)}\cdot \frac{1}{\cos^2 x} = \int \frac{\sin x}{(2+3\sin x)}\cdot \sec^2 xdx$$
again Using Integration parts
$$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x - \int \frac{d}{dx}\left\{\frac{\sin x}{(2+3\sin x)}\right\}\cdot \tan xdx$$
So $$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x-2\int\frac{\sin x}{(2+3\sin x)^2}dx$$
$$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x-\frac{2}{3}\int \frac{[(2\sin x+3)-3]}{(2\sin x+3)^2}dx$$
So $$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x- \frac{2}{3}\int\frac{1}{(2\sin x+3)}dx+4I$$
Now Let $$\displaystyle K = \int\frac{1}{2\sin x+3}dx = \int\frac{2\sin x-3}{4\sin^2 x-9}dx = -2\int\frac{\sin x}{5+4\cos^2 x}dx+3\int\frac{\sec^2 x}{5\tan^2 x+9}dx$$
My Question is can we solve it Without Using Integration by parts
If Yes, The plz explain here
thanks
| Hint: Let $~I(b)=\displaystyle\int\frac{dx}{a\sin x+b},~$ and solve it using the Weierstrass substitution. Then notice
that your original integral is nothing more than $-I'(b)$.
| {
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Evaluation of a class of continued fractions Is there a closed-form way of writing the continued fraction:
$$
1 + \frac{2}{3+ \frac{4}{5 + \frac{6}{7 + ...}}}
$$
EDIT: Since the above has been determined as $\frac{1}{\sqrt{e}-1}$, is there a similar expression for:
$$
2 + \frac{3}{4+ \frac{5}{6 + \frac{7}{8 + ...}}}
$$
More generally, are there general closed-form expressions for all continued fractions of the form:
$$
a_n = n + \frac{n+1}{(n+2) + \frac{n+3}{(n+4) + ...}} \\
f(x) = x + \frac{x+1}{(x+2) + \frac{x+3}{(x+4) + \cdots}} = x + \frac{x+1}{f(x+2)} \\
f(x) f(x+2) = xf(x+2) + x+1
$$
And can said closed form be extended to all real numbers? For example, I experimented with extending the sequence to negative values of n and found that for all negative odd $n, a_n = -1$.
| If $$z_1 = 1 + 2/(3 + 4/(5 + 6/( ... = {1\over \exp(1/2)-1 }$$ (as given in the answer before) and $$z_3 =3 + 4/(5 + 6/( ... $$ then $$ z_1 = 1 + 2/z_3$$ or $$z_3 = { 2 \over z_1-1} = { 2 \over {1 \over \exp(1/2)-1}-1} = { 2 \exp(1/2)-2 \over 2 - \exp(1/2) } $$
From this $z_5,z_7,...$ follow analoguously, and they are all rational compositions of the "magic" constant $\beta= \frac 1{\sqrt{ e}}$ in the form $$ { 1\over z_{2k+1} +1} = a_{2k+1} + b_{2k+1} \beta $$.
For the even indexed $z_{2k}$ the rule is the same, however I did not yet find their "magic constant", say $\gamma$ .
Just for the visualization, I made a table focusing on the cases, where the continued fraction becomes rational. The grey-shaded entries indicate evaluation of the cont-fractions to irrational numbers. The two yellow entried are the two known irrational numbers:
| {
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Proving a second derivative Given that $$y = \sin^3 x + \cos^3 x$$
prove that $$\frac{d^2 y}{dx^2} = \frac{3}{2} (\cos x + \sin x)(3 \sin 2x - 2)$$
I began with differentiating the equation as it is and it took me around twelve steps to reach the answer, and I'm guessing that factoring the sum of the cubes before differentiating wouldn't make it any easier. Is there a simpler approach to this problem?
Thanks in advance.
| Note that $$y=\sin^3 x+ \cos^3 x=(\sin x+\cos x)(\sin^2 x - \sin x\cos x+ \cos^2 x)=(\sin x+\cos x)(1- \frac12\sin 2x)$$
Then $$y'=-(\sin x+\cos x)\cos 2x+(\cos x-\sin x)(1- \frac12\sin 2x)$$ and $$y''=2(\sin x+\cos x)\sin 2x-2(\cos x-\sin x)\cos 2x-(\sin x+\cos x)(1- \frac12\sin 2x)$$
For the middle term $\cos 2x= \cos^2 x-\sin^2 x=(\cos x +\sin x)(\cos x-\sin x)$ and $(\cos x- \sin x)^2=\cos^2 x-2\cos x\sin x+ \sin^2 x=1-\sin 2x$
This gets you a factor $\cos x+ \sin x$ and the remaining factor in terms of $\sin 2x$, so it remains to simplify.
| {
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Prove $a^2+b^2+c^2=\frac{6}{5}$ if $a+b+c=0$ and $a^3+b^3+c^3=a^5+b^5+c^5$
if $a,b,c$ are real numbers that $a\neq0,b\neq0,c\neq0$ and $a+b+c=0$ and $$a^3+b^3+c^3=a^5+b^5+c^5$$
Prove that $a^2+b^2+c^2=\frac{6}{5}$.
Things I have done: $a+b+c=0$ So $$a^3+b^3+c^3=a^5+b^5+c^5=3abc$$
also $$a^2+b^2+c^2=-2ab-2bc-2ca$$
I tried multiplying $(a^3+b^3+c^3)$ and $a^2+b^2+c^2$ but I was not able to reach a useful result from that.
| There are probably cleverer methods for this problem, but here is an approach using the general theory of what are called symmetric polynomials. We use the Newton-Girard identities in order to write power-sum symmetric polynomials in terms of elementary symmetric polynomials. Here,
$$p_k:=x_1^k+\cdots+x_n^k \qquad e_k=\sum_{1\le i_1<\cdots<i_k\le n} x_{i_1}\cdots x_{i_k}$$
In particular, with three variables $a,b,c$ we have
$$p_1=a+b+c, \quad p_2=a^2+b^2+c^2, \quad p_3=a^3+b^3+c^3, \quad p_5=a^5+b^5+c^5 $$
$$e_1=a+b+c, \quad e_2=ab+bc+ac, \quad e_3=abc.$$
We know that $p_1=e_1$ which in our case $=0$, and that $p_3=p_5$. Using the formulas for $p_k$ in terms of the other $e_k$s given in the link, and setting $e_1=0$, we have
$$p_2=-2e_2, \quad p_3=3e_3, \quad p_5=-5e_3e_2.$$
Since $p_3=p_5\implies 3e_3=-5e_3e_2\implies e_2=-\frac{3}{5}$, we can plug into $p_2=-2e_2$ to get $p_2=\frac{6}{5}$.
| {
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Prove $\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=0$ if $\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0$
if $a,b,c$ are real numbers and $$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0$$
Prove $$\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=0$$
things i have done: using the assumption i deduced that $$\frac{a}{b-c}=-\frac{(b-c)(a-b-c)}{(c-a)(a-b)}\rightarrow\frac{a}{(b-c)^2}=\frac{(c+b-a)}{(c-a)(a-b)}$$
So some rewriting $$\frac{(c+b-a)}{(c-a)(a-b)}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2} =\frac{(c+b-a)(c-a)(a-b)+b(a-b)^2+a(c-a)^2}{(c-a)^2(a-b)^2}$$
I tried to write numerator in expanded form but it was to large and it seemed like there was not going to be something useful after factoring. so is there any better way than my brute force ? can my solution get continued?
| $$\left(\frac a{b-c}+\frac b{c-a}+\frac c{a-b}\right)\left(\frac1{b-c}+\frac1{c-a}+\frac1{a-b}\right)$$
$$=\sum\frac a{(b-c)^2}+\sum\frac b{c-a}\left(\frac1{b-c}+\frac1{a-b}\right)$$
$$=\sum\frac a{(b-c)^2}-\sum\frac b{c-a}\left(\frac{c-a}{(b-c)(a-b)}\right)$$
$$=\sum\frac a{(b-c)^2}-\frac{\sum b(c-a)}{(b-c)(a-b)(c-a)}$$
Now, $\displaystyle\frac{\sum b(c-a)}{(b-c)(a-b)(c-a)}=0$
| {
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Find the derivative of the function $f(x)=(5-x^2)(\sqrt{x})$ I am stuck on this question, not sure what I am doing wrong:
\begin{align*}
f(x) &=(5-x^2)(\sqrt{x}) \\
\\
f'(x) &=(5-x^2)(x)^{1/2} \\
&=(-2x)(x)^{1/2}+(-x^2+5)(\dfrac{1}{2})(x)^{-1/2}\\
&=-2x^{3/2}+\dfrac{1}{2}(-x^{3/2}+5x^{-1/2})\\
&=\dfrac{1}{2}x^{-1/2}\bigg(x^2-2x^2+10\bigg)\\
&=\dfrac{1}{2}x^{-1/2}\bigg(-x^2+10\bigg)
\end{align*}
| Answer
In your fourth step:
$f'(x) = \frac{-5}{2}x^{\frac{3}{2}} + \frac{5}{2}x^{\frac{-1}{2}}$
$f'(x) =-\frac{5}{2}\left(\frac{x^2 -1}{\sqrt{x}}\right)$
| {
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Indefinite integral of a rational function: $\int\frac{6x+4}{x^2+4}\,dx$ Find $\displaystyle\int\frac{6x+4}{x^2+4}\,dx$
The question asks to find integral of the expression so I divided them into two parts:
$$
\int\frac{6x}{x^2+4}\,dx
$$
and
$$\int\frac{4}{x^2+4}.
$$
So, for the first integral, I set $u=x^2+4$ and got $\int\frac{3}{u}\,du$ which is $3\ln|x^2+4|+c$. But I don't know how to integrate the second part.
| We have
\begin{align}
\int\frac{6x+4}{x^2+4}dx&=\int\frac{6x}{x^2+4}dx+\int\frac{4}{x^2+4}dx\\&=3\ln(x^2+4)+2\arctan \frac{x}{2}+c
\end{align}
| {
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Turning $2\le x$ into $\sqrt{1+\frac{4}{x^6}}\le \sqrt{5}$? I am supposed to turn $2\le x$ into $\sqrt{1+\frac{4}{x^6}}\le \sqrt{5}$, and I have no idea on how to approach this. I'll post my steps, even though I don't think they'll be of much help.
$$2\le x \rightarrow \frac{1}{x} \ge \frac{1}{2} \rightarrow\frac{4}{x}\ge2$$
and that is pretty much all I got. Thanks for all the help in advance!
| In fact, we only need that $x \geq 1$:
\begin{align*}
x \geq 2
&\implies x \geq 1 \\
&\implies x^6 \geq 1^6 \\
&\implies \frac{1}{x^6} \leq \frac{1}{1^6} \\
&\implies \frac{4}{x^6} \leq \frac{4}{1^6} \\
&\implies 1 + \frac{4}{x^6} \leq 1 + \frac{4}{1^6} \\
&\implies \sqrt{1 + \frac{4}{x^6}} \leq \sqrt{1 + \frac{4}{1^6}} = \sqrt{5} \\
\end{align*}
| {
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Prove the sum to n terms of the series Prove that the sum to $n$ terms of a sequence:
$$\frac{1^2}{1\times 3}+\frac{2^2}{3\times 5}+\frac{3^2}{5\times 7}+\cdots $$
is
$$
\frac{n(n+1)}{2(2n+1)}
$$
| If you don't notice the telescoping sum, since you already know the answer in advance you can take a more 'dumbass' approach.
Another way of writing this is $$\sum_{i=1}^n\frac{i^2}{(2i-1)(2i+1)}=\frac{n(n+1)}{2(2n+1)}$$
We can prove this is true by induction. The formula holds when $n=1$, because
$$\frac{1^2}{1\times3}=\frac{1(1+1)}{2(2\cdot1+1)}$$
Suppose the formula holds when $n=k$, for some positive integer $k$. Then:
\begin{align*}
\sum_{i=1}^{k+1}\frac{i^2}{(2i-1)(2i+1)} &= \left(\sum_{i=1}^k\frac{i^2}{(2i-1)(2i+1)}\right)+\frac{(k+1)^2}{(2(k+1)-1)(2(k+1)+1)} \\
&= \frac{k(k+1)}{2(2k+1)}+\frac{(k+1)^2}{(2k+1)(2k+3)} \\
&= \frac{k+1}{2k+1}\left(\frac{k}{2}+\frac{k+1}{2k+3}\right) \\
&= \frac{k+1}{2k+1}\left(\frac{k(2k+3)+2(k+1)}{2(2k+3)}\right) \\
&= \frac{k+1}{2k+1}\left(\frac{(2k+1)(k+2)}{2(2k+3)}\right) \\
&= \frac{(k+1)((k+1)+1)}{2(2(k+1)+1)}
\end{align*}
Which is exactly what the formula should be when $n=k+1$, so by induction the formula holds for all $n\in\mathbb{N}$.
| {
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Solve the equation: $1+2^x+4^x+8^x+16^x+32^x=3(1+2^x+4^x)$ I am doing some math repetition and am a bit stuck on this exercise:
Solve the equation: $1+2^x+4^x+8^x+16^x+32^x=3(1+2^x+4^x)$.
Now, this is a geometric sum on both the $LHS$ and $RHS$, which I guess is something that I should use to solve the equation...
Another way is to simply start to eliminate terms:
$$1+2^x+4^x+8^x+16^x+32^x=3+3 \times 2^x+3\times4^x$$
$$-2 -2\times 2^x -2\times4^x+8^x+16^x+32^x = 0$$
$$8^x+16^x+32^x = 2 +2\times 2^x +2\times4^x$$
$$8^x+16^x+32^x = 2(1 + 2^x + 4^x)$$
But I am stuck here...
| Setting $2^x=a,$
We have $$1+a+a^2+a^3(1+a+a^2)=3(1+a+a^2)$$
$$\iff (1+a+a^2)(a^3-2)=0$$
If $x$ is real $2^x>0\implies1+a+a^2>0$
So, we have $2=(2^x)^3\iff2^{3x-1}=1$
Now if $b^m=1$
either $m=0,b\ne0$
or $b=1$
or $b=-1,m$ is even
| {
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problem solving logarithmic equation and reaching an equivalence ok so i've had a problem trying to simplify the $\ln\left[ \sqrt{1+\frac{u^2}{a^2}} + \frac{u}{a} \right]$ and this is supposed to be equal to : $\ln [ \sqrt{a^2+u^2} + u ]$
how is this posible ?? i've tried to solve this for more than 2 hours and couldn't get to this equivalence. any suggestions ?
| \begin{align}
& \ln\left( \sqrt{1+\frac{u^2}{a^2}} + \frac{u}{a} \right) = \ln\left(\sqrt{\frac{a^2}{a^2}+\frac{u^2}{a^2}} + \frac u a\right) = \ln\left(\sqrt{\frac{a^2+u^2}{a^2}} + \frac u a\right) \\[10pt]
= {} & \ln\left(\frac {\sqrt{a^2+u^2}} a + \frac u a\right) = \ln\left(\frac{\sqrt{a^2+u^2}+u} a\right) = \ln\left(\sqrt{a^2+u^2} + u\right) - \ln a
\end{align}
If this is viewed as a function of $u$, then $a$ is constant, so it's
$$
\ln\left(\sqrt{a^2+u^2} + u\right) + \text{constant}
$$
| {
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For the given quadratic equation find the value of p For the equation 3x^2 + px + 3 = 0 , p>0, if one of the roots is square of the other, then p is equal to?
Solving the equation, i get the value of p as -6 but the question states that p>0. Is there another way of solving this or is the question wrong?
| The value is $p=3$.
Proof: Let $a, a^2$ be the roots of the equation. Then you get
$$3x^2+px+3 = 3(x-a)(x-a^2) = 3x^2-3(a^2+a)x+a^3,$$
So you first have to solve $a^3=1$. This gives the three solutions (the three cube-roots of unity)
$$a_1=1,\quad a_2=-\frac{1}{2}+\frac{i}{2}\sqrt{3},\quad a_3=-\frac{1}{2}-\frac{i}{2}\sqrt{3}$$
which in turn give the possible values for $p$
$$p_1 = -3(a_1^2+a_1) = -6$$
$$p_2 = -3(a_2^2+a_2) = 3$$
$$p_3 = -3(a_3^2+a_3) = 3$$
$p_1\;$ is your negative value, but $p_2=p_3=3\;$ is positive and therefore $p=3\;$ is the answer to the question.
| {
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symetric inequality for a rational function of three variables $a,b,c$ are positive real numbers such that $a+b+c=1$. Prove that: $$\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2} \geqslant \dfrac{1}{2} $$ I have tried with Cauchy-Schwarz inequality in Engel form...
| $$\sum_{cyc}\frac{a^3}{a^2+b^2}-\frac{1}{2}=\sum_{cyc}\frac{a^3}{a^2+b^2}-\frac{a+b+c}{2}=\sum_{cyc}\left(\frac{a^3}{a^2+b^2}-\frac{a}{2}\right)=$$
$$=\sum_{cyc}\frac{a^3-ab^2}{2(a^2+b^2)}=\sum_{cyc}\left(\frac{a^3-ab^2}{2(a^2+b^2)}-\frac{a-b}{2}\right)=\sum_{cyc}\frac{(a-b)^2b}{2(a^2+b^2)}\geq0.$$
Done!
| {
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Evaluating $\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}$ I did this:
$$\begin{align}
\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x} &\sim \lim_{x \to 0} \frac{(1 + x + x^2)^{1/x} - (1 + x)^{1/x}}{x} = \\
&= \lim_{x \to 0} \left [ (1+x)^{1/x} \frac{\left ( \frac{1 + x + x^2}{1 + x} \right )^{1/x} - 1}{x} \right ] =\\
&= \lim_{x \to 0} (1+x)^{1/x} \cdot \lim_{x \to 0} \frac{\left ( 1 + \frac{x^2}{1 + x} \right )^{1/x} - 1}{x} =\\
&= e \cdot \lim_{x \to 0} \frac{e^{\frac{1}{x} \cdot \ln \left ( 1 + \frac{x^2}{1+x} \right )} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{e^{\frac{x}{1 + x}} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{1}{1 + x} = e
\end{align}$$
Is it right? If it is, how to evaluate the limit faster? It was pretty long the way I did it.
| By applying Lagrange's Mean Value Theorem to the function $f(t) = e^t$ we have that
$$\begin{align}
\frac{1}{x} \left( (1+ \sin x + \sin^2 x)^{1/x} - (1+ \sin x)^{1/x} \right)
&= \frac{1}{x} \left( e^{ \frac{ \log(1+ \sin x + \sin^2 x) }{x} } - e^{ \frac{ \log(1+ \sin x) }{x} } \right)
\\&= \frac{1}{x} e^{\theta_x} \left( \frac{ \log(1+ \sin x + \sin^2 x)}{x} - \frac{ \log(1+ \sin x)}{x} \right)
\\ &= e^{\theta_x} \left( \frac{ \log(1+ \sin x + \sin^2 x) - \log(1+ \sin x)}{x^2} \right)
\end{align}$$
where $\theta_x \in \left( \frac{ \log(1+ \sin x) }{x}, \frac{ \log(1+ \sin x + \sin^2 x) }{x} \right)$. Then by squeezing, it is easy to see that $\lim_{x\to0} \theta_x = 1$.
Furthermore we note that by Taylor Series
$$\begin{align}
\frac{ \log(1+ \sin x + \sin^2 x) - \log(1 + \sin x)}{x^2} &= \frac{ \log(1+ x + x^2 + O(x^3)) - \log(1 + x + O(x^3))}{x^2} \\&= \frac{ x + x^2/2 - ( x - x^2/2) + O(x^3)}{x^2} \\&= 1 + O(x)
\end{align}$$
Now passing to the limit as $x\to0$ in the upper expression yields the desired result.
| {
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Inequality for a rational function of three variables $x,y,z$ are positive real numbers such that $$x^2+y^2+z^2=1$$ Prove that $\dfrac{x^2}{1+2yz}+\dfrac{y^2}{1+2xz}+\dfrac{z^2}{1+2xy} \geqslant \dfrac{3}{5}$.Again, I try with Engel form of Cauchy inequality...
|
$$Z=\dfrac{x^2}{1+2yz}+\dfrac{y^2}{1+2xz}+\dfrac{z^2}{1+2xy} $$
By Cauchy-Schwarz:
$$\left(\dfrac{x^2}{1+2yz}+\dfrac{y^2}{1+2xz}+\dfrac{z^2}{1+2xy} \right)(x^2(1+2yz)+y^2(1+2xz)+z^2(1+2xy))\geq(x^2+y^2+z^2)^2=1$$
So:
$$Z\geq\frac1{x^2+y^2+z^2+2xyz(x+y+z)}=\frac1{1+2xyz(x+y+z)}$$
Now $x+y+z\geq3(xyz)^{1/3}$, but also $1=x^2+y^2+z^2\geq3(xyz)^{2/3}\implies (xyz)^{4/3}\leq1/{9}$
So:
$$Z\geq\frac1{1+2xyz(x+y+z)}\geq\frac1{1+6(xyz)^{4/3}}\geq\frac1{1+6(1/9)}=\frac35$$
| {
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Solve a PDE: $ x(y^2+z)p-y(x^2+z)q=(x^2-y^2)z$ Solve the PDE $$ x(y^2+z)p-y(x^2+z)q=(x^2-y^2)z$$
where, $ p=\displaystyle \frac{\partial z}{\partial x}$ and $ q=\displaystyle \frac{\partial z}{\partial y}$
My attempt:
I start with Lagrange's auxiliary equation
$$\frac{dx}{x(y^2+z)}=\frac{dy}{-y(x^2+z)}=\frac{dz}{z(x^2-y^2)}$$
Relation 1: $$\frac{x \, dx+y \, dy- dz}{x^2y^2+x^2z-y^2x^2-y^2z-x^2z+y^2z} = \frac{x \, dx+y \, dy-dz}{0} $$
Integrating $x \, dx+y \, dy-dz=0$
$$\frac{x^2}{2}+\frac{y^2}{2}-z=c \implies x^2+y^2-2z=c_1 $$
I can't see/find a second relation for me to solve the problem. Please help.
| Using multipliers $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$,
Relation 2: $$\frac{\frac{1}{x} \, dx+\frac{1}{y} \, dy +\frac{1}{z} dz}{y^2+z-x^2-z+x^2-y^2} = \frac{\frac{1}{x} \, dx+\frac{1}{y} \, dy +\frac{1}{z} dz}{0} $$
Integrating $\frac{1}{x} \, dx+\frac{1}{y} \, dy +\frac{1}{z} dz=0$
$$\log x+\log y+\log z=c \implies xyz=e^c = c_2 $$
| {
"language": "en",
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$P(X^2+Y^2<1)$ of two independent n(0,1) random variables Suppose that X and Y are independent n(0,1) random variables.
a) Find $P(X^2+Y^2<1)$
Attempt:
a) Let $U = X^2 + Y^2$, $V = Y$.
Then $X = \sqrt{V^2 -U}$, $Y = V$.
$J = \left| \begin{array}{ccc}
\frac{-1}{\sqrt{V^2-U}} & \frac{V}{V^2-U} \\
0 & 1\\
\end{array} \right| $
Then the joint distribution of $f_{u,v}(u,v)$ is:
$$f_{u,v}(u,v)= \frac{1}{2\pi}e^{\frac{-\sqrt{v^2-u}}{2}}e^{\frac{-u^2}{2}}\frac{1}{\sqrt{v^2-u}}$$
Then $P(X^2 +Y^2 <1)$ is:
$$\int_0^\infty \int_0^{v^2-u} \frac{1}{2\pi}e^{\frac{-\sqrt{v^2-u}}{2}}e^{\frac{-u^2}{2}}\frac{1}{\sqrt{v^2-u}}dudv$$
However, at this point I simply do not know how any tricks to complete this integration.
| Since $X$ and $Y$ are independent, the joint density of $X$ and $Y$ is the product of the individual densities, so it is
$$\frac{1}{2\pi}e^{-(x^2+y^2)/2}$$
on the entire $x$-$y$ plane. We want to find
$$\int_D \frac{x^2+y^2}{2\pi}e^{-(x^2+y^2)/2}\,dy\,dx,$$
where $D$ is the unit disk. Switch to polar coordinates. We want to find
$$\int_{r=0}^{1}\int_{\theta=0}^{2\pi} \frac{r^3}{2\pi}e^{-r^2/2}\,d\theta\,dr.$$
The integration with respect to $\theta$ gets rid of the $2\pi$ in the denominator.
So we want
$$\int_0^1 r^3 e^{-r^2/2}\,dr.$$
This can be done by integration by parts. To make things more familiar, you may wish to first make the substitution $t=r^2/2$. Our integral becomes
$$\int_0^{1/2} 2te^{-t}\,dt.$$
| {
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"answer_id": 3
} |
Limits without L'Hopitals Rule Evaluate the limit without using L'hopital's rule
a)$$\lim_{x \to 0} \frac {(1+2x)^{1/3}-1}{x} $$
I got the answer as $l=\frac 23$... but I used L'hopitals rule for that... How can I do it another way?
b)$$\lim_{x \to 5^-} \frac {e^x}{(x-5)^3}$$
$l=-\infty$
c)$$\lim_{x \to \frac {\pi} 2} \frac{\sin x}{\cos^2x} - \tan^2 x$$
I don't know how to work with this at all
So basically I was able to find most of the limits through L'Hopitals Rule... BUT how do I find the limits without using his rule?
| a) $$\lim_{x\to0}{\frac{\big(1+2x\big)^{\frac{1}{3}}-1}{x}}$$
First, let $f(x)=\big(1+2x\big)^{\frac{1}{3}}$, we know that $f(0)=1$ and $f'(x)=\frac{2}{3\big(1+2x\big)^{\frac{2}{3}}}$ $\Rightarrow$ $f'(0)=\frac{2}{3}$
so $$\lim_{x\to0}{\frac{\big(1+2x\big)^{\frac{1}{3}}-1}{x}}$$
$$=\lim_{x\to0}{\frac{f(x)-f(0)}{x-0}}=f'(0)=\frac{2}{3}$$
b) $$\lim_{x\to5^-}{\frac{e^x}{\big(x-5\big)^3}}$$
$$=\lim_{x\to5^-}{e^x}\cdot\lim_{x\to5^-}{\frac{1}{\big(x-5\big)^3}}$$
$$=e^5\cdot-\infty=-\infty$$
c) $$\lim_{x\to\frac{\pi}{2}}{\frac{\sin x}{\cos^2 x}-\tan^2 x}=L$$
Let's multiply by the conjugate of $\frac{\sin x}{\cos^2 x}-\tan^2 x$:
$$L=\lim_{x\to\frac{\pi}{2}}{\frac{\big(\frac{\sin x}{\cos^2 x}-\tan^2 x\big)\big(\frac{\sin x}{\cos^2 x}+\tan^2 x\big)}{\frac{\sin x}{\cos^2 x}+\tan^2 x}}$$
By smplifying the numerator we get:
$\big(\frac{\sin x}{\cos^2 x}-\tan^2 x\big)\big(\frac{\sin x}{\cos^2 x}+\tan^2 x\big)=\frac{\sin^2 x}{\cos^4 x}-\tan^4 x$
$=\tan^2 x\big(\frac{1-\sin^2 x}{\cos^2}\big)=\tan^2 x$
$$\Rightarrow L=\lim_{x\to\frac{\pi}{2}}{\frac{\tan^2 x}{\frac{\sin x}{\cos^2 x}+\tan^2 x}}$$
$$=\lim_{x\to\frac{\pi}{2}}{\frac{\sin x}{1+\sin x}}=\frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$
Integrate $$I=\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$$
Let $$\begin{align}u^2=\frac{\sin(x-a)}{\sin(x+a)}\implies 2udu&=\frac{\sin(x+a)\cos(x-a)-\sin(x-a)\cos(x+a)}{\sin^2(x+a)}dx\\2udu&=\frac{\sin((x+a)-(x-a))}{\sin^2(x+a)}dx\\
2udu&=\frac{\sin(2a)}{\sin^2(x+a)}dx\end{align}$$
Now:
$$\begin{align}u^2&=\frac{\sin(x+a-2a)}{\sin(x+a)}
\\u^2&=\frac{\sin(x+a)\cos(2a)-\cos(x+a)\sin(2a)}{\sin(x+a)}
\\u^2&=\cos(2a)-\sin(2a)\cot(x+a)
\\\cot(x+a)&=(\cos(2a)-u^2)\csc(2a)
\\\csc^2(x+a)=\cot^2(x+a)+1&=(\cos(2a)-u^2)^2\csc^2(2a)+1
\\\csc^2(x+a)&=\frac{\cos^2(2a)+u^4-2u^2\cos2(2a)+\sin^2(2a)}{\csc^2(2a)}
\\\sin^2(x+a)&=\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}\end{align}$$
Now:
$$\begin{align}
I&=\int u.\frac{2udu\sin^2(x+a)}{\sin(2a)}\\
I&=\int\frac2{\sin(2a)}.u^2.\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}du
\\I&=2\sin(2a)\int\frac{u^2}{u^4-2ku^2+1}du\quad k:=\cos 2a
\\\frac If&=\int\frac{2+u^{-2}-u^{-2}}{u^2-2k+u^{-2}}du=\int\frac{1+u^{-2}}{u^2-2k+u^{-2}}du+\int\frac{1-u^{-2}}{u^2-2k+u^{-2}}du\quad \\f:=\sin(2a)
\\&=\int\frac{d(u-u^{-1})}{(u-u^{-1})^2+2-2k}+\int\frac{d(u+u^{-1})}{(u+u^{-1})^2-2-2k}\end{align}$$
Now: $2-2k=2(1-\cos 2a)=4\sin^24a,2+2k=2(1+\cos 2a)=4\cos^24a$
So:
$$I=\sin2a\left(\frac1{2\sin4a}\arctan\left(\frac{u-u^{-1}}{2\sin(4a)}\right)+\frac1{4\cos4a}\ln\left|\frac{u+u^{-1}-2\cos 4a}{u+u^{-1}+2\cos 4a}\right|\right)+C$$
Or:
$$I=\frac1{4\cos2a}\arctan\left(\frac{-\sin a\cos x}{\sin4a\sqrt{\sin(x+a)\sin(x-a)}}\right)+\frac{\sin2a}{4\cos 4a}\ln\left|\frac{\sin x\cos a-\cos4a\sqrt{\sin(x+a)\sin(x-a)}}{\sin x\cos a+\cos4a\sqrt{\sin(x+a)\sin(x-a)}}\right|+C$$
But the textbook answer is:
$$\cos a\arccos\left(\frac{\cos x}{\cos a}\right)-\sin a\ln(\sin x+\sqrt{\sin^2x-\sin^2a})+c$$
| Let us write $$I_a=\int\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}\mathrm dx$$
and compute $I_+=I_a+I_{-a}$ and $I_-=I_a-I_{-a}$. We get
$$I_\pm=I_a\pm I_{-a}=\int\frac{\sin(x-a)\pm\sin(x+a)}{\sqrt{\sin(x+a)\sin(x-a)}}\mathrm dx.$$
Using trigonometric formulas, one obtains
$$I_+=2\cos a\int\frac{\sin x}{\sqrt{\cos^2a-\cos^2x}}\mathrm dx\\
I_-=2\sin a\int\frac{\cos x}{\sqrt{\sin^2x-\sin^2a}}\mathrm dx.$$
Now just apply the adequate changes of variables
$$I_+=2\cos a \arccos\left(\frac{\cos x}{\cos a}\right)+c_+\\
I_-=-2\sin a \ln\left(\sin x+\sqrt{\sin^2x-\sin^2a}\right)+c_-$$
and the result is $I_a=\frac12\left(I_++I_-\right)$.
| {
"language": "en",
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"source": "stackexchange",
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} |
Finding limits of two functions of two variables Show using the definition of limit that
$$\lim _{(x,y)\to(0,1)}\frac{x^2-y^2}{x^2+y^2} = -1$$
and
$$\lim_{ (x,y)\to(0,0)}\frac{ (1-\cos(xy))\sin y}{(x^2+y^2) }= 0$$
Definition of limit:
$\lim_{(x,y)\to(a,b)} f(x,y) =L$ if and only if for every $\epsilon >0$ exist $\delta>0$ such that if $((x-a)^2+(y-b)^2)^{1/2}<\delta$ then $|f(x,y)-L|<\epsilon$.
| We give more or less full detail for the first problem, and treat the second problem more casually. Let $\epsilon\gt 0$ be given. Note that
$$\left|\frac{x^2-y^2}{x^2+y^2}-(-1)\right|=\frac{2x^2}{x^2+y^2}.\tag{1}$$
Suppose that $\delta\le \frac{1}{2}$. We will need to put additional conditions on $\delta$ later.
If $\sqrt{(x-0)^2+(y-1)^2}\lt \delta$, then $(y-1)^2\lt \frac{1}{4}$ and therefore $|y-1|\lt \frac{1}{2}$. It follows that $y\gt \frac{1}{2}$. Thus $y^2\gt \frac{1}{4}$, and therefore $\frac{1}{x^2+y^2}\lt 4$. It follows that $\frac{2x^2}{x^2+y^2}\lt 8x^2$.
We want to make sure that $8x^2\lt \epsilon$. Note that $\sqrt{x^2+(y-1)^2}\gt |x|$. So if $\sqrt{x^2+(y-1)^2}\lt \frac{\sqrt{\epsilon}}{\sqrt{8}}$, then $x^2\lt \frac{\epsilon}{8}$, and therefore $8x^2\lt \epsilon$.
So if we set $\delta=\min(1/2,\sqrt{\epsilon}/\sqrt{8})$, then if $\sqrt{x^2+(y-1)^2}\lt \delta$ then the expression on the left side of (1) will be less than $\epsilon$.
For the second problem, we just mention the basic strategy. We are interested in the absolute value of $\frac{(1-\cos(xy))\sin y}{x^2+y^2}$. Multiply top and bottm by $1+\cos(xy)$, and use the fact that $|\sin t|\le t$. We get that
$$\left|\frac{(1-\cos(xy))\sin y}{x^2+y^2}\right|\le \frac{(x^2y^2)|y|}{(|1+\cos(xy)|)(x^2+y^2)}.$$
Now if $\sqrt{x^2+y^2}$ is small enough (and it doesn't have to be very small!) we have $1+\cos(xy)\gt 1$. Also, since $(x-y)^2\ge 0$, we have $\frac{|xy|}{x^2+y^2}\le \frac{1}{2}$ unless $(x,y)=(0,0)$. The $\epsilon$ stuff should now be straightforward. There is a lot of slack. Even if $1-\cos(xy)$ is replaced by its square root, the limit is $0$.
| {
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Clearing the confusion of the area In the example 1:
They are finding the area as $8$ $\sqrt[2]{30}$. But, I am finding $8$ $\sqrt[2]{15}$.
I have done $\sqrt[2]{16\times5\times8\times3}$
Then,
Breaking it up,
$2\times2\times2\times\sqrt[2]{15}$
Please help me clear my confusion.
| I think you're just forgetting to multiply by two.
Note that
$$
\begin{align}
\sqrt{16\cdot 5\cdot 8\cdot 3} &= \sqrt{4 \cdot 4 \cdot 5 \cdot 4 \cdot 2 \cdot 3}\\
&= \sqrt{4}\sqrt{4}\sqrt{4}\sqrt{5 \cdot 2 \cdot 3}\\
&= 2 \cdot 2 \cdot 2 \sqrt{5 \cdot 2 \cdot 3}\\
&= 8 \sqrt{5 \cdot 2 \cdot 3}\\
&= 8 \sqrt{30}\\
\end{align}
$$
| {
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What algebraic manipulation is used to express the solution to this integral? According to WolframAlpha, $$
\int \frac{2}{\sqrt{x^2 + 4}} dx = 2 \sinh^{-1}\left(\frac{x}{2}\right) + c.$$
I'm wondering how this was obtained, as I got the following:
Let $x = 2\tan\theta$. Then $dx = 2\sec^2\theta$, so \begin{align*}\int\frac{2}{\sqrt{x^2 + 4}}dx &= \int\frac{4\sec^2\theta}{\sqrt{4\tan^2\theta + 4}}d\theta = \int\frac{2\sec^2\theta}{\sqrt{\tan^2\theta + 1}}d\theta = 2\int\frac{\sec^2\theta}{\sec\theta}d\theta\\ &= 2\int\sec\theta{}d\theta = 2\ln\mid{\tan\theta + \sec\theta}\mid + c.\end{align*}
The substitution $x = 2\tan\theta$ implies that $\sec\theta = \frac{\sqrt{x^2 + 4}}{2}$ and $\tan\theta = \frac{x}{2}$. Hence $$
\int \frac{2}{\sqrt{x^2 + 4}} dx = 2\ln\left|{\frac{x}{2} + \frac{\sqrt{x^2 + 4}}{2}}\right| + c.$$
Either I made a mistake or I don't know the right algebraic manipulation.
| We could use hyperbolic trig substitution like Andre Nicolas mentioned.
Also, $$\sinh^{-1}(u) = \ln\left(u+\sqrt{1+u^2}\right).$$
| {
"language": "en",
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How to derive the hyperbola giving the foci and the fixed differene Given the two foci coordinates $(x_1,y_1)$ and $(x_2,y_2)$ of the hyperbola and the fixed difference distance, how can I derive its function to be able to draw it.
| Let $f$ denote the distance between the foci, and let $d$ denote the constant difference between the distances from any point on the given hyperbola $H$ and the foci, which coincides with the distance between the vertices. The shape of the hyperbola (but not its size) is determined precisely by the ratio of these,
$$\epsilon := \frac{f}{d},$$
a parameter we call the eccentricity of the hyperbola. Now, one branch of an west-east facing hyperbola with eccentricity $\epsilon$ and distance $d$ between the vertices can be parameterized in polar coordinates (with the focus "inside" the parameterized branch at the origin) by
$$r(\theta) = \frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)},$$
which by construction is congruent $H$ via a rotation and a translation. Now, converting to rectangular coordinates gives the parameterization
$$\left\{\begin{array}{l} x(\theta) = \frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \cos \theta \\ y(\theta) = \frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \sin \theta\end{array}\right.$$
Next, we shift the hyperbola so that its center is at the origin. In this case, the distance to the focus is $\frac{1}{2}f$, so the shifted hyperbola is
$$\left\{\begin{array}{l} x(\theta) = \frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \cos \theta - \frac{f}{2}\\ y(\theta) = \frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \sin \theta\end{array}\right.$$
Now, to orient the hyperbola in the same direction, we need to apply a rotation transformation. Let $X := x_2 - x_1$ and $Y := y_2 - y_1$. We need to rotate the horizontal direction to the direction given by the vector $(X, Y)$ without scaling, and we can do that by applying the rotation
$$\frac{1}{X^2 + Y^2} \left(\begin{array}{cc} X & Y \\ -Y & X\end{array}\right).$$
Note that, by the Pythagorean theorem, $X^2 + Y^2 = f^2$. The result of this transformation is
$$\left\{\begin{array}{l} x(\theta) = \frac{1}{f^2}\left[X\left(\frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \cos \theta - \frac{f}{2}\right) + Y\left(\frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \sin \theta\right)\right]\\ y(\theta) = \frac{1}{f^2}\left[X\left(\frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \sin \theta\right) - Y\left(\frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \cos \theta - \frac{f}{2}\right)\right] \end{array}\right.$$
All that remains is to translate the hyperbola to coincide with $H$. We need to send the center $0$ of our hyperbola to the center of $H$, which we denote by $(\bar{x}, \bar{y})$, where $\bar{x} = \frac{1}{2}(x_1 + x_2)$ and $\bar{y} = \frac{1}{2}(y_1 + y_2)$, so the translation we want is $(x, y) \mapsto (x + \bar{x}, y + \bar{y})$. Composing with our parameterization gives
$$\left\{\begin{array}{l} x(\theta) = \frac{1}{f^2}\left[X\left(\frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \cos \theta - \frac{f}{2}\right) + Y\left(\frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \sin \theta\right)\right] + \bar{x}\\ y(\theta) = \frac{1}{f^2}\left[X\left(\frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \sin \theta\right) - Y\left(\frac{d(\epsilon^2 - 1)}{2(1 + \epsilon \cos \theta)} \cos \theta - \frac{f}{2}\right)\right] + \bar{y}\end{array}\right. .$$
It's possible to rearrange these steps some, and in particular I don't claim that this procedure is optimized for anything except (very arguably) geometric intuition.
| {
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How to prove this inequality using am gm? Is it true that
$$\frac{1}{3k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}>\frac{1}{2k+1}+\frac{1}{2k+2}$$
for all natural k? Is bashing going to work?
| The LHS is
$$\frac{1}{6k+2}+\frac{1}{6k+2}+\frac{1}{6k+4}+\frac{1}{6k+4}+\frac{1}{6k+6}+\frac{1}{6k+6}$$
while the RHS is
$$\frac{1}{6k+3}+\frac{1}{6k+3}+\frac{1}{6k+3}+\frac{1}{6k+6}+\frac{1}{6k+6}+\frac{1}{6k+6}$$
hence we just need to prove that:
$$\frac{1}{6k+2}+\frac{1}{6k+2}+\frac{1}{6k+4}+\frac{1}{6k+4}>\frac{1}{6k+3}+\frac{1}{6k+3}+\frac{1}{6k+3}+\frac{1}{6k+6}$$
that follows from:
$$\frac{1}{6k+2}-\frac{1}{6k+3}=\frac{1}{36k^2+30k+6},$$
$$\frac{1}{6k+4}-\frac{1}{6k+6}=\frac{2}{36k^2+60k+24},$$
$$\frac{1}{6k+3}-\frac{1}{6k+4}=\frac{1}{36k^2+42k+12},$$
$$\frac{2}{36k^2+30k+6}+\frac{2}{36k^2+60k+24}>\frac{1}{36k^2+42k+12}.\tag{3}$$
To prove $(3)$, you can use AM-GM, AM-HM or many other methods, it is not a tight inequality.
Another possibility is given by writing both terms as geometric series. Since:
$$\frac{1}{6(k+1)-\eta}=\frac{1}{6(k+1)}+\sum_{j=1}^{+\infty}\frac{\eta^j}{(6(k+1))^{j+1}}$$
and so on, we just have to prove that:
$$2\sum_{j=1}^{+\infty}\frac{1+2^{j}+4^j}{(6(k+1))^{j+1}}>3\sum_{j=1}^{+\infty}\frac{1+3^{j}}{(6(k+1))^{j+1}}$$
that is trivial since $2(1+2^j+4^j)> 3(1+3^j)$ holds by induction for any $j\geq 1$.
| {
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Integration without substitution of $\frac{x^2+3}{x^6\left(x^2+1\right)}$ This is a repost of a question i had written incorrectly earlier.
How do I integrate this without substitutions ?
$$
\frac{x^2+3}{x^6\left(x^2+1\right)}
$$
I got:
$$
\frac{1}{x^6}+\frac{2}{x^6\left(x^2+1\right)},
$$
but wasn't able to eliminate the 2.
| Once you have the expression equal to $\frac{1}{x^6}+\frac{2}{x^6\left(x^2+1\right)}$ (as you stated in the question), you can get to the partial fraction decomposition without partial fractions. Using the formula for an infinite geometric series, expand $\frac{1}{x^2+1}=1-x^2+x^4-x^6\dots$, which allows you to write your expression as
$$\frac{1}{x^6}+\frac{2}{x^6}-\frac{2}{x^4}+\frac{2}{x^2}-2(1-x^2+x^4-x^6\dots)=\frac{3}{x^6}-\frac{2}{x^4}+\frac{2}{x^2}-\frac{2}{1+x^2}.$$
| {
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Find the function and domain for $ (f\circ f)$ when $ f(x) = x+ \frac {1} {x} \ $ Find the function and domain for $ (f\circ f)$
My answer is $ \frac {x^4+3x^2+1x} {(x^2+1)(x)}?$
However, the program I am using states I am wrong. What have I done incorrectly?
| $$f(x) = x + \frac 1x = \frac {x^2+1}{x}$$
$$f(f(x)) = \frac{x^2 + 1}{x} + \frac{1}{\frac{x^2+1}x} = \frac{x^2+1}{x} + \frac{x}{x^2+1} = \frac{(x^2+1)^2 + x^2}{x(x^2+1)} = \frac{x^4 + 3x^2 + 1}{x(x^2 + 1)}$$
Domain: All $x \in \mathbb R$ such that $x\neq 0.$ (The denominator is undefined only when $x = 0.)$ $$(-\infty, 0)\cup (0, +\infty)$$
Note that $x^2 + 1 \gt 0\;\; \forall x \in \mathbb R$.
| {
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Proof for $\log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log 2$ How can I prove that
$\displaystyle \log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log2$
using $\displaystyle \frac{x}{1+x}<\log(1+x)<x$
| As shown in $(1)$ here,
$$
\begin{align}
H(2n)-H(n)
&=\sum_{k=n+1}^{2n}\frac1k\\
&=\sum_{k=1}^{2n}\color{#C00000}{\frac1k}-\sum_{k=1}^n\color{#00A000}{\frac1k}\\
&=\sum_{k=1}^n\left(\color{#C00000}{\frac1{2k-1}+\frac1{2k}}-\color{#00A000}{\frac1k}\right)\\
&=\sum_{k=1}^n\left(\frac1{2k-1}-\frac1{2k}\right)\\
&=\sum_{k=1}^{2n}(-1)^{k-1}\frac1k\tag{1}
\end{align}$$
which is a series of positive terms, $\left(\frac1{2k-1}-\frac1{2k}\right)$, whose limit is $\log(2)=\sum\limits_{k=1}^\infty(-1)^{k-1}\frac1k$. Thus, each partial sum is less than $\log(2)$. That is,
$$
\sum_{k=n+1}^{2n}\frac1k\lt\log(2)\tag{2}
$$
Furthermore,
$$
\begin{align}
e^{\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}}
&\gt\left(1+\frac1{n+1}\right)\left(1+\frac1{n+2}\right)\cdots\left(1+\frac1{2n}\right)\\
&=\frac{2n+1}{n+1}\tag{3}
\end{align}
$$
Taking the log of $(3)$, we get
$$
\sum_{k=n+1}^{2n}\frac1k\gt\log\left(\frac{2n+1}{n+1}\right)\tag{4}
$$
Putting together $(2)$ and $(4)$ yields
$$
\log\left(\frac{2n+1}{n+1}\right)\lt\sum_{k=n+1}^{2n}\frac1k\lt\log(2)\tag{5}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/957977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Mathematical Induction: Sum of first n even perfect squares So the series is $$P_k: 2^2 + 4^2 + 6^2 + ... + (2k)^2 = \frac{2k(k+1)(2k+1)}3$$
and i have to replace $P_k$ with $P_{k+1}$ to prove the series.
I have to show that $$\frac{2k(k+1)(2k+1)}3 + [2(k+1)]^2 = \frac{2(k+1)(k+2)(2k+3)}3$$ but I don't know how.
| $$\frac{2k(k+1)(2k+1)}{3}+(2(k+1))^2=\frac{2k(k+1)(2k+1)}{3}+\frac{3(2(k+1))^2}{3}=\\=
\frac{2k(k+1)(2k+1)+3 \cdot 4(k+1)^2}{3}=\frac{2(k+1)(k(2k+1)+6(k+1))}{3}=\\=\frac{2(k+1)(2k^2+k+6k+6)}{3}=\frac{2(k+1)((2k+3)(k+2))}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Number of ways to choose a sequence of three letters from the letters of MISSISSIPPI How many ways can a sequence of three letters be chosen from the letters of MISSISSIPPI?
I'm just a little confused how to go about this since so many letters repeat
I=4
S=4
P=2
So ultimately there are 4 different letters.
| Our desired $3$-tuples, will consist of $x$ M's, $y$ I's, $z$ S's and $w$ P's.
These should be picked among the letters of the word MISSISSIPI and must sum to $3$:
\begin{equation}
x+y+z+w=3
\end{equation}
*
*$x$ -> # of M's, thus: $x=0,1$
*$y$ -> # of I's, thus: $y=0,1,2,3,4$ (note that $y=4$, is obviously excluded)
*$z$ -> # of S's, thus: $z=0,1,2,3,4$ (note that $z=4$, is obviously excluded)
*$w$ -> # of P's, thus: $w=0,1$
The possible solutions to the equation $$x+y+z+w=3$$ under the above restrictions on the values of $x,y,z,w$, are displayed in the following table:
$$
\begin{array}{c|c|c|c}
x & y & z & w \\ \hline
0 & 1 & 2 & 0 \\
0 & 2 & 1 & 0 \\
0 & 3 & 0 & 0 \\
0 & 0 & 3 & 0 \\
1 & 2 & 0 & 0 \\
1 & 0 & 2 & 0 \\
1 & 1 & 1 & 0 \\
0 & 2 & 0 & 1 \\
0 & 0 & 2 & 1 \\
0 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1
\end{array}
$$
What remains to do, is to perform the following summation:
$$\sum_{x+y+z+w=3}\frac{3!}{x!y!z!w!}$$
on the rows of the table above. Consequently we get:
$$\frac{3!}{1!2!}+\frac{3!}{2!1!}+\frac{3!}{3!}+\frac{3!}{3!}+\frac{3!}{1!2!}+\frac{3!}{1!2!}+\frac{3!}{1!1!1!}+\frac{3!}{2!1!}+\frac{3!}{2!1!}+\frac{3!}{1!1!1!}+\frac{3!}{1!1!1!}+\frac{3!}{1!1!1!}= \\ =3+3+1+1+3+3+6+3+3+6+6+6=44$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/960046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Deriving equality with algebra? Can someone walk me through computation of the following equality?
$$\frac{1}{i}\int \frac{dz}{z(a + \frac{z + z^{-1}}{2})^2} = \frac{4}{i}\int \frac{zdz}{(z^2 + 2az + 1)^2}$$
I've gotten as far as:
$$\frac{4}{i}\int \frac{dz}{4z(a^2 + a(z + z^{-1}) + z^2 + z^{-2} + 2)}$$
but am unsure of how to proceed after this.
Thanks.
| $$\frac{1}{i}\int\frac{dz}{z(a+\frac{z+z^{-1}}{2})^2}=\frac{1}{i}\int\frac{4dz}{z(2a+z+z^{-1})^2}=\frac{1}{i}\int\frac{4dz}{z\frac{(2az+z^2+1)^2}{z^2}}=\frac{4}{i}\int\frac{zdz}{(2az+z^2+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/961337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence and estimation of the rests of $\sum\limits_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} - \sum\limits_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}} $ For $x\in ]-1,1[$ is required to evaluate the expression
$$
S(x) = \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} - \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}}
$$
with absolute error $\epsilon= 3\times 10^{-10}.$
*
*I need to prove that the series $\displaystyle\sum_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} $ converges and $\displaystyle \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}}$ also.
*How many terms are required to evaluate the first series in order to have an absolute error lower than $\epsilon$?
*How (if possible) you can re-write $S(x)$ in order to be evaluated faster.
Sorry if that question is very simple, but for me it is hard for me.
Thanks.
| Rewrite $$\frac{1}{\sqrt{k^3 \pm x}}=\frac{1}{k^{3/2}}\frac{1}{\sqrt{1 \pm y}}$$ where $y=\frac{x}{k^3}$. Now, consider that $y$ is small compared to $1$ and expand $$\sqrt{1 \pm y}=1\pm\frac{y}{2}+\frac{3 y^2}{8}\pm\frac{5 y^3}{16}+O\left(y^4\right)$$ and replace $y$ by $\frac{x}{k^3}$. You then end, after simplifications, with $$\frac{1}{\sqrt{k^3+x}} - \frac{1}{\sqrt{k^3 - x}}=-\left(\frac{1}{k}\right)^{9/2} x-\frac{5}{8} \left(\frac{1}{k}\right)^{21/2}
x^3+O\left(\left(\frac{1}{k}\right)^{31/2}\right)$$ which, after summation, write $$S(x) = \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3+x}} - \sum_{k=1} ^\infty \frac{1}{\sqrt{k^3 - x}}\approx-\zeta \left(\frac{9}{2}\right)x-\frac{5 \zeta \left(\frac{21}{2}\right)}{8}x^3$$ $$S(x) \approx -1.05471x-0.625438 x^3$$
Let us try with $x=\frac{1}{4}$; the approximation leads to $-0.273449$ while the whole summation would lead to $-0.273957$; for sure, if you increase the value of $x$, more terms should be required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/961887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A Sine integral: problem I Is it possible to demonstrate a solution for the integral
\begin{align}
\int_{0}^{\infty} x^{n} \, \sin\left( a x^{2} + \frac{b}{x^{2}} \right) \, dx
\end{align}
| Part 1
Lets assume that $n$ is even ($n=2m$), and the parameters are chosen that the integral converges. Call our integral of interest $J(a,b)$.
Now observe that
\begin{align}x^{2m}\sin(ax^2+b/x^2)&= (-1)^{(m+3)/2}\frac{d^m}{da^m}\cos(ax^2+b/x^2),\qquad
\text{if} \; m=1,3,5,\ldots
\\
x^{2m}\sin(ax^2+b/x^2)&= (-1)^{m/2}\frac{d^m}{da^m}\sin(ax^2+b/x^2),\qquad\;\,\quad
\text{if}\;m=2,4,6...
\end{align}
So essentially we only have to calculate
$\displaystyle\int_0^{\infty}\cos(ax^2+b/x^2)\mathrm{d}x$ and $\displaystyle\int_0^{\infty}\sin(ax^2+b/x^2)\mathrm{d}x$.
This can be done as follows (we only do the sine integral the other cosine one goes along the same lines):
*
*First complete the square in the Integrands
\begin{align}
a/x^2+b/x^2= (\sqrt{a}x-\sqrt{b}/x)^2+2\sqrt{ab},
\end{align}
*Use the trigonometric identity
\begin{align}
\sin(x+y)=\sin x\cos y+\cos x\sin y,
\end{align}
to obtain
\begin{align}
_{e}I(a,b)&=\int_0^{\infty}\sin(ax^2+b/x^2)\mathrm{d}x \\
&= \cos(2\sqrt{ab})\int_0^{\infty}\sin(\sqrt{a}x-\sqrt{b}/x)^2\mathrm{d}x+\sin(2\sqrt{ab})\int_0^{\infty}\cos(\sqrt{a}x-\sqrt{b}/x)^2\mathrm{d}x
\end{align}
*Now we apply the inversion $\displaystyle x \rightarrow \frac{\sqrt{b}}{\sqrt{a}x}$ to both integrals and add them together with their non-inverted versions. We end up with
\begin{align}
2 _{e}I(a,b) &=
\cos(2\sqrt{ab})\int_0^{\infty}\sin(\sqrt{a}x-\sqrt{b}/x)^2 \left(1+\frac{\sqrt{b}}{\sqrt{a}x^2}\right)\mathrm{d}x \\
& \hspace{10mm} +\sin(2\sqrt{ab})\int_0^{\infty}\cos(\sqrt{a}x-\sqrt{b}/x)^2\left(1+\frac{\sqrt{b}}{\sqrt{a}x^2}\right)\mathrm{d}x.
\end{align}
*Substitute $\sqrt{a}x-\sqrt{b}/x=y$ and $\displaystyle\frac{dy}{\sqrt{a}}=\left(1+\frac{\sqrt{b}}{\sqrt{a} x^2}\right)\mathrm{d}x$ to get
\begin{align}
2 _{e}I(a,b) &=\frac{\cos(2\sqrt{ab})}{\sqrt{a}} \int_{-\infty}^{\infty} \sin y^2\mathrm{d}y + \frac{\sin(2\sqrt{ab})}{\sqrt{a}} \int_{-\infty}^{\infty}\cos y^2\mathrm{d}y \\
&= \frac{2 \, \cos(2\sqrt{ab})}{\sqrt{a}} \int_{0}^{\infty} \sin y^2\mathrm{d}y + \frac{2\, \sin(2\sqrt{ab})}{\sqrt{a}} \int_{0}^{\infty}\cos y^2 \mathrm{d}y.
\end{align}
*Use the well known values $\displaystyle\int_0^{\infty}\sin y^2\mathrm{d}y=\int_0^{\infty}\cos y^2\mathrm{d}y=\sqrt{\frac{\pi}{8}}$.
So in the end we have
\begin{align}
_{e}I(a,b) = \sqrt{\frac{\pi}{8 a}}\left(\cos(2\sqrt{ab})+\sin(2\sqrt{ab})\right)
= \frac{1}{2} \sqrt{\frac{\pi}{a}} \, \sin\left( 2 \sqrt{ab} + \frac{\pi}{4} \right).
\end{align}
The final result is now
\begin{align}
J(a,b)=\int_0^{\infty}x^{2m}\sin(ax^2+b/x^2)\mathrm{d}x=(-1)^{(m+3)/2}\frac{d^m}{da^m}_{e}I(a,b),
\end{align}
if $m=1,3,5\ldots$
Edit:
For $m=2,4,6\ldots$
we get
\begin{align}
_{e}K(a,b) &= \int_0^{\infty}\cos(ax^2+b/x^2)\mathrm{d}x = \sqrt{\frac{\pi}{8 a}}\left(\cos(2\sqrt{ab})-\sin(2\sqrt{ab})\right) \\
&= \frac{1}{2} \sqrt{ \frac{\pi}{a} } \, \cos\left( 2 \sqrt{ab} + \frac{\pi}{4} \right),
\end{align}
where we used
$\cos(x+y)=\cos x\cos y-\sin x \sin y$ and
\begin{align}
J(a,b)=\int_0^{\infty}x^{2m}\cos(ax^2+b/x^2)\mathrm{d}x=(-1)^{m/2}\frac{d^m}{da^m}{_{e}K(a,b)}.
\end{align}
Part 2
I finally managed managed to get an answer for $n=2m+1$
We again observe the differentiation trick works again:
\begin{align}x^{2m+1}\sin(ax^2+b/x^2)&= (-1)^{(m+3)/2}\frac{d^m}{da^m}x\cos(ax^2+b/x^2),
\qquad \text{if}\; m=1,3,5\ldots
\\
x^{2m+1}\sin(ax^2+b/x^2)&= (-1)^{m/2}\frac{d^m}{da^m}x
\sin(ax^2+b/x^2),\qquad\quad\;\,
\text{if} \; m=2,4,6\ldots
\end{align}
So we only need $\displaystyle\int_0^{\infty}x\cos(ax^2+b/x^2)\mathrm{d}x$ and $\displaystyle\int_0^{\infty}x\sin(ax^2+b/x^2)\mathrm{d}x$.
We do this as follows (only the sine case in detail...)
*
*Substitute $x\rightarrow\sqrt{y}$,$dx=dy/\sqrt{y}$ and luckily the intergal simplifies a lot:
\begin{align}
_oI(a,b)=\int_0^{\infty}\sin(ay+b/y)\mathrm{d}y.
\end{align}
*Observe that our Integral can after rescaling $y=\sqrt\frac{b}{a}y$ be written as
\begin{align}
_oI(a,b)=-\sqrt{\frac{b}{a}}\Im\int_0^{\infty}e^{-\sqrt{ab}i(y+1/y)}\mathrm{d}y
\end{align}
*Referring to the the beautiful answers in this link and analytically (which is as far as I can see legitimate) continuing them to purely imaginary parameters ($a-\rightarrow ia$,$b\rightarrow ib$)
we can write
\begin{align}
_oI(a,b)=-2\sqrt{\frac{b}{a}}\Im K_1(2i\sqrt{ab})
\end{align}
where $K_1(z)$ is a modified Bessel function
*Using furthermore the Relations $K_1(z)=\frac{i \pi}{2}e^{i \pi/2}H_1(iz)$ and $H_1(z)=J_1(z)+iN_1(z)$, with $H_1(z)$ an Hankel function and $J_1(z)$/$N_1(z)$ Bessel functions of first/second kind, see here
for example, we finally get
\begin{align}
_oI(a,b)=-\pi\sqrt{\frac{b}{a}}N_1(2\sqrt{ab})
\end{align}
*Repeating this procedure (with $\Re$ instead of $\Im$) for the Cosine case we get
\begin{align}
_oK(a,b)=-\pi\sqrt{\frac{b}{a}}J_1(2\sqrt{ab}).
\end{align}
The whole Integral is then given by
\begin{align}
J(a,b)=(-1)^{(m+3)/2}\frac{d^m}{da^m}{_{o}I(a,b)},
\end{align}
if $m=1,3,5\ldots$ or
\begin{align}
J(a,b)=(-1)^{m/2}\frac{d^m}{da^m}{_{o}K(a,b)},
\end{align}
if $m=2,4,6\ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/963861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How to derive $J_v(x)$ I've seen many sources say that $\frac{\text{d}}{\text{d}x}J_v(x) = J_{v-1}(x) - \frac{v}{x}J_v(x)$, but every time I try to derive it, I get the conjugate, $\frac{v}{x}J_v(x) - J_{v-1}(x)$. Could someone walk me through the steps to arrive at this derivative, so I may see where I went wrong?
$$
J_v(x) = \sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v}}\\
\begin{align}
\frac{\text{d}}{\text{d}x}J_v(x)
& = \sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\frac{\text{d}}{\text{d}x}\left(\frac{x}{2}\right)^{2n+v}} \\
& = \sum_{n=0}^\infty{\frac{(-1)^n(2n+v)}{2\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v-1}} \\
& = \sum_{n=0}^\infty{\frac{(-1)^nn}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v-1}} + \frac{v}{2}\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v-1}} \\
& = \left(\frac{x}{2}\right)^{-1}\left[\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)(n-1)!}\left(\frac{x}{2}\right)^{2n+v}} + \frac{v}{2}\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v}}\right] \\
& = \left(\frac{2}{x}\right)\left[\sum_{n=1}^\infty{\frac{(-1)^{n+1}}{\Pi(v+n-1)n!}\left(\frac{x}{2}\right)^{2(n-1)+v}} + \frac{v}{2}\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v}}\right] \\
& = -\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi((v-1)+n)n!}\left(\frac{x}{2}\right)^{2n+(v-1)}} + \frac{v}{x}\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v}} \\
& = \frac{v}{x}J_v(x) - J_{v-1}(x)
\end{align}
$$
| We have
$$\frac{dJ_{\nu}}{dx} = \sum_{m=0} \frac{(-1)^m}{m!\Gamma(m+\nu+1)} (2m +\nu)\frac{x^{2m+\nu-1}}{2^{2m+\nu}} = \sum_{m=0} \frac{(-1)^m}{m!\Gamma(m+\nu)} \frac{2m +\nu}{2m+2\nu}\left({x\over 2}\right)^{2m+\nu-1}$$
where we have used $\Gamma(m+\nu+1)=(m+\nu)\Gamma(m+\nu)$. Now use $$\frac{2m+\nu}{2m+2\nu} = 1 -\frac{\nu}{2(m+\nu)}$$
The sum splits into two sums. The first sum is
$$\sum_{m=0} \frac{(-1)^m}{m!\Gamma(m+\nu)}\cdot 1 \cdot\left({x\over 2}\right)^{2m+\nu-1} \equiv J_{\nu -1}$$
and the second sum is
$$~~~~~~\sum_{m=0} \frac{(-1)^m}{m!\Gamma(m+\nu)}\cdot \left(-\frac{\nu}{2(m+\nu)} \right)\cdot\left({x\over 2}\right)^{2m+\nu-1} \\= -\frac{\nu}{x}\sum_{m=0} \frac{(-1)^m}{m!\Gamma(m+\nu+1)} \cdot\left({x\over 2}\right)^{2m+\nu} \equiv -\frac{\nu}{x}J_\nu$$
where we have used $\Gamma(m+\nu)(m+\nu) = \Gamma(m+\nu+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/965360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the limit as $x$ approaches $5$ $f(x) = \dfrac{\sin(x-5)}{x^2-2x-15}$
Find the limit as $x$ approaches $5$.
I got up to : $\dfrac{\sin}{ ( x+3)}$.
I know the answer is $\frac18$ but I just don't know how to get it.
Unfortunately, I did cancel out the (x-5) =(. Is it because the the numerator (x-5) is considered an angle? like sin theta? and is not similar to the one in the denominator?
| $$\frac{\sin(x-5)}{x^2-2x-15}=\frac{\sin(x-5)}{x^2-2x+1-16}=\frac{\sin(x-5)}{(x-1)^2-4^2}=$$
$$=\frac{\sin(x-5)}{(x-1-4)(x-1+4)}=\frac{\sin(x-5)}{(x-5)}\frac{1}{(x+3)}\to\frac{1}{8}\text{if}x\to5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$5$ digit no. in which at least $3$ digits are identical. The number of $5$-digit numbers that can be made with digits $\left\{1,2,3,4,5,6\right\}$
in which at least $3$ digits are identical is.
$\bf{My\; Try::}$
No.,s in which at least $3$ digits are Identical is $ = $ Total -no digit are identical - exactly $2$ digit are
identical.
So Total no. of ways is $ = 6\times 6 \times 6 \times 6 \times 6=6^5$
and no digit are identical is $ = 6\times 5 \times 4 \times 3 \times 2 = 6!$
and I did not understand how can i find the cases in which exactly $2$ digits are identical.
Help me
Thanks
| There are 3 cases:
1. 3 digits identical: $aaabc$
There are $4+3+2+1$ ways of arranging $b$ and $c$ such that $b$ always comes before $c$. Then there are $6$ choices for $b$, $5$ for $c$ and $4$ for $a$. Thus there are $10\cdot6\cdot5\cdot4=1200$ numbers.
For $aaabb$, we have $10$ ways of arrangement and therefore a total of $10\cdot6\cdot5=300$ numbers.
2. 4 digits are identical: $aaaab$
There are $5$ positions for $b$. So there $5\cdot6\cdot5=150$ numbers.
3. 5 digits are identical: $aaaaa$
There are clearly $6$ such numbers.
So there are $\boxed{1656}$ such numbers.
To count the ones with $2$ identical digits, we have $2$ cases.
1. $aabcd$
There are ${5\choose 2}$ ways of placing the $a$'s. We will not arrange $bcd$ since it will cause over counting. So we have $10\cdot6\cdot5\cdot4\cdot3=3600$ such numbers.
2. $aabbc$
There are $5$ ways for placing the $c$. For the remaining $4$ positions we have $4\choose 2$ ways of placing the $a$'s and $1$ way of placing the $b$'s. So there are $5\cdot 6$ ways of arranging. But we divide this by $2$ since $a$ and $b$ are interchangeable. So we have $15$ ways of arranging these. Thus there are $15\cdot6\cdot5\cdot4=1800$ numbers.
So there are $5400$ numbers with $2$ digits identical.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/969786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What is the minimum value of $(\tan^2)(A/2)+(\tan^2)(B/2)+(\tan^2)(C/2)$, where $A$, $B$ and $C$ are angles of a triangle
What is the minimum value of $(\tan^2)(A/2)+(\tan^2)(B/2)+(\tan^2)(C/2)$, where $A$, $B$ and $C$ are angles of a triangle?
I know that the sum of the angles is $\pi$, but I am unable to find the minimum value of the above expression.
| Just another way: noting $\tan^2 \frac x 2$ is convex, by Jensen's inequality you have
$$\tan^2\frac A 2 + \tan^2 \frac B 2 + \tan^2 \frac C 2 \ge 3\tan^2\frac{A+B+C}{2\cdot 3 } = 1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Factorize $6x^2 -5x -14 = 0$ I'm throwing a bit of a blank on the best way to factor this :
$$6x^2 -5x -14 = 0$$
I know that I could multiply $6$ by $14$ and then find a pair of factors that add to $-5$ (b), but this feels a bit convoluted and I'm sure that I've gone about this a bit of an easier way before, I just can't remember how...
Does anyone know the simplest way?
cheers!
| \begin{align}
& 6x^2 - 5x - 14 \\[8pt]
= {} & 6\left( x^2 - \frac 5 6 x \right) - 14 \\[8pt]
= {} & 6\left( x^2 - \frac 5 6 x + \frac{25}{144} \right) - 14 - 6 \left(\frac{25}{144}\right) \quad\text{(completing the square)} \\[8pt]
= {} & 6 \left(\left( x - \frac 5 {12} \right)^2 - \frac{361}{144}\right) \\[8pt]
= {} & 6 \left(\left( x - \frac 5 {12} \right)^2 - \left(\frac{19}{12}\right)^2 \right) \\[8pt]
= {} & 6\left(x - \frac5{12} - \frac{19}{12}\right)\left(x - \frac5{12} + \frac{19}{12}\right) \quad\text{(factoring a difference of two squares)} \\[8pt]
= {} & 6(x-2)\left(x + \frac 7 6\right) \\[8pt]
= {} & (x-2)(6x+7).
\end{align}
If one says
$$
x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a} = \frac{5\pm\sqrt{5^2-4\cdot6\cdot(-14)}}{2\cdot6} = \frac{5\pm19}{12} = \left(\frac{24}{12}\text{ or }\frac{-14}{12}\right) = \left(2\text{ or }\frac{-7}6\right)
$$
then one knows that $x-2$ and $x-\left(\dfrac{-7}6\right)$ must be factors, and one must have $6$ as the leading coefficient, so one gets the same result.
The usual way to prove that
$$
x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}
$$
if $ax^2+bx+c=0$ is by completing the square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving a sequence is convergent Let $\
(x_n )_{n \ge 0}
$ be a convergent sequence .
Prove that another sequence $\
(y_n )_{n \ge 0}
\
$ defined as $
x_n = y_n + 2y_{n + 1}
$ is convergent as well . I tried mixing the $\
\varepsilon ,N(\varepsilon )
$ definition of a limit with the Bolzano-Weierstrass theorem but I didn't get anything useful .
Any ideas ?
| Suppose that $x_n$ converges to $L$ and for the sake of argument assume that $y_n$ converges to $A$. Then $L = A + 2 A = 3A$ and $A = L/3$.
Now let us examine the sequence
$x_0 = y_0 + 2y_1$ so $\frac{x_0 - y_0}{2} = y_1$, and $$y_2 = \frac{x_1 - y_1}{2} = \frac{x_1}{2} - \frac{x_0}{4} + \frac{y_0}{4}$$
Then $$y_3 = \frac{x_2 - y_2}{2} = \frac{x_2}{2} - \frac{x_1}{4} + \frac{x_0}{8} - \frac{y_0}{8}$$
And $$y_4 = \frac{x_3}{2} - \frac{x_2}{4} + \frac{x_1}{8} - \frac{x_0}{16} + \frac{y_0}{16}$$
By induction we have:
$$y_{n+1} = \sum_{k=0}^n (-1)^{n-k} \frac{x_k}{2^{n-k+1}} + \frac{y_0}{2^n}$$
Now let $\epsilon > 0$, and suppose $N \in \mathbb{N}$ is chosen so that $|x_n-L| < \epsilon$ when $n>N$.
For $n>N$ write $$M_n = \sum_{k=1}^N (-1)^{n-k} \frac{x_k}{2^{n-k+1}} = \frac{1}{2^{n-N}} \sum_{k=1}^N (-1)^{n-k} \frac{x_k}{2^{N-k+1}}.$$ Pick $n$ large enough so that $M_n < \epsilon$ and $y_0/2^n < \epsilon$.
Then show that $$\left|y_{n+1} - \sum_{k=N}^n(-1)^{n-k}\frac{L}{2^{n-k+1}}\right|$$ is bounded by $4\epsilon.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find $\left\lfloor\sum_{r=1}^{80}\int_0^1x^{\sqrt r-1}dx\right\rfloor$
$$\left\lfloor\sum_{r=1}^{80}\int_0^1x^{\sqrt r-1}dx\right\rfloor$$
My try:
$$K=\left\lfloor\sum_{r=1}^{80}\int_0^1x^{\sqrt r-1}dx\right\rfloor=\left\lfloor\sum_{r=1}^{80}\frac1{\sqrt r}\right\rfloor=\left\lfloor\frac1{\sqrt 1}+\frac1{\sqrt 2}+\ldots+\frac1{\sqrt{80}}\right\rfloor$$
I can only say:
$$\left\lfloor\sum_{r=1}^{80}\frac1{\sqrt {80}}\right\rfloor<K<\left\lfloor\sum_{r=1}^{80}\frac1{\sqrt 1}\right\rfloor\\
\lfloor\sqrt{80}\rfloor<K<\lfloor80\rfloor\\
8<K<80$$
So I don't know how to find K?
| Hint: You need tighter bounds on $\dfrac{1}{\sqrt{r}}$.
Note that $\sqrt{r+1} - \sqrt{r} = \dfrac{1}{\sqrt{r+1}+\sqrt{r}} < \dfrac{1}{2\sqrt{r}} < \dfrac{1}{\sqrt{r}+\sqrt{r-1}} = \sqrt{r} - \sqrt{r-1}$.
Hence, $2(\sqrt{r+1} - \sqrt{r}) < \dfrac{1}{\sqrt{r}} < 2(\sqrt{r} - \sqrt{r-1})$.
Using these bounds will give you two telescoping sums, which are easy to evaluate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/976237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove that for all positive integers $x$, $\left\lfloor \frac{x^2 +2x + 2}{4}\right\rfloor =\left\lfloor \frac{x^2 + 2x + 1}{4}\right\rfloor$. Title says it all, basically. I believe it to be true that
$$\left\lfloor \dfrac{x^2 + 2x + 2}{4} \right\rfloor=\left\lfloor \dfrac{x^2 + 2x + 1}{4} \right\rfloor$$ for all positive integers $x$. I am, however, having a difficult time proving this. My current proof reads along the lines of the fact that, when adding $2$ or $1$, it is impossible to cause a large enough difference in the two numbers that, when divided by four and floored, they evaluate to different numbers. This basically comes down to proving that $x^2 + 2x \neq 4k + 2$ for some integer $k$, but I'm not sure if this is a good way of proving it.
Could anyone shine some light on this? Thanks.
| The only time the two floors are different is when $x^2+2x+2$ is a multiple of $4$ (or, equivalently, looking at the right-hand side, $x^2+2x+1 \equiv 3\mod{4}$.
But $x^2+2x+2 = (x+1)^2+1$, and $(x+1)^2\equiv 0, 1\mod{4}$, so we cannot have $x^2+2x+2\equiv 0\mod{4}$. Almost identically, considering the RHS instead, $x^2+2x+1 = (x+1)^2\equiv 0, 1\mod{4}$, so it cannot be $3\mod{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
How to Solve : $ A =\frac{1}{6}\left((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2(24))^3\right) $ $ A =\frac{1}{6}\left((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2(24))^3\right).$
Solve for $2^A.$
(no calculators or graphs are permitted)
The way I went about solving this problem was using the properties of base $2$. For example, $2^x = 8$ , $x$ is obviously $3$, and $2^y = 16$, $y = 4 = x+1$, so adding $1$ to the exponent will give twice the answer. Since the arguments are twice each other, I used this pattern:
$ A =\frac{1}{6}\left((x)^3-(x+1)^3-(x+2)^3+(x+3)^3\right)$
$= \frac{1}{6}\left((x^3-(x^3 + 3x^2 + 3x + 1)-(x^3 + 6x^2 + 12x + 8)+(x^3 + 9x^2 + 27x + 27)\right)$
Simplifying this we get $A = (12x + 18)/6 = 2x + 3.$
So now I solved for $2^{2x+3}$ using $2^{2x} =2^x * 2^x = 3^2 = 9.$ Now, since we add $3$ to the exponent, we have $2^{2x+3} = (2^3)(9) = 72.$
Would this be the correct answer, and if not, what would be a good way of going about it?
| If it were to be asked in the exam, you could save some time in the simplification exploiting the fact that $a^3-b^3 = (a-b).(a^2+ab+b^2)$
Rearanging in the following way
$$\frac{1}{6}\left((x+3)^3 - (x+2)^3 - \left((x+1)^3 - x^3)\right)\right)$$
$$\frac{1}{6}\left((x+3)^2 + (x+2)^2+(x+3)(x+2) - \left((x+1)^2 + x^2+x(x+1))\right)\right)$$
$$\frac{1}{6}\left((x^2+6x+9+x^2+4x+4+x^2+5x+6) - \left((x^2+2x+1+x^2+x^2+x)\right)\right)$$
$$\frac{1}{6}\left((3x^2+15x+19) - \left((3x^2+3x+1)\right)\right)$$
$$\frac{1}{6}\left(12x+18\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/978930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Inequalities with more than one absolute value I saw a question which asked to find all the solutions to:
$|x+2|+|x-5|=7$
For $x\leq -2$, the answer is $-2$.
For $-2< x <5$, the answer is $R$.
For $x>5$, the answer is $5$.
First I thought the answer should be $(-2, 5)$, but then I realized $-2$ and $5$ solve it too! What would you do?
| We have
$$|x+2|+|x-5| = \begin{cases}
-(x+2) - (x-5) & \text{ if } x \leq -2 \\
+(x+2) - (x-5) & \text{ if } 2 < x \leq 5 \\
+(x+2) + (x-5) & \text{ if } x > 5
\end{cases}
$$
Simplifying,
$$|x+2|+|x-5| = \begin{cases}
-2x+3 & \text{ if } x \leq -2 \\
7 & \text{ if } 2 < x \leq 5 \\
2x-3 & \text{ if } x > 5
\end{cases}
$$
So what does this tell us?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given the frame $B=\{(1,1,0),(0,1,1),(1,1,1),(0,0,1),(0,1,-1)\}$, find (if possible) a (i) $(2,1)$ surgery, Given the frame $B=\{(1,1,0),(0,1,1),(1,1,1),(0,0,1),(0,1,-1)\}$, find (if possible) a (i) $(2,1)$ surgery, and a $(1,2)$ surgery that produce tight frames.
A frame is tight if and and only if the frame operator $B^TB$ is a multiple of identity. So, the off-diagonal entries must be zero and the diagonal entries must be equal to each other.
$\textbf{Part i:}$
$\begin{pmatrix} 1 & 0 & 1 & a \\ 1 & 1 & 1 & b \\ 0 & 1 & 1 & c \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ a & b & c \end{pmatrix}=\begin{pmatrix} 2+a^2 & 2+ab & 1+ac \\ 2+ab & 3+b^2 & 2+bc \\ 1+ac & 2+bc & 2+c^2 \end{pmatrix}$
So to get matrix to produce a tight frame, we have the following equations:
\begin{cases}
2+a^2=3+b^2=2+c^2 \\
2+ab=0 \\
1+ac=0 \\
2+bc=0 \\
\end{cases}
From the first statement we have $$ab+2=bc+2 \iff ab=bc \iff a=c.$$ So $ac+1=0 \iff a^2=1=0$ but then there are no real values that will satisfy the equation. Hence there are no values of $(a,b,c)$ to produce a tight frame.
$\textbf{Part ii:}$
$\begin{pmatrix} 1 & 0 & 1 & 0 & a & d \\ 1 & 1 & 1 & 0 & b & e \\ 0 & 1 & 1 & 1 & c & f \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ a & b & c \\d & e & f \end{pmatrix}=\begin{pmatrix} 2+a^2+d^2 & 2+ab+de & 1+ac+df \\ 2+ab+de & 3+b^2+e^2 & 2+bc+ef \\ 1+ac+df & 2+bc+ef & 2+c^2+f^2 \end{pmatrix}$
So to get matrix to produce a tight frame, we have the following equations:
\begin{cases}
2+a^2+d^2=3+b^2+e^2=2+c^2+f^2 \\
2+ab+de=0 \\
1+ac+ef=0 \\
2+bc+df=0 \\
\end{cases}
What is a simple way of finding solutions for these equations?
| So to get matrix to produce a tight frame, we have the following equations:
\begin{cases}
1+a^2+d^2=3+b^2+e^2=3+c^2+f^2 \\
1+ab+de=0 \\
ac+df=0 \\
bc+ef=0 \\
\end{cases}
One way is to follow the mathematics of wishful thinking. Here, I wishfully think that I can choose many things to be zero. For instance, if $d = e = c = 0$, then the last two equations hold trivially and we are instead left with the system
$$\begin{cases} 1 + a^2 = 3 + b^2 = 3 + f^2 \\
1 + ab = 0
\end{cases}$$
If $f = b$, then $f$ is covered. So really, we just have the pair $1 + ab = 0$ and $1 + a^2 = 3 + b^2$. Fortunately, the first means that $a = -1/b$. Plugging into the other gives
$$1 + \frac{1}{b^2} = 3 + b^2.$$
Rearranging and calling $B = b^2$ gives
$$B^2 + 2B - 1 = 0,$$
with solutions $B = -1 \pm \sqrt 2$. Let's choose the positive square root, $B = 1 + \sqrt 2$. Then $b = \sqrt{1 + \sqrt 2}$, and $a = -1/b$.
So in total,
$$\begin{cases}
a = \frac{-1}{\sqrt{1 + \sqrt 2}}\\
b = \sqrt{1 + \sqrt{2}} \\
c = 0 \\
d = 0 \\
e = 0 \\
f = \sqrt{1 + \sqrt{2}}
\end{cases}$$
is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/983219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $ (0.3)^n\sum_{m=0}^n \left(\frac{0.8}{0.3}\right)^m $? Can someone please help me solve this sum:
$$
(0.3)^n\sum_{m=0}^n \left(\frac{0.8}{0.3}\right)^m u[n]
$$
where $u[n]$ means just that $n \ge 0$.
I keep getting $$-2(0.3^n -(0.8^(n+1))/(0.3))$$ but according to my textbook it is wrong.
I used the $$\sum_{i=m}^{n-1} a^i = \frac{a^m-a^n}{1-a}$$ forumla from which m becomes n+1.
Thanks guys
I get this: $$ 2*(0.8)^(n+1) - (3/10)^n *3/5$$, and now I try to transform $$(3/10)^n$$ to
$$(3/5)^n *(1/2)^n$$ and after that I can't do anything with $$(1/2)^n$$
| $$
\sum_0^n a^n = \frac{1-a^{n+1}}{1-a}
$$
Substitute $a = \frac{0.8}{0.3} = \frac{8}{3} $ giving
$$\sum_0^n \left( \frac{8}{3} \right) ^m = \frac
{ 1-\frac{8^{n+1}} {3^{n+1}}}
{1-\frac{8}{3}} =
\frac{3}{5} \left( \frac{8^{n+1}} {3^{n+1}} -1 \right)
$$
Now multiply by $\left( \frac{3}{10} \right) ^ n$. The powers of 3 cancel in the first term, and some powers of 2 cancel as well, and you are left with
$$
\frac{ 2^{2n+3} }{5^{n+1}} -
\frac{3^{n+1}} { 2^n \cdot 5^{n+1} } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/984621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\sin 3x}{1+2\sin x}\right)dx$ and $\int_{0}^{2} \left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\right)dx$ Evaluation of Some Integrals::
$\displaystyle (a)\;\;\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\sin 3x}{1+2\sin x}\right)dx\;\;\;\;\;\;(b)\;\; \int_{0}^{2} \left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\right)dx\;\;\;\;\;\;$
$\displaystyle (c)\;\;\int_{0}^{1}\frac{4x^3\cdot \left(1+(x^4)^{2010}\right)}{\left(1+x^4\right)^{2012}}dx\;\;\;\;\;\; (d)\;\; $
$\bf{My\; Try::}$I have tried for third one:: Let $\displaystyle I = \int_{0}^{1}\frac{4x^3\cdot \left(1+(x^4)^{2010}\right)}{\left(1+x^4\right)^{2012}}dx$
Now Put $\displaystyle x^4=t\;,$ Then $\displaystyle 4x^3dx=dt$ and changing Limit, we Get
$\displaystyle I = \int_{0}^{1}\frac{1+t^{2010}}{(1+t)^{2012}}dt$
Now How can I solve after that , Help me
and i did not understand how can i solve $(a)$ and $(b)\;,$ Help me
Thanks
| Despite its appearance, there is actually a surprisely simple way to evaluate integral $(b)$.
Consider following change of variables,
$$y = \sqrt{1+x^3}-1 \quad\iff\quad x = \sqrt[3]{(y+1)^2-1} = \sqrt[3]{y^2 + 2y}$$
When $x$ varies from $0$ to $2$, the corresponding $y$ varies from $0$ to $2$ too.
As a result,
$$\int_0^2 \left(\sqrt{1+x^3} + \sqrt[3]{x^2+2x}\right) dx
= \int_0^2 \left( 1 + \sqrt[3]{x^2+2x} \right) dx + \int_0^2 y\, d\sqrt[3]{y^2+2y}\\
= \left[ x + x\sqrt[3]{x^2+2x}\right]_0^2
= 2 + 2\sqrt[3]{2^2+2\cdot 2} = 6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_0^{\frac{\pi}{2}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta$ Prove
$$\int_0^{\frac{\pi}{2}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta = \frac{\pi^2}{4}$$
| I am also a proponent of the opinion that the proposed rules are against the way how we communicate ideas in this community. At the same time, however, the mathematical part of OP is something worth it to be dealt with. So here is a solution:
1. Preliminary
Before the calculation we make some preliminary results:
Lemma 1. For any $u > 0$ and $n > 0$, we have
$$\frac{1}{n^{2}} \log \left(1 + \frac{4\pi^{2}n^{2}}{u^{2}} \right) = \pi^{2} \int_{u/2}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} \, \frac{ds}{s}.$$
Proof. Differentiating both sides with respect to $u$, we check that they must equal up to a constant. Taking $u \to \infty$, we find that this constant should equal zero. ////
Lemma 2. For any real $x$, we have
$$ \sum_{n=1}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} = \frac{s \coth s - 1}{s^{2}}. $$
Although non-trivial, this is a standard result in complex analysis. So we omit the proof.
Lemma 3. Let $f(s) = (1 - e^{-2s})(s\coth s - 1)$. Then
*
*$f(s) = (s-1) + (s+1)e^{-2s}$ and hence $f''(s) = 4s e^{-2s}$.
*$f(s)/s^{2}$ and $f'(s)/s$ converges to $0$ as $s \to 0$ and $s \to +\infty$.
Proof. The first assertion is just a simple calculation. To prove the second assertion, it suffices to look into the McLaurin series expansion $f(s) = \frac{2}{3}s^{3} - \frac{2}{3}s^{4} + \cdots$. ////
2. Calculation
Now we are ready to calculate the integral. Let $I$ denote the integral. Then with the substitution $\sin^{2}\theta = e^{-t}$ (so that $d\theta/\tan\theta = -dt/2t$), we have
\begin{align*}
I
&= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\log^{2} \sin^{2}\theta}{4\pi^{2} + \log^{2}\log^{2}\theta} \right) \frac{\log^{2}\cos^{2}\theta}{\tan\theta} \, d\theta \\
&= \frac{1}{4} \int_{0}^{\infty} \log(1 - e^{-t}) \log\left( \frac{t^{2}}{4\pi^{2} + t^{2}} \right) \, dt\\
&= \frac{1}{4} \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{e^{-nt}}{n} \log \left(1 + \frac{4\pi^{2}}{t^{2}} \right) \, dt.
\end{align*}
Now we utilize the Tonelli's theorem to interchange the summation and integral. Then
\begin{align*}
I
&= \frac{1}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} \frac{e^{-nt}}{n} \log \left(1 + \frac{4\pi^{2}}{t^{2}} \right) \, dt \\
&= \frac{1}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} \frac{e^{-u}}{n^{2}} \log \left(1 + \frac{4\pi^{2}n^{2}}{u^{2}} \right) \, du, \quad (u = nt) \\
&= \frac{\pi^{2}}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} e^{-u} \left( \int_{u/2}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} \, \frac{ds}{s} \right) \, du,
\end{align*}
where the last equality follows from Lemma 1. Applying the Tonelli's theorem again, Lemma 2 shows that
\begin{align*}
I
&= \frac{\pi^{2}}{4} \int_{0}^{\infty} e^{-u} \left( \int_{u/2}^{\infty} \frac{s \coth s - 1}{s^{2}} \, \frac{ds}{s} \right) \, du \\
&= \frac{\pi^{2}}{4} \int_{0}^{\infty} \left( \int_{0}^{2s} e^{-u} \, du \right) \frac{s \coth s - 1}{s^{3}} \, ds \\
&= \frac{\pi^{2}}{4} \int_{0}^{\infty} \frac{f(s)}{s^{3}} \, ds,
\end{align*}
where we applied Tonelli's theorem again in the second line, and $f(s)$ denotes the function in Lemma 3. So it suffices to prove that the last integral, without the constant $\pi^{2}/4$, equals 1. Indeed, Lemma 3 shows that
\begin{align*}
\int_{0}^{\infty} \frac{f(s)}{s^{3}} \, ds
&= \left[ -\frac{f(s)}{2s^{2}} \right]_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} \frac{f'(s)}{s^{2}} \, ds \\
&= \left[ -\frac{\smash{f'}(s)}{2s} \right]_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} \frac{f''(s)}{s} \, ds \\
&= \int_{0}^{\infty} 2e^{-2s} \, ds
= 1
\end{align*}
and therefore we get $I = \pi^{2}/4$ as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Prove that $2\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)}\ge 2ax+2by-ay-bx$? Let $a,b,x,y$ be real numbers , then is it true that
$$2\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)}\ge 2ax+2by-ay-bx?$$ Actually I am trying to prove the triangle inequality for the norm of numbers in $K(\rho)$ , where $\rho$ is the imaginary cube-root of unity , $|a+b\rho|:=\sqrt{(a+b\rho)(a+b \rho^2)}=a^2-ab+b^2$
| $$
\begin{align}
(u-v)^2&\ge0\tag{1}\\[6pt]
u^2v^2+3u^2+3v^2+9&\ge u^2v^2+6uv+9\tag{2}\\[6pt]
(u^2+3)(v^2+3)&\ge(uv+3)^2\tag{3}\\[3pt]
\sqrt{(2s-1)^2+3}\,\sqrt{(2t-1)^2+3}&\ge\left|\,(2s-1)(2t-1)+3\tag{4}\,\right|\\[3pt]
2\sqrt{s^2-s+1}\,\sqrt{t^2-t+1}&\ge\left|\,2st-s-t+2\,\right|\tag{5}\\
2\sqrt{\left(\frac ab\right)^2-\frac ab+1}\,\sqrt{\left(\frac xy\right)^2-\frac xy+1}&\ge\left|\,2\frac ab\frac xy-\frac ab-\frac xy+2\,\right|\tag{6}\\
2\sqrt{a^2-ab+b^2}\,\sqrt{x^2-xy+y^2}&\ge\left|\,2ax-ay-bx+2by\,\right|\tag{7}\\
\end{align}
$$
Explanation:
$\begin{array}{l}
(1):&\text{square of a real is non-negative}\\
(2):&\text{multiply by $3$ and add $u^2v^2+6uv+9$}\\
(3):&\text{factor}\\
(4):&\text{square root, $u=2s-1$ and $v=2t-1$}\\
(5):&\text{expand and divide by $2$}\\
(6):&\text{$s=\frac ab$ and $t=\frac xy$}\\
(7):&\text{multiply by $|by|$}
\end{array}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/989274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find $ \lim_{x\rightarrow 0}\frac{\tan^2 x+2x}{x+x^2} $ without L'Hôpital's Rule I need to find $$\lim_{x\to 0}\frac{\tan^2 x+2x}{x+x^2}$$
This is what I did:
Let $f(x)=\frac{\tan^2(x)+2x}{x+x^2}=\frac{\frac{\sin^2(x)}{\cos^2(x)}+2x}{x+x^2}=\frac{\sin^2(x)+2x\cos^2(x)}{x(x+1)\cos^2(x)}=\frac{\sin^2(x)}{x(x+1)\cos^2(x)}+\frac{2}{x+1}=\frac{\sin (x)}{x}\cdot\frac{\sin(x)}{(x+1)\cos^2(x)}+\frac{2}{x+1}$
$\lim_{x\to0}\left(\frac{\sin(x)}{x}\cdot\frac{\sin(x)}{(x+1)\cos^2(x)}+\frac{2}{x+1}\right)=\underbrace{\left(\lim_{x\to0}\frac{\sin(x)}{x}\right)}_{=1}\cdot\left(\lim_{x\to0}\frac{\sin(x)}{(x+1)\cos^2(x)}\right)+\underbrace{\lim_{x\to0}\left(\frac{2}{x+1}\right)}_{=2}=\lim_{x\to0}\frac{\sin(x)}{(x+1)\cos^2(x)}+2=\frac{\lim_{x\to0}\sin(x)}{\underbrace{\left(\lim_{x\to0}x+1\right)}_{=1}\cdot\underbrace{\left(\lim_{x\to0}\cos^2(x)\right)}_{=1^2=1}}+2=\lim_{x\to0}\sin(x)+2=0+2=2$
Is this correct? I feel like I have assumed the some limits rather than proving it.
We are allowed to use the common result $\frac{\sin x}{x}$ is $1$ as $x$ tends to $0$.
| $$\frac{\tan^2 x+2x}{x+x^2}$$ is equivalent to$$\frac{\tan(x)\frac{\tan(x)}{x}+2}{1+x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/990775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculus - Help with this series $S_n=\sum_{k=1}^\infty \frac{1}{9k^2+3k-2}$ $$S_n=\sum_{k=1}^\infty \frac{1}{9k^2+3k-2}$$
My solution
$$S_n=\sum_{k=1}^\infty \frac{1}{9k^2+3k-2} = \sum_{k=1}^\infty \frac{1}{3k(3k+1)-2}=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{3k(3k+1)-2}$$
I understand that it converge (since r<1), but I need to find the sum.
Any ideas?
Thanks!
| As has been pointed out in hints the following holds:
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{9k^2 + 3k-2} &= \frac{1}{3} \sum_{n=1}^{\infty} \left(\frac{1}{3k-1}-\frac{1}{3k+2}\right) \\
&= \frac{1}{3} \left[ \frac{1}{2} + \sum_{n=1}^{\infty} \frac{1}{3n+2} - \sum_{k=1}^{\infty} \frac{1}{3k+2} \right] \\
&= \frac{1}{6}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integration Using Trigonometric Substitutions with $x^2$ and $\sqrt x$ in denominator How can I integrate a function in this form:
$$\int\frac{1}{x^2\sqrt{a+x^2}}dx$$
I know I need to use trigonometric substitution but don't see how to apply it for this form of function. What should I look for when integrating these types of functions?
| Let $$x = \sqrt a\tan\theta \implies dx = \sqrt a\sec^2 \theta$$ and $x^2 = a \tan^2\theta$
$$\int\frac{1}{x^2\sqrt{a+x^2}}\,dx = \int \frac {\sqrt a \sec^2 \theta}{a\tan^2\theta\sqrt{a + a \tan^2 \theta}}\,d\theta$$
$$ = \int \dfrac{\sec^2\theta}{a \tan^2\theta \sqrt{\underbrace{1+ \tan^2}_{\sec^2 \theta}\theta}}\,d\theta$$
$$ = \frac 1a\int \dfrac{\sec^2\theta}{\tan^2\theta\sec\theta}\,d\theta$$
$$=\frac 1a \int \dfrac{\cos \theta\,d\theta}{\sin^2\theta} = \frac{-1}{a\sin\theta} + C $$ Now, since $$x = \sqrt a \tan\theta, \tan\theta = \frac x{\sqrt a}\implies \sin\theta = \dfrac{x}{\sqrt{a+x^2}}$$
So $$\frac{-1}{a\sin\theta} + C = -\dfrac{\sqrt{a+x^2}}{ax} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $ \int_0^{\sqrt{\pi / 2}}\left(\int_x^{\sqrt{\pi / 2} }\sin(y^2) dy \right)dx$ I'm trying to solve this:
$$
\int_0^{\large\sqrt{\frac{\pi}{2}}}\left(\int_x^{\large\sqrt{\frac{\pi}{2}}}\, \sin y^2\, dy \right)dx
$$
But I'm having trouble with finding an primitive to $\sin(y^2)$. I got the tips to invert the order of the integrals but I fail to do that. (My try was to simply replace $x$ with $y$ and do the calculations). How to solve this?
| $\int_0^{\sqrt{\frac{\pi}{2}}}\int_x^{\sqrt{\frac{\pi}{2}}}\sin(y^2)~dy~dx$
$=\left[x\int_x^{\sqrt{\frac{\pi}{2}}}\sin(y^2)~dy\right]_0^{\sqrt{\frac{\pi}{2}}}-\int_0^{\sqrt{\frac{\pi}{2}}}x~d\left(\int_x^{\sqrt{\frac{\pi}{2}}}\sin(y^2)~dy\right)$
$=\left[x\int_x^{\sqrt{\frac{\pi}{2}}}\sum\limits_{n=0}^\infty\dfrac{(-1)^ny^{4n+2}}{(2n+1)!}dy\right]_0^{\sqrt{\frac{\pi}{2}}}+\int_0^{\sqrt{\frac{\pi}{2}}}x\sin(x^2)~dx$
$=\left[x\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^ny^{4n+3}}{(2n+1)!(4n+3)}\right]_x^{\sqrt{\frac{\pi}{2}}}\right]_0^{\sqrt{\frac{\pi}{2}}}+\int_0^{\sqrt{\frac{\pi}{2}}}\dfrac{\sin(x^2)}{2}d(x^2)$
$=\left[x\left(\sum\limits_{n=0}^\infty\dfrac{(-1)^n\pi^{2n+\frac{3}{2}}}{2^{2n+\frac{3}{2}}(2n+1)!(4n+3)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{4n+3}}{(2n+1)!(4n+3)}\right)\right]_0^{\sqrt{\frac{\pi}{2}}}-\left[\dfrac{\cos(x^2)}{2}\right]_0^{\sqrt{\frac{\pi}{2}}})$
$=\dfrac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Computing the Laplacian of $\frac{\mu\ \cdot\ \mathbf{r}}{\|\mathbf{r}\|^3}$ How does one compute the Laplacian of
$$\dfrac{(\mathbf \mu \cdot \mathbf r)}{r^3} \;\; \text{where} \;\; r = \Vert \mathbf r \Vert?$$
I am aware that the Laplacian is defined to be $\Delta f=\sum_i \partial_i^2f$ but am a little confused about this computationally in this case.
I have got as far as saying $\partial_i^2[1/r^3(\mu_1r_1+\mu_2r_2+...+\mu_nr_n)]$, but don't know how to compute this.
| WARNING: GRAPHIC CONTENT! It's not going to be pretty as Joonas' answer (btw thank you for the correction!)
Let's do it in cartesian coordinates. The function is
\begin{equation}
u(x_1,\dots,x_n)=\frac{\sum_i\mu_ix_i}{\left(\sum_j x_j^2\right)^{3/2}}=\frac{\mu\cdot\mathbf{r}}{r^3}
\end{equation}
We have that
\begin{equation}
\frac{\partial\ r^s}{\partial x_k}=sx_k\ r^{s-2}
\end{equation}
and also
\begin{equation}
\frac{\partial\ (\mu\cdot\mathbf{r})}{\partial x_k}=\mu_k
\end{equation}
so,
\begin{equation}
\frac{\partial u}{\partial x_k}=\frac{\mu_kr^3-3x_kr(\mu\cdot\mathbf{r})}{r^6}=\frac{1}{r^5}\left(\mu_kr^2-3(\mu\cdot\mathbf{r})x_k\right)
\end{equation}
and
\begin{equation}
\frac{\partial^2 u}{\partial x_k^2}=
\frac{1}{r^{10}}\left[(2\mu_kx_k-3\mu_kx_k-3(\mu\cdot\mathbf{r}))r^5-5x_kr^3(\mu_kr^2-3(\mu\cdot\mathbf{r})x_k)\right]=\\
=\frac{1}{r^7}\left[-\mu_kx_kr^2-3(\mu\cdot\mathbf{r})r^2-5\mu_kx_kr^2+15(\mu\cdot\mathbf{r})x_k^2\right]=\\
=\frac{3}{r^7}\left[(\mu\cdot\mathbf{r})(5x_k^2-r^2)-2\mu_kx_kr^2\right]
\end{equation}
Summing everything over $k$ we get
\begin{equation}
\Delta u = \sum_k\frac{\partial^2u}{\partial x_k^2}=\frac{3}{r^7}\sum_k\left[(\mu\cdot\mathbf{r})(5x_k^2-r^2)-2\mu_kx_kr^2\right]=\\
=\frac{3}{r^7}\left[(\mu\cdot\mathbf{r})\sum_k(5x_k^2-r^2)-2r^2\sum_k\mu_kx_k\right]=\\
=\frac{3}{r^7}\left[(\mu\cdot\mathbf{r})(5r^2-nr^2)-2r^2(\mu\cdot\mathbf{r})\right]=\\
=3(3-n)\frac{(\mu\cdot\mathbf{r})}{r^5}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/995131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve equation for $0^\circ < x < 360^\circ$ Solve the following equation for $0^\circ < x < 360^\circ$
$$\cos(2x - 15^\circ) = -0.145$$
By finding out the cos inverse, I get $81.7^\circ$. Because $-0.145$ is negative, it lies on the 2nd and 3rd quadrant. That is we can find $x$ by $(180 + \theta)$ and $(180 - \theta)$.
So this was what I've thought:
$$2x-15^\circ = 81.7^\circ$$
From here according to me, $$x=(81.7^\circ + 15^\circ)/2$$ gives $$x=48.4$$ which is not the answer in my book.
Then comes $$2x-15^\circ=(180+81.7^\circ),(180-81.7^\circ)$$
From here I find $$x=138.35^\circ, 56.7^\circ$$ these answers are correct.
Now, the first answer $48.4$ is wrong, and there is one more answer in my book, i.e, there are $4$ values in the answer, where I got $3$ answers and $1$ is wrong.
Could someone help help me to find the two other answers?
| \begin{align}
\cos( 2x - 15^\circ ) &= -0.145 \\
2x - 15^\circ &= \cos^{-1}(-0.145) + 360^\circ k \quad\quad \ k\in \ldots -2, -1, 0, 1, 2, \ldots \\
x &= \frac{1}{2} \left( 15^\circ + \cos^{-1}(-0.145) + 360^\circ k \right) \quad\quad \ k\in \ldots -2, -1, 0, 1, 2, \ldots \\
x &= \frac{1}{2} \left( 15^\circ \pm 98.3372791889057^\circ + 360^\circ k \right) \quad\quad \ k\in \ldots -2, -1, 0, 1, 2, \ldots \\
x &= 56.6686395944528^\circ, 138.331360405547^\circ, 236.668639594453^\circ, 318.331360405547^\circ
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/995307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.