Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Elementary proof of a cotangent inequality
Let $0<x<\pi/2$. Then
$$ \cot{x} > \frac{1}{x}+\frac{1}{x-\pi}. $$
(This is still true for $-\pi<x<0$, but the given range is the one I'm concerned about.)
Is there an elementary proof of this? The local inequality $\cot{x}<1/x$ is easy to prove since it is equivalent to $x<\tan{x}$, which even has a simple geometric proof.
One can massage the Mittag-Leffler formula
$$ \cot{x} = \frac{1}{x} + \sum_{n \neq 0} \frac{1}{x-n\pi} + \frac{1}{n\pi} $$
into
$$ \cot{x} = \frac{1}{x} + \frac{1}{x-\pi} + \sum_{n=1}^{\infty} \frac{1}{z- (n+1)\pi}+\frac{1}{z+\pi n}, $$
and the terms in the second sum are all positive in the range considered since
$$ \frac{1}{z- (n+1)\pi}+\frac{1}{z+\pi n} = \frac{\pi-2z}{n(n+1)\pi^2+z(\pi-z)}>0, $$
but this is rather heavyweight for such a simple-looking inequality.
An equivalent formulation is
$$ \tan{y} > \frac{1}{\pi/2-y}-\frac{1}{\pi/2+y} = \frac{8y}{\pi^2-4y^2} $$
for $0<y<\pi/2$, if one desires more symmetry.
| We need to prove that
$$\frac{1}{\tan{x}}>\frac{\pi-2x}{x(\pi-x)}$$ or
$$\pi x-x^2>(\pi-2x)\tan{x}$$ or
$$x^2-(2\tan{x}+\pi)x+\pi\tan{x}<0$$ or
$$\tan{x}+\frac{\pi}{2}-\sqrt{\tan^2x+\frac{\pi^2}{4}}<x<\tan{x}+\frac{\pi}{2}+\sqrt{\tan^2x+\frac{\pi^2}{4}},$$
for which it's enough to prove that $f(x)>0$, where
$$f(x)=x-\tan{x}-\frac{\pi}{2}+\sqrt{\tan^2x+\frac{\pi^2}{4}},$$
which is easy.
Indeed,
$$f'(x)=1-\frac{1}{\cos^2x}+\frac{\tan{x}}{\cos^2x\sqrt{\tan^2x+\frac{\pi^2}{4}}}=$$
$$=\frac{\sin{x}\left(1-\sin{x}\sqrt{\sin^2x+\frac{\pi^2}{4}\cos^2x}\right)}{\cos^3x\sqrt{\tan^2x+\frac{\pi^2}{4}}}=$$
$$=\frac{\sin{x}\left(1-\sin^4x-\frac{\pi^2}{4}\sin^2x\cos^2x\right)}{\cos^3x\sqrt{\tan^2x+\frac{\pi^2}{4}}\left(1+\sin{x}\sqrt{\sin^2x+\frac{\pi^2}{4}\cos^2x}\right)}=$$
$$=\frac{\sin{x}\cos^2x\left(1+\sin^2x-\frac{\pi^2}{4}\sin^2x\right)}{\cos^3x\sqrt{\tan^2x+\frac{\pi^2}{4}}\left(1+\sin{x}\sqrt{\sin^2x+\frac{\pi^2}{4}\cos^2x}\right)}=$$
$$=\frac{\left(\frac{\pi^2}{4}-1\right)\sin{x}\left(\frac{2}{\sqrt{\pi^2-4}}-\sin{x}\right)\left(\frac{2}{\sqrt{\pi^2-4}}+\sin{x}\right)}{\sqrt{\sin^2x+\frac{\pi^2}{4}\cos^2x}\left(1+\sin{x}\sqrt{\sin^2x+\frac{\pi^2}{4}\cos^2x}\right)},$$
which gives $x_{max}=\arcsin\frac{2}{\sqrt{\pi^2-4}}$ and since $$\lim_{x\rightarrow0^+}f(x)=\lim_{x\rightarrow\frac{\pi}{2}^-}f(x)=0,$$
we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Volume from iterated integrals and two regions The prompt is to find the volume of the solid which is described the equations and is bounded.
$$x^2+y^2+z^2=9 $$
$$x^2-3x+y^2=0 $$
The first one is a sphere with radius 3, the shadow is on the y-x plane.
For the second on I tried using completing the squares.
$$x^2-3x + y^2 =0 $$
$$x^2-3x+ 1/25 + y^2 = 1/25 $$
$$(x-1/5)^2 + y^2 = 1/25 $$
i dont know how to procede now. I also tried...
$$x^2+y^2-3x = 0$$
$$r^2-3x = 0 $$
$$ r^2 = 3x$$
$$ r^2 = 3cos\theta$$
$$ r = \sqrt{3cos\theta} $$
$$ \int_0^3\int_0^{2\pi} \int_0^{\sqrt{3cos\theta}}x^2+y^2+z^9rdrd\theta dz$$
Please correct me if the method to get the radius, if its wrong? Im kinda new to calculus.
| If you want integrate in cartesian coordinates, note that:
substituting $x^2+y^2=3x$ in the first equation we find that the limiting values for $z$ are given by $3x+z^2=9 $ that gives
$$
-\sqrt{3(3-x)}<z<\sqrt{3(3-x)}
$$
The circle in $x-y$ plane $x^2+y^2-3x=0$ gives: $y=\pm \sqrt{3x-x^2}$
so the limits for $y$ are:
$$
-\sqrt{x(3-x)} <y<\sqrt{x(3-x)}
$$
and, since $y$ must be real we have also the limits ofr $x$
$$
0<x<3
$$
So, using the symmetry withe respect to the $x-y$ plane, the volume can be calculates as:
$$
V=2\int_0^3\int_{-\sqrt{x(3-x)}}^{\sqrt{x(3-x)}}\int _0^{\sqrt{3(3-x)}}dzdydx
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Write Taylor Expansion Of $\frac{1}{(1-z)^2}$
Write Taylor Expansion Of $\frac{1}{(1-z)^2}$ around $z_0=0$
I need to "get rid" of the 2 power and I can use $\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$ but how can I do that?
| Since$$\frac1{1-z}=1+z+z^2+z^3+\cdots\text,$$then $\frac1{(1-z)^2}$ is the Cauchy product of the geometric series by itself. So,\begin{align*}\frac1{(1-z)^2}&=(1+z+z^2+z^3+\cdots)(1+z+z^2+z^3+\cdots)\\&=1+2z+3z^2+4z^3+\cdots\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337360",
"timestamp": "2023-03-29T00:00:00",
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Simplified expression of the ratio of this series I need to simplify the expression of this series:
$$
\sum\limits_{n=1}^{\infty}\frac{3^{n}(4n)!}{1 · 4 · 7 · ⋯ ·3n+1}
$$
As the ratio (for the test): $$\frac{a_{n+1}}{a_n}$$
First I multiplied the reciprocal of a(n) by a(n+1):
$$
\frac{a_{n+1}}{a_n}=\frac{3^{n+1}(4(n+1))!}{(3(n+1)+1)}\cdot\frac{3n+1}{3^{n}(4n)!}
$$
Then I expanded the terms:
$$
\frac{3^{n+1}\cdot4n!(4n+4)(4n+3)(4n+2)(4n+1)(3n+1)}{3^{n}\cdot4n!(3n+4)(3n+3)(3n+2)(3n+1)}
$$
Simplified down to:
$$
\frac{a_{n+1}}{a_n}=\frac{3(4n+4)(4n+3)(4n+2)(4n+1)}{(3n+4)(3n+3)(3n+2)}
$$
Am I missing a step or performing any incorrectly?
This question is from a practice test, which is multiple choice, but I don't have an answer key:
See choices here
| \begin{align}
\frac{a_{n+1}}{a_n}&=
\frac{3^{n+1}\cdot4n!(4n+4)(4n+3)(4n+2)(4n+1)(3n+1)}{3^{n}\cdot 4n!(3n+4)}\\
&=
\frac{3 (4n+4)(4n+3)(4n+2)(4n+1)(3n+1)}{(3n+4)}.
\end{align}
Answer F.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Correct answer of an indefinite integral Find the value of
$$ \int{\frac{dx}{x\sqrt{1-x^3}}} $$
I assumed $x^3 = \sin^2\theta$ and found the solution as
$$\frac{2}{3} \log\left|\frac{1}{x\sqrt{x}} - \frac{\sqrt{1-x^3}}{x\sqrt{x}} \right| + c$$
but the solution is given as
$$\frac{1}{3} \log{\left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|} + c$$
Any help to reach to this provided solution will be appreciated.
| You have found the correct answer already. Both solutions are the same, they are just written in a different fashion.
If you use the property of the logarithm you can rewrite your initial results by
$$
\frac{2}{3} \log |\cdots| = \frac{1}{3} \log |\cdots|^2 = \frac{1}{3} \log \left|\frac{2 - x^3 - 2 \sqrt{1-x^3}}{x^3} \right|
$$
In the case of the given answer you can rewrite it by
$$
\frac{1}{3} \log \left| \frac{\sqrt{1-x^3} -1}{\sqrt{1-x^3} +1} \cdot\frac{\sqrt{1-x^3} -1}{\sqrt{1-x^3} -1}\right|
$$
and work out numerator and denominator. This gives the same expression.
Normally you would need to be a bit careful with the sign of the argument in the logarithm, but here with the absolute values that is automatically taken care of.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I solve this fraction question? If $a = -1/5$, how do I calculate:
$$3 a + 2 a^2$$
I did $3\times(-1/5) + (-1/5) \times (-1/5) \times 2$, but can't figure out what the right way to solve this is.
| It might help you to add a denominator of $1$ to integers: $$\frac{3}{1} \times \frac{-1}{5} = \frac{-3}{5}.$$
So far so good, right? Next, $$\frac{2}{1} \times \left(\frac{-1}{5}\right)^2 = \frac{2}{1} \times \frac{1}{25} = \frac{2}{25}.$$
And then $$\frac{-3}{5} + \frac{2}{25} = \frac{-15}{25} + \frac{2}{25} = ?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ax^2+bx+c = 0$ and $bx^2+cx + a = 0$ have a common root and $a\neq 0$, then find $\frac{a^3+b^3+c^3}{abc}$ If $ax^2+bx+c = 0$ and $bx^2+cx + a = 0$ have a common root and $a\neq 0$, then find $$\frac{a^3+b^3+c^3}{abc}$$
I tried that for both equations to have a common root, the expression on left hand sides must be equal, ie $$ax^2+bx+c = bx^2+cx + a$$
for this we must have $x=1$ (i cannot prove this, but it appears to be true). Also both of these must be equal to $0$, so we have:
$$a+b+c=0$$
So using this we say
$$\frac{a^3+b^3+c^3}{abc} = \frac{a^3+b^3+c^3-3abc+3abc}{abc}=\frac{(a+b+c)(...)}{abc}+3$$
So we get the answer as $3$.
How do we say that $x =1$ is the commmon root? Thanks!
| HINT:
Let $t$ be the common root
So, we have $$at^2+bt+c=0\ \ \ \ (1)$$ and $$bt^2+ct+a=0\ \ \ \ (2)$$
So, we have two simultaneous equations in $t^2,t$
Solve for $t,t^2$ and use the identity $$t^2=(t)^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Calculating Imaginary Number Is this step-by-step computation correct?
\begin{align}
\frac {3i^{30}-i^{19}}{2i-1} &= \frac {3(i^{2})^{15}-(i^{2})^{9}i}{2i-1} \\
&= \frac {3(-1)^{15}-(-1)^{9}i}{2i-1} \\
&= \frac {-3+i}{2i-1} \\
&= \frac {-3+i}{2i-1} \times \frac {2i+1}{2i+1} \\
&= \frac {-6i-5+i}{-5} \\
&= \frac {-5i-5}{-5} \\
&= i+1.
\end{align}
| An alternate method is to take advantage of $i^{4} = 1$ in such a way that $i^{19} = i^{4 \cdot 4 + 3} = (i^{4})^{4} \, i^{3} = i^{3} = - i$ and $i^{30} = i^{4 \cdot 7 + 2} = i^{2} =-1$. This leads to
\begin{align}
\frac{3i^{30}-i^{19}}{2i-1} &= \frac{i - 3}{2i -1} = \frac{(i-3)(2i +1)}{4 i^{2} - 1} = i+1.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the expression $\cos(\frac{1}{2}\tan^{-1}(-\frac{4}{3}))$ I can prove that $\cos(\tan^{-1}(x)) = +\dfrac{1}{\sqrt{1+x^2}}$, set $y=\tan^{-1}(x)$ and therefore $y=\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$
Dividing both sides by $2$:
$\dfrac{y}{2}=\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$
$\cos(\dfrac{1}{2}\tan^{-1}(x))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}}))$
$\cos(\dfrac{1}{2}\tan^{-1}(-\frac{4}{3}))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{3}{5}))$
This is my approach to the problem, are there any better ways to simplify the initial expression and perhaps find an approximation even without using a calculator?
| Consider the following :
$$\begin{align}
&\implies\frac{\arctan\left(\frac{-4}{3}\right)}{2}=y \\
&\implies2y=\arctan(\frac{-4}{3})\\
&\implies\tan(2y)=-4/3\\
&\implies\frac{2\tan y}{1-\tan^2y}=-4/3\\
&\implies2\tan^2y-3\tan y-2=0\\
\end{align}$$
Solving this equation we get-
$$\tan y=-1/2$$and $$\tan y=2$$
Using this we can find the value of $\cos y$ which will give the value of the expression. It comes out to be:
$$\cos y=\frac{1}{\sqrt{5}}$$
Hope my answer is helpful. Sorry for any errors and please let me know if there are some.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
What is the value of the expression $[1 + \cos(\frac{\pi}{8})][1 + \cos(\frac{3\pi}{8})][1 + \cos(\frac{5\pi}{8})][1 + \cos(\frac{7\pi}{8})]$? This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.
What is the value of the following expression?
$$\left( 1+\cos { \left( \frac { \pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 5\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 7\pi }{ 8 } \right) } \right) $$
Edit: Do I manually put the values of $\cos\frac{\pi}{8}$ and all other cosine terms? Is there a better and shorter way?
| The answer is $\dfrac{1}{8}$
As $\cos(\pi-x)=-\cos x$ we can write
$$\left(1+\cos\frac{\pi}{8}
\right)\left(1+\cos\frac{3\pi}{8}
\right)\left(1+\cos\frac{5\pi}{8}
\right)\left(1+\cos\frac{7\pi}{8}
\right)=\left(1+\cos\frac{\pi}{8}
\right)\left(1+\cos\frac{3\pi}{8}
\right)\left(1-\cos\frac{3\pi}{8}
\right)\left(1-\cos\frac{\pi}{8}
\right)=
\left(1-\cos^2\frac{\pi}{8}
\right)\left(1-\cos^2\frac{3\pi}{8}
\right)=\sin^2\frac{\pi}{8}\,\sin^2\frac{3\pi}{8}=\left(\sin\frac{\pi}{8}\,\sin\frac{3\pi}{8}\right)^2$$
For the Werner's formula
$$\sin x\sin y=\frac{1}{2} (\cos (x-y)-\cos (x+y))$$
so
$$\sin\frac{\pi}{8}\,\sin\frac{3\pi}{8}=\frac{1}{2} \left(\cos \left(\frac{\pi}{8}-\frac{3\pi}{8}\right)-\cos \left(\frac{\pi}{8}+\frac{3\pi}{8}\right)\right)=\frac{1}{2}\left(\cos\frac{-\pi}{4}-\cos\frac{\pi}{2}\right)=\frac{\sqrt{2}}{4}$$
Remember that
$\cos\dfrac{-\pi}{4}=\cos\dfrac{\pi}{4}=\dfrac{\sqrt 2}{2}$
So the result is $\left(\dfrac{\sqrt{2}}{4}\right)^2=\dfrac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352508",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find the remainder when $7^{7^{7}}$ is divided by 1000 I need help with this problem please
Find the remainder when $7^{7^{7}}$ is divided by $1000$
My try follow
$1000=8×125$ , now
$7 \equiv -1 \;\bmod\; (8)$
$\to$ $7^{7^{7}} \equiv -1 \;\bmod\; (8)$ and
$7^{100}\equiv 1 \;\bmod\; (125)$
Any help to complete this solution?
| $(7^{100} \mod 125)= 1$ so $7^{7^7} = 7^{(7^7 \bmod 100)} \mod 125 = 7^{43} \mod 125 = 93$
Now you can use Chinese remainder theorem with $x = 93 \pmod {125}$ and $x = -1 \pmod 8$, fastest way by hand is to check the $8$ values of $125 + 93k$ in the range of $0$ to $1000$ to see which one is $-1$ of $\pmod 8$.
Alternatively
$$\begin{align}
%
7 \text{^} (7 \text{^} 7) \mod 1000 & =
%
7 \text{^} (7 \text{^} 7 \mod \Phi 1000) \mod 1000 \\
%
&= 7 \text{^} (7 \text{^} 7 \mod 400) \mod 1000 \\
%
&= 7 \text{^} 343 \mod 1000 \\
%
&= 343 \\
%
\end{align}$$
Just calculating $7 \text{^} 823543 \pmod {1000}$ is trivial with modular exponentiation as well.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\sum\limits_{n\geq 1}\frac{1}{1+2+\ldots+n} $ Calculate $\displaystyle \sum_{n\geq 1}\dfrac{1}{1+2+\ldots+n} $
My attempts :
\begin{aligned}
\sum_{n\geq 1}\dfrac{1}{1+2+\ldots+n} &= \sum_{n\geq 1}\dfrac{2}{n(n+1)}=2\sum_{n\geq 1}\left( \dfrac{1}{n}-\dfrac{1}{n+1}\right)=2\left( \dfrac{1}{1}-\dfrac{1}{+\infty}\right)=2(1-0)=2?
\end{aligned}
| \begin{aligned}
\sum_{n\geq 1}\dfrac{2}{n(n+1)}=2\sum_{n\geq 1}\dfrac{1}{n}-2\sum_{n\geq 1}\dfrac{1}{n+1}
\end{aligned}
False. The equality does not hold because the two sums on the right are divergent, while the sum on the left is convergent.
Better go at it this way:
$$\sum_{n=1}^\infty\frac{1}{n(n+1)} = \sum_{n=1}^\infty\frac{1}{n}-\frac{1}{n+1} = \left(\frac 11 - \frac12\right) + \left(\frac12-\frac13\right) + \left(\frac13-\frac14\right) + \cdots$$
Now, remember that $$\sum_{n=1}^\infty a_n = \lim_{N\to\infty} \sum_{n=1}^N a_n$$
In your case, $$\sum_{n=1}^N \frac 1n-\frac{1}{n+1}$$ can be calculated relatively easily.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Help with summation induction Prove that
$$\sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i } } = } 2-\frac { n+2 }{ { 2 }^{ n } } \quad { for\quad all}\quad n\in \mathbb N $$
My attempt:
For my inductive step I tried the following:
$$\left(2 − \left(\frac{k + 2}{2^k}\right)\right) + \left(\frac{k + 1}{2^{k+1}}\right) = 2 − \left(\frac{k + 3}{2^{k+1}}\right)$$
But they never equal, this is where I'm stuck.
| Given $ \sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i } } = } 2-\frac { n+2 }{ { 2 }^{ n } } \quad for\quad all\quad n\in\mathbb N$
you should show
$$\sum _{ i=1 }^{ n+1 }{ \frac { i }{ { 2 }^{ i } } = } 2-\frac { n+3 }{ { 2 }^{ n+1 } } \quad $$
$$\sum _{ i=1 }^{ n+1 }{ \frac { i }{ { 2 }^{ i } } = } \sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i } } +\frac { n+1 }{ { 2 }^{ n+1 } } } =2-\frac { n+2 }{ { 2 }^{ n } } +\frac { n+1 }{ { 2 }^{ n+1 } } =2+\frac { n+1-2n-4 }{ { 2 }^{ n+1 } } =2-\frac { n+3 }{ { 2 }^{ n+1 } } \\ $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Telescoping(?) an infinite series Find the value of the sum $\displaystyle \sum\limits_{n=1}^{\infty} \frac{(7n+32) \cdot3^n}{n(n+2) \cdot 4^n}.$
Using partial fraction decomposition, I found the above expression is equivalent to $\displaystyle \sum\limits_{n=1}^{\infty} \frac{25}{n} \cdot \left(\frac{3}{4}\right)^n - \sum\limits_{n=1}^{\infty} \frac{18}{n+2} \cdot \left(\frac{3}{4}\right)^n,$ where I got couldn't find a closed form of either expression because of the $\left(\dfrac{3}{4}\right)^n.$
Similarly, trying to telescope one of the the sums with $\displaystyle \sum\limits_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+2}\right) \cdot \left(\frac{3}{4}\right)^n$ fails for the same reason. How can I further simplify the above expression? Thanks.
| HINT:
$$7n+32=4^2(n+2)-3^2\cdot n$$
$$\implies\dfrac{7n+32}{n(n+2)}\left(\dfrac34\right)^n=\dfrac{4^2(n+2)-3^2\cdot n}{n(n+2)}\left(\dfrac34\right)^n=\dfrac{16\left(\dfrac34\right)^n}n-\dfrac{16\left(\dfrac34\right)^{n+2}}{(n+2)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all $x$ such that $11\mid 3x+7$ I found this question in Beachy and Blair: Abstract algebra book, they even have a solution to this but its not satisfactory for me. They only say "$x\equiv 5 \pmod{11}$ ". Which one can "feel" simply by trial and error. I would like to know what is the proper approach. Thank you in advance!
| $11\mid 3x+7$ is the same thing as saying there exists an integer $y$ such that $3x + 7 = 11y$.
Trying $y=0, 1$ and finally $y=2$, we find that $x=5$ and $y=2$ is a solution. So we consider solutions of the form $x = 5+u$ and $y=2+v$.
\begin{align}
3x + 7 &= 11y \\
3(5+u) + 7 &= 11(2+v) \\
15 + 3u + 7 &= 22 + 11v \\
3u + 22 &= 11v + 22 \\
3u &= 11v \\
3 &\mid 11v \\
3 &\mid v
\end{align}
So $v = 3t$ and $y=2+3t$ for some integer $t$. The "for some" part is a problem. It suggests that, sometimes, $y=2+3t$ may not be a solution. It is very easy to show $y=2+3t$ is always going to be a solution.
Start with $3x+7=11y$ and let $y = 2+3t$.
\begin{align}
3x+7 &= 11y \\
3x + 7 &= 11(2+3t) \\
3x + 7 &= 22 + 33t \\
3x &= 33t + 15 \\
x &= 11t + 5
\end{align}
But, of course, saying $x = 11t+5$ for any integer $t$ means that $x \equiv 5 \pmod{11}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
If $\sqrt{1-\cos^2x}-\sqrt{1+\sin^2x}=k$, find $\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x}.$ Additional data added
If $\sqrt{1-\cos^2x}-\sqrt{1+\sin^2x}=k$, find $\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x}$
I raised $\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x}$ to the second power in order to express $k$:
\begin{align}&\color{white}=1-\cos^2x+2\sqrt{1-\cos^2x}\sqrt{1+\sin^2x}+1+\sin^2x\\&=1-\cos^2x+\sqrt{1+\sin^2x}(2\sqrt{1-\cos^2x}-\sqrt{1+\sin^2x})\\&=1-\cos^2x+\sqrt{1+\sin^2x}2k.\end{align}
Since it doesn't give the answer, I also raised the second expression equalling $k$ to the second power. That didn't work out. The variants are:
A)$1.5k$ B)$2k$ C)$\frac{2}{k}$ D)$-k$ E)$-\frac{1}{k}$
I have solved it in the way Fred did (having learnt from him) but as is obvious there is NOT such an answer in the variants above.
The correct answer is E)$-\frac{1}{k}$ (according to the answer tables in the book).
How to get to that correct answer E)???
| It's obvious that $k<0$.
Let $\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x}=p$.
Thus, $\sqrt{1+\sin^2x}=\frac{p-k}{2}$ and $\sqrt{1-\cos^2x}=\frac{p-k}{2}$.
Id est, $$\left(\frac{p-k}{2}\right)^2-\left(\frac{p+k}{2}\right)^2=1+\sin^2x-1+\cos^2x$$ or
$$-pk=1$$ or
$$p=-\frac{1}{k}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Evaluating $\lim_{x \rightarrow +\infty} \frac{\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\log(1+{e}^{3\sqrt x})}$ I'm trying to calculate the following limit:
$$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log(1+{e}^{3\sqrt x})}$$
For WolframAlpha the result is: $\frac 13$, I've seen its step by step process but I didn't understand the background logic.
Before I got stuck, I did this step:
$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log[{e}^{3\sqrt x}({e}^{-3\sqrt x}+1)]}$ and then
$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log({e}^{3\sqrt x})+\log({e}^{-3\sqrt x}+1)}$
so $$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle {3\sqrt x}}$$
Someone could give me a hint for solve it?
| HINT: Break it up as $$\frac{\sqrt{x}}{3\sqrt{x}} + \frac{x^3 - \sqrt{x^6 - 2x^4} - x}{3\sqrt{x}}.$$
To deal with the numerator of the second term, write it as $$x^3\left(1 - \sqrt{1 - 2/x^2} - 1/x^2\right)$$
and use the Taylor expansion of $\sqrt{1 - y}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =(x-i)(x+i)$, I don't think it is possible to show using partial fractions.
Secondly, decided to use differentiation
$y=\frac{2x}{x^2 +1}$
$\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }$
For stationary points:
$\frac {dy}{dx} = 0$
$\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0$
$x=-1$ or $x=1$
When $x=-1,y=-1$
When $x=1,y=1$
Therefore, this implies that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
^I wonder if this is the correct method or did I leave out something?
The third way was using discriminant
Assume that $y=\frac{2x}{x^2 +1}$ intersects with $y=-1$ and $y=1$
For $\frac{2x}{x^2 +1} = 1$,
$x^2 -2x+1 = 0$
Discriminant = $ (-2)^2 -4(1)(1) = 0 $
For $\frac{2x}{x^2 +1} = -1$,
$x^2 +2x+1 = 0$
Discriminant = $ (2)^2 -4(1)(1) = 0 $
So, since $y=\frac{2x}{x^2 +1}$ touches $y=-1$ and $y=1$, $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
Is the methods listed correct?Is there any other ways to do it?
| Squares are non-negative so
$$-(x+1)^2\le 0\le (x-1)^2$$
so
$$-x^2-2x-1\le 0\le x^2-2x+1$$
so
$$-x^2-1\le {2x}\le x^2+1$$
so
$$-1\le \frac{2x}{x^2+1}\le 1$$
since this is division by $x^2+1>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 6
} |
How can I convert a general ellipsoid equation into a vector form in MATLAB? What function can I use in MATLAB or how by hand can I convert the ellipsoid:
$$\frac{(x+12t-11)^2}{4}+y^2+z^2-1=0$$
into matrix form, from the form $X^\top {\rm A} X = 0$
$$X=\pmatrix{x \\ y \\ z \\ 1} $$
For correction and testing see the following
| I got
$$ \pmatrix{x \\ y \\ z \\ 1}^\top
\begin{bmatrix} \frac{1}{4} &0 & 0 & \frac{12 t-11}{4} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \frac{12 t-11}{4} & 0 & 0 & \frac{3 (48 t^2-88 t+39)}{4} \end{bmatrix} \pmatrix{x \\ y \\ z \\1} =0 $$
How?
I matched coefficients with $$\pmatrix{x \\ y \\ z \\ 1}^\top \begin{bmatrix} A_{11} & & & A_{14} \\ & A_{22} & & & \\ & & A_{33} & \\ A_{14} & & & A_{44} \end{bmatrix} \pmatrix{x \\ y \\ z \\ 1} =$$
$$ \frac{(x+12t-11)^2}{4}+y^2+z^2-1 = A_{11} x^2 +2 A_{14} x + A_{22} y^2 + A_{33} z^2 + A_{44} = 0$$
The reason I picked those elements of the coefficient matrix ${\rm A}$ is because the expression only contains terms of $x^2$, $x$, $y^2$, $z^2$ and a constant term.
In General
$$ \begin{bmatrix} A_{11} & 0 & 0 & A_{14} \\ 0 & A_{22} & 0 & A_{24} \\ 0 & 0 & A_{33} & A_{34} \\ A_{14} & A_{24} & A_{34} & A_{44} \end{bmatrix} \Longrightarrow A_{11} x^2 + A_{22} y^2 + A_{33} z^2 + 2 A_{14} x + 2 A_{24} y + 2 A_{34} z + A_{44} = 0 $$
$$ \frac{(x-x_c)^2}{a^2} + \frac{(y-y_c)^2}{b^2} + \frac{(z-z_c)^2}{c^2} = 1 \Longrightarrow \begin{bmatrix}
\frac{1}{a^2} & 0 & 0 & -\frac{x_c}{a^2} \\
0 & \frac{1}{b^2} & 0 & -\frac{y_c}{b^2} \\
0 & 0 & \frac{1}{c^2} & -\frac{z_c}{c^2} \\
-\frac{x_c}{a^2} & -\frac{y_c}{b^2} & -\frac{z_c}{c^2} & \kappa^2-1
\end{bmatrix} $$
where $\kappa^2 = \frac{x_c^2}{a^2}+\frac{y_c^2}{b^2}+\frac{z_c^2}{c^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \frac {\pi}{4}}$
Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}}$.
My Attempt:
\begin{align}
\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta }{\theta - \dfrac {\pi}{4}}
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \cos \dfrac {\pi}{4} + \sin \dfrac {\pi}{4} - \sin \theta}{\theta - \dfrac {\pi}{4}}
\\
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {2\sin \dfrac {\pi-4\theta }{8}\cos \dfrac {\pi+4\theta}{8} - 2\sin \dfrac {4\theta + \pi}{8}\sin \dfrac {4\theta -\pi}{8}}{\theta - \dfrac {\pi}{4}}.
\end{align}
How do I proceed?
| Note: $cos(π/4) = sin(π/4) = 1/√2$.
$\frac{cos(\theta) - sin(\theta)}{\theta - π/4} =$
√2 $\frac{sin(π/4) cos(\theta) - cos(π/4)sin(\theta)}{\theta - π/4} = $
√2$ \frac{sin(π/4 - \theta)}{\theta - π/4} = $
-√2$ \frac{sin(x)}{x}$ , where x: = $\theta - π/4$.
$\lim_{x \rightarrow 0}$ (-√2 $\frac{sin(x)}{x}) =$ -√2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
What is the simplest way to factor the following polynomial $x^4-8x^2+x+12$?
What is the simplest way to factor the following polynomial $$x^4-8x^2+x+12$$ ?
Note : I already knew the factorization which is
$$(x^2-x-3)(x^2+x-4)$$
I need the way to get that
Thank you for your help.
| I like the following way.
$$x^4-8x^2+x+12=(x^2+k)^2-((2k+8)x^2-x+k^2-12).$$
Now, we need to find a value of $k$, for which $(2k+8)x^2-x+k^2-12=(ax+b)^2$,
for which we need $$1-4(2k+8)(k^2-12)=0$$ or
$$(2k+7)(4k^2+2k-55)=0.$$
Easy to see that $k=-3.5$ is valid.
Indeed, $$x^4-8x^2+x+12=(x^2-3.5)^2-(x^2-x+0.25)=$$
$$=(x^2-3.5)^2-(x-0.5)^2=(x^2-x-3)(x^2+x-4)$$
and we are done!
In the general case we'll get a cubic equation with the variable $k$ and one of this equation roots always gives difference of squares and possibility of the factorization.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Decomposing a matrix into binary ones Let $A$ denote the set of all 3×2 binary matrices (those containing only 0's and 1's) in which the sum of each column adds up to 2. Can I decompose
$$B=\begin{bmatrix}
3/4 & 1/2 \\
3/4 & 3/4 \\
1/2 & 3/4 \end{bmatrix}$$
into a linear combination of matrices in $A$? For example,
$$B=1/2 \begin{bmatrix}1 & 1 \\ 0 & 0 \\ 1 & 1\end{bmatrix} + 1/2 \begin{bmatrix}1 & 0 \\ 0 & 1 \\ 1 & 1\end{bmatrix}$$
or something like that.
EDIT: I want to show there is at least a matrix that cannot be decomposed. How about
$$C=\begin{bmatrix}
3/4 & 2/3 \\
3/4 & 2/3 \\
1/2 & 2/3 \end{bmatrix}$$
| There are exactly $9$ matrices in $A$. The dimension of the vector space $V$ of $3\times 2$ matrices is $6$.
The rank of the matrix (formed putting each matrix of $A$ in a row)
$$\begin{pmatrix}
1&1&0&1&1&0\\
1&1&0&1&0&1\\
1&1&0&0&1&1\\
1&0&1&1&1&0\\
1&0&1&1&0&1\\
1&0&1&0&1&1\\
0&1&1&1&1&0\\
0&1&1&1&0&1\\
0&1&1&0&1&1
\end{pmatrix}$$
is $5$, and this means that $A$ does not generate $V$. The other answer shows that, nevertheless, the particular matrix $B$ is generated by $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Estimate the error that results when $\sqrt{1 + x}$ is replaced by $1 + \frac{1}{2}x$ if $|x| < 0.01$ Question
Estimate the error that results when $\sqrt{1 + x}$ is replaced by $1 + \frac{1}{2}x$ if $|x| < 0.01$
Definition
Taylors formula is $f(x) = P_n(x) + R_n(x)$ where $P_n(x)$ is
\begin{equation}
\begin{aligned}
P_n(x) = f(a)
+ \frac{f'(a)}{1!}(x - a)
+ \frac{f''(a)}{2!}(x - a)^2
& + \ldots
+ \frac{f^{(n)}(a)}{n!}(x - a)^n \\
\end{aligned}
\end{equation}
And $R_n (x) $ is (\emph{where $\xi$ is between $a$ and $x$ })
\begin{equation}
\begin{aligned}
R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - a)^{(n + 1)}
\end{aligned}
\end{equation}
Working
I'm not sure how to go about this, would I say that this is a first order
approximation as
\begin{equation}
\begin{aligned}
P(x) & = 1 - \frac{1}{2}x \\
P'(x) &= - \frac{1}{2} \\
P''(x) &= 0
\end{aligned}
\end{equation}
Then the remainder term would be
\begin{equation}
\begin{aligned}
R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - a)^{(n + 1)}
\end{aligned}
\end{equation}
Where $n + 1 = 2$. For $f(x) = \sqrt{1 + x}$ this would be
\begin{equation}
\begin{aligned}
R_n(x) & = \frac{- \frac{1}{4} (1 + \xi)^{-3/2}}{(3)!}
\end{aligned}
\end{equation}
And $\xi $ is between $-0.01$ and $0.01$
This would give the maximum error as
\begin{equation}
\begin{aligned}
R_n(x) & = \frac{- \frac{1}{4} (1 \pm 0.01 )^{-3/2}}{(3)!} \approx -0.0423
\end{aligned}
\end{equation}
The error is greatest when $\xi = -0.01$.
| You have that
$$f(x)=f(0)+\dfrac{f'(0)}{1!}x+R_1(x).$$ Since $f(x)=\sqrt{1+x}$ it is $f(0)=1$ and $f'(0)=\dfrac 12.$ Thus
$$\sqrt{1+x}=1+\dfrac{1}{2}x+R_1(x).$$
Now, $$R_1(x)=\dfrac{f''(\xi)}{2!}x^2.$$ Since $$f''(\xi)=-\dfrac14 (1+\xi)^{-3/2}$$ we get
$$\sqrt{1+x}-\left(1+\dfrac{1}{2}x\right)=-\dfrac18 (1+\xi)^{-3/2}x^2.$$ Now, if $|x|\le 0.01$ we have to get a bound for the RHT. We have that
$$\dfrac18 (1+\xi)^{-3/2}x^2\le \dfrac{0.01^2}{8},$$ where we have used that $$(1+\xi)^{-3/2}\le 1.$$
| {
"language": "en",
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"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Integration by Partial Fractions: Numerator Should Be One Degree Lower than Denominator At 1:36:13 (on https://www.youtube.com/watch?v=KJGp0pyPoVo&list=PLDesaqWTN6EQ2J4vgsN1HyBeRADEh4Cw-&index=11), the professor says:
Whatever factor you have, your numerator should be one degree lower than that. So for linears, one degree lower than a linear is a constant. One degree lower than a quadratic is a linear.
However, this doesn't seem to hold true in some of the worked examples in Stewart's Early Transcendentals (8th Edition) where you have:
$$\frac{x^3-x+1}{x^2(x-1)^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} + \frac{E}{(x-1)^3}$$
Another example shows:
$$\frac{4x}{(x-1)^2(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}$$
With D and E in the first equation, and B in the second equation being constants, don't these examples disprove the statement above?
| It's just an alternative way of writing it, since
$$\frac{A}{x} + \frac{B}{x^2} = \frac{Ax + B}{x^2}$$
and the polynomial $Ax+B$ is one degree lower than $x^2$.
If the polynomial isn't $x^n$, then the constants change a little, since
$$\frac{C}{x-1} + \frac{D}{(x-1)^2} + \frac{E}{(x-1)^3} = \frac{C(x-1)^2 + D(x-1) + E}{(x-1)^3} = \frac{C'x^2 + D'x + E'}{(x-1)^3}$$
for some constants $C', D', E'$, so it's sort of up to you. You can either go calculate $C,D,E$ or $C',D',E'$.
| {
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"url": "https://math.stackexchange.com/questions/2371071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of recursive sequences: $x_{n+1}=\frac 1 3 (x_n+2y_n), y_{n+1}=\frac 1 3 \left( \frac 1 {x_n}+ \frac 1 {2y_n} \right).$ Let $\left(\,x_{n}\,\right)_{\ n\ \geq\ 1}$ and
$\left(\,y_{n}\,\right)_{\ n\ \geq\ 1}$ be two sequences defined as follows
$\left(\,x_{1},y_{1} > 0\,\right)$:
$$
\left\{\begin{array}{rcl}
{\displaystyle x_{n + 1}} & {\displaystyle =} &
{\displaystyle{1 \over 3}\left(\,x_{n} + 2y_{n}\,\right)}
\\[2mm]
{\displaystyle y_{n + 1}} & {\displaystyle =} &
{\displaystyle{1 \over 3}\left(\,{1 \over x_{n}} + {1 \over 2y_n}\,\right)}
\end{array}\right.
$$
Show that the sequences are convergent and find their limits.
MY TRY:
*
*I think about using the Weierstrass' Theorem ( every bounded and monotone sequence is convergent ), but I can't find the monotonity for any sequence.
*Afterwards, I can think of denoting the two limits as $L_{1}$ and $L_{2}$ and replacing them in the reccurence relationship.
| Let us consider the two sequences
$$a_n = \frac{x_n}{y_n}$$
and
$$b_n = x_n y_n$$ instead
Dividing the two given recurrences (of $x_{n+1}$ and $y_{n+1}$) gives us
$$a_{n+1} = 2 b_n$$
Note that it is enough to show that $b_n$ is convergent, as that
implies $a_n$ is convergent, which in turn implies $x_n^2 = a_nb_n$ is
convergent, which implies $y_n$ is too.
Now multiplying the two give recurrences(of $x_{n+1}$ and $y_{n+1}$) in the question gives us
$$9 b_{n+1} = 2 + \frac{a_n}{2} + \frac{2}{a_n} = 2 + b_{n-1} + \frac{1}{b_{n-1}}$$
To simplify the algebra, Let $$s_n = \sqrt{b_n}$$
Thus
$$(3s_{n+1})^2 = \left(s_{n-1} + \frac{1}{s_n}\right)^2$$
Since $s_n \gt 0$, we have
$$3s_{n+1} = s_{n-1} + \frac{1}{s_{n-1}}$$
Now by considering the even $n$ and odd $n$ separately, we just need to analyze the sequence
$$3c_{n+1} = c_n + \frac{1}{c_n}$$
(for eg by setting $c_k = s_{2k}$ or $c_k = s_{2k+1}$)
Let $$L = \frac{1}{\sqrt{2}}$$
Let $$e_n = c_n -L$$
Note that $$e_n = 0 \implies e_{n+1} = 0 \implies c_n \to L$$
So assume $e_n \ne 0$
We have that $$3L = L + \frac{1}{L}$$ and so
$$3c_{n+1} -3L = c_n - L + \frac{1}{c_n} - \frac{1}{L}$$
i.e.
$$ 3 e_{n+1} = e_n - \frac{e_n}{Lc_n}$$
Thus
$$\left|\frac{e_{n+1}}{e_n}\right| = \frac{|1 - \frac{1}{Lc_n}|}{3}$$
Let us now show that $$\frac{|1 - \frac{1}{Lc_n}|}{3} \le d$$ for some constant $d \lt 1$ which implies $e_n \to 0$.
Now we have that $$c_n \ge \frac{2}{3}$$
using $x + \frac{1}{x} \ge 2$ (put $x = c_{n-1}$).
$$ \implies Lc_n \ge \frac{\sqrt{2}}{3}$$
And so
$$\frac{3Lc_n}{\sqrt{2}} \ge 1$$
$$\frac{(3-\sqrt{2})Lc_n}{\sqrt{2}} \ge 1 - Lc_n$$
Thus
$$\frac{3-\sqrt{2}}{3\sqrt{2}} \ge \frac{1 - Lc_n}{3Lc_n}$$
Now if $Lc_n \gt 1$, then $$\frac{\left|1 - \frac{1}{Lc_n}\right|}{3} = \frac{1 - \frac{1}{Lc_n}}{3} \le \frac{1}{3}$$
if $Lc_n \le 1$ then
$$\frac{\left|1 - \frac{1}{Lc_n}\right|}{3} = \frac{1 -Lc_n}{3Lc_n} \le \frac{3-\sqrt{2}}{3\sqrt{2}}$$
Thus $e_n \to 0$ and so $c_n \to L$. Thus $s_{2n}, s_{2n+1} \to L$ and so $b_n$ is convergent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the set of values of $a$ for which the inequality $(x-3a)(x-3-a)<0$ is satisfied for all $x$ in the interval $1\le x\le3$. Find the set of values of $a$ for which the inequality $(x-3a)(x-3-a)<0$ is satisfied for all $x$ in the interval $1\le x\le3$.
$$(x-3a)(x-3-a)<0\implies a+3<x<3a$$ and we are given that $1\le x\le3$.
So $a+3>1$ and $3a<3$ ,so $a\in (-2,1)$ but the answer given is $a\in(0,\frac{1}{3})$.
| The function $f(x):=(x-3a)(x-3-a)$ has two zeros $3a$ and $3+a$ which coincide at ${9\over2}$ when $a={3\over2}$. Note that both zeros move to the right when $a$ is increasing. Since the coefficient of $x^2$ in $f$ is positive the function is negative between the two zeros. As we want it to be negative in the $x$-interval $[1,3]$ we obtain the condition
$$\min\{3a,3+a\}<1\qquad\wedge\qquad\max\{3a,3+a\}>3\ .\tag{1}$$
When $a\geq{3\over2}$ then both zeros are $\geq{9\over2}>3$, hence such values of $a$ are a priori forbidden. When $a<{3\over2}$ then $\min\{3a,3+a\}=3a$ and $\max\{3a,3+a\}=3+a$. The condition $(1)$ then translates into
$$3a<1\qquad\wedge\qquad3+a>3\ ,$$
which is fulfilled iff $0<a<{1\over3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cosecants of half angles
The tangent of an angle is $2.4$. Find the cosecant of half the angle.
My tries:
As $\tan A=2.4=\dfrac{12}{5}\implies \sin A=\pm\dfrac{12}{13}=\dfrac{1}{\csc A}$
Also:
$\sin \frac{A}{2}+\cos \frac{A}{2}=\pm\sqrt{1+\sin A}=\pm\dfrac{5}{{\sqrt{13}}}$ , considering $\sin A=\dfrac{12}{13}$
$\sin \frac{A}{2}-\cos \frac{A}{2}=\pm\sqrt{1-\sin A}=\pm\dfrac{1}{{\sqrt{13}}}$, considering $\sin A=\dfrac{13}{13}$
Adding them gives: $\sin \frac{A}{2}=\pm\dfrac{3}{\sqrt{13}}=\dfrac{1}{\csc{\frac A2}}$
I did same by considering $\sin A=-\dfrac{12}{13}$ then also it gave the same result.
But answer provided by the author is $\dfrac{\pm\sqrt{13}}{2}$ and $\dfrac{\pm\sqrt{13}}{3}$.
So what did I miss? please help.
| $$|\sin\frac{A}{2}|=\sqrt{\frac{1-\cos{A}}{2}}$$ and
$$\frac{1}{\cos^2A}=1+\tan^2A=6.76.$$
Thus, $|\cos{A}|=\frac{1}{2.6}=\frac{5}{13}$, which says that $\frac{1}{\sin\frac{A}{2}}$ can get four values:
$\sqrt{\frac{2}{1-\frac{5}{13}}}=\frac{\sqrt{13}}{2},$ $-\frac{\sqrt{13}}{2},$ $\sqrt{\frac{2}{1+\frac{5}{13}}}=\frac{\sqrt{13}}{3}$ and $-\frac{\sqrt{13}}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the number of rearrangements of the string 12345 in which none of the sequences 12, 23, 34, and 45 occur. My attempt: Let $A_{1}$ denote where 12 occurs, $A_{2}$ denote where 23 occurs, $A_{3}$ denote where 34 occurs, $A_{4}$ denote where 45 occurs.
|$A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$|= |$A_{1}$|+|$A_{2}$|+|$A_{3}$|+|$A_{4}$|-(|$A_{1}A_{2}$|+|$A_{1}A_{3}$|+|$A_{1}A_{4}$|+|$A_{2}A_{3}$|+|$A_{2}A_{4}$|+|$A_{3}A_{4}$|)+|$A_{1}A_{2}A_{3}$|+|$A_{1}A_{2}A_{4}$|+|$A_{1}A_{3}A_{4}$|+|$A_{2}A_{3}A_{4}$|-|$A_{1}A_{2}A_{3}A_{4}$|
I don't know if I'm doing this write or not. I tried to glue the 12, 23, 34, and 45 together but I don't think this is correct. The total ways this can be done with no restrictions is |U|=5!
The actual answer is: 5!-[96-3+8-1]=53.
If someone can tell me how to get this answer that would be excellent.
| Yes, your approach is correct. Note that $|A_i|=4\cdot 3!=24$, $|A_iA_j|=6$, $|A_iA_j A_k|=2$ and $|A_1A_2A_3A_4|=1$. Hence
$|A_{1} \cup A_{2} \cup A_{3} \cup A_{4}|$ is equal to
$$\binom{4}{1}\cdot |A_i|-\binom{4}{2}\cdot |A_iA_j|+\binom{4}{3}\cdot |A_iA_j A_k|-\binom{4}{4}\cdot |A_1A_2A_3A_4|.$$
which implies that the number of those permutations is $|U|-|A_{1} \cup A_{2} \cup A_{3} \cup A_{4}|$:
$$5!-(4\cdot 24-6\cdot 6+4\cdot 2-1)=120-(96-36+8-1)=53.$$
Alternative approach. Let $p(n)$ be the number of permutations of $[1,\dots,n]$ having no substring $[k,k+1]$ with $1\leq k\leq n-1$. Then $p(1) = 1$,
$p(2) =1$ and, for $n\geq 2$ it satisfies the recurrence
$$p(n+1) = n p(n) + (n-1)p(n-1).$$
Hence $p(3)=2p(2)+p(1)=3$, $p(4)=3p(3)+2p(2)=11$ and $$p(5)=4p(4)+3p(3)=44+9=53.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding range of an unknown in a biquadratic equation If the equation $x^4- ax^3+ 11x^2 -ax+1=0$ has four distinct positive roots then the range of $a$ is $(m, M) $. Find the value of $m+2M$.
| Rewrite it as $x^2 +\dfrac{1}{x^2} - a\left(x+\dfrac{1}{x}\right) + 11 = 0$, and put $y = x+\dfrac{1}{x}\implies y^2- ay+9=0$, and this equation has $2$ positive distinct roots. Thus $\triangle > 0 \implies a^2-36 > 0\implies a > 6 = m$. Thus $y = \dfrac{a \pm \sqrt{a^2-36}}{2}> 2\implies a - \sqrt{a^2-36} > 4\implies (a-4)^2 > a^2-36 \implies 8a < 52 \implies a < \dfrac{13}{2} = M \implies m+2M = 6 + 2\cdot \dfrac{13}{2} = 19$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to take the square root of a number in a negative base? For example, in negadecimal (base -10): how to take the square root of $185_{-10}=25_{10}$?
Or in negabinary (base -2): the square root of $1100100_{-2}=36_{10}$?
Converting to another base, taking the square root and converting back to the negative base is not an option. I'm looking for an algorithm fully done in the negative base.
I take the square root of a binary number by dividing a number like $11001_2=25_{10}$ as follows:
$01|10|01$
And then conditionally subtracting $01$ from each segment from left to right, adding a $1$ to the answer and appending the answer to the $01$ subtraction.
| The standard positive base $b$ square root algorithm is:
*
*Split the input digits into groups of two (such that the decimal point is one of the split points).
*Initialize $\mathit{remainder}$ to $0$ and $\mathit{output}$ to $0$.
*Append the next digit group to the end of $\mathit{remainder}$.
*Find the largest digit $d$ such that $(2 \cdot b \cdot \mathit{output} + d) \cdot d \le \mathit{remainder}$.
*Subtract $(2 \cdot b \cdot \mathit{output} + d) \cdot d$ from $\mathit{remainder}$.
*Append $d$ to $\mathit{output}$.
*While more digits are desired, repeat from step 3.
The reason this fails in a negative base is that step 4 is really looking for the first digit of the real-number solution to
$$\begin{align*}
(b \cdot \mathit{output} + x)^2 &= b^{2n} \cdot \mathit{input} \\
\iff (2 \cdot b \cdot \mathit{output} + x) \cdot x &= b^{2n} \cdot \mathit{input} - (b \cdot \mathit{output})^2,
\end{align*}$$
where $\mathit{remainder}$ is maintained as the integer part of the right side. A negative base presents four difficulties:
*
*Because we’re outputting one digit at a time, on any given iteration, output may be of either sign, which means that $(2 \cdot b \cdot \mathit{output} + x) \cdot x$ may be either increasing or decreasing in $x$.
*The first digit of $x$ may now be greater than $x$. The set of numbers starting with a digit $d$ is no longer $[d.000000\cdots, d.999999\cdots) = [d, d + 1)$, but rather $[d.909090\cdots, d.090909\cdots) = \left[d + \frac{b}{1 - b}, d + \frac{1}{1 - b}\right)$.
*For the same reason, we might sometimes need more than two input digits to produce one output digit: $\sqrt{02.9921_{-10}} = 1.09_{-10}$ starts with $1$, but $\sqrt{02.81_{-10}} = 2.9_{-10}$ starts with $2$.
*Without taking into account whether we’re outputting a digit of positive or negative place value, it’s easy to inadvertently produce the negative square root instead of the positive square root.
One strategy for addressing these difficulties involves building $\mathit{remainder}$ from the digits of $\mathit{input} \cdot (1 - b)$ instead of $\mathit{input}$. The points at which the first digit of $x$ changes are described by $x = d + \frac{1}{1 - b}$:
$$\left(2 \cdot b \cdot \mathit{output} + d + \frac{1}{1 - b}\right) \cdot \left(d + \frac{1}{1 - b}\right) = b^{2n} \cdot \mathit{input} - (b \cdot \mathit{output})^2 \\
\begin{multline}
\iff (2 \cdot b \cdot \mathit{output} + d) \cdot d \cdot (1 - b) + 2 \cdot b \cdot \mathit{output} + 2 \cdot d + \frac{1}{1 - b} \\
= b^{2n} \cdot \mathit{input} \cdot (1 - b) - (b \cdot \mathit{output})^2 \cdot (1 - b).
\end{multline}$$
The pre-decimal part of the right side also changes here (since the right side equals an integer plus $\frac{1}{1 - b}$), so maintaining that pre-decimal part as $\mathit{remainder}$ guarantees sufficient precision to pin down the first digit of $x$ correctly.
Taking this all into account, we can update the algorithm for negative base $b$ as follows:
*
*Multiply the input by $1 - b$.
*Split the resulting digits into groups of two (such that the decimal point is one of the split points).
*Initialize $\mathit{remainder}$ to $0$ and $\mathit{output}$ to $0$.
*Append the next digit group to the end of $\mathit{remainder}$.
*
*
*If $\mathit{output} = 0$ and we’re outputting a digit of negative place value, set $d = 0$.
*Otherwise if $\mathit{output} \le 0$, find the smallest digit $d$ such that $$(2 \cdot b \cdot \mathit{output} + d) \cdot d \cdot (1 - b) + 2 \cdot b \cdot \mathit{output} + 2 \cdot d \ge \mathit{remainder}.$$
*Otherwise, find the smallest digit d such that $$(2 \cdot b \cdot \mathit{output} + d) \cdot d \cdot (1 - b) + 2 \cdot b \cdot \mathit{output} + 2 \cdot d < \mathit{remainder}.$$
*Subtract $(2 \cdot b \cdot \mathit{output} + d) \cdot d \cdot (1 - b)$ from $\mathit{remainder}$.
*Append $d$ to $\mathit{output}$.
*While more digits are desired, repeat from step 4.
| {
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"url": "https://math.stackexchange.com/questions/2381721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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What is the general term formula of this recurrence relation, if it exists? $$a_n=-\frac{(a_{n-1} + 1)^2}{a_{n-1}+2}\quad \quad a_1 = -\frac 12$$
I transformed to:
$$a_n = -a_{n-1} - \frac{1}{a_{n-1} + 2}$$
$$a_{n-1} = -a_{n-2} - \frac{1}{a_{n-2} + 2}$$
...
And then
$$a_n - a_{n-1} + a_{n-2} -\cdots + \cdots - a_1$$
$$=$$
$$- a_{n-1} + a_{n-2} - ... + ... - \frac{1}{a_{n-1} + 2} + \frac{1}{a_{n-2} + 2} - ... + ... $$
Many terms cancel each other and:
$$a_n = - \frac{1}{a_{n-1} + 2} + \frac{1}{a_{n-2} + 2} - ... + ... + \frac{1}{a_{1} + 2} + a_1$$ (Here need to divide cases odd vs even)
And then I'm forever stuck. I can't find analytical form of the sum above
I try to get the value of the first few dozen terms to observe. I find nothing other than the terms are negative fractions that satisfy the recurrence relation.
Question is, is there a general term formula? How to find it?
| This is an incomplete answer.
The recurrence is given by:
$$
a_n=-\frac{(a_{n-1}+1)^2}{a_{n-1}+2}
$$
Here we can separate both numerator and denominator as independent series. Let's create two new series, for them $c_n$ and $d_n$ respectively:
$$
a_{n-1}=\frac{c_{n-1}}{d_{n-1}}
$$
So:
$$
a_n=-\frac{(\frac{c_{n-1}}{d_{n-1}}+1)^2}{\frac{c_{n-1}}{d_{n-1}}+2}\\
=\frac{-(c_{n-1}+d_{n-1})^2}{c_{n-1}+2d_{n-1}}
=\frac{-c_{n-1}^2-2c_{n-1}d_{n-1}-d_{n-1}^2}{c_{n-1}+2d_{n-1}}\frac{1}{d_{n-1}}\\
$$
Hence we can define the new sequences as:
$$
c_n=-c_{n-1}^2-2c_{n-1}d_{n-1}-d_{n-1}^2\\
d_n=c_{n-1}d_{n-1}+2d_{n-1}^2
$$
Both are quadratic forms, very similar:
$$
c_n=
\begin{bmatrix}
c_{n-1} & d_{n-1}
\end{bmatrix}
\begin{bmatrix}
p_{111} & p_{112}\\
p_{121} & p_{122}
\end{bmatrix}
\begin{bmatrix}
c_{n-1}\\
d_{n-1}
\end{bmatrix} \\
d_n=
\begin{bmatrix}
c_{n-1} & d_{n-1}
\end{bmatrix}
\begin{bmatrix}
p_{211} & p_{212}\\
p_{221} & p_{222}
\end{bmatrix}
\begin{bmatrix}
c_{n-1}\\
d_{n-1}
\end{bmatrix}
$$
with:
$$
\mathbf{p}_{1\cdot\cdot}=\begin{bmatrix}
-1 & -1\\
-1 & -1
\end{bmatrix} \\
\mathbf{p}_{2\cdot\cdot}=\begin{bmatrix}
0 & 1/2\\
1/2 & 2
\end{bmatrix}
$$
Which can be summarized by renaming the numerator and denominator series as $\mathbf{x}_{1|n}$ and $\mathbf{x}_{2|n}$, and by using tensorial notation with implicit summation over repeated indexes, excepting the recurrence index:
$$
\mathbf{x}_{k_n}=\mathbf{x}_{i_{n-1}}\mathbf{p}_{k_ni_{n-1}j_{n-1}}\mathbf{x}_{j_{n-1}}
$$
This recurrence, though tensorial, can be expressed up to the first term $\mathbf{x}_{\cdot|1}=[-1,2]$, where the tensor $\mathbf{p}$ appears $2^{n-1}-1$ times:
$$
\mathbf{x}_{k_n}=\mathbf{x}_{i_{1}} \mathbf{p}_{i_{2}i_{1}j_{1}} \mathbf{x}_{j_{1}}
\cdots
\mathbf{p}_{k_ni_{n-1}j_{n-1}}
\cdots
\mathbf{x}_{i_{1}} \mathbf{p}_{j_{2}i_{1}j_{1}} \mathbf{x}_{j_{1}}
$$
This could be depicted as a repetitive product of matrices, with $P$ appearing $2^{n-1}-1$ times:
$$
x_n=x_1 P x_1\cdots P \cdots x_1 P x_1
$$
Though this expression is not closed and surely very far from what the OP wanted, this is not a recurrence, because it depends only on the first value of the sequence, and is a repetitive aplication of multidimensional products over the initial value.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to deal with cubic root equations If x,y,z are real positive number, and the conditions are:
\begin{cases}
\begin{array}{ll}
1995x^3=1996y^3 \\
1996y^3=1997z^3 \\
\sqrt[3]{1995x^2+1996y^2+1997z^2}=\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997}
\end{array}
\end{cases}
What's the result of:
\begin{equation}
\begin{array}{ll}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
\end{array}
\end{equation}
I can do below transfer,but don't know how to get rid of the x:
\begin{equation}
\begin{array}{ll}
\sqrt[3]{\frac{1995x^3}{x}+\frac{1996y^3}{y}+\frac{1997z^3}{z}}=\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997} \\
\sqrt[3]{1995x^3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}=\sqrt[3]{1995}+\sqrt[3]{1996} +\sqrt[3]{1997} \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{(\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997})^3}{1995x^3}
\end{array}
\end{equation}
| Call $1995x^3=A$, $1996y^3=B$ and $1997z^3=C$.
Then $A=B=C$ and
$(1995^{1/3}+1996^{1/3}+1997^{1/3})^3=1995\frac{A^{2/3}}{1995^{2/3}}+1996\frac{B^{2/3}}{1996^{2/3}}+1997\frac{C^{2/3}}{1997^{2/3}}=A^{2/3}(1995/1995^{2/3}+1997/1997^{2/3}+1997/1997^{2/3})$
Solve for $A=(\frac{(1995^{1/3}+1996^{1/3}+1997^{1/3})^3}{1995/1995^{2/3}+1997/1997^{2/3}+1997/1997^{2/3}})^{3/2}$ and you got it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382861",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{n \to \infty} \frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\cdots+\frac{n}{n^2+n^2}$ Evaluate
$$ \lim_{n \to \infty} \frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\cdots+\frac{n}{n^2+n^2}$$
I used definite integral as a limit of a sum as:
$$S= \lim_{ n \to \infty}\frac{1}{n} \sum_{r=1}^{n} \frac{\left(\frac{r}{n}\right)}{1+\left(\frac{r}{n}\right)^2}$$
So
$$S=\int_{0}^{1}\frac{ x \:dx}{1+x^2}=\frac{1}{2} \log 2$$
Is there any other approach?
| HINT:
$$\frac{n(n+1)}{2}\frac{1}{n+n^2}\le \sum_{k=1}^n \frac{k}{k+n^2}\le \frac{n(n+1)}{2}\frac{1}{1+n^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Polynomial divisibility exercise How do I prove that $(X+1)^{6n+1}-X^{6n+1}-1$ is divisible by $(X^2+X+1)^2$ without derivative functions? (I am 10th grade).
| One way is to use the binomial theorem.
Assume $a$ is a root of $x^2+x+1=0$. In particular $a^3=1$ and $a+1=-a^2$.
We want to prove that we can divide two times by $y=x-a$ the polynomial $$p(x)=(x+1)^{6n+1}-x^{6n+1}-1$$
Replacing $x=y+a$ we get
$$(y+(a+1))^{6n+1}-(y+a)^{6n+1}-1$$
Using the binomial theorem this is
$$\sum_{k=0}^{6n+1}\binom{6n+1}{k}y^k(a+1)^{6n+1-k}-\sum_{k=0}^{6n+1}\binom{6n+1}{k}y^ka^{6n+1-k}-1$$
The constant term is $$(a+1)^{6n+1}-a^{6n+1}-1=(-a^2)^{6n}(-a^2)-a^{6n}\cdot a-1=-a^2-a-1=0$$
The coefficient of the term of degree $1$ is
$$(6n+1)(a+1)^{6n}-(6n+1)a^{6n}=(6n+1)[(-a^2)^{6n}-a^{6n}]=(6n+1)[1-1]=0$$
Therefore, $y^2=(x-a)^2$ divides $p(x)$. Since the computation didn't depend on which root of $x^2+x+1$ we used, it follows that $(x^2+x+1)^2$ divides $p(x)$.
In the realm of polynomials, everything that can be proved using derivatives can be proved using the binomial theorem instead.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving complex number by geometrical method. Let C1
and C2
are concentric circles of radius $1$ and $8/3$ respectively having centre at $(3, 0)$ on the argand
plane. If the complex number $z$ satisfies the inequality, $\log_{1/3}{\frac
{|z-3|^2+2}{11|z-3 |-2}}> 1$ then :
(A) $z$ lies outside C1
but inside C2
(B) $z$ lies inside of both C1
and C2
(C) $z$ lies outside both of C1
and C2
(D) none of these
I tried to do it like this.${\frac
{|z-3|^2+2}{11|z-3 |-2}}<\frac{1}{3}$.So by simplifying I got ${\frac
{3|z-3|^2+8-11|z-3 |}{11|z-3 |-2}}<0$ .Hence by substituting $|z-3|$ by $t$ I get $t=1$ and $t=8/3$.So I get the negative value between $1$ and $8/3$.But I don't know how to proceed and check which option is correct.
| $z$ is in the annulus formed by the two circles. Answer (A).
Indeed set $p=|z-3|$
$\log_{1/3}{\dfrac
{p^2+2}{11p-2}}> 1$
is verified if
$0<\dfrac
{p^2+2}{11p-2}<\dfrac{1}{3}$
Let's solve
$\dfrac
{p^2+2}{11p-2}>0$ numerator is always positive so must be $11p-2>0\to p>\dfrac{2}{11}$
and
$\dfrac
{p^2+2}{11p-2}<\dfrac{1}{3}\to \dfrac
{p^2+2}{11p-2}-\dfrac{1}{3}<0 \to \dfrac{3p^2+6-11p+2}{3(11p-2)}<0\to \dfrac{3p^2-11p+8}{3(11p-2)}<0$
As the denominator must be positive for the previous condition, then the numerator must be negative, that is
$3p^2-11p+8<0\to 1<p<\dfrac{8}{3}$
Thus $1<|z-3|<\dfrac{8}{3}$
which means that $z$ is outside $C_1$ but inside $C_2$ as stated above.
Hope this helps
| {
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Hints to prove $\tan^2{\theta}=\tan{A}\tan{B}$, given $\frac{\sin{(\theta + A)}}{\sin{(\theta + B)}} = \sqrt{\frac{\sin{2A}}{\sin{2B}}}$ I need some hints on solving this trigonometry problem.
Problem
If $\dfrac{\sin{(\theta + A)}}{\sin{(\theta + B)}} = \sqrt{\dfrac{\sin{2A}}{\sin{2B}}}$, then prove that $\tan^2{\theta}=\tan{A}\tan{B}$.
I tried to expand the left hand side of the equation, but no clue what to do next. I also tried to use $\sin{2\alpha} = \dfrac{2\tan{\alpha}}{1 + \tan^2{\alpha}}$ for the right hand side of the equation with no result.
I appreciate for any help. Thank you.
| $$\begin{align*}
\frac{\sin{(\theta + A)}}{\sin{(\theta + B)}} &= \sqrt{\frac{\sin{2A}}{\sin{2B}}}\\
\frac{\sin \theta\cos A + \cos\theta\sin A}{\sin \theta\cos B + \cos\theta\sin B} &= \sqrt\frac{\sin A \cos A}{\sin B \cos B}\\
\frac{\tan \theta\cos A + \sin A}{\tan \theta\cos B + \sin B} &= \sqrt\frac{\sin A \cos A}{\sin B \cos B}\\
(\tan \theta\cos A + \sin A)\sqrt{\sin B \cos B}&=(\tan \theta\cos B + \sin B)\sqrt{\sin A\cos A}\\
\tan\theta(\cos A\sqrt{\sin B \cos B} -\cos B\sqrt{\sin A\cos A}) &= -\sin A\sqrt{\sin B \cos B} + \sin B \sqrt{\sin A\cos A}\\
\tan\theta\sqrt{\cos A\cos B}(\sqrt{\cos A\sin B}-\sqrt{\cos B\sin A}) &= \sqrt{\sin A \sin B}(-\sqrt{\sin A \cos B} + \sqrt{\sin B\cos A})\\
\tan\theta &= \sqrt{\frac{\sin A \sin B}{\cos A\cos B}}\\
\tan^2\theta &= \tan A \tan B
\end{align*}$$
Assuming $\sqrt{\cos A\sin B}-\sqrt{\cos B\sin A} \ne 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to calculate a vec-permutation matrix? I'm trying to follow the maths in a paper on modelling population dynamics and have come unstuck on the calculation of the vec-permutation matrix.
Let $p$ be the number of patches and $s$ be the number of stages. For my example, both $s$ and $p$ are $= 2$.
The vec-permutation matrix has dimension $(sp × sp)$ and is given by:
$P(s,p) = \sum_{i=1}^{s}\sum_{j=1}^{p}E_{ij}\otimes E_{ij}^T$
The authors state $E_{ij}$ is an $s × p$ matrix with a $1$ in the $(i, j)$
position and zeros elsewhere and $\otimes$ denotes the Kronecker
matrix product.
From the paper, it should be the $4$ x $4$ matrix:
$P(2,2)= \begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1
\end{pmatrix} $
I don't understand the $E_{ij}$ matrix. I thought every matrix position could be considered $ij$.
https://www.researchgate.net/profile/Christine_Hunter5/publication/228754096_The_use_of_the_vec-permutation_matrix_in_spatial_matrix_population_models/links/5551497808ae956a5d25ed48.pdf
| Maybe looking at the $3 \times 3$ case will be helpful. Note that
$$
\sum_{i,j} E_{ij} \otimes E_{ij} =
\left[\begin{array}{ccc|ccc|ccc}
1&0&0&0&1&0&0&0&1\\
0&0&0&0&0&0&0&0&0\\
0&0&0&0&0&0&0&0&0\\
\hline
0&0&0&0&0&0&0&0&0\\
1&0&0&0&1&0&0&0&1\\
0&0&0&0&0&0&0&0&0\\
\hline
0&0&0&0&0&0&0&0&0\\
0&0&0&0&0&0&0&0&0\\
1&0&0&0&1&0&0&0&1
\end{array}
\right]
$$
If we take the transpose of each $3 \times 3$ block, we get the desired matrix
$$
\sum_{i,j} E_{ij} \otimes E_{ij}^T =
\left[\begin{array}{ccc|ccc|ccc}
1&0&0&0&0&0&0&0&0\\
0&0&0&1&0&0&0&0&0\\
0&0&0&0&0&0&1&0&0\\
\hline
0&1&0&0&0&0&0&0&0\\
0&0&0&0&1&0&0&0&0\\
0&0&0&0&0&0&0&1&0\\
\hline
0&0&1&0&0&0&0&0&0\\
0&0&0&0&0&1&0&0&0\\
0&0&0&0&0&0&0&0&1
\end{array}
\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2386989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a recurrence equation with constant term so I have this recurrence equation:
$$G(n) = 2G(n-1) + 100 $$
I believe this a linear non-homogeneous equation of the form $G(n) = LH + F(n) $ where LH is the associated linear homogeneous equation which is G(n-1).
If $F(n)$ is $2n$ we can take a trial solution as $Cn+D$ and solve or if F(n) is say $7^n$ we can take $C\cdot7^n$ and try to solve for constants $C$ and $D$. What do I do if $F(n)$ is a constant like $100$. The initial condition is G($0$) = $2$.
How do I go about solving this?
Thanks in advance!
| Another approach using generating functions:
$\begin{align}
F(x) &= \sum_{n=0}^{\infty} G(n) x^n \\
F(x) &= 2x^0 + \sum_{n=1}^{\infty} (2G(n-1) + 100) x^n \\
F(x) &= 2 + 2\sum_{n=1}^{\infty} G(n-1)x^n + 100\sum_{n=1}^{\infty} x^n \\
F(x) &= 2 + 2x\sum_{n=1}^{\infty} G(n-1)x^{n-1} + 100(-x^0 + \sum_{n=0}^{\infty} x^n) \\
F(x) &= 2 + 2x\sum_{n=0}^{\infty} G(n)x^{n} + 100(-1 + \sum_{n=0}^{\infty} x^n) \\
F(x) &= 2 + 2xF(x) + 100\left(-1 + \frac{1}{1-x}\right) \\
F(x) &= 102 \cdot \frac{1}{1 - 2 x} - 100 \cdot \frac{1}{1 - x}\\
\end{align}$
Take the $n$th coefficient of this generating function and we get:
$G(n) = 102 \cdot 2^n - 100$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
integral of $\int_0^\infty \frac{ x^3 \cos(x)}{x^6 + 1} \mathrm{d}x $ How does someone solve
$$\int_0^\infty \frac{ x^3 \cos(x)}{x^6 + 1} \mathrm{d}x $$
Is this analytically possible? I have no idea and I think if it is, it must be probably be solved via the Fourier/residue theorem or something like that.
| As Sangchul Lee commented, we could use partial fractions to get
$$\frac{ x^3 \cos(x)}{x^6 + 1}=\sum_{i=1}^6 a_i \frac{\cos(x)}{x-b_i}$$ and then use $$\int_0^\infty \frac{\cos(x)}{x-c}\,dx=-\text{Ci}(-c) \cos (c)-\text{Si}(c) \sin (c)-\frac{1}{2} \pi \sin (c)$$ (provided that $\Im(c)\neq 0\lor \Re(c)\leq 0$). This would be quite tedious but doable.
To my surprise, a CAS gave
$$\int_0^\infty \frac{ x^3 \cos(x)}{x^6 + 1} \,dx=\frac{1}{2} \sqrt{\frac{\pi }{3}} G_{1,7}^{4,1}\left(\frac{1}{46656}|
\begin{array}{c}
\frac{1}{3} \\
0,\frac{1}{3},\frac{1}{3},\frac{2}{3},\frac{1}{6},\frac{1}{2},\frac{5}{6}
\end{array}
\right)$$ where appears the Meijer G function (see here and here).
Its numerical value is $0.118802427933651$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding error in evaluation of triple integral The following problem is from lecture 26 of MIT OCW's multivariable calculus:
Evaluate $$\int_0^{2\pi} \int_{0}^{\frac{\pi}{4}} \int_{\frac{\sec \phi}{\sqrt{2}}}^1 \rho^2 \sin \phi\; d\rho\; d\phi\; d\theta$$
For some reason, I keep getting a different answer to the lecturer. Furthermore, I get a negative answer although the integral is supposed to describe the volume of a segment of a sphere. I have been unable to figure out what I am doing wrong. This is how I have done the problem:
Inner integral:
$$\int_{\frac{\sec \phi}{\sqrt{2}}}^1 \rho^2 \sin \phi\; d\rho=\sin \phi \cdot [\frac{\rho^3}{3}]_{\frac{\sec \phi}{\sqrt{2}}}^1=\frac{\sin \phi}{3}-\frac{\sin \phi \cdot \sec^3 \phi}{6\sqrt{2}}$$
Middle Integral:
$$\int_{0}^{\frac{\pi}{4}}\frac{\sin \phi}{3}-\frac{\sin \phi \cdot \sec^3 \phi}{6\sqrt{2}}\;d\phi=[-\frac{\cos \phi}{3}-\frac{\tan^2 \phi}{12\sqrt{2}}]_0^{\frac{\pi}{4}}=-\frac{\sqrt{2}}{6}-\frac{1}{12\sqrt{2}}$$
Outer Integral:
$$\int_0^{2\pi}-\frac{\sqrt{2}}{6}-\frac{1}{12\sqrt{2}} \; d\theta=-\frac{2\pi\sqrt{2}}{6}-\frac{2\pi}{12\sqrt{2}}$$
Which simplifies to $$-\frac{\pi \sqrt{2}}{3}-\frac{\pi}{6\sqrt{2}}=-(\frac{12\pi+3\pi}{18\sqrt{2}})=-\frac{5\pi}{6\sqrt{2}}$$
The answer that the lecturer gets is $$\frac{2\pi}{3}-\frac{5\pi}{6\sqrt{2}}$$
Could anyone tell me what I have done wrong? Any help would be greatly appreciated.
| In the middle integral: $\cos 0 = 1$. This gives you a $+ \frac{1}{3}$, which integrated in the next step adds $\frac{2\pi}{3}$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Green's function and derivatives of Dirac's Delta In my textbook in a section about electromagnetism I came across the following lines:
\begin{align}
\left[\left(\frac{\partial}{\partial x^0}\right)^2 - \Delta\right] \frac{\delta(x^0-|\mathbf{x}|)}{|\mathbf{x}|} &= \frac{1}{|\mathbf{x}|}\delta'' - \Delta \left(\frac{1}{|\mathbf{x}|}\right) \delta(x^0 - |\mathbf{x}|) - \frac{1}{|\mathbf{x}|} \Delta \delta(x^0-|\mathbf{x}|) - 2 \nabla \left(\frac{1}{|\mathbf{x}|}\right) \cdot \nabla \delta(x^0-|\mathbf{x}|)\\
&= \frac{1}{4\pi} \delta(\mathbf{x}) \delta(x^0-|\mathbf{x}|).
\end{align}
I'm having problems seeing the second equation; I know that
\begin{align}
\nabla^2 \frac{1}{|\mathbf{x}|} = -4\pi \delta(\mathbf{x}),
\end{align}
but I don't know how to deal with the derivatives of the delta distribution.
Can anyone explain, how to see that the equation above is correct?
Thanks in advance!
| Let's look at the a more general case: a function of the form $f(x^0 - |\mathbf{x}|)/|\mathbf{x}|$. For the sake of compactness, I'll switch to spherical polar coordinates where $|\mathbf{x}| = r$. We therefore have
$$
\begin{multline}
\left[\left(\frac{\partial}{\partial x^0}\right)^2 - \nabla^2 \right] \frac{f(x^0-|\mathbf{x}|)}{|\mathbf{x}|} \\= \frac{f''(x^0 - r)}{r} - f(x^0 - r) \nabla^2 \left( \frac{1}{r} \right) - \frac{1}{r} \nabla^2 f(x^0 - r) - 2 \left[ \nabla \left(\frac{1}{r}\right) \right] \cdot \left[\nabla (f(x^0 - r)) \right]
\end{multline}
$$In spherical coordinates, we have
$$
\frac{1}{r} \nabla^2 f(x^0 - r) = \frac{1}{r} \left[ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial f(x^0 - r)}{\partial r} \right) \right] = \frac{1}{r} f''(x^0 - r) - \frac{2}{r^2} f'(x^0 - r)
$$
and
$$
2 \left[ \nabla \left(\frac{1}{r}\right) \right] \cdot \left[\nabla (f(x^0 - r)) \right] = 2 \left( - \frac{1}{r^2} \hat{r} \right) \cdot \left( - f'(x^0 - r) \hat{r} \right) = \frac{2}{r^2} f'(x^0 - r).
$$
(If you wish to check these, note that
$$
\frac{\partial f(x^0 - r)}{\partial r} = - f'(x^0 - r) \qquad \frac{\partial^2 f(x^0 - r)}{\partial r^2} = f''(x^0 - r)
$$
due to the chain rule.) Putting all of these together, we find that
$$
\frac{f''(x^0 - r)}{r} - \frac{1}{r} \nabla^2 f(x^0 - r) - 2 \left[ \nabla \left(\frac{1}{r}\right) \right] \cdot \left[\nabla (f(x^0 - r)) \right] = 0,
$$
and so for any function $f$, we have
$$
\left[\left(\frac{\partial}{\partial x^0}\right)^2 - \nabla^2 \right] \frac{f(x^0-|\mathbf{x}|)}{|\mathbf{x}|} = - f(x^0 - r) \nabla^2 \left( \frac{1}{r} \right) = 4 \pi f(x^0 - |\mathbf{x}|) \delta^3(\mathbf{x}).
$$
The above derivation works for any function of $x^0 - r$ alone, including $\delta(x^0 - r)$.
The reason this works for any function, by the way, is that the general solution for the wave equation in three dimensions under the assumption of spherical symmetry (and excluding the origin) is
$$
u(r, t) = \frac{1}{r} \left[ f(r - t) + g(r + t) \right]
$$
for any two functions $f$ and $g$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
partial fractions decomposition issue How do you decompose the following fraction?
$$\frac{2}{2x^2 (x+1)} = \frac{a}{2x^2} + \frac{b}{x+1}$$
i just want to know if I am on the right track or I am missing something(based on the above beginning).
| Notice that you can cancel the common factor $2$ first.
Since $x=0$ is a double root, you need to take:
$$\frac{1}{x^2 (x+1)} = \frac{a}{x^2} + \frac{b}{x} + \frac{c}{x+1}$$
Can you take it from there?
Sometimes, you can avoid having to solve a system (in your case of three linear equations in three unknowns) to find the coefficients/numerators with a few handy manipulations.
$$\begin{align}
\frac{1}{x^2 (x+1)}
& = \frac{1\color{blue}{+x-x}}{x^2 (x+1)} \\[5pt]
& = \frac{1+x}{x^2 (x+1)} - \frac{x}{x^2 (x+1)} \\[5pt]
& = \frac{1}{x^2} - \frac{1}{x (x+1)}\\[5pt]
& = \frac{1}{x^2} - \frac{1\color{blue}{+x-x}}{x (x+1)}\\[5pt]
& = \frac{1}{x^2} - \frac{1+x}{x (x+1)}+\frac{x}{x (x+1)}\\[5pt]
& = \frac{1}{x^2} - \frac{1}{x}+\frac{1}{x+1}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does $(a + b)^3$ expand to $a^3 + 3ab^2 + 3a^2b + b^3$
Why does $(a + b)^3$ expand to $a^3 + 3ab^2 + 3a^2b + b^3$
Why does this work? I am confused as to why this happens.
| Let's break the algebra down, one step at a time. Trying to use colors to keep track of what comes from what.
You have that $$(a+b)^3 = \color{green}{(a+b)^2}\cdot (\color{#AA33FF}{a}+\color{#11AAFF}{b})$$
I assume you know $(a+b)^2 = a^2+2ab+b^2$, from which
$$
(a+b)^3 = \color{green}{(a^2+2ab+b^2)}\cdot (\color{#AA33FF}{a}+\color{#11AAFF}{b})
$$
Now, distributing (i.e., $c(a+b)=ca+cb$),
$$
(a+b)^3 = \color{green}{(a^2+2ab+b^2)}\cdot \color{#AA33FF}{a}+\color{green}{(a^2+2ab+b^2)}\cdot \color{#11AAFF}{b}
$$
Distributing again,
$$
(a+b)^3 = a^2\cdot \color{#AA33FF}{a}+2ab\cdot \color{#AA33FF}{a}+b^2\cdot \color{#AA33FF}{a} + a^2\cdot \color{#11AAFF}{b}+2ab\cdot \color{#11AAFF}{b}+b^2\cdot \color{#11AAFF}{b}
$$
that is
$$
(a+b)^3 = a^3+\color{red}{ 2a^2b}+\color{blue}{ a b^2} + \color{red} {a^2b}+\color{blue}{ 2ab^2}+b^3
$$
and regrouping the blue and red terms together,
$$
(a+b)^3 = a^3+\color{red}{3a^2b}+\color{blue}{3ab^2}+b^3
$$
giving the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Maclurin series for $e^{\sqrt{x^2+1}}$ i am giving $e^{\sqrt{x^2+1}}$ and asked to find the Maclaurin series for this term.
Here is my solution:
let $u=\sqrt{x^2+1}$, and given that we know that Maclaurin series for $e^x= 1+x+\frac{x^2}{2!} ...$
then: $$e^u= 1+u+\frac{u^2}{2}+...$$
hence:
$$e^{\sqrt{x^2+1}}=1+\sqrt{x^2+1}+\frac{x^2+1}{2}+...$$
Am I doing it right? please help
| You can get the first terms of the series computing the derivatives of $f(x)=e^{\sqrt{x+1}}$. You get:$$\begin{align}f(x)&=e^{\sqrt{x+1}}\\f'(x)&=\frac{e^{\sqrt{x+1}}}{2\sqrt{x+1}}\\f''(x)&=\frac{e^{\sqrt{x+1}} \left(\sqrt{x+1}-1\right)}{4(x+1)^{3/2}}\\f^{(3)}(x)&=\frac{e^{\sqrt{x+1}} \left(x-3 \sqrt{x+1}+4\right)}{8(x+1)^{5/2}}\\f^{(4)}(x)&=\frac{e^{\sqrt{x+1}} \left(x \left(\sqrt{x+1}-6\right)+16\sqrt{x+1}-21\right)}{16(x+1)^{7/2}}\\f^{(5)}(x)&=\frac{e^{\sqrt{x+1}}
\left(x^2+\left(47-10 \sqrt{x+1}\right) x-115
\sqrt{x+1}+151\right)}{32(x+1)^{9/2}}\end{align}$$Therefore$$\begin{align*}f(0)&=e\\f'(0)&=\frac e2\\\frac{f'(0)}{2!}&=0\\\frac{f^{(3)}(0)}{3!}&=\frac e{48}\\\frac{f^{(4)}(0)}{4!}&=-\frac{5e}{384}\\\frac{f^{(5)}(0)}{5!}&=\frac{3e}{320}\end{align*}$$So, the first terms of the Maclaurin series of $f(x)$ are$$e+\frac e2x+\frac e{48}x^3-\frac{5e}{384}x^4+\frac{3e}{320}x^5$$and therefore the first terms of the Maclaurin series of $e^{\sqrt{x^2+1}}$ are$$e+\frac e2x^2+\frac e{48}x^6-\frac{5e}{384}x^8+\frac{3e}{320}x^{10}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
When is $\sqrt[3]{a+\sqrt b}+\sqrt[3]{a-\sqrt b}$ an integer? I saw a Youtube video in which it was shown that
$$(7+50^{1/2})^{1/3}+(7-50^{1/2})^{1/3}=2$$
Since there are multiple values we can choose for the $3$rd root of a number, it would also make more sense to declare the value of this expression to be one of $2, 1 + \sqrt{-6},$ or $1 - \sqrt{-6}$
We may examine this more generally. If we declare $x$ such that
$$x=(a+b^{1/2})^{1/3}+(a-b^{1/2})^{1/3}$$
$$\text{(supposing } a \text{ and } b \text{ to be integers here)}$$
one can show that
$$x^3+3(b-a^2)^{1/3}x-2a=0$$
Which indeed has $3$ roots.
We now ask
For what integer values of $a$ and $b$ is this polynomial solved by an integer?
I attempted this by assuming that $n$ is a root of the polynomial. We then have
$$x^3+3(b-a^2)^{1/3}x-2a$$
$$||$$
$$(x-n)(x^2+cx+d)$$
$$||$$
$$x^3+(c-n)x^2+(d-nc)x-nd$$
Since $(c-n)x^2=0$ we conclude that $c=n$ and we have
$$x^3+3(b-a^2)^{1/3}x-2a=x^3+(d-c^2)x-cd$$
And - to continue our chain of conclusions - we conclude that
$$3(b-a^2)^{1/3}=d-c^2 \quad\text{and}\quad 2a=cd$$
At this point I tried creating a single equation and got
$$108b=4d^3+15c^2d^2+12c^4d-c^6$$
This is as far as I went.
| One choice is where $a-b^{1/2}=0$ so $a=4k^3$ for integer $k$ and $b=16k^6$, then
$$(a+b^{1/2})^{1/3}+(a-b^{1/2})^{1/3}=2k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A $3\times 3$ matrix to the power of $n$. I can't find a formula for :
$$
A =\begin{pmatrix}
1 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 3 \\
\end{pmatrix}^n
$$
I tried to separate $A = I + J$ with $J$ nilpotent but I didn't success.
Can you give me a hint?
Thanks.
| After computing a few terms, we see that
$$
\begin{pmatrix}
1 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 3 \\
\end{pmatrix}^n
=
\begin{pmatrix}
1 & 2^n-1 & a_n \\
0 & 2^n & 3^n-2^n \\
0 & 0 & 3^n \\
\end{pmatrix}
$$
where, $a_n=S(n+1,3)$, Stirling numbers of second kind (A000392).
In this case, as noted by @Joffan,
$$
a_n = \dfrac{(3^n-2^n)-(2^n-1)}{2}= \dfrac{3^n-2^{n+1}+1}{2}
$$
All this comes from expanding
$$
\begin{pmatrix}
1 & a & b \\
0 & c & d \\
0 & 0 & e \\
\end{pmatrix}
\begin{pmatrix}
1 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 3 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 2a+1 & a+3b \\
0 & 2c & c+3d \\
0 & 0 & 3e \\
\end{pmatrix}
$$
and noting that the relations $a=c-1, d=e-c, 2b=d-a$ are preserved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is this a valid formula for the tribonacci numbers? After about 5 pages of derivation, I derived a formula for the Tribonacci series:
$$
T(x) = \left( \frac{a-b-c + (b-c)r_1 + cr_1^2}{k(r_1-r_2)(r_1-r_3)} \right)r_1^n
+ \left( \frac{a-b-c + (b-c)2_1 + cr_2^2}{k(r_2-r_1)(r_2-r_3)} \right)r_2^n
+ \left(\frac{a-b-c + (b-c)r_3 + cr_3^2}{k(r_3-r_1)(r_3-r_2)} \right)r_3^n
$$
Where $a = T_2,b=T_1,c=T_0$, and:
$$
1 - x - x^2 - x^3 = k(1-r_1)(1-r_2)(1-r_3)
$$
In this case:
$$
r_1 = \frac{1}{\frac13\left(\sqrt[3]{3\sqrt{33}-17}-\dfrac2{\sqrt[3]{3\sqrt{33}-17}}-1\right)}
$$
and(approx)
$$
r_2 = \frac{1}{1.11514i - 0.771845}
$$
$$
r_3 = \frac{1}{1.11514i + 0.771845}
$$
$$
k = \frac{1}{r_2r_3}
$$
Is this a valid formula?
(I'll edit this answer to include the derivation, if it's really needed. It's a bit long.)
| Just to make your formulae looking simpler.
I must precise that I did not check the calculations.
Define $$t=\sqrt[3]{17+3 \sqrt{33}}$$ this makes the roots to be such that
$$\frac 1{r_1}=-\frac{1}{3}+\frac{t}{3}-\frac{2}{3 t}$$
$$\frac 1{r_2}=-\frac{1}{3}-\frac{t}{6}+\frac{1}{3 t}+\frac i {\sqrt 3}\left(\frac 1 t +\frac t2\right)$$
$$\frac 1{r_3}=-\frac{1}{3}-\frac{t}{6}+\frac{1}{3 t}-\frac i {\sqrt 3}\left(\frac 1 t +\frac t2\right)$$
$$k=\frac 1{r_1\,r_2}=\frac{4}{9 t^2}-\frac{2}{9
t}+\frac{1}{3}+\frac{t}{9}+\frac{t^2}{9}$$
| {
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"answer_id": 0
} |
Finding the Derivative of $\arctan \frac{x}{a - \sqrt{a^2 - x^2}}$
I am trying to simplify the derivative of $\arctan \frac{x}{a - \sqrt{a^2 - x^2}}$.
My work:
Everybody knows that $\frac{d}{dx} (\arctan \space u) = \frac{1}{1 + u^2} \frac{du}{dx}$
We let $u = \frac{x}{a - \sqrt{a^2 - x^2}} .$ To get the $du,$ I remember that $\frac{d}{dx} \left( \frac{u}{v} \right)= \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}.$
So:
$$\frac{d}{dx} \left ( \frac{x}{a - \sqrt{a^2 - x^2}} \right) = \frac{a-\sqrt{a^2 - x^2} \frac{d}{dx} (x) - x \frac{d}{dx} (a-\sqrt{a^2 - x^2)}}{(a-\sqrt{a^2 - x^2)^2}} $$
$$ = \frac{a - \sqrt{a^2 - x^2} -x \left(\frac{x}{(\sqrt{a^2 - x^2})}\right)}{a^2 - 2\sqrt{a^2 - x^2} +a^2 - x^2}$$
$$ = \frac{ \frac{a - \sqrt{a^2 - x^2}}{1} + \frac{-x^2}{ \sqrt{a^2 - x^2} }}{2a^2 - 2\sqrt{a^2 - x^2} - x^2}$$
$$ = \frac{\frac{(a^2 - x^2)^{\frac{1}{2}} (a - (a^2 - x^2)^\frac{1}{2}) - x^2}{(a^2 - x^2)^{\frac{1}{2}}}}{2a^2 - x^2 - 2\sqrt{a^2 - x^2} }$$
$$ \frac{d}{dx} \left( \frac{x}{a - \sqrt{a^2 - x^2}}\right) = \frac{a(a^2- x^2)^\frac{1}{2} - (a^2- x^2) - x^2}{2a^2 - x^2 - 2\sqrt{a^2 - x^2} }$$
Then:
$$\frac{d}{dx} \left( \arctan \frac{x}{a - \sqrt{a^2 - x^2}} \right) = \frac{1}{1 + \left( \frac{x}{a - \sqrt{a^2 - x^2}} \right)^2} \left( \frac{a(a^2- x^2)^{\frac{1}{2}} - (a^2- x^2) - x^2}{2a^2 - x^2 - 2\sqrt{a^2 - x^2} } \right) $$
At this point, simplifying it is difficult. How do you get the derivative of $\arctan \frac{x}{a - \sqrt{a^2 - x^2}}?$
| Using $$u = \frac{x}{a-\sqrt{a^2-x^2}} = \frac{a+\sqrt{a^2-x^2}}{x}$$ results in $$\frac{du}{dx} = -\frac{a^2 + a \sqrt{a^2-x^2}}{x^2\sqrt{a^2-x^2}},$$ and $$\frac{d}{dx}\tan^{-1}(u) = -\frac{1}{2\sqrt{a^2-x^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof writing for polynomials Let the polynomial $f(x)=x^n+a_1 x^{n-1}+a_2 x^{n-2}+.....a_{n-1}x+a_n$ have integral coefficients. If there exists four distinct integers $a,b,c$ and $d$ such that $f(a)=f(b)=f(c)=f(d)=5$ show that there is no integer $k$ such that $f(k)=8$
I have tried to prove it but I somehow feel that my proof is incorrect please point out the errors and suggest some other way to do the question.
This how I proceeded
$\frac{x^{n+1}-1}{x-1}=x^n+a_1 x^{n-1}+a_2 x^{n-2}+.....a_{n-1}x+a_n $
$\frac{a^{n+1}-1}{a-1}=5$
hence $\frac{a^{n+1}-1}{5}=a-1$
Since $a$ is an integer $a-1$ will also be an integer hence $5|a^{n+1}-1$
Using fermat's little theorem we know that $a^4-1\equiv_5 0$
Since $f(a)=5$ we get $n+1=4$
Let k be an integer such that $f(k)=8$
$\frac{a^4-1}{8}=a-1$ Since $a-1$ is an integer $8|a^4-1$
We will consider two cases
*
*when $a$ is an even integer
*when $ a$ is an odd integer
If $a$ is even then $a^4-1$ will be odd and it not be divisible by $8$
If $a$ is odd then it will be relatively prime to $8$ hence $8$ will be Carmichael number
$a^7-1\equiv0(mod8)$ which proves that $a^4-1$ is not divisible by 8.
$\therefore$ there is no integer k such that $f(k)=8$
| I think the first part of your solution is wrong ($\frac{x^{n+1}-1}{x-1}=x^n+x^{n-1}+\cdots+x+1$, not $x_n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$).
Since $f$ is a polynomial with integer coefficients, we have $a-b|f(a)-f(b)$ for all $a, b\in \mathbb{Z}, a\neq b$. Thus if we suppose there exists integer $k$ with $f(k)=8$, we have $k-x|f(k)-f(x)=3$, where $x=a,b,c,d$. Since $a,b,c,d$ are all distinct we have $\{a,b,c,d\}=\{k-3,k-1,k+1,k+3\}$.
Then, consider $g(x)=f(x-k)-5$. $g$ clearly is a monic polynomial with integer coefficients such that $g(0)=3, g(-3)=g(-1)=g(1)=g(3)=0$. If we set $g(x)=x^n+b_1x^{n-1}+\cdots+b_{n-1}x+b_n$, by conditions we have $b_n=3$ and following:
$3^n+3^{n-1}b_1+\cdots+3b_{n-1}+3=0 \rightarrow 3^{n-1}+3^{n-2}b_1+\cdots+b_{n-1}+1=0$
$(-3)^n+(-3)^{n-1}b_1+\cdots+(-3)b_{n-1}+3=0 \rightarrow (-3)^{n-1}+(-3)^{n-2}b_1+\cdots+b_{n-1}-1=0$
From the first equation we have $b_{n-1}\equiv-1 \mod 3$, while from the second equation we have $b_{n-1}\equiv 1 \mod 3$, a contradiction. Thus there exists no such $f$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
find limit of a sequence I have a sequence:
$$ X_n\:=\:\cos\left(\left(\frac{3^n+\pi^n}{3^n+\left(\pi-\frac{1}{4} \right)^n} \right)^{1/n}\right) $$
I have to find the limit when $n \to \infty $ where $n \in \mathbb{N}$.
Which is the best way to find the answer ? Can I reduce or use the Squeeze theorem in that case ?
| $$a_n=\sqrt[n]{\frac{3^n+ \pi^n}{3^n+ (\pi -1/4)^n}}\leq \sqrt[n]{\frac{2 \pi^n}{3^n}}=\sqrt[n]{2} \frac{\pi}{3} \to \frac{\pi}{3}$$
Also $$a_n=\sqrt[n]{\frac{3^n+ \pi^n}{3^n+ (\pi -1/4)^n}} \geq \sqrt[n]{ \frac{\pi^n}{3^n+3^n}}=\frac{1}{\sqrt[n]{2}} \frac{\pi}{3} \to \frac{\pi}{3}$$
So from squeeze theorem $a_n \to \frac{\pi}{3}$.
Now using the continuity of $\cos{x}$ we have that $$\cos{a_n} \to \cos{\frac{\pi}{3}}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Quartic equation with integer solutions I am trying to find the solutions of the following quartic equation given that all the solutions are integers.
$$x^4+22x^3+172x^2+552x+576=0$$
Below is the original phrasing of the problem with hints building up to this equation. I can prove all of the results it asks for before the final part of the question, but I am struggling to actually find the solutions to the equation and require help with explaining the process as well.
I then know that:
(1) $k_1k_2k_3k_4 = 576 = (1)(2^6)(3^2)$
(2) $(k_1+1)(k_2+1)(k_3+1)(k_4+1) = 1323=(1)(3^3)(7^2)$
(3) $(k_1-1)(k_2-1)(k_3-1)(k_4-1) = 175 = (1)(5^2)(7)$
However, I do not know how to proceed from here and the solution to this problem didn't explain in enough detail for me to either understand the solution, or the approach.
| For $$f(x) = x^4+22x^3+172x^2+552x+576,$$
notice that if $x\geq0$ then $f(x) \geq 576 > 0,$
so all roots of the equation $f(x) = 0$ are negative.
Since the roots are $-k_1, -k_2, -k_3, -k_4,$
this tells us that $k_1, k_2, k_3, k_4$ are all positive.
That leaves only a few possible values of $k_i$ that could occur
in the equation
$$(k_1-1)(k_2-1)(k_3-1)(k_4-1) = 175 = 5^2 \cdot 7.$$
The factors of $175$ are $1, 5, 5^2 = 35, 7, 5\cdot7 = 35,$
and $5^2 \cdot 7 = 175.$
(I chose this equation to start with because $175$ has fewer factors to consider than either $576$ or $1323$.)
Since each factor $k_i - 1$ must be a factor of $175,$
the possible values of any $k_i$ can only be among the numbers
$2, 6, 26, 8, 36,$ and $176.$
From
$$k_1k_2k_3k_4 = 576 = 2^6 \cdot 3^2, \tag1$$
we know any of the $k_i$ can have only $2$ or $3$ as prime factors.
So $k_i \neq 26 = 2 \cdot 13$ and $k_i \neq 176 = 16 \cdot 11.$
From
$$(k_1+1)(k_2+1)(k_3+1)(k_4+1) = 1323 = 3^3 \cdot 7^2,$$
we know $k_i + 1$ must be divisible by $3$ or by $7.$
So $k_i + 1 \neq 37,$ and therefore $k_i \neq 36.$
Therefore the only possible values of any of the $k_i$ are $2,$ $6,$ or $8.$
But Equation $(1)$ implies either that one of the $k_i$ is divisible by $9$
(which we now know cannot be true) or that two of the $k_i$ are each divisible by $3.$ If $k_i$ is divisible by $3,$ its only possible value is $6$ (since neither $2$ and $8$ is divisible by $3$).
Without loss of generality, we can set $k_1 = k_2 = 6.$
Dividing both sides of Equation $(1)$ by $k_1k_2 = 36,$
we now have $k_3k_4 = 2^4.$ Each of $k_3$ or $k_4$ can then only be $2$ or $8$ (since $6$ is not a power of $2$), but if one is $2$ then the other is $8$ and vice versa. So without loss of generality we can set $k_3 = 2$ and
$k_4 = 8.$
The four roots therefore are
$-k_1 = -6,$ $-k_2 = -6,$ $-k_3 = -2,$ and $-k_4 = -8.$
In ascending order they are $-8, -6, -6, -2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Proving that if $x,y \in \Bbb R$ and $|x|=|y|$ then $x^2=y^2$ I've seen that if $|x|=|y|$ for reals $x$ and $y$ that $x^2$ is also equal to $y^2$. Generally that makes handling the absolute values much easier algebraically. I was wondering if there was any more need proof beyond If
$x,y \in \Bbb R$ then $|x|=\sqrt{x^2}$ so if $|x|=|y|$ then $\sqrt{x^2}=\sqrt{y^2}$ square both sides: $x^2=y^2$.
I was also wondering if it would be valid to say if $x,y \in \Bbb R$ and $|x|=y$ then $x=\pm y$ and $\pm x=y$, and in extension $x^2=y$ then $x=\pm \sqrt{y}$.
| Thats a good question.
I think it is more consequential than direct though.
For real numbers,
We have as an axiom for $a > 0$ and $x < y$ than $ax < ay$. This gives us: if $x > 0$ then $x^2 = x*x > x*0 = 0$.
We also have as an axiom that for ever real $b$ there is a unique $-b$ so that $b + (-b) = 0$. And we have an axiom that if $x < y$ than $x + c < y +c$ for any $c$. From this we have $b > 0 \iff b +(-b) > 0 + (-b) \iff 0 > -b$.
And there therefore we have a proposition that for $a < 0$ and $x < y$ than $ax > ay$. (Because $a < 0 \implies -a > 0 \implies -ax < -ay \implies ax > ay$.) [Okay, I skipped proving that $(a)(-b) = -(ab)$ but... that's details...].
Thus we have. For any $x \in \mathbb R$ exactly one of the following are true:
i) $x > 0$ and $x^2 > 0$.
ii) $x = 0$ and $x^2 = 0$.
iii) $x < 0$ and $x^2 = x*x > x*0 = 0$.
So all squares are positive or zero (and only zero if it is zero squared).
.... Hmm.... okay, we do need to prove that $(a)(-b) = -(ab)$ and that $-(-a) = a$ and that $(-a)(-b) = ab$, after all.
We have an axiom that $a(b+c) = ab+bc$. So $ab+ (a)(-b) = a(b + (-b)) = a*0 = 0$. Now we have an axiom that there is one unique $-ab$ so that $ab + (-ab) =0$ and $ab + (a)(-b) =0$.... so that mean $(a)(-b)$ must be equal to that unique $-ab$.
We also have $a + (-a) = -a + a = 0$ so there is a unique $-(-a)$ so that $-a + (-(-a)) = 0$. So that must be $-(-a) = a$.
And finally $(-a)(-b) = -(-a)(b) = --(ab) = ab$.
So .... given all that....
AND defining that $|a| = a$ if $a > 0$ and $|a| = 0$ if $a = 0$ and that $|a| = -a$ (which is positive) if $a < 0$.
We can have:
$|a| = |b|$ means either.
1) $a = b$ and $a\ge 0 ; b\ge 0$ or $a < 0; b < 0$ and $a^2 = b^2$.
2) $a = -b$ and either $a \ge 0; b \le 0$ or $a < 0; b> 0$ and then $a^2 = (-b)^2 = b^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Definite integral of the product of powers of the sine and cosine When integration limits were from $0$ to $2\pi$ and the function was $\sin x \cdot \cos^2 x$ and also when the function was $\cos x \cdot \sin^2 x$,answer is $0$, so is there a standard formula when limits are $0$ to $2\pi$ and function is product of higher powers of $\sin x$ and $\cos x$? I have searched about it but got no such information..
| Here's an answer not just for higher powers of $\sin x$ and $\cos x,$
but indeed for any non-negative integer powers of $\sin x$ and $\cos x.$
In this answer, assume at all times that $m$ and $n$ are non-negative integers.
If $m$ and $n$ both are even, then
$$
\int_0^{2\pi} \sin^m t \cos^n t\,dt
$$
evaluates to a positive real number according to procedures shown later in this answer.
In any other case ($m$ odd, $n$ odd, or $m$ and $n$ both odd),
$$
\int_0^{2\pi} \sin^m t \cos^n t\,dt = 0.
$$
Consider three cases that cover all possibilities:
Case 1: $m$ is odd.
Since $\sin(-t) = -\sin t$ and $m$ is odd, $\sin^m(-t) = -\sin^m t,$
that is, $\sin$ is an odd function.
But $\cos(-t) = \cos(t)$ and therefore $\cos^n(-t) = \cos^n t.$
It follows that
\begin{align}
\sin^m(-t)\cos^n(-t) &= -\sin^m t \cos^n t; \\
\int_{-\pi}^0 \sin^m t \cos^n t\,dt &= - \int_0^\pi \sin^m t \cos^n t\,dt; \\
\int_{-\pi}^\pi \sin^m t \cos^n t\,dt &= 0. \\
\end{align}
Now observe that
$\sin(t - 2\pi) = \sin t$
and $\cos(t - 2\pi) = \cos t,$
so $\sin^m (t - 2\pi) \cos^n (t - 2\pi) = \sin^m t \cos^n t$
and
\begin{align}
\int_0^{2\pi} \sin^m t \cos^n t\,dt
& = \int_0^\pi \sin^m t \cos^n t\,dt
+ \int_\pi^{2\pi} \sin^m t \cos^n t\,dt \\
& = \int_0^\pi \sin^m t \cos^n t\,dt
+ \int_{-\pi}^0 \sin^m t \cos^n t\,dt \\
&= \int_{-\pi}^{\pi} \sin^m t \cos^n t\,dt.
\end{align}
Therefore
$$ \int_0^{2\pi} \sin^m t \cos^n t\,dt = 0. $$
Case 2: $n$ is odd.
Since (as we already know)
$\sin^m (t - 2\pi) \cos^n (t - 2\pi) = \sin^m t \cos^n t,$
it follows that
\begin{align}
\int_0^{2\pi} \sin^m t \cos^n t\,dt
& = \int_0^{3\pi/2} \sin^m t \cos^n t\,dt
+ \int_{3\pi/2}^{2\pi} \sin^m t \cos^n t\,dt \\
& = \int_0^{3\pi/2} \sin^m t \cos^n t\,dt
+ \int_{-\pi/2}^0 \sin^m t \cos^n t\,dt \\
&= \int_{-\pi/2}^{3\pi/2} \sin^m t \cos^n t\,dt \\
&= \int_0^{2\pi} \sin^m \left(t-\frac\pi2\right)
\cos^n \left(t-\frac\pi2\right)\,dt \\
&= \int_0^{2\pi} \left(-\cos t\right)^m \sin^n t\,dt \\
&= \pm \int_0^{2\pi} \cos^m t \sin^n t\,dt \\
\end{align}
(where the $\pm$ sign depends on whether $m$ is even or odd).
But we now have $\sin t$ raised to an odd power, so Case 1 shows that
$$ \int_0^{2\pi} \cos^m t \sin^n t\,dt = 0.$$
Therefore
$$ \int_0^{2\pi} \sin^m t \cos^n t\,dt = 0.$$
Case 3: $m$ and $n$ are both even.
In this case we can use the identity $\sin^2 t = 1 - \cos^2 t$ to show that
$$
\int_0^{2\pi} \sin^m t \cos^n t\,dt
= \int_0^{2\pi} (1 - \cos^2 t)^{m/2} \cos^n t\,dt,
$$
and since $\frac m2$ is an integer we can expand $(1 - \cos^2 t)^{m/2}$
to get a polynomial in $\cos^2 t.$
Since $n$ is even, $\cos^n t$ is a power of $\cos^2 t$
and therefore the entire integrand
$$ (1 - \cos^2 t)^{m/2} \cos^n t $$
is a polynomial in $\cos^2 t.$
We can integrate each term of this polynomial separately;
the integral for any non-constant term can be evaluated using the formula
$$
\int_0^{2\pi} \cos^{2p} t\,dt
= 2\pi \left(\frac{1 \cdot 3 \cdot 5 \cdots(2p-1)}{2 \cdot 4 \cdot 6 \cdots 2p}\right)
$$
where $p$ is a positive integer
(see Definite integral of even powers of Cosine.,
integration of $\int_0^{2\pi} cos^{2n}(t)dt$,
and the answers to those questions for details).
In the case $n=0$ there is also a constant term whose integral
(of course) is $\int_0^{2\pi} 1\,dt = 2\pi.$
Example: Integrate $\sin^4 t \cos^2 t.$
We have
\begin{align}
\int_0^{2\pi} \sin^4 t \cos^2 t \,dt
&= \int_0^{2\pi} (1 - \cos^2 t)^2 \cos^2 t \,dt \\
&= \int_0^{2\pi} (\cos^2 t - 2\cos^4 t + \cos^6 t)\,dt \\
&= \int_0^{2\pi} \cos^2 t\,dt - \int_0^{2\pi} 2\cos^4 t\,dt + \int_0^{2\pi} \cos^6 t\,dt \\
&= \int_0^{2\pi} \cos^2 t\,dt - \int_0^{2\pi} 2\cos^4 t\,dt + \int_0^{2\pi} \cos^6 t\,dt \\
&= 2\pi\left(\frac12\right) - 2\left(2\pi\left(\frac{1\cdot3}{2\cdot4}\right)\right)
+ 2\pi\left(\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\right) \\
&= 2\pi\left(\frac{24 - 2(18) + 15}{48}\right) \\
&= \frac18 \pi.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
proof of $\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$ without solving $\int\frac{1}{x^2+1}dx$ So I know $\int\frac{1}{x^2+1}dx=\arctan(x)+c$
Now when i tried to integral $\frac{1}{x^2+1}$ in a different way i got this:
$$\frac{1}{x^2+1}=\frac{1}{(x-i)(x+i)}=\frac{A}{x-i}+\frac{B}{x+i}\\\implies1=Ax+Ai+Bx-Bi=x(A+B)+i(A-B)\\\implies
\begin{cases}
A+B=0
\\[2ex]
A-B=\frac{1}{i}=-i
\end{cases}\\\implies2A=-i\implies A=\frac{-i}{2}\implies B=\frac{i}{2}$$
So in the end i get:$$\int\frac{1}{x^2+1}dx=\int\frac{i}{2(x+i)}-\frac{i}{2(x-i)}dx=\frac{i}{2}\int\frac{1}{(x+i)}-\frac{1}{(x-i)}dx$$
After integrating this i get $\frac{i}{2}(\ln(x+i)-\ln(x-i)+C)$ after comparing this to $\arctan(x)$ at $x=0$ i find $C=-i\pi$
$\therefore~\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$
now my question is how can i prove that without using the integral
| Here is another way to look at this: consider that $z=|z|e^{i\theta}$ and that $z^*=|z^*|e^{-i\theta}$. Then
$$
\frac{|z|}{|z^*|}=e^{2i\theta}\\
\theta=\frac{1}{2i}(\ln |z|-\ln |z^*|)
$$
Now, let $z=x+i$, then
$$
\theta=\cot^{-1}x=\frac{1}{2i}(\ln |x+i|-\ln |x-i|)
$$
Now,
$$
\cot^{-1}x+\tan^{-1}x=\frac{\pi}{2}
$$
so that
$$
\begin{align}
\tan^{-1}x
&=-\frac{1}{2i}(\ln |x+i|-\ln |x-i|)+\frac{\pi}{2}\\
&=\frac{i}{2}(\ln |x+i|-\ln |x-i|-i\pi)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Difficult Telescoping Series Finding the explicit sum of a telescoping series with two factors in the denominator is not a problem: we split the fractions in the difference of two pieces.
But what about 2+ factors
\begin{equation}
\lim_{n\to\infty} \sum \frac{1}{k(k+2)(k+4)}
\end{equation}
| hint: we can do telescoping inside teloscoping as follows:
$\dfrac{1}{n(n+2)(n+4)}= \dfrac{1}{4}\left(\dfrac{1}{n(n+2)} - \dfrac{1}{(n+2)(n+4)}\right)$, and split again: $\dfrac{1}{n(n+2)} = \dfrac{1}{2}\left(\left(\dfrac{1}{n} - \dfrac{1}{n+1}\right) + \left(\dfrac{1}{n+1} - \dfrac{1}{n+2}\right)\right)$. Do the same for the other term and there are a total of $4$ "mini" telescoping sums to be evaluated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the sum of this series? Show that the series $\displaystyle \sum_{n=0}^{\infty}\frac{x}{(1+x^2)^n} $ converges $\forall x\in \mathbb{R}$ and find the sum $\forall x\in \mathbb{R}$
So, I've shown that the series converges $\forall x\in \mathbb{R}$ and, through testing, suspect that the sum is equal to $\frac{x}{1+x^2}$. Heres what I did:
Let $s =\displaystyle \sum_{n=0}^{k}\frac{x}{(1+x^2)^n}$ = $\frac{x}{(1+x^2)}+ \frac{x}{(1+x^2)^2} + ... +\frac{x}{(1+x^2)^k} \Rightarrow ... \Rightarrow\frac{(x^2+1)^k}{x}s = (1+x^2)^{k-1}+(1+x^2)^{k-2} + ...+1 $.
Not quite sure what to do from here. Any tips ?
| Let $x$ be a non-zero real number.
Then given sum is $x\cdot(1+\frac 1{1+x^2}+\frac 1{{(1+x^2)}^2}+\frac 1{{(1+x^2)}^3}+\cdot\cdot\cdot).$
Observe that $1+\frac 1{1+x^2}+\frac 1{{(1+x^2)}^2}+\frac 1{{(1+x^2)}^3}+\cdot\cdot\cdot$ is a geometric series with first term $a=1$ and common ratio $r=\frac 1{1+x^2} \implies |r|=\frac 1{1+x^2} \lt 1$.
Thus this geometric series converges to limit $\frac a{1-r}=\frac 1{1-\frac 1{1+x^2}}=\frac {1+x^2}{x^2}.$
Thus given sum is $x\cdot(1+\frac 1{1+x^2}+\frac 1{{(1+x^2)}^2}+\frac 1{{(1+x^2)}^3}+\cdot\cdot\cdot) =x \cdot \frac {1+x^2}{x^2}=\frac 1x + x.$
In case of $x=0$, every term of the series is zero. Thus whole sum is also zero.
| {
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"url": "https://math.stackexchange.com/questions/2417825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int {\sqrt{x^4+1}}dx$ Evaluate $$\int \sqrt{x^4+1}dx$$
My Try: I used parts as follows:
$$I=x{\sqrt{x^4+1}}-4 \int \frac{x^4 dx}{\sqrt{x^4+1}}$$ $\implies$
$$I=\frac{x}{\sqrt{x^4+1}}-4I+4 \int \frac{dx}{\sqrt{x^4+1}}$$ $\implies$
$$5I=\frac{x}{\sqrt{x^4+1}}+4J$$ where
$$J=\int \frac{dx}{\sqrt{x^4+1}}$$
any clue here?
| To get
$$I = \int\sqrt{x^4+1}\,\mathrm{d}x=\int\frac{x^4+1}{\sqrt{x^4+1}}\,\mathrm{d}x.$$
First use partial fractions to get
$$
I=x{\sqrt{x^4+1}}-2\int x \mathrm{d}\sqrt{x^4+1}=x{\sqrt{x^4+1}}-2 \int \frac{x^4+1-1}{\sqrt{x^4+1}}\mathrm{d}x
$$
and rewrite as
$$
I=x{\sqrt{x^4+1}}-2I+2 J,\quad \mathrm{where}\quad J=\int \frac{1}{\sqrt{x^4+1}}\mathrm{d}x
$$
let $t^4=-x^4$, solved
$$
\begin{aligned}
t &= e^{\pi i / 4} x\\
x &= e^{-\pi i / 4} t\\
1-x^4&=1 - t^4= (1-t^2)(1- i^2t^2)\\
\end{aligned}
$$
So
$$
J= \int \frac{1}{\sqrt{1+x^4}}\mathrm{d}x = \int \frac{e^{-\pi i/4}}{\sqrt{1-t^4}}\mathrm{d}x=e^{-\pi i/4}F(\arcsin t; i^2)
$$
where F(x;k) is Elliptic Integral of the First Kind.
$$
F(x;k^2)=\int _{0}^{\sin x}{\frac {{\rm {d}}t}{\sqrt {(1-t^{2})(1-k^{2}t^{2})}}}
$$
Then
$$
\begin{aligned}
I
&=\frac{x}{3}{\sqrt{x^4+1}}+\frac{2}{3}J\\
&=\frac{x}{3}{\sqrt{x^4+1}}+\frac{2e^{-\pi i/4}}{3}F(\arcsin e^{\pi i/4} x; -1)\\
\end{aligned}
$$
Finally, take the derivation and verify that the result is correct
D[x Sqrt[x^4 + 1] / 3 + 2 Exp[-Pi I / 4] EllipticF[ArcSin[Exp[Pi I / 4] x], -1] / 3, x] // FullSimplify
| {
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Given a positive number $b$, under what conditions does there exist a rectangle with perimeter $2b$ and area $\frac{b}{2}$? Given a positive number $b$, under what conditions does there exist a rectangle with perimeter $2b$ and area $\frac{b}{2}$?
$$2l + 2w = 2b$$ $$l+w=b$$
and $$lw = \frac{b}{2}$$
I tried to use AGM inequality
$$lw = \frac{l+w}{2}$$
For that to be a AGM inequality the right side had to be squared and the equality will only hold if $l = w$. I don't know how to get the conditions from here.
| $2w + 2l = 2b$
$l = b-w$.
$wl = w(b-w) = bw -w^2 = \frac b2$
$w^2 - bw + \frac b2 = 0$
$w = \frac {b \pm\sqrt{b^2 - 4*\frac b2}}2 = \frac {b \pm \sqrt {b^2 - 2b}}2$
This can be done whenever $b^2 \ge 2b$ which (as $b > 0$) mean $b\ge 2$.
If $b = 2$ this is possible with square of $w=l = 1$ and $P = 4=2*2$ and $A = 1^2 = 1 = \frac 22$.
If $b > 2$ then this is possible with $w = \frac {b + \sqrt {b^2 - 2b}}2$ and $l = \frac {b - \sqrt {b^2 - 2b}}2$. The $P= 2w + 2l = b + \sqrt {b^2 - 2b} + b - \sqrt {b^2 - 2b} = 2b$ and $A = wl = \frac {b + \sqrt {b^2 - 2b}}2*\frac {b - \sqrt {b^2 - 2b}}2 = \frac {b^2 - (b^2 - 2b)}{4} = \frac b2$.
If $b < 2$ well....
$\frac {l + w}2 \ge \sqrt {lw}$ by AM-GM
But $\frac {l + w}2 = $ perimeter/$4$ $= \frac b2$
$\sqrt {lw} = \sqrt{\text {area}} = \sqrt {\frac b2}$
So $\frac b2 \ge \sqrt {\frac b2}$
So $\frac b2 \ge 1$ and $b \ge 2$.
So it can't be done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality : $\frac{a^{n+1}}{x}+\frac{b^{n+1}}{y}+\frac{c^{n+1}}{z}\geq1 $
Prove that if $a, b, c, x, y, z >0$ and $n$ is positive integer such that
$(a^n+b^n+c^n)^{n+1}=x^n+y^n+z^n$ , then $$\frac{a^{n+1}}{x}+\frac{b^{n+1}}{y}+\frac{c^{n+1}}{z}\geq1 $$
My attempt :
By Holder Inequality,
$$(a^{n+1}+b^{n+1}+c^{n+1})^n(1+1+1) \geq (a^n+b^n+c^n)^{n+1}$$
$$(a^{n+1}+b^{n+1}+c^{n+1})^n \geq \frac{x^n+y^n+z^n}{3}$$
I don't know how to proceed.
| Hint: Holder:
$$
(x^n+y^n+z^n)^{1/n+1}(\frac{a^{n+1}}x+\frac{b^{n+1}}y+\frac{c^{n+1}}z)^{n/n+1}>?
$$
| {
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"url": "https://math.stackexchange.com/questions/2421500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove that if $1 \le m \le n$ then $\left(1+\frac1n\right)^m < 1+\frac mn + \left(\frac mn\right)^2$ We know that $\dfrac1n>0$, $\dfrac1n<\dfrac mn<1$, $\left(\dfrac1n+1\right)^m < \left(\dfrac mn+1\right)^m$, so we have to prove that $\left(\dfrac mn+1\right)^m<\dfrac mn+1+\left(\dfrac mn\right)^2$.
I tried using Newton's binomial expression of $\left(\dfrac mn+1\right)^m$ and calculating the difference, but I didn't get there.
| For integer $m$ we can use induction.
Indeed, for $m=1$ we have $$1+\frac{1}{n}<1+\frac{1}{n}+\frac{1}{n^2}.$$
Let $$\left(1+\frac{1}{n}\right)^m<1+\frac{m}{n}+\frac{m^2}{n^2}.$$
Thus, $$\left(1+\frac{1}{n}\right)^{m+1}<\left(1+\frac{m}{n}+\frac{m^2}{n^2}\right)\left(1+\frac{1}{n}\right).$$
Thus, it remains to prove that
$$\left(1+\frac{m}{n}+\frac{m^2}{n^2}\right)\left(1+\frac{1}{n}\right)<1+\frac{m+1}{n}+\frac{(m+1)^2}{n^2}$$ or
$$\frac{2m+1}{n^2}>\frac{m^2}{n^3},$$
which is true because $n\geq m$.
Finally, we need to check what happens for $m=n$:
$$\left(1+\frac{1}{n}\right)^n<1+1+1$$ or
$$\left(1+\frac{1}{n}\right)^n<3,$$
which is known.
Done!
| {
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Limit of a sequence : $x_n = \frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \ldots + \frac{1}{n(n+1)(n+2)}$ Can someone help me with this problem?
Finding the limit $\lim_{n \to \infty}\ x_n$ where
$$x_n = \frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \ldots + \frac{1}{n(n+1)(n+2)},\quad n\in\mathbb{N}.$$
I don't have a clue how to do this.
| From the question here we find that
$$\sum_{r=1}^\infty \frac 1{r^\overline{m}}=\frac 1{(m-1)(m-1)!}$$
Putting $m=3$ gives
$$\sum_{r=1}^\infty \frac 1{r^\overline{3}}=\sum_{r=1}^\infty \frac 1{r(r+1)(r+2)}=\frac 1{2\cdot 2!}=\color{red}{\frac 14}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that for any $x\geq0$ and $y \ge 0$ we have: $|\sqrt{x} − \sqrt{y}| \le \sqrt{|x − y|}.$
Prove that for any $x\geq0$ and $y \ge 0$ we have: $$|\sqrt{x} − \sqrt{y}| \le \sqrt{|x − y|}.$$
$$|\sqrt(x) - \sqrt(y)|^2 \le \sqrt{|x-y|}^2$$
$$(\sqrt{x}-\sqrt{y})^2 \le |x-y|$$
$$x - 2\sqrt{xy} + y \le |x-y|$$
How do I take the absolute value separately to prove it?
| Consider $2$ cases:
$$\begin{align} & x<y \Rightarrow \sqrt{x}-\sqrt{y}<0< \sqrt{|x-y|}, \\
& x\ge y \Rightarrow \sqrt{x}-\sqrt{y}\le \sqrt{x-y} \Rightarrow x-2\sqrt{xy}+y\le x-y \Rightarrow y\le x. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\sum_{n=1}^{2017}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$
What's
$$\sum_{n=1}^{2017}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$$
What have I tried?
$$(n+2)^4=n^4+8n^3+24n^2+32n+16$$
$$(n+1)^4=n^4+4n^3+6n^2+4n+1$$
Remainder:
$$4n^3+18n^2+28n+15$$
mod:
$$2n^2-1\pmod{4}$$
I can compute $\sum x^2$ but I don't know what to do with $$\sum_{n=1}^{2017}\left(2n^2-1\mod{4}\right)$$
| For odd $n$, $(n+1)^4$ is a multiple of $4$, so in that case $$(X\bmod(n+1)^4)\bmod 4 = X \bmod 4$$ and we just have to sum $(n+2)^4\bmod 4$. But $(n+2)^4$ is an odd square, and all odd squares are $1$ modulo $4$.
For even $n$ your calculation of $(n+2)^2 \bmod (n+1)^4 = 4n^3+18n^2+28n+15$ works as long as $(n+1)^4 < 2(n+1)^4$ which happens when $n>\frac{1}{\sqrt[4]2-1}-1 \approx 4.3$. So the cases $n=2,4$ need to be treated as special cases.
Otherwise you're summing (as you say) $2n^2-1$ modulo $4$. But since $n$ is now even, $2n^2$ will be a multiple of $4$, so it disappears and only the $-1$ is left.
So for $n\ge 4$ the $+1$ from an odd $n$ cancels out the $-1$ from the even $n$ next to it, and all that is left is
$$\sum_{n=1}^{3}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$$
which can be done by hand easily.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Logarithm-Based Sum Find the value of $x$ satisfying
$18^{4x-3}=(54\sqrt{2})^{3x-4}$.
The given options are $2,6,3,4$.
I don't know how to do this.
| Notice that, following Math Lover's hint, $\displaystyle 54\sqrt{2}=18^\frac{3}{2}$, because $54=(3\sqrt{2})^3$ and $18=(3\sqrt{2})^2$.
Therefore, $18^{4x-3}=(18^\frac{3}{2})^{3x-4}$, so $18^{4x-3}=18^{\frac{3}{2}(3x-4)}$.
Taking the base $18$ logarithm of each side gives
$ 4x-3=\frac{3}{2}(3x-4)$
$8x-6=9x-12$
$\boxed{x=6}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x$: $\sqrt{1-(\frac{\sin x}{n})^2}\sqrt{1-(n\sin (A-\arcsin (\frac{\sin x}{n})))^2}=\cos x \cos (A-\arcsin (\frac{\sin x}{n}))$ Solve for $x$ in terms of the constants $n$ and $A$.
$$\sqrt{1-(\frac{\sin x}{n})^2}\sqrt{1-(n\sin (A- \arcsin (\frac{\sin x}{n})))^2}=\cos x \cos (A- \arcsin (\frac{\sin x}{n}))$$
This gets really messy when I try and simplify and manipulate.
I notice that there is a pair of difference-of-two-squares on the left. But multiplying out just makes things even messier.
| Define $s := \sin x$ and $r := n^2-s^2$.
Then,
$$\sin\left(A -\operatorname{asin}\frac{s}{n}\right) = \sin A \cos\operatorname{asin}\frac{s}{n}-\cos A \sin \operatorname{asin}\frac{s}{n} = \frac{1}{n}\left(r \sin A - s \cos A\right)$$
$$\cos\left(A -\operatorname{asin}\frac{s}{n}\right) = \cdots = \frac{1}{n}\left(r \cos A + s \sin A\right)$$
Squaring your equation, and multiplying-through by $n^2$, gives
$$\left(n^2-s^2\right)\left(1 - \left( r \sin A - s \cos A \right)^2\right) = \left( 1 - s^2 \right)( r \cos A + s \sin A )^2$$
With a bit of symbol-crunching help from Mathematica, this simplifies to
$$(n^2-1) \sin A \cdot \left(-n^2 \sin A + 2 s^2 \sin A + 2 r s \cos A\;\right) = 0$$
For $\sin A \neq 0$ and $n \neq \pm 1$, we can eliminate $r$ to get
$$\left(\;n^2 ( 1 - \cos A ) - 2 s^2\;\right)\; \left(\;n^2 ( 1 + \cos A ) - 2 s^2\;\right) = 0$$
which leads to
$$\begin{align}
s^2 &= n^2 \frac{1-\cos A}{2} = n^2 \sin^2 \frac{A}{2} \quad\to\quad \sin x = \pm n \sin\frac{A}{2}\\[6pt]
s^2 &= n^2 \frac{1+\cos A}{2} = n^2 \cos^2 \frac{A}{2} \quad\to\quad \sin x = \pm n \cos\frac{A}{2}
\end{align}$$
Sanity checking for extraneous solutions to the original equation is left as an exercise to the reader.
| {
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If $\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$, then $\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $.
If $$\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$$ then how can I show that the sum of cosines of each angle ($x$, $y$, $z$) and sines of each angle sum up to zero? i.e. $$\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $$
I tried:
• Expanded using $\cos(A-B) = \cos A\cos B+\sin A\sin B $, but it did nothing.
After spending one hour to this problem, I thought that there must be a shorter and ideal way.
| Let $x-y=\alpha$, $y-z=\beta$. Thus, $z-x=-\alpha-\beta$ and we have
$$\cos\alpha+\cos\beta+\cos(\alpha+\beta)+\frac{3}{2}=0$$ or
$$2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+2\cos^2\frac{\alpha+\beta}{2}-1+\frac{3}{2}=0$$ or
$$4\cos^2\frac{\alpha+\beta}{2}+4\cos\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}+1=0,$$
which gives
$$4\cos^2\frac{\alpha-\beta}{2}-4\geq0$$ or
$$\sin\frac{\alpha-\beta}{2}=0.$$
Thus, also $$\left|\cos\frac{\alpha+\beta}{2}\right|=\frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometric Identity question
$x=2+\csc(\theta)$
$y=\dfrac14\tan(\theta)$
Eliminate $\theta$.
I tried extensively doing $x^2$ and $y^2$ and tried to equate but cannot manage to do it. This is the right method, however.
| Do it the hard way:
$x = 2 + \csc \theta = 2 + \frac 1 {\sin \theta}$
$x -2 = \frac 1{\sin \theta}$. (We can conclude that $\sin \theta \ne 0$ and that $x-2 \ge 1$ or $x - 2 \le -1$ and in any event, $x-2 \ne 0$.)
$\sin \theta =\frac 1{x-2}$
$y = \frac 14{\tan \theta} = \frac 14 \frac {\sin \theta}{\cos \theta}=\frac 14 \frac {\sin \theta}{\pm\sqrt {1-\sin^2 \theta}}$ (We can conclude $\sin \theta \ne \pm 1$).
$=\frac 14 \frac 1{(x-2)\sqrt{1 - \frac 1{(x-2)^2}}} $
$= \frac 14 \frac 1{\pm\sqrt{(x-2)^2 -1}}$
$16y^2 = \frac 1{x^2 -4x +3}$
| {
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Limit of $a_1=\sqrt{3}$, $a_2=\sqrt{3\sqrt{3}}$, $a_3=\sqrt{3\sqrt{3\sqrt{3}}}$... Find the limit of:
$a_1=\sqrt{3}$
$a_2=\sqrt{3\sqrt{3}}$
$a_3=\sqrt{3\sqrt{3\sqrt{3}}}$
...
By using induction, I found that the limit is 3 but it seems pretty strange to me, I thought it would go to a further number.
I just want to know if I'm right (proof-verification).
| 1)$3 > a_{i+1} > a_i > 1$.
Pf: by induction.
Base step: $a_1 = \sqrt {3} > 1; a_1 = \sqrt{3} < 3$.
Induction: If $3> a_i > 1$ then $a_{i+1} = \sqrt{3*a_i} = \sqrt 3\sqrt{a_i}$
So $3 = \sqrt{3}\sqrt{3} > \sqrt{3}\sqrt{a_i} =a_{i+1}=\sqrt{3}\sqrt{a_i} > \sqrt{3}*1 > 1$.
And $a_{i+1} = \sqrt{3}\sqrt{a_i} > \sqrt{a_i}\sqrt{a_i} = a_i$.
So $3 > a_{i+1} > a+i > 1$.
2) So $a_i$ is increasing and bounded so $\lim\limits_{n\to\infty} a_n$ exists and is less than or equal to $3$.
I honestly have no idea whatsoever why you would have possibly thought the limit would be more than $3$.
Let $\lim\limits_{n\to\infty} a_n = A$
Then $\lim\limits_{n\to \infty} \sqrt{3a_n} = \sqrt{3A}$
$\lim\limits_{n\to\infty}a_{n+1} = \sqrt{3A}$
$\lim\limits_{n+1\to\infty}a_{n+1} = \sqrt{3A}$
$A = \sqrt{3A}$
$A^2 = 3A$
$A =3$ or $A = 0$ but as $A > 1$ we have $A=3$.
| {
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Prove that $\lim\limits_{x \to 1} \frac{x+2}{x^2+1} = \frac{3}{2}$ I am required to prove the following limit using the epsilon-delta definition:
$$\lim_{x\to 1}\frac{x+2}{x^2+1}=\frac{3}{2}$$
So, ($\forall\epsilon>0)(\exists\delta>0)[0<\vert x-1\vert<\delta \implies\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert<\epsilon]$
Below is my working on getting the $\delta$ that I need:
$\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert\le\vert\frac{x+2}{x^2+1}\vert+\frac{3}{2}=\frac{\vert x+2 \vert}{\vert x^2+1 \vert}+\frac{3}{2}$ (Triangle Inequality)
But I know that $x^2+1\ge1$ for all values of $x$, so I can say that $\frac{\vert x+2 \vert}{\vert x^2+1 \vert}+\frac{3}{2}\le\vert x-1+3\vert+\frac{3}{2}\le \vert x-1\vert +3 + \frac{3}{2}<\delta+\frac{9}{2}$
So for every $\epsilon$, I choose $\delta=\epsilon-\frac{9}{2}$, then:$$0<\vert x-1\vert<\delta \implies\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert \le \vert\frac{x+2}{x^2+1}\vert+\frac{3}{2}\le\vert x+2\vert+\frac{3}{2}\le\vert x-1+3\vert+\frac{3}{2}\le \vert x-1\vert +3 + \frac{3}{2}<\delta+\frac{9}{2}=\epsilon$$
Am I doing this right? Is there something wrong to assume that $x^2+1\ge1$ for all values of $x$?
| $\large{\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert=\vert\frac{2x+4-3x^2-3}{2(x^2+1)}\vert=\vert\frac{(x-1)(-3x-1)}{2(x^2+1)}\vert}$ Can you take it further?
| {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite integral $\int \arctan^2 x dx$ in terms of the dilogarithm function I read about the integral
$$\int \arctan^2 x dx$$ in this old post: Evaluation of $\int (\arctan x)^2 dx$
By replacing
$$\arctan x = -\frac{i}{2}\left[\log(1+ix) - \log(i-ix)\right],$$
as suggested there, I ended up with this solution
$$\int\arctan^2 x dx = x\arctan^2x - \frac{1}{2}\log(1+x^2)\arctan x -\log 2 \arctan x + \mbox{Im}\left\{\mbox{Li}_2\left(\frac{1+ix}{2}\right)\right\} + K, \tag{1}\label{uno}$$
where, as usual, $\mbox{Li}_2(z)$ is the dilogarithm function
$$\mbox{Li}_2(z) = -\int_0^z \frac{\log(1-u)}{u}du=\sum_{k=1}^{+\infty}\frac{z^k}{k^2}.$$
Is this a correct development?
In that case, if I determine, using \eqref{uno}, the definite integral $\int_0^1 \arctan^2xdx$ I get the result
$$\int_0^1 \arctan^2xdx=\frac{\pi^2}{16}-\frac{3\pi}{8}\log 2 + \mbox{Im}\left\{\mbox{Li}_2\left(\frac{1+i}{2}\right)\right\}.$$
If I now compare this result with the one given in Definite Integral of $\arctan(x)^2$, i.e.
$$\int_0^1 \arctan^2xdx=\frac{\pi^2}{16}+\frac{\pi}{4}\log 2 - C,$$
where $C$ is the Catalan constant
$$C = \sum_{k=0}^{+\infty} \frac{(-1)^k}{(2k+1)^2},$$
I get the following expression:
$$\mbox{Im}\left\{\mbox{Li}_2\left(\frac{1+i}{2}\right)\right\} = \frac{5\pi}{8}\log 2 - C.$$
Is that reasonable?
| Landen's identity states that if $z \notin (1,\infty)$, then
$$
\operatorname{Li}_2(z) = -\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac12\ln^2(1-z).
$$
For $z:=(1+i)/2$, we have $z/(z-1) = -i$, thus
$$
\operatorname{Li}_2\left(\frac{i+1}{2}\right) = -\operatorname{Li}_2(-i)-\frac12\ln^2\left(\frac{1-i}{2}\right).
$$
It is well known that
$$
\operatorname{Li}_2(-i) = -iC -\frac{\pi^2}{48},
$$
where $C$ is Catalan's constant. For the logarithmic term we have
\begin{align}
\frac12\ln^2\left(\frac{1-i}{2}\right) &= -\frac{1}{32}\left(\pi - 2\ln(2)\cdot i\right)^2 \\
&= \frac{\ln^2 2}{8} - \frac{\pi^2}{32} + \frac{\pi}{8}\ln(2) \cdot i.
\end{align}
Finally
$$
\operatorname{Li}_2\left(\frac{i+1}{2}\right) = \frac{5\pi^2}{96} - \frac{\ln^2 2}{8} + \left(C - \frac{\pi}{8}\ln 2\right)\cdot i.
$$
You could find a more general approach in this answer, where there is a solution for all $z \in \mathbb{C}$, such that $\left|z/(z-1)\right|=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Show that the prime must be $13$ if $p\mid n^2+3$ and $p\mid (n+1)^2+3$ Let $p$ be a prime for which it exists a $n\in \mathbb{Z}$ with $p\mid n^2+3$ and $p\mid (n+1)^2+3$. I want to show that it must hold that $p=13$ and that there are infinitely many integers $n$ so that the above relations hold.
$$$$
We have that $p\mid n^2+3$ and $p\mid (n+1)^2+3 \Rightarrow p\mid n^2+2n+1+3$, so we get $p\mid (n^2+2n+1+3)-(n^2+3) \Rightarrow p\mid 2n+1$, right?
How could we continue?
| Keep using all the expressions you know that $p$ divides the same way you have to make more and simpler numbers you know that $p$ divides until you end up somewhere.
Next is that $p$ divides $n(2n+1) - 2(n^2 + 3) = n-6$. Finally, $p$ must divide $2n+1 - 2(n-6) = 13$.
Then it remains to show that there are infinitely many $n$ that actually make $13$ divide both $n^2 + 3$ and $(n+1)^2 + 3$. First, find some $n<13$ which works. We find $n = 6$, since $3\cdot 13 = 6^2 + 3$ and $4\cdot 13 = 7^2 + 3$. Then note that if $n$ works, so does $n+13$:
$$
(n+13)^2 + 3 = n^2 + 26n + 169 + 3 = 13(2n + 13) + n^2 + 3
$$
and
$$
(n+13 + 1)^2 + 3 = n^2 + 26n+2n + 169 + 26 + 1 + 3\\
= 13(2n + 13 + 2) + n^2 + 2n + 1 + 3 = 13(2n + 13 + 2) + (n+1)^2 + 3
$$
PS: Using modular arithmetic, this final $n+13$-statement is trivial. Modular arithmetic also helps speed up the search for the initial $n = 6$: We want $n^2\equiv (n+1)^2$ ($\equiv -3$, but that's irrelevant here). This becomes $0\equiv 2n+1$, which immediately tells us that $n \equiv 6$ is the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
AGM inequality for four variables Prove that for any $x,y,z,w \ge 0$
$$\sqrt[4]{xyzw} \le \frac{x+y+z+w}{4}$$
I tried using AGM and doing $xy \cdot zw = (\frac{xy + zw}{2})^2$ but couldn't get the proof
| $(\sqrt{x}-\sqrt{y})^2\ge0$ => $x+y-2\sqrt{x}\sqrt{y}\ge0$ => $\frac{x+y}{2}\ge\sqrt{xy}$
Using this, we get
$x+y\ge2\sqrt{xy}$ and $z+w\ge2\sqrt{zw}$
Adding both, we get
$x+y+z+w\ge2\sqrt{xy}+2\sqrt{zw}$ => $\frac{x+y+z+w}{2}\ge\sqrt{xy}+\sqrt{zw}$
Now using formula again on RHS, we get
$\sqrt{xy}+\sqrt{zw}\ge2(xyzw)^{\frac{1}{4}}$
Combining last two equations, we get the required answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate this $\int_{-1}^1 (\frac{1}{x}+1-\frac{\sqrt{1-x^2}}{x})\arctan\frac{2}{x^2}dx$ I dont know how to evaluate this improper integral:
$$I=\int_{-1}^1 \left(\frac{1}{x}+1-\frac{\sqrt{1-x^2}}{x}\right )\arctan\frac{2}{x^2}dx$$
At first I tried to do it in trigonometric substitution：$x=\sin t,dx=\cos tdt$
$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\csc t+1-\cot t\right)\arctan(2\csc ^2t)\cos tdt$$
$$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cot t+\cos t-\cot t\cos t\right)\arctan(2\csc ^2t)dt$$
I don't know what to do next,any help will be appreciated.
| Let me write a complete answer：
\begin{align}
I&=\int_{-1}^1\left(\frac{1}{x}+1-\frac{\sqrt{1-x^2}}{x}\right)\arctan\left(\frac{2}{x^2}\right)dx\\
&=\int_{-1}^1\arctan\left(\frac{2}{x^2}\right)dx\qquad(\text odd function)\\
&=\int_0^2\arctan\frac{2}{(x-1)^2}dx\\
&=\int_0^2\arctan\frac{x+(2-x)}{1-x(2-x)}dx\\
&=\int_0^2(\arctan x+\arctan(2-x))dx\\
&=2\int_0^2\arctan xdx\\
&=2(x\arctan x|_0^2-\int_0^2\frac{x}{1+x^2}dx)\\
&=4\arctan 2-\ln (1+x^2)|_0^2\\
&=4\arctan 2-\ln 5
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Concrete Mathematics Section 3.5: Floor / Ceiling sum manipulation Problem: To find the closed form of $\sum_{0 \le k \lt n}\lfloor \sqrt{k} \rfloor$
The text goes on solving as follows
$$
\begin{align}
\sum_{0 \le k \lt n}\lfloor \sqrt{k} \rfloor
&= \sum_{k,m \ge 0}m[k < n][m = \lfloor \sqrt{k} \rfloor]\\
&= \sum_{k,m \ge 0}m[k < n][m \le \sqrt{k} < m + 1]\\
&= \sum_{k,m \ge 0}m[k < n][m^2 \le k < (m + 1)^2]\\
&= \sum_{k,m \ge 0}m[m^2 \le k < (m + 1)^2 \le n] + \sum_{k,m \ge 0}m[m^2 \le k < n < (m + 1)^2]
\end{align}
$$
Let's assume that $n = a^2$ is a perfect square. Then the second sum is zsero and the first can be evaluated as
$$
\newcommand{\fallingfactorial}[1]{%
^{\underline{#1}}%
}
\begin{align}
\sum_{k,m \ge 0}m[m^2 \le k < (m + 1)^2 \le a^2]
&= \sum_{m \ge 0}m((m + 1)^2 - m^2)[m + 1 \le a]\\
&= \sum_{m \ge 0}m(2m + 1)[m < a]\\
&= \sum_{m \ge 0}(2m\fallingfactorial{2} + 3m\fallingfactorial{1})[m < a]\\
&= \sum_0^a (2m\fallingfactorial{2} + 3m\fallingfactorial{1}) \delta m\\
&= \frac{1}{6}(4a + 1)a(a - 1)
\end{align}
$$
The next part I don't understand
It says: In general we can let a = $\lfloor \sqrt{n} \rfloor$; then we merely need to add the term $a^2 \le k < n$, which are all equal to a, so they sum to $(n - a^2)a$. This gives the desired closed form,
$$
\sum_{0 \le k < n} \lfloor \sqrt{k} \rfloor = na - \frac{1}{3}a^3 - \frac{1}{2}a^2 - \frac{1}{6}a, \text{where } a = \lfloor \sqrt{n} \rfloor
$$
Why we are considering $a^2 \le k \lt n$, shouldn't we consider $a \le k \lt n$ as we have already calculated for $0 \le k \lt a$ before ?
| We consider the second sum in somewhat more detail which might help to clarify the situation.
We obtain
\begin{align*}
\color{blue}{\sum_{k,m\geq 0}}&\color{blue}{m[m^2\leq k<n<(m+1)^2]}\tag{1}\\
&=\sum_{k\geq 0}a[a^2\leq k<n<(a+1)^2]\tag{2}\\
&=\sum_{{k\geq 0}\atop{a^2\leq k <n}}a\tag{3}\\
&=\sum_{a^2\leq k <n}a=\sum_{k=a^2}^{n-1}a\tag{4}\\
&\color{blue}{=(n-a^2)a}
\end{align*}
Comment:
*
*In (1) we observe that if $n$ is fixed there is precisely one integer $m$ which fulfills the condition
$$[m^2\leq k<n<(m+1)^2]$$
So, we sum in fact only over $k$.
*In (2) we sum over $k$ only and denote the single value $m$ with $a$. We have
$$m=a=\lfloor \sqrt n\rfloor$$
The condition $[\color{blue}{a^2\leq k<n}<(a+1)^2]$ tells us that we have a non-zero contribution only if $a^2\leq k<n$.
*In (3) we rewrite the condition from (2) as index range.
*In (4) we do a small simplification.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int\frac1{x^2+bx+c}dx$ where $b$ and $c$ are real numbers and $d=4c-b^2$ Let's assume $d>0$ as well.
$$\int\frac1{x^2+bx+c}dx$$
where $b$ and $c$ are real numbers and $d=4c-b^2$
So my observation here is that this looks mightily similar to the standard integral that results in $\arctan u$, that integral usually solves from some form like $\frac1{u^2+1}$ where u is whatever has been substituted.
When the conditions are like this with the real numbers $b,c$. And $d$ being an expression. I'm not really sure where to even begin. Throws my integral intuition out the window!
Any ideas on how a solution is reached on this one? Much appreciated. :)
| Complete the square, we have
$$x^2 + bx + c = \left(x - \frac b 2\right)^2-\left(\frac b 2\right)^2 + c$$
$$= \left(x - \frac b 2\right)^2-\frac {b^2} 4 + c$$
Applying $d = 4c - b^2$,
$$\left(x-\frac b 2\right)^2 + \frac 1 4 \left(4c - b^2\right)=\left(x - \frac b 2\right)^2 +\frac d 4$$
So we have,
$$\int \frac 1 {\left(x - \frac b 2\right)^2 + \frac d 4} \mathrm dx$$
Now we can apply a substitution of
$$x - \frac b 2 = \frac{\sqrt d} 2\tan\theta$$
To evaluate the integral. You can probably take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Laplace transform of a normal distribution Let $X$ be a normal distribution with mean $0$ and variance $\sigma^2$. Then the Laplace transform of $X$ should be
$\int_0^\infty e^{-tx} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\{-\frac{x^2}{2\sigma^2}\}dx$
Then I try to do integral on
$\int_0^\infty \exp\{-\frac{x^2}{2\sigma^2}-tx\}dx=\int_0^\infty \exp\{-\frac{1}{2\sigma^2}[x^2+2\sigma^2tx]\}dx$
finishing the square we can get
$\int_0^\infty \exp\{-\frac{1}{2\sigma^2}(x+\sigma^2 t)^2\}dx$
Then I replace $x+\sigma^2 t$ by $u$
$\int_{\sigma^2t}^\infty \exp\{-\frac{1}{2\sigma^2}u^2\}du$
Then I am stuck at this step.
| Well, in the general case you're looking at:
$$\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right):=\int_0^\text{m}\frac{\text{n}}{\exp\left(\text{s}\cdot x+\pi\cdot\text{n}^2\cdot x^2\right)}\space\text{d}x\tag1$$
Substitute:
$$\text{u}:=\frac{2\cdot\pi\cdot\text{n}^2\cdot x+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\tag2$$
So, for the integral we get:
$$\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right):=\text{n}\cdot\frac{\exp\left(\frac{\text{s}^2}{4\cdot\pi\cdot\text{n}^2}\right)}{2\cdot\text{n}}\int_{\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}^{\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}\frac{2\cdot\exp\left(-\text{u}^2\right)}{\sqrt{\pi}}\space\text{d}\text{u}\tag2$$
This is a special integral:
$$\int_{\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}^{\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}\frac{2\cdot\exp\left(-\text{u}^2\right)}{\sqrt{\pi}}\space\text{d}\text{u}=\left[\text{erf}\left(\text{u}\right)\right]_{\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}^{\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}=$$
$$\text{erf}\left(\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)-\text{erf}\left(\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)\tag3$$
So, we end up with:
$$\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right)=\text{n}\cdot\frac{\exp\left(\frac{\text{s}^2}{4\cdot\pi\cdot\text{n}^2}\right)}{2\cdot\text{n}}\cdot\left\{\text{erf}\left(\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)-\text{erf}\left(\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)\right\}\tag4$$
Now, your task is to simplify and to take:
$$\lim_{\text{m}\to\infty}\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right)\tag5$$
Try to prove:
$$\lim_{\text{m}\to\infty}\text{erf}\left(\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)=1\tag6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that the roots of the quadratic equation
Show that the roots of the quadratic equation $$(b-c)x^2+2(c-a)x+(a-b)=0$$
are always reals.
My Attempt:
$$(b-c)x^2+2(c-a)x+(a-b)=0$$
Comparing above equation with $Ax^2+Bx+C=0$
$$A=b-c$$
$$B=2(c-a)$$
$$C=(a-b)$$
Now,
$$B^2-4AC=[2(c-a)]^2-4(b-c)(a-b)$$
$$=4(c-a)^2 - 4(ab-b^2-ac+bc)$$
$$=4c^2-8ac+4a^2-4ab+4b^2+4ac-4bc$$
$$=4(a^2+b^2+c^2-ab-bc-ca)$$
How do I proceed further?
| Certainly, using the AM-GM is easier. An alternative method is to consider cases. We want to prove the condition $D\ge 0$ for all $a,b,c\in R$:
$$D=(c-a)^2-(c-b)(a-c)\ge 0 \Rightarrow \\ (c-a)(c-a)\ge(a-b)(b-c)=(b-a)(c-b).$$
Case 1: $a\le b\le c:$
$$c-a\ge b-a; \ c-a\ge c-b.$$
Case 2: $a\le c\le b:$
$$(c-a)^2\ge 0 \ge (a-b)(b-c).$$
Case 3: $c\le a\le b:$
$$(c-a)^2\ge 0 \ge (a-b)(b-c).$$
Case 4: $c\le b\le a:$
$$(c-a)^2=(a-c)^2 \ge (a-b)(b-c).$$
Case 5: $b\le a\le c:$
$$(c-a)^2\ge 0 \ge (a-b)(b-c).$$
Case 6: $b\le c\le a:$
$$(c-a)^2\ge 0 \ge (a-b)(b-c).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Two consecutive numbers are removed from the progression 1, 2, 3..n. The arithmetic mean of the remaining numbers is 105/4. What is the value of n? Two consecutive numbers are removed from the progression 1, 2, 3..n. The arithmetic mean of the remaining numbers is 105/4. What is the smallest value of n?
?What are the two numbers removed?
Now many answers are given but they are given in a hit and trial method.
Is there any other method to solve this problem?
| $\sum _{k=1}^n k= \dfrac{1}{2} n (n+1)$
Subtract $x+(x+1)$. The sum is then $\dfrac{1}{2} n (n+1) -2x-1$
The mean is then $\dfrac{\frac{1}{2} n (n+1) -2x-1}{n-2}=\dfrac{105}{4}$
$x=\dfrac{1}{8} \left(2 n^2-103 n+206\right)$ and $0<x<n$
$x>0 \to 2 n^2-103 n+206>0$ which means $n\ge 50$
And $x<n \to \dfrac{1}{8} \left(2 n^2-103 n+206\right)<n $ which means $2\le n\le 53$
so we have $50\le n\le 53$ but the only value that gives an integer $x$ is $n=50$
Finally the solution is $x=7$ when $n=50$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to prove that for all $|z|<1$, $|(1-z)e^z-1|\leq |z|^2$?
For all $|z|<1$, $$|(1-z)e^z-1|\leq |z|^2.$$
Applying Taylor's expansion (does it hold for all $z$?), we have
$$\begin{aligned}|(1-z)e^z-1|&=\left|(1-z)\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots \right)-1~\right|\\&=\left|\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots \right)-\left(z+z^2+\frac{z^3}{2!}+\frac{z^4}{3!}+\cdots\right)-1~\right|\\
&=\left|~\left(\frac{1}{2!}-\frac{1}{1!}\right)z^2+\left(\frac{1}{3!}-\frac{1}{2!}\right)z^3+\cdots~\right|\end{aligned}$$
How should I go on?
| In this particular case, one does not need to use Weierstrass's canonical factors. By the triangle inequality,
$$
\left|~
\left(\frac{1}{2!}-\frac{1}{1!}\right)z^2
+\left(\frac{1}{3!}-\frac{1}{2!}\right)z^3
+\left(\frac{1}{4!}-\frac{1}{3!}\right)z^4
+\cdots\ \right|\\
\leqslant\frac12|z|^2+
\left|\left(\frac{1}{3!}-\frac{1}{2!}\right)z^3
+\left(\frac{1}{4!}-\frac{1}{3!}\right)z^4
+\cdots\right|
$$
Thus it suffices to show that
$$
\left|\left(\frac{1}{3!}-\frac{1}{2!}\right)z^3
+\left(\frac{1}{4!}-\frac{1}{3!}\right)z^4
+\cdots\right|\leqslant\frac12|z|^2
$$
Factoring out $z^3$ on the left hand side, applying the triangle inequality again, one has
$$
|z|^3\left[\left|\frac{1}{3!}-\frac{1}{2!}\right|
+\left|
\frac{1}{4!}-\frac{1}{3!}\right|
+\cdots\right]=|z|^3\cdot\frac{1}{2}
$$
Since $|z|<1$, we are done.
One can justify the "$\cdots$" part by a limiting argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Confused on how to solve $\lim_{x\to 0} \frac{(1+mx)^n-(1+nx)^m}{x^2}$ and $\lim_{x\to 1} (\frac{m}{1-x^m}-\frac{n}{1-x^n})$ I'm having trouble solving these two problems using only elementary math. Find:
$$\lim_{x\to 0} \frac{(1+mx)^n-(1+nx)^m}{x^2}$$
and
$$\lim_{x\to 1} (\frac{m}{1-x^m}-\frac{n}{1-x^n})$$
My confusion may be due to the fact that I don't remember encountering these type of problems before. Do any of you recommend books that can help me improve my problem solving skills via elementary math or on different methods of problem solving (like the two I posted)?
| Expanding the numerator by Binomial theorem, we get
\begin{align*}
(1+mx)^n-(1+nx)^m &= \left(1+ nmx + \binom{n}{2}m^2x^2 + \cdots \right) - \left(1+ mnx + \binom{m}{2}n^2x^2 + \cdots \right) \\
&= \frac{nm(n-m)}{2}x^2 + \cdots
\end{align*}
Hence the first limit is $\frac{nm(n-m)}{2}$
For the second limit, we need
\begin{align*}
\lim_{x \rightarrow 1} \frac{m(1-x^n)-n(1-x^m)}{(1-x^n)(1-x^m)} &= \lim_{x\rightarrow 1} \frac{(1-x)(m(1+x+\cdots +x^{n-1}) - n(1+x+\cdots+x^{m-1}))}{(1-x)^2(1+x+\cdots+x^{m-1})(1+x+\cdots+x^{n-1})}\\
&= \lim_{x \rightarrow 1} \frac{(m(1+x+\cdots +x^{n-1}) - n(1+x+\cdots+x^{m-1}))}{(1-x)(1+x+\cdots+x^{m-1})(1+x+\cdots+x^{n-1})}\\
&= \frac{1}{mn}\lim_{x \rightarrow 1} \frac{(m(1+x+\cdots +x^{n-1}) - n(1+x+\cdots+x^{m-1}))}{(1-x)}\\
&= \frac{1}{mn}\lim_{x \rightarrow 1}\frac{m[(1-x)(-x^{n-2}-2x^{n-3} - \cdots - (n-1)) + n] - n[(1-x)(-x^{m-2}-2x^{m-3} - \cdots - (m-1)) + m]}{1-x} \\
&= \frac{1}{mn}\left[m\left(-\frac{n(n-1)}{2}+n\right) - n\left(-\frac{m(m-1)}{2}+m\right) \right]\\
&=\frac{m-n}{2}
\end{align*}
In the above, we have used this:
$$1+x+x^2+\cdots +x^{n-1} = (1-x)(-x^{n-2}-2x^{n-3}-3x^{n-4} - \cdots - (n-2)x - (n-1)) + n $$
obtained by the usual division of $1+x+ \cdots +x^{n-1}$ by $1-x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find k if the two polynomials have one common root The equations $x^2−4x+k=0$ and $x^2+kx−4=0$, where $k$ is a real number, have exactly one common root. What is the value of $k$?
I know the answer but can it be done with the relation of roots.
$a$ and $b$ are roots of equation 1 and $a$ and $c$ are roots of equation 2. So the relations are :-
*
*$a + b = 4$
*$ab = k$
*$a + c = (-k)$
*$ac= (-4)$
I tried doing all the stuff but couldn't get it. Can we find it using these 4 equations?
| If $a+b=4$, then $a^2 + ab = 4a$. So
$$a^2-4a+k=0. \tag 1$$
If $a+c=-k$, then $a^2+ac=-ka$. So
$$a^2 + ka -4 = 0. \tag 2$$
So $0 = (a^2 + ka -4)-(a^2-4a+k)=(k+4)a-(k+4)=(k+4)(a-1)$
So $k=-4$ or $a=1$
If $k=-4$, then both trinomials become $x^2-4k-4$ and so they have two roots in common: $2+\sqrt 8$ and $2-\sqrt 8$.
If $a=1$, then plugging $a=1$ into equations $(1)$ and $(2)$ yields the equations
$$1-4+k=0 \qquad \text{and} \qquad 1+k-4=0$$
and both equations have the single solution $k=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An equation relating sine of angles in a triangle I wanted to solve the following problem.
In $\triangle ABC$ we have $$\sin^2 A + \sin^2 B = \sin^2 C + \sin A \sin B \sin C.$$ Compute $\sin C$.
Since it's an equation for a triangle, I assumed that $\pi = A + B + C$ would be important to consider.
I've tried solving for $\sin C$ as a quadratic, rewriting $\sin C = \sin (\pi - A - B)$, but nothing seemed to work.
How does one approach this problem? Any help would be appreciated.
The answer is
$$\frac{2\sqrt{5}}{5}$$
| using the Theorem of sines we get
$$\sin^2(C)\left(\frac{a^2+b^2}{c^2}\right)=\sin^2(C)+\frac{ab}{c^2}\sin^3(C)$$
since $$\sin(C)\neq 0$$ we obtain
$$\sin(C)=\frac{a^2+b^2-c^2}{ab}$$ using $$2\cos(C)=\frac{a^2+b^2-c^2}{ab}$$ we get $$\tan(C)=2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I solve this equation for real numbers? How can I solve this equation for real numbers?
$$(x+2)^4+x^4=82.$$
I tried $(x+2)^4+x^4-82=
2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$
It is very difficult to continue.
| $$u=x+1$$
$$(u+1)^4+(u-1)^4=82$$
$$u^4+4u^3+6u^2+4u+1+u^4-4u^3+6u^2-4u+1=82$$
$$2u^4+12u^2=80$$
$$u^4+6u^2-40=0$$
$$(u^2+10)(u^2-4)=0$$
$$u^2=4$$
$$u=\pm2$$
$$x=-3\lor x=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to simplify this fraction using algebraic logic? Well, I need to simplify this fraction using algebraic logic
$$\frac{(2^4+2^2+1)(4^4+4^2+1)(6^4+6^2+1)(8^4+8^2+1)(10^4+10^2+1)}{(3^4+3^2+1)(5^4+5^2+1)(7^4+7^2+1)(9^4+9^2+1)(11^4+11^2+1)}$$
For resolve this, I think I will use this concept.
If you have:
$$\frac{(2^2)(4^2)(6^2)(8^2)(10^2)}{(3^2)(5^2)(7^2)(9^2)(11^2)}$$
Simplifying, you must get:
$$\frac{3840^2}{10395^2}$$
Because all are terms with same exponent, like:
$$(2^2)(2^2) = 4^2 $$
Aplying this, I get:
$$\frac{3840^4+3840^2+1}{10395^4+10395^2+1}$$
But, its not the same, I resolve the first fraction and I get: 0.0225539...
The second one result in: 0.0186220...
This is embarrassing, probably i´m complety lost, but I need help
| Factoring this equation can allow you to simplify significantly.
$$n^4+n^2+1=(n^2-n+1)(n^2+n+1)$$
See if you recognize a pattern after applying this for $n=2,3,4...$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How would I find the 3×3 matrix A that satisfies the following? $$A\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}2\\-5\\ 1\end{pmatrix}$$
$$A\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}-1\\2\\-5\end{pmatrix}$$
$$A\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}-4\\-1\\5\end{pmatrix}$$
I find this problem particularly confusing as you can't find the inverse of any of these 3×1 matrices.
| Since $A$ is being multiplied by the three standard basis vectors of $\Bbb R^3$, just concatenate the result column vectors together:
$$A=\begin{bmatrix}2&-1&-4\\-5&2&-1\\1&-5&5\end{bmatrix}$$
Verify that this satisfies the three given equations.
| {
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"url": "https://math.stackexchange.com/questions/2442751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the remainder when $f(x)$ divided by $(x^2 + x + 1)(x+1)$. When a polynomial $f(x)$ is divided by $x^2 + x + 1$ and $(x+1)^2$, the remainder are $x+5$ and $x-1$ respectively. Find the remainder when $f(x)$ is divided by $(x^2 + x + 1)(x+1)$.
First, I let the remainder be $Ax^2 + Bx +C$, then I try to find the values of $A$, $B$ and $C$. There is $3$ unknowns so we need three equations but I can only get two equations.
| Write $$ f(x)=k(x)(x^2+x+1)+x+5$$
and $$ f(x)=q(x)(x+1)^2+x-1 \Longrightarrow f(-1) = -2$$
So $$-2 = f(-1) = k(-1)\cdot 1 +4 \Longrightarrow k(-1) = -6$$
thus
$$ k(x) = p(x)(x+1)-6 $$ and finally
\begin{eqnarray}
f(x) &= &(p(x)(x+1)-6)(x^2+x+1)+x+5 \\
&=& p(x)(x+1)(x^2+x+1)-6x^2-5x-1
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the only integer solution to the equation $x^2+y^2=3z^2$ is $x=y=z=0$. I started by saying that $x^2$ and $y^2$ must be divisible by 9 because they are perfect squares and must add to a multiple of three. $z^2$ also must be a multiple of 9, otherwise the equation wouldn't be correct. $3z^2$ will have an extra multiple of $3$, so the equation is not true for values other than $x=y=z=0$.
Is my reasoning correct? And would this logic apply for $x^2+y^2=5z^2$?
| Suppose $x^2+y^2=3z^2$ is a minimal non zero solution. Now consider $x$ and $y$ modulo $3$ ($1^2 \equiv 2^2 \equiv 1 \pmod{3}) $. Both must be multiples of $3$ so $x=3a$ and $y=3b$. this give $3a^2+3b^2=z^2$ so $z$ must be a multiple of $3$. $z=3c$ and we have $a^2+b^2=3c^2$ which contradict the hypothesis that $x^2+y^2=3z^2$ is a minimal non zero solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right)$ I was studying for some quizzes when wild question appears. It looks like this:
Find $\prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right)$
My work
I think it's a repeated multiplication of the expression $1-\frac{1}{k^2}$. It looks like this:
$$\prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right) = \left(1-\frac{1}{(2)^2}\right)\left(1-\frac{1}{(3)^2}\right)\left(1-\frac{1}{(3)^2}\right)\left(1-\frac{1}{(4)^2}\right).....$$
I barely had any experience evaluating these new summation...How do evaluate $\prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right)$?
| \begin{eqnarray*}
&=&\prod_{k=2}^{+\infty}\left(1-\frac1{k^2}\right)\\
&=&\prod_{k=2}^{+\infty}\frac{(k-1)(k+1)}{k^2}\\
&=&\frac{1\cdot3}{2\cdot 2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot 4}\cdot\ldots\\
&=&\lim_{k\to+\infty}\frac{k+1}{2\cdot k}\\
&=&\frac12
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Tricky dice problem We simultaneously roll two dice in one round until at least one of them shows 3 once. What is the expected number of rolls needed?
I've calculated probability of getting 3 on the 1st roll
I am really bad in math, sorry
$= (1/6)^2 + 2(5/6 * 1/6) = 11/36$
Then on the second
$= (1 - p(\text{on the 1st roll})) \cdot (11/36)$
Third
$= (1 - p(\text{on the 2nd})) \cdot (11/36)$
And so on, then I've just calculated the expected number of this distribution and reached the conclusion that I'll need at least $4$ rolls, since we operate only with integers.
I am really confused, if this approach is correct?
Or should we use the geometric distribution and to get the expected number of rolls we just have to divide $1$ over probability of success, i.e. $11/36$?
And consequently how can we find the minimum number of dice needed to get $3$ at least once on at least one dice if we can simultaneously roll them no more than $2$ times?
Thank you in advance.
| $\frac {11}{36} + (2)(\frac {25}{36})(\frac {11}{36}) + (3)(\frac{25}{36})^2(\frac {11}{36})+ \cdots +(n+1)(\frac {25}{6})^n(\frac {11}{36})\cdots$
That is a $3$ on the first roll, or no $3$ followed by a $3,$ up to $n$ no 3's followed by a $3.$
You have something that resembles a geometric series. It is not quite, but I forget what it is called.
$E[Y] = \frac {11}{36}\sum_\limits{n=0}^{\infty} (n+1)(\frac{25}{36})^n$
Now look what happens when we subtract $\frac {25}{36} E[Y]$ from $E[Y]$
$E[Y] -\frac {25}{36} E[Y]= \frac {11}{36} + (2)(\frac {25}{36})(\frac {11}{36})-(\frac {25}{36})(\frac {11}{36}) + (3)(\frac{25}{36})^2(\frac {11}{36}) - (2)(\frac{25}{36})^2(\frac {11}{36})+ \cdots +(n+1)(\frac {25}{6})^n(\frac {11}{36})-(n)(\frac {25}{6})^n(\frac {11}{36})\cdots$
$E[Y](1 -\frac {25}{36}) = \frac {11}{36}\sum_\limits{n=0}^{\infty}(\frac{25}{36})^n$
and that is a more ordinary geometric series. And we perform a similar operation and we get.
$E[Y](1 -\frac {25}{36})^2 = \frac {11}{36}\\
E[Y] = \frac {36}{11}$
Slightly anti-climactic, isn't it?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$ Prove that $\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$
My steps
A= $\frac{a}{b+c}$,B= $\frac{b}{c+a}$ &C=$\frac{c}{a+b}$
A.M$\ge$H.M
$\frac{A+B+C}{3} \ge \frac{3ABC}{AB+BC+AC}$
I am struck after this step
${A+B+C} \ge \frac{9}{\frac{a}{c}+\frac{c}{a}+\frac{b}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}}$
| set $$x=b+c,y=c+a$$ and $$z=a+b$$ then we get
$$\frac{-x+y+z}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\geq \frac{3}{2}$$
and this is
$$\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}-3\geq 3
$$
and then use $$\frac{A}{B}+\frac{B}{A}\geq 2$$ for $$A,B>0$$
| {
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"url": "https://math.stackexchange.com/questions/2451455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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$\lim_{x\to\infty} x^2\operatorname{cos}\left(\frac{3x+2}{x^2}-1\right)$ without using L’Hôpital. Can I calculate the following limit without applying L'Hôpital?
$$\lim_{x\to\infty} x^2\operatorname{cos}\left(\frac{3x+2}{x^2}-1\right)$$
With L'Hôpital it gives me as a result $\frac{-9}{2}$.
| I think he want to type and solve this $$\lim_{x\to\infty} x^2(cos(\frac{3x+2}{x^2})-1)=\frac{-9}{2}$$ this can be solve by multiplying conjugate
$$\quad{\lim_{x\to\infty} x^2(cos(\frac{3x+2}{x^2})-1)=\\
\lim_{x\to\infty} x^2(cos(\frac{3x+2}{x^2})-1)\times \frac{cos(\frac{3x+2}{x^2})+1}{cos(\frac{3x+2}{x^2})+1}=\\
\lim_{x\to\infty} x^2(-\sin^2(\frac{3x+2}{x^2}))\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\}$$
as $x\to \infty \implies \sin(\frac{3x+2}{x^2})\to 0 \implies \sin(\frac{3x+2}{x^2})\sim \frac{3x+2}{x^2}\\$so
$$\quad{\lim_{x\to\infty} x^2(-\sin^2(\frac{3x+2}{x^2}))\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\
\lim_{x\to\infty} -x^2(\frac{3x+2}{x^2})^2\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\
\lim_{x\to\infty} -(\frac{(3x+2)^2}{x^2})\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\\lim_{x\to\infty} -(\frac{9x^2+12x+4}{x^2})\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\
\lim_{x\to\infty} -(\frac92)\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\
-(\frac92)\times \frac{1}{cos(0)+1}=\\\frac{-9}{2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the Galois group of $x^4-5x^2+6$ over $\mathbb{Q}$
Find the Galois group of $x^4-5x^2+6$ over $\mathbb{Q}$
$$x^4-5x^2+6 = a^2-5a+6 = (a-3)(a-2) = (x^2-3)(x^2-2)$$
The roots are $\pm \sqrt{3}, \pm \sqrt{2}$, and the Galois group is made of the automorphisms that mixes the roots of $x^2-3$ and $x^2-2$. The possibilities of automorphisms:
$$e: \sqrt{2}\to\sqrt{2}, \sqrt{3}\to\sqrt{3}\\a: \sqrt{2}\to-\sqrt{2}, \sqrt{3}\to\sqrt{3}\\b: \sqrt{2}\to\sqrt{2}, \sqrt{3}\to-\sqrt{3}\\c: \sqrt{2}\to-\sqrt{2}, \sqrt{3}\to-\sqrt{3}$$
Is this the Galois group?
| The roots of $x^4-5x^2+6$ are $\pm\sqrt2$ and $\pm\sqrt3$ as you have pointed out.
Let $r$ be an automorphism of $\Bbb Q(\sqrt2,\sqrt3)$ that fixes $\Bbb Q$.
Then, $r(\sqrt2)^2 = r(2)=2$, so $r(\sqrt2)=\pm\sqrt2$.
Similarly, $r(\sqrt3)=\pm\sqrt3$.
Since an automorphism of a field is completely determiend by its generators, there are only $4$ possible automorphisms. There are three involutions, i.e. three elements with order $2$, so the group is $V_4$.
| {
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Recurrence relation problem: Given $a_n = \left \lfloor \sqrt{a_1 + a_2 + \cdots + a_{n-1}} \right \rfloor$, find $a_{1000}$. I've been struggling with this problem for some time.
Let there be a recurrent sequence ${a_n}$ such that $a_1 = 1$ and $a_n = \left \lfloor\sqrt{a_1 + a_2 + \cdots + a_{n-1}}\right \rfloor$ for all $n > 1$. Find $a_{1000}$.
I've tried to plug in different $n$ and have noticed that the square root makes it grow slowly, and with bigger $n$ the terms become more distinct (different integers), but that still hasn't gotten me close to solving for $a_{1000}$.
Update: I've solved it. I will post my solution in a bit. The hints you guys have given me have proved valuable, and I appreciate the caution with which you guys go about giving me a full solution based on my lack of steps taken.
| If we write down some of the first numbers, we get: $(1,1,1,1,2,2,2,3,3,4,4,4,5,5,6,6,7,7,8,8,8,9,9,…)$. From this it is visible that number $1$ appears four times, while all other numbers appear twice, thrice at most. We will further define that numbers that appear thrice are the powers of $2$, starting with $2^1$. All other numbers therefore appear twice.
Suppose we wrote down the beginning of the sequence until the first occurrence of the number $n \ (n>1)$, and that the sequence visibly behaves as defined.
Let $k$ be the largest integer satisfying $2^{k}$. Then the sum of all the values is:
$s_1 = (1+2+...+n) + (1+2+...+n-1) + (1+2+2^2+...+2^k) +1 = n^2 + 2^{k+1}$
since $2^{k+1} = 2 \times 2^k < 2n < 2n +1 = (n+1)^2 - n^2$. We have $s_1 < (n+1)^2$ and the following member is therefore $\lfloor\sqrt{s_1}\rfloor = n$.
Now we define the next member in this order, and the sum is now:
$s_2 = s_1 + n = n^2 + n + 2^{k+1}$,
so if $2^{k+1} < n+1$, then $s_2 < (n+1)^2$ and the next member is $n$.
However, $k$ is the largest integer satisfying $2^k < n$, so it's true that $n\le 2^{k+1}$. The previous situation therefore occurs iff $2^{k+1} = n$.
When $n$ isn't a power of two, the next member is $n+1$, because $n+1\le 2^{k+1} < 2n < 3n +4$, which follows from $(n +1)^2 \le n^2 + n + 2^{k+1} < (n+2)^2$.
Now we only need to show that if $n = 2^{k+1}$, then after three occurrences of $n$, an $n+1$ will follow.
Now the sum is:
$s_3 = s_2 + n = n^2 + 2n + 2^{k+1} = n^2 + 3n$, and we immediately get the inequality $(n+1)^2 < s_3 < (n+2)^2$.
Now we finally see that $500 = a_{1010} = a_{1009}$. Thus, $a_{1000} = \boxed{495}$.
| {
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Sum of series $\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$
The problem is to find the sum of series $$\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$$
to infinite terms, if $|x|<1$
I tried taking $\frac{x}{1-x^2}$ outside but could not solve it.
How to proceed further?
| If you combine the first two terms, you should end up with $$\frac{x(1+x^2)}{1-x^4}+\frac{x^2}{1-x^4}=\frac{x+x^2+x^3}{1-x^4}$$
If you combine the first three terms, you should end up with $$\frac{(x+x^2+x^3)(1+x^4)}{1-x^8}+\frac{x^4}{1-x^8}=\frac{x+x^2+x^3+x^4+x^5+x^6+x^7}{1-x^8}$$
Following this pattern, you will find out that the sum of the first $n$ terms is $$\frac{x}{1-x^{2^n}}\sum_{i=0}^{2^n-2}x^i$$
And you should be able to evaluate this sum, because it is the finite sum of a geometric progression. Taking the limit as $n\to\infty$ should yield your final answer of $$\frac{x}{1-x}$$
| {
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Prove $\binom{2n+2}{n+1} = \binom{2n}{n+1} + 2\binom{2n}{n} + \binom{2n}{n-1}$ I need some help showing that these are equivalent. I made a couple attempts to get this right but so far the following work is as far as I've gotten.
Here is the question in its entirety:
Let n be a natural number. Give a combinatorial proof of the following:
$\binom{2n+2}{n+1} = \binom{2n}{n+1} + 2\binom{2n}{n} + \binom{2n}{n-1}$
My first impression was that I could use the $\binom{n}{k} = \binom{n}{n-k}$ identity to make the term "$\binom{2n}{n-1}$" equal "$\binom{2n}{n+1}$" so I could simplify the equation a bit. I now have:
$\binom{2n+2}{n+1} = 2\binom{2n}{n+1} + 2\binom{2n}{n} = 2[\binom{2n}{n+1} + \binom{2n}{n}]$
Afterwards i figured I could use pascals identity $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ to get rid of yet more terms resulting in:
$\binom{2n +2}{n+1} = 2\binom{2n+1}{n+1}$
This is where I realize I made a pretty big mistake and got stuck, I'm fairly sure I made a maths error somewhere but I am unsure where after my third attempt at this.
| The question asks for combinatorial proof. But going by your way, ie using Pascal's identity it can be done as follows:
$$\binom{2n}{n} + \binom{2n}{ n-1} = \binom{2n+1}{n}$$
And $$\binom{2n}{n} + \binom{2n}{n+1} = \binom{2n+1}{n+1}$$
so adding both the equations,
$$\begin{align}
\binom{2n}{ n-1} +2\binom{2n}{n}+ \binom{2n+1}{n} &= \binom{2n+1}{n} + \binom{2n+1}{n+1}\\
&= \binom{2n+2}{n+1}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Trying to solve $2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)=3$ without seeing the obvious solution
Determine all real values of $x$ such that $$\log_{5x+9}(x^2+6x+9)+\log_{x+3}(5x^2+24x+27)=4$$
Taken from Waterloo 2012: Link
What I tried:
$$2\log_{5x+9}(x+3)+\log_{x+3}\left((x+3)(5x+9)\right)=4$$
$$2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)=3$$
Then I was stuck and I graphed it:
It was clearly $0$:
$$2\log_{9}(3)+\log_{3}(9)=3$$
But how could I do it without seeing the "zero"? Is there a way?
| To complete @MrYouMath
$2u+\frac{1}{u} = 3$
where $u = \log_{5x+9} x+3$
$u = 1$ or $u = \frac{1}{2}$
Case 1: $u = 1$
$x+3 = 5x + 9$
$x = -\frac{3}{2}$
Case 2 : $u = \frac{1}{2}$
$(x+3)^2 = 5x+9$
$ x^2+x = 0$
$x = 0$ or $x= -1$
All three values are greater than $x = -3$ and $x = -\frac{9}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Value of $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$ if $z^4-2z^3+z^2+z-7=0$ for $z=\alpha$, $\beta$, $\gamma$, $\delta$ Let $\alpha$, $\beta$, $\gamma$, $\delta$ be the roots of $$z^4-2z^3+z^2+z-7=0$$ then find value of $$(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$$
Are Vieta's formulas appropriate?
| There is no need to use Vieta's formulas. Let
$$f(z)=z^4-2z^3+z^2+z-7=(z-\alpha)(z-\beta)(z-\gamma)(z-\delta).$$
Then, since $(i-a)(-i-a)=-i^2+a^2=1+a^2$, it follows that
$$(\alpha^2+1)(\beta ^2+1)(\gamma^2+1)(\delta^2+1)=f(i)f(-i)=|f(i)|^2=|-7+3i|^2=49+9=58.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 1
} |
Use delta-epsilon to prove continuity of a function $$f(x,y) =\begin{cases} x\cos(1/y) & \text{ if }y\neq 0,\\ 0 &\text{ if }x= 0.\end{cases}$$
How would I show this function is continuous (using delta-epsilon) on any point $(x,y)$ given $y$ is not $0$, as well as at the origin?
So far I have that for all $\epsilon$ > 0 there exist $\delta$ > 0 s.t. $$|f(x,y)-0| < \epsilon$$
whenever $0 < \|(x,y)-(0,0)\| < \delta$.
Now, when $y$ is not $0$ we have $|x\cos(1/y)| \le |x||cos(1/y)|\cdots$
As mentioned $|\cos(1/y)|\le 1$ but this is the point at which I get stuck and have no idea how to proceed further.
Any pointers will be greatly appreciated!
| denote the point as $(x_0,y_0)$, obviously, you need to consider two cases
Case1: $x_0=0$.
then for any $(x, y)$:
$$
|f(x,y)-f(x_0,y_0)| = \mid x\cos \dfrac {1}{y} -0\mid=\mid x\cos \dfrac {1}{y}\mid\leq \mid x\mid \cdot \mid\cos \dfrac {1}{y}\mid \leq \mid x \mid
$$
because $\mid\cos \dfrac {1}{y}\mid\leq 1$. choose $\delta=\epsilon $, for all $\|(x,y)-(x_0,y_0)\|<\delta$, since $x_0=0$, we get
$$
\mid x \mid<\delta =\epsilon
$$
which means $|f(x,y)-f(x_0,y_0)|<\epsilon$.
Case2: $x_0\neq 0$.
Then
$$
\begin{align}
|f(x,y)-f(x_0,y_0)| &= \mid x\cos \dfrac {1}{y} -x_0\cos \dfrac {1}{y_0}\mid\\
&=\mid (x-x_0)(\cos \dfrac {1}{y} -\cos \dfrac {1}{y_0})+x_0 (\cos \dfrac {1}{y} -\cos \dfrac {1}{y_0})+(x-x_0)\cos \dfrac {1}{y_0}\mid\\
&\leq \mid x-x_0\mid \mid \cos \dfrac {1}{y} -\cos \dfrac {1}{y_0}\mid +\mid x_0 \mid \mid \cos \dfrac {1}{y} -\cos \dfrac {1}{y_0}\mid +\mid x-x_0\mid \mid \cos\dfrac {1}{y_0}\mid\\
&\leq \mid x-x_0\mid \mid \cos \dfrac {1}{y} -\cos \dfrac {1}{y_0}\mid +\mid x_0 \mid \mid \cos \dfrac {1}{y} -\cos \dfrac {1}{y_0}\mid +\mid x-x_0\mid\\
\end{align}
$$
the second line uses a simple identity, while the third line uses triangular inequality.
Now, by trigonometric identities
$$
\begin{align}
\mid \cos \dfrac {1}{y} -\cos \dfrac {1}{y_0} \mid &= \mid -2\sin (\dfrac {1}{2y}-\dfrac {1}{2y_0})\sin (\dfrac {1}{2y}+ \dfrac {1}{2y_0})\mid \\
&= 2 \mid \sin (\dfrac {1}{2y}- \dfrac {1}{2y_0})\mid \mid \sin (\dfrac {1}{2y}+ \dfrac {1}{2y_0})\mid\\
&\leq 2 \mid \sin (\dfrac {1}{2y}- \dfrac {1}{2y_0})\mid \\
&\leq 2 \mid \dfrac {1}{2y}- \dfrac {1}{2y_0}\mid\\
&=\mid \dfrac {y-y_0}{yy_0} \mid
\end{align}
$$
The third line comes from $\mid \sin (\dfrac {1}{2y}+ \dfrac {1}{2y_0})\mid\leq 1$, while the fourth line uses the relationship that $\mid \sin(x)\mid < \mid x\mid $ for all $x$.
choose $\delta<\dfrac {\mid y_0 \mid }{2}$, then if $\|(x,y)-(x_0,y_0)\|<\delta$, we would get $\mid y-y_0\mid<\delta$ and $\mid y\mid>\dfrac {\mid y_0 \mid }{2}$. plug it in, then
$$
\mid \cos \dfrac {1}{y} -\cos \dfrac {1}{y_0} \mid\leq \dfrac {2\delta}{y^2_0}
$$
note that $\|(x,y)-(x_0,y_0)\|<\delta$ also implies $\mid x-x_0\mid<\delta$, now we can handle
$$
\begin{align}
|f(x,y)-f(x_0,y_0)| &\leq \mid x-x_0\mid \mid \cos \dfrac {1}{y} -\cos \dfrac {1}{y_0}\mid +\mid x_0 \mid \mid \cos \dfrac {1}{y} -\cos \dfrac {1}{y_0}\mid +\mid x-x_0\mid\\
&\leq \dfrac {2\delta^2}{y^2_0}+\dfrac {2\delta \mid x_0\mid}{y^2_0}+\delta\\
&\leq \dfrac {\delta \mid y_0\mid}{y^2_0}+\dfrac {2\delta \mid x_0\mid}{y^2_0}+\delta\\
&=\delta (\dfrac {1}{\mid y_0\mid}+\dfrac {\mid x_0\mid}{y_0^2} +1)\\
&< \epsilon
\end{align}
$$
choose $\delta=\min \{\dfrac {\mid y_0\mid }{2}, \dfrac {\epsilon}{\mid y_0\mid+\dfrac {\mid x_0\mid}{y_0^2} +1} \}$, done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ for $x,y,z \in \Bbb{R}$ and $xyz > 0 $.
I know that i can use the axioms of the real numbers, but i can't finde an usefull equivalent transformation
| since $x,y,z\in\mathbb R$
$$(x-y)^2+(y-z)^2+(z-x)^2\geq0\\
x^2+y^2+z^2\geq xy+yz+zx\\
\frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Basic Algebra Help \begin{align}
x^2-2x+1 &= x^2-2x+1\\
(1-x)^2 &= (x-1)^2\\
\sqrt{(1-x)^2} &= \sqrt{(x-1)^2}\\
1-x &= x-1\\
1-x &= -1\cdot(1-x)\\
1 &= -1
\end{align}
I definitely did something wrong.
| \begin{align}
x^2-2x+1 &= x^2-2x+1\\
(1-x)^2 &= (x-1)^2\\
\sqrt{(1-x)^2} &= \sqrt{(x-1)^2}\\
\end{align}
OK, as square root is unambiguously defined as non-negative value.
First (very common) error is that from the last equation follows
\begin{align}
1-x &= x-1\\
\end{align}
That is not correct, as you may see from this example:
\begin{align}
4^2 &= (-4)^2\\
\end{align}
so
\begin{align}
\sqrt{4^2} &= \sqrt{(-4)^2}\\
\end{align}
in spite of
\begin{align}
4 \neq -4
\end{align}
(Remember, square root is a non-negative value, so right-hand side is incorrect.)
Second, also a common error is dividing both sides of an equation with the same number without a consideration if it is not zero:
\begin{align}
1-x &= -1\cdot(1-x)\\
1 &= -1
\end{align}
The correct method for solution of the equation
\begin{align}
1-x &= -1\cdot(1-x)\\
\end{align}
is put all members with $x$ to one side of the equal sign and others to the opposite one:
\begin{align}
-2x &= -2\\
x &= 1\\
\end{align}
From this (single) solution you may immediately see why dividing both sides by $(1-x)$ was incorrect - we divided them by zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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