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Convergence of Cauchy sequence $(x_n)$, where $x_n=1/n^2$. I'm just looking at an example where we're asked to prove that the sequence $(x_n)$ is Cauchy, where $x_n=1/n^2$. (the example is from here at time 7:15 minutes; this requires registration but is free) Anyway, the proof goes exactly like this:
Let $\varepsilon>0$ and let $N=\lfloor 1/(2\varepsilon)\rfloor+1$. Then $q>p>N$ implies $q>p>1/(2\varepsilon)$ and
$$\left(\frac{1}{q}-\frac{1}{p}\right)\left(\frac{1}{q}+\frac{1}{p}\right)\leq 2\left(\frac{1}{q}-\frac{1}{p}\right)<2\frac{1}{2\varepsilon}.\tag{1}$$
Hence $$\frac{1}{q^2}-\frac{1}{p^2} < \varepsilon,\tag{2}$$ as required.
My questions are:
(i) the $\varepsilon$ in equation (2) does not follow from the $2\times 1/(2\varepsilon)$ in equation (1), so is this just a mistake in the given proof?
(ii) Shouldn't the proof be using absolute notation, e.g.
$$\left|\left(\frac{1}{q}-\frac{1}{p}\right)\left(\frac{1}{q}+\frac{1}{p}\right)\right|\leq 2\left|\frac{1}{q}-\frac{1}{p}\right|<2\frac{1}{2\varepsilon},$$ because $q>p$ implies that $$\frac{1}{q}-\frac{1}{p}<0$$ is negative, which is not something we should allow. In particular, equation (2) should not be negative because we are trying to show that $$0<\frac{1}{q^2}-\frac{1}{p^2}<\varepsilon,$$ i.e. goes to zero as $\varepsilon\to 0$.
| (i) Yes. This is a mistake.
(ii) The absolute value is necessary, OR they could have said $p > q$. However, they did define it wrong for their argument to work.
Here's how I believe that proof is supposed to go. (It is similar, but there appear to be typos in the above.
Let $\{x_n\}$ be a sequence defined by $x_n = \frac{1}{n^2}, \forall n \in \mathbb{N}$. Show $\{x_n\}$ is Cauchy.
Fix $\epsilon > 0$. Then, there exists an $N = \lceil\frac{4}{\epsilon} \rceil$ such that $\forall n,m > N$ we know that
$$ \left| \frac{1}{n^2} - \frac{1}{m^2} \right| = \left| (\frac{1}{n} - \frac{1}{m} ) (\frac{1}{n} + \frac{1}{m})\right| \le 2 \left(\frac{1}{n} + \frac{1}{m}\right) < 2 \left(\frac{\epsilon}{4} + \frac{\epsilon}{4}\right) = \epsilon,$$
where the second to last inequality follows since $\left(\frac{1}{n} + \frac{1}{m} \right) \ge \left| \frac{1}{n} - \frac{1}{m} \right|$, and the last inequality follows since $\frac{1}{m}$ and $\frac{1}{n}$ are both less than $\frac{1}{N} < \frac{\epsilon}{4}$.
Moreover, here is another method that is more straightforward.
Fix $\epsilon > 0$. Let $N > \left\lceil \sqrt{\frac{2}{\epsilon}} \right\rceil $, or equivalently, $\frac{1}{N^2} < \frac{\epsilon}{2}$. Then,$\forall n,m \ge N$
$$ \left| \frac{1}{n^2} - \frac{1}{m^2} \right| \le \frac{1}{n^2} + \frac{1}{m^2} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Cauchy sequence $a_n = 1 + \frac12 + \frac14 + ... + \frac{1}{2^n}$ For the sequence $a_n = 1 + \frac12 + \frac14 + ... + \frac{1}{2^n}$, $n \ge 1$, find a formula $N = N(\epsilon)$ such that
for all $\epsilon > 0$ and for all $m,n \ge N(\epsilon)$, $|a_m - a_n| < \epsilon$.
I have tried many things but it's just not working. I know that the condition is the definition of Cauchy sequences. But that didn't help.
| Hint:
\begin{align*}
a_n
&= 1 + \frac12 + \frac14 + \cdots + \frac{1}{2^n} \\
&= \frac{1}{2^n} (2^n + 2^{n-1} + 2^{n-2} + \cdots + 1) \\
&= \frac{2^{n+1} - 1}{2^n} = 2 - \frac{1}{2^n}
\end{align*}
so
\begin{align*}
|a_n - a_m| &= \left| \left( 2 - \frac{1}{2^n} \right) - \left( 2 - \frac{1}{2^m} \right) \right| \\
&= \left| \frac{1}{2^n} - \frac{1}{2^m} \right| \\
&< \frac{1}{2^m} + \frac{1}{2^n} \\
&\le \frac{1}{2^N} + \frac{1}{2^N} = \frac{2}{2^N}
\end{align*}
Pick $N(\epsilon)$ so that $\frac{2}{2^N} < \epsilon$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$? Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$?
So far I have:
$a^2+(a+2)^2+(a+4)^2+1$
$=a^2+a^2+4a+4+a^2+8a+16+1 $
$=3a^2+12a+21$
$=3(a^2+4a+7) $
where do I go from here.. the solution I have is divisible by $3$ not $12$...
| Hint:
If $a$ is odd then by the Division algorithm $a = 2k + 1$ for some integer $k$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Ways of writing $n=2a+b$ with $a$ and $b$ are non-negative integers For a non-negative number $n$, let $r_n$ be the number of ways of writing
$n = a + 2b$, where $a$ and $b$ are non-negative integers.
For example,
$5 = 1 \cdot 5 + 2 \cdot 0 = 3 \cdot 1 + 1 \cdot 2 = 1 \cdot 1 + 2 \cdot 2$,
so $r_5 = 3$.
My question is I have found the rule that:
$r_0=r_1=1$,
$r_2=r_3=2$,
$r_4=r_5=3$,
$r_6=r_7=4$,
......
$r_{n-1}=r_{n}=\frac{n+1}{2}$ where $n$ is odd.
But how can we prove this rule?
| Note that $a$ is determined by $n$ and $b$ via $a=n-2b$, so $r_n$ is the number of valid choices for $b$. Since $a$ and $b$ must be nonnegative, we get $b\geq 0$ and $n-2b\geq 0$, so $\frac n2\geq b \geq 0$. If $n=2k$ or $2k+1$, then the only integer values $b$ can take are $0, 1, \ldots, k$ so there are $k+1$ choices. In particular, in the case $n=2k+1$, $$k+1=\frac{n-1+2}{2}=\frac{n+1}{2}$$ which is exactly the formula you found.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$? How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$?
Hmmm... So I didn't know what to do here so I tested some cases for a rule.
If a number had the factors $3^2$ and $2^1$, you can make $5$ distinct factors: $2^1$, $3^1$, $3^2$, $2^1 \cdot 3^1$, $2^1 \cdot 3^2$... I don't see a pattern yet.
How does one go about this?
I don't think the answer is $5!$....
| Your approach with $3^2$ and $2^1$ is very close. You only forgot to account for the factor $1 = 3^0\cdot2^0$. This brings us to $6$ distinct factors for this smaller example. You can calculate this by multiplying the number of all the different exponents used for each factor. $3$ can occur either $2$, $1$ or $0$ times, while $2$ can only occur $1$ or $0$ times. This brings us to $3\cdot2 = 6$ combinations of these exponents.
Now, the "real" question has another little trick to it because the factors aren't all that nice just yet. $2^5\cdot3^4\cdot4^3\cdot3^2\cdot2^1 = 2^6\cdot3^6\cdot4^3$
Much better.
Now, we can have between $0$ and $6$ occurences of $2$, making it $7$ possibilities.
Another $7$ possibilities for occurences of $3$ and finally $4$ possibilities for $4$.
This makes a total of $7\cdot7\cdot4 = 196$ possible distinct factors of $2^6\cdot3^6\cdot4^3 = 2^5\cdot3^4\cdot4^3\cdot3^2\cdot2^1$.
Edit
Alright, I didn't pay attention either. As Konstantinos Gaitanas stated, $4^3 = 2^6$, making the whole thing $2^{12}\cdot3^6$.
So the solution should indeed be $13\cdot7$ = 91.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why aren't these two integration methods yielding the same answer? I'm trying to solve this (not homework, if it matters), and both u-substitution and integration by parts are both yielding two different answers. Where am I going wrong?
Equation: $$\int \frac{(4x^3)}{(x^4+7)}dx$$
u-substitution answer: $$=\ln\big|(x^4+7)\big|+C$$
integration by parts answer:
$$=\int4x^3*(x^4+7)^{-1}dx$$
$$=4x^3*\ln\big|x^4+7\big|-\int 12x^2*(x^4+7)^{-1}dx$$
$$=4x^3*\ln\big|x^4+7\big|-(12x^2*ln\big|x^4+7\big|-\int 24x*(x^4+7)^{-1}dx)$$
$$=4x^3*\ln\big|x^4+7\big|-(12x^2*ln\big|x^4+7\big|-24x*ln\big|x^4+7\big|-\int 24(x^4+7)^{-1}dx)$$
$$= 4x^3*\ln\big|x^4+7\big|-(12x^2*\ln\big|x^4+7\big|-(24x\ln\big|x^4+7\big|-24\ln\big|x^4+7\big|))$$
$$=(4x^3-12x^2+24x-24)(\ln\big|x^4+7\big|)$$
| The correct result is:
$$\int\frac{4x^3}{x^4+7}dx=\ln|x^4+7|+c$$.
With substitution: $u=x^4+7$ from which $du=4x^3dx$ hence the integral become:
$$\int\frac{du}{u}=\ln|u|+c$$.
Integration by parts it's not raccomanded in this case.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I put $\sqrt{x+1}$ into exponential notation? I think $\sqrt{x+1} = x^{1/2} + 1^{1/2}$. Is this incorrect? Why or why not?
| $$
9^{1/2} + 16^{1/2} = 3 + 4 = 7 \ne 5 = 25^{1/2} = (9+16)^{1/2}.
$$
So
$$
\left(\frac{9}{16}\right)^{1/2} + 1^{1/2} = \frac 3 4 + 1 = \frac 7 4 \ne \frac 5 4 = \left(\frac{25}{16}\right)^{1/2} = \left(\frac{9}{16}+1\right)^{1/2}
$$
In other words, if $x=\dfrac{9}{16}$ then $x^{1/2}+1^{1/2}$ is not at all the same as $(x+1)^{1/2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676634",
"timestamp": "2023-03-29T00:00:00",
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A good site documenting approximations of irrationals I'm thinking of Sloane here but I believe that only takes sequences/series into account.
Basically I've derived an interesting, appealing formula for e and want to know if it's already been discovered.
| We want to show equality
$$ e={2 \over 3} \left( {1 \over 1!} +{1 +2\over 2!} + {1 +2+3\over 3!} + {1 +2+3+4\over 4!} + ... \right) $$
Let's define a more general expression $f$ using the indeterminate $x$ at each summand inside the parenthese and derive the equality by finally inserting $x=1$:
$$ \begin{array}{}
f(x)&= {1 \over 1!} +{3\over 2!}x + {6\over 3!}x^2 + {10\over 4!}x^3 + ... \\
&={1 \over 2}\left( {2 \over 1!} +{6\over 2!}x + {12\over 3!}x^2 + {20\over 4!}x^3 + ... \right) \qquad \text{//cancel biggest factors}\\
&={1 \over 2}\left( {2 \over 0!} +{3\over 1!}x + {4\over 2!}x^2 + {5\over 3!}x^3 + ... \right) \qquad \text{//separate in two series} \\
&={1 \over 2}\left( {1 \over 0!} +{2\over 1!}x + {3\over 2!}x^2 + {4\over 3!}x^3 + ... \right) + {1 \over 2}\left( {1 \over 0!} +{1\over 1!}x + {1\over 2!}x^2 + {1\over 3!}x^3 + ... \right) \\
&={1 \over 2}\left( {1 \over 0!}x +{1\over 1!}x^2 + {1\over 2!}x^3 + {1\over 3!}x^4 + ... \right)'+ {1 \over 2}e^x \qquad \text{//note the first derivative!}\\
&={1 \over 2}\left( x e^x \right)'+ {1 \over 2}e^x\\
&={1 \over 2}\left( 1+ x \right)e^x+ {1 \over 2}e^x\\
&={1 \over 2}(2+x)e^x\\ \end{array}
$$
and if we insert $1$ for $x$ in $f(x)$ we get equality
$$ e = {2\over 3} f(1) ={2\over 3} \left( {1 \over 2} (2+1) \right)e^1=e $$
| {
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"url": "https://math.stackexchange.com/questions/680189",
"timestamp": "2023-03-29T00:00:00",
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how to find this line integral and what is its answer evaluate the line integral $$\int_C (xy^2 dy-x^2y dx), $$ taken in the counter-clockwise sense along the cardioid $$r= a(1+\cos\theta)$$
here putting the parametric form of cardioid $x=a(2\cos t-\cos2t), y= a(2\sin t-\sin2t) $ and taking $\theta$ , $0 $ to $2\pi $ tried to solve but then it became complicated, so is there any other method and in this method please solve these as well ?
| If $x=a(2\cos t-\cos 2t)$ and $y=a(2\sin t-\sin 2t)$, then
$$
xy=a^2\big(2\sin 2t-2\sin3t+\tfrac{1}{2}\sin 4t\big),
$$
$$
x^2-y^2=a^2\big(4\cos 2t-4\cos 3t+\cos4t\big),
$$
and
$$
x\,dx-y\,dy=a^2\big(-4\sin 2t+6\sin 3t-2\sin4t\big)\,dt.
$$
Hence
$$
xy^2\,dy-x^2y\,dx=a^4\big(2\sin 2t-2\sin3t+\tfrac{1}{2}\sin 4t\big)
\big(4\sin 2t-6\sin 3t+2\sin4t\big)\,dt,
$$
and
\begin{align}
\int_C xy^2\,dy-x^2y\,dx &=\int_0^{2\pi}a^4\big(2\sin 2t-2\sin3t+\tfrac{1}{2}\sin 4t\big)
\big(4\sin 2t-6\sin 3t+2\sin4t\big)\,dt\\ &=a^{4}\int_0^{2\pi}\big(8\sin^22t+12\sin^23t+\sin^44t\big)\,dt=\pi a^4 (8+12+1)=21\pi a^4.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/680787",
"timestamp": "2023-03-29T00:00:00",
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What does $\prod_{n\geq2}\frac{n^4-1}{n^4+1}$ converge to?
What does $\prod_{n\geq2}\frac{n^4-1}{n^4+1}$ converge to?
As far as I can tell, this has no closed-form solution (not saying much, I don't know much math), but a friend of mine swears he saw a closed-form solution to this in some text he doesn't remember.
Running it through WolframAlpha gives me an approximation which, entered through an inverse symbolic calculator, gets no results. This is satisfactory enough for me to believe there is no closed-form solution, but my friend really does insist there is one.
Edit
Okay, not meaning to sound greedy, but I'd also be interested in seeing how I'd derive a closed-form solution algebraically. I do appreciate the fast answers, though.
| Start with the infinite product expansion of $\sin x$,
$$\sin x = x \prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2\pi^2}\right)$$
We get
$$\prod_{n=2}^\infty \left(1 - \frac{\alpha^2}{n^2}\right) = \frac{\sin(\alpha\pi)}{\alpha\pi(1-\alpha^2)}$$
In particular,
$$\begin{align}
\prod_{n=2}^\infty \left(1 - \frac{1}{n^2}\right)
&= \lim_{\alpha\to 1}\frac{\sin(\alpha\pi)}{\alpha\pi(1-\alpha^2)}
= \lim_{a\to1} \frac{\pi\cos(\pi\alpha)}{\pi\alpha(-2\alpha)}
= \frac{1}{2}\\
\prod_{n=2}^\infty \left(1 + \frac{1}{n^2}\right)
&=\frac{\sin(i\pi)}{i\pi(1-i^2)}
= \frac{\sinh\pi}{2\pi}\\
\prod_{n=2}^\infty \left(1 \mp \frac{i}{n^2}\right)
&= \frac{\sin(\frac{1\pm i}{\sqrt{2}}\pi)}{\frac{1\pm i}{\sqrt{2}}\pi(1 \mp i)}
= \frac{
\sin(\frac{\pi}{\sqrt{2}})\cosh(\frac{\pi}{\sqrt{2}})
\pm i \cos(\frac{\pi}{\sqrt{2}})\sinh(\frac{\pi}{\sqrt{2}})
}{\frac{1\pm i}{\sqrt{2}}\pi(1 \mp i)}
\end{align}$$
From this, we get
$$\begin{align}
\prod_{n=2}^\infty\frac{n^4-1}{n^4+1}
&= \frac{\displaystyle
\prod_{n=2}^\infty \left(1 - \frac{1}{n^2}\right)
\prod_{n=2}^\infty \left(1 + \frac{1}{n^2}\right)
}{\displaystyle
\prod_{n=2}^\infty \left(1 - \frac{i}{n^2}\right)
\prod_{n=2}^\infty \left(1 + \frac{i}{n^2}\right)
}\\
&= \frac{\displaystyle
\frac{\sinh\pi}{4\pi}
}{\displaystyle\left(
\frac{\sin(\frac{\pi}{\sqrt{2}})^2\cosh(\frac{\pi}{\sqrt{2}})^2
+\cos(\frac{\pi}{\sqrt{2}})^2\sinh(\frac{\pi}{\sqrt{2}})^2}{2\pi^2}\right)
}\\
&= \frac{\pi\sinh\pi}{\cosh(\sqrt{2}\pi) - \cos(\sqrt{2}\pi)}\\
\\
&\approx 0.84805404935290039212965018340500770584798748\ldots
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the following system of equations. $2x^2-3xy+2y^2=2\frac{3}{4}\\x^2-4xy+y^2+\frac{1}{2}=0$
I tried all the methods that I know, but I could't isolate $x$ or $y$ to form one equation.
| Multiply second equation by $2$,
$$2x^2-8xy+2y^2+1=0$$
Substract the first equation by the second,
$$2x^2-3xy+2y^2-\frac{11}{4}-(2x^2-8xy+2y^2+1)=0\\2x^2-3xy-2y^2-\frac{11}{4}-2x^2+8xy-2y^2-1=0\\5xy=\frac{15}{4}\\x=\frac{3}{4y}$$
Substitute $x=\frac{3}{4y}$ in the first equation,
$$2x^2-3xy+2y^2=\frac{11}{4}\\2(\frac{3}{4y})^2-3(\frac{3}{4y})y+2y^2-\frac{11}{4}=0\\\frac{9}{8y^2}+2y^2-5=0\\16y^4-40y^2+9=0$$Now use the quadratic Formula to solve $16y^4-40y^2+9=0$,$$y^2=\frac{-(-40)\pm\sqrt{(-40)^2-4.16.9}}{2.16}\\y^2=\frac{40\pm\sqrt{1600-4.16.9}}{2.16}\\y^2=\frac{40\pm\sqrt{16(100-4.9)}}{2.16}\\y^2=\frac{40\pm4\sqrt{64}}{2.16}=\frac{40\pm4.8}{2.16}=\frac{5\pm4}{4}\\y^2=\frac{5+4}{4}=\frac{9}{4}\Rightarrow y=\pm\frac{3}{2}\\or,~~y^2=\frac{5-4}{4}=\frac{1}{4}\Rightarrow y=\pm\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/681976",
"timestamp": "2023-03-29T00:00:00",
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How find this integral $I=\int_{-\infty}^{+\infty}\frac{x^3\sin{x}}{x^4+x^2+1}dx$ Find this integral
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
my idea:
$$I=2\int_{0}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
because
$$\dfrac{x^3\sin{x}}{x^4+x^2+1}\approx\dfrac{\sin{x}}{x},x\to\infty$$
so
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
converges
then I can't,Thank you
| Just like the comments given by @heropup and @Random Variable, this problem can be solved by using residue theorem and Jordan's lemma.
Note that the denominator of the integrand can be factorize by
\begin{equation}
z^4+z^2+1=(z^2-z+1)(z^2+z+1)
\end{equation}
It have four poles that
\begin{equation}
z_{1,2}=\pm\frac{1}{2}+\frac{\sqrt{3}}{2}i\\
z_{3,4}=\pm\frac{1}{2}-\frac{\sqrt{3}}{2}i\\
\end{equation}
Only $z_1$ and $z_2$ are in upper half plane. Then, from residue theorem and Jordan's lemma, we have:
\begin{equation}
I=\int_{-\infty}^{+\infty} \frac{x^3\sin(x)}{x^4+x^2+1}dx\\
=\Im(2\pi i\sum_{k=1,2}\mathrm{Res}_{z=z_k}\frac{z^3e^{iz}}{z^4+z^2+1})\\
=-\frac{1}{3}e^{-\sqrt{3}/2}\sin(\frac{1}{2})\pi\sqrt{3}+e^{-\sqrt{3}/2}\cos(\frac{1}{2})\pi
\end{equation}
The approximation of the result is $0.7938888330$. You can check it by some numerical method.
| {
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Newbie: Find the intersection of a line and a circle and interpret geometrically Find the points of intersection of the line $x+y+k=0$ and the circle $x^2+y^2=2x$. Show that there are two points of intersection if: $-1-\sqrt2<k<-1+\sqrt2$, one point of intersection if: $k=-1\pm\sqrt2$, and none otherwise. Interpret the results geometrically. What is the shortest distance between the point (1,0) and either of the lines $x+y-1\pm\sqrt2=0$
My attempt:
$x + y=0$
$y=-x-k$
$x^2+y^2=2x$
$x^2+(-x-k)^2=2x$
$x^2+(x^2+2xk=k^2)=2x$
$2x^2 -2x+2xk+k^2=0$
$2x^2+(-2+2k)x +k^2=0$
Following what I read here: How do I calculate the intersection(s) of a straight line and a circle? I substituted into the quadratic equation and made the assumption that if $b^2 -4ac=0$ then there would be a single point of intersection.
$0=b^2-4ac$
$0=(-2+2k)^2-4(2)(k^2)$
$0=-4k^2-8k+4$
$0=k^2+2k-1$
$0=(k-1)^2$
The parameters for k, as stated in the question, are clearly not met so I'm not sure what else to try.
| Everything is correct up until your last two lines. You seem to be assuming that $k^2 + 2k -1$ is equal to $(k-1)^2$, which is not true.
Your equation $k^2 + 2k -1 = 0$ is correct, and it has roots $k = -1 \pm \sqrt2$, which is probably what you were expecting.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculating $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $ If $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\sqrt{4-\sqrt{ 4+\sqrt{4-\dots}}}}}} $
then find value of 2x-1
I tried the usual strategy of squaring and substituting the rest of series by x again but could not solve.
| Since $$x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\cdots}}}},$$ you have $$x^2=4+\sqrt{4-\sqrt{4+\sqrt{4-\cdots}}},$$ so $$(x^2-4)^2=4-\sqrt{4+\sqrt{4-\cdots}}.$$ Hence, $$(x^2-4)^2=4-x,$$ which you can try to solve.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Difference of consecutive cubes never divisible by 5. This is homework from my number theory course.
Since $(x+1)^3-x^3=3x^2+3x+1$ and $x^3-(x+1)^3=-3x^2-3x-1$, to say that the difference of two cubes is divisible by 5 is the same as saying that $3x^2+3x+1\equiv 0\mod 5$ or $-3x^2-3x-1\equiv 0\mod 5$. Both of these statements imply that $x(x+1)\equiv 3\mod 5$. Thus I can finish this by showing that there are no such integers which satisfy $x(x+1)\equiv 3\mod 5$.
I want to say that it is sufficient to check by hand for the values 0,1,2,3, and 4 (for which it is not true), but other than following this by a messy induction I was wondering if there is an easier way to show that there are no integers such that $x(x+1)\equiv 3\mod 5$?
| $x^3 - y^3 = (x - y)(x^2 + x y + y^2)$. Now
$x^2 + x y + y^2 \equiv (x + 3 y)^2 - 3 y^2 \mod 5$, and since $3$ is not a quadratic residue mod $5$ we find that there are no solutions to
$x^2 + x y + y^2 \equiv 0 \mod 5$ other than the trivial $(0,0)$. Thus
$x^3 \equiv y^3 \mod 5$ only when $x \equiv y \mod 5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Differentiation wrt parameter $\int_0^\infty \sin^2(x)\cdot(x^2(x^2+1))^{-1}dx$ Use differentiation with respect to parameter obtaining a differential equation to solve
$$
\int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)}dx
$$
No complex variables, only this approach. Interesting integral and it should have a nice ODE. I have not found the right way yet. we have singularities at $x=\pm i$.
| Well, as an alternative (like Mr. Ron Gordon did).
\begin{align}
\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}dx&=\int_0^\infty\left[\frac{\sin^2x}{x^2}-\frac{\sin^2x}{1+x^2}\right]dx\\
&=\int_0^\infty\frac{\sin^2x}{x^2}dx-\frac{1}{2}\int_0^\infty\frac{1-\cos2x}{1+x^2}dx\\
&=\frac{\pi}{2}-\frac{1}{2}\int_0^\infty\frac{1}{1+x^2}dx+\frac{1}{2}\int_0^\infty\frac{\cos2x}{1+x^2}dx\\
&=\frac{\pi}{2}-\frac{1}{2}\frac{\pi}{2}+\frac{1}{2}\frac{\pi}{2e^2}\\
&=\frac{\pi}{4}+\frac{\pi}{4e^2}
\end{align}
where I use these links: $\displaystyle\int_0^\infty\frac{\sin^2x}{x^2}dx$ and $\displaystyle\int_0^\infty\frac{\cos2x}{1+x^2}dx$ to help me out.
Unfortunately, this is not differentiation with respect to parameter method (the Feynman way) but I still love this method. (>‿◠)✌
| {
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Solving $\arcsin(1-x)-2\arcsin(x)=\pi/2$ \begin{eqnarray*}
\arcsin(1-x)-2\arcsin(x) & = & \frac{\pi}{2}\\
1-x & = & \sin\left(\frac{\pi}{2}+2\arcsin(x)\right)\\
& = & \cos\left(2\arcsin(x)\right)\\
& = & 1-2\left(\sin\left(\arcsin(x)\right)\right)^{2}\\
& = & 1-2x^{2}\\
x & = & 2x^{2}\\
x\left(x-\frac{1}{2}\right) & = & 0
\end{eqnarray*}
So $x=0$ or $x=\frac{1}{2}$
But puttig $x=\frac{1}{2}$ in the original expression gives $-\frac {\pi} 4 \ne \frac \pi 2$
So, why do we get $x=-1/2$ as an answer?
| As $\displaystyle \sin\left(\frac\pi2\pm A\right)=\cos A,$
$\displaystyle\sin\left(\frac\pi2\pm2\arcsin x\right)=\cos(2\arcsin x)=1-2\left[\sin(\arcsin x)\right]^2=1-2x^2$
So, $\displaystyle x=\frac12$ corresponds to $\displaystyle\arcsin(1-x)+2\arcsin x=\frac\pi2$ as $\displaystyle\arcsin\frac12=\frac\pi6$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Pre Calculus Expression The questions is:
$$\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}$$
My answer is: $$\dfrac{3(x+2)^2 + 6x^2-4}{(x-3)^2}$$
Am I right? If not, where have I failed?
| $$ \begin{align}\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}&=
\dfrac{(x-3)(3(x-3)(x+2)^2-2(x+2)^3)}
{(x-3)^4}\\
&= \dfrac{3(x-3)(x+2)^2-2(x+2)^3}{(x-3)^3}\\
&= \dfrac{(x+2)^2(2(x-3)-2(x+2))}{(x-3)^3} \\
&= \dfrac{(x+2)^2(3x-9-2x-4)}{(x-3)^3}\\
&= \dfrac{(x+2)^2(x-13)}{(x-3)^3}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Interpretation of a homogeneous transformation matrix of the plane I have the transformation matrix $\begin{pmatrix} 1&0&0\\0&1&0\\0&-1&1\end{pmatrix}$. This $3\times 3$ matrix is a homogeneous transformation matrix in $2-D$ space.
My book says that this matrix translates the line $y=x+1$ to $y=x$. I don't see how. Let us take the point $(a,b)$. After the translation by $-1$ along the $y$-axis, the point should become $(a,b-1)$. Now let us determine $\begin{pmatrix} 1&0&0\\0&1&0\\0&-1&1\end{pmatrix}\begin{pmatrix} a\\b\\1\end{pmatrix}$. We get $\begin{pmatrix} a\\b\\1-b\end{pmatrix}$. After transforming this resultant matrix to 2-D form, we get $\begin{pmatrix} \frac{a}{1-b}&\frac{b}{1-b}\end{pmatrix}$. How is this equal to $(a,b-1)$?
| It appears you've either put the translation elements in the wrong position or you're using column vectors instead of row vectors:
$$
\begin{pmatrix}a&b&1\end{pmatrix}
\begin{pmatrix}1&0&\color{#00A000}{0}\\0&1&\color{#00A000}{0}\\\color{#C00000}{0}&\color{#C00000}{-1}&1\end{pmatrix}
=\begin{pmatrix}a&b-1&1\end{pmatrix}
$$
or
$$
\begin{pmatrix}1&0&\color{#C00000}{0}\\0&1&\color{#C00000}{-1}\\\color{#00A000}{0}&\color{#00A000}{0}&1\end{pmatrix}
\begin{pmatrix}a\\b\\1\end{pmatrix}
=\begin{pmatrix}a\\b-1\\1\end{pmatrix}
$$
translation elements in $\color{#C00000}{\text{red}}$, perspective elements in $\color{#00A000}{\text{green}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/696389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trigonometric Substitution
I am having trouble with this problem even though everything I did seemed right to me since we went over a similar one in my class. I used the method of setting up a triangle, my hypotenuse is $\sqrt{54+9x^2}$ and my sides are $\sqrt{54}$ and $3x$. I got $\tan(t)=3x/\sqrt{54}$ so
$$x=\sqrt{54} \tan(t) \frac{1}{3}$$
which left
$$\sec(t) = \frac{\sqrt{54+9x^2}}{\frac{\sqrt{54}}{3}}$$
and then
$$\frac{\sqrt{54}}{3} \sec(t) = \sqrt{54+9x^2}.$$
This left me with a simplified $6 \int \sec^3(t) \, dt$. After using the reduction formula my answer was
$$3 \tan(t) \sec(t) +3 \ln |\sec(t) + \tan(t)| +C$$
and then I plugged back in with my $x$ values. If anyone can help it would be greatly appreciated!
| Sorry, I misunderstood your question the first time round.
First, I'm getting
$$9 \sec \theta \tan \theta + 9 \ln | \sec \theta + \tan \theta | + C $$
as my intermediate answer. From there, you need to take out all the thetas and put in $x$. From what you wrote, I'm assuming you see how we get $\tan \theta = x / \sqrt{6}$ and $\sec \theta = \sqrt{(x^2 + 6) / 6}$. (If not, let me know.) Therefore, we rewrite the above expression as
$$
\\ 9 \cdot \frac{\sqrt{x^2 + 6}}{\sqrt{6}} \cdot \frac{x}{\sqrt{6}} + 9 \ln \left| \frac{\sqrt{x^2 + 6}}{\sqrt{6}} + \frac{x}{\sqrt{6}} \right| + C = \frac{3}{2}x\sqrt{x^2 + 6} + 9 \ln \left| \frac{\sqrt{x^2 + 6} + x}{\sqrt{6}} \right| + C \
$$
A lot of times with these kinds of integrals the trick is with the $\ln$. Remember that $\ln A / B = \ln A - \ln B$; this lets us simplify the expression like so:
$$\frac{3}{2}x\sqrt{x^2 + 6} + 9 \ln \left| \sqrt{x^2 + 6} + x \right| + C$$
Doing this always seems weird to me, but $-9 \ln \sqrt{6}$ is a constant, after all...
| {
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Fourier Series for $|\cos(x)|$ I'm having trouble figuring out the Fourier series of $|\cos(x)|$ from $-\pi$ to $\pi$.
I understand its an even function, so all the $b_n$s are $0$
$$a_0 = \frac 2 \pi \int_0^\pi |\cos(x)|\,dx = 0$$
$$a_n = \frac 2 \pi \int _0^\pi |\cos(x)| \cos(nx) \, dx = \frac 2 \pi \int_0^\pi \cos^2(x)\,dx.$$
since for all $j,k$ not equal the integral is zero.
so only $a_1$ remains. is this correct?
How would I evaluate $\sum_{n=1}^\infty (-1)^{n-1} /(4n^2 - 1)\ {}$?
| Although $ \int_0^\pi \cos(x)\,dx = 0$, $a_0\ne 0$ because $$\int_0^{\pi/2} |\cos(x)|\,dx=\int_{\pi/2}^{\pi} |\cos(x)|\,dx. $$
We can evaluate it as follows, as can be seen in the plot below
$$a_0 = \frac 1 \pi \int_{-\pi}^\pi |\cos(x)|\,dx=\frac 2 \pi \int_0^\pi |\cos(x)|\,dx=\frac 4 \pi \int_0^{\pi/2} |\cos(x)|\,dx = \frac 4 \pi \int_0^{\pi/2} \cos(x)\,dx=\frac 4 \pi.$$
$$\tag{1}$$
Plot of $\cos x$ (doted line) and $|\cos x|$ (solid line) in the interval $[-\pi,\pi]$.
The coefficients $b_n=0$ as you concluded. As for the $a_n$ coefficients only the odd ones are equal to $0$ (see below). The functions $\cos(x)$ and $\cos(nx)$ are orthogonal in the interval $[-\pi,\pi]$, but $|\cos(x)|$ and $\cos(nx)$ are not. Since
\begin{equation*}
\left\vert \cos (x)\right\vert =\left\{
\begin{array}{c}
\cos (x) \\
-\cos (x)
\end{array}
\begin{array}{c}
\text{if} \\
\text{if}
\end{array}
\begin{array}{c}
0\leq x\leq \pi /2 \\
\pi /2\leq x\leq \pi,
\end{array}
\right. \tag{2}
\end{equation*}
we have that
\begin{eqnarray*}
a_{n} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }\left\vert \cos (x)\right\vert\cos (nx)\,dx=\frac{2}{\pi }\int_{0}^{\pi }\left\vert \cos (x)\right\vert
\cos (nx)\,dx \\
&=&\frac{2}{\pi }\int_{0}^{\pi /2}\left\vert \cos (x)\right\vert \cos
(nx)\,dx+\frac{2}{\pi }\int_{\pi /2}^{\pi }\left\vert \cos (x)\right\vert
\cos (nx)\,dx \\
&=&\frac{2}{\pi }\int_{0}^{\pi /2}\cos (x)\cos (nx)\,dx-\frac{2}{\pi }
\int_{\pi /2}^{\pi }\cos (x)\cos (nx)\,dx. \\
a_{1} &=&\frac{2}{\pi }\int_{0}^{\pi /2}\cos ^{2}(x)\,dx-\frac{2}{\pi }\int_{\pi /2}^{\pi }\cos ^{2}(x)\,dx=0.
\end{eqnarray*}
Using the following trigonometric identity, with $a=x,b=nx$,
\begin{equation*}
\cos (a)\cos (b)=\frac{\cos (a+b)+\cos (a-b)}{2},\tag{3}
\end{equation*}
we find
\begin{eqnarray*}
a_{2m} &=&\frac{4}{\pi \left( 1-4m^{2}\right) }\cos (\frac{2m\pi }{2})=\frac{
4}{\pi \left( 1-4m^{2}\right) }(-1)^{m} \\
a_{2m+1} &=&\frac{4}{\pi ( 1-4(2m+1)^{2}) }\cos (\frac{(2m+1)\pi
}{2})=0,\qquad m=1,2,3,\ldots.\tag{4}
\end{eqnarray*}
The expansion of $\left\vert \cos (x)\right\vert $ into a trigonometric
Fourier series in the interval $[-\pi ,\pi ]$ is thus
\begin{equation*}
\left\vert \cos x\right\vert =\frac{a_{0}}{2}+\sum_{n=1}^{\infty }\left(
a_{n}\cos (nx)+b_{n}\sin (nx)\right) =\frac{2}{\pi }+\frac{4}{\pi }
\sum_{m=1}^{\infty }\frac{(-1)^{m}}{1-4m^{2}}\cos (2mx)\tag{5}
\end{equation*}
$$|\sin(x)|\ \text{(blue) and the partial sum }\frac{2}{\pi }+\frac{4}{\pi }
\sum_{m=1}^{5 }\frac{(-1)^{m}}{1-4m^{2}}\cos (2mx) \ \text{(red) in }[-\pi,\pi]$$
Setting $x=0$ in $(5)$, we obtain
\begin{equation*}
1=\frac{2}{\pi }+\frac{4}{\pi }\sum_{m=1}^{\infty }\frac{(-1)^{m}}{1-4m^{2}}=\frac{2}{\pi }-\frac{4}{\pi }\sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{1-4n^{2}}.\tag{6}
\end{equation*}
Hence
\begin{equation*}
\sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{1-4n^{2}}=\frac{1}{2}-\frac{\pi }{4}.\tag{7}
\end{equation*}
| {
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Proof: $a^2 - b^2 = (a-b)(a+b)$ holds $\forall a,b \in R$ iff R is commutative We want to show that for some ring $R$, the equality $a^2 - b^2 = (a-b)(a+b)$ holds $\forall a,b \in R$ if and only if $R$ is commutative.
Here's my proof --- I'm not sure if the first part stands up to examination. I'd be grateful if someone could take a look.
Forward: $a^2 -b^2 = (a-b)(a+b) \forall a,b \in R$ implies $R$ is commutative
Let $x = (a-b)$. Then \begin{align}
x(a+b) &= xa+xb\\
&= (a-b)a + (a-b)b\\
&= a^2 -ba + ab - b^2\end{align}
Then we note that $a^2 - ba + ab - b^2 = a^2 - b^2$ iff $-ba + ab = 0$ if and only if $ab=ba$ iff $R$ is commutative.
Backwards: $R$ is commutative implies $a^2 - b^2 = (a-b)(a+b) \forall a,b \in R$.
Let $x = (a+b)$. Then $(a-b)x = ax - bx = a(a+b) - b(a+b) = a^2 + ab - ba - b^2$. $R$ is commutative, so $ab-ba = 0$, so $a^2 + ab - ba - b^2 = a^2 - b^2$.
| It looks good! The only holes are the following typos: $x(a+b)=xa+b$ and $-ba=ab=0$
In fact, you don't even need the substitution $x=a-b$, since you can simply multiply out the right-hand side normally assuming distributivity for the ring!
| {
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"timestamp": "2023-03-29T00:00:00",
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Fourier transform of $\operatorname{erfc}^2\left|x\right|$ Could you please help me to find the Fourier transform of
$$f(x)=\operatorname{erfc}^2\left|x\right|,$$
where $\operatorname{erfc}z$ denotes the the complementary error function.
| We have
$$ \int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|x|) e^{-i\xi x} \, dx = \frac{4}{\xi}e^{-\xi^{2}/4} \left\{ \operatorname{erfi}\left( \frac{\xi}{2} \right) - \operatorname{erfi}\left( \frac{\xi}{2\sqrt{2}} \right) \right\}, \tag{1} $$
where $\operatorname{erfi}$ is the imaginary error function defined by
$$\operatorname{erfi}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{t^{2}} \, dt. $$
So here goes my solution. Notice that we can write
$$ \operatorname{erfc}(|t|) = \frac{2}{\sqrt{\pi}} \int_{1}^{\infty} \left| t \right| e^{-t^{2}x^{2}} \, dx. $$
Using this, we can write
\begin{align*}
\int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|t|) e^{-i\xi t} \, dt
&= \frac{4}{\pi} \int_{-\infty}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} t^{2} e^{-(x^{2}+y^{2})t^{2}} e^{-i\xi t} \, dxdydt \\
{\scriptsize(\because \text{ Fubini})}
&= \frac{4}{\pi} \int_{1}^{\infty} \int_{1}^{\infty} \left( \int_{-\infty}^{\infty} t^{2} e^{-r^{2}t^{2}} e^{-i\xi t} \, dt \right) \, dxdy. \tag{2}
\end{align*}
Using some standard complex analysis technique, we can show that
$$ \int_{-\infty}^{\infty} t^{2} e^{-r^{2}t^{2}} e^{-i\xi t} \, dt = \frac{\sqrt{\pi}}{4} \frac{2r^{2} - \xi^{2}}{r^{5}} e^{-\xi^{2} / 4r^{2}}. \tag{3} $$
Indeed, we have
\begin{align*}
\int_{-\infty}^{\infty} t^{2} e^{-r^{2}t^{2}} e^{-i\xi t} \, dt
&= e^{-\xi^{2}/4r^{2}} \int_{-\infty}^{\infty} t^{2} \exp \left\{ - r^{2} \left( t + \frac{i\xi}{2r^{2}} \right)^{2} \right\} \, dt \\
{\scriptsize(\because \text{ contour shift})}
&= e^{-\xi^{2}/4r^{2}} \int_{-\infty}^{\infty} \left( t - \frac{i\xi}{2r^{2}} \right)^{2} e^{-r^{2}t^{2}} \, dt \\
&= e^{-\xi^{2}/4r^{2}} \int_{0}^{\infty} \frac{4r^{4}t^{2} - \xi^{2}}{2r^{4}} e^{-r^{2}t^{2}} \, dt,
\end{align*}
which immediately yields $\text{(3)}$ by exploiting the gamma function. Plugging $\text{(3)}$ back to our calculation $\text{(2)}$, we get
\begin{align*}
\int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|t|) e^{-i\xi t} \, dt
&= \frac{1}{\sqrt{\pi}} \int_{1}^{\infty} \int_{1}^{\infty} \frac{2r^{2} - \xi^{2}}{r^{5}} e^{-\xi^{2} / 4r^{2}} \, dxdy \\
{\scriptsize(\because \text{ symmetry})}
&= \frac{2}{\sqrt{\pi}} \iint_{1\leq y\leq x} \frac{2r^{2} - \xi^{2}}{r^{5}} e^{-\xi^{2} / 4r^{2}} \, dxdy \\
{\scriptsize(\because \text{ polar coordinate})}
&= \frac{2}{\sqrt{\pi}} \int_{0}^{\frac{\pi}{4}} \int_{\csc\theta}^{\infty} \frac{2r^{2} - \xi^{2}}{r^{4}} e^{-\xi^{2} / 4r^{2}} \, drd\theta.
\end{align*}
Using the substitution $u = \xi / 2r$, we get
\begin{align*}
\int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|t|) e^{-i\xi t} \, dt
&= \frac{8}{\xi\sqrt{\pi}} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{\xi}{2}\sin\theta} (1 - 2u^{2}) e^{-u^{2}} \, dud\theta \\
&= \frac{8}{\xi\sqrt{\pi}} \int_{0}^{\frac{\pi}{4}} \left[ u e^{-u^{2}} \right]_{0}^{\frac{\xi}{2}\sin\theta} \, d\theta \\
&= \frac{4}{\sqrt{\pi}} \int_{0}^{\frac{\pi}{4}} \sin \theta \exp\left( -\frac{\xi^{2}}{4} \sin^{2}\theta \right) \, d\theta.
\end{align*}
Finally, using the substitution $v = \frac{1}{2}\xi \cos\theta$, we get
\begin{align*}
\int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|t|) e^{-i\xi t} \, dt
&= \frac{8}{\xi \sqrt{\pi}} e^{-\frac{1}{4}\xi^{2}} \int_{\xi/2\sqrt{2}}^{\xi/2} e^{v^{2}} \, dv
= \frac{4}{\xi} e^{-\frac{1}{4}\xi^{2}} \left[ \operatorname{erfi}(v) \right]_{\xi/2\sqrt{2}}^{\xi/2}.
\end{align*}
This proves $\text{(1)}$ as desired.
| {
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Find the greatest common divisor (gcd) of $n^2 - 3n - 1$ and $2$ Find the greatest common divisor (gcd) of $n^2 - 3n - 1$ and $2$ considering that $n$ is an integer. Thanks.
| $2$ only has two divisors, 0 and 1. So $gcd(a,2)=1$ or $2, \forall a \in \mathbb{Z}$.
Case 1: $n$ is even. Then $n^2 \equiv 0 \pmod{2}, 3n\equiv 0 \pmod{2}.$ So then $gcd(n^2-3n-1,2)=1$ because $n^2-3n-1 \equiv 0+0-1 \equiv 1 \pmod{2}$.
Case 2: $n$ is odd. Then $n^2 \equiv 1 \pmod{2}, 3n \equiv 1 \pmod{2}.$ So then $gcd(n^2-3n-1,2)=1$ again because $n^2-3n-1 \equiv 1+1-1 \equiv 1 \pmod{2}$.
$$\therefore \forall n \in \mathbb{Z}, gcd(n^2-3n-1,2)=1.$$
Alternatively, by Fermat's Little Theorem, $a^2 \equiv a \pmod{2},$ so$$ n^2-3n-1 \equiv n-3n-1 \equiv -2n-1 \equiv 1 \pmod{2} \implies gcd(n^2-3n-1,2)=1.$$
| {
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Prove $ \left |\sin(x) - x + \frac{x^3}{3!} \right | < \frac{4}{15}$ Prove $ \left |\sin(x) - x + \dfrac{x^3}{3!} \right | < \dfrac{4}{15}$ $\forall x \in [-2,2]$
By Maclaurin's formula and Lagrange's remainder we have $\sin(x) = x - \dfrac{x^3}{3!} + \dfrac{\sin(\xi)}{5!}x^5$ for some $0<\xi<2$
subbing this in we get $\left|\dfrac{\sin(\xi)}{5!}x^5 \right| \leq \left |\dfrac{x^5}{5!} \right| \leq \dfrac{2^5}{5!} = \dfrac{4}{15}$, but the question has $<$ rather than $\leq$ - where have I done wrong?
edit: thinking the $\cos(\xi)$ should be there rather than $\sin(\xi)$
| From the Leibniz rule it is known that, if the sequence of $\frac{x^4}{5!},\frac{x^6}{7!},\frac{x^8}{9!},...$ is decreasing, which is the case for $x^2<6\cdot7=42$, $|x|\le6$ to get a round number, then
$$
0\le\frac{x^4}{5!}-\frac{x^6}{7!}
\le \frac{\sin x}{x}-1+\frac{x^2}{3!}
\le\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}
=\frac{x^4}{5!}-\frac{x^6}{7!}\left(1-\frac{x^2}{72}\right).
$$
Since under the assumed restrictions $1-\frac{x^2}{72}\ge\frac12$, we get
$$
\left|\sin x-x+\frac{x^3}{3!}\right|\le\frac{|x|^5}{5!}\left(1-\frac{x^2}{84}\right)<\frac{|x|^5}{5!}
$$
for $0<|x|<6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Matrices and determinants question. Establish that if A is the matrix
\begin{bmatrix}
b+c & a^2 & a \\
c+a & b^2 & b \\
a+b & c^2 & c \\
\end{bmatrix}
then $|A| = -(a-b)(b-c)(c-a)(a+b+c)$.
| using linearity with respect to the first column, those matrix have the same determinant:
\begin{bmatrix}
b+c & a^2 & a \\
c+a & b^2 & b \\
a+b & c^2 & c \\
\end{bmatrix}
\begin{bmatrix}
b+c + a& a^2 & a \\
c+a + b& b^2 & b \\
a+b +c & c^2 & c \\
\end{bmatrix} whose determinant is $a+b +c$ multiplied by
\begin{bmatrix}
1 & a^2 & a \\
1 & b^2 & b \\
1 & c^2 & c \\
\end{bmatrix} which is $-1 \times$ the Vandermonde determinant.
| {
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How to Prove This Trigonometry Identity? I have to prove that:
$$\tan^2\theta \sin^2\theta = \tan^2\theta - \sin^2 \theta$$
Here is what I have tried
$$\tan^2\theta \sin^2\theta$$
$$=\left(\frac{\sin^2\theta}{\cos^2\theta}\right)\left(\sin^2\theta\right)$$
$$=\frac{\sin^4\theta}{\cos^2\theta}$$
Not much of an attempt, but now I am stuck. What should I do next? Thanks in advance for your answers ;)
| $$\tan^2\theta \sin^2\theta+\sin^2 \theta=\sin^2\theta(\tan^2\theta +1) = \sin^2\theta\cdot\sec^2\theta=\frac{\sin^2\theta}{\cos^2\theta}$$
Alternatively, $$\frac1{ \sin^2\theta}-\frac1{\tan^2\theta}=\csc^2\theta-\cot^2\theta=1$$
| {
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"url": "https://math.stackexchange.com/questions/711580",
"timestamp": "2023-03-29T00:00:00",
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Integral $\int_0^1 \left(\arctan x \right)^2\,dx$ Evaluate
$$\int_0^1 \left(\arctan x \right)^2\,dx$$
The answer should be
$${\pi^2\over16} + \frac{\pi\ln(2)}{4} -C$$
where $C$ is Catalan's constant.
How do I proceed?
I tried doing integration by parts twice and got stuck at $$\int_0^1{\frac{\log\left(\frac1{x}+x\right)}{1+x^2}}\,dx$$
| If one know that $\displaystyle \int_0^{\dfrac{\pi}{2}}\log (\sin (x))dx=-\dfrac{\pi}{2}\log(2)$
so then :
$\displaystyle \int_0^\dfrac{\pi}{2}\dfrac{x}{\tan x }dx=\dfrac{\pi}{2}\log(2)$ (by integration by parts)
Let $I=\displaystyle \int_0^\dfrac{\pi}{2}\dfrac{x}{\tan x }dx$
Change of variable $u=\tan x$ :
$I=\displaystyle \int_0^{+\infty} \dfrac{\arctan x}{x(1+x^2)}dx$
$I=\displaystyle \int_0^1 \dfrac{\arctan x}{x(1+x^2)}dx+\int_1^{+\infty} \dfrac{\arctan x}{x(1+x^2)}dx$
in the second integral, in the right member make the change of variable $u=\dfrac{1}{x }$
$I= \displaystyle\int_0^1 \dfrac{\arctan x}{x(1+x^2)}dx+ \int_0^1 \dfrac{x\arctan
\Big(\dfrac{1}{x}\Big) }{1+x^2}dx$
For $x>0$ one have $\arctan\Big(\dfrac{1}{x}\Big)+\arctan x=\dfrac{\pi}{2}$
$I= \displaystyle\int_0^1 \dfrac{\arctan x}{x(1+x^2)}dx+\dfrac{\pi}{2}\int_0^1\dfrac{x}{1+x^2}dx-\int_0^1\dfrac{x\arctan x}{1+x^2}dx$
But:
$\dfrac{1}{x(1+x^2)}=\dfrac{1}{x}-\dfrac{x}{1+x^2}$
So then:
$I=\displaystyle\int_0^1 \dfrac{\arctan x}{x}dx-\displaystyle\int_0^1 \dfrac{x\arctan x}{1+x^2}dx+\dfrac{\pi}{2}\int_0^1\dfrac{x}{1+x^2}dx-\int_0^1\dfrac{x\arctan x}{1+x^2}dx$
Hence:
$I=\displaystyle\int_0^1 \dfrac{\arctan x}{x}dx-2\displaystyle\int_0^1 \dfrac{x\arctan x}{1+x^2}dx+\dfrac{\pi}{4}\Big[\log(1+x^2)\Big]_0^1$
and so:
$I=\displaystyle\int_0^1 \dfrac{\arctan x}{x}dx-2\displaystyle\int_0^1 \dfrac{x\arctan x
}{1+x^2}dx+\dfrac{\pi}{4}\log(2)$
But:
the derivative of $x\rightarrow(\arctan x)^2$ is $x\rightarrow\dfrac{2\arctan x }{1+x^2}$
So then:
$I=\displaystyle\int_0^1 \dfrac{\arctan x}{x}dx-\Big(\big[ x(\arctan x)^2\big]_0^1-\int_0^1(\arctan x)^2 dx\Big)+\dfrac{\pi}{4}\log(2)$
$I=\displaystyle\int_0^1\dfrac{\arctan x }{x}dx-\dfrac{\pi^2}{16}+\int_0^1(\arctan x)^2 dx$
But $I=\dfrac{\pi}{2}\log(2)$ so then:
$$\displaystyle\int_0^1 (\arctan x)^2
dx=\dfrac{\pi^2}{16}-G+\dfrac{\pi}{4}\log(2)$$
where $\displaystyle G=\int_0^1 \dfrac{\arctan x}{x}dx$ is the Catalan's constant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Product of two number $a$ and $b$ How to prove that if $a, b$ are two positive integers so that $a^2+b^2=4100$ and $a<b<2a$ then $ab=2000$.
| As pointed by another user, $a,b$ must be positive integers, otherwise it's easy to find a counterexample.
Let's find some limitation for $a$. From the condition we have:
$$4100 = a^2 + b^2 > 2a^2 \iff a^2 < 2050 \iff a < \sqrt{2050} \approx 45.27$$
Since $a$ is integer we can conclude $a\le 45$
Now for the lower bound:
$$4100 = a^2 + b^2 < a^2 + 4a^2 = 5a^2 \iff a^2 > 820 \iff a^2 > \sqrt{820} \approx 28.63$$
Now we found out that $29 \le a \le 45$
Now some modular arithemtic would come in use. Working modulo 8 we have:
$$a^2 + b^2 = 4100 \equiv 4 \pmod 8$$
Since $0,1,4$ are the only quadratic residues we can only make a sum of residues $4$ if $a^2 \equiv 2 \pmod 8$ and $b^2 \equiv 0 \pmod 8$ or other way around.
For the first case let: $a=2k$ and $b=4n$, where $k$ is an odd integer. Now back to the limitation we have:
$$29 \le a = 2k \le 45 \iff 15 \le k \le 22$$
Now since $k$ is odd integer integer we have: $k \in \{15,17,19,21\}$
Checking all possibilities for $k$ wouldn't give us any solution.
Now check the case when $a=4n$ and $b=2k$, where $k$ is odd integer.
You'll get $a=40, b=50$ as the only integer solution satisfying the requirements. So we have: $ab = 40 \cdot 50 = 2000$.
Hence the proof.
| {
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Do either of the following series converge: $\sum_{n = 1}^\infty \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots{2n}}$ Does $$\sum_{n = 1}^\infty \frac{1\cdot3\cdot5\cdots{2n-1}}{2\cdot4\cdot6\cdots{2n}}$$ converge and additionally does the following converge $$\sum_{n = 1}^\infty \frac{1^2\cdot3^2\cdot5^2\cdots{(2n-1)}^2}{2^2\cdot4^2\cdot6^2\cdots{(2n)}^2}$$
| Hint:
$$ \frac{1\cdot3\cdot5\dots{2n-1}}{2\cdot4\cdot6\dots{2n-2}}= \frac{3}{2} \frac{5}{4}... \frac{2n-1}{2n-2} >1$$
| {
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Another combinatorics problem: $\sum\limits_{k = 0}^n (-1)^k \binom{2n-k}k2^{2n-2k}=2n+1$ I am dealing with problem 10F in the textbook A Course in Combinatorics by J. H. van Lint and R. M. Wilson.
Problem 10F: Prove directly the equality
$\sum\limits_{k = 0}^n {{{\left( { - 1} \right)}^k}\left( {\begin{array}{*{20}{c}}
{2n - k}\\k\end{array}} \right){2^{2n - 2k}} = } 2n + 1$.
Sketch of the proof: Evaluate
$\sum\limits_{n = 0}^\infty {\sum\limits_{k = 0}^n {{{\left( { - 1} \right)}^k}\left( {\begin{array}{*{20}{c}}{2n - k}\\k\end{array}} \right){2^{2n - 2k}}{x^{2n}}} } $. To do this, use Hint 1. Then, use Hint 2.
Hint 1: $\sum\limits_{j = 0}^\infty {\left( {\begin{array}{*{20}{c}}{a + j}\\j
\end{array}} \right){x^j} = {{\left( {1 - x} \right)}^{ - a - 1}}} ,\,\,a \in {\mathbb{Z}^ + }$.
Hint 2:
${\sum\limits_{i = 0}^\infty {\left( {2i + 1} \right){x^{2i}} = \left( {\frac{x}{{1 - {x^2}}}} \right)} ^\prime }$.
Attempt: It is enough to show that
$\sum\limits_{n = 0}^\infty {\sum\limits_{k = 0}^n {{{\left( { - 1} \right)}^k}\left( {\begin{array}{*{20}{c}}
{2n - k}\\
k
\end{array}} \right){2^{2n - 2k}}{x^{2n}}} } = {\left( {\frac{x}{{1 - {x^2}}}} \right)^\prime }$. But I can not use Hint 1.
| Finally got it. First, note that we can take the inner sum to be $\sum_{k=0}^\infty$ since $\binom{2n-k}{k} = 0$ when $k > n$. Next, observe that
$$
\binom{2n-k}{k} = \binom{2(n-k) + k}{k} = \binom{2j + k}{k}
$$
where $j = n - k$. Then
\begin{align*}
\sum_{n \geq 0} \sum_{k \geq 0} (-1)^k \binom{2n - k}{k}2^{2(n-k)} x^{2n} &= \sum_{j \geq 0} \sum_{k \geq 0}(-1)^k \binom{2j + k}{k} 2^{2j} x^{2(j+k)}\\
&= \sum_{j \geq 0} (2x)^{2j} \sum_{k \geq 0} \binom{2j + k}{k} (-x^2)^k \, .
\end{align*}
By Hint 1, the inner sum is $\frac{1}{(1+x^2)^{2j+1}}$, so we find
\begin{align*}
\sum_{j \geq 0} (2x)^{2j} \sum_{k \geq 0} \binom{2j + k}{k} (-x^2)^k &= \sum_{j \geq 0} (2x)^{2j} \frac{1}{(1+x^2)^{2j+1}} = \frac{1}{1+x^2}\sum_{j \geq 0} \left(\frac{(2x)^2}{(1 + x^2)^2}\right)^j\\
&=\frac{1}{1+x^2} \frac{1}{1 - \frac{(2x)^2}{(1 + x^2)^2}} = \frac{1 + x^2}{(1 + x^2)^2 - (2x)^2}\\
&= \frac{1+x^2}{(1 + x^2 - 2x)(1 + x^2 + 2x)} = \frac{1+x^2}{(1 - x)^2(1 +x)^2} = \frac{1+x^2}{(1 - x^2)^2} \, .
\end{align*}
You can easily check using the quotient rule that this last function is indeed $\frac{d}{dx} \frac{x}{1 - x^2}$. The result then follows from Hint 2.
| {
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evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\frac{\pi}{4}+x\right)
$$
This changes the integral to
$$
\int \cos (2x)\cdot \ln \left(\tan \left(\frac{\pi}{4}+x\right)\right)\,dx
$$
Now let $t=\left(\frac{\pi}{4}+x\right)$ such that $dx = dt$. Then the integral with changed variables becomes
$$
\begin{align}
\int \cos \left(2t-\frac{\pi}{2}\right)\cdot \ln (\tan t)dt &= \int \sin (2t)\cdot \ln (\tan t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \frac{\sec^2(t)}{\tan t}\cdot \cos (2t)\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \cot (2t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\ln \left|\sin (2t)\right|
\end{align}
$$
where $t=\displaystyle \left(\frac{\pi}{4}+x\right)$.
Is this solution correct? Is there another method for finding the solution?
| Integrate by parts: $\int udv=uv-\int v du$, where
$$u=\ln\frac{\cos x+\sin x}{\cos x-\sin x}\Rightarrow du=\frac{\frac{(\cos x-\sin x)(-\sin x+\cos x)-(\cos x+\sin x)(-\sin x +\cos x) }{(\cos x-\sin x)^2}}{\frac{\cos x+\sin x}{\cos x-\sin x}}=...=\frac{2}{\cos 2x}dx$$
and
$$ dv=\cos 2x dx \Rightarrow v=\frac{1}{2}\sin 2x.$$
Then,
$$\int \cos 2x \ln(\frac{\cos x+\sin x}{\cos x-\sin x}) dx=\frac{1}{2}\sin 2x \ln\frac{\cos x+\sin x}{\cos x-\sin x}-\int \frac{1}{2}\sin 2x \frac{2}{\cos 2x} dx=$$
$$=\frac{1}{2}\sin 2x\ln \frac{\cos x+\sin x}{\cos x-\sin x}-\int \tan 2x dx= $$
$$=\frac{1}{2}\sin 2x \cdot\ln\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)-\frac{1}{2}\ln|\sec 2x|+c. $$
| {
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Integrate $\int\frac{5x-7}{x^2-3x+2}$ I want to integrate $\int\frac{5x-7}{x^2-3x+2}$ but my result differs from the one on Wolframalpha http://www.wolframalpha.com/input/?i=integrate+%285x-7%29%2F%28x%5E2-3x%2B2%29
I did the following steps:
$$\frac{5x-7}{(x-2)(x-1)} = \frac{A}{x-2}+\frac{B}{x-1}$$
$$5x-7 = A(x-1)+B(x-2)$$
$$5 = A + B$$
$$-7 = -A-2B$$
$$A=5-B\\
-7= -(5-B)-2B\\
-2 = -B\\
2 = B$$
Therefore $A = 3$ and $B=2$
$$\int\frac{3}{x-2}+\int\frac{2}{x-1}=3\ln(x-2)+2\ln(x-1) + C$$
While on Wolframalpha it is $3\ln(2-x)+2\ln(1-x)$ Where did I do the error?
| They are both valid. $$\frac{d}{dx}\ln(a-x)=\frac{-1}{a-x}=\frac{1}{x-a}=\frac{d}{dx}\ln(x-a)$$
| {
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Find the value of $k$ such that $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$. $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$.
(a) -1 (b) 1 (c) 2 (d) -2
I am getting the value of k: $-17/29$ after equating the remainders.
$p(x)= kx^3 + 4x^2 + 3x - 4/(x – 3)$: remainder= $(30k+32)$
$q(x)= x^3 - 4x + k/(x – 3)$: remainder= $(k+15)$
So,
$30k+32=k+15$
=> $k=(-17/29)$
Any help would be appreciated. :)
| By the polynomial remainder theorem, the remainder of $f(x)$ when divided by $x - a$ is equal to $f(a)$. So we have
$$p(3) = q(3)$$
$$27k + 41 = 15 + k$$
$$k = -1$$
| {
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If $\sum x_k= \sqrt{n}$, then $\sum\frac{x_k^2}{(x_k^2+1)^2}\leq\frac{n^2}{(n+1)^2}$ A new question has emerged after this one was successfully answered by r9m: If $a+b+c+d = 2$, then $\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}+\frac{c^2}{(c^2+1)^2}+\frac{d^2}{(d^2+1)^2}\le \frac{16}{25}$. I thought of this generalization. Does it hold?
$$\dfrac{x_1^2}{(x_1^2+1)^2}+\dfrac{x_2^2}{(x_2^2+1)^2}+\cdot\cdot\cdot+\dfrac{x_n^2}{(x_n^2+1)^2}\le \dfrac{n^2}{(n+1)^2}$$ with $$x_1+x_2+\cdot\cdot\cdot+x_n= \sqrt{n}$$ $$x_1,x_2,\cdot\cdot\cdot,x_n \ge0$$ $$ n \in \mathbb{N}$$
| Alternative way to reach the result, that is already proposed in the other answers:
Consider the non-linear maximization problem $$\max_{x_i}\sum_{i=1}^{n}\frac{x_i^2}{(x_i^2+1)^2}$$ subject to $x_1+x_2+\ldots+x_n=\sqrt{n}$ and $x_i\ge0$. In that case the Lagrange function is defined as $$Λ(x, λ)=\sum_{i=1}^{n}\frac{x_i^2}{(x_i^2+1)^2}+λ\left(\sum_{i=1}^{n}x_i-\sqrt{n}\right)$$ The equation $\nablaΛ=0$ yields $$λ=\frac{2x_i(1-x_i^2)}{(1+x_i^2)^3}$$ for all $1\le x_i \le n$. Obviously the point $x^0$ with $$x^0_i=\begin{cases}1, & 1\le i \le \lfloor \sqrt{n}\rfloor \\ \\0, & \lfloor \sqrt{n}\rfloor \le i \le n \end{cases}, \qquad λ=0$$ solves the above system (both sides are equal to zero) and is thus a stationary point. The value of the objective function in $x^0$ is equal to $$\lfloor \sqrt{n}\rfloor\frac14$$
It is easy to see that for $n \ge 16$ this values is greater than $1$ and thus larger than the RHS of the given inequality. For $n<16$ this point does not yield a value larger than the RHS.
Perhaps there are other stationary points that yield a higher value for the objective function, even for $n<16$.
| {
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Integrate with square root in square $\int \left(1 + \sqrt{\frac{x-1}{x+1}}\right)^2 dx$
How would you attack this? I've tried variable substitution with $t = \sqrt{\frac{x-1}{x+1}}$
| Expand the square
$$\int {\left( {1 + \frac{{x - 1}}{{x + 1}}} \right)dx + 2\int {\sqrt {\frac{{x - 1}}{{x + 1}}} } } dx$$
Rearrange the first integral and collect $2$
$$2\left[ {\int {\left( {1 - \frac{1}{{x + 1}}} \right)} \;dx + \int {\sqrt {\frac{{x - 1}}{{x + 1}}} dx} } \right]$$
First integral is easy; for the second substitute $\dfrac{x-1}{x+1}=t$
Solve for $x$ and substitute
$$x = \frac{{t + 1}}{{1 - t}};\quad dx = \frac{{2\,dt}}{{{{\left( {t - 1} \right)}^2}}}$$
$$\int {\sqrt {\frac{{x - 1}}{{x + 1}}} \,dx} = \int {\frac{{2\sqrt t }}{{{{\left( {t - 1} \right)}^2}}}} \;dt$$
Substitute again $u=\sqrt{t};\;t=u^2;\;dt=2u\,du$
$$\int {\frac{{2u \cdot 2u\,du}}{{{{\left( {{u^2} - 1} \right)}^2}}}} = \int {\frac{{4{u^2}\,du}}{{{{\left( {{u^2} - 1} \right)}^2}}}} $$
use partial fraction to get
$$\frac{{4{u^2}}}{{{{\left( {{u^2} - 1} \right)}^2}}} = \frac{1}{{{{\left( {u + 1} \right)}^2}}} + \frac{1}{{{{\left( {u - 1} \right)}^2}}} + \frac{1}{{u - 1}} - \frac{1}{{u + 1}}$$
and integrate
$$\int {\frac{{4{u^2}}}{{{{\left( {{u^2} - 1} \right)}^2}}}} \,du = \int {\frac{{du}}{{{{\left( {u + 1} \right)}^2}}}} + \int {\frac{{du}}{{{{\left( {u - 1} \right)}^2}}}} + \int {\frac{{du}}{{u - 1}}} - \int {\frac{{du}}{{u + 1}}} $$
$$ - \frac{1}{{u + 1}} + \frac{1}{{1 - u}} + \ln \left( {u - 1} \right) - \ln \left( {u + 1} \right) + C=$$
$$=\frac{{2u}}{{1 - {u^2}}} + \ln \frac{{u - 1}}{{u + 1}} + C = \frac{{2u}}{{1 - {u^2}}} + \ln \left( {1 - \frac{2}{{u + 1}}} \right) + C=$$
$$=\frac{{2\sqrt {\frac{{x - 1}}{{x + 1}}} }}{{1 - \frac{{x - 1}}{{x + 1}}}} + \ln \frac{{\sqrt {\frac{{x - 1}}{{x + 1}}} - 1}}{{\sqrt {\frac{{x - 1}}{{x + 1}}} + 1}} + C$$
Simplified result should be
$$2\left( {x+\ln \left( {\sqrt {\frac{{x - 1}}{{x + 1}}} - \frac{x}{{x + 1}}} \right) + \left( {x + 1} \right)\sqrt {\frac{{x - 1}}{{x + 1}}} } \right) + C$$
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Is there a general relation between $a/b$ and $(a+c)/(b+c)$ where $a,b,c > 0 $? Is there a general relation between $a/b$ and $(a+c)/(b+c)$ where $a,b> 0$ and $c\geq 0$ ?
Is there a general proof for that relation ?
| Suppose $a,b,c \in \mathbb N$, $a < b$.
$$\color{red}{\dfrac ab}
= \dfrac{a(b+c)}{b(b+c)}
= \dfrac{ab+ac}{b(b+c)}
\color{red}{<} \dfrac{ab+bc}{b(b+c)}
= \dfrac{b(a+c)}{b(b+c)}
= \color{red}{\dfrac{a+c}{b+c}}
\color{red}{<} \dfrac{b+c}{b+c} = \color{red}1$$
Suppose $a,b,c \in \mathbb N$, $a > b$.
$$\color{red}{\dfrac ab}
= \dfrac{a(b+c)}{b(b+c)}
= \dfrac{ab+ac}{b(b+c)}
\color{red}{>} \dfrac{ab+bc}{b(b+c)}
= \dfrac{b(a+c)}{b(b+c)}
= \color{red}{\dfrac{a+c}{b+c}}
\color{red}{>} \dfrac{b+c}{b+c} = \color{red}1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/734128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to prove it? Let $y_0\geqslant 2$, $y_n=y_{n-1}^2-2$, $n\in\mathbb{N}_+$, set $\displaystyle S_n=\sum_{k=0}^{n}\frac{1}{y_0\cdots y_k}$, how to prove
$$\lim_{n\to\infty}S_n=\frac{y_0-\sqrt{y_0^2-4}}{2}.$$
Do you have some idea?
I can only get that
$$y_n=\begin{cases}y_0, & n=0 \\ \left(\frac{y_0^2-2+\sqrt{y_0^4-4y_0^2}}{2}\right)^{2^{n-1}}+\left(\frac{y_0^2-2+\sqrt{y_0^4-4y_0^2}}{2}\right)^{-2^{n-1}}, & n\ge1\end{cases}.$$
| Since, $y_n=y_{n-1}^2-2$, we have $y_n^2=(y_{n-1}^2-2)^2=y_{n-1}^2(y_{n-1}^2-4)+4$.
or, $y_n^2-4=y_{n-1}^2(y_{n-1}^2-4)=y_{n-1}^2y_{n-2}^2(y_{n-2}^2-4)=\ldots=(\prod\limits_{i=0}^{n-1}y_i^2)(y_0^2-4)$.
Thus, $\dfrac{y_n^2-4}{\prod\limits_{i=1}^{n-1}y_i^2}=y_0^2-4$ for $n\ge1$.
$\implies \dfrac{\sqrt{y_n^2-4}}{\prod\limits_{i=0}^{n-1}y_i}=\sqrt{y_0^2-4}$
Thus, $\lim\limits_{n \to \infty} \dfrac{\sqrt{y_n^2-4}}{\prod\limits_{i=0}^{n-1}y_i} = \lim\limits_{n \to \infty} \dfrac{y_n}{\prod\limits_{i=0}^{n-1}y_i} = \sqrt{y_0^2-4}$
Now, $\displaystyle \frac{1}{y_0\cdots y_k} = \frac{1}{2}\bigg(\frac{y_k}{y_0\cdots y_{k-1}}-\frac{y_{k+1}}{y_0\cdots y_k}\bigg)$, for each $k\ge 1$.
Thus, it undergoes telescopic cancellation to yield,
$\displaystyle \lim\limits_{n \to \infty} S_n = \frac{1}{y_0} + \frac{y_1}{2y_0} - \frac{\sqrt{y_0^2-4}}{2} = \frac{y_0^2-\sqrt{y_0^2-4}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/735029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\sum_{n=1}^\infty \frac{1}{n! \sin(n)}$ diverge or converge? Does the series $$ \sum_{n=1}^\infty \frac 1 {n!\sin(n)}$$ converge or diverge? Even the necessary condition of the convergence is difficult to verify.
| Since at least one person seemed uncomfortable with the details of Hagen von Eitzen's answer, I thought it would be a nice exercise to flesh it out.
As he says, the irrationality measure says that there are only finitely many positive integers $p,q$ such that $|\pi - \frac{p}{q}| < q^{-8}$. Since the left side is never 0 ($\pi$ is irrational), we may find a constant $c$ such that for all positive integers $p,q$, we have $|\pi - \frac{p}{q}| \ge c q^{-8}$.
To elaborate, let $\{(p_1, q_1), \dots, (p_m, q_m)\}$ be all the pairs of positive integers such that $|\pi - \frac{p_i}{q_i}| < q_i^{-8}$. Then set $$c := \min\left\{1, q_i^8 \left|\pi - \frac{p_i}{q_i}\right| : i=1, \dots, m\right\}.$$ Since $c$ is the minimum of finitely many strictly positive numbers, we have $c>0$.
Now given a positive integer $n \ge 3$, let $k_n$ be the nearest integer to $\frac{n}{\pi}$. Then $|n - k_n \pi| \le \frac{\pi}{2}$. Noting that for any $|x| < \frac{\pi}{2}$, we have $|\sin x| \ge \frac{|x|}{2}$, we thus have
$$|\sin (n)| = |\sin(n-k_n\pi)| \ge \frac{|n-k_n\pi|}{2} = \frac{k_n}{2} \left| \frac{n}{k_n} - \pi \right| \ge \frac{k_n}{2} \cdot c k_n^{-8} = \frac{c}{2 k_n^7}. $$
I claim for sufficiently large $n$ we have $k_n \le \frac{n}{3}$. Specifically, since $k_n$ is the nearest integer to $\frac{n}{\pi}$, we have $$k_n \le \frac{n}{\pi} + \frac{1}{2} = \frac{n}{3} + \frac{1}{2} - \left(\frac{1}{3}-\frac{1}{\pi}\right)n.$$ Therefore, we will have $k_n \le \frac n 3$ for any $n$ such that $\frac{1}{2} - \left(\frac{1}{3}-\frac{1}{\pi}\right)n \le 0$. Solving this inequality for $n$, we find that we get $k_n \le \frac n 3$ if $$n \ge \frac{1}{2 \left(\frac{1}{3} - \frac{1}{\pi}\right)} \approx 33.28;$$ in particular, for all $n \ge 34$.
Hence for all $n \ge 34$ we have
$$|\sin(n)| \ge c' n^{-7}$$
where $c' = \frac{3^7 c}{2}$, and therefore
$$\left|\frac{1}{n! \sin(n)} \right| \le c' \frac{n^7}{n!}.$$
Since $\sum \frac{n^7}{n!}$ converges, we have that $\sum \frac{1}{n! \sin(n)}$ converges absolutely.
| {
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"url": "https://math.stackexchange.com/questions/736361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
Algebra Question ( possible values) If
$$k = \dfrac{a}{b+c} = \dfrac{b}{a+c} = \dfrac{c}{b+a}$$
How many possible values of $k$ are there?
| Add like this:
$$(b+c)k+(a+c)k+(b+a)k=a+b+c$$
$$k=\frac{a+b+c}{b+c+a+c+b+a}=\frac12$$
if $a+b+c\ne0$, else $a=-(b+c)\Longrightarrow\dfrac a{b+c}=-1$.
So there are $2$ possible values for $k$, because first is e.g. for $a=b=c=1$ and the second e.g. for $a=b=1, c=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/738282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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relation between sides of a triangle and circumradious prove that $a.b.c=4R. S_{\bigtriangleup ABC}$, ($S$ stands for surface) and ($R$ stands for circumradious)
any response would be appreciated
| A good starting point is to use the formula for area $S$ of $\bigtriangleup ABC$:
$S = \dfrac{1}{2} \cdot ab \cdot sinC$. We now prove this formula:Let $H$ be the point on side $BC$ such that $AH$ is perpendicular to $BC$, and let $h = AH$. So $S = \dfrac{1}{2}\cdot AH \cdot BC = \dfrac{1}{2} \cdot h \cdot BC = \dfrac{1}{2} \cdot (b\cdot sinC)\cdot a = \dfrac{1}{2} \cdot ab \cdot sinC$. Thus: $4S = 2\cdot ab \cdot sinC$ , and use the law
of sine for the same triangle: $sinC = \dfrac{c}{2R}$ and get: $4S = 2 \cdot ab \cdot \dfrac{c}{2R} = \dfrac{abc}{R}$. So $4R\cdot S_{\bigtriangleup ABC} = abc$.
Note: As for a quick proof of the law of sines in triangle $\bigtriangleup ABC$:Let $O$ be the center of the circumscribing circle of the $\bigtriangleup ABC$, and let $K$ be midpoint of side $AB$, then $\bigtriangleup AOB$ is isosceles since $OA = OB = R$. So $OK$ is the altitude of $\bigtriangleup ABC$. So $\angle C = \dfrac{\angle AOB}{2} = \angle AOK$, and then $sinC = sin\angle AOK = \dfrac{\frac{c}{2}}{R} = \dfrac{c}{2R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/739673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding an asymptotic expansion for $\sum_{k=0}^{n} \frac{1}{1+\frac{k}{n}}$ It is well known that an asymptotic expansion of the n-th harmonic number is $$H_{n}= \sum_{k=1}^{n} \frac{1}{k} \sim \ln(n) + \gamma + \frac{1}{2n} -\frac{1}{12n^{2}} + O(n^{-4}).$$
How could we find an asymptotic expansion for the sum $ \displaystyle \sum_{k=0}^n \frac{1}{1+\frac{k}{n}}$ to a similar order?
| $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\dd}{{\rm d}}
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\newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,{\rm Li}_{#1}}
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Since the series has a closed expression in terms of the Digamma Functions, it's interesting to check the expansion by means of the asymptotic behavior of those functions:
\begin{align}&\color{#66f}{\large\sum_{k=0}^{n}\frac{1}{1 + k/n}}
=n\sum_{k=0}^{n}\frac{1}{k + n}
=n\sum_{k=0}^{\infty}\pars{\frac{1}{k + n} - \frac{1}{k + 2n + 1}}
\\[5mm]&=n\bracks{\Psi\pars{2n + 1} - \Psi\pars{n}}\tag{1}
\end{align}
where $\ds{\Psi}$ is the Digamma Function and we used
${\bf 6.3.16}$. With
${\bf 6.3.5}$, expression $\pars{1}$ becomes:
\begin{align}&\color{#66f}{\large\sum_{k=0}^{n}\frac{1}{1 + k/n}}
=n\bracks{\Psi\pars{2n} - \Psi\pars{n}} + \half
\end{align}
With the
Digamma Asymptotic Expansion
${\bf 6.3.18}$:
\begin{align}&\color{#66f}{\large\sum_{k=0}^{n}\frac{1}{1 + k/n}}
\sim n\times
\\[5mm]&\braces{\!\!\bracks{%
\ln\pars{2n} - \frac{1}{4n} - \frac{1}{48n^{2}} + \frac{1}{1920n^{4}}
-\frac{1}{16128n^{6}}}\!\!
-\!\!\bracks{%
\ln\pars{n} - \frac{1}{2n} - \frac{1}{12n^{2}} + \frac{1}{120n^{4}}
-\frac{1}{252n^{6}}}}
\\[5mm]&-\half
\\[1cm]&=n\bracks{\ln\pars{2} + \frac{1}{4n} + \frac{1}{16n^{2}} - \frac{1}{128n^{4}} + \frac{1}{256n^{6}}} + \half
\\[5mm]&=\color{#66f}{\large n\ln\pars{2} + \frac{3}{4}
+ \frac{1}{16n} - \frac{1}{128n^{3}} + \frac{1}{256n^{5}}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/740088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Convergence of series $A_n = \sqrt{\sum_{k=n}^{\infty} a_k} - \sqrt{\sum_{k=n+1}^{\infty}a_k} $ if series $a_n$ converges I must show that if a series $\sum_{n=1}^{\infty} a_n$ with positive terms converges, then the series $\sum_{n=1}^{\infty} A_n$, where $A_n = \sqrt{\sum_{k=n}^{\infty} a_k} - \ \sqrt{\sum_{k=n+1}^{\infty} a_k},$ also converges, and $a_n = o(A_n)$ as $n\to\infty$.
Here is my (rough) attempt:
Since $\sum_{n=1}^{\infty}a_n$ is positive and finite, $\sqrt{\sum_{k=n}^{\infty}a_k} $ is defined $\forall n \in \Bbb N$.
By the Cauchy criterion, for all $\frac{\varepsilon^2}{4} > 0,$ there is a $N \in \Bbb N$ such that $|\ a_n+\ ...\ +a_m \ | < \frac{\varepsilon^2}{4}$ for all $m \geq n > N$.
Then for $| \ A_n+\ ...\ +A_m | = \left|\sqrt{\sum_{k=n}^{\infty} a_k} - \ \sqrt{\sum_{k=m+1}^{\infty} a_k} \ \right| \leq \left|\sqrt{\sum_{k=n}^{\infty} a_k}\right| + \left|\sqrt{\sum_{k=m+1}^{\infty} a_k}\right| $=
$ \sqrt{ \left | \sum_{k=n}^{\infty} a_n \right| }$ + $\sqrt{ \left | \sum_{k=m+1}^{\infty} a_n \right| }$ $ < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $, where $m \geq n > N$. Thus $A_n$ converges.
For $$\lim \limits_{n \to \infty} \frac{a_n}{ \left(\sqrt{\sum_{k=n}^{\infty} a_k} - \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)} = \lim \limits_{n \to \infty} \frac{a_n \cdot \left( \sqrt{\sum_{k=n}^{\infty} a_k} + \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)}{\left( \sqrt{\sum_{k=n}^{\infty} a_k} - \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)\cdot\left( \sqrt{\sum_{k=n}^{\infty} a_k} + \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)} =
\lim \limits_{n \to \infty} \frac{a_n \cdot \left( \sqrt{\sum_{k=n}^{\infty} a_k} + \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)}{ \left(\sum_{k=n}^{\infty} a_k - \ \sum_{k=n+1}^{\infty} a_k\right)} =
\lim \limits_{n \to \infty} \frac{a_n \cdot \left( \sqrt{\sum_{k=n}^{\infty} a_k} + \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)}{ a_n} =
\lim \limits_{n \to \infty} \sqrt{\sum_{k=n}^{\infty} a_k} + \ \sqrt{\sum_{k=n+1}^{\infty} a_k} = 0 + 0 = 0 $$
| Fix some sequence $(b_k)_{k\geqslant1}$ and define $c_k=b_k-b_{k+1}$ for every $k\geqslant1$. Then, for every $n\geqslant0$,
$$
\sum_{k=1}^nc_k=b_1-b_{n+1}.
$$
If the sequence $(b_k)$ converges, this implies that the series $\sum\limits_kc_k$ converges and that its sum is $b_1-b_\infty$, where $b_\infty=\lim\limits_{k\to\infty}b_k$.
Applying this to
$$
b_k=\sqrt{\sum_{i=k}^\infty a_i},
$$
one sees that the series $\sum\limits_kA_k$ converges and that its sum is
$$
\sum_{k=1}^\infty A_k=\sqrt{\sum_{i=1}^\infty a_i}.
$$
To prove that $a_n\ll A_n$, note that
$$
a_n=A_n\cdot\left(b_n+b_{n+1}\right),\qquad b_n+b_{n+1}\to0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/741156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Integrate $\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}dx$. HW Since this is a homework problem, a hint would be appreciated to help me get this started, since I have no idea how to start this. Thanks
Here's the problem:
Compute the improper integral:
$$\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}dx$$
| Here is the full solution
Using the substitution $x=\sin \theta$, the limit of the integrand changes from $x=0$ to $\theta =0$ and from $x=1$ to $\theta = \frac{\pi}{2}$ since $\theta=\arcsin(x)$.
Also, $dx=\cos \theta d\theta\Leftrightarrow dx=\sqrt {1 - {{\sin }^2}\theta } d\theta \Leftrightarrow dx=\sqrt {1 - {x^2}} d\theta \Leftrightarrow d\theta=\frac{{dx}}{{\sqrt {1 - {x^2}} }}$.
So we have,
$$I = \int\limits_0^1 {\frac{{\ln x}}{{\sqrt {1 - {x^2}} }}} dx = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin x)dx} $$
from the properties of definite integral, recall that
$$\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)dx} $$
$$ \Rightarrow I=\int\limits_0^{\frac{\pi }{2}} {\ln (\sin x)dx} = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {\frac{\pi }{2} - x} \right))dx} = \int\limits_0^{\frac{\pi }{2}} {\ln (\cos x)dx} $$
$$ \Rightarrow 2I = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin x) + \ln \left( {\cos x} \right)dx} = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin x\cos x)dx} $$
$$ \Rightarrow 2I = \int\limits_0^{\frac{\pi }{2}} {\ln (\frac{{\sin \left( {2x} \right)}}{2})dx} = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx} - \int\limits_0^{\frac{\pi }{2}} {\ln (2)dx} $$
$$ \Rightarrow 2I = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx} - \int\limits_0^{\frac{\pi }{2}} {\ln (2)dx} = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx} - \frac{\pi }{2}\ln \left( 2 \right)$$
In the $\int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx} $, if we let $2x = t \Rightarrow dx = \frac{{dt}}{2}$ and the limit of the integrand becomes $t = 0$ and $t=\pi$. That is, $\int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx} = \frac{1}{2}\int\limits_0^\pi {\ln (\sin \left( t \right))dt}$
Since
$$\int\limits_0^a {f\left( x \right)} dx = 2\int\limits_0^{{\textstyle{a \over 2}}} {f\left( x \right)} dx \quad\mbox{ if } f\left( {a - x} \right) = f\left( x \right)$$
$$ \Rightarrow \frac{1}{2}\int\limits_0^\pi {\ln (\sin \left( t \right))dt} = \frac{2}{2}\int\limits_0^{{\textstyle{\pi \over 2}}} {\ln (\sin \left( t \right))dt} = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( x \right))dx} \quad \mbox{ since } \sin \left( {\pi - x} \right) = \sin \left( x \right) \mbox{ and $t$, $x$ are dummy variables}$$
$$ \Rightarrow 2I = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( {2x} \right))dx} - \frac{\pi }{2}\ln \left( 2 \right) = I - \frac{\pi }{2}\ln \left( 2 \right)$$
$$ \Rightarrow I = - \frac{\pi }{2}\ln \left( 2 \right) = \frac{\pi }{2}\ln \left( {\frac{1}{2}} \right)$$
$$ \Rightarrow \int\limits_0^{\frac{\pi }{2}} {\ln (\sin \left( x \right))dx} = \frac{\pi }{2}\ln \left( {\frac{1}{2}} \right)$$
Any Question?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/741404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Trig Reduction with Pythagoras If $\sin 10 = p$, then determine $$\tan^2 30^\circ \times \tan^2190^\circ$$ in terms of p.
| $\tan^2 30^\circ$ is easy because it's a common angle; we know that it is equal to $\frac{1}{3}$
Note that $\tan^2 190^\circ = \tan^2 10^\circ = \frac{\sin^2 10^\circ}{\cos^2 10^\circ}$
Since $\cos^2 \theta = 1 - \sin^2 \theta$, we find that $\tan^2 190^\circ = \frac{p^2}{1 - p^2}$
Thus, the final answer is $\frac{p^2}{3(1 - p^2)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/742040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the last two digits of the number $9^{9^9}$
Find the last two digits of the number $9^{9^9}$ .
[Hint: $9^9 \equiv 9 \pmod {10} $; hence, $9^{9^9}$ = $9^9+10k$ ;now use the fact that $9^9 \equiv 89 \pmod {100}$]
| $$9^{9^9}=(10-1)^{9^9}\equiv(-1)^{9^9}+9^9\cdot10^1\cdot(-1)^{9^9-1}\equiv-1+10\cdot9^9\pmod{100}$$
Now, $\displaystyle9^9=(10-1)^9\equiv-1\pmod{10}\implies10\cdot9^9\equiv-10\pmod{10\cdot10}$
$$\implies9^{9^9}\equiv-1-10\pmod{100}\equiv100-11$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this $\cos{x}+\cos{y}+\cos{z}=1$ Question:
let $x,y,z\in R$ and such $x+y+z=\pi$,and such
$$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$$
show that
$$\cos{x}+\cos{y}+\cos{z}=1$$
My idea: let $$x+y-z=a,x+z-y=b,y+z-x=c$$
then
$$a+b+c=\pi$$
and
$$\tan{\dfrac{a}{4}}+\tan{\dfrac{b}{4}}+\tan{\dfrac{c}{4}}=1$$
we only prove
$$\cos{\dfrac{b+c}{2}}+\cos{\dfrac{a+c}{2}}+\cos{\dfrac{a+b}{2}}=1$$
Use
$$\cos{\dfrac{\pi-x}{2}}=\sin{\dfrac{x}{2}}$$
$$\Longleftrightarrow \sin{\dfrac{a}{2}}+\sin{\dfrac{b}{2}}+\sin{\dfrac{c}{2}}=1$$
let
$$\tan{\dfrac{a}{4}}=A,\tan{\dfrac{b}{4}}=B,\tan{\dfrac{\pi}{4}}=C$$
then
$$A+B+C=1$$
and use $$\sin{2x}=\dfrac{2\tan{x}}{1+\tan^2{x}}$$
so we only prove
$$\dfrac{2A}{1+A^2}+\dfrac{2B}{1+B^2}+\dfrac{2C}{1+C^2}=1$$
other idea:let
$$\dfrac{y+z-x}{4}=a,\dfrac{x+z-y}{4}=b,\dfrac{x+y-z}{4}=c$$
then we have
$$a+b+c=\dfrac{\pi}{4},\tan{a}+\tan{b}+\tan{c}=1$$
we only prove
$$\cos{(2(b+c)}+\cos{2(a+c)}+\cos{2(a+b)}=\sin{(2a)}+\sin{(2b)}+\sin{(2c)}=1$$
then I fell very ugly, can you some can help?
Thank you very much!
|
Looking down at the positive octant , ( arrow tips are coordinate axes).
$x+y+z=\pi$ ( the cyan colored plane. )
$\cos{x}+\cos{y}+\cos{z}=1$ , ( the pink colored area. )
$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$ , (the light gray colored area. )
I can see three solutions where the gray central area, surrounded by pink triangle meets the cyan plane.
Just a picture!
| {
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"url": "https://math.stackexchange.com/questions/749758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to find $\lim_{x\to\ln 2} \frac{2e^{3x}-16}{3e^{2x}-12}$? Need to find without using L'Hopital's rule or derivatives.
I know the answer is 2, but how can I find this analytically, without using limit tables?
Thanks!
| Let $u = e^x$. Then
\begin{align}
\lim_{x \rightarrow \ln 2} \frac{2e^{3x}-16}{3e^{2x}-12} &= \lim_{u \rightarrow 2} \frac{2u^3-16}{3u^2-12} &\text{substitute } u = e^{2x} \\
&= \lim_{u \rightarrow 2} \frac{2(u^3-8)}{3(u^2-4)} & \text{factor a 2 from top and a 3 from bottom} \\
&= \lim_{u \rightarrow 2} \frac{2(u-2)(u^2+2u+4)}{3(u+2)(u-2)} & \text{numerator: }(a^3-b^3)=(a-b)(a^2+ab+b^2)\\
&= \lim_{u \rightarrow 2} \frac{2(u^2+2u+4)}{3(u+2)} & (u-2)\text{'s cancel} \\
&= \frac{2((2)^2+2(2)+4)}{3((2)+2)} & \text{plug in $u=2$} \\
&= \frac{24}{12} = 2 \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/750312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Multiplying Adjacent Matrices? My teacher hasn't explained it too well, so i'm looking for an explanation:
$$A =
\begin{pmatrix}
0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{pmatrix}$$
$$A^2 =
\begin{pmatrix}
2 & 1 & 1\\
1 & 2 & 1\\
1 & 1 & 2
\end{pmatrix}$$
I would have thought the first number in $A^2$ would be $0$, as $0 \cdot 0 = 0$, and am struggling to see where the $2$ comes from.
| $A^2$ is simply $A\cdot A$, where the $\cdot$ denotes matrix multiplication.
Matrices are not multiplied elementwise (other than for addition), but as follows:
The matrix element of the resulting matrix which sits on row $i$ and column $j$ is calculated by multiplying each element of $i$-th row of the first matrix with the corresponding element of the $j$-th column of the second matrix, and add up all those products. For your specific case, when interested in the upper left element of
$$\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}\cdot\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}$$
you have to look at the first row of the first matrix,
$$\begin{pmatrix}0&1&1\end{pmatrix}$$
and the first column of the second matrix,
$$\begin{pmatrix}0\\1\\1\end{pmatrix}$$
and then find $0\cdot 0+1\cdot 1+1\cdot 1 = 2$. Therefore the upper left matrix element of the resulting matrix is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/751723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How find this system $a^2+b^2=3,a^2+c^2+ac=4,b^2+c^2+\sqrt{3}bc=7$ Find the this system real solution
$$\begin{cases}
a^2+b^2=3\\
a^2+c^2+ac=4\\
b^2+c^2+\sqrt{3}bc=7
\end{cases}$$
I think that one can use Geometry to solve this system. Maybe there exist an algebraic method.
$$a^2+b^2=\sqrt{3}^2$$
$$a^2+c^2-2ac\cos{(120°)}=2^2$$
$$b^2+c^2-2bc\cos{150°}=\sqrt{7}$$
and note $$150°+120°+90°=360°$$
This is Mathematics competition of Zhejiang province today.
| Assuming $a,b,c>0$, then as you noted the equalities are just cosine laws for $3$ triangles which form a larger triangle with sides $2,\sqrt3,\sqrt7$, because the angles add up to $2\pi$. That triangle is right, because $4+3=7$, so we can find the lengths analytically if we draw it like this:
Here we set $A=(0,0)$, $B=\left(0,\sqrt3\right)$ and $C=(2,0)$. The angle $APB$ is right, so $P$ lies on the circle with the center $S_1=\left(0,\frac{\sqrt3}2\right)$ and radius $\frac{\sqrt3}2$. The size of the angle $APC$ is $\frac{2\pi}3$, so $ACX$ is an equilateral triangle and $P$ lies on the circle with the center $S_2=\left(1,-\frac{\sqrt3}3\right)$ and radius $\frac{2\sqrt3}3$.
So if $P=(x,y)$, then
\begin{alignat*}{5}x^2+\left(y-\tfrac{\sqrt3}2\right)^2\ &=\left(\tfrac{\sqrt3}2\right)^2&&=\tfrac34&\ \Longleftrightarrow\ &&0\ &=x^2+y^2-\sqrt3\ y\\
(x-1)^2+\left(y+\tfrac{\sqrt3}3\right)^2\ &=\left(\tfrac{2\sqrt3}3\right)^2&&=\tfrac43&\ \Longleftrightarrow\ &&2x\ &=x^2+y^2+\tfrac{2\sqrt3}3y\end{alignat*}
So $x=\frac12\left(\tfrac{2\sqrt3}3+\sqrt3\right)y=\tfrac{5\sqrt3}6y$ and $0=\left(\tfrac{5\sqrt3}6y\right)^2+y^2-\sqrt3\ y\Longleftrightarrow0=y(111y-36\sqrt3)$, so $y=\frac{12\sqrt3}{37}$ and $x=\frac{30}{37}$, because we want the nonzero solution.
Therefore $a=\sqrt{\left(\frac{30}{37}\right)^2+\left(\frac{12\sqrt3}{37}\right)^2}=\dfrac6{\sqrt{37}}$, $b=\sqrt{\left(\frac{30}{37}\right)^2+\left(\sqrt3-\frac{12\sqrt3}{37}\right)^2}=\dfrac{5\sqrt3}{\sqrt{37}}$ and $c=\sqrt{\left(2-\frac{30}{37}\right)^2+\left(\frac{12\sqrt3}{37}\right)^2}=\dfrac8{\sqrt{37}}$.
If you want all real solutions, then it can be solved geometrically too. Just consider different signs before $ac$ and $bc$. For example $-ac=-2ac\cos\frac\pi3$ and $\sqrt3\,bc=-2bc\cos\frac{5\pi}6$. We see that $\frac\pi2+\frac\pi3=\frac{5\pi}6$, so we can draw the following picture:
The triangle $ACX$ is again equilateral and it can be solved in a similar fashion. This gives $a=\dfrac6{\sqrt{13}}, b=\dfrac{\sqrt3}{\sqrt{13}}, c=\dfrac8{\sqrt{13}}$ and if we negate $a$ we get a solution to the original problem.
Similar configurations can be made for the other $2$ possibilities, but there are no solutions, so these two we found together with their negations are all $4$ real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/751807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Finding the limit of a sequence by diagonalising a matrix Consider the sequence described by:
$\frac11 , \frac32 , \frac75 , ... ,\frac {a_{n}}{b_{n}}$
where $ a_{n+1} = a_n +2b_n $ and $b_{n+1} = a_n+b_n$
Find a matrix $A$ such that
$$\begin{bmatrix} a_{n+1} \\b_{n+1} \end{bmatrix} = A \begin{bmatrix} a_{n} \\b_{n} \end{bmatrix}$$
By diagonalising $A$ find explicit formulae for $a_n \text{ and } b_n $ and hence show that
$$\lim_{n \to \infty} \frac {a_n} {b_n} = \sqrt{2}$$
My progress so far:
$A= \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}$
with eigenvalues $ \lambda_1 = 1 + \sqrt{2} \text{ and } \lambda_2 = 1 -\sqrt{2}$ that correspond to eigenvectors $(\sqrt{2},1) , (-\sqrt{2},1)$.
So the matrix $P=\begin{bmatrix} \sqrt{2} & -\sqrt{2} \\ 1 & 1 \end{bmatrix}$ is such that $P^{-1}A P =\begin{bmatrix} 1+ \sqrt{2} & 0 \\ 0 & 1 - \sqrt{2} \end{bmatrix}$
I am unsure how this helps find explicit formulae for $ a_n \text{ and } b_n$.
Update:
The Solution I was given defined a coordinate system ,such that $\begin{bmatrix} 1 \\1 \end{bmatrix}$ becomes $P^{-1} \begin{bmatrix} 1 \\1 \end{bmatrix} = \begin{bmatrix} p_1 \\p_2 \end{bmatrix}$ where $$p_1 = \frac {\sqrt{2}+1} {2 \sqrt{2}} , p_2 = \frac {\sqrt{2}-
1} {2 \sqrt{2}} $$
then
$a_n = p_1 \lambda_1^{n-1} \sqrt{2} - p_2 \lambda_2^{n-1} \sqrt{2}$ and $b_n = p_1 \lambda_1^{n-1} + p_2 \lambda_2^{n-1} $
therefore $$\frac {a_n} {b_n} = \frac { p_1 \lambda_1^{n-1} \sqrt{2} - p_2 \lambda_2^{n-1} \sqrt{2}} {p_1 \lambda_1^{n-1} + p_2 \lambda_2^{n-1}}$$
$$ = \frac {\sqrt{2} -(p_2/p_1)(\lambda_2/ \lambda_1)^{n-1} \sqrt{2}} {1+(p_2/p_1)(\lambda_2/ \lambda_1)^{n-1}}$$
$$= \sqrt{2} ( \frac {1-(p_2/p_1)(\lambda_2/ \lambda_1)^{n-1} } {1 + (p_2/p_1)(\lambda_2/ \lambda_1)^{n-1}})$$
Since $ 0 <|\lambda_2 /\lambda_1| <1 $ we deduce that
$$\lim_{n \to \infty} \frac {a_n} {b_n} = \sqrt{2}$$
My Solution:
Upon following the advice of Git Gud and Mark Bennet, I was able to find:
$$ a_n = \frac12 (1 +\sqrt{2}) \lambda_1^{n-1}- \frac12 (\sqrt{2} -1) \lambda_2^{n-1} $$
and
$$ b_n = \frac {(1 +\sqrt{2}) \lambda_1^{n-1} - (\sqrt{2}-1 ) \lambda_2^{n-1}}{2\sqrt{2}}$$
so $ \frac {a_n} {b_n}$ simplifies to:
$$ \frac {a_n} {b_n} = \frac {\sqrt2 ((1 + \sqrt2)\lambda_1^{n-1} - (\sqrt2 -1 )\lambda_2^{n-1})} { (1 +\sqrt{2}) \lambda_1^{n-1} - (\sqrt{2}-1 ) \lambda_2^{n-1}} $$
*
*How and why are they able to define such a coordinate system?
*Why does this coordinate system simplify the question?
*Is there any discrepancy between my solution and their solution?
| Hint: From $\begin{bmatrix} a_{n+1} \\b_{n+1} \end{bmatrix} = A \begin{bmatrix} a_{n} \\b_{n} \end{bmatrix}$ find $\begin{bmatrix} a_{n+1} \\b_{n+1} \end{bmatrix} = A^k \begin{bmatrix} a_{0} \\b_{0} \end{bmatrix}$, for a suitable $k\in \mathbb N$.
Edit: Mark has already answered 1. and 2. As for 3., your solution is wrong:$$b_n = \frac {(1 +\sqrt{2}) \lambda_1^{n-1} \color{red}- (\sqrt{2}-1 ) \lambda_2^{n-1}}{2\sqrt{2}}.$$
The red minus sign should be a plus.
Also note that $1+\sqrt 2=\lambda _1$ and $\lambda _2=1-\sqrt 2$, this allows you to simplify the final expression a lot.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/751907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove $\sum\limits_{k=0}^n\ (3k^2+2k+1) = n^3 + 5 \begin{pmatrix}n+1\\2\end{pmatrix}+1$ $\sum\limits_{k=0}^n\ (3k^2+2k+1) = n^3 + 5
\begin{pmatrix}n+1\\2\end{pmatrix}+1$
How would you go on proving this equation? Doesn't have to be induction..
| $$(k+1)^3=k^3+3k^2+3k+1\quad=>\quad(k+1)^3-k^3=3k^2+3k+1\quad=>$$
$$=>\quad\sum_0^n(3k^2+2k+1)=\sum_0^n\Big[(3k^2+3k+1)-k\Big]=\sum_0^n\underbrace{(3k^2+2k+1)}_{\large(k+1)^3-k^3}-\sum_0^nk=$$
$$=(n+1)^3-\frac{n(n+1)}2=n^3+\underbrace{3n^2+3n}_{6\tfrac{n(n+1)}2}+1-\frac{n(n+1)}2=n^3+5\cdot\underbrace{\frac{n(n+1)}2}_{\Large{n+1\choose2}}+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/752610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Problem with trigonometric equation I am having trouble solving this equation
$$4\cdot \sin \theta + 2 \cdot \sin 2\theta =5$$
Thank you for your help.
| \begin{align}
4\sin \theta+2\sin 2\theta&=5\\
4\sin \theta+2\cdot2\sin \theta\cos\theta&=5\\
4\sin \theta+4\sin \theta\cos\theta&=5\\
4\sin \theta(1+\cos\theta)&=5
\end{align}
Take square both sides.
\begin{align}
4^2\sin^2 \theta(1+\cos\theta)^2&=5^2\\
16(1-\cos^2\theta)(1+\cos\theta)^2&=25
\end{align}
Let $x=\cos\theta$.
\begin{align}
16(1-x^2)(1+x)^2&=25\\
x^4+2x^3-2x-1&=-\frac{25}{16}\\
x^4+2x^3-2x&=-\frac{9}{16}\\
16x^4+32x^3-32x+9&=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/754542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the System of Equations in $x$ and $y$ \begin{equation}
x+\frac{3\,x-y}{x^2+y^2}=3 \tag{1}
\end{equation}
\begin{equation}
y=\frac{x+3\,y}{x^2+y^2} \tag{2}
\end{equation}
| I am assuming that you are solving this equation over the reals only. This approach is motivated because you have $x^2 + y^2$, and 'anti-symmetric' coefficients.
Set $z = x+iy $.
Take the first equation, add it to $i$ times the second equation, we get
$$ (x+iy) + \frac{ (3-i) ( x-iy) } { x^2 + y^2} = 3.$$
Converting this to $z$, we get
$$ z + \frac{ (3-i) \bar{z} } { |z|^2 } = 3.$$
Recall that $ z \bar{z} = |z|^2$. Multiplying throughout by $z$, we get
$$ z^2 - 3z + (3-i) = 0.$$
This has solutions $ z= 1-i, z = 2+i.$ This yields $(x,y) = (1,-1), (2,1) $.
If you want more problems along the line of using complex substitution, try Advanced System Of Equations and Making An Equation Complex Doesn't Make It Harder.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/755174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Showing $\lim_{n\to\infty} \left( \frac{n}{n^2+1^2} + \frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) = \frac{\pi}{4}$ How could I go about proving the following limit:
$$
\lim_{n\to\infty} \left( \frac{n}{n^2+1^2} + \frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) = \frac{\pi}{4}
$$
| Hint: Write each term individually as
$$\frac{n}{n^2 + k^2} = \frac{1}{n} \frac{1}{1 + (k/n)^2}$$
and try to recognize a Riemann sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/755905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder I am in my pre-academic year. We recently studied the Remainder sentence (at least that's what I think it translates) which states that any polynomial can be written as $P = Q\cdot L + R$
I am unable to solve the following:
Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder.
| I am going to prove it by modulo. In modulo $$
x^{2}+x+1 \equiv 0\left(\bmod x^{2}+x+1\right) \Rightarrow x+1\equiv -x^2\textrm{ and } x^3 \equiv 1 \left(\bmod x^{2}+x+1\right).
$$
In modulo $x^{2}+x+1$, we have
$$
\begin{aligned}
(x+1)^{2 n+1}+x^{n+2} \equiv &\left(-x^{2}\right)^{2 n+1}+x^{n+2} \\
\equiv &-x^{4 n+2}+x^{n+2} \\
\equiv & x^{n+2}\left(1-x^{3 n}\right) \\
\equiv & x^{n+2}\left(1-1\right) \\
\equiv & 0
\end{aligned}
$$
Therefore it has no remainder when divided by $x^{2}+x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Does the series $\sum_{n=1}^\infty \left|\frac{\cos2^n}{n}\right|$ converge or not? $$\sum_{n=1}^\infty~\left|\frac{\cos2^n}{n}\right|$$
I just confused what to do.
I am sorry, i don't understand. But this task is important for me. Can you give full solution?
| I guess the homework tag has now become irrelevant and it's about understanding the argument. So let's spell out what user2566092 hinted.
The addition theorem for the cosine is
$$\cos (x+y) = \cos x \cos y - \sin x \sin y.$$
Specialising $y$ to $x$, we obtain the double-angle formula
$$\cos (2x) = \cos^2 x - \sin^2 x.$$
With the identity $\cos^2 x + \sin^2 x \equiv 1$, we can write that as
$$\cos (2x) = \cos^2 x - (1-\cos^2 x) = 2\cos^2 x - 1.$$
If $\lvert \cos x\rvert$ is small, then $\cos^2 x$ is very small, and $2\cos^2 x - 1$ is close to $-1$, and therefore $\lvert \cos (2x)\rvert$ is then not small.
We can quantify that: If $\lvert\cos x\rvert \leqslant \frac{1}{2}$, then $0 \leqslant \cos^2 x \leqslant \frac{1}{4}$, and $\cos (2x) \leqslant 2\cdot\frac{1}{4} - 1 = -\frac{1}{2}$, so then $\lvert \cos (2x)\rvert \geqslant \frac{1}{2}$. Thus in the sequence $a_n = \lvert\cos 2^n\rvert$, of every two consecutive terms, at least one is $\frac{1}{2}$ or greater. Therefore, grouping two consecutive terms, we see that
$$\sum_{n=1}^{2k} \left\lvert \frac{\cos 2^n}{n}\right\rvert \geqslant \sum_{m=1}^k \frac{\lvert\cos 2^{2m-1}\rvert + \lvert\cos 2^{2m}\rvert}{2m} \geqslant \sum_{m=1}^k \frac{1/2}{2m} = \frac{1}{4}\sum_{m=1}^k \frac{1}{m} > \frac{1}{4}\log k,$$
which shows that the series does not converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/758340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Interesting association between tangent lines of slope one and ellipses Why is it that a tangent line with slope $1$ to an ellipse centered at the origin will have a transformation of $\pm \sqrt{a^2 +b^2}$ where $a$ and $b$ are the major and minor axis of the ellipse?
For example: The tangent line of slope one to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is $y=x+\sqrt{13}$.
| It's actually quite simple. An ellipse centered at the origin with semimajor and semiminor axes $a$ and $b$ can be transformed to a circle by a suitable scaling of the horizontal axis by a factor of $b/a$. Since such a linear transformation preserves tangency, the resulting line has a slope that is also scaled by the same factor; i.e., if $y = x + \beta$ is the tangent line, then the transformed line after the scaling has the form $y' = \frac{a}{b}x' + \beta$, where $(x',y') = (\frac{b}{a}x, \frac{b}{a}y)$. Now, such a line is tangent to the circle $(x')^2 + (y')^2 = b^2$, so we have by similarity of right triangles $$\frac{a}{b} = \frac{\sqrt{\beta^2 - b^2}}{b},$$ or equivalently, $$\beta^2 = a^2 + b^2,$$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/758638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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eigenvectors of a matrix Good Day,
I have a matrix of
\begin{bmatrix}
28 & 10\\
10 & 19
\end{bmatrix}
I have found the eigenvalues...
first eigenvalue (v1) : 24 + 5sqrt5 = rd off to 35.18
second eigenvalue (v2) : 24 - 5sqrt5 = rd off to -12.81
Following the page above,
if eigenvalue = 35.18, -6.18x+10y = 0 and 10x-16.18y=0
if eigenvalue = 12.82, 16.18x+10y = 0 and 10x+6.18y=0
How should I proceed from here to get the Eigenvector?
With egreg's help, I am able to derive to [(sqrt{5}+1)/2 1] and [(sqrt5-1)/2 1].
There's another part of the question as shown below:
This is my working currently:
Is my current working correct (before getting matrix of abcd)?
I have expanded to the following...
$\frac{Yn}{Xn} = \frac{[[(\frac{\sqrt{5}+1}{2})(24+5\sqrt{5})^{n}+(\frac{\sqrt{5}-1}{2})(24+5\sqrt{5})^{n}]+90[(\frac{\sqrt{5}+1}{2})(\frac{1-\sqrt{5}}{2})(24+5\sqrt{5})^{n}+(\frac{\sqrt{5}-1}{2})(\frac{\sqrt{5}+1}{2})(24-5\sqrt{5})^{n}]]}{[(24+5\sqrt{5})^{n}+(24+5\sqrt{5})^{n}]+90[(\frac{1-5\sqrt{5}}{2})(24+5\sqrt{5})^{n}+(\frac{\sqrt{5}+1}{2})(24-5\sqrt{5})^{n}]}$
How can I simplify it?
Thanks.
| You shouldn't round, if you can avoid it. Your matrix is
\begin{bmatrix}
29 & 10 \\
10 & 19
\end{bmatrix}
whose characteristic polynomial is
$$
X^2 - 48X + 451
$$
The roots are given by the formula
$$
\frac{48\pm\sqrt{48^2-4\cdot 451}}{2}
$$
so they are $24+5\sqrt{5}$ and $25-5\sqrt{5}$, so you computed correctly. An eigenvector relative to $25+5\sqrt{5}$ is a non zero solution of
$$
\begin{bmatrix}
29-(24+5\sqrt{5}) & 10 \\
10 & 19-(24+5\sqrt{5})
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=
\begin{bmatrix}
0\\
0
\end{bmatrix}
$$
However, since we know that only one equation is sufficient, because by definition of eigenvalue the matrix has rank less than $2$, we can simply solve
$$
(5-5\sqrt{5})x_1+10x_2=0
$$
or
$$
(1-\sqrt{5})x_1+2x_2=0
$$
We can obviously give any (non zero) value to $x_2$, so we set $x_2=1$ and so
$$
x_1=\frac{2}{\sqrt{5}-1}=\frac{2}{\sqrt{5}-1}\frac{\sqrt{5}+1}{\sqrt{5}+1}
=\frac{2(\sqrt{5}+1)}{5-1}=\frac{\sqrt{5}+1}{2}
$$
and an eigenvector is
\begin{bmatrix}
\frac{\sqrt{5}+1}{2}\\[2ex]
1
\end{bmatrix}
Similarly for the other eigenvector.
| {
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"url": "https://math.stackexchange.com/questions/759646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Integration by reduction I have learnt how to integrate by reduction formula but this one seems to give me hell someone to lift me by telling me what to do or simply to solve it.
\begin{equation}
I_n=\int\sec^n x\,dx
\end{equation}
| Write the integral as
\begin{align}
\int\sec^n x\,dx=\int\sec^{n-2}x\sec^2 x\,dx
\end{align}
Then use integration by parts. Let
\begin{align}
u&=\sec^{n-2}x\\
du&=(n-2)\sec^{n-3}x\sec x\tan x\,dx=(n-2)\sec^{n-2}x\tan x\,dx\\
\end{align}
and
\begin{align}
dv&=\sec^{2}x\\
v&=\int\sec^{2}x\,dx=\int\,d(\tan x)=\tan x\\
\end{align}
Hence
\begin{align}
\int\sec^{n-2}x\sec^2 x\,dx&=\sec^{n-2}x\tan x-(n-2)\int\tan x\sec^{n-2}x\tan x\,dx\\
&=\sec^{n-2}x\tan x-(n-2)\int\sec^{n-2}x\tan^2x\,dx\\
&=\sec^{n-2}x\tan x-(n-2)\int\sec^{n-2}x(\sec^2 x-1)\,dx\\
&=\sec^{n-2}x\tan x-(n-2)\int\sec^{n}x\,dx+(n-2)\int\sec^{n-2}x\,dx\\
I_n&=\sec^{n-2}x\tan x-(n-2)I_n+(n-2)I_{n-2}\\
I_n+(n-2)I_n&=\sec^{n-2}x\tan x+(n-2)I_{n-2}\\
(n-1)I_n&=\sec^{n-2}x\tan x+(n-2)I_{n-2}\\
I_n&=\frac{\sec^{n-2}x\tan x+(n-2)I_{n-2}}{n-1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/760583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
is $324+455^n$ ever prime Another question that I can only solve in part.
Is there an $n$ such that $324+455^n$ is prime?
When $n$ is odd, this is false since
$$ 324+455^n = (2\cdot 3^2)^2+(5\cdot 91)^n \equiv (-1)^2+(5\cdot 15)^n \equiv 1+(6\cdot 3)^n \equiv 1+(-1)^n\pmod{19}$$
so that $19\mid 324+455^n$ whenever $n$ is odd. Now, what about $n$ even?
| Let's analyse three possible cases:
If $n\equiv 1\pmod 2$, we have $455^n+324 \equiv 0\pmod {19}$ because
$$\begin{eqnarray}
(455^1+324) & = & 19\times 41 \\
(455^2-1) & = & 19\times 10896\\
\end{eqnarray}$$
If $n\equiv 2\pmod 4$, we have $455^n+324 \equiv 0\pmod {17}$ because
$$\begin{eqnarray}
(455^2+324) & = & 17\times 12197\\
(455^4-1) & = & 17\times 2521138272\\
\end{eqnarray}$$
Finally, if $n=4k$, we can apply the equality $$x^4+4y^4 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)$$ to $x=455^k$ and $y=3$ to obtain $$455^{4k}+4\cdot3^4 = (455^{2k} + 6\cdot 455^k + 18)(455^{2k}-6\cdot 455^k + 18)$$
Since both factors are strictly greater than $1$, their product is certainly a composite number.
Thus, the quantity $455^n+324$ is not a prime for any natural number $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/760966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to evaluate this expression: $∫\frac{\sin2x}{(1 - \cos2x)^4}\:\mathrm{d}x$?
How to evaluate: $$∫\frac{\sin2x}{(1 - \cos2x)^4}\:\mathrm{d}x$$ If anyone knows the answer, please help.
| $$
\int \frac{\sin 2x dx}{(1 - \cos 2x)^4} = \frac{1}{2}\int\frac{\sin 2x d(2x)}{(1 - \cos 2x)^4} = \frac{1}{2}\int \frac{d(1 - \cos 2x)}{(1 - \cos 2x)^4} = \frac{1}{2}\int u^{-4}du
$$
$$
= -\frac{1}{6}(\cos 2x)^{-3} = -\frac{1}{6\cos^3(2x)} + C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/761695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Trigonometric problem: Elevation angle The elevation of the top of a tower $KT$ from a point $A$ is $27^\circ$. At another point $B$, $50$ meters nearer to the foot of the tower where $ABK$ is a straight line, the angle of elevation is $40^\circ$. Find the height of the tower $KT$.
| Let $AK=d$, $BK=d-50$, and $KT=h$. Then
$$
\tan 27^\circ=\frac{h}{d}\qquad\rightarrow\qquad d=\frac{h}{\tan 27^\circ}\tag1
$$
and
$$
\tan 40^\circ=\frac{h}{d-50}\qquad\rightarrow\qquad d-50=\frac{h}{\tan 40^\circ}\tag2
$$
Substitute $(1)$ to $(2)$, yield
\begin{align}
\frac{h}{\tan 27^\circ}-50&=\frac{h}{\tan 40^\circ}\\
\frac{h}{\tan 27^\circ}-\frac{h}{\tan 40^\circ}&=50\\
h\left(\frac{1}{\tan 27^\circ}-\frac{1}{\tan 40^\circ}\right)&=50\\
h\left(\frac{\tan 40^\circ-\tan 27^\circ}{\tan 27^\circ\tan 40^\circ}\right)&=50\\
h&=50\left(\frac{\tan 27^\circ\tan 40^\circ}{\tan 40^\circ-\tan 27^\circ}\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/763055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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When is the sum of consecutive squares a prime? For what integers $x$ do there exist $x$ consecutives integers, the sum of whose squares is prime?
I tried use $$1^2+2^2+...+n^2=\frac {n(n+1)(2n+1)}{6}$$
| The sum of the squares of the first $x$ consecutive integers, starting from $n+1$, equals
$$(n+1)^2+(n+2)^2+\ldots+(n+x)^2=\frac{(n+x)(n+x+1)(2(n+x)+1)}{6}-\frac{n(n+1)(2n+1)}{6}$$ $$=\frac{1}{6}\cdot x\cdot(6n^2+6nx+2x^2+6n+3x+1).$$
For this to be prime we must have $x\mid 6$ or
$$6n^2+6nx+2x^2+6n+3x+1\mid6.$$
Solving the quadratic equation
$$6n^2+6nx+2x^2+6n+3x+1=c,$$
for $n$ where $c\mid 6$ yields the solutions
$$n=-\frac{1}{2}(x+1)\pm\frac{1}{2}\sqrt{3+6c-3x^2},$$
which shows that $c\geq-\tfrac{1}{2}$ and $x\leq\sqrt{1+2c}$, which reduces to $c\in\{2,3,6\}$ and $x\in\{2,3\}$ as $x=1$ is impossible. The only values that yield integral solutions for $n$ are $c=3$ and $x=2$, with solutions $n=0$ and $n=3$ corresponding to the sums of squares
$$1^2+2^2=5\qquad\text{ and }\qquad 4^2+5^2=41.$$
Otherwise $x\mid 6$. For $x=2$, $x=3$ and $x=6$ we may take $n=1$ to find
$$2^2+3^2=13,\qquad 2^2+3^2+4^2=29,\qquad 2^2+3^2+4^2+5^2+6^2+7^2=139,$$
which are all prime, and $x=1$ is still impossible. Hence the values of $x$ for which there exists $x$ consecutive integers, the sum of whose squares is prime, are $2$, $3$ and $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/763773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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system of linear congruences How can I solve this system of linear congruences? $$2x\equiv 0\, \text{mod 3}\\ 3x\equiv 2\, \text{mod 5}\\ 5x\equiv 4\, \text{mod 7}$$
I don´t know where to start, I am having a lot of troubles, can you lend me a hand please?
| Let's solve this equation without using the CRT. so $2x = 3k$, and this requires that $ 2 | k$. So $k = 2t$. So $x = 3t$, and $3(3t) = 5m + 2$. So $9t = 5m + 2 = 9m + 2 - 4m$, and $t = m + \dfrac{2 - 4m}{9}$. So $2 - 4m = 9p$, and $m = \dfrac{2 -9p}{4} = -2p + \dfrac{2 - p}{4}$. So $p = 2 - 4q$. So $t = -2(2 - 4q) + q + 2 - 4q = -2 + 5q$. So $x = 3(-2 + 5q) = -6 + 15q$. So $5(-6 + 15q) = 7w + 4$ gives: $-30 + 75q = 7w + 4$ or $75q = 7w + 34 = 75w + 34 - 68w = 75w + 34(1 - 2w)$. So $q = w + \dfrac{34(1 - 2w)}{75}$. So $1 - 2w = 75u$. So $w = \dfrac{1 - 75u}{2} = -38u + \dfrac{u+1}{2}$. So $u = 2z - 1$. So $w = -38(2z - 1) + z = -75z + 38$. So $q = -75z + 38 + 34(2z - 1) = -7z + 4$. So $x = -6 + 15q = -6 + 15(-7z + 4) = -105z + 54$, with $z \in \mathbb{Z}$. So the answer is: $x = -105z + 54$, with $z \in \mathbb{Z}$. Check each equation holds.
| {
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"url": "https://math.stackexchange.com/questions/764014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Odd series convergence Prove that we have following inequality:
$1+ \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{397} > \frac{9}{4}$
Anybody can help me to figure it out?
| Note that $\frac{1}{2k-1}+\frac{1}{2k+1}=\frac{4k}{4k^2-1}>\frac{4k}{4k^2}=\frac{1}{k}$.
Denote $1+\frac{1}{3}+\ldots+\frac{1}{2k-1}$ by $S_k$, then $S_{199}=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\ldots+\frac{1}{397}$, so
\begin{align}
S_{199}&>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots\frac{1}{198}\\
&>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\ldots+\underbrace{\frac{1}{128}+\ldots+\frac{1}{128}}_{32\;\text{terms}}+\frac{1}{130}+\ldots+\frac{1}{198} \\
&>1+\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\ldots+\frac{32}{128}\\
&=1+\frac{1}{2}+\underbrace{\frac{1}{4}+\frac{1}{4}+\ldots\frac{1}{4}}_{6\;\text{terms}} \\
&=1+\frac{1}{2}+\frac{6}{4} \\
&=3
\end{align}
This means $1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\ldots+\frac{1}{397}>3>\frac{9}{4}$.
| {
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"url": "https://math.stackexchange.com/questions/764846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Are there complex solutions for $z^3=\bar z$ I'm asked to solve $z^3=\bar z$. I got $z=0, 1, -1$. Are there any complex solutions $a+bi$ to this though?
| $(a+bi)^3 = (a^3 - 3ab^2) + (3a^2b - b^3)i = a - bi$
$a^3 - 3ab^2 = a, 3a^2b - b^3 = -b$
If $a = 0$, then $-b^3 = -b$, and the only reals which statisfy that is $b = 0, 1, -1$
If $b = 0$, then similarly $a^3 = a$ so $a = 0, 1, -1$
So far this gives $0, -1, 1, i, -i$
If $a, b \neq 0$,
$a^2 - 3b^2 - 1 = 0, b^2 - 3a^2 - 1 = 0$. This implies $a = b$, so $a^2 - 3a^2 - 1 = 0 \rightarrow 2a^2 = -1$ which is impossible in the reals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/766599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Prove the inequality $\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$
If $a,b,c\in\mathbb R^+$ prove that:
$$\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$$
| By Cauchy Schwarz Lemma:
$$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\geqslant \frac{(a+b+c)^2}{2(a+b+c)}$$
Also By $HM\leqslant AM$
$$\frac 3 {\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}}\leqslant \frac{2(a+b+c)}{3}$$
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\geqslant\frac9 {2(a+b+c)}$$
Add both of them. Then apply $AM\geqslant GM$ on RHS
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Maximum value of $abc$ for $a, b, c > 0$ and $ab + bc + ca = 12$ $a,b,c$ are three positive real numbers such that $ab+bc+ca=12$.
Then find the maximum value of $abc$
| Applying the inequality $AM \ge GM $ on {$ab,bc,ca$}, we get:
$$\frac{ab+bc+ca}{3} \ge (a^2.b^2.c^2)^{\frac13}$$
$\Rightarrow$ $(abc)^{\frac23} \le \frac{12}{3} = 4$
i.e. the maximum value of abc is $4^{\frac32}$ = $8$.
Also, maximum value of $8$ is attained for $a=b=c=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Integral: $\int \frac{dx}{\sqrt{x^{2}-x+1}}$ How do I integrate this? $$\int \frac{dx}{\sqrt{x^{2}-x+1}}$$
I tried solving it, and I came up with $\ln\left | \frac{2\sqrt{x^{2}-x+1}+2x-1}{\sqrt{3}} \right |+C$. But the answer key says that the answer should be $\sinh^{-1}\left ( \frac{2x-1}{\sqrt{3}} \right )+C$. Any answer is very appreciated.
| Note that these are actually the same answer, since:
$$ \sinh^{-1} x = \ln \left(x + \sqrt{x^2+1} \right) $$
and
$$ \frac{2x-1}{\sqrt{3}} + \sqrt{\left(\frac{2x-1}{\sqrt{3}}\right)^2 + 1} = \frac{2x - 1 + 2\sqrt{x^2 - x + 1}}{\sqrt{3}}$$
So the answer you got is also correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Prove that $\frac{2^{4n + 2} + 1}{5}$ is composed for all $n \ge 2$ The problem is to prove that $$\frac{2^{4n + 2} +1 }{5}$$ is composed for every $n \ge 2$.
I've tried this way: write $2^{4n + 2} + 1$ as $1 - (-4)^{2n + 1}$ and $5 = 1 - (-4)$, then $$\frac{2^{4n + 2} + 1}{5} = \frac{1 - (-4)^{2n + 1}}{1 - (-4)} = 1 - 4 + 16 - 64 + \cdots + (-4)^{2n}$$
How to proceed from this?
Thanks!
| One approach. First observe that we must have
$$2^{4n + 2} = - 1 \pmod 5$$
This is true because it is equivalent to
$$4 * 16^n = -1 \pmod 5$$
And there are various ways of showing this holds for all $n$.
So we know the expression is an integer. As for compositeness, a hint is as follows. Note that
$$4x^4 + 1 = (2x^2 + 2x + 1)(2x^2 - 2x + 1)$$
What does this say about the numerator?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding primes $p$ such that $ \dfrac {p+1}2$ and $\dfrac{p-1}4$ are primes How many odd primes $p$ are there such that both $ \dfrac {p+1}2$ and $\dfrac{p-1}4$ are primes ?
| Think modulo 12. Either $p = 3$ or $p \cong \{1,5,7,11\} \mod 12$. $3$ doesn't work because $(3-1)/4$ isn't an integer. $\frac{\{1,5,7,11\}+1}{2} \cong \{1,3,4,6\} \mod 12$ so your $\frac{p+1}{2}$ prime requirement eliminates any $p \not \cong 1 \mod 12$. If $p \cong 1 \mod 12$ and $p>1$, then $12$ divides $p-1$, so $3$ divides $\frac{p-1}{4}$. If $\frac{p-1}{3\cdot4} = 1$ then we get the solution $p=13$. Otherwise some multiple of $3$ (other than just $3$) divides $\frac{p-1}{4}$, so it is not prime.
Therefore the only solution is $p=13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/772447",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Tricky Triangle Area Problem This was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\sqrt{33}$ in it. How could I have done this? The answer was $2\sqrt{6}$ if it helps.
Edited to add that it was a multiple choice question, with possible answers:
a. $2\sqrt{6}$
b. $5$
c. $3\sqrt{6}$
d. $5\sqrt{6}$
| You could've used Heron's formula straight away, actually.
$$\begin{align}
T & = \tfrac{1}{4} \sqrt{(a+b-c)(a-b+c)(-a+b+c)(a+b+c)} \\
& = \tfrac{1}{4} \sqrt{(2+5-\sqrt{33})(2-5+\sqrt{33})(-2+5+\sqrt{33})(2+5+\sqrt{33})} \\
& = \tfrac{1}{4} \sqrt{(7-\sqrt{33})(-3+\sqrt{33})(3+\sqrt{33})(7+\sqrt{33})} \\
& = \tfrac{1}{4} \sqrt{(7+\sqrt{33})(7-\sqrt{33})(\sqrt{33}+3)(\sqrt{33}-3)} \\
& = \tfrac{1}{4} \sqrt{(49-33)(33-9)} \\
& = \tfrac{1}{4} \sqrt{16 \cdot 24} \\
& = \tfrac{1}{4} \sqrt{64 \cdot 6} \\
& = \tfrac{8}{4} \sqrt{6} \\
& = 2 \sqrt{6}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/773504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
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How to solve this limit without using L'Hôspital? This limit wildly appeared in a test and, well, I could not solve it without using L'Hôspital rule - which were not allowed in the test. Can anyone help me?
$$\lim_{x \to +\infty}\sqrt{x-\sqrt{x}}-\sqrt{x-1}$$
Answer, thanks to @user71352.
$$
\lim_{x\to\infty}\frac{\sqrt{x-\sqrt{x}}-\sqrt{x-1}}{1}\cdot \frac{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}
$$
Multiply by the conjugate:
$$
\lim_{x\to\infty}\frac{(\sqrt{x-\sqrt{x}})^2-(\sqrt{x-1})^2}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}
$$
Multiply by $\frac{1}{\sqrt{x}}$:
$$
\lim_{x\to\infty}\frac{1-\sqrt{x}}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}\cdot\frac{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x}}}
$$
Knowing that $\lim_{x\to\infty}\frac{1}{x} = \lim_{x\to\infty}\frac{1}{\sqrt{x}} = 0$:
$$
\lim_{x\to\infty}\frac{\frac{1}{\sqrt{x}}-1}{\sqrt{1-\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{x}}}=\frac{0-1}{\sqrt{1-0}+\sqrt{1-0}}=-\frac{1}{2}
$$
| $$
\sqrt{x-\sqrt{x}}-\sqrt{x-1}=\frac{x-\sqrt{x}-x+1}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}=\frac{1-\sqrt{x}}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}=\frac{\frac{1}{\sqrt{x}}-1}{\sqrt{1-\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{x}}}
$$
| {
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"url": "https://math.stackexchange.com/questions/776523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Use logarithmic differentiation to find the derivative of $y = (1 +\frac1x)^{2x}$ Can someone guide me through solving this problem?
$$\dfrac{\mathrm d}{\mathrm dx}\left(1 +\dfrac1x\right)^{2x}$$
| Take logs on both sides and differentiate,
\begin{align}
y &=\left(1+\frac{1}{x}\right)^{2x}\\
\ln y &=2x \ln \left(1+\frac{1}{x}\right)\\
\frac{\frac{dy}{dx}}{ y} &=2 \ln \left(1+\frac{1}{x}\right)-2 x\frac{\frac{1}{x^2}}{1+\frac{1}{x}} \\
\frac{dy}{dx} &=2 y\ln \left(1+\frac{1}{x}\right)-2 yx\frac{\frac{1}{x^2}}{1+\frac{1}{x}}
\end{align}
and you can substitute in for $y$ and simplify.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Riemann-Stieltjes integral Compute the following Riemann-Stieltjes integral:
The integral from 0 to 3 of $dg(x)/(sqrt(1 + x^2))$ where $g(x) = |2x - 2| + |3x - 6| - |x - 3|$
What I have tried:
Using formula: sum of $f(c_i)*(g(x_(i+1)) - g(x-i))$.
--> $1/\sqrt(2)*(1 - 5) + 1/\sqrt(5)*(1 - 1) = -2\sqrt(2)$
I'm not sure that this is the correct formula to use, but I saw that another person used it on this site solving a similar question. I'm not sure if my answer is correct or not.
| Redefining the function g as follows
\begin{align}
g\left( x \right) &= \left| {2x - 2} \right| + \left| {3x - 6} \right| - \left| {x - 3} \right|
\\
&= \left\{ \begin{array}{l}
2x - 2,\,\,\,\,\,x \ge 1 \\
2 - 2x,\,\,\,\,\,0 < x < 1 \\
\end{array} \right. + \left\{ \begin{array}{l}
3x - 6,\,\,\,\,\,x \ge 2 \\
6 - 3x,\,\,\,\,\,0 < x < 2 \\
\end{array} \right. - \left\{ \begin{array}{l}
x - 3,\,\,\,\,\,x \ge 3 \\
3 - x,\,\,\,\,\,0 < x < 3 \\
\end{array} \right.
\\
&= \left\{ \begin{array}{l}
2x - 2 + 3x - 6 - x + 3,\,\,\,\,\,\,\,x > 3 \\
2x - 2 + 3x - 6 - 3 + x,\,\,\,\,\,\,\,2 < x \le 3 \\
2x - 2 + 6 - 3x - 3 + x,\,\,\,\,\,\,\,1 < x \le 2 \\
2 - 2x + 6 - 3x - 3 + x,\,\,\,\,\,\,\,0 \le x \le 1 \\
\end{array} \right.
\\
&= \left\{ \begin{array}{l}
4x - 5,\,\,\,\,\,\,\,x > 3 \\
6x - 11,\,\,\,\,\,\,2 < x \le 3 \\
2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 < x \le 2 \\
5 - 4x,\,\,\,\,\,\,\,0 \le x \le 1 \\
\end{array} \right.
\end{align}
then
\begin{align}\int_0^3 {\frac{{dg\left( x \right)}}{{\sqrt {1 + {x^2}} }}} &= \int_0^1 {\frac{{dg\left( x \right)}}{{\sqrt {1 + {x^2}} }}} + \int_1^2 {\frac{{dg\left( x \right)}}{{\sqrt {1 + {x^2}} }}} + \int_2^3 {\frac{{dg\left( x \right)}}{{\sqrt {1 + {x^2}} }}}
\\
&=\int_0^1 {\frac{{d\left( {5 - 4x} \right)}}{{\sqrt {1 + {x^2}} }}} + \int_1^2 {\frac{{d\left( 2 \right)}}{{\sqrt {1 + {x^2}} }}} + \int_2^3 {\frac{{d\left( {6x - 11} \right)}}{{\sqrt {1 + {x^2}} }}}
\\
&= - \int_0^1 {\frac{{4dx}}{{\sqrt {1 + {x^2}} }}} + 0 + \int_2^3 {\frac{{6dx}}{{\sqrt {1 + {x^2}} }}}
\end{align}
Finally using the substitution $x = \tan u$, then we get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/781966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How find the equation $\cot x=\frac{\sin 20^\circ - \sin 80^\circ \cos 20^\circ}{\sin 80^\circ \sin 20^\circ}$ let $x\in R$, and such
$$\cot x =\frac{\sin 20^\circ -\sin 80^\circ \cos 20^\circ}{\sin 80^\circ \sin 20^\circ}$$
Find $x$
my idea:
$$\cot x=\csc 80^\circ - \cot 20^\circ$$
then I can't
| Let $\theta=20^\circ$, and defin
$\Delta=\sin\theta-\sin4\theta\cos\theta+\sqrt{3}\sin4\theta\sin\theta
$.
Then
$$\eqalign{2\Delta&=2\sin\theta-(\sin5\theta+\sin3\theta)+\sqrt{3}(\cos3\theta-\cos5\theta)\cr
&=2\sin\theta-\sin5\theta-\sqrt{3}\cos5\theta\cr
&=2\sin\theta-2(\sin5\theta\cos60^\circ+\cos5\theta\sin60^\circ)\cr
&=2\sin\theta-2\sin8\theta=0
}
$$
From $\Delta=0$ we conclude that $cot x=-\sqrt{3}$ thus $x\in\left\{-\frac{\pi}{6}+\pi k:k\in\Bbb{Z}\right\}$.$\qquad\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/782921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How did Euler realize $x^4-4x^3+2x^2+4x+4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$? How did Euler find this factorization?
$$\small x^4 − 4x^3 + 2x^2 + 4x + 4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$$
where $\alpha = \sqrt{4+2\sqrt{7}}$
I know that he had some super powers, like he was sent to us from a super intelligent alien universe just to humiliate our intelligence, but how the hell did he do that three centuries ago? :|
| Once it is noticed that
$$
x^4-4x^3+2x^2+4x+4=(x-1)^4-4(x-1)^2+7
$$
the square can be completed to get
$$
\left((x-1)^2-2\right)^2+3=\color{#00A000}{\left((x-1)^2-2-i\sqrt3\right)}\color{#0000FF}{\left((x-1)^2-2+i\sqrt3\right)}
$$
Solving for $(\alpha+i\beta)^2=2+i\sqrt3$, we get $\alpha^2-\beta^2=2$ and $2\alpha\beta=\sqrt3$. Adding the squares and taking the square root gives $\alpha^2+\beta^2=\sqrt7$. Solving for $\alpha$ and $\beta$ yields
$$
\color{#00A000}{\left({\small\sqrt{\frac{2+\sqrt7}2}}+i\,{\small\sqrt{\frac{-2+\sqrt7}2}}\right)^2=2+i\sqrt3}
$$
$$
\color{#0000FF}{\left({\small\sqrt{\frac{2+\sqrt7}2}}-i\,{\small\sqrt{\frac{-2+\sqrt7}2}}\right)^2=2-i\sqrt3}
$$
The full factorization is
$$
\overset{\underbrace{\color{#0000FF}{\left[x-1-\sqrt{\frac{2+\sqrt7}2}+i\,\sqrt{\frac{-2+\sqrt7}2}\right]}\color{#00A000}{\left[x-1-\sqrt{\frac{2+\sqrt7}2}-i\,\sqrt{\frac{-2+\sqrt7}2}\right]}}_{}}
{\left[\left(x-1-\sqrt{\frac{2+\sqrt7}2}\right)^2+\frac{-2+\sqrt7}2\right]}
\overset{\underbrace{\color{#00A000}{\left[x-1+\sqrt{\frac{2+\sqrt7}2}+i\,\sqrt{\frac{-2+\sqrt7}2}\right]}\color{#0000FF}{\left[x-1+\sqrt{\frac{2+\sqrt7}2}-i\,\sqrt{\frac{-2+\sqrt7}2}\right]}}_{}}
{\left[\left(x-1+\sqrt{\frac{2+\sqrt7}2}\right)^2+\frac{-2+\sqrt7}2\right]}
$$
This gives the factorization sought.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 2
} |
To prove or disprove $abc=1$ $\implies$ $ \sqrt {a^4+a^2b^2+b^4}+\sqrt {b^4+b^2c^2+c^4}+\sqrt {c^4+c^2a^2+a^4} > \dfrac92$ Let $a,b,c$ be positive real numbers such that $abc=1$ , then is it true that
$ \sqrt {a^4+a^2b^2+b^4}+\sqrt {b^4+b^2c^2+c^4}+\sqrt {c^4+c^2a^2+a^4} > \dfrac92$ ?
| With the hope that the problem remains unchanged;
First see that:
$$
\sqrt{a^4+a^2b^2+b^4}\geq \sqrt 3ab
$$
Hence:
$$
\sqrt {a^4+a^2b^2+b^4}+\sqrt {b^4+b^2c^2+c^4}+\sqrt {c^4+c^2a^2+a^4}\\\geq \sqrt 3(ab+bc+ac)\geq 3\sqrt 3> \dfrac92
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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About a combinatorics problem I'm dealing with this problem. Show that $\sum\limits_{j = 1}^n {{{\left( { - 1} \right)}^{j + k}}\left( {\begin{array}{*{20}{c}}i\\j\end{array}} \right)} \left( {\begin{array}{*{20}{c}}j\\
k\end{array}} \right) = 0$ for $1 \le i,k \le n$ and $i \ne k$.
Attempt: Since
$\left( {\begin{array}{*{20}{c}}i\\j
\end{array}} \right) = 0$ for
$j > i$ and
$\left( {\begin{array}{*{20}{c}}
j\\
k
\end{array}} \right) = 0$ for
$j < k$, it can be written as
$\sum\limits_{j = k}^i {{{\left( { - 1} \right)}^{j + k}}\left( {\begin{array}{*{20}{c}}
i\\
j
\end{array}} \right)} \left( {\begin{array}{*{20}{c}}
j\\
k
\end{array}} \right)$. Also, we can let
$i > k$ and so
$i = k + a$ for some
$1 \le a \le n$. But I can not continue.
| The solution for odd values of $i + k$ is here.
\begin{equation}
\sum\limits_{j = k}^i {{{\left( { - 1} \right)}^{j + k}}\left( {\begin{array}{*{20}{c}}
i\\
j
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
j\\
k
\end{array}} \right) = \sum\limits_{a = 0}^{i - k} {{{\left( { - 1} \right)}^a}\left( {\begin{array}{*{20}{c}}
i\\
{k + a}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{k + a}\\
k
\end{array}} \right)} } .
\end{equation}
Since $i + k$ is odd, $i-k$ is odd. Then, there are even number terms in this sum. Observe that
\begin{equation}
\left( {\begin{array}{*{20}{c}}
i\\
{k + a}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{k + a}\\
k
\end{array}} \right) = \frac{{i!}}{{\left( {i - \left( {k + a} \right)} \right)!\left( {k + a} \right)!}}.\frac{{\left( {k + a} \right)!}}{{k!a!}}
\end{equation}
\begin{equation}
= \frac{{i!}}{{a!}}.\frac{1}{{\left( {i - \left( {k + a} \right)} \right)!k!}}
\end{equation}
Multiplying and dividing by $\left( {i - a} \right)!$, we get
\begin{equation}
\left( {\begin{array}{*{20}{c}}
i\\
{k + a}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{k + a}\\
k
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
i\\
{i - a}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{i - a}\\
k
\end{array}} \right).
\end{equation}
Observe that the sign of
${\left( { - 1} \right)^a}$ is different from the sign of
${\left( { - 1} \right)^{i - k - a}} = {\left( { - 1} \right)^{a + 1}}$ since
$i-a = k + \left( {i - k - a} \right)$. Hence, this sum consists of equal terms having opposite sign and so is equal to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/786172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\rightarrow 1}\frac{\sin{(x^2-1)}}{x-1}$ My problem is to calculate
$$\lim_{x\rightarrow 1}\frac{\sin{(x^2-1)}}{x-1}$$
I evaluated to $\frac{\sin(x+1)(x-1)}{x-1}$ and then to $\sin(x+1)$ but I wonder about the result of limit as x approach 1 of $\sin{(x+1)}$. I cannot think of any result I cannot connect to anything.
| Notice
$$ \frac{\sin ( x^2 - 1 )}{x-1} = \frac{\sin ( x^2 - 1 )}{x-1} \cdot \frac{ x+1}{x+1} = \frac{\sin ( x^2 - 1 )}{x^2-1} \cdot (x+1) \to_{x \to 1} 2 $$
since
$$ \frac{ \sin y}{y} \to_{y \to 0} 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/788048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
calc the length of $y = x^2-1$ I want to know the lenght of the curce $y = x^2-1$. To do this I use the integral method and get the integral $$\int \sqrt{(2x)^2+1)}dx$$. Substitute $2x = u$ and then $u+\sqrt{u^2+1} = s$ gives $$\frac{s^4-1}{8s^2}+\frac{1}{2}\ln(s)$$
The answer should be $\sqrt{5}+\ln(2+\sqrt{5})$. So the above primitive is correct if we integrate from 0 to 1.
However I've two problems with this:
Why is the integration from 0 to 1 and not from -1 to 1, or 0 to 1 and times 2?
I fail to simplify $\frac{s^4-1}{8s^2}$ to $\sqrt{5}$ and the only way I know that it is correct is by numerically solve it with a calculator. This is not okay...
=== Edit ===
I'm looking for the length of the curve below the x-axis
| I wrote about this topic together with two colleagues some years ago. We also derived almost the same formula as you gave which can be found as $L_{kurve}$ at the beginning of page 2 in the following (Danish) article:
LMFK-article on parabola length
As you can see, your answer is off by a factor $2$. This accounts for both problems! The limits $x=-1$ and $x=1$ works and $\sqrt 5$ 'pops out'. The primitive function should be
$$
\frac{s^4-1}{16s^2}+\frac{\ln(s)}{4}
$$
Plugging in $s=2+\sqrt 5$ the first fraction above becomes (noting that $s^2=9+4\sqrt 5$)
$$
\frac{s^4-1}{16s^2}=\frac{s^2}{16}-\frac{1}{16s^2}=\frac{9+4\sqrt 5}{16}-\frac{1}{16(9+4\sqrt 5)}
$$
then if we scale the last fraction by $(9-4\sqrt 5)$ noting that $(9+4\sqrt 5)(9-4\sqrt 5)=1$ we obtain
$$
\frac{s^4-1}{16s^2}=\frac{9+4\sqrt 5}{16}-\frac{9-4\sqrt 5}{16}=\frac{\sqrt 5}{2}
$$
Doing the same steps for $s=-2+\sqrt 5$ multiplying the last fraction by $(9+4\sqrt 5)$ this time will lead to a result of $-\frac{\sqrt 5}{2}$ and subtracting these two results we have the $\sqrt 5$ you asked for. There is one minor detail left. Subtracting the fractions with $\ln(s)$ one gets
$$
\frac{\ln(2+\sqrt 5)}{4}-\frac{\ln(-2+\sqrt 5)}{4}=\frac{\ln\left((2+\sqrt 5)/(-2+\sqrt 5)\right)}{4}
$$
where one can rewrite $(2+\sqrt 5)/(-2+\sqrt 5)=(2+\sqrt 5)^2$. Then dividing $\ln\left((2+\sqrt 5)^2\right)$ by $2$ to take the square root inside the logarithm we get the final result (that differs from yours!!!)
$$
\sqrt 5+\frac{\ln(2+\sqrt 5)}{2}
$$
This result can be verified by Wolfram Alpha Computation 1 and Wolfram Alpha Computation 2. Note that in computation 2 you should see the Alternative Form nr. 2 where the expression I gave is found.
| {
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"url": "https://math.stackexchange.com/questions/790740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Verify a vector is an eigenvector of a matrix I have been asked to verify whether $v = \begin{bmatrix}1\\4\end{bmatrix}$ is an eigenvector of $A = \begin{bmatrix}-3&1\\-3&8\end{bmatrix}$? If yes, find the eigenvalue.
The way that I approached this question is to find eigenvalues, then use eigenvalues to verify whether $v$ is an eigenvector of the matrix. Here is how I find the eigenvalues:
$$\begin{align*}
&det(A - \lambda I)\\
&=det(\begin{bmatrix}-3&1\\-3&8\end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix})\\
&=det(\begin{bmatrix} -3-\lambda & 1 \\ -3 & 8 - \lambda \end{bmatrix})\\
&=(-3-\lambda)(8-\lambda) - (-3)\\
\therefore \lambda &= \frac{1}{2} (5 \pm \sqrt{109})
\end{align*}
$$
To verify:
$$
\begin{align*}
Av = \lambda v\\
\begin{bmatrix}-3&1\\-3&8\end{bmatrix} \begin{bmatrix}1\\4\end{bmatrix} = \begin{bmatrix}1\\4\end{bmatrix}\\
\begin{bmatrix}1\\29\end{bmatrix} = \frac{1}{2} (5 \pm \sqrt{109}) \begin{bmatrix}1\\ 4 \end{bmatrix}
\end{align*}
$$
I am not sure this is the right approach. Appreciate your comment, insight, etc.
| This is the long way!
Recall that $v$ is an eigenvector of $A$ if $Av=\lambda v$ for some $\lambda$.
Here we have
$$
\begin{bmatrix}
-3 & 1 \\ -3 & 8
\end{bmatrix}
\begin{bmatrix}
1\\ 4
\end{bmatrix}=
\begin{bmatrix}
1\\ 29
\end{bmatrix}
$$
But is $\begin{bmatrix}
1\\ 29
\end{bmatrix}$ a scalar multiple of $\begin{bmatrix}
1\\ 4
\end{bmatrix}$?
No! So $\begin{bmatrix}1\\ 4\end{bmatrix}$ is not an eigenvector of $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/794396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Finding eigenvectors of a matrix I want to find all eigenvalues and eigenvectors of the matrix $\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix}$.
Here is how I find eigenvalues:
$$\begin{align*}
\det(A - \lambda I) &= \det \Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}\lambda&0&0\\0&\lambda&0\\0&0&\lambda \end{bmatrix} \Bigg)\\
&= \det \Bigg(\begin{bmatrix} -\lambda&1&0 \\ 0&-\lambda&1 \\ -1&0&-\lambda \end{bmatrix} \Bigg)\\
&= -\lambda^3 - 1\\
\therefore \lambda =& -1
\end{align*}$$
Using eigenvalue that I found ($-1$), I want to find eigenvectors:
$$\begin{align*}
(A - \lambda I)\vec{V} =& 0\\
\Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\Bigg) \begin{bmatrix}x\\y\\z \end{bmatrix} =& \begin{bmatrix}0\\0\\0\end{bmatrix}\\
\begin{bmatrix}1&1&0\\0&1&1\\-1&0&1\end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} = & \begin{bmatrix}0\\0\\0\end{bmatrix}\\
\begin{bmatrix} x+y \\ y+z \\ -x+z \end{bmatrix} = & \begin{bmatrix}0\\0\\0\end{bmatrix}\\
\end{align*}$$
But what I should do from now? What is really the eigenvectors? Does this means that I have unlimited eigenvectors and any number that satisfies three equations can be eigenvectors?
| Well, the polynomial $\lambda^3+1$ has two more (complex) roots, which means a rotation in a $2$ dimensional subspace.
In your last equation substitute $\lambda=-1$ and, say, $x=1$ to find one eigenvector.
(You are right: there are infinitely many eigenvectors if there is one as they always form a subspace.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Algebraic solution for the intersection point(s) of two parabolas I recently ran through an algebraic solution for the intersection point(s) of two parabolas $ax^2 + bx + c$ and $dx^2 + ex + f$ so that I could write a program that solved for them. The math goes like this:
$$
ax^2 - dx^2 + bx - ex + c - f = 0 \\
x^2(a - d) + x(b - e) = f - c \\
x^2(a - d) + x(b - e) + \frac{(b - e)^2}{4(a - d)} = f - c + \frac{(b - e)^2}{4(a - d)} \\
(x\sqrt{a - d} + \frac{b - e}{2\sqrt{a - d}})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\
(a - d)(x + \frac{b - e}{2(a - d)})^2 = f - c + \frac{(b - e)^2}{4(a - d)} \\
x + \frac{b - e}{2(a - d)} = \sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} \\
x = \pm\sqrt{\frac{f - c + \frac{(b - e)^2}{a - d}}{a - d}} - \frac{b - e}{2(a - d)} \\
$$
Then solving for $y$ is as simple as plugging $x$ into one of the equations.
$$
y = ax^2 + bx + c
$$
Is my solution for $x$ and $y$ correct? Is there a better way to solve for the intersection points?
| All the other answers are fine from a mathematical point of view, but they ignore the fact that using the quadratic formula is a very bad way to solve quadratic equations in computer code (using floating point arithmetic).
The problem arises when one of the roots is near zero. In this case, either the "$+$" or the "$-$" formula of the $\pm$ will cause you to subtract two numbers that are nearly equal, and this will result in catastrophic cancellation error. The problem is illustrated here, and the well-known solutions are described.
For other interesting examples, see George Forsythe's famous paper "Why a Math Book is not Enough".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/799519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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} |
How to prove $\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$ Question:
let $a,b,c>0$,show that:
$$\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$$
maybe this inequality can Use Holder inequality to solve it
$$\left(\sum_{cyc}\sqrt{\dfrac{ab}{2a^2+bc+ca}}\right)^2\sum_{cyc}(2a^2+bc+ca)\ge(\sum_{cyc}\sqrt[3]{ab})^3$$
then I can't prove it
| By Hölder's inequality we have $$\left(\sum_{cyc}\sqrt{\frac{a}{2bc + ca + abc}}\right)^2 \cdot \left(\sum_{cyc}\frac{2bc + ca + abc}{a}\right) \geq 27$$ Hence it suffices to prove that $$27 \geq \frac{81}{2} \cdot \frac{abc}{(a + b + c)^3} \cdot \left(\sum_{cyc}\frac{2bc + ca + abc}{a}\right)$$ which is obvious after homogenization and expansion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/800082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$ Hi how can we prove this integral below?
$$
I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}
$$
I tried to use
$$
I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx
$$
and now tried changing variables to $y=x(1-x)$ in order to write
$$
I\propto \int_0^1 \sum_{n=0}^\infty y^n
$$
however I do not know how to manipulate the $\log^2 x$ term when doing this procedure when doing this substitution. If we can do this the integral would be trivial from here.
Complex methods are okay also, if you want to use this method we have complex roots at $x=(-1)^{1/3}$. But what contour can we use suitable for the $\log^2x $ term?
Thanks
| Note
$$ \frac{1}{x^2-x+1}=\frac{1+x}{1+x^3}. $$
So
\begin{eqnarray*}
I&=&\int_0^1 \frac{\ln^2x}{x^2-x+1}dx=\int_0^1 \frac{(1+x)\ln^2x}{1+x^3}dx\\
&=&\sum_{n=0}^{\infty}(-1)^n \int_0^1(1+x)x^{3n}\ln^2xdx\\
&=&\sum_{n=0}^{\infty}(-1)^n \int_0^1(x^{3n}+x^{3n+1})\ln^2xdx\\
&=&2\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{(3n+1)^3}+\frac{1}{(3n+2)^3}\right)\\
&=&2\sum_{n=-\infty}^{\infty}(-1)^n\frac{1}{(3n+1)^3}\\
&=&\frac{10\pi^3}{81\sqrt3}.
\end{eqnarray*}
Here we use the following theorem
$$ \sum_{n=-\infty}^\infty (-1)^nf(n)=-\pi \sum_{k=1}^m\text{Re}(\frac{f(z)}{\sin\pi z},a_k) $$
where $a_1,a_2,\cdots,a_m$ are poles of $f(z)$. For $f(z)=\frac{1}{(3z+1)^3}$, $z=-\frac{1}{3}$ is the only pole and
$$ \text{Re}(\frac{f(z)}{\sin\pi z},-\frac{1}{3})=-\frac{5\pi^2}{81\sqrt3}. $$
Thus we have the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 8,
"answer_id": 4
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Calculus Question: $\int_2^\infty\frac{\log^3(x-1)}{x^2}dx$ I have just taken calculus quiz but I could not find $\displaystyle \int_2^\infty\frac{\log^3(x-1)}{x^2}dx$? Any help would be appreciated. Thanks in advance.
EDIT:
Forgot to mention, my tutor gave us hints about this question.
*
*Use Taylor series
*$\displaystyle \zeta(3)=\sum_{n=1}^{\infty}\frac{1}{n^3}$
Those the hints that she gave to us.
| By substitutions, the following integrals are equivalent:
\begin{align*}
\int_{2}^{\infty} \, \frac{\log^3(x-1)}{x^2}\, dx &= \int_{1}^{\infty} \, \frac{\log^3(x)}{(1+x)^2} \, dx\\
&= -\int_{0}^{1} \, \frac{\log^3(x)}{(1+x)^2}\, dx \tag 1
\end{align*}
$(1)$ can be written as a sum, consider:
\begin{align*}
\int_{0}^{1} \, \frac{x^a}{(1+x)^2} dx &= \int_{0}^{1} \, \sum_{n\ge 0} (-1)^n (n+1)\, x^{a+n} \, dx\\
&= \sum_{n\ge 0} \int_{0}^{1} \, (-1)^n (n+1)\, x^{a+n} \, dx\\
&= \sum_{n\ge 0} (-1)^n \frac{(n+1)}{a+n+1}\tag 2
\end{align*}
Differentiate $(2)$ w.r.t a thrice and set $a=0$, and from $(1)$,
\begin{align*}
\int_{2}^{\infty} \, \frac{\log^3(x-1)}{x^2}\, dx &= 6\, \sum_{n\ge 0}\frac{(-1)^n}{(n+1)^3}\\
&= \frac{9}{2} \zeta{(3)} \approx 5.40925606421817428
\end{align*}
So, a general result looks like:
\begin{align*}
\int_{2}^{\infty} \, \frac{\log^n(x-1)}{x^2}\, dx &= \left(1-\frac{1}{2^{n-1}}\right)n!\, \zeta{(n)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Summation with Ceilinged Logarithmic Function According to Johann Blieberger's paper - "Discrete Loops and Worst Case Performance" (1994):
$$
\sum_{i = 1}^{n}\left \lceil \log_2{(i)} \right \rceil = n\left \lceil \log_2{(n)} \right \rceil - 2^{\left \lceil \log_2{(n)} \right \rceil} + 1
$$
Now, I was wondering if someone knows what the following may equal?
$$
\sum_{i = 1}^{n}i\left \lceil \log_2{(i)} \right \rceil = ?
$$
| Based on @vonbrand's answer, using integration by parts is a good approximation:
Concretely,
$$
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = n.n(\left \lceil \log_a{(n)} \right \rceil - 1) - 1(\left \lceil \log_a{(1)} \right \rceil - 1) - \int_{1}^{n}i(\left \lceil \log_a{(i)} \right \rceil - 1) di
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = n^2(\left \lceil \log_a{(n)} \right \rceil - 1) + 1 - \left [ \frac{i^2(2\left \lceil \log_a{(i)} \right \rceil - 3)}{4}\right ]_{1}^{n}
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = n^2(\left \lceil \log_a{(n)} \right \rceil - 1) + 1 - \left [ \frac{n^2(2\left \lceil \log_a{(n)} \right \rceil - 3)}{4} - \frac{1(2\left \lceil \log_a{(1)} \right \rceil - 3)}{4}\right]
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = n^2(\left \lceil \log_a{(n)} \right \rceil - 1) + 1 - \left [ \frac{n^2(2\left \lceil \log_a{(n)} \right \rceil - 3)}{4} + \frac{3}{4}\right]
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = n^2\left \lceil \log_a{(n)} \right \rceil - n^2 + \frac{1}{4} - \frac{n^2\left \lceil \log_a{(n)} \right \rceil}{2} +\frac{3n^2}{4}
\\
\int_{1}^{n}i\left \lceil \log_a{(i)} \right \rceil di = \frac{n^2\left \lceil \log_a{(n)} \right \rceil}{2} - \frac{n^2}{4} + \frac{1}{4}
\\\\
\text{Therefore, } \sum_{i = 1}^{n}i\left \lceil \log_a(i) \right \rceil \approx \left \lfloor \frac{n^2\left \lceil \log_a{(n)} \right \rceil}{2} - \frac{n^2}{4} + \frac{1}{4} \right \rfloor
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/805479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A logarithmic integral $\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$ How to prove the following
$$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$
I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are not best ideas to approach the problem.
Any other ideas ?
| Let $-1\le a \le 1$ and:
\begin{align*}
I(a) &= \int_{0}^{1} \, \log\left(\frac{1+a\,x}{1-a\, x}\right)\frac{1}{x\sqrt{1-x^2}}\, dx \tag 1\\
\frac{\partial}{\partial a}I(a) &= \int_{0}^{1} \, \frac{1}{(1+a\, x)\sqrt{1-x^2}} + \frac{1}{(1-a\, x)\sqrt{1-x^2}} \, dx\\
&= \frac{1}{\sqrt{1-a^2}}\, \left(\arcsin\left(\frac{x+a}{1+a\, x}\right)+\arcsin\left(\frac{x-a}{1-a\, x}\right) \right) \Big|_0^1\\
&= \frac{\pi}{\sqrt{1-a^2}}\\
\therefore I(a) &= \pi\, \arcsin{a} + C \tag 2\\
\end{align*}
Putting $a=0$, in $(1)$ and $(2)$, we see that $C=0$
Hence,
\begin{align*}
I(a) &= \int_{0}^{1} \, \log\left(\frac{1+a\,x}{1-a\, x}\right)\frac{1}{x\sqrt{1-x^2}}\, dx = \pi\, \arcsin{a}
\end{align*}
and for this problem
$$I(1)=\frac{\pi^2}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/805893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 10,
"answer_id": 7
} |
Linear Algebra determinant reduction Prove, without expanding, that
\begin{vmatrix}
1 &a &a^2-bc \\
1 &b &b^2-ca \\
1 &c &c^2-ab
\end{vmatrix} vanishes.
Any hints ?
| \begin{align}
\begin{vmatrix}
1 & a & a^2-bc \\
1 & b & b^2-ca \\
1 & c & c^2-ab
\end{vmatrix}
&=\begin{vmatrix}
1 & a & a^2-bc \\
0 & b-a & b^2-a^2+bc-ca \\
0 & c-b & c^2-b^2+ca-ab
\end{vmatrix} \\
&=\begin{vmatrix}
b-a & b^2-a^2+bc-ca \\
c-b & c^2-b^2+ca-ab
\end{vmatrix} \\
&=\begin{vmatrix}
b-a & (b-a)(b+a+c) \\
c-b & (c-b)(c+b+a)
\end{vmatrix}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/806779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Prove that $x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq 2xy(x+y+xy)$ Prove that $x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq
2xy(x+y+xy)$ for $x,y \in \mathbb{R}^+$.
I started by multiplying everything out on the RHS to get the equivalent statement
\begin{align*}
x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq
2 x^2 y + 2 x y^2 + 2 x^2 y^2.
\end{align*}
I've tried numerous methods of attacking this problem, but I haven't had much luck yet. The closest I seem to have been able to get is to use the identity $(x-y)^2\geq 0$ to deduce that
\begin{align*}
x^3+x^3y^2\geq x^3+xy^2=x(x^2+y^2)\geq 2x^2y,
\end{align*}
and similarly,
\begin{align*}
y^3x^2+y^3\geq x^2y+y^3=y(x^2+y^2)\geq 2xy^2.
\end{align*}
But now I am left with having to prove that $x^2+y^2\geq 2x^2y^2$, which is obviously false for $x,y \in \mathbb{R}^+$. I'm thinking I'm giving up too much trying to make the identity $(x-y)^2\geq 0$ work for this problem.
I also tried writing everything as a polynomial in $x$ and showing that the resulting polynomial is positive for $x>0$, but this gets extremely messy. Does anyone have any ideas? Thanks in advance!
| By AM-GM, we have $$x^3+y^3+x^2y^3+x^3y^2 \geq 4 \sqrt[4]{x^3y^3x^2y^3x^3y^2} = 4x^2y^2.$$
Similarly, we have $$x^3y^2 + y^2 + x^2 + x^3 \geq 4 \sqrt[4]{x^3y^2y^2x^2x^3} = 4x^2y$$ and $$x^2y^3+ x^2 + y^2 + y^3 \geq 4 \sqrt[4]{x^2y^3x^2y^2y^3} = 4xy^2.$$ Adding these inequalities yields the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/807654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$p(x)=0$ with real coefficient has purely Imaginary roots.Then the equation $p(p(x)) = 0$ has If the Quadratic equation $p(x)=0$ with real coefficient has purely Imaginary roots.Then the
equation $p(p(x)) = 0$ has
$\bf{OPTIONS::}$
$(a)\;\; $ Only purely Inaginary Roots. $\;\;\;\;\;\;(b)$ all real roots.
$(c)$ Two real and Two purely imaginary roots.$\;\;\;\;\;\;(d)$ neither real nor purely imaginary roots.
$\bf{My\; Try}::$ Let $p(x) = x^2+1.$ Then $p(p(x)) = (x^2+1)^2+1 = x^4+2x^2+2$.
Now Roots of $\displaystyle p(p(x))=0\Rightarrow x^4+2x^2+2=0\Rightarrow x^2 = \frac{-2\pm\sqrt{4-8}}{2} = \frac{-2\pm 2i}{2} = -1\pm i$
So $x^2=-1+i\;\;,-1-i$. Means $x$ must be in the form of $a+ib\;,$ where $a,b\neq 0$ and $i=\sqrt{-1}$
So option $(d)$ mut be Correct.
But my question is how can we solve the given question in Analytical method. means not by guessing.
Help me
Thanks
| If a quadratic equation has purely imaginary roots, then it can be written in the form $p(x)=k(x-ai)(x+ai)=k(x^2+a^2)$ for some $k,a\ne0$.
Hence $p(p(x))=k[k^2(x^2+a^2)^2+a^2]=k^3x^4+2k^3a^2x^2+k^3a^4+ka^2$. By the quadratic equation, this gives us
$$x^2=\frac{-2k^3a^2\pm\sqrt{4k^6a^4-4k^3(k^3a^4+ka^2)}}{2k^3}=\frac{-2k^3a^2\pm\sqrt{-4k^4a^2}}{2k^3}=\frac{-ka^2\pm ia}{k}$$
This means that $x$ can be neither purely real or purely imaginary, since if it were, then $x^2$ would be real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Find a formula for $1 + 3 + 5 + ... +(2n - 1)$, for $n \ge 1$, and prove that your formula is correct. I think the formula is $n^2$.
Define $p(n): 1 + 3 + 5 + \ldots +(2n − 1) = n^2$
Then $p(n + 1): 1 + 3 + 5 + \ldots +(2n − 1) + 2n = (n + 1)^2$
So $p(n + 1): n^2 + 2n = (n + 1)^2$
The equality above is incorrect, so either my formula is wrong or my proof of the implication is wrong or both.
Can you elaborate?
Thanks.
| $p(n): 1 + 3 + 5 + \ldots +(2n − 1) = n^2$
$p(n + 1): 1 + 3 + 5 + \ldots +(2n − 1) + 2n+1 =n^2+2n+1= (n + 1)^2$
next term after $2n-1$ is $2n+1$ is not $2n$ as you mean
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Elliptic integral $\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk$ Question:
Prove that
$$\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk=\frac{1}{16\pi}\Gamma^4\left(
\frac{1}{4}\right)$$
My attempt
Start by the transformation
$$k \to \frac{2\sqrt{k}}{1+k}$$
Hence we have
$$\int^{1}_0 K\left(\frac{2\sqrt{k}}{1+k}\right)\,\frac{1}{\sqrt{k}(1+k)}dk$$
Now we use that
$$K(k)=\frac{1}{k+1}K\left( \frac{2\sqrt{k}}{1+k} \right)$$
So we have
$$\int^1_0 \frac{K(k)}{\sqrt{k}}\,dk=2\int^1_0 K(k^2)\,dk$$
[1] I have no idea how to solve the last integral?
[2] Should I use another approach to solve the integral ?
By definition we have
$$K(k) = \int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2\,x^2}}\,$$
| Consider the integral
\begin{align}
I = \int_{0}^{1} \frac{K(x)}{\sqrt{1-x^2}} \ dx
\end{align}
where $K(x)$ is the complete elliptic integral of the first kind. It can be shown that the hypergeometric form is
\begin{align}
K(x) = \frac{\pi}{2} \ {}_{2}F_{1}(\frac{1}{2}, \frac{1}{2}; 1; x^{2}).
\end{align}
By using the series form the integral becomes
\begin{align}
I &= \frac{\pi}{2} \ \sum_{r=0}^{\infty} \frac{(1/2)_{r} (1/2)_{r}}{r! (1)_{r}} \ \int_{0}^{1} x^{2r} (1-x^{2})^{-1/2} \ dx.
\end{align}
By making the substitution $x = \sqrt{t}$ the integral can be cast into Beta function form and yields
\begin{align}
I &= \frac{\pi}{4} \ \sum_{r=0}^{\infty} \frac{(1/2)_{r} (1/2)_{r}}{r! (1)_{r}} \ B(r+1/2, 1/2) \\
&= \left( \frac{\pi}{2} \right)^{2} \ {}_{3}F_{2}(a,a,a; 1,1; 1)
\end{align}
where $a = 1/2$. Now, by using the identity
\begin{align}
{}_{3}F_{2}(a,a,a; 1,1; 1) = \frac{\Gamma\left(1 - \frac{3a}{2}\right) \ \cos\left( \frac{a \pi}{2} \right)}{\Gamma^{3}\left( 1 - \frac{a}{2}\right)}
\end{align}
which is valid for $Re(a) < 2/3$, the integral value is seen to be
\begin{align}
I = \left( \frac{\pi}{2} \right)^{2} \ \frac{\Gamma\left(\frac{1}{4}\right) \ \cos\left( \frac{\pi}{4} \right)}{\Gamma^{3}\left(\frac{3}{4}\right)}.
\end{align}
Now using the relation
\begin{align}
\Gamma\left(\frac{3}{4}\right) = \frac{\sqrt{2} \pi}{\Gamma\left(\frac{1}{4}\right)}
\end{align}
the final value is obtained, namely,
\begin{align}
I = \frac{1}{16 \pi} \Gamma^{4}\left(\frac{1}{4}\right).
\end{align}
Hence
\begin{align}
\int_{0}^{1} \frac{K(x)}{\sqrt{1-x^2}} \ dx = \frac{1}{16 \pi} \Gamma^{4}\left(\frac{1}{4}\right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
solving a system of equations dealing with Lorentz transformations Can anyone help me to find the solutions of this system of equations:
$$c^2x^2-v^2y^2=c^2$$
$$y^2-c^2z^2=1$$
$$vy^2+c^2zx=0$$
I know the answer:
$$x= \frac{1}{ \sqrt{1- \frac{ v^{2} }{ c^{2} } } } $$
$$y= \frac{1}{ \sqrt{1- \frac{ v^{2} }{ c^{2} } } } $$
$$z= -\frac{v}{c^2} \sqrt{1- \frac{ v^{2} }{ c^{2} } } $$
But I can't follow the steps. If any one could say me how to solve this kind of problem that would be helpful for me. Thank you in advance.
| The second equality says that
$$y^2=c^2z^2+1$$
Replace $y^2$ in the first and third equalities. You get:
$$c^2x^2-v^2(1+c^2z^2)=c^2,$$
$$v(1+c^2z^2)+c^2zx=0,$$
Therefore:
$$c^2x^2-v^2c^2z^2=c^2+v^2,\;\;(E_1)$$
$$vc^2z^2+c^2zx=-v,\;\;(E_2)$$
Now, you get from the first equality $(E_1)$:
$$x^2=\dfrac{c^2+v^2+c^2v^2z^2}{c^2},\;\;(S_1)$$
The previous second equality $(E_2)$ (squared) says that :
$$(c^2zx)^2=(-v-vc^2z^2)^2,\;\;(S_2)$$
Use $(S_1)$ and $(S_2)$ to get $z^2$:
$$z^2=\dfrac{v^2}{c^2(c^2-v^2)}$$
Now you get $$y^2=\dfrac{c^2}{c^2-v^2}$$
And $$x^2=y^2$$
P.S. I suppose that $c\neq v$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Calculate a sum $1+4e+9e^2+...+n^2e^{n-1}$, where e is nth root of unity Its similar to Calculate a sum involving nth root of unity
I dont get how to do it when $e$ is not $1$.
Please help!
| Start with: $1 + x + x^2 +...+ x^n = \dfrac{1 - x^{n+1}}{1-x}$, differentiate both sides with respect to $x$:
$1 + 2x + 3x^2 + ...+ nx^{n-1} = \dfrac{nx^{n+1} - (n+1)x^n + 1}{(1 - x)^2}$, multiply $x$ both sides again:
$x + 2x^2 + 3x^3 + ...+ nx^n = \dfrac{nx^{n+2} - (n+1)x^{n+1} + x}{(1 - x)^2}$, and differentiate again both sides with respect to $x$, and substitute $x = e$, and $e^n = 1$ to get:
$1 + 4e + 9e^2 +...+ n^2e^{n-1} = n^2e - 2ne^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/811729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the jordan form the matrix let characteristic polynomial $P_A(x)=(x+2)^4(x-3)^2$ and
minimal polynomial $m_A=(x+2)^3(x-3)$ find the jordan form that possible.
we know $q_6=\frac{f_6}{f_5}$ ($f_i$ is gcd{det of i x i submatrices which isnt equal to 0})
$q_6=\frac{f_6}{f_5}=\frac{(x+2)^4(x-3)^2}{f_5}=(x+2)^3(x-3)^2$ so $f_5=(x+2)(x-3)$
I know $q_5.q_4.q_3.q_2.q_1=(x+2)(x-3)$
how do we find the rest?$ (q_5,q_4,q_3...) $
can we choose them arbitrary like $q_5=x+2,q_4=x-3,q_3=q_2=q_1=1$ or $q_5=(x-3)(x+2),q_4=q_3=q_2=q_1=1$ ?
| We are given that the characteristic and minimal polynomial of a matrix $A$ are
\begin{align*}
p_A(x) &= (x+2)^4(x-3)^2 & m_A(x) &= (x+2)^3(x-3)
\end{align*}
This tells us that $A$ is a $6\times 6$ matrix whose eigenvalues are $\lambda_1=-2$ and $\lambda_2=3$. The Jordan form of $A$ is thus of the form
$$
\begin{bmatrix}
-2 & * & 0 & 0 & 0 & 0 \\
0 & -2 & * & 0 & 0 & 0 \\
0 & 0 & -2 & * & 0 & 0 \\
0 & 0 & 0 & -2 & 0 & 0 \\
0 & 0 & 0 & 0 & 3 & * \\
0 & 0 & 0 & 0 & 0 & 3
\end{bmatrix}
$$
The minimal polynomial of $A$ tells us that the largest Jordan block associated to $\lambda_1=-2$ has size $3$ and the largest Jordan block associated to $\lambda_2=3$ has size $1$. This gives one possible Jordan form (up to re-ordering):
$$
\begin{bmatrix}
-2 & 1 & 0 & 0 & 0 & 0 \\
0 & -2 & 1 & 0 & 0 & 0 \\
0 & 0 & -2 & 0 & 0 & 0 \\
0 & 0 & 0 & -2 & 0 & 0 \\
0 & 0 & 0 & 0 & 3 & 0 \\
0 & 0 & 0 & 0 & 0 & 3
\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/814234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int_0^\pi\frac{\cos nx}{a^2-2ab\cos x+b^2}\, dx$
How to compute the following integral
\begin{equation}
\int_0^\pi\frac{\cos nx}{a^2-2ab\cos x+b^2}\, dx
\end{equation}
I have been given two integral questions by my teacher. I cannot answer this one. I have also searched the similar question here but it looks like nothing is similar so I think this is not a duplicate. I could compute the integral if
\begin{equation}
\int_0^\pi\frac{dx}{a^2-2ab\cos x+b^2}
\end{equation}
The $\cos nx$ part makes the integral is really difficult. I want to use the result to compute this integral (the real question given by my teacher)
\begin{equation}
\int_0^\pi\frac{x^2\cos nx}{a^2-2ab\cos x+b^2}\, dx
\end{equation}
My question is how to compute the first integral (in the grey-shaded part) preferably with elementary ways (high school methods)?
| Let $I_n(a,b)$ be the desired integral. Note that $I_n(a,b)=I_n(b,a)$, and $I_n(a,b)=I_n(-a,-b)$. So, we may suppose that $|b|< a$ Note that
$$\eqalign{
\frac{a^2-b^2}{a^2-2ab\cos x+b^2}&=\frac{a}{a-e^{ix}b}+\frac{be^{-ix}}{a-e^{-ix}b}\cr
&=\sum_{n=0}^\infty \left(\frac{b}{a}\right)^ne^{inx}+\frac{be^{-ix}}{a}\sum_{n=0}^\infty
\left(\frac{b}{a}\right)^ne^{-inx}\cr
&=1+\sum_{n=1}^\infty \left(\frac{b}{a}\right)^ne^{inx}+ \sum_{n=1}^{\infty}
\left(\frac{b}{a}\right)^{n}e^{-inx}\cr
&=1+2\sum_{n=1}^\infty \left(\frac{b}{a}\right)^n\cos(n x)
}
$$
It follows, using the uniform convergence of the series on $[0,\pi]$, that
$$
\int_0^\pi\frac{(a^2-b^2)\cos(mx)}{a^2-2ab\cos x+b^2}dx
=\int_0^\pi\cos(mx)dx+2\sum_{n=1}^\infty \left(\frac{b}{a}\right)^n\int_0^\pi\cos(n x)\cos(mx)dx
$$
But $\int_0^\pi\cos(n x)\cos(mx)dx=0$ if $n\ne m$, and
$\int_0^\pi\cos^2(n x)dx=\pi/2$ if $n\ne0$. So
$$\eqalign{I_m(a,b)=
\int_0^\pi\frac{\cos(mx)}{a^2-2ab\cos x+b^2}dx
&=\left\{\matrix{\frac{\pi}{a^2-b^2}&\hbox{if}&m=0\cr
\frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^m&\hbox{if}&m\ne0
}
\right.}
$$
which is the desired formula for $|b|<a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/815855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
"answer_id": 1
} |
Kernel of linear transformation in $\Bbb R^3$ Let $T: \Bbb R^3 \rightarrow \Bbb R^3$ be a linear transformation satisfying
\begin{align*}
T(0,1,1) =& (-1,1,1) \\
T(1,0,1) =& (1,-1,1) \\
T(1,1,0) =& (1,-1,0) .
\end{align*}
Is it necessary true that $\ker(T) = \operatorname{Sp}\{(1,-1,1)\}$ ?
Well, I tried to say that we know that $\operatorname{Im}(T) = \operatorname{Sp}\{T(0,1,1),\,T(1,0,1),\,T(1,1,1)\}$
So, $\operatorname{Im}(T) = \operatorname{Sp}\{(-1,1,1),\,(1,-1,1),\,(1,-1,0)\}$ which means $\operatorname{Sp}\{(1,-1,1)\} \in \operatorname{Im}(T)$
and also $(1,1,1)$ is linearly independent by $2$ other vectors which are in $\operatorname{Im}(T)$.
Now, how can I prove that $\operatorname{Sp}\{(1,1,1)\}$ not inside $\ker(T)$? or maybe $\operatorname{Sp}\{(1,1,1)\} \in \ker (T)$ which makes it $\operatorname{Sp}\{(1,1,1)\} = \ker (T)$?
| We can simply put our vectors in the matrix and do row operations in the following way - we are trying to get a basis in the right part:
$$\left(
\begin{array}{ccc|ccc}
0 & 1 & 1 &-1 & 1 & 1\\
1 & 0 & 1 & 1 &-1 & 1\\
1 & 1 & 0 & 1 &-1 & 0
\end{array}
\right)\sim
\left(
\begin{array}{ccc|ccc}
0 & 1 & 1 &-1 & 1 & 1\\
0 &-1 & 1 & 0 & 0 & 1\\
1 & 1 & 0 & 1 &-1 & 0
\end{array}
\right)\sim
\left(
\begin{array}{ccc|ccc}
1 & 2 & 1 & 0 & 0 & 1\\
0 &-1 & 1 & 0 & 0 & 1\\
1 & 1 & 0 & 1 &-1 & 0
\end{array}
\right)\sim
\left(
\begin{array}{ccc|ccc}
1 & 1 & 0 & 1 &-1 & 0\\
0 &-1 & 1 & 0 & 0 & 1\\
1 & 2 & 1 & 0 & 0 & 1
\end{array}
\right)\sim
\left(
\begin{array}{ccc|ccc}
1 & 1 & 0 & 1 &-1 & 0\\
0 &-1 & 1 & 0 & 0 & 1\\
1 & 3 & 0 & 0 & 0 & 0
\end{array}
\right)$$
What have we found out by doing this?
Notice that at the beginning we have $a|b$, where $T(a)=b$ in each row. This property is not changed using row operations.
So we see that $T(1,3,0)=(0,0,0)$. This means that $(1,3,0)\in\operatorname{Ker} T$.
We have also found out that the image is generated by the vectors $(1,-1,0)$ and $(0,0,1)$. (If we look only on the right part, then we have tried to row reduce the matrix consisting of images of basis vectors.) Since these vectors are linearly independent, we get that $\dim\operatorname{Im} T =2$. By rank-nullity theorem we know that $\dim\operatorname{Ker} T=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
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