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Calculating a sum in the coefficient of some generating function I tried to calculate the coefficient of a generating function
\begin{align*}
\frac{1}{n} [u^{n-1}] e^{un} \frac{1}{(1-u)^2}
\end{align*}
and got to
\begin{align*}
\frac{1}{n} [u^{n-1}] e^{un} \frac{1}{(1-u)^2} &= \frac{1}{n} [u^{n-1}] \Bigg(\sum_{k\geq 0}\frac{1}{k!} (un)^k\Bigg) \Bigg(\sum_{k\geq 0} \underbrace{\binom{n+2-1}{2-1}}_{=~n+1} u^n\Bigg) \\
&= \frac{1}{n} [u^{n-1}] \sum_{k\geq 0} \sum_{l=0}^k \frac{n^l}{l!} (k-l+1) u^{k} = \frac{1}{n} \sum_{l=0}^{n-1} \frac{n^l}{l!} (n-l).
\end{align*}
According to wolfram alpha, this sum equals
\begin{align*}
\frac{1}{n} \sum_{l=0}^{n-1} \frac{n^l}{l!} (n-l) = \frac{n^n}{n!}.
\end{align*}
Is there a quick and simple way to show this or could I have done a different calculation in the beginning to arrive at this solution?
| One could also work with geometric series instead of using the general binomial theorem:
\begin{align*}
[z^n]g(f(z)) &= \frac{1}{n}[u^{n-1}]\phi(u)^n g'(u) \\
&= \frac{1}{n}[u^{n-1}]e^{nu} \left( \frac{1}{1-u}\right)' \\
&= \frac{1}{n}[u^{n-1}]e^{nu} \left( \sum_{k \geq 0}u^k\right)' \\
&= \frac{1}{n}[u^{n-1}]\sum_{k \geq 0}n^k\frac{u^k}{k!} \sum_{k \geq 1}ku^{k-1} \\
&= \frac{1}{n}[u^{n-1}]\sum_{k \geq 0}n^k\frac{u^k}{k!} \sum_{k \geq 0}(k+1)u^{k} \\
&= \frac{1}{n}[u^{n-1}]\sum_{k \geq 0}\sum^{k}_{l = 0}{(k-l+1) \frac{n^l}{l!}} u^k \\
&= \frac{1}{n} \sum^{n-1}_{l = 0}{(n-1-l+1) \frac{n^l}{l!}} \\
&= \frac{1}{n} \sum^{n-1}_{l = 0}{(n-l) \frac{n^l}{l!}} \\
&= \frac{n^n}{n!}.
\end{align*}
where at the end we used
\begin{align}
\frac{1}{n} \sum^{n-1}_{l=0}{\frac{n^l}{l!}(n-l)} &= \frac{1}{n}\sum^{n-1}_{l=1}\frac{n^l}{l!}(n-l) + 1 \\
&= \sum^{n-1}_{l=1}\frac{n^l}{l!} - \sum^{n-1}_{l=1}{\frac{n^{l-1}}{(l-1)!}} + 1\\
&= \sum^{n-1}_{l=1}\frac{n^l}{l!} - \sum^{n-2}_{l=0}{\frac{n^{l}}{l!}} + 1 \\
&= \frac{n^{n-1}}{(n-1)!} \\
&= \frac{n^{n-1}n }{n(n-1)!} \\
&= \frac{n^n}{n!}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Does the pattern of identities for $(x+y+z)^{2n-1}$ continue past $2 n - 1 = 5$? We have the following identities.
\begin{align}
(x+y+z)^3&=x^3+y^3+z^3+3(x+y)(y+z)(z+x)\\
(x+y+z)^5&=x^5+y^5+z^5+5(x+y)(y+z)(z+x)(x^2+y^2+z^2+xy+yz+zx)
\end{align}
I'm wondering is there any pattern like above in the expansion of $(x+y+z)^{2n-1}$ where $n\in\mathbb{Z}^+$? The pattern I see so far is appearance of the coefficient $2n-1$ and $(x+y)(y+z)(z+x)$.
The above expansions can be found by factoring $(x+y+z)^{t}-x^{t}-y^{t}-z^{t}$ by using Cyclic Polynomials. For example for $t=5$ the factorization is provided on Brilliant.
I tried factoring $(x+y+z)^{7}-x^{7}-y^{7}-z^{7}$ using the method I got:
$$(x+y)(y+z)(z+x) [A(x^4+y^4+z^4)+B(x^3y+y^3z+z^3x)+C(x^2y^2+y^2z^2+z^2x^2)]$$
Here I'm not sure whether I should include $D(xy^3+yz^3+zx^3)$ too or not.
| Yes you do keep the factors $x+y,x+z,y+z$. We can prove that, for instance, $x+z$ divides $(x^k+y^k+z^k)-(x+y+z)^k$ for all odd $k$ by observing that the difference goes to zero simultaneously with the claimed factor $x+z$ when we specify $z=-x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving by induction that $\frac{1}{n} \ge \frac{n!}{n^n}$ I have been trying to prove by induction that for all $n \ge 0, \frac{1}{n} \ge \frac{n!}{n^n} $
Here is my proof so far:
Prove that for $ n = 1, \frac{1}{n} \ge \frac{n!}{n^n} $
$ \frac{1}{1} = 1 $
$ \frac{1!}{1^1} = 1 \le 1 $
Next, assume there exists $ n \ge 0 $ such that $ \frac{n!}{n^n} \le \frac{1}{n}. $.
We'll prove that $ \frac{1}{n+1} \ge $ $\frac{(n+1)!}{(n+1)^{n+1}} $.
$\frac{(n+1)!}{(n+1)^{n+1}} = \frac{(n!)(n+1)}{(n+1)^{n+1}} = \frac{n!}{(n+1)^n} \lt \frac{n!}{n^n} \le \frac{1}{n}$
However, I don't know how to show now that $\frac{n!}{(n+1)^n} \le \frac{1}{n +1}$
| Looks like a trivial proof holds. $n!$ has $n$ terms, each $\le n$. So $\frac{n!}{n^n}=\frac{1}{n}\times C\le \frac{1}{n}$, since $C=\frac{2}{n}\times...\frac{n}{n}\le 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Help with $\frac{\sin\theta+\cos\theta}{-\sin\theta+\cos\theta}=?$ Assume that $0<\theta<\frac{\pi}{4}$. If $\sin2\theta=\frac{1}{4}$, then $\frac{\sin\theta+\cos\theta}{-\sin\theta+\cos\theta}=?$
My approach with this problem is to rationalize the denominator in the fraction to get this:
$$
\frac{\sin2\theta+1}{\cos2\theta}=\frac{\frac{1}{4}+1}{\cos2\theta}
$$
After that i got stuck with $\cos2\theta$ any advice how to attack the problem?
| Alternative approach:
$$\frac{1}{4} = \sin(2\theta) = 2\sin(\theta)\cos(\theta). \tag1 $$
Also, since $0 < \theta < (\pi/4)$, you have that
$$\cos(2\theta) = \sqrt{1 - \left(\frac{1}{4}\right)^2} = \frac{\sqrt{15}}{4}. \tag2 $$
Finally, you have that
$$\cos^2(\theta) - \sin^2(\theta) = \cos(2\theta) = \frac{\sqrt{15}}{4}. \tag3 $$
You have that:
$$\frac{\cos(\theta) + \sin(\theta)}{\cos(\theta) - \sin(\theta)} \times
\frac{\cos(\theta) - \sin(\theta)}{\cos(\theta) - \sin(\theta)}$$
$$= ~\frac{\cos^2(\theta) - \sin^2(\theta)}{\cos^2(\theta) - 2\sin(\theta)\cos(\theta) + \sin^2(\theta)}. \tag4 $$
Using (1), (2), and (3), the expression in (4) equals
$$\frac{\frac{\sqrt{15}}{4}}{1 - \frac{1}{4}} = \frac{\sqrt{15}}{3}.$$
| {
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To find the function $g(x)$ in terms of $f(x)$. For each point $(a,b)$ on the graph $y = f(x)$ the point $(3a-1,\frac{b}{2})$ is plotted forming the graph of another function $y = g(x)$. The problem asks us to find the function $g(x)$ in terms of $f(x)$.
Also, we need to describe the transformation.
Since $a$ goes to $3a-1$, the inverse of the function will be $x \to \frac{x+1}{3}$.
Hence the required function is $$g(x) = \frac{1}{2}f\left( \frac{x+1}{3}\right).$$
Note that $$g(3a-1) = \frac{1}{2}f\left( \frac{3a-1+1}{3}\right) = \frac{1}{2}f(a) = \frac{b}{2}.$$
Description of the transformation:
$\bullet$ Horizontal stretch by a factor of $3$.
$\bullet$ Vertical stretch by a factor of $\frac{1}{2}$.
$\bullet$ Horizontal translation by $-1$.
Is the solution correct?
| Let $$X=3a-1\;\text{ and }\; g(X)=\frac b2$$
with
$$b=f(a) $$
or
$$2g(X)=f(\frac{X+1}{3})$$
So
$$g(X)=\frac 12f(\frac{X+1}{3})$$
| {
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Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality case, for example, when $a=x^4,b=x^3,c=x^2,d=x$, and as $x$ tends to $\infty$, the LHS tends to $3$, but this means that the inequality is very unlikely to be solved with traditional methods, such as Cauchy-Schwartz (my starting idea), so I got stuck.
| Again partial hint :
Using the formula :
$$\cos\left(\operatorname{arccot}\left(\frac{a}{b}\right)\right)-\sqrt{\frac{a^{2}}{a^{2}+b^{2}}}=0$$
We use Karamata's inequality with $-\cos(x)$:
$$\frac{a}{b}\leq \frac{b}{c}\leq \frac{c}{d}\leq \frac{d}{a}$$
And $$3=4\cos\left(\operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right)\right)$$
With constraint :
$$f(a,b)\geq \operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right),\quad,f(a,b)+f(b,c)\geq 2 \operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right),f(a,b)+f(b,c)+f(c,d)\geq 3\operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right),\quad f\left(a,b\right)+f\left(b,c\right)+f\left(c,d\right)+f\left(d,a\right)\geq4\operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right)$$
Where :
$$f\left(x,y\right)=\operatorname{arccot}\left(\frac{x}{y}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the integral by Riemann $\int_{0}^{1}\sqrt{x}\,dx$ Calculate the integral by Riemann $\displaystyle \int _{0}^{1}\sqrt{x} \, dx$
$$\displaystyle a_{k} =0+k\cdot \frac{1}{n} =\frac{k}{n}$$
$$\displaystyle \Delta x=\frac{1-0}{n} =\frac{1}{n}$$
\begin{align*}
\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\sqrt{\frac{k}{n}}\right) \cdotp \frac{1}{n} & = \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{\sqrt{k}}{\sqrt{n}}\right) \cdotp \frac{1}{n} \\[1ex]
& = \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{1}{n\sqrt{n}} \cdotp \sqrt{k}\right) \\[1ex]
& = \lim_{n\rightarrow \infty }\frac{1}{n \sqrt{n}} \cdotp \sum_{k=1}^{n}\left(\sqrt{k}\right)
\end{align*}
I am stuck on the sum of $\sqrt k$ maybe have another way to solve this question?
| Here is a direct way to look at this if you would like to stick with a Riemann sum that uses equally spaced subintervals. The strategy is, in small doses, to replace the irrational $\sqrt{k}$ values with nearby integers.
$$\begin{align}
\lim_{n\to\infty}\frac{1}{n\sqrt{n}}\sum_{k=1}^n\sqrt{k}
&=\lim_{n\to\infty}\frac{1}{n^2\sqrt{n^2}}\sum_{k=1}^{n^2}\sqrt{k}
\end{align}$$
Already this might give you pause. But for one thing, as $n\to\infty$, so does $n^2$. The sum on the right is like the sum on the left but between one value of $n$ and the next, you add
$$\frac{1}{n^2\sqrt{n^2}}\left(\sqrt{(n-1)^2+1}+\sqrt{(n-1)^2+2}+\cdots+\sqrt{(n-1)^2+(2n-1)}\right)$$
That is, you add $2n-1$ terms that are each less than or equal to $n$, and the result is divided by $n^3$. So this tail that you add with each step, all by itself, converges to $0$. So the two sums have the same limit. Starting again:
$$\begin{align}
\lim_{n\to\infty}\frac{1}{n\sqrt{n}}\sum_{k=1}^n\sqrt{k}
&=\lim_{n\to\infty}\frac{1}{n^2\sqrt{n^2}}\sum_{k=1}^{n^2}\sqrt{k}\\
&=\lim_{n\to\infty}\frac{1}{n^3}\left(\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n^2}\right)\\
&\leq\lim_{n\to\infty}\frac{1}{n^3}\left(\sqrt{1}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{9}+\sqrt{9}+\cdots+\sqrt{n^2}\right)\\
\end{align}$$
Here we have one $\sqrt{1}$s, then three $\sqrt{4}$s, then five $\sqrt{9}$s, and so on.
$$\begin{align}
&=\lim_{n\to\infty}\frac{1}{n^3}\left(1+3(2)+5(3)+\cdots+(2n-1)n\right)\\
&=\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^n(2k-1)k\\
&=\lim_{n\to\infty}\frac{1}{n^3}\left(2\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}2\right)\\
&=\lim_{n\to\infty}\left(\frac13(1+1/n)(2+1/n)-\frac{(1+1/n)}{2n}\right)\\
&=\frac23
\end{align}$$
So we have an upper bound and the limit you are after is at most $\frac23$. (Of course we know independently that the limit actually is $\frac23$.)
Do you see how to modify this to get a lower bound? Hint: $1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,\ldots$.
| {
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A number theory question that is probably wrong Prove that $a^3-b^3 = 2011$ has no integer solutions.
I think the question is wrong as
$a^3-b^3 = 2011$
$(a-b)(a^2+ab+b^2) = 2011$
As $2011$ is prime so the only factors $2011$ has are $1$ and $2011$ itself.
So if $a-b = 1$ and $a^2+ab+b^2 = 2011$ , then we can say that
$a^2-2ab+b^2 = 1$
So , $a^2+b^2 = 2ab+1$
Putting the value in the second equation
$3ab +1 = 2011$
$3ab = 2010$
And as $2010$ is divisible by $3$ it has integer solutions , which is contradictory to the original question. Can anybody help me with this?
| The possible values or $x^3\pmod 9$ are $\{0,1,-1\}$ only.
So their difference can only take the values $\{0,1,2,-1,-2\}$, yet $2011\equiv 4\pmod 9$ so it is not reachable.
| {
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$f(x)=x^{3}-3 x+a$. Given that it has $3$ integer roots $\alpha, \beta, \gamma$. Find all possible values of $a$. So, I tried vieta and with some algebra, I got to the point $0=\alpha^{2}+\beta^{2}+\gamma^{2}+2(\alpha \gamma+\gamma \beta+\alpha \beta)$ where $(\alpha \gamma+\gamma \beta+\alpha \beta)=a$. I don't know how to progress from here
| $$f(x) = (x - \alpha)(x - \beta)(x - \gamma) = x^3 - 3x + a$$
$$f(x) = x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \alpha\gamma + \beta\gamma)x - \alpha\beta\gamma = x^3 - 3x + a$$
Matching up the coefficients gives you the system of equations:
$$\alpha + \beta + \gamma = 0$$
$$\alpha\beta + \alpha\gamma + \beta\gamma = -3$$
$$\alpha\beta\gamma = -a$$
From the first one, we get $\gamma = -(\alpha + \beta)$. Substitute into the other two.
$$\alpha\beta - (\alpha + \beta)(\alpha + \beta) = -3 \implies - \alpha^2 - \alpha\beta - \beta^2 = -3 \implies \alpha^2 + \beta^2 + \alpha\beta = 3$$
$$\alpha\beta(-(\alpha + \beta)) = -a \implies \alpha^2\beta + \alpha\beta^2 = a$$
Expressing the first one of these as a quadratic in terms of $\beta$, we get:
$$\beta^2 + \alpha\beta + (\alpha^2 - 3) = 0 \implies \beta = \frac{-\alpha \pm \sqrt{\alpha^2 - 4(\alpha^2 - 3)}}{2} = \frac{-\alpha \pm \sqrt{12 - 3\alpha^2}}{2}$$
This means that in order for $\beta$ to be a real number, we must have $\alpha^2 \le 4$, or $|\alpha| \le 2$. Since we're given that $\alpha$ is an integer, it follows that $\alpha \in \{-2, -1, 0, 1, 2\}$. But we also need $\beta$ to be an integer, so $\alpha = 0$ is not valid, as then we'd be dealing with $\sqrt{12} = 2\sqrt{3}$.
Since the equations are symmetric in $\alpha$ and $\beta$, we must also have $\beta \in \{-2, -1, 1, 2\}$.
Now, for each of the possible combinations of ($\alpha$, $\beta$), we can calculate $\gamma = -(\alpha + \beta)$, and then check if the ($\alpha$, $\beta$, $\gamma$) triple follows the constraint $\alpha\beta + \alpha\gamma + \beta\gamma = -3$. It turns out that there are six valid combinations:
*
*$(-2, 1, 1)$
*$(-1, -1, 2)$
*$(-1, 2, -1)$
*$(1, -2, 1)$
*$(1, 1, -2)$
*$(2, -1, -1)$
From each one, we can calculate $a = -\alpha\beta\gamma$. And it turns out that there are only two possibilities, $a = \pm 2$.
| {
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How do you finish the equations for $S^2 + \frac{7}{4}S= \frac{1}{2}$? Complete the Square:
$$S^2 + \frac{7}{4}S = \frac{1}{2}$$
$$S^2 + \frac{7}{4}S \_\_\_\_\_\_\_ = \frac{1}{2}$$
$$S^2 + \frac{7}{4}S + \frac{49}{64} = \frac{81}{64}$$
THIS is where I get stuck. I know that the Square Root of $\frac{81}{64}$ is $\frac{9}{8}$, but how do I find the Square Root of $S^2 + \frac{7}{4}S + \frac{49}{64}$?
Quadratic Equation:
$$A=1,\ \ B=\frac{7}{4},\ \ C= \frac{-1}{2}$$
$$\frac{\frac{-7}{4} \pm
\sqrt{\frac{49}{16}} + 2}{2}$$
THIS is where I get stuck because I don't know what to do with the 2 in the discriminant.
| I'm thinking that with the algebra/precalculus tag, the equation you're asking about is $$s^2+\frac {7}{4} s = \frac {1}{2}$$
Completing the square is correct; when you have
$$s^2 + \frac {7}{4}s + \frac {49}{64} = \frac {1}{2} + \frac {49}{64}$$ you can factor the left hand side; then you'll have $$\left(s + \frac {7}{8} \right)^2=\frac {81}{64}$$ Take the square roots of both sides, subtract $\dfrac {7}{8}$, and you're finished. (You should get $\dfrac {1}{4}$ and $-2$ as your final solutions.)
Note: If you had intended to solve the equation $$s^2+\frac {7}{4s} = \frac {1}{2}$$ you can multiply by $4s$ and then you'll have an cubic equation $$4s^3-2s+7 = 0$$ The solution to this is not nice; the best way to solve this is by numerical methods.
| {
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What is the maximum and minimum value of this function? Find the maximum and minimum values of the $f(z) = \vert 1 + z\vert + \vert 1-z+z^2\vert$ on the $\{z\in \mathbb{C} \vert$ $\vert z \vert =1, Re(z) \geq 0\}$ (Here the $\mathbb{C}$ is the set of
complex numbers.)
Let me show you my solution.
Put $z = e^{i\theta}$ with $cos\theta \geq 0$ (I.e. $0 \leq \theta \leq {\pi \over2}$ or ${3\pi \over2} \leq \theta \leq 2\pi$ )
Since $\vert 1+z \vert = 2\vert \cos({\theta \over 2})\vert $ and $\vert 1-z+z^2 \vert =\vert \frac{z^3+1}{z+1}\vert = \vert 4cos^2(\frac{\theta}{2})-3\vert$
$f(z)$ would be $ f(\theta) = 2\vert \cos({\theta \over 2})\vert +\vert 4cos^2(\frac{\theta}{2})-3\vert$
Firstly , I put the $u = \cos({\theta \over 2})$. And now the range of the $\frac{1}{\sqrt2} \leq u \leq 1 $ or $ -1\leq u \leq \frac{-1}{\sqrt2} $ from the $cos^2\frac{\theta}{2} = \frac{1+cos\theta}{2} \geq \frac{1}{2}$
Second, only just we find the either max or min of the $f(u)=2\vert u \vert + \vert 4u^2 -3 \vert$ over the $\frac{1}{\sqrt2} \leq u \leq 1 $ or $ -1\leq u \leq \frac{-1}{\sqrt2} $.
Third, I divided the possible cases.
$(1)$ $u \leq 0, 4u^2-3 \leq0 $ ; Find the value of the $f(u)=-4u^2+3-2u$ on the $ \frac{-\sqrt3}{2} \leq u \leq \frac{-1}{\sqrt 2} $ ; The min and max are $\sqrt 3 ,3$
$(2)$ $u \leq0, 4u^2-3 \geq0 $ ; Find the value of the $f(u)=4u^2-3-2u$ over the $ -1\leq u \leq \frac{-\sqrt3}{2} ; $ The min and max are $\sqrt 3 ,3$
$(3)$ $u \geq 0, 4u^2-3 \leq0 $ ; Find the value of the $f(u)=-4u^2+3+2u$ over the $ \frac{1}{\sqrt 2} \leq u \leq \frac{\sqrt3}{2} $ ; The min and max are $ \sqrt3, 1+\sqrt 2$
$(4)$ $u \geq 0, 4u^2-3 \geq0 $ ; Find the value of the $f(u)=4u^2-3+2u$ over the $ \frac{\sqrt3}{2} \leq u \leq 1 $ ; The min and max are $\sqrt3 ,3$
So my answer is $\sqrt3$ and $3$
But In my Teaching assistance's note the answer was $\frac{13}{4}, \sqrt3$. I can't understand why the answer like those. Are my processes right? Which one is right between him and mine? Plus, Would you tell me the error and its reason if my answer is not correct.
| For your given function, your answer is correct. The maximum is attained when $\theta = 0$, giving $z = 1$, hence $$f(1) = |1+1| + |1^2-1+1| = 3.$$ This is clearly greater than $\sqrt{3}$ claimed by the note. The minimum is attained when $\theta \in \pm \pi/3$, giving $z \in \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$, hence $$f(e^{i \pi/3}) = |1 + e^{i \pi /3}| + |e^{2\pi i/3} - e^{i \pi/3} + 1| = \sqrt{3}.$$ The claimed answer $13/4$ doesn't even make sense because it is larger than $3$.
In terms of $\theta$, the function $f$ is equivalent to
$$g(\theta) = 2 \cos \frac{\theta}{2} + |1 - 2 \cos \theta| = \begin{cases}
2 \cos \frac{\theta}{2} - 1 + 2 \cos \theta, & 0 \le |\theta| \le \pi/3 \\
2 \cos \frac{\theta}{2} + 1 - 2 \cos \theta, & \pi/3 < |\theta| \le \pi/2. \end{cases}$$ Since $g$ is an even function we may consider just the two cases $0 \le \theta \le \pi/3$ and $\pi/3 < \theta \le \pi/2$, and the rest of the analysis is straightforward as you have already done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A Clarkson-type inequality $|a+b+c|^q + \sum\limits_{\mathrm{cyc}} |a + b - c|^q \ge \sum\limits_{\mathrm{cyc}} (|a + b|^q + |a - b|^q)$ Show that, for $a,b,c \in \mathbb{R}$ and $q \geq 2$:
$$ |a+b|^{q} + |a-b|^{q} + |a+c|^q + |a-c|^q + |b+c|^q + |b-c|^q \\\leq \\|a+b+c|^q + |a+b-c|^q + |a-b+c|^q + |a-b-c|^q $$
When $c=0$, this follows from the result here.
Is some generalization of Clarkson's inequality in the real case for more than 2 variables known?
Edit: Numerical experiments indicate that $\frac{lhs}{rhs}$ is maximized when one of $a, b, c$ is non-zero and the other two are zero. Many inequalities involving convex functions (for example: for $p\geq 1$, $(x_1^2 + x_2^2 + ... +x_n^2)^p \geq ( (x_1^2)^{p} + (x_2^2)^{p} + ... +(x_n^2)^{p})$ satisfy this property (the inequality being tight when one variable is non-zero and rest all is zero). This gives the impression that the inequality in question follows from some convexity arguments.
| WLOG, assume that $a\ge b \ge c \ge 0$.
Fact 1: It holds that
\begin{align*}
&(a+b-c)^q + (a-b+c)^q + |b+c-a|^q \\
\ge\,& a^q + b^q + c^q
+ (a - b)^q + (b - c)^q + (a - c)^q.
\end{align*}
(The proof is given at the end.)
Fact 2: It holds that
$$(a + b + c)^q + a^q + b^q + c^q
\ge (a + b)^q + (b + c)^q + (c + a)^q.$$
(Note: Take derivative.)
Using Facts 1-2, the desired result follows.
Proof of Fact 1:
Let
\begin{align*}
&x_1 = (a + b - c)^2,
~ x_2 = (a - b + c)^2, ~ x_3 = (b + c - a)^2,\\
&y_1 = a^2, ~ y_2 = b^2, ~ y_3 = c^2, ~ y_4 = (a-b)^2, ~ y_5 = (b-c)^2, ~ y_6 = (a-c)^2.
\end{align*}
Let $p = (x_1, x_2, x_3, 0, 0, 0)$ and $q = (y_1, y_2, y_3, y_4, y_5, y_6)$.
First, we have $x_1 \ge x_2 \ge x_3$ and $y_1 = \max(y_1, y_2, y_3, y_4, y_5, y_6)$ and $y_2 \ge y_3$ and $y_6 - y_4 \ge 0$ and $y_6 - y_5 \ge 0$ and
\begin{align*}
x_1 - y_1 &\ge 0,\\
x_1 + x_2 - y_1 - y_6 &= 2ac + 2b^2 - 4bc + c^2 \\
&= 2(a-b)c + 2b(b-c) + c^2\\
&\ge 0, \\
x_1 + x_2 - y_1 - y_2 &= a^2 + b^2 - 4bc + 2c^2\\
&= 2c(a - b) + 2(b-c)^2 + 2(a-b)^2 + 2(a-b)(b-c)\\
&\ge 0.
\end{align*}
Second, we have
$$x_1 + x_2 + x_3 = y_1 + y_2 + y_3 + y_4 + y_5 + y_6.$$
Thus, $q$ is majorized by $p$.
Note that $u\mapsto u^{q/2}$ is convex on $u \ge 0$.
Using Karamata inequality, the desired result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
Attempt $1:$ $A=I+B$, where $B=\begin{bmatrix}1&1&1\\0&0&0\\1&1&1\end{bmatrix}$
$B^n$ comes out to be $2^{n-1}B$
After that I tried doing $A^8=(I+B)^8$ and then opening RHS with binomial. But it didn't help.
Attempt $2:$ Finding characteristic equation and putting $A$ in it. It didn't help either.
The equation I got here is $A^3-5A^2+7A-3I=0$
Note: Concept of minimal polynomial is not in syllabus. Also, if we can do this without characteristic polynomial, that would be great.
| Following your attempt 1 and using
$$\begin{cases}
A-I &= B\\
B^2 &= 2B
\end{cases}$$ you get that
$A^2-4A + 3I=0$, i.e. that $q(A)=0$ where $q(x) = x^2-4x +3$. You want to get $p(A)$ where $p(x)=x^8-5x^7+7x^6-3x^5+x^4-5x^3+8x^2-2x+1$.
Performing the long division of $p$ by $q$ you get
$$p(x) = (x^6-x^5+x^2-x+1)q(x) + 5x-2.$$
Hence $$p(A)= 5A-2I=5B + 3I=\begin{bmatrix}8&5&5\\0&3&0\\5&5&8\end{bmatrix}$$
Note: when you'll learn the minimum polynomial, you'll be able to see that $q$ is the minimal polynomial of $A$. However, the notion is not required for the question.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can we know how many times a number is divisible by 2? For example 36:
36/2=18
18/2=9
So for 36 it is divisible by 2 2 times.
For 2ⁿ×(2k-1) it is divisible by 2 n times.
But how do I know for 3k-2?
Is there a formula to find how many times a number is divisible by 2?
| For $k = 1,2,3,\ldots$ and so $3k-2 = 1,4,7,\ldots$, the highest power of $2$ dividing $3k-2$ are:
$$1,4,1,2,1,16,1,2,1,4,1,2,1,8,1,\underbrace2_{k=16},\ldots$$
This looks like some translated or transformed version of the highest power of $2$ dividing $k$ (https://oeis.org/A006519), but instead term $4^n=2^{2n}$ first appears earlier than $4^n/2 = 2^{2n-1}$.
If $3k-2$ is divisible by up to $2^n$ (but not higher), then
$$3k-2 \equiv 2^n \pmod{2^{n+1}}$$
One form of $3^{-1} \pmod{2^{n+1}}$ is
$$\begin{align*}
3^{-1} &\equiv 1-2+2^2-\cdots + (-2)^n \pmod{2^{n+1}}\\
3\left[1-2+2^2-\cdots + (-2)^n\right] &= 3\cdot\frac{1-(-2)^{n+1} }{1-(-2)}\\
&= 1-(-2)^{n+1}\\
&\equiv 1
\end{align*}$$
Back to the mod equation,
$$\begin{align*}
3k-2&\equiv 2^n \pmod{2^{n+1}}\\
k&\equiv 2^n\left[1-2+2^2-\cdots + (-2)^n\right]+2\left[1-2+2^2-\cdots + (-2)^n\right]\\
&\equiv 2^n + \underbrace{\left[2-2^2+2^3-\cdots-(-2)^{n-1} -(-2)^n\right]}_{n\text{ terms}}\\
&\equiv \begin{cases}
2^n+0\equiv 1&& n=0\\
2 - 2^2+2^3 - \cdots -(-2)^{n-1} && n\ge 1
\end{cases}
\end{align*}$$
From this form of $k$, if $n\ge 2$, $k$ is divisible by $2$ but not by $4$. Writing $k$ explicitly for some small $n$s, i.e. if $2^n\mid (3k+1)$ but $2^{n+1}\not\mid (3k+1)$,
$$\begin{align*}
n&=0, &k&\equiv 1 \pmod 2\\
n&=1, &k&\equiv 0 \pmod 4\\
n&=2, &k&\equiv 2 \pmod 8\\
n&=3, &k&\equiv -2\equiv 14 \pmod{16}\\
n&=4, &k&\equiv 6 \pmod{32}\\
n&=5, &k&\equiv -10\equiv54 \pmod{64}
\end{align*}$$
So if the divisibility of $k$ by powers of $2$ is known, without performing further division,
*
*If $1\mid k$ but $2\not \mid k$, then $1\mid (3k-2)$ but $2\not\mid (3k-2)$;
*If $2\mid k$ but $4\not\mid k$, then $4\mid (3k-2)$, but no further conclusions for higher powers of $2$;
*If $4\mid k$ (and possibly by higher powers of $2$), then $2\mid (3k-2)$ but $4\not\mid(3k-2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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problem canceling in multiplication of fractions now that i have a grasp on formatting mathematical fractions and whatnot i present you with my latest confusion
$$3\frac15 \times 1\frac23 \times 2\frac 34$$
converted into improper fractions
$$\frac{16}5 \times \frac 53 \times \frac{11}4$$
this much i understand. the following bit is whats driving me insane
$$\frac41 \times \frac13 \times \frac{11}1$$
i simply dont get how it goes from the improper fractions to the next set of fractions. apparently its called canceling but i cant see any sense in it
| We know that $\dfrac{a}{b}\times\dfrac{c}{d}=\dfrac{a\times c}{b\times d}$ and that we can commute $\dfrac{a}{b}\times\dfrac{c}{d}=\dfrac{c}{b}\times\dfrac{a}{d}=\dfrac{c}{d}\times\dfrac{a}{b}$ and so on, of course in the nominator and denominatior. Then, in your exercise:
$$\frac{16}5 \times \frac 53 \times \frac{11}4=\dfrac{16}{4}\times\dfrac{5}{5}\times\dfrac{11}{3}=\dfrac{4}{1}\times\dfrac{1}{1}\times\dfrac{11}{3}=\dfrac{4}{1}\times\dfrac{1}{3}\times\dfrac{11}{1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Why are the two calculations of $ \int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x $ give two distinct answers? One of the calculations:
$$
\begin{aligned}
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x &=\int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{1}{\sqrt{x}} \mathrm{~d} x \\
&=2 \int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \mathrm{~d}(\sqrt{x})=2 \arcsin \sqrt{x}+C .
\end{aligned}
$$
The other:
$$
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x = \int \frac{d\left(x-\frac{1}{2}\right)}{\sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}} = \arcsin \frac{x-1 / 2}{1 / 2}+C=\arcsin(2x-1)+C
$$
| They are both correct. Let $y=2 \arcsin \sqrt x$. Then $x=(\sin (\frac y 2))^{2}$. So $2x-1=2(\sin (\frac y 2))^{2}-1=\sin (y+\frac {\pi} 2)$. So $y=\sin (2x-1)-\frac {\pi} 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the integer solution: $a+b+c=3d$, $\: a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$ Find the integer solutions:
$$a+b+c=3d$$
$$a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$$
Attempt:
Notice that $a=b=c=d=1$ is a solution.
Other facts:
Notice that $a^{2} + b^{2} + c^{2} > 0$, so $4d^{2}-2d+1>0$. Notice that $4d^{2}-2d+1$ has negative discriminant: $D = -12$, and so it is always one sign, which is positive in this case. So we cannot narrow $d$-solution this way.
Next, since $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab+ac+bc)$, we get
$$ 9d^{2} - 2(ab+ac+bc) = 4d^{2}-2d+1 $$
$$ 5d^{2} + 2d -1 = 2(ab+ac+bc) $$
$5d^{2} + 2d - 1 = 0$ has solutions:
$$ d_{1,2} = \frac{-2 \pm \sqrt{24}}{10} = \frac{-1 \pm \sqrt{6}}{5}$$
and when $d > (-1 + \sqrt{6})/5$, OR, $d < (-1 - \sqrt{6})/5$ we have
$5d^{2}+2d-1 > 0$. And when $d$ is between the 2 roots we have $5d^{2}+2d-1<0$.
Then also nocitce that $(a+b+c)d= 3d^{2}$, and so
$$a^{2}+b^{2}+c^{2}-(a+b+c)d = (d-1)^{2}$$
Now if $d \ne 1$ we must have
$$ a^{2}+b^{2}+c^{2}-(a+b+c)d > 0$$
Another fact:
$$d = \frac{a+b+c}{3} \ge (abc)^{1/3} $$
by AM-GM.
Also notice $d < \max(a,b,c)$, because
$$a^{2}+b^{2}+c^{2} =d^{2} + d^{2} + d^{2} + (d-1)^{2}$$
| Your equation has no non-zero solutions.
Lemma. The Diophantine equation $3x^2+y^2=2z^2$ has no non-zero solutions in integers. Proof. Let $x_0$ is smallest positive integer such that there are integers $y_0$ and $z_0$ such that $3x_0^2+y_0^2=2z_0^2$. If $y_0 \not \equiv 0 \pmod 3$ then $z_0 \not \equiv 0 \pmod 3$. Then $y_0^2 \equiv 1 \pmod 3$ and $2z_0^2 \equiv 2 \pmod 3$. Then $2z_0^2-y_0^2 \equiv 1 \pmod 3$. But $2z_0^2-y_0^2=3x_0^2 \equiv 0 \pmod 3$. Therefore $y_0 \equiv 0 \pmod 3$. Then $z_0 \equiv 0 \pmod 3$. Let $y_0=3k$, $z_0=3n$. Then $3x_0^2+9k^2=18n^2$. Then $x_0^2+3k^2=6n^2$. Then $x_0 \equiv 0 \pmod 3$. Let $x_0=3p$. Then $9p^2+3k^2=6n^2$. Then $3p^2+k^2=2n^2$.We have come to a contradiction because $0<p<x_0$ and $p$ is integer solution of $3x^2+y^2=2z^2$.
Now solve your equation. Let $z=a+c$, $y=c-a$. Then $c=\frac{z+y}{2}$, $a=\frac{z-y}{2}$. Then from $a+b+c=3d$ we have $z+b=3d$. Then $b=3d-z$. Then $\left( \frac{z-y}{2}\right)^2+\left( \frac{z+y}{2}\right)^2+(3d-z)^2=4d^2-2d+1$. Then $z^2+y^2+2(3d-z)^2=8d^2-4d+2$. Then $3z^2-12zd+y^2=-10d^2-4d+2$. Then $3(z-2d)^2+y^2=2d^2-4d+2$. Then $3(z-2d)^2+y^2=2(d-1)^2$.The last equation has no non-zero solutions by the lemma.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the solution to this non-linear second order differential equation? I'm trying to solve the following non-linear second order differential equation:
$$\tag{1}
\frac{d\, }{dx} \Bigl( \frac{1}{y^2} \, \frac{dy}{dx} \Bigr) = -\, \frac{2}{y^3},
$$
where $y(x)$ is an unknown function on the real axis. I already know the "trivial" solution $y(x) = y_0 \pm x$. The solution I'm looking may involve trigonometric functions, but I'm not sure. Take note that we may pose
$$\tag{2}
u = \frac{1}{y},
$$
so that (1) takes another form:
$$\tag{3}
\frac{d^2 u}{dx^2} - 2 \, u^3 = 0.
$$
So what is the non-linear solution $y(x)$? Or $u(x)$?
| Solve
\begin{align*}
u^{\prime \prime}&=2 u^{3}
\end{align*}
Multiplying both sides by $u^{\prime}$ gives
\begin{align*}
u^{\prime} u^{\prime \prime}&=2 u^{3} u^{\prime}
\end{align*}
Integrating both sides w.r.t. $x$ gives
\begin{align*}
\int{u^{\prime} u^{\prime \prime}\, \mathrm{d}x} &=\int{2 u^{3} u^{\prime}\, \mathrm{d}x}\\
\int{u^{\prime} u^{\prime \prime}\, \mathrm{d}x} &=\int{2 u^{3}\, \mathrm{d}u} \tag{1}
\end{align*}
But
$$
\int{u^{\prime} u^{\prime \prime}\, \mathrm{d}x} = \frac{1}{2} \left(u^{\prime}\right)^2
$$
Hence equation (1) becomes
\begin{align*}
\frac{1}{2} \left(u^{\prime}\right)^2 &=\int{2 u^{3}\, \mathrm{d}u} \tag{2}
\end{align*}
But
$$
\int{2 u^{3}\, \mathrm{d}u} = \frac{u^{4}}{2}
$$
Therefore equation (2) becomes
\begin{align*}
\frac{1}{2} \left(u^{\prime}\right)^2 &=\frac{u^{4}}{2} + c_2
\end{align*}
Where $c_2$ is an arbitrary constant of integration.
This is first order ODE which is now solved for $u$.
Solving for $u^{\prime}$ gives
\begin{align*}
u^{\prime}&=\sqrt{u^{4}+2 c_{2}}\tag{1} \\
u^{\prime}&=-\sqrt{u^{4}+2 c_{2}}\tag{2}
\end{align*}
Let just solve (1) as (2) is similar.
\begin{align*}
\frac{1}{\sqrt{u^{4}+2 c_{2}}}\mathop{\mathrm{d}u} &= \mathop{\mathrm{d}x}\\
\int \frac{1}{\sqrt{u^{4}+2 c_{2}}}\mathop{\mathrm{d}u} &= \int \mathop{\mathrm{d}x}\\
\int \frac{1}{\sqrt{u^{4}+c_{3}}}\mathop{\mathrm{d}u} &= x +c_{1}
\end{align*}
Using the computer, integrating the left side above gives the solution
\begin{align*}
\frac{\sqrt{1-\frac{i u^{2}}{\sqrt{c_{3}}}}\, \sqrt{1+\frac{i u^{2}}{\sqrt{c_{3}}}}\, \operatorname{EllipticF}\left(u \sqrt{\frac{i}{\sqrt{c_{3}}}}, i\right)}{\sqrt{\frac{i}{\sqrt{c_{3}}}}\, \sqrt{u^{4}+c_{3}}} = x +c_{1}
\end{align*}
Similar solution for the second ode.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding Solutions to a Symmetric Non-Linear Equation (for Some Cases Beside $x = y = z$)
Find all values $x,y,z$ (whether real or complex) such that :
$2x-2y+z^{-1} = 2022^{-1}\\ 2z-2x+y^{-1}=2022^{-1}\\ 2y-2z + x^{-1} = 2022^{-1}$
I know that for case when $x=y=z$ i can easily find : $x=y=z = 2022$ (which seems to be the only real solution), What i'm struggling with is finding $x,y,z $ for cases beside $x = y = z$.
Attempt :
*
*$\displaystyle B_1+B_2+B_3 : x^{-1}+y^{-1}+z^{-1} = 3\cdot 2022^{-1}$
*$\displaystyle B_1-B_2+B_3 : 4(x-y) + 2z^{-1}-y^{-1}=2022^{-1} $
*$\displaystyle B_1+B_2 : 2(z-y) =2\cdot2022^{-1} $
It seems like no matter which row i'm operating i can never find equation in form of :
$ F_1(x,y,z)\cdot (x-y) = F_2(x,y,z) \cdot (x-y) \\ F_3(x,y,z)\cdot (x-z) = F_4(x,y,z) \cdot (x-z) \\ F_5(x,y,z)\cdot (y-z) = F_6(x,y,z)\cdot (y-z) $
My question : how do i turn the early equation into such equation above
so that i wouldn't need to find for $x/y/z$ manually (in a longer & standard way).
I need an assistance with this one, any help is appreciated
| To simplify the calculations, doing the same as @Dietrich Burde did in his answer, let $a=\frac 1{2022}$ and you end with the cubic
$$12 a x^3-36 x^2-3 a^3 x+a^2=0$$
which has three real roots since
$$\Delta=1296 a^2 \left(a^4+12\right)^2 \quad > 0~~\forall a$$ Then the three roots are given by
$$x_k=\frac 1 a +\frac{\sqrt{3\left(a^4+12\right) }}{3 a}\cos \left(\frac{1}{3} \left(2 \pi k-\cos
^{-1}\left(\sqrt{\frac{12}{a^4+12}}\right)\right)\right) \qquad (k=0,1,2)\tag 1$$ Using $a=\frac 1{2022}$, you obtain the results already given by @Dietrich Burde but the two small roots are not identical in absolute value; in this case, they differ by $8.96\times 10^{-12}$.
In fact, if $a$ is small, a Taylor expansion gives
$$x_1+x_2=-\frac{2 a^3}{27}+O\left(a^7\right)$$
Edit
Using $(1)$, for small values of $a$, the series expansions are
$$x_0=\frac{3}{a}+\frac{2 a^3}{27}-\frac{7 a^7}{4374}+O\left(a^{11}\right)$$
$$x_1=\frac{a}{6}-\frac{a^3}{27}+\frac{a^5}{486}+\frac{7 a^7}{8748}-\frac{5
a^9}{52488}+O\left(a^{11}\right)$$
$$x_2=-\frac{a}{6}-\frac{a^3}{27}-\frac{a^5}{486}+\frac{7 a^7}{8748}+\frac{5a^9}{52488}+O\left(a^{11}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability of exactly $7$ unreplaced draws from $\{1,...,10\}$ to get $1,2,3$? A box contains tickets numbered $1$ to $10$. Tickets are drawn at random without replacement until the numbers $1, 2$ and $3$ are drawn. Find the probability that exactly $7$ draws are required.
I am a little confused with this question. My first thought was to define the numbers $1,2,$ and $3$ as successes and the rest of the numbers as failures. Then,
$$P(\text{in $6$ draws you get $2$ successes}) = \frac{\binom{3}{2} \binom{7}{4}}{\binom{10}{6}}.$$ But I don't know where to go next or if this is even needed.
Can anyone help me? Thanks.
| Probability $1,2,3$ are in the first $7$ draws is the probability none in last $3$ are $1,2,3$: $\frac{7 \choose 3}{10 \choose 3}$
Probability $1,2,3$ are in the first $6$ draws is the probability none in last $4$ are $1,2,3$: $\frac{7 \choose 4}{10 \choose 4}$
Probability exactly $7$ draws are required is the difference $\frac{7 \choose 3}{10 \choose 3}-\frac{7 \choose 4}{10 \choose 4} = \frac18$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Properties of Continued fractions of $\sqrt n$ I got to know from this Brilliant article (no pun intended) that the continued fraction expansion expression of $\sqrt n$ is of the form $[a_0,\overline{a_1,a_2,\dots ,a_k}]$ for integers $a_i$ where
*
*$a_0=\lfloor\sqrt n\rfloor$ and $a_k=2a_0$
*$a_1,a_2\dots ,a_{k-1}$ is a palindrome, i.e., $a_i=a_{k-i}$
Unfortunately, these properties were presented without proof, and a quick google search didn't yield much. A suggested question seems to be similar to a part of my question, although the answers seem to use too advanced machineries for me to understand.
I would like to have simple proofs or intuitive explanations to understand why these beautiful patters hold. This table listing some continued fractions may be helpful. This paper suggested by mathcounterexamples.net is quite informative.
| here I give an example fo converting the square root to continuous fraction:
We want to find $\sqrt{41}$. The greatest perfect square less than $41$ is $36=6^2$ , so we may write:
$\sqrt{41}\approx 6 +\frac 1 x$
We have:
$\frac 1 x=\sqrt {41}-6\Rightarrow x=\frac 1{\sqrt{41}-6}\space\space\space\space(1)$
Multiplying numerator and denominator be conjugate $\sqrt {41}+6$ we get:
$x=\frac {\sqrt{41}+6} 5$
$\sqrt{41}+6>10 \Rightarrow x>2$
So we may write:
$x=2+\frac 1y=\frac{\sqrt{41}+6}5\Rightarrow \frac 1y=\frac{\sqrt{41}-4}5\Rightarrow y=\frac 5{\sqrt{41}-4}$
Multiplying numerator and denominator be conjugate $\sqrt {41}+4$ we get:
$y=\frac {\sqrt{41}+4}5>2$
Suppose:
$\frac 1y=2+\frac1z$
We have:
$\frac 1z=\frac{\sqrt{41}+4}5-2=\frac{\sqrt{41}-6}5$
$\Rightarrow z=\frac 5{\sqrt{41}-6}$
Multiplying numerator and denominator be conjugate $\sqrt {41}+6$ we get:
$z=\sqrt {41}+6>12$
Let $z=12+\frac 1u$, we can write:
$\frac 1 u=\sqrt {41}+6-12=\sqrt{41}-6\space\space\space\space(2)$
Comparing relation (2) with (1) we find that $u=x$, s0 our continuous fraction is:
$$\sqrt {41}\approx 6+\frac1 {2+\frac1{2+\frac1{12+\frac 1{2+\frac1{...}}}}}$$
Or $\sqrt {41}\approx(6, 2, 2 ,12, 2, 2, 12, ...)$
Now we convert one part of this to ordinary fraction:
$2+\frac 1{12}=\frac {25}{12}: \frac 1{\frac{25}{12}}=\frac {12}{25} $
$2+\frac{12}{25}=\frac{62}{25}: \frac 1{\frac{62}{25}}=\frac {25}{62}$
finally :
$\sqrt{41}\approx [6+\frac{25}{62}=\frac{397}{62}]\approx 6.403$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Finding a closed form of $f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$. I'm trying to find the closed form of
$f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$.
Empirically, it turns out the answer is simply $f(n)=1+\frac{n-4}{5}\ ,\ n \geq 4$, but I'm having a hard time getting there. I've tried two ways but neither is very successful. Could someone please suggest a way through?
Attempt 1: I try to set up a generating function
$$G(x)=f(4)x^4+f(5)x^5+\dots=\sum_{k=4}^\infty f(k)x^k$$
Substituting in the recurrence (edit: noticed I forgot to add the +1)
$$\begin{align}G(x)&=x^4+\sum_{k=5}^\infty(1-\frac{4}{k})f(k-1)x^k \\
&=x^4+x\sum_{k=5}^\infty (1-\frac{4}{k}) f(k-1)x^{k-1} \\
&= x^4+x\sum_{k=4}^\infty (1-\frac{4}{k+1})f(k)x^k \\
&= x^4+xG(x)-x\sum_{k=4}^\infty \frac{4}{k+1}f(k)x^k
\end{align} $$
But I don't know how to express the last part in terms of G(x).
Attempt 2: Just brute force it:
With arbitrary $k$,
$$\begin{align}f(n) &= \frac{(n-4)(n-5) \dots (n-(k+3))}{n(n-1)\dots(n-(k-1))}f(n-k)\\
&\ \ + \left(1+\frac{n-4}{n}+\frac{(n-4)(n-5)}{n(n-1)}+\dots+\frac{(n-4)\dots(n-(k+2))}{n\dots(n-(k-2))}\right)\end{align}$$
Setting $n-k=4 \implies k = n -4$ and using $f(4)=1$,
$$\begin{align} f(n) &= \frac{(n-4)!}{n!/4!}+\left(1+\frac{n-4}{n}+\frac{(n-4)(n-5)}{n(n-1)}+\dots+\frac{(n-4)\dots2}{n \dots 6}\right) \\
&= {n \choose 4}^{-1}+\left(1+\frac{(n-4)!}{n!}\left(\frac{(n-1)!}{(n-5)!} + \frac{(n-2)!}{(n-6)!}+\dots +\frac{5!}{1!}\right)\right)
\\
&= {n \choose 4}^{-1}+\frac{(n-4)!}{n!}\sum_{k=1}^{n-5}\frac{(k+4)!}{k!} \\
&= {n \choose 4}^{-1}+\frac{1}{n(n-1)(n-2)(n-3)}\sum_{k=1}^{n-5}(k+4)(k+3)(k+2)(k+1) \\
&= {n \choose 4}^{-1}+\frac{1}{n(n-1)(n-2)(n-3)}\sum_{k=5}^{n-1}k(k-1)(k-2)(k-3)\end{align}$$
Empirically this seems to be correct, but even Mathematica refuses to simplify it all the way down. Surely there is a better way?
| Alt. hint: write it as $\lambda_nf(n) = \lambda_{n-1}f(n-1) + \mu_n$ where $\lambda_n=\mu_n=n(n-1)(n-2)(n-3)$.
[ EDIT ] After telescoping $\,\lambda_nf(n) = \lambda_4 f(4) + \sum_{k=5}^n \mu_k\,$ where $\,\lambda_4 = 24\,$, $\,f(4)=1\,$, and the sum of falling factorials can be calculated as shown at $\sum r(r+1)(r+2)(r+3)$ is equal to?, or in the general case Partial sums of falling factorials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the maximum value of $M$ with given constraints.
Let $a, b \in \mathbb{R}$ such that $n^{4a-\log_5{n^2}} \leq 25^{40-b^2}$ for all positive real number $n$. Find the maximum value of $M=a^2+b^2+a-3b$.
What I have tried:
Taking the logarithm base 5 of both sides from the inequality, we get $(4a-2\log_5{n})\log_5n\leq2(40-b^2)$
Now, let $x=\log_5 n$, we get $x(4a-2x)\leq2(40-b^2)$, and this is equivalent to $x^2+2ax-b^2+40
\geq0$
The new inequality is true if and only if the discriminant $4a^2-4(40-b^2)\leq0$, this is equivalent to $a^2+b^2\leq40$, so we have $\max(a^2+b^2)=40$
This gave me hope of finding the maximum value of $M$ by finding the maximum value of $a-3b$ but I have no idea how to do it.
I'd like to call for some help. Thank you!
|
$x^2+2ax-b^2+40
\geq0$
You dropped a sign: should be
$x^2-2ax-b^2+40
\geq0.$
This won't matter to the rest of your analysis, which is good. You reached the point of realizing that
you must have
$$a^2 + b^2 = 40 \implies $$
$$ -\sqrt{40} \leq a \leq \sqrt{40} ~~\text{and}~~
-\sqrt{40} \leq b \leq \sqrt{40}. \tag1 $$
So, in effect, your problem reduces to maximizing $(a - 3b)$, subject to the constraint in (1) above.
Note that for any fixed value of $a$ such that $~\displaystyle -\sqrt{40} < a < \sqrt{40}$,
you can have that $~\displaystyle b = \pm \sqrt{40 - a^2}.$
Setting
$\displaystyle f(a) = a \pm 3\left[40 - a^2\right]^{(1/2)} \implies $
$\displaystyle f'(a) = 1 \pm \left\{3 \times (1/2) \times \left[40 - a^2\right]^{(-1/2)} \times (-2a)\right\} \implies $
$$f'(a) = 1 \pm (3a) \times \left[40 - a^2\right])^{(-1/2)} ~: ~-\sqrt{40} < a < \sqrt{40}. \tag2 $$
Since the second derivative, $f''(a),$ is messy, I will examine the behavior of $f(a)$ solely by considering $f'(a)$ as well as the behavior of $f(a)$ at $~\displaystyle a = \pm \sqrt{40}.$
$\displaystyle f'(a) = 0 \implies
1 \pm\frac{3a}{\sqrt{40 - a^2}} = 0.$
This implies that
$$1 = \frac{9a^2}{40 - a^2} \implies 40 - a^2 = 9a^2 \implies a = \pm 2 \implies b = \pm 6.$$
Then:
*
*$\displaystyle (a,b) = \left(-\sqrt{40},0\right) \implies (a - 3b) = -\sqrt{40}.$
*$\displaystyle (a,b) = \left(+\sqrt{40},0\right) \implies (a - 3b) = \sqrt{40}.$
*$\displaystyle (a,b) = \left(2,-6\right) \implies (a - 3b) = 20.$
So, the maximum value of $M$ is $(40 + 20) = 60.$
Edit
I just realized that my analysis can reasonably be considered to be out-of-bounds. This is because I computed the derivative, $f'(a)$, and your problem specified the algebra-precalculus tag.
Unfortunately, I never really studied the behavior of the circle $a^2 + b^2 = 40$, from the perspective of maximizing $(a - 3b)$ solely through Analytical Geometry. If this is really the problem composer's intent, perhaps someone else may provide an answer that does not depend on examining the derivative, $f'(a).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the differential equation $L^2 y=0$ where $L=x^2\frac{d^2}{dx^2}+x\frac{d}{dx}-(x^2+1)\text{id}$ Consider the linear differential equation defined by
$$
L^2y=0,\quad \text{where}\ L:=x^2\frac{d}{dx^2}+x\frac{d}{dx}-(x^2+1)\text{id}. \tag{1}$$
Here $\text{id}$ denotes the identity operator, and $L^2$ denotes the operator $L$ applied twice. Written out explicitly, (1) is of the form
$$x^4\frac{d^4 y}{dx^4}+6 x^3\frac{d^3y}{dx^3}+(5x^2-2x^4)\frac{d^2y}{dx^2}-(x+6x^3)\frac{dy}{dx}+(x^2-1)^2y=0. \tag{2}$$
I am interested in the general solution to (1) which is bounded at $x=0$.
Since the solution that we are looking for is bounded at $x=0$, we may write it as
$$y(x)=\sum_{n=0}^\infty a_n x^n,$$
near the origin. Substituting the above solution into (2) and equating the coefficient of $x^n$ to be zero, we obtain the following recurrence relation
$$ a_2=\frac{2}{9}a_0,\quad a_3=\frac{1}{8}a_1,\quad a_4=\frac{1}{75}a_0,\quad a_5=\frac{1}{192}{a_1},\quad a_6=\frac{4}{11025}a_0,\quad\dots.$$
Setting $(a_0,a_1)=(1,0)$ and $(a_0,a_1)=(0,1)$, we obtain two linearly independent solutions whose linear combination gives the general solution to (1) bounded at $x=0$.
The solution to with $(a_0,a_1)=(0,1)$ can be written as
$$
y_1(x)=x+\frac{1}{8}x^3+\frac{1}{192}x^5+\cdots=2I_1(x),$$
where $I_1(x)$ is the modified Bessel function of the first kind, while the solution with $(a_0,a_1)=(1,0)$ can be represented as
$$
y_2(x)=1+\frac{2}{9}x^2+\frac{1}{75}x^4+\frac{4}{11025}x^6+\cdots$$
My question is that: can $y_2(x)$ be represented in terms of any special function (like $y_1(x))$?
Update: As pointed out in the comments, the $x^0$ coefficient relation gives $a_0=0$, so the solution $y_2(x)$ doest not exist.
| Per a c.a.s. the general solution is
\begin{split}y(x) = A I_1(x) + B K_1(x) + C \left[-I_1(x) \int_{x_0}^x \frac{K_1(t)^2 \,dt}{t} + K_1(x) \int_{x_0}^x \frac{I_1(x) K_1(x)}{x}\,dt \right] \qquad \qquad \\+ D \left[\frac{1}{2} K_1(x) \left(I_0(x)^2 - I_1(x)^2\right) - I_1(x) \int_{x_0}^x \frac{I_1(x) K_1(x)}{x}\,dt\right] ,\end{split} where $I_\alpha, K_\alpha$ are the modified Bessel functions of the First and Second Kinds, respectively.
Expanding in a series shows that as $x \searrow 0$,
$$y(x) = \frac{1}{2} C \frac{\log x}{x} + o\left(\frac{\log x}{x}\right),$$
so for any solution bounded near $x = 0$ we must have $C = 0$. Expanding $y(x) \vert_{C = 0}$ around $x = 0$ now gives
$$y(x) = \left(B + \frac{D}{2}\right) \frac{1}{x} + o\left(\frac{1}{x}\right),$$
so boundedness near $0$ implies that $D = -2 B ,$ and expanding $y(x) \vert_{C = 0, D = -2B}$ gives a $2$-parameter family of solutions bounded near $0$; n.b. all of these functions are odd. Now the series expansion as $x \searrow 0$ is
$$\frac{B}{2} x \log x + \left(\frac{A}{2} - \frac{B}{4}\right) x + o(x) ,$$ so the solutions are bounded exactly for such choices of coefficient. Rewriting per the below comment of Maxim gives a simple expression for the general bounded solution:
$$y(x) = a I_1(x) + b \left(\frac{I_0(x)}{x} - K_1(x)\right) .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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why does $1/((x-1)(x^2 - 1)) = (x^4-1)/(x-1) \cdot (x^8 - 1)/(x^2 - 1) \cdot (x^{16} - 1)/(x^4 - 1) \cdots$
Why does $1/((x-1)(x^2 - 1)) = (x^4-1)/(x-1) \cdot (x^8 - 1)/(x^2 - 1) \cdot (x^{16} - 1)/(x^4 - 1) \cdots$?
I think the equality holds as formal power series. Expanding the LHS, one should get $(1+x+x^2 +\cdots )(1+x^2 + x^4+\cdots).$ But doing cancellation for the RHS gives that it equals $(x^2+1)/(x-1) \cdot (x^4 + 1)(x^8 + 1)\cdots,$ which clearly contradicts the equality. If however, one multiplies both sides by $x^2 - 1$, one gets an obvious equality after cancellation.
I was wondering if someone could explain what I'm doing wrong here (i.e. why I get two different "answers")?
| "But doing cancellation for the RHS gives that it equals $(x^2+1)/(x−1)\cdot(x^4+1)(x^8+1)\cdots$, which clearly contradicts the equality."
In fact $(1 + x^2)(1 + x^4)(1 + x^8)(1 + x^{16}) \cdots = 1 + x^2 + x^4 + x^6 + x^8 + x^{10} + \ldots = \frac{1}{1 - x^2}$ -- the first equality is just the generating-functional version of binary expansion, see e.g. this recent question: Demonstrate recursively that $\prod_{k = 0}^\infty (1 + x^{2^k}) = \frac{1}{1-x}$. (Or maybe you mean the fact that you've lost a minus sign? That's because the leftover factor is $x^{2^n} - 1$, which is approaching $-1$ as $n \to \infty$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For a triangle, prove that $\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\frac{1}{r^2} $
Context: I made this question and am contemplating about submitting it as a contest question. The contest is not a large one; its scope does not comprise even our whole class. Its just a friendly, small contest within a small group. This originally occurred to me when we were asked to prove $\tan \frac A2\tan\frac B2 +\tan \frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1$ in class. I tried a geometrical proof but failed, but I got the following for my efforts.
Derivation: Given $\triangle ABC$ and its incircle $\Im$ with inradius $r$, we have the known (and easily provable) formula $$\tan \frac A2\tan\frac B2 +\tan \frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1.\tag{1}\label{1}$$ We know, $IE=IF=IG=r$. Now, $$\tan \frac A2=\frac{r}{AE}=\frac{r}{AF}$$$$\tan \frac B2=\frac{r}{BE}=\frac{r}{BG}$$$$\tan \frac C2=\frac{r}{CG}=\frac{r}{CF}$$ Putting appropriate values in the formula $(1)$, we get $$\frac{r^2}{AE\cdot BE}+ \frac{r^2}{CG\cdot BG}+ \frac{r^2}{AF\cdot CF}=1$$
or $$\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\frac{1}{r^2}$$ Finally, using the identities $\Delta=rs, s=\dfrac{a+b+c}{2}, \Delta=\dfrac{abc}{4R}$ (where R is the circumradius) and the sine law we get $$\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\dfrac{s^2}{\Delta^2}=\dfrac{(a+b+c)^2}{4\Delta^2}=\dfrac{(a+b+c)^2\cdot 16R^2}{4a^2b^2c^2} $$ Thus, $$\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=2R\frac{a+b+c}{abc}$$ $$=2R\frac{2R(\sin A+\sin B+\sin C)}{8R^3\sin A\sin B\sin C}$$$$=\frac{1}{2R}\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$ so that $$2R\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$
Question: Prove that for a triangle ABC, $$2R\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$
The result looks pretty symmetrical and nice, and also reminds one of Ceva’s theorem because of the segments taken consecutively.
If you were given this question and did not know the above derivation, how would you approach it? Basically, I am looking for alternative solutions to this problem just for reference.
If the question has any flaws, please tell me. Also, I would really like to know how you would rate this question for a high school level informal contest. If any of you can come up with variants of this, especially as inequalities, I would be grateful. Thanks in advance.
| By the result $1$ found in my post, $$\displaystyle x yz=r^{2}(x+y+z),\tag*{(*)} $$
where $r$ is the in-radius of the triangle ABC of sides $a,b$ and $c$ and $a=y+z, b=z+x $ and $c=x+y$.
Dividing the equation (*) by $xyzr^2$ yields the required equation
$$
\frac{1}{y z}+\frac{1}{x z}+\frac{1}{x y}=\frac{1}{r^{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Integrate $\ln^3(\sin(x))$ using Fourier series
Show that $$\int_0^{\frac\pi2} \ln^3(\sin(x)) \, dx = -\frac\pi2 \ln^3(2) - \frac{\pi^3}8 \ln(2) - \frac{3\pi}4 \zeta(3)$$
I have seen a method for this elsewhere, but I would specifically like to reproduce this result in the same way I've computed the similar integrals of $\ln(\cos(x))\ln(\sin(x))$ and $\ln^2(\sin(x))$, as shown here and here - which involves a "complicated series", as Jack puts it.
In particular, I want to use the Fourier series
$$f(x) = \ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k$$
Expanding the integrand yields
$$\begin{align*}
-f(x)^3 &= \ln^3(2) + 3 \ln^2(2) \sum_{a=1}^\infty \frac{\cos(2ax)}a \\ &\quad + 3\ln(2) \left(\sum_{a=1}^\infty \frac{\cos^2(2ax)}{a^2} + 2 \sum_{a\neq b} \frac{\cos(2ax) \cos(2bx)}{ab}\right) \\
& \quad + \left(\sum_{a=1}^\infty \frac{\cos^3(2ax)}{a^3} + 3 \sum_{a\neq b} \frac{\cos^2(2ax) \cos(2bx)}{a^2b} + 6 \sum_{a\neq b\neq c} \frac{\cos(2ax) \cos(2bx) \cos(2cx)}{abc}\right)
\end{align*}$$
In the integral, the series with $\cos(2ax)$ vanishes; by orthogonality, $\cos(2ax)\cos(2bx)$ vanishes; in combination of both of these facts, $\cos^2(2ax)\cos(2bx)$ and $\cos^3(2ax)$ also vanish. So the integral reduces to
$$\int_0^{\frac\pi2} f(x)^3 \, dx = -\frac\pi2 \ln^3(2) - \frac{\pi^3}8 \ln(2) - 6 \sum_{a\neq b\neq c} \frac1{abc} \int_0^{\frac\pi2} \cos(2ax) \cos(2bx) \cos(2cx) \, dx$$
In the remaining integral, I'm fairly sure that most of the terms integrate to $0$ using the orthogonality argument, except in the case of $a+b=c$,
$$\int_0^{\frac\pi2} \cos(2ax) \cos(2bx) \cos(2(a+b)x) \, dx \\ = \int_0^{\frac\pi2} \frac{\cos(2(a-b)x)\cos(2(a+b)x) + \cos^2(2(a+b)x)}2 \, dx = \frac\pi8$$
ETA: If there are no other triples, then I should end up with
$$-6 \sum_{a\neq b\neq c} \frac1{abc} \int_0^{\frac\pi2} \cos(2ax) \cos(2bx) \cos(2cx) \, dx = -\frac{3\pi}4 \sum_{a+b=c} \frac1{abc} = -\frac{3\pi}4 \zeta(3) \\ \implies \sum_{a\neq b} \frac1{ab(a+b)} = \zeta(3)$$
Now,
$$\begin{align*}
\sum_{a\neq b} \frac1{ab(a+b)} &= \sum_{(a,b)\in\Bbb N^2} \frac1{ab(a+b)} - \frac12 \sum_{a=1}^\infty \frac1{a^3} \\[1ex]
&= 2 \sum_{a<b} \frac1{ab(a+b)} - \frac12 \zeta(3) \\[2ex]
\sum_{a<b} \frac1{ab(a+b)} &= \sum_{b=2}^\infty \frac1{b(b+1)} + \sum_{b=3}^\infty \frac1{2b(b+2)} + \sum_{b=4}^\infty \frac1{3b(b+3)} + \cdots \\[1ex]
&= \sum_{b=2}^\infty \left(\frac1b - \frac1{b+1}\right) + \frac14 \sum_{b=3}^\infty \left(\frac1b - \frac1{b+2}\right) + \frac19 \sum_{b=4}^\infty \left(\frac1b - \frac1{b+3}\right) + \cdots \\[1ex]
&= (H_2 - H_1) + \frac{H_4 - H_2}4 + \frac{H_6 - H_3}9 + \cdots \\[1ex]
&= \sum_{n=1}^\infty \frac{H_{2n} - H_n}{n^2} \\[1ex]
&= \sum_{n=1}^\infty \frac{H_{2n}}{n^2} - 2\zeta(3)
\end{align*}$$
where the last equality is due to (31), and it remains to show
$$\sum_{n=1}^\infty \frac{H_{2n}}{n^2} = \frac{11}4 \zeta(3)$$
Rewriting the sum as follows leads me to think there may be a hidden Cauchy product, but I have not been able to find a decomposition.
$$H_{2n}-H_n = \sum_{k=1}^{2n} \frac1k - \sum_{k=1}^n \frac1k = \sum_{k=n+1}^{2n} \frac1k = \sum_{k=1}^n \frac1{n+k} \\ \implies\sum_{n=1}^\infty \frac{H_{2n}-H_n}{n^2} = \sum_{n=1}^\infty \sum_{m=1}^n \frac1{n^2(n+m)}$$
| $$\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n^2-H_n^{(2)}}{n+1}\cos(2(n+1)x)\\
=\Re\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n+1}(-e^{2ix})^{n+1}$$
$$=\Re\left\{-\frac13 \ln^3(1+e^{2ix})\right\}$$
$$=x^2\ln(2\cos x)-\frac13\ln^3(2\cos x)\tag{*}$$
where we used the generating function
$$\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n+1}x^{n+1}=-\frac13\ln^3(1-x)$$
which follows from integrating
$$\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^{n}=\frac{\ln^2(1-x)}{1-x}.$$
Integrate both sides of $(*)$ from $x=0\to \pi/2$,
$$\int_0^{\pi/2}\left[x^2\ln(2\cos x)-\frac13\ln^3(2\cos x)\right]dx$$
$$=\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n^2-H_n^{(2)}}{n+1}\int_0^{\pi/2}\cos(2(n+1)x)dx$$
$$=\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n^2-H_n^{(2)}}{n+1}\left(-\frac{\sin(n\pi)}{2(n+1)}\right)=0$$
so
$$\int_0^{\pi/2}\ln^3(2\cos x)dx=3\int_0^{\pi/2}x^2\ln(2\cos x)dx=-\frac34\pi\zeta(3).$$
Note that $\int_0^{\pi/2}\ln^3(2\sin x)dx=\int_0^{\pi/2}\ln^3(2\cos x)dx.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
How do I handle equations like $f(x)-x = f'(x)$? How do I handle equations like $f(x)-x = f'(x)$?
Integrating both sides gets me nowhere... I have been trying to guess an example, but with no avail. I cannot think of any other approaches ...
Edit: ok i noticed $f(x) =x+1$ ... can someone guide be to how i can solve for the other solutions?
| I know that the below solution doesn't work in many situations, but I like to share its idea.
Suppose $f(x) = \sum_{n \geq 0}a_nx^n$ is the power series of $f$. Because $f(x) - x = f'(x)$, term by term differentiating of the power series gives us
\begin{align*}
\sum_{n \geq 0}a_nx^n - x = \sum_{n \geq 1}na_nx^{n-1} \implies a_0 + (a_1 - 1)x + a_2x^2 + \cdots = a_1 + 2a_2x + 3a_3x^2 + \cdots
\end{align*}
Because the above equation is correct for all $x$, so we have:
\begin{align*}
a_1 &= a_0\\
a_1 - 1 = 2a_2 \implies a_2 &= \frac{a_1 - 1}{2} = \frac{a_0 - 1}{2}\\
a_2 = 3a_3 \implies a_3 &= \frac{a_2}{3} = \frac{a_0 - 1}{3!}
\end{align*}
Computing in this way shows that for all $n \geq 2$ we have $a_n = \frac{a_0 - 1}{n!}$. Therefore
\begin{align*}
f(x) = \sum_{n \geq 0}a_nx^n &= a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots\\
&= a_0 + a_0x + (a_0 - 1)\frac{x^2}{2!} + (a_0 - 1)\frac{x^3}{3!} + \cdots\\
&= 1 + x + (a_0 - 1) + (a_0 - 1)x + (a_0 - 1)\frac{x^2}{2!} + (a_0 - 1)\frac{x^3}{3!} + \cdots\\
&= 1 + x + \underbrace{(a_0 - 1)}_Ce^x
\end{align*}
For an arbitrary constant $C \in \mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Finding Equation of tangent to the curve $xy^2=25$ at point $(5,5)$
Find equation of the tangent to the curve $xy^2=25$ at point $(5,5)$
My Working
I tried this sum using two different approaches. But I'm getting two different answers. I wonder if any is correct. Can you please guide me on this
First approach;
use product rule to differentiate expression as it is to find the $\frac{dy}{dx}$ gives me,
$x.2y\frac{dy}{dx}+y^2.1=0$
$\frac{dy}{dx}=-\frac{y}{2x}$
Finally Im getting the equation as, $y-5=-\frac{1}{2}(x-5)$
Second approach;
differentiate function explicitly after making $y^2$ the subject,
$y^2=\frac{25}{x}$
$2y\frac{dy}{dx}=-\frac{25}{x^2}$
$\frac{dy}{dx}=-\frac{25}{2yx^2}$
$y-5=-\frac{1}{10}(x-5)$
Thank you in advance!
| You did the differentiation right.
$$\frac{dy}{dx} = - \frac{y}{2x} = - \frac{25}{x^2y}$$
Note that if $xy^2 = 25$, then $- \frac{25}{2x^2y} = - \frac{25y}{2x^2y^2} = - \frac{25y}{2x(xy^2)} = - \frac{25y}{2x(25)} = - \frac{y}{2x}$, so the two expressions are equivalent.
The problem is that the question is ill-formed: If $x = y = 5$, then $xy^2 = 125 \ne 25$. So with two contradictory values of $xy^2$, of course the math won't be consistent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Evaluate the integral $I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$ To evaluate the integral
$$I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$$
I define $t=-1+\sqrt{\frac{4}{x}-3}$, then we have $x=\frac{4}{(t+1)^2+3}$ and
$$dx=\frac{-8(t+1)}{(t^2+2t+4)^2}dt$$
So we get
$$\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx=8\int_0^\infty \frac{\sqrt{t}(t+1)}{(t^2+2t+4)^2}\ dt$$
Again, let $s=\sqrt{t}$, then $dt=2sds$ and the integral becomes
$$8\int_0^\infty \frac{s(s^2+1) 2s}{(s^4+2s^2+4)^2}\ ds=16\int_0^\infty \frac{s^2(s^2+1)}{(s^4+2s^2+4)^2}\ ds$$
Next, I write the integrand as
$$\frac{s^2(s^2+1)}{(s^4+2s^2+4)^2}=\frac{1}{s^4+2s^2+4}-\frac{s^2+4}{(s^4+2s^2+4)^2}$$
so I have to evaluate
$$I_1=\int_0^\infty \frac{1}{s^4+2s^2+4}\ ds\ and\ I_2=\int_0^\infty \frac{s^2+4}{(s^4+2s^2+4)^2}\ ds$$
But when I evaluate the integral $I_1$, I get a weird result as follows:
$$I_1=\int_0^\infty \frac{1}{(s^2+2)^2-(\sqrt{2}s)^2}\ ds
=\int_0^\infty \frac{1}{(s^2+\sqrt{2}s+2)(s^2-\sqrt{2}s+2)}\ ds=\int_0^\infty \left(\frac{\frac{1}{4\sqrt{2}}s+\frac{1}{4}}{s^2+\sqrt{2}s+2}+\frac{\frac{-1}{4\sqrt{2}}s+\frac{1}{4}}{s^2-\sqrt{2}s+2}\right)\ ds$$
and if we separate two parts and evaluate the integrals, we would get two divergent improper integrals. So how can I find $I_1$?(Hope that the method is elementary and without complex analysis.)
Another question: I'm a beginner at learning complex analysis. I conjecture that we can evaluate the integral $I$ in complex analysis (or maybe not worked). Hope everybody can give me some hints or solutions with the method in complex analysis.
| $ \text {Let } y=\sqrt{-1+\sqrt{\frac{4}{x}-3}}\textrm{ then ,}$
$ \displaystyle \begin{aligned}I&=16 \int_{0}^{\infty} \frac{y^{2}\left(y^{2}+1\right) d y}{\left(y^{4}+2 y^{2}+4\right)^{2}}\\&=4\left[3 \underbrace{\int_{0}^{\infty} \frac{y^{2}\left(y^{2}+2\right)}{\left(y^{4}+2 y^{2}+4\right)^{2}} d y}_{J}+\underbrace{\int_{0}^{\infty} \frac{y^{2}\left(y^{2}-2\right)}{\left(y^{4}+2 y^{2}+4\right)^{2}}}_{K} d y\right] \end{aligned}\tag*{} $
Now let’s play a little trick on the integral $ J$.
$\displaystyle \begin{aligned}J &=\int_{0}^{\infty} \frac{1+\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y \\&=\int_{0}^{\infty} \frac{d\left(y-\frac{2}{y}\right)}{\left[\left(y-\frac{2}{y}\right)^{2}+6\right]^{2}} \\&=\int_{-\infty}^{\infty} \frac{d u}{\left(u^{2}+6\right)^{2}}\\ &\stackrel{u=\sqrt6 \tan \theta}{=}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sqrt{6} \sec ^{2} \theta d \theta}{\left(6 \sec ^{2} \theta\right)^{2}}\\&=\frac{\pi}{12 \sqrt{6}} \end{aligned} \tag*{} $
For the integral $ K$ , we first split the interval into two.
$ \displaystyle \begin{aligned}K &=\int_{0}^{\infty} \frac{1-\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y \\&=\int_{0}^{1} \frac{1-\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y+\int_{1}^{\infty} \frac{1-\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y}+2\right)^{2}} d y \\&=\int_{0}^{1} \frac{d\left(y+\frac{2}{y}\right)}{\left[\left(y+\frac{2}{y}\right)^{2}-2\right]^{2}}+\int_{1}^{\infty} \frac{d\left(y+\frac{2}{y}\right)}{\left[\left(y+\frac{2}{y}\right)^{2}-2\right]^{2}} d y \\&=\int_{\infty}^{3} \frac{d u}{\left(u^{2}-2\right)^{2}}+\int_{3}^{\infty} \frac{d v}{\left(v^{2}-2\right)^{2}} \\&=0 \end{aligned} \tag*{} $
Now we can conclude that
$\displaystyle \boxed{I=4\left(3 \cdot \frac{\pi}{12 \sqrt{6}}\right)=\frac{\pi}{\sqrt{6}}}\tag*{} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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$\lim\limits_{x\to 0} \frac{x^2\cos x - 6 \ln(1+x^2) +5x^2}{(e^{\sqrt[4]{1+4x^3+8x^4}}-e)\arcsin(x)}$ without L'Hosptial rule. I have a belief that it is possible to evaluate any limit without using L'Hospital rule or series expansions. However the limit $$\lim\limits_{x\to 0} \frac{x^2\cos x - 6 \ln(1+x^2) +5x^2}{(e^{\sqrt[4]{1+4x^3+8x^4}}-e)\arcsin(x)}$$ stumps me.
I thought at the beginning that is it nothing but dividing top and bottom by $x^2$ and then using the limits: $$\lim\limits_{x\to 0} \frac{\ln(1+x^2)}{x^2} = 1, \quad \quad \lim \limits_{x\to 0} \frac{\arcsin x}{x} = 1$$
And then I found $$\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}}-e}{x} = e\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}-1}-1}{\sqrt[4]{1+4x^3+8x^4}-1} \cdot \frac{\sqrt[4]{1+4x^3+8x^4}-1}{(1+4x^3+8x^4)-1} \cdot \frac{4x^3+8x^4}{x} = 0$$
Which made the limit $\frac{0}{0}$. I have no clue how to split the limit into a product or a sum of existent limits. Is it possible to solve this without L'Hosptial? Is there is possible usage of the squeeze theorem or a geometric interpretation to compute this limit? The answer according to WolframAlpha is $\frac{5}{2e}$ and it is clear how to find the $\frac{1}{e}$ but not the $\frac{5}{2}$.
| With the hint of eyeballfrog, I was able to arrive at the following solution.
We write the nuemreator as $x^2(\cos x - 1) - 6(\ln(1+x^2)-x^2)$ then dividing top and bottom by $x^4$ the limit becomes $$\lim\limits_{x\to 0} \frac{\frac{\cos x - 1}{x^2}-\frac{6}{x^4}(\ln(1+x^2)-x^2)}{\frac{\arcsin x}{x} \cdot \frac{e^{\sqrt{1+4x^3+8x^4}-1}-1}{x^3}} $$
The numerator can be computed using the limits $\lim\limits_{x\to 0}\frac{\cos x -1}{x^2} = -\frac{1}{2}$ and $\lim\limits_{x\to 0} \frac{\ln(1+x^2)-x^2}{x^4} = \frac{-1}{2}$ where the latter is proved in Are all limits solvable without L'Hôpital Rule or Series Expansion
For the bottom, we use the limits in the original post except now it is more divided by $x^3$
$$\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}}-e}{x^3} = e\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}-1}-1}{\sqrt[4]{1+4x^3+8x^4}-1} \cdot \frac{\sqrt[4]{1+4x^3+8x^4}-1}{(1+4x^3+8x^4)-1} \cdot \frac{4x^3+8x^4}{x^3} = e \cdot \frac{1}{4} \cdot 4$$
Combining all of this we arrive at the desired result $\frac{5}{2e}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Evaluating $\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$. Why doesn't $x^2$ overpower $2x$ when $x\to\infty$, so that the answer is $0$ (instead of $1$)?
Evaluate $$\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$$
Hello!
I was solving this problem, and here is my approach:
Inside the square root $x^2$ overpowers $2x$ as $x \to \infty$, so $$\lim_{x \to \infty} \sqrt{x^2 + 2x} - x = \lim_{x \to \infty} \sqrt{x^2} - x = \lim_{x \to \infty} |x| - x = 0$$ which is wrong answer. Actual answer is $1$.
This is the way I have solved so many problems when $x \to \infty$, especially when the terms are in denominator.
For example, consider $$\lim_{x \to \infty} \frac{x^2 + x}{2x^2 + 4x + 10}$$
Now, we ignore the smaller power terms, and answer is $\frac{1}{2}$. Why can we ignore smaller powers here but not in the original question on top?
Sorry if this is a stupid question.
| $\lim_{x \to \infty} \sqrt{x^2+2x}-x =$
$\lim_{x \to \infty} (\sqrt{x^2+2x}-x) \cdot \frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{(\sqrt{x^2+2x}-x)\cdot(\sqrt{x^2+2x}+x)}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{|x^2+2x|-x^2}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{x^2+2x-x^2}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{2x}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{2}{\frac{\sqrt{x^2+2x}}{x}+1} =$
$\lim_{x \to \infty} \frac{2}{\sqrt{1+\frac{2}{x}}+1} =$
$\frac{2}{\sqrt{1+\lim_{x \to \infty} \frac{2}{x}}+1} = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Calculate the sum of $\left\lfloor \sqrt{k} \right\rfloor$ I'm trying to calculate for $n\in \mathbb{N}$ the following sum :
$\sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor$.
I tried putting in the first terms, which gave me
$\sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor=(1+2+3+\cdots+n)+\left\lfloor \sqrt{2} \right\rfloor+\left\lfloor \sqrt{3} \right\rfloor+\left\lfloor \sqrt{5} \right\rfloor+\cdots+\left\lfloor \sqrt{n^2-1} \right\rfloor$
$\iff \sum_{k=1}^{n^2}\left\lfloor \sqrt{k} \right\rfloor=\frac{n(n+1)}{2}+\left\lfloor \sqrt{2} \right\rfloor+\left\lfloor \sqrt{3} \right\rfloor+\left\lfloor \sqrt{5} \right\rfloor+\cdots+\left\lfloor \sqrt{n^2-1} \right\rfloor$.
I've been trying to somehow find a pattern between the different integer parts of the irrational numbers just like I did with the integers but I fail to success.
Is there a trick to use here or is my take wrong ?
Thank you.
| The sum can be calculated as follows:
$$
\sum_{k=1}^{n^2} \lfloor \sqrt k \rfloor = 1 + 1 + 1 + \overbrace{2 + .. + 2}^{5 \space \text{terms}}+ \overbrace{3 + .. + 3}^{7 \space \text{terms}} + \overbrace{4 + .. + 4}^{9 \space \text{terms}} + .. + n
$$
That reduces to:
$$
3 \times 1 + 5 \times 2 + 7 \times 4 + .. + (2n-1) \times (n-1) + n
$$
The above sum is:
$$
n + \sum_{k=1}^{n-1} (2n+1)\times n = \frac{n(4 n^2 - 3n + 5)}{6}
$$
Which is gotten from the OEIS found by @lulu. I used the method given in the proof at the OEIS (given by J. M. Bergot), with insight from @Thomas Andrews and I believe the last equality is straightforward to prove by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Spivak: How do we divide $1$ by $1+t^2$, to obtain $\frac{1}{1+t^2}=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$? In Chapter 20, "Approximation by Polynomial Functions" in Spivak's Calculus, there is the following snippet on page $420$
The equation
$$\arctan{x}=\int_0^x \frac{1}{1+t^2}dt$$
suggests a promising method of finding a polynomial close to $\arctan$
*
*divide $1$ by $1+t^2$, to obtain a polynomial plus a remainder:
$$\frac{1}{1+t^2}=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}\tag{1}$$
How exactly do we perform this division of $1$ by $1+t^2$ to obtain $(1)$?
| Use geometric sum:
$$1-t^2+t^4-t^6+...+(-1)^nt^{2n}=\frac{1-(-t^2)^{n+1}}{1-(-t^2)}=\frac{1}{1+t^2}+\frac{(-1)^nt^{2n+2}}{1+t^2}$$
Move terms and we get:
$$\begin{align}\frac{1}{1+t^2}&=1-t^2+t^4-t^6+...+(-1)^nt^{2n}-\frac{(-1)^{n}t^{2n+2}}{1+t^2}\\
\\\frac{1}{1+t^2}&=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Given $p(x)=x^2+ax+b$, where $a,b\in \mathbb{Z}$. If $p(2020 + p(2020))=1$, find $a,b$
Given $p(x)=x^2+ax+b$, where $a,b\in \mathbb{Z}$. If $p(2020 + p(2020))=1$,
find $a,b.$
My attempt:
Given $p(x) = x^2+ax+b$, then $p(x+p(x))=p(x^2+(a+1)x+b)=(x^2+(a+1)x+b)^2+a(x^2+(a+1)x+b)+b=1$
After a few pages of algebra, I get...
$(a x + b + x^2) (a x + a + b + x^2 + 2 x + 1) =1$
Substituting $x=2020.$
$(2020a + b + 2020^2) (2020a + a + b + 2020^2 + 2(2020) + 1) =1$
Equating the factors of 1 yields 2 systems of equations.
$$\begin{cases} (2020a + b + 2020^2)=1 \\ (2020a + a + b + 2020^2 + 2(2020) + 1) =1\\ \end{cases}\implies \boxed{a=-4041, b=4082421}$$
$$\begin{cases} (2020a + b + 2020^2)=-1 \\ (2020a + a + b + 2020^2 + 2(2020) + 1) =-1\\ \end{cases}\implies \boxed{a=-4041, b=4082419}$$
This was a competition question. I felt my solution isn't elegant since it required a lot of algebra. Any thoughts on a better solution?
| weird fact about parabolas:
$$p(x)=x^2+ax+b\iff \boxed{p(n)p(n+1)}=p\Big(p(n)+n\Big)$$
$$\begin{cases}2020a+b+2020^2&=\pm1\\2021a+b+2021^2&=\pm1\end{cases}$$
$0=2021^2-2020^2+a \iff a=-4041$
$b=\pm 1-2021(-4041)-2021^2=4082420 \pm 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is there a closed form for $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(\sin x+\cos x)^{2 n+1}} d x$? When I encountered the integral $$I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(\sin x+\cos x)^{2 n+1}} d x,$$
where n is a non-negative integer, I, as usual, used the substitution $x\mapsto \frac{\pi}{2}-x$ to transforms
$$I=
\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(\sin x+\cos x)^{5}} d x
$$
Then $$
\begin{aligned}
2 I &=\int_{0}^{\frac{\pi}{2}} \frac{d x}{(\sin x+\cos x)^{4}} \\
&=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{d x}{\cos ^{4}\left(\frac{\pi}{4}-x\right)} \\
&=\frac{1}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d(\tan x)\\&= \frac{1}{3}\\ \therefore I&= \frac{1}{6}
\end{aligned}
$$
Similarly, we can generalise $I$ to
$$I_n =\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(\sin x+\cos x)^{2 n+1}} d x=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(\sin x+\cos x)^{2 n+1}} $$
$$
\begin{aligned}
2 I_{n} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin x+\cos x}{(\sin x+\cos x)^{2 n+1}} d x \\
&=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left[\sqrt{2} \cos \left(\frac{\pi}{4}-x\right)\right]^{2 n}}\\&=\frac{1}{2^{n}} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2 n} x d x
\end{aligned}
$$
Letting $t=\tan x$ yields
$$
\begin{aligned}
I_{n} &=\frac{1}{2^{n}} \int_{0}^{1}\left(1+t^{2}\right)^{n-1} d t \\
&=\frac{1}{2^{n}} \sum_{k=0}^{n-1}\left(\begin{array}{l}
n-1 \\
\quad k
\end{array}\right) \int_{0}^{1} t^{2 k} d t \\
&=\frac{1}{2^{n}} \sum_{k=0}^{n-1}\left(\begin{array}{c}
n-1 \\
k
\end{array}\right) \frac{1}{2 k+1}
\end{aligned}
$$
My question is how to simplify the last sum?
| Something which could be interesting
$$I_n=\frac 1 {2^n} \sum_{k=0}^{n-1}\binom{n-1}{k}\frac{1}{2 k+1}=2^{-n} \, \,_2F_1\left(\frac{1}{2},1-n;\frac{3}{2};-1\right)$$ generates the unknown (?) sequence
$$\left\{\frac{1}{2},\frac{1}{3},\frac{7}{30},\frac{6}{35},\frac{83}{630},\frac{73}{6
93},\frac{523}{6006},\frac{476}{6435},\frac{14051}{218790},\frac{13103}{230945},
\frac{32867}{646646},\frac{31130}{676039},\cdots\right\}$$ but
$$\lim_{n\to \infty } \,( n\,I_n)=\frac 12$$
Let $n=10^k$ and compute
$$\left(
\begin{array}{cc}
k & 10^k\, I_{10^k} \\
1 & 0.5673645240 \\
2 & 0.5051036873 \\
3 & 0.5005010035\\
4 & 0.5000500100 \\
5 & 0.5000050001 \\
6 & 0.5000005000
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit of $u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k !$ I encountered a sequence $(u_n)_{n \in \mathbb{N}} $ defined as
$$ u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k ! $$
And I wonder what is the limit. It seems to be 1 but even Wolfram Alpha cannot figure it out.
My first idea was to write $$ \frac{k!}{n!} = \frac{1}{(k+1)...(n-1)n}$$ and
$$ (k+1)^{n-k} \leq (k+1)...(n-1)n \leq n^{n-k} $$
EDIT : I found a solution but Yanko found something more elementary.
My solution :
\begin{align}
u_n &= 1 + \frac{1}{n} + \frac{1}{n(n-1)} + \frac{1}{n(n-1)(n-2)} + ... + \frac{1}{n!} \\
&= 1 + \frac{1}{n} + (n-1) \times o(\frac{1}{n^2}) \\
&= 1 + \frac{1}{n} + o(\frac{1}{n}) \\
\end{align}
Thus all terms converge towards 0 except the first one, and the limit is 1.
| To go beyond the limit itself, using the same approach as @Yanko, let
$$a_p=\frac 1{\prod_{i=0}^p (n-i)}$$ and consider
$$1+\sum_{p=0}^\infty a_p=1+\sum_{k=0}^\infty \frac {B_k}{n^{k+1}}=1+\frac{1}{n}+\frac{1}{n^2}+\frac{2}{n^3}+\frac{5}{n^4}+O\left(\frac{1}{n^5}
\right)$$ where appear Bell numbers.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Prove $\int_{- \infty}^{\infty} \frac{\mathrm{dx}}{({x}^{2} + {b}^{2}) {({x}^{2} + {a}^{2})}^{2}} = \frac{\pi (2 a + b)}{2 {a}^{3} b {(a + b)}^{2}}$ I am in chapter 6 of Whittaker and Watson, and I am currently confused attempting to find the mistake I made in computing the following integral. The problem asks me to prove the following relationship
\begin{align}\int_{- \infty}^{\infty} \frac{\mathrm{dx}}{\left({x}^{2} + {b}^{2}\right) {\left({x}^{2} + {a}^{2}\right)}^{2}} = \frac{\pi \left(2 a + b\right)}{2 {a}^{3} b {\left(a + b\right)}^{2}}\end{align}
Attempted solution: I thought it best to try an evaluate this integral using Jordan's lemma. I already feel comfortable with my proof for demonstrating that the integral along a semicircle in the upper half plane will collapse to $0$ as its radius trends towards infinity. For simplicity, I am examining the case $a > 0$ and $b > 0$. (I believe that the case when one of them is zero could be solved by producing another semicircular domain around the real axis.)
I assert
\begin{align}
I & = \int_{- \infty}^{\infty} \frac{\mathrm{dx}}{\left({x}^{2} + {b}^{2}\right) {\left({x}^{2} + {a}^{2}\right)}^{2}} \\
& = \int_{- \infty}^{\infty} \frac{\mathrm{dx}}{\left(x - i b\right) \left(x + i b\right) {\left(x - i a\right)}^{2} {\left(x + i a\right)}^{2}} \\
& = \int_{C} \frac{\mathrm{dz}}{\left(z - i b\right) \left(z + i b\right) {\left(z - i a\right)}^{2} {\left(z + i a\right)}^{2}} \\
\end{align}
where $C$ is a semicircular contour in the upper half plane formed by the line from $- R$ to $R$ and a semicircle $\Gamma$.
For concision, I define
\begin{align}
f \left(z\right) = \frac{1}{\left(z - i b\right) \left(z + i b\right) {\left(z - i a\right)}^{2} {\left(z + i a\right)}^{2}} \\
\end{align}
such that
\begin{align}
\int_{C} f \left(z\right) \mathrm{dz} = 2 \pi i \left[{\text{Res}}_{z = i b} f \left(z\right) + {\text{Res}}_{z = i a} f \left(z\right)\right] \\
\end{align}
The point $i b$ is an order $1$ pole, so I use the formula
\begin{align}
{\text{Res}}_{z = i b} f \left(z\right) & = \lim_{z \to i b} \left(z - i b\right) f \left(z\right) \\
& = \lim_{z \to i b} \frac{1}{\left(z + i b\right) {\left({z}^{2} + {a}^{2}\right)}^{2}} \\
& = \frac{1}{2 i b \left({a}^{2} - {b}^{2}\right)} \\
\end{align}
Likewise, $i a$ is an order $2$ pole, so
\begin{align}
{\text{Res}}_{z = i a} f \left(z\right) & = \lim_{z \to i a} \frac{d}{\mathrm{dz}} \left({\left(z - i a\right)}^{2} f \left(z\right)\right) \\
& = \lim_{z \to i a} \frac{d}{\mathrm{dz}} \left(\frac{1}{\left({z}^{2} + {b}^{2}\right) {\left(z + i a\right)}^{2}}\right) \\
& = \lim_{z \to i a} \frac{- 2 z {\left(z + i a\right)}^{2} - 2 \left({z}^{2} + {b}^{2}\right) \left(z + i a\right)}{{\left({z}^{2} + {b}^{2}\right)}^{2} {\left(z + i a\right)}^{2}} \\
& = \frac{8 i {a}^{3} - 4 i a {b}^{2} + 4 i {a}^{3}}{\left({b}^{2} - {a}^{2}\right) \left(- 4 {a}^{2}\right)} \\
\end{align}
Then,
\begin{align}
I & = \frac{2 \pi i}{{\left({a}^{2} - {b}^{2}\right)}^{2}} \left(\frac{1}{2 i b} + \frac{8 {a}^{3} i - 4 i a {b}^{2} + 4 i {a}^{3}}{\left(- 4 {a}^{2}\right)}\right) \\
& = \pi \left(\frac{{a}^{2} + 4 {a}^{3} b - 2 a {b}^{3} + 2 b {a}^{3}}{{a}^{2} b {\left({a}^{2} - {b}^{2}\right)}^{2}}\right) \\
\end{align}
However, I believe that I have made a mistake as using the identity, $\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$ and multiplying by $\frac{a}{a}$, I find
\begin{align}
I = \pi \left(\frac{{a}^{4} + 4 {a}^{4} b - 2 {a}^{2} {b}^{3} + 2 b {a}^{4}}{{a}^{2} b {\left({a}^{2} - {b}^{2}\right)}^{2}}\right) \\
\end{align}
with the contradiction that
\begin{align}
\left(2 a + b\right) {\left(a - b\right)}^{2} & = \left(2 a + b\right) \left({a}^{2} + {b}^{2} - 2 a b\right) \\
& = 2 {a}^{3} + 2 a {b}^{2} - 4 {a}^{2} b + b {a}^{2} + {b}^{3} - 2 a {b}^{2} \\
& = 2 {a}^{3} - 3 {a}^{2} b + {b}^{3} \\
& \ne {a}^{4} + 4 {a}^{4} b - 2 {a}^{2} {b}^{3} + 2 b {a}^{4} \\
\end{align}
Question: Where is the mistake in my above reasoning? Is my computation of the residues incorrect? (I hope that it was not a simple algebraic oversight given how long it took to type my question.)
| Without residues but partial fractions, the integrand is
$$-\frac{i}{2 b \left(b^2-a^2\right)^2 (z-i b)}+\frac{i}{2 b \left(b^2-a^2\right)^2
(z+i b)}+$$ $$\frac{1}{4 a^2 \left(a^2-b^2\right) (z-i a)^2}+\frac{1}{4 a^2
\left(a^2-b^2\right) (z+i a)^2}+$$ $$\frac{i \left(3 a^2-b^2\right)}{4 a^3
\left(a^2-b^2\right)^2 (z-i a)}-\frac{i \left(3 a^2-b^2\right)}{4 a^3
\left(a^2-b^2\right)^2 (z+i a)}$$
Integrated between $-t$ and $+t$, it gives
$$\frac{a b t \left(b^2-a^2\right)+\left(a^2+t^2\right) \left(2 a^3 \tan
^{-1}\left(\frac{t}{b}\right)+\left(b^3-3 a^2 b\right) \tan
^{-1}\left(\frac{t}{a}\right)\right)}{a^3 b \left(a^2-b^2\right)^2
\left(a^2+t^2\right)}$$ Expanded as a series
$$I=\int_{- t}^{+t} \frac{\mathrm{dx}}{({x}^{2} + {b}^{2}) {({x}^{2} + {a}^{2})}^{2}} =\frac{\pi (2 a+b)}{2 a^3 b (a+b)^2}-\frac{2}{5
t^5}+O\left(\frac{1}{t^7}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Can we write square roots in a fraction separately? eg, $\sqrt{\frac{9-x^2}{x-2}}$ vs $\frac{\sqrt{9-x^2}}{\sqrt{x-2}}$ I was doing some questions related to functions when I came across this question where they gave us two functions as
$$\sqrt{\frac{9-x^2}{x-2}}\quad\text{and}\quad\frac{\sqrt{9-x^2}}{\sqrt{x-2}}$$
I can see that those two are a little different, but I have learned that we write $$\sqrt\frac{a}{b}=\frac{\sqrt a}{\sqrt b}$$
I am a little confused: is there a specific condition in which we can apply this?
| Hint: $\sqrt{\frac{-3}{-2}}$ is defined, while $\frac{\sqrt{-3}}{\sqrt{-2}}$ is not (in case we work only with real numbers).
In your case, for $\sqrt{\frac{9-x^2}{x-2}}$ to be defined, you need $\frac{9-x^2}{x-2} \geq 0 \Leftrightarrow x \in (-\infty, -3] \cup [2, 3]$ and for $\frac{\sqrt{9-x^2}}{\sqrt{x-2}}$ you need both $9 - x^2 \geq 0$ and $x-2\geq 0$, which means $x \in [2, 3]$.
So, for $x = -4$ you have $\sqrt{\frac{9-x^2}{x-2}} = \sqrt{\frac{-7}{-6}} = \sqrt{\frac{7}{6}}$ and $\frac{\sqrt{9-x^2}}{\sqrt{x-2}} = \frac{\sqrt{-7}}{\sqrt{-6}}$ which is undefined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $a_{n+1} = a_{n} + a_{n}^{\frac{1}{3}}$ satisfies $\lim_{n\to\infty}\frac{a_n}{pn^q}=1$ for some real valued $p, q$. Show that $a_{n+1} = a_{n} + a_{n}^{\frac{1}{3}}$ satisfies $\lim_{n\to\infty}\frac{a_n}{pn^q}=1$ for some real valued $p, q$.
I had hoped that I could use a strategy similar to Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$, wherein the author found that $\lim_{n\to\infty}\sqrt{a_{n+1}}-\sqrt{a_n}=\frac{1}{2}$ and used the telescoping sum $\sum_{k=1}^{n-1}\sqrt{a_{k+1}}-\sqrt{a_k}\approx\frac{1}{2}n$ to approximate $\sqrt{a_n}$. However, $a_{n+1}^{\frac{1}{3}}-a_{n}^{\frac{1}{3}}=\frac{a_{n+1}-a_{n}}{a_{n+1}^{\frac{2}{3}}+a_{n+1}^{\frac{1}{3}}a_{n}^{\frac{1}{3}}+a_{n}^{\frac{2}{3}}}=
\frac{a_{n}^\frac{1}{3}}{a_{n+1}^{\frac{2}{3}}+a_{n+1}^{\frac{1}{3}}a_{n}^{\frac{1}{3}}+a_{n}^{\frac{2}{3}}}=
\frac{1}{\frac{a_{n+1}^{\frac{2}{3}}}{a_n^{\frac{1}{3}}}+a_{n+1}^{\frac{1}{3}}+a_{n}^{\frac{1}{3}}}
$ which converges to $0$. I also tried $a_{n+1}^q-a_n^q$ for different positive and negative values of $q$ to no avail.
If anyone has any hints, I'd appreciate it. Thanks.
| If $a_0=0$ then the result is not true.
I'm assuming that $a_0>0$.
It's easy to check that $\{a_n\}$ is non-decreasing. So it either diverges to $+\infty$ or converges to a finite number $l>0$.
If the latter, then $l$ must verify $l=l+l^{\frac 1 3}$, which is a contradiction. Thus the sequence diverges to $+\infty$.
Let $\alpha$ be a positive number. Then
$$\begin{split}
a_{n+1}^\alpha &= \left(a_n + a_n^{\frac 1 3} \right)^\alpha\\
&= a_n^\alpha \left( 1+a_n^{-\frac 2 3}\right)^\alpha\\
&=a_n^\alpha\left( 1+\alpha a_n^{-\frac 2 3}+o\left ( a_n^{-\frac 2 3}\right)\right)\\
&=a_n^\alpha + \alpha a_n^{\alpha -\frac 2 3}+o\left ( a_n^{\alpha-\frac 2 3}\right)
\end{split}$$
Thus, selecting $\alpha=\frac 2 3$ yields
$$a_{n+1}^{\frac 2 3} = a_n^{\frac 2 3} + \frac 2 3 +o(1)$$
Summing this gives
$$a_n^{\frac 2 3}=\frac{2n}3+o(n)$$
Equivalently,
$$a_n=\left(\frac {2n}3\right)^{\frac 3 2}+o\left(n^{\frac 3 2}\right)$$
$$\boxed{\lim_{n\rightarrow+\infty} \frac{a_n}{\left(\frac {2n}3\right)^{\frac 3 2}}=1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Two types of days, expected number of days of one type Say I have two types of days, A and B. We have that:
*
*A is followed by A with probability $p$.
*A is followed by B with probability $1 - p$.
*B is followed by B with probability $q$.
*B is followed by A with probability $1 - q$.
Over a year with 365 days, what’s the expected number of A days?
Edit: I’m not sure how to even start this problem. It’s not even specified if the sequence begins with A or B!
This problem however reminds me of a similar problem that is simpler, which is to calculate the expected number of runs given however many coin flips. That’s going to be the total number of coin flips + 1, divided by 2. But I’m wondering how to generalize this to the problem at hand here.
| Here is the brute force method:
Notation: Let us give the state $A$ the value $1$ and the state $B$ the value $0$. Let us denote the state in time $n$ as $X_n$. $X_0$ will now denote the starting state. Now when we are in state $A$ at time $n$, we say $X_n = 1$ and when we are in state $B$ at that time we say $X_n = 0$. Now let us define $S_n := X_0 + \dots + X_{n-1}$. Now the advantage of giving $A$ the value $1$ becomes apparent: $S_n$ counts the number of times the state $A$ was visited up to time $n$ since $1$ is added for every time we are in state $A$ and nothing is added when we are in state $B$.
How does the process start?: Since we do not know how the system is started, the best thing we can do is assume it starts with a starting distribution vector
$$B = \begin{pmatrix} B_1, & B_2 \end{pmatrix},$$
where $B_1 = \mathbb{P}\left(X_0 = A \right) = \mathbb{P}\left(X_0 = 1 \right)$ and $B_2 = \mathbb{P}\left(X_0 = B \right) = \mathbb{P}\left(X_0 = 0\right)$ (I choose the letter $B$ for Begin), thus $B_1 + B_2 = 1$. Note that $B = \begin{pmatrix} 1 & 0 \end{pmatrix}$ corresponds to starting in $A$ with probability 1 and likewise $\begin{pmatrix} 0 & 1 \end{pmatrix}$ is starting in $B$ with probability 1, but for now we take general $B_1$ and $B_2$.
Probability to be in $A$ at time $n$: Now let us calculate the probability to be in state $A$ at time $n$. For that we first write down the transition matrix. For that, see the following figure.
So now we have the transition matrix
$$P = \begin{pmatrix} p & p-1 \\ 1-q & q \end{pmatrix}.$$
Now to determine the probability that we are in state $A$ at time $n$, we need the probability transition matrix for $n$ time steps, which is simply the $n$-th power of the probability transition matrix. I leave it up to the OP to verify (either via induction of diagonalization) that
$$P^n = \frac{1}{2 - p - q} \begin{pmatrix} 1 - q + (1- p)(p + q - 1)^n & 1 - p - (1- p)(p + q - 1)^n \\ 1 - q - (1- q)(p + q - 1)^n & 1 - p + (1- q)(p + q - 1)^n \end{pmatrix}.$$
Now the probability vector at time $n$ given that we start with the probability vector $B$ will be
\begin{align*}
BP^n &= \begin{pmatrix} B_1 & B_2 \end{pmatrix}
\frac{1}{2 - p - q}\begin{pmatrix} 1 - q + (1- p)(p + q - 1)^n & 1 - p - (1- p)(p + q - 1)^n \\ 1 - q - (1- q)(p + q - 1)^n & 1 - p + (1- q)(p + q - 1)^n \end{pmatrix}\\
&= \begin{pmatrix} B_1 & B_2 \end{pmatrix}
\frac{1}{2 - p - q}\left[ \begin{pmatrix} 1 - q & 1 - p \\ 1 - q & 1- p \end{pmatrix} + (p + q - 1)^n\begin{pmatrix} 1 - p & p -1 \\ q - 1 & 1 - q \end{pmatrix}\right]\\
&=\frac{1}{2 - p - q}\left[ \begin{pmatrix}B_1(1 - q) + B_2(1-p), & B_1(1 - q) + B_2(1-p)\end{pmatrix}\\ + (p + q -1)^n \begin{pmatrix}B_1(1-p) - B_2(1-p), & -B_1(1-q) + B_2(1-q)\end{pmatrix} \right],
\end{align*}
where the last expression is a sum of two row matrices again. Now since the first coördinate corresponds to state $A$, we find that
$$\mathbb{P}\left(X_n = A \right) = \frac{1}{2 - p - q}\left( B_1(1 - q) + B_2(1-p) + (p + q -1)^n\cdot (B_1(p-1) - B_2(1-p)) \right).$$
We will later denote the above probability again by $\mathbb{P}\left(X_n = 1\right)$.
Expected number of $A$ days:
Now we are ready to calculate the expected number of $A$ days up to day $n = 365$. That is, we determine $\mathbb{E}\left[S_n\right]$. Here you have to remember that we give $X_n$ value $1$ if it is in state $A$ and value $0$ if it is in state $B$. Using linearity of the expectation we can calculate
\begin{align*}
\mathbb{E}\left[S_n\right] &= \mathbb{E}\left[ X_0 + \dots + X_{n-1} \right]\\
&= \sum_{k = 0}^{n-1}\mathbb{E}\left[X_n\right]\\
&= \sum_{k = 0}^{n-1}\left( 1\cdot \mathbb{P}\left( X_n = 1\right) + 0\cdot \mathbb{P}\left( X_n = 0\right) \right) \\
&= \sum_{k = 0}^{n-1} \frac{1}{2 - p - q}\left[ B_1(1 - q) + B_2(1-p) + (p + q -1)^n\cdot (B_1(1-p) - B_2(1-p)) \right] \\
&= \frac{n(B_1(1-q) B_2(1-p))}{2 - p - q} + \frac{B_1(1-p) - B_2(1-p)}{2 - p - q}\sum_{k = 0}^{n-1}(p + q - 1)^k\\
&= \frac{n(B_1(1-q) B_2(1-p))}{2 - p - q} + \frac{B_1(1-p) - B_2(1-p)}{2 - p - q}\cdot \frac{1 - (p + q - 1)^n}{2 - p - q}.
\end{align*}
Now filling in $n = 365$ we get that the expected number of $A$ days in a year of 365 days will be:
$$\mathbb{E}\left[S_{365}\right] = \frac{365(B_1(1-q) B_2(1-p))}{2 - p - q} + \frac{B_1(1-p) - B_2(1-p)}{2 - p - q}\cdot \frac{1 - (p + q - 1)^{365}}{2 - p - q}.$$
If you see any typos / mistakes, please let me know.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the derivative of $y=\ln^3(3-x+x^2)$ Find the derivative of $$y=\ln^3(3-x+x^2)$$ We have to use the chain rule, but I am not sure how to think about it when we have more complex functions as in this problem. We can rewrite the function as $$y=\left[\ln(3-x+x^2)\right]^3,$$ so the outermost function is $g(x)=x^3$, right? And its derivative is $g'(x)=3x^2$. Then we have $t(x)=\ln(3-x+x^2)$ and $t'(x)=\dfrac{1}{3-x+x^2}(3-x+x^2)^{'}$ and finally $v(x)=(3-x+x^2)$ with $v'(x)=2x-1$. How do we use that in order to find $y'$? I would be grateful if you explained it in detail.
| Another possible approach:
\begin{align*}
y = \ln^{3}(3 - x + x^{2}) & \Longleftrightarrow y^{1/3} = \ln(3 - x + x^{2})\\\\
& \Longleftrightarrow \frac{y'}{3y^{2/3}} = \frac{2x - 1}{3 - x + x^{2}}\\\\
& \Longleftrightarrow y' = \frac{3(2x - 1)\ln^{2}(3 - x + x^{2})}{3 - x + x^{2}}
\end{align*}
Hopefully this helps!
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determinant of block matrices. $$
X = \begin{pmatrix}
1+b_1 & 1 & 0 & 0 & 0 & \frac{1}{a_{6}} \\
1+b_2 & 1 & 1 & 0 & 0 & -\frac{a_1}{a_6} \\
b_3 & 1 & 1 & 1 & 0 & -\frac{a_2}{a_6} \\
b_4 & 0 & 1 & 1 & 1 & -\frac{a_3}{a_6} \\
b_5 & 0 & 0 & 1 & 1 & 1-\frac{a_4}{a_6} \\
b_6 & 0 & 0 & 0 & 1 & 1-\frac{a_5}{a_6}
\end{pmatrix}$$
The Schur complement w.r.t. the first and last row/column gives
$$S = \begin{pmatrix}
1+b_1 & \frac{1}{a_6}\\
b_6 & 1 - \frac{a_5}{a_6}
\end{pmatrix}
-\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 &1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 0 &1 &1&1\\0&0&1&1 \end{pmatrix}^{-1}
\begin{pmatrix}1+b_2 & -\frac{a_1}{a_6} \\ b_3 & -\frac{a_2}{a_6}\\ b_4 & -\frac{a_3}{a_6}\\ b_5 & 1-\frac{a_4}{a_6}\end{pmatrix}.$$
$$S = \begin{pmatrix} b_1 - b_2 + b_4 - b_5 & - \frac{-1-a_1+a_3-a_4+a_6}{a_6}\\-1-b_2+b_3-b_5+b_6 & \frac{a_1-a_2 + a_4 - a_5}{a_6}\end{pmatrix}$$
Then $\det(X) = \det\Biggr(\begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 0 &1 &1&1\\0&0&1&1 \end{pmatrix}\Biggl). \det(S)$.
How matrix $S$ is obtained? I am not sure why and how to take the blocks here. How $S$ is derived? I can see that in matrix $M$ but how to approach it in matrix $X$?
Suppose $M = \begin{pmatrix} A & B \\ C & D\end{pmatrix}$. The Schur complement of $D
$ w.r.t $M$ is given by $M/D = A - B D^{-1} C$. It is easy to see when there are contiguous blocks. But when they are not contiguous, how we apply the formula? Like how we get matrix $S$. But I am not sure how to connect this with matrix $X$.
How matrix $X$ will be different from $Y$ below, by clubbing the blocks together the blocks used to construct matrix $S$.
$$
Y = \begin{pmatrix}
1+b_1 & \frac{1}{a_6} & 1 & 0 & 0 & 0 \\
b_6 & 1-\frac{a_5}{a_6} & 0 & 0 & 0 & 1 \\
1+b_2 & -\frac{a_1}{a_6} & 1 & 1 & 0 & 0 \\
b_3 & -\frac{a_2}{a_6} & 1 & 1 & 1 & 0 \\
b_4 & -\frac{a_3}{a_6} & 0 & 1 & 1 & 1\\
b_6 & 1-\frac{a_4}{a_6} & 0 & 0 & 1 & 1
\end{pmatrix}$$
| I think you have a partition as follows
$$ X=\left(\begin{array}{c|cccc|c}
1+b_{1} & 1 & 0 & 0 & 0 & \frac{1}{a_{6}}\\
\hline 1+b_{2} & 1 & 1 & 0 & 0 & -\frac{a_{1}}{a_{6}}\\
b_{3} & 1 & 1 & 1 & 0 & -\frac{a_{2}}{a_{6}}\\
b_{4} & 0 & 1 & 1 & 1 & -\frac{a_{3}}{a_{6}}\\
b_{5} & 0 & 0 & 1 & 1 & 1-\frac{a_{4}}{a_{6}}\\
\hline b_{6} & 0 & 0 & 0 & 1 & 1-\frac{a_{5}}{a_{6}}
\end{array}\right) $$
Consider these two projection matrics, $P$ a 6×2 matrix, and its orthogonal $Q$ a 6×4 matrix.
$$\begin{aligned}P & =\left(\begin{array}{cc}
1\\
\hline \\
\\
\\
\\
\hline & 1
\end{array}\right) & Q & =\left(\begin{array}{cccc}
\\
\hline 1\\
& 1\\
& & 1\\
& & & 1\\
\hline \\
\end{array}\right)\end{aligned}$$
then you can extract the submatrices as follows
$$\begin{array}{cc}
A=P^{\top}XP & B=P^{\top}XQ\\
C=Q^{\top}XP & D=Q^{\top}XQ
\end{array}$$
The result is
$$\begin{array}{cc}
A=\begin{bmatrix}1+b_{1} & \frac{1}{a_{6}}\\
b_{6} & 1-\frac{a_{5}}{a_{6}}
\end{bmatrix} & B=\begin{bmatrix}1 & 0 & 0 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}\\
C=\begin{bmatrix}1+b_{2} & -\frac{a_{1}}{a_{6}}\\
b_{3} & -\frac{a_{2}}{a_{6}}\\
b_{4} & -\frac{a_{3}}{a_{6}}\\
b_{5} & 1-\frac{a_{4}}{a_{6}}
\end{bmatrix} & D=\begin{bmatrix}1 & 1 & 0 & 0\\
1 & 1 & 1 & 0\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 1
\end{bmatrix}
\end{array}$$
and the composition of $X$ from the submatrices is
$$X=PAP^{\top}+PBQ^{\top}+QCP^{\top}+QDQ^{\top}$$
Considering the above decomposition, the Shur complement is
$$\begin{aligned}S & =A-B\,D^{-1}C\\
& =\left(P^{\top}XP\right)-\left(P^{\top}XQ\right)\left(Q^{\top}XQ\right)^{-1}\left(Q^{\top}XP\right)\\
& =P^{\top}\left(X-XQ\left(Q^{\top}XQ\right)^{-1}Q^{\top}X\right)P
\end{aligned}$$
Now for $Y$ the partition is contiguous I think
$$Y=\left(\begin{array}{cc|cccc}
1+b_{1} & \frac{1}{a_{6}} & 1 & 0 & 0 & 0\\
b_{6} & 1-\frac{a_{5}}{a_{6}} & 0 & 0 & 0 & 1\\
\hline 1+b_{2} & -\frac{a_{1}}{a_{6}} & 1 & 1 & 0 & 0\\
b_{3} & -\frac{a_{2}}{a_{6}} & 1 & 1 & 1 & 0\\
b_{4} & -\frac{a_{3}}{a_{6}} & 0 & 1 & 1 & 1\\
b_{6} & 1-\frac{a_{4}}{a_{6}} & 0 & 0 & 1 & 1
\end{array}\right)$$
which makes the extraction of the sub-matrices trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of $(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$ Find $a,b$ of:
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
I can't use L'hopital, I tried multiplying by the conjugate, and solving it,
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-(ax+b))\cdot \frac{\sqrt{4x^2+2x+1}+(ax+b)}{\sqrt{4x^2+2x+1}+(ax+b)}$$
$$=\lim_{x \to \infty}\frac{4x^2+2x+1-(ax+b)^2}{\sqrt{4x^2+2x+1}+(ax+b)} = \lim_{x \to \infty}\frac{x^2\left(4+\frac{2}{x}+\frac{1}{x^2}-a^2-\frac{2ab}{x}-\frac{b^2}{x^2}\right)}{x^2\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
Applying limit on the numerator and denominator
$$\frac{\lim_{x \to \infty}\left(4-a^2+\frac{2-2ab}{x}+\frac{1-b^2}{x^2}\right)}{\lim_{x \to \infty}\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
It can be seen that the denominator tends to $0$
| $$ L=\lim_{x\to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
Let $x=\frac{1}{t}$, then
$$L=\lim_{t\to 0}\left( \frac{2}{t} (1+\frac{t}{2}+{\frac{t^2}{4}})^{\frac{1}{2}}-\frac{a}{t}-b\right)$$
Use $(1+z)^k=1+kz+O(z^2)$. Then
$$L=\lim_{t\to 0}\frac{2(1+\frac{t}{4}+O(t^2))-a-bt}{t}=\lim_{t\to 0} \frac{(2-a)+(\frac{1}{2}-b)t+O(t^2)}{t}.$$ For $L$ to be finite, we demand $2-a=0$, then $L=(\frac{1}{2}-b)=\frac{-1}{2}.$ Thus we get $a=2$ and $b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
The equation with integer solution: $x^4+x-y^4+y=2x^2y^2(x^4-y^4+1)^3$ I am trying to solve this equation:
Given $x,y\in \text {Z}$: Solve : $x^4+x-y^4+y=2x^2y^2(x^4-y^4+1)^3$
Obviously, when $x=0$, it is easy to find that $y=0,y=1$ and $y=0$ then $x=0, x=-1$.
My question is how to find another solutions.
My try is factorization, but the power of $x,y$ is too big, i guess there is relation between $x,y$ so i tried to divide polynomial $x^4+x-y^4+y-2x^2y^2(x^4-y^4+1)^3$ by $xy$ or $xy-1$ (base on CAS gives solutions $x,y$ and $xy=0,xy=1$) but i failed.
I am bad at inequalities so i can't find another approach.
Every help is precious to me,thank you for reading.
| Write $a=x+y$, $b=x-y$, $c=x^2+y^2$ then the equation becomes
$$2a(bc+1)=(c-ab)(c+ab)(abc+1)^3.$$
Now if $|a|>1$ and $|bc|>1$ then by the fundamental theorem of arithmetic either $a$ or $bc+1$ must contain three copies of all the prime factors of $abc+1$. However $a$ is coprime with $abc+1$, so $bc+1$ must contain three copies of all the prime factors of $abc+1$, an impossibility. Hence there are no integer solutions other than the trivial cases.
★ Note that
$$
\begin{align}
c-ab&=x^2+y^2-(x^2-y^2)=2y^2\\
c+ab&=x^2+y^2+(x^2-y^2)=2x^2\\
abc&=x^4-y^4\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Geometric proof for derivative of $f(x)=\sqrt{x}$ Using a square of area $x$,
we obtain
$$dx = (2\sqrt{x})d\sqrt{x}+(d\sqrt{x})^2$$
$$\frac{d\sqrt{x}}{dx} = \frac{1}{2 \sqrt{x}} - \frac{(d\sqrt{x})^2}{2\sqrt{x}dx}$$
How do I show that $\frac{(d\sqrt{x})^2}{2\sqrt{x}dx} \rightarrow 0$ as $dx\rightarrow0$?
| \begin{eqnarray}
\Delta x &=& (2\sqrt{x})\Delta\sqrt{x}+(\Delta\sqrt{x})^2\\
\Delta x&=&(2\sqrt{x}+\Delta\sqrt{x})\Delta\sqrt{x}\\
(2\sqrt{x}+\Delta\sqrt{x})\Delta\sqrt{x}&=&\Delta x\\
\frac{\Delta\sqrt{x}}{\Delta x}&=&\frac{1}{2\sqrt{x}+\Delta\sqrt{x}}
\end{eqnarray}
As $\Delta x\to0$, then $\Delta\sqrt{x}\to0$.
So, as $\Delta x\to0$ we have
$$ \frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4524342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is this limit at infinity negative or positive 1
$\lim_{x \to -∞} \frac{x}{\sqrt{x^2+4}}$
I solved it:
$\lim_{x \to -∞} \frac{x}{\sqrt{x^2+4}} = \frac{x}{\sqrt{x^2}} = \frac{x}{x} = 1 , -1$ ?
Since its $x/x$, a negative infinity number divided by another negative infinity number is a positive infinity. But I am told that this limit is $-1$
| Note that
$$
\frac{x}{\sqrt{x^2+4}} = \frac{x}{\sqrt{x^2(1+4x^{-2}})} = \frac{x}{|x|}\frac{1}{\sqrt{1+4x^{-2}}}
$$
so
$$
\lim_{x\to -\infty}\frac{x}{\sqrt{x^2+4}} = \lim_{x\to -\infty}\frac{x}{|x|} \times\lim_{x\to -\infty}\frac{1}{\sqrt{1+4x^{-2}}} = -1\times 1 = -1
$$
since
$$
\frac{x}{|x|} = -1
$$
for $x<0$ and $(1+4x^{-2})^{-1/2}\to 1$ as $x\to -\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
prove $\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{n}-\frac{1}{n^2}$
Let $x_1,x_2,\cdots,x_n$ be $n$ non-negative numbers such that
$x_1+x_2+\cdots+x_n=1$. Show $$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j}\ge \frac{2}{n}-\frac{1}{n^2}.$$
We may consider using Cauchy. That is
$$\sum_{i=1}^n\sum_{j=1}^i x_j\cdot \sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j}\ge \left(\sum_{i=1}^nx_i\right)^2=1.$$
How to go on?
| Base Case:- $n=1$
$$\sum_{i=1}^1 \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{1}-\frac{1}{1^2}$$
$$1 \ge 1$$
Thus the base case holds.
Hypothesis:-
$$\sum_{i=1}^{n+1} \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{n+1}-\frac{1}{{(n+1)}^2}$$
And now for the step:-
$$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j} + \frac{x_{n+1}^2}{\sum_{j=1}^{n+1} x_j}\ge \frac{2}{n}-\frac{1}{n^2} + \frac{x_{n+1}^2}{\sum_{j=1}^{n+1} x_j}$$
$$\sum_{i=1}^{n+1} \frac{x_i^2}{\sum_{j=1}^i x_j} \ge \frac{2}{n}-\frac{1}{n^2} + \frac{x_{n+1}^2}{1}$$
This is as $x_1+x_2+\cdots+x_n+x_{n+1}=1$
As $n \ge 1$ so:-
$$2n^2\ge1$$
$$\Rightarrow 2n(n+1)\ge2n+1$$
$$\Rightarrow 2 \ge \frac{2n+1}{n(n+1)}$$
$$\Rightarrow 2\cdot\left[\frac{1}{n(n+1)}\right] \ge \frac{2n+1}{n^2(n+1)^2}$$
$$\Rightarrow 2\cdot\left[\frac{1}{n} - \frac{1}{n+1}\right]\ge \frac{1}{n^2}-\frac{1}{(n+1)^2}$$
$$\Rightarrow\frac{2}{n}-\frac{1}{n^2}\ge \frac{2}{n+1} -\frac{2}{(n+1)^2} $$
$$\Rightarrow\frac{2}{n}-\frac{1}{n^2} + x_{n+1}^2\ge \frac{2}{n+1} -\frac{1}{(n+1)^2} $$
This is as $x_{n+1}$ is non-negative.
Thus we finally get:-
$$\sum_{i=1}^{n+1} \frac{x_i^2}{\sum_{j=1}^i x_j} \ge \frac{2}{n}-\frac{1}{n^2} + x_{n+1}^2\ge \frac{2}{n+1} -\frac{1}{(n+1)^2} $$
completing the step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4537840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find all real values of x and y such that $\left(x+yi\right)^3$ is real and greater than $8$, represent these values in the xy-plane My solution
$\left(x+yi\right)^3>8$
$\left(x^3-3xy^2-8\right)+\left(3x^2y-y^3\right)i>0$
and now I'm stuck with these equations
$\left(x^3-3xy^2-8\right)>0,\left(3x^2y-y^3\right)>0$
| I don't know if my answer is correct or not but this is my idea:
$(x+yi)^3= r^3(\cos\phi+i\sin\phi)^3,$
where $r=\sqrt{x^2+y^2}$ and $\phi \in [-\pi,\pi)$.
By de Moivre's formula, we have
$(x+yi)^3= r^3(\cos\phi+i\sin\phi)^3 = r^3(\cos(3\phi)+i\sin(3\phi)).$
Since this is real, then $\sin(3\phi)=0 \Leftrightarrow \phi = \dfrac{k\pi}{3}$, where $k\in \lbrace -1, 0, 1\rbrace$.
Then $(x+yi)^3= r^3 = \left(\sqrt{x^2+y^2} \right)^3$.
So $ \left(\sqrt{x^2+y^2} \right)^3 >8 \Leftrightarrow x^2+y^2 > 4$. Then in the $xy$-plane, this is the domain outside a circle with $(0,0)$ as a center and radius $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Computing the exact value of the integral $∫_0^∞ \tanh(2x)\ln(\tanh x)dx.$ LATEST EDIT
Thanks to @FDP’s alternative method and @Claude Leibovici’s generalisation on the integral. Meanwhile, I had found a formula for the integral with power n in general.
$$\boxed{I_n=∫_0^∞ \tanh(2x)\ln^n(\tanh x)dx =\frac{(-1)^n n !}{2^n}\left(1-\frac{1}{2^{n+1}}\right)\zeta(n+1)} $$
and its proof is shown below.
Recently, I investigate the integral
$$∫_0^∞ \tanh(2x)\ln (\tanh x)dx,$$
using the substitution $y=\tanh x$.
$$
\begin{aligned}
I &=\int_0^{\infty} \frac{2 \tanh x}{1+\tanh ^2 x} \ln (\tanh x) d x \\
&=\int_0^{\infty} \frac{2 y \ln y}{1+y^2} \cdot \frac{d y}{2\left(1-y^2\right)} \\
&=\int_0^{\infty} \frac{y \ln y d y}{1-y^4} \\
&\stackrel{y^2\mapsto y}{=} \frac{1}{4} \int_0^{\infty} \frac{\ln y}{1-y^2} d y
\end{aligned}
$$
By my post,
$$
\begin{aligned}
&\int_0^{\infty} \frac{\ln y}{1-y^2} d y=-\frac{\pi^2}{4}
\end{aligned}
$$
We now conclude that
$$\boxed{∫_0^∞ \tanh(2x)\ln (\tanh x)dx= -\frac{\pi^2}{16}}$$
Is there any other method to evaluate the integral? Your comments and alternative methods are highly appreciated.
| \begin{align}J&=\int_0^\infty \tanh(2x)\ln (\tanh x)dx\\
&=\int_0^\infty \frac{\text{e}^{2x}-\text{e}^{-2x}}{\text{e}^{2x}+\text{e}^{-2x}}\ln\left(\frac{\text{e}^{x}-\text{e}^{-x}}{\text{e}^{x}+\text{e}^{-x}}\right)dx\\
&=\int_0^\infty \frac{1-\text{e}^{-4x}}{1+\text{e}^{-4x}}\ln\left(\frac{1-\text{e}^{-2x}}{1+\text{e}^{-2x}}\right)dx\\
&\overset{u=-\text{e}^{-2x}}=\frac{1}{2}\int_0^1 \frac{1-u^2}{(1+u^2)u}\ln\left(\frac{1-u}{1+u}\right)du\\
&\overset{z=\frac{1-u}{1+u}}=\int_0^1 \frac{2z\ln z}{1-z^4}dz\\
&\overset{t=z^2}=\frac{1}{2}\int_0^1 \frac{\ln t}{1-t^2}dt\\
&=\frac{1}{2}\int_0^1 \frac{\ln t}{1-t}dt-\frac{1}{2}\underbrace{\int_0^1 \frac{t\ln t}{1-t^2}dt}_{y=t^2}\\
&=\frac{3}{8}\int_0^1 \frac{\ln t}{1-t}dt\\
&=\frac{3}{8}\times -\frac{\pi^2}{6}=\boxed{-\frac{\pi^2}{16}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
complex numbers $z^n = 1$ complex plane I would like to know if I answered the question below correctly or if I need to change something.
Question: Take $n ∈ \mathbb N$. Find all solutions for $z^n=1$. Where can we find the solutions in the complex plane? Draw them in the complex plane for some values of $n$.**
$z^n = 1$ and $ n ∈ \mathbb N$. Assumption: $ 1 = e^{i0} = e^{i2π} = e^{i4π} = e^{i2(2π)} $.
$z^n = 1$ and $ n ∈ \mathbb N$.
= $ \cos (0) + i \sin (0) $, (as $r = 1$)
= $ \cos (0+2kπ) + i \sin (0+2kπ)$
= $ \cos (2kπ) + i \sin (2kπ) $, this gives us $ e^{i(2kπ)} $ with $k ∈ \mathbb Z$.
$z^n = 1$, so $z=\sqrt[n]{1 }$ and $z = 1^{1/n}$.
Using deMoivre's theorem, we get $ z = [e^{i(2kπ)}]^{1/n} $ =
$ z = e^{i2π(k/n)} , k = 0,1,2,...,n-1 $ with $ n ∈ \mathbb N$ .
Using deMoivre's theorem again, we get $ z = [\cos(2kπ) + i \sin (2kπ)]^{1/n} = \cos (2kπ/n) + i \sin (2kπ/n) $.
Now, we draw the solutions for some values of $n$ in the complex plane.
$ n = 1 \to z = 1 $ gives $ \cos (2kπ) + i \sin (2kπ) $, so $ z = 1 $.
$ n = 2 \to z = 1^{1/2} $ gives $ \cos (2kπ/2) + i \sin (2kπ/2) = \cos (kπ/2) + i \sin (kπ/2) $ , so $ z = 1, -1 $.
$ n = 3 \to z = 1^{1/3} $ gives $ \cos (2kπ/3) + i \sin (2kπ/3) $ , so $ z = -1/2 + 1/2√3 i, -1/2 - 1/2√3 i, 1 $ .
$ n = 4 \to z = 1^{1/4} $ gives $ \cos (2kπ/4) + i \sin (2kπ/4) = \cos (kπ/2) + i \sin (kπ/2) $, so $z = i, -1, -i, 1$.
Did I do it right? Should I change something in my notation?
| An easier way to solve would be:
$$
\begin{align*}
z^{n} &= 1 \Rightarrow z^{n} = |z^{n}|\\
z &= |z| \cdot \mathrm{e}^{\arg(z) \cdot \mathrm{i}}
(|z| \cdot \mathrm{e}^{\arg(z) \cdot \mathrm{i}})^{n} &= 1\\
|z|^{n} \cdot (\mathrm{e}^{\arg(z) \cdot \mathrm{i}})^{n} &= 1\\
|z|^{n} \cdot \mathrm{e}^{n \cdot \arg(z) \cdot \mathrm{i}} &= 1\\
\mathrm{e}^{n \cdot \arg(z) \cdot \mathrm{i}} &= 1 \quad\mid\quad (\text{
})^{\frac{1}{2}}\\
\mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= 1^{\frac{1}{2}}\\
\mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= \pm \sqrt{1}\\
\mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= \pm 1\\
&\Rightarrow \mathrm{e}^{\arg(z) \cdot \mathrm{i}} = \Re(\mathrm{e}^{\arg(z) \cdot \mathrm{i}})\\
&=\Rightarrow \arg{z} = 2 \cdot \frac{k_{k \in \mathbb{Z}} + n}{n} \cdot \pi
\end{align*}
$$
That means that there are n solutions with $$ k = 1, ~2, ~..., ~n - 1, $$
wich means that there are mure solutios for z then $$ z = \left\{1, ~i, ~-1, ~-i\right\}, $$ aka you should change this.
e.g.
$$
\begin{align*}
z &= \frac{\mathrm{1}}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}}\\
z^{4} &= (\frac{\mathrm{1}}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}})^{4}\\
z^{4} &= (1^{\frac{1}{4}})^{4}\\
z^{4} &= 1
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \sqrt{x^2 + a^2}\mathrm{d}x$ I tried using integration by parts and could get to this point:
\begin{align}
\int \sqrt{x^2 + a^2}\mathrm{d}x &= x\sqrt{x^2+a^2} - \int x\frac{2x}{\sqrt{x^2 + a^2}}\mathrm{d}x
\\\\ &= x\sqrt{x^2+a^2} -2\int \sqrt{x^2 + a^2}\mathrm{d}x + \int \frac{2a^2}{\sqrt{x^2 + a^2}}\mathrm{d}x
\end{align}
then, I don't know how to get:
$$\int \frac{2a^2}{\sqrt{x^2 + a^2}}\mathrm{d}x$$
| You have a little mistake in the integration by parts, $\displaystyle(\sqrt{x^2 + a^2})' = \frac{x}{x^2 + a^2}$ therefore
$$
\int \sqrt{x^2 + a^2}\,dx = x\sqrt{x^2 + a^2} - \int \sqrt{x^2 + a^2}\,dx + a^2\int\frac{1}{\sqrt{x^2 + a^2}}.
$$
The latter integral is simply a can be calculated using the hyperbolic trig substitution or you can recall that $\displaystyle\left(\sinh^{-1}(x)\right)' = \frac{1}{x^2 + 1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculate the limit $\lim_{n\to \infty }\int_{1}^{n}(n^2+x^3)^{-1/2} dx$ \begin{align}
\lim_{n\to \infty }\int_{1}^{n}\frac{dx}{\sqrt{n^2+x^3}}& \overset{x=nt}{=}\int_{1/n}^{1}\frac{dt}{\sqrt{1+nt^3}} \\
& \overset{t^3=p}{=}\int_{1/n^3}^{1}\frac{1/3p^{-2/3}}{\sqrt{1+np}}dp\\
& \overset{pn=q}{=}\frac{n^{-1/3}}{3}\int_{1/n^2}^{n}\frac{t^{-2/3}}{\sqrt{t+1}}dt \\
& \overset{t=\mathrm{tg}^2\theta }{=} 2\int_{0}^{\pi/2}\sin^{-1/3}\theta\cos^{-2/3}\theta d\theta\\
& =\mathrm{B}\left ( \frac{1}{3},\frac{1}{6} \right ) \\
& =\frac{\Gamma \left ( \frac{1}{3} \right )\Gamma \left ( \frac{1}{6} \right )}{\sqrt{\pi}}
\end{align}
Question: Have I calculated the limit correctly?
Why am I asking? After replacing $x=nt$, the limit and the integral can be interchanged and the limit will tend towards zero and the answer will be $0$.
Am I solving it correctly?
| Your mistake is saying
$$\frac{n^{-\frac{1}{3}}}{3}\int_{\frac{1}{n^{2}}}^{n}\frac{t^{-\frac{2}{3}}}{\sqrt{t+1}}dt = 2\int_{0}^{\frac{\pi}{2}}\sin\left(\theta\right)^{-\frac{1}{3}}\cos\left(\theta\right)^{-\frac{2}{3}}d\theta.$$
I'm not too sure how you tried using the substitution $t=tg^2\theta$ since that second integral does not depend on $n$. In fact, I'm not sure what $g$ is supposed to be.
The previous steps are correct, though, assuming you meant to keep writing $\displaystyle \lim_{n\to\infty}$ for each step.
(Answer) Going off $\frac{n^{-1/3}}{3}\int_{\frac{1}{n^{2}}}^{n}\frac{t^{-\frac{2}{3}}}{\sqrt{t+1}}dt$, an antiderivative of the integrand is $$\frac{n^{-1/3}}{3} \cdot 3t^{1/3} \cdot \left(_2F_1 \left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-t\right)\right),$$ where we use the hypergeometric function. Using the Fundamental Theorem of Calculus and doing some simplifying, we get the integral to equal $$\frac{n^{-1/3}}{3}\left(\frac{3\left(-_{2}F_1 \left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{n^2}\right)+n\left(_2F_1 \left(\frac{1}{3}, \frac{1}{2};\frac{4}{3};-n\right)\right)\right)}{n^{2/3}}\right),$$ where we assumed $n>0$. Applying $n\to\infty$, we get
$$
\eqalign{
&\displaystyle \lim_{n\to\infty} \frac{n^{-1/3}}{3}\left(\frac{3\left(-_{2}F_1 \left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{n^2}\right)+n\left(_2F_1 \left(\frac{1}{3}, \frac{1}{2};\frac{4}{3};-n\right)\right)\right)}{n^{2/3}}\right) \cr
=& \left(\lim_{n\to\infty} \frac{n^{-1/3}}{3}\right)\left(\lim_{n\to\infty}\frac{3\left(-_{2}F_1 \left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{n^2}\right)+n\left(_2F_1 \left(\frac{1}{3}, \frac{1}{2};\frac{4}{3};-n\right)\right)\right)}{n^{2/3}}\right) \cr
=& 0 \cdot \frac{\Gamma{\left(1/6\right)\Gamma{\left(1/3\right)}}}{\sqrt{\pi}} \cr
=& 0. \cr
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve initial value problem using Laplace Transform Use Laplace transform to solve the following initial–value problems.
I'm confused with $f(t)$ defined by cases and both $x,y$ being differentiated as shown below.
*
*$(\frac{d^2}{dt^2}-4\frac{d}{dt}+4)y=f(t)$, subject to $y(0)=-2,y'(0)=1$, with
$$f(t)=\begin{cases}
t & \text{if }0\leq t\leq 3,\\
t+2 & \text{if }t\geq 3.
\end{cases}$$
If we solve the two cases separately, we will get
$$y=\frac{1}{4}+\frac{t}{4}-\frac{9}{4}e^{2t}+\frac{21e^{2t}t}{4}$$
and
$$y=\frac{3}{4}+\frac{t}{4}-\frac{11}{4}e^{2t}+\frac{25e^{2t}t}{4}.$$
But, how can we proceed?
*$\frac{d^2x}{dt^2} = -2x+y,\frac{d^2y}{dt^2} = x − 2y$, subject to the initial conditions
$$x(0) = y(0) = 1,\quad x'(0) = y'(0) = 0.$$
| The Laplace transform of $f(t)$ is given by
\begin{align}
f(t) &\doteqdot \int_{0}^{\infty} e^{-s t} \, f(t) \, dt \\
&\doteqdot \int_{0}^{3} e^{- s t} \, t \, dt + \int_{3}^{\infty} e^{-s t} \, (t + 2) \, dt \\
&\doteqdot \frac{1}{s^2} + \frac{2}{s} \, e^{- 3 s}.
\end{align}
Now the differential equation, when transformed, takes the form:
\begin{align}
y^{''} - 4 \, y^{'} + 4 \, y &= f(t) \\
(s-2)^2 \, \overline{y} &= 4 \, y(0) + y^{'}(0) - y(0) \, s + \frac{1}{s^2} + \frac{2}{s} \, e^{- 3 s} \\
\overline{y} &= \frac{4 \, y(0) + y^{'}(0)}{(s-2)^2} - \frac{y(0) \, s}{(s-2)^2} + \frac{1}{s^2 \, (s-2)^2} + \frac{2 \, e^{-3 s}}{s \, (s-2)^2}
\end{align}
which, when inverted, leads to
$$ y(t) = \frac{1}{4} \, \left[ 1 + t + ((1 + 8 \, y(0) + 4 \, y^{'}(0)) \, t - (1 + 4 \, y(0)) ) \, e^{2 t} \right] + \frac{1}{2} \, \left( 1 + (2 t -7) \, e^{2(t -3)} \right) \, H(t-3), $$
where $H(t)$ is the Heaviside step function. Applying the conditions gives
$$ y(t) = \frac{1}{4} \, \left[ 1 + t + (7 - 11 \, t) \, e^{2 t} \right] + \frac{1}{2} \, \left( 1 + (2 t -7) \, e^{2(t -3)} \right) \, H(t-3) $$
which is the same as:
$$ y(t) = \begin{cases} \frac{1}{4} \, \left[ 1 + t + (7 - 11 \, t) \, e^{2 t} \right] & t \leq 3 \\ \frac{1}{4} \, \left[ 1 + t + (7 - 11 \, t) \, e^{2 t} - 2 \, (7 - 2 \, t) \, e^{2 (t-3)} \right] & t > 3 \end{cases}. $$
| {
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"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$ by partial fractions
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$
I know that we can use substitution to make the expression simplier before solving it, but I am trying to solve this by using partial fractions only.
$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2} = \frac{A(x+2)^2+B(x-3)(x+2) + C(x-3)}{(x-3)(x+2)} = \frac{Ax^2 + 4Ax + 4A + Bx^2 - Bx - 6B +Cx - 3C}{(x-3)(x+2)}$
Looking at the numerator: $x^2(A+B) + x(4A-B+C) + (4A-6B-3C)$
So, comparing coefficients:
$A+B=1$,
$4A-B+C=0$
$4A-6B-3C=0$
I am struggling to solve these 3 equations to find A,B,C
| From
$$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$$
we evaluate the coefficients directly as follows
\begin{align}
&A=\left[\frac{x^2}{(x+2)^2} - \frac{B(x-3)}{x+2} + \frac{C(x-3)}{(x+2)^2} \right]_{x=3}=\frac9{25}\\
&B=\left[\frac{x^2}{(x-3)(x+2)} - \frac{A(x+2)}{x-3} -{\frac C{x+2}}\right]_{x\to\infty}=1-A=\frac{16}{25}\\
&C=\left[\frac{x^2}{x-3} - \frac{A(x+2)^2}{x-3} -{B(x+2)}\right]_{x=-2}=-\frac4{5}
\end{align}
| {
"language": "en",
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"answer_count": 5,
"answer_id": 3
} |
How is ${n \choose n/2}$ approximated in this manner? In this question, the author asked how to get an asymptotic growth of ${n \choose n/2}$, to which the answer is as follows. First:
$$ {n \choose n/2} = \frac{n!}{(n/2)!(n- n/2)!} = \frac{n!}{(\tfrac{n}{2}!)^2} $$
Now, to find an upper bound on this, take the upper bound of Stirling's approximation for $n!$ and the lower bound for $\tfrac{n}{2}!$ which if I'm not mistaking gives the following:
$$ {n \choose n/2} \leq \frac{\sqrt{2 \pi n} (\frac{n}{e})^n e^{\tfrac{1}{12n}}}{\Big( \sqrt{2 \pi \tfrac{n}{2}} (\frac{\tfrac{n}{2}}{e})^\tfrac{n}{2} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} $$
However, the answer states this should give:
$$ {n \choose n/2} \leq \frac{e \cdot n^{n + \tfrac{1}{2}} e^{-n}}{\Big( \sqrt{2 \pi} \cdot \tfrac{n}{2}^\tfrac{n+1}{2} e^{-\tfrac{n}{2}}\Big )^2} $$
I'm confused as to how one would get from the former to the latter. Up to now I get stuck around here:
$$ {n \choose n/2} \leq \frac{\sqrt{2 \pi n} (\frac{n}{e})^n e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi \tfrac{n}{2}} (\frac{\tfrac{n}{2}}{e})^\tfrac{n}{2} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} \\
\leq \frac{\sqrt{2 \pi n} \cdot n^n e^{-n} e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi \tfrac{n}{2}} \cdot \tfrac{n}{2}^\tfrac{n}{2} e^{-\tfrac{n}{2}} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} \\
\leq \frac{\sqrt{2 \pi} \cdot n^{n + \tfrac{1}{2}} e^{-n} e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi} \cdot \tfrac{n}{2}^\tfrac{n+1}{2} e^{-\tfrac{n}{2}} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2}
$$
Now, do the latter terms just disappear due to $\lim_{n \rightarrow \infty} \frac{e^{\tfrac{1}{12n}}}{\Big ( e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} = 0$? But in this case, why should there still be an $e$ term in the numerator? Also what happened to the $\sqrt{2 \pi}$ in the numerator? It cannot have canceled out with the one in the denominator since the latter is still there...
| Using,as you did,
$$\sqrt{2 \pi n}\ \left(\frac{n}{e}\right)^n e^{\frac{1}{12n + 1}} < n! < \sqrt{2 \pi n}\ \left(\frac{n}{e}\right)^n e^{\frac{1}{12n}}$$
$$\frac{2^{n+\frac{1}{2}} e^{\frac{1}{12 n+1}-\frac{1}{3 n}}}{\sqrt{\pi n}
}\lt \binom{n}{\frac{n}{2}} \lt \frac{2^{n+\frac{1}{2}} e^{\frac{1}{12 n}-\frac{2}{6 n+1}}}{\sqrt{\pi n} }$$
You could use
$$e^{\frac{1}{12 n+1}-\frac{1}{3 n}}=1-\frac{1}{4 n}+\frac{7}{288 n^2}-\frac{1}{3456 n^3}+O\left(\frac{1}{n^4}\right)$$
$$e^{\frac{1}{12 n}-\frac{2}{6 n+1}}=1-\frac{1}{4 n}+\frac{25}{288 n^2}-\frac{89}{3456
n^3}+O\left(\frac{1}{n^4}\right)$$ which give very tight bounds.
For $n=10$, this gives
$$ 251.972 < 252 < 252.127 $$
Edit
I use the opportunity of the post to put an unpublised result from one of my past students, the late Dr. Bois,
$$\color{blue}{\binom{n}{\frac{n}{2}} > \frac{2^{n+\frac{1}{2}}\, e^{-\frac 1 {4n}}}{\sqrt{\pi n} }}$$ which is, for any $n$, better than the above left bound. For $n=10$, it gives $251.990$.
| {
"language": "en",
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} |
Proving $\frac1{2^1}+\frac2{2^2}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$ by induction I need to prove that
$$\frac1{2^1}+\frac2{2^2}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$$ by induction.
The base case was fine, and after that we have the induction hypothesis:
$$\frac1{2^1}+\frac2{2^2}+\cdots+\frac{k}{2^k}=2-\frac{k+2}{2^k} \tag1$$
holds for any given $n=k$.
The induction step is where I get stuck. We have
$$\frac1{2^1}+\frac2{2^2}+\cdots+\frac{k+1}{2^{k+1}} \tag2$$ and by the induction hypothesis we get
$$\begin{align}
2-\frac{k+2}{2^{k+1}} + \frac{k + 1}{2^{k+1}} &=2-\frac{2(k+2)}{2^{k+1}} +\frac{k+1}{2^{k+1}} \tag3 \\[6pt]
&= 2- \frac{3k+5}{2^{k+1}} \tag4
\end{align}$$ I certainly have made a mistake somewhere, but I can’t for the life of me figure out where.
| You forgot to take into account the fact that the
$\dfrac{2(k + 2)}{2^{k + 1}}$
term had a negative coefficient.
Observe that
\begin{align*}
2 - \frac{2(k + 2)}{2^{k + 1}} + \frac{k + 1}{2^{k + 1}} & = 2 - \left[\frac{2(k + 2)}{2^{k + 1}} - \frac{k + 1}{2^{k + 1}}\right]\\
& = 2 - \left[\frac{2k + 4}{2^{k + 1}} + \frac{k + 1}{2^{k + 1}}\right]\\
& = 2 - \frac{k + 3}{2^{k + 1}}\\
& = 2 - \frac{(k + 1) + 2}{2^{k + 1}}
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/4559625",
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"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\frac{1-4\cos80^\circ}{\tan20^\circ}$ I am working on simplifying the value
$$\frac{1-4\cos80^\circ}{\tan20^\circ}$$
I am looking for it to be expressed in terms of tangent, because the value seems to be $\tan40^\circ$. However, I can't find any obvious way to do so. How would simplify it to express like such?
| Let $\theta = \pi/9 = 20^\circ$. We wish to show that $$1 - 4 \cos 4\theta = \tan \theta \tan 2\theta. \tag{1}$$ This suggests the tangent angle addition identity $$\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \tag{2}$$ for the choice $$\alpha = \theta, \beta = 2\theta$$ which yields
$$(\tan 3\theta)(1 - \tan \theta \tan 2\theta) = \tan \theta + \tan 2\theta. \tag{3}$$ Since $3\theta = \pi/3 = 60^\circ$ hence $\tan 3\theta = \sqrt{3}$, we find
$$\begin{align}
1 - \tan \theta \tan 2\theta
&= \frac{1}{\sqrt{3}}(\tan \theta + \tan 2\theta) \\
&= \frac{\sin \theta \cos 2\theta + \sin 2\theta \cos \theta}{\sqrt{3} \cos \theta \cos 2\theta} \\
&= \frac{\sin 3\theta}{\sqrt{3} \cos \theta \cos 2\theta} \\
&= \frac{1}{2\cos \theta \cos 2\theta}. \tag{4}
\end{align}$$
We now claim that $$8 \cos \theta \cos 2\theta \cos 4\theta = 1. \tag{5}$$ To this end, write
$$\begin{align}
8 \sin \theta \cos \theta \cos 2\theta \cos 4 \theta
&= 4 \sin 2\theta \cos 2\theta \cos 4\theta \\
&= 2 \sin 4\theta \cos 4\theta \\
&= \sin 8\theta \\
&= \sin(\pi - \theta) \\
&= \sin \theta,
\end{align}$$
and now dividing both sides by $\sin \theta$ yields the proof of $(5)$. Therefore, $(4)$ is equivalent to
$$1 - \tan \theta \tan 2\theta = 4 \cos 4\theta,$$ which in turn is equivalent to $(1)$.
Something unsettled me about my solution above; I had the feeling it was unnecessarily complicated. After some thought I noticed that it could be simplified as follows: to show $(1)$, consider
$$\begin{align}
1 - \tan \theta \tan 2\theta
&= \frac{\cos \theta \cos 2\theta - \sin \theta \sin 2\theta}{\cos \theta \cos 2\theta} \\
&= \frac{\cos 3\theta}{\cos \theta \cos 2\theta} \\
&= \frac{1}{2 \cos \theta \cos 2\theta},
\end{align}$$
which takes us directly to $(4)$ without the tangent addition identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Prove that $(\sin x + \cos x)(6 - \sin x)<9$ Is there any elementary way to prove that $(\sin x + \cos x)(6 - \sin x)<9$?
I've noticed that $(\sin x + \cos x)$ has to be positive so $x \in\left(-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right)$ and then $(6 - \sin x)\in\left(6-\dfrac{\sqrt2}{2},6+ \dfrac{\sqrt2}{2}\right)$ but since $(\sin x + \cos x) \in\left(0, \sqrt2\right]$ the maximum possible value of $(\sin x + \cos x)(6 - \sin x)$ can be $(6+ \dfrac{\sqrt2}{2}) * \sqrt2 = 1+6\sqrt2$ which is greater than $9$.
My second attempt was to find the derivative, but I couldn't find its roots.
| Using the product-to-sum identity $\sin\theta\sin\phi = \frac{1}{2}(\cos(\theta - \phi) - \cos(\theta + \phi))$ to simplify:
\begin{align*}
& (\sin x + \cos x)(6 - \sin x) \\
= & \sqrt{2}\sin(x + \pi/4)(6 - \sin x) \\
= & 6\sqrt{2}\sin(x + \pi/4) - \sqrt{2}\sin x\sin(x + \pi/4) \\
= & 6\sqrt{2}\sin(x + \pi/4) - \sqrt2\times \frac{\cos(\pi/4) - \cos(2x+\pi/4)}{2} \\
= & 6\sqrt{2}\sin(x + \pi/4) + \frac{\sqrt{2}}{2}\cos(2x + \pi/4) - \frac{1}{2} \\
\leq & 6\sqrt{2} + \frac{\sqrt{2}}{2} - \frac{1}{2} \\
= & 8.69\ldots < 9.
\end{align*}
| {
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"answer_count": 3,
"answer_id": 0
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Using complex analysis, show that $\int_0^\infty\frac{\arctan(x)}{1+x^2}\,dx=\frac{\pi^2}8$ The result is of course confirmed by substituting $x\mapsto\tan(x)$:
$$I = \int_0^\infty \frac{\arctan(x)}{1+x^2} \, dx = \int_0^{\frac\pi2} x \, dx = \frac{\pi^2}8$$
Or integrating by parts:
$$I = \lim_{x\to\infty} \arctan^2(x) - I \implies 2I = \left(\frac\pi2\right)^2 \implies I = \frac{\pi^2}8$$
Or splitting the integral at $x=1$ and substituting $x\mapsto\frac1x$ on the integral over $[1,\infty)$:
$$I= \int_0^1 \frac{\arctan(x) + \arctan\left(\frac1x\right)}{1 + x^2} \, dx = \frac\pi2 \int_0^1 \frac{dx}{1+x^2} = \frac{\pi^2}8$$
Or differentiating under the integral sign:
$$I(a) = \int_0^\infty \frac{\arctan(ax)}{1+x^2} \, dx \implies I'(a) = \int_0^\infty \frac x{(1+x^2)(1+a^2x^2)} \, dx = \frac{\ln(a)}{a^2-1} \\ I(0) = 0 \implies I(1) = \int_0^1 \frac{\ln(x)}{x^2-1} \, dx = \frac{\pi^2}8$$
Or getting the same integral of $\frac{\ln(x)}{x^2-1}$ by converting $\arctan(x)$ to an integral representation and computing the resulting double integral (per @Dr.WolfgangHintze's suggestion) using the same substitution as in the third method above:
$$\begin{align*}
I &= \int_0^\infty \int_0^x \frac x{(1+x^2)(1+x^2y^2)} \, dy \, dx \\[1ex]
&= \int_0^\infty \int_y^\infty \frac x{(1+x^2)(1+x^2y^2)} \, dx \, dy \\[1ex]
&= \frac12 \int_0^\infty \frac{\ln\left(\frac{y^2+y^4}{1+y^4}\right)}{y^2-1} \, dy \\[1ex]
&= \int_0^\infty \frac{\ln(y)}{y^2-1} \, dy + \frac12 \int_0^\infty \frac{\ln(1 + y^2)}{y^2-1} \, dy - \frac12 \int_0^\infty \frac{\ln(1+y^4)}{y^2-1} \, dy \\[1ex]
&= (1 + 2 - 2) \int_0^1 \frac{\ln(y)}{y^2-1} \, dy = \frac{\pi^2}8
\end{align*}$$
I was wondering how, if at all possible, one might approach it with the residue theorem? I see that $z=\pm i$ are simple poles of $\frac1{1+z^2}$, but they're also the branch points of $\arctan(z)$, since
$$\arctan(z) = -\frac i2 \log\left(\frac{i-z}{i+z}\right)$$
so I don't believe the theorem can be readily applied here. The integrand is odd so I don't think there's much to infer from symmetry. Maybe there's a way to massage the integrand to get closer to something with which we can use a contour integral.
| Complex integration can also be used to evaluate the integral; often, this is a very power tool for finding general solutions - if the system has an appropriate symmetry.
We will consider a general case:
$$I(a,b)=\int_0^\infty\frac{x^b\arctan x}{a^2+x^2}dx;\,b\in[0;1);\,a>0\tag{0}$$
We will also choose $a<1$ - it allows us to separate poles from branch points; as the integrand is a continuous function of $a$, our solution will be valid for all $a>0$ (after an appropriate rearrangement if $a>1$).
We define $\arctan z$ in a standart way: $\displaystyle\arctan z=-\frac{i}{2}\ln\frac{i-z}{i+z}$, and the regular cuts are $[-i;-i\infty);\,[i;i\infty)$.
The modified keyhole contour looks
To define the integrand on the banks of the cut (upper half-plane), we choose $z=r<1$ on the axis $X$, then make a turn counter-clockwise around $\displaystyle z=0$: $\displaystyle\,r\to re ^\frac{\pi i}{2}; \ln z\to-\frac{i}{2}\ln\frac{i-ir}{i+ir}=-\frac{i}{2}\ln\frac{1-r}{1+r}$. To move on the bank of the cut (area A), we make another turn (angle $\pi$ around $z=i$; counter-clockwise): $\displaystyle 1-r\to (1-r)e ^{\pi i};\,-\frac{i}{2}\ln\frac{1-r}{1+r}\to -\frac{i}{2}\ln\Big(\frac{1-r}{1+r}e ^{\pi i}\Big)=-\frac{i}{2}\ln\frac{1-r}{1+r}+\frac{\pi}{2}$
If we move to the opposite bank (region C), we will get $-\frac{i}{2}\ln\frac{1-r}{1+r}-\frac{\pi}{2}$. On the both banks of the cut $z= re ^\frac{\pi i}{2}$ and $z^b= re^\frac{\pi ib}{2}$.
Integral along our close contour
$$\oint=I(a,b)(1-e^{2\pi ib})+I_A+I_C+I_D+I_E+I_R+I_{r_k}$$
It is not difficult to show that the integrals along a big circle and small circles (around $z=0,\pm i$) $I_R$ and $I_{r_k}$ tend to zero at $R\to\infty$ and $r_k\to0$. Making the similar analysis for the bottom part of the contour (as was done above for the regions A, C) and bearing in mind that in this case $z= re ^\frac{3\pi i}{2}$ (our contour goes counter-clockwise)
$$I(a,b)(1-e^{2\pi ib})+I_A+I_C+I_D+I_E=2\pi i\underset {z=ae ^\frac{\pi i}{2}, \,ae ^\frac{3\pi i}{2}}{\operatorname{Res}}-\frac{i}{2}\frac{z^b}{z^2+a^2}\ln\frac{i-z}{i+z}\tag{1}$$
To evaluate $\displaystyle I_{A, C, D, T}$ we notice that the terms with $\ln\frac{i-z}{i+z}$ cancel (we integrate in the opposite directions), and we are left, for example
$$I_A+I_C=-\frac{\pi}{2}e ^\frac{\pi i}{2}e ^\frac{\pi ib}{2}\int_1^\infty\frac{x^b}{a^2-x^2}dx+\frac{\pi}{2}e ^\frac{\pi i}{2}e ^\frac{\pi ib}{2}\int_\infty^1\frac{x^b}{a^2-x^2}dx$$
$$=-\pi ie ^\frac{\pi ib}{2}\int_1^\infty\frac{x^b}{a^2-x^2}dx$$
In the similar way,
$$I_D+I_E=-\pi ie ^\frac{3\pi ib}{2}\int_1^\infty\frac{x^b}{a^2-x^2}dx$$
The residues evaluation gives
$$2\pi i\sum\operatorname {Res}=-\frac{\pi i}{2}a^{b-1}\ln\frac{1-a}{1+a}\Big(e ^\frac{\pi ib}{2}+e ^\frac{3\pi ib}{2}\Big)$$
Putting all in (1)
$$I(a,b)=\frac{\pi}{2\sin\frac{\pi b}{2}}\int_1^\infty\frac{x^b}{x^2-a^2}dx-\frac{\pi}{4\sin\frac{\pi b}{2}}a^{b-1}\ln\frac{1+a}{1-a}$$
Integrating the first term by part
$$\boxed{\,\,I(a,b)=\frac{\pi}{4a\sin\frac{\pi b}{2}}\bigg((1-a^b)\ln\frac{1+a}{1-a}+b\int_1^\infty x^{b-1}\ln\frac{x+a}{x-a}dx\bigg);\, a\in(0;1]\,\,}\tag{2}$$
Now we can consider different cases.
*
*Leading $\displaystyle b\to 0$
$$I(a,0)=\int_0^\infty\frac{\arctan x}{a^2+x^2}dx=\int_1^\infty\frac{\ln x}{x^2-a^2}dx-\frac{\ln a}{2a}\ln\frac{1+a}{1-a}\tag{a}$$
*Putting in turn $\displaystyle a=1$
$$I(1,0)=\int_0^\infty\frac{\arctan x}{1+x^2}dx=\int_1^\infty\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}\tag{b}$$
*Taking $a\to1$ (the first term in (2) is zero), integrating the second term by part, using expressions for gamma (duplication formula) and digamma functions, we can get an interesting result
$$I(1,b)=\int_0^\infty\frac{x^b\arctan x}{1+x^2}dx=\frac{\pi}{4\sin\frac{\pi b}{2}}\bigg(\psi\Big(\frac{1}{2}\Big)-\psi\Big(\frac{1-b}{2}\Big)\bigg);\,b\in[0;1)\tag{c}$$
When $b\to 0$, we get again
$$I(1;0)=\int_0^\infty\frac{\arctan x}{1+x^2}=\frac{1}{4}\psi'\Big(\frac{1}{2}\Big)=\sum_{k=1}^\infty\frac{1}{(2k-1)^2}=\frac{\pi^2}{8}\tag{d}$$
*The integral has also a closed form for $\displaystyle b=\frac{1}{2}$ (and arbitrary $a>0$):
$$I\Big(a, \frac{1}{2}\Big)=\frac{\pi}{2\sqrt2 a}\bigg((1-\sqrt a)\ln\frac{1+a}{1-a}+\frac{1}{2}\int_1^\infty\ln\frac{x+a}{x-a}\frac{dx}{\sqrt x}\bigg)$$
Making in the second term the substitution $t=\sqrt x$, integrating by part and using $\arctan\frac{1}{s}=\frac{\pi}{2}-\arctan s$
$$= \frac{\pi}{2\sqrt2 a}\bigg((1-\sqrt a)\ln\frac{1+a}{1-a}+\ln\frac{1-a}{1+a}+\sqrt a\ln\frac{1+\sqrt a}{1-\sqrt a}+2\sqrt a\arctan\sqrt a\bigg)$$
and we get the known result:
$$I\Big(a, \frac{1}{2}\Big)=\int_0^\infty\frac{\sqrt x\arctan x}{a^2+x^2}dx=\frac{\pi}{\sqrt {2a}}\bigg(\arctan\sqrt a+\ln\frac{1+\sqrt a}{\sqrt{1+a}}\,\bigg),\,a>0\tag{e}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
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} |
Evaluate the limit of a sequence by Riemann sums and mean value theorem Calculate:
$\displaystyle \lim_{n \rightarrow \infty} \left( \frac{n \pi}{4} - \left( \frac{n^2}{n^2+1^2} + \frac{n^2}{n^2+2^2} + \cdots \frac{n^2}{n^2+n^2} \right) \right)$.
I solved it by taking into account that $\displaystyle \int_0^{1} \frac{1}{1+x^2} \mathrm{d}x = \frac{\pi}{4}$ and let the given sequence be:
$a_n= \displaystyle \frac{n \pi}{4} - \left( \frac{n^2}{n^2+1^2} + \frac{n^2}{n^2+2^2} + \cdots + \frac{n^2}{n^2+n^2} \right)$
Let $f(x) = \frac{1}{1+x^2}$, then:
$a_n = \displaystyle \frac{n \pi}{4} - \sum_{i=1}^n \frac{1}{1+\left( \frac{i}{n} \right)^2} = n \int_0^{1} f(x) \mathrm{d}x - \sum_{i=1}^n f\left( \frac{i}{n} \right) = n \sum_{i=1}^n \int_{\frac{i-1}{n}}^{\frac{i}{n}} f(x) \mathrm{d}x - n \sum_{i=1}^n \int_{\frac{i-1}{n}}^{\frac{i}{n}} f\left( \frac{i}{n} \right) \mathrm{d}x = n \sum_{i=1}^n \int_{\frac{i-1}{n}}^{\frac{i}{n}} \left( f(x)- f\left( \frac{i}{n} \right) \right) \mathrm{d}x$
Using Mean Value Theorem and doing a lot of calculations, I finally get that the limit is $\displaystyle \frac{1}{4}$.
Is it correct? Is there an easier method to solve the problem?
| Using
$$ \sum_{k=1}^{n} \frac{1}{k^2 +n^2} = \frac{i}{2 n} \, \left( \psi(1 - i n) - \psi(1 + i n) + \psi(1 + (1+i) \, n) - \psi(1 + (1-i) \, n) \right), $$
where $\psi(x)$ is the digamma function, and
$$ \psi(x) \approx \ln(x) + \frac{1}{2 x} - \sum_{j=1}^{\infty} \frac{B_{2j}}{2 \, j \, x^{2 j}}, $$
where $B_{n}$ are the Bernoulli numbers, then
$$ \sum_{k=1}^{n} \frac{n^2}{k^2 + n^2} \approx \frac{n \, \pi}{4} - \frac{1}{4} - \sum_{j=1}^{\infty} \frac{B_{2 j}}{2^{j+1} \, j \, n^{2 j -1}} \, \sin\left(\frac{j \pi}{2}\right). $$
Now the limit in question takes the form
$$ \lim_{n \to \infty} \left( \frac{n \pi}{4} - \sum_{k=1}^{n} \frac{n^2}{k^2 + n^2} \right) = \lim_{n \to \infty} \left( \frac{1}{4} + \sum_{j=1}^{\infty} \frac{b_{j}}{n^{2 j -1}} \right) = \frac{1}{4}. $$
| {
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"url": "https://math.stackexchange.com/questions/4564940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Telescoping a finite sum Working through an upper-school pre-calculus book which starts with a bit of revision. Problem is that one of the questions in the development section of exercises is solvable using a technique they’ve not yet taught called ‘telescoping’.
If I process the equation by hand, I see the telescoping mechanism in action: all middle terms cancel, leaving only the first and last. However, the answer in the back of the book as well as the answer my CAS calculator generates has the opposite polarity to my own (theirs is positive where as mine is negative).
Had a look at some examples of telescoping online but I’m still confused as to why my handworked example is wrong.
The calculation is:
$$\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\ldots +\dfrac{1}{\sqrt{15}+\sqrt{16}}$$
The correct answer is $3$, but when I work it out I get $-3$.
I begin by rationalising the denominators by multiplying each term (both the numerator and the denominator) by the denominator with the operator inverted, such that the difference of squares is formed in the denominator, cancelling the radical and the terms in all of the denominators equal one, like so:
$$\dfrac{\sqrt{1}-\sqrt{2}}{\left( \sqrt{1}\right) ^{2}-\left( \sqrt{2}\right) ^{2}}+\dfrac{\sqrt{2}-\sqrt{3}}{\left( \sqrt{2}\right) ^{2}-\left( \sqrt{3}\right) ^{2}}+\dfrac{\sqrt{3}-\sqrt{4}}{\left( \sqrt{3}\right) ^{2}-\left( \sqrt{4}\right) ^{2}}+\ldots +\dfrac{\sqrt{15}-\sqrt{16}}{\left( \sqrt{15}\right) ^{2}-\left( \sqrt{16}\right) ^{2}}$$
Simplified:
$$\dfrac{\sqrt{1}-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}-\sqrt{4}}{3-4}+\ldots +\dfrac{\sqrt{15}-\sqrt{16}}{15-16}$$
The penultimate operation looks like this:
$$\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+\ldots \sqrt{15}-\sqrt{16}$$
So all of the middle terms are cancelling (telescoping) except the first and last, leaving:
$$\sqrt{1}-\sqrt{16}=1-4=-3$$
Clearly my reasoning is flawed. Where does my error(s) creep in?
| $$\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+...+\frac{\sqrt{15}-\sqrt{16}}{15-16}$$
Your work till here is correct. Let's evaluate each of this in turn. We notice that the deominantor of all the fractions is $-1$, so they evaluate to the following:
$$-(\sqrt{1}-\sqrt{2})-(\sqrt{2}-\sqrt{3})-(\sqrt{3}-\sqrt{4})-...-(\sqrt{15}-\sqrt{16}) =\\ \sqrt{2} - \sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{16}-\sqrt{15}$$
Since all the terms ranging from $\sqrt{2}$ to $\sqrt{15}$ cancel out, the answer is $$\sqrt{16}-\sqrt{1} = 4-1 = \boxed{3}.$$
I think you forgot the $-1$ when expanding the terms, and that's why you got $-3$ as the final answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating double integral on different domains $D$ $$\iint\limits_{D} \left(x^2+y^2\right)\mathrm{d}x \mathrm{d}y$$
where $D$ is given each time by
$D=x^2-y^2=1,\hspace{0.5cm} x^2-y^2=9,\hspace{0.5cm} xy=2,\hspace{0.5cm} xy=4$
I try to use Polar coordinate transformation
\begin{cases}
x &= \rho \cos \theta \\
y &= \rho \sin \theta
\end{cases}
but I don’t know how to find the range of $\rho$ and $\theta$
| $D$ is made of two disconnected but symmetric regions. $xy$ is positive, so either both $x,y$ are positive or they are both negative. Let $D_+$ be the part of $D$ in the first quadrant. The other part is mirrored in the fourth quadrant. The integrand $x^2+y^2$ is symmetric about the origin, so its integral over $D$ is exactly twice the integral over $D_+$.
Using polar coordinates:
First rewrite the boundaries of $D_+$.
$$x^2 - y^2 = a \implies r^2 \left(\cos^2(\theta) - \sin^2(\theta)\right) = a \implies r = \sqrt a \sqrt{\sec(2\theta)} \\
xy = b \implies r^2 \cos(\theta) \sin(\theta) = b \implies r = \sqrt{2b} \sqrt{\csc(2\theta)}$$
Split up $D_+$ into three regions:
$$D_+ = D_1 \cup D_2 \cup D_3 \\
D_1 = \left\{(r,\theta) \mid 2 \sqrt{\csc(2\theta)} \le r \le 3\sqrt{\sec(2\theta)} \text{ and } \theta_1 \le \theta \le \theta_2\right\} \\
D_2 = \left\{(r,\theta) \mid 2 \sqrt{\csc(2\theta)} \le r \le 2\sqrt2 \sqrt{\csc(2\theta)} \text{ and } \theta_2 \le \theta \le \theta_3\right\} \\
D_3 = \left\{(r,\theta) \mid \sqrt{\sec(2\theta)} \le r \le 2\sqrt2 \sqrt{\csc(2\theta)} \text{ and } \theta_3 \le \theta \le \theta_4\right\}$$
where $\theta_1,\theta_2,\theta_3,\theta_4$ are the values of $\theta$ corresponding to the vertices of $D_+$, represented by the rays in the plot in counter-clockwise order:
The specific coordinates are given in the table below.
$$\begin{array}{c|cr}
(x,y) & (r,\theta) \\
\hline
\left(\sqrt{\frac{9+\sqrt{97}}2}, \frac{\sqrt{97}-9}4 \sqrt{\frac{9+\sqrt{97}}2}\right) & \left(\sqrt[4]{97}, \tan^{-1}\left(\frac{\sqrt{97}-9}4\right)\right) & (\theta_1) \\
\left(\sqrt{\frac{9+\sqrt{145}}2}, \frac{\sqrt{145}-9}8 \sqrt{\frac{9+\sqrt{145}}2}\right) & \left(\sqrt[4]{145}, \tan^{-1}\left(\frac{\sqrt{145}-9}4\right)\right) & (\theta_2) \\
\left(\sqrt{\frac{1+\sqrt{17}}2}, \frac{\sqrt{17}-1}8 \sqrt{\frac{1+\sqrt{17}}2}\right) & \left(\sqrt[4]{17}, \tan^{-1}\left(\frac{\sqrt{17}-1}4\right)\right) & (\theta_3) \\
\left(\sqrt{\frac{1+\sqrt{65}}2}, \frac{\sqrt{65}-1}8 \sqrt{\frac{1+\sqrt{65}}2}\right) & \left(\sqrt[4]{65}, \tan^{-1}\left(\frac{\sqrt{65}-1}8\right)\right) & (\theta_4)
\end{array}$$
Then the integral is
$$\begin{align*}
\iint_D (x^2+y^2) \,dx\,dy &= 2 \iint_{D_+} r^3 \, dr \, d\theta \\[1ex]
&= 2 \left\{\int_{\theta=\theta_1}^{\theta_2} \int_{r=2\sqrt{\csc{2\theta}}}^{3\sqrt{\sec(2\theta)}} + \int_{\theta=\theta_2}^{\theta_3} \int_{r=2\sqrt{\csc{2\theta}}}^{2\sqrt2\sqrt{\csc(2\theta)}} + \int_{\theta=\theta_3}^{\theta_4} \int_{r=\sqrt{\sec{2\theta}}}^{2\sqrt2\sqrt{\csc(2\theta)}}\right\} r^3 \, dr \, d\theta
\end{align*}$$
Doable, yes, but you're much better off using Cartesian coordinates or making the substitution suggested in comments and the other answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$ If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$
My approach is as follow
$\sqrt 3 \sin C = 2\sec A - \tan A \Rightarrow \sqrt 3 \sin C = \sec A + \sec A - \tan A$
$\sqrt 3 \sin C = \sec A + \left( {\frac{{{{\sec }^2}A - {{\tan }^2}A}}{{\sec A + \tan A}}} \right) \Rightarrow \sqrt 3 \sin C = \sec A + \left( {\frac{1}{{\sec A + \tan A}}} \right)$
$\sqrt 3 \sin C = \sec A + \left( {\frac{{\cos A}}{{1 + \sin A}}} \right) \Rightarrow \sqrt 3 \sin C = \frac{1}{{\cos A}} + \left( {\frac{{\cos A}}{{1 + \sin A}}} \right) \Rightarrow \sqrt 3 \sin C = \frac{{1 + \sin A + 1 - {{\sin }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}$
$\sqrt 3 \sin C = \frac{{2 + \sin A - {{\sin }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}$
Not able to proceed further
| Hint:
We have $$\sqrt3\sin C\cos A+\sin A=2$$
But $$\sqrt3\sin C\cos A+\sin A\le\sqrt{(\sqrt3\sin C)^2+1^2}=\sqrt{4-3\cos^2C}\le2$$
The equality occurs if $\cos C=0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Determine the value of $\theta$. Given two random variables $X_1$ and $X_2$ with pdf
$$
f\left(x_i\right)= \begin{cases}\frac{1}{2 \theta}, & -\theta<x_i<\theta \\ 0, & x_i \text { otherwise }\end{cases}
$$
If it is known that $X_1$ and $X_2$ are independent and $\operatorname{Var}\left(X_1 X_2\right)=\frac{64}{9}$, determine the value of $\theta$.
My working:
If the random variable X comes from continuous data with pdf
$$\operatorname{Var}\left(X_1 X_2\right)=\frac{64}{9}$$
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\mu)^2f(x) dx$$
for $\mu$ is the expected value. For example
$$\mu=\int xf(x) dx$$
if subtitute $f(x)=\frac{1}{2 \theta}$, thus
$$\mu=\int x\frac{1}{2 \theta} dx$$
$$\mu=\frac{x^2}{4 \theta}+C$$
So
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\mu)^2f(x) dx$$
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\frac{x^2}{4 \theta})^2\frac{1}{2 \theta} dx$$
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x^2-\frac{x^3}{2 \theta}+\frac{x^4}{16 \theta})\frac{1}{2 \theta} dx$$
$$\frac{64}{9}=(\frac{x^3}{3}-\frac{2x^4}{\theta}+\frac{5x^5}{16 \theta})\frac{1}{2 \theta}$$
$$\frac{64}{9}=(\frac{x^3}{6 \theta}-\frac{x^4}{\theta^2}+\frac{5x^5}{32 \theta})$$
How the next step and please correct if i'm wrong for my work. Thank u
| Note that $EX_1 = EX_2 = 0$, hence
$$
\begin{align}
Var (X_1 X_2) &= E(X_1X_2)^2 - \left(E\left(X_1X_2\right)\right)^2 \\
&= EX_1^2 EX_2^2 - (EX_1 EX_2)^2 \\
&=(EX_1^2)^2 \\
&= \left(\int_{-\theta}^\theta\frac{x^2}{2\theta} dx\right)^2 \\
&= \left(\frac{\theta^2}{3}\right)^2 \\
\end{align}
$$
Solve equatioin $\left(\frac{\theta^2}{3}\right)^2 = \frac{64}9$,
derive $\theta = 2\sqrt2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Struggles solving : $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
Solve $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
(Hint: the equation can be boiled down to a homogeneous one.)
My attempt:
I didn't know how to obtain a homogeneous equation, so I'm going to show what I did instead.
We can write $\frac23xyy'=|x|^3\sqrt{1-\frac{y^4}{x^6}}+y^2.$ Let $z=\frac{y^2}{x^3}=y^2x^{-3}.$ Then
$$\begin{aligned}z'&=2yy'x^{-3}-3y^2x^{-4}\\ \implies x^4z'&=2xyy'-3y^2\\ \implies\frac23xyy'&=\frac{x^4z'+3y^2}3=\frac{x^4z'}3+y^2\\ \implies\frac{x^4z'}3+y^2&=|x|^3\sqrt{1-z^2}+y^2\\ \implies\frac{|x|z'}3&=\sqrt{1-z^2}\\ \implies\frac{dz}{\sqrt{1-z^2}}&=3\frac{dx}{|x|}\\ \implies\arcsin z&=3\ln|x|+C, C\in\Bbb R\\ \implies z&=\sin\ln(C|x|^3), C>0\\ \implies y^2&=x^3\sin\ln(C|x|^3)C>0.\end{aligned}$$
I'm not sure my answer is correct. I didn't get far plugging the result into the equation.
How do we solve the given ODE? Or should I better ask, how do we obtain a homogeneous ODE from it?
| To make it homogeneous, we use the following variable change
$y=t^{a}$ so $~dy=at^{a-1}~dt$
$\frac{2}{3}ax t^{2a-1}~dt=(\sqrt{x^{6}-t^{4a}}+t^{2a})~dx$
To be homogeneous : $2a=3$ ,so $a=\frac{3}{2}$
$xt^{2}~dt=(\sqrt{x^{6}-t^{6}}+t^{3})~dx$ , $t=zx$
$z'x+z=\frac{\sqrt{1-z^{6}}+z^{3}}{z^{2}}$
$\int\frac{z^2}{\sqrt{1-z^{6}}}~dz=\int\frac{1}{x}~dx$
$arcsin(z^{3}) =3\ln(x)+C$
we had : $z=\frac{t}{x}$ , $t=y^{\frac{2}{3}}$
$$y^{2}=x^{3}sin(3\ln(x)+C)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$ Solve the equation $$\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$$
$x=5$ is the only real solution
We can note that $x^2-9x+24>0$ and $6x^2-59x+149>0$ for all $x$. The first thing I decided to try: as $$|5-x|=\begin{cases}5-x,x\le5\\x-5,x>5\end{cases},$$ we can look at two different cases based on the sign of $5-x$: let's $x>5$. Then the equation becomes $$\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=x-5\\\sqrt{x^2-9x+24}=x-5+\sqrt{6x^2-59x+149}\\x^2-9x+24=(x-5)^2+6x^2-59x+149+2(x-5)\sqrt{6x^2-59x+149}\\30x-3x^2-7=(x-5)\sqrt{6x^2-59x+149}$$ We have to raise both sides to the power of 2 again, so I decided to stop here.
Something else we can try is to raise both sides of the initial equation to the power of 2, as $|5-x|^2=(5-x)^2,$ so we have $$7x^2-68x+173-2\sqrt{(x^2-9x+24)(6x^2-59x+149)}=x^2-10x+25\\3x^2-29x+72=\sqrt{(x^2-9x+24)(6x^2-59x+149)}$$ I think it's obvious I am missing something.
| Well we must have the following conditions.
$x^2 -9x +24 \ge 0$ (other wise the square root doesn't exist)
$6x^2 - 59x + 149 \ge 0$. (ditto)
$\sqrt{x^2 - 9x + 24} \ge \sqrt{6x^2 - 59x + 149}$ (because their difference is $|5-x|$ which is $\ge 0$)
$x^2 - 9x + 24 \ge 6x^2 - 59x + 149$ (because $0 \le a \le b \implies a^2 \le b^2$)
So $5x^2 - 50x + 125 \le 0$. We can factor to get
$5(x-5)^2 \le 0$. and we'll.. that actually throws me throw a loop... I was only trying to find a valid range that $x$ can be in but this is ... big.
$(x-5)^2$ is a perfect square and so $(x-5)^2 \ge 0$ so the only way we can have $5(x-5)^2 \le 0$ is if $(x-5)^2 = 0$ and $x = 5$.
So if there is any answer the only possible answer is $x=5$.
But that doesn't mean that there is any answer. After all anyone can write down anything. That doesn't mean it has an answer.
So we have to check that if $x=5$ the equation holds.
If $x =5$ then $x^2 -9x + 24 = 25-45 + 24 = 4$ and $\sqrt{x^2 -9x+24} = 2$.
And $6x^2 - 59x + 149= 150 -295 + 149= 4$ and $\sqrt{6x^2 - 59x + 149} = 2$.
And $|5-x| = |5-5|=0$ and $\sqrt{x^2 -9x+24} - \sqrt{6x^2 - 59x + 149}=2-2=0=|5-x|$ and thus $x = 5$ is a solution.
| {
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"url": "https://math.stackexchange.com/questions/4579852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why $\lim\limits_{n \to \infty} \frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim\limits_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$? I just need clarification about answer to this question:
$$\lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n}} = \lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n+1} \frac{1}{\sqrt{i}} - \sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n+1} - \sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} = 2$$
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n}} \sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}\right)^k = 2^k$$
In this case I don't understand why
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$$
My reasoning. $$\frac{ \frac{1}{a} }{b - c} = \frac{1}{a(b-c)}$$Hence, we get:
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}=\lim_{n \to \infty} \frac{1}{n+1 - \sqrt{n}\sqrt{n+1}} = \lim_{n \to \infty} \frac{ \frac{1}{n} }{ \frac{n}{n} + \frac{1}{n} + \sqrt{\frac{n^2}{n^2}+\frac{n}{n^2} } } = \lim_{n \to \infty}\frac{0}{1+0+\sqrt{1+0}}=0$$
While $\lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} = \lim_{n \to \infty}\frac{ \sqrt{\frac{n}{n} + \frac{1}{n}} + \sqrt{\frac{n}{n}} }{ \sqrt{\frac{n}{n} + \frac{1}{n} } } = \lim_{n \to \infty}\frac{\sqrt{1+0} + \sqrt{1} }{\sqrt{1+0}} = 2$
Any explanation, what's going on? Thank you
| This equality is purely algebraic. Forget about the limits.
$$\frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}=\frac{\frac1{\sqrt{n+1}}\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}.$$The numerator equals $\frac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}}$ and the denominator equals $\sqrt{n+1}^2-\sqrt n^2=n+1-n=1.$
Your mistake is in "$\lim_{n\to\infty}\frac1{n+1-\sqrt n\sqrt{n+1}}=\lim_{n\to\infty}\frac{\frac1n}{\frac nn+\frac1n+\sqrt{\frac{n^2}{n^2}+\frac n{n^2}}}$" (the minus sign in the denominator has become a plus sign).
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
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Prove $\frac{1}{(y+z) x^4} + \frac{1}{(x+z) y^4} + \frac{1}{(y+x) z^4}\geq\frac{3}{2}$ for $x, y, z>0$, with $xyz=1$ How to prove that $\frac{1}{(y+z) x^4} + \cfrac{1}{(x+z) y^4} + \cfrac{1}{(y+x) z^4}\geq\frac{3}{2}$ for $x, y, z>0$...
I've tried substituting $a=1/x$, $b=1/y,$ $c=1/z$, and also writing the 1 that is being divided into $(xyz)^4.$ In both of those cases I tried applying Titu's lemma, but didn't manage to do anything. In the first case, I've observed that it is similar to Nesbitt's inequality, just with an extra $a^2$,$b^2$ and $c^2$ multiplied
| By arithmetic mean-geometric mean inequality: $y^2z^2 + z^2x^2 \geq 2z^2xy = 2z$. Also, we can use Sedrakyan's Form of Cauchy-Schwarz inequality and arithmetic mean-geometric mean inequality we get
$S= \cfrac{1}{(y+z) x^4} + \cfrac{1}{(x+z) y^4} + \cfrac{1}{(y+x) z^4} = \cfrac{(xyz)^4}{(y+z) x^4} + \cfrac{(xyz)^4}{(x+z) y^4} + \cfrac{(xyz)^4}{(y+x) z^4}\\
= \cfrac{(yz)^4}{(y+z)} + \cfrac{(xz)^4}{(x+z)} + \cfrac{(xy)^4}{(y+x)} \geq \dfrac{(y^2z^2 + z^2x^2 + x^2y^2)^2}{2(x+y+z)} \geq \dfrac{(x+y+z)^2}{2(x+y+z)} = \dfrac{1}{2}(x+y+z) \geq \dfrac{3}{2}\sqrt[3]{xyz} = \dfrac{3}{2}.$
Equality holds when $x=y=z=1$.
| {
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Verify that $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots As the title states, the goal is to verify that the quadratic equation: $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots. This problem comes from an interschool mathematics contest for High-schoolers, here's my (brute force) attempt:
Since we need to verify that the roots are rational, we need to prove that the discriminant of this equation (which I'll denote with $D$) is greater than $0$. In other words:
$D=B^2-4AC>0$ where $B=c(3a^2+b^2)$, $A=abc^2$ and $C=3a^2-ab+b^2$ Therefore:
$D=[c(3a^2+b^2)]^2-4(abc^2)(3a^2-ab+b^2)$
$D=[c^2(9a^4+6a^2b^2+b^4)]+[c^2(-12a^3b+4a^2b^2-4ab^3)]$
$D=c^2(9a^4+10a^2b^2+b^4-12a^3b-4ab^3)$
I'm not quite sure where to proceed from here. And this certainly does not prove that the roots are real as this expression can be negative. Is there a way to proceed from here? Are there any better or alternative ways to answer this? Please share your approaches!
| It seems your approach is the most straightforward, and you are almost done. Observe that WLOG we may assume $a, b \ge0$, and we, by AM-GM, have:
$$9a^4+10a^2b^2+b^4 \ge (6a^4+6a^2b^2)+(4a^2b^2+b^4) \ge 12a^3b +4ab^3.$$
EDIT: Another easy way is: $D=c^2(3a^4+6(a^2-ab)^2+(b^2-2ab)^2),$ not using the AM-GM inequality.
| {
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Trig and de Moivre's theorem
A) Use de Moivre's theorem to prove that $\cos^4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$
B) Therefore deduce that $\cos(\pi/8) = \left(\frac{2 + \sqrt{2}}{4}\right)^{1/2}$
C) and write down an expression for $\cos(3\pi/8)$.
I have proved the first part of the question but I am not sure where to go from there.
My attempt
de Moivre's theorem states that $(\cos x +i\sin x)^n = \cos nx + i\sin nx$
using this and $\cos ^2θ + \sin ^2θ =1$, I did the following:
\begin{align}
\cos 4θ + i\sin 4θ &= (\cos θ +i\sin θ)^4 + \cos 4θ + 4i\cos ^3θ\sin θ - 6\cos ^2θ\sin ^2θ-4i\cos θ\sin 3θ+\sin ^4θ\\
\cos 4θ &= \cos ^4θ -6\cos ^2θ\sin ^2θ +\sin ^4θ\\
\cos 4θ &= \cos ^4θ - 6\cos ^2θ(1-\cos ^2θ) + (1-\cos ^2θ)^2\\
\cos 4θ &= \cos ^4θ - 6\cos ^2θ + 6\cos ^4θ +1 - 2\cos ^2θ + \cos ^θ\\
\cos 4θ &= 8\cos ^4θ -8\cos ^2θ +1
\end{align}
| As noted, the firt proposition should be
$\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$
What you show is fine.
What is $\cos \frac {\pi}{8}$?
$\cos 4\frac {\pi}{8} = 8\cos^4\frac {\pi}{8} - 8\cos^2\frac {\pi}{8} +1\\
\cos \frac {\pi}{2} = 0 =8\cos^4\frac {\pi}{8} - 8\cos^2\frac {\pi}{8} +1$
Let $u = \cos \frac{\pi}{8}$
$8u^4 - 8u^2 + 1 = 0$
Apply the quadratic formula.
$u^2 = \frac {8\pm\sqrt{64-32}}{16} = \frac 12 \pm \frac {\sqrt{2}}{4}$
$u = \pm \sqrt {\frac 12 \pm \frac {\sqrt{2}}{4}}$
But this gives 4 answers... how do we make sense of that?
All values of $\cos(\frac {(2n+1)\pi}{8})$ will solve $8u^4-8u^2 + 1$
$\cos \frac {\pi}{8} = \sqrt {\frac 12 + \frac {\sqrt{2}}{4}}\\
\cos \frac {3\pi}{8} = \sqrt {\frac 12 - \frac {\sqrt{2}}{4}}\\
\cos \frac {5\pi}{8} = -\sqrt {\frac 12 - \frac {\sqrt{2}}{4}}\\
\cos \frac {7\pi}{8} = -\sqrt {\frac 12 + \frac {\sqrt{2}}{4}}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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compute $\int_0^1 f(x)dx$
A continuous function $f:[0,1)\to [0,\infty)$ satisfies $f(x/2+1/2) = f(x)+1$ for $x\in[0,1)$ and $f(1-x) = 1/f(x)$ for all $x\in (0,1)$. Compute $\int_0^1 f(x)dx$.
This is question 11 from this problem set.
$f(x)$ cannot be zero for any $x>0$ since otherwise $f(1-x) = 1/f(x)$ would not hold. My first thought is to use some sort of substitution (e.g. $x\mapsto 1-x$, which gives $\int_0^1 f(x)dx=\int_0^1 \dfrac{1}{f(x)}$ ). It might be true that $f(x)+f(1/2-x)=1$ for $0\leq x\leq 1/2$ (or we could extend $f$ to negative $x$ if necessary). $f(0)+f(1/2) = 1$ follows from $f(1/2) = f(0) + 1$ and $f(1/2) = 1/f(1/2)\Rightarrow f(1/2)^2 = 1\Rightarrow f(1/2) = 1\Rightarrow f(0)=0$.
We know $f(1/2-x)=f(1-(1/2+x)) = 1/f(1/2+x) = 1/(f(2x)+1).$ So we need to equivalently show that $f(x)+1/f(x+1/2) = 1,$ or $f(x+1/2) = 1/(1-f(x)).$
However, I'm not sure how to proceed from here.
Then it might be useful to come up with recursive formulas for an integral and use that to find $\int_0^1 f(x)dx$. It might be useful to split $[0,1]$ into carefully chosen subintervals.
| The first thing to notice is that $f(x)+f(1/2-x)=1$ whenever $0<x<\frac{1}{2}$:
\begin{align*}
f(x)+f\left(\frac{1}{2}-x\right)&=\frac{1}{f(1-x)}+\frac{1}{f\left(\frac{1}{2}+x\right)}\\
&=\frac{1}{f(1-2x)+1}+\frac{1}{f(2x)+1}\\
&=\frac{1}{\frac{1}{f(2x)}+1}+\frac{1}{f(2x)+1}\\
&=\frac{f(2x)}{1+f(2x)}+\frac{1}{f(2x)+1}\\
&=1 \, .
\end{align*}
It then follows that
\begin{align*}\int_0^{1/2} f(x) \, dx &= \frac{1}{2} \int_0^{1/2} \left[f(x)+f\left(\frac{1}{2}-x\right)\right] \, dx\\
&=\frac{1}{2}\int_0^{1/2} 1 \, dx\\
&=\frac{1}{4}
\end{align*}
and so
\begin{align*}
\int_0^1 f(x) \, dx &= \int_0^{1/2} f(x) \, dx +\int_{1/2}^1 f(x) \, dx\\
&=\frac{1}{4} + \int_{1/2}^1 f(x) \, dx\\
&=\frac{1}{4} + \frac{1}{2}\int_0^1 f\left(\frac{u+1}{2}\right) \, du\\
&=\frac{1}{4} + \frac{1}{2}\int_0^1 \left[f(u)+1\right] \, du\\
&=\frac{3}{4} + \frac{1}{2}\int_0^1 f(u) \, du
\end{align*}
meaning that we have $\displaystyle\int_0^1 f(x) \, dx = \frac{3}{2}$.
Incidentally, $f$ is uniquely defined, and is a rather interesting function: e.g., it maps the dyadic rationals in $[0,1)$ surjectively to all rationals in $\left[0,\infty\right)$.
| {
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Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. What is its general solution formula? Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. For examples,
$\begin{array}{l}
{1^2}+ {3^2} + {4^2} = {1^2} + {5^2}\\
{1^2} + {2^2} + {6^2} = {4^2} + {5^2}\\
{322^2} + {562^2} + {597^2} = {601^2} + {644^2}\\
{938^2} + {1063^2} + {4722^2} = {536^2} + {4901^2}\\
\vdots
\end{array}$
What is its general solution formula? How to get the general solution formula?
| A very simple equation. You can write various combinations.
https://artofproblemsolving.com/community/c3046h1055645_quadratic_diophantine_equation
For the equation:
$$X_1^2+X_2^2=Y_1^2+Y_2^2+Y_3^2$$
Solutions have the form:
$$X_1=t^2+2(p+s-k)t+2k^2+2p^2+4ps-4pk-2sk$$
$$X_2=t^2+2(p+s-k)t+2k^2+2s^2+4ps-2pk-4sk$$
$$Y_1=t^2+2(p+s-k)t+2k^2+2ps-2pk-2sk$$
$$Y_2=t^2+2(p+s-k)t+2ps$$
$$Y_3=2(p+s-k)(t+p+s-k)$$
| {
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How to find the shape of $x^3 + y^3 - 3xy = 0$ using polar coordinates? There is a curve known as the Folium of Descartes, and it has the equation $x^3 + y^3 - 3axy = 0$. I wanted to know whether it was possible to determine the shape of the case a = 1, using polar coordinates. I have only determined its polar coordinates representation, which is given by
$r = \frac{3 \sin \theta \cos \theta}{\sin^3 \theta + \cos^3 \theta}$
However, I am not too sure how to proceed from here. Do I use calculus or find the values at specific values of $\theta$, and if so, how? Any help would be greatly appreciated. Thank you!
| You can write $r^2$ in terms of $\sin 2\theta$, which shows some of the symmetries of the curve: $$(s^3+c^3)^2 = (s^2+c^2)^3 +2s^3c^3 - 3s^2c^2(s^2+c^2)$$so $$r^2=\dfrac{\tfrac{9}{4}\sin^2 2\theta}{1+\tfrac{1}{4}\sin^3 2\theta-\tfrac{3}{4}\sin^2 2\theta}=\dfrac{9\sin^2 2\theta}{4+\sin^3 2\theta-3\sin^2 2\theta}=\dfrac{9\sin^2 2\theta}{(1+\sin 2\theta)(2-\sin 2\theta)^2}.$$ But this curve just isn't very nice in polar co-ordinates: it's much nicer in co-ordinates $X=\dfrac{x+y}{\sqrt{2}}$, $Y=\dfrac{x-y}{\sqrt{2}}$. Then $$3Y^2=\frac{X^2(3-\sqrt{2}X)}{1+\sqrt{2}X}$$ or, if you don't care about scale, (replacing $Y$ with $3Y$, $X$ with $3\sqrt{2}X$), $$Y^2=\frac{2X^2(1-2X)}{1+6X}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Right triangle with its perimeter and median and altitude A right triangle $ABC$ is given with $\measuredangle ACB=90^\circ$. If the perimeter of the triangle is $72$ and the difference between the lengths of the median and the altitude to the hypotenuse is $7$, find the area.
Let $CD=x$ $(x>0)$, then $CM=x+7$. By the Pythagorean theorem, it follows that $$MD=\sqrt{(x+7)^2-x^2}=\sqrt{14x+49}.$$
For the perimeter as $AM=MB=CM=x+7$ we have $$P_{ABC}=2x+14+a+b=72\Rightarrow a+b=58-2x,$$ where $BC=a,AC=b$.
We have to find $S_{ABC}=\dfrac{2(x+7)x}{2}=x(x+7)=x^2+7x$. I can't see how to do that.
On the other hand, if the sides of the triangle are $a,b$ and $c$, then $$\begin{cases}a+b+c=72\\\dfrac{c}{2}-\dfrac{ab}{c}=7\\a^2+b^2=c^2\end{cases}$$ and we only need to solve it for $\dfrac{ab}{2}$. I cannot do that as well. Thanks!
| For the last three equations, there is a way to solve them.
$$a+b=72-c$$
$$a^2+2ab+b^2=5184-144c+c^2$$
$$2ab=5184-144c$$
Now, we know that:
$$\frac{c}{2}-\frac{ab}{c}=7$$
$$2ab=c^2-14c$$
$$c^2-14c=5184-144c$$
And that yields $c=32$ and $c=-162$, and obviously we'll use the positive value. Since we now have $c$, its very easy to calculate the rest
| {
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Is it possible to simplify $n \equiv - \frac d 4\bigg(\frac 3 2\bigg)^{m-2} - 2 \pmod {3^m}$ further? Is it possible to simplify this further:
$$
n \equiv - \frac d 4\bigg(\frac 3 2\bigg)^{m-2} - 2 \pmod {3^m} \qquad \text{where } n, d \geq 0 \text{ and } m \geq 2 \text{ are integers}
$$
or to glean some info on what $n$ should be to satisfy the equivalence?
I obtained the above as follows:
I want to know when
$$
\frac 1 {3^m} \big(2^m(n+2) - (2 - \Delta) 3^m \big)
\qquad
\text{where } \Delta = \frac d 9 \text{ for some integer } d \geq 0
$$
is an integer. We have
$$
\begin{align}
2^m (n+2) &\equiv (2 - \Delta) 3^m &\pmod {3^m}\\
2^m (n+2) &\equiv (2 - \frac d 9) 3^m &\pmod {3^m}\\
2^m (n+2) &\equiv 2 \cdot 3^m - \frac d {3^2} \cdot 3^m &\pmod {3^m}\\
2^m (n+2) &\equiv - d \cdot 3^{m-2} &\pmod {3^m}\\
4 \cdot 2^{m-2} (n+2) &\equiv - d \cdot 3^{m-2} &\pmod {3^m}\\
n+2 &\equiv - \frac d 4 \bigg(\frac 3 2\bigg)^{m-2} &\pmod {3^m}\\
n &\equiv - \frac d 4 \bigg(\frac 3 2\bigg)^{m-2} - 2 &\pmod {3^m}
\end{align}
$$
From $2^m (n+2) \equiv - d \cdot 3^{m-2} \pmod {3^m}$, I know that if $d \equiv 0 \pmod 9$, then $n \equiv -2 \pmod {3^m}$.
Also, by writing $d = 9k + (d \bmod 9)$, we have
$$
\begin{align}
2^m (n+2) &\equiv - (9k + (d \bmod 9)) \cdot 3^{m-2} &\pmod {3^m}\\
2^m (n+2) &\equiv -9k \cdot 3^{m-2} + -(d \bmod 9) \cdot 3^{m-2} &\pmod {3^m}\\
2^m (n+2) &\equiv -k \cdot 3^m + -(d \bmod 9) \cdot 3^{m-2} &\pmod {3^m}\\
&\boxed{
2^m (n+2) \equiv -(d \bmod 9) \cdot 3^{m-2} \pmod {3^m}
}
\end{align}
$$
| Simplfy $n \equiv - \dfrac d 4\bigg(\dfrac 3 2\bigg)^{m-2} - 2 \pmod {3^m} \qquad \text{where } n, d \geq 0 \text{ and } m \geq 2 \text{ are integers}$.
Multiplying by $\dfrac94$ we get
$$\dfrac94n\equiv -\dfrac d4 \left(\dfrac 32\right)^m-\dfrac 92\pmod{3^m}\Rightarrow 9n\equiv \left(\dfrac{-d}{2^m}\right)3^m-18\pmod{3^m}$$ Then $$\boxed{n\equiv-2\mod{3^m}}$$
| {
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Show that the sums are equal to: a) If $\sum_{k=1}^∞ \frac{1}{k^2} = \frac{π^2}{6}$ show that $\sum_{k=1}^∞ \frac{1}{(2k+1)^2} = \frac{π^2}{8}$
b) Show that $\sum_{k=1}^∞ \frac{1}{k(k+1)(k+2)} = \frac{1}{4}$
Hint for b): Find A, B, C so that for ∀k:
$\frac{1}{k(k+1)(k+2)}$= $\frac{A}{k}$ +$\frac{B}{k+1}$+ $\frac{C}{k+2}$
I managed to do a) but don't know how to do b)
a) $\sum_{k=1}^∞ \frac{1}{k^2}$ - $\sum_{k=1}^∞ \frac{1}{(2k)^2}$ = $\sum_{k=1}^∞ \frac{1}{(2k+1)^2}$
$\sum_{k=1}^∞ \frac{1}{k^2}$ - $\sum_{k=1}^∞ \frac{1}{4}\frac{1}{k^2}$ = $\frac{π^2}{6}$ - $\frac{1}{4}$$\frac{π^2}{6}$=$\frac{π^2}{8}$
| If you find the constants $A,B,C$, they are $\frac{1}{2},-1,\frac{1}{2}$ respectively, then you term
$\frac{1}{k(k+1)(k+2)}=\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2(k+2)}=\frac{1}{2}(\frac{1}{k}-\frac{1}{k+1})+\frac{1}{2}(\frac{1}{k+2}-\frac{1}{k+1})$ if you take the series (Harmonic series) you limit is $\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$.
Recall that $\sum_{1}^n (a_{k+1}-a_{k})=a_{n+1}-a_1$ then $\sum_{K=1}^{\infty}(a_{k+1}-a_{k})=\lim_{n\rightarrow \infty}\sum_{1}^n (a_{k+1}-a_{k})=\lim_{n\rightarrow \infty}a_{n+1}-a_1$
| {
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Describe the set of real numbers $x$ which satisfy $2\log_{2x+3}x<1$. Describe the set of real numbers $x$ which satisfy $2\log_{2x+3}x<1$.
Here, I am really stuck . In this problem, I tried to compute by writing this expression as $(2x+3)^a=x^2$, where $a<1$. But, I don't have a clue what to do ? I am not quite getting this...
| General approach.
Notice that, the logarithm $\log_{2x+3}x$ is defined only by the following condition:
$$x>0\wedge 2x+3>0\wedge 2x+3≠1$$
This is equivalent to the condition $x>0$.
Remember that,
If $a>1$ and $b,c>0$ then the inequality $\log_a b<\log_a c$ is equivalent to
$b<c$ and if $0<a<1$ and $b,c>0$ then the inequality $\log_a b<\log_a c$ is equivalent to $b>c.$
Therefore, if $2x+3>1\wedge x>0$, then we have:
$$\begin{align}&\log_{2x+3}x^2<\log_{2x+3}(2x+3)\\
\iff &x^2<2x+3\end{align}$$
If $0<2x+3<1\wedge x>0$, then we have:
$$\begin{align}&\log_{2x+3}x^2<\log_{2x+3}(2x+3)\\
\iff &x^2>2x+3.\end{align}$$
However, we see that the inequality $0<2x+3<1\wedge x>0$ implies $x\in\emptyset$, therefore in this case we doesn't have a solution.
Thus, the original problem is reduced to the inequality $x^2<2x+3$ under the restriction $x>0$.
Finally we conclude:
$$
\begin{align}&x^2-2x-3<0\wedge x>0\\
\implies &x>0\wedge (x+1)(x-3)<0\\
\implies &x\in(0,3).\end{align}
$$
| {
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Help w/$(\frac{a}{b})^4+(\frac{b}{c})^4+(\frac{c}{d})^4+(\frac{d}{e})^4+(\frac{e}{a})^4\ge\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{e}{d}+\frac{a}{e}$ How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II)
\begin{equation*}
\bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e}
\end{equation*}
There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS?
| Applying the AM-GM
$$LHS - \bigl(\frac{e}{a}\bigr)^4 = \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 \ge4 \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot\frac{d}{e} = 4\cdot\frac{a}{e} $$
Do the same thing for these 4 others terms, and make the sum
$$5 LHS - LHS \ge 4 RHS$$
$$\Longleftrightarrow LHS \ge RHS$$
The equality occurs when $a=b=c=d=e$
| {
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The sum $\frac{1}{2\sqrt1+1\sqrt 2}+\frac{1}{3\sqrt2+2\sqrt 3}+\cdots +\frac{1}{100\sqrt{99}+99\sqrt{100}}$ is equal to: From an exercise of a textbook of an high school of 15 years old, I have this partial sum
$$S_n=\frac{1}{2\sqrt1+1\sqrt 2}+\frac{1}{3\sqrt2+2\sqrt 3}+\cdots +\frac{1}{100\sqrt{99}+99\sqrt{100}}$$
Considering that every fraction is $<1$ I have seen that the general term is $$\frac{1}{100\sqrt{99}+99\sqrt{100}}=\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}$$$$=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}=\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+1}}$$
$$A) \frac{999}{1000}, \quad B) \frac{99}{100}, \quad C) \frac{9}{10}, \quad D) 9, \quad E) 1$$
How should I arrive at the result without to use the calculator? My students not have studies the $\sum$.
| We can split the terms into two that can be eliminated between adjacent terms.
\begin{align}
\sum_{i=1}^{99}{\frac{1}{\left( i+1 \right) \sqrt{i}+i\sqrt{i+1}}}&=\sum_{i=1}^{99}{\frac{\left( i+1 \right) \sqrt{i}-i\sqrt{i+1}}{\left( i+1 \right) ^2i-i^2\left( i+1 \right)}}=\sum_{i=1}^{99}{\frac{\left( i+1 \right) \sqrt{i}-i\sqrt{i+1}}{i^2+i}}
\\
&=\sum_{i=1}^{99}{\left( \frac{\sqrt{i}}{i}-\frac{\sqrt{i+1}}{i+1} \right)}
\\
&=1-\frac{\sqrt{100}}{100}=\frac{9}{10}
\end{align}
So the answer is C.
P.S. To write it without the symbol $\sum$, analyzing each term and then writing the equation termwise would help.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Need advice: what should be my next step for solving the derivative of $f(z)$ using the definition? What should be my next step for solving the derivative of $f(z)$ using the definition?
$$f(z) = (2{z^2} + 1) \cdot ({z^3} - \sqrt {z}) $$
$$f'(z) = \mathop {\lim }\limits_{\vartriangle z \to 0} \frac{{f(z + \vartriangle z) - f(z)}}{{\vartriangle z}} \\= \mathop {\lim }\limits_{\vartriangle z \to 0} \frac{{[2{{(z + \vartriangle z)}^2} + 1] \cdot [{{(z + \vartriangle z)}^3} - \sqrt {(z + \vartriangle z)} ] - (2{z^2} + 1) \cdot ({z^3} - \sqrt {z}) }}{{\vartriangle z}}$$
At this point I don't know how to proceed: when I try to expand it, I can't isolate the ∆z to simplify and avoid the division by zero. What am I missing here?
| After distributing the terms we have...
$$ f(z) = (2{z^2} + 1) \cdot ({z^3} - \sqrt {z}) = 2z^5 - 2z^{
\frac{5}{2}} + z^3 - z^{\frac{1}{2}} $$
by definition of derivative...
$$ f'(z) = \lim _{\Delta z \to 0} \frac{f(z+\Delta z) - f(z)}{\Delta z} $$
$$ = \lim _{\Delta z \to 0} \frac{2{(z+\Delta z)}^5 - 2(z+\Delta z)^{\frac{5}{2}} + (z+\Delta z)^3 - (z+\Delta z)^{\frac{1}{2}} - \Big[ 2z^5 - 2z^{
\frac{5}{2}} + z^3 - z^{\frac{1}{2}} \Big]}{\Delta z} $$
$$ = \lim _{\Delta z \to 0} \frac{2{(z+\Delta z)}^5 - 2(z+\Delta z)^{\frac{5}{2}} + (z+\Delta z)^3 - (z+\Delta z)^{\frac{1}{2}} - 2z^5 + 2z^{
\frac{5}{2}} - z^3 + z^{\frac{1}{2}}}{\Delta z} $$
You can proceed by factoring out the larger powers; solving for limits; and then factoring a $\Delta z$ from the remaining terms in the numerator. At that point, the $\Delta z$ in the numerator will cancel the one in the denominator, and you will be left with the answer. However, it is much easier to obtain the result using the standard differentiation rules, all of which can be proven using the definition of a limit. Otherwise, you are faced with the tedious task of computing an enormous polynomial.
For example, using standard differentation rules...
$$ f'(z) = \frac{d}{dz} f(z) = \frac{d}{dz} \Big[ 2z^5 - 2z^{
\frac{5}{2}} + z^3 - z^{\frac{1}{2}} \Big] $$
by the sum and difference rules we have
$$ = \frac{d}{dz} 2z^5 - \frac{d}{dz} 2z^{
\frac{5}{2}} + \frac{d}{dz} z^3 - \frac{d}{dz} z^{\frac{1}{2}} $$
by the constant multiple rule we have
$$ = 2 \frac{d}{dz} z^5 - 2 \frac{d}{dz} z^{
\frac{5}{2}} + \frac{d}{dz} z^3 - \frac{d}{dz} z^{\frac{1}{2}} $$
and by the power rule we have
$$ = 2 \cdot 5z^{(5-1)} - 2 \cdot \frac{5}{2} z^{(
\frac{5}{2} - 1)} + 3 \cdot z^{(3-1)} - \frac{1}{2} \cdot z^{(\frac{1}{2}-1)} $$
and then we clean things up...
$$ = 10z^4 - 5z^{\frac{3}{2}} + 3z^2 - \frac{1}{2} z^{-\frac{1}{2}} $$
$$ = 10z^4 - 5z\sqrt{z} + 3z^2 - \frac{1}{2\sqrt{z}} $$
If you have any doubts about all of this, then I suggest reviewing the proofs of the standard differentiation rules. In particular, prove the sum/difference rule, the constant multiple rule, and power rule for any arbitrary function, using the definition of derivative. Then you will have confidence in this result without having to explicitly prove it from the ground up. You can find proof of said rules here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does cancelling by $\sin x$ when solving $4\tan x = 5\sin x$ for $0\leq x < 2\pi $ miss solutions? So the solution to the problem is:
$$4 \tan x = 5 \sin x$$
$$4 \frac{\sin x}{\cos x} = 5 \sin x$$
$$4 \sin x = 5 \sin x \cos x$$
So, either $\sin x = 0$ or $\cos x = 4/5$. I'm fine with the logic of this. But is there some deeper reason why cancelling the $\sin x$ straight away loses solutions?
| We can’t cancel terms on both sides without knowing whether it is zero.
Just like solving
$x^2=x \Leftrightarrow x^2-x=0 \Leftrightarrow x(x-1)=0 \Leftrightarrow x=0 \textrm{ or }x=1, \tag*{} $
we shall miss the solution $x=0.$
Back to our equation, we have
$$
\begin{aligned}
& 4 \tan x=5 \sin x \\
\Leftrightarrow \quad & \frac{4 \sin x}{\cos x}=5 \sin x \\
\Leftrightarrow \quad& 4 \sin x=5 \sin x \cos x \\
\Leftrightarrow \quad& \sin x(5 \cos x-4)=0 \\
\Leftrightarrow \quad& \sin x=0 \text { or } 5 \cos x-4=0 \\
\Leftrightarrow \quad& x=n \pi \quad \text { or }\quad \cos x=\frac{4}{5}\\
\Leftrightarrow \quad & x=n \pi \quad \textrm{ or } \quad 2 n \pi \pm \cos ^{-1}\left(\frac{4}{5}\right)
\end{aligned}
$$
where $n\in Z$.
If we cancel $\sin x$ on both sides, we shall miss the solutions $x=n\pi$.
| {
"language": "en",
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Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which is not easily solved. Any suggestions would be helpful.
| Another variation.
We observe that, $\sqrt {1-x^2}≥0$ and $1+\sqrt {1-x^2}>0$ imply $x≥0$.
Letting $\sqrt {1-x^2}=u,~u≥0$ we have:
$$\begin{cases}
ux=u-x\\u^2+x^2=1\end{cases}$$
Then setting $u=a+b,~ x=a-b$ under the restriction $a≥b≥0$, we have:
$$\begin{align}&\begin{cases}a^2-b^2=2b\\a^2+b^2=\frac 12\end{cases}\\
\implies &4b^2+4b-1=0\\
\implies &x=\frac{\sqrt {4b+1}}{2}-b.\end{align}$$
This completes the answer.
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
deriving pdf with two methods, with two different results I have this problem to determine a density function $f_{Y}(y)$ of $Y$ given $X$ as a random variable with uniform distribution on $[−1, 1]$ and $Y = X^2$. So: $f_X(x) = \frac{1}{2}$ for $x \in [-1, 1]$. I approached this problem with two different ways, 1) through determining first the $F_Y$ as the $CDF$, then derive from it $PDF$ 2) by transformation.
For the first method I got $f_Y(y) = \frac{1}{2\sqrt(y)}$ for $y \in ]0, 1]$ This seems correct. However, for transformation method, I tried the following:
$g(x) = x^2$ and $g'(x) = 2x$
$x_1 = g^{-1}(y) = \sqrt(y)$
so $f_Y(y) = \frac{f_X(x_1)}{|g'(x_1)|} = \frac{1/2}{2\sqrt(y)} = \frac{1}{4\sqrt(y)}$
Also, there is the example of $Y = X^3$, then:
$g(x) = x^3$ and $g'(x) = 3x^2$
$x_1 = g^{-1}(y) = \sqrt[3](y)$
so $f_Y(y) = \frac{f_X(x_1)}{|g'(x_1)|} = \frac{1/2}{3\sqrt[3](y)^2} = \frac{1}{6\sqrt(y)^2}$ but the correct answer is $\frac{1}{3\sqrt(y)^2}$
| This idea that you proposed only works if the transform $g(\cdot)$ is one to one.
https://www.math.arizona.edu/~jwatkins/f-transform.pdf
This is why the following reasoning is incorrect, for instance, if $g(x)=x^2$
\begin{align}
f_Y(y)&=f_X(g^{-1}(y))\Big| \frac{d}{dy}g^{-1}(y) \Big| \\
&= \frac{1}{2} \Big| \frac{1}{2} y^{-1/2} \Big|
\end{align}
The correct would be
\begin{align}
F_Y(y) &= P(Y \leq y) = F_X(\sqrt{y})- F_X(-\sqrt{y}) \\
f_Y(y)&= \frac{1}{2\sqrt{y}}(f_X(\sqrt{y})+f_X(-\sqrt{y})) = \frac{1}{2\sqrt{y}}
\end{align}
| {
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"source": "stackexchange",
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} |
Integrate expression with 3 multiplications $\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)}dx$ Specifically I want to integrate the following:
$$\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)}$$
I do know that
$$\frac4{\pi}\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)} = \dfrac{-16\cos(n\pi)}{(4n^2-1)^2{\pi}}$$
I just have no idea how to integrate it. I tried using trig identities to simplify the expression something by parts could be used for, but I got a very wrong answer.
| Using prostapheresis formulas:
$$\sin(p)+\sin(q)=2\cdot\sin\left(\frac{p+q}{2}\right)\cdot\cos\left(\frac{p-q}{2}\right)$$
One can write:
$$\left\{\begin{matrix}
\frac{p+q}{2}=2nx
\\\frac{p-q}{2}=x
\end{matrix}\right.$$
Which leads to:
$$\left\{\begin{matrix}
q=(2n-1)x
\\p=(2n+1)x
\end{matrix}\right.$$
So:
$$\int_0^{\pi/4}{x\cos(x)\sin(2nx)}=\frac{1}{2}\int_{0}^{\pi/4}x\cdot\sin((2n+1)x)dx+\frac{1}{2}\int_{0}^{\pi/4}x\cdot\sin((2n-1)x)dx$$
Evaluating, one get:
$$\frac{1}{2}\cdot\left[-\frac{x\cdot\cos((2n+1)x)}{2n+1}+\frac{\sin((2n+1)x)}{(2n+1)^2}\right]_{0}^{\pi/4}+\frac{1}{2}\cdot\left[-\frac{x\cdot\cos((2n-1)x)}{2n-1}+\frac{\sin((2n-1)x)}{(2n-1)^2}\right]_{0}^{\pi/4}$$
Now, $(2n+1)x=(2n-1)x+2x$. For $x=\frac\pi4$, one have:
$$\sin\left(\frac{(2n+1)\pi}4\right)=\sin\left(\frac{(2n-1)\pi}4+\frac\pi2\right)=\sin\left(\frac{(2n-1)\pi}4\right)$$
And:
$$\cos\left(\frac{(2n+1)\pi}4\right)=\cos\left(\frac{(2n-1)\pi}4+\frac\pi2\right)=-\cos\left(\frac{(2n-1)\pi}4\right)$$
Can you take it from here?
| {
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"question_score": "2",
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Proof by induction that $17n^3+103n$ is a multiple of 6 Problem use induction to prove that $17n^3+103n$ is a multiple of 6.
Im new to learning proof by induction and was wondering if my proof would be acceptable for my maths test coming up.
Proof. base case $n=1 = 120$ $17k^3+103k=6a$ for some $a$. then we prove $k+1$ $$17(k+1)^3+103(k+1)=6b$$ for some b. we expand and get $17k^3+51k^2+51k+17+103k+103=6b$ combine like terms and substitute $17k^3+103$ for $6a$ then we get $$6a+51k^2+51k+120=6b$$ simplify to $51n(n+1)$ then we see that $n(n+1)$ has to multiply to become an even number so 51n(n+1) is also a multiple of six QED.
would this proof work or are there any better methods.
| Your base case is correct: $17(1)^3+103(1)=120$ which is divisible by $6$.
For the inductive step we assume $17k^3+103k$ is divisible by $6$ and consider $17(k+1)^3+103(k+1)$.
Expanding this we get $$17(k+1)^3+103(k+1)=17(k^3+3k^2+3k+1)+103k +103=(17k^3+103k)+51k^2+51k+120$$
We know that $17k^3+103k$ is divisible by $6$ because of the induction hypothesis. We also know that $120$ is divisible by $6$ so what we really have to show is that $51k^2+51k$ is divisible by $6$.
$51k^2+51k=51(k^2+k)$ and $51$ is divisible by $3$ so we only need to show that $k^2+k$ is even.
Now, $k$ and $k^2$ are either both even or both odd. Since the sum of two even numbers and the sum of two odd numbers are even, $k^2+k$ is even so $51k^2+51k$ is divisible by $6$.
Therefore, by induction, $17n^3+103n$ is divisible by $6$ for all $n$.
| {
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Evaluation of $\iint_{x^2-xy+y^2\leq1}(x-y)^2\mathrm{d}x\mathrm{d}y$ $$
I:=\iint_{x^2-xy+y^2\leq1}(x-y)^2\mathrm{d}x\mathrm{d}y
$$
I've got the following 2 colored equations but neither seems useful for evaluating this double integral.
$$\begin{align}
x^2-xy+y^2&= \color{blue}{\left(x- {1 \over 2 }y \right)^2+ {3 \over 4}y^2} \\
&= \color{green}{(x-y)^2+xy}
\end{align}$$
I want your wisdom.
| Observe that $a(x+y)^2+b(x-y)^2=x^2+y^2-xy$ if and only if $a=\frac1{4}$ and $b=\frac{3}{4}$. Now let the change of variable $T(x,y):=\frac1{2}(x+y,\sqrt{3}(x-y))=(z,t)$ and notice that $T=\left[\begin{smallmatrix}1/2&1/2\\\sqrt{3}/2&-\sqrt{3}/2\end{smallmatrix}\right]$ is an invertible linear map, therefore
$$
\begin{align*}
\int_{\left\{(x,y):\frac1{4}(x+y)^2+\frac{3}{4}(x-y)^2\leqslant 1\right\}}(x-y)^2 d(x,y)&=\int_{\{(z,t):z^2+t^2\leqslant 1\}}\frac{4}{3}t^2 d(T^{-1}(z,t))\\
&=\int_{\{(z,t):z^2+t^2\leqslant 1\}}\frac{4}{3}t^2|\det T^{-1}| d(z,t)\\
&=\frac4{3|\det T|}\int_{[0,1]\times [0,2\pi]}r^3(\sin \theta)^2\,d (r,\theta)\\
&=\frac2{3\sqrt{3}}\int_{0}^{2\pi}(\sin \theta )^2 \,d \theta \\
&=\frac{2\pi}{3\sqrt{3}}
\end{align*}
$$
where we used the change of variable $(z,t)=(r\cos \theta ,r \sin \theta )$.∎
| {
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Determinant of $A_{2n}$ Determine the determinant of $$A_{2n}:=\begin{pmatrix}
a &0 &\cdots & \cdots &0 &b \\
0& a & & & b&0 \\
\vdots &\vdots &\ddots &\ddots &\vdots &\vdots \\
\vdots&\vdots & \ddots & \ddots &\vdots &\vdots \\
0& b & & & a&0 \\
b&0 &\cdots &\cdots &0 &a
\end{pmatrix}\in \mathbb{R}^{2n \times 2n}$$
My initial thought is if we assume that it is $2 \times 2$, then we will get $a^2-b^2$. If it is a $4 \times 4$ one then we would get $(a^2-b^2)^2$. So would the determinant for a $2n \times 2n$ is $(a^2-b^2)^n$? But how to prove it mathematically?
| We expand the determinant along the first row:
$$
\det A_{2n}=aM_{11}+(-1)^{2n+1}bM_{1,2n}
$$
Now we expand $M_{11}$ and $M_{1,2n}$ along the last row:
$$
\det A_{2n}=a^2\det A_{2n-2}-b^2\det A_{2n-2}=(a^2-b^2)\det A_{2n-2}.
$$
Now induction works:
$$
\det A_{2n}=(a^2-b^2)(a^2-b^2)^{n-1}=(a^2-b^2)^n.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find a permutation $X \in S_6$ which satisfies the equation $\begin{pmatrix}1&2&3&4&5&6\\ 4&3&1&5&2&6\end{pmatrix} \circ X = \begin{pmatrix}1&2&3&4&5&6\\ 3&2&4&5&1&6\end{pmatrix}$
I've tried this solution and checked this website, but none of them were helpful or my problem is little bit different and may require more steps to get the solution. Could anybody help me sort this out?
| Hint:
$$X=\begin{pmatrix}1&2&3&4&5&6\\ 4&3&1&5&2&6\end{pmatrix}^{-1}\circ \begin{pmatrix}1&2&3&4&5&6\\ 3&2&4&5&1&6\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4598172",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Integrals of the form $\int_0^1\left(\frac{1}{x+1}-\ln\left|\frac{x+1}{2}\right|-\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx$
How could we evaluate the following integrals? (Integrands pictures above)
Blue:
\begin{align}
\int_0^1 \left(\frac{1}{x+1}-\ln\left|\frac{x+1}{2}\right|-\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx &= \gamma - \ln\pi + \ln 2 + 1 \\
\int_0^\infty \left(\frac{1}{x+1}-\ln\left|\frac{x+1}{2}\right|-\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx &= -1 + \ln 2
\end{align}
Red:
$$\begin{align}
\int_0^1 \left(\frac{1}{x+1}+\ln\left|\frac{x-1}{2}\right|+\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx &= -\gamma + \ln\pi - \ln 2 - 1 \\
\int_0^\infty \left(\frac{1}{x+1}+\ln\left|\frac{x-1}{2}\right|+\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx &= -1 - \ln 2
\end{align}$$
| A solution using Riemann $\zeta$, with $\zeta(0)=-1/2$ and $\zeta'(0)/\zeta(0)=\ln2\pi$. Let \begin{align}
F(a,b)&=\int_a^b\left(\frac1{x+1}+\ln\left|\frac{x-1}{2}\right|+\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx,\\
G(a,b)&=\int_a^b\left(\frac1{x+1}-\ln\left|\frac{x+1}{2}\right|-\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx,\\
S(a,b)&=F(a,b)+G(a,b)=\int_a^b\left(\frac2{x+1}-\ln\left|\frac{x+1}{x-1}\right|\right)dx,\\
D(a,b)&=F(a,b)-G(a,b)=\int_a^b\left(\ln\left|\frac{x^2-1}4\right|+2\ln\ln\left|\frac{x+1}{x-1}\right|\right)dx.
\end{align}
Here $S(a,b)$ is elementary, and we easily find $S(0,1)=0$ and $S(1,\infty)=-2$.
Similarly, after elementary integration, we get $D(0,1)=2I-2$, where $$I=\int_0^1\ln\ln\frac{1+x}{1-x}\,dx=2\int_0^\infty\frac{e^t\ln t\,dt}{(e^t+1)^2}$$ after $x=(e^t-1)/(e^t+1)$. Thus $I=2f'(0)=f'(0)/f(0)$ where, for $\Re s>-1$, $$f(s)=\int_0^\infty\frac{t^s e^t\,dt}{(e^t+1)^2}=\Gamma(s+1)\eta(s)$$ with $\eta(s)=(1-2^{1-s})\zeta(s)$; this can be obtained for $\Re s>0$ using integration by parts, and then for $\Re s>-1$ by analytic continuation. With $\zeta(0)$ and $\zeta'(0)$ at hand, we obtain $I=\ln(\pi/2)-\gamma$.
For $D(1,\infty)$, we substitute $x=(e^t+1)/(e^t-1)$ and integrate by parts: \begin{align}
D(1,\infty)&=2\int_0^\infty\ln\frac{t^2 e^t}{(e^t-1)^2}\,d\frac{-1}{e^t-1}=2g(0),\\g(s)&=\int_0^\infty\left(1+\frac2t-\frac{2e^t}{e^t-1}\right)\frac{t^s\,dt}{e^t-1}.
\end{align}
This time we get $g(s)=\Gamma(s)\big(s\zeta(s+1)+2(1-s)\zeta(s)\big)$, for $\Re s>1$ via integration by parts, then for $\Re s>-1$ by analytic continuation, and finally $g(0)=\gamma+1-\ln2\pi$.
Now we have all the pieces to gather the results from.
| {
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Evaluate $\int_0^\infty\frac{\sqrt[5]{x}}{(x^2+1)(x+1)^2}dx$ using Gamma and Beta functions I need to evaluate the following integral, using Gamma and Beta functions:
$$\int_0^\infty\frac{\sqrt[5]{x}}{(x^2+1)(x+1)^2}dx$$
I tried to use partial fractions, but I am not sure if that is correct, since doing that I get that the integral is divergent.
Since
$\frac{1}{\left(x^2+1\right)\left(x+1\right)^2}=\frac{1}{2}\left(\frac{1}{x+1}\right)+\frac{1}{2}\left(\frac{1}{\left(x+1\right)^2}\right)-\frac{1}{2}\left(\frac{x}{x^2+1}\right)$
Then,
$\int_{0}^{\infty}{\frac{\sqrt[5]{x}}{\left(x^2+1\right)\left(x+1\right)^2}dx=\frac{1}{2}\int_{0}^{\infty}{\frac{\sqrt[5]{x}}{x+1}dx}+\frac{1}{2}\int_{0}^{\infty}{\frac{\sqrt[5]{x}}{\left(x+1\right)^2}dx}-\frac{1}{2}\int_{0}^{\infty}{\frac{x\sqrt[5]{x}}{x^2+1}dx}}$
But you can see that the first of this new integrals is divergent, and we only need one integral to be divergent to say that the whole integral diverges, but it is not true since the original integral is convergent, so there must be something wrong with this procedure.
| We can make your approach work by modifying it slightly. Consider the integral
$$I(s)=\int_0^\infty\frac{x^{s-1}}{(1+x^2)(1+x)^2}\,\mathrm dx$$
instead. Applying your partial fraction expansion, an integral representation of the beta function $\text{B}(x,\,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}\,\mathrm dt$, and the reflection formula for the gamma function $\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)}$ tells us that
\begin{align*}
I(s) &= \frac{1}{2}\int_0^\infty\frac{x^{s-1}}{1+x}\,\mathrm dx+\frac{1}{2}\int_0^\infty\frac{x^{s-1}}{(1+x)^2}\,\mathrm dx-\frac{1}{2}\int_0^\infty\frac{x^s}{1+x^2}\,\mathrm dx \\
&= \frac{1}{2}\text{B}(s,\,1-s)+\frac{1}{2}\text{B}(s,\,2-s)-\frac{1}{4}\text{B}\left(\frac{s+1}{2},\,\frac{1-s}{2}\right) \\
&= \frac{\pi}{2}(2-s)\csc(\pi s)-\frac{\pi}{4}\sec\left(\frac{\pi s}{2}\right).
\end{align*}
This last expression agrees with I(s) for $0<\Re(s)<4$. Setting $s=6/5$ yields
$$I\left(\frac{6}{5}\right)=\frac{\pi\phi}{2}-\frac{4\pi}{5}\sqrt{\frac{2}{5}+\frac{\phi}{5}}$$
where $\phi$ is the golden ratio.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Two linear equations with three unknowns and a magnitude constraint Given $u,v,w,x,y,z$ is there a way to solve the following for $a,b,c$, assuming that there is a solution (e.g. $(u,v,w)$ and $(x,y,z)$ are not proportional):
$$\begin{aligned}
au+bv &= cw\\
ax+by &= cz\\
a^2 + b^2 &= 1
\end{aligned}$$
An equivalent problem is solving for $c$ and $\theta$:
$$\begin{aligned}
u\cos\theta + v\sin\theta &= cw\\
x\cos\theta + y\sin\theta &= cz\\
\end{aligned}$$
There must be some straightforward way to solve this, but I've got a mental block. (This is not homework; I'm trying to fit a curve to some data.)
| The system should be digestible by a Grobner basis computation. I ran it in Mathematica V13 with respect to the monomial order $\{a, b, c\}$, with the variables $b$ and $c$ eliminated
GroebnerBasis[{a u + b v - c w, a x + b y - c z, a^2 + b^2 - 1}, {a, b, c}, {b, c}]
and obtained the following polynomial for $a$
$$a^2 w^2 x^2 - w^2 y^2 + a^2 w^2 y^2 - 2 a^2 u w x z + 2 v w y z -
2 a^2 v w y z + a^2 u^2 z^2 - v^2 z^2 + a^2 v^2 z^2$$
Similarly, by eliminating $a$ and $c$ we get a polynomial for $b$:
$$-w^2 x^2 + b^2 w^2 x^2 + b^2 w^2 y^2 + 2 u w x z - 2 b^2 u w x z -
2 b^2 v w y z - u^2 z^2 + b^2 u^2 z^2 + b^2 v^2 z^2$$
and by eliminating $a$ and $b$ we get a polynomial for $c$
$$-v^2 x^2 + c^2 w^2 x^2 + 2 u v x y - u^2 y^2 + c^2 w^2 y^2 -
2 c^2 u w x z - 2 c^2 v w y z + c^2 u^2 z^2 + c^2 v^2 z^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Frullani like Trig integral $$\int_{0}^{\infty}\frac{\left|\cos\left ( x-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( x+\frac{\pi}{4} \right ) \right| }{x}dx=\sqrt{2}\ln(1+\sqrt{2})$$ The above integral seems to look like a frullani type integral and has a closed form in terms of natural log. I tried to indefinitely integrate it. But the closed form are in terms $\text{Si}(x)$ and $\text{Ci}(x)$. I would highly appreciate if there's any method or unique substitution to evaluate this Integral.
| Let $$I = \int_{0}^{\infty} \frac{\left|\cos\left(x-\frac{\pi}{4}\right) \right|-\left|\cos\left(x+\frac{\pi}{4}\right) \right|}{x} \, \mathrm dx .$$
Then
$ \require{cancel} \begin{align}I &= \int_{0}^{\infty}\frac{\left|\cos\left(x-\frac{\pi}{4}\right) \right|-\left|\sin\left(x-\frac{\pi}{4} \right) \right|}{x} \, \mathrm dx \\ &\overset{(1)}{=} \int_{0}^{\infty} \mathcal{L} \left\{\left|\cos\left(x-\frac{\pi}{4}\right) \right|-\left|\sin\left(x-\frac{\pi}{4} \right) \right| \right\}(t) \, \mathrm dt \\ &= \int_{0}^{\infty} \left[\int_{0}^{\infty} \left(\left|\cos\left(x-\frac{\pi}{4}\right) \right| -\left|\sin\left(x-\frac{\pi}{4}\right) \right| \right) e^{-tx} \, \mathrm dx \right]\, \mathrm dt \\ &= \int_{0}^{\infty} e^{-\pi t/4}\int_{-\pi/4}^{\infty} \left(\left|\cos(u) \right|-\left|\sin(u) \right| \right) e^{-tu} \, \mathrm du \, \mathrm dt \\ &= \small\int_{0}^{\infty} e^{- \pi t/4} \left( \int_{- \pi/4}^{0} \cos(u) e^{-tu} \, \mathrm du + \int_{-\pi/4}^{0} \sin(u) e^{-tu} \, \mathrm du + \int_{0}^{\infty} \left|\cos(u) \right| e^{-tu} \, \mathrm du -\int_{0}^{\infty} \left|\sin(u) \right| e^{-tu} \, \mathrm du \right) \, \mathrm dt \\ & \overset{(2)}{=} \small\int_{0}^{\infty} e^{- \pi t/4} \left(\frac{e^{\pi t/4} (1+t )-\sqrt{2}t}{\sqrt{2}(1+t^{2})} + \frac{e^{\pi t/4}(1-t)-\sqrt{2}}{\sqrt{2}(1+t^{2})} + \frac{1}{1+t^{2}} \left(t+ \frac{1}{\sinh \left(\frac{\pi t}{2} \right)} \right) -\frac{1}{(1+t^{2})\tanh \left(\frac{\pi t}{2} \right)} \right) \, \mathrm dt \\ &\overset{(3)}{=} \int_{0}^{\infty} \left(\frac{\sqrt{2}}{1+t^{2}} - \frac{1}{(1+t^{2})\cosh\left(\frac{\pi t}{4} \right)} \right) \, \mathrm dt \\ & \overset{(4)}{=} \frac{\sqrt{2} \pi}{2} - \frac{1}{2} \left(\psi \left(\frac{7}{8} \right) - \psi \left(\frac{3}{8} \right) \right) \\ & \overset{(5)}{=} \frac{\sqrt{2} \pi}{2} -\frac{1}{2} \left[- \frac{\pi}{2} \, \cot \left(\frac{7 \pi}{8} \right) + \frac{\pi}{2} \, \cot \left(\frac{3 \pi}{8} \right) + 2 \sqrt{2} \log \left(\sin \left(\frac{\pi}{8} \right) \right) - 2 \sqrt{2} \ln \left(\sin \left(\frac{3\pi}{8} \right) \right) \right] \\ &= \frac{\sqrt{2} \pi}{2} -\frac{1}{2} \left[\frac{\pi}{2} \cot \left(\frac{\pi}{8} \right) + \frac{\pi}{2} \tan \left(\frac{\pi}{8} \right)+ 2 \sqrt{2} \log \left(\sin \left(\frac{\pi}{8} \right) \right) - 2 \sqrt{2} \ln \left(\cos \left(\frac{\pi}{8} \right) \right) \right]\\ &=\frac{\sqrt{2} \pi}{2} -\frac{1}{2} \left[\pi \csc\left(\frac{\pi}{4} \right)- 2 \sqrt{2} \log \left(\cot \left(\frac{\pi}{8} \right) \right) \right] \\ &= \sqrt{2} \ln(1+\sqrt{2}). \end{align}$
$(1)$ If $f(x)$ is a continuous function and $\int_{0}^{\infty} f(x) e^{-s_{0}x} \, \mathrm dx$ converges, we have the identity $$ \int_{0}^{\infty} \frac{f(x)}{x} e^{-sx} \, \mathrm dx = \int_{s}^{\infty}\mathcal{L} \left\{f(x) \right\} (t) \, \mathrm dt $$ for $s> s_{0}$.
But if $\int_{0}^{\infty} \frac{f(x)}{x} \, \mathrm dx$ converges, then $$\lim_{s \to 0^{+}} \int_{0}^{\infty} \frac{f(x)}{x} e^{-sx} \, \mathrm dx = \int_{0}^{\infty} \frac{f(x)}{x} \, \mathrm dx = \int_{0}^{\infty} \mathcal{L} \left\{f(x) \right\}(t) \, \mathrm dt $$ by Abel's test for uniform convergence..
$(2)$ An evaluation of the Laplace transform of $|\sin(u)|$ can be found here.
For the Laplace transform of $|\cos(u)|$, express the integral as $$\int_{0}^{\pi/2} \cos(u)e^{-tu} \, \mathrm du + \sum_{k=1}^{\infty} \int_{(2n-1)\pi/2}^{(2n+1)\pi/2}|\cos(u)| e^{-tu} \, \mathrm du. $$
$(3)$ $e^{-x/2} \left(\frac{1}{\sinh(x)} - \frac{1}{\tanh(x)}-1 \right) = - \frac{1}{\cosh(x/2)}$
$(4)$ See the answers here.
$(5)$ Gauss' Digamma Theorem
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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In a triangle $\triangle ABC$, the sum of two sides is $x$, and their product is $y$. If $x^2-c^2=y$, find $\frac{r}{R}$ in terms of $x, c, y$ This problem comes from a previous JEE Advanced examination, the problem is as follows:
In a triangle $\triangle ABC$, the sum of two sides is $x$, and their product is $y$. If $x^2-c^2=y$ where $c$ is the third side, find the ratio of the inradius to the circumradius in terms of $x, c, y$
This is indeed a pretty challenging and tricky problem. Upon initial examination, spamming algebra seems to not lead anywhere. I found a solution using some angle chasing, which I'll share below, please share your own approaches as well!
Let $a+b=x$ and $ab=y$
We know that:
$$x^2-c^2=y$$
$$(a+b)^2-c^2=ab$$
$$a^2+b^2-c^2=-ab$$
Dividing by $2ab$ and exploiting the Law of Cosines, we can say that:
$$\frac{a^2+b^2-c^2}{2ab}=-\frac{1}{2}$$
Now, this implies that $\angle C=120^\circ$
Now, using the Law of Sines, we get:
$$\frac{1}{2R}=\frac{\sin{C}}{c}$$
$$\frac{c}{2R}=\frac{\sqrt3}{2}$$
$$R=\frac{c}{\sqrt3}$$
We also know that:
$$\Delta= \frac{1}{2}ab\sin{C}$$
$$\Delta=\frac{\sqrt{3}y}{4}$$
Further, we can note that:
$$s=\frac{a+b+c}{2}$$
$$s=\frac{x+c}{2}$$
Also, $r=\frac{\Delta}{s}$, therefore:
$$\frac{2\sqrt{3}y}{4(x+c)}=r$$
$$r=\frac{\sqrt{3}y}{2(x+c)}$$
Finally:
$$\frac{r}{R}=(\frac{\sqrt{3}y}{2(x+c)})(\frac{\sqrt{3}}{c})$$
$$\frac{r}{R}=\frac{3y}{2c(x+c)}$$
| Let’s assume $S$ is the Area of the triangle and $P$ is half of the perimeter.
If $x=a+b$ and $y=ab$ $\\$ then $(a-b)^2=(a+b)^2-4ab=x^2-4y$.
We know that $S=\frac{abc}{4S}$ and $r=\frac{S}{P}$ then using the Heron’s formula we have:$$\frac{r}{R}=\frac{\frac{S}{P}}{\frac{abc}{4S}}=\frac{4S^2}{Pabc}=\frac{4(P-a)(P-b)(P-c)}{abc}$$
And also, $(P-a)(P-b)(P-c)=\frac{c-(a-b)}{2}\times\frac{c+(a-b)}{2}\times\frac{x-c}{2}=\frac{c^2-(a-b)^2}{4}\times\frac{x-c}{2}$ $$=\frac{c^2-(x^2-4y)}{4}\times\frac{x-c}{2}=\frac{3y(x-c)}{8}$$
Therefore $$\frac{r}{R}=\frac{\frac{3y(x-c)}{2}}{yc}=\frac{{3(x-c)}}{2c}$$
Which is consistent with Goku カカロット
's final answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Help Solve $\int \sqrt{x^{2}+x-2}\,dx$ I have this integral $\int \sqrt{x^{2}+x-2}\,dx=\int |x-1|\sqrt{\frac{x+2}{x-1}}\,dx$.
My textbook advices to use the substitution $t^{2}=\frac{x+2}{x-1}$.
Observation: This integral should be solvable without using integral of modules as it is placed before moduled function integrals chapter.
I did the substitution and i obtaind $-18\int \frac{{t^{2}}}{|(t^{2}-1)^{3}|}\,dt$ . At this point i thought that since the substitution has to be an invertible function, i may consider a restrinction to cancel the absolute value.
I checked on wolfram alpha both integrals but it gives different results. Wolfram solution:
$$\int \sqrt{(x^2 + x - 2)}\,dx = \frac14 (2 x + 1) \sqrt{(x^2 + x - 2)} - \frac98 \log\left(2 \sqrt{(x^2 + x - 2)} + 2 x + 1\right) + c$$
Any help?
The purpose of the exercise is to trasform the irrational function into a rational one and use the Hermite's equation to solve it
| $$
\begin{aligned}
I &= \int{\sqrt{x^2 + x - 2}dx} = \int{\underbrace{\sqrt{x^2 + x - 2}}_{u}\underbrace{dx}_{dv}} = |\text{ integrating by parts }| = \\
&= x\sqrt{x^2+x-2}-\int{xd\sqrt{x^2+ x-2}} = \\
&= x\sqrt{x^2+x-2}-\int{x\frac{2x+1}{2\sqrt{x^2+x-2}}dx} = \\
&= x\sqrt{x^2+x-2}-\int{\frac{2\left(x^2+x-2\right)-\left(x-4\right)}{2\sqrt{x^2+x-2}}dx} = \\
&= x\sqrt{x^2+x-2}-\underbrace{\int{\sqrt{x^2+x-2}dx}}_{I} + \int{\frac{x-4}{2\sqrt{x^2+ x-2}}dx} \Leftrightarrow \\
2I &= x\sqrt{x^2+x-2} + \frac{1}{2}\underbrace{\int{\frac{2x+1}{2\sqrt{x^2+ x - 2}}dx}}_{\int{d\sqrt{x^2+ x-2}}=\sqrt{x^2+x-2}}-\frac{9}{4}\int{\frac{dx}{\sqrt{x^2+x-2}}} = \\
&= \left(x+\frac{1}{2}\right)\sqrt{x^2+x-2}-\frac{9}{4}\int{\frac{dx}{\sqrt{\left(x+\frac{1}{2}\right)^2-\frac{9}{4}}}} = \left|\text{ note that }dx = d\left(x+\frac{1}{2}\right)\right|=\\
&= \left(x+\frac{1}{2}\right)\sqrt{x^2+x-2}-\frac{9}{4}\int{\frac{d\left(x+\frac{1}{2}\right)}{\sqrt{\left(x+\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2}}} = \left|\int{\frac{du}{\sqrt{u^2-a^2}}} = \ln\left|u+\sqrt{u^2-a^2}\right|\right|=\\
&= \left(x+\frac{1}{2}\right)\sqrt{x^2+x-2}-\frac{9}{4}\ln\left|\left(x+\frac{1}{2}\right)+\sqrt{\left(x+\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2}\right| + C \Leftrightarrow \\
I &=\frac{1}{2}\left(x+\frac{1}{2}\right)\sqrt{x^2+x-2}-\frac{9}{8}\ln\left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x-2}\right| + C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Any better way to find $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
Find the value of $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
My Method:
I used the following Identities:
\begin{aligned}
& \sin A \cos B-\cos A \sin B=\sin (A-B) \\
& 2 \sin A \sin B=\cos (A-B)-\cos (A+B) \\
& \cos (2 A)=2 \cos ^2 A-1 \\
& \cos (3 A)=4 \cos ^3 A-3 \cos A \\
& \sin (2 A)=2 \sin A \cos A \\
& 2 \sin A \cos B=\sin (A+B)+\sin (A-B) \\
&
\end{aligned}
$$\begin{aligned}
S_1 & =\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)=\cot(10^{\circ})-\cot \left(50^{\circ}\right)+\tan(20^{\circ}) \\
\\
\Rightarrow S_1 & =\frac{\cos \left(10^{\circ}\right)}{\sin \left(10^{\circ}\right)}-\frac{\cos \left(50^{\circ}\right)}{\sin \left(50^{\circ}\right)}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\
\\
\\
\Rightarrow S_1 & =\frac{2 \sin \left(40^{\circ}\right)}{2 \sin \left(10^{\circ}\right) \sin \left(50^{\circ}\right)}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\
\\
\Rightarrow S_1 & =\frac{2 \sin \left(40^{\circ}\right)}{\cos \left(40^{\circ}\right)-\frac{1}{2}}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\ \\
\Rightarrow S_1 & =\frac{4 \sin 40^{\circ}}{4 \cos ^2\left(20^{\circ}\right)-3}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)}=\frac{6 \sin \left(40^{\circ}\right) \cos \left(20^{\circ}\right)-3 \sin \left(20^{\circ}\right)}{0.5}
\end{aligned}$$
$$\begin{aligned}
& \Rightarrow S_1=\frac{3}{0.5}\left(2 \sin \left(40^{\circ}\right) \cos \left(20^{\circ}\right)-\sin \left(20^{\circ}\right)\right) \\ \\
& \Rightarrow S_1=6\left(\sin \left(60^{\circ}\right)\right)=3 \sqrt{3}
\end{aligned}$$
| Yet an other way to proceed, which arguably explains why we get a "nice value" for the sum. Let $c_1,c_2,c_3$ be the three different values of the cotangent function computed in $x_1,x_2,x_3$, which are respectively
$10^\circ$, $70^\circ$, $130^\circ$. These angles are of the shape $10^\circ+k\cdot 60^\circ$, so taking the cotangent of the triple value we get
$\cot 30^\circ=\sqrt3$. There is a formula for $\cot (3x)$ in terms of $c:=\cot x$ obtained from $\displaystyle\cot(x+y)=(\cot x\cot y-1)/(\cot x+\cot y)$, namely
$$
\cot(3x)=\frac{c^3-3c}{3c^2-1}=:f(c)\ .
$$
Plugging in $c_1,c_2,c_3$ into $f$ we thus obtain every time $\cot 30 ^\circ=\sqrt 3$. So $c_1,c_2,c_3$ are three (thus all) roots of the polynomial equation:
$$
c^3-3c=\sqrt3(3c^2-1)\ ,\qquad\text{ i.e. }\qquad
c^3-3\sqrt 3c^2-3c+\sqrt 3=0\ .
$$
From here we obtain immediately relations like:
$$
\begin{aligned}
\color{blue}{3\sqrt 3} &= \color{blue}{c_1+c_2+c_3}\ ,\\
-3 &= c_1c_2 + c_3c_3+c_3c_1\ ,\\
-\sqrt 3 &= c_1c_2c_3\ ,\\[3mm]
33 &=c_1^2+c_2^2+c_3^2\ ,\\
105\sqrt3 &=c_1^3+c_2^3+c_3^3\ .
\end{aligned}
$$
The blue relation is the needed one.
$\square$
Computer check:
sage: c1, c2, c3 = cot(pi/18), cot(7*pi/18), cot(13*pi/18)
....: (c1 + c2 + c3).minpoly()
....: (c1 + c2 + c3).numerical_approx()
x^2 - 27
5.19615242270663
So $c_1+c_2+c_3$ is the positive root of the polynomial $x^2-27$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the concrete mathematical argument why $a^3+a=b^3+b\Rightarrow a=b$ is true in $\mathbb{R}$? The question arose from the following proof:
$f(x)=x^3+x+1$ is injective:
Let $a,b\in \mathbb{R}$ and $f(a)=f(b)$:
$f(a)=f(b)\Leftrightarrow a^3+a+1=b^3+b+1\Leftrightarrow a^3+a=b^3+b \Rightarrow a=b$
Is $a^3+a=b^3+b\Rightarrow a=b$ mathematically rigorous enough? I don't know the concrete mathematical argument why $a^3+a=b^3+b\Rightarrow a=b$ is true?
| If $a^3+a=b^3+b$, then $$(a^3-b^3)+(a-b)=0 \quad \Leftrightarrow $$
$$(a-b)(a^2+ab+b^2+1)=0.$$
However, $a^2+ab+b^2+1=(a+\frac{b}{2})^2+\frac{3b^2}{4}+1>0$. This immediately implies that $a-b=0$, i.e. $a=b$. Please let me know does it make sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find the symmetric matrix given its eigenvalues and eigenvector.
$A$ is a $3 \times 3$ symmetric matrix. It has the eigenvalue $\lambda_1 = 3$ with the eigenvector
$$
\begin{bmatrix}
1 \\ 0 \\ -1
\end{bmatrix}
$$
and a double eigenvalue $\lambda_2 = -1$.
Find the matrix $A$.
Since $A$ is symmetric, it can be diagonalized orthogonally, with an orthogonal matrix $P$ as $A = PDP^T$. The matrix is made of the eigenvectors that are orthogonal to each other. What I then did was with the help of inner product, I got a vector that is orthogonal to $v_1$ and with the help of cross product I got another orthogonal vector. Those vectors are
$$
\begin{bmatrix}
1 \\ 1 \\ 1
\end{bmatrix}
\quad\text{and}\quad
\begin{bmatrix}
1 \\ -2 \\ 1
\end{bmatrix}
$$
so
$$
A = PDP^{-1}
= \begin{bmatrix}
1 & 1 & -5 \\
1 & -5 & 1 \\
-5 & 1 & -5
\end{bmatrix}.
$$
But my answer is wrong obviously since $Av \neq \lambda v$. The answer is
$$
\begin{bmatrix}
1 & 0 & -2 \\
0 & -1 & 0 \\
-2 & 0 & 1
\end{bmatrix}
$$
where the vectors are
$v_2 = \begin{bmatrix} 1 & 0 & 1 \end{bmatrix}^T$ and
$\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}^T$. The answer sheet says that these vectors are orthogonal to $v_1$. Also they didn't do $A = PDP^T$ instead they did $A = PDP^{-1}$. Can anyone help me understand?
| Since $3, -1, -1$ are all eigenvalues of $A$, $4, 0, 0$ are all eigenvalues of the matrix $A + I$, which means the rank of the symmetric matrix $A + I$ is $1$, hence $A + I = vv^T$ for some length-$3$ vector $v$.
Denote $\begin{bmatrix} 1 & 0 & -1 \end{bmatrix}^T$ by $w$, $Aw = 3w$ and $A + I = vv^T$ implies that
\begin{align}
Aw + w = 3w + w = 4w = vv^Tw,
\end{align}
i.e.,
\begin{align}
vv^Tw = 4w. \tag{1}
\end{align}
In other words:
\begin{align}
(v^Tw)v = 4w. \tag{2}
\end{align}
Since $w \neq 0$, $(2)$ implies that $v^Tw \neq 0$, hence $v$ is a multiple of $w$, say $v = \lambda w$. Multiply both sides in equation $(1)$ by $v^T$, we have
\begin{align}
v^Tvv^Tw = 4v^Tw, \text{ or } v^Tw(v^Tv - 4) = 0.
\end{align}
Since $v^Tw \neq 0$, it follows that $v^Tv = \lambda^2w^Tw = 4$. Because $w^Tw = 1^2 + 0^2 + (-1)^2 = 2$, it follows that $\lambda^2 = 4/2 = 2$. Therefore,
\begin{align}
& A = vv^T - I = \lambda^2ww^T - I = \\
=& 2\begin{bmatrix} 1 \\ 0 \\ - 1\end{bmatrix}\begin{bmatrix} 1 & 0 & - 1\end{bmatrix} - \operatorname{diag}(1, 1, 1) \\
=& \begin{bmatrix}
2 & 0 & -2 \\
0 & 0 & 0 \\
-2 & 0 & 2
\end{bmatrix} -
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} \\
=& \begin{bmatrix}
1 & 0 & -2 \\
0 & -1 & 0 \\
-2 & 0 & 1
\end{bmatrix}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$, prove $abcd \geq 3$. Given four positive real numbers $a,b,c,d$. It is given that $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$. Prove $abcd \geq 3$.
I've applied AM-HM inequality and got the result, $a^4+b^4+c^4+d^4 \geq 12$. Then by AM-GM, $a^4+b^4+c^4+d^4\geq 4abcd; 12\geq 4abcd \implies 3\geq abcd$.
But this is contradictory. Then I realized I was doing all wrong. Does anyone have method for this one?
| Let
$$w=\frac{1}{1+a^4}, x=\frac{1}{1+b^4}$$$$ y=\frac{1}{1+c^4}, z=\frac{1}{1+b^4}$$
So, $$a^4=\frac{1-w}{w}=\frac{x+y+z}{w}$$$$ b^4=\frac{1-x}{x}=\frac{w+y+z}{x}$$$$ c^4=\frac{1-y}{y}=\frac{w+x+z}{y}$$$$ d^4=\frac{1-z}{z}=\frac{w+x+y}{z}$$
$$a^4b^4c^4d^4=(\frac{x+y+z}{w}) (\frac{w+y+z}{x})( \frac{w+x+z}{y} )(\frac{w+x+y}{z})-(1)$$
Apply AM-GM for these four terms on RHS individually in eqation $(1)$,
RHS$\geq$$(\frac{3(xyz)^{\frac{1}{3}}}{w})$$(\frac{3(wyz)^{\frac{1}{3}}}{x})$$(\frac{3(wxz)^{\frac{1}{3}}}{y})$$(\frac{3(wxy)^{\frac{1}{3}}}{z})$
$$⇒RHS\geq 81$$
RHS is also equal to $a^4b^4c^4d^4$.
$$a^4b^4c^4d^4\geq81 ⇒ abcd\geq3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2- \frac{2}{\sqrt{n+1}}$
Let $i\in\mathbb{N}$ $$\sum_{i=1}^{n} \frac{1}{(i²+i)^\frac{3}{4}}>2-\frac{2}{\sqrt{n+1}}$$
A.M.-G.M. inequality makes this problem complex to solve.
It was true for $n=1,2,3,4$ then solving this became difficult. How can I solve this?
| Note that since
$$
(i^2 + i)^{\frac34} = (i^2 + i)^{\frac12} \cdot i^\frac14 \cdot (i+1)^\frac14 < \sqrt{i} \cdot \sqrt{i+1} \cdot \frac{\sqrt{i} + \sqrt{i+1} }{2} = \frac12 \cdot \frac{\sqrt{i} \cdot \sqrt{i+1}}{\sqrt{i+1} - \sqrt{i}},
$$
we have
$$
\frac{1}{(i^2 + i)^{\frac34}} > \frac{2(\sqrt{i+1} - \sqrt{i})}{\sqrt{i}\cdot\sqrt{i+1} \cdot} = 2 \left(\frac{1}{\sqrt{i}} - \frac{1}{\sqrt{i+1}}\right).
$$
Therefore,
$$
\sum_{i = 1}^n \frac{1}{(i^2 + i)^{\frac34}} > 2\left(1 - \frac{1}{\sqrt{n+1}}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.