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How can I prove/disprove that if n is an integer which is not divisible by 3, then $n^3+n^2+2n+1 ≡ 2 \pmod 3$? By doing a few examples, it seems as if the statement is true. However, I'm having trouble proving (or disproving) it. I know that the numerical value of some $N$ can be written as $n_{0}10^{0}+n_{1}10^{1}+n_{2}10^{2}+\dotsb$ and $10\equiv 1 \pmod{3}$, but this just shows the divisibility rule for $3$. Can I use this somehow?
| There are three options $n \equiv 0 \mod 3$ (which you are told is not true) or $n \equiv 1 \mod 3$ or $n\equiv -1 \mod 3$.
So $n = 3m \pm 1$ for some integer $m$.
So $n^3 + n^2 + 2n + 1= (3m\pm 1)^3 + (3m\pm 1)^2 + 2(3m \pm 1) + 1$
$= 27m^3 \pm 27 m^2 + 9m \pm 1 + 9m^2 \pm 6m + 1 + 6m \pm 2 + 1=$
$3K \pm 3 + 2 \equiv 1 \mod 3$ where $K = $... one third of the sum of all those terms that are multiples of three.
.... or more sophisticatedly.
$n \equiv \pm 1 \mod 3$
So $n^2 \equiv (\pm 1)^2 \equiv 1 \mod 3$
$n^3 \equiv (\pm 1)^3 \equiv - 1 \mod 3$
So $n^3 + n^2 + 2n + 1 \equiv \pm 1 + 1 \pm 2 +1\equiv \pm 3 + 2\equiv 2 \mod 3$.
| {
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Complex modulus of $\left|\frac{-3z+2i}{2iz+1}\right|$ given that $\left|z\right|=\frac{1}{\sqrt3}$ The following question was on my first year algebra exam way back in 1989.
If $\left|z\right|=\frac{1}{\sqrt3}$, then find $\left|\frac{-3z+2i}{2iz+1}\right|$.
I couldn't figure it out then, and 28 years on, I still can't. It was only worth 4 marks, so the solution must be simpler than all the things I have tried over the years to no avail.
| Since $ \mid Z \mid = \dfrac{1}{\sqrt{3}}$, considering $Z > 0$ then it follows that $Z = \dfrac{1}{\sqrt{3}}$. \ \
Now by the absolute value properties $ | \dfrac{-3z + 2i}{2iz + 1} | = \dfrac{| -3z + 2i |}{| 2iz + 1 |} $ \ \
Since $| -3z +2i | = \sqrt{(-3z+2i) (-3z - 2i)} = \sqrt{9z^2 +4} = \sqrt{9( \dfrac{1}{ \sqrt{3}})^2 + 4} = \sqrt{7}$
and $|2iz + 1 | = \sqrt{(1 +2iz)(1 - 2iz)} = \sqrt{1 + 4z^2} = \sqrt{\dfrac{7}{3}} $ \ \
Hence $ | \dfrac{ -3z + 2i}{2iz + 1} | = \dfrac{| -3z + 2i|}{|2iz + 1|} = \dfrac{\sqrt{7}}{\sqrt{\dfrac{7}{3}}} = \sqrt{3} $
| {
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Find all functions $f:\mathbb R \rightarrow \mathbb R$, such that: $(x^2 − y^2)\cdot f(xy) = x\cdot f(x^2y) − y\cdot f(xy^2)$ Find all functions $f:\mathbb R \rightarrow \mathbb R$, such that:
$$(x^2 − y^2)\cdot f(xy) = x\cdot f(x^2y) − y\cdot f(xy^2)$$
for all $x,y \in \mathbb R$
My work so far:
1) $f(0)=0$
2) $y=1; y=\frac1x; y=-\frac1x; y=kx$
| The assumption of continuity at $ x = 1 $ is not necessary. As @Hyperplane has shown, we know that:
*
*$ f ( 0 ) = 0 $.
*For $ x \ne 0 $ we have $ f ( - x ) = - f ( x ) $ and thus it suffices to find $ f ( x ) $ for $ x > 0 $.
*$ f \big( x ^ 2 \big) = x f ( x ) $.
Now let $ g ( x ) = f ( x ) - a x $ where $ a $ is a constant real number. Using the original functional equation, it's easy to see that $ g $ satisfies the same equation, i.e.
$$ \big( x ^ 2 - y ^ 2 \big) g ( x y ) = x g \big( x ^ 2 y \big) - y g \big( x y ^ 2 \big) \text . \tag 0 \label 0 $$
Thus $ g $ satisfies the three properties above that hold for $ f $. We can also let $ a = f ( 1 ) $ and have the additional property $ g ( 1 ) = 0 $. Now multiplying \eqref{0} by $ x y $ and using the third property, we get
$$ \big( x ^ 2 - y ^ 2 \big) g \big( x ^ 2 y ^ 2 \big) = g \big( x ^ 4 y ^ 2 \big) - g \big( x ^ 2 y ^ 4 \big) $$
and hence for $ x , y > 0 $ we have
$$ ( x - y ) g ( x y ) = g \big( x ^ 2 y \big) - g \big( x y ^ 2 \big) \text . \tag 1 \label 1 $$
Now letting $ y = \frac 1 x $ in \eqref{1} we get
$$ g \bigg( \frac 1 x \bigg) = g ( x ) \text . \tag 2 \label 2 $$
Again, letting $ y = \frac 1 { x ^ 2 } $ in \eqref{1} and using \eqref{2}, we have
$$ \bigg( x - \frac 1 { x ^ 2 } \bigg) g ( x ) = - g \big( x ^ 3 \big) \tag 3 \label 3 $$
and substituting $ \frac 1 x $ for $ x $ in \eqref{3} and using \eqref{2} we also have
$$ \bigg( \frac 1 x - x ^ 2 \bigg) g ( x ) = - g \big( x ^ 3 \big) \text . \tag 4 \label 4 $$
Subtracting \eqref{3} from \eqref{4} yields
$$ \bigg( x + \frac 1 x + 1 \bigg) \bigg( x - \frac 1 x \bigg) g ( x ) = 0 $$
which shows $ g ( x ) = 0 $ for $ 1 \ne x > 0 $. Therefore $ g $ is the constant zero function and thus $ f ( x ) = a x $.
| {
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Find the value of minimize $a+b$ If $a,b$ are positive integers and $$\frac{1}{2a}+\frac{1}{3a}+\frac{1}{4a}=\frac{1}{b^2-2b}$$. Find the value of minimize $a+b$
From $\frac{1}{2a}+\frac{1}{3a}+\frac{1}{4a}=\frac{1}{b^2-2b}\Leftrightarrow13b^2-26b-12a=0$
$\Leftrightarrow 12(a+b)=13b^2-14b$
$\Leftrightarrow a+b=\frac{13b^2-14b}{12}$
$\Leftrightarrow a+b=b^2-b+\frac{b^2-2b}{12}=b^2-b+\frac{b\left(b-2\right)}{12}$
We have: $b$ must even number $(1)$
For $\frac{b\left(b-2\right)}{2.2.3}\in Z\Rightarrow b⋮3$ or $b-2⋮3$ $(2)$
From $(1);(2)$ $\Rightarrow b=6k$ or $b-2=6k$ $(k\ge1)$
*)With $b=6k\Rightarrow a+b=\frac{13\left(6k\right)^2-14\cdot 6k}{12}=3k^2-7k$
We have function $f(k)=39k^2-7k$ is function covariate withg $k\ge1$
So $a+b$ have minimize when k have minimize or $k=1$ $\Rightarrow b=6;a=26$
*)With $b-2=6k$.....
this method is inconvenient with me, so i can a new method
| Hint:
$12a/13=b(b-2)$
$\implies13|a, a=13c$(say)
$(b-2)b=12c\implies$
$(i) b(b-2)\ge12\implies b>4$
$(ii)b$ must be even $b=2d, d>2$
$d(d-1)=3c$
Now $(d,d-1)=1$
| {
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Prove that $\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2} \text{ when } abc=1$ So I have a possible proof but I'm not certain it's right:
$\text{As } \frac{a_1}{x_1} + \frac{a_2}{x_2} + \frac{a_3}{x_3} \geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}\text{, we obtain } \frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{9}{a^3(b+c) + b^3(a+c) + c^3(a+b)}$
$\begin{align}
a^3(b+c) + b^3(a+c) + c^3(a+b) & = a^2(\frac{1}{b} + \frac{1}{c}) + b^2(\frac{1}{a} + \frac{1}{c}) c^2(\frac{1}{a} + \frac{1}{b}) \\
& \geq \frac{4a^2}{b+c} + \frac{4b^2}{a+c} + \frac{4c^2}{a+b} \text{ by the same inequality as above} \\
& \geq 4\frac{(a+b+c)^2}{2(a+b+c)} \text{ (same reason)} \\
& \geq 2(a+b+c)
\end{align}$
Now, we have:
$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{9}{2(a+b+c)}$
I'm not sure if this is a logical progression to make though. Because if $a \geq b$ then $\frac{1}{a} \leq \frac{1}{b}$, so is this not true? I feel like it must be as I can get the desired result, but I'm confused about it.
However, if this is true it is then sufficient to prove that $a+b+c \geq 3$.
$a+b+c = a+b + \frac{1}{ab} \geq 2\sqrt{ab} + \frac{1}{ab}$
It can be proved with calculus that $\forall x\ge 0,f(x) = 2\sqrt{x} + \frac{1}{x} \geq 3$. Thus the inequality is true, and the proof is finished.
Can someone please explain either why my proof is true (namely what I am confused about) or why this false proof yields the right result?
| I think your first step gives a wrong inequality.
For the proof we can make the following thinks.
By Holder and AM-GM we obtain:
$$\sum_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{b^3c^3}{b+c}\geq\frac{(ab+ac+bc)^3}{3\sum\limits_{cyc}(b+c)}=$$
$$=\frac{(ab+ac+bc)(ab+ac+bc)^2}{6(a+b+c)}\geq\frac{3(ab+ac+bc)^2}{6(a+b+c)}=\frac{(ab+ac+bc)^2}{2abc(a+b+c)}\geq\frac{3}{2},$$
where the last inequality it's just $\sum\limits_{cyc}c^2(a-b)^2\geq0$.
Done!
| {
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Solution of differential equation which is quadratic in $\frac{dy}{dx}$ Consider the following differential equation
$$ x\frac{dy}{dx} + y = x^4(\frac{dy}{dx})^2$$
I used the quadratic formula and got $\frac{dy}{dx} = \frac{x \pm \sqrt {x^2 + 4yx^4}}{2x^4}$. Now how to proceed or is there any other method?
Edit:
I proceeded further this way
$\frac{dy}{dx} = \frac{1 \pm \sqrt {1 +4yx^2}}{2x^3}$
Let $1 + 4yx^4 = t^2$
then $8yx + 4x^2\frac{dy}{dx} = 2t\frac{dt}{dx}$
multiplying throughout with $x$ I got $4yx^3 +2x^3\frac{dy}{dx} = tx\frac{dt}{dx}$
now putting the value of $2x^3\frac{dy}{dx}$ and $t^2$ in 1st equation I got
$tx\frac{dt}{dx} - t^2 = \pm t$
which gave 2 solutions $$t = 0$$ $$and$$ $$x\frac{dt}{dx} = t \pm 1$$
Using $t = 0$ I got one solution $$1 + 4x^2y = 0$$ and the other by solving the differential equation $x\frac{dt}{dx} = t \pm 1$ as $$c^2x^2 + 4y^2x^2 = 4cyx^2 + 2c$$
Now the answer to this question which I have got is $$4x^2y + 1 = 0$$ $$and$$ $$xy - c^2x + c = 0$$
Why I couldn't get the second solution right?
| Let $v = 4yx^2$ then we can write
$$
\frac{d}{dx}\frac{v}{4x^2} = \frac{x\pm \sqrt{x^2+vx^2}}{2x^4} = \frac{1\pm\sqrt{1+v}}{2x^3} = \frac{1}{4x^2}v' -\frac{1}{2x^3}v =\frac{1\pm\sqrt{1+v}}{2x^3}
$$
we can re-arrange
$$
xv'-2v = 2\pm\sqrt{1+v} \implies xv' = 2(v+1)\pm 2\sqrt{1+v}
$$
we can clean up
$$
xu ' = 2u\pm 2\sqrt{u}
$$
where $u=v+1$.
then set $u=t^2$ to find
$$
x2tt' = 2t^2\pm 2t\implies xt' = t \pm 1
$$
can you take it from here and transform back to your original variables?
| {
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Evaluate $\sum_{k=0}^{n} {3n \choose 3k}$ Evaluate $\sum_{k=0}^{n} {3n \choose 3k}$.
It's not hard to evaluate by putting roots of unity.
Therefore, I would like to see some solutions using elementary mathematical stuffs but not roots of unity.
And the answer is $\frac{1}{3}(2^{3n}+2(-1)^n)$
| First, observe that
$$\sum\limits_{k=0}^n \binom{3n}{3k} = \sum\limits_{k=0}^n \binom{3n-3}{3k} + 3\binom{3n-3}{3k-1} + 3\binom{3n-3}{3k-2} + \binom{3n-3}{3k-3}$$
by repeated expansion using Pascal's recurrence. Also, due to the limits on $k$ and the fact that $\binom{n}{m} = 0$ whenever $m \leq 0$ or $m > n$, we can write this
$$\sum\limits_{k=0}^n \binom{3n}{3k} = 3\sum\limits_{k=0}^n \bigg{(} \binom{3n-3}{3k-1} + \binom{3n-3}{3k-2} \bigg{)} + 2\sum\limits_{k=0}^n \binom{3n-3}{3k-3}$$
Now, let $F_n = \sum\limits_{k=0}^n \binom{3n}{3k}$. We use this substitution and the fact that the given sum over two binomial coefficients spans all binomial coefficients except those with bottom part a multiple of 3 to obtain;
$$F_n = 3\bigg{(} 2^{3n-3} - F_{n-1} \bigg{)} + 2F_{n-1} \implies F_n + F_{n-1} = 3*2^{3n-3}$$
To solve a linear nonhomogeneous recurrence like this, we let $F_n = h_n + p_n$ where $h_n$ is the homogeneous solution and $p_n$ is a particular solution to the nonhomogeneous form.
Solving $h_n + h_{n-1} = 0$ yields $h_n = A*(-1)^n$. We assume $p_n$ is of the form $p_n = B*2^{3n}$ and we solve;
$$p_n + p_{n-1} = 3*2^{3n-3} \implies B2^{3n-3}(2^{3} + 1) = 3*2^{3n-3}$$
which simplifies to $B = \dfrac{1}{3}$, so that $p_n = \dfrac{1}{3} 2^{3n}$.
Using an initial value to solve for $A$ suffices to prove the same formula the author claims.
| {
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How can I simplify $\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$? This problem got me thinking hard:
$$\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$$
Considering the fact that the common base is '$3$', I decided to do this:
$$\log_3\left(\frac{(n+1)(n-1)}{(n+2)(n)}\right)$$
$$\log_3\left(\frac{n^2-n+n+1}{n^2+2n}\right)$$
$$\log_3\left(\frac{n^2 + 1}{n^2 + 2n}\right)$$
$$\log_3\left(\frac{1}{2n}\right)$$
I am not 100% sure if have simplified the problem. I need some clarification.
| $$\frac{3^{n+1}+3^{n-1}}{3^{n+2}+3^n}=\frac{3^{n-1}(3^2+1)}{3^{n-1}(3^3+3)}=\frac{10}{30}=\frac 13$$
This shows your simplification is incorrect for a number of reasons:
*
*$\log(3^{n+1}+3^{n-1})\neq \log(3^{n+1})\log(3^{n-1})$. The $\log$ rule works the other way: $\log(a)+\log(b)=\log(ab)$
*You cannot simplify $\frac{n^2+1}{n^2+2n}$ to $\frac{1}{2n}$: big no-no. You can simplify multiplied expressions, not $summed$ expressions. For instance $\frac{n(n+1)}{(n+3)(n+1)}=\frac{n}{n+3}$ is fine.
| {
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Show that for any integer $n\ge5$, the inequality ${2n \choose n}>3^n$ holds. I used induction, but got stuck at the last step. By cancelling common factors in the expanded $2n$ choose $n$, I got a simpler equation that didn't make sense.How do i prove this inequality?
| The main step is to prove for $n+1$ then
$${2(n+1)\choose n+1}=\frac{(2n+2)!}{(n+1)!(n+1)!}=\frac{(2n+2)(2n+1)}{(n+1)^2}{2n\choose n}=\frac{2(2n+1)}{(n+1)}{2n\choose n}$$
Using the hypothesis
$${2(n+1)\choose n+1}>\frac{2(2n+1)}{(n+1)}\cdot 3^n$$
but
$$\frac{2(2n+1)}{n+1}=\frac{4n+2}{n+1}=\frac{3(n+1)+(n-1)}{n+1}=3+\frac{n-1}{n+1}>3$$
so,
$${2(n+1)\choose n+1}>\frac{2(2n+1)}{(n+1)}\cdot 3^n>3\cdot 3^n=3^{n+1}$$
| {
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Problem with calculation this integral: $\int_0^\pi \frac{dx}{1+3\sin^2x}$ Question
Calculate this integral: $$\displaystyle\int_0^\pi \frac{dx}{1+3\sin^2x}$$
Solution
$$I=\displaystyle\int \frac{dx}{1+3\sin^2x}=\displaystyle\int \frac{dx}{\cos^2x+4\sin^2x}=\displaystyle\int \frac{\sec^2x\;dx}{1+4\tan^2x}$$ Let's apply substitution $u=2\tan x$, so $du=2\sec^2x\;dx$ $$I=\dfrac12\displaystyle\int \frac{du}{1+u^2}=\frac12\arctan(u)+c=\frac12\arctan(2\tan x)+c$$ In this case; $$\displaystyle\int_0^\pi \frac{dx}{1+3\sin^2x}=\left[\frac12\arctan(2\tan x)\right]_{x=0}^{x=\pi}=0$$.
But we know that: $$\frac{1}{1+3\sin^2 x} \ge \frac14$$ Therefore; $$\displaystyle\int_0^\pi \frac{dx}{1+3\sin^2x} \ge \displaystyle\int_0^\pi \frac{dx}{4}=\frac{\pi}4$$ On the other hand; $$0 \ge \frac{\pi}4$$ which is not true.
1) Where is the problem? (Why?)
2) How we can correct this mistake by applying the same substitution?
| HINT: write
$$
\int_0^{\pi}\frac{dx}{1+3\sin^2x}=\frac{1}{2}\int_0^{\pi/2}\frac{dx}{1+3\sin^2x}.
$$
Then, the substitution you suggest is legitimate because $\tan x$ is monotone on $(0,\pi/2)$.
| {
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Prove that for every prime number $p>3$, $4p^2+1$ can be written as the sum of three square numbers Given that $p>3$ prove that $4p^2+1$ can be written as the sum of three distinct positive square numbers.
Plugging in $5$ I get $101=49+36+16=7^2+6^2+4^2$
I also know that all primes greater than $3$ can be written in the form $3k+1$ and $3k+2$ but plugging those values in I get:
$36k^2+24k+5$, $36k^2+48k+17$ and the solution probably lies in arranging these numbers in such a way that we get the desired squares, but I can't come up with a combination ,or is my "idea" not even in the right direction?
| If $p=6k\pm 1$ then $4p^2+1 = 144k^2\pm 48k +5$.
So you might want to write $$4p^2+1 = (ak+1)^2+(bk+2)^2+(ck)^2=(a^2+b^2+c^2)k^2+(2a+4b)k+5$$
So you need to find $a,b,c$ so that $a^2+b^2+c^2=144, a+2b=\pm 24$, and $c\neq 0$.
I'll leave it to you to solve for $a,b,c$.
But you might get a quess of what they are by looking at the case $p=7$ along with your solution for $p=5:$
$$4\cdot 7^2+1 = 9^2+10^2+4^4$$
| {
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Calculus - Infinite Series Decide whether the infinite series converges of diverges.
$$\sum_{n=1}^\infty \frac{2^n+3^n}{3^n+4^n}$$
My thought process:
The nth term test doesn't seem viable after initial use.
The comparison function I derived for DCT/LCT is: $\frac{3^n}{4^n}$.
I cannot create an appropriate inequality between the given function and my comparison function I found, so I tried LCT. I am aware that
$\sum_{i=1}^\infty \frac{3^n}{4^n}$ is geometric and because the common ratio (r) is $ \frac{3}{4}$, $-1<r<1$ the series converges, however I cannot compute the limit LCT requires.
Application of the root test would seem unbeneficial because roots do not split over sums.
As for the ratio test I was also unsuccessful in proceeding with the computation of the limit.
The above are the only tests I have learned as of right now. My question is what test should I be looking to utilize to determine convergence or divergence?
| Besides the already provided solutions, I'd like to point out that both the LCT and the Root Test do work here too.
If we're applying the LCT to $\sum_{n=1}^\infty \frac{2^n+3^n}{3^n+4^n}$ and $\sum_{n=1}^\infty \frac{3^n}{4^n}=\sum_{n=1}^\infty \left(\frac{3}{4}\right)^n$:
$$\frac{\frac{2^n+3^n}{3^n+4^n}}{\frac{3^n}{4^n}}=\frac{2^n+3^n}{3^n+4^n}\cdot\frac{4^n}{3^n}=\frac{8^n+12^n}{9^n+12^n}=\frac{(8^n+12^n)\color{red}{\div12^n}}{(9^n+12^n)\color{red}{\div12^n}}=\frac{\left(\frac{8}{12}\right)^n+1}{\left(\frac{9}{12}\right)^n+1}\to\frac{0+1}{0+1}=1,$$
so the LCT applies to these two series and they behave the same way.
A similar trick works with the Root Test too. We can factor our $3^n$ and $4^n$ in the numerator and the denominator, respectively, to find that
$$\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}=\sqrt[n]{\frac{\color{red}{3^n\cdot}\left(\left(\frac{2}{3}\right)^n+1\right)}{\color{red}{4^n\cdot}\left(\left(\frac{3}{4}\right)^n+1\right)}}=\color{red}{\frac{3}{4}\cdot}\sqrt[n]{\frac{\left(\frac{2}{3}\right)^n+1}{\left(\frac{3}{4}\right)^n+1}}\to\frac{3}{4}\cdot1=\frac{3}{4}.$$
UPDATE: I've just noticed that the OP mentioned the Ratio Test too. I'm not going to type one more solution, but the limit of the fraction in the Ratio Test can be simplified here using a trick similar to the one used in the LCT calculation above.
| {
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} |
If $p\mid (3^n+1)$, then $p\equiv 1\pmod{3}$
Show that if $p> 2$ is a prime, $n > 1$ is odd and $p\mid (3^n+1)$,
then $p\equiv 1\pmod{3}$.
Since $n$ is odd, we have $3^{n+1} \equiv -3 \pmod{p}$ is a quadratic residue. Then I thought about using Quadratic Reciprocity but didn't see how to apply it.
We have $x^2 \equiv -3 \pmod{p}$ for some integer $x$. By Fermat's Little Theorem we have $$x^{p-1} \equiv x^2 \cdot x^{p-3} \equiv -3 \cdot x^{p-3} \equiv 1 \pmod{p}.$$ Thus, $x^{p-3} \equiv -3^{-1} \pmod{p}$.
I didn't see how to get a contradiction if $p \equiv 2 \pmod{3}$.
| You have shown that $(\frac{-3}{p}) = 1$ in this case. The truth is, in fact, that $(\frac{-3}{p}) = 1 \iff p \equiv 1 \pmod{3}$, which is what you are trying to prove.
$(\frac{-3}{p}) = (\frac{-1}{p})(\frac{3}{p})$ due to multiplicity of Legendre symbol.
From Euler's criterion, $(\frac{-1}{p}) = (-1)^{\frac{p-1}{2}}$.
From QR, $(\frac{3}{p}) = (-1)^{\frac{p-1}{2}}(\frac{p}{3})$.
Hence $(\frac{-3}{p}) = (-1)^{p-1}(\frac{p}{3}) = (\frac{p}{3}) = 1 \iff p \equiv 1 \pmod{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Show that $\int_{-\infty}^{+\infty}\left|\frac{\sin x}{x}\right|\mathrm dx$ diverges to infinity Show that integration of $$\int_{-\infty}^{+\infty}\left|\dfrac{\sin x}{x}\right|\mathrm dx$$ is equal to infinity.
Is the limit $$\lim_{x\to\infty}\frac{\sin x}{x} =1$$ useful here?
| Let $$\mathcal{A}=\left\{ x\in\mathbb{R}_{>0}:|\sin x| \ge \frac{1}{2} \right\} \subset \bigcup_{n\in\mathbb{N}}\left[ \pi n+ \frac{\pi }{6} , \pi n+ \frac{5 \pi }{6}\right]$$ then $$\int_{\mathbb{R}}\left| \frac{\sin x}{x} \right| \text{d}x \ge \int_{\mathbb{R}_{>0}}\left| \frac{\sin x}{x} \right| \text{d}x \ge \int_{\mathcal{A}}\left| \frac{\sin x}{x} \right| \text{d}x \ge \frac{1}{2} \int_{\mathcal{A}} \frac{1}{x}\text{d}x \ge \frac{1}{2} \sum_{n\in\mathbb{N}} \int_{\pi n+ \frac{\pi }{6}}^{\pi n+ \frac{5\pi }{6}} \frac{1}{x}\text{d}x $$
but
$$ \frac{1}{2} \sum_{n\in\mathbb{N}} \int_{\pi n+ \frac{\pi }{6}}^{\pi n+ \frac{5\pi }{6}} \frac{1}{x}\text{d}x=\frac{1}{2}\sum_{n\in\mathbb{N}} \ln \left( \frac{\pi n+ \frac{5\pi }{6}}{\pi n+ \frac{\pi }{6}} \right)$$
and $$\ln \left( \frac{\pi n+ \frac{5\pi }{6}}{\pi n+ \frac{\pi }{6}} \right)\sim \frac{\pi}{\pi n+ \frac{\pi }{6}}$$
and $$\sum_{n\in\mathbb{N}} \frac{\pi}{\pi n+ \frac{\pi }{6}}$$
is divergent so the integral too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2234138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Summation of alternating series, Mercator series: $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}$ I am struggling with solving sum of this alternate series:
$$
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}\
$$
I know that:
$$
\log(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \cdot x^n\
$$
But It seems that I can't find a way to get to this form.
Thanks.
| It is well-known fact that
$${ \sum _{ n=1 }^{ \infty }{ \frac { (-1)^{ n+1 } }{ n } } =\ln { 2 } . }$$
So
$$\sum _{ n=1 }^{ \infty } \frac { (-1)^{ n+1 } }{ n(n+1) } =\sum _{ n=1 }^{ \infty } \left( \frac { (-1)^{ n+1 } }{ n } -\frac { (-1)^{ n+1 } }{ n+1 } \right) =\sum _{ n=1 }^{ \infty } \frac { (-1)^{ n+1 } }{ n } -\sum _{ n=1 }^{ \infty } \frac { (-1)^{ n+1 } }{ n+1 } =\left( 1-\frac { 1 }{ 2 } +\frac { 1 }{ 3 } -\frac { 1 }{ 4 } +.. \right) -\left( \frac { 1 }{ 2 } -\frac { 1 }{ 3 } +\frac { 1 }{ 4 } +.. \right) =\\ =\ln { 2 } -\left( -\ln { 2 } +1 \right) =2\ln { 2 } -1\\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral of alternating series I was trying to answer a question about a random walk when I came across the integral
$$
\int_0^\infty \sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\left(1-e^{-(2m+1)^2/x^2}\right)\mathrm{d}x.
$$
For probabilistic reasons, I think it has a finite value. Is there a simple proof of this? Is there a way to compute or simplify the expression? If one could exchange the $\int$ with the $\sum$, then one could use that
$$\int_0^\infty 1-e^{-(2m + 1)^2 / x^2}\mathrm{d}x = (2m+1)\sqrt{\pi}.$$
Thanks.
| The integral converges to $\sqrt{\pi}/2$. Indeed, let
$$ F(x) = \int_{0}^{x} (1- e^{-1/t^2}) \, dt. $$
By the Abel's test, the series
$$ \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) $$
converges uniformly on $[0,\infty)$ (with the convention $e^{-\infty} = 0$). So we can switch the integration and summation in the following computation:
\begin{align*}
I_R
&:= \int_{0}^{R} \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) \, dx \\
&\hspace{3em}= \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} \int_{0}^{R} (1 - e^{-(2m+1)^2/x^2}) \, dx\\
&\hspace{6em}= \sum_{m=0}^{\infty} (-1)^m F\left(\frac{R}{2m+1}\right).
\end{align*}
Proceeding,
\begin{align*}
I_R
&= \sum_{m=0}^{\infty} \left\{ F\left(\frac{R}{4m+1}\right) - F\left(\frac{R}{4m+3}\right) \right\} \\
&\hspace{3em}= \sum_{m=0}^{\infty} \int_{\frac{R}{4m+3}}^{\frac{R}{4m+1}} (1 - e^{-1/t^2}) \, dt \\
&\hspace{6em}= \sum_{m=0}^{\infty} \int_{\frac{4m+1}{R}}^{\frac{4m+3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx,
\end{align*}
where we applied the substitution $x = 1/t$ in the last line. (This substitution is not essential for our argument, but I adopted this step to make clear how monotonicity works.)
Now using the fact that the integrand is decreasing, we can bound $2I_R$ from below by
$$ 2I_R
\geq \sum_{m=0}^{\infty} \left( \int_{\frac{4m+1}{R}}^{\frac{4m+3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx + \int_{\frac{4m+3}{R}}^{\frac{4m+5}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx \right)
= \int_{\frac{1}{R}}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx. $$
Similar idea shows that
$$ 2I_R \leq \int_{\frac{1}{R}}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx + \int_{\frac{1}{R}}^{\frac{3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx. $$
Finally, taking $R \to \infty$ proves
$$ \int_{0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) \, dx
= \lim_{R\to\infty} I_R = \frac{1}{2} \int_{0}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx = \frac{\sqrt{\pi}}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
How to integrate this : $\int \frac{x^2+\cos^2 x}{1+x^2}\csc^2xdx$ How to integrate :
$\int \frac{x^2+\cos^2x}{1+x^2}\csc^2xdx$
I am not getting any clue how to move further, request you to please provide hint, will be of great help.
Thanks..
| Before integrating, let's do some changes
$$\frac{(x^2+\cos^2 x)\csc^2x}{1+x^2}=\frac{x^2\csc^2x+\cos^2 x\csc^2x}{1+x^2}=\frac{x^2\csc^2x+\cot^2 x}{1+x^2}=\frac{x^2\csc^2x+\csc^2 x-1}{1+x^2}=\frac{-1+(1+x^2)\csc^2x}{1+x^2}=-\frac{1}{1+x^2}+\csc^2x.$$
But
$$\int -\frac{1}{1+x^2}dx=-\arctan x+C$$
and
$$\int \csc^2xdx=-\cot x+C.$$
So
$$\int \frac{(x^2+\cos^2 x)\csc^2x}{1+x^2}dx=-\arctan x-\cot x+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to integrate $\int_0^{2\pi} \frac{dx}{(2+\cos(x))(3+\cos(x))}$ The integral is
$$\int_0^{2\pi} \frac{dx}{(2+\cos(x))(3+\cos(x))}$$
I did try Tangent half-angle substitution but only got:
$$2\int^{2\pi}_0 \frac{dt}{9t^2-t^4-12}$$
which does not seem to be easier... am I wrong on the way? Or is there easier way?
| You can see that $\cos (2\pi - x) = \cos x$ and hence the integral from $0$ to $2\pi$ is twice the integral from $0$ to $\pi$. Thus we have
\begin{align}
I &= 2\int_{0}^{\pi}\frac{dx}{(2 + \cos x)(3 + \cos x)}\notag\\
&= 2\int_{0}^{\pi}\frac{dx}{2 + \cos x} - 2\int_{0}^{\pi}\frac{dx}{3 + \cos x}\notag\\
&= 2\frac{\pi}{\sqrt{4 - 1}} - 2\frac{\pi}{\sqrt{9 - 1}}\notag\\
&= 2\pi\left(\frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{2}}\right)\notag
\end{align}
The integral $$\int_{0}^{\pi}\frac{dx}{a + b\cos x}$$ for $a > |b|$ is calculated using the substitution $$(a + b\cos x)(a - b\cos y) = a^{2} - b^{2}$$ and it easily evaluates to $\pi/\sqrt{a^{2} - b^{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
If $(A-B)^2=AB$, prove that $\det(AB-BA)=0$.
Let $A,B\in M_{n}(\mathbb{Q})$. If $(A-B)^2=AB$, prove that $\det(AB-BA)=0$.
I considered the function $f:\mathbb{Q}\rightarrow \mathbb{Q}$, $f(x)=\det(A^2+B^2-BA-xAB)$ and I obtained that:
$$f(0)=\det(A^2+B^2-BA)=\det(2AB)=2^n\det(AB)$$
$$f(1)=\det(A^2+B^2-BA-AB)=\det((A-B)^2)=\det(AB)$$
$$f(2)=\det(A^2+B^2-BA-2AB)=\det((A-B)^2-AB)=\det(AB-AB)=0$$
I don't have any other idea.
| The idea is to use $(-3\pm\sqrt 5)/2$. We begin with
$$\begin{align}
(A+xB)(A+x'B) &= A^2 + x BA + x' AB + xx' B^2 \\ &= A^2 + (x' + x) AB+x(BA-AB)+xx'B^2.\end{align}
$$
Now we find numbers $x$, $x'$ such that $x'+x = -3$, $xx'=1$.
These numbers are the roots of the quadratic equation
$$
\lambda^2 +3\lambda +1 = 0.
$$
Thus, we have $x = \frac{-3+\sqrt 5}2$ and $x'= \frac{-3-\sqrt 5}2$.
With these, we have
$$\begin{align}
(A+xB)(A+x'B) &= A^2 -3AB + B^2 + x(BA-AB) \\ &= BA-AB + x(BA-AB) = (BA-AB)(1+x).\end{align}
$$
We now have that
$$
\det(A+xB)(A+x'B) =q \in \mathbb{Q},
$$
and
$$
q=(1+x)^n \det(BA-AB).
$$
Then it follows by $A, B\in M_n(\mathbb{Q})$ and $(1+x)^n\in\mathbb{R} \backslash \mathbb{Q}$ that $\det(BA-AB)=0$.
An example with $AB\neq BA$
For the example that @Jonas Meyer requested, here it is:
$$
A=\begin{pmatrix} 1 & 1 \\ 0 & \frac{-3+\sqrt 5}2\end{pmatrix}, \ \ B=\begin{pmatrix} \frac{3+\sqrt 5}2 & 1 \\ 0 & -1\end{pmatrix}.
$$
Then
$$
A-B= \begin{pmatrix} \frac{-1-\sqrt 5}2 & 0 \\ 0 & \frac{-1+\sqrt 5}2\end{pmatrix},$$
$$
(A-B)^2 = \begin{pmatrix} \frac{3+\sqrt 5}2 & 0 \\ 0 & \frac{3-\sqrt 5}2\end{pmatrix} = AB.
$$
But,
$$
BA =\begin{pmatrix} \frac{3+\sqrt 5}2 & \sqrt 5 \\ 0 & \frac{3-\sqrt 5}2\end{pmatrix} \neq AB.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 3,
"answer_id": 1
} |
Find the sum of the following series: $ \frac{1}{1 \cdot 2}\,+\,\frac{1}{2 \cdot 3}\,+\,\frac{1}{3 \cdot 4} +\cdots\,+\frac{1}{100 \cdot 101}$
Find the sum of the following series
$$ \frac{1}{1 \cdot 2}\,+\,\frac{1}{2 \cdot 3}\,+\,\frac{1}{3 \cdot 4} +\cdots\,+\frac{1}{100 \cdot 101}$$
My Attempt:
$$\frac{1}{2 \cdot 3} =\frac{1}{2}-\frac{1}{3}.$$
So we can write question as:
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\\
=\frac{1}{1}-\frac{1}{101}\\
=\frac{100}{101}.$$
Am i right?
| _Sum up of everybody else's comments: _Yes, it is correct. The only thing I'd say is that as Theophile said, you need to make your prove more convincing, and back up every single hole in. A proof is never at an optimal proving state, but you can maximise this, by filling in every single statement you make with a law or rule, or in other cases, another statement. In the end, all the loose bits would be tied up, and your proof will be complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Least squares for rational function Simple question: If I want to fit data points to a function of the form $y=a+\frac{b}{x}$, is there any reason why I can't use a least-squares approach?
I want to minimize $$E=\sum_{i=1}^{n}(y-a-\frac{b}{x})^2$$
So just set $\frac{\partial E}{\partial a}=\frac{\partial E}{\partial b}=0$ and solve the linear system?
| Problem statement
Start with a set of $m$ measurements $\left\{ x_{k}, y_{k} \right\}$ and the model function
$$
y(x) = a + \frac{b}{x}.
$$
Assume the data are distinct to the point where the system matrix will have full column rank. Assume none of the $x$ values are $0$.
Least squares definition
The least squares solution is defined as
$$
(a,b)_{LS} = \left\{
(a,b)\in \mathbb{R}^{2} \colon
\sum_{k=1}^{m} \left( y_{k} - a - \frac{b}{x_{k}} \right)^{2}
\text{ is minimized}
\right\}
$$
Linear system
$$
\begin{align}
\mathbf{A} a &= y \\
%
\left[
\begin{array}{cc}
1 & \frac{1}{x_{1}} \\
\vdots & \vdots \\
1 & \frac{1}{x_{m}} \\
\end{array}
\right]
%
\left[
\begin{array}{c}
a \\
b \\
\end{array}
\right] &=
%
\left[
\begin{array}{cc}
y_{1} \\
\vdots \\
y_{m} \\
\end{array}
\right]
%
\end{align}
$$
Linear system solution
The least squares solution for the full column rank problem is
$$
%
\left[
\begin{array}{c}
a \\
b \\
\end{array}
\right]_{LS}
=
\mathbf{A}^{+} y
$$
One path to this solution is to use the normal equations:
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} \, a &= \mathbf{A}^{*}y \\
%
\left[
\begin{array}{cc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot \frac{1}{x} \\
\frac{1}{x} \cdot\mathbf{1} & \frac{1}{x} \cdot \frac{1}{x} \\
\end{array}
\right]
%
\left[
\begin{array}{c}
a \\
b \\
\end{array}
\right] &=
%
\left[
\begin{array}{cc}
\mathbf{1} \cdot y \\
\frac{1}{x} \cdot y
\end{array}
\right]
%
\end{align}
$$
These are exactly the same equations you would get by the minimization process you outlined:
$$
\begin{align}
%
& \frac{\partial}{\partial a} \sum_{k=1}^{m} \left( y_{k} - a - \frac{b}{x_{k}} \right)^{2} = 0, \\
%
& \frac{\partial}{\partial b} \sum_{k=1}^{m} \left( y_{k} - a - \frac{b}{x_{k}} \right)^{2} = 0. \\
%
%
%
\end{align}
$$
The solution is then
$$
\left[
\begin{array}{c}
a \\
b \\
\end{array}
\right]_{LS}
=
\left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} y
$$
(These remarks are an elaboration of the points made by @David.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find an inverse in a quotient of a Polynomial Ring I know this question is so easy that should be a first year algebra, but It has been a long time since I last studied this argument.
I have to find the inverse of $x+2$ in the ring $\mathbb{Q}[x]/\left<x^6 + x^4 + x^2 + 1\right>$.
I tried to divide $(x^6 + x^4 + x^2 + 1)+1$ for $x + 2$ but I obtained a reminder of $86$. Then I tried to solve
$$(x + 2)(ax^5+bx^4+cx^3+dx^2+ex+f)=q(x^6 + x^4 + x^2 + 1)+1$$
and obtained that q must be a complex number. So now I'm wondering if there's a better way to proceed. Can anybody give me an hint?
Thanks
| You can write $x^6+x^4+x^2+1=(x^5-2x^4+5x^3-10x^2+21x-42)(x+2)+85$
So, from here, $85=(x^6+x^4+x^2+1)-(x^5-2x^4+5x^3-10x^2+21x-42)(x+2)$, i.e.,
$85\equiv -(x^5-2x^4+5x^3-10x^2+21x-42)(x+2) \pmod{x^6+x^4+x^2+1}$
So, $(x+2)^{-1}=-\frac1{85}(x^5-2x^4+5x^3-10x^2+21x-42)$ in the quotient ring.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Polynomials with four distinct common rational roots Let $P(x) = x^5 + a_1x^4 + a_2x^3+ a_3x^2+ a_4x + 14$
and $Q(x) = x^5 + b_1x^4 + b_2x^3+ b_3x^2+ b_4x + 42$
be polynomials with integral coefficients.
If $P(x)$ and $Q(x)$ have four distinct common rational roots.
Find all possible $Q(x)$.
Please check my answer.
$P(x) = (x-1)(x+1)(x+2)(x+7)(x-1)\Rightarrow Q(x) = (x-1)(x+1)(x+2)(x+7)(x-3)$
$P(x) = (x-1)(x+1)(x-2)(x-7)(x-1)\Rightarrow Q(x) = (x-1)(x+1)(x-2)(x-7)(x-3)$
$P(x) = (x-1)(x+1)(x-2)(x+7)(x+1)\Rightarrow Q(x) = (x-1)(x+1)(x-2)(x+7)(x+3)$
$P(x) = (x-1)(x+1)(x+2)(x-7)(x+1)\Rightarrow Q(x) = (x-1)(x+1)(x+2)(x-7)(x+3)$
| Since the product of the roots of $P$ is an integer, if $P$ has four different rational roots, it has five rational roots. The same holds for $Q$.
By the rational root theorem, all roots of $P$ and $Q$ are integers and:
*
*the roots of $P$ are integers that divide $14$,
*the roots of $Q$ are integers that divide $42$.
Moreover, the product of the roots of $P$ is $14$ and the product of the roots of $Q$ is $42$. Therefore:
*
*for $P$, exactly one root is $\pm 2$ and exactly one root is $\pm 7$. The others are $\pm 1$.
*for $Q$, exactly one root is $\pm 2$, exactly one root is $\pm 3$, and exactly one root is $\pm 7$. The others are $\pm 1$.
The possible combinations follow from this.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve a simple equation: $x+x\sqrt{(2x+2)}=3$
$x+x\sqrt{(2x+2)}=3$
I must solve this, but I always get to a point where I don't know what to do. The answer is 1.
Here is what I did:
$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1) \\
\frac{9-6x+x^{2}}{2x^{2}}&=x+1 \\
\frac{9-6x+x^{2}-2x^{2}}{2x^{2}}&=x \\
\frac{9-6x+x^{2}-2x^{2}-2x^{3}}{2x^{2}}&=0
\end{align}$$
Then I got:
$-2x^{3}-x^{2}-6x+9=0$
| $f(x)=x+x\sqrt{2x+2}$
We are trying to find $f(x)=3$ but notice that for $-1\le x\le 0$ then $f(x)\le 0$.
$f'(x)=1+\sqrt{2x+2}+\frac {x}{\sqrt{2x+2}}>0$ for $x>0$
so $f$ is increasing from $f(0)=0$ to $\lim\limits_{x\to+\infty}f(x)=+\infty$
Thus there is an unique $x$ such that $f(x)=3$ and since $f(1)=3$ we are done.
It's not fordidden to have a look at the graph, before trying more complicated stuff.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find the exact value of $\sin(\pi+\alpha)+\cos(\frac{3\pi}{2}+\alpha)+\tan(-\frac{\pi}{2}+\alpha)$ Given that $\tan(\alpha) = -3$ find the exact value.
I tried:
$$\sin(\pi+\alpha)+\cos(\frac{3\pi}{2}+\alpha)+\tan(-\frac{\pi}{2}+\alpha) = \\
(\sin(\pi)\cos(\alpha)+\cos(\pi)\sin(\alpha))+(\cos(\frac{3\pi}{2})\cos(\alpha)-\sin(\frac{3\pi}{2})\sin(\alpha))+\tan(-\frac{\pi}{2}+\alpha) = \\
-\sin(\alpha)+\sin(\alpha)+\frac{\tan(-\frac{\pi}{2})+\tan(\alpha)}{1-\tan(-\frac{\pi}{2})\tan(\alpha)} = \\
\frac{\tan(-\frac{\pi}{2})+\tan(\alpha)}{1-\tan(-\frac{\pi}{2})\tan(\alpha)} = \\
\frac{-\tan(\frac{\pi}{2})-3}{1-(-\tan(\frac{\pi}{2})(-3)} = \\
\frac{-\tan(\frac{\pi}{2})-3}{1-3\tan(\frac{\pi}{2})} = \\
???$$
What do I do next?
| Simply,
$$\sin(\pi + \alpha)=-\sin \alpha$$
$$\cos(\frac{3\pi}{2} + \alpha)=\sin\alpha$$
$$\tan(-\frac{\pi}{2} + \alpha)=-\cot \alpha$$
Now, By drawing triangle. $-\sin \alpha=\frac{-3}{\sqrt{10}}$,$\sin\alpha=\frac{3}{\sqrt{10}}$ and $\cot \alpha=\frac{1}{3}$. Rest is simple arithmatics.
Assumptions: $\tan \alpha=3$
| {
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"url": "https://math.stackexchange.com/questions/2245733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that if $x$ is odd then $x^2 + (x + 2)^2$ is divisible by $2$ but not for $4$ This is what I tried:
$$x^2 + x^2 + 4x + 4=2x^2 + 4(x + 1),$$
so it's divisible by $2$, since this expression is a sum of a multiple of $2$ and a multiple of $4$.
Therefore, for the expression not to be a multiple of $4$, $2x^2$ can't be multiple of $4$, that's what I can't prove.
Brazilian student, sorry for my English.
| Perhaps the easiest way is to put $x=2k-1$, $k$ an integer, since $x$ is odd. Then
$$ x^2+(x+2)^2 = (2k-1)^2+(2k+1)^2 = 8k^2+2, $$
which is clearly $2$ more than a multiple of $4$, and hence not a multiple of $4$ (but is $2(4k^2+1)$, so obviously a multiple of $2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find locus with circle and intersection of lines
$M(x_0,y_0), P(x,y)$
The equation of the circle is $x^2+y^2=25$.
MN is a chord in the circle perpendicular to the axis $x$. P is the intersection of BM and NA.
I found that
$$MB: y = \frac{y_0}{x_0+5}x+\frac{5y_0}{x_0+5}$$
$$NA: y = \frac{y_0}{5-x_0}x-\frac{5y_0}{5-x_0}$$
I need to find the locus of the points P.
| So I let the equation of the cord MN be $x = t$ where $t\in (-5,5)$. Then its is clear that the line BP has the form $y = \frac{\sqrt{25-t^2}}{5+t} x + c$. Substituting $(-5,0)$ I find that $c = \frac{5 \sqrt{25-t^2}}{5+t}$. Similarly, we get the equation of the line NP: $y = \frac{\sqrt{25-t^2}}{5-t}[ x -5]$. To find the point P we set these two equations to be equal and calculate: $$\frac{\sqrt{25-t^2}}{5+t} [x+5] = \frac{\sqrt{25-t^2}}{5-t} [x-5]$$ we can then cancel the $\sqrt{25-t^2}$ because $t\not\in\{-5,5\}$. Rearranging we get $x^2 -25 = 25-t^2$ and so when $t>0$ we have $x = \sqrt{50-t^2}$ and when $t< 0$ we have $x = -\sqrt{50-t^2}$. We can then find $y$ by substitution into one of our original equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos\left (\theta -\frac{\pi}{4}\right)$ If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos \left(\theta -\frac{\pi}{4}\right).$
I could not get any idea to solve. However I tried by using $\theta =0^\circ $. But could not get the answer.
| $$\begin{align}
\tan (\pi \cos \theta) &= \cot (\pi \sin \theta) \\
\implies \frac{\sin(\pi \cos \theta)}{\cos(\pi \cos \theta)} &= \frac{\cos(\pi \sin \theta)}{\sin(\pi \sin \theta)} \\
\implies \sin(\pi \cos \theta)\sin(\pi \sin \theta) &= \cos(\pi \sin \theta)\cos(\pi \cos \theta) \\
\end{align}$$
and then
$$\begin{align}
0 &= \cos(\pi \sin \theta)\cos(\pi \cos \theta) - \sin(\pi \cos \theta)\sin(\pi \sin \theta) \\
&= \cos(\pi\sin\theta + \pi\cos\theta) \\
&= \cos(\pi(\sin\theta + \cos\theta)) \\
\implies \frac{\pi}{2} &= \pi(\sin\theta + \cos\theta) \\
\implies \frac{1}{2} &= \sin\theta + \cos\theta \\
\end{align}$$
Now we use the fact that $\sin \pi/4 = \cos \pi/4 = \frac{1}{\sqrt{2}}$.
$$\begin{align}
\sin\theta + \cos\theta &= \sqrt{2}\bigg(\frac{1}{\sqrt{2}}\sin\theta + \frac{1}{\sqrt{2}}\cos\theta\bigg) \\
&= \sqrt{2}(\sin\theta\cos \pi/4 + \cos\theta\sin \pi/4) \\
&= \sqrt{2}\sin(\theta + \pi/4) \\
&= \frac{1}{2}
\end{align}$$
Finally, $\cos(\theta - \pi/4) = \sin(\theta + \pi/4) = \frac{1}{2\sqrt{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\int {\sin \theta + \cos \theta \over \sqrt{\sin 2\theta} }d\theta$
$$\int {\sin \theta + \cos \theta \over \sqrt{\sin 2\theta} }d\theta$$
After simplifying I get,
$${1\over \sqrt{2}}\int {\tan \theta + 1 \over \sqrt{\tan \theta}} d\theta$$
Substituting $u = \tan \theta$
$${1\over \sqrt{2}}\int {u + 1\over \sqrt{u}(1 + u^2)} du$$
Substituting $t^2 = u$
$${\sqrt{2}}\int {t^2 + 1\over (1 + t^4)} dt = \sqrt{2} \int {1 + 1/t^2 \over(t - 1/t)^2 + 2 }dt$$
Substituting $z = t - 1/t$
$$\sqrt{2} \int {1\over z^2 + 2 }dz = \arctan\left(\tan \theta - 1\over \sqrt{2\tan \theta} \right) + C = {\arcsin(2\sqrt{\sin2\theta} (\sin\theta - \cos \theta))\over 2} + C$$
Which is far from the right answer of $\arcsin(\sin \theta - \cos \theta) + C$.
Where did I go wrong ?
Edit :
I would like to know where I am wrong rather than knowing how to solve the problem.
| Hint:
Try the substitution $u=\sin 2\theta$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Solving an equation containing cube roots: $\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$ I'm trying to figure out a way to solve this equation:
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1.$$
I tried to cube both sides, but I ended up with an equation looking like this:
$$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1}-\sqrt[3]{5x+7})=-6.$$
At this point I'm out of stuff to do. Any help would be appreciated. Thanks in advance.
| $$a=\sqrt[3]{5x+7},b=\sqrt[3]{5x-12}\\a-b=1\\(a-b)^3=1\\a^3-3a^2b+3ab^2-b^3=1\\a^3-b^3-3ab(a-b)=1\\a^3-b^3-3ab=1\\5x+7-5x+12-3\sqrt[3]{5x+7}\sqrt[3]{5x-12}=1\\-3\sqrt[3]{5x+7}\sqrt[3]{5x-12}=-18\\\sqrt[3]{5x+7}\sqrt[3]{5x-12}=6\\(5x+7)(5x-12)=216\\25x^2-60x+35x-84=216\\25x^2-25x-300=0\\x^2-x-12=0\\x_1=4,x_2=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
} |
Find Big-O Estimate of $f(n) = n^3 + 4n^2 \log(n) + n + 1$ So to show the big-O complexity I'm trying to find a C and a K such that $f(n) \le C g(n)$ for all $n > k$
To start solving $f(n) = n^3 + 4n^2 \log(n) + n + 1$
Fix $k = 1$, so $n > 1$
$f(n) = n^3 + 4n^2 \log(n) + n + 1 \le n^3 + 4n^3 \log(n) + n^3 + n^3$
$f(n) = n^3 + 4n^2 \log(n) + n + 1 \le 3n^3 + 4n^3 \log(n)$
But I'm not sure what to do next or if I even did that last step right. Any help is appreciated, thanks in advance.
| Hint
$$f(n) = n^3 + 4n^2 \log(n) + n + 1 \leq n^3 +4n^3+n^3+n^3$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Rounding a number $a$ is a positive whole number. $k$ is a real number greater than one. $\operatorname{round}(x)=\left\lfloor x+\frac12\right\rfloor$.
$b(a)=\operatorname{round}(k\cdot\operatorname{round}(a/k))$.
Can $b(a)$ be:
*
*greater than $a$?
*less than $a$?
*can $b(\dots (b(a)))$ tend to infinity?
The problem is practical: It is about rounding errors when calculating paid time period ($a$ or $b(a)$) on a Web site dependently on changes of the price (price is changed $k$ times).
| Both inequalities are possible. For $1 < k < 2$:
*
*For $0 < x < \frac{1}{2} (k + 1)$, we have $b(x) = 0 < x$.
*For $\frac{1}{2}(k + 1) < x < 1$, we have $b(x) = 1 > x$.
Now, for any $k > 1$, the operation $b$ actually satisfies $b(b(x)) = b(x)$ for all $x$, that is, $b$ is idempotent. In particular, the series $b(x), b(b(x)), \ldots$ cannot escape to infinity.
Hint To see this, first observe that it follows from the definition and the fact that $u - 1 < \lfloor u \rfloor < u$,
$$k r(x) - \frac{1}{2} < b(x) < k r(x) + \frac{1}{2} .$$
Dividing by $k$, adding $\frac{1}{2}$, and applying the floor function gives $$\left\lfloor r(x) + \frac{1}{2} \left(1 - \frac{1}{k} \right) \right\rfloor \leq r(b(x)) \leq \left\lfloor r(x) + \frac{1}{2} \left(1 + \frac{1}{k} \right) \right\rfloor .$$ Now, since $k > 1$, $0 < \frac{1}{2} \left(1 \pm \frac{1}{k} \right) < 1$, and since $r(x)$ is an integer, $$\left\lfloor r(x) + \frac{1}{2} \left(1 \pm \frac{1}{k} \right) \right\rfloor = \lfloor r(x) \rfloor = r(x),$$ so $r(b(x)) = r(x)$, from which the claim follows quickly.
| {
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"url": "https://math.stackexchange.com/questions/2249656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the first derivative and nth derivative of the following function $ y = \sqrt {2 +\sqrt {3 + \sqrt{x}}}$ What is the first derivative and nth derivative of the following function $ y = \sqrt {2 +\sqrt {3 + \sqrt {x}}}$
I think taking the ln for both sides will remove the first square root only?
Could anyone give me a hint ?
| First of all. Use chain rule. I'd recommend learning something new instead of relying on app.(personal opinion)
$$\frac{dy}{dx}=\frac{1}{2\sqrt{2+\sqrt{3+\sqrt{x}}}}.\frac{d\sqrt{3+\sqrt{x}}}{dx}$$
$$\frac{dy}{dx}=\frac{1}{2\sqrt{2+\sqrt{3+\sqrt{x}}}}.\frac{1}{2\sqrt{3+\sqrt{x}}}\frac{d\sqrt{x}}{dx}$$
$$\frac{dy}{dx}=\frac{1}{2\sqrt{2+\sqrt{3+\sqrt{x}}}}.\frac{1}{2\sqrt{3+\sqrt{x}}}\frac{1}{2\sqrt{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If the limit exists (prove) and find the value
Let $$f(x, y) = \frac{x^2(x - 1) - y^2}{x^2 + y^2}, (x, y) \ne (0, 0)$$ Determine if $\lim_{(x, y) \to (0, 0)} f(x, y)$ exists, and if so, find the value.
Let $y = mx$, then
$$ = \lim_{x \to 0} \frac{x^2(x - 1) - m^2x^2}{x^2 + m^2x^2} = \lim_{x \to 0} \frac{x - 1 - m^2}{m^2 + 1} = -1 $$
Then we apply the squeeze theorem to prove $|f(x,y) - (-1)| < B(x)$ where $B(x) \to 0$
Main question: if we let $y = mx$, then as $y \to 0$, how can we be sure that $x \to 0$?
Since $x = y/m$ we see that if $m \ne 0$ then $x \to 0$ as $y \to 0$.
So, in the hypothesis when we say let $y = mx$, is it for non zero $m$?
| Use polar coordinates: $x = r\cos\theta, \, y = r\sin\theta$ and let $r \to 0$.
EDIT
$$\begin{align}
\lim_{(x,y)\to(0,0)} \frac{x^2(x - 1) - y^2}{x^2 + y^2}
& = \lim_{r \to 0} \frac{r^2 \cos^2\theta \, (r \cos\theta - 1) - r^2\sin^2\theta}{r^2} \\
& = \lim_{r \to 0} \left(\cos^2\theta \, (r \cos\theta - 1) - \sin^2\theta\right) \\
& = \cos^2\theta \cdot (-1) - \sin^2\theta \\
& = -1
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2253351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $a = \log_{\sqrt{6}}{\sqrt[3]{4}}$ find $\log_{\sqrt{3}}{\sqrt[3]{2}}$ I have attempted to solve this and got the result of $$\frac{2a}{4-3a}$$
Whereas the correct answer:
$$\frac{10a}{12-15a}$$
I have attached my calculations. Could you tell me where they are inaccurate?
$$a=\log_{\sqrt{6}}\sqrt[3]{4}$$
$$\log_{\sqrt{3}}\sqrt[3]{2}=\frac{1}{3}\log_{\sqrt{3}}2=\frac{1}{3}\log_{\sqrt{3}}4^{1/2}=\frac{1}{6}\log_{\sqrt{3}}4=\frac{1}{2}\cdot \frac{1}{3}\log_{\sqrt{3}}4$$
$$=\frac{1}{2}\log_{\sqrt{3}}\sqrt[3]{4}=\frac{1}{2}\cdot \frac{\log_{\sqrt{6}}\sqrt[3]{4}}{\log_{\sqrt{6}}\left(\frac{\sqrt{6}}{\sqrt{2}}\right)}=\frac{1}{2}\cdot \frac{a}{1-\log_{\sqrt{6}}\sqrt{2}}$$
$$=\frac{a}{2(1-\frac{1}{2}\log_{\sqrt{6}}2)}=\frac{a}{2(1-\frac{1}{4}\log_{\sqrt{6}}4)}=\frac{\frac{1}{3}a}{\frac{2}{3}-\frac{2}{12}\log_{\sqrt{6}}4}$$
$$=\frac{\frac{1}{3}a}{\frac{2}{3}-\frac{2}{4}\cdot \frac{1}{3}\cdot \log_{\sqrt{6}} 4}=\frac{\frac{1}{3}a}{\frac{2}{3}-\frac{1}{2}a}=\frac{\frac{1}{3}a}{\frac{4-3a}{6}}=\frac{2a}{4-3a}$$
Hence, the final answer is:
$$\bbox[5px,border:2px solid #C0A000]{\log_{\sqrt{3}}\sqrt[3]{2}=\frac{2a}{4-3a}}$$
| I followed your method, and every step was correct. Checking the answer in a calculator I get that $$\log_{\sqrt{3}}{\sqrt[3]{2}} = 0.42061...$$
and $$\frac{2a}{4-3a} = 0.42061...=\log_{\sqrt{3}}{\sqrt[3]{2}}\quad\text{as required}$$
whereas $$\frac{10a}{12-15a} = 1.20997...\quad\text{which is clearly incorrect}$$
| {
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"url": "https://math.stackexchange.com/questions/2255326",
"timestamp": "2023-03-29T00:00:00",
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Find minimum value of the function If $g(x) = \max|y^2 - xy|$ for $0\leq y\leq 1$. Then the minimum value of $g(x)$ is?
I am not being able to proceed. Tried drawing the graph.
| Computing the function $g(x) = \max{|y^2 - xy|}$ is equivalent to solving for the maximum value of the expression $|y^2 - xy| = f(y)$ with a fixed parameter $x$. Let's then imagine that $x$ is a constant. In general, the maximum value of a function (of one variable) can be found in either
*
*Where the derivative is zero;
*Where the derivative is not defined; or
*At the boundaries of the domain.
Let's go through these cases one by one. Firstly, we want to find $\frac{d f(y)}{dy}=0$. For $y^2 - xy < 0$ the result is $y = \frac{x}{2}$, and for $y^2 -xy > 0$ there is no solution.
The derivative is not defined when $y^2 - xy = 0 \Rightarrow y= 0$ or $y = x$.
The boundaries of the domain were defined to be $y = 0$ and $y = 1$.
Inserting these results into $f(y)$ gives us $f(\frac{x}{2}) = \frac{x^2}{4}$; $f(0)=f(x)=0$; $f(1)=|1-x|$. We need the largest of these, which is
$g(x) = \begin{cases}
\left| 1 - x \right| , \text{if} -2(1+\sqrt{2}) \leq x \leq 2(\sqrt{2}-1)\\
x^2/4 , \text{otherwise}
\end{cases}$
This was found out by solving the equation $|1-x|>\frac{x^2}{4}$.
The minimum of $g(x)$ can be found at $x= 2(\sqrt{2}-1)$, which is $g(2(\sqrt{2}-1)) = 1- 2(\sqrt{2}-1) = 3-2\sqrt{2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Number Theory: Prove that $p|m-n$
Assume $p$ is an odd prime number, $\displaystyle q = \frac{3p-5}{2}$ and $$ S_q = \frac{1}{2.3.4} + \frac {1}{5.6.7} + ... + \frac{1}{q(q+1)(q+2)}$$.
If $\displaystyle \frac{1}{p} - 2S_q = \frac{m}{n}$ and $m$ and $n$ are two integers, Prove that $p\ | \ m-n$.
Any hints how to start the proof?
| You could start with the identity
$$
\frac{1}{(3n-1)(3n)(3n+1)}=\frac{1}{2}\left(\frac{1}{3n-1}+\frac{1}{3n}+\frac{1}{3n+1}\right)-\frac{1}{2}\frac{1}{n}.
$$
Using this:
$$
\begin{align*}
S_q &= \sum_{n=1}^{(p-1)/2}\frac{1}{2}\bigg[\left(\frac{1}{3n-1}+\frac{1}{3n}+\frac{1}{3n+1}\right)-\frac{1}{n}\bigg]\\
&=\frac{1}{2}\left(\sum_{k=2}^{(3p-1)/2}\frac{1}{k}-\sum_{n=1}^{(p-1)/2}\frac{1}{n}\right)=\frac{1}{2}\left(-1+\sum_{k=(p-1)/2}^{(3p-1)/2}\frac{1}{k}\right).
\end{align*}
$$
Can you take it from there? You might find Wolstenholme's Theorem useful.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the derivatives of $\frac{d}{dx}\int _2^{x^3}\sin \left(t^2\right)dt$ and $\frac{d}{dx}\int _{\tan x}^{\sqrt{x+2}}\sqrt[3]{1+t^2}dt$ I know that $\frac{d}{dx}\int _a^xf\left(t\right)dt=f\left(x\right)\:$ and $\frac{d}{dx}\int _a^{g\left(x\right)}f\left(t\right)dt=f\left(g\left(x\right)\right)\cdot g'\left(x\right)$
I think I know what to do on the first problem, but I'm stumped on the second one. $a=\tan \left(x\right)$ rather than some constant, so I feel like I did not do it correctly.
a. $\frac{d}{dx}\int _2^{x^3}\sin \left(t^2\right)dt$
$$=\sin \left(\left(x^3\right)^2\right)\cdot 3x^2$$
$$=\sin \left(x^6\right)\cdot 3x^2$$
b. $\frac{d}{dx}\int _{\tan x}^{\sqrt{x+2}}\sqrt[3]{1+t^2}dt$
$$=\sqrt[3]{1+\sqrt{x+2}^2}\cdot \frac{1}{2\sqrt{x+2}}$$
$$=\sqrt[3]{x+3}\cdot \frac{1}{2\sqrt{x+2}}$$
$$=\frac{\sqrt[3]{x+3}}{2\sqrt{x+2}}$$
| This is a standard application of Fundamental Theorem of Calculus.
Since you seen to know what to do for (a), I'll show you the process for (b), although they're near identical.
Let $$F(t) = \int (1+t^2)^{1/3}dt $$
By FTC,
$$F(\sqrt{x+3}) - F(\tan(x)) = \int_{\tan(x)}^{\sqrt{x+3}} (1+t^2)^{(1/3)}dt$$
Taking the derivatives of both sides, we get
$$\frac{d}{dx} \int_{\tan(x)}^{\sqrt{x+3}} (1+t^2)^{(1/3)} = F'(\sqrt{x+3})\left(\frac{1}{2 \sqrt{x+3}}\right)- F'(\tan(x))(\sec^2(x))$$
And $F'(x) = (1 + x^2)^{1/3}$
I trust you can plug in the rest
| {
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"timestamp": "2023-03-29T00:00:00",
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Differentiation to tangent and normals The equation of the curve is given by
$ y = \frac{6}{2x^2+1} $
Find the equation of the tangent and the normal at the point when
$ x = -1$
My workings
When $x= -1, y= 2 $
$\frac{d}{dx} (Y) = \frac{-24}{(2x^2+1)^2} $
Sub $x=-1$ to the above expression to find the gradient ...
$\frac{d}{dx} (Y) = \frac{-8}{3} $
Now , to find the equation of the tangent , what can I do ?
Is it $y=mx+c$ ?
Sub in $x=-1 , y=2$ to find the y intercept and then continue to find the equation of the tangent ?
Which is
$y= \frac{-8}{3} x - \frac{2}{3} $
But that don't feel right to me .. so im here to ask .
| HINT: write your function in the form $$y=6(2x^2+1)^{-1}$$ then the derivative is given by
$$y'=6(-1)(2x^2+1)^{-2}\cdot 4x=\frac{-24x}{(2x^2+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Which numbers is bigger $80^{105}$ or $28^{140}$
Which numbers is bigger $80^{105}$ or $28^{140}$
My try follows
$80^{105}$ = $(80^{3})^{35}$=$(512000)^{35}$
$28^{140}$=$(28^{4})^{35}$=$ (614656)^{35}$
So $28^{140}$ is bigger
Is there another idea for this problem involving relations between $80$ and $81$ and $27$ and $28$ since $81$ and $27$ have common base $3$
Thank you for your help
| As $(105,140)=35$
we can check for $80^3,28^4$
$80^3=(5\cdot2^4)^3=5^32^{12},28^4=(7\cdot2^2)^4=7^4\cdot2^8$
$80^3>=<28^4\iff5^32^{12}>=<7^4\cdot2^8\iff5^32^4>=<7^4$
Now $7^4=2401,5^32^4=2\cdot10^3=2000$
| {
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"url": "https://math.stackexchange.com/questions/2260561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdot...\cdot\frac{2n-1}{2n}\le \frac{1}{\sqrt{3n+1}}$ We want to prove by induction that
$$\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdots\frac{2n-1}{2n}\le \frac{1}{\sqrt{3n+1}}$$ for all $n,k \in \mathbb{Z^+}$
For k=1 : $\frac1{2}\le\frac{1}{2}$ which it is true.
Assume the induction hypothesis $\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdots\frac{2k-1}{2k}\le \frac{1}{\sqrt{3k+1}}$
We need to prove $\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdots\frac{2(k+1)-1}{2(k+1)}\le \frac{1}{\sqrt{3(k+1)+1}}$
We can take the induction hypothesis and multiply by $\frac{2(k+1)-1}{2(k+1)}$ on both sides.
Then, we need to prove that $\frac{1}{\sqrt{3k+1}}\cdot\frac{2(k+1)-1}{2(k+1)}\le \frac{1}{\sqrt{3(k+1)+1}}$
$\frac{(2k+1)^2}{4(3k+1)(k+1)^2}\le \frac{1}{(3k+4)} \Leftrightarrow$
$(3k+4)(2k+1)^2\le 4(3k+1)(k+1)^2 \Leftrightarrow$
$12k^3+28k^2+19k+4\le 12k^3+16k^2+20k+4 \Leftrightarrow$
$12k^2-k\le 0$ which is not true!
Where is the mistake??
| for the right-hand side i have got $$12k^3+28k^2+20k+4$$
and the left-Hand side is given by $$12\,{k}^{3}+28\,{k}^{2}+19\,k+4$$
| {
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"url": "https://math.stackexchange.com/questions/2260778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving the lemma when the intersection is inside. In our book we had a lemma that says if $G$ and $H$ are tangent points and $F$ and $E$ are mid points then $GH$,$FE$ and $CD$ intersect in the same point and the angle $BIC=90$.
This is the lemma when the intersection is outside.To prove this lemma we can connect $D$ to $H$ and get a right angle then we have to prove $DHIB$ is cyclic which can easily proved by showing that angles $DBH=HID$.When the intersection is inside proving $BIC=90$ is a bit different but possible using cyclic quadrilaterals but then I don't know how to prove these three lines intersect in the same point.
| *
*Let the line $CD$ intersect $EF$ at point $I_1$. $\, EF$ is a midsegment so $EF || AC$ and $EF = \frac{1}{2} \, CA$. Thus $EI_1 || CA$. Since $CD$ is an angle bisector and $EI_1 || CA$
$$ \angle \, ECI_1 = \angle \, ACI_1 = \angle \, EI_1C$$ which means that triangle $CEI_1$ is isosceles so $$EI_1 = CE = BE$$ which by the way means that triangle $CBI_1$ is right-angled.
$$ $$
*Let line $GH$ intersect $EF$ at point $I_2$. Since $EI_2 || CA$ triangles $AGH$ and $FI_1H$ are similar and since $AG = AH$, also $FI_1 = FH$. Our goal is, given that $BC = a, \, CA = b, \, AB = c$ to calculate the length of $$EI_2 = EF + FI_1 = \frac{1}{2} \, CA + FI_1 = \frac{b}{2} + FI_1 = \frac{b}{2} + FH$$ Thus, we need to calculate $FH$. But $$FH = FA - AH = \frac{1}{2} \, AB - AH = \frac{c}{2} - AH$$ Moreover,
$$AH = \frac{b+c - a}{2}$$ so
$$FH =\frac{c}{2} - AH = \frac{c}{2} - \frac{b+c- a}{2} = \frac{a-b}{2}$$ Therefore, $$EI_2 = EF + FI_1 = EF + FH = \frac{b}{2} + \frac{a-b}{2} = \frac{a}{2} = \frac{1}{2} \, BC = CE$$ However, we proved already that $EI_1 = CE$. Therefore $EI_1 = EI_2$ which is possible if and only if $I_1 \equiv I_2 $. Thus the tree lines $GH, \, CD$ and $EF$ intersect at a common point we call $I \equiv I_1 \equiv I_2$.
We already proved that $\angle \, BIC \equiv \angle \, BI_1C = 90$
| {
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"timestamp": "2023-03-29T00:00:00",
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A differential equation with homogeneous coefficients: $(x+y) dx - (x-y) dy = 0$. I have a differential equation with homogeneous coefficient:
$$ (x+y) dx - (x-y) dy = 0 $$
Functions $ P(x) = x+y$ and $ Q(x) = - x-y $ are both homoegeneous functions of order 1. So I can use substitusion:
$$ y = ux, dy = udx + xdu $$
(because it will lead to a differential equation in which the variables are separable)
So after this substitution I'm obtaining result:
$$ (u+ux)dx - (x-ux)(udx + xdu) = 0 $$
After simplification I have:
$$ x^{2} (u-1)du + x(1+u^{2})dx = 0 $$
Now I'm dividing both sides by $x^{2} \cdot (1+u^{2})$ and I'm obtaining:
$$ \frac{u-1}{1+u^{2}}du + \frac{dx}{x} = 0 $$
Next, because:
$$ \int \frac{u-1}{1+u^{2}}du = \frac{1}{2} \ln|u^{2} +1 | - \arctan(u) +C $$
and:
$$ \int \frac{dx}{x} = \ln| x| + C$$
I have a solution:
$$ \frac{1}{2} \ln| u^{2} + 1| - \arctan(u) + \ln|x| = C$$
$$ \frac{1}{2} \ln|\frac{y^{2}}{x^{2}} +1| - \arctan(\frac{y}{x}) + \ln|x| = C $$
After simplification, I can obtain result:
$$ \frac{1}{2} \ln( x^{2} + y^{2}) - \arctan(\frac{y}{x}) = C $$
In book from this exercise from there is a little diffrent answer :
$$ \arctan(\frac{y}{x}) - \frac{1}{2} \ln(x^2 + y^2) = C $$
Is my solution incorrect? I will be grateful for an explanation
| Both solutions are correct. Note that $C$ is an arbitrary constant. Therefore, one may define $C=-k$ on your solution:
$$\frac{1}{2}\ln(x^2+y^2)-\arctan\left(\frac{y}{x}\right)=-k$$
Multiplying both sides by $-1$, we obtain the same solution the book obtained:
$$\arctan\left(\frac{y}{x}\right)-\frac{1}{2}\ln(x^2+y^2)=k$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that the product of eight consecutive numbers can't be a number raised to exponent 4? How to prove this? I tried something like $$P(n,8)=\frac{n!}{(n-8)!} = b^4$$ but I can't proceed to a solution.
| Here's a proof that there is no positive integer $n$ such that
$$f(n) = (n)\cdots (n+7)$$
is a perfect $4$-th power.
The proof is along the same lines as the proof I gave showing that there is no positive integrer $n$ such that $f(n)$ is a perfect square, but the proof for the case of $4$-th powers is easier, and can be done by hand.
Thus, suppose $n$ is a positive integer such that $f(n)$ is a perfect $4$-th power.
We start by getting a quick lower bound on $n$ . . .
Considering factors of $7$, it's clear that $f(n)$ must be a multiple of $7$, hence must be a multiple of $7^4$.
But at most two of the $8$ factors can be multiples of $7$, and at most one of them can be a multiple of $7^2$. Thus, in order for $f(n)$ to be a multiple of $7^4$, one of the $8$ factors must be a multiple of $7^3$.
It follows that $7^3 \le n + 7$, so $n \ge 336$.
Next we pair up the $8$ factors . . .
Since
\begin{align*}
n(n+7) &= n^2 + 7n\\[4pt]
(n+1)(n+6) &= n^2 + 7n + 6\\[4pt]
(n+2)(n+5) &= n^2 + 7n + 10\\[4pt]
(n+3)(n+4) &= n^2 + 7n + 12\\[4pt]
\end{align*}
it follows that
$$(n^2 + 7n)^4 < f(n) < (n^2 + 7n + 12)^4$$
hence we must have
$$f(n) = (n^2 + 7n + c)^4$$
for some positive integer $c < 12$.
From the definition of $f(n)$, we have $\bigl((n)(n+7)\bigr) \mid f(n)$.
\begin{align*}
\text{Then}\;\;&\bigl((n)(n+7)\bigr) \mid f(n)\\[4pt]
\implies\; &f(n) \equiv 0 \pmod {n^2 + 7n}\\[4pt]
\implies\; &(n^2 + 7n + c)^4 \equiv 0 \pmod {n^2 + 7n}\\[4pt]
\implies\; &c^4 \equiv 0 \pmod {n^2 + 7n}\\[4pt]
\implies\; &n^2 + 7n \le c^4\\[4pt]
\implies\; &n^2 < c^4\\[4pt]
\implies\; &n < c^2\\[4pt]
\implies\; &n < 144\qquad\text{[since $c < 12$]}\\[4pt]
\end{align*}
contrary to our lower bound, $n \ge 336$.
Hence there is no positive integer $n$ such that $f(n)$ is a perfect $4$-th power.
| {
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"url": "https://math.stackexchange.com/questions/2271343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Ratio test involving factorials: $a_{n} = \frac{n-2}{(n+1)!}$; finding $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|$ The correct answer is apparently $0$ however I end up with $1$. I do the following:
$$a_{n+1} = \frac{(n+1)-2}{((n+1)+1)!} = \frac{n-1}{(n+2)!}$$
$$
\begin{split}
\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|
&= \lim_{n \to \infty} \left|\frac{n-1}{(n+2)(n!)}
\times \frac{(n+1)(n!)}{n-2}\right| \\
&= \lim_{n \to \infty} \left|\frac{(n-1)(n+1)}{(n+2)(n-2)}\right| \\
&= \lim_{n \to \infty} \left|\frac{n^2-1}{n^2-2}\right| \\
&= \lim_{n \to \infty} \frac{n^2}{n^2} \\
&= 1
\end{split}
$$
I assumed as $n$ approaches $\infty$ the '$-1$' and '$-2$' become meaningless, however I'm pretty certain that is now incorrect unless I've made an error earlier on.
| Note that $(n+2)! = ((n+1)!)(n+2)$
I think that is the mistake.
$$\lim_{n \rightarrow \infty} \frac{n-1}{(n+1)!(n+2)} \frac{(n+1)!}{n-2}=\lim_{n \rightarrow \infty} \frac{n-1}{(n+2)(n-2)}=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of $\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}$ Let $x,y,z>0$ and $x+y+z=xyz$. What is the minimum value of $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}?$$
In the case when $x=y=z$, the equation $x+y+z=xyz$ translates to $3x=x^3$, or $x=\sqrt{3}$. If $A$ denotes the quantity that we want to minimize, then $A=\sqrt{3}$ as well.
If we use the inequality of arithmetic and geometric means, we get
$$A\geq \frac{3}{\sqrt[3]{xyz}}=\frac{3}{\sqrt[3]{x+y+z}}.$$
| From our constraint $x + y + z = xyz$, we obtain that $z = \frac{x+y}{xy-1}$, plugging that into
$$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}$$
we obtain
$$\frac{x}{y^2} + \frac{y(xy-1)^2}{(x+y)^2} + \frac{x+y}{x^2 (xy-1)}$$
Call this $f(x,y)$. We can then find a minimum value by doing the second derivative test. First, we need to find the critical points by setting
$$f(x,y) = \frac{x}{y^2} + \frac{y(xy-1)^2}{(x+y)^2} + \frac{x+y}{x^2 (xy-1)} = 0$$
We would obtain some ordered pair $(a, b)$ (there may be more than one set of critical points - there may be infinitely many even!). From this we calculate the Hessian,
$$H = \det \begin{pmatrix} f_{xx}(a,b) & f_{xy}(a,b) \\ f_{yx}(a,b) & f_{yy}(a,b)\end{pmatrix}$$
For a minimum we want $H > 0$ and $f_{xx}(a,b) > 0$, once we find that culprit, we have found our minimum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Primes dividing sum of binomial coefficients. Given that $p>3$ is a prime, we have $k=\lfloor \frac {2p}{3} \rfloor$ then prove or disprove that $\sum_{i=1}^k\binom{p}{i}\equiv 0\pmod{p^2}$.
Its easy to see how $p$ divides each of the binomial coefficients, but I tried using maximal powers of $p$ in each coefficient but was stuck can someone help.
| We want to prove that $\sum \limits_{i=1}^{\lfloor \frac{2p}{3} \rfloor} \binom{p}{i} = 0 \mod p^2$ for $p$- primes.
$\sum \limits_{i=1}^{\lfloor \frac{2p}{3} \rfloor} \binom{p}{i} = 0 \mod p^2$ imply that $\sum \limits_{i=1}^{\lfloor \frac{2p}{3} \rfloor} \frac{1}{p}\binom{p}{i} = \sum \limits_{i=1}^{\lfloor \frac{2p}{3} \rfloor} \frac{(-1)^{i-1}}{i} = 0 \mod p $ (as mentioned in the comments).
For $p=3k+1$ => $\sum \limits_{i=1}^{2k} \frac{(-1)^{i-1}}{i} = 1+\sum \limits_{i=1}^{k-1}\frac{1}{2i+1}-\sum \limits_{i=1}^{k} \frac{1}{2i} = \frac{1}{2} \left(H_{k-\frac{1}{2}}-H_k+\log (4)\right)$
and since $H_{2k}=\frac{1}{2}(H_k+H_{k-1/2}+\ln 4)$
we can get that
$\frac{1}{2}(H_{k-1/2}-H_k+\ln 4)-\frac{1}{2}(H_{k-1/2}+H_k+\ln 4) = -H_{2k} \mod (3k+1)$.
Which is just $-H_k = -H_{2k} \mod (3k+1)$
Rearranging that we arrive at $H_{2k}-H_k = 0 \mod (3k+1)$ when $p=3k+1$ is prime.
For $p=3k+2$ => $\sum \limits_{i=1}^{2k+1} \frac{(-1)^{i-1}}{i} = 1+\sum \limits_{i=1}^{k-1}\frac{1}{2i+1}-\sum \limits_{i=1}^{k} \frac{1}{2i}+\frac{1}{2k+1} = \frac{1}{2} \left(H_{k-\frac{1}{2}}-H_k+\log (4)\right)+\frac{1}{2k+1}$
and since $H_{2k}=\frac{1}{2}(H_k+H_{k-1/2}+\ln 4)$
we can get that
$\frac{1}{2}(H_{k-1/2}-H_k+\ln 4)-\frac{1}{2}(H_{k-1/2}+H_k+\ln 4)+\frac{1}{2k+1} = -H_{2k} \mod (3k+2)$.
Which is just $-H_k+\frac{1}{2k+1} = -H_{2k} \mod (3k+2)$
Rearranging that we arrive at $H_{2k}-H_k+\frac{1}{2k+1} = 0 \mod (3k+2)$ when $p=3k+2$ is prime.
Note : $H_k = \sum \limits_{i=1}^{k} \frac{1}{i}$ is the Harmonic number and also the highlighted part,are the statements that i did not prove.
| {
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show that $11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $
Let $n$ be postive integer,show that
$$11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $$
I have use ugly metods to solve consider this case,$n\equiv 0,\pm 1,\pm 2,\pm 3,\pm 4,\pm 5\pmod {11}$,have other more simple methods it?
| Starting from @Crostul's comment, we have
$$
f(n)=n^6-2n^5+4n^4-8n^3+16n^2-32n+64 = \frac{n^7+2^7}{n+2}
$$
and so $f(n)(n+2)=n^7+2^7$.
Therefore, if $11$ divides $f(n)$, then $11$ divides $n^7+2^7$.
Now $n \mapsto n^7$ is an injective map mod $11$ because $\gcd(7,\phi(11))=1$.
Therefore, if $11$ divides $f(n)$, then $n^7 \equiv -2^7 = (-2)^7 \bmod 11$ and so $n \equiv -2 \bmod 11$.
So you only have to check that $11$ does not divide $f(n)$ for $n=-2$.
| {
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Which solutions to use for the following hypergeometric equation? I am currently trying to understand the solution to a hypergeometric equation given in a paper on scalar fields and rotating black holes by S. Detweiler (https://journals.aps.org/prd/abstract/10.1103/PhysRevD.22.2323).
The equation I have trouble with is (17):
$$z(z+1)\frac{d}{dz}[z(z+1)\frac{dR}{dz}]+[P^2-l(l+1)z(z+1)]R=0$$
When I try to solve it in mathematica I get associated Legendre polynomials:
$$R(z) = C[1]LegendreP[l, 2 I P, 1 + 2 z] + C[2] LegendreQ[l, 2 I P, 1 + 2 z]$$
I know these can be written in terms of hypergeometric functions but they don't seem to match up when I try.
In the paper, the author provides a solution of the following form:
$$R(z)=(\frac{z}{z+1})^{iP}G(-l,l+1;1-2iP;z+1)$$
where G is any solution to the hypergeometric equation.
From here he uses two independent hypergeometric functions $U_3$ and $U_4$ (which are in turn linear combinations of two other solutions $U_1$ and $U_5$):
$$U_3 = (-z)^lF(-l,-l-2iP;-2l;-z^{-1})$$
$$U_4 = (-z)^{-l-1}F(l+1,l+1-2iP;2l+1;-z^{-1})$$
I am unsure how, given his solution in terms of G, the author knew which independent hypergeometric functions to use in the following steps to give the correct solution.
Thank you in advance for any advice you can give me!
EDIT: As I was unable to find them in the paper, I have calculated what I think are the correct boundary conditions. The requirement is that at the horizon, $r \rightarrow r_+$, there is only an ingoing wave solution. It is useful to know that in the above equations:
$$ P = (am - 2Mr_+ \omega)/(r_+ - r_{-}) $$
$$ z = (r-r_+)/(r_+ - r_-)$$
then the boundary condition at $r \rightarrow r_+$ is
$$R \sim (r-r_+)^{-i\alpha}$$
where $$ \alpha = \frac{Mr_+\omega-ma/2}{\sqrt{M^2-a^2}} $$
| $z(z+1)\dfrac{d}{dz}\left(z(z+1)\dfrac{dR}{dz}\right)+(P^2-l(l+1)z(z+1))R=0$
$z(z+1)\dfrac{d^2R}{dz^2}+(2z+1)\dfrac{dR}{dz}+\left(\dfrac{P^2}{z(z+1)}-l(l+1)\right)R=0$
$\dfrac{d^2R}{dz^2}+\dfrac{2z+1}{z(z+1)}\dfrac{dR}{dz}+\dfrac{1}{z(z+1)}\left(\dfrac{P^2}{z(z+1)}-l(l+1)\right)R=0$
$\dfrac{d^2R}{dz^2}+\left(\dfrac{1}{z}+\dfrac{1}{z+1}\right)\dfrac{dR}{dz}+\dfrac{1}{z(z+1)}\left(\dfrac{P^2}{z}-\dfrac{P^2}{z+1}-l(l+1)\right)R=0$
Which relates to Riemann's differential equation.
Let $R=z^a(z+1)^bS$ ,
Then $\dfrac{dR}{dz}=z^a(z+1)^b\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)S$
$\dfrac{d^2R}{dz^2}=z^a(z+1)^b\dfrac{d^2S}{dz^2}+z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a(a-1)}{z^2}+\dfrac{2ab}{z(z+1)}+\dfrac{b(b-1)}{(z+1)^2}\right)S=z^a(z+1)^b\dfrac{d^2S}{dz^2}+2z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a(a-1)}{z^2}+\dfrac{2ab}{z(z+1)}+\dfrac{b(b-1)}{(z+1)^2}\right)S$
$\therefore z^a(z+1)^b\dfrac{d^2S}{dz^2}+2z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a(a-1)}{z^2}+\dfrac{2ab}{z(z+1)}+\dfrac{b(b-1)}{(z+1)^2}\right)S+\left(\dfrac{1}{z}+\dfrac{1}{z+1}\right)\left(z^a(z+1)^b\dfrac{dS}{dz}+z^a(z+1)^b\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)S\right)+\dfrac{1}{z(z+1)}\left(\dfrac{P^2}{z}-\dfrac{P^2}{z+1}-l(l+1)\right)z^a(z+1)^bS=0$
$\dfrac{d^2S}{dz^2}+\left(\dfrac{2a}{z}+\dfrac{2b}{z+1}\right)\dfrac{dS}{dz}+\left(\dfrac{a(a-1)}{z^2}+\dfrac{2ab}{z(z+1)}+\dfrac{b(b-1)}{(z+1)^2}\right)S+\left(\dfrac{1}{z}+\dfrac{1}{z+1}\right)\dfrac{dS}{dz}+\left(\dfrac{1}{z}+\dfrac{1}{z+1}\right)\left(\dfrac{a}{z}+\dfrac{b}{z+1}\right)S+\left(\left(\dfrac{1}{z}-\dfrac{1}{z+1}\right)\left(\dfrac{P^2}{z}-\dfrac{P^2}{z+1}\right)-\dfrac{l(l+1)}{z(z+1)}\right)S=0$
$\dfrac{d^2S}{dz^2}+\left(\dfrac{2a+1}{z}+\dfrac{2b+1}{z+1}\right)\dfrac{dS}{dz}+\left(\dfrac{a^2+P^2}{z^2}+\dfrac{2ab+a+b-P^2-l(l+1)}{z(z+1)}+\dfrac{b^2+P^2}{(z+1)^2}\right)S=0$
Choose $a=iP$ and $b=-iP$ , the ODE becomes
$\dfrac{d^2S}{dz^2}+\left(\dfrac{2iP+1}{z}-\dfrac{2iP-1}{z+1}\right)\dfrac{dS}{dz}-\dfrac{3P^2+l(l+1)}{z(z+1)}S=0$
Let $z=-t$ ,
Then $\dfrac{d^2S}{dt^2}-\left(\dfrac{2iP+1}{-t}-\dfrac{2iP-1}{-t+1}\right)\dfrac{dS}{dt}-\dfrac{3P^2+l(l+1)}{-t(-t+1)}S=0$
$\dfrac{d^2S}{dt^2}+\left(\dfrac{2iP+1}{t}-\dfrac{2iP-1}{t-1}\right)\dfrac{dS}{dt}-\dfrac{3P^2+l(l+1)}{t(t-1)}S=0$
Which relates to Gaussian hypergeometric equation.
| {
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Log of A to the base 9=Log of B to the base 12=Log of (A+B) to the base of 16, then what is B/A? I tried,
$9^k$=A, $12^k$=B, $16^k$=A+B
Now, $16^k$*$9^k$=$(12^k$)^2
That means, (A+B)*A=$B^2$,
therefore, $A^2$-$B^2$=-AB
and, (A-B)(A+B)=-AB
and, (B-A)(B+A)=AB.
I don't know what to do further.
| Let $\log_9A=\log_{12}B=\log_{16}(a+B)=k$.
Hence, $A=9^k$, $B=12^k$ and $A+B=16^k$, which gives
$$\left(3^k\right)^2+3^k\cdot4^k-\left(4^k\right)^2=0$$ or
$$\left(\frac{3}{4}\right)^k=\frac{\sqrt5-1}{2}.$$
Thus,
$$\frac{B}{A}=\frac{12^k}{9^k}=\left(\frac{4}{3}\right)^k=\left(\frac{4}{3}\right)^{\log_{\frac{3}{4}}\frac{\sqrt5-1}{2}}=\frac{2}{\sqrt5-1}=\frac{\sqrt5+1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $\sum_{k=N+1}^\infty \frac {1}{k!} < \frac {1}{N!}$ I was constructing a proof through inequalities, but I am having a bit of problem showing the following step: $$\sum_{k=N+1}^\infty \frac {1}{k!} < \frac {1}{N!}$$
Is there any quick way to show this?
| All we need to prove is :
$$\sum_{k=N+1}^\infty \frac {N!}{k!} < 1$$
Now, since
$$\frac{1}{N+1}+\frac{1}{(N+1)(N+2)}+\frac{1}{(N+1)(N+2)(N+3)} \ldots <\frac{1}{(N+1)}+\frac{1}{(N+1)(N+2)}+\frac{1}{(N+2)(N+3)}$$
Can you see the telescoping sum now, doing that we get
$$\frac{1}{N+1}+\frac{1}{(N+1)(N+2)}+\frac{1}{(N+1)(N+2)(N+3)} \ldots <\frac{2}{N+1} < 1 ~\forall ~N \ge 2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Asymptotics of a double integral I want to calculate the asymptotic form as $x\to 0$ of the following integral.
\begin{alignat}{2}
I_2(x) = \int_0^{\infty}du\int_0^{\infty}dv\, \frac{1}{(u+v)^{\frac{3}{2}}}\exp\left(-\frac{x}{u+v}\right)
\end{alignat}
How can we solve?
This question is related with this post (Asymptotics of a double integral: $ \int_0^{\infty}du\int_0^{\infty}dv\, \frac{1}{(u+v)^2}\exp\left(-\frac{x}{u+v}\right)$),
or the solutions in three-dimensional space for this post (https://physics.stackexchange.com/questions/61498/a-four-dimensional-integral-in-peskin-schroeder#)
Thank you so much
| $$\int \frac{1}{(u+v)^{\frac{3}{2}}}\exp\left(-\frac{x}{u+v}\right)\,dv=-\frac{\sqrt{\pi } }{\sqrt{x}}\,\text{erf}\left(\frac{\sqrt{x}}{\sqrt{u+v}}\right)$$
$$\int_0^{\infty} \frac{1}{(u+v)^{\frac{3}{2}}}\exp\left(-\frac{x}{u+v}\right)\,dv=\frac{\sqrt{\pi }}{\sqrt{x}}\,\text{erf}\left(\frac{\sqrt{x}}{\sqrt{u}}\right)$$
$$\int \text{erf}\left(\frac{\sqrt{x}}{\sqrt{u}}\right)\,du=(u+2 x)\, \text{erf}\left(\frac{\sqrt{x}}{\sqrt{u}}\right)+\frac{2 \sqrt{u} \sqrt{x}
}{\sqrt{\pi }} e^{-\frac{x}{u}}$$
$$\int_0^M \text{erf}\left(\frac{\sqrt{x}}{\sqrt{u}}\right)\,du=(M+2 x) \,\text{erf}\left(\frac{\sqrt{x}}{\sqrt{M}}\right)+\frac{2 \sqrt{M} \sqrt{x}
}{\sqrt{\pi }}e^{-\frac{x}{M}}-2 x$$ which does not converge if $M\to \infty$.
$$I_2(x)=\frac{\sqrt{\pi }}{\sqrt{x}}\int_0^M \text{erf}\left(\frac{\sqrt{x}}{\sqrt{u}}\right)\,du=2 \sqrt{M} e^{-\frac{x}{M}}-2 \sqrt{\pi } \sqrt{x}+\frac{\sqrt{\pi } (M+2 x)
}{\sqrt{x}}\text{erf}\left(\frac{\sqrt{x}}{\sqrt{M}}\right)\tag 1$$ Using series, we should end with $$I_2(x)=4 \sqrt{M}-2 \sqrt{\pi } \sqrt{x}+\frac{4 x}{3 \sqrt{M}}-\frac{2 x^2}{15
M\sqrt{M}}+O\left(x^3\right)$$
Edit
Concerning the derivatives of $I_2(x)$, using $(1)$, we should have
$$\frac{d}{dx} I_2(x)=\frac{\sqrt{M} e^{-\frac{x}{M}}}{x}+\frac{\sqrt{\pi } \left((M-2 x)
\text{erfc}\left(\frac{\sqrt{x}}{\sqrt{M}}\right)-M\right)}{2 x^{3/2}}$$
$$\frac{d^2}{dx^2} I_2(x)=-\frac{3 \sqrt{M} e^{-\frac{x}{M}}}{2 x^2}+\frac{\sqrt{\pi } \left((3 M-2 x)
\text{erf}\left(\frac{\sqrt{x}}{\sqrt{M}}\right)+2 x\right)}{4 x^{5/2}}$$
Expanding as series around $x=0$, this would lead to
$$\frac{d}{dx} I_2(x)=-\frac{\sqrt{\pi }}{\sqrt{x}}+\frac{4}{3 \sqrt{M}}-\frac{4 x}{15
M^{3/2}}+O\left(x^{2}\right)$$
$$\frac{d^2}{dx^2} I_2(x)=\frac{\sqrt{\pi }}{2 x^{3/2}}-\frac{4}{15 M^{3/2}}+\frac{4 x}{35
M^{5/2}}+O\left(x^{2}\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the number of $2$ inversions in $S_5$? How would I find the number of permutations in $S_5$ that have exactly $2$ inversions?
By inversion I mean if $i<j$ but $\sigma(i)>\sigma(j)$ i.e the order in which an element appears is switched in the permutation.
I have found the following $6$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
1 & 2 & 5 & 3 &4
\end{pmatrix}$$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
1 & 4 & 2 & 3 &5
\end{pmatrix}$$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 1 & 2 & 4 &5
\end{pmatrix}$$
simply by pulling the element $5$ back two places and then $4$ back two places and then the same for $3$. This no longer works with $2$ because we have to move it to $5$ and this gives more than two inversions however we can reverse it and move $1$ forward two places and then move $2$ forward two places and likewise for $3$ this gives.
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
2 & 3 & 1 & 4 &5
\end{pmatrix}$$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
1 & 3 & 4 & 2 &5
\end{pmatrix}$$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
1 & 2 & 4 & 5 &3
\end{pmatrix}$$
this gives us $6$ permutations with only $2$ inversions.
Is this all? Can anyone show me a better way that just trying to figure them out?
Thanks!
| The generating function for inversions is the factorial $q$-analogue
$$\left[n\right]_q!$$
The $q$-analogue of $k$ is
$$\left[k\right]=1+q+q^2+q^3+\cdots +q^{k-1}$$
so the factorial $q$-analogue is
$$\left[n\right]_q!=1(1+q)(1+q+q^2)(1+q+q^2+q^3)\cdots (1+q+q^2+q^3+\cdots +q^{n-1})$$
The number of permutations with $k$ inversions is given by the $x^k$ coefficient $A_{n,k}$ of
$$\left[n\right]_q! = \sum_{k=0}^{\binom{n}{2}}A_{n,k}x^k$$
in your case you want $A_{5,2}$.
There is a simple proof of this inversion generating function in Theorem 1.2 page 3 of this book
Multiplying these polynomials is either a job for a computer algebra system like sage or we can do it manually.
In sage if we input
show(expand(1*(1+x)*(1+x+x^2)*(1+x+x^2+x^3)*(1+x+x^2+x^3+x^4)))
we get the output
$$x^{10} + 4 \, x^{9} + 9 \, x^{8} + 15 \, x^{7} + 20 \, x^{6} + 22 \, x^{5} + 20 \, x^{4} + 15 \, x^{3} + 9 \, x^{2} + 4 \, x + 1$$
which gives 9 as your answer.
Edit: The full list of permutations with $2$ inversions. I have highlighted all occurrences of a larger number occurring to the left of a smaller. The ones accounted for in the question are checked $\checkmark$.
$$\begin{array}{c}\text{Permutations}\\
\begin{array}{|ccccc|c|}\hline
\boxed{2} & \boxed{3} & 1 & 4 & 5 &\checkmark\\
\boxed{2} & 1 & \boxed{4} & 3 & 5&\\
\boxed{2} & 1 & 3 & \boxed{5} & 4&\\
1 & \boxed{3} & \boxed{4} & 2 & 5&\checkmark\\
1 & \boxed{3} & 2 & \boxed{5} & 4&\\
1 & \boxed{4} & 2 & 3 & 5&\checkmark\\
1 & 2 & \boxed{4} & \boxed{5} & 3&\checkmark\\
1 & 2 & \boxed{5} & \boxed{3} & 4&\checkmark\\
\boxed{3} & 1 & 2 & 4 & 5&\checkmark\\\hline
\end{array}\end{array}$$
| {
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"question_score": "1",
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Solve $\sqrt{3+\sqrt{3+x}}=x$
Solve
$$\sqrt{3+\sqrt{3+x}}=x$$
My try:
$$\sqrt{3+\sqrt{3+x}}=x \\
3+\sqrt{3+x}=x^2\\\sqrt{3+x}=x^2-3\\3+x=(x^2-3)^2$$
$$x^4-6x^2+9=x+3\\x^4-6x^2-x+6=0$$
Now ?
| First state the conditions that $x$ has to satisfy:
*
*The initial equation requires $x\ge 0$.
*$\sqrt{3+x}=x^2-3$ requires $x^2\ge 3$, i.e., taking into account the previous condition, $x\ge \sqrt3$.
The resulting equation is easy to factorise: rewrite it as
\begin{align}
x^4-x-6x^2+6&=x(x^3-1)-6(x^2-1)=(x-1)\bigl(x(x^2+x+1)-6(x+1)\bigr)\\&=(x-1)(x^3+x^2-5x-6).
\end{align}
One tests the existence of rational roots for the second factor, among $\pm1, \pm2,\pm3,\pm6$, one finds $-2$, so dividing by $x+2$:
$$x^4-x-6x^2+6=(x-1)(x+2)(x^2-x-3).$$
Set $p(x)=x^2-x-3$. Its discriminant is $\Delta=13$. As $p(\sqrt 3)=-\sqrt 3<0$, $\sqrt 3$ separates the roots, so the only root $\ge\sqrt3$ is
$$x=\frac{1+\sqrt{13}}2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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PreCalc Roots of Unity Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}$$
So far, I have simplified that down to $\frac{-2}{(1+\omega)^2}$, but I don't know how to continue. Thank you!
| We can simplify the individual terms by reducing the numerators to $1$. For example, for the first term we have
$$
\frac{\omega}{1+\omega^2} = \frac{\omega \omega^4}{(1+\omega^2)\omega^4} = \frac{1}{\omega^4 + \omega^6} = \frac{1}{\omega + \omega^4}
$$
and we can do something similar with the other terms. After simplification we find
$$f(\omega) = \frac{2}{\omega+\omega^4} + \frac{2}{\omega^3 + \omega^2}$$
Now it's straightforward to compute the sum of these two terms:
$$f(\omega) = 2 \frac{(\omega^3 + \omega^2) + (\omega + \omega^4)}{(\omega^3 + \omega^2) (\omega + \omega^4)} = 2 \frac{\omega + \omega^2 + \omega^3 + \omega^4}{\omega^3 + \omega^4 +\omega^6 + \omega^7}= 2 \frac{\omega + \omega^2 + \omega^3 + \omega^4}{\omega^3 + \omega^4 +\omega + \omega^2}=2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve this partial fraction decomposition? Please help me to solve the following partial fraction decomposition:
$$\frac{1-v^2}{v+v^3} = \frac{A}{v}+\frac{Bv+C}{1+v^2}$$
| If you play around up- and downstairs you find:
\begin{align*}
\frac{1-v^2}{v + v^3} & = \frac{1-v^2 + v^2 - v^2}{v \, (1+v^2)} \\
& = \frac{1+v^2}{v(1+v^2)} + \frac{-2v^2}{v (1+v^2)} \\
& = \frac{1}{v} - \frac{2 v}{1+v^2}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
An Unconventional Elliptic Integral? I came across the following integral recently:
$$ I = \int_a^b d \lambda \sqrt{(\lambda^2-a^2)(b^2 - \lambda^2)} $$
The author of the paper claims that this integral can be transformed into an elliptic integral, giving the answer:
$$ I = b [(a^2 + b^2) E(k) - 2a^2 K(k)] \quad k^2 = \frac{b^2 - a^2}{b^2} $$
I tried to manipulate $I$ into one of the standard elliptic integrals, but I couldn't quite get it right. The most promising form I have obtained is via the substitution:
$$ \lambda = \sqrt{\frac{a^2 + b^2}{2}} \sin \theta $$
Which gives the integral:
$$ I = \sqrt{\frac{a^2 +b^2}{2}} \int d \theta \, \cos \theta \sqrt{\left(\frac{a^2-b^2} 2 \right)^2 - \left(\frac{a^2+b^2} 2\right)^2 \cos^4 \theta } $$
Am I on the right track? Is there some nice trick to evaluate this integral?
| You have a typo in your expression, your expression is missing an overall factor $\frac13$.
You can verify that by comparing the integral and expression you have at $a \to 0$ and $b = 1$.
At that limit, $I \to \frac13$ while your expression
$$b ((a^2 + b^2) E(k) - 2a^2 K(k))\quad\to\quad 1((0^2+1^2)\times 1 - 2\times 0 ) = 1$$
Let $c^2 = b^2 - a^2$ and $\displaystyle\;k = \frac{c}{b}$. Change variable to
$$u = \frac1c \sqrt{b^2 - \lambda^2}
\quad\iff\quad \lambda = \sqrt{b^2 - c^2u^2} = b\sqrt{1-k^2u^2}$$
Notice
$$\lambda^2 - a^2 = (b^2 - a^2) - (b^2 - \lambda^2) = c^2(1-u^2)
\quad\text{ and }\quad
d\lambda = -\frac{bk^2u du}{\sqrt{1-k^2u^2}}$$
The integral at hand equals to
$$I = \int_1^0 c^2u\sqrt{1-u^2}\left(-\frac{bk^2u du}{\sqrt{1-k^2u^2}}\right)
= b^3k^4\int_0^1 \frac{u^2(1-u^2)}{\sqrt{(1-u^2)(1-k^2u^2)}}du
$$
To evaluate this integral, one can use the fact (trick?)
$$\frac{d}{du}\left[u\sqrt{(1-u^2)(1-k^2u^2)}\right]
= \frac{1 - 2u^2 + k^2u^2 - 3k^2u^2(1-u^2)}{\sqrt{(1-u^2)(1-k^2u^2)}}$$
and
$u\sqrt{(1-u^2)(1-k^2u^2)}$ vanishes at $u = 0$ and $1$. With this, one can transform the integral to
$$I = \frac{b^3k^2}{3}\int_0^1\frac{1 - 2u^2 + k^2u^2}{\sqrt{(1-u^2)(1-k^2u^2)}} du$$
Notice $$1 - 2u^2 + k^2u^2 = 1 + (k^2-2)\frac{1 - (1-k^2u^2)}{k^2} =
\frac{1}{k^2}\left((2-k^2)(1-k^2u^2) - (2-2k^2)\right)$$
We can simplify the integral to
$$\begin{align} I = &\frac{b^3}{3}\left[(2-k^2)\int_0^1\sqrt{\frac{1-k^2u^2}{1-u^2}}du - (2-2k^2)\int_0^1\frac{1}{\sqrt{(1-u^2)(1-k^2u^2)}}du\right]\\
= & \frac{b^3}{3}\left[(2-k^2)E(k) - (2-2k^2)K(k)\right]\\
= & \frac{b}{3}\left[(b^2+a^2)E(k) - 2a^2K(k)\right]
\end{align}
$$
Up to a factor $\frac13$, this is the expression you have.
| {
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"timestamp": "2023-03-29T00:00:00",
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what's the approach used in this integral? in : $\displaystyle \int (\sin 3x)(\cos 5x)\, dx $ ( which recurred as hyperbolic sin and cosine as well )
*
*ive noted the resemblance to the sum of $ \sin 8x$ but didnt know how to use it
*ive noted that $ \cos 5x $ can be expanded to $ \cos (2x+3x ) $ but again failed to simplify the resulted form
| *
*Someone already suggested using the identity $$\sin A \cos B = \frac{ \sin(A-B) + \sin(A+B)} 2.$$ That's one way.
*Another way is this:
\begin{align}
& \int \sin(3x)\cos(5x)\,dx = \int \frac{e^{3ix} - e^{-3ix}}{2i} \cdot \frac{e^{5ix}+ e^{-5ix}} 2 \,dx \\[10pt]
= {} & \frac 1 {4i} \int (e^{8ix} - e^{2ix} - e^{-2ix} + e^{-8ix}) \,dx
\end{align}
then integrate term by term and then use $e^{i\theta} = \cos\theta + i\sin\theta$ to get a function of sines and cosines.
*And here's another way:
\begin{align}
& \int \sin(3x)\Big(\cos(5x)\,dx\Big) = \overbrace{\int u\,dv = uv - \int v\,du}^\text{integration by parts} \\[10pt]
= {} & \sin(3x) \frac{\sin(5x)} 5 - \int \frac {\sin(5x)} 5 \cdot 3\cos(3x)\,dx \\[10pt]
= {} & \frac 1 5 \sin(3x)\sin(5x) - \frac 3 5 \int \cos(3x) \Big(\sin(5x)\,dx\Big) \\[10pt]
= {} & \frac 1 5 \sin(3x)\sin(5x) - \frac 3 5 \left( \int s\,dt \right) \\[10pt]
= {} & \frac 1 5 \sin(3x)\sin(5x) - \frac 3 5 \left( st - \int t \,ds \right) \\[10pt]
= {} & \frac 1 5 \sin(3x)\sin(5x) - \frac 3 5 \left( \frac{-1} 5 \cos(3x)\cos(5x) - \int \frac 3 5 \sin(3x)\cos(5x)\,dx \right) \\[10pt]
= {} & \frac 1 5 \sin(3x)\sin(5x) + \frac 3 {25} \cos(3x)\cos(5x) + \frac 9 {25} \int \sin(3x)\cos(5x)\, dx \\[10pt]
& \text{and so we have:} \\
I = {} & \frac 1 5 \sin(3x)\sin(5x) + \frac 3 {25} \cos(3x)\cos(5x) + \frac 9 {25} I. \\[10pt]
& \text{Adding } \frac 9 {25} I \text{ to both sides, we get:} \\
\frac{34}{25} I = {} & \frac 1 5 \sin(3x)\sin(5x) + \frac 3 {25} \cos(3x)\cos(5x) + \text{constant} \\[10pt]
& \text{and then multiply both sides by } 25/34.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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why isn't $\int \ln\left((x+1)(x+2)\right)\ dx = 1/(x+1) + 1/(x+2) + C$? by the product rule for Logarithm,
$$\ln(xy) = \ln(x) + \ln(y) $$
the question is, $\int \ln\left(x^2 + 3x + 2\right)\ dx = \int \ln\left((x+1)(x+2)\right)\ dx$
so, I thought I could divide them into $\int\ln(x+1)\ dx + \int\ln(x+2)\ dx$ and the answer will be $\1/(x+1) + 1(x+2) + C$,
however, the system gave me that that is wrong answer and system showed another answer with using another formular.
So I want to know why my answer is wrong.
Thank you!
| This is because the derivative of the natural logarithm is the reciprocal function. The antiderivative of the natural logarithm function is:
$$ \int \ln x \, dx = x \ln x - x + C $$
So:
\begin{align*}
\int \ln(x^2 + 3x + 2) \, dx &= \int \ln[(x+1)(x+2)] \, dx \\
&= \int \ln[(x+1)(x+2)] \, dx \\
&= \int [\ln(x+1) + \ln(x+2)] \, dx \\
&= (x+1)\ln(x+1) - (x+1) + (x+2)\ln(x+2) - (x+2) + C \\[5pt]
&= (x+1)\ln(x+1) + (x+2)\ln(x+2) - 2x + C \\[5pt]
&= x\ln(x^2+3x+2) + \ln(x+1) + 2\ln(x+2) - 2x + C \\
\end{align*}
| {
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Using digits $0,1,2,3,4$ and $5$ only once, how many $5$ digit numbers can be formed which are divisible by $4$? Using digits $0,1,2,3,4$ and $5$ only once, how many $5$ digit numbers can be formed which are divisible by $4$?
I worked on this problem and I have the solution as:
The number of 5 digit numbers formed will be $5 \cdot 5 \cdot 4 \cdot 3 \cdot 2$, which comes to $600$. Now, to get the numbers divisible by $4$, what I did was $600/4$, which comes out to be $150$, but the answer to this problem is $144$.
Please tell me where I went wrong, and if there is a better approach, please provide details about it.
| We can think to this "scheme": you have five ordered empty "boxes" and each of them has to be filled with one digit from $0,\ldots,5$ (which are $6$ "objects") with no repetitions, in order to get a $5$-digits number. This means that the last two digits must form a number which is divisible by $4$; then the possible last two digits are as follows: $04$, $12$, $20$, $24$, $32$, $40$, $52$. Now, take for instance $04$; how many numbers (strings) of five distinct digits ending with $04$ are there? Since the last two digits are taken, we have to choose the first three among the remaining ones, that is we have to choose and order $3$ objects out of the $4$ left: this means we have $D_{4,3} = 4 \cdot 3 \cdot 2$ such strings. The same reasoning applies to $20$ and $40$, but not to $12$, $24$, $32$ and $52$: in these cases, in facts, for the remaining three digits $0$ is available but cannot be chosen as first digit, since we want $5$-digits numbers. Thus, in these four cases, the first digit can be filled in $3$ ways (and not $4$, because $0$ is also excluded), the second in $3$ ways and the third in $2$, giving $3\cdot 3\cdot 2$ possible numbers for each couple of final two non-zero digits. Hence, since for $04$, $20$ and $40$ (which are $3$ "cases") we apply the first reasoning, and for $12$, $24$, $32$ and $52$ ($4$ "cases") the second one, the total amount of numbers of five distinct digits (chosen among $0,\ldots,5$) that are divisible by $4$ is $$3 \cdot (4 \cdot 3 \cdot 2) + 4 \cdot (3 \cdot 3 \cdot 2) = 144.$$
| {
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Prove that matrix can be square of matrix with real entries Prove that matrix
\begin{bmatrix}1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}
can be square of matrix with all real entries.
I have found one such matrix to be
\begin{bmatrix}1&0&0\\0&1&-1\\0&2&-1\end{bmatrix} but is there an elegant way to do it without any trial and error?
| I will try to present a solution "without any trial and error".
Here we have a matrix $F$ in the form
$\begin{bmatrix}
1 & 0_{1 \times 2} \\ 0_{2 \times 1} & A_{2 \times 2} \end{bmatrix}$ so if $G^2=F$ then $G$ can be in the form $\begin{bmatrix}
\pm 1 & 0_{1 \times 2} \\ 0_{2 \times 1} & B_{2 \times 2} \end{bmatrix}$ where $B^2=A$.
Now let's concentrate on $B^2=\begin{bmatrix}
-1 & 0 \\ 0 & -1 \end{bmatrix} $.
We have two matrix equations
$B^2=-I$ i.e.
$B^2+I=0$
and general equation from Cayley-Hamilton theorem for $ 2 \times 2$ matrices
$B^2-\text{tr}(B)B+\det(B)I=0$.
Comparing both equations we obtain
$\text{tr(B)}=0$ , $ \det(B)=1$.
So if we denote $ B=\begin{bmatrix} a & b \\c & d\end{bmatrix} $ then $d=-a$ and consequently $-a^2-bc=1$.
These conditions are sufficient to obtain a solution not only with real numbers, but even with integer values.
For example
if $a=0$ then $b=1$, $c=-1$
($b,c$ should have always opposite signs because -$a^2-bc$ has to be positive - possible other solution $-1,1$, also $a$ and $d$ can be switched)
if $a=1$ then $b=2$, $c=-1$
if $a=2$ then $b=5$, $c=-1$
if $a=3$ then $b=5$, $c=-2$
...
if $a=8$ then $b=5$, $d=-13$ , etc... infinite number of solutions - for every integer $a$ we can find proper integer values of $b$ and $c$ from $-bc=a^2+1$..
Let's check the last listed solution.
Indeed
$ \begin{bmatrix} 8 & 5 \\-13 & -8\end{bmatrix} \begin{bmatrix} 8 & 5 \\-13 & -8\end{bmatrix} = \begin{bmatrix} 8 \cdot {8} - 5\cdot 13 & 8\cdot 5 - 5 \cdot 8 \\ -13\cdot 8 + 8 \cdot 13 & -13\cdot 5 +8\cdot 8\end{bmatrix} = \begin{bmatrix} -1 & 0 \\0 & -1\end{bmatrix} $ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Different Representations of GF(8) Can anyone point me in the right direction with the following problem?
Given that
$$GF(8)=\frac{Z_{2}}{x^3+x^2+1}= \frac{Z_{2}}{x^3+x+1}$$
Find $\beta$ as a function of $\alpha$ , where $\alpha$ is a root of $x^3+x+1$ and $\beta$ is a root of $x^3+x^2+1$
| Brief answer:
As one polynomial is just the reverse of the other, the roots of one are the inverses of the roots of the other. Hence we may take $\beta=\alpha^{-1}$
Here is the complete story:
One can find the three possibilities this way.
Neither $0$ nor $1$ is a root of either $X^3+X^2+1$ or $X^3+X+1$; and it is trivial to see that these polynomials are coprime and so have no common root.
The order of the multiplicative group of $\mathbb{F}_8$ is seven and so is cyclic of that order.
If $\alpha$ is a root of $X^3+X^2+1$ so is $\alpha^2$ since $\alpha^6+\alpha^4+1=(\alpha^3+\alpha^2+1)^2=0$. Hence the roots of this polynomial are $\alpha, \alpha^2, \alpha^4(=(\alpha^{2})^2)$ and these are distinct as the multiplicative order of $\alpha$ is $7$.
In the same way the roots of $X^3+X+1$ are $\beta,\beta^2,\beta^4$.
That accounts for all the elements of $\mathbb{F}_8$.
Therefore $\beta,\beta^2,\beta^4$ must be the "missing" powers of $\alpha$, namely $\alpha^3,\alpha^5,\alpha^6$: or more elegantly $\alpha^{-1}, \alpha^{-2}, \alpha^{-4}$.
So $\beta$ is one of $\alpha^{-1}, \alpha^{-2}, \alpha^{-4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Closed-form expression of sum of series $\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\frac{x^k}{(k+2n)!}$
I would like a closed-form expression of the following series:
$\left(\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\right)+\left(\frac{x}{3!}+\frac{x^2}{4!}+\frac{x^3}{5!}+\frac{x^4}{6!}+\cdots\right)+\left(\frac{x}{5!}+\frac{x^2}{6!}+\frac{x^3}{7!}+\frac{x^4}{8!}+\cdots\right)+\cdots$
Clearly the first bracket is $e^x-1$, the second bracket is $x^{-2}(e^x-1-x-\frac{x^2}{2!})$, the third bracket is $x^{-4}(e^x-1-x-\frac{x^2}{2!}-\frac{x^3}{3!}-\frac{x^4}{4!})$, and so on. Is there a way to combine these infinitely many terms into a closed form expression?
| For any formal Laurent series in $t$, $f(t) = \sum\limits_{\ell=-\infty}^\infty \alpha_\ell t^\ell$, we will use the notation $[t^\ell] f(t)$ to denote the coefficient $\alpha_\ell$ in front of the monomial $t^\ell$. When $f(t)$ defines a
function holomorphic over an annular region $\mathcal{A} \subset \mathbb{C}$, $[t^\ell] f(t)$ can be reinterpreted as a contour integral over some suitably chosen circle $C_R = \{ t : |t| = R \}$ lying within $\mathcal{A}$.
$$[t^\ell]f(t) \quad\longleftrightarrow\quad\frac{1}{2\pi i}\oint_{C_R} \frac{f(t)}{t^{\ell+1}} dt$$
The series at hand can be rewritten as
$$\begin{align}
\mathcal{S} \stackrel{def}{=}\sum_{\ell=0}^\infty \sum_{k=1}^\infty \frac{x^k}{(k+2\ell)!}
&= \sum_{\ell=0}^\infty \sum_{k=1}^\infty \sum_{m=0}^\infty \delta_{m,k+2\ell} \frac{x^k}{m!}
= \sum_{\ell=0}^\infty \sum_{k=1}^\infty \sum_{m=0}^\infty \delta_{m-k,2\ell} \frac{x^k}{m!}\\
&= \sum_{\ell=0}^\infty \left\{ \sum_{k=1}^\infty \sum_{m=0}^\infty [t^{2\ell}]\left(\frac{x}{t}\right)^k \frac{t^m}{m!}\right\}\tag{*1}
\end{align}
$$
Over the complex plane, the sum
$\displaystyle\;\sum\limits_{k=1}^\infty \sum\limits_{m=0}^\infty \left(\frac{x}{t}\right)^k \frac{t^m}{m!}\;$
converges absolutely for $|t| > |x|$.
If we interpret the expression inside curly braces of $(*1)$ as contour integrals over circle $C_R$ with $R > |x|$, we can switch the order of summation and $[\ldots]$.
We find
$$\mathcal{S} = \sum_{\ell=0}^\infty [t^{2\ell}]
\sum\limits_{k=1}^\infty \sum\limits_{m=0}^\infty \left(\frac{x}{t}\right)^k \frac{t^m}{m!}
= \sum_{\ell=0}^\infty [t^{2\ell}] \frac{xe^t}{t-x}
= \sum_{\ell=0}^\infty [t^0] t^{-2\ell} \frac{xe^t}{t-x}
$$
Over the complex plane, the term $\sum\limits_{\ell=0}^\infty t^{-2\ell}$ converges absolutely when $|t| > 1$. If we choose $R > \max\{ |x|, 1 \}$,
we can change the order of summation and $[\cdots]$ again and get
$$\mathcal{S}
= [t^0]\left[\left(\sum_{\ell=0}^\infty t^{-2\ell}\right)\frac{xe^t}{t-x}\right]
= [t^0] \left(\frac{t^2}{t^2-1}\frac{xe^t}{t-x}\right)
= \frac{1}{2\pi i}\oint_{C_R} \frac{tx e^t}{(t^2-1)(t-x)} dt
$$
Since $R > \max\{ |x|, 1 \}$, there are 3 poles $x, \pm 1$ inside $C_R$. If one
sums over the contributions from these poles, one obtains
$$\begin{align}
\mathcal{S}
&= \frac{x^2 e^x}{(x^2-1)} + \frac{xe}{2(1-x)} + \frac{-xe^{-1}}{(-2)(-1-x)}\\
&= \frac{x}{2(x^2-1)}\left[ 2xe^x - e (x+1) - e^{-1}(x-1)\right]\\
&= \frac{x}{2(x^2-1)}\left[ x(2e^x - (e+e^{-1})) - (e-e^{-1})\right]
\end{align}
$$
Reproducing what Did mentioned in hir comment.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Lagrange reduction of quadratic form to canonical form How can I normalize quadratic form using Lagrange method?
$f=x_1^2+x_2^2+4x_3^2+x_4^2+2x_1x_2+4x_1x_3-2x_1x_4+4x_2x_3-6x_2x_4$
Any kind of help is appreciated.
Lagrange's reduction: http://www.solitaryroad.com/c138.html
| Changing to $\;x,y,z,w\;$ :
$$\begin{align*}&\color{red}{x^2}+\color{red}{y^2}+\color{green}{4z^2}+w^2+\color{red}{2xy}+\color{green}{4xz}-2xw+4yz-6yw=\\{}\\
&=(x+y)^2+4\left(z+\frac12x\right)^2-x^2+w^2-2xw+4yz-6yw=\\{}\\
&=(x+y)^2+4\left(z+\frac12x\right)^2-(x+w)^2+2w^2+4yz-6yw=\\{}\\
&=(x+y)^2+4\left(z+\frac12x\right)^2-(x+w)^2+2\left(w-\frac32y\right)^2-\frac92y^2+4yz=\\{}\\
&=(x+y)^2+4\left(z+\frac12x\right)^2-(x+w)^2+2\left(w-\frac32y\right)^2-\frac92\left(y-\frac49z\right)^2+\frac89z^2\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}$ I am trying to find the limit as $x\to 8$ of the following function. What follows is the function and then the work I've done on it.
$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}$$
\begin{align}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} &= \frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} \times \frac{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}} \\\\
& = \frac{\frac{1}{x+1}-\frac{1}{9}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac{8-x}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac {-1}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}\end{align}
At this point I try direct substitution and get:
$$ = \frac{-1}{\frac{2}{3}}$$
This is not the answer. Could someone please help me figure out where I've gone wrong?
| $$\lim\limits_{x\rightarrow8}\frac{\frac{1}{\sqrt{x+1}}-\frac{1}{3}}{x-8}=\lim\limits_{x\rightarrow8}\frac{8-x}{3(x-8)\sqrt{x+1}\left(3+\sqrt{x+1}\right)}=-\frac{1}{3\cdot3\cdot6}=-\frac{1}{54}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$ then $x+\dfrac{1}{x}=?$
let :
$$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$$
then :
$$x+\dfrac{1}{x}=?$$
My try :
$$\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18}=\sqrt3+\sqrt 3\times\sqrt2+4+3\sqrt2 \\ \sqrt3(1+\sqrt2+3)+4$$
Now what ?
| Note that
$$\begin{align}
(1+\sqrt2)(2+\sqrt2+\sqrt3)
&=(2+\sqrt2+\sqrt3)+(2\sqrt2+2+\sqrt6)\\\\
&=4+3\sqrt2+\sqrt3+\sqrt6\\\\
&=\sqrt{16}+\sqrt{18}+\sqrt3+\sqrt6
\end{align}$$
Thus
$$x={\sqrt3+\sqrt6+\sqrt{16}+\sqrt{18}\over\sqrt2+\sqrt3+\sqrt4}={(1+\sqrt2)(2+\sqrt2+\sqrt3)\over\sqrt2+\sqrt3+2}=1+\sqrt2$$
so that
$${1\over x}={1\over1+\sqrt2}={1-\sqrt2\over1-2}=\sqrt2-1$$
and thus
$$x+{1\over x}=2\sqrt2$$
Remark: The first part of the derivation, starting with the factorization, is, needless to say, presented without all the finagling that went into finding it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Derive a recurrence relation for $(a_n)=\int_0^{{\pi / 3}}(\cos{x})^{-n}dx$ I am trying to derive a recurrence relation for
$$(a_n)=\int_0^{{\pi\over 3}}(\cos{x})^{-n}dx$$
$n=0,1,2,\dots$
It is pretty obvious $a_0={\pi \over 3}$. I get stuck on $a_1=\int_0^{{\pi / 3}}{1\over\cos{x}}dx$. I am able to transform it to $a_1=\int_0^{{\sqrt{3} / 2}}{1\over1-y^2}dy$ but not further. Also, this is still quite far from a recurrence equation. Do you have any suggestions?
| One can just power through and discover
\begin{align}
a_{n} &= \int_{0}^{\pi/3} \left(\frac{1}{\cos(x)}\right)^{n} \, dx \\
&= \left[ \sin(x) \, {}_{2}F_{1}\left(\frac{1}{2}, \frac{n+1}{2}; \frac{3}{2}; \sin^{2}(x) \right) \right]_{0}^{\pi/3} \\
&= \frac{\sqrt{3}}{2} \, {}_{2}F_{1}\left(\frac{1}{2}, \frac{n+1}{2}; \frac{3}{2}; \frac{3}{4}\right).
\end{align}
It is of interest to note that it becomes evident that
$$ {}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; \frac{3}{2}; \frac{3}{4}\right) = \frac{2 \, \pi}{3 \, \sqrt{3}}. $$
There are several recurrence relations for the ${}_{2}F_{1}$-hypergeometric functions for which relations can be developed. An example is to use:
\begin{align}
(b-c+1) \, {}_{2}F_{1}\left(a,b; c; x\right) &= (2b - c+2 +(a-b-1) \, x) \, {}_{2}F_{1}(a,b+1; c; x) \\
& \hspace{20mm}+ (b+1)(x-1) \cdot \, {}_{2}F_{1}(a,b+2; c; x)
\end{align}
which leads to
\begin{align}
(n+3) \, a_{n+4} = (5 n +6) \, a_{n+2} - 4 n \, a_{n}.
\end{align}
Since $a_{0} = \pi/3$ and $a_{1} = \ln(2+\sqrt{3})$ then the remaining terms can be developed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{n\to\infty}\sin^4x+\frac{1}{4}\sin^42x+.....+\frac{1}{4^n}\sin^42^nx$ $\lim_{n\to\infty}\sin^4x+\frac{1}{4}\sin^42x+.....+\frac{1}{4^n}\sin^42^nx$
I feel that this question will be solved by telescoping series but i cannot express it as telescoping series.
$\sin^4x=(\frac{1-\cos2x}{2})^2=\frac{1+\cos^22x-2\cos2x}{4}$
I am not able to solve it further.
The options are $(A)\sin^4x,(B)\sin^2x,(C)\cos^2x$(D)does not exist
| Using the hint given by Almot1960 we have $\sin^4 x =\frac{3}{8} \color{red}{-\frac{1}{2} \cos(2x)} \color{blue}{ +\frac{1}{8} \cos (4x) }$ ...
\begin{eqnarray*}
\sin^4 x +\frac{1}{4} \sin^4 2x + \frac{1}{4^2} \sin^4 4x +\frac{1}{4^3} \sin^4 8x + \cdots \\
=\frac{3}{8} \color{red}{-\frac{1}{2} \cos(2x)} \color{blue}{ +\frac{1}{8} \cos (4x) } +\frac{3}{8} \frac{1}{4} \color{blue}{-\frac{1}{8} \cos(4x)} \color{red}{ +\frac{1}{32} \cos (8x) } +\frac{3}{8}\frac{1}{4^2} \color{red}{-\frac{1}{32} \cos(8x)} \color{blue}{ +\frac{1}{128} \cos (16x) } \cdots
\end{eqnarray*}
So you were absolutely right , it is a telescoping sum & lots of the terms cancel. The terms in black are simply a geometric series
\begin{eqnarray*}
\frac{3}{8} \left( 1+ \frac{1}{4}+\frac{1}{4^2}+ \cdots \right)= \frac{1}{2}
\end{eqnarray*}
So the sum is $ \frac{1- \cos(2x)}{2} =\sin^2 (x)$ and the answer is $\color{red}{(B)}$.
| {
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Prove by induction that $ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$
Prove by induction that
$$ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$ with $n \geq 1$
*
*Testing n=1:
$$\sin(x)= \frac {1-\cos(2x)}{2\sin(x)}$$
$$2\sin^2(x)=1-\cos(2x)$$
$$2\sin^2(x)=1-[\cos^2(x)-\sin^2(x)]$$
$$\sin^2(x)=1-\cos^2(x)$$
$$\sin^2(x)+\cos^2(x) =1$$
It shows that n=1 yields a true identity (Pythagorean identity)
*Let's assume that $P_n$ is true:
$$\ sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$
*Let's consider adding $\sin(2n+1)x$ to $P_n$:
$$ \sin(x) +\sin(3x) +...+ \sin (2n-1)x+ \sin(2n+1)x= \frac{1-\cos(2nx)}{2\sin x} + \sin(2n+1)x$$
Considering the RHS:
$$\frac{1-\cos(2nx)}{2\sin x} + \sin (2n+1)x$$
$$\frac{1-\cos(2nx)+ \sin(2n+1)x \cdot 2\sin x}{2 \sin x}$$
$$\frac{1-\cos(2nx)+ 2[\sin(2n+1)x \cdot sin x]}{2sinx}$$
$$\frac{1-\cos(2nx)+ 2 \cdot \frac{1}{2} [\cos[(2n+1)x-x]-\cos[(2n+1)x+x]]}{2\sin x}$$
$$\frac{1-\cos(2nx)+ \cos(2nx)-\cos(2n+2)x}{2\sin x}$$
$$\frac{1-\cos(2n+2)x}{2\sin x}$$
It follows that
$$\sin(x) +\sin(3x) +...+ \sin [(2(n+1)-1)x]= \frac{1-\cos(2(n+1)x)}{2\sin x}$$
Therefore,
$$ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$ is true
Any input is much appreciated.
| You can also use $$\sin(kx)=\frac{e^{ikx}-e^{-ikx}}{2i}$$and geometric series for a direct proof.
| {
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Complex: $z^2 = 1+2i$ (find real- and imaginary part of $z$) these are my toughs:
$$z^2 = 1 + 2i \Longrightarrow (x+yi)(x+yi) = 1 + 2i$$
so: $x^2-y^2 = 1$ and $2xy = 2$
then i got that $x = 1/y$ but i cant continue to find the real- and imaginary part of z anymore. Appriciated any help
| In general, we have this
Lemma: If $z=a+ib \in \mathbb{C}$, with $a,b\in\mathbb{R}$, then
$$ w = \sqrt{\frac{|z|+a}{2}} + i\epsilon\sqrt{\frac{|z|-a}{2}},$$
where $\epsilon =\pm 1$ according to $b=\epsilon|b|$, satisfies $w^2 = z$.
Proof: Let $w=x+iy$ satisfying $w^2=z$. Then $x^2 - y^2+2xyi =a+bi$.
This equation is equivalent to the system
$$ x^2 - y^2 = a \text{ ; } 2xy = b.$$
Since $w^2 = z$, we have $x^2+y^2 = |z|$ too, and we can conclude that
$$x^2 = \frac{|z|+a}{2} \text{ ; } y^2 = \frac{|z|-a}{2}.$$
Choosing the positive square roots, we can write
$$x = \sqrt{\frac{|z|+a}{2}} \text{ ; }
y = \epsilon\sqrt{\frac{|z|-a}{2}}$$
satisfying $2xy = b$, i.e., $\epsilon = 1$ if $b > 0$ and $\epsilon = -1$ if $b < 0$.
| {
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An integral and series to prove that $\log(5)>\frac{8}{5}$ A relationship between $\log(5) \approx 1.6094$ and $\dfrac{3}{2}=1.5$ can be justified by the harmonic approximation $$\log(5) \approx H_2=1+\frac{1}{2}=\frac{3}{2}$$
that can be obtained by regrouping Lehmer's logarithm
$$\log(5) = \sum_{k=0}^\infty \left(\frac{1}{5k+1}+\frac{1}{5k+2}+\frac{1}{5k+3}+\frac{1}{5k+4}-\frac{4}{5k+5}\right)$$
symmetrically around the negative term
$$\log(5)=\frac{3}{2}+\sum_{k=1}^\infty \left( \frac{1}{5k-2}+\frac{1}{5k-1}-\frac{4}{5k}+\frac{1}{5k+1}+\frac{1}{5k+2} \right)$$
The corresponding integral is
$$\log(5)-\frac{3}{2}=\int_0^1 \frac{x^2(1-x)(1+3x+x^2)}{1+x+x^2+x^3+x^4}\:dx$$
(answer https://math.stackexchange.com/a/1656356/134791 by Olivier Oloa)
This is a direct proof that $\log(5)>\dfrac{3}{2}$ because the integrand is non-negative in $(0,1)$.
However, $\dfrac{8}{5}=1.6$ would be a closer approximation using small numbers, so the natural question is:
What are Dalzell-type integral and series for $\log(5)-\dfrac{8}{5}$?
| I'm not quite sure what you're looking for, but I get
$$ \log(5) = \frac{8}{5} -\frac{1}{5}\int_0^1 {\frac {x \left( 12\,{x}^{5}+10\,{x}^{4}+15\,{x}^{3}-5\,{x}^{2}-
7 \right) }{{x}^{4}+{x}^{3}+{x}^{2}+x+1}}
\; dx $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A Question about solution of Laplace's equation in p.d.e Let $u:\mathbb{R}^2\setminus\{(0,0)\}\rightarrow \mathbb{R}$ be a $C^2$ function satisfying $\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}=0$,for all $(x,y)\neq (0,0)$ suppose $u$ is the form $u(x,y)=f(\sqrt{x^2+y^2})$, where $f:(0,\infty)\rightarrow \mathbb{R}$ is non constant function , then
$1. \lim _{x^2+y^2\rightarrow 0}|u(x,y)|=\infty$
$2. \lim _{x^2+y^2\rightarrow 0}|u(x,y)|=0$
$3. \lim _{x^2+y^2\rightarrow \infty}|u(x,y)|=\infty$
$4. \lim _{x^2+y^2\rightarrow \infty}|u(x,y)|=0$
My attempt:
solution of the Laplace's equation with no boundary conditions:
so general solution of the this laplace's equation
but how to solve this question
| $u(x,y)=f(\sqrt{x^2+y^2})\\
u_x = \frac{x}{\sqrt{x^2+y^2}}f'(\sqrt{x^2+y^2})\\u_{xx} = \frac{\left[\sqrt{x^2+y^2}\left[\frac{x^2}{\sqrt{x^2+y^2}}f''(\sqrt{x^2+y^2})+f'(\sqrt{x^2+y^2}) \right]-\frac{x^2}{\sqrt{x^2+y^2}}f'(\sqrt{x^2+y^2})\right]}{(\sqrt{x^2+y^2})^2}\\u_{xx} = \frac{x^2 f''(\sqrt{x^2+y^2})}{(\sqrt{x^2+y^2})^2}+\frac{f'(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}-\frac{x^2 f'(\sqrt{x^2+y^2})}{(\sqrt{x^2+y^2})^3} \\ Let \; r=\sqrt{x^2+y^2}\;\; then \\u_{xx}=\frac{x^2f''(r)}{r^2}+\frac{f'(r)}{r}-\frac{x^2f'(r)}{r^3}\\ similarly \;\; u_{yy}=\frac{y^2f''(r)}{r^2}+\frac{f'(r)}{r}-\frac{y^2f'(r)}{r^3}\\$
hence the given Laplace equation $u_{xx}+u_{yy}=0$ becomes
$\frac{(x^2+y^2)f''(r)}{r^2}+\frac{2f'(r)}{r}-\frac{(x^2+y^2)f'(r)}{r^3}=0\\ \frac{r^2f''(r)}{r^2}+\frac{2f'(r)}{r}-\frac{r^2f'(r)}{r^3}=0\\ f''(r)+\frac{2f'(r)}{r}-\frac{f'(r)}{r}\\ f''(r)+\frac{f'(r)}{r}=0\\ \frac{f''(r)}{f'(r)}=-\frac{1}{r}$
on integration we get
$ \;\;log(f'(r))=-logr+loga \\ f'(r)=\frac{a}{r}$
on integrating again we get $\;\; f(r)=a\ logr+b $
f(r) is nonconstant if $a\ne 0 \\ as \;\;x^2+y^2\to 0, \;\;r^2\to 0 \Rightarrow r\to 0 \\ hence \;\; lim_{x^2+y^2\to 0} |u(x,y)|=lim_{r\to 0} |f(r)|=lim_{r\to 0}( a\ logr+b)=\infty\\ also \;\;lim_{x^2+y^2\to \infty} |u(x,y)|=lim_{r\to \infty} |f(r)|=lim_{r\to \infty}( a\ logr+b)=\infty $
option 1 and 3 are correct
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2301623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Derivative of oscillatory integral with absolute value For the function $F(x) = \int_0^x |\cos(1/u)| du$, I want to determine if the derivative exists at $x=0$ and find $F'(0)$.
I figured this out when the absolute value is removed using integration by parts:
$$\frac{F(h) - F(0)}{h} = \frac{1}{h}\int_0^h\cos(1/u)du = \frac{1}{h}\int_{1/h}^\infty\frac{\cos(x)}{x^2}dx \\= - h \sin(1/h) + \frac{1}{h}\int_{1/h}^\infty\frac{2\sin(x)}{x^3}dx = -h \sin(h) + \frac{1}{h}O(h^2)$$
So $F'(0) = 0$ here.
This breaks down for $F(x) = \int_0^x \cos(1/u)du$.
Thank you for any help.
| As you already observed, $F(h) = \int_{1/h}^\infty \frac{|\cos x|}{x^2}\,dx$. Now consider one of the pieces of the transformed integral between sign switches, on $[(n-1/2)\pi, (n+1/2)\pi]$. On this interval, $\frac{1}{(n+1/2)^2\pi^2} |\cos x| \le \frac{|\cos x|}{x^2} \le \frac{1}{(n-1/2)^2\pi^2} |\cos x|$. Therefore,
$$\frac{2}{(n+1/2)^2\pi^2} \le \int_{(n-1/2)\pi}^{(n+1/2)\pi} \frac{|\cos x|}{x^2}\,dx \le \frac{2}{(n-1/2)^2\pi^2}.$$
From this, we conclude:
$$\sum_{n=N}^\infty \frac{2}{(n+1/2)^2\pi^2} \le \int_{(N-1/2)\pi}^{\infty} \frac{|\cos x|}{x^2}\,dx \le \sum_{n=N}^\infty \frac{2}{(n-1/2)^2\pi^2}.$$
Now, since $\sum_{n=N}^\infty \frac{1}{(n+a)^2} \sim \frac{1}{N}$ as $N \to \infty$ for any constant $a$, we can conclude:
$$\int_{(N-1/2)\pi}^\infty \frac{|\cos x|}{x^2}\,dx \sim \frac{2}{N \pi^2}~\mathrm{as}~N\to \infty.$$
Now, for $(N-3/2)\pi \le X \le (N-1/2)\pi$, $\int_X^{(N-1/2)\pi} \frac{|\cos x|}{x^2}\,dx \le \frac{2}{(N-3/2)^2 \pi^2} = O(1/N^2)$. Therefore, in general,
$$\int_X^\infty \frac{|\cos x|}{x^2} \, dx \sim \frac{2}{X \pi}~\mathrm{as}~X \to \infty.$$
From this, we can conclude $\frac{F(h)}{h} \to \frac{2}{\pi}$ as $h \to 0^+$. Now, since $F$ is an odd function, $\frac{F(h)}{h}$ is even as a function of $h$, so in fact $\frac{F(h)}{h} \to \frac{2}{\pi}$ as $h \to 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that $\int_{0}^{\pi/2}{\arctan(2\cos^2 x)\over \cos^2 x}\mathrm dx={\pi\over \sqrt{\phi}}?$ Given that:
Where $\phi={\sqrt{5}+1\over 2}$
$$\int_{0}^{\pi/2}{\arctan(2\cos^2 x)\over \cos^2 x}\mathrm dx={\pi\over \sqrt{\phi}}\tag1$$
$t=2\cos^2 x\implies dt=-4\sin x\cos xdx$
$${1\over 4}\int_{0}^{1}{\arctan t\over t\sqrt{t-t^2}}\mathrm dt\tag2$$
$t=\tan v\implies dt=\sec^2 v dv$
$${1\over 4}\int_{0}^{\pi/4}{v\sec^2 v\over \tan v\sqrt{\tan v-\tan^2 v}}\mathrm dv\tag3$$
$${1\over 2}\int_{0}^{\pi/4}{v\over \sin(2v)\sqrt{\tan v-\tan^2 v}}\mathrm dv\tag4$$
Or we leave it in terms of $\tan t$
$${1\over 4}\int_{0}^{\pi/4}{v(1+\tan^2 v)\over \tan v\sqrt{\tan v-\tan^2 v}}\mathrm dv\tag5$$
| Let
$$I(a)=\int_{0}^{\pi/2}{\arctan(a\cos^2 x)\over \cos^2 x}\mathrm dx.$$
Then $I(0)=0,I(2)=I$ and
\begin{eqnarray}
I'(a)&=&\int_{0}^{\pi/2}{1\over{1+a^2\cos^4 x}}\mathrm dx\\
&=&\int_{0}^{\pi/2}{\sec^2x\over{a^2+\sec^4 x}}\sec^2x\mathrm dx\\
&=&\int_{0}^{\infty}{1+u^2\over{a^2+(1+u^2)^2}}\mathrm du\\
&=&\int_{0}^{\infty}{1+u^2\over{(u^2+1+ai)(u^2+1-ai)}}\mathrm du\\
&=&\frac12\int_{0}^{\infty}\left({1\over{u^2+1+ai}}+{1\over{u^2+1-ai}}\right)\mathrm du\\
&=&\frac12\bigg[\frac{1}{\sqrt{1+ai}}\arctan\frac{u}{\sqrt{1+ai}}+\frac{1}{\sqrt{1-ai}}\arctan\frac{u}{\sqrt{1-ai}}\bigg]\bigg|_{0}^{\infty}\\
&=&\frac{\pi}{4}\bigg(\frac{1}{\sqrt{1-ai}}+\frac{1}{\sqrt{1+ai}}\bigg)
\end{eqnarray}
So
\begin{eqnarray}
I(2)&=&\frac{\pi}{4}\int_{0}^{2}\bigg(\frac{1}{\sqrt{1-ai}}+\frac{1}{\sqrt{1+ai}}\bigg)\\
&=&\frac{\pi}{4}\bigg(\frac{\sqrt{1+ai}}{i}+\frac{\sqrt{1-ai}}{-i}\bigg)\bigg|_0^2\\
&=&\frac{\pi}{4}\frac{\sqrt{1+2i}-\sqrt{1-2i}}{i}\\
&=&\frac{\sqrt{2\sqrt5-2}}{4}\pi=\frac{\pi}{\sqrt\phi}
\end{eqnarray}
| {
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"url": "https://math.stackexchange.com/questions/2304028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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There don't exist $a, b$ positive integers such that $a^2 + b^2$ and $a^2 - b^2$ are perfect squares I need to prove that there don't exist $a, b$ positive integers such that $a^2 + b^2$ and $a^2 - b^2$ are perfect squares.
I suopose that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ with $c, d$ positive integers, so
$$(a^2 + b^2)(a^2 - b^2) = c^2d^2$$
Therefore
$$(a^2 + b^2)(a^2 - b^2) = (cd)^2$$
and
$$a^4 - b^4 = (cd)^2$$
but this equation doesn't have solution in positive integers.
Is that right?
| If $a^4-b^4=(cd)^2$ then that yields $b^4+(cd)^2=a^4$ . @awllower proved here that there can never exist a right triangle with integer sides such that a leg and the hypotenuse can be perfect squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integration Using Trig Sub and Partial Fractions $\displaystyle\int\frac{1}{x^4 + x^2 + 1}\mathrm{d}x$ using the trig substitution.
My attempt: I got $\left(x^2 + \frac{1}{2}\right)^2+\frac{3}{4}$ and did $x= \frac{1}{\sqrt2}\tan \theta$ but did not know how to proceed because I got $\displaystyle\int\frac{\sec^2\theta}{\sec^4\theta+ 3}\mathrm{d}\theta$ after factoring out the constant
| HINT:
$$\dfrac2{x^4+x^2+1}=\dfrac{x^2+1}{x^4+x^2+1}-\dfrac{x^2-1}{x^4+x^2+1}$$
Now $\dfrac{x^2+1}{x^4+x^2+1}=\dfrac{1+\dfrac1{x^2}}{x^2+1+\dfrac1{x^2}}$
As $\displaystyle\int\left(1+\dfrac1{x^2}\right)dx=x-\dfrac1x,$
write $x^2+1+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the initial value problem $y' = \frac{xy^3}{\sqrt{1+x^2}}$ with $y(0)=-1$
Solve the initial value problem $y' = \frac{xy^3}{\sqrt{1+x^2}}$ with
$y(0)=-1$
I'd like to know if my solution is correct. To keep it short, just assume that all integration steps I did are right.
The differential equation isn't in its standard form $y'+P(x)y=Q(x)$, so we bring it to that form:
$$y'-\frac{x}{\sqrt{1+x^2}}y^3=0$$
And at this step is my first doubt; is it really the standard form? It doesn't have to be $y$ instead of $y^3$?
Continuing, the integration factor is $e^{\int{P(x)dx}}=e^{\int{\frac{x}{\sqrt{1+x^2}}}dx}=e^{\sqrt{1+x^2}}$, in next step we will multiply the differential equation with $\sqrt{1+x^2}$.
$$\Rightarrow \frac{dy}{dx}-\frac{x}{\sqrt{1+x^2}}y^3=0 \Leftrightarrow \sqrt{1+x^2} \cdot\frac{dy}{dx}- \sqrt{1+x^2} \cdot \frac{x}{\sqrt{1+x^2}}y^3=0 \Leftrightarrow \frac{d}{dx}(\sqrt{1+x^2} \cdot y)-x \cdot y^3=0 \Leftrightarrow$$
$$\Leftrightarrow \int{\frac{d}{dx}(\sqrt{1+x^2} \cdot y)-x \cdot y^3} = \int{0} \text{ }dx \Leftrightarrow \sqrt{1+x^2} \cdot y = c \Leftrightarrow y = \frac{1}{\sqrt{1+x^2}}c$$
We know that $y(0)=-1 \Leftrightarrow -1=\frac{1}{\sqrt{1+0^2}}c \Leftrightarrow c=-1$
$\Rightarrow y = -\frac{1}{\sqrt{1+x^2}}$
| You are making it yourself way too hard here.
We have
$$\frac{dy}{dx} = \frac{xy^3}{\sqrt{1+x^2}}$$
or equivalently:
$$\frac{dy}{y^3} = \frac{xdx}{\sqrt{1+x^2}}$$
Now integrate both sides, to obtain:
$$\frac{-1}{2y^2} = \sqrt{1+x^2} + c$$
Now solve for $y$ and fill in the initial values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve this limit without using L'Hopital's Rule: $\lim_{x\to0^-}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}$?
Find:
$$\lim_{x\to0^-}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}\cdot$$
I tried to simplify the expression, but I kept getting stuck. I also tried to make a substitution $u=\frac{1}{x}$, but I got stuck again. Please give me hint or explain what to do. Thanks.
| Note that, as $x \to 0$, we have
$$
\sin x \sim x \implies\sin^2(2x)\approx4x^2
\\ \cos x \sim1 \implies\cos^2(x)\approx1
$$
We can use these to greatly simplify the limit at hand:
$$\lim_{x \to 0}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}=\lim_{x \to 0}\frac{\sqrt{8x^4}}{x^2} = \color{red}{\sqrt 8}$$
If you want a mor rigorous proof, we can instead use squeeze theorem or $O$ notation. Using the latter, we can write $\sin^2(2x) \in 4x^2 + O(x^6)$ and $\cos^2(x) \in 1+O(x^4)$ as $x \to 0$, giving us
$$\begin{align}
\lim_{x \to 0}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2} &=\lim_{x \to 0}\frac{\sqrt{x^4[1+O(x^4)]+2x^2[4x^2+O(x^6)]-x^4}}{x^2}
\\&=\lim_{x \to 0}\frac{\sqrt{x^4+8x^4+O(x^8)-x^4}}{x^2}
\\&=\lim_{x \to 0}\sqrt{\frac{8x^4+O(x^8)}{x^4}}
\\&=\lim_{x \to 0}\sqrt{8+O(x^4)}
\\&=\sqrt{8+\lim_{x \to 0}O(x^4)}
\\&= \color{red}{\sqrt 8}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Tiling a 7×7 rectangle with trominoes We are given a $7 \times 7$ array. We want to remove a $1 \times 1$ square, such that the remaining shape can be covered with $1 \times 3$ triomino.
What are the possible positions of the eliminated square?
| You can color the array as
$$
\pmatrix{
0 & 1 & 2 & 0 & 1 & 2 & 0 \\
1 & 2 & 0 & 1 & 2 & 0 & 1 \\
2 & 0 & 1 & 2 & 0 & 1 & 2 \\
0 & 1 & 2 & 0 & 1 & 2 & 0 \\
1 & 2 & 0 & 1 & 2 & 0 & 1 \\
2 & 0 & 1 & 2 & 0 & 1 & 2 \\
0 & 1 & 2 & 0 & 1 & 2 & 0 \\
}
$$
and since there are $17$ zeros and $16$ of each other number, the missing square has to be a $0$.
But you can also color it as
$$
\pmatrix{
0 & 1 & 2 & 0 & 1 & 2 & 0 \\
2 & 0 & 1 & 2 & 0 & 1 & 2 \\
1 & 2 & 0 & 1 & 2 & 0 & 1 \\
0 & 1 & 2 & 0 & 1 & 2 & 0 \\
2 & 0 & 1 & 2 & 0 & 1 & 2 \\
1 & 2 & 0 & 1 & 2 & 0 & 1 \\
0 & 1 & 2 & 0 & 1 & 2 & 0 \\
}
$$
so the missing quare has to be $0$ in both colorings, that is, it has to be a corner or a midpoint of a side, or the center point.
It is easy to see how to do the tiling in each of these cases:
*
*If a corner is missing, say the upper left corner, then place to triminoes horizontally covering the top, and the rest vertically, two in each column.
*If a midpoint of a side is missing, do the same, only now the two horizontal triminoes are separated by the missing space.
*if the center square is missing, then place two triminoes horizontally across the center row, separated by the center space. You are left ith two 3x7 horizontal bands, which can be tiled vertically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find equation of tangent to the circle Coming back to this after about 6 months I now know how to solve it.
First I found the gradient of the radius $\frac{changeiny}{changeinx}$ >> $\frac{-8}{6}$ >> $1.33333333$
Then I found the negative reciprocal of the radius gradient to get the gradient of the tangent because it is at $90^\circ$ to the radius which becomes $\frac{1}{1.33333333}$ >> $0.75$
I then substituted the $x$ and $y$ values where the tangent touches the circle into the $y=mx+c$ equation >> $-8=0.75*6+c$ >> $-8=4.5+c$ >> $c=-12.5$
Which means that the equation of the tangent turned out to be ..... $y=0.75x-12.5$
ORIGINAL QUESTION:
The circle has the equation $x^2+y^2=100$
Find the equation of the tangent to the circle at the point $A(6,-8)$.
Any clues on how to solve this?
| Alternatively if you haven't learnt calculus you can say there a line in the form $y=mx+c$ passing through the point $(6,-8)$ so $-8= 6m+c \implies c=-8-6m \implies y=mx+(-8-6m) $
Then substitute this into the circle equation:
$$ x^2+y^2=100 \implies x^2+(mx+(8-6m))^2=100 $$
$$ \Leftrightarrow x^2+m^2x^2+2mx(8-6m)+(8-6m)^2=100 \implies m^2 x^2 - 12 m^2 x + 36 m^2 + 16 m x - 96 m + x^2 + 64=100$$
$$ \Leftrightarrow x^2(m^2+1) +x(-12m^2+16m)+(36m^2-96m-36)=0$$
For tangency the discriminant of this equation must equal zero so
$$ (-12m^2+16m)^2-4(m^2+1)(36m^2 - 96m-36) =0$$
$$ \Leftrightarrow 16(4m+3)^2=0 \implies m=\frac{-3}{4}$$
Hence the equation of the tangent line is
$$ y=\frac{-3}{4}x - \frac{7}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $S = \frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b}$ if values of $a+b+c$ and $\frac1{a+b}+\frac1{b+c}+\frac1{a+c}$ are given I just stumbled upon a contest question from last year's city olympiad math contest:
Question: For the real numbers $a,b,c$ such that: $a+b+c = 6, \dfrac{1}{a+b}+\dfrac{1}{b+c} + \dfrac{1}{c+a} = \dfrac{47}{60}$, find the value of $S = \dfrac{a}{b+c}+\dfrac{b}{c+a} + \dfrac{c}{a+b}$.
Since I just saw it from an online forum "elsewhere", I thought I'd want to hear from other more skilled and experienced MSE members about your tactics and approaches to the solution of this interesting question.
| Multiplying the given expressions together:
\begin{align}
\frac{47}{10} &= (a+b+c)\bigg(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\bigg) \\ \\
&= \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a} \\ \\
&=3+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} \\ \\
\end{align}
$$\Longrightarrow \frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} = \frac{17}{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 2
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Find closed form for $a_{1}=2, a_{n}=a_{n-1}+n+6$ I have determined that $a_{2} = 10, a_{3} = 19, a_{4} = 29, a_{5} = 40, a_{6} = 52,$ and $a_{7} = 65$. I can see that there is a pattern in that each value increases by 8, then 9, then 10, then 11, then 12, etc. but I am having difficulty making an equation for it.
I thought I had it when I realized that $a_{2} = 9+1, a_{3} = 16 + 4, a_{4} = 25 + 4, a_{5} = 36 + 4,$ and so on, but then I realized it was not very consistent. Also, the difference between $a_{n} - (n-1)^2$ starts to get smaller as the value of n increases, and then begins to increase again later.
Am I going about this completely wrong? Is there a way to find a closed form for $a_{n}=a_{n-1}+n+6$ when $a_{1}=2$?
| Hint:
reorder the terms:
$$
a_1=2
$$
$$
a_2=a_1+2+6=2+2+6
$$
$$
a_3=a_2+3+6=2+2+6+3+6=2+(2+3)+2\cdot 6
$$
$$
a_4=a_3+4+6=2+(2+3)+2\cdot 6+4+6=2+(2+3+4)+3\cdot 6
$$
this suggests:
$$
a_n=2+\frac{(n-1)(2+n)}{2}+6(n-1)
$$
Now prove that it works giving $a_n=a_{n-1}+n+6$
| {
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"url": "https://math.stackexchange.com/questions/2315846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ramanujan and his problem Following was proposed by Ramanujan:
$ \sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11+\cdots}}}}=1+4\sin(10^o)$
Working on this I got the radical on the left equal to $(1+2\sqrt{2})$ implying that $\sin(10^o)=1/\sqrt{2}$
How is this possible? What is wrong here?
| Actually you have written the identity incorrectly. The original, by Ramanujan, stated:
$$ 1 + 4\sin 10^\circ = \sqrt{11 - 2\sqrt{11 + 2\sqrt{11 - 2(1 + 4\sin 10^\circ) \dots }}} $$
Edit: note the period of signs is $-, +, -, -, +, -, -, +, -, \dots$
or letting $ x = 1 + 4\sin 10^\circ $
$$ x = \sqrt{11 - 2\sqrt{11 + 2\sqrt{11 - 2x \dots }}} $$
According to WA, we have $x \approx 1.695$ and again $1 + 4 \sin 10^\circ \approx 1.695$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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System of equations with 4 unknowns. I'm trying to solve this system of equations but I'm reaching a dead end.
$$\begin{array}{lcl}
xyz & = & x+y+z \ \ \ \ \ \ \ (1)\\
xyt & = & x+y+t \ \ \ \ \ \ \ (2)\\
xzt & = & x+z+t \ \ \ \ \ \ \ (3)\\
yzt & = & y+z+t \ \ \ \ \ \ \ (4)\\
\end{array}$$
So, (1)-(2) gives $$xyz-zyt=xy(z-t)=z-t\Rightarrow xy=1.$$
(3)-(4) gives $$xzt-yzt=zt(x-y)=x-y\Rightarrow zt=1.$$
This system reduces to $$\begin{array}{lcl}
xy & = & 1 \ \ \ \ \ \ \ (5)\\
zt & = & 1 \ \ \ \ \ \ \ (6)\\
\end{array}$$
One trivial solution is $x=y=z=t=\pm1,$ but this can't be because then I'd have divided by zero earlier in my simplifications.
How do I solve this one?
| We will find all solutions using only (1), (2), and (3).
Making the difference between the first two equations you get $xy=1$ or $x=y$. Let us suppose that $xy=1$. Then $x,y$ have the same sign and are both $\neq 0$. This implies that
$$
z=xyz=x+y+z=x+\frac{1}{x}+z \implies x+\frac{1}{x}=0,
$$
which is impossibile since $|x+\frac{1}{x}|\ge 2$ for all non-zero reals $x$ (by AM-GM). Therefore $x=y$ ($\neq 1$ and $\neq -1$, otherwise $xy$ would still be $1$).
The first and second equations become (*)
$$
z(x^2-1)=2x \text{ and }t(x^2-1)=2x.
$$
In particular, $z=t$. This means the only two first equations imply $x=y \notin \{\pm 1\}$ and $z=t$.
Lastly, using (3), we get $xz^2=x+2z$, hecen we reduce to
$$
x(z^2-1)=2z\text{ and }z(x^2-1)=2x.
$$
If $x=0$ then also $z=0$ and viceversa. Otherwise substituing the first into the second we get
$$
z=\frac{2x}{x^2-1}=\frac{4z}{\left(z^2-1\right) \left(\left(\frac{2z}{z^2-1}\right)^2-1\right)}=\frac{4z(z^2-1)}{4z^2-(z^2-1)^2}.
$$
Since $z\neq 0$ then
$$
4z^2-(z^2-1)^2=4(z^2-1) \Leftrightarrow (z^2-3)(z^2+1)=0.
$$
Hence $|z|=\sqrt{3}$. Thus we have only the trivial solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Evaluate $\int_1^{\infty} {\frac{\ln{x}}{(x-1)(2x-1)}\,dx}$ Evaluate:
$$\int_1^{\infty} {\left(\frac{\ln{x}}{\left(x-1\right)\left(2x-1\right)}\,dx\right)}$$
It turns out that the value of the integral is exactly:
$$\frac{1}{12}\left(\pi^2+6\ln^2{2}\right)$$
as found by WolframAlpha, but Wolfram gives no indication of how it arrives at this curious result.
How would one solve the integral analytically? A first step might be the $u$-sub, $u=\frac{1}{x}$, which gives:
$$\int_1^{\infty} {\left(\frac{\ln{x}}{\left(x-1\right)\left(2x-1\right)}\,dx\right)}=-\int_0^1 {\left(\frac{\ln{x}}{\left(1-x\right)\left(2-x\right)}\,dx\right)}$$
Thanks!
| Start by performing the substitution $u=\frac 1x$, $\text dx=-\frac {1}{u^2}\text du$:
\begin{align}\int_1^\infty\left(\frac{\ln x}{(x-1)(2x-1)}\right)\text dx&=\int_1^0\left(\left(\frac{\ln\left(\frac 1u\right)}{\left(\frac 1u-1\right)\left(2\frac 1u-1\right)}\right)\times \left(-\frac {1}{u^2}\right)\right)\text du\\
&=-\int_0^1\left(\frac{-\ln\left(\frac 1u\right)}{(u-2)(u-1)}\right) du\\
\end{align}
We can now write the integral as follows, using partial fraction decomposition:
\begin{align}-\int_0^1\left(\frac{-\ln\left(\frac 1u\right)}{(u-2)(u-1)}\right) du&=-\int_0^1\left(-\ln\left(\frac 1u\right)\left(\frac{1}{(u-2)(u-1)}\right)\right)\text du\\
&=-\int_0^1\left(-\ln\left(\frac 1u\right)\left(\frac{1}{u-2}-\frac{1}{u-1}\right)\right)\text du\\
&=-\int_0^1\left(\ln\left(\frac 1u\right)\left(\frac{1}{u-1}-\frac{1}{u-2}\right)\right)\text du\end{align}
This now leaves us with:
\begin{align}-\int_0^1\left(\ln\left(\frac 1u\right)\left(\frac{1}{u-1}-\frac{1}{u-2}\right)\right)\text du&=-\left(\left(\int_0^1\ln\left(\frac 1u\right)\frac1{u-1}\right)\text du-\left(\int_0^1\ln\left(\frac 1u\right)\frac1{u-2}\right)\text du\right)\\
&=\left(\int_0^1\ln\left(\frac 1u\right)\frac1{u-2}\text du\right)-\left(\int_0^1\ln\left(\frac 1u\right)\frac1{u-1}\text du\right)\\
&=\left(\int_0^1-\ln(u)\frac1{u-2}\text du\right)-\left(\int_0^1-\ln(u)\frac1{u-1}\text du\right)\\
&=\left(-\int_0^1\frac{\ln u}{u-2}\text du\right)-\left(-\int_0^1\frac{\ln u}{u-1}\text du\right)\\
&=\left(\int_0^1\frac{\ln u}{u-1}\text du\right)-\left(\int_0^1\frac{\ln u}{u-2}\text du\right)\end{align}
Can you continue from here? At a glance, I would recommend trying integration by parts.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is my mistake in this integral: $\int \tan^5(x) dx$? I'm doing an exercise from Stewart's Calculus textbook in which I have to evaluate the following integral:
$\int \tan^5(x) dx$
I start by rewriting the integral this way:
$\int \tan^5(x) dx = \int \frac{\sin^5(x)}{\cos^5(x)} dx = \int \frac{\sin^4(x)}{\cos^5(x)} \sin(x) dx$
And here I make a substitution:
$u = \cos(x)$
$-du = \sin(x)dx$
So the integral with the substitution becomes:
$\int \frac{(1-u^2)^2}{u^5} (-1) du$
$\int \frac{(1-2u^2+u^4)}{u^5} (-1) du$
$\int \frac{(2u^2-u^4-1)}{u^5} du$
$\int \frac{(2u^2)}{u^5} - \frac{(u^4)}{u^5} - \frac{1}{u^5} du = 2 \frac{u^{-2}}{(-2)} - \ln\vert u \vert - \frac{u^{-4}}{(-4)} + constant$
$= \frac{-1}{u^2} - \ln\vert u \vert + \frac{1}{4u^4} + constant $
$= \frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} + constant $
This antiderivative seems to be correct, because if I differentiate it I get:
$\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} + constant] $
$-(-2)\cos^{-3}(x)(-\sin(x))-\frac{1}{\cos(x)}(-\sin(x))+\frac{1}{4}(-4)\cos^{-5}(x)(-\sin(x))$
$ \frac{-2\sin(x)}{\cos^3(x)} +\tan(x)+\frac{\sin(x)}{\cos^5(x)}$
$\tan(x)[1-\frac{2}{\cos^2(x)} + \frac{1}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-2\cos^2(x)+1}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-\cos^2(x)+1-\cos^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-\cos^2(x)+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^2(x)(\cos^2(x)-1)+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^2(x)(-\sin^2(x))+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^2(x)(1-\cos^2(x))}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^2(x)(\sin^2(x))}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^4(x)}{\cos^4(x)}] = \tan^5(x)$
However, when I graph $\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} ] $ and $\tan^5(x)$ at a first glance they look like they are the same, but when I start to zoom in I notice that these curves are actually not equal:
The magenta curve is $\tan^5(x)$ and the black curve is $\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} ] $
The answer that Stewart provides at the end of the book is the following:
$ \int \tan^5(x) dx = \frac{1}{4} \sec^4(x)-\tan^2(x)+\ln\vert\sec(x)\vert + C$
Which is confusing me even more, because if I try to differentiate his answer I get:
$\frac{d}{dx}[\frac{1}{4} \sec^4(x)-\tan^2(x)+\ln\vert\sec(x)\vert + C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)+\ln\vert1/\cos(x)\vert + C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)+\ln(1) - \ln\vert\cos(x)\vert+ C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)- \ln\vert\cos(x)\vert+ C]$
$= \frac{1}{4}(-4)\cos^{-5}(x)-2\tan(x)\sec^2(x)-\frac{1}{\cos(x)}(-\sin(x))$
$= \frac{-1}{\cos^5(x)}-2\tan(x)\sec^2(x)+\tan(x)$
$= \frac{-1}{\cos^5(x)}+\tan(x)(1-2\sec^2(x))$
$= \frac{-1}{\cos^5(x)}+\tan(x)(1-\sec^2(x) -\sec^2(x))$
$= \frac{-1}{\cos^5(x)}+\tan(x)(-\tan^2(x) -\sec^2(x))$
$= \frac{-1}{\cos^5(x)}-\tan^3(x)-\tan(x)\sec^2(x)$
$= \frac{-1}{\cos^5(x)}-\frac{\sin^3(x)}{\cos^3(x)}-\frac{\sin(x)}{\cos^3(x)}$
$=\frac{-1-\cos^2(x)\sin^3(x)-\cos^2(x)\sin(x)}{\cos^5(x)}$
$=\frac{-1-(1-\sin^2(x))\sin^3(x)-(1-\sin^2(x))\sin(x)}{\cos^5(x)}$
$=\frac{-1-(\sin^3(x)-\sin^5(x))-(\sin(x)-\sin^3(x))}{\cos^5(x)}$
$=\frac{-1-\sin^3(x)+\sin^5(x)-\sin(x)+\sin^3(x)}{\cos^5(x)}$
$=\frac{-1-\sin(x)+\sin^5(x)}{\cos^5(x)}$
$=\tan^5(x)-\frac{(1+\sin(x))}{\cos^5(x)}$
If I plot the derivative of Stewart's answer along the derivative of my answer and the original function $\tan^5(x)$ this is how the the graph looks like:
*
*Magenta curve: $\tan^5(x)$
*Black curve: $\frac{d}{dx}[\text{my answer}]$
*Blue curve: $\frac{d}{dx}[\text{Stewart's answer}]$
What's going on? Is my answer correct? If it isn't, what am I doing wrong? And what about Stewart's answer? Am I making a mistake when I differentiate it or is it correct?
Thanks for your help!
| When you are differentiating his answer, you took the derivative of $\frac{\sec^4x}{4}$ wrong.
What you need to do is -
$$\left(\frac{\sec^4 x}{4}\right)'=\left(\frac{\cos^{-4} x}{4}\right)'=\frac{-1}{\cos^5 x} \color{red}{\cdot (\cos x)'}=\frac{\sin x}{\cos^5 x}$$
Your answer is absolutely correct. Even Wolfram Alpha says that.
Your and the books answer are both the same, but seem to be different because they differ by a constant. $$-\sec^2 x=-1-\tan^2 x$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Find sum of series $\frac{1}{6} +\frac{5}{6\cdot12} +\frac{5\cdot8}{6\cdot12\cdot18} +\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...$ How to find sum of above series
$$\frac{1}{6} +\frac{5}{6\cdot12} +\frac{5\cdot8}{6\cdot12\cdot18} +\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...$$
How to find sum of series I can find its convergence but not sum of series.
Can anyone explain?
| Using binomial expansion
\begin{eqnarray*}
(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots
\end{eqnarray*}
with $x=-\frac{1}{2}$ and $n=-\frac{2}{3}$.
\begin{eqnarray*}
(1-\frac{1}{2})^{-\frac{2}{3}}=1+\left(-\frac{2}{3}\right)\left(-\frac{1}{2}\right)+\frac{\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{2}\left(-\frac{1}{2}\right)^2+\frac{\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)\left(-\frac{8}{3}\right)}{3!}\left(-\frac{1}{2}\right)^3+\cdots
\end{eqnarray*}
Rearrange a bit & we have
\begin{eqnarray*}
\color{red}{\frac{2^{\frac{2}{3}}-1}{2}}=\frac{1}{6} +\frac{5}{6\cdot12} +\frac{5\cdot8}{6\cdot12\cdot18} +\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\cdots
\end{eqnarray*}
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Area under the graph - integration
The region $P$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the line $x=a$ . The region $Q$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the lines $x=2a$ and $x=a$. Given that the area of $Q$ is twice the area of $P$, find the value of $a$ .
Firstly , on the first step , in already stuck ...
I used definite integral to find the area of $P$ -
$$\int^a_0\ (3x-x^2)dx=\frac{9a^2-2a^3}{6}$$
However when I calculate area of $Q$ , it's the same as Area of $P$ - $$\frac{9a^2-2a^3}{6}$$
Then since
$Q= 2P$
$9a^2 - 2a^3 = 18a^2 - 4a^3 $
From here, I definitely can't find the value of $a$ ... where have I gone wrong or misunderstood ?
| Area of $Q$ certainly is not $$\frac{9a^2-2a^3}{6}$$
It is:
$$Q = \int_a^{2a}\ (3x-x^2)dx=\left .\frac{9x^2-2x^3}6\right |_{x=a}^{x=2a}$$
$$= \frac{9(2a)^2-2(2a)^3}6 - \frac{9a^2-2a^3}6 $$
$$= \frac {36a^2-16a^3}6 - \frac{9a^2-2a^3}6$$
$$= \frac {27a^2-14a^3}6$$
so $Q=2P$ yields
$$27a^2-14a^3 = 2(9a^2-2a^3)$$
$$27a^2-14a^3 = 18a^2-4a^3$$
which others solved already.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329685",
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"source": "stackexchange",
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All functions satisfying $f(x+(f(y))^2)=(f(x+y))^2$.
Let $f: \mathbb{R} \to \mathbb{R} $ be a function such that $f(0)$ is rational and for real numbers $x$ and $y$
$$f \big(x+(f(y))^2 \big)=(f(x+y))^2$$
Find all functions satisfying above conditions.
My try:
Let $a \in \mathbb{R}$ be a number such that $f(a)=0$ and let $x=0$, $y=a$ in the equation, then $$f(0)=f(a)^2=0$$
Now let $y=0$, then
$$f (x)=(f(x))^2$$
Thus $f(x)=0, 1$.
Is this correct?
Edit: Considering @Travis's comment the above argument is not correct.
Let $f(0)=a$ and let $x=y=0$ in the equation, then $f(a^2)=a^2$. Now let $x=a^2$ and $y=0$, then $$f(2a^2)=(f(a^2))^2=a^4$$Once again let $x=2a^2$ and $y=0$, then $$f(3a^2)=(f(2a^2))^2=(a^4)^2=a^8$$Hence induction will show that $$f(na^2)=a^{2^n}$$For $n \in \mathbb{N}$. Any ideas how to proceed?
| This answer is now complete: the only such $f$ are the constant $0$ and the constant $1$.
Things in block quotes will be true regardless of the assumptions made in the surrounding text.
As you point out,
If $f$ has a root, then $f(0) = 0$, and $f(x) \in \{0,1\}$ for all $x$, and $$f(x+f(y)) = f(x+y)$$
Clearly $f=0$ is a solution. Otherwise, suppose $f(y)=1$. Then for all $x$, $f(x+1) = f(x+y)$, so letting $x=1$ we have $f(y-1) = 0$. [Call this property DECREMENT.]
On the other hand, whenever $f(y) = 0$ we have $f(x) = f(x+y)$ for all $x$. Therefore $f(x) = f(x+yn)$ for all $x$ and for all $n \in \mathbb{Z}$. [Call this property TRANSLATE, and if $y$ is fixed with $f(y)=0$, let TRANSLATE(n) denote the fact that $f(x) = f(x+yn)$ for all $x$.]
So suppose $f(0) = f(1) = 0$, and $f(y) = 1$. Then $f(y-1) = 0$ (DECREMENT) but also $f(y-1) = f(y-1+1)$ (TRANSLATE(1)) which is $f(y) = 1$, a contradiction.
If $f(1) = 0$, then $f$ is the constant $0$ function.
So if $f(0) = 0$ but $f$ is nonzero, then $f(1) = 1$. Hence $f(2) = 0$ (since if not, it would be $1$ and hence we would have $f(2-1) = 0$ by DECREMENT), and so $f(x) = f(x+2n)$ for all $x$ and all $n \in \mathbb{Z}$. That is, $f$ is defined by its values on $(-1, 1]$.
Moreover, in that case $f\left(\frac{1}{n}\right) = 1$ for all $n>1$ because if $f\left(\frac{1}{n}\right) = 0$ then $0 = f(0) = f\left(0+\frac{1}{n} \times n \right)$ by TRANSLATE(n), which is $f(1) = 1$, a contradiction.
This means that $f\left(\frac{1}{n}-1\right) = 0$ for $n>1$, so for all $x \in \mathbb{R}, m \in \mathbb{Z}, n > 1$ we have $f(x) = f(x+m(\frac{1}{n}-1))$ by TRANSLATE(m); in particular, letting $x=m$ we have $f(m) = f(\frac{m}{n})$ for all $m \in \mathbb{Z}, n > 1$. So letting $m=3n$, we have $f(3) = f(3n)$ for all $n > 1$; but $f(3) = f(1) = 1$ by TRANSLATE(2) and so $f(3n) = 1$. Letting $n=2$, we obtain $f(6) = 1$, but we already know $f(2k) = 0$ for all $k$ by TRANSLATE(2), so we obtain a contradiction.
If $f$ has a root, then $f$ is the constant $0$ function.
(Actually I think @lulu's trick works to show this much faster.)
If ever $y = f(y)^2$, then $f(x+y) = f(x+y)^2$ for all $x$, so $f(x) \in \{0,1\}$ for all $x$.
So it has a root unless it's the constant $1$ function.
If $f$ is not the constant $1$ function, and there is some $y$ with $y=f(y)^2$, then $f$ is the constant $0$ function.
In particular,
If $f(1) = 1$, then $f$ is the constant $1$ function. $f(1)$ cannot be $-1$.
If $y$ is such that $f(y) = 1$, then $f(x+1) = f(x+y)^2$ for all $x$, so letting $x=y-1$ we obtain $1 = f(y) = f(1)^2$, so $f(1) = 1$ or $f(1) = -1$ (the latter being a contradiction). That is:
If $f$ ever hits $1$, then $f$ is the constant $1$ function.
Note that by letting $x=0$, we see that $f$ has a fixed point at $f(y)^2$ for every $y$.
So, letting $y=0$, we obtain $f(x+f(0)^2) = f(x)^2$, which means that $f(x+f(0)^2)$ is a fixed point of $f$ for every $x$.
In particular, letting $x=-f(0)^2+z$,
$f(z)$ is a fixed point of $f$ for every $z$. (That is, $f$ is idempotent: $y$ is a fixed point of $f$ iff $y$ is in the range of $f$.)
Also let $x = y-f(y)^2$ to obtain
$f(y) \geq 0$ for all $y$.
And let $x=0$ to obtain $f(f(y)^2) = f(y)^2$; that is, the square of anything in the range of $f$ is a fixed point of $f$, or
The square of a fixed point is a fixed point.
Now, let $x=f(y)^2$ and $y=0$ to obtain $$f(2a^2) = f(a^2)^2$$ where $a=f(0)$; so $f(2a^2) = a^4$ since the square of a fixed point $a$ is a fixed point $a^2$.
Let $x=\bar{x}, y=\bar{x}$ where $\bar{x}$ is a fixed point, to obtain $f(\bar{x} + f(\bar{x})^2) = f(2\bar{x})^2$, or $f(\bar{x} + \bar{x}^2) = f(2 \bar{x})^2$.
Let $x = \bar{x}^2, y=\bar{x}$ to obtain $f(\bar{x}^2+f(\bar{x})^2) = f(\bar{x}+\bar{x}^2)^2$.
Combining those facts, obtain:
If $\bar{x}$ is a fixed point of $f$, then $f(2\bar{x}^2) = f(2\bar{x})^4$.
And let $x=a^2, y=0$ to obtain $f(2a^2) = f(a^2)^2$, so $f(2a^2) = a^4$ since $a^2$ is a fixed point, so by the previous fact, $a^4 = f(2a)^4$; since $a=f(0)$ is in the range of $f$, and since $f$ is always positive, we have:
$a = f(2a)$.
Now, we prove by induction that $f(na) = a$.
Indeed, let $y=na$ and $x=a$.
Then $$f(a+f(na)^2) = f(a(n+1))^2$$ which by inductive hypothesis means $$f(a+a^2) = f(a(n+1))^2$$
But we already know that $f(a+a^2) = f(2a)^2$ because that's just the definition of $f$ with $x=a, y=a$; so $f(a(n+1))^2 = f(2a)^2$, which we already know is $a$.
So
$f(na) = a$ for every positive integer $n$.
Let $n = m a$ to obtain:
$f(ma^2) = a$ whenever $ma$ is integer.
In particular, let $a=\frac{p}{q}$, and let $m=q$, to obtain $f(qa^2) = a$.
But you have already shown that $f(ma^2) = a^{2^m}$ for every $m$, so we have $a = a^{2^q}$. That is, $a=0$ or $a=1$ (since $m \not = 0$); so $f$ hits $0$ or $1$.
| {
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"url": "https://math.stackexchange.com/questions/2329771",
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"question_score": "9",
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An infinite series for $\ln 2$ and taking limits of summations Example $1$.$$\tfrac 14\text{Log }2=\sum\limits_{k=1}^\infty\frac 1{\left\{2(2k-1)\right\}^3-2(2k-1)}\tag1$$
Proof. The right side above is equal to $\tfrac 12\{\varphi(2)-\varphi(4)\}$. Hence, the result follows from the two identities$$\text{Log }2=\tfrac 12\varphi(2)$$$$\tfrac 32\text{Log }2=\varphi(4)$$
Question: How do you prove $(1)$?
I started off with $\varphi(2,n)$ and $\varphi(4,n)$; subtracted them, and tried to show that together, the limit was equal to $\tfrac 12\text{Log }2$. However, this is all I have$$\begin{align*}\varphi(2,n)-\varphi(4,n) & =\sum\limits_{k=1}^n\left\{\frac 1{2k-1}+\frac 1{2k+1}-\frac 1k\right\}-\sum\limits_{k=1}^n\left\{\frac 1{4k-1}+\frac 1{4k+1}-\frac 1{2k}\right\}\\ & \\ & =\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}-\sum\limits_{k=1}^n\frac 1k-\sum\limits_{k=1}^n\frac 1{4k-1}-\sum\limits_{k=1}^n\frac 1{4k+1}+\sum\limits_{k=1}^n\frac 1{2k}\\ & \\ & =\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}-\sum\limits_{k=1}^n\frac 1{4k-1}-\sum\limits_{k=1}^n\frac 1{4k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k\end{align*}$$However, it's after this step that I'm not sure what to do. If you split up twice the RHS of $(1)$, you get a $1/(4k-3)$ term, a $1/(4k-1)$ term and a $1/(2k-1)$ term. I'm puzzled on how they got there.
Also, I should probably add the definitions of the notations$$\lim\limits_{n\to\infty}\varphi(a,n)=\varphi(a)$$$$\varphi(a,n)=1+2\sum\limits_{k=1}^n\left\{\frac 1{(ak)^3-ak}\right\}$$Where $a$ is greater than one.
Sorry for the increase in summation questions, I'm analyzing Ramanujan's works on Harmonic Series and Inverse Tangent Functions.
| By partial fraction decomposition
$$ \frac{1}{8x^3-2x} = \frac{1}{(2x-1)2x(2x+1)} = \frac{1}{2}\cdot\frac{1}{2x-1}-\frac{1}{2x}+\frac{1}{2}\cdot\frac{1}{2x+1}\tag{1} $$
hence by replacing $x$ with $2k-1$ we get:
$$ \frac{1}{8(2k-1)^3-2(2k-1)}=\frac{1}{2}\cdot\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{2}\cdot\frac{1}{4k-1}\tag{2} $$
or:
$$ \frac{1}{8(2k-1)^3-2(2k-1)}=\frac{1}{2}\int_{0}^{1}\left(x^{4k-4}-2x^{4k-3}+x^{4k-2}\right)\,dx\tag{3} $$
and by summing both sides of $(3)$ over $k\geq 1$:
$$ \sum_{k\geq 1}\frac{1}{8(2k-1)^3-2(2k-1)}=\frac{1}{2}\int_{0}^{1}\frac{(1-x)^2}{1-x^4}\,dx = \color{red}{\frac{1}{4}\log 2} \tag{4}$$
always by partial fraction decomposition, or through:
$$ \int_{0}^{1}\frac{(1-x)}{(1+x)(1+x^2)}\,dx\stackrel{x\mapsto\tan\theta}{=} \int_{0}^{\pi/4}\frac{\cos\theta-\sin\theta}{\sin\theta+\cos\theta}\,d\theta = \left[\log(\sin\theta+\cos\theta)\right]_{0}^{\pi/4}.\tag{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Log with $\sqrt x$ base I'd like to know how this simplification happened:
$$\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)=\log _2\left(x-2\right)$$
$$
\begin{array}{l}
\color{red}{2 \log _{2} x+\log _{\frac{1}{2}}(1-\sqrt{x})=\frac{1}{2} \log _{\sqrt{2}}(x-2 \sqrt{x}+2) \quad } \color{blue}{0<x<1} \\
\Leftrightarrow 2 \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2) \\
\Leftrightarrow \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2)-\log _{2} x
\end{array}
$$
| $\log_2(x-2)$ is the value $a$ such that $2^a = x-2$. This means $(\sqrt{2})^{2a} = x-2$, so $\log_{\sqrt{2}}(x-2) = 2a$, or $$\frac{1}{2}\log_{\sqrt{2}}(x-2) = a = \log_2(x-2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
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Solving $\dfrac{(3^x-4^x)\cdot \ln (x+2)}{x^2-3x-4}\leq 0$
The solution set of the inequality $\dfrac{(3^x-4^x)\cdot \ln (x+2)}{x^2-3x-4}\leq 0$ is.
My attempts:
For $x>0\implies 3^x-4^x<0$
$\implies \dfrac{\ln (x+2)}{x^2-3x-4}\geq 0$. Now root of numerator is $x=-1$ and of denominator is $x=4,-1$
By wavy curve method:
$\implies x\epsilon(4,\infty)$
Now for $x\leq 0\implies 3^x-4^x\geq 0$
Hence:
$\dfrac{\ln (x+2)}{x^2-3x-4}\leq 0$
$\implies x\epsilon(-1,0]$
My answer $(-1,0]\cup (4,\infty)$
But book's answer said I missed one interval, please help.
I observed something that $-1$ are common roots in both numerator and denominator, so should we put $-ve$ on LHS of $-1$? If we do this then my answer will be correct, but don't know whether it's correct thinking or not, if it is then why?
| The term $3^x-4^x$ is positive for $x<0$ and negative for $x>0$.
The term $\ln(x+2)$ is positive for $x>-1$ and negative for $-2<x<-1$ (undefined for $x\le-2$).
The term $x^2-3x-4$ is positive for $x<-1$ or $x>4$ and negative for $-1<x<4$.
$$
\begin{array}{l|ccccccccc}
&& -2 & & -1 && 0 && 4 \\
\hline
3^x-4^x & <0 &<0 & <0 & <0 & <0 & =0 & >0 & >0 & >0 \\
\ln(x+2) & \text{u} & \text{u} & <0 & =0 & >0 & >0 & >0 & >0 & >0 \\
x^2-3x-4 & >0 & >0 & >0 & =0 & <0 & <0 & <0 & =0 & >0 \\
\hline
f(x) & \text{u} & \text{u} & >0 & \text{u} & >0 & =0 & <0 & \text{u} & >0
\end{array}
$$
where “u” means “undefined”.
The last row is obtained by using the “rule of signs” on the factors. The function is undefined where the denominator vanishes.
Hence you get, for the solutions, $(-2,-1)\cup(-1,0]\cup(4,\infty)$.
The problem with your solution is that two factors vanish at $-1$, so the global sign doesn't change when we cross it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Elements with order $100$ in $\mathbb{Q}/\mathbb{Z}$. How many elements with order $100$ are in $\mathbb{Q}/\mathbb{Z}$?
| Consider the orders of $26/100$ and $27/100$ in this group:
$$
\begin{array}{clclc}
& 26\times 1 = 26 & & 27\times1=27 \\
& 26\times 2 = 52 & & 27\times2=54 \\
& 26\times 3 = 78 & & 27 \times 3 = 81 \\
& 26\times 4 = 104 \equiv 4 & & 27 \times 4 = 108 \equiv 8 \\
& 26 \times 5 = 130 \equiv 30 & & 27\times 5 = 135 \equiv 35 \\
& \qquad\vdots & & \qquad \vdots \\
\text{back to 0 after 50 steps }\longrightarrow & 26 \times 50 = 1300 \equiv0 & & 27 \times 50 = 1350 \equiv 50 \longleftarrow \text{not yet back to 0} \\
& & & 27 \times 51 = 1377 \equiv 77 \\
& & & 27 \times 52 = 1406 \equiv 6 \\
& & & 27 \times 53 = 1433 \equiv 33 \\
& & & \qquad \vdots \\
& & & 27 \times 100 = 2700 \equiv 0
\end{array}
$$
Every number will get you back to $0$ after $100$ steps; the reason $26$ gets you back there earlier is that it has the factor $2$ in common with $100:$
$$
\frac{26}{100} = \frac{2\times13}{2\times 50} = \frac{13}{50},
$$
so $26\times50 = (13\times2)\times50 = 13\times(2\times50) = 13\times100.$
Thus the ones that take the full $100$ steps to get back to $0$ are those that have no factors in common with $100$ except the trivial factor that all numbers share, which is $1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Taylor expansion of $\frac{1}{(x^2-2x+17)^2}$ at point $c=1$ Taylor expansion of $$\frac{1}{(x^2-2x+17)^2}$$ at point $c=1$ ?
My attempt: $y = x -c = x-1$ so $x = y +1 $.
When we substitute this we get the expression $\frac{1}{(x^2+16)^2}$. Now, I have the formula $(1+x)^\alpha = \sum_{n=0}^{\infty}{\binom{\alpha}{n}x^n},\lvert x \rvert < 1$
I was wondering if I could write my expression as $4((\frac{x}{4})^2+1)^{-2}$ and apply the formula, or should I take a different approach? Am I even allowed to do that?
| Let $z=(x-1)$
\begin{eqnarray*}
\frac{1}{(x^2-2x+17)^2}=\frac{1}{(z^2+16)^2} =\frac{1}{16^2} \left(1+\frac{z^2}{16} \right)^{-2}
\end{eqnarray*}
$\left(1+\frac{z^2}{16} \right)^{-2}$ is now ripe to apply the binomial expansion
\begin{eqnarray*}
\left(1+\frac{z^2}{16} \right)^{-2}= 1+(-2) \frac{z^2}{16}+\frac{(-2)(-2-1)}{2!}\left(\frac{z^2}{16} \right)^{-2}+\cdots
\end{eqnarray*}
and so we have
\begin{eqnarray*}
\frac{1}{(x^2-2x+17)^2}=\frac{1}{16^2} \left(1-\frac{(x-1)^2}{8} +\frac{3(x-1)^4}{16^2} +\cdots \right)
\end{eqnarray*}
and this series is valid provided $ \mid x-1 \mid <4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is $\sum_{k=1}^n{{n}\choose{k}}\left (1+\frac{c}{2n}\right)^{n-k}o\left(\frac{1}{n}\right)^k$ equal to $o(1)$? where $c,n$ are some finite, positive real numbers, and the $o$ is little oh.
My attempt is that $\left (1+\frac{c}{2n}\right)^{n-k}o\left(\frac{1}{n}\right)^k$ is just a constant times $o\left(\frac{1}{n}^k\right)$ and is thus $o\left(\frac{1}{n}\right)^k$, and all
$o\left(\frac1n\right)^K$ terms are dominated by the ${{n}\choose{1}} o\left(\frac1n\right)$ term, which is $o(1)$?
| Try binomial theorem
$$S=\sum_{k=1}^{n}\binom{n}{k}\left(1+\frac{c}{2n}\right)^{n-k}o\left(\frac{1}{n}\right)^k=\sum_{k=0}^{n}\binom{n}{k}\left(1+\frac{c}{2n}\right)^{n-k}o\left(\frac{1}{n}\right)^k - \left(1+\frac{c}{2n}\right)^{n}=\\
=\left(1+\frac{c}{2n}+ o\left(\frac{1}{n}\right)\right)^{n} - \left(1+\frac{c}{2n}\right)^{n}=$$
which is
$$=o\left(\frac{1}{n}\right)\left(\sum_{k=0}^{n-1} \left(1+\frac{c}{2n}+ o\left(\frac{1}{n}\right)\right)^{n-k}\left(1+\frac{c}{2n}\right)^{k} \right)\tag{1}$$
From $e^x\geq x+1$ for $\forall x>-1$ we have
$$o\left(\frac{1}{n}\right)\left(\sum_{k=0}^{n-1} 1\right)<S<o\left(\frac{1}{n}\right)\left(\sum_{k=0}^{n-1} e^{\frac{c(n-k)}{2n}+ o\left(\frac{1}{n}\right)(n-k)} \cdot e^{\frac{ck}{2n}}\right)=\\
=o\left(\frac{1}{n}\right)\left(\sum_{k=0}^{n-1} e^{\frac{c}{2}+ o\left(\frac{1}{n}\right)(n-k)}\right)<o\left(\frac{1}{n}\right)\left(\sum_{k=0}^{n-1} e^{\frac{c}{2}+ o\left(1\right)}\right)$$
or
$$n\cdot o\left(\frac{1}{n}\right)<S<n \cdot o\left(\frac{1}{n}\right)e^{\frac{c}{2}+ o\left(1\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the sum $\sum\limits_{n=2}^{\infty}5^{-(n+1)}\ln\big(\frac{n^5}{n+1}\big)$
I want to find the sum of this series:
$$\sum_{n=2}^{\infty}5^{-(n+1)}\ln\bigg(\frac{n^5}{n+1}\bigg)$$
I have went through theses steps:
$$\sum_{n=2}^{\infty}\frac{1}{5^{n+1}}\ln\bigg(\frac{n^5}{n+1}\bigg)=\sum_{n=2}^{\infty}\frac{1}{5^{n+1}}[\ln(n^5)-\ln(n+1)]=\sum_{n=2}^{\infty}\frac{1}{5^{n+1}}[5\ln(n)-\ln(n+1)]$$
But I can't get past this.
| First, note that the series converges.
$$\sum_{n=2}^{\infty}5^{-(n+1)}\ln\bigg(\frac{n^5}{n+1}\bigg)$$
To see the convergence, observe that $5^{(n+1)}$ grows much faster than $\ln\bigg(\frac{n^5}{n+1}\bigg)$.
$5^{(n+1)}$ is in the denominator and $\ln\bigg(\frac{n^5}{n+1}\bigg)$ in the numerator.
$$
\begin{array}{rcl|l}
\sum_{n=2}^{\infty}5^{-(n+1)}\ln\bigg(\frac{n^5}{n+1}\bigg) & = & \sum_{n=2}^{\infty}\frac{1}{5^{n+1}}[5\ln(n)-\ln(n+1) & \text{basic properties of logorithms} \\
& = & \sum_{n=2}^{\infty}\frac{1}{5^{n+1}}*5\ln(n)-\sum_{n=2}^{\infty}\frac{1}{5^{n+1}}*\ln(n+1) & \text{distribute} \\
& = & \sum_{n=2}^{\infty}\frac{1}{5^{n+1}}*5\ln(n)-\sum_{k=3}^{\infty}\frac{1}{5^{k}}*\ln(k) & \text{substitute k = n + 1 } \\
& & & \text{for index of summation in rightmost sum} \\
& = & \sum_{n=2}^{\infty}\frac{1}{5^{n}}*\ln(n)-\sum_{k=3}^{\infty}\frac{1}{5^{k}}*\ln(k) & \frac{1}{5^{n+1}} = \frac{1}{5}*\frac{1}{5^{n}} \\
& = & \frac{1}{5^{2}}*\ln(2) + \sum_{n=3}^{\infty}\frac{1}{5^{n}}*\ln(n)-\sum_{k=3}^{\infty}\frac{1}{5^{k}}*\ln(k) & \text{pull $n=2$ term out of the leftmost sum} \\
& = & \frac{1}{5^{2}}*\ln(2) + 0 & \text{the two series are identical} \\
& = & \frac{\ln(2)}{25} & \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
find the value $\prod_{k=0}^{9}(2+z^k)$ Let $$z=\cos\left( \frac{2\pi}{10}\right)+i\sin\left(\frac{2\pi}{10}\right)$$
Find the value
$$\prod_{k=0}^{9}(2+z^k)$$
| Clearly, $z^k,0\le k\le9$ are the roots of $$z^{10}-1=0$$
Let $y=2+z^k\iff z^k=y-2\implies$ $$0=(y-2)^{10}-1=y^{10}-\binom{10}1y^9+\cdots+2^{10}-1$$
$$\implies\prod_{k=0}^9(2+z^k)=\dfrac{2^{10}-1}1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The set of values of $p$ for which the points of extremum of the function $f(x)=x^3-3px^2+3(p^2-1)x+1$ lie in the interval $(-2,4)$ The set of values of $p$ for which the points of extremum of the function $f(x)=x^3-3px^2+3(p^2-1)x+1$ lie in the interval $(-2,4)$ is
$(A)(-3,5)$
$(B)(-3,3)$
$(C)(-1,3)$
$(D)(-1,5)$
$$f'(x)=3x^2-6px+3(p^2-1)=3[(p-x)^2-1]$$
For the points of extremum to lie between $(-2,4)$
The vertex of the parabola $f'(x)=3x^2-6px+3(p^2-1)$ should lie between $(-2,4)$
The vertex of the parabola $f'(x)=3x^2-6px+3(p^2-1)$ is $2p$
$$-2<2p<4\implies -1<p<2$$
But the answer is $(-1,3)$
| Because you need $-3<p<3$ and $-1<p<5$, which gives $-1<p<3$
Derivative gives $(x-p)^2-1=0$ and $x_{max}=p-1$, $x_{min}=p+1$.
$-2<p+1<4$ with $-2<p-1<4$ gives $-1<p<3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integrating $ \int\frac{1}{\sqrt{x}+\sqrt[3]{x}}\, dx$ I'm stuck on this integration with substitution problem:
$$
\int\frac{1}{\sqrt{x}+\sqrt[3]{x}}\,dx
$$
It tells me that I have to substitute with $\displaystyle u={x}^{\frac{1}{6}}$ but no matter what I try, I seem to either be getting further away from the answer of it is too long to compute analytically. I've tried multiplying by the conjugate, or factorising with the square root of $x$, but its not turning out how the answer is supposed to be.
Any help is much appreciated!!
| Let
\begin{eqnarray}
u &=& x^\frac{1}{6}\\
\Rightarrow x&=&u^6\\
\Rightarrow dx&=&6u^5 \cdot du
\end{eqnarray}
Hence
\begin{eqnarray}
\int\frac{1}{\sqrt{x}+\sqrt[3]{x}}dx&=& 6 \cdot \int\frac{u^5}{u^3+u^2}du\\
&=& 6 \cdot \int\frac{u^3}{u+1}du\\
&=& 6 \cdot \int\left(\frac{(u^3 + 1)}{u+1} - \frac{1}{u+1}\right)du\\
&=&6 \cdot \int(u^2 - u + 1) \cdot du - 6 \cdot \int\frac{1}{u+1} \cdot du\\
&=& 6 \cdot \left(\frac{u^3}{3} - \frac{u^2}{2} + u\right) - 6 \cdot \ln({u+1}) + C\\
&=& 6 \cdot \left(\frac{\sqrt{x}}{3} - \frac{\sqrt[3]{x}}{2} + \sqrt[6]{x} - \ln({\sqrt[6]{x}+1})\right) + C
\end{eqnarray}
$$\int\left(\frac{1}{\sqrt{x}+\sqrt[3]{x}}\right) \cdot dx = \boxed{\left(2 \sqrt{x} - 3 \sqrt[3]{x} + 6 \sqrt[6]{x} - 6 \ln({\sqrt[6]{x}+1})\right) + C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Correct answer of an indefinite integral Find the value of
$$ \int{\frac{dx}{x\sqrt{1-x^3}}} $$
I assumed $x^3 = \sin^2\theta$ and found the solution as
$$\frac{2}{3} \log\left|\frac{1}{x\sqrt{x}} - \frac{\sqrt{1-x^3}}{x\sqrt{x}} \right| + c$$
but the solution is given as
$$\frac{1}{3} \log{\left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|} + c$$
Any help to reach to this provided solution will be appreciated.
| Please multiply numerator and denominator by $x^2$ and then put $t=x^3$ with $dt=3x^2dx $.
you will get $$I=\frac {1}{3}\int \frac {dt}{t\sqrt {1-t}} $$
to finish, put $$u=\sqrt {1-t }.$$
to find
$$I=\frac {1}{3}\int \frac {2du}{u^2-1} $$
$$=\frac {1}{3}\int (\frac {1}{u-1}-\frac {1}{u+1})du $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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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.