Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Eigenfunction of the Fourier transform I want to show that $$ \frac{1}{ \sqrt{ 2 \pi}} \int_{-\infty}^{\infty} \frac{e^{-iwx}}{\cosh{ (x \sqrt{\frac{\pi}{2}}} ) } = \frac{1}{\cosh{ (w \sqrt{\frac{\pi}{2}}} ) } .$$
My attempt is to first make a substitution $y = x \sqrt{ \frac{\pi}{2} }$, and write $\cosh$ as exponentials which yeilds
$$ \frac{1}{ \sqrt{ 2 \pi}} \int_{-\infty}^{\infty} \frac{e^{-iwx}}{\cosh{ (x \sqrt{\frac{\pi}{2}}} ) }dx = \frac{2}{ \pi} \int_{-\infty}^{\infty} \frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{e^{y} + e^{-y} }dy = \frac{2}{ \pi} \int_{-\infty}^{\infty} e^{-y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{-2y}) }dy$$
Now I can write it as a convergent geometric series
$$\frac{2}{ \pi} \int_{-\infty}^{\infty} e^{-y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{-2y}) }dy = \frac{2}{ \pi} \int_{-\infty}^{\infty} \sum_{k = 0}^{\infty} -e^{-2yk} e^{-y}e^{-iwy\sqrt{\frac{2}{\pi}}}dy$$ $$ = \frac{2}{ \pi} \sum_{k = 0}^{\infty}\int_{-\infty}^{\infty} -e^{-y(2k+1+iw\sqrt{\frac{2}{\pi}})}dy$$
However this integral diverges. Does anyone have any hint on what to do next, or another strategy to show this?
Since this would mean that $\frac{1}{\cosh( x \sqrt{\frac{\pi}{2}})}$ would be an eigenfunctions of the Fourier transforms, should you be able to write it as a linear combination of gaussians?
Thanks!
$\textbf{Added:}$
So before writing it as a geometric series, I split the integration in two parts
$$\frac{2}{ \pi} \int_{-\infty}^{\infty} \frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{e^{y} + e^{-y} }dy = \frac{2}{ \pi} \int_{0}^{\infty} e^{-y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{-2y}) }dy + \frac{2}{ \pi} \int_{-\infty}^{0} e^{y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{2y}) }dy$$
$$= \frac{2}{ \pi} \sum_{k = 0}^{\infty}\int_{0}^{\infty} (-1)^k e^{-y(2k+1+iw\sqrt{\frac{2}{\pi}})}dy + \frac{2}{ \pi} \sum_{k = 0}^{\infty}\int_{-\infty}^{0} (-1)^k e^{y(2k+1-iw\sqrt{\frac{2}{\pi}})}dy$$
$$ = \frac{4}{\pi} \sum_{k=0}^{\infty} (-1)^k \frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2} $$
Now in a previous exercise I proved this identity:
$$ \frac{\cosh(a( \pi -x))}{\sinh(a \pi)} = \frac{1}{a \pi} + \frac{2}{\pi} \sum_{n = 1}^{\infty}\frac{a}{a^2+n^2}\cos(nx). $$ for $ 0 \leq x \leq \pi$, which look very similar with $x = \pi$. However I'm not sure how to proceed.
| You are just about there. One trick that I find helps over this hump is to recognize that your sum may be rewritten as
$$ \frac{2}{\pi} \sum_{k=-\infty}^{\infty} (-1)^k \frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2}$$
Now you can apply the residue theorem to this infinite sum. I will state the following result without (much) proof: the following convergent sum satisfies
$$\sum_{k=-\infty}^{\infty} (-1)^k \, f(k) = -\sum_m \text{Res}_{z=z_m} \pi \csc{(\pi z)} f(z)$$
where $z_m$ are the poles of $f$ that are not real integers. You may prove this by integrating the function $\pi \csc{(\pi z)} f(z)$ along a rectangular contour about the real line interval $x \in [-N,N]$, and taking the limits as $N \to \infty$. Now in this case,
$$f(z) = \frac{2}{\pi}\frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2}$$
Your poles are at $z_{\pm} = -(1/2) \pm i w \sqrt{\frac{1}{2 \pi}} $. Now simply take these poles and plug into the above residue formula:
$$\frac{2}{\pi}\sum_{k=-\infty}^{\infty} (-1)^k \frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2} = - \left [\csc{\left(-\frac{\pi}{2}+i w \sqrt{\frac{\pi}{2}}\right)} + \csc{\left(-\frac{\pi}{2}-i w \sqrt{\frac{\pi}{2}}\right)} \right ]$$
After simplification, I get as your FT:
$$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} dx \, \frac{e^{-i w x}}{\cosh{[x \sqrt{(\pi/2)}]}} = \frac{2}{\cosh{[w \sqrt{(\pi/2)}]}}$$
which would certainly serve as an eigenfunction of the FT.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/434334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Trigonometry - Finding the range of the function Problem :
$$f(\theta)=(2\sqrt{3}+4)\sin\theta +4\cos \theta $$
I have studied if the function is in the form : $f(\theta)=a\cos\theta + b\sin\theta$ then the range of this function can be given as $$[-\sqrt{a^2+b^2} ,\sqrt{a^2+b^2}]$$
So, here the range will be [-$44+16\sqrt{3} , 44+16\sqrt{3}]$ But this is the wrong answer.. please guide.. thanks..
| The answer is indeed $ [-\sqrt{a^2+b^2},\, \sqrt{a^2+b^2}]$.
You made a calculation mistake somewhere. The range is $[-\sqrt{44+16\sqrt{3}},\sqrt{44+16\sqrt{3}}]$
Here's a proof:
Let $a=2\sqrt{3}+4$ and $b= 4$
Choose $y \in [0,2\pi]$ such that $\sin y = \dfrac{b}{\sqrt{a^2+b^2}}$ and $\cos y = \dfrac{a}{\sqrt{a^2+b^2}}$
Then $f(\theta) = a\sin\theta + b\cos\theta = \sqrt{a^2+b^2}(\dfrac{a}{\sqrt{a^2+b^2}}\sin\theta + \dfrac{b}{\sqrt{a^2+b^2}}\cos\theta) = \sqrt{a^2+b^2}(\cos y\,\sin\theta+\sin y\,\cos \theta) = \sqrt{a^2+b^2}\,\sin(\theta+y)$
which lies in the range $ [-\sqrt{a^2+b^2},\, \sqrt{a^2+b^2}]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/435819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Using trig substitution to evaluate $\int \frac{dt}{( t^2 + 9)^2}$ $$\int \frac{\mathrm{d}t}{( t^2 + 9)^2} = \frac {1}{81} \int \frac{\mathrm{d}t}{\left( \frac{t^2}{9} + 1\right)^2}$$
$t = 3\tan\theta\;\implies \; dt = 3 \sec^2 \theta \, \mathrm{d}\theta$
$$\frac {1}{81} \int \frac{3\sec^2\theta \mathrm{ d}\theta}{ \sec^4\theta} = \frac {1}{27} \int \frac{ \mathrm{ d}\theta}{ \sec^2\theta} = \dfrac 1{27}\int \cos^2 \theta\mathrm{ d}\theta $$
$$ =\frac 1{27}\left( \frac{1}{2} \theta + 2(\cos\theta \sin\theta)\right) + C$$
$\arctan \frac{t}{3} = \theta \;\implies$
$$\frac{1}{27}\left(\frac{1}{2} \arctan \frac{t}{3} + 2 \left(\frac{\sqrt{9 - x^2}}{3} \frac{t}{3}\right)\right) + C$$
This is a mess, and it is also the wrong answer.
I have done it four times, where am I going wrong?
| I would like to handle the integral by integration by parts. $$
\begin{aligned}
\int \frac{d t}{\left(t^{2}+9\right)^{2}} &=-\int \frac{1}{2 t} d\left(\frac{1}{t^{2}+9}\right) \\
&=-\frac{1}{2 t\left(t^{2}+9\right)}-\frac{1}{2} \int \frac{1}{t^{2}\left(t^{2}+9\right)} d t \\
&=-\frac{1}{2 t\left(t^{2}+9\right)}-\frac{1}{18} \int\left(\frac{1}{t^{2}}-\frac{1}{t^{2}+9}\right) d t \\
&=-\frac{1}{2 t\left(t^{2}+9\right)}+\frac{1}{18 t}+\frac{1}{54} \arctan \left(\frac{t}{3}\right)+C \\
&=\frac{1}{54}\left[\frac{3 t}{t^{2}+9}+\arctan \left(\frac{t}{3}\right)\right]+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Does $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ converge or diverge? I have to show whether $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ is convergent or divergent. I tried using the squeeze theorem to prove it was convergent. So what I did was
bound ${y_n}$ in between $\frac{-1}{n}$ and $\frac{1}{n}$. That is $\frac{-1}{n} \leq \frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n} \leq \frac{1}{n}$ Since we also proved earlier that $lim\frac{1}{n}=0$ and $lim\frac{-1}{n}= -lim\frac{-1}{n}=0$ It follows by the squeeze theorem that $lim(y_n)=0$. Would this be correct?
Edit: This what I have now.
Well this what I get so far that $ \sum_{k=n+1}^{2n}\frac{1}{k}≥ \sum_{k=n+1}^{2n}\frac{1}{k+1}$ Hence its divergent. I think it works since $ \sum_{k=n+1}^{2n}\frac{1}{k+1}=\frac{1}{(n+1)+1}+\frac{1}{(n+2)+1}+...+\frac{1}{2n+1}$ ≤ $ \sum_{k=n+1}^{2n}\frac{1}{k}= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} $.
| If you know that
$$\gamma_n=\frac{1}{1}+\frac{1}{2}+..+\frac{1}{n}- \ln(n) \,,$$ is convergent then your sequence is exactly
$$\gamma_{n^2}-\gamma_n+ \ln(n) \,.$$
As $\gamma_{n^2}-\gamma_n \to 0$ and $\ln(n) \to \infty$ it follows that your sequence diverges to $\infty$.
Edit With the change, your sequence is
$$\gamma_{2n}-\gamma_n+\ln(2n)-\ln(n)= \gamma_{2n}-\gamma_n+\ln(2)$$
which is convergent.
A simpler solution
your sequence is also
$$\frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}}$$
which is the Riemann Sum associated to $f(x)=\frac{1}{x}$ on $[1,2]$ with $x_k=x_k^*=1+\frac{k}{n}$.
Both solutions also Yield $\ln(2)$ as the limit.
Third solution
$$\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}=\frac{1}{1}\frac{1}{2}+..+\frac{1}{2n-1}-\frac{1}{2n}$$
is a well known identity, pretty standard induction problem. Then you can use the Alternating Series Test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/437244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Analytic solution of $ \int_t^{t+r} \frac{x^2}{(t+r-\sqrt{2rx-x^2})^4}dx $ Could anyone help me to solve this defined integral analytically?
$$
\int_t^{t+r} \dfrac{x^2}{(t+r-\sqrt{2rx-x^2})^4}dx
$$
| The antiderivative of the integrand of the integral under consideration can be found by the Euler changes $\sqrt{2rx-x^2}=tx$ or/and $\sqrt{2rx-x^2}=t(x-2r)$. The output done with Maple by the command $int(x^2/(r+t-2*sqrt(2*r*x-x^2))^4, x)$ is huge. It can be seen in the worksheet exported as a pdf file (20MB). The definite integral under consideration depends on the two parameters $t$ and $r$ so parametric analysis is necessary.
I give only its results:$ t \le r, t \ge r/3$ under the assumption $t \ge 0$. In other cases the integral may diverge because of the singularity caused by the zeros of the denominator of the integrand $r+t-2\sqrt{2rx-x^2}=0$:$$x_1=r+\frac 1 2\,\sqrt {3\,{r}^{2}-2\,rt-{t}^{2}},x_2=r-\frac 1 2\,\sqrt {3\,{r}^{2}-2\,rt-{t}^{2}}.
$$
For concrete values of the parameters the integral under consideration can be found with Maple by the command
$$J:=(t,r)->int(x^2/(r+t-2*sqrt(2*r*x-x^2))^4,x=t..t+r).$$ For example,
$$J(1,2)= {\frac {1440}{2401}}\,{\it arctanh} \left( 2/7\,\sqrt {7} \right)
\sqrt {7}+{\frac {360}{2401}}\,\sqrt {7}\ln \left( \sqrt {7}\sqrt {3}
+8\,\sqrt {3}-9 \right) -$$ $${\frac {360}{2401}}\,\sqrt {7}\ln \left(
\sqrt {7}\sqrt {3}+8\,\sqrt {3}+9 \right) +$$ $${\frac {360}{2401}}\,\sqrt
{7}\ln \left( -\sqrt {7}\sqrt {3}+8\,\sqrt {3}+9 \right) -$$ $${\frac {360
}{2401}}\,\sqrt {7}\ln \left( -\sqrt {7}\sqrt {3}+8\,\sqrt {3}-9
\right) +{\frac {27404}{3087}}\,\sqrt {3}+{\frac {15494}{1029}}
$$
and
$$J\left(\frac 1 {10},2\right) = \infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/437364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Sum of one, two, and three squares If a square $n^2$ can be written as the sum of two nonzero squares as well as the sum of three nonzero squares, then can we conclude that it can be written as the sum of any number of nonzero squares up to $n^2 - 14$ nonzero squares?
Example: $13^2 = 12^2 + 5^2 = 12^2 + 4^2 + 3^2$. But also $13^2 = 11^2 + 4^2 + 4^2 + 4^2$ and $13^2 = 12^2 + 4^2 + 2^2 + 2^2 + 1^2$ etc up to $13^2 = 3^2 + 2^2 + 2^2 + 1^2 + .. + 1^2$.
| Yes, because all numbers can be written as the sum of four squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/437919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Summing Finitely Many Terms of Harmonic Series: $\sum_{k=a}^{b} \frac{1}{k}$ How do I calculate sum of a finite harmonic series of the following form?
$$\sum_{k=a}^{b} \frac{1}{k} = \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \cdots + \frac{1}{b}$$
Is there a general formula for this? How can we approach this if not?
| The formula for sum of H.P. remained unknown for many years, but now I have found the formula as an infinite polynomial.
The formula or the infinite polynomial which is
equal to the sum of th H.P. $\frac{1}{a} + \frac{1}{a+d} + \frac{1}{a+2d} + \frac{1}{a+3d} + ........$ is :-
$Sum of H.P. = (1/a) [1 + \frac{1}{1×(1+b)}(x-1) - \frac{b}{(2×(1+b)(2+b)}(x-1)(x-2) + \frac{b^2}{3×(1+b)(2+b)(3+b)}(x-1)(x-2)(x-3) - \frac{b^3}{4×(1+b)(2+b)(3+b)(4+b)}(x-1)(x-2)(x-3)(x-4) + \frac{b^4}{5×(1+b)(2+b)(3+b)(4+b)(5+b)}(x-1)(x-2)(x-3)(x-4)(x-5) - .....]$
Here, $b=d/a$ and $x$ is the number of terms upto
which you want to find the sum of the H.P.
Substitute any natural number in place of x and
get the sum of the harmonic series.
For more information, visit https://facebook.com/ElementaryResearchesinMathematics.
I have not provided any proof here since that will be a very difficult task at this platform.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/439177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 6
} |
Proving $\frac{a^4}{b^3} +\frac{b^4}{c^3} +\frac{c^4}{a^3} \ge 3$ For each $a,b,c > 0$ and $a^5+b^5+c^5=3 $ .How to prove that :
$$\frac{a^4}{b^3} +\frac{b^4}{c^3} +\frac{c^4}{a^3} \ge 3$$
| Since $a, b, c$ are positive, we can set
$$
a = x^2\quad b = y^2 \quad c = z^2
$$
The inequality becomes
$$
\frac {x^8} {y^6} + \frac {y^8} {z^6} + \frac {z^8} {x^6} \geq 3
$$
with the condition $x^{10} + y^{10} + z^ {10} = 3$.
Applying Hölder's inequality we get:
$$
3 = (x^{10} + y^{10} + z^{10})^{9/10} \cdot (1 + 1 + 1)^{1/10} \geq x^9 + y^9 + z^9
$$
By means of AM - GM inequality we can write
$$
\begin{align}
3 &\geq \frac {(x^9 + y^9 + z^9)^2} {3} \\
&= \sum_{cyc} x^9 \frac {x^9 + 2y^9} {3} \\
&\geq \sum_{cyc} x^9 \sqrt[3] {x^9 y^{18}} \\
&= x^{12} y^6 + y^{12} z^6 + z^{12} x^6
\end{align}
$$
Applying Cauchy-Schwarz inequality we conclude
$$
\left(\frac {x^8} {y^6} + \frac {y^8} {z^6} + \frac {z^8} {x^6} \right) \cdot
(x^{12} y^6 + y^{12} z^6 + z^{12} x^6) \geq (x^{10} + y^{10} + z^{10})^2 = 9
$$
and finally
$$
\frac {x^8} {y^6} + \frac {y^8} {z^6} + \frac {z^8} {x^6} \geq \frac 9 {x^{12} y^6 + y^{12} z^6 + z^{12} x^6} \geq 3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/439508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$
Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$
example1: $|x-1|+|x-2|\geq 1$
my solution：(substitution)
$x-1=t,x-2=t-1,|t|+|t-1|\geq 1,|t-1|\geq 1-|t|,$
square,
$t^2-2t+1\geq 1-2|t|+t^2,\text{Since} -t\leq -|t|,$
so proved.
question1 : Is my proof right? Alternatives?
one reference answer:
$1-|x-1|\leq |1-(x-1)|=|1-x+1|=|x-2|$
question2 : prove:
$|x-1|+|x-2|+|x-3|\geq 2$
So I guess:( I think there is a name about this, what's that? wiki item?)
$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$
How to prove this? This is question3. I doubt whether the two methods I used above may suit for this general case.
Of course, welcome any interesting answers and good comments.
| You want
$S(x)
=\sum_{i=1}^n |x-i|$.
I will show
$\begin{cases}
\text{if } x \le 1&S(x) \ge \dfrac{n(n-1)}{2}\\
\text{if } x \ge n &S(x) \ge \dfrac{n(n-1)}{2}\\
\text{otherwise} &S(x) \ge \dfrac{n^2-1}{4}
\end{cases}
$
Exact results are
$\begin{cases}
\text{if } x \le 1, & S(x) = n(1-x) + \dfrac{n(n-1)}{2}\\
\text{if } x \ge n, & S(x) = n(x-n)+ \dfrac{n(n-1)}{2}\\
\text{if } j\le x < j+1 \text{ where } 1 \le j < n, &S(x) = \dfrac{n^2+(n-2x+1)^2-(1-2y)^2}{4}\text{ where } y=x-j
\end{cases}
$
As a check, these are correct for
$x=0, 1, n$,
and
$n+1$.
We consider three cases:
$x \le 1$,
$x \ge n$,
and $1 < x < n$.
If $x \le 1$,
then
$|x-i| = -x+i$,
so
$\begin{align}
S(x) &= \sum_{i=1}^n (-x+i)\\
&=-nx + \frac{n(n+1)}{2}\\
&=n(1-x)-n+ \frac{n(n+1)}{2}\\
&=n(1-x) +\frac{n(n-1)}{2}\\
\end{align}
$
If $x \ge n$,
then
$|x-i| = x-i$,
so
$\begin{align}
S(x) = \sum_{i=1}^n (x-i)
&=nx - \frac{n(n+1)}{2}\\
&=n(x-n)+n^2 - \frac{n(n+1)}{2}\\
&=n(x-n) \frac{2n^2-n(n+1)}{2}\\
&= n(x-n)+\frac{n^2-n}{2}\\
&= n(x-n)+\frac{n(n-1)}{2}\\
\end{align}
$
If $j \le x < j+1$
where $1 \le j < n$,
let $y = x-j$,
so $0 \le y < 1$.
Then
$\begin{align}
S(x)
&=\sum_{i=1}^n |x-i|\\
&=\sum_{i=1}^n |j+y-i|\\
&=\sum_{i=1}^j |j+y-i|+\sum_{i=j+1}^n |j+y-i|\\
&=\sum_{i=1}^j (j+y-i)+\sum_{i=j+1}^n -(j+y-i)\\
&=\sum_{i=1}^j (j+y-i)+\sum_{i=j+1}^n (i-j-y)\\
&=jy+\sum_{i=0}^{j-1} (i)-(n-j)y+\sum_{i=1}^{n-j} (i)\\
&=(2j-n)y+\frac{j(j-1)+(n-j)(n-j+1)}{2}\\
&=(2j-n)y+\frac{j^2-j+(n-j)^2+(n-j)}{2}\\
&=(2j-n)y+\frac{(n^2+(n-2j)^2)/2+(n-2j)}{2}\quad (*)\\
&=(2j-n)y+\frac{n^2+(n-2j)^2+2(n-2j)}{4}\\
&=\frac{n^2+(n-2j)^2+2(n-2j)-4y(n-2j)}{4}\\
&=\frac{n^2+(n-2j)^2+2(1-2y)(n-2j)}{4}\\
&=\frac{n^2+(n-2j+1-2y)^2-(1-2y)^2}{4}\\
&=\frac{n^2+(n-2x+1)^2-(1-2y)^2}{4}\\
&\ge \frac{n^2-1}{4}\\
\end{align}
$
Note: The line marked "(*)" used
$a^2+b^2 = \dfrac{(a+b)^2+(a-b)^2}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/439745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Calculating this Riemann sum limit
Calculate the limit $$\lim_{n\to \infty} {\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}}$$
How exactly do we calculate this limit of the Riemann sum? I am never able to find what is the partition. I know that our $f(x)$ is $\sin(x)$.
| Recall that if $f$ is integrable on $[a,b]$, then:
$$
\int_a^b f(x)~dx = \lim_{n\to \infty} \dfrac{b-a}{n}\sum_{k=1}^n f \left(a + k \left(\dfrac{b-a}{n}\right) \right)
$$
Notice that:
$$
\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}
= \sum_{k=1}^{n} {\left(\dfrac{k}{n^2} - \dfrac{1}{n^3}\right) \sin\frac{k}{n}}
= \dfrac{1}{n}\sum_{k=1}^{n} \dfrac{k}{n}\sin\frac{k}{n} - \dfrac{1}{n^3}\sum_{k=1}^{n} \sin\frac{k}{n}
$$
Hence, by letting $a=0$ and $b=1$ and considering the functions $f(x)=x \sin x$ and $g(x) = \sin x$, we obtain:
$$ \begin{align*}
\lim_{n\to \infty} {\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}}
&= \lim_{n\to \infty} \left[ \dfrac{1}{n}\sum_{k=1}^{n} \dfrac{k}{n}\sin\frac{k}{n} - \dfrac{1}{n^3}\sum_{k=1}^{n} \sin\frac{k}{n} \right] \\
&= \lim_{n\to \infty} \left[ \dfrac{1}{n}\sum_{k=1}^{n} \dfrac{k}{n}\sin\frac{k}{n} \right] - \lim_{n\to \infty}\left[\dfrac{1}{n^2} \right] \cdot \lim_{n\to \infty} \left[\dfrac{1}{n}\sum_{k=1}^{n} \sin\frac{k}{n} \right] \\
&= \int_0^1 x \sin x~dx - 0 \cdot \int_0^1 \sin x~dx \\
&= \int_0^1 x \sin x~dx\\
&= \left[\sin x - x\cos x \right]_0^1\\
&= \sin 1 - \cos 1\\
\end{align*} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/439889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
For a positive integer $n$, $p_n$ denotes the product of the digits of n, and $s_n$ denotes the sum of the digits of n. For a positive integer $n$, $p_n$ denotes the product of the digits of $n$, and $s_n$ denotes the sum of the digits of $n$. What is the number of integers between 10 and 1000 for which $p_n + s_n = n$ ?
Let $n = xy$ be the two digit number satisfying the given condition.
Given, the product of the digits of $n + \text{sum of the digits of $n$} = n$,
\begin{align}
p_n + s_n &= n\\
xy + x + y &= 10x + y\\
9x – xy &= 0\\
x (9 - y) &= 0.
\end{align}
But $x$ is not zero, because $xy$ is a two digit number.
So, $9 - y = 0 ⇒ y = 9$.
So, $xy$ can be 19, 29, 39, 49, 59, 69, 79, 89, and 99. i.e, 9 numbers.
Let $n = xyz$ be the three digit number satisfying the given condition.
Since $p_n + s_n = n$
\begin{align}
xyz + x + y + z &= 100x + 10y + z.\\
99x + 9y – xyz &= 0.\\
xyz &= 99x + 9y.
\end{align}
| Given the conditions, your last equation doesn't have any solution. Let's see.
As $x \neq 0$ let's divide the equation by $x$:
$$yz=99+ \frac{y}{x}$$
Therefore $x |y$.
But for $y=0$ there's no solution, and for $y \neq 0$:
$$yz >99 \Rightarrow y>10 \, \mathrm{or} \, z>10.$$
Which is impossible since $y$ and $z$ are digits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/440417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove inequality by induction Once again, I'm stuck in a demonstration by induction, this time, it's really proving that an inequality is valid. So, here is the inequality:
Prove that $\binom{2n}{n} \geq (n+5)^2 \ \forall n \geq 5, n \in \mathbb{N} $
Then, what I wanted to prove is that:
$\binom{2n+2}{n+1} \geq (n+6)^2$
For $n=5$:
$\binom{10}{5} = 252 \geq 100$
The inductive step would be:
$\binom{2n+2}{n+1} \geq (n+6)^2$
$\binom{2n+2}{n+1} = \frac{(2n + 2)(2n+1)(2n)!}{(n+1)(n+1)n!n!} = \frac{2(2n+1)}{n+1}\frac{(2n)!}{n!n!}$
By Inductive Hypothesis, I know that:
$\frac{2(2n+1)}{n+1}\frac{(2n)!}{n!n!} \geq \frac{2(2n+1)}{n+1} (n+5)^2 $
I want to prove, indeed, that:
$\frac{2(2n+1)}{n+1} (n+5)^2 \geq (n+6)^2$
And there I'm stuck. I've tried doing this:
$ 2(2n+1)(n+5)^2) \geq (n+6)^2(n+1) $
$\Rightarrow 4n^3 + 40 n^2 + 120 n + 50 \geq n^3 + 13n^2+38n+36$
$\Rightarrow n(4n^2 + 42n + 120) + 50 \geq n(n^2 + 13n + 38) + 36$
Because $50 \geq 36 \Rightarrow n(4n^2+42n+120) \geq n(n^2+13n+38) $
But they told me here that this inequality is not necessarily true. So.. any ideas how should I follow?
Thanks for your help!
| What you have is right, you want to show that if $n \geq 5$, then
$$2(2n+1)(n+5)^2 \geq (n+6)^2(n+1) $$
This can be shown in various ways. For example, if you just expand everything, you get that
$$2(2n+1)(n+5)^2 \geq (n+6)^2(n+1) \Leftrightarrow 3n^3 + 29n^2 + 72n + 14 \geq 0 $$
which is obviously true for the region that we want.
Note, what they are telling you, is that just because $50 \geq 36$, does not necessary imply that $2(2n+1)(n+5)^2 \geq (n+6)^2(n+1) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/443735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find an equation of each tangent plane to a cone that is parallel to a given plane. Is the solution correct? Please check it and tell me my mistakes. Thank you.
Find an equation for each plane tangent to $K$ which is parallel to the plane $x-y+z=5$, where $K$ is the half-cone $z=\sqrt{x^2+y^2}$
Let $f(x,y,z) = z-\sqrt{x^2+y^2}$.
To get the normal vector to the cone at the $(x,y,z)$, calculate the gradient of $f$:
\begin{align*}
\nabla f(x,y,z) &= \frac {2x}{-2\sqrt{x^2 + y^2}}\mathbf i+\frac{2y}{-2\sqrt{x^2+y^2}}\mathbf j+\mathbf k \\
&=-\frac x {\sqrt{x^2+y^2}}\mathbf i - \frac y {\sqrt{x^2+y^2}}\mathbf j + \mathbf k.
\end{align*}
Since we want the tangent plane to be parallel to the plane $x-y+z=5$, the normal vector has to be parallel to the vector $\mathbf i-\mathbf j+\mathbf k$ since this is normal to the plane ($x-y+z=5$). That is,
\begin{gather}
\require{cancel}
-\frac x {\sqrt{x^2+y^2}}\mathbf i
- \frac y {\sqrt{x^2+y^2}}\mathbf j
+ \cancel{\mathbf k} = c(\mathbf i - \mathbf j + \cancel{\mathbf k})
\qquad \text{constant $c = 1$} \\
\left.\begin{gathered}
-\frac x z = 1 \implies \bbox[2.5pt,border:1pt solid black]{x = -z}\\
-\frac y z = 1 \implies \bbox[2.5pt,border:1pt solid black]{y = z}
\end{gathered}\right\}
\:
\begin{gathered}
\land z = \sqrt{x^2+y^2} \implies z = \sqrt{2z^2} \implies z=0 \\
\bbox[2.5pt,border:1pt solid black]{x=0,\,y=0}
\end{gathered}
\end{gather}
So $K$ has no tangent plane at the origin, which is parallel to $x-y+z=5$.
| This solution looks correct. However, there is one careless error towards the very end. Where you have:
$$-\frac{y}{z} = 1$$
It should read:
$$-\frac{y}{z} = -1$$
But, it appears to be a careless error, because you ended up correcting for it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/447665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $p =\frac{4\sin\theta \cos\theta}{\sin\theta +\cos\theta}$ Find the value of $\frac{p+2\sin\theta}{p-2\sin\theta}$ Problem :
If $\displaystyle p =\frac{4\sin\theta\cos\theta}{\sin\theta +\cos\theta}$, find the value of $\displaystyle \frac{p+2\sin\theta}{p-2\sin\theta} + \frac{p+2\cos\theta}{p-2\cos\theta}$.
Please help how to proceed in such problem..Thanks..
| You can directly substitute the value P in the given term
We get:
$\large \large \frac {3\sinθ + \cosθ}{\sinθ - \cosθ} + \frac {3\cosθ + \sinθ}{\cosθ - \sinθ}$
which simplifies to $2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/447978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$ How to solve problems involving roots. If we square them they may go to fourth degree.There must be some technique to solve this.
$$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$
| $$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$
Put x-1 = t^2
so the equation becomes
$$\sqrt{t^2+4-4t} + \sqrt{t^2+9-6t} =1$$
complete the squares .
EDIT:
My answer was incorrect. Thanks to egreg and Samatix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/449300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$\det(I+A) = 1 + tr(A) + \det(A)$ for $n=2$ and for $n>2$? Let $I$ the identity matrix and $A$ another general square matrix. In the case $n=2$ one can easily verifies that
\begin{equation}
\det(I+A) = 1 + tr(A) + \det(A)
\end{equation}
or
\begin{equation}
\det(I+tA) = 1 + t\ tr(A) + t^2\det(A)
\end{equation}
for some scalar $t \in \mathbb{R}$.
I have tried to see if there exists a similar formula for $n>3$. This is a natural question. But the calculations are very big and difficulty to see. Then I do the answer. Is there a similar formula for $n>2$?
| $$
\left|
\begin{array}{ccc}
1+a_{11} & a_{12} & a_{13} \\
a_{21} & 1+a_{22} & a_{23} \\
a_{31} & a_{32} & 1+a_{33} \\
\end{array}
\right|
=
\left|
\begin{array}{ccc}
1 & a_{12} & a_{13} \\
0 & 1+a_{22} & a_{23} \\
0 & a_{32} & 1+a_{33} \\
\end{array}
\right|
+
\left|
\begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
a_{21} & 1+a_{22} & a_{23} \\
a_{31} & a_{32} & 1+a_{33} \\
\end{array}
\right|=
$$
$$
=
\left|
\begin{array}{cc}
1+a_{22} & a_{23} \\
a_{32} & 1+a_{33} \\
\end{array}
\right|
+
\left|
\begin{array}{ccc}
a_{11} & 0 & a_{13} \\
a_{21} & 1 & a_{23} \\
a_{31} & 0 & 1+a_{33} \\
\end{array}
\right|
+
\left|
\begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & 1+a_{33} \\
\end{array}
\right|
=\ldots
$$
$$
=
1+{\rm tr}A + M_{11} + M_{22} + M_{33}+\det A
$$
and further by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 0
} |
Use the addition or subtraction formula for cosine to compute $\cos(-5\pi/12)$ Use the addition or subtraction formula for cosine to compute $\cos(-5\pi/12)$ (Leave your answer in exact form.)
I have $$\begin{align}\cos(-5\pi/12)&=\cos((\pi/4)-(5\pi/6))\\
&=\cos(\pi/4)\cos(5\pi/6)+\sin(\pi/4)\sin(5\pi/6)\\
&=\frac{\sqrt2}2\cdot\frac{\sqrt3}2+\frac{\sqrt2}2\cdot\frac 12\end{align}$$
Is this right?
| Alternative, you can use the half-angle formula (I am aware that this was not specified):
$$\cos{\left(-\frac{5 \pi}{12}\right)} = \sqrt{\frac{1+\cos{(-5 \pi/6)}}{2}} = \frac{\sqrt{2-\sqrt{3}}}{2}$$
Now, $2-\sqrt{3} = (\sqrt{3}-1)^2/2$
so that
$$\cos{\left(-\frac{5 \pi}{12}\right)} = \frac{\sqrt{3}-1}{2 \sqrt{2}} = \frac{\sqrt{6}-\sqrt{2}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
An inequality involving arctan of complex argument I have the following conjecture:
\begin{equation}
\text{Re}\left[(1+\text{i}y)\arctan\left(\frac{t}{1+\text{i}y}\right)\right] \ge \arctan(t), \qquad \forall y,t\ge0.
\end{equation}
Which seems to be true numerically.
Can anyone offer some advice on how to approach proving (or disproving) this?
It originates from a question involving the (complex) Hilbert transform of a symmetric non-increasing probability distribution:
\begin{equation}
h(y) = (1+\text{i}y)\int_{-\infty}^\infty \frac{1}{1 + \text{i}(y-t)}\text{d}G(t)
\end{equation}
and attempting to show $\text{Re}[h(y)]$ takes its minimum at $y=0$.
| I start at the beginning, with a symmetric non-increasing density $\varphi$ on $\mathbb R$. The layer cake representation reduces the matter to $\varphi=\chi_{[-a,a]}$. Therefore, the goal is to show that
$$\int_{-a}^a \operatorname{Re}\frac{1+iy}{1+i(y-t)}\,dt > \int_{-a}^a \operatorname{Re}\frac{1 }{1-i t}\,dt \tag1$$
for all $y>0$ and $a>0$. Let's find this real part:
$$
\operatorname{Re}\frac{1+iy}{1+i(y-t)} = \frac{1+y^2-ty}{1+(y-t)^2}
\tag2$$
and add the value at $-t$ so that we integrate over $[0,a]$ only:
$$
\frac{1+y^2-ty}{1+(y-t)^2}+ \frac{1+y^2+ty}{1+(y+t)^2} = 2\,\frac{1+2y^2+t^2+y^4-y^2t^2}{(1+(y-t)^2)(1+(y+t)^2)}
\tag3$$
Looks bad, but we have to keep going. The goal is to prove that
$$
\int_0^a 2\,\frac{1+2y^2+t^2+y^4-y^2t^2}{(1+(y-t)^2)(1+(y+t)^2)}\,dt >
\int_0^a \frac{2}{1+t^2}\,dt
\tag4$$
Take the difference of two sides, and it simplifies!
$$
\int_0^a \frac{2 y^2 t^2 (3+y^2-t^2)}{(1+t^2)(1+(y-t)^2)(1+(y+t)^2)}\,dt > 0
\tag5$$
Sweet. The denominator is always positive. The numerator changes sign only once, from plus to minus.
Therefore, as a function of $a$, the integral in (5) increases from $0$ (at $a=0$) to some positive
value, and then decreases to its limit as $a\to\infty$. It turns out that the limit as $a\to\infty$ is $0$:
$$
\int_0^\infty \frac{2 y^2 t^2 (3+y^2-t^2)}{(1+t^2)(1+(y-t)^2)(1+(y+t)^2)}\,dt = 0
\tag6$$
which completes the proof.
Well, I ought to prove (6) instead of trusting my computer. Let
$$
R(z) = \frac{2 y^2 z^2 (3+y^2-z^2)}{(1+z^2)(1+(y-z)^2)(1+(y+z)^2)},\quad z\in\mathbb C
\tag7$$
This is an even rational function which is $O(|z|^{-2})$ at infinity. Hence, $\int_0^\infty R(z)\,dz$
is $\pi i$ times the sum of residues of $R$ in the upper halfplane. For computing the
residues, it makes sense to go back to where $R$ came from:
$$
R(z) = \frac{1+y^2-zy}{1+(y-z)^2}+ \frac{1+y^2+zy}{1+(y+z)^2} -\frac{2}{1+z^2}
\tag8$$
From (8) we easily get
$$
\begin{split}
\operatorname{res}\limits_{z=i} R(z) &= -\frac{2}{2i} = i \\
\operatorname{res}\limits_{z=i-y} R(z) &= \frac{1+y^2+(i-y)y}{2(i-y)} =\frac{1+iy}{2(i-y)} =-\frac{i}{2}\\
\operatorname{res}\limits_{z=i+y} R(z) &= \frac{1+y^2-(i+y)y}{2(i+y)} = \frac{1-iy}{2(i+y)}= -\frac{i}{2}
\end{split} \tag{9}$$
which indeed sum to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Fractions in Ancient Egypt In ancient Egypt, fractions were written as sums of fractions with numerator 1. For instance,$ \frac{3}{5}=\frac{1}{2}+\frac{1}{10}$. Consider the following algorithm for writing a fraction $\frac{m}{n}$ in this form$(1\leq m < n)$: write the fraction $\frac{1}{\lceil n/m\rceil}$ , calculate the fraction $\frac{m}{n}-\frac{1}{\lceil n/m \rceil}$ , and if it is nonzero repeat the same step. Prove that this algorithm always finishes in a finite number of steps.
Note:if $ n\in \mathbf{Z} $ and $n-1<x\leq n , \lceil x\rceil=n$
| We proceed by induction on $m\in \Bbb{N}$.
Base Case: When $m=1$, $\dfrac{m}{n}$ is already in the desired form, so terminate.
Induction Hypothesis: Assume that for all $m \in \{1,...,k\}$ (where $k \geq 1$), the algorithm will terminate when its input is $\dfrac{m}{n}$.
It remains to prove the claim true for $m=k+1$. After one step of the algorithm, we obtain the new fraction:
$$
\dfrac{k+1}{n}-\dfrac{1}{\left\lceil \frac{n}{k+1} \right\rceil}
= \dfrac{\left\lceil \frac{n}{k+1} \right\rceil(k+1)-n}{\left\lceil \frac{n}{k+1} \right\rceil n}
$$
Now observe that:
$$ \begin{align*}
\left\lceil \frac{n}{k+1} \right\rceil(k+1)-n
&= \left( \left\lfloor \frac{n-1}{k+1} \right\rfloor+1 \right)(k+1) -n \\
&\leq \left( \frac{n-1}{k+1} +1 \right)(k+1) -n \\
&= (n-1)+(k+1) -n \\
&= k
\end{align*} $$
Hence, by the induction hypothesis, the algorithm must terminate. This completes the induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
If $a,b,c$ are in Geometric Progression and$ a-b,c-a,b-c$ are in Harmonic Progression, then the value of $a+4b+c$ is equal to? If $a,b,c$ are in Geometric Progression and $a-b$,$c-a$,$b-c$ are in Harmonic Progression, then the value of $a+4b+c$ is equal to?
The statements I could get are
1)$b^2=ac$
2)$(c-a)^2=2(b-a)(b-c)$
I get that $(a+4b+c)^2=24ac+6ab+6bc$.
how to proceed and get to the answer??Options given are
a)1
b)0
3)2abc
d)b^2+ac
| Let $b=ra$ and $c=r^2a$. Then $c-a=\frac{2(a-b)(b-c)}{a-c}$, so $(c-a)^2=2(b-a)(b-c)$ and therefore $(r^2-1)^2 a^2=2(r-1)ar(1-r)a=-2r(r-1)^2a^2$. Dividing by
$(r-1)^2a^2$ gives $(r+1)^2=-2r$ and therefore $r^2+4r+1=0$. Thus $a+4b+c=a+4ra+r^2 a=a(r^2+4r+1)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral of normal distribution How does one calculate the indefinite integral?
$$\int\frac1{\sigma\sqrt{2\pi}}\exp\left(-\frac{x^2}{2\sigma^2}\right)dx$$
Where $\sigma$ is some constant.
Work so far:
Integrating from exp as rest is constant.
$$\begin{align}
\int\exp\left(-\frac{x^2}{2\sigma^2}\right)dx&=\sum_{n=0}^\infty\frac{\left(-\frac{x^2}{2\sigma^2}\right)^n}{n!}=-\sum_{n=0}^\infty n!^{-1}2^{-1}\sigma^{-2n}\int x^{2n}dx\\
&=-\sum_{n=0}^\infty n!^{-1}\sigma^{-2n}x^{2n}x^{-1}\\
\end{align}$$
I pulled it apart, integrated it, now I cant put it back together.
| It's been a while since you've asked, but this is the indefinite integral:
[\begin{array}{l}
\int {\left( {\frac{{{e^{ - \frac{1}{2}\;\cdot\;{{\left( {\frac{x}{\sigma }} \right)}^2}}}}}{{\sigma \sqrt {2\pi } }}} \right)\;dx = } \frac{{\int {\left( {\frac{{{e^{ - \frac{1}{2}\;\cdot\;{{\left( {\frac{x}{\sigma }} \right)}^2}}}}}{\sigma }} \right)} \;dx}}{{\sqrt {2\pi } }}\\
u = \frac{x}{\sigma }\\
du = \frac{1}{\sigma }\;dx\\
= \frac{{\int {\left( {{e^{ - \frac{{{u^2}}}{2}}}} \right)} \;du}}{{\sqrt {2\pi } }}\\
= \frac{{\int {{e^{ - \frac{{{u^2}}}{2}}}} \;du}}{{\sqrt {2\pi } }}\\
= \frac{{\int {{e^{ - \frac{{{u^2}}}{2}}}} \cdot1\;du}}{{\sqrt {2\pi } }}\\
= \frac{{{e^{ - {u^2}}}\cdot\int 1 \;du - \int {\left( { - u{e^{ - \frac{{{u^2}}}{2}}}\cdot\int 1 \;du} \right)} \;du}}{{\sqrt {2\pi } }}\\
= \frac{{\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1} - \int {\left( {\frac{{ - {u^2}{e^{ - \frac{{{u^2}}}{2}}}}}{1}} \right)\;} \;du}}{{\sqrt {2\pi } }}\\
= \frac{{\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1}\; - \frac{{ - 2{e^{ - \frac{{{u^2}}}{2}}}}}{1}\;\cdot\;\int {{u^2}} \;du + \int {\left( {\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1}\;\cdot\int {{u^2}} \;du} \right)} \;du}}{{\sqrt {2\pi } }}\\
= \frac{{\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1}\; + \frac{{{u^3}{e^{ - \frac{{{u^2}}}{2}}}}}{3} + \int {\left( {\frac{{{u^4}{e^{ - \frac{{{u^2}}}{2}}}}}{3}} \right)} \;du}}{{\sqrt {2\pi } }}\\
= \frac{{\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1}\; + \frac{{{u^3}{e^{ - \frac{{{u^2}}}{2}}}}}{3} + \frac{{{e^{ - \frac{{{u^2}}}{2}}}}}{3}\;\cdot\int {\left( {{u^4}} \right)} \;du - \int {\left( {\frac{{ - u{e^{ - \frac{{{u^2}}}{2}}}}}{3}\;\cdot\int {{u^4}} \;du} \right)} \;du}}{{\sqrt {2\pi } }}\\
= \frac{{\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1}\; + \frac{{{u^3}{e^{ - \frac{{{u^2}}}{2}}}}}{3} + \frac{{{u^5}{e^{ - \frac{{{u^2}}}{2}}}}}{{15}} - \int {\left( {\frac{{ - {u^6}{e^{ - \frac{{{u^2}}}{2}}}}}{{15}}} \right)} \;du}}{{\sqrt {2\pi } }}\\
= \frac{{\sum\limits_{n = 1}^\infty {\left( {\frac{{{u^{2n - 1}}\cdot{e^{ - \frac{{{u^2}}}{2}}}}}{{\prod\limits_{i = 1}^n {\left( {2i - 1} \right)} }}} \right)} }}{{\sqrt {2\pi } }}\\
= \sum\limits_{n = 1}^\infty {\left( {\frac{{{{\left( {\frac{x}{\sigma }} \right)}^{2n - 1}}\cdot{e^{ - \frac{1}{2}\;\cdot\;{{\left( {\frac{x}{\sigma }} \right)}^2}}}}}{{\sqrt {2\pi } \;\cdot\prod\limits_{i = 1}^n {\left( {2i - 1} \right)} }}} \right)}
\end{array}]
$$ \int \frac{e^{-\frac{x^2}{2\sigma^2}}}{\sigma \sqrt{2\pi}} ~dx =
\sum_{n=1}^\infty \left( \frac{\left(\frac{x^2}{\sigma^2}\right)^{2n-1} e^{-\frac{x^2}{2\sigma^2}}}{\sqrt{2\pi}\cdot \prod_{i=1}^n (2i-1)} \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/461433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is wrong with my solution of finding remainder of $50^{(51^{52})}$ when divided by 11? I used the following method using remainder theorem. (I used method from here: Find the remainder of $128^{1000}/153$.)
$$\begin{align}
(50^{{51}^{52}})/11 & = (50^{2652})/11 \implies \\
(6^{2652})/11 & = (36^{1326})/11 \implies \\
(3^{1326})/11 &= (27^{442})/11 \implies \\
(5^{442})/11 & = ({25}^221)/11 \implies \\
(3^{221})/11
\end{align}$$
Now,
3^1/11 implies remainder 3
3^2/11 implies remainder 9
3^3/11 implies remainder 5
3^4/11 implies remainder 4
3^5/11 implies remainder 1
3^6/11 implies remainder 3
So modulo 11, powers of 3 repeat with period 5
so 221 is 1 modulo 5
so $50^{{51}^{52}}/11$ should become finally $3^1/11$ implies remainder 3 (my answer). Actual answer is 6.
What is wrong with this method?
| Since $50\equiv 6\pmod{11}$, you should start by looking at powers of $6$ modulo $11$. If you do, you'll find that $50^{10}\equiv6^{10}\equiv 1\pmod{11}$. That means that when computing $50^x\pmod{11}$ we can discard all multiples of $10$ in $x$ and just concentrate on the remainder mod $10$. For example, if we needed to know $50^{672}\pmod{11}$ we could write
$$
672=67\cdot10+2
$$
and so
$$
50^{672}= 50^{67\cdot 10+2}= (50^{67\cdot 10})(50^2)= (50^{10})^{67}(50^2)\equiv(1)^{67}(50^2)\equiv 50^2\pmod{11}
$$
To recap, we now know
$$
50^{\,x}\equiv 50^{\,x\ \bmod{10}}\pmod{11}
$$
In this problem, we have $x=51^{52}$ which is good news since modulo $10$ we have
$$
51^{52}\equiv 1^{52} \equiv 1 \pmod{10}
$$
Putting the two display lines above together, we have
$$
50^{51^{52}} \equiv 50^{\,51^{52}\ \bmod {10}}\equiv50^1\equiv 6^1\equiv 6\pmod{11}
$$
As Thomas suggested in his comment to your post, there's a cleaner way to do this if you know Fermat's Little Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/462028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using the Legendre Symbol Which have solutions? $x^{2} \equiv7 \mod{53}$, $x^{2} \equiv53 \mod{7}$, $x^{2} \equiv 14 \mod{31}$, $x^{2} \equiv 25\mod{997}$?
I have all of these properties for this Legendre symbol and no idea what to do with it to find whether these have a soltion, let alone solve them if they do.
| They all have solutions. Since $53 \equiv 1 \pmod{4}$, we have
$$\left(\frac{7}{53}\right) = \left(\frac{53}{7}\right) = \left(\frac{4}{7}\right) = \left(\frac{2}{7}\right)^2 = 1,$$
which settles the first two. $31 \equiv 7 \pmod{8}$, so
$$\left(\frac{14}{31}\right) = \left( \frac{2}{31}\right) \left(\frac{7}{31}\right) = \left(\frac{7}{31}\right) = - \left(\frac{31}{7}\right) = - \left(\frac{3}{7}\right) = \left(\frac{7}{3}\right) = \left(\frac13\right) = 1.$$
And finally
$$\left(\frac{25}{997}\right) = \left(\frac{5}{997}\right)^2 = 1.$$
Finding solutions is a different matter, but
$$22^2 = 484 = 9\cdot 53 + 7;\quad 2^2 = 4 \equiv 53 \pmod{7};\quad 13^2 = 169 = 5\cdot 31 + 14; \quad 5^2 = 25.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving an equation involving binomial coefficients and complex numbers Question:
Solve the following equation for $x$:
$$\sum_{k=0}^{n}\binom{n}{k}x^{k}\cos(k\theta )=0$$
Attempt:
I think this equation come from:
$$(x\cos\theta+ix\sin\theta)^{k}$$
Is that right?
I don't know what to do after that.
| $$\sum_{0\le r\le n}\binom nkx^ke^{ik\theta}=(1+xe^{i\theta})^n=(1+x\cos\theta+ix\sin\theta)^n$$
Let $1+x\cos\theta=R\cos\alpha, x\sin\theta=R\sin\alpha$
$$R^2=(1+x\cos\theta)^2+(x\sin\theta)^2=1+x^2+2x\cos\theta$$
$$\implies \cos\alpha=\frac{1+x\cos\theta}{\sqrt{1+x^2+2x\cos\theta}}$$
$$\implies (1+x\cos\theta+ix\sin\theta)^n=\{R(\cos\alpha+i\sin\alpha)\}^n=R^n(\cos n\alpha+i\sin n\alpha)$$
So, we need either $R=0$ or $\cos n\alpha=0$
If $R=0, x\sin\theta=1+x\cos\theta=0$
Case $1:$ If $x=0, 1+x\cos\theta=1\ne0\implies x\ne0$
Case $2:$ If $\sin\theta=0,\cos\theta=\pm1,1\pm x=0\implies x=\mp1 $
Else $R\ne0 ,\cos n\alpha=0\implies n\alpha=(2m+1)\frac{\pi}2$ where $m$ is any integer
Now we need to solve for $x,\theta$ from $$\cos (2m+1)\frac{\pi}{2n}=\frac{1+x\cos\theta}{\sqrt{1+x^2+2x\cos\theta}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\cos^4 \theta −\sin^4 \theta = x$. Find $\cos^6 \theta − \sin^6 \theta $ in terms of $x$. Given $\cos^4 \theta −\sin^4 \theta = x$ , I've to find the value of $\cos^6 \theta − \sin^6 \theta $ .
Here is what I did:
$\cos^4 \theta −\sin^4 \theta = x$.
($\cos^2 \theta −\sin^2 \theta)(\cos^2 \theta +\sin^2 \theta) = x$
Thus
($\cos^2 \theta −\sin^2 \theta)=x$ ,
so $\cos 2\theta=x$ .
Now $x^3=(\cos^2 \theta −\sin^2 \theta)^3=\cos^6 \theta-\sin^6 \theta +3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $
So if I can find the value of $3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $ in terms of $x$ , the question is solved. But how to do that ?
| As already found $\cos^2\theta-\sin^2\theta=x,\implies \cos2\theta=x$
Using $a^3-b^3=(a-b)^3+3ab(a-b),$
$$\cos^6\theta-\sin^6\theta=(\cos^2\theta)^3-(\sin^2\theta)^3$$
$$=(\cos^2\theta-\sin^2\theta)^3+3\cos^2\theta\sin^2\theta(\cos^2\theta-\sin^2\theta) $$
$$=x^3+3x\cos^2\theta\sin^2\theta$$
$$=x^3+\frac{3x}4\sin^22\theta\text{ as }\sin2A=2\sin A\cos A$$
$$=x^3+\frac{3x}4(1-\cos^22\theta)$$
$$=x^3+\frac{3x}4(1-x^2)=\frac{4x^3+3x-3x^3}4=\frac{x(x^2+3)}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
For $a$, $b$, $c$, $d$ the sides of a quadrilateral, show $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. (A generalization of IMO 1983 problem 6)
Let $a$, $b$, $c$, and $d$ be the lengths of the sides of a quadrilateral. Show that
$$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0 \tag{$\star$}$$
Background: The well known 1983 IMO Problem 6 is the following:
IMO 1983 #6. Let $a$, $b$ and $c$ be the lengths of the sides of a triangle.
Prove that $$a^{2}b(a - b) + b^{2}c(b - c) +c^{2}a(c - a)\ge 0. $$
See: here.
A lot of people have discussed this problem. So this problem (IMO 1983) has a lot of nice methods.
Now I found for quadrilaterals a similar inequality. Are there some nice methods for inequality $(\star)$?
| Let $\;$ $LHS=f(a,b,c,d)$. Note that $f(a,b,c,d)>f(a-k,b-k,c-k,d-k)$ for $0<k\le\min\{a,b,c,d\}$. So, WLOG we can take $d=0$ to prove the inequality. In that case we should show $ab^2(b-c)+bc^3\ge0$. If $b\ge c$ obviously we are done. For $c\ge b \ge a$ and $c\ge a \ge b$ cases arranging the inequality shows us $ab^3+bc(c^2-ab)\ge0$. Now, $a\ge c \ge b$ case remained. We will use the inequality $a\ge c$. Let's arrange the inequality again. Then, $ab^2(b-c)+bc^3\ge cb^2(b-c)+bc^3=b^3c-b^2c^2+bc^3\ge0$ $\;$ (by AM-GM $b^3c+bc^3\ge 2b^2c^2$ )
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/466271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 4,
"answer_id": 3
} |
$7^{10}$ by $51$ what rest? How to find the remains of Division $7^{10}$ by $51$ using arithmetic debris?
$$$$
$$7\equiv51\;\text{mod}11$$
| $7^3=343=17\cdot20+3$, so $7^3\equiv3\pmod{17}$, and $7^{10}\equiv3^3\cdot7\equiv27\cdot7\equiv10\cdot7\equiv2\pmod{17}$, since $4\cdot17=68$. Of course $7^{10}\equiv1^{10}\equiv1\pmod3$. Thus, we want an integer $n\in\{0,1,\dots,50\}$ such that $n\equiv2\pmod{17}$ and $n\equiv1\pmod3$; $2+17=19$ clearly does the trick.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/466796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Inequality with square roots: $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$
Let $x$ and $y$ be nonnegative real numbers such that $x+y=1$. How do I show that $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$?
How do I deal with square roots inside the inequality?
| Assuming the contrary, for some $0\le x \le 1$, and $y=1-x$,
$$\begin{align}
\sqrt{x^2+1}+\sqrt{y^2+1} <& \sqrt{5}\\
\left( \sqrt{x^2+1} + \sqrt{y^2+1} \right)^2 <& 5 &\text{(Square both sides)}\\
x^2 +1 + 2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} + y^2 + 1 <& 5\\
2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} <& 3 - x^2 -y^2\\
2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} <& 3 - \left(x+y\right)^2 + 2xy\\
2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} <& 3 - 1 + 2xy &\text{(Substitute }x+y=1\text{)}\\
\sqrt{\left( x^2+1 \right) \left( y^2 + 1 \right)} <& 1+xy\\
\left( x^2+1 \right) \left( y^2 + 1 \right) <& \left( 1+xy \right)^2 &\text{(Square both sides)}\\
x^2 y^2 + x^2 + y^2 + 1 <& 1 + 2xy + x^2 y^2\\
x^2 -2xy+y^2 <& 0\\
\left( x - y \right)^2 <& 0\\
\end{align}$$
which contradicts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 3
} |
Why is $\frac{6\sin(2\pi/3)}2 = \frac{3\sqrt3}{2}$ The source of my problem is that I have an integral:
$$\int_{\pi/3}^\pi{(6\cos 2x)dx}$$
For anyone having the same problem as me, to see the boundaries, They are $\pi/3$ and $\pi$
The primitive function of $\,6\cos 2x\,$ is, if I'm not mistaken, $\dfrac{6\sin2x}2$
I haven't figured out how to do large square brackets so I've skipped the enclosed formal primitive function below (Maybe someone could help me with this?)
$$\int_{\pi/3}^\pi{(6\cos 2x)dx} = \frac{6\sin(2\pi)}2 - \frac{6\sin(2\pi/3)}2$$
Well... So far, so good I think. $2\pi$ should be possible to replace with 0, right?, and $2\pi/3$ could be replaced with 120? In that case I would end up with:
$\dfrac{3\sin (120)}2$
Wolfram alpha says:
$$\frac{6\sin(2\pi/3)}2 = \frac{3\sqrt3}{2}$$
So... the question(s) in short:
*
*How is that last transformation based on?
*Is the calculation correct overall? I tried to figure out how to input it all to wolfram alpha for an answer but weren't able to format it correctly. The reason I'm asking is that I have a couple of sample exam questions, but no answers, and I don't want to learn this wrong from the beginning.
| $$\dfrac {6\sin\left(\frac{2\pi}{3}\right)}{2} = 3 \sin \left(\frac {2\pi}{3}\right) = 3\sin \left(\frac{\pi}{3}\right) = 3\left(\frac{\sqrt 3}{2}\right)$$
Re: your question in the comment below your question: $\sin\left(\frac{\pi}{3}\right) = \sin\left(\pi - \frac {\pi}{3}\right) = \sin \left(\frac{2\pi}{3} \right)$, which can be confirmed using the unit circle. If it is more intuitive for you to visualize angles in degrees, then the equivalent identity can be stated as follows:
$$\sin\left(60^\circ\right) = \sin\left(180^\circ - 60^\circ\right) = \sin \left(120^\circ \right) = \frac{\sqrt 3}{2}$$
$(2)$ Your evaluation of the integral is correct. Note that $$\begin{align}\int_\frac\pi3^\pi{(6\cos 2x)dx} & = \frac{6\sin(2\pi)}2 - \frac{6\sin(2\pi/3)}2\\ \\ & = {3\sin(2\pi)} - {3\sin(2\pi/3)} \\ \\ & = 0 - 3\left(\frac{\sqrt 3}{2}\right)\\ \\ & = - 3\left(\frac{\sqrt 3}{2}\right)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/468204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\lim_{x\to+\infty}(\sqrt{x^2-3x}-x)$ Let $f(x) = \sqrt{x^2-3x}$ and $g(x) = x$. Calculate the following limits, showing all working.
I've done the first two -
$$\lim_{x\to0}f(x)\\
=\lim_{x\to0}\sqrt{x^2-3x}=0$$
$$\lim_{x\to-\infty}\frac{f(x)}{g(x)}\\
=\lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{x}=\sqrt{1-\frac{3}{x}}=1$$
How do I calculate this one?
$$\lim_{x\to+\infty}(f(x)-g(x))$$
| Your second one is incorrect, since $\sqrt{x^2}=-x$ for all $x<0$. Hence:
$$
\lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{x}=\lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{-\sqrt{x^2}}=\lim_{x\to-\infty}-\sqrt{1-\frac{3}{x}}=-1
$$
For the last one, we multiply by the conjugate:
\begin{align*}
\lim_{x \to \infty} [f(x)-g(x)]
&= \lim_{x \to \infty} \left[ \sqrt{x^2-3x}-x \right] \\
&= \lim_{x \to \infty} \left[\left(\sqrt{x^2-3x}-x\right) \cdot \dfrac{\sqrt{x^2-3x}+x}{\sqrt{x^2-3x}+x} \right] \\
&= \lim_{x \to \infty} \dfrac{(x^2-3x)-x^2}{\sqrt{x^2-3x}+x} \\
&= \lim_{x \to \infty} \dfrac{-3x}{\sqrt{x^2-3x}+x} \\
&= \lim_{x \to \infty} \dfrac{-3x}{\sqrt{x^2(1-\frac3x)}+x} \\
&= \lim_{x \to \infty} \dfrac{-3x}{x\sqrt{1-\frac3x}+x} \qquad \text{since }\sqrt{x^2}=x \text{ for all }x>0\\
&= \lim_{x \to \infty} \dfrac{-3}{\sqrt{1-\frac3x}+1} \\
&= \dfrac{-3}{\sqrt{1-0}+1} \\
&= \dfrac{-3}{2} \\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to factor $2x^2-x-3$? I know its:
$$(x+1)(2x-3)$$
But how do you come to that conclusion?
| Here's a trick: multiply the first and last coefficients and see which factor pairs add to the middle coefficient.
In this case, $2\times (-3) = -6$. The factor pairs of $-6$ are $1,-6$, $2,-3$, $-1,6$, and $-2,3$. Since $2+(-3) = -1$, we can break up the middle coefficient and factor by grouping:
\begin{align*}
2x^2 - x - 3 &= 2x^2 + (2-3)x - 3 \\ &= 2x^2 + 2x - 3x - 3 \\ &= 2x(x + 1) - 3(x + 1) \\ &= (2x - 3)(x+1).
\end{align*}
The rule comes from the distributive rule: $(ax + b)(cx + d) = (ac)x^2 + (ad + bc)x + bd$. The first coefficient is $ac$ and the last is $bd$, they multiply together to $abcd$, and the middle coefficient is $ad + bc$, the sum of a factor pair of $abcd$.
Another way to do it is to use the quadratic formula to find the roots of the polynomial: $a = 2$, $b=-1$, and $c = -3$, so
$$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm\sqrt{1 - -24}}{4} = \frac{1\pm 5}{4} = \frac{3}{2}, -1,$$
which gives a factorization into the monic polynomials $x+1$ and $2x - 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Congruence Equation $3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}$ How to solve the following congruence equation?
$$3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}$$
If $t_n$ be the $n$th triangular number, then
$$t_1^2+t_2^2+...+t_n^2=\frac{t_n(3n^3+12n^2+13n+2)}{30}$$
so if we solve this congruence equation we find the values of $n$ that $t_n$ divides $t_1^2+t_2^2+...+t_n^2$.
| We solve it by looking at different primes one at a time:
modulo 2:
The equation becomes $0\equiv0\pmod2$, since $x^n\equiv x\pmod2$ for every integer $x\gt0$. Hence this is always true.
modulo 3:
We obtain now: $n+2\equiv0\pmod3$, i.e. $n\equiv1\pmod3$.
modulo 5:
The equation is seen to be $(1+n^2)(3n+2)\equiv0\pmod5,$ hence $\begin{cases}n^2\equiv4\pmod5&\text{or}\\n\equiv1\pmod5\end{cases}.$ Thus the solution is $n\equiv1, 2, -2\pmod5.$
To sum up, we only have to solve for the set of congruence equations: $$\begin{cases}n\equiv1\pmod3\\n\equiv1, 2, -2\pmod5\end{cases}.$$
We find the solution by noticing that $$\begin{cases}6\equiv0\pmod3&10\equiv1\pmod3\\6\equiv1\pmod5&10\equiv0\pmod5\end{cases}.$$
Consequently, the solutions are the linear combinations: $n\equiv10+6a\pmod{15},$ where $a=1, 2, -2.$ Namely, $$n\equiv1, 7, 13\pmod{15}.$$
Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/475526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is there any solution to the following system of equations? Is there any solution to the following system of diophantine equations?
$$
\left\{\begin{array}{l}
2.a^2 = b^2+c^2+d^2 \\
a^2 = e^2+f^2+g^2 , & \mbox{with }((a,b,c,d,e,f,g)>2)\in N\mbox{ and differents among them}
\end{array}
\right.
$$
| In the system of equations:
$\left\{\begin{aligned}&X^2+Y^2+Z^2=2R^2\\&F^2+G^2+T^2=R^2\end{aligned}\right.$
Solutions have the form:
$X=4(c^2+f^2+k^2)k^2-(2c^2-2d^2+f^2)f^2-(d^2-c^2)^2$
$Y=4kd(2k^2+d^2-c^2-f^2)$
$Z=4(f^2+c^2-k^2)k^2+(f^2+2c^2-2d^2)f^2+(d^2-c^2)^2$
$F=4kc(2k^2+d^2-c^2-f^2)$
$G=4kf(2k^2+d^2-c^2-f^2)$
$T=(2k^2+d^2)^2+(c^2+f^2)^2-2(d^2+4k^2)(c^2+f^2)$
$R=4(k^2+d^2)k^2+(2c^2+f^2-2d^2)f^2+(d^2-c^2)^2$
$k,c,d,f$ - integers and sets us.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/476245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Is there another simpler method to solve this elementary school math problem? I am teaching an elementary student. He has a homework as follows.
There are $16$ students who use either bicycles or tricycles. The total
number of wheels is $38$. Find the number of students using bicycles.
I have $3$ solutions as follows.
Using a single variable.
Let $x$ be the number of students in question. The number of students using tricycles is $16-x$. The total number of wheels is the sum of the total number of bicycles times $2$ and the total number of tricycles times $3$.
$$
2\times x + 3 \times (16-x) = 38
$$
The solution is $x=10$.
Using $2$ variables.
Let $x$ and $y$ be the number of students using bicycles and tricycles, respectively.
It implies that
\begin{align}
x+y&=16\\
2x+3y&=38
\end{align}
The solution is $x=10$ and $y=6$.
Using multiples
The multiples of $2$ are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22,24, 26,28,30,32,\dotsc$
The multiples of $3$ are $3,6,9,12,15,18,21,24,27,30,33,36,\dotsc$
The possible wheel combinations with format $(\#\text{bicycle wheels}, \#\text{tricycle wheels})$:
$\hspace{6cm} (32,6)$ but there are $18$ students
$\hspace{6cm} (26,12)$ but there are $17$ students
$\hspace{6cm} (20,18)$ there are $16$ students
$\hspace{6cm} (14, 24)$ there are $15$ students
$\hspace{6cm} (8, 30)$ there are $14$ students
$\hspace{6cm} (2,36)$ there are $13$ students
Thus the correct combination is $10$ bicycles and $6$ tricycles.
My question
Is there any other simpler method?
| Consider the total wheel function $f$ of two variables $x$ the number of bicyclists and $y$ the number of "tricyclists". Since every time we add a bicyclist we increase the number of wheels by two, we have
$$
f_x = 2.
$$
Similarly for case of tricylists,
$$
f_y = 3.
$$
So we have the solutions
$$
f(x,y) = 2x + A(y) \\
f(x,y) = 3y + B(x)
$$
which give
$$
f(x,y) = 2x + 3y + C.
$$
However, $f(0, 0) = 0$ as we have no wheels when we have no riders, so $C = 0$.
Since the total number of cyclists is $16$, we are only interest in the portion of $f$ that lies on the line $y = 16 - x$, that is $f(x, 16 - x)$. Morever, we what to find when the total number of wheels on this line is $38$, so
$$
\begin{aligned}f(x, 16 - x) & = 38 \\
2x + 3(16-x) & = 38 \\
-x + 48 & = 38 \\
-x & = -10 \\
x & = 10.
\end{aligned}
$$
Thus the solution is the point $(x,y) = (10, 6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/478212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "158",
"answer_count": 19,
"answer_id": 6
} |
Modified Pell equation: $x^2-D y^2 = m$, $m\neq1$. How does one solve the Diophantine equation
$$ x^2-Dy^2=m, $$
where $m$ is some fixed arbitrary integer?
I understand that given the fundamental solution to $r^2-D s^2=1$, and any solution to the above I can generate an infinite sequence of solutions. But how do I find all fundamental solutions of the equation with $m\neq1$? Mathworld says it is complicated and several may exist. I have been unable to find a clear explanation of this, unfortunately.
What is the range $0<y\leq Y$ that all fundamental solutions are guaranteed to belong to?
| Here I analyse the equation $x^2 -D y^2 = z^2$.Generally these type of equations can have several solutions if x, y, D and z are particular function of a certain parameter like m.
Suppose $x=2m^2 + p$; plugging in equation we get:
$(2m^2 +p)^2 - z^2 = D y^2 $ ⇒ $4 m^4 + 4p m^2+p^2 -z^2=Dy^2 $
If $4m^2 | p^2 - z^2$ then we have:
$p^2 - z^2 = 4 k m^2 $ or $4m^2(m^2 +p+k)=D y^2$
So we have following system of equations:
$y^2 = 4m^2$ ⇒ $ y = ± 2m$
$ D= m^2 +p+k$
Therefore with certain values for m, p, and k equation $x^2 -D y^2 = z^2$ is valid. In fact equation $p^2 = z^2 + 4 k m^2 $ can have several solutions if $p^2$, $z^2$ and $4km^2$ make a Pythagorean triple$ a^2 +b^2= c^2$ in which c and b are two consecutive numbers, that is $4km= p-1$ and $4km^2$is a perfect square. For example Consider triple $25^2 = 24^2 +7^2$ and compare it with above equation, we get:
$p=25$, $z=7$ and:
$4km^2=24^2= 4. 4. 6^2$ ⇒ $(k=4 , m=6)$ or $( k=36 , m=2)$
For example for $m=2, k=36, p=25 and z=7$ we have:
$x=2m^2 +p =2. 2^2 +25 = 33$
$y = 2m= 2.2=4$
$D=m^2 +p+k=2^2 +25+36=65$ and the equation has the form:
$x^2 -65y^2=z^2$
Since there are infinitely many such triples, then there are infinitely many forms of equation$x^2 -D y^2 = z^2$, that is there are several solution. Using this algorithm you can analyse the equation when RHS of equation is not perfect square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/478570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
| A probability approach.
Let $\{C_n\}_{n=1}^{\infty}$ be independent random bits such that $P(C_n=1)=\frac{1}{2(n+1)}$.
Then $$\prod_{k=1}^{n}P(C_k=0)=\prod_{k=1}^{n}\left(1-\frac{1}{2(k+1)}\right)\to 0\text{ as }n\to\infty$$
So we can define a random variable $X$ to be the least $n$ such that $C_n=1$, and you get that $$\sum_{k=1}^{\infty} P(X=n) = 1,$$ since $P(X>n)\to 0$.
But a quick calculation shows:
$$P(X=n)=P(C_n=1)\prod_{k=1}^{n-1}P(C_k=0)=\frac{1}{2(n+2)}\prod_{k=1}^{n-1} \frac{2k+1}{2k+2} = \frac{1\cdot 3\cdot \cdots \cdot (2n-1)}{4\cdot 6\cdot \cdots \cdot 2(n+2)}$$
Now your sum is $4\sum_{n=1}^{\infty} P(X=n) = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/479610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 6
} |
Limit of $\lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}$ Find the limit of:
$$\lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}$$
| $$\lim_{x \to \infty} \frac{\cos \frac1x - 1}{\cos \frac2x - 1} = \lim_{x \to \infty} \frac{-2\sin^2 \frac{1}{2x}}{-2\sin^2 \frac{1}{x}} = \lim_{x \to \infty} \frac{\sin^2 \frac{1}{2x}}{\sin^2 \frac{1}{x}} = \lim_{x \to \infty} \frac14 \frac{\sin^2 \frac{1}{2x}}{\frac{1}{(2x)^2}} \frac{\frac{1}{x^2}}{\sin^2 \frac{1}{x}} = \frac14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/481421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 3
} |
Solving for $x$ in an inequality $$\frac{x^2-3x-2}{x^2+5x+6}<\frac{2-x}{x^2-4}$$
First, I got the things it couldn't be, which were: -3, -2, and 2
I thought to factor everything I could, and I got
$$\frac{x^2-3x-2}{(x+3)(x+2)}<\frac{2-x}{(x+2)(x-2)}$$
I also cancelled out one value on the right side
$$\frac{x^2-3x-2}{(x+3)(x+2)}<\frac{-1}{x+2}$$
but now, I'm stuck.
| Starting from where you stopped, we have
$$\dfrac{x^2-3x-2}{(x+3)(x+2)}+\dfrac{1}{x+2}<0$$
Adding, we have
$$\dfrac{(x-1)^2}{(x+3)(x+2)}<0$$
As the numerator is non-negative, we can ignore it for now. So we need solutions to
$$(x+3)(x+2)<0$$
You should be able to do this now simply by splitting the number line into three regions, viz. $x < -3, -3 < x < -2, x > -2$ and checking in each.
Dont forget to exclude $x=1$ if it is in the feasible region!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/482340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = x^2$ and $y = 4$ into two regions with equal area. Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = x^2$ and $y = 4$ into two regions with equal area.
I could find the area of this region but i have no clue how to split it parallel to the $x$ axis perfectly... this is too abstract for me...
I found the whole area and got $\dfrac{32}{3}$... I tried plugging that back into the definite integral but that definitely wasn't right...
| If $y=b$,
then $y=x^2$ intersects that
at $x=\sqrt{b}$.
That whole area is a rectangle
with vertices
$
( \sqrt{b}, 0),
( \sqrt{b}, b)
( -\sqrt{b}, b),
( -\sqrt{b}, b)
$.
Its area is
$2b\sqrt{b}
=2b^{3/2}
$.
The area below the parabola is
$2\int_0^{\sqrt{b}} x^2 dx
=2 \frac{x^3}{3}\big|_0^{\sqrt{b}}
=\frac23 b^{3/2}
$.
Therefore,
the area between the parabola
and the line $y=b$
is
$2b^{3/2}-\frac23 b^{3/2}
=\frac43 b^{3/2}
$.
If we put $b=4$,
the area between the parabola
and the line $y=4$
is
$\frac43 4^{3/2}
=\frac43 8
=\frac{32}{3}
$.
We want the area between
the parabola and the line
$y=b$ to be half of this,
which is
$\frac{16}{3}
$.
Therefore,
$\frac43 b^{3/2}
=\frac{16}{3}
$
or
$b^{3/2}
=4
$
or
$b = 4^{2/3}
=\sqrt[3]{16}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/482557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Proving that $\frac{\pi}{4}$$=1-\frac{\eta(1)}{2}+\frac{\eta(2)}{4}-\frac{\eta(3)}{8}+\cdots$ After some calculations with WolframAlfa, it seems that
$$
\frac{\pi}{4}=1+\sum_{k=1}^{\infty}(-1)^{k}\frac{\eta(k)}{2^{k}}
$$
Where
$$
\eta(n)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{n}}
$$
is the Dirichlet Eta function.
Could it be proved that this is true, or false?
Thanks.
ADDED:
If we consider the Dirichlet Beta function
$$
\beta(z)=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{z}}
$$
We can write this as
$$
\beta(1)=1+\sum_{k=1}^{\infty}(-1)^{k}\frac{\eta(k)}{2^{k}}
$$
ADDED:
I recently also noted that
$$
\frac{\pi}{4}=\sum_{k=1}^{\infty}\frac{\eta(k)}{2^{k}}
$$
So summing both of the expressions
Whe have that
$$
\frac{\pi}{2}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{2^{2k}}
$$
and
$$
\frac{1}{2}=\sum_{k=1}^{\infty}\frac{\eta(2k-1)}{2^{2k-1}}
$$
| If we can change the order of summation, we obtain
$$\begin{align}
1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\eta(k) &= 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^k} \\
&= 1 + \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^\infty \frac{(-1)^k}{(2n)^k}\\
&= 1 + \sum_{n=1}^\infty (-1)^{n+1} \left(-\frac{1}{2n}\right)\frac{1}{1 + \frac{1}{2n}}\\
&= 1 + \sum_{n=1}^\infty \frac{(-1)^n}{2n+1},
\end{align}$$
which is the Leibniz series for $\frac{\pi}{4}$.
The convergence of the double sum is not absolute, so the change of summation requires a justification. We obtain that by a slightly more circumspect computation:
$$\begin{align}
1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\eta(k) &= 1 - \frac{\eta(1)}{2} + \sum_{k=2}^\infty \frac{(-1)^k}{2^k}\eta(k)\\
&= 1 - \frac{\eta(1)}{2} + \sum_{k=2}^\infty\frac{(-1)^k}{2^k}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^k}\\
&= 1 - \frac{\eta(1)}{2} + \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=2}^\infty \frac{(-1)^k}{(2n)^k}\\
&= 1 - \frac{\eta(1)}{2} + \sum_{n=1}^\infty (-1)^{n+1}\left(-\frac{1}{2n}\right)^2\frac{1}{1+\frac{1}{2n}}\\
&= 1 - \frac{\eta(1)}{2} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n(2n+1)}\\
&= 1 - \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n(2n+1)}\\
&= 1 + \sum_{n=1}^\infty (-1)^{n+1}\left(\frac{1}{2n(2n+1)} -\frac{1}{2n}\right)\\
&= 1 + \sum_{n=1}^\infty \frac{(-1)^n}{2n+1}.
\end{align}$$
Here the change of order of summation is unproblematic, since
$$\sum_{n=1}^\infty\sum_{k=2}^\infty \frac{1}{(2n)^k} = \sum_{n=1}^\infty \frac{1}{2n(2n-1)} < \infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/483061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 0
} |
Help with minimization problem help me, if $x$ and $y$ are real such that $3x-4y = 12$, determine the minimum value of $z = x ^ 2 + y ^ 2$?$$$$I thought of $$3x-4y = 12\Longrightarrow x=4\frac{y+3}{3}\\z = x ^ 2 + y ^ 2\Longrightarrow z = \left(4\frac{y+3}{3}\right) ^ 2 + y ^ 2$$ and then?
| $$
{\rm F} \equiv x^{2} + y^{2} -\mu\left(3x - 4y - 12\right)\,,
\quad
\left\vert
\begin{array}{rcl}
{\partial F \over \partial x} = 0
&\Longrightarrow&
2x - 3\mu = 0\ \Longrightarrow\ x = {3 \over 2}\,\mu
\\[1mm]
{\partial F \over \partial y} = 0
&\Longrightarrow&
2y + 4\mu = 0\ \Longrightarrow\ y = -2\mu
\end{array}\right.
$$
$$
12 = 3x - 4y = {9 \over 2}\,\mu + 8\mu = {25 \over 2}\,\mu\ \Longrightarrow\
\mu = {24 \over 25}
\quad\Longrightarrow\quad
\left\vert%
\begin{array}{rcl}
x & = &{3 \over 2}\,{24 \over 25} = {36 \over 25}
\\[2mm]
y & = & -2\,{24 \over 25} = -\,{48 \over 25}
\end{array}\right.
$$
$$
\begin{array}{|c|}\hline\\
{\large\quad%
z_{\rm min}
=
\left(36 \over 25\right)^{2} + \left(-\,{48 \over 25}\right)^{2}
=
\color{#0000ff}{144}\quad}
\\ \\ \hline
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove that $1^3 + 2^3 + 3^3 +\cdots+ n^3 = \frac14n^4 + \frac12n^3 + \frac14n^2$ I have to prove that this is true using mathematical induction.
I have this:
for every $n \in \mathbb N$: $1^3 + 2^3 + 3^3 + ... + n^3 = \frac 14n^4 + \frac 12n^3 + \frac 14n^2$
for $n = 1: 1^3 = 1/4 + 1/2 + 1/4$, hence $P(1)$ is true.
Let $N \in \mathbb N$ be given and assume that $P(N)$ is true, that is $$1^3 + 2^3 + 3^3 + ... + N^3 = \frac 14N^4 + \frac 12N^3 + \frac 14N^2$$
For n = $N$ + 1:
And now what? I just couldn't solve it.
| Here's the idea. First note that your right hand side is actually $\frac{1}{4}N^2(N+1)^2$. Let's add $(N+1)^3$ to both sides of the inductive step to see if we can get $\frac{1}{4}(N+1)^2(N+2)^2$.
Doing this, we have
$$1^3 + 2^3 + \cdots + (N+1)^3 = \frac{1}{4}N^2(N+1)^2 + (N+1)^3 = (N+1)^2\left(\frac{1}{4}N^2 + N + 1\right) = \frac{1}{4}(N+1)^2(N^2+4N+4) = \frac{1}{4}(N+1)^2(N+2)^2.$$
From this, you clearly have that the truth of $P(N)$ implies the truth of $P(N+1)$. You should be able to handle write-up from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ If $a, b$ are positive real numbers and $a+b = 1$, prove that :
$$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$
I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.
Thank you.
| First
$$(a+1/a)^2 + (b+1/b)^2 \geq\frac{1}{2} (a+b+1/a+1/b)^2=\frac{1}{2}(1+1/(ab))^2.$$
Then note that
$$ab\le(a+b)^2/4=1/4.$$
Take it into the first one, you may get your inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/487486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 10,
"answer_id": 4
} |
Limit $\lim\limits_{x\to\infty} \frac{5x^2}{\sqrt{7x^2-3}}$ Evaluate the following limit: $$\lim_{x\to\infty} \frac{5x^2}{\sqrt{7x^2-3}}$$
I'm not really sure what to do when there is a square root for an infinity limit.
Please Help!
| You can consider $$\lim_{x\to\infty}\frac{5x^2}{\sqrt{7x^2-3}}=\lim_{x\to\infty}\frac{5x^2}{\sqrt{7x^2-3}}\frac{\sqrt{7x^2-3}}{\sqrt{7x^2-3}}=\lim_{x\to\infty}\frac{5x^2\sqrt{7x^2-3}}{7x^2-3}=
\lim_{x\to\infty}\frac{5\sqrt{7x^2-3}}{7-\frac{3}{x^2}}$$
Now for $x\to\infty$ we have $\sqrt{7x^2-3}\to\infty,\frac{3}{x^2}\to{}0$ so by using the arithmetics of limits rule, we get
$$\lim_{x\to\infty}\frac{5\sqrt{7x^2-3}}{7-\frac{3}{x^2}}=\frac{5\cdot\infty}{7-0}=\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/487752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
A difficult differential equation $ y(2x^4+y)\frac{dy}{dx} = (1-4xy^2)x^2$ How to solve the following differential equation?
$$ y(2x^4+y)\dfrac{dy}{dx} = (1-4xy^2)x^2$$
No clue as to how to even begin. Hints?
| $$y(2x^4+y)\frac{dy}{dx}=(1-4xy^2)x^2$$
$$2x^4y\frac{dy}{dx}+y^2\frac{dy}{dx}=x^2-4x^3 y^2$$
$$2x^4y\frac{dy}{dx}+4x^3y^2+y^2\frac{dy}{dx}=x^2$$
$$\frac{d}{dx}(x^4 y^2)+y^2\frac{dy}{dx}=x^2$$
$$x^4y^2+\frac{y^3}{3}=\frac{x^3}{3}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int \frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}} \mathrm dx$ I am having real trouble evaluating this indefinite integral
$$\int{\frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}}}dx$$
Need I mention that I already tried WolframAlpha with little success? It returned a complicated expression full of many $\arctan$ terms.
Here is what I have already tried:
*
*Seperating denominator as $(1-\sqrt{2}\sin{x}\cos{x})(1-\sqrt{2}\sin{x}\cos{x})$ and applying partial fractions.
*Multiplying and dividing by $\sec^4{x}$
Please provide any help. It would be appreciated if you give a hint rather than the whole solution.
| We first split the integral into two.
$$
\begin{aligned}
I:=& \int \frac{\sin x+\cos x}{\sin ^{4} x+\cos ^{4} x} d x \\
=&-\int \frac{d(\cos x)}{\left(1-\cos ^{2} x\right)^{2}+\cos ^{4} x}+\int \frac{d(\sin x)}{\sin ^{4} x+\left(1-\sin ^{2} x\right)^{2}} \\
=& \frac{1}{2}\left[\underbrace{-\int \frac{d y}{y^{4}-y^{2}+\frac{1}{2}}}_{J}+\underbrace{\int \frac{d z}{z^{4}-z^{2}+\frac{1}{2}}}_{K}\right]
\end{aligned}
$$
where $y=\cos x$ and $z=\sin x$
We need to find one of them, say $J$. We are going to play a little trick on $J$.
\begin{aligned}J &=\int \frac{\frac{1}{y^{2}} d y}{y^{2}+\frac{1}{2 y^{2}}-1} \\
&=\frac{1}{\sqrt 2} \int \frac{1+\frac{1}{y^{2}}-\left(1-\frac{1}{y^{2}}\right)}{y^{2}+\frac{1}{2 y^{2}}-1} d y \\
&=\frac{1}{\sqrt2} \int \frac{d\left(y-\frac{1}{y}\right)}{\left(y-\frac{1}{y}\right)^{2}+1}-\frac{1}{\sqrt2} \int \frac{d\left(y+\frac{1}{y}\right)}{\left(y+\frac{1}{y}\right)^{2}-3}\\
&=\frac{1}{\sqrt2} \tan ^{-1}\left(y-\frac{1}{\sqrt2y}\right)-\frac{1}{2 \sqrt{6}} \ln \left|\frac{y+\frac{1}{\sqrt2y}-\sqrt{3}}{y+\frac{1}{\sqrt2y}+\sqrt{3}}\right|+C_{1} \\
&=\frac{1}{\sqrt2} \tan ^{-1}(\cos x-\frac{\sec x}{\sqrt2})-\frac{1}{2 \sqrt{6}} \ln \left|\frac{\sqrt2\cos ^{2} x-\sqrt{6} \cos x+1}{\sqrt2\cos ^{2} x+\sqrt{6} \cos x+1}\right|+C_{1}\end{aligned}
Replacing y by z yields
$$
K=\frac{1}{2} \tan ^{-1}(\sin x-\frac{\csc x}{\sqrt2})-\frac{1}{2 \sqrt{6}} \ln \left|\frac{\sqrt2\sin ^{2} x-\sqrt{6} \sin x+1}{\sqrt2\sin ^{2} x+\sqrt{6} \sin x+1}\right|+C_{2}
$$
We can now conclude that
$$
\begin{aligned}
I=& \frac{1}{2\sqrt2}\left[\tan ^{-1}(\sin x-\frac{\csc x}{\sqrt2})-\tan ^{-1}(\cos x-\frac{\sec x}{\sqrt2})\right] \\
&+\frac{1}{4 \sqrt{6}} \ln \left| \frac{\left(\sqrt2\cos ^{2} x-\sqrt{6} \cos x+1\right)\left(\sqrt2\sin ^{2} x+\sqrt{6} \sin x+1\right)}{\left(\sqrt2\cos ^{2} x+\sqrt{6} \cos x+1\right)\left(\sqrt2\sin ^{2} x-\sqrt{6} \sin x+1\right)}\right|+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How find this maximum and minimum of$|x+1|+|x-1|+\sqrt{4-x^2}$
show that
$$2+\sqrt{3}\le|x+1|+|x-1|+\sqrt{4-x^2}\le2\sqrt{5}$$
This problem have nice methods? Thank you
my ugly methods,
since $-2\le x\le 2$,and $f(x)=|x-1|+|x+1|+\sqrt{4-x^2}\Longrightarrow f(x)=f(-x)$
so we only find $x\in [0,2]$ $f(x)_{\max},f(x)_{\min}$
so when
(1):
$$0\le x\le 1\Longrightarrow f(x)=2+\sqrt{4-x^2}\le 4$$
(2): when $1\le x\le 2$, then
$$f(x)=2x+\sqrt{4-x^2}\Longrightarrow f'(x)=0\Longrightarrow x=\dfrac{4}{\sqrt{5}}$$
so $\cdots\cdots$
My question This problem have nice methods?Thank you
| The function is defined only for $|x| \le 2$, and is even. Hence let $x = 2 \sin A$, for $A \in [0, \frac{\pi}{2}]$. Then:
$F = |x+1| + |x-1| + \sqrt{4-x^2} = 2 \sin A + 1 + |2\sin A - 1|+ 2\cos A$
For $A \in [0, \frac{\pi}{6}]$, $F = 2 + 2 \cos A $ has the obvious range $[2 + 2 \cos \frac{\pi}{6}, 2 + 2 \cos 0] = [2+\sqrt{3}, 4]$
and for $A \in [\frac{\pi}{6}, \frac{\pi}{2}]$, $F = 4\sin A + 2 \cos A = 2\sqrt{5} \cos (A - \alpha)$ where $\tan \alpha = 2$. Noting $\alpha \in [\frac{\pi}{6}, \frac{\pi}{2}]$ we now have a range $[2\sqrt{5} \cos(\frac{\pi}{6} - \alpha), 2\sqrt{5} \cos 0] = [2 + \sqrt{3}, 2\sqrt{5}]$
Thus $2 + \sqrt{3} \le F \le 2 \sqrt{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Relationships between the elements $(a,b,c,d)$ of a solution to $A^2+B^2+4=C^2+D^2$ I have reduced a certain equation (in positive integers) to the equation
$$A^2 + B^2 + 4 = C^2 + D^2. \quad(\star)$$
Assume the positive integers $(a,b,c,d)$ are any solution to $(\star)$. Are there any algebraic restrictions on [i.e., relationships between] the elements which are well-known or easy to prove? I'm thinking of things like Pythagorean means, or other relative size or congruence restrictions.
Thank you,
Kieren.
| For the equation: $X^2+Y^2+4=Z^2+D^2$
You can draw a simple formula.
$X=(2c+b+2)c+b$
$Y=(2c+b+2)c$
$Z=2c^2+bc-2$
$D=(2c+b+4)c+b$
more:
$X=\frac{(a^2-q^2)b}{2}+(b-2)(q-1)$
$Y=ab$
$Z=bq+2-b$
$D=\frac{(a^2-q^2)b}{2}+(b-2)q+2$
more:
$X=\frac{(a^2-q^2)b}{2}-(b+2)(q+1)$
$Y=ab$
$Z=b+bq+2$
$D=\frac{(a^2-q^2)b}{2}-(b+2)q-2$
$c,b,q,a$ - what some integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Roots of $ x^3-3x+1$ How can I find the roots of $x^3-3x+1$ using Cardano's formula?
So far, I found that
$$x = \sqrt[3]{\dfrac{-1+\sqrt{-3}}{2}} + \sqrt[3]{\dfrac{-1-\sqrt{-3}}{2}}$$
$$x = \sqrt[3]{\dfrac{-1+i\sqrt{3}}{2}} + \sqrt[3]{\dfrac{-1-i\sqrt{3}}{2}}$$
$$x = \sqrt[3]{\frac{-1}{2}+\frac{i\sqrt{-3}}{2}} + \sqrt[3]{\frac{-1}{2}-\frac{i\sqrt{-3}}{2}}$$
I am now trying to express each cubic radicand in their exponential form.
Euler's formula : $e^{i\theta} = \cos \theta + i\sin \theta$.
I have $r = |z| = 1$.
However, when I try to find the angle, I get
$$\theta = \arccos \frac{-1}{2} = 2\pi/3$$
but at the same time
$$\theta = \arcsin \frac{\sqrt{-3}}{2} = \pi/3$$
Shouldn't the two angles be the same for one radicand?
Plus, once I will have expressed the two radicands in their exponential form, what is the next step?
$$x = \sqrt[3]{e^{i2\pi/3}} + \sqrt[3]{e^{i4\pi/3}}$$
$$x = e^{i2\pi/9} + e^{i4\pi/9}$$
Am I on the right track?
| Use the Cardano approach:
Let
$$ x=u+v $$ then we will get
$$u^3+v^3+3(uv-1)(u+v)+1=0$$
by setting $uv-1=0$ we get $$uv=1$$ and $$u^3+v^3=-1$$ by cubing the first equation we obtain a simple quadratic system in terms of $u^3$ and $v^3$ which you can be reduced to a quadratic equation in terms of $u^3$ for example: $$u^3+\frac{1}{u^3}+1=0$$
Then use the Quadratic formula to find $u^3$ and $v^3=-1-u^3$
After extracting cubic roots of the obtained solution you need to take into account that $uv$ is real. This will give the appropriate pairs of $u$ and $v$ whose sums will yield all three roots of the equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/498636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
condition for a cubic polynomial to have a real root Let $a,b \in R$ and assume that $x=1$ is a root of the polynomial
$p(x)= x^4+ax^3+bx^2+ax+1$. Find the ranges of values of $a$ for which $p$ has a complex root which is not real.
Here first I factored out $x-1$ which left me with a cubic polynomial and then i thought of using the discriminant of a cubic polynomial, $\Delta= 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$ $< 0$ to get the condition on a.
But I don't know how this discriminant is derived. And I want to know if there is another method where we don't need to use the discriminant.
| There is a well known method to solve equations of the type
$$x^4+ax^3+bx^2+ax+1=0 \tag{1}$$
If $x \ne 0$ we can transform it to
$$x^2(x^2+\frac{1}{x^2})+ax^2(x+\frac{1}{x})+bx^2=0 \tag{2}$$
Now we observe that
$$(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2$$
Using the substitution $y=x+\frac{1}{x}$ from $(2)$ a quadratic equation of $y$ can be derived that is easier to analyze than $(1)$.
The quadratic equation in $y$ is
$$y^2+ay+b-2=0 \tag{3}$$
$x=1$ is a solution of $(1)$ therefore $b=-2a-2$.
Substituing this in $(3)$ gives
$$y^2+ay-2a-4=0$$
This equation has two roots $y=2$ and $y=-a-2$. So the much simpler question to solve is when does have
$$x+\frac{1}{x}=-a-2 \tag{4}$$
a complex root that is not real.
A simpler method:
If $x=1$ is a solution we get $b=−2a−2$ by substituting $x$ in $(1)$ and therefore
$$ x^4+ax^3-2(a+1)x^2+ax+1=0$$
Therefore we can divide this polynomial by $x-1$ and get
$$x^3+\left(a+1\right)\,x^2+\left(-a-1\right)\,x-1$$
$x=-1$ is a solution of this polynomial to and so finally $(1)$ can be transformed to
$$(x-1)^2(x^2+(a+2)x+1)=0$$
So we have to investigate
$$x^2+(a+2)x+1=0$$
which is equivalent to $(4)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Real solutions of $x$ in $\{x\} = \{x^2\} = \{x^3\}$
Calculate real values of $x$ in $\{x\} = \{x^2\} = \{x^3\}$ , where $\{x\}$ is the fractional part of $x$.
My Attempt:
Let $\{x\} = \{x^2\} = \{x^3\} = k$. Because the fractional part of $X$ is given by $\{X\} = X-\lfloor X \rfloor$, we know the following to be true:
$$
0\leq \{x\} <1\\
0\leq \{x^2\} <1\\
0\leq \{x^3\} <1\\
$$
Using the definition of the fractional part, our equation becomes
$$
x-\lfloor x \rfloor = x^2-\lfloor x^2 \rfloor = x^3-\lfloor x^3 \rfloor = k
$$
I'm not sure how to proceed from here.
| For any $x$ such $\{x\} = \{x^2\}$, we have
$$x^2 - x = \lfloor x^2\rfloor + \{x^2\} - \lfloor x\rfloor - \{x\} = \lfloor x^2\rfloor - \lfloor x\rfloor \in \mathbb{Z}.$$
Likewise $x^3 - x \in \mathbb{Z}$ and $x^3 - x^2 \in \mathbb{Z}$. If $x^2 - x \neq 0$ (that is, $x \neq 0, 1$) then
$$x = \frac{x^3 - x^2}{x^2 - x} \in \mathbb{Q}.$$
Let $x = \frac{p}{q}$ where $(p, q) = 1$. Then $x^2 - x = \frac{p^2 - pq}{q^2} = \frac{p(p-q)}{q^2}$. As $x^2 - x \in \mathbb{Z}$, $q^2 \mid p(p-q)$, but $(p, q) = 1$, so $q^2 \mid p - q$. We can therefore write $p$ in the form $p = kq^2 + q$ where $k \in \mathbb{Z}\setminus\{0\}$. It follows that $(p, q) = q$, so we must have $q = 1$; hence $x \in \mathbb{Z}\setminus\{0, 1\}$.
Checking $x = 0$ and $x = 1$ separately, we find they are also solutions.
Therefore, $\{x\} = \{x^2\} = \{x^3\}$ if and only if $x$ is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Common tangent lines of two quadratic functions Find all such lines that are tangent to the following curves:
$$y=x^2$$ and $$y=-x^2+2x-2$$
I have been pounding my head against the wall on this. I used the derivatives and assumed that their derivatives must be equal at those tangent point but could not figure out the equations. An explanation will be appreciated.
| If $y=x^2$ we have that $\frac {dy}{dx} = 2x$
This gives the gradient of the tangent at the point $(x, y)=(a, a^2)$
If the tangent line is $y=mx + c$ we therefore have $a^2=(2a)\cdot a+c$ whence $c=-a^2$ and the general tangent line to $y=x^2$ is $$y=2ax-a^2$$
If $y=-x^2+2x-2$ we have $\frac {dy}{dx} = -2x+2$, so that the tangent line $y=mx +c$ at $(x, y)=(b, -b^2+2b-2)$ is $-b^2+2b-2=(-2b+2)b+c$ whence $c=b^2-2$ and the general tangent to the parabola is $$y=(-2b+2)x+b^2-2$$
If the two tangents are to be the same line we equate coefficients to give:$$a=1-b$$ and $$a^2=2-b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Proof of $(a+b)^{n+1}$ I have to do proof of $(a+b)^n$ and $(a+b)^{n+1}$ with mathematical induction.
I finished the first one $(a+b)^n = a^n + na^{n-1}b+\dots+b^n$.
I however have trouble with the second one, I don't know what to start exactly. Any help/hints?
| Note that $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$, so
$$\begin{align}
(a+b)^{n+1}&=(a+b)^n(a+b)\\&=\left(\sum_{k=0}^{n}\binom{n}{k}a^kb^{n-k}\right)(a+b)\\
&=\sum_{k=0}^{n}\binom{n}{k}a^{k+1}b^{n-k}+\sum_{k=0}^{n}\binom{n}{k}a^kb^{n-k+1}\\
&=a^{n+1}+\sum_{k=1}^{n}\binom{n}{k-1}a^kb^{n-k+1}+\sum_{k=1}^{n}\binom{n}{k}a^kb^{n-k+1}+b^{n+1}\\
&=a^{n+1}+\sum_{k=1}^{n}\left(\binom{n}{k-1}+\binom{n}{k}\right)a^kb^{n-k+1}+b^{n+1}\\
&=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^kb^{n-k+1}+b^{n+1}\\&=\sum_{k=0}^{n+1}\binom{n+1}{k}a^kb^{n-k+1}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integrating $\int \frac{x^2}{\sqrt{2+3x}}$ using u-sub and integration by parts I'm supposed to integrate this problem by two methods and show that they are the same or that they differ by a constant.
Problem: $$\int \frac{x^2}{\sqrt{2+3x}} dx$$
First method: u-sub with $u = 2+3x$
Second method: integration by parts $u = x^2$ and $dv = \frac{1}{\sqrt{2+3x}}$
Method 1)
$$ u = 2 + 3x $$
$$ \frac{1}{3} du = dx$$
$$\frac{1}{3} \int \frac{x^2}{\sqrt{u}} du $$
Am I able to finish this up with tabular integration? Or do I need to do an integration by parts? The $u^\frac{-1}{2}$ seems to be where it will likely get complicated which is why I ask about tabular integration.
Method 2)
$$ u = x^2$$
$$ \frac{1}{2} du = dx $$
$$ dv = \frac{1}{\sqrt{2+3x}} $$
Now for v, it looks like I'll have to u-sub...
$$ w = 2+3x $$
$$ \frac{1}{3} dw = dx $$
$$ v = \frac{1}{3} \int \frac{dw}{w} $$
$$ v = \frac{1}{3} ln | 2+3x | $$
Now we put it together with integration by parts
$$ (x^2)(\frac{1}{3} ln |2+3x| - \int ln |2+3x| (\frac{1}{2}) $$
Now I'll have to integrate the second part
$$ \frac{1}{2} ln |2 + 3x | $$
u-sub
$$ u = 2+3x $$
$$ \frac{1}{3} du = dx $$
New equation is
$$ \frac{1}{6} \int \ln | u | du $$
Integration by parts
$$ w = u $$
$$ dw = \frac{1}{u} $$
$$ dv = du $$
$$ v = u $$
So that'll give us
$$ \frac{1}{6} [w*w - \int u*\frac{1}{w}] dw$$
which is
$$ \frac{1}{6} [w^2 - \int \frac{w}{w}] dw$$
reduce the w and integrate it
$$ \frac{1}{6} [ w^2 - w ] dw $$
Now we need to plug in our u for w. ($w^2 = \sqrt{u}$)
$$ \frac{1}{6} [ \sqrt{u} - u ] du $$
Now lets plug in 2 + 3x for our u
$$ \frac{1}{6} [ \sqrt{2+3x} - 2+3x $$
Put our integrals together...
$$ (x^2)(\frac{1}{3} ln |2+3x| - \frac{1}{6} [ \sqrt{2+3x} - 2+3x ] + c $$
I'm not sure if I did this correctly so I would appreciate it if anyone can check it and also hopefully advise me on the first method as well.
Edited Answers...
Method 1:
$$\frac{1}{27} [ \frac{2}{5}(2+3x)^{5/3} - \frac{8}{3}(2+3x)^{3/2} - 8(2+3x)^{1/2}] + c$$
Method 2:
$$ x^2 \sqrt{2+3x} - \frac{2}{3} [ 2x(\frac{2}{3}(2+3x)^{3/2}) - \frac{1}{3}(\frac{2}{5}(2+3x)^{5/2} ] + c $$
| In the first case, you need to express $x^2$ in terms of $u$, by noting $$u = 2 + 3x \iff x =\frac 13(u -2).$$ So $$x^2 = \dfrac 19 (u - 2)^2$$
In the second case, $u = x^2 \implies du = 2x dx \iff \dfrac 12 u\,du = x\,dx$
Also, for $dv = \dfrac 1{\sqrt{2 + 3x}}$, and $w = 2 + 3x \implies \dfrac 13 dw = dx$, then $$dv = \frac 1{\sqrt w}\cdot \,dw \implies v = \frac 13 \int w^{-1/2} \,dw = \dfrac{{\frac 13 w^{\frac 12}}}{\frac 12} + C = \dfrac 23w^\frac 12 + C = \frac 23\sqrt{2 + 3x} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/507190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How can I calculate summated math questions without using some kind of computer script or program? For example, how can I find the sum of $\frac 1 {(2 + n) \cdot (3 + n)}$ + ... for $n = 0$ to $n = 97$
i.e.
$$\frac{1}{2 \cdot 3} + \frac 1 {3 \cdot 4} + ... + \frac 1 {99 + 100}$$
| Note that $$\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$$
Hence, your second series can be rewritten as
$$\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + ...$$
terminating, of course, at an appropriate point.
Similarly,
$$\frac{1}{(2 + n)(3 + n)} = \frac{1}{2 + n} - \frac{1}{3 + n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/507565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How find this interesting sum $\sum_{n=1}^\infty\ln(f(n))+\ln(g(n))$ How find this following interesting sum
$$\sum\limits_{n = 1}^\infty {\left\{ {{{\ln }^2}\left( {n - \frac{1}{4}} \right) + {{\ln }^2}\left( {n{\rm{ + }}\frac{1}{4}} \right) - {{\ln }^2}\left( {n - \frac{1}{2}} \right) - {{\ln }^2}\left( {n + \frac{1}{2}} \right)} \right\}} = \ln 2\ln \pi - \frac{5}{2}{\ln ^2}2$$
I think we can use
$$n!\approx\left(\dfrac{n}{e}\right)^n\sqrt{2n\pi}$$
and before I have solve following this
$$\sum_{n=1}^{\infty}\left(n\ln{(\dfrac{2n+1}{2n-1})}-1\right)=\dfrac{1-\ln{2}}{2}$$
| Here is my solution to the problem. The idea behind it is not difficult. It's the same thing that one can do to evaluate series like
$$\sum_{n=1}^{\infty} \frac{n^2}{2^n} = 6$$
using generating functions. Note that
$$\frac{d}{ds} \frac{1}{(n+a)^s} = \ln^2(n+a) \frac{1}{(n+a)^s}$$
We can view the series in question as the second derivative of function evaluated at 0.
So we're going to consider the following function
$$F(s) = \sum_{n=1}^{\infty} \frac{1}{(n+1/4)^s}+\frac{1}{(n-1/4)^s}-\frac{1}{(n+1/2)^s}-\frac{1}{(n-1/2)^s}$$
For $s$ with real part greater than 1 we can split the sum into a sum of Hurwitz zeta functions but we're not going to actually consider them explicitly since we'll calculate what the series are directly. Let's take care of each one.
$$\sum_{n=1}^{\infty} \frac{1}{(n+1/4)^s} = 4^s \sum_{n=1}^{\infty} \frac{1}{(4n+1)^s} = 4^s\left(\sum_{n=0}^{\infty} \frac{1}{(4n+1)^s}-1\right) \tag{1}$$
$$\sum_{n=1}^{\infty} \frac{1}{(n-1/4)^s} = 4^s \sum_{n=0}^{\infty} \frac{1}{(4n+3)^s} \tag{2}$$
$$\sum_{n=1}^{\infty} \frac{1}{(n+1/2)^s} = 2^s \sum_{n=1}^{\infty} \frac{1}{(2n+1)^s} = 2^s \left(\sum_{n=0}^{\infty} \frac{1}{(2n+1)^s} - 1\right) \tag{3}$$
$$\sum_{n=1}^{\infty} \frac{1}{(n-1/2)^s} = 2^s \sum_{n=1}^{\infty} \frac{1}{(2n-1)^s} = 2^s \sum_{n=0}^{\infty} \frac{1}{(2n+1)^s} \tag{4}$$
We combine $(1)$ and $(2)$
$$\sum_{n=1}^{\infty} \frac{1}{(n+1/4)^s} + \sum_{n=1}^{\infty} \frac{1}{(n-1/4)^s} = -4^s + 4^s\left(\sum_{n=0}^{\infty} \frac{1}{(4n+1)^s}+\sum_{n=0}^{\infty} \frac{1}{(4n+3)^s}\right) \\= -4^s + 4^s \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}$$
We combine this with $(3)$ and $(4)$ to get
$$F(s) = 2^s-4^s + (4^s-2\cdot2^s)\sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}$$
The last series is not difficult to find a closed form of. We have
$$\zeta(s) = \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s} + \sum_{n=0}^{\infty}\frac{1}{(2n)^s} = \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s} + 2^{-s}\zeta(s)$$
so
$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^s} = (1-2^{-s})\zeta(s)$$
Finally,
$$F(s) =2^s-4^s + (4^s - 2\cdot 2^s)(1-2^{-s})\zeta(s) = (2^s-1)((2^s-2)\zeta(s)-2^s)$$
Using
$$\zeta(0) = -\frac{1}{2}$$
$$\zeta'(0) = -\frac{1}{2}\ln 2$$
we find that
$$F''(0) = \ln^2 2(-\zeta(0)-1)+2\ln 2(-\ln2+\ln2 \zeta(0)-\zeta'(0)) \\= -\frac{1}{2}\ln^2 2 + 2\ln 2\left(-\frac{3}{2}\ln 2 + \frac{1}{2}\ln(2\pi)\right)\\= \ln2 \ln \pi - \frac{5}{2}\ln^22$$
and we're done.
Note: We are actually doing things that are not explicitly allowed. For example, we're implicitly assuming that we can interchange the infinite summation with term by term differentiation and some other complex analytic phenomena. Someone with better analytic knowledge than me can hopefully point out justification as to why we're allowed to do all that.
Note 2: I'm thinking that we're really using something equivalent to Abel's theorem but with Dirichlet generating functions. Perhaps one could view this as using Perron's formula to convert to an ordinary generating function and then using Abel's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/508525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
How to prove the identity $3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$? I'm trying to prove a trigonometric identity but I can't. I've been trying a lot but I can't prove it. The identity says like this:
$$3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$$
The identity would be easy if $1-\cos^4x=\sin^4x$ and $1-\cos^6x=\sin^6x$ but we know that $\sin^4x+\cos^4x$ isn't equal with $1$ and $\sin^6x+\cos^6x$ isn't equal with $1$.
Can anybody help me?!
Thank you!
| $$3\sin^4 x - 2\sin^6 x = 3 (\sin^2 x)^2 - 2(\sin^2x)^3 = (\sin^2 x)^2(3 - 2\sin^2 x)\cdots$$
Now we can use the Pythagorean Identity: $$\sin^2x + \cos^2x = 1 \iff \sin^2 x = 1 -\cos^2 x$$
$$\begin{align} 3\sin^4 x - 2\sin^6 x & = 3 (\sin^2 x)^2 - 2(\sin^2 x)^3 \\ \\
& = (\sin^2x)^2 (3 - 2\sin^2 x) \\ \\
& = (1 - \cos^2 x)^2\Big(3 - 2(1 - \cos ^2 x)\Big)\\ \\
& = (1- 2\cos^2 x + \cos^4 x)(1 + 2 \cos^2 x) \\ \\
& = 1 - 3 \cos^4 x + 2 \cos^6 x\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/508763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
solving trigonometric equation $3^{\sin^2(x)}+3^{\cos^2(x)}=4$ Please help me to solve this trigonometric equation.
$$3^{\sin^2(x)}+3^{\cos^2(x)}=4.$$
| As $\sin^2x+\cos^2x=1$
Let $3^{\cos^2x}=a$
$\displaystyle 3^{\sin^2x}=3^{1-\cos^2x}=\frac3{3^{\cos^2x}}=\frac3a$
So,we have $\displaystyle \frac3a+a=4\iff a^2-4a+3=0\iff a=1,3$
If $a=1,$ $3^{\cos^2x}=1=3^0\iff \cos^2x=0\iff \cos x=0,x=(2n+1)\frac\pi2$
If $a=3,$ $3^{\cos^2x}=3=3^1\iff \cos^2x=1\iff \sin x=0,x=n\pi=2n\frac\pi2$ where $n$ is any integer
So, the answer reduces to any [integral] multiple of $\frac\pi2,$ right?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/508853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Prime trios of the form $p$,$p+2$,$p+4$ My question is about number theory about the prime numbers.
Here is the question:
Prove that there is only one prime trio of the form $p$,$p+2$,$p+4$.
I mean,we can only find the primes 3,5,7
That satisfies the condition.
| Let $n$ be an integer. Then one of $n$, $n+2$, $n+4$ is divisible by $3$.
For the remainder when $n$ is divided by $3$ is $0$, $1$, or $2$.
If it is $0$, then $n$ is divisible by $3$.
If it is $1$, then $n+2$ is divisible by $3$.
If it is $2$, then $n+4$ is divisible by $3$.
A number $\gt 3$ which is divisible by $3$ cannot be prime.
Now suppose that $n,n+2,n+4$ are all prime.
Then $n$ cannot be $1$, since $1$ is not prime.
Also, $n$ cannot be $2$, since $4$ (and $6$) are not prime.
Certainly $n$ can be $3$, giving the primes $3,5,7$ of the question.
And $n$ cannot be greater than $3$, for then all of $n,n+2,n+4$ are $\gt 3$, and one of them is divisible by $3$, so not prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/509674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How prove this inequality $\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)}\ge 1$ let $a,b,c>0$,and such
$a+b+c=3$,
show that
$$\dfrac{2}{(a+b)(4-ab)}+\dfrac{2}{(b+c)(4-bc)}+\dfrac{2}{(a+c)(4-ac)}\ge 1$$
I think this inequality use this
$$ab\le\dfrac{(a+b)^2}{4}$$
| By C-S $$\sum_{cyc}\frac{1}{(a+b)(4-ab)}=\sum_{cyc}\frac{(c+6)^2}{(a+b)(4-ab)(c+6)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(c+6)\right)^2}{\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2}=\frac{441}{\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2}.$$
Thus, it remains to prove that
$$882\geq\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2.$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, it's obvious that the last inequality is a linear inequality of $w^3$,
which says that it's enough to prove the last inequality for the extremal value of $w^3$,
which happens in te following cases.
*
*$w^3\rightarrow0^+$.
Let $c\rightarrow0^+$ and $b=3-a$.
After this substitution we obtain:
$$882-\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2=18>0;$$
*$b=a$ and $c=3-2a$.
In this case we get $(a-1)^2(2a^3-3a^2+6a+3)\geq0$, which is obvious.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 0
} |
Trying to decompose Partial Fraction. Is this the method? Decompose $\frac{(x^3+x+1)}{(x^2+1)^2}$
Based on my understanding so far:
The partials of the denominator are $(x^2+1)$ and $(x^2+1)^2$
$\frac{(x^3+x+1)}{(x^2+1)^2}$ can be decomposed into partial fractions as below:
$\frac{(x^3+x+1)}{(x^2+1)^2} = \frac{A}{x^2+1} +\frac{B}{(x^2+1)^2}$
multiply both sides by LCD which is $(x^2+1)^2$ we get:
$$x^3+x+1=A(x^2+1)+B$$
removing brackets
$$x^3+x+1=Ax^2+A+B$$
simplify
$$x(x^2+1)+1=A(x^2+1)+B$$
$\therefore$ looking at coefficients, $B=1$ and $A=x$.
Looking at the above, I don't seem to have a solution. Where am I going wrong? Am I on right path?
| By inspection, the partial fractions decomposition is $\dfrac{x}{x^2+1}+\dfrac{1}{(x^2+1)^2}$. This is because $x^3+x=x(x^2+1)$.
Remark: If the top were less simple, we would look for constants $A, B, C, D$ such that our expression is identically equal to $\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{z}{(z-1)(z-3)} = -3\sum\limits_{n=0}^{\infty}\frac{(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$. Show that $\frac{z}{(z-1)(z-3)} = -3\sum\limits_{n=0}^{\infty}\frac{(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$.
I tried breaking it up using partial fractions but it did not seem to get me any closer to what I want.
| $$\frac z{(z-1)(z-3)}=-\frac1{2(z-1)}+\frac3{2(z-3)}=-\frac1{2(z-1)}-\frac34\frac1{1-\frac{z-1}2}\;\;(**)$$
Now, we have that
$$|z-1|<2\implies \frac{|z-1|}2<1\;,\;\;\text{so}$$
$$(**)=-\frac1{2(z-1)}-\frac34\sum_{n=1}^\infty\frac{(z-1)^n}{2^n}=-3\sum_{n=0}^\infty\frac{(z-1)^n}{2^{n+2}}-\frac1{2(z-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/512443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove an equation in complex numbers Prove an equation ($n \in N$, $z \in C$, $a \in R$):
$$ z^n + \frac1{z^n} = 2 \cos\alpha n$$
if $$ z + \frac1{z} = 2 \cos\alpha$$
I tried math induction, but did not solve.
| Using de Moivre's formula $(\cos \alpha + i \sin \alpha)^n = \cos n\alpha + i \sin n\alpha$, we see that setting $z = r(\cos \alpha + i\sin \alpha)$, we have $z^{-1} = r^{-1}(\cos \alpha - i \sin \alpha)$; thus
$z + z^{-1} = r(\cos \alpha + i \sin \alpha) + r^{-1}(\cos \alpha - i \sin \alpha) = 2 \cos \alpha. \tag{1}$
This implies
$ r\cos \alpha + r^{-1}\cos \alpha = 2 \cos \alpha, \tag{2}$
as well as
$i(r\sin \alpha -r^{-1}\sin \alpha) = 0, \tag{3}$
this latter equation (3) since $2 \cos \alpha$ is real.
If $\cos \alpha \ne 0$, (2) yields
$r + r^{-1} = 2, \tag{4}$
or
$(r - 1)^2 = r^2 - 2r + 1 = 0; \tag{5}$
$r = 1$ and $z = \cos \alpha + \sin \alpha$; then $z^n = \cos (n\alpha) + i\sin (n\alpha)$, $z^{-n} = \cos (-n\alpha) + i\sin (-n\alpha) = \cos (n\alpha) - i\sin (n\alpha)$, so
$z^n + z^{-n} = 2 \cos n \alpha \tag{6}$
in this case.
In the event that $\cos \alpha = 0$, we have $\sin \alpha \ne 0$, and from (3) we infer
$r \sin \alpha = r^{-1} \sin \alpha, \tag{7}$
whence
$r^2 = 1, \tag{8}$
implying $r = 1$ since $r > 0$. Thus we still have $z = \cos \alpha + i\sin \alpha$ and (6) still follows as in the case $\cos \alpha \ne 0$. I credit Daniel Littlewood's comment for making clear to me the utility of equation (3) in this context.
Well, I hope that helps. Cheers,
and as always,
Fiat Lux!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/512589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ So, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ means $1=ax+by$, and want to show $a^2x+b^2y=1$.
By squaring $1=ax+by$ both sides, I get, $1=(ax)^2+2(ax)(by)+(by)^2$, but this doesn't help my proof.
| suppose $gcd(a^2, b^2) = d$ and $d \ne 1$
Then $a^2 = dm, b^2 = dn \implies a^2b^2 = d^2mn$, where $gcd(m,n) = 1$
This implies both $m$ and $n$ are perfect squares. $m = p^2, n = q^2$
We have $a^2 = dp^2, b^2 = dq^2 \implies d$ is a perfect square. Let $d = x^2$
Now we have $a^2 = x^2p^2, b^2 = x^2q^2 \implies a = xp, b = xq \implies gcd(a,b) = x = 1$
Hence $d = x^2 = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/512924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 4
} |
A basic doubt on a infinite series problem I see in Rudin the following statement is claimed for the following convergent series:
$$1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$$
If $s$ is the sum of this series then $$s \lt 1 -\frac{1}{2} + \frac{1}{3}$$. How is that possible to tell without knowing $s$ ?
| Notice that
$$ 1-\frac{1}{2}+\frac{1}{3}-\left( \frac{1}{4}-\dots \right)=1-\frac{1}{2}+\frac{1}{3}-a< 1-\frac{1}{2}+\frac{1}{3}$$
Since $a>0$.
Note:
$$a=\frac{1}{4}-\frac{1}{5}+\dots.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/513336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Verify this identity: $\sin x/(1 - \cos x) = \csc x + \cot x$ Verify this identity $\frac{\sin {x}} {1 - \cos {x}}\ = \csc x + \cot x$ I got the left side to $\frac{1-\cos x} {\csc x}$, but I can't get any farther. Am I on the right path? Can I get some help?
| The right-hand side can be rewritten as
$$\csc{x} + \cot{x} = \frac{1}{\sin{x}} + \frac{\cos{x}}{\sin{x}} = \frac{1 + \cos{x}}{\sin{x}}$$
Multiplying the left side top and bottom by $1 + \cos{x}$, we find that
$$\frac{\sin{x}}{1 - \cos{x}} = \frac{\sin{x}(1 + \cos{x})}{1 - \cos^2{x}} = \frac{\sin{x}(1 + \cos{x})}{\sin^2{x}} = \frac{1 + \cos{x}}{\sin{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/514229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Limits of integral: $\iiint_{D} \frac {\mathrm{d}x\mathrm{d}y\mathrm{d}z}{(x + y + z + 1)^3}$ , where $D =\{ x > 0 , y > 0 , z > 0 , x + y + z < 2\}$ What are the limits of the integral:
$$\iiint_{D} \frac {\mathrm{d}x\mathrm{d}y\mathrm{d}z}{(x + y + z + 1)^3}$$ where
$$ D =\{ x > 0 , y > 0 , z > 0 , x + y + z < 2\}$$
I have previously done double integrals and was wondering if someone could help me derive the limits of this triple one.
Help is appreciated!
| Hint: $(x,y,z)\in D$ if and only if $0<x<2$, $0<y<2-x$, and $0<z<2-x-y$. Then, $$\underset{D}{\iiint}\frac{\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z}{(x+y+z+1)^3}=\int_{x=0}^{x=2}\left[\int_{y=0}^{y=2-x}\left(\int_{z=0}^{z=2-x-y}\frac{1}{(x+y+z+1)^3}\mathrm{d}z\right)\mathrm{d}y\right]\mathrm{d}x.$$ Be careful: since the limits depend on other variables in some of the integrals on the right-hand side, the order of integration must not be changed! That this trick works is a consequence of Fubini's famous theorem.
Detailed calculations: $$\int_{z=0}^{z=2-x-y}\frac{1}{(x+y+z+1)^3}\mathrm{d}z=\left[-\frac{1}{2(x+y+z+1)^2}\right]_{z=0}^{z=2-x-y}=-\frac{1}{18}+\frac{1}{2(x+y+1)^2}.$$
\begin{align*}\int_{y=0}^{y=2-x}\left(-\frac{1}{18}+\frac{1}{2(x+y+1)^2}\right)\mathrm{d}y=&\left[-\frac{y}{18}-\frac{1}{2(1+x+y)}\right]_{y=0}^{y=2-x}\\=&\,\frac{x-2}{18}-\frac{1}{6}+\frac{1}{2(1+x)}\end{align*}
\begin{align*}\int_{x=0}^{x=2}\left(\frac{x-2}{18}-\frac{1}{6}+\frac{1}{2(1+x)}\right)\mathrm{d}x=&\left[\frac{\dfrac{x^2}{2}-2x}{18}-\frac{x}{6}+\frac{\log (1+x)}{2}\right]_{x=0}^{x=2}\\=&\,-\frac{2}{18}-\frac{2}{6}+\frac{\log 3}{2}-0=-\frac{4}{9}+\frac{\log3}{2}.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find lim:$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$ Find lim: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$$.
You can use L'Hospitale, or Maclaurin, etc
| One way without MacLaurin or L'Hospital: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}=\lim_{x\to0} \frac{\tan(\tan x) - \tan(\sin x)+\tan(\sin x)-\sin(\sin x)}{\tan x -\sin x} = \lim_{x\to0} \frac{[\tan(\tan x -\sin x)][1-\tan(\tan x)\tan(\sin x)]}{\tan x -\sin x}+\lim_{x\to0} \frac{\sin(\sin x)[1 - \cos(\sin x)]}{\cos(\sin x)(\tan x -\sin x)}=$$ $$=1+\lim_{x\to0} \frac{\sin(\sin x)}{\sin x}\lim_{x\to0} \frac{1 - \cos(\sin x)}{\sin^2 x}\lim_{x\to0}\frac{x^2}{1-\cos x}\lim_{x\to0}\frac{\sin^2x}{x^2} = 1+1.\frac{1}{2}.2.1 =2 $$.
We used $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b}$$ $$\lim_{x\to0}\frac{\sin x}{x}= \lim_{x\to0}\frac{\tan x}{x}=\lim_{x\to0}\cos x=1, \lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/516483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Complex Numbers.... Suppose a is a complex number such that:
$$a^2+a+\frac{1}{a}+\frac{1}{a^2}+1=0$$
If m is a positive integer, find the value of:
$$(a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}$$
My Approach:
After I could not solve it using the usual methods I tried a bit crazier approach. I thought that as the question suggests that the value of the expression does not depend upon the value of m, provided it is positive, hence the graph of the expression on Y-axis and m on X-axis would be parallel to X-axis and thus the slope be zero. So I differentiated it with respect to m and equated it to zero and after factorizing and solving I got $a^m=-1$ or $a^m=1$ or $a^m=\left(\frac{-1}{4}+i\frac{\sqrt15}{4}\right)$ or $a^m=\left(\frac{-1}{4}-i\frac{\sqrt15}{4}\right)$. But if $a^m=1$ then $a=1$ which does not sattisfy the first equaion. Thus the value of $$(a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}=\frac{-9}{4}$$
OR
$$(a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}=0$$
What other approach would you suggest? What are the flaws in my approach (if any)?
| If we multiply
$$a^2 + a + \frac1a + \frac{1}{a^2} + 1 = 0$$
with $a^2$, we obtain (since evidently $a \neq 1$)
$$0 = a^4 + a^3 + a^2 + a + 1 = \frac{a^5-1}{a-1},$$
so $a^5 = 1$,
$$a = e^{(2\pi ik)/5},\quad k \in \{1,2,3,4\}.$$
Then, if $m$ is a multiple of $5$, we have $$(a^2)^m + a^m + \frac1{a^m} + \frac{1}{(a^2)^m} = 1+1+1+1 = 4,$$
and if $m$ is not a multiple of $5$, the four numbers
$$a^{2m},\, a^m,\, a^{-m}\, a^{-2m}$$
are the numbers $a^2,\, a,\, a^{-1},\,a^{-2}$, possibly in a different order, then the sum is $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/518060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
if $f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$ If $\displaystyle f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$
$\underline{\bf{My\;\; Try}}::$ Given $\displaystyle f(x) = x-\frac{1}{x} = \frac{x^2-1}{x}.$ Now Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get.
$\displaystyle f(f(x)) = x-\frac{1}{x}-\frac{x}{x^2-1} = \frac{x^2-1}{x}-\frac{x}{x^2-1} = \frac{(x^2-1)^2-x^2}{x.(x^2-1)} = \frac{x^4-3x^2+1}{x.(x^2-1)}$
again Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get
$\displaystyle f(f(f(x))) = \frac{\left(\frac{x^2-1}{x}\right)^4-3\left(\frac{x^2-1}{x}\right)^2+1}{\left(\frac{x^2-1}{x}\right).\left\{\left(\frac{x^2-1}{x}\right)^2-1\right\}}$
Now I did not understand how can I solve This $8^{th}$ Degree equation,
So Help Required,
Thanks
| The following method may be unnecessarily time intensive, but will definitely work. Call $f(x) = y$ and $f(y) = f(f(x)) = z$. The equality you need to solve is $f(z) = 1$. This gives a quadratic equation for $z$. Solve it for $z$; this gives you two answers. For these two answers $z_1, z_2$, solve $f(y) = z_1, z_2$, giving you at most 4 answers. Repeat, to solve $f(x) = y_1, ... ,y_4$, giving you at most 8 answers. Eliminate duplicate answers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/518683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Taylor Series for $f(z)=\frac{z}{2}+\frac{z}{e^z-1}$. The question reads as follows:
Let the function $f$ be given by $f(z)=\frac{z}{2}+\frac{z}{e^z-1}$ if
$z\neq0$ and $f(z)=1$ if $z=0$. Show that $f$ is analytic at $z=0$ and
that $f(z)=f(-z)$. Deduce that there is a Taylor series
$f(z)=a_0+a_2x^2+a_4x^4+...$ valid for $|z|<2\pi$. Find $a_0,a_2,a_4$.
Clearly the function is analytic at $z=0$, and $f(z)$ is even for all $z$. Now, the singularities occur where $z=2ni\pi$ where $n\in\mathbb{Z}$. Given $|z|<2\pi$, we have that $f(z)$ is analytic within this disk. Hence a Taylor series exists for $f(z)$ such that $z_0=0$. Thus $f(z)=a_0+a_1z+a_2x^2+a_3z^3+a_4x^4+....$ in $|z|<2\pi$ where $a_n=\frac{f^{n}(0)}{n!}$ where $n\in\mathbb{Z^+}$.
Since $f(z)$ is even this implies that only the even powers exist. Thus $f(z)=a_0+a_2x^2+a_4x^4+...$. I am having trouble finding the coefficients, $a_0,a_2,a_4$.
Thank you in advanced for you help.
| To find the first terms of the Taylor series of $f$, there are various possibilities. One can simply differentiate the function and use $a_n = \frac{1}{n!}f^{(n)}(0)$ to obtain the coefficients, but that soon becomes unwieldy. Still, for the coefficients up to order $4$, it's doable.
Another standard way, when one has such a quotient, is to expand the numerator (trivial here) and the denominator, and divide both by a suitable power of $z$ times a constant, so that the denominator has the form $1 - h(z)$, and expand $\frac{1}{1-h(z)}$ in a geometric series. That leads to (ignoring the $\frac{z}{2}$ summand)
$$\begin{align}
\frac{z}{e^z-1} &= \frac{z}{z + \frac{z^2}{2} + \frac{z^3}{6} + \dotsb}\\
&= \frac{1}{1 + \left(\frac{z}{2} + \frac{z^2}{6} + \dotsb\right)}\\
&= 1 - \left(\frac{z}{2} + \frac{z^2}{6} + \dotsb\right) + \left(\frac{z}{2} + \frac{z^2}{6} + \dotsb\right)^2 - \dotsb\\
&= 1 - \frac{z}{2} +\left(\frac14 - \frac16\right)z^2 + \dotsb\\
&= 1 - \frac{z}{2} + \frac{z^2}{12} + \dotsb
\end{align}$$
The coefficients $a_0$ and $a_2$ are found in this way without much computation, but finding $a_4$ involves some more computation. It's still not too hard.
Yet another way is to multiply the unknown expansion with the known expansion of the denominator to obtain a recurrence for the coefficients:
$$\begin{align}
z &= \frac{z}{e^z-1}(e^z-1)\\
&= \left(\sum_{k=0}^\infty \frac{b_k}{k!}z^k\right)\cdot\left(\sum_{m=1}^\infty \frac{z^m}{m!}\right)\\
&= \sum_{n=1}^\infty \left(\sum_{k=0}^{n-1}\frac{b_k}{k!(n-k)!}\right)z^n\\
&= \sum_{n=1}^\infty \left(\sum_{k=0}^{n-1} \binom{n}{k}b_k\right)\frac{z^n}{n!}.
\end{align}$$
Equating the coefficients yields $1 = \binom{1}{0}b_0$, so $b_0 = 1$, and for $n > 1$, the recurrence
$$\sum_{k=0}^{n-1}\binom{n}{k}b_k = 0.\tag{1}$$
For $n = 2$, we obtain $0 = \binom{2}{0}b_0 + \binom{2}{1}b_1 = 1 + 2b_1$, so $b_1 = -\frac12$. For $n = 3$, it becomes
$$0 = \binom{3}{0}b_0 + \binom{3}{1}b_1 + \binom{3}{2}b_2 = 1 - \frac{3}{2} + 3b_2 \Rightarrow b_2 = \frac16.$$
Note that we have $a_k = \frac{b_k}{k!}$, so this agrees with the result above.
Using the fact that $b_{2k+1} = 0$ for $k > 0$ by the evenness of $f$, we need not compute $b_3$ and further odd coefficients, which saves some work. Computing $a_4$ with the recurrence $(1)$ is then not much work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/519963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$x,y,z$ positive real numbers , $x+y+z=3$ $\implies x^4y^4z^4(x^3+y^3+z^3)≤3$ If $x,y,z$ are positive real numbers with $x+y+z=3$ then how to prove (without using calculus) that $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$ ?
| Let's use the arithmetic-geometric inequality.
That gives us $(xyz)^{\frac{1}{4}}≤\frac{1}{4}(x+y+z)\leq\frac{3}{4}$
thus $(xyz)^4\leq(\frac{3}{4})^{16}$
Now all you have to prove is that $x^3+y^3+z^3\leq3*(\frac{4}{3})^{16}$
But $x^3+y^3+z^3\leq 3^3+3^3+3^3=81$ because $x\leq3$ and $y\leq3$ and $z\leq3$.
Thus you have to prove that $3^{15+4}\leq4^{16}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/520129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
Apollonius circle, its radius and center I've got the following set:
$\{|z-a|=k|z-b|\}$, where z is a complex number, a an b are fixed, and $k>0$,$k \ne 1$.
I need to prove that this is a circle (called Apollonius circle).
I also have to prove that this circle's radius is equal to $k|a-b||1-k^2|^{-1}$ and it's centre is $(a-k^2b)(1-k^2)^{-1}$.
I don't know what to do. I've tried to work with the analytic circle equation ($|z-c|^2=r^2$), substituting given radius and center, but it didn't work. I also tried to square both sides of the first given equation ($|z-a|=k|z-b|$), which usually works with complex number, but also didn't get any result... Can somebody show me how to solve this problem? I will be very grateful.
| This is an old post, but I wanted to see if the calculation could be illuminated by unpacking to real coordinates later, and by solving a simpler equation. After doing this I show how to do the calculation without unpacking to real coordinates.
We want to show that the set of $z$ satisfying $|z - a| = k |z - b|$ forms a circle.
We can simplify our calculation by noting that translation preserves circles. We can translate to $w = z - b$ and solve the equation $|w - A| = k|w|$, where $A = a - b$, and then translate our solution back to $z$.
Note that $$|w-A|^2 = (w-A) (\overline{w-A}) = |w|^2 - 2 \mathrm{Re}(\bar{A}w) + |A|^2.$$
This is a version of the polarization identity, noting that $\mathrm{Re}(\bar{A} z)$ is the Euclidean dot product between $A$ and $z$.
Then we see
\begin{align*}
|w - A| = k |w| &\Longrightarrow |w - A|^2 = k^2 |w|^2 \\
&\Longrightarrow |w|^2 - 2 \mathrm{Re}(\bar{A}w) + |A|^2 = k^2 |w|^2 \\
&\Longrightarrow (1 - k^2)|w|^2 - 2\mathrm{Re}(\bar{A} w) + |A|^2 = 0 \ \ \ (\star)
\end{align*}
Let $w = x + i y$ and $A = A_x + i A_y$. If $k \neq 1$, then
\begin{align*}
(\star) &\Longrightarrow |w|^2 - \frac{2}{1 - k^2} \mathrm{Re}(\bar{A}w) + \frac{|A|^2}{1 - k^2} = 0 \\
&\Longrightarrow x^2 + y^2 - \frac{2 A_x}{1 - k^2}x - \frac{2 A_y}{1 - k^2}y + \frac{|A|^2}{1 - k^2} = 0
\end{align*}
Completing the square shows that sets determined by an equation in the following form are always (possibly degenerate) circles.
\begin{align*}
x^2 + y^2 + Ex + Fy + G = 0 \Rightarrow (x + \frac{E}{2})^2 + (y + \frac{F}{2})^2 = \frac{E^2 + F^2}{4} - G.
\end{align*}
Replacing coefficients in this formula with those of our previous expression, we find
\begin{align*}
\left(x - \frac{A_x}{1 - k^2} \right)^2 + \left(y - \frac{A_y}{1 - k^2}\right)^2 = \frac{A_x^2 + A_y^2}{(1 - k^2)^2} - \frac{|A|^2}{1 - k^2}
= \left( \frac{k |A|}{|1 - k^2|} \right)^2
\end{align*}
Since $A = a - b$, we confirm that the radius is $r = k\frac{|a - b|}{|1 - k^2|}$.
In complex form, we find $$|w - \frac{a - b}{1 - k^2}| = \frac{k|a - b|}{|1 - k^2|},$$
Translating back to $z$, we find
$$|z - \frac{a - b k^2}{1 - k^2}| = \frac{k|a - b|}{|1 - k^2|},$$
which shows that the center is $c = \frac{a - b k^2}{1 - k^2}$.
There's a slicker calculation which doesn't unpack to real coordinates. It uses the polarization identity again, which serves as a complex substitute for completing the square: $$|z|^2 - 2 \mathrm{Re}(\bar{d}z) = |z - d|^2 - |d|^2$$
From $(\star)$,
\begin{align*}
0 &= |w|^2 - 2\mathrm{Re}(\frac{\bar{A}}{1 - k^2}w) + \frac{|A|^2}{1 - k^2} \\
&= | w - \frac{A}{1 - k^2} |^2 - |\frac{A}{1 - k^2}|^2 + \frac{|A|^2}{1 - k^2},
\end{align*}
which implies
\begin{align*}
| w - \frac{A}{1 - k^2} |^2 = \left( \frac{k |A|}{1 - k^2} \right)^2.
\end{align*}
Then we can finish as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/520469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Parametrize the intersection of 2 planes. Parametrize the intersection of $\frac {x^2} {3}+y^2+\frac {z^2} {10} = 1$ with $z=2$ (level curve) plane.
Here is what I did.
Plugged in $z=2$ into the plane $\frac {x^2} {3}+y^2+\frac 25=1$. I got $y^2=9-5x^2$ Then I substituted $y^2$ into the plane $\frac {x^2} {3}+9-5x^2+\frac25=1$ to solve for $x^2$. I got $x^2=1.8$ and then got $y=0$. How should I parametrize the intersection from here onward?
$\mathbf Error$ in my calculation, $y^2=\frac 35-\frac{x^2}{3}$
| I'm afraid you have make something wrong.Plugged in $z=2$ into the plane $\frac {x^2} {3}+y^2+\frac 25=1$. You should get $y^2=\frac{3}{5}-\frac{x^2}{3}$, rather than $y^2=9-5x^2$ this is the equation of the curve while $z=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/520920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How find the $(x^2_{1}+x_{1}x_{2}+x^2_{2})(x^2_{2}+x_{2}x_{3}+x^3_{3})(x^2_{3}+x_{3}x_{1}+x^2_{1})$ Let $x_1$, $x_2$, and $x_3$ be the roots of the equation
$$4x^3-6x^2+7x-8=0.$$
Find this value:
$$(x_1^2+x_1x_2+x_2^2)(x_2^2+x_2x_3+x_3^3)(x_3^2+x_3x_1+x_1^2)$$
My try:
$$x_1+x_2+x_3=\frac 3 2$$
$$x_1 x_2+x_2 x_3+x_1 x_3=\frac 7 4$$
$$x_1 x_2 x_3=2$$
Now I have
$$x_1^2+x_1x_2+x_2^2=(x_1+x_2)^2-x_1x_2=(\frac 3 2-x_3)^2-\frac 2 {x_3}.$$
But this is very ugly, and I think this problem should have a cleaner solution. Thanks.
| Assuming that we need to calculate $$(x^2_1+x_1x_2+x^2_2)(x^2_2+x_2x_3+x^2_3)(x^2_3+x_3x_1+x^2_1)$$
Let $\displaystyle x^2_1+x_1x_2+x^2_2=y_3,x^2_2+x_2x_3+x^2_3=y_1, x^2_3+x_3x_1+x^2_1=y_2$
So, $\displaystyle y_3=x^2_1+x_1x_2+x^2_2=(x_1+x_2)^2-x_1x_2$
$=\left(\frac32-x_3\right)^2-\frac2{x_3}$ as $x_1+x_2+x_3=\frac32$ and $x_1 x_2 x_3=2$
On re-arrangement we have, $\displaystyle 4x_3^3-12x_3^2+x_3(9-4y_3)-8=0\ \ \ \ (1)$
Again as $x_3$ is a root of the given equation, $4x_3^3-6x_3^2+7x_3-8=0\ \ \ \ (2)$
$(2)-(1)\implies 6x_3^2+x_3\{7-(9-4y_3)\}=0\iff 2x_3(3x_3+2y_3-1)=0$
As $0$ does not satisfy the given equation $x_3\ne0$
$\displaystyle \implies 3x_3+2y_3-1=0\iff x_3=\frac{1-2y_3}3$
Putting the value of $x_3$ in $(2)$ we get,
$$4\left(\frac{1-2y_3}3\right)^3-6\left(\frac{1-2y_3}3\right)^2+7\left(\frac{1-2y_3}3\right)-8=0$$
On re-arrangement we have,
$\displaystyle 4(2y_3)^3+(\cdots)y_3^2+(\cdots)y_3+3^3\cdot8-3^3\cdot7+3\cdot6-4=0$
$\displaystyle \implies 32y_3^3+(\cdots)y_3^2+(\cdots)y_3+167=0\ \ \ \ (3)$
As the values of $y_1,y_2,y_3$ are symmetric, we shall reach at the same equation $(3)$ if we start with $y_1$ or $y_2$
$\displaystyle \implies y_1,y_2,y_3$ are the roots equation $(3)$
Using Vieta's Formula, $\displaystyle y_1y_2y_3=-\frac{167}{32}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/521225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Generating functions word problem(balloons) Find the generating function for the number of ways to create a bunch of n balloons selected from white, gold, and blue balloons so that the bunch contains
at least one white balloon, at least one gold balloon, and at most two blue balloons. How many ways are there to create a bunch of 10 balloons subject to these
requirements?
So I have $$\frac{x^2}{(1-x)^2} \cdot \frac{(1-x^3)}{1-x}$$ and I have no idea where to go from here. Have a test on this material in the morning so any help would be greatly appreciated.
| You have $$(x^2-x^5)\frac{1}{(1-x)^3}$$
We have the generating function $$\frac{1}{(1-x)^3}=\sum_{n\ge 0} {n+2\choose 2}x^n$$
Multiply by $x^k$ gives $$x^k\frac{1}{(1-x)^3}=\sum_{n\ge 0}{n+2\choose 2}x^{n+k}$$
We reindex using $m=n+k$ to get $$x^k\frac{1}{(1-x)^3}=\sum_{m\ge k}{m-k+2\choose 2}x^m$$
We do this twice, for $k=2$ and $k=5$, then subtract. The result is
$$(x^2-x^5)\frac{1}{(1-x)^3}=\sum_{m\ge 5}{m-2+2\choose 2}-{m-5+2\choose 2}x^m + \sum_{m=2}^4{m-2+2\choose 2}x^m$$
Hence your answer (for $m\ge 5$) is ${m\choose 2}-{m-3\choose 2}$. You want $m=10$, so ${10\choose 2}-{7\choose 2}=24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/522116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
integrate $ \frac {7x^2 + 2x − 7}{ x^3 − x}$ dx I keep messing up with the integration part of this I think. Before that I have to factor out the $x$ on the bottom, and then set up the $A$ and $B$ right?
Evaluate the integral. (Remember to use $\ln|u|$ where appropriate. Use $C$ for the constant of integration.)
$$\int\frac{7x^2+2x-7}{x^3-x}\,dx$$
| HINT:
As $x^3-x=x(x^2-1)=x(x+1)(x-1),$
Using Partial fraction decomposition
$$\frac{7x^2+2x-7}{x(x-1)(x+1)}=\frac A{x+1}+\frac Bx+\frac C{x-1}$$
Multiply either sides by $x^3-x=x(x^2-1)=x(x+1)(x-1)$
and compare the coefficients of the different power of $x$ to find $A,B,C$
Finally use $\displaystyle \int\frac{dx}{x+m}=\ln|x+m|+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/523253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Constant.
I was able to express it in terms of Euler Sums but it does not seem to be of any use.
\begin{align}
&\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x
\\[3mm] = &\
\frac{1}{16}\sum_{n = 1}^{\infty}
\frac{\psi_{1}\left(1/4 + n\right) -
\psi_{1}\left(3/4 + n\right)}{n} \tag{2}
\end{align}
Here $\psi_{n}\left(z\right)$ denotes the polygamma function.
Can you help me solve this problem $?$.
| The following is a proof of the formula $$S= \sum_{k=1}^{\infty} \frac{H_{k}}{ (k+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$
This formula is mentioned in a comment under Bennett Gardiner's answer.
(For $a=0$, the right side of the equation should be interpreted as a limit).
$$ \begin{align} S &= \sum_{k=1}^{\infty} \frac{H_{k}}{(k+a)^{2}} \\ &= \sum_{k=1}^{\infty} \frac{1}{(k+a)^{2}} \sum_{n=1}^{k} \frac{1}{n} \\& = \sum_{n=1}^{\infty} \frac{1}{n} \sum_{k=n}^{\infty} \frac{1}{(k+a)^2} \\ &= \sum_{n=1}^{\infty} \frac{\psi_{1}(a+n)}{n} \\ &= - \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} \frac{x^{a+n-1} \ln x}{1-x} \, dx \tag{1} \\ &= - \int_{0}^{1} \frac{x^{a-1} \ln x}{1-x} \sum_{n=1}^{\infty} \frac{x^{n}}{n} \, dx \\ &= \int_{0}^{1} \frac{x^{a-1} \ln x \ln(1-x)}{1-x} \, dx \\ &= \lim_{b \to 0^{+}} \frac{\partial }{\partial a \, \partial b} B(a,b) \\ &= \small \lim_{b \to 0^{+}} \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} \left( \psi(a) \psi(b) - \psi(a)\psi(a+b) - \psi(b) \psi(a+b) + \psi^{2}(a+b) - \psi_{1}(a+b) \right) \tag{2} \\ &= \lim_{b \to 0^{+}} \frac{\Gamma(a)}{\Gamma(a+b)} \left( \frac{1}{b} - \gamma + \mathcal{O}(b) \right)\left( \left( \gamma \psi_{1}(a) + \psi(a) \psi_{1} (a) - \frac{\psi_{2}(a)}{2} \right)b + \mathcal{O}(b^{2}) \right) \\ &= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \end{align}$$
$(1)$ https://en.wikipedia.org/wiki/Trigamma_function#Calculation
$(2)$ http://mathworld.wolfram.com/BetaFunction.html (26)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 5
} |
Coordinate vectors for different size matrices Hi am having trouble with this question
Let $A = \left[\begin{array}{cc}-1&3&-4&-3\\2&-1&0&-3\end{array}\right]$ Find the coordinate vector for the matrix $A \cdot A^{T}$
with respect to the standard basis for $R^{2 \times 2}$
I figured out the $A \cdot A^{T} = \left[\begin{array}{cc}5&-5&-4&-3\\-5&10&12&-6\\-4&12&16&-12\\-3&-6&12&18\end{array}\right]$ and the standard basis for $R^{2 \times 2}$ is $\left[\begin{array}{cc}1&0\\0&0\end{array}\right],
\left[\begin{array}{cc}0&1\\0&0\end{array}\right],
\left[\begin{array}{cc}0&0\\1&0\end{array}\right],
\left[\begin{array}{cc}0&0\\0&1\end{array}\right]$but that is it I get stuck at this point because i cant figure out how $2\times2$ matrices can result to a $4 \times 4$ matrix
| $$AA^t=\begin{pmatrix}-1&\;\;3&-4&-3\\\;\;2&-1&\;\;0&-3\end{pmatrix}\begin{pmatrix}-1&\;\;2\\\;\;3&-1\\-4&\;\;0\\-3&-3\end{pmatrix}=\begin{pmatrix}35&4\\4&14\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/525400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the probability of $a>b+c$, where $a$, $b$, $c$ are $U(0,1)$ What is the probability that $a > b + c$?
$a, b, c$ are picked independently and uniformly at random from bounded interval [0,1] of $\mathbb{R}$.
| Let $d = b + c$.
$d$ has the Irwin-Hall distribution with $n=2$ independent variables.
Its CDF is equal to $F_d(x)=\frac{1}{n!}\sum_{k=0}^{\lfloor x\rfloor}(-1)^k\binom{n}{k}(x-k)^n$.
Now,
$$P(a<d) = \int_{-\infty}^{\infty} F_a(x)\, F'_d(x)\,dx.$$
Explanation of this equality. $a$ and $d$ are independent and continuous random variables with cumulative distributions $F_a(x)$ and $F_d(x)$.
So, we can write:
$$P(a > b + c) = P(a > d) = 1 - P(a < d) = 1 - \int_{-\infty}^{\infty} F_a(x)\, F'_d(x)\,dx.$$
$F'_d(x)=0$ for $x\in(-\infty,0)\cup[2, \infty)$. Thus, integral is non-zero only between two intervals:
*
*$x \in [0,1). \lfloor x\rfloor = 0.$
$$F_a(x) = x$$
$$F_d(x) = \frac{1}{2}\sum_{k=0}^{0}(-1)^k\binom{2}{k}(x-k)^2 = \frac{1}{2}x^2$$
$$F_d'(x) = x$$
*$x \in [1,2). \lfloor x\rfloor = 1.$
$$F_a(x) = 1$$
$$F_d(x) = \frac{1}{2}\sum_{k=0}^{1}(-1)^k\binom{2}{k}(x-k)^2 = \frac{1}{2}x^2 - \frac{1}{2}\cdot2(x-1)^2 =-\frac{1}{2}x^2+2x-1$$
$$F_d'(x) = -x+2$$
Therefore:
$$P(a > b + c) = 1 - \int_{0}^{1}F_a(x)\, F'_d(x)\,dx - \int_{1}^{2}F_a(x)\, F'_d(x)\,dx =$$
$$1 - \int_{0}^{1}x\, x\,dx - \int_{1}^{2}1\cdot(-x+2)\,dx = 1 - \frac{1}{3} - \frac{1}{2} = \frac{1}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/525648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Why does $(3\sqrt3)^2 = 27$? How does $(3\sqrt3)^2 = 27$?
I've tried to solve this using binomial expansion and using the FOIL method from which I obtain $9 + 3\sqrt3 +3\sqrt3 + 3$.
it has been a while since I've done this kind of thing so it may be something obvious that I can't see.
| Other people have basically already said it, but the FOIL is only something that you can apply when you have expressions of exactly this form:
$$
(a + b)(c+d).
$$
In this case you do: (F)irst (O)utside (I)nside (L)ast: The First part of $(a + b)^2$ is $ac$. The Outside is $ad$, the Inside if $bc$ and the Last is $bd$. In all you get
$$
ac + ad + bc + bd.
$$
Now consider what happens when you have
$$
(a+b)^2 = (a+b)(a+b).
$$
In this case you get
$$
aa+ ab + ba + bb = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2.
$$
Now, you are asking about what to do with
$$
(3\sqrt{3})^2
$$
Note that this is not of exactly the form $(a+b)(c+d)$ or $(a+b)^2$. You have the product of $3$ and $\sqrt{3}$ and to use FOIL you would need the sum $3+\sqrt{3}$. Therefore it doesn't make sense to use FOIL here. It simply isn't applicable (unless you rewrite the expression first as explain in one of the other answers).
So what do you do in this case? Your expression is exactly of the form
$$
(ab)^2
$$
with $a=3$ and $b = \sqrt{3}$. And then we have a rule that says that
$$
(ab)^2 = a^2b^2.
$$
So you get
$$
(3\sqrt{3})^2 = 3^2(\sqrt{3})^2 = 9\cdot 3 = 27.
$$
And that is it.
You could also do it slightly differently by using the rule/definition
$$
(ab)^2 = abab
$$
So
$$
(3\sqrt{3})^2 = 3\sqrt{3}\cdot3\sqrt{3} = 3\cdot 3\cdot \sqrt{3}\sqrt{3} = 9 \sqrt{3\cdot 3} = 9\sqrt{9} = 9\cdot 3 = 27.
$$
Here we also used that the rule $\sqrt{c}\sqrt{d} = \sqrt{cd}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 1
} |
$\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^3}\right)^{-1}\right\rfloor=2n(n-1)$ is true for any $n\in\mathbb N$? Question : Is the following true for any $n\in\mathbb N$?
$$\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^3}\right)^{-1}\right\rfloor=2n(n-1).$$
Note that $\lfloor x\rfloor$ is the largest integer not greater than $x$.
Motivation : I've been able to prove the following $(\star)$:
"For any $n\in\mathbb N$,
$$\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^2}\right)^{-1}\right\rfloor=n-1."$$
Then, I reached the above expectation. However, I can't prove that this expectation is true. Can anyone help?
Proof for $(\star)$ : The $n=1$ case is obvious. In the following, let $n\ge 2$. First, since
$$\sum_{k=n}^{\infty}\frac {1}{k^2}\gt\sum_{k=n}^{\infty}\frac{1}{k(k+1)}=\sum_{k=n}^{\infty}\left(\frac 1k-\frac{1}{k+1}\right)=\frac 1n,$$
we get
$$\begin{align}\sum_{k=n}^{\infty}\frac{1}{k^2}\gt \frac 1n\qquad(1)\end{align}$$
Next, since
$$\sum_{k=n}^{\infty}\frac {1}{k^2}\lt\sum_{k=n}^{\infty}\frac{1}{k(k-1)}=\sum_{k=n}^{\infty}\left(\frac 1{k-1}-\frac{1}{k}\right)=\frac 1{n-1},$$
we get
$$\begin{align}\sum_{k=n}^{\infty}\frac{1}{k^2}\lt \frac 1{n-1}\qquad(2)\end{align}$$
$(1)(2)$ give us
$$\frac 1n\lt\sum_{k=n}^{\infty}\frac{1}{k^2}\lt\frac{1}{n-1}\rightarrow n-1\lt\left(\sum_{k=n}^{\infty}\frac{1}{k^2}\right)^{-1}\lt n.$$
Hence, $$\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^2}\right)^{-1}\right\rfloor=n-1$$
as desired.
| Yes, it is true:
$$\sum_{k=n}^{\infty} \frac{1}{k^3} < \sum_{k=n}^{\infty} \frac{1}{k(k^2-1)} = \sum_{k=n}^{\infty} \Big( \frac{1}{2k(k-1)} - \frac{1}{2k(k+1)}\Big) = \frac{1}{2n(n-1)}$$ and $$\sum_{k=n}^{\infty} \frac{1}{k^3} > \sum_{k=n}^{\infty} \frac{k}{k^4 + \frac{1}{4}} = \sum_{k=n}^{\infty} \Big( \frac{1}{2k^2 -2k + 1} - \frac{1}{2(k+1)^2 - 2(k+1) + 1}\Big)$$ $$=\frac{1}{2n^2 - 2n + 1} = \frac{1}{2n(n-1) + 1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/527214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Existence of $2^{1/2}$ I'm trying to prove, that the square root of 2 exists in $\mathbb{R}$.
I'm looking at the set $A:=\{y\in\mathbb{R}: y\geq 0, y^2\geq 2\}$. I have already proven that the greatest lower bound $b$ of $A$ exists. However, now I don't know how to proceed. I know that I should consider the cases $b^2 > 2$ and $b^2 < 2$, and bring them to a contradiction, so only the possibility $b^2=2$ is left. I have been given the tip to use the quantities $a:=\frac{1}{2}(1 - \frac{2}{b^2}),\; a^\prime := \frac{1}{2}(1-\frac{b^2}{2}),\; b(1-a),\; \frac{b}{1-a^\prime}$.
I believe I should get to the contradiction that $b$ can't be the greatest lower bound in both cases, but I don't see how to get there.
Any ideas?
| Suppose that the infimum is $b$ and there is no real number $c$ such that $c^2=2$. Then if $b^2> 2$, then we have $0<a<1$ and moreover:
$$
b(1-a)< b\\
b^2(1-a)^2=b^2\left(\frac 12+\frac 1{b^2}\right)^2\geq b^2\left(\frac {\sqrt 2}{b}\right)^2=2
$$
where the last inequality comes from simple AM-GM inequality. But this is a contradiction because $b(1-a)< b$ which means that we found $b(1-a)$ less than the infimum $b$ but still bigger than or equal to 2.
On the other hand, if $b^2<2$ then we also have $0<a'<1$, hence $b/(1-a')> b$ and because $b$ is infimum then $\frac{b^2}{(1-a')^2}> 2$ (if $\frac{b^2}{(1-a')^2}=2$ then we found $c$ such that $c^2=2$). But:
$$
\frac{b^2}{(1-a')^2}=\frac{b^2}{\left(\frac 12+\frac {b^2}4\right)^2} \leq \frac{b^2}{\left(\frac { 2 b}{\sqrt 8}\right)^2}=2
$$
which is again a contradition.
Therefore $b^2=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/528590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Calculate volumes using triple integrals I have to calculate these two volumes using triple integrals:
volume of $A = \{(x,y,z) \in \Bbb R^3 : {x^2\over a^2} + {y^2\over b^2} \leq z \leq 1 \}$
volume of $B = \{(x,y,z) \in \Bbb R^3 : x^2+y^2+z^2 \leq a^2, x^2+y^2-ax \geq 0,x^2+y^2+ax \geq 0 \}$
So I want to calculate the integral of the function 1 over A but I'm having a hard time finding the limits of integration. Thanks in advance.
| Volume A is an elliptic paraboloid with its "vertex" at the origin, truncated by the plane $ \ z \ = \ 1 \ $ . Because it has a symmetry axis along the $ \ z-$ axis, it is actually convenient to integrate using a single variable, since every "slice" parallel to the $ \ xy-$ plane is an ellipse, the area of which has a simple dependence upon $ \ z \ $ .
If it is insisted upon that the volume be found using triple integration, we can work out some of the integration limits from the equation for the surface. At any specific value $ \ z \ \ge \ 0 \ $ , we have
$$ \ z \ = \ {x^2\over a^2} + {y^2\over b^2} \ \Rightarrow \ y \ = \ \pm \ b \sqrt{z} \ \cdot \ \sqrt{1 - \frac{x^2}{a^2 z}} \ \ . $$
The elliptical cross-section at a particular $ \ z \ $ has a "semi-horizontal" axis of length $ \ a \ \sqrt{z} \ $ and a "semi-vertical" axis $ \ b \ \sqrt{z} \ $ . Because of the four-fold symmetry of these cross-sections, we can just integrate over the first quadrant and multiply the result by four. The volume integral in Cartesian coordinates is then
$$ V_A \ = \ 4 \ \int_0^1 \int_0^{a \sqrt{z}} \int_0^{b \sqrt{z} \ \cdot \ \sqrt{1 - \frac{x^2}{a^2 z}}} \ \ dy \ dx \ dz \ \ = \ \ 4b \ \int_0^1 \int_0^{a \sqrt{z}} \ \sqrt{z} \ \cdot \ \sqrt{1 - \frac{x^2}{a^2 z}} \ \ dx \ dz $$
[the integration over $ \ y \ $ simply gives us $ \ y \ $ evaluated at the upper integration limit]
$$ = \ \ 4b \ \int_0^1 \ \frac{1}{2} \sqrt{z} \ \left[ \ a \sqrt{z} \ \arcsin \left(\frac{x}{a \sqrt{z}} \right) \ + \ \frac{1}{a \sqrt{z}} \cdot x \ \sqrt{1 - \frac{x^2}{a^2 z}} \ \right]\vert_0^{a \sqrt{z}} \ \ dz $$
[the integration over $ \ x \ $ is accomplished using a sine-substitution: the anti-derivative appears in many applications, so I won't derive it here]
$$ = \ \ 2b \ \int_0^1 \ \sqrt{z} \ \left( \ a \sqrt{z} \ \arcsin \ 1 \ + \ 0 \ - \ 0 \ + \ 0 \ \right) \ \ dz \ \ = \ \ 2ab \ \int_0^1 \ z \ \cdot \ \frac{\pi}{2} \ \ dz $$
[pulling the constants together shows us that each "horizontal slice" is indeed an ellipse with the expected area $ \ \pi \ ab \ z \ $ ]
$$ = \ \pi \ ab \ \left( \ \frac{1}{2} z^2 \ \right) \vert_0^1 \ = \ \frac{\pi \ ab}{2} \ \ , $$
confirming Felix Marin's result using cylindrical coordinates. Since the "height" of the volume is one unit, the volume expression gives the appearance of only having two dimensions, but in fact has three.
$$ \ \ $$
Volume B is that of a sphere of radius $ \ a \ $ centered on the origin, less the volume of two circular cylinders of diameter $ \ a \ $ with their symmetry axes parallel to the $ \ z-$ axis, these axes passing through $ \ ( \ \frac{a}{2} , 0, 0 \ ) \ $ and $ \ ( \ -\frac{a}{2} , 0, 0 \ ) \ $ , respectively. (We want the volume "inside the sphere, but outside of the cylinders".)
What we will compute is the volume of the portion of the sphere within the two cylinders, and subtract the result from the familiar expression for the volume of a sphere. The symmetry of this geometrical arrangement is eight-fold, so we can consider integration in the first octant. In the $ \ xy-$ plane, we will integrate over a semi-circle of radius $ \ \frac{a}{2} \ $ with its center at $ \ ( \ \frac{a}{2} , 0 \ ) $ , and integrate in the $ \ z-$ direction from that plane "upward" to the surface of the sphere.
In rectangular coordinates, the "bounding function" for the semi-circle is given by
$$ \ ( x - \frac{a}{2} )^2 \ + \ y^2 \ = \ \left( \frac{a}{2} \right)^2 \ \Rightarrow \ y \ = \ \sqrt{ax \ - \ x^2} \ \ , $$
and that for the spherical surface is
$$ x^2 \ + \ y^2 \ + \ z^2 \ = \ a^2 \ \Rightarrow \ z \ = \ \sqrt{a^2 \ - \ x^2 \ - \ y^2} \ \ . $$
The integral for the volume of the sphere contained by the cylinder is therefore
$$ 8 \ \int_0^a \int_0^{\sqrt{ax \ - \ x^2}} \int_0^{\sqrt{a^2 \ - \ x^2 \ - \ y^2}} \ \ dz \ dy \ dx \ \ . $$
However, this soon produces a rather daunting calculation. It proves to be much easier to work in cylindrical coordinates; setting up the integral can be done using our Cartesian-coordinate integral as a guide. The integration in $ \ z \ $ now "runs" from $ \ 0 \ $ to $ \ \sqrt{a^2 \ - \ r^2 } \ $ . The semi-circle expressed in polar coordinates is half of the "one-petal rosette" $ \ r \ = \ a \ \cos \ \theta \ $ , with the angle "running" from $ \ 0 \ $ to $ \ \frac{\pi}{2} \ $ . So we will instead calculate
$$ 8 \ \int_0^{\pi / 2} \int_0^{a \ \cos \ \theta} \int_0^{\sqrt{a^2 \ - \ r^2 }} \ \ dz \ r \ dr \ d\theta \ \ = \ \ 8 \ \int_0^{\pi / 2} \int_0^{a \ \cos \ \theta} \ r \ \sqrt{a^2 \ - \ r^2 } \ \ dr \ d\theta $$
$$ = \ \ 8 \ \int_0^{\pi / 2} \ \left[ \ -\frac{1}{3} \cdot \ (a^2 \ - \ r^2)^{3/2} \ \right] \vert_0^{a \ \cos \ \theta} \ \ d\theta $$
$$ = \ \ \frac{8}{3} \ \int_0^{\pi / 2} \ a^3 \ - \ ( \ a^2 \ - \ a^2 \ \cos^2 \ \theta \ )^{3/2} \ \ d\theta \ \ = \ \ \frac{8}{3} a^3 \ \int_0^{\pi / 2} \ 1 \ - \ \sin^3 \ \theta \ \ d\theta $$
$$ = \ \frac{8}{3} a^3 \ \left[ \ \theta \ \vert_0^{\pi / 2} \ - \ \left( \ u \ - \ \frac{1}{3} u^3 \ \right) \vert_{-1}^0 \ \right] $$
[using, in the second integrand term, the substitution $ \ du \ = \ \sin \ \theta \ d\theta \ , \ u \ = \ -\cos \ \theta \ $ ]
$$ = \ \frac{8}{3} a^3 \ \left[ \ \frac{\pi}{2} \ - \ \left( \ 0 \ - \ 0 \ - \ [\ -1 \ ] \ + \ [ \ -\frac{1}{3} \ ] \ \right) \ \right] $$
$$ = \ \frac{8}{3} a^3 \ \left( \ \frac{\pi}{2} \ - \ \frac{2}{3} \ \right) \ = \ \frac{4 \pi}{3} a^3 \ - \ \frac{16}{9} a^3 \ \ . $$
Since this is the volume of the portion of the sphere within the cylinders, it is to be removed from the total volume of the sphere, which is $ \ \frac{4 \pi}{3} a^3 \ $ . Hence, the volume of region B is
$$ V_B \ = \ \frac{16}{9} a^3 \ \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/530361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A triangle has side lengths 4,6,8. A tangent is drawn to incircle parallel to side 4 cutting ..... Problem :
A triangle has side lengths $4,6,8$. A tangent is drawn to incircle parallel to side $4$ cutting other two sides at M and N, than length of MN is
(a) $\frac{10}{9}$
(b) $\frac{20}{ 9}$
(c) $\frac{5}{3}$
(d) $\frac{4}{3}$
I have no clue how to proceed such problem however, please help on this I will be greatful to you thanks...
| Let the triangle be $ABC$, such that $M$ is on $AC$, and $N$ is on $AB$ and $x=MN$.
As $ABC$ and $AMN$ are similar we want to derive the scaling factor $k$ by calculating their areas. Let the games begin!
Let $F_1$ be the area of $ABC$, $F_2$ the area of $AMN$ and $F_3$ the area of $NMBC$. From Heron we know $F_1=3\sqrt{15}$ and from $F_1=s\cdot r$ (where $s=(a+b+c)/2$ and $r$ is the radius of the inscribed circle) we get $r=\sqrt{15}/3$. Now $F_3$ is the area of a trapezoid, so $F_3=\dfrac{x+4}{2}\cdot2r=(x+4)\dfrac{\sqrt{15}}{3}$. A short computation -- recalling that $F_2=F_1-F_3$ -- reveals that
$$k^2=\frac{F_2}{F_1}=\frac{5-x}{9}\quad\text{hence}\quad k=\frac{\sqrt{5-x}}{3}.$$
Grand finale: From $x=4k$ we'll happily derive $x=20/9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/533397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\gcd(c^a + 1, c^b + 1)$ for even $a$ and $b$? Following on this question, what is the Greatest Common Denominator of $c^a + 1$ and $c^b + 1$, where $a, b, c \in N$. I know that for odd a and b, we have $\gcd(c^a + 1, c^b + 1) = c^{\gcd(a, b)} + 1$ Thanks, Aleks Vlasev
I also found that for a odd and b even (or vice versa), the result is 1 or 0 depending on odd or even c.
The last case remains: For both $a$ and $b$ even: How can $gcd(c^a +1, c^b+1)$ be simplified such that it can be computed more quickly?
I tried to adapt this answer to my case but got nowhere*. Or maybe I should be using Fermat's little theorem?
Edit:
I did use the last linked answer to simplify this. Let a > b, $(c^b + 1, c^a + 1) = (c^b + 1, -c^{a-b} + 1)$.
| If $b=a$, then we have the trivial case $c^b+1 = c^a+1$.
Now take $b>a$ (without loss of generality). Note that $\gcd(r,s) \mid \gcd(r+s,r-s)$ for all integers $r,s$. Hence here we need consider
\begin{align}
&\gcd(c^b+c^a+2,c^b-c^a)\\
&\qquad= \gcd(c^b+c^a+2, c^a(c^{b-a}-1)) \\
&\qquad= \gcd\left(c^b+c^a+2, c^a(c-1)(c^{b-a-1}+c^{b-a-2}+\dotsb+1)\right).
\end{align}
Now you can do a factor-by-factor examination, e.g., $\gcd(c^a,c^b+2)$.
EDIT: Notice that right away you can eliminate any odd prime factors of $c-1$, since $\gcd(c-1,c^a+1)=1$ or $2$ because $a$ is even. Hence you only really have to consider
\begin{align}
\gcd\left(c^b+c^a+2, c^a(c^{b-a-1}+c^{b-a-2}+\dotsb+1)\right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/533769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$ This is for homework, and I could use a little help. The question asks
Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$.
Here is what I have done so far. Since $f$ is entire, I wrote
$$ f(z) = \sum_{n=0}^{\infty} a_n z^n = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 + \dotsb $$
for some $z \in \mathbb{C}$. Then
$$ f(2z) = a_0 + 2a_1z + 4a_2z^2 + 8a_3z^3 + 16a_4z^4 + \dotsb $$
and
$$ (1-2z)f(z) = a_0 + (a_1-2a_0)z + (a_2-2a_1)z^2+(a_3-2a_2)z^3 + (a_4-2a_3)z^4 + \dotsb. $$
Comparing coefficients, I find that
\begin{align*}
a_0 &= a_0 \\
a_1 &= -2a_0 \\
a_2 &= \frac{2^2}{3}a_0 \\
a_3 &= -\frac{2^3}{7 \cdot 3}a_0 \\
a_4 &= \frac{2^4}{15 \cdot 7 \cdot 3}a_0 \\
a_5 &= \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3} a_0 \\
&\vdots
\end{align*}
Now $f$ looks like
$$ f(z) = a_0 \left( 1 - 2z + \frac{2^2}{3}z^2 - \frac{2^3}{7 \cdot 3}z^3 + \frac{2^4}{15 \cdot 7 \cdot 3}z^4 - \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3}z^5 + \dotsb \right). $$
Does the series in parenthesis represent any elementary function? Besides the denominators, it looks like the Taylor expansion of $e^{-2z}$.
| Your function can be expressed as a convergent infinite product
$$f(z) = (1-z)(1-z/2)(1-z/4)(1-z/8)\dots$$
It's easy to see directly from this expression that it satisfies the functional equation $f(2z) = (1-2z)f(z)$.
I don't think that it can be expressed in terms of elementary functions, but I may be wrong.
Its values at the points $z=2^{-n}$ are related to the values of the Dedekind eta function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/534517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
About the inscribed sphere and the exspheres of a $n$-dimensional simplex Let us consider $n$-dimensional simplex $K$ in $n$-dimensional Euclidean space. Let $r_0$ be the radius of the inscribed sphere of $K$, and let be $r_1, r_2, \cdots, r_{n+1}$ be each radius of the exsphere of $K$. Then, here is my question.
Question : How can we represent $r_0$ by $r_1, r_2, \cdots, r_{n+1}$?
Motivation : I've known the followings :
In the $n=2$ case,
$$r_0=\frac{1}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}}.$$
In the $n=3$ case,
$$r_0=\frac{2}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\frac{1}{r_4}}.$$
Then, I reached the following conjecture (of course, this is just a conjecture) :
$$r_0=\frac{n-1}{\sum_{i=1}^{n+1}\frac{1}{r_i}}.$$
I don't have any good idea for $n$ in general. Can anyone help?
Update : I crossposted to MO.
| I'll prove the formula for a triangle, and it'll be "obvious" that the argument extends to any-dimensional space.
Consider $\triangle ABC$ with incenter $O$ and inradius $r$. Recall the fundamental observation that point $O$ helps dissect the triangle into sub-triangles of equal height ($r$), so that
$$\begin{align}
|\triangle ABC| &= |\triangle OBC| + |\triangle OCA|+ |\triangle OAB| \\[6pt]
&= \frac{1}{2}r\;|BC| + \frac{1}{2}r\;|CA| + \frac{1}{2}r\;|AB|\\[6pt]
&=\frac{r}{2}\left( a + b + c \right)
\end{align}$$
Similarly, if $O_A$ is the excenter on the opposite side of $\overline{BC}$ from $A$, and if $r_A$ is the corresponding exradius, then we have this formula:
$$\begin{align}
|\triangle ABC| &= -|\triangle O_A BC| + |\triangle O_A CA| + |\triangle O_A AB| \\[6pt]
&= \frac{r_A}{2}\left(-a + b + c \right)
\end{align}$$
Likewise, for exradii $r_B$ and $r_C$:
$$|\triangle ABC| = \frac{r_B}{2}\left( a - b + c \right) = \frac{r_C}{2}\left( a + b - c \right)$$
Thus,
$$\frac{1}{r_A}+\frac{1}{r_B}+\frac{1}{r_C} = \frac{(-a+b+c)+(a-b+c)+(a+b-c)}{2|\triangle ABC|} = \frac{a+b+c}{r \left(a+b+c\right)} = \frac{1}{r}$$
as expected.
The higher-dimensional cases proceed according to the same idea: the incenter and each excenter provides a dissection of the simplex into equal-height sub-simplices. The resulting "content" formulas are identical, except for a change in sign (and the radius involved). The extra coefficient appears as a natural consequence of the increasing failure of negative components to cancel positive ones.
Here's a run-through of the case of tetrahedron $ABCD$, where I'll write $w$, $x$, $y$, $z$ for the areas of faces opposite $A$, $B$, $C$, $D$. For incenter $r$, we have
$$\begin{align}|\operatorname{tet}ABCD| &= |\operatorname{tet}OBCD|+|\operatorname{tet}OCDA|+|\operatorname{tet}ODAB|+|\operatorname{tet}OABC| \\
&= \frac{1}{3} r \;\left( w+x+y+z \right)
\end{align}$$
And for excenter $r_A$ opposing $A$, we have:
$$\begin{align}|\operatorname{tet}ABCD| &= -|\operatorname{tet}O_A BCD|+|\operatorname{tet}O_A CDA|+|\operatorname{tet}O_A DAB|+|\operatorname{tet}O_A ABC| \\[6pt]
&= \frac{r_A}{3} \;\left(-w+x+y+z \right)
\end{align}$$
Also,
$$|\operatorname{tet}ABCD| = \frac{r_B}{3} \;\left(w-x+y+z \right) = \frac{r_C}{3} \;\left(w+x-y+z \right) = \frac{r_D}{3} \;\left(w+x+y-z \right)$$
whence
$$\begin{align}
\frac{1}{r_A} + \frac{1}{r_B} + \frac{1}{r_C} + \frac{1}{r_D} &= \frac{(-w+x+y+z)+\cdots+(w+x+y-z)}{3|\operatorname{tet}ABCD|} = \frac{2(w+x+y+z)}{r(w+x+y+z)} \\[6pt]
&= \frac{2}{r}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/535692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Laurent series of $z^{-3}$ at $z_0 = i$. Is there a way to do this by hand or is the question just evil? I have to find the two Laurent series expansions of $\frac{1}{z^3}$ about $i$. The only approach I can think of is to do:
$$\frac{1}{z^3} = \frac{1}{(z-i)^3} \left( \frac{z-i}{z} \right) ^3 = \frac{1}{(z-i)^3} \left( 1 - \frac{i}{i+(z-i)} \right) ^3 = \frac{1}{(z-i)^3} \left( 1 - \frac{1}{1-i(z-i)} \right) ^3 $$
This expansion will work in the disk $|z-i|<1:$
$$=\frac{1}{(z-i)^3} \left( 1 - \sum_{n=0}^{\infty}i^n(z-i)^n \right) ^3 =i \left( \sum_{n=0}^{\infty}i^{n}(z-i)^{n} \right) ^3 .$$
But this is nasty, because I don't know how to cube the series! And I don't think anything nice would come out of it, anyway.
The second expansion is in the annulus $|z-i|>1:$
$$\frac{1}{z^3} = ... = \frac{1}{(z-i)^3} \left( 1 - \frac{1}{z-i}\frac{1}{ \frac{1}{z-i}-i} \right) ^3 = \frac{1}{(z-i)^3} \left( 1 - \frac{1}{z-i}\frac{i}{ 1-\frac{-i}{z-i}} \right) ^3 =$$
$$= \left( \frac{1}{z-i} - i \sum_{n=0}^{\infty} \frac{(-i)^n}{(z-i)^{n-2}} \right) ^3$$
Same problem, except now it's even worse... How do I deal with this?
| I'm sure you have been taught about this before. $\frac{1}{z^3}$ is regular at
$z = i$, so the Laurent series there is simply the ordinary Taylor series expansion (in the vicinity of $i$).
For any $\alpha > 0$ and $a \in \mathbb{C}$ not on the negative real axis, we have
$$\begin{align}
\frac{1}{z^\alpha}
= & \sum_{k=0}^{\infty} \frac{1}{k!} \left( \left.\frac{d^k}{dz^k} \frac{1}{z^{\alpha}}\right|_{z=a}\right) (z-a)^k\\
= & \sum_{k=0}^{\infty} \frac{1}{k!} \left( \frac{(-\alpha)(-\alpha-1)\cdots(-\alpha-k+1)}{a^{\alpha+k}}\right) (z-a)^k\\
= & \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \left( \frac{\alpha(\alpha+1)\cdots(\alpha+k-1)}{a^{\alpha+k}}\right) (z-a)^k\\
\end{align}$$
When $\alpha$ is a positive integer $n$, this reduces to
$$
\frac{1}{z^{n}}
= \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k} \frac{(z-a)^k}{a^{n+k}}
= \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{n-1} \frac{(z-a)^k}{a^{n+k}}
\tag{*1}$$
In particular, when $n = 3$ and $a = i$, one has
$$\frac{1}{z^3} = \sum_{k=0}^{\infty} (-1)^k \binom{k+2}{2} \frac{(z-i)^k}{i^{3+k}}
=\sum_{k=0}^{\infty} i^{k+1}\frac{(k+1)(k+2)}{2} (z-i)^k
\tag{*2}$$
Update
To derive the other Laurent series (i.e, the one in the vicinity of $\infty$),
one can let $u = \frac{1}{z-i}$ and rewrite
$$\frac{1}{z^3} \quad\text{ as }\quad\frac{1}{(i + \frac{1}{u})^3} = \frac{i u^3}{(u-i)^3}.$$
One then consider the Taylor series expansion of it at $u = 0$. Compare this with $(*1)$ and $(*2)$, one find $u$ is now taking the role of $z-i$ and $-i$ is taking the role of $a$ there. As a result,
$$\frac{1}{z^3} = i u^3 \sum_{k=0}^{\infty}(-1)^k\binom{k+2}{k} \frac{u^k}{(-i)^{3+k}}
= \sum_{k=0}^{\infty}(-i)^k \frac{(k+1)(k+2)}{2}\frac{1}{(z-i)^{k+3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/536121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Improper Integral $\int\limits_0^1\frac{\ln(x)}{x^2-1}\,dx$ How can I prove that?
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$
I know that
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\sum_{n=0}^{\infty}\int_0^1-x^{2n}\ln(x)\,dx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
but I want another method.
| $$\begin{align}\int_0^1\frac{\ln(x)}{x^2-1}dx&=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^1\frac{\ln(v)}{v^2-1}dv\right)\\
&=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_1^{\infty}\frac{\ln(v)}{v^2-1}dv\right)\\
&=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^{\infty}\frac{\ln(u)}{u^2-1}du-\int_0^{1}\frac{\ln(u)}{u^2-1}du\right)\\
&=\frac{1}{2}\int_0^{\infty}\frac{\ln(u)}{u^2-1}du=\frac{1}{2}\times\frac{1}{2}\int_0^{\infty}\frac{\ln(u^2)}{u^2-1}du
=\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\ln\left(\frac{1}{u^2}\right)du\\
&=\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\left[\ln\left(\frac{1+v}{1+u^2v}\right)\right]_{v=0}^{v=\infty}du\\
&=\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\left(\int_0^{\infty}\left(\frac{1}{1+v}-\frac{u^2}{1+u^2v}\right)dv\right)du\\
&=\frac{1}{4}\int_0^{\infty}\int_0^{\infty}\frac{1}{(1+v)(1+u^2v)}dv\,du=\frac{1}{4}\int_0^{\infty}\left(\frac{1}{1+v}\int_0^{\infty}\frac{1}{1+u^2v}du\right)dv\\
&=\frac{1}{4}\int_0^{\infty}\left(\frac{1}{1+v}\left[\frac{\tan^{-1}(\sqrt{v}u)}{\sqrt{v}}\right]_{u=0}^{u=\infty}\right)dv=\frac{1}{4}\times\frac{\pi}{2}\int_0^{\infty}\frac{1}{\sqrt{v}(1+v)}dv\\
&=\frac{\pi}{8}\int_0^{\infty}\frac{2w}{w(1+w^2)}dw=\frac{\pi}{8}\times2\left[\tan^{-1}(w)\right]_0^{\infty}=\frac{\pi}{8}\times2\times\frac{\pi}{2}=\frac{\pi^2}{8}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/537903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 8,
"answer_id": 5
} |
How to solve system equation $x^3+y^3+z^3=x+y+z, x^2+y^2+z^2=xyz$ for $x,y,z\in\mathbb{R}$? How to solve system equation $\left\{\begin{matrix}x^3+y^3+z^3=x+y+z&\\x^2+y^2+z^2=xyz&\end{matrix}\right.$ , $x,y,z\in\mathbb{R}$ ?
| We can show that there is no solution for $x,y,z>0$. Firstly we have:
$$
x^{2}+y^{2}+z^{2}=xyz \implies xyz\geq 3(xyz)^{\frac 23}\\
\implies xyz\geq 27.
$$
On the other hand we have:
$$
x^{2}+y^{2}+z^{2}=xyz\geq \frac 13(x+y+z)^{2}=\frac 13(x^3+y^3+z^3)^{2}\geq \frac 13(3xyz)^{2}=3(xyz)^2\\
xyz\leq \frac 13
$$
So we have $xyz\leq \frac 13$ and $xyz\geq 27$ which is a contradiction and hence there is no solution for positive $x,y,z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/538252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proofs of the properties of Jacobi symbol The definition and properties of Jacobi symbol are stated in this article.
I don't have a textbook handy containing the proofs of the following properties of Jacobi symbol.
It seems to me that not many textbooks on elementary number theory contain them.
So I think it is nice not only for me but also for other users to have them here assuming the corresponding properties of Legendre symbol.
1.
Let $n$ be a positive odd integer.
Let $a, b$ be integers such that $a \equiv b$ (mod $n$).
Then
$$\left(\frac{a}{n}\right) = \left(\frac{b}{n}\right)$$
2.
Let $a$ be an integer.
Let $m, n$ be positive odd integers.
Then
$$\left(\frac{a}{mn}\right) = \left(\frac{a}{m}\right)\left(\frac{a}{n}\right)$$
3.
Let $a, b$ be integers.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{ab}{n}\right) = \left(\frac{a}{n}\right)\left(\frac{b}{n}\right)$$
4.
Let $m, n$ be positive odd integers such that gcd$(m,n) = 1$.
Then
$$\left(\frac{m}{n}\right) \left(\frac{n}{m}\right) = (-1)^{\frac{m-1}{2}\frac{n-1}{2}}$$.
5.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$$.
6.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{2}{n}\right) = (-1)^{\frac{n^2-1}{8}}$$.
| 1.
Let $n$ be a positive odd integer.
Let $a, b$ be integers such that $a \equiv b$ (mod $n$).
Then
$$\left(\frac{a}{n}\right) = \left(\frac{b}{n}\right)$$
Proof:
Let $n = \prod p$ be the prime decomposition of $n$ with every prime factor repeated according to its multiplicity.
Since $a \equiv b$ (mod $p$) for every prime factor of $n$, $\left(\frac{a}{p}\right) = \left(\frac{b}{p}\right)$.
Hence $\prod \left(\frac{a}{p}\right) = \prod \left(\frac{b}{p}\right)$.
Hence $\prod \left(\frac{a}{n}\right) = \prod \left(\frac{b}{n}\right)$.
2.
Let $a$ be integers.
Let $m, n$ be positive odd integers.
Then
$$\left(\frac{a}{mn}\right) = \left(\frac{a}{m}\right)\left(\frac{a}{n}\right)$$
Proof:
Clear from the definition of Jacobi symbol.
3.
Let $a, b$ be integers.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{ab}{n}\right) = \left(\frac{a}{n}\right)\left(\frac{b}{n}\right)$$
Proof:
Let $n = \prod p$ be the prime decomposition of $n$ as in the proof of $1.$
$\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$ for every prime factor $p$ of $n$.
Hence $\prod \left(\frac{ab}{p}\right) = \prod \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$.
Hence $\left(\frac{ab}{n}\right) = \left(\frac{a}{n}\right)\left(\frac{b}{n}\right)$.
Lemma 1
Let $a, b$ be odd integers.
Then $(ab - 1)/2 \equiv (a - 1)/2 + (b - 1)/2$ (mod $2$).
Proof:
Since $a - 1$ and $b - 1$ are even,
$(a - 1)(b - 1) \equiv 0$ (mod $4$).
Hence $ab - a - b + 1 \equiv 0$ (mod $4$).
Hence $ab - 1 \equiv (a - 1) + (b - 1)$ (mod $4$).
Hence $(ab - 1)/2 \equiv (a - 1)/2 + (b - 1)/2$ (mod $2$).
Lemma 2
Let $a, b$ be odd integers.
Then $(a^2b^2 - 1)/8 \equiv (a^2 - 1)/8 + (b^2 - 1)/8$ (mod $2$).
Proof:
$a^2 - 1 \equiv 0$ (mod $4$).
$b^2 - 1 \equiv 0$ (mod $4$).
Hence
$(a^2 - 1)(b^2 - 1) \equiv 0$ (mod $16$).
Hence
$a^2b^2 - a^2 - b^2 + 1 \equiv 0$ (mod $16$).
Hence $a^2b^2 - 1 \equiv (a^2 - 1) + (b^2 - 1)$ (mod $16$).
Hence
$(a^2b^2 - 1)/8 \equiv (a^2 - 1)/8 + (b^2 - 1)/8$ (mod $2$).
Lemma 3
Let $a, b, c$ be positive odd integers.
Suppose
$\left(\frac{a}{c}\right) \left(\frac{c}{a}\right) = (-1)^{\frac{a-1}{2}\frac{c-1}{2}}$.
$\left(\frac{b}{c}\right) \left(\frac{c}{b}\right) = (-1)^{\frac{b-1}{2}\frac{c-1}{2}}$.
Then
$\left(\frac{ab}{c}\right) \left(\frac{c}{ab}\right) = (-1)^{\frac{ab-1}{2}\frac{c-1}{2}}$.
Proof:
$\left(\frac{ab}{c}\right) \left(\frac{c}{ab}\right) =\left(\frac{a}{c}\right) \left(\frac{c}{a}\right)\left(\frac{b}{c}\right) \left(\frac{c}{b}\right) = (-1)^{\frac{a-1}{2}\frac{c-1}{2} + \frac{b-1}{2}\frac{c-1}{2}}
= (-1)^{(\frac{a-1}{2} + \frac{b-1}{2})\frac{c-1}{2}}
= (-1)^{\frac{ab-1}{2}\frac{c-1}{2}}$.
The first equality follows from $2.$ and $3.$
The last equality follows from Lemma 1.
4.
Let $m, n$ be positive odd integers such that gcd$(m,n) = 1$.
Then
$$\left(\frac{m}{n}\right) \left(\frac{n}{m}\right) = (-1)^{\frac{m-1}{2}\frac{n-1}{2}}$$.
Proof:
This follows immediately from Quadratic Reciprocity Theorem using Legendre symbol and Lemma 3.
5.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$$.
Proof:
This follows immediately from the first supplementary law of quadratic reciprocity using Legendre symbol and $2.$ and Lemma 1.
6.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{2}{n}\right) = (-1)^{\frac{n^2-1}{8}}$$.
Proof:
This follows immediately from the second supplementary law of quadratic reciprocity using Legendre symbol and $2.$ and Lemma 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/539006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Smallest such $n \in \mathbb{N}$ that $2^{n} \equiv 1 \pmod{5\cdot 7\cdot 9\cdot 11\cdot 13}$ Can anybody give me a hint about how to find smallest such $n \in \mathbb{N}$ that $2^{n} \equiv 1 \pmod{5\cdot 7\cdot 9\cdot 11\cdot 13}$?
I thought that I will find it piece by piece with help from my friend Fermat's Little Theorem, so :
*
*$2^{n_{1}} \equiv 1 \mod 5$, so $n_{1}=4$.
*$2^{n_{2}} \equiv 1 \mod 7$, so $n_{2}=6$.
*$2^{n_{3}} \equiv 1 \mod 3$, so $n_{3}=2$.
*$2^{n_{4}} \equiv 1 \mod 11$, so $n_{4}=10$.
*$2^{n_{5}} \equiv 1 \mod 13$, so $n_{5}=12$.
So, since I know that $x \equiv y \mod mz \Leftrightarrow x \equiv y \mod m$ when $z \neq 0$, so my lucky was to take the lowest common multiple of $4,6,2,10,12$, which is $60$ and lo and behold it fits the bill. But is it the smallest such $n$? If so, how to explain it?
| First of all, $p-1$ is not necessarily the Multiplicative Order ord$_pa$ for prime $p$ and $(a,p)=1$
For example, ord$_72=3\ne 7-1$
Again, $9=3^2$ is not prime, we need Euler's Totient Theorem here
Now as ord$_{11}2=10$ and ord$_{13}2=12,$ there can be no number $<$lcm
$(10,12)=60$ to satisfy $2^n\equiv1\pmod {5\cdot7\cdot9\cdot11\cdot13}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/540477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find the Minimum value of this expression Let $x,y,z>0$ such that: $x+y+z=1$. Find the minimum value of this expression.
$P=(\frac{x+1}{x})^3.\frac{z^2}{z^2+1}+(\frac{y+1}{y})^3.\frac{x^2}{x^2+1}+(\frac{z+1}{z})^3.\frac{y^2}{y^2+1}$
| Using rearrangement inequality:
$$P=\sum_{cyc} \left[\left(\frac{x+1}{x}\right)^3 \frac{z^2}{z^2+1}\right] \ge \sum \left[\left(\frac{x+1}{x}\right)^3 \frac{x^2}{x^2+1}\right] = \sum \frac {(1+x)^3}{x \, (1+x^2)}$$
Noting $f(x) = \dfrac{(1+x)^3}{x \, (1+x^2)}$ is convex in $(0, 1)$, by Jensen's inequality:
$$\frac{P}{3} \ge \frac {(1+\frac13)^3}{\frac13 \, (1+\frac19)} = \frac{32}5$$
So the minimum of $P$ is $\frac{96}5$, achieved when $x=y=z=\frac13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$xy=22$ and $yz=26$: What is $x+y+z $ equal to? Given the following: $$xy=22,\qquad yz=26,$$ where $x,y,z\in\mathbb{N}$. Which of the following is a possible value of $ x + y + z $?
$ \textbf {(A) } 22 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 26 \qquad \textbf {(D) } 48 $
| Look at $xy$ divisors and $yz$ divisors. (Because we have that $x,y,z \in \mathbb{N}$.)
$$
\begin{array}{|c|lcr|}
\hline xy \text{ and } yz & \text{} \\
\hline
22 & 1 & 2 & 11 & 22 \\
26 & 1 & 2 & 13 & 26 \\
\hline
\end{array}
$$
So if $xy=22=\color{red}{2}\cdot11 = \color{green}{1} \cdot 22$, and $yz=26=\color{red}{2}\cdot13 = \color{green}{1} \cdot 26$
then $y$ is either $\color{red}{2}$ or $\color{green}{1}$ and nothing else. We can do an easy shortcut: In the possibilities, we only have even numbers. $(1)$
So $y$ cannot be equal to $1$ since it would imply that $x+y+z$ is odd in contradiction with $(1)$.
But $x+y+z$ will be even if $y = 2$, so: $x+y+z=11+2+13=26$
Therefore the right answer is: $\boxed{\textbf{(C) } 26}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/548410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Pre Calculus Synthetic Division two of the roots of the equation $2x^3-3x^2+px+q=0$ are $3$ and $-2$. Find the third root of the equation.
| The cubic must factor as $2(x-r_1)(x-r_2)(x-r_3)$ and the coefficient of $x^2$ is $2 \cdot (-3/2).$ So the sum of the roots is $3/2$ and you know two roots are $3,-2$ with sum $1$, forcing the third root to be $1/2$.
Note that we do not have to find $p,q$ to use this method, since the sum of the roots of a monic cubic gives the negative of the coefficient of $x^2$, and the given equation may be divided by $2$ to be
$$x^3-(3/2)x^2+(p/2)x+(q/2)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/550873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\sin 2\theta +\sin \theta =1$ I tried Wolframalpha to solve this equation. The solution is $\theta\approx 0.355$. Since once a wise guy at MSE told me not to trust machines, I would like to know what methods can be used to solve this equation. I would appreciate for a brief explanation.
By the way $0\leq\theta\leq\pi/4$.
My second question is for any $x>0$ and $x\in\mathbb{R}$, Does $$x\sin 2\theta +\sin \theta =1$$ have a real valued solution for $0\leq\theta\leq\pi/4$?
| Well, note that if we put $t=\tan\frac\theta 2,$ then we have $$\begin{align}\sin\theta &= \sin\left(2\cdot\frac\theta2\right)\\ &= 2\sin\frac\theta2\cos\frac\theta2\\ &= 2\tan\frac\theta2\cos^2\frac\theta2\\ &= \cfrac{2t}{\sec^2\frac\theta2}\\ &= \cfrac{2t}{\tan^2\frac\theta2+1}\\ &= \frac{2t}{t^2+1},\end{align}$$ and in a similar fashion, we can show that $$\cos\theta=\frac{1-t^2}{t^2+1}.$$
Now, use the double-angle formula:
$$x\sin 2\theta+\sin\theta=1\\2x\sin\theta\cos\theta+\sin\theta=1\\2x\cdot\frac{2t}{t^2+1}\cdot\frac{1-t^2}{t^2+1}+\frac{2t}{t^2+1}=1\\\frac{4xt-4xt^3}{(t^2+1)^2}+\frac{2t}{t^2+1}=1\\0=1-\frac{4xt-4xt^3}{(t^2+1)^2}-\frac{2t}{t^2+1}\\0=1+\frac{4xt^3-4xt}{(t^2+1)^2}=\frac{-2t}{t^2+1}\\0=\frac{(t^2+1)^2}{(t^2+1)^2}+\frac{4xt^3-4xt}{(t^2+1)^2}+\frac{-2t(t^2+1)}{(t^2+1)^2}\\0=\frac{t^4+2t^2+1}{(t^2+1)^2}+\frac{4xt^3-4xt}{(t^2+1)^2}+\frac{-2t^3-2t}{(t^2+1)^2}\\0=\frac{t^4+(4x-2)t^3+2t^2-(4x+2)t+1}{(t^2+1)^2}$$
Since $t^2+1$ is nonzero for all real $t,$ then
$$0=t^4+(4x-2)t^3+2t^2-(4x+2)t+1.$$ Note that the coefficients of the above polynomial sum to $0,$ so $t=1$ is a root of the polynomial, and so $t-1$ is a factor. In particular, $$0=(t-1)\bigl(t^3+(4x-1)t^2+(4x+1)t-1\bigr),$$ so $$t=1\qquad\text{or}\qquad t^3+(4x-1)t^2+(4x+1)t-1=0.$$ The cubic equation can be solved using the general formula for roots of a cubic. (Warning: The roots thus found will probably be ugly-looking even if they happen to be rational numbers, and may look non-real even when they are actually real.)
Once you've found all your values of $t=\tan\frac\theta2,$ the values of $\theta$ will be those values of $2\arctan(t)$ lying in the appropriate interval, if there are any such values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/554933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How prove this inequality $\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\ge \frac{3}{2}$ let $x,y,z>0$,and such $x^n+y^n+z^n=3(n\ge 1),n\in N^*$,
show that:
$$\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy}\ge \dfrac{3}{2}$$
My try: if $n=1$ ,
since $x+y+z=3$,then
use Cauchy-Schwarz inequality
$$\left(\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy} \right)(x^2+y^2+z^2+3xyz)\ge (x+y+z)^2$$
then we only prove
$$\dfrac{9}{x^2+y^2+z^2+3xyz}\ge\dfrac{3}{2}$$
$$\Longleftrightarrow x^2+y^2+z^2+3xyz\le 6$$
Then I can't,and for $n$ how prove it?
| By Holder $$3=x^n+y^n+z^n=\frac{1}{3^{n-1}}(1+1+1)^{n-1}(x^n+y^n+z^n)\geq\frac{1}{3^{n-1}}(x+y+z)^n,$$ which gives $$x+y+z\leq3.$$
Now, let $x=ka$, $y=kb$ and $z=kc$ such that $k>0$ and $a+b+c=3$.
Thus, $k(a+b+c)\leq3$ gives $$0<k\leq1$$ and
$$\sum_{cyc}\frac{x}{x+yz}=\sum_{cyc}\frac{ka}{ka+k^2bc}=\sum_{cyc}\frac{a}{a+kbc}\geq\sum_{cyc}\frac{a}{a+bc}.$$
Id est, it's enough to prove that
$$\sum_{cyc}\frac{a}{a+bc}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\left(\frac{a}{a+bc}-\frac{1}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{a-bc}{a+bc}\geq0$$ or
$$\sum_{cyc}\frac{a(a+b+c)-3bc}{a+bc}\geq0$$ or
$$\sum_{cyc}\frac{(a-b)(a+3c)-(c-a)(a+3b)}{a+bc}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{a+3c}{a+bc}-\frac{b+3c}{b+ac}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2c(a+b+3c-3)}{(a+bc)(b+ac)}\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2c^2}{(a+bc)(b+ac)}\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/556519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to prove the following inequality of logarithm? Let $x,y,z\in\mathbb{C}.$ Suppose $$z=\frac{1}{2}(xy\pm\sqrt{x^2y^2-4(x^2+y^2)} ).$$
Show that $$log^+|z|\leq log^+|x|+log^+|y|+log 2.$$
Where $log^+\phi=max\{0,log\phi\}.$ Here we are also considering the complex root function with respect to the principle branch of logarithm.
| To prove what you want, we basically just need to apply the triangle inequality to your formula. I'll look at cases separately depending whether $|x|,|y|$ are greater than or smaller than unit magnitude.
*
*If $|x| \le 1$, $|y| \le 1$, then $|x^2y^2-4x^2-4y^2| \le |x^2y^2|+|4x^2|+|4y^2|\le 9$,so:
$$2z=xy\pm\left(x^2y^2-4x^2-4y^2\right)^{1/2},$$
$$\begin{aligned}\left|2z\right|&\le|xy|+|x^2y^2-4x^2-4y^2|^{1/2}\\
&\le 1+ 9^{1/2}=4 \end{aligned} $$
Hence $\log\left|z\right| \le \log 2$.
*If $|x| \ge 1$, $|y| \le 1$, then
$$\frac{z}{x}=\frac{y}{2}\pm\left(\frac{y^2}{4}-\frac{y^2}{x^2}-1\right)^{1/2},$$
$$\begin{aligned}\left|\frac{z}{x}\right|&\le\frac{|y|}{2}+\left|\frac{y^2}{4}-\frac{y^2}{x^2}-1\right|^{1/2}\\
&\le \frac{|y|}{2}+ \left(\frac{|y|^2}{4}+\left|\frac{y^2}{x^2}\right|+1\right)^{1/2} \\
&\le \frac{1}{2}+ \left|\frac{1}{4}+2\right|^{1/2}= \frac{1}{2}+ \left|\frac{9}{4}\right|^{1/2}=2 \end{aligned} $$
Hence $\log\left|\frac{z}{x}\right| \le \log 2$.
*If $|y| \ge 1$, $|x| \le 1$, then by symmetry with $x$ and $y$ and case 2 we have $\log\left|\frac{z}{y}\right| \le \log 2$.
*If $|x| \ge 1$, $|y| \ge 1$, then $|1/x^2+1/y^2| \le 2$ so:
$$\frac{z}{xy}=\frac{1}{2}\pm\left(\frac{1}{4}-\frac{1}{x^2}-\frac{1}{y^2}\right)^{1/2},$$
$$\begin{aligned}\left|\frac{z}{xy}\right|&\le\frac{1}{2}+\left|\frac{1}{4}-\frac{1}{x^2}-\frac{1}{y^2}\right|^{1/2}\\
&\le \frac{1}{2}+ \left(\frac{1}{4}+\left|\frac{1}{x^2}+\frac{1}{y^2}\right|\right)^{1/2} \\
&\le \frac{1}{2}+ \left|\frac{1}{4}+2\right|^{1/2}= \frac{1}{2}+ \left|\frac{9}{4}\right|^{1/2}=2 \end{aligned} $$
Hence $\log\left|\frac{z}{xy}\right| \le \log 2$.
In summary,
$$
\log|z| \le \begin{cases} \log 2 & \text{ if } |x| \le 1, |y| \le 1\\
\log|x| + \log 2 & \text{ if } |x| \ge 1, |y| \le 1\\
\log|y| + \log 2 & \text{ if } |x| \le 1, |y| \ge 1\\
\log|x| + \log|y| + \log 2 & \text{ if } |x| \ge 1, |y| \ge 1\\
\end{cases}.
$$
Then since $\log^+|x|=\begin{cases}0 & \text{ if } |x| \le 1\\ \log|x| & \text{ if }|x| \ge 1\end{cases}$, this is equivalent to:
$$
\log|z| \le \log^+|x| + \log^+|y| + \log 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/561999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Third-degree cosine inequality for obtuse triangle Suppose $\triangle ABC$ is an obtuse triangle with side lengths $a=BC, b=CA, c=AB$. I want to show that $$a^3\cos A+b^3\cos B+c^3\cos C<abc.$$
My idea is to use the cosine rule. I have $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, etc. Plugging into the inequality I get
$$a^4b^2+a^4c^2-a^6+b^4a^2+b^4c^2-b^6+c^4a^2+c^4b^2-c^6<2a^2b^2c^2.$$
How can I show this?
| Using $\sin2x,\cos2x$ formula
$$a^3\cos A=(2R\sin A)^3\cos A= 2R^3(2\sin^2A)(2\sin A\cos A)$$
$$=2R^3(1-\cos2A)\sin2A=R^3(2\sin2A-2\sin2A\cos2A)=R^3(2\sin2A-\sin4A)$$
Using this, $\sum \sin2A=4\prod \sin A$
Now, $\sin4A+\sin4B+\sin4C=2\sin(2A+2B)\cos(2A-2B)+2\sin2C\cos2C$
Now, $\cos2C=\cos\{2\pi-2(A+B)\}=\cos2(A+B)$ and $\sin2(A+B)=\sin(2\pi-2C0=-\sin2C$
$\implies\sin4A+\sin4B+\sin4C=-2\sin2C\cos(2A-2B)+2\sin2C\cos2(A+B)$
$=-\sin2C\{\cos(2A-2B)-\cos(2A+2B)\}=-\sin2C\cdot2\sin2A\sin2B$
$=-2(2\sin C\cos C)(2\sin A\cos A)(2\sin B\cos B)$
For an obtuse triangle, only one angle is between $(\frac\pi2,\pi)$ so exactly one of the cosine ratio $<0$ and all the sine ratios are $>0$
$\implies\sin4A+\sin4B+\sin4C>0$
$\implies\sum a^3\cos A <2R^3(\sum\sin2A)=2R^3(4\sin A\sin B\sin C)=\prod(2R\sin A)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/564276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.