Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $A+B+C=0$, then prove that the value of the determinant is $0$. I'll state the question from my textbook below:
If $A+B+C=0$, the prove that $\begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0$.
This is how I tried solving the problem:
$LHS = \begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix}$
$= 1(1- \cos^2 A) - \cos C (\cos C - \cos A \cos B) + \cos B (\cos A \cos C - \cos B)$
$= 1 + 2 \cos A \cos B \cos C - (\cos^2 A +\cos^2 B + \cos^2 C)$
I don't know how to proceed further. I tried using the fact that $A+B+C=0$ but it didn't lead to anything I could solve. I don't know where is it supposed to be used.
Also, I read a solution to this problem somewhere in which the term $(\cos^2 A +\cos^2 B + \cos^2 C)$ was replaced by $1 + 2 \cos A \cos B \cos C$ as $A+B+C=0$. Are these two terms equal for the given condition? Also, is there a way to prove the statement without using this fact?
Any help would be appreciated.
| When at least one of $A,B$ or $C$ is not an integer multiple of $\pi$, the vector $(\sin A,\sin B,\sin C)$ is nonzero. However, as $A+B+C=0$, the sum-of-angle formula implies that
$$
\pmatrix{1&\cos C&\cos B \\ \cos C&1&\cos A\\ \cos B&\cos A& 1}
\pmatrix{\sin A\\ \sin B\\ \sin C}
=0
$$
and hence the matrix is singular.
In case all of $A,B,C$ are integer multiples of $\pi$, you may complete the proof by a continuity argument, or by noticing that the matrix must have at least two identical rows, because up to a permutation, $(\cos A,\cos B,\cos C)$ is either $(1,1,1)$ or $(1,-1,-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Jordan basis unique? Is Jordan basis unique? I have a 4x4 matrix, find eigenvectors and one generalized eigenvector, also trying different linearly independent eigenvectors but the matrix P so that PJP^-1 =A only works for certain vector.
\begin{pmatrix}-\frac{1}{2}&0&-\frac{1}{2}&0\\ \frac{3}{2}&0&\frac{1}{2}&0\\ -\frac{3}{2}&0&-\frac{3}{2}&-1\\ 1&1&1&1\end{pmatrix}\begin{pmatrix}1&-1&0&-1\\ 0&2&0&1\\ -2&1&-1&1\\ 2&-1&2&0\end{pmatrix}\begin{pmatrix}1&1&0&0\\ -1&0&1&1\\ -3&-1&0&0\\ 3&0&-1&0\end{pmatrix}=\begin{pmatrix}-1&0&0&0\\ 0&1&0&-1\\ 0&0&1&1\\ 0&0&0&1\end{pmatrix}
| It is not: the Jordan matrix is unique, but there are several basis which will yield that matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Let $a^2<2, b=2(a+1)/(a+2)$. Show $b^2<2$ (assignment) It is a part of my assignment.
$$ \text {Let }a^2<2, \quad b=2\frac {(a+1)}{(a+2)}\quad \text{ Show } b^2<2$$
I already proved that a
But, I am struggling to prove $b^2<2$.
My lecturer said that I need to manipulate $b^2$, which is larger than $b^2$ but less than 2. That is $4(a+1)^2/(a+2)^2$ < some number < 2.
I was trying this for hours, but I couldn't find the way to solve. Thanks for helping in advance.
| same thing but with more details.
$b = \frac{(2a+1)}{a+2}$
$\frac{b}{2} = \frac{a+1}{a+2} = \frac{a+2-1}{a+2} = = 1 - \frac{1}{a+2}$
since
$a^2 < 2$
$-\sqrt{2} < a < \sqrt{2}$
$2 -\sqrt{2} < 2 + a < 2 + \sqrt{2}$
both sides are positive so
$\frac{1}{2 + \sqrt{2}} < \frac{1}{2+a} < \frac{1}{2 - \sqrt{2}}$
$-\frac{1}{2 - \sqrt{2}} < -\frac{1}{2+a} < -\frac{1}{2 + \sqrt{2}}$
$1-\frac{1}{2 - \sqrt{2}} < 1-\frac{1}{2+a} < 1-\frac{1}{2 + \sqrt{2}}$
$1-\frac{1}{2 - \sqrt{2}} < \frac{b}{2} < 1-\frac{1}{2 + \sqrt{2}}$
$2(1-\frac{1}{2 - \sqrt{2}}) < b < 2(1-\frac{1}{2 + \sqrt{2}})$
$(1-\frac{1}{2 + \sqrt{2}})^2$
so you need to show this is less than 2 and it's QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $\cos2\theta=0$, then $\Delta^2=$? I'll state the question from my textbook here:
If $\cos2\theta=0$, then $\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2=$?
This is how I solved the problem:
$\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2$
$= (\cos^3 \theta + \sin^3 \theta)^2$
$= (\cos \theta + \sin \theta)^2(\cos^2 \theta - \cos \theta \sin \theta + \sin^2 \theta)^2$
$= (1+ \sin2\theta)(1-\sin2\theta + \sin^2 \theta \cos^2 \theta)$
$= (1+ \sin2\theta)(1-\sin2\theta) + (1 + \sin2\theta)\sin^2\theta \cos^2 \theta$
$= \cos^2 2\theta + \frac 14 (1 + \sin2\theta)\sin^2 2\theta$
$= \frac 14 (1 + \sin2\theta)\sin^2 2\theta$
Now since $\cos2\theta=0$, $\sin2\theta = \pm 1$.
Therefore the above expression can take the values 0 and $\frac12$.
My textbook gives the answer as $\frac12$. I don't see any grounds on rejecting the other answer of 0. Have I made a mistake somewhere? Or am I forgetting something?
| When in doubt, use the relations in the original problem. Let $\theta=\frac{3\pi}4$. Then $\cos2\theta=0$, while $\sin\theta=\sqrt2/2=a$ and $\cos\theta=-\sqrt2/2=-a$.
The expression in the question is now
$$\begin{vmatrix}0&-a&a\\-a&a&0\\a&0&-a\end{vmatrix}^2$$
Clearly, adding up all three rows of the matrix produces the zero vector, so the whole expression evaluates to zero. Your working is entirely correct: the book is wrong to omit 0 as an answer.
The result of $\frac12$ is obtained with the other principal solution to $\cos2\theta=0$, $\theta=\frac\pi4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Using polar co-ordinates to solve a dynamical system I have the system
$x^\prime = x-y-x(x^{2}+y^{2})$
$y^\prime = x+y-y(4x^{2}+y^{2})$.
I want to show with polar coordinates that I eventually end up with
$r^\prime = r-r^{3}(1+ \frac{3}{4}$sin$^{2}(2 \theta))$.
I have used the following substitutions:
$x = r$cos$(\theta)$, $y=r$sin$(\theta)$, $x^{2} + y^{2} = r^{2}$ and $rr^\prime = xx^\prime + yy^\prime $.
I have made a few attempts but my recent one is as follows:
\begin{align}
rr^\prime &= xx^\prime + yy^\prime
\\
&=x(x-y-x^{3} -xy^{2}) + y(x+y-4x^{2}y^{2}+y^{3})
\\
&=x^{2} -xy + x^{4} + x^{2}y^{2} + yx + y^{2} - 4x^{2}y^{2} + y^{4}
\\
&=r^{2} + (x^{4} + y^{4}) - 3x^{2}y^{2} .
\end{align}
I am not sure if I am heading in the right direction. I also have noted that $\sin(2\theta) = 2\sin(\theta)\cos(\theta) $.
Any help would be much appreciated.
| Actually you are on the right track but you've just made a few slips with your algebra.
You should have $$rr'=x(x-y-x^{3} -xy^{2}) + y(x+y-4x^{2}y^{\color{red}{1}}\color{red}{-}y^{3})$$
this will simplify to $$x^2+y^2-(x^4+y^4)-5x^2y^2$$
$$=r^2-(r^4-2x^2y^2)-5x^2y^2$$
$$=r^2-r^4-3x^2y^2$$
now use the double angle formula for $\sin2\theta$ and you are done.
I hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Limit of a function tending to a finite number If
$$\lim_{x\to 0} \frac{ae^x - b\cos x +ce^{-x}}{x\sin x} = 2$$
then find the value of $a+b+c$.
My book has given the following solution to the above problem :-
We observe that as $x$ tends to zero , numerator tends to $a-b+c$ whereas the denominator tends to zero. Therefore for the limit to exist , we must have ,$a-b+c=0$
Now I am really confused at this point. Why would we want the numerator to attain the value of $0$ . Wouldn’t that give us an indeterminate answer? But actually it’s suposed to be two . Can you please explain ? Thank you for your help.
| If $a-b+c \neq 0$, the limit would be either $+\infty$ or $-\infty$.
Note, as written as Robert Z, that
$$
\frac{ae^x-b\cos x+cx^{-x}}{x\sin x}=\frac{(a-b+c)+x(a-c)+x^2(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}))+O(x^3)}{x^2+O(x^3)}
$$
as $x\to 0$. Since the limit exists, you need $a-b+c=0$, $a=c$ and $\frac{a}{2}+\frac{b}{2}+\frac{c}{2}=2$. Therefore $$a+b+c=4$$ (and you also know that $a=1, b=2, c=1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2676038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How do you solve this equation $\sqrt{1+log_{0.004}x}+\sqrt{3+log_{0.2}x}=1$? This is a problem from Prilepko's book on Mathematics for High School, I have tried to solve this equation but cannot succeed. The hint is to put $t =$ $\sqrt{3+log_{0.2}x}$. But then it reduces to some quite complicated expression. If I've also tried $t=$ $\sqrt{1+log_{0.004}x}$ but that doesn't simplify anything. Could you help me on this?
$\sqrt{1+log_{0.004}x}+\sqrt{3+log_{0.2}x}=1$
| Let $t = \sqrt{3+log_{0.2}x} \implies \frac{\ln x}{\ln 0.2} = t^2 - 3$
Now $\sqrt{1+log_{0.04}x} = \sqrt{1 + (t^2 - 3)\ln{0.2}/\ln{0.04}} = \sqrt{1 + \frac{t^2 - 3}{2}} = \sqrt{\frac{t^2 - 1}{2}} $
So we get $\sqrt{\frac{t^2 - 1}{2}} + t = 1 \implies t^2 -1 = 2(t-1)^2$
$ t^2 -1 = 2t^2 - 4t + 2 \implies t^2 - 4t + 3 = 0 \implies t = 3, 1$
solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2678846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Bounded sequences. We have the following sequences:
$a_n=n+sin(n+3^n)$
$b_n=\frac{{(-1)^n}(n+1)}{3\sqrt{n^2+7}}$
Show that $a_n$ is non bounded and $b_n$ is bounded. Also show that the subsequences $b_{2n}$ and $b_{2n-1}$ are convergent.
My work: $a_n<n+1$ and as $n\rightarrow∞$ $a_n<$∞ , so $a_n$ is non bounded
For $b_n$ , $|\frac{{(-1)^n}(n+1)}{3\sqrt{n^2+7}}|=
\frac{n+1}{3\sqrt{n^2+7}}\rightarrow\frac{1}{3}$ as $n\rightarrow∞$
so we get $0<|b_n|<\frac{1}{3}$.
I don't know how to show that the subsequences are convergent. Maybe Weierstrass but i'm not sure.
| Note that
$$a_n=n+\sin(n+3^n)>n-100 \to \infty$$
thus $a_n$ is not bounded and
$$-\frac{(n+1)}{3\sqrt{n^2+7}}\le b_n=\frac{{(-1)^n}(n+1)}{3\sqrt{n^2+7}}\le \frac{(n+1)}{3\sqrt{n^2+7}}$$
with $\frac{(n+1)}{3\sqrt{n^2+7}}$ bounded since $\frac{(x+1)}{3\sqrt{x^2+7}}$ is continuous and
$$\frac{(n+1)}{3\sqrt{n^2+7}}\to\frac13$$
thus for Weierstrass EVT $b_n$ is bounded.
Finally note that
$$b_{2n}=\frac{{(-1)^{2n}}(2n+1)}{3\sqrt{4n^2+7}}=\frac{(2n+1)}{3\sqrt{4n^2+7}}\to \frac13$$
$$b_{2n-1}=\frac{{(-1)^{2n-1}}(2n)}{3\sqrt{4n^2-4n+8}}=-\frac{2n}{3\sqrt{4n^2-4n+8}}\to -\frac13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
how to find the value of the $k$ to make $\lim_{x\to 0} \frac{\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}}{x^{k}}=A$ exist? the question described as follow:
$$\lim_{x\to 0} \frac{\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}}{x^{k}}=A$$
the $A$ is constant and $A\not=0$
and find the $k$ to make this limit exist.
and I did this:
$$let\space t\space be\space x^{1/15}\space, then \space x=t^{15}.$$
$$then \space \lim_{x\to 0} \frac{\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}}{x^{k}}=\lim_{x\to 0} \frac{\sqrt[5]{t^{15}+t^{5}}-\sqrt[3]{t^{15}+t^{3}}}{t^{15k}}.$$
$$then \space \lim_{x\to 0} \frac{\sqrt[5]{t^{15}+t^{5}}-\sqrt[3]{t^{15}+t^{3}}}{t^{15k}}=\lim_{x\to 0} \frac{t\sqrt[5]{t^{10}+1}-t\sqrt[3]{t^{12}+1}}{t^{15k}}.$$
use taylor expantion:
$$then \space \lim_{x\to 0} \frac{t\sqrt[5]{t^{10}+1}-t\sqrt[3]{t^{12}+1}}{t^{15k}}=\lim_{x\to 0} \frac{t(1+\frac{1}{5}t^{10}+o(t^{10}))-t(1+\frac{1}{3}t^{12}+o(t^{12}))}{t^{15k}}.$$
$$then \space\lim_{x\to 0} \frac{t(1+\frac{1}{5}t^{10}+o(t^{10}))-t(1+\frac{1}{3}t^{12}+o(t^{12}))}{t^{15k}}=\lim_{x\to 0} \frac{\frac{1}{5}t^{11}+o(t^{11})}{t^{15k}}.$$
$$then \space\lim_{x\to 0} \frac{\frac{1}{5}t^{11}+o(t^{11})}{t^{15k}}=\lim_{x\to 0} \frac{\frac{1}{5}t^{11}}{t^{15k}}=\lim_{x\to 0} \frac{1}{5t^{15k-11}}=A.$$
since A is a non-zero constant, so the $t^{15k-11}$ should be $t^{0}=1$.
then we get $15k-11=0$ and finally, we find $k=\frac{11}{15}$.
Am I right?
suppose I was right.
but unfortunately I found the image of $f(x)=\frac{\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}}{x^{\frac{11}{15}}}$ in quick graph app.
the value of $f(0)$ goes to $\infty$ instead any constant.
which is wrong, the app or me?
if I was wrong, how to find the right k?
| It seems correct indeed
$$\sqrt[5]{x+\sqrt[3]{x}}\sim x^\frac1{15} \left(1+ \frac{x^\frac23}5 \right)=x^\frac1{15} + \frac{x^\frac{11}{15}}5 $$
$$\sqrt[3]{x+\sqrt[5]{x}}\sim x^\frac1{15} \left(1+\frac{x^\frac45}3\right)=x^\frac1{15} +\frac{x^\frac{13}{15}}3$$
thus
$$\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}\sim \frac{x^\frac{11}{15}}5 $$
and
$$\lim_{x\to 0} \frac{ \sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}} }{x^\frac{11}{15}}=\frac15$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is it wrong to factor in this way? I found an exercise for factoring:
$\frac{4x^2-8x+4}{x^2-1} * \frac{x+1}{12}$
I resolve like this:
= $\frac{(2x -2)^2}{(x-1)(x+1)} * \frac{(x+1)}{12}$
= $\frac{(2x -2)^2}{12(x-1)}$
Well up here, because here comes my problem:
$1-.$ If I factor like this:
= $\frac{2(x -1)^2}{12(x-1)}$
= $\frac{(x-1)}{6}$ <--- First answer
$2-.$ If I factor with this other way:
= $\frac{(2x-2)(2x-2)}{12(x-1)}$
= $\frac{4(x-1)(x-1)}{12(x-1)}$
= $\frac{(x-1)}{3}$ <-- Second answer
then, what is the wrong answer ? and why ?
| When you factor something out of an expression which is squared, be wary of where the brackets go
$$(2x-2)^2=[2(x-1)]^2=[2]^2[x-1]^2=4(x-1)^2$$
So the first answer is wrong, the second is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
polynomial long division Integration $$
\int \frac{3x^2 +6x +2}{x^2 + 3x +2}dx
$$
I've been trying to solve this integral, my gut tells me to use polynomial long division so I factored and got
$$
\int \frac{3x^2 +6x +2}{(2+x)(1+x)}dx
$$
I have never done a division with two factors in the denominator, would anybody be able to help out?
| You do polynomial long division just like you do long division on integers.
$
\require{enclose}
\begin{array}{r}
3 \\
x^2 + 3x + 2 \enclose{longdiv}{3x^2+6x+2} \\
\underline{3x^2 + 9x + 6} \\
-3x - 4
\end{array}
$
$\frac {3x+6x + 2}{x^2+3x + 2} = 3 - \frac {3x+ 4}{x^2+3x+2}$
But I think what you are really looking for is something like
$\frac {3x^2+6x + 2}{x^2+3x + 2} = A + \frac {B}{x+1} + \frac {C}{x+2}\\
3x^2+6x + 2 = A(x^2 + 3x + 2) + B(x+2) + C(x+1)\\
3x^2 + 6x +2 = Ax^2 +(3A + B + C)x + (2A + 2B + C)$
The two polynomial are equal if all of the coefficients are equal
$A = 3\\
3A + B + C = 6\\
2A + 2B + C = 2$
And solve for $A,B,C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Compute $I=\int\frac{\sin x+\cos x}{\sin^2x+\cos^4x}\mathrm d x$
$$I=\int\frac{\sin x+\cos x}{\sin^2x+\cos^4x}\mathrm d x=?$$
I saw the related problem but that didn't help much. I attempted this question by splitting the numerator into $$I_1=\int\frac{\sin x}{\sin^2x+\cos^4x}\mathrm d x \text { and } I_2=\int\frac{\cos x}{\sin^2x+\cos^4x}\mathrm d x$$
For $I_1$, I set $t=\sin x$ and solved it the straightforward way. For $I_2$, I set $t=\cos x$ and again solved it the straightforward way.
My method got extremely lengthy in the end, while the question is supposed be decently okay. The answer that my book got is:
$$I=\frac 1{2\sqrt3}\ln\left|\frac{\sqrt3+t}{\sqrt3-t}\right|+\arctan t+C$$
for $t=\sin x-\cos x$. I have absolutely no clue how to get to such an easy result.
|
Here is a sketch only that should facilitate leading to the coveted result. To that end, we proceed.
First, noting that
$$\sin^2(x)+\cos^4(x)=\cos^2(x)+\sin^4(x)$$
we can write
$$\begin{align}
\int\frac{\sin(x)+\cos(x)}{\sin^2(x)+\cos^4(x)}\,dx&=\int \frac{\cos(x)}{1-\sin^2(x)+\sin^4(x)}\,dx+\int \frac{\sin(x)}{1-\cos^2(x)+\cos^4(x)}\,dx\\\\
&=\left.\left(\int \frac1{v^4-v^2+1}\,dv\right)\right|_{u=\sin(x)}-\left.\left(\int \frac1{u^4-u^2+1}\,du\right)\right|_{u=\cos(x)}
\end{align}$$
Next, we use partial fraction expansion to write
$$\begin{align}
\frac{1}{x^4-x^2+1}&=\frac{1}{(x^2+\sqrt 3 x +1)(x^2-\sqrt 3 x+1)}\\\\
&=\frac{1}{2\sqrt3}\left(\frac{x+\sqrt3}{x^2+\sqrt 3x+1}-\frac{x-\sqrt3}{x^2-\sqrt 3x+1}\right)\\\\
&=\frac{1}{2\sqrt3}\left(\frac{x+\sqrt3/2+\sqrt3/2}{x^2+\sqrt 3x+1}-\frac{x-\sqrt3/2-\sqrt3/2}{x^2-\sqrt 3x+1}\right)\\\\
&=\frac{1}{4\sqrt3}\left(\frac{2x+\sqrt3}{x^2+\sqrt 3x+1}-\frac{2x-\sqrt3}{x^2-\sqrt 3x+1}\right)\\\\
&+\frac{1}{2}\left(\frac{1}{x^2+\sqrt 3x+1}-\frac{1}{x^2-\sqrt 3x+1}\right)\tag1
\end{align}$$
To finish, observe that the first two terms on the right-hand side of $(1)$ are perfect differentials and integration leads to logarithm terms. Integration of the last two terms lead to arctangent terms.
This way forward isn't pretty, but it is effective and tractable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Suppose $A \in M_{3,3}$ with eigenvalues $1,2,3$, eigenvectors $b_1, b_2, b_3$. Let $v = b_1 - 4b_2 + 3b_3$. Compute $A^5 v$. I think I did this right, but someone skim and double check?
Q: Suppose $A$ is a $3 \times 3$ matrix, with eigenvalues $1,2,3$ and corresponding eigenvectors $b_1, b_2, b_3$. Suppose that $v = b_1 - 4b_2 + 3b_3$. Compute $A^5 v$.
Let $\mathcal{B}$ be the eigenvector basis and $D$ be the diagonal eigenvalue matrix:
\begin{align*}
D &= \begin{pmatrix}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 \\
\end{pmatrix} \\
P_{\mathcal{BE}} &= \begin{pmatrix} \begin{pmatrix} b_1 \end{pmatrix} & \begin{pmatrix} b_2 \end{pmatrix} & \begin{pmatrix} b_3 \end{pmatrix} \end{pmatrix} \\
[v]_{\mathcal{B}} &= (1,-4,3)^T \\
A &= P_{\mathcal{BE}}^{-1} D P_{\mathcal{BE}} \\
A^5 &= P_{\mathcal{BE}}^{-1} D^5 P_{\mathcal{BE}} \\
A^5 v &= P_{\mathcal{BE}}^{-1} D^5 P_{\mathcal{BE}} [v]_{\mathcal{E}} \\
[A^5 v]_{\mathcal{B}} &= D^5 [v]_{\mathcal{B}} \\
[A^5 v]_{\mathcal{B}} &= (1^5 \cdot 1, 2^5 \cdot -4, 3^5 \cdot 3)^T \\
[A^5 v]_{\mathcal{B}} &= (1, -128, 729)^T \\
A^5 v &= b_1 - 128 b_2 + 729 b_3 \\
\end{align*}
| That's fine, to check note that for each eigenvectors
$$A\vec v=\lambda \vec v\implies A^n(c\vec v)=c\lambda^n \vec v$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find all possible values of $p$, $q$ and $r$ such that this matrix is orthogonal
Find all possible values of $p$, $q$ and $r$ such that the following matrix is orthogonal.
$$B= \begin{pmatrix}
\frac1{3}&\frac2{3}&\frac2{3} \\
\frac2{3}&\frac1{3}&-\frac2{3} \\
p&q&r
\end{pmatrix}$$
| If $B$ is your matrix, then$$B^T.B=\begin{pmatrix}p^2+\frac{5}{9} & p q+\frac{4}{9} & p r-\frac{2}{9} \\ p q+\frac{4}{9} & q^2+\frac{5}{9} & q r+\frac{2}{9} \\ p r-\frac{2}{9} & q r+\frac{2}{9} & r^2+\frac{8}{9}\end{pmatrix}$$and you want this to be the identity matrix. So, since you only want to have $1$'s in the main diagonal, $p=q=\pm\frac23$ and $r=\pm\frac13$. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $\sin(x)=3\cos(x)$, compute $\sin(x)*\cos(x)$ I drew a triangle and was instructed to use a property unknown to me. Apparently the answer is $\frac{3}{\sqrt{10}}*\frac{1}{\sqrt{10}}=\frac{3}{10}$. Is this answer correct, and what theorem/formula is this?
| We have the identity
$1 + \tan^2 \theta = \sec^2 \theta = \dfrac{1}{\cos^2 \theta}; \tag 1$
so with
$\sin x = 3\cos x, \tag 2$
we infer
$\tan x = \dfrac{\sin x}{\cos x} = 3, \tag 3$
whence
$\dfrac{1}{\cos^2 x} = 1 + \tan^2 x = 1 + 3^2 = 10, \tag 4$
yielding
$\cos x = \pm \dfrac{1}{\sqrt{10}}; \tag 5$
since
$\sin^2 x + \cos^2 x = 1, \tag 6$
we infer
$\sin^2 x = 1 - \dfrac{1}{10} = \dfrac{9}{10}, \tag 7$
whence
$\sin x = \pm \dfrac{3}{\sqrt{10}}; \tag 8$
now, as has been pointed out by Foobaz John in his answer, $\tan x = 3$ implies the sign of $\sin x$ and $\cos x$ must be the same; thus
$\sin x \cos x = \dfrac{3}{\sqrt{10}} \dfrac{1}{\sqrt{10}} = \dfrac{3}{10}. \tag 9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
reverse three digit subtraction I cannot get this. How does it work? Like what is the concept in this?
Imagine $895 - 598$:
*
*Step 1. $8 - 5 = 3$ (I got $8$ and $5$ from the hundreths place in the numbers I am subtracting)
*Step 2. $3 × 100 = 300$ ($100$ is because $8$ and $5$ were originally from the hundreths)
*Step 3. $300 - 3 = 297$ (Subtract the answer from step 2 from the answer from step 1.)
*Final result. $297$ is the answer to $895 - 598$.
| Simply,
$$895-598=\left(8\times10^2+9\times10^1+5\times10^0\right)-\left(5\times10^2+9\times10^1+8\times10^0\right)=\left(8\times10^2-5\times10^2\right)+\left(9\times10^1-9\times10^1\right)+\left(5\times10^0-8\times10^0\right)=3\times10^2+0\times10^1+\left(-3\times10^0\right)=300-3=297$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Number of ways 6 things can be selected from a group of identical things
In how many ways can a party of 6 men be selected out of 10 Hindus, 8 Muslims, and 6 Christians if the party consists of at least one person of each religion( Consider only the religion of the person?
Now, I know this problem can be solved by finding the non-negative integral solution to the given equation,
$$x+y+z=6$$
Which the answer turns out to be $\binom{6-1}{3-1}$
But I want to know is there any other method of solving this problem?
Any help would be appreciated.
| Stars and Bars
Once we pre-select one of each religion, this is the same as asking "how many ways we can choose $3$ men from $9$ Hindus, $7$ Muslims, and $5$ Christians?" This is the same as asking how many ways to solve
$$
x+y+z=3
$$
with $x,y,z\ge0$. Stars and Bars say $\binom{5}{2}=10$.
Generating Functions
$$
\begin{align}
&\left[x^n\right]
\overbrace{\left(x+\cdots+x^{10}\right)}^{1\dots10\text{ Hindus}}
\overbrace{\left(x+\cdots+x^8\right)}^{1\dots8\text{ Muslims}}
\overbrace{\left(x+\cdots+x^6\right)}^{1\dots6\text{ Christians}}\\
&=\left[x^n\right]x^3\frac{1-x^{10}}{1-x}\frac{1-x^8}{1-x}\frac{1-x^6}{1-x}\\
&=\left[x^{n-3}\right]\frac{1-x^6-x^8-x^{10}+x^{14}+x^{16}+x^{18}-x^{24}}{(1-x)^3}\\
&=\left(\left[x^{n-3}\right]-\left[x^{n-9}\right]-\left[x^{n-11}\right]-\left[x^{n-13}\right]+\left[x^{n-17}\right]+\left[x^{n-19}\right]+\left[x^{n-21}\right]-\left[x^{n-27}\right]\right)\frac1{(1-x)^3}\\
&=\textstyle(-1)^{n-3}\left[\binom{-3}{n-3}-\binom{-3}{n-9}-\binom{-3}{n-11}-\binom{-3}{n-13}+\binom{-3}{n-17}+\binom{-3}{n-19}+\binom{-3}{n-21}-\binom{-3}{n-27}\right]\\[6pt]
&=\bbox[5px,border:2px solid #C0A000]{\textstyle\binom{n-1}{n-3}-\binom{n-7}{n-9}-\binom{n-9}{n-11}-\binom{n-11}{n-13}+\binom{n-15}{n-17}+\binom{n-17}{n-19}+\binom{n-19}{n-21}-\binom{n-25}{n-27}}
\end{align}
$$
This is the number of ways to pick a group of $n$ people from a pool of $10$ Hindus, $8$ Muslims, and $6$ Christians, where there is at least $1$ of each.
The only one of these that is not $0$ for $n=6$ is $\binom{n-1}{n-3}=\binom{5}{3}$.
Note that we could reduce the last line to
$$
\textstyle\binom{n-1}{2}-\binom{n-7}{2}-\binom{n-9}{2}-\binom{n-11}{2}+\binom{n-15}{2}+\binom{n-17}{2}+\binom{n-19}{2}-\binom{n-25}{2}
$$
except that the equation $\binom{n}{k}=\binom{n}{n-k}$ only holds when $k$ and $n-k$ are non-negative integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2689421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Fourier Transform of $ e^ {t/2} \operatorname{sech}(t) $ My question is what is the Fourier transform of the function
$f(t)=\exp(t/2)\text{sech}(t)$
Is there any closed form available? One possibility is that
$f(t)^2=\frac{1}{2} \text{sech}(t) + \frac{1}{4} (\dfrac{d}{dt}\operatorname{sech}t)$
and then the Fourier transform at frequency f is
$F(f(t)^2)=\frac{1}{2} F(\operatorname{sech}t)(1 + j\pi f )$
and $F(\operatorname{sech}t)=\pi \operatorname{sech}(\pi^2 f)$. But I am interested in $F(f(t))$ itself.
| The best way to attack the problem is to use the definition of the hyperbolic secant, together with the Geometric Series.
$$\text{sech}(t) = \frac{2}{e^t + e^{-t}}$$
We then have:
$$\mathcal{F}\left(e^{t/2}\text{sech}(t)\right)(k) = \frac{2}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{t/2} e^{-i kt }\frac{\text{d}t}{e^t + e^{-t}}$$
Now we have to split the integral into the two ranges $(-\infty, 0]\cup[0, +\infty)$ in order to use geometric series.
Geometric series will be used for the denominator as follows:
$$\frac{1}{e^t + e^{-t}} = \frac{1}{e^t(1 + e^{-2t})} = \frac{e^{-t}}{1 + e^{-2t}} = e^{-t}\sum_{j = 0}^{+\infty}(-1)^j e^{-2jt} ~~~~~~~ \text{for} ~~ [0, +\infty)$$
$$\frac{1}{e^t + e^{-t}} = \frac{1}{e^{-t}(1 + e^{2t})} = \frac{e^{t}}{1 + e^{2t}} = e^{t}\sum_{j = 0}^{+\infty}(-1)^j e^{2jt} ~~~~~~~ \text{for} ~~ (-\infty, 0]$$
Hence the two integrals we need to calculate are:
I
$$\frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j\int_0^{+\infty} e^{t/2} e^{-ikt} e^{-t}e^{-2jt}\ \text{d}t = \frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{2}{1 + 4j + 2ik}$$
Provided that $4 \Re(j)+1>2 \Im(k)$ which should be the case in question.
The sum can be obtained by some knowledge of Special Functions, and it sums:
$$\frac{4}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{1}{1 + 4j + 2ik} = \frac{\Phi \left(-1,1,\frac{1}{4} (2 i k+1)\right)}{\sqrt{2 \pi }}$$
The Hurwitz function that obtained can be expanded into "more elementary" functions, like the Polygamma Function:
$$\frac{\Phi \left(-1,1,\frac{1}{4} (2 i k+1)\right)}{\sqrt{2 \pi }} \equiv \frac{\psi ^{(0)}\left(\frac{i k}{4}+\frac{5}{8}\right)-\psi ^{(0)}\left(\frac{i k}{4}+\frac{1}{8}\right)}{2 \sqrt{2 \pi }}$$
Here $\Phi$ denotes the Hurwitz function and $\Psi$ denotes the Polygamma.
II
Part two is similar to the first one, except for the calculations. In any case, similarly, we get:
$$\frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j\int_0^{+\infty} e^{t/2} e^{-ikt} e^{t}e^{2jt}\ \text{d}t = \frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{2}{4 j-2 i k+3}$$
Provided that $\Im(k)+2 \Re(j)>-\frac{3}{2}$ which again holds.
The sum can be evaluated in the same manner:
$$\frac{4}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{1}{4 j-2 i k+3} = \frac{\Phi \left(-1,1,\frac{1}{4} (3-2 i k)\right)}{\sqrt{2 \pi }}$$
And again:
$$\frac{\Phi \left(-1,1,\frac{1}{4} (3-2 i k)\right)}{\sqrt{2 \pi }} \equiv \frac{\psi ^{(0)}\left(\frac{7}{8}-\frac{i k}{4}\right)-\psi ^{(0)}\left(\frac{3}{8}-\frac{i k}{4}\right)}{2 \sqrt{2 \pi }}$$
The final result is thence the sum of the two previous results, which means:
Final Result
$$\mathcal{F}\left(e^{t/2}\text{sech}(t)\right)(k) = \frac{\psi ^{(0)}\left(\frac{i k}{4}+\frac{5}{8}\right)-\psi ^{(0)}\left(\frac{i k}{4}+\frac{1}{8}\right)}{2 \sqrt{2 \pi }} + \frac{\psi ^{(0)}\left(\frac{7}{8}-\frac{i k}{4}\right)-\psi ^{(0)}\left(\frac{3}{8}-\frac{i k}{4}\right)}{2 \sqrt{2 \pi }} = $$
$$ = \frac{-\psi ^{(0)}\left(\frac{i k}{4}+\frac{1}{8}\right)+\psi ^{(0)}\left(\frac{i k}{4}+\frac{5}{8}\right)-\psi ^{(0)}\left(\frac{3}{8}-\frac{i k}{4}\right)+\psi ^{(0)}\left(\frac{7}{8}-\frac{i k}{4}\right)}{2 \sqrt{2 \pi }}$$
FINAL BEAUTY
Through definitions of the Polygamma, and few algebra, we can transform that pretty ugly expression into something cute and simple. The "true" final result is then:
$$\color{red}{\frac{\pi \tan \left(\frac{\pi i k}{4}+\frac{\pi }{8}\right)+\pi \cot \left(\frac{\pi i k}{4}+\frac{\pi }{8}\right)}{2 \sqrt{2 \pi }}}$$
Which can be written also as
$$\color{blue}{\frac{\pi \tan \left(\frac{1}{8} (2 \pi i k+\pi )\right)+\pi \cot \left(\frac{1}{8} (2 \pi i k+\pi )\right)}{2 \sqrt{2 \pi }}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2692041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Summation of Fourier Coefficients in Fourier Series Problem First poster!
I have been posed with a problem that is to find the Fourier Series of the periodic function
$$f(x) =
\begin{cases}
0, & -\pi < x < 0 \\
1, & 0<x<\frac{\pi}{2}\\
0, &\frac{\pi}{2}<x<\pi
\end{cases}$$
$$f(x) = f(x+2\pi)$$
I have found
$$
a_0 = \frac{1}{4},
$$
$$
a_n = \frac{1}{n\pi}\sin\left({n\frac{\pi}{2}}\right),
$$
$$
b_n = \frac{-1}{n\pi}\cos\left({n\frac{\pi}{2}}\right)-1
$$
At the moment my final answer for the problem is
$$
F.S f = \frac{1}{4}+\sum_{n=1}^\infty \left(\left(\frac{1}{n\pi}\sin\left({n\frac{\pi}{2}}\right)\right)\cos(nx)+\left(\frac{-1}{n\pi}\cos\left({n\frac{\pi}{2}}\right)-1\right)\sin(nx)\right)
$$
I realize, for example, for $a_n$:
For every odd integer of $n$, $\sin\left({n\frac{\pi}{2}}\right) = 1$, but every second odd integer results in $-1$. My question is how would I represent this pattern in the Fourier Series without including
$\sin\left({n\frac{\pi}{2}}\right)$ in the $a_n$ coefficient, an the same for the $b_n$ coefficient.
How do I simplify this?
Is this a valid answer whether there is or is not a way to simplify it?
Thanks everyone
| What you have looks fine, and I likely wouldn't bother trying to simplify it at all (unless you want to take Spencer's suggestion and work with complex numbers). That being said, one might note that the angle addition formula for the sine function might allow some moderate simplification. Recall that
$$ \sin(\alpha + \beta) = \cos(\alpha)\sin(\beta) + \cos(\beta)\sin(\alpha) $$
for all $\alpha$ and $\beta$. With $\alpha = -nx$ and $\beta = \frac{n\pi}{2}$, we obtain
$$ \cos(nx) \sin\left( \frac{n\pi}{2} \right) - \cos\left(\frac{n\pi}{2}\right) \sin(nx)
= \sin\left( -nx + \frac{n\pi}{2} \right)
= \sin\left(n\left(\frac{\pi}{2}-x\right)\right).$$
Your series then becomes
\begin{align}
&\frac{1}{4}+\sum_{n=1}^\infty \left[\left(\frac{1}{n\pi}\sin\left({n\frac{\pi}{2}}\right)\right)\cos(nx) + \left(\frac{-1}{n\pi}\cos\left( {n\frac{\pi}{2}}\right) - 1 \right)\sin(nx)\right] \\
&\qquad\qquad= \frac{1}{4}+\sum_{n=1}^\infty \left[ \frac{1}{n\pi} \left( \cos(nx)\sin\left({n\frac{\pi}{2}}\right) - \cos\left(\frac{n\pi}{2}\right) \sin(nx) \right) - \sin(nx) \right]\\
&\qquad\qquad= \frac{1}{4} + \sum_{n=1}^{\infty} \left[ \frac{1}{n\pi} \sin\left( n\left(\frac{\pi}{2}-x\right) \right) - \sin(nx) \right].
\end{align}
I'm not sure that this is any simpler, really, but it is correct (at least, assuming that your derivation tis correct) and it eats up slightly less ink.
On the other hand, I am slightly worried about your derivation. Are you sure that you completed the Fourier coefficients correctly? When I plot the 50th partial sum, I get the following:
Your function is piecewise continuous, hence the Fourier series converges pointwise, and should converge to $\chi_{[0,\frac{\pi}{2}]}$ (i.e. the characteristic function of the interval $[0,\frac{\pi}{2}]$, i.e. the "box" you initially described) almost everywhere. I don't see the sums converging nicely... Indeed, you have a bunch of terms that, taken together, look like $\sum \sin(nx)$. Since $\sin(nx)$ is 1 quite a lot, this is going to cause some major problems for you as $n$ goes to infinity. I would encourage you to check your work.
There is definitely an error in your derivation of the Fourier coefficients. For $n=0$ I get $c_0 = \frac{1}{4}$, and for $n \ne 0$ I get the (complex) Fourier coefficients
\begin{align}
c_n
&= \frac{1}{2\pi} \int_{0}^{2\pi} \chi_{[0,\frac{\pi}{2}]}(x) \mathrm{e}^{-inx}\,\mathrm{d}x \\
&= \frac{1}{2\pi} \int_{0}^{\frac{\pi}{2}} \mathrm{e}^{-inx}\,\mathrm{d}x \\
&= \frac{1}{2\pi in} \left( 1 - \mathrm{e}^{-\frac{in\pi}{2}} \right).
\end{align}
This then gives us
\begin{align}
f(x)
\sim \frac{1}{4} + \sum_{n=-\infty}^{\infty} \frac{1}{2\pi in} \left( 1 - \mathrm{e}^{-\frac{in\pi}{2}} \right) \mathrm{e}^{inx}
&= \frac{1}{4} + \sum_{n=-\infty}^{\infty} \frac{1}{2\pi in} \left( \mathrm{e}^{inx} - \mathrm{e}^{inx-\frac{in\pi}{2}} \right) \\
&= \frac{1}{4} + \sum_{n=1}^{\infty} \frac{1}{2\pi i n} \left( \left(\mathrm{e}^{inx} - \mathrm{e}^{-inx} \right) - \left( \mathrm{e}^{in(x-\frac{\pi}{2})} - \mathrm{e}^{in(x-\frac{\pi}{2})} \right) \right) \\
&= \frac{1}{4} + \sum_{n=1}^{\infty} \frac{1}{n\pi} \left( \sin(nx) - \sin\left( n\left(x-\frac{\pi}{2}\right) \right) \right).
\end{align}
Modulo the (expected) Gibbs phenomenon, I'd say this looks pretty good:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the limit (proof). (assignment) This is my assignment question.
Find $\lim_{n \to \infty} (2n^2+10n+5)/n^2$.
My attempt: suppose $\lim_{n \to \infty} (2n^2+10n+5)/n^2=2$
Then, $|\frac {2n^2+10n+5}{n^2}-2|= \frac {10n+5}{n^2}\le \frac {15n}{n^2}= \frac {15}{n}$.
Then, let $M \in N$ and $\frac {15}M< \varepsilon$. Then, for $\varepsilon >0$, there exists $M$ such that $|\frac {2n^2+10n+5}{n^2}-2|= \frac {10n+5}{n^2}\le \frac {15n}{n^2}= \frac {15}{n}\le\frac {15}{M}<\varepsilon$ for $n\ge M$.
My question is (1) Can I say "suppose $\lim_{n \to \infty} (2n^2+10n+5)/n^2=2$" before finding the limit ?
(2)$\frac {10n+5}{n^2}\le \frac {15n}{n^2}$. I am also not sure whether it is true.
Thank you in advance.
| You could just do $$\lim_{n \to \infty}\frac{2n^2+10n+5}{n^2}=\lim_{n\to\infty}\frac{2+\frac{10}n+\frac5{n^2}}{1}=\frac{2+0+0}{1}=2$$ since both $\dfrac1n$ and $\dfrac1{n^2}$ tend to $0$ as $n\to\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find derivative of $y=\sin^{-1}\frac{2x}{1+x^2}$
Find $\frac{dy}{dx}$ if $y=\sin^{-1}\frac{2x}{1+x^2}$
The solution is given as $\frac{2}{1+x^2}$. But is it a complete solution ?
My Attempt
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2},\quad |x|\leq 1\\
\pi-\sin^{-1}\frac{2x}{1+x^2},\quad |x|>1 \;\&\; x>0\\
-\pi-\sin^{-1}\frac{2x}{1+x^2},\quad |x|>1 \;\&\;x>0\\
\end{cases}\\
\sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x,\quad |x|\leq 1\\
\pi-2\tan^{-1}x,\quad |x|>1 \;\&\; x>0\\
-\pi-2\tan^{-1}x,\quad |x|>1 \;\&\;x>0\\
\end{cases}\\
$$
Thus,
$$
\frac{dy}{dx}=\frac{d}{dx}\bigg[\sin^{-1}\frac{2x}{1+x^2}\bigg]=\begin{cases}
\frac{d}{dx}[2\tan^{-1}x]=\frac{2}{1+x^2},\quad |x|\leq 1\\
\frac{d}{dx}[\pm\pi-2\tan^{-1}x]=\frac{-2}{1+x^2},|x|>1
\end{cases}
$$
Is it correct ?
| Hint
Your answer seems correct to me even I don't really understand the need of tan function
$$y=\sin^{-1}\frac{2x}{1+x^2} \implies \sin y=\frac{2x}{1+x^2}$$
Differentiate
$$y'\cos(y)=\frac {-2(x^2-1)}{(x^2+1)^2}$$
use the fact that $$\sin^2(y)+\cos^2(y)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Alternative for calculating the nth of quadratic sequence Given the quadratic sequence
$$f(n)=1, 7, 19, 37, \cdots$$
To calculate the $f(n)$ for $n\ge1$. $$f(n)=an^2+bn+c$$
We start with the general quadratic function, then sub in for $n:=1,2$ and $3$
$$f(1)=a+b+c$$
$$f(2)=4a+2b+c$$
$$f(3)=9a+3b+c$$
Now solve the simultaneous equations
$$a+b+c=1\tag1$$
$$4a+2b+c=7\tag2$$
$$9a+3b+c=19\tag3$$
$(2)-(1)$ and $(3)-(2)$
$$3a+b=6\tag4$$
$$5a+b=12\tag5$$
$(5)-(4)$
$$a=3$$
$$b=-3$$
$$c=1$$
$$f(n)=3n^2-3n+1$$
This method is very long. Is there another easy of calculating the $f(n)$?
| Yes!
You know $f(1)= 1 \implies f(1) - 1 = 0 \implies f(x) - 1$ has a root at x=1
Now,
$f(x) - 1 = a(x-1)(x-b)$
Put $x = 2$, $a(2-b) = 6$
Put x = 3, $a*2*(3-b) = 18$
Divide both equation, we get $b = 0 $ and a = 3
$f(x) - 1 = 3*(x-1)x \implies f(x) = 3x^2 - 3x + 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\lim _{t\to \infty}\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}$ I am having difficulties evaluating this limit:
$$\lim _{t\to \infty \:}\left(\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}\right)$$
I have tried to divide out by $\frac{\sqrt{t}}{\sqrt{t+1}}$ in the numerator and denominator but I run into problems and I also tried to divide through with $\sqrt{t}$ but I still get 0/0.
I've been stumped for hours and need a heads up on this.
| What I interpret Dr. Graubner's directions to mean.
\begin{align*}
\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}
&= \frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}} \cdot \frac{\sqrt{t+1}\sqrt{t+2}}{\sqrt{t+1}\sqrt{t+2}} \\
&= \frac{\sqrt{t+1}-\sqrt{t} }{2 \sqrt{t+2}-\sqrt{4t+1}}
\cdot \frac{\sqrt{t+2}}{\sqrt{t+1}} \\
&= \frac{\sqrt{t+1}-\sqrt{t} }{2 \sqrt{t+2}-\sqrt{4t+1}}\cdot \frac{2 \sqrt{t+2}+\sqrt{4t+1}}{2 \sqrt{t+2}+\sqrt{4t+1}} \cdot \frac{\sqrt{t+1}+\sqrt{t}}{\sqrt{t+1}+\sqrt{t}}
\cdot \frac{\sqrt{t+2}}{\sqrt{t+1}} \\
&= \frac{ t+1-t}{4 (t+2)-(4t+1)} \cdot \frac{2 \sqrt{t+2}+\sqrt{4t+1}}{\sqrt{t+1} + \sqrt{t}} \cdot \frac{\sqrt{t+2}}{\sqrt{t+1}} \\
&= \frac{1}{7} \cdot \frac{2 \sqrt{t+2}+\sqrt{4t+1}}{\sqrt{t+1} + \sqrt{t}} \cdot \frac{\sqrt{t+2}}{\sqrt{t+1}} \\
&= \frac{1}{7} \cdot \frac{\sqrt{t}(2\sqrt{1+2/t}+\sqrt{4+1/t})}{\sqrt{t}(\sqrt{1+1/t} + 1)} \cdot \frac{\sqrt{t} \sqrt{1+2/t}}{\sqrt{t}\sqrt{1+1/t}} \\
&= \frac{1}{7} \cdot \frac{2\sqrt{1+2/t}+\sqrt{4+1/t}}{\sqrt{1+1/t} + 1} \cdot \frac{\sqrt{1+2/t}}{\sqrt{1+1/t}} \\
&\rightarrow \frac{1}{7} \frac{2\sqrt{1}+\sqrt{4}}{\sqrt{1} + 1} \cdot \frac{\sqrt{1}}{\sqrt{1}} \\
&= \frac{2}{7} \text{.}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove that the general term of this multinomial expansion is as given
If $(6-12x+12x^2)^n = \displaystyle \sum_{r=0}^{2n} T_r x^r$, prove $ T_r = (-2)^r 3^n \bigg[\binom{2n}{r} + \binom{2n-2}{r}\binom{n}{1} + \binom{2n-4}{r}\binom{n}{2} + \dots \bigg]$
I tried by writing $$(6−12x+12x^2)^n= 6^n(1-2x+2x^2)^n = 6^n\displaystyle \sum_{a,b,c}^{a+b+c = n} \frac{n!}{a!b!c!}(-2x)^b(2x^2)^c = 6^n \sum_{a,b,c}^{a+b+c = n} \frac{n!}{a!b!c!}(-1)^b 2^{b+c} x^{b+2c}$$ which gives two conditions, $a+b+c = n$ and $b+2c =r$
So the (a,b,c)'s possible here are (n-r,r,0) ; (n-r+1,r-2,1) ; (n-r+2,r-4,2) ... (0,0,n)
So substituting those a,b,c's I get the coefficient of $x^r$ as $$6^n \bigg[\frac{n!}{(n-r)!r!0!}(-1)^r 2^{r} + \frac{n!}{(n-r+1)!(r-2)!1!}(-1)^{r-2} 2^{r-1} + \frac{n!}{(n-r+2)!(r-4)!2!}(-1)^{r-4} 2^{r-2} + \cdots \bigg] =$$
$$\displaystyle 6^n(-1)^r2^{r} \bigg[\frac{n!}{(n-r)!r!0!} + \frac{n!}{(n-r+1)!(r-2)!1!} 2^{-1} + \frac{n!}{(n-r+2)!(r-4)!2!} 2^{-2} + \cdots \bigg]$$
$ = \displaystyle (-2)^r 3^n \bigg[\frac{n!}{(n-r)!r!0!}2^n + \frac{n!}{(n-r+1)!(r-2)!1!} 2^{n-1} + \frac{n!}{(n-r+2)!(r-4)!2!} 2^{n-2} + \cdots \bigg]$
I guess I have to prove that $$\displaystyle \bigg[\frac{n!}{(n-r)!r!0!}2^n + \frac{n!}{(n-r+1)!(r-2)!1!} 2^{n-1} + \frac{n!}{(n-r+2)!(r-4)!2!} 2^{n-2} + \cdots \bigg] = \bigg[\binom{2n}{r} + \binom{2n-2}{r}\binom{n}{1} + \binom{2n-4}{r}\binom{n}{2} + \dots \bigg] $$
now, which I'm not able to..
WolframAlpha doesn't help with that last part. ..or is there a better method to do the whole question?
| Working with the innermost sum we have for
$$\sum_{q=0}^n {2n-2q\choose r} {n\choose q}$$
that it is
$$\sum_{q=0}^n [w^r] (1+w)^{2n-2q} [z^{n-q}] (1+z)^n
\\ = [w^r] (1+w)^{2n} \sum_{q=0}^n (1+w)^{-2q}
[z^{n}] z^q (1+z)^n.$$
Now we may extend $q$ to infinity as there is no contribution
to the coefficient extractor $[z^n]$ when $q\gt n,$ getting
$$[w^r] (1+w)^{2n} [z^n] (1+z)^n \sum_{q\ge 0} (1+w)^{-2q} z^q
\\ = [w^r] (1+w)^{2n} [z^n] (1+z)^n \frac{1}{1-z/(1+w)^2}
\\ = [w^r] (1+w)^{2n+2} [z^n] (1+z)^n \frac{1}{1+2w-z+w^2}.$$
Observe that
$$\sum_{r=0}^{2n} T_r x^r = \sum_{r\ge 0} T_r x^r$$
because we may extend $r$ beyond $2n$ with the coefficient extractor
$[w^r]$ returning zero for all $q$ in this case (first line). We
obtain
$$\sum_{r\ge 0} x^r
(-2)^r 3^n [w^r] (1+w)^{2n+2} [z^n] (1+z)^n
\frac{1}{1+2w-z+w^2}
\\ = 3^n [z^n] (1+z)^n
\sum_{r\ge 0} x^r
(-2)^r [w^r] (1+w)^{2n+2}
\frac{1}{1+2w-z+w^2}
\\ = 3^n [z^n] (1+z)^n
(1-2x)^{2n+2}
\frac{1}{1-4x-z+4x^2}
\\ = 3^n (1-2x)^{2n+2}
[z^n] (1+z)^n \frac{1}{1-4x-z+4x^2}.$$
This is
$$\frac{3^n (1-2x)^{2n+2}}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
(1+z)^n \frac{1}{1-4x-z+4x^2} \; dz.$$
Introducing $z/(1+z) = v$ we get $z = v/(1-v)$ and $dz = 1/(1-v)^2 \; dv$
and hence
$$\frac{3^n (1-2x)^{2n+2}}{2\pi i}
\int_{|v|=\gamma} \frac{1}{v^n} \frac{1-v}{v}
\frac{1}{1-4x-v/(1-v)+4x^2} \; \frac{1}{(1-v)^2} \; dv
\\ = \frac{3^n (1-2x)^{2n+2}}{2\pi i}
\int_{|v|=\gamma} \frac{1}{v^{n+1}}
\frac{1}{(1-4x)(1-v) -v +4x^2 (1-v)} \; dv
\\ = \frac{3^n (1-2x)^{2n+2}}{2\pi i}
\int_{|v|=\gamma} \frac{1}{v^{n+1}}
\frac{1}{(1-4x+4x^2) - v (2-4x+4x^2 ) } \; dv.$$
With $x$ a formal parameter this evaluates to
$$3^n (1-2x)^{2n+2} \frac{1}{1-4x+4x^2}
\frac{(2-4x+4x^2)^n}{(1-4x+4x^2)^n}
\\ = 3^n (1-2x)^{2n+2} \frac{1}{(1-2x)^2}
\frac{(2-4x+4x^2)^n}{(1-2x)^{2n}}
\\ = (6-12x+12x^2)^n.$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trace of matrices with primes as elements I noticed this curious pattern that matrices composed of primes seem to have traces equal to or close to square numbers:
\begin{align}tr\begin{bmatrix}2&3\\5&7\end{bmatrix}=9=3^2\end{align}
\begin{align}tr\begin{bmatrix}2&3&5\\7&11&13\\17&19&23\end{bmatrix}=36=6^2\end{align}
\begin{align}tr\begin{bmatrix}2&3&5&7\\11&13&17&19\\23&29&31&37\\41&43&47&53\end{bmatrix}=99=10^2-1\end{align}
\begin{align}tr\begin{bmatrix}2&3&5&7&11\\13&17&19&23&29\\31&37&41&43&47\\53&59&61&67&71\\73&79&83&89&97\end{bmatrix}=224=15^2-1\end{align}
This can be generally written as:
$$\sum_{i=0}^{n-1} {p_{1+i(n+1)}}\leq (T_n)^2$$
Where n is the dimension of the matrix, $n\geq2$, p are the prime numbers, and $T_n$ are the triangular numbers.
Can someone please explain why this pattern exists?
| I think this pattern exist, because these matrices are relatively small. If you take for example matrix of dimension $n=7$, then sum is equal to $724$, which is $3$ away from $27^2$.
Interesting thing happens when for $n=10$ it gives diagonal value of: $\sum=2,451$ which is in "the middle" of $49^2=2401$ and $50^2=2500$, however the pattern is not regular as the distance grow in size.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric series sum involving tangents
$$\frac{1}{4}\tan\bigg(\frac{\pi}{8}\bigg)+\frac{1}{8}\tan\bigg(\frac{\pi}{16}\bigg)+\frac{1}{16}\tan\bigg(\frac{\pi}{32}\bigg)+\cdots\cdots \infty$$
Try: $$\cos x\cdot \cos(x/2)\cdot\cos(x/2^2)\cdots\cdots \cos(x/2^{n-1})=\frac{1}{2^{n-1}}\frac{\cos(x)}{\sin(x/2^{n-1})}$$
Could some help me how to solve ahead , thanks
| Using the identity:
$$
\cot x-\tan x = 2\cot 2x \Rightarrow \tan x=\cot x -2\cot 2x,
$$
one obtains a telescoping sum:
$$
S(x):=\sum_{n=0}^\infty \frac{\tan(x/2^n)}{2^n}=\sum_{n=0}^\infty\left( \frac{\cot(x/2^n)}{2^n}-\frac{\cot(x/2^{n-1})}{2^{n-1}}\right)\\=
-2\cot(2x)+\lim_{n\rightarrow\infty}\frac{\cot(x/2^n)}{2^n}=\frac{1}{x}-2\cot (2x).
$$
Applying this to your case one obtains
$$\frac{1}{4}S\left(\frac{\pi}{8}\right)=\frac{2}{\pi}-\frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to simplify arc length problems: expressions under the root I am having trouble during my studies of arc length by integration in Calculus.
Arc Length formula: $\int\sqrt{1+('y)^2}dx$
My problem values:
$y=\frac{1}{10}x^5+\frac{1}{6}x^-3$
$y'=\frac{5}{10}x^5+(-\frac{3}{6}x^{-4})$ = $\frac{1}{2}x^4-\frac{1}{2}x^-4$ = $\frac{1}{2}(x^4-x^{-4})$
$1+('y)^2=1+(\frac{1}{2}(x^4-x^{-4})^2$
My question is: How does $1+(\frac{1}{2}(x^4-x^{-4})^2$ become $(\frac{1}{2}(x^4+x^{-4})^2$ as seen in the solution?
Is this an obvious algebra step that I am missing? I understand that the soln completes the square, does addition, and achieves sign change, but I don't understand the motive or method. The steps in question can be seen in the linked solution at the bottom of this post. Thank you.
Link to solution
| $$\begin {align}1+\frac12\left(x^4-x^{-4}\right)^2&=1+\frac12\left(x^8-2\cdot x^4 \cdot x^{-4}+x^{-8}\right)\\&=1+\frac12(x^8-2+x^{-8})\\
&=-1+\frac12(x^8+2+x^{-8})\\
&=-1+\frac12\left(x^4+x^{-4}\right)^2\end {align}$$
Your expression is missing the $-1$ out front.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve: $5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$ Solve to find the general value of $x$:
$$5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$$
My Attempt:
$$5(1-\sin^2 (x))-4\sin (x)\cos (x)+3\sin^2 (x)=2$$
$$5-5\sin^2 (x)-4\sin (x)\cos (x)+3\sin^2 (x)=2$$
$$2\sin^2 (x)+4\sin (x)\cos (x)=3$$
| You can divide by $\cos^2 x \quad $ (Since $x=\pi/2$ does not work)
$$\implies 5-4 \tan x+3 \tan^2 x=2 \sec^2 x$$
yet $\sec^2 x=1+\tan^2 x .....$
Continuing you have a quadratic in $\tan x$ which I'm sure you know how to solve...
$$\implies z^2-4z+3=(z-3)(z-1)=0 \quad |\quad z=\tan x $$
So for any $n \in \mathbb{Z}$
$$x=\arctan 1 +n\pi=\frac{\pi(1+4n)}{4}$$
$$\land$$
$$x=\arctan 3 +n\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\sum_{k=1}^\infty \frac{k-1}{2^{k+1}}$ as a telescoping series? Okay, so my book shows (with absolutely no explanation) an equivalency to
$\lim\limits_{x \to \infty} (\frac{1}{2} - \frac{x+2}{2^x})$
, and I have no idea where this came from. From my understanding, telescoping series generally telescope down as a result of combining positives and negatives and may lead to a result of the first term minus the last term in the series ($\infty$ in such as case). I do see a pattern in the sequence once I split into
$\frac{k}{2^{k+1}}$ and $\frac{-1}{2^{k+1}}$ but I'm not sure of there being any telescoping in the way the lesson examples show.
-Can someone tell me how $\lim\limits_{x \to \infty} (\frac{1}{2} - \frac{x+2}{2^x})$ was obtained, since I have no explanation in the chapter or solution?
| Apparently there was a typo somewhere: “$\frac{x + 2}{2^x}$” should have being replaced by “$\frac{x}{2^x}$”. That being said, let $s_x \mathop{:=} 1/2 - x \mathbin/ 2^x$ for $x \in \mathbf{N}^*$. Observe that
$$
s_{x + 1} - s_x = \frac{1}{2} - \frac{x + 1}{2^{x + 1}} - \biggl(\frac{1}{2} - \frac{x}{2^x}\biggr) = \frac{x}{2^x} - \frac{x + 1}{2^{x + 1}} = \frac{2 x}{2 \cdot 2^x} - \frac{x + 1}{2 \cdot 2^x} = \frac{2x - (x + 1)}{2^{x + 1}}
\text,$$
which is finally $(x - 1) \mathbin/ 2^{x + 1}$. Let us denote that value $u_x$; then the initial problem is to compute
$$
\sum_{k = 1}^{\infty} u_k = \sum_{k = 1}^{\infty} (-s_k + s_{k + 1})
\text.$$
If we consider the truncated sum for $1 \leq k \leq x - 1$, we get
$$
(-s_1 + s_2) + (-s_2 + s_3) + \cdots + (-s_{x - 2} + s_{x - 1}) + (-s_{x - 1} + s_x)
\text,$$
where almost all the terms “telescope” by appearing once positively and once negatively, finally yielding
$$
\sum_{k = 1}^x u_k = -s_1 + s_x
\text.$$
As $s_1 = 0$, this further simplifies into “$s_x$”; so that finally, letting $x$ tend to infinity, the sum of the infinite series is the limit of $s_x$ when $x \to \infty$, namely $1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How many integers in the range satisfy a given congruence? Specifically how many integers $x$ in the range $1 ≤ x ≤ 60$ satisfy $21x ≡ 24\ (mod\ 60)$? And how many integers $a$ are there in the range $1 ≤ a ≤ 74$ for which the equation $x^2 ≡ a\ (mod\ 37)$ is soluble?
I have the solutions to be $3$ and $38$ respectively but not too sure how they came up with those answers.
For the first one, I believe it's because $21x ≡ 24\ (mod\ 60) \implies 3\cdot7x \equiv 3\cdot8\ (mod\ 60) \implies 7\cdot8x\ (mod\ 20) \implies x \equiv 4\ (mod\ 20)$
Then looking in the range from $1\ to\ 60$, $60/20 = 3$ to give us 3 integers.
Not sure about the second..
| \begin{align}
21x &\equiv 24 \pmod{60} \\
7x &\equiv 8 \pmod{20} \\
x &\equiv 4 \pmod{20} \\
x &\in \{4, 24, 44 \}
\end{align}
TOTAL = 3
\begin{array}{|c|c|}
\hline
x & x^2 \pmod{37}\\
\hline
0,37 & 0 \\
1,36 & 1 \\
2,35 & 4 \\
3,34 & 9 \\
4,33 & 16 \\
5,32 & 25 \\
6,31 & 36 \\
7,30 & 12 \\
8,29 & 27 \\
9,28 & 7 \\
10,27 & 26 \\
11,26 & 10 \\
12,25 & 33 \\
13,24 & 21 \\
14,23 & 11 \\
15,22 & 3 \\
16,21 & 34 \\
17,20 & 30 \\
18,19 & 28 \\
\hline
\end{array}
TOTAL = $2 \times 19 = 38$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proof by Induction $\frac{n+1}{2n} = \left(1-\frac{1}{2^2}\right)\dotsc \left(1-\frac{1}{n^2}\right)$ proof by induction. I'm pretty close to done with this proof but got stuck at the end
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{n^2}\right) = \frac {n+1}{2n}$$
for all integers $n \geq 2$
*
*basis step
$$1-\frac{1}{2^2} = .75 \qquad \frac{2+1}{2\times2} = 3/4= .75$$
*set $n = k$ assumed to be true
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) = \frac {k+1}{2k}$$
*set $n = k+1$
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) \left(1-\frac{1}{(k+1)^2}\right) = \frac{(k+1)+1}{2k+1}$$
from step 2 we know that
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) = \frac {k+1}{2k}$$
so we are going to plug that in
$$\frac {k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right) = \frac{(k+1)+1}{2k+1} $$
now to get both sides equal we multiply by $2k$ to get common denominators
$$\frac{k+1}{2k} \times {\left(1-{1\over(k+1)^2}\right) \times 2k \over 2k}$$ This is the left hand side
now to combine it
$${\left((k+1) \left(1- {1\over(k+1)^2}\right) \times 2k \right)\over 2k}$$ - this is where I got stuck...
I wanted to cancel the $2k$'s on the top and bottom but then there would be no way to get it to equal the $(k+1)+1\over2(k+1)$ because the denominator required a $2k$. Any help is appreciated thank you.
| You are close.
You want to show that
$\frac {k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right)
= \frac{(k+1)+1}{2k+2}
$.
Note that the denominator on the right is
$2(k+1) = 2k+2$,
not the $2k+1$ you have.
That is why you were unable
to complete the proof -
you were trying to prove
something that was false.
$\begin{array}\\
\frac {k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right)
&=\frac {k+1}{2k} \times \left(\frac{(k+1)^2-1}{(k+1)^2}\right)\\
&=\frac {1}{2k} \times \left(\frac{k^2+2k+1-1}{k+1}\right)\\
&=\frac {1}{2k} \times \left(\frac{k^2+2k}{k+1}\right)\\
&=\frac {1}{2k} \times \left(\frac{k(k+2)}{k+1}\right)\\
&=\frac{k+2}{2(k+1)}\\
\end{array}
$
which is what you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Explaining $\int_{-1}^1\frac{1}{1+x^2}\,dx = \frac{\pi}{2}$. How would you explain to a student that
$$ \int_{-1}^1\frac{1}{1+x^2}\,dx = \arctan(1) - \arctan(-1) = \frac{\pi}{2} $$
and not
$$ \int_{-1}^1\frac{1}{1+x^2}\,dx = \arctan(1) - \arctan(-1) \neq \frac{\pi}{4} - \frac{3\pi}{4} = -\frac{\pi}{2}$$
besides the obvious fact that $\arctan x$ cannot map to two distinct values?
| We are actually using the substitution $x=\tan\theta$.
$$ \int_{-1}^1\frac{1}{1+x^2}dx=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \left(\frac{1}{1+\tan^2\theta}\right)(\sec^2\theta)d\theta=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} d\theta=\frac{\pi}{2}$$
We take the $\displaystyle \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ as $\tan\theta$ is differentiable in the range.
If we want to take $\displaystyle \theta=\frac{3\pi}{4}$, the interval cannot be $\displaystyle \left(\frac{\pi}{4},\frac{3\pi}{4}\right)$ as $\displaystyle \tan\frac{\pi}{2}$ does not exist.
We have to break down the interval as $\displaystyle \left(\frac{3\pi}{4},2\pi\right)$ and $\displaystyle \left(0,\frac{\pi}{4}\right)$ to avoid the trouble.
\begin{align*}
\int_{-1}^1\frac{1}{1+x^2}dx&=\int_{-1}^0\frac{1}{1+x^2}dx+\int_0^1\frac{1}{1+x^2}dx\\
&=\int_{\frac{3\pi}{4}}^{2\pi}d\theta+\int_0^{\frac{\pi}{4}}d\theta\\
&=\frac{\pi}{4}+\frac{\pi}{4}\\
&=\frac{\pi}{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Number theory and set theory problem
Let $A$ be a finite set. For $0\leq i\leq 2$, let $a_i$ be the number of subsets $B$ of $A$ such that
$$|B|\equiv i \pmod{3}.$$
Prove that $$|a_i - a_j| \leq 1,$$ for all $0\leq i,j\leq 2$.
Firstly I assumed that $|A|=n$. Then $a_0 = {^nC_0} + {^nC_3} + {^nC_6} +\cdots$ and $a_1 = {^nC_1}+ {^nC_4}+{^nC_7}+\cdots$ and $a_2$ in a similar manner but I am unable to show the main part.
Could anyone give a hint!!
| Let $a_0(n),a_1(n),a_2(n)$ be given by
$$
\begin{cases}
a_0(n) = \binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \cdots\\[4pt]
a_1(n) = \binom{n}{1} + \binom{n}{4} + \binom{n}{7} + \cdots\\[4pt]
a_2(n) = \binom{n}{2} + \binom{n}{5} + \binom{n}{8} + \cdots\\
\end{cases}
$$
We want to show that for all nonnegative integers $n$, we have $|a_i(n)-a_j(n)|\le 1$, for $0 \le i < j \le 2$.
Proceed by induction on $n$.
By direct evaluation, we have
$$
\begin{cases}
a_0(0) = 1\\[4pt]
a_1(0) = 0\\[4pt]
a_2(0) = 0\\
\end{cases}
$$
so the claim holds for the base case, $n=0$.
Next, suppose the claim holds for some nonnegative integer $n$.
Then we get
\begin{align*}
a_0(n) &= \;\;\binom{n}{0}\;\;+\;\;\binom{n}{3}\;\;+\;\;\binom{n}{6}\;\;+\;\;\cdots\\[4pt]
a_1(n) &= \;\;\binom{n}{1}\;\;+\;\;\binom{n}{4}\;\;+\;\;\binom{n}{7}\;\;+\;\;\cdots\\[4pt]
\implies\;a_0(n) + a_1(n) &= {\small{\binom{n+1}{1}}}+{\small{\binom{n+1}{4}}}+ {\small{\binom{n+1}{7}}}+\;\;\cdots\\[4pt]
\implies\;a_0(n)+a_1(n) &= a_1(n+1)\\[30pt]
a_1(n) &= \;\;\binom{n}{1}\;\;+\;\;\binom{n}{4}\;\;+\;\;\binom{n}{7}\;\;+\;\;\cdots\\[4pt]
a_2(n) &= \;\;\binom{n}{2}\;\;+\;\;\binom{n}{5}\;\;+\;\;\binom{n}{8}\;\;+\;\;\cdots\\[4pt]
\implies\;a_1(n) + a_2(n) &= {\small{\binom{n+1}{2}}}+{\small{\binom{n+1}{5}}}+ {\small{\binom{n+1}{8}}}+\;\;\cdots\\[4pt]
\implies\;a_1(n)+a_2(n) &= a_2(n+1)\\[30pt]
a_2(n) &= \;\;{\phantom{\binom{n}{0}\;\;+\;\;}}\binom{n}{2}\;\;+\;\;\binom{n}{5}\;\;+\;\;\cdots\\[4pt]
a_0(n) &= \;\;\binom{n}{0}\;\;+\;\;\binom{n}{3}\;\;+\;\;\binom{n}{6}\;\;+\;\;\cdots\\[4pt]
\implies\;a_2(n) + a_0(n) &= {\small{\binom{n+1}{0}}}+{\small{\binom{n+1}{3}}}+ {\small{\binom{n+1}{6}}}+\;\;\cdots\\[4pt]
\implies\;a_2(n)+a_0(n) &= a_0(n+1)\\[4pt]
\end{align*}
Thus, we have
\begin{align*}
a_0(n)+a_1(n) &=\; a_1(n+1)
\qquad\qquad\qquad\qquad
\\[4pt]
a_1(n)+a_2(n) &=\; a_2(n+1)\\[4pt]
a_2(n)+a_0(n) &=\; a_0(n+1)\\[4pt]
\end{align*}
hence, subtracting the above equations in pairs, we get
\begin{align*}
a_2(n+1)-a_1(n+1) &=\;a_2(n)-a_0(n)
\qquad\qquad\qquad\qquad\qquad
\\[4pt]
a_0(n+1)-a_2(n+1) &=\;a_0(n)-a_1(n)\\[4pt]
a_1(n+1)-a_0(n+1) &=\;a_1(n)-a_2(n)\\[4pt]
\end{align*}
Then applying the inductive hypothesis, it follows that the claim holds for the case $n+1$.
This completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\sin(z) =\frac{e^{iz}-e^{-iz}}{2i} =\sin(x)\cosh(y) +i\cos(x)\sinh(y)$ For complex $z=x+iy$.
How to prove that
$$\sin(z) =\frac{e^{iz}-e^{-iz}}{2i} =\sin(x)\cosh(y) +i\cos(x)\sinh(y)$$
by using the power series definition
$$\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots$$
and Euler's formula $$e^z =e^x (\cos (y) +i\sin(y))?$$
| Is it really necessary to use the power series definition of sine? The left equality is easier to prove right-to-left with the exp. power expansion.
\begin{align}
\sin z &=\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!} z^{2n+1} =\sum_{n=0}^\infty \frac{i^{2n}}{(2n+1)!} z^{2n+1}\\
&=\frac{1}{i} \sum_{n=0}^\infty \frac{i^{2n+1}}{(2n+1)!} z^{2n+1} =\frac{1}{i} \sum_{n=0}^\infty \frac{(iz)^{2n+1}}{(2n+1)!}.
\end{align}
Consider $\dfrac{1-(-1)^k}{2}$ which equals $1$ for odd $k$ and $0$ for even $k$.
\begin{align}
\sin z &=\frac{1}{i} \sum_{k=0}^\infty \left(\frac{1-(-1)^k}{2}\right) \frac{(iz)^k}{k!}
=\frac{1}{2i} \sum_{k=0}^\infty \frac{1-(-1)^k}{k!} (iz)^k\\
&=\frac{1}{2i} \sum_{k=0}^\infty \frac{(iz)^k}{k!} -\frac{1}{2i} \sum_{k=0}^\infty \frac{(-iz)^k}{k!} \\
&=\frac{e^{zi} -e^{-zi}}{2i}\\
&=\frac{e^{(x+yi)i} -e^{-(x+yi)i}}{2i}\\
&=\frac{e^{-y+xi} -e^{y-xi}}{2i}\\
&=\frac{e^{-y} e^{xi} -e^{y} e^{-xi}}{2i}\\
&=\frac{e^{-y} (\cos x +i\sin x) -e^{y} (\cos x -i\sin x)}{2i}\\
&=\frac{i}{i} \cdot \sin x \cdot \frac{e^{y} +e^{-y}}{2} -\frac{1}{i} \cdot \cos x \cdot \frac{e^{y} -e^{-y}}{2}\\
&=\sin x \cosh y +i \cos x \sinh y.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the limit of sequence $\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}$ without using of derivatives and etc. I need to find limit of sequence
$$
\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}\right)
$$
I tried to solve it and stopped here
$$
f(n+1) = \frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}+\frac{2n+1}{2^{n+1}}
$$
$$
2f(n+1) = 1+\frac{3}{2}+\frac{5}{2^{2}}+\cdots+\frac{2n-1}{2^{n-1}}+\frac{2n+1}{2^{n}}
$$
$$
2f(n+1) -f(n) = 1+ \left(1 + \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\right) = 1 + g(n)
$$
I can find the limit of $g$, but what to do with the other parts?
| Use summation by parts. Given a sequence $f(n)$ define $(\Delta f)(n)=f(n+1)-f(n)$ and note that
$$
\sum_{n=a}^b(\Delta f)(n)=f(b+1)-f(a).
$$
Further given two sequences $f(n)$ and $g(n)$, note that
$$
(\Delta fg)(n)=f(n+1)\Delta g(n)+g(n)\Delta f(n)
$$
whence
$$
\sum_{n=a}^bg(n)\Delta f(n)=[f(b+1)g(b+1)-g(a)f(a)]-\sum_{n=a}^bf(n+1)\Delta g(n)\tag{1}
$$
Let $h(n)=-2/2^n$, and note that $\Delta h(n)=2^{-n}$. So by (1)
$$
\sum_{n=1}^k\frac{2n-1}{2^n}=[(2k+1)(-2^{-k})-1(-1)]+2\sum_{k=1}^n2^{-k}.\tag{2}
$$
Let $k\to \infty$ in (2) and use the formula for a geometric series to deduce that
$$
\sum_{n=1}^\infty\frac{2n-1}{2^n}=1+2=3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Possible values of $x+y+z$ Given that $xyz=11(x+y+z)$, find the possible values of $x+y+z$ such that $x\leq y$ satisifies $x^2+y^2=2018$
I know that $x,y$, or $z$ should be divisible by $11$. I am having trouble going to the next step
| I think the problem also states that $x,y,z$ must be integers.
Assume that $x$ is divisible by $11$, then $x=11k$ ($k$ is an integer) and $(11k)^2+y^2=2018$.
We will have $y^2=2018-121k^2$. Because $y^2\ge 0$, we can prove that $k^2$ must be in this set of numbers: $(1;4;9;16)$.
We have four cases:
*
*$k^2=1\Rightarrow y=43.55$, result eliminated
*$k^2=4\Rightarrow y=39.17$, result eliminated
*$k^2=9\Rightarrow y=30.48$, result eliminated
*$k^2=16\Rightarrow y=9.05$, result eliminated
So there are no cases available for "$x$ is divisible by $11$" that satisfy $x^2+y^2=2018$, this means "$y$ is divisible by $11$" cannot be true as well.
Because of this, someone should attempt it by solving the equation $x^2+y^2=2018$ for integers $x,y$ instead, you can make a table that consists of $45$ cases from $1$ to $45$ for $x$ to find $y$ (or fewer cases if you use the condition $x\le y$ to eliminate some cases). This method is really tedious, but you should get $x=13$ and $y=43$, then you can solve for $z$.
However, $z=1.12$ which means $z$ is not an integer, this means there are no three integers $x,y,z$ satisfy all the conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Polar form from a complex number? How do I solve for the polar form of $-3\sqrt{2} - 3\sqrt{3}\,i\; ?$
I think I solved for $r$ which is $3\sqrt{5}$ from using $r=\sqrt{a^2+b^2}.$
When I look for theta I use $\tan\theta = \frac ab,$ but when doing so I get $\frac{\sqrt{6}}{ 2}.$ Anyway I looked in the answer key for $\theta$ and it's in degrees. Around $230.8^{\circ}$. How do I figure that out from $\frac{\sqrt{6}}{ 2}?$ Did I even do it right ?
| Same way always:
$z = -3\sqrt{2} + 3\sqrt 3 i = re^{i\theta}$ where $r = |-3\sqrt{2} + 3\sqrt 3 i| = \sqrt{(-3\sqrt{2})^2 + (3\sqrt 3)^2} = \sqrt{9*2 + 9*3}=\sqrt {45} = 3\sqrt 5$.
And $\theta$ is so that $\cos \theta = \frac {-3\sqrt{2}}r=\frac {-3\sqrt{2}}{3\sqrt 5}=\frac {-\sqrt 2}{\sqrt 5}; \sin\theta = \frac {3\sqrt{3}}r= \frac {3\sqrt{3}}{3\sqrt 5}=\frac {\sqrt 3}{\sqrt 5}; \tan \theta = \frac {3\sqrt{3}}{-3\sqrt {2}}= -\frac{\sqrt{3}}{\sqrt{2}}$.
As $\cos \theta <0$ and $\sin \theta > 0$ we know $\frac \pi 2 < \theta < \pi$.
$\theta = \arctan \frac {-\sqrt 2}{\sqrt 5} = 2.45687$
So $z = 3\sqrt 5e^{2.45687i}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the remainder when $ 6^{65} $ is divided by 80? In the solutions it used Chinese remainder theorem in the following way: $6^{65}\equiv 1 \pmod 5 $ and $ 6^{65} \equiv 0 \pmod{16} $. I can see how they got the first congruence but am really struggling to see it for the second one. Any help would be much appreciated!
| Note:
$$\begin{align} 6 &\equiv 1\pmod 5 \\
6^{61} &\equiv 1 \pmod 5 \\
6^{61}\cdot 2^4 &\equiv 1\cdot 2^4 \pmod {5\cdot 2^4}\\
3^4 &\equiv 1\pmod{80}\\
6^{61}\cdot 2^4\cdot 3^4 &\equiv 1\cdot 2^4\cdot 1 \pmod{80}\\
6^{65} &\equiv 16 \pmod{80}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
The square of the solution of the equation... The square of the solution of the equation
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
is equal to: ...
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
$\implies x\sqrt{7} = \sqrt{8 + 3\sqrt{7}} - \sqrt{8 - 3\sqrt{7}}$
$\implies 7x^2 = \left(8 + 3\sqrt{7}\right) + \left(8 - 3\sqrt{7}\right) - \underline{2\sqrt{\left(8+3\sqrt{7}\right)\cdot\left(8-3\sqrt{7}\right)}}$
$\implies 7x^2 = 16 - 2\cdot\sqrt{64-63} = 14$
$\implies x^2 = 2$
I know how to do all the steps up to the part where the underlined section comes into play, can someone please explain where does this come from. Thanks!
| Alt. hint: show first that $\;\sqrt{8 \pm 3\sqrt{7}} = \dfrac{1}{\sqrt{2}}\left(3 \pm \sqrt{7}\right)\,$, then no squaring is needed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What is $x$, if $\cot ^{-1} \left(3x+\frac{2}{x}\right)+\cot ^{-1} \left(6x+\frac{2}{x}\right)+\cot ^{-1} \left(10x+\frac{2}{x}\right)+\cdots = 1$?
Let
$$S_{n}=\cot ^{-1} \left(3x+\frac{2}{x}\right)+\cot ^{-1} \left(6x+\frac{2}{x}\right)+\cot ^{-1} \left(10x+\frac{2}{x}\right)+\cdots \quad\text{($n$ terms)}$$
where $x>0$. If $\lim _{n \to \infty} S_{n}=1$, then find the value of $x$.
Can the given series be converted to telescopic series? I converted in into $\tan^{-1}$ but in every term $x$ is in numerator? Could someone please given some hint?
| We have
\begin{eqnarray*}
\cot^{-1}(A)-\cot^{-1}(B)=\cot^{-1}\left(\frac{AB+1}{B-A} \right).
\end{eqnarray*}
In your case this gives
\begin{eqnarray*}
\cot^{-1}\left(\frac{i(i+1)x}{2}+\frac{2}{x} \right)=\cot^{-1}\left(\frac{ix}{2} \right)-\cot^{-1}\left(\frac{(i+1)x}{2} \right).
\end{eqnarray*}
So the sum is telescopic and we get $\color{red}{x=\cot(1)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Finding a tight lower bound for $\left(\frac{1+x}{(1+x/2)^2}\right)^n$ I am trying to find a tight lower bound for $\left(\frac{1+x}{(1+x/2)^2}\right)^n$ as a function of $x$ and $n$ and for large $n$,
where $x$ changes with $n$ such that $\lim_{n\to\infty}x=0$.
I am not sure wether my approach to solve this is right or not, but this is what I did:
\begin{align*}\left(\frac{1+x}{(1+x/2)^2}\right)^n&=e^{n(\ln({1+x})-2\ln{(1+x/2)})}\\
&=e^{n(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots-2(\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{24}-\cdots))}\\
&=e^{n(-\frac{x^2}{4}+\frac{x^3}{4}-\frac{15x^4}{64}+\cdots)}\\
&\geq e^{n(-\frac{x^2}{4})}\\
&=1-(n\frac{x^2}{4})+(n\frac{x^2}{4})^2-\cdots \\
&\geq 1-(n\frac{x^2}{4})
\end{align*}
We know $\lim_{n\to\infty}x=0$, but we don't know whether $\lim_{n\to\infty}nx^2=0$ . Hence, the last inequality is not necessarily correct,
because the sum of the terms after $1-(n\frac{x^2}{4}) $ may not be greater than zero.
| Start considering $$y=\frac{1+x}{1+\left(\frac x2 \right)^2}$$ and use Taylor around $x=0$; this will give
$$y=1+x-\frac{x^2}{4}-\frac{x^3}{4}+O\left(x^4\right)$$ Now, use the binomial expansion and get
$$y^n=1+nx+\frac n4 (2n-3)x^2+\frac n{12}(2 n^2-9 n+4)x^3+O\left(x^4\right)$$ Define $t=nx$ and you can write as
$$y^n=1+t+\left(\frac{1}{2}-\frac{3}{4 n}\right) t^2+\left(\frac{1}{6}-\frac{3}{4 n}+\frac{1}{3 n^2}\right)t^3+O\left(t^5\right)\tag 1$$ So, for large $n$
$$y^n <1+t+\frac 12 t^2+\frac 16 t^3\tag 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What does this connection between addition and multiplication mean? I noticing that some numbers when multiplied together give the same product as their sum. After some trial and error I came up with a simple rule for this.
$$n\times\frac{n}{n-1} = n+\frac{n}{n-1}$$
For example
$$7\times\frac{7}{6} = \frac{49}{6}$$ $$ 7+\frac{7}{6} = \frac{49}{6}$$
I was wondering a few things,
*
*Is there any significance to this "theorem"?
*If it has already been discovered is there a name for it?
*Is there a similar theorem for division and subtraction?
| Notice $1 + \frac 1{n-1} = \frac {n-1}{n-1} + \frac 1{n-1} = \frac n{n-1}$
So $n\times \frac n{n-1} = n(1 + \frac 1{n-1}) = n + \frac n{n-1}$
Also note: If we were asked to solve $m*n = m + n$ then $m*n -m = n$ and $m(n-1) = n$.
Then if $n -1 = 0$ then $n = 1$ and $m*0 = 1$ which is impossible. So $n\ne 1$ and $n -1 \ne 0$ and so $m = \frac n{n-1}$.
So the only solutions are if $n \frac n{n-1}=n + \frac n{n-1}$ and we must solve for $n$.
If $n = 0$ then $n \frac n{n-1}=0$ and $n + \frac n{n-1} = 0$ so $n = 0$ is a solution.
If $n \ne 0$ then $n = \frac {n-1}n (n + \frac n{n-1}) = (n-1) + 1 = n$ so an number (integer or not) will be a solution.
So $nm = n + m \iff (n,m) = (0,0) $ or $m = \frac n{n-1}$ (and $m = \frac n{n-1} \iff n = \frac m{m-1}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
General formula of $\sum_{n\ge1}{\binom{2n}{n}^{-k}}$. If $k=1$, we have
\begin{align}
\sum_{n\ge1}\frac{1}{\binom{2n}{n}}&=\frac12\sum_{n\ge1}n\mathrm{B}(n,n)\\
&=\frac12\int _{0}^{1}\sum_{n\ge1}n(t-t^2)^{{n-1}}{\mathrm d}t\\
&=\frac12\int _{0}^{1}\frac{1}{(t-t^2-1)^2}{\mathrm d}t\\
&=\frac13+\frac{2\sqrt3\pi}{27}
\end{align}
I wonder if there exists a general formula of $\sum_{n\ge1}{\binom{2n}{n}^{-k}}$ with $k\in\Bbb N^*$. This may involve some ${}_pF_q$ functions.
| Chappers already wrote these series as hypergeometric functions, so I will just outline how to deal with the case $k=2$, exhibiting a relation with elliptic integrals. The key identities are
$$\frac{4^n}{(2n+1)\binom{2n}{n}}=\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx,\qquad \arcsin^2(x)=\frac{1}{2}\sum_{n\geq 1}\frac{(4x^2)^n}{n^2 \binom{2n}{n}} $$
which have been deeply exploited also here.
The case $k=2$.
$$ \sum_{n\geq 0}\binom{2n}{n}^{-2}=\phantom{}_2 F_1\left(1,1,1;\tfrac{1}{2},\tfrac{1}{2};\tfrac{1}{16}\right) $$
$$ 2\arcsin^2\left(\frac{x}{4}\right)=\sum_{n\geq 1}\frac{x^{2n}}{4^n n^2 \binom{2n}{n}},\quad \frac{x\arcsin(x/4)}{\sqrt{1-x^2/16}}=\sum_{n\geq 1}\frac{2 x^{2n}}{4^n n \binom{2n}{n}}$$
$$\frac{x^3}{4-\frac{x^2}{4}}+\frac{x^4\arcsin(x/4)}{16\left(1-x^2/16\right)^{3/2}}+\frac{x^2\arcsin(x/4)}{\sqrt{1-x^2/16}}= \sum_{n\geq 1}\frac{4 x^{2n+1}}{4^n \binom{2n}{n}} $$
$$ \frac{x^2 \left(x \sqrt{16-x^2} \left(64-x^2\right)+16 \left(32+x^2\right) \arcsin\left(\frac{x}{4}\right)\right)}{\left(16-x^2\right)^{5/2}}= \sum_{n\geq 1}\frac{(2n+1) x^{2n+1}}{4^n \binom{2n}{n}} $$
$$ \sum_{n\geq 1}\binom{2n}{n}^{-2}=\\=\int_{0}^{\pi/2}\frac{\sin^2(\theta) \left(\sin(\theta) \sqrt{16-\sin^2(\theta)} \left(64-\sin^2(\theta)\right)+16 \left(32+\sin^2(\theta)\right) \arcsin\left(\frac{\sin\theta}{4}\right)\right)}{\left(16-\sin^2\theta\right)^{5/2}}\,d\theta\\ = \frac{3}{5}-\frac{32}{5\sqrt{15}}\arcsin\left(\frac{1}{4}\right)+ 16\int_{0}^{1/4}\frac{u^2 \left(2+u^2\right) \arcsin\left(u\right)}{\sqrt{1-16 u^2}\left(1-u^2\right)^{5/2}}\,du $$
where the last integral is related to $E\left(\frac{1}{4}\right)$ and $K\left(\frac{1}{4}\right)$, which can be computed through algorithms with quadratic convergence (Brent-Salamin and the like).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solving a system of equations: $\begin{cases}\frac xy-\frac yx=\frac{15}4\\2x-5y=9\end{cases}$ How should I approach this type of a system equation?
$$
\begin{cases}
\dfrac xy-\dfrac yx=\dfrac{15}4\\
2x-5y=9
\end{cases}
$$
I tried to multiply the first equation by $4xy$ and divide the second one by $2$. After that I got this system:
$$
\begin{cases}
4x^2 - 4y^2 = 15xy\\
x - 2.5y = 4.5 \Longrightarrow x = 4.5 + 2.5y
\end{cases}
$$
Then I put $x$ from the second equation in the first one:
$$4(4.5 + 2.5y)^2 - 4y^2 = 15y(4.5 + 2.5y)$$
When I solved it I got these results:
*
*$$x_1 = \frac{189}{22},\ y_1 = \frac{18}{11}.$$
*$$x_2 = -3,\ y_2 = -3.$$
But these results are incorrect.
These are the answers from my book:
*
*$$x_1 = 12,\ y_1 = 3.$$
*$$x_2 = \frac{9}{22},\ y_2 = -\frac{18}{11}.$$
| Put $x/y =t$ then $$t-{1\over t} = {15\over 4} \implies 4t^2-15t-4 =0$$
Since $(4t+1)(t-4)=0$ so $t=-1/4$ or $t=4$.
*
*case $y=-4x$ put in to $2x+20x=9$ so $x=9/22$ and $y=-18/11$.
*case $x=4y$ and $y=3$ and $x=12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Finding the Equation of the Ellipse By Completing the Square I have the following equation of an ellipse:
$$5x^2+9y^2+40x=100$$
I need to put it in the form:
$$\frac{(x-h)^2}{m^2} + \frac{(y-k)^2}{n^2} =1 $$
I was trying to complete the square with the coefficients that have an $x$ variable.
$$5x^2 + 40x + 9y^2 = 100$$
$$5(x^2 + 8x) + 9y^2 = 100$$
$$5(x^2 + 8x + 16) + 9y^2 = 100 + 16$$
$$5(x+4)^2 + 9y^2 = 116$$
I then divided both sides of the equation by $116$.
$$\frac{5(x + 4)^2}{116} + \frac{9y^2}{116} = 1$$
However, when I graph the equation I do not get the same ellipse that was represented with the original equation of $5x^2 + 9y^2 + 40x = 100$. From the graph, it seems like they are two similar ellipses. Where am I making the mistake? Any help will be appreciated.
| You did a good job just you made a mistake here
$$5(x^2 + 8x + 16) + 9y^2 = 100 + 16$$
Should be
$$5(x^2 + 8x + 16) + 9y^2 = 100 + 80$$
$$5(x^2 +4)^2 + 9y^2 = 180$$
$$\frac {(x^2 +4)^2}{36} + \frac {y^2}{20} = 1$$
$$.....$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate without L'Hopital: $\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$ Evaluate the following limit without using L'Hospital method
$$\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$$
My turn is
$$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{x^{14}-x^9-x^5+1}\right\}$$
$$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{(x^{14}-1)-(x^9-1)-(x^5-1)}\right\}$$
then divide the numerator and the denominator by $$x-1$$
But I got again $$\frac{0}{0}$$
| Observe that, by the standard binomial expansion,
$$
(1+h)^9-1=9h+36h^2+O(h^3),\qquad (1+h)^5-1=5h+10h^2+O(h^3)
$$ then, by setting $x=1+h$, as $x \to 1$ we have $h \to 0$, we get
$$
\begin{align}
\lim_{x\to1}\left(\frac{9}{x^9-1}-\frac{5}{x^5-1}\right)&=\lim_{h\to0}\left(\frac{9}{(1+h)^9-1}-\frac{5}{(1+h)^5-1}\right)
\\\\&=\lim_{h\to0}\left(\frac{9}{9h+36h^2+O(h^3)}-\frac{5}{5h+10h^2+O(h^3)}\right)
\\\\&=\lim_{h\to0}\left(\frac{1}{h+4h^2+O(h^3)}-\frac{1}{h+2h^2+O(h^3)}\right)
\\\\&=\lim_{h\to0}\frac{-2h^2+O(h^3)}{(h+4h^2+O(h^3))(h+2h^2+O(h^3))}
\\\\&=-2.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Probability of forming a triangle given sides to choose from
What is the probability of being able to form a triangle from three segments chosen at random from five line segments of lengths $1,3,5,7$ and $9$?
Denote with $A$ the event of forming a triangle from three segments chosen from the list above. Then, the segments
$$\begin{aligned}
&1)\,\,3-5-7\\
&2)\,\,5-7-9\\
&3)\,\,3-7-9\\
\end{aligned}$$
are the only ones that can be used to create a triangle. Each case can be arranged in $3!$ ways, thus $N(A)=3!\cdot3$. On the other hand, $N=5\cdot4\cdot3\,$ and so
$$P(A)=\frac{N(A)}{N}=\frac{3!\cdot3}{5\cdot4\cdot3}=0.3$$
Would this look right? I am assuming that a side can be chosen only once, if that's not the case, then the following wold be how I would do it:
The segments to choose from are now
$$\begin{aligned}
&1)\,\,3-5-7\\
&2)\,\,5-7-9\\
&3)\,\,3-7-9\\
&4)\,\,1-1-1\\
&5)\,\,3-3-3\\
&6)\,\,5-5-5\\
&7)\,\,7-7-7\\
&8)\,\,9-9-9\\
\end{aligned}$$
The first three cases can be arranged in $3!$ ways and the other five only in $1$ way. thus $N(A)=3!\cdot3 +5$. On the other hand, $N=5^3$ and so
$$P(A)=\frac{N(A)}{N}=\frac{3!\cdot3+5}{5^3}=0.184$$
For these $2$ solutions I am not sure if the order matters or not. I know that the sides $3-5-7$ and $7-5-3$ form the same triangle, so would this mean that order does not matter? The way I argued is that if we set $A_1=3$ for the first and $A_1=7$ for the second, it is easy to see that $A_1\neq A_2$, so by this logic order matters.
| You can choose $3$ segments from those $5$ on ${5\choose 3} =10$ ways, but only $3$, you listed, are OK.- So the answer is $0.3$
Order is here irrelevant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\cot(x)=2$ and $\cot(y)=3$ show that $x+y=\frac{\pi}{4}$. If $\cot(x)=2$ and $\cot(y)=3$ show that $x+y=\frac{\pi}{4}$.
I was trying to find x and y but is some weird thing. Also note that $x,y\:\in (0,\pi/2)$
Second idea was to write : $\frac{\cos x}{\sin x}=2$ and $\frac{\cos y}{\sin y}=3$ and adding this two I'll get $\frac{\cos(x+y)}{\cos x \cos y}=5$. Than what?
| From this wikipedia page on trigonmetric identities, we see that
$\tan(\alpha + \beta) = \dfrac{(\tan \alpha + \tan \beta)}{1 - (\tan \alpha)(\tan \beta)}; \tag 1$
with
$\cot x = 2, \; \cot y = 3, \tag 2$
we find
$\tan x= \dfrac{1}{2} = 2^{-1}, \; \tan y = \dfrac{1}{3} = 3^{-1}; \tag 3$
thus
$\tan(x + y) = \dfrac{2^{-1} + 3^{-1}}{1 - 2^{-1} \cdot 3^{-1}} = \dfrac{5 \cdot 6^{-1}}{1 - 6^{-1}} = \dfrac{5 \cdot 6^{-1}}{5 \cdot 6^{-1}} = 1; \tag 4$
with $x, y \in (0, \pi / 2)$, we have $x + y \in (0, \pi)$; the only value
of $x + y$ in this range satisfying (4) is
$x + y = \dfrac{\pi}{4}. \tag 5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Taking common $2$ from $R_1$
$$2\left|
\begin{matrix}
(b+c)&(a+c)&(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
How do I proceed further?
| In the case $3 \times 3$ you might want to use the rule of Sarrus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\cos(x_n)$ we get
$$\cos(x_{n+1})=\cos(x_n-\theta)$$ where $\cos(\theta)=\frac{4}{5}$
Now Since $f_0=0$ we have $b_0=0$ and $x_0=\frac{\pi}{2}$
Now we have
$$x_{n+1}=x_n-\theta$$
Putting $n=0,1,2,3 \cdots 10$ and adding all we get
$$x_{10}=\frac{\pi}{2}-10\theta$$
Hence
$$b_{10}=\cos\left(\frac{\pi}{2}-10\theta\right)=\sin(10\theta)=\sin\left(10\arcsin\left(\frac{3}{5}\right)\right)$$
How to proceed further?
| De Moivre's theorem says
$(\cosθ+i\sinθ)^{10}=\cos{10θ}+i\sin{10θ}$
$$\sin{10θ}=10\sinθ\cos^9θ-_{10}C_3\sin^3θ\cos^7θ+_{10}C_5\cos^5θ\sin^5θ
-_{10}C_7\cos^3θ\sin^7θ+10\cosθ\sin^9θ$$
$=\dfrac{10*3*4^9-10*3*4*3^3*4^7+9*2*7*3^5*4^5-10*3*4*3^7*4^3+10*4*3^9}{5^{10}}$
$$=\dfrac{1476984}{9765625}.$$ The result of wolfram alpha give same value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Calculating $3^{m-n}=?$ $$9^m + 9^n = 52$$
$$9^m -4 = 2 \cdot 9^n$$
$$3^{m-n}=?$$
Let me show what I've tried
Simpifyling the both equalities.
$$3^{2^m} + 3^{2n} = 2 \cdot 13 \tag{1}$$
$$3^{2m} -2^2 = 2 \cdot 3^{2n} \tag{2}$$
Diving the second equality by $2$ and we have
$$\frac{3^{2m} -2^2}{2} =3^{2n} \tag{3}$$
Here is where I'm stuck.
My Kindest Regards!
| Putting
$$
x=9^{m}, \quad y=9^{n},
$$
we have
\begin{eqnarray*}
x+y&=&52\\
x-2y&=&4
\end{eqnarray*}
It follows that
$$
9^{m}=x=\frac{3x}{3}=\frac{108}{3}=36, \quad 9^{n}=y=\frac{3y}{3}=\frac{48}{3}=16,
$$
i.e.
$$
3^{m}=\sqrt{9^{m}}=\sqrt{36}=6,\quad 3^{n}=\sqrt{9^{n}}=\sqrt{16}=4.
$$
Hence
$$
3^{m-n}=\frac{3^{m}}{3^{n}}=\frac{6}{4}=\frac{3}{2}.
$$
Remark: The numbers $m$ and $n$ are obviously not integers, in fact
$$
m=\log_3(6)=\frac{\ln(6)}{\ln(3)}\approx 1.63092975\ldots ,\quad n=\log_3(4)=\frac{\ln(4)}{\ln(3)}\approx 1.261859507\ldots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is the function is differentiable at $0$? Is the function given by
$\displaystyle f(x) = \begin{cases} \dfrac{1}{x\log(2)} - \dfrac{1}{2^x -1}, \quad &x \neq 0, \\ \dfrac{1}{2} , &x = 0 \end{cases}$
differentiable at $0$ ?
My attempt : $f'(0) = \frac {f(x) -f(0)}{x-0}$
$$\lim _{x\to 0}f'(0) = \lim _{x\to 0}\frac {\frac{1}{x\log(2)} - \frac{1}{2^x -1} - \frac{1}{0\log(2)} - \frac{1}{2^0 -1}}{x- 0}$$
$$\lim _{x\to 0}\frac {\frac{1}{x\log(2)} - \frac{1}{2^x -1} - \frac{1}{0} - \frac{1}{0}}{x- 0}$$
Now I am not able to proceed further.
Any hints /solution will be appreciated.
Thanks in advance
| You approach is correct, but the substitution is wrong! You should have proceeded in this manner:-
$$f'\left( 0 \right) = \lim_{x \rightarrow 0}\dfrac{f \left( x \right) - f \left( 0 \right)}{x - 0}$$
$$= \lim_{x \rightarrow 0}{\dfrac{ \left( \dfrac{1}{x \log 2} - \dfrac{1}{2^{x} - 1} \right) - \left( \dfrac{1}{2} \right)}{x}}$$
$$= \lim_{x \rightarrow 0} \dfrac{2 \left( 2^{x} - 1 - x \log 2 \right) - \left( 2^x - 1 \right)x \log 2}{2 x^2 \log 2 \left( 2^x - 1 \right)}$$
Use L'Hospital's Rule to get the answer!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solutions to parametric polynomial equations The equation $ax^3-x^2+1=0$ has three real solutions if $a^2<4/27$. The three solutions are plotted below against parameter $a$
Is there some intuitive reason why instead of being two continuous functions, both the yellow branch and green branch are discontinuous at $a=0$? This leads to the consequence that $x^2-1=0$ can't be solved by solving $ax^3-x^2+1=0$ and bring $a\to 0$.
P.S. The solutions are shown below, if anyone's interested.
$$
\cases{x_1 = \frac{1}{3} \left(\frac{\sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}{\sqrt[3]{2} a}+\frac{\sqrt[3]{2}}{a \sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}+\frac{1}{a}\right)\\
x_2=-\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}{6 \sqrt[3]{2} a}-\frac{1+i \sqrt{3}}{3\ 2^{2/3} a \sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}+\frac{1}{3 a}\\
x_3=-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}{6 \sqrt[3]{2} a}-\frac{1-i \sqrt{3}}{3\ 2^{2/3} a \sqrt[3]{-27 a^2+3 \sqrt{3} \sqrt{27 a^4-4 a^2}+2}}+\frac{1}{3 a}}
$$
| One obvious thing is that the three functions given are all odd functions: the contents of the cube root is an even function, so if we write it as $b = \frac{1}{2}(2-27a^2-3\sqrt{3}\sqrt{27a^4-4a^2})$, we have three functions of the form
$$ \frac{1}{3a} \left(1+\omega b^{-1/3}+\omega^{-1} b^{1/3} \right), $$
where $\omega$ is one of the cube roots of $1$ (namely $1$ and $(-1\pm i\sqrt{3})/2$), and all three of these are the product of an odd and even function and so must be odd.
If you replace $b$ by $\frac{1}{2}(2-27a^2-3\sqrt{3}a\sqrt{27a^2-4})$ (i.e. pull an $a$ out of the square root, so choosing a branch so that $\sqrt{27a^4-4a^2}$ is instead odd), this is no longer the case, and you'll find that both finite solutions are now continuous at zero:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that the following inequality is true: $(\frac{1}{a} + \frac{1}{bc}) (\frac{1}{b} + \frac{1}{ca})(\frac{1}{c} + \frac{1}{ab}) \geq 1728$ This is a question from a past Olympiad paper:
Three positive real numbers $a, b, c$ satisfy the following constraint:
$a+b+c = 1$. Show that the following inequality is true:
$\left(\dfrac{1}{a} + \dfrac{1}{bc}\right) \left(\dfrac{1}{b} + \dfrac{1}{ca}\right)\left(\dfrac{1}{c} + \dfrac{1}{ab}\right) \geq 1728$.
Starting from the L.H.S, I end up with:
$\dfrac{abc(abc+a^2 + b^2 + c^2 + 1) + a^2b^2 + a^2c^2+b^2c^2}{(abc)^3}$.
From the numerator I suspect $a+b+c$ will be factorised, but this is the furthest I have got to.
Using the hint given by timon92:
$\dfrac{1}{a} + \dfrac{1}{bc} = \dfrac{(a+b)(a+c)}{abc}$
and likewise for other two, I end up with:
$(a+b)^2(b+c)^2(a+c)^2 \geq 1728(abc)^3$
$\left((a+b)(b+c)(a+c)\right)^\frac{2}{3} \geq 8abc$
Many thanks in advance.
| Alt. hint: by AM-GM:
$$
\frac{1}{a}+\frac{1}{bc}=\frac{1}{a}+\frac{a+b+c}{bc} = \frac{1}{a} + \frac{1}{b}+\frac{1}{c}+\frac{a}{bc}\ge 4 \sqrt[4]{\frac{1}{b^2c^2}} = 4 \sqrt{\frac{1}{bc}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Partial fractions integral. Compute $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$
Evaluate $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$.
I directly approached this with partial fractions and rewritten $x^6+1=(x^2)^3+1=(x^2+1)(x^4+x^2+1)$.
Therefore the integral is:
$$\int_{0}^{1} \frac {x^4+1}{(x^2+1)(x^4+x^2+1)}dx=-2\int_{0}^{1} \frac {x}{1+x^2}dx+2\int_{0}^{1}\frac 1{1+x^2}dx-\int_{0}^1\frac {(x-1)^2}{(x^2+1)^2+(\frac {\sqrt{3}}2)^2}dx$$
But I don't know how to solve the last term.
I answered the question... I realized my silly mistake, sorry for bothering...
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{1}{x^{4} + 1 \over x^{6} + 1}\,\dd x & =
\int_{0}^{1}{x^{4} + 1 - x^{10} - x^{6} \over 1 - x^{12}}\,\dd x =
{1 \over 12}\int_{0}^{1}{x^{-7/12} + x^{-11/12} - x^{-1/12} - x^{-5/12} \over 1 - x}\,\dd x
\\[5mm] & =
{1 \over 12}\pars{\int_{0}^{1}{1 - x^{-5/12} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{-7/12} \over 1 - x}\,\dd x}
\\[2mm] & +
{1 \over 12}\pars{\int_{0}^{1}{1 - x^{-1/12} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{-11/12} \over 1 - x}\,\dd x}
\\[5mm] & =
{\pars{H_{-5/12} - H_{-7/12}} +
\pars{H_{-1/12} - H_{-11/12}}\over 12}\qquad\pars{~H_{z}:\ Harmonic\ Number~}
\\[5mm] & =
{\pi\cot\pars{5\pi/12} + \pi\cot\pars{\pi/12}\over 12}
\qquad\pars{~Euler\ Reflection\ Formula~}
\\[5mm] & =
{\pi \over 12}\,{\sin\pars{\pi/2} \over \sin\pars{5\pi/12}\sin\pars{\pi/12}} =
{\pi \over 12}\,{1 \over \bracks{\cos\pars{\pi/3} - \cos\pars{\pi/2}}/2} =
{\pi \over 6}\,{1 \over 1/2 - 0}
\\[5mm] & = \bbx{\pi \over 3} \approx 1.0472
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
If $x+y+z=2$ prove that $\sum_{cyc}\frac{1}{\sqrt{x^2+y^2}}\ge2+\frac{1}{\sqrt{2}}$ Let $x,y,z$ be non-negative reals whose sum is $2$. Prove that
$\frac{1}{\sqrt{x^2+y^2}}+\frac{1}{\sqrt{y^2+z^2}}+\frac{1}{\sqrt{z^2+x^2}}\ge2+\frac{1}{\sqrt{2}}$
I have tried bounding them up (assuming that $a\le b\le c$), many inequalities (AM-GM, QM-AM, Cauchy) but nothing has worked. I know that equality is achieved when one of $x,y,z$ is $0$ and the other two are $1$ but that doesn't help.
| Without loss of generality, assume that $x \geq y \geq z$. It is easy to verify that using $(x + \frac{z}{2}, y + \frac{z}{2}, 0)$ to replace $(x, y, z)$ would make the LHS smaller, i.e.,
$$
\frac{1}{\sqrt{x^2 + y^2}} +
\frac{1}{\sqrt{y^2 + z^2}} +
\frac{1}{\sqrt{x^2 + z^2}}
\geq
\frac{1}{\sqrt{\left(x + \frac{z}{2}\right)^2 + \left(y + \frac{z}{2}\right)^2}} +
\frac{1}{y + \frac{z}{2}} +
\frac{1}{x + \frac{z}{2}}.
$$
This is obvious as $\left( y + \frac{z}{2}\right)^2 \geq (y + z)^2, \left( x + \frac{z}{2}\right)^2 \geq (x + z)^2$.
As a consequence, we only need to prove the case where $z = 0$, i.e., for nonnegative $x$ and $y$ with $x + y = 2$,
$$
\frac{1}{\sqrt{x^2 + y^2}} + \frac{1}{x} + \frac{1}{y} \geq 2 + \frac{1}{\sqrt{2}}.\tag{*}
$$
Note that the LHS in (*) could be written as
$$
\begin{align}
\frac{1}{\sqrt{(x+y)^2 - 2xy}} + \frac{x+y}{xy}
&= \frac{1}{\sqrt{4 - 2xy}} + \frac{2}{xy}\\
&= 2 + \frac{1}{\sqrt{2}} + (1 - xy) \left[\frac{2}{xy} - \frac{1}{\sqrt{2}(2 - xy + \sqrt{2 - xy})}\right]\\
&\geq 2 + \frac{1}{\sqrt{2}} + (1 - xy)\left(2 - \frac{1}{2\sqrt{2}}\right)\\
&\geq 2 + \frac{1}{\sqrt{2}},
\end{align}
$$
where we have used the fact that $xy \leq \left(\frac{x + y}{2}\right)^2 = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate this binomial sum? I come across the following binomial sum when studying the average-case time complexity of an algorithm:
$$\sum_{i = 1}^{n-k+1} i \binom{n-i}{k-1}$$
How to evaluate this sum?
| \begin{align*}
\sum_{i=1}^{n-k+1} i \binom{n-i}{k-1} &= \sum_{i=0}^{n-k} (i+1) \binom{n-i-1}{k-1} \\
&= \sum_{i=0}^{n-k} \big((n+1) - (n-i)\big) \binom{n-i-1}{k-1} \\
&= \sum_{i=0}^{n-k} (n+1) \binom{n-i-1}{k-1} - \sum_{i=0}^{n-k} (n-i) \binom{n-i-1}{k-1} \\
&= (n+1) \sum_{i=0}^{n-k}\binom{n-i-1}{k-1} - k \sum_{i=0}^{n-k} \binom{n-i}{k} \\
&= (n+1) \sum_{m=k-1}^{n-1}\binom{m}{k-1} - k \sum_{m=k}^{n} \binom{m}{k} \\
&= (n+1) \binom{n}{k} - k \binom{n+1}{k+1} \\
&= \binom{n+1}{k+1}
\end{align*}
It uses
$$ r \binom{r-1}{k-1} = k \binom{r}{k} $$
and
$$ \sum_{0 \le k \le n} \binom{k}{m} = \binom{n+1}{m+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant? Why is the difference between these two functions a constant?
$$f(x)=\frac{2x^2-x}{x^2-x+1}$$
$$g(x)=\frac{x-2}{x^2-x+1}$$
Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant.
Of course I can calculate it is true: the difference is $2$, but my intuition is still completely off here. So, who can provide some intuitive explanation of what is going on here? Perhaps using a graph of some kind that shows what's special in this particular case?
Thanks!
BACKGROUND: The background of this question is that I tried to find this integral:
$$\int\frac{x dx}{(x^2-x+1)^2}$$
As a solution I found:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{2x^2-x}{3\left(x^2-x+1\right)}+C$$
Whereas my calculusbook gave as the solution:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{x-2}{3\left(x^2-x+1\right)}+C$$
I thought I made a mistake but as it turned out, their difference was constant, so both are valid solutions.
| Would you be surprised that the difference of $\dfrac{2x^2+x+1}{x^2}$ and $\dfrac{x+1}{x^2}$ is $2$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 16,
"answer_id": 0
} |
Solve differential equation $(1+x^2) \frac{dy}{dx} - 2xy = x$ Solve differential equation $$(1+x^2) \frac{dy}{dx} - 2xy = x$$
I simplified it to $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$ \int \frac{1}{1+2y} dy =\int \frac{x}{1+x^2} dx $$
$$\frac{1}{2} \int \frac{2}{1+2y} dy = \frac{1}{2} \int \frac{2x}{1+x^2} dx$$
$$\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + C $$
From here, I got stuck. I have to remove y from here to solve it.
The answer in the textbook gave $ y= (k(1+x^2) - 1)/{2}$ I believe $k$ is the integration constant. How do I remove the $\ln$ from both sides?
| $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + \frac {1}{2} \ln C$$
$$ | 1 + 2y | =C(1+x^2)$$
$$ 2y = -1 +C(1+x^2)$$
$$ y = \frac {-1 +C(1+x^2)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Solve the differential equation: $\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} } $ $$\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} }$$
Simplified it to: $$ \int \frac{1}{\sqrt{1-y^2}} dy = \int \frac{1}{\sqrt{1-x^2}} dx$$
$$\implies\sin^{-1} y = ( \sin^{-1} x + C)$$
$$y = (\sin (\sin^{-1} x) \cos C_1) + \cos(\sin^{-1} x) (\sin C_1)$$
How to I simplify $\cos(\sin^{-1} x)$ to get the final answer ?
| You can simplify it like this
$$ y = \sin(\sin^{-1} x + C) = \cos (C) \sin(\sin^{-1}x) + \sin (C)\cos(\sin^{-1}x) \\
= \cos (C)x + \sin(C)\sqrt{1-x^2} $$
The explanation is
$$ \cos(\sin^{-1}x) = \sqrt{1-\sin^2(\sin^{-1}x)} = \sqrt{1-x^2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove $3^{2n+1} + 2^{n+2} = 7m$ I'm trying to prove that $3^{2n+1} + 2^{n+2}$ is a multiple of 7 by using induction.
So I started to prove it for $n=1$: $3^{2(1)+1}+2^{1+2}=3^3+2^3=27+8=35=7(5)$.
Next, try to prove that the statement being true $n=k$ implies it being true for $n=k+1$. Thus:
$3^{2(k+1)+1}+2^{(k+1)+2} = 3^{2k+3}+2^{k+3} = (3)(3)(3^{2k+1})+(2)(2^{k+2}) = 9(3^{2k+1})+2(2^{k+2})$
I feel like I'm almost there, if I could've factor 9 and 2 somehow I could say that $3^{2k+1}+2^{k+2}=7m$ for some integer $m$, but I can't find a way to do it. What am I missing? Or did I do a blunder somewhere along the road?
Thanks in advance.
| $$9(3^{2k+1}) + 2(2^{k+2}) = 7(3^{2k+1}) + 2(3^{2k+1}) + 2(2^{k+2}) = 7(3^{2k+1}) + 2(3^{2k+1} + 2^{k+2}) = 7(3^{2k+1} + 2m)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2769913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
For $n \ge 4$, does it follow that ${{3n} \choose {n}} > 4^n$ I believe that the answer is yes. Here's my thinking:
(1) For $n \ge 4, { {2n} \choose {n}} \ge \frac{4^n}{n}$
By induction: $64 = \frac{4^4}{4} \le {{8}\choose{4}}= 70$. Assume it is true up to $n-1 \ge 4$. Then, ${ {2n} \choose n} = 2\left(\frac{2n-1}{n}\right){{2n-1}\choose{n-1}} > 2\left(\frac{2n-1}{n}\right)\frac{4^{n-1}}{n-1} > 2 \cdot 2\cdot \frac{4^{n-1}}{n} = \frac{4^n}{n}$
Note: This argument was taken from Wikipedia.
(2) $(3n-1) > \left(\frac{3}{2}\right)(2n-1), (3n-2) > \left(\frac{3}{2}\right)(2n-2), \dots, (2n-n+1) > \left(\frac{3}{2}\right)(2n-n+1)$
If $2a = 3b$, then $2a - 2 > 3b - 3$ and $2(a-1) > 3(b-1)$ so that $(a-1) > \frac{3}{2}(b-1)$
(3) For $n\ge 2$, ${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n} \choose {n}}$
From (2) above: ${{3n} \choose{n}} = \frac{(3n)(3n-1)\dots(3n-n+2)(3n-n+1)}{n!} > \left(\frac{3}{2}^n\right)\frac{(2n)(2n-1)\dots(2n-n+2)(2n-n+1)}{n!} = \left(\frac{3}{2}^n\right){{2n} \choose {n}}$
(4) Since for $n \ge 4, \left(\frac{3}{2}\right)^n > n$, it follows that:
${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n}\choose{n}} > n\left(\frac{4^n}{n}\right) = 4^n $
Note: The argument that for $n \ge 2, \left(\frac{3}{2}\right)^n > n$ can be found here.
| It is even true for $n\ge3$. Indeed $\;\dbinom 93=\dfrac{9!}{3!\,6!}=84 >64$.
\begin{align}
&\text{Next, we have }\qquad\qquad
&\binom{3(n+1)}{n+1}&=\binom{3n}n\,\frac{(3n+1)(3n+2)(3n+3)}{(n+1)(2n+1)(2n+2)},&&\qquad\qquad\end{align}
so, by induction, all we have to prove is that
$$\frac{3(3n+1)(3n+2)}{2(n+1)(2n+1)}\ge4$$
This inequality is equivalent to
$$ 27n^2+27n+6\ge 16n^2+24n+8\iff 11n^2+3n\ge 2, $$
which is obviously true if $n \ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Computing the coefficient of the term of a certain degree in a polynomial Given the polynomial
${1\over8}((1+z)^9 + 3(1-z)^4(1+z)^5 + (1-z)^6(1+z)^3)$
(which is the weight enumerator of a code)
how do I find out the coefficient of $z^2$?
The solution given is ${1 \over 8}(36-12+0) = 3$.
I got $36$ for the $z^2$ coefficient of $(1+z)^9$ using the Binomial Theorem, but I don't know how to get $-12$ for the $z^2$ coefficient of $3(1-z)^4(1+z)^5$. By using the Binomial Theorem separately on $(1-z)^4$ and $(1+z)^5$ I get the following two polynomials, repsectively:
$z^4-4z^3+6z^2-4z+1$
$z^5+5z^4+10z^3+10z^2+5z+1$
I am unsure what to do next, or even if this is going in the right direction.
| Notice that the coefficient of $z^2$ in $(a+bz+cz^2+\cdots)(d+ez+fz^2+\cdots)$ is $af+be+cd$.
After expansion of the powers,
$$1+9z+36z^2+\cdots+\\
3(1-4z+6z^2-\cdots)(1+5z+10z^2+\cdots)+\\
(1-6z+15z^2-\cdots)(1+3z+3z^2+\cdots)
$$
Which gives
$$\frac{36+3(10-20+6)+(3-18+15)}8=\frac{24}8.$$
Using Lord Shark the Unknown's shortcuts, and the rule $(a+bz+cz^2+\cdots)(d+ez^2+\cdots)$ that gives $ae+cd$,
$$1+9z+36z^2+\cdots+\\
3(1-4z^2+\cdots)(1+z)+\\
(1-3z^2+\cdots)(1-3z+3z^2+\cdots)
$$
Now
$$\frac{36+3(-4)+(3-3)}8=\frac{24}8.$$
Using a CAS,
$$\frac{5z^9+9z^8+24z^7+80z^6+138z^5+138z^4+80z^3+24z^2+9z+5}8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
quadratic system of three equations with three unknowns If $x,y,z$ are real and positive and they verify the system
$$\begin{cases}
x^2+xy+y^2=36\\
y^2+yz+z^2=64\\
z^2+zx+x^2=100
\end{cases}$$
Find the value of the sum $S=xy+yz+zx$
Of course I tried the obvious, to sum them all up, to bring each equation to the canonical form, to guess. I can see that the system is symmetric, but I don't know how does it help. Someone who already solved it told me that $S$ is proportional to $\sqrt{3}$. How to proceed?
| If you are willing to use the Wolfram Cloud Sandbox then the answer comes out as
In[1]:= (x y+y z+z x)^2 /. Solve[{x^2+x y+y^2 == 36,
y^2+y z+z^2 == 64, z^2+z x+x^2 == 100}] //
Simplify//Sqrt//InputForm
Out[1]//InputForm=
32*Sqrt[3], 32*Sqrt[3], 32*Sqrt[3], 32*Sqrt[3]}
and so the answer is $\;32\sqrt{3}.$
Alternatively, define
$$ q_1 := x^2+xy+y^2,\; q_2 := y^2+yz+z^2,\; q_3 := z^2+zx+x^2 $$
and also let
$$ e_1 := x+y+z,\; e_2 := xy+yz+zx,\; e_3 := xyz. $$
Given that
$\;q_1=36, q_2=64, q_3=100,\;$ then
$\;q_1+q_2+q_3 = 2e_1^2-3e_2 = 200\;$ and $\;q_1q_2+q_2q_3+q_3q_1 = e_1^4 + 3e_2^2 -3e_2e_1^2 = 12304.\;$ Now from the first equation $\;e_2=(2e_1^2-200)/3\;$ and when substituted into the second equation gives
$\;(40000-200e_1^2+e_1^4)/3 = 12304\;$ with solution $\;e_1^2 = 100+48\sqrt{3},\;$ and now $\;e_2 = 32\sqrt{3}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is there a fast way to prove a tridiagonal matrix is positive definite? I' m trying to prove that
$$A=\begin{pmatrix}
4 & 2 & 0 & 0 & 0 \\
2 & 5 & 2 & 0 & 0 \\
0 & 2 & 5 & 2 & 0 \\
0 & 0 & 2 & 5 & 2 \\
0 & 0 & 0 & 2 & 5 \\
\end{pmatrix}$$
admits a Cholesky decomposition.
$A$ is symmetric, so it admits a Cholesky decomposition iff it is positive definite. The only methods I know for checking this are:
*
*$X^tAX > 0, \quad \forall X \in \mathbb{K}^n- \{0\}$.
*If $\lambda$ is an eigenvalue of $A$, then $\lambda>0.$
I have failed to prove it using 1 and 2 is taking me so much time. Is there any easier way to do this, given that $A$ is tridiagonal?
| It is well-known that the eigenvalues of
$$
S=\begin{bmatrix} 0&1&0&0&0\\ 1&0&1&0&0\\ 0&1&0&1&0\\ 0&0&1&0&1\\ 0&0&0&1&0
\end{bmatrix}
$$
are
$$
2\cos\frac{k\pi}{n+1},\ \ \ k=1,\ldots,n.
$$
So the eigenvalues of $4I+2S$ are
$$
4+4\cos\frac{k\pi}{6}>0,\ \ \ k=1,\ldots,5.
$$
Being symmetric, this tells us that $$4I+4S=\begin{bmatrix} 4&2&0&0&0\\ 2&4&2&0&0\\ 0&2&4&2&0\\ 0&0&2&4&2\\ 0&0&0&2&4
\end{bmatrix}$$
is positive definite. Now $A$ is also positive definite, as $A=B+4I+2S$, where
$$
B=\begin{bmatrix} 0&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\end{bmatrix}
$$
is positive semi definite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 8,
"answer_id": 7
} |
Prove that the level curves of $\log\frac{|z-\sqrt{z^2-1}|}{2}$ are ellipses In a book it was mentioned that the level curves of the complex function $\phi (z) = \log\frac{|z-\sqrt{z^2-1}|}{2}$ is an ellipse with foci $1$ and $-1$. I can not establish a relation between the ellipse equation in the complex plane $|z-a|+|z-b|=c$ and the function $\phi(z)$. Is the change of variable possible here? (for example I tried $z = \sinh\omega$ but it did not work).
Can anyone help me to prove it?
Thanks in advance.
| You can just ignore $\log$ and $2$ in the fraction, and concentrate on $|z-\sqrt{z^2-1}| = r$ for $r>0$.
Let $w = z-\sqrt{z^2-1}$. Then $z - w = \sqrt{z^2-1}$ and $(z-w)^2 = z^2 - 1$. It follows that $z = \frac{w^2+1}{2w}.$ We have
\begin{align}
\left| \frac{w^2+1}{2w} + 1 \right| + \left| \frac{w^2+1}{2w} - 1 \right| &= \left| \frac{(w+1)^2}{2w}\right| + \left| \frac{(w-1)^2}{2w}\right|\\
&= \frac{(w+1)(\overline w + 1)}{2|w|} + \frac{(w-1)(\overline w - 1)}{2|w|}\\
&= \frac{w\overline w + w + \overline w + 1}{2|w|} + \frac{w\overline w - w - \overline w + 1}{2|w|}\\
&= \frac{|w|^2 + 1}{|w|}
\end{align}
and thus if $|w|$ is constant, so is $\left| \frac{w^2+1}{2w} + 1 \right| + \left| \frac{w^2+1}{2w} - 1 \right|$. Therefore, solutions of $|z-\sqrt{z^2-1}|=r$ lie on an ellipse.
However, it seems the contours are actually not the whole ellipses.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
prove this inequality with five variables Let $$\begin{cases}S=a^2+b^2+c^2+d^2+e^2\\
ab+bc+cd+de+ea=T_{1}\\
ac+ce+eb+bd+da=T_{2}
\end{cases}$$
Find the range $A$,such any postive real numbers have
$$S\ge AT_{1}+(1-A)T_{2}$$
This problem is creat by Wang yong xi .since I'm using the AM-GM inequality.
$$S=a^2+b^2+c^2+d^2+e^2\ge ab+bc+cd+de+ea=T_{1}$$
and
$$S=a^2+b^2+c^2+d^2+e^2\ge T_{2}$$
so $$S\ge \max{(T_{1},T_{2})}\ge AT_{1}+(1-A)T_{2}$$
so $A\in [0,1]?$
It is said that this is not the correct answer.
| Indeed, the condition is satisfied for $-0.62 \simeq \frac{1-5^{1/2} }{2} < A< \frac43$ as will be shown below. Two examples are given.
We need to establish $S - AT_{1}-(1-A)T_{2} \ge 0 $ and we have that
$$
2[S - AT_{1}-(1-A)T_{2}] = A \sum_{cyc} (a-b)^2 + (1-A) \sum_{cyc} (a-c)^2
$$
where $\sum_{cyc}$ says that the argument is to be taken over all 5 cyclic shifts.
So as long as $0 \le A \le 1$ this is always nonnegative.
Case 1: Now consider $A<0$ and write $B = -A > 0$. Then we need to establish
$$
\sum_{cyc} (a-c)^2 \ge \frac{B}{1+B} \sum_{cyc} (a-b)^2
$$
Case 2: Consider $A>1$ and write $C = A-1 > 0$. Then we need to establish
$$
\sum_{cyc} (a-c)^2 \le \frac{1+C}{C} \sum_{cyc} (a-b)^2
$$
It remains to be discussed whether and for which $B,C$ these cases hold true.
Let $a-b = x$, $b-c = y$, etc. with variables $x,y,z,w,v$. Then $\sum_{cyc} x =0$ and in case 1,
$$
\sum_{cyc} (x+y)^2 - \frac{B}{1+B} \sum_{cyc} x^2 \ge 0
$$
Now this is an expression which is unrestricted in the variables $x,y,z,w,v$, which means we can apply calculus under the extra condition (enforced with Lagrangian $\lambda$) $\sum_{cyc} x =0$ . So let $q = \frac{B}{1+B}$ and consider
$$
f(x,y,z,w,v) = \sum_{cyc} (x+y)^2 - q \sum_{cyc} x^2 + \lambda \sum_{cyc} x
$$
The function is quadratic, so if we can establish global convexity, a local minimum will also be a global minimum. Global convexity is established if the Hessian is positive definite everywhere. We easily compute the Hessian
$$
H(f) = 2 \left(
\begin{matrix}
-q + 2& 1& 0& 0& 1\\
1& -q + 2& 1& 0& 0\\
0& 1& -q + 2& 1& 0\\
0& 0& 1& -q + 2& 1\\
1& 0& 0& 1& -q + 2\\
\end{matrix}
\right)
$$
For global convexity, we need all eigenvalues of $H$ be positive. (Note that positivity of the determinant of $H$ is necessary but not sufficient.) The eigenvalues are $$\lambda_1 =
8 - 2q \\
\lambda_2 = 5^{1/2} - 2q + 3\\
\lambda_3 =
3 - 5^{1/2} - 2q
$$
where $\lambda_2$ and $\lambda_3$ are double. We obtain $q < 4$ and $q < \frac{5^{1/2} + 3}{2} \simeq 2.61$ and $q < \frac{-5^{1/2} +3}{2} \simeq 0.38$ e.g. in total $q = \frac{B}{1+B} < \frac{-5^{1/2} +3}{2} $ or $B < \frac{q}{1-q} = \frac{5^{1/2} -1}{2} \simeq 0.62$
Case 2 can be treated in the very same fashion. Setting $q = \frac{1+C}{C} $ we need that $
f(x,y,z,w,v) \le 0$, hence we need to establish concavity which leads to requiring that all eigenvalues of $H$ are negative, which is satisfied with $q = \frac{1+C}{C} > 4 $ or $C< \frac13$.
Combining the two cases gives $-0.62 \simeq \frac{1-5^{1/2} }{2} < A< \frac43$ as the solution.
For illustration, let's consider two special settings.
1) Let $(a,b,c,d,e) = (1,1,2,2,2)$. Then $
A \sum_{cyc} (a-b)^2 + (1-A) \sum_{cyc} (a-c)^2 = 2 A + 4 (1-A) = 4 - 2A \ge 0
$
which is true for $A<2$, and indeed $A< \frac43$ is the sharper condition.
2) Let $(a,b,c,d,e) = (1,2,1,2,1)$. Then $
A \sum_{cyc} (a-b)^2 + (1-A) \sum_{cyc} (a-c)^2 = 4 A + 2 (1-A) = 2 + 2A \ge 0
$
which is true for $-1<A$, and indeed $-0.62 \simeq \frac{1-5^{1/2} }{2} < A$ is the sharper condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\sqrt[3]{2}$ is irrational without using prime factorization Prove that $\sqrt[3]{2}$ is irrational without using prime factorization.
The standard proof that $\sqrt[3]{2}$ is irrational uses prime factorization in an essential way. So I wondered if there is a proof that does not use it.
This was inspired by the fact that I know two proofs that $\sqrt{2}$ is irrational that do not use prime factorization.
The first uses
$$\sqrt{2}=\sqrt{2}\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{2-\sqrt{2}}{\sqrt{2}-1}
$$
to show that if $\sqrt{2} = \dfrac{a}{b}$ then
$$\sqrt{2}=\frac{2-\dfrac{a}{b}}{\dfrac{a}{b}-1}=\frac{2b-a}{a-b}
$$
is a rational $\sqrt{2}$ with a smaller denominator.
The second uses
$$(x^2-2y^2)^2=(x^2+2y^2)^2-2(2xy)^2
$$
and $3^2-2\cdot 2^2 = 1$ to show that $x^2-2y^2=1$ has arbitrarily large solutions and this contradicts $\sqrt{2}$ being rational.
I have not been able to extend either of these proofs to $\sqrt[3]{2}$. Results that I do not consider "legal" in solving this problem include Fermat's Last Theorem (which definitely uses unique factorization) and the rational root theorem (which uses unique factorization
in its proof).
| You can do it using the concept of odds and evens which is sort of a poor mans version of unique prime factorization but surely acceptable. After all the classic proof of the irrationality of $\sqrt 2$ used even and odds without assuming unique prime factorization.
First note: If $b \in \mathbb Z$ is $b = 2m$ is even, then $b^3 = 8m^3$ is even and if $b = 2m + 1$ is odd then $b^3 = 8m^3 + 12m^2 + 6m + 1$ is odd.
So if $\frac ab; b\ne 0$ and $a$ and $b$ in "lowest terms" (have no factors in common, in particular are not both even), and if $(\frac ab)^3 = 2$ then....
$2a^3 = b^3$ and so $b^3$ is even and so $b$ is even and so $b = 2m$ and so $2a^3 = 8m^3$ and so $a^3 = 4m^3$ and so $a^3$ is even and so $a$ is even and so $a$ and $b$ are both even but we said that wasn't the case so that's impossible and nyah nyah nyah nyah nyah.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Without resorting to induction show that $L_n^2=L_{n+1}L_{n-1}+5(-1)^n$,Where $L_n$ is $n^{th}$ Lucas number.
Without resorting to induction show that
$L_n^2=L_{n+1}L_{n-1}+5(-1)^n$,Where $L_n$ is $n^{th}$ Lucas number.
By definition of Lucas number $L_n=L_{n-1}+L_{n-2}\implies L_{n-1}=L_n-L_{n-2}$ and $L_{n+1}=L_{n}+L_{n-2}.$
Now,$L_{n+1}L_{n-1}=L_n^2+L_{n}L_{n-1}-L_{n}L_{n-2}-L_{n-2}L_{n-1}$
From,here i'm not getting how to proceed further...
| The identity
$$
L_n^2=L_{n+1}L_{n-1}+5(-1)^n
$$
for the Lucas numbers seems to be closed related to Cassini's identity for the Fibonacci numbers:
$$
F_n^2 - F_{n+1}F_{n-1} = (-1)^{n-1}
$$
which follows from taking determinants in
$$
\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}
=
\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n
$$
Indeed, the identity in question has a closely related proof:
it follows from taking determinants in
$$
\begin{pmatrix}L_{n+1}&L_n\\L_n&L_{n-1}\end{pmatrix}
=
\begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1}
\begin{pmatrix}L_{2}&L_1\\L_1&L_{0}\end{pmatrix}
=
\begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1}
\begin{pmatrix}3&1\\1&2\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2779542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Eliminate $x,y,z$ from equations Eliminate $x,y,z$ from equations
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=a$$
$$\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=b$$
$$\left(\frac{x}{y}+\frac{y}{z}\right)\left(\frac{y}{z}+\frac{z}{x}\right)\left(\frac{z}{x}+\frac{x}{y}\right) =c$$
I try this problem with this method.
$$ab=\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)$$
$\implies$$$ab=3+\sum_{cyclic}\frac{x^2}{yz}+\sum_{cyclic}\frac{xy}{z^2}$$
$$c=\left(\frac{x}{y}+\frac{y}{z}\right)\left(\frac{y}{z}+\frac{z}{x}\right)\left(\frac{z}{x}+\frac{x}{y}\right)=2+\sum_{cyclic}\frac{xy}{z^2}+\sum_{cyclic}\frac{x^2}{yz} $$
from above two equations I get,
$ab-3=c-2$
$ab=c+1$.
Is it correct ?or not correct ?
is there exists another way to do it? i want to know solution
| Following is an alternate way to derive $c = ab - 1$.
Let $\lambda_1 = \frac{x}{y}$, $\lambda_2 = \frac{y}{z}$ and $\lambda_3 = \frac{z}{x}$. It is trivial to see $\lambda_1\lambda_2\lambda_3 = 1$.
In terms of $\lambda_1,\lambda_2,\lambda_3$, we have
$$\begin{align}
a &= \frac{x}{y} + \frac{y}{z} + \frac{z}{x} = \lambda_1 + \lambda_2 + \lambda_3\\
b &= \frac{y}{x} + \frac{z}{y} + \frac{x}{z} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3} = \lambda_2\lambda_3 + \lambda_3\lambda_1 + \lambda_1\lambda_2
\end{align}$$
Using Vieta's formula, we obtain
$$(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3) = \lambda^3 - a\lambda^2 + b\lambda - 1$$
Notice the three factors in $c$ are simply $a - \lambda_i$ for $i = 1,2,3$. As a result,
$$c = (a-\lambda_1)(a-\lambda_2)(a-\lambda_3) = a^3 - a a^2 + b a - 1 = ba - 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational.
If $x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational numbers. What does $x^3 - 5x$ equals to?
| Since $x\ne 0$ and $x\ne 2$ we have $${x^3-5x\over x^2-2x} = {x^2-5\over x-2}= q\in \mathbb{Q}$$
So $$x^2-5 = xq-2q \;\;\;\;{\rm and }\;\;\;\;x^2 =2x+r\in \mathbb{Q}$$
so $$ 2x+r-5 = xq-2q\implies (q-2)x =-2q-r+5$$ so if $q\ne 2$ we get $$x={-2q-r+5\over q-2}\;\;\;$$
but this can't be since $x$ is irational and left side is rational so $q=2$ and $r= 1$:
$$ x^3-5x = (x^2-2x)q = qr = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Solve $\int \frac{2x}{(x^2+x+1)^2}dx$ Solve
$$\int \frac{2x}{\left(x^2+x+1\right)^2}dx$$
I tried using integrating in parts by using the partial fraction method where the rational function is split into
$$\frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{(x^2+x+1)^2}$$
But then I just get the original formula with A=$0$ B=$0$ C=2 and D=$0$...
Please help!
| Hint:
$\dfrac{d(x^2+x+1)}{dx}=?$
$$\dfrac{2x}{(x^2+x+1)^2}=\dfrac{2x+1}{(x^2+x+1)^2}-\dfrac1{(x^2+x+1)^2}$$
As $4(x^2+x+1)=(2x+1)^2+3$
how about starting with $2x+1=\sqrt3\tan y$
See also : Trigonometric substitutions, wiki
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to find the numerically greatest term in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$ How to find the numerically greatest term in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$?
My attempt
$$(3x+5y)^{12}=\left(3x\left(1+\frac{5y}{3x}\right)\right)^{12}$$
$$=3^{12}x^{12}\left(1+\frac{5y}{3x}\right)^{12}$$
I then compared $\left(1+\frac{5y}{3x}\right)^{12}$ with $(1+x)^n$ and got $n=17,\space x=(\frac53)(\frac yx)=(\frac53)\frac {\frac43}{\frac12} =\frac{40}{9}$.
Then I used $$\frac{(n+1)|x|}{1+|x|}=\frac{(12+1)\left(\frac{40}{9}\right)}{1+\frac {40}{9}}$$
After solving, I got
$$=10\frac{30}{49}$$ which is not an integer.
I am stuck here, can anyone explain how to solve this problem.
| The $n^{th}$ term of $\left(\frac 32 + \frac{20}{3} \right)^{12}$ is
$t_n =\binom{12}{n}\left(\frac 32\right)^{12-n}\left(\frac{20}{3}\right)^n$ for $n = 0,1,\dots,12$. The ratio of consecutive terms is
$\rho_n = \dfrac{t_{n+1}}{t_n} = \dfrac{40}{9}\left(\dfrac{12-n}{n+1}\right)$.
These ratios are easy to compute
\begin{array}{|c|cccccccccccc|}
\hline
\text{n} & 0 & 1 & 2 & 3
& 4 & 5 & 6 & 7
& 8 & 9 & 10 & 11\\
\hline
r_n & \frac{160}{3} & \frac{220}{9} & \frac{400}{27} & 10
& \frac{64}{9} & \frac{140}{27} & \frac{80}{21} & \frac{25}{9}
& \frac{160}{81} & \frac{4}{3} & \frac{80}{99} & \frac{10}{27} \\
\hline
\end{array}
The last occurence of $r_n>1$ happens at $n=9$. So
$t_{10} =\binom{12}{10}\left(\frac 32\right)^2\left(\frac{20}{3}\right)^{10}$ will be the largest term.
As @BarryCipra has pointed out. It is not necessary to compute the above table. We can solve the following inequality.
\begin{align}
\dfrac{t_{n+1}}{t_n} &\ge 1 \\
\dfrac{40}{9}\left(\dfrac{12-n}{n+1}\right) &\ge 1 \\
480-40n &\ge 9n + 9 \\
49n &\le 472 \\
n &\le \dfrac{472}{49} \\
n &\le 9 \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2784469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $\frac{a}{b} = 1-\frac{1}{2}+\frac{1}{3}-\dots+\frac{1}{67},$ such that $\gcd(a,b)=1$. Show that $101\mid a.$ Let $\frac{a}{b} = 1-\frac{1}{2}+\frac{1}{3}-\dots+\frac{1}{67},$ such that $\gcd(a,b)=1$. Show that $101\mid a.$
i got this problem in olympiad mathematics .
$\frac{a}{b}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{67}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{66}\right)$
$\frac{a}{b}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{67}-\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{33}\right)$
$\frac{a}{b}=\frac{1}{34}+\frac{1}{35}+...+\frac{1}{67}$
Then what will I do?
| Hint:
You're on the right track. In the new sum you have, you can pair up the terms $1/n$ and $1/(101-n)$:
$$\sum_{n=34}^{50} \frac{1}{n}+\frac{1}{101-n}.$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaulating the trigonometric integral $\int \frac{1}{(x^2+1)^2} \, dx$ Problem:
Evaluate the following integral:
\begin{eqnarray*}
\int \frac{1}{(x^2+1)^2} \, dx \\
\end{eqnarray*}
Answer:
To do this, I let $x = \tan u$. Now we have $dx = \sec^2 u du$.
\begin{eqnarray*} \int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\sec^2{u} \, du}{(\tan^2{u} + 1)^2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{1}{\sec^2{u}} \, du
= \int \cos^2{u} \, du \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\cos{(2u)} + 1}{2} \, du
= \frac{\sin(u)}{4} + \frac{u}{2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \frac{\sqrt{1 - \cos^2{u}}}{4} + \frac{u}{2} \\
\end{eqnarray*}
Now, I think I am right so far but I do not know have to get rid of the $u$ in
the $\cos^2(u)$ term. Please help.
Thanks
Bob
| $$\int(1+\cos2u)du=u+\dfrac{\sin2u}2+C$$
$$\sin2u=\dfrac{2\tan u}{1+\tan^2u}=?$$
Another way: $$\int\dfrac{dx}{(x^2+1)^n}=\int\dfrac1{2x}\dfrac{2x}{(x^2+1)^n}dx$$
$$=\dfrac1{2x}\int\dfrac{2x}{(x^2+1)^n}dx-\int\left(\dfrac{d(1/2x)}{dx}\int\dfrac{2x}{(x^2+1)^n}dx\right)dx=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 1
} |
How to show that $\ln\left(\sin(\pi x)/(\pi x)\right)=\sum_{n=1}^{+\infty}\ln\left(1-x^2/n^2\right)$ I wanted to show for $x \in \left]-1,1\right]$ the beautiful relation
$$\ln\left(\frac{\sin\left(\pi x\right)}{\left(\pi x\right)}\right)=\sum_{n=1}^{+\infty}\ln\left(1-\frac{x^2}{n^2}\right)$$
I want to use it in order to prove the complement formula about $\Gamma$ function, so I would like to know if I could show this with differential equation, or partial sums but without fourier series. I know that it is a well known product result if we take the exponential.
| It's pretty straightforward if you are familiar with the product expansion of $\sin\theta$
$$\frac {\sin\theta}{\theta}=\prod\limits_{n\geq1}\left(1-\frac {\theta^2}{n^2\pi^2}\right)$$
Take the natural log of both sides to immediately see that$$\log\left(\frac {\sin\theta}{\theta}\right)=\sum\limits_{n\geq1}\log\left(1-\frac {\theta^2}{n^2\pi^2}\right)$$
Proof: The proof isn't very simple. Start off with the basic identity$$\sin \theta=2\sin\left(\frac {\theta}2\right)\cos\left(\frac {\theta}2\right)=2\sin\left(\frac {\theta}2\right)\sin\left(\frac {\pi}2+\frac {\theta}2\right)\tag1$$
Now, let $\theta\mapsto\frac {\theta}2$ to derive
$$\sin\left(\frac {\theta}2\right)=2\sin\left(\frac {\theta}4\right)\sin\left(\frac {2\pi}4+\frac {\theta}4\right)$$
And similarly, let $\theta\mapsto\frac {\pi}2+\frac {\theta}2$ so that
$$\sin\left(\frac {\pi}2+\frac {\theta}2\right)=2\sin\left(\frac {\pi}4+\frac {\theta}4\right)\sin\left(\frac {3\pi}4+\frac {\theta}4\right)$$Replacing the right-hand side of (1) gives us$$\sin\theta=2^3\sin\left(\frac {\theta}{4}\right)\sin\left(\frac {\pi}4+\frac {\theta}4\right)\sin\left(\frac {2\pi}{4}+\frac {\theta}4\right)+\cdots\tag2$$In fact, continuing indefinitely (try it yourself!), gives us$$\sin\theta=2^{n-1}\sin\left(\frac {\theta}n\right)\sin\left(\frac {\pi}n+\frac {\theta}n\right)\sin\left(\frac {2\pi}n+\frac {\theta}n\right)+\cdots+\sin\left(\frac {\pi(n-1)}n+\frac {\theta}n\right)\tag3$$The last factor of (3) can be simplified to give us what we desire. Using the basic identity for sine that we all learned in first grade, most notably$$\sin(\pi-\theta)=\sin\theta$$we get$$\sin\theta=2^{n-1}\sin\left(\frac {\theta}n\right)\sin\left(\frac {\pi}n+\frac {\theta}n\right)\sin\left(\frac {2\pi}n+\frac {\theta}n\right)\cdots\sin\left(\frac {2\pi}n-\frac {\theta}n\right)\sin\left(\frac {\pi}n-\frac {\theta}n\right)$$And thus, selectivly multiplying the "conjugates" leaves$$\begin{multline}\sin\theta=2^{n-1}\sin\left(\frac {\theta}n\right)\cos\left(\frac {\theta}n\right)\left[\sin^2\left(\frac {\pi}n\right)-\sin^2\left(\frac {\theta}n\right)\right]\left[\sin^2\left(\frac {2\pi}n\right)-\sin^2\left(\frac {\theta}n\right)\right]\\\cdots\left[\sin^2\left(\frac {\pi(\tfrac n2-1)}n\right)-\sin^2\left(\frac {\theta}n\right)\right]\end{multline}\tag4$$Finally, starting with (4), divide both sides by $\sin\left(\frac {\theta}n\right)$ and take the limit as $n\to0$ to isolate $n$ on the left-hand side. Hence$$n=2^{n-1}\sin^2\left(\frac {\pi}n\right)\sin^2\left(\frac {2\pi}n\right)\cdots\sin^2\left(\frac {\pi\left(\tfrac n2-1\right)}n\right)\tag5$$Dividing (4) by (5) and taking the limit as $n\to\infty$ gives$$\begin{align*}\sin\theta & =n\sin\left(\frac {\theta}n\right)\left[1-\frac {\sin^2\left(\frac {\theta}n\right)}{\sin^2\left(\frac {\pi}n\right)}\right]\left[1-\frac {\sin^2\left(\frac {\theta}n\right)}{\sin^2\left(\frac {2\pi}n\right)}\right]\cdots\left[1-\frac {\sin^2\left(\frac {\theta}n\right)}{\sin^2\left(\frac {\pi\left(\frac {n}2-1\right)}n\right)}\right]\end{align*}$$As $n\to\infty$, we are left with$$\sin\theta=\theta\prod\limits_{k\geq1}\left(1-\frac {\theta^2}{k^2\pi^2}\right)$$Hence, the identity has been proven. Man this was a long post!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \frac {dx}{1+\sin x}$ Integrate $\int \dfrac {dx}{1+\sin x}$
My attempt:
$$\begin{align}&=\int \dfrac {dx}{1+\sin x}\\
&=\int{\dfrac{dx}{1+\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1+\tan ^2\left( \dfrac{x}{2} \right)}}}\\
&=\int{\dfrac{1+\tan ^2\left( \dfrac{x}{2} \right)}{\left( 1+\tan \left( \dfrac{x}{2} \right) \right) ^2}}\,dx\end{align}$$
| $$\int\frac{1}{1+sin(x)}dx$$
Apply u-substitution:$u=tan(\frac x2)$
$$=\int \frac{2}{u^2+1+2u}$$
$$=2\int \frac{1}{u^2+1+2u}$$
$$=2\int\frac{1}{(u+1)^2}$$
Again apply u-substitution:$v=u+1$
$$=2\int\frac{1}{v^2}dv$$
$$=2\int\ v^{-2}dv$$
$$=2\frac{v^{-2+1}}{-2+1}$$
Substitute back $v=u+1,u=tan\frac {x}{2}$
$$=-\frac{2}{tan\frac x2+1}$$
$$\int\frac{1}{1+sin(x)}dx=-\frac{2}{tan\frac x2+1}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find $\int\arcsin(\sqrt{x})dx$
Find $\displaystyle\int\arcsin(\sqrt{x})dx$
My Attempt
Put $y=\sqrt{x}\implies dy=\frac{1}{2\sqrt{x}}dx\implies dx=2ydx$
$$
\int\arcsin(\sqrt{x})dx=2\int \arcsin(y)\,y\,dy=2\bigg[\frac{y^2}{2}\arcsin(y)-\int\frac{1}{\sqrt{1-y^2}}\frac{y^2}{2}dy\bigg]\\
=y^2\arcsin(y)-\int\frac{y^2}{\sqrt{1-y^2}}dy.
$$
How do I proceed further and find the solution or is there any easier way ?
| $$\int\frac{-y^2}{\sqrt{1-y^2}}dy=\int\frac{1-y^2-1}{\sqrt{1-y^2}}dy
=\int\frac{1-y^2}{\sqrt{1-y^2}}dy-\int\frac{1}{\sqrt{1-y^2}}dy=$$ $$=
\int\sqrt{1-y^2}dy-\arcsin(y)=\left[y=\sin{t}, dy=\cos{t}\ dt\right]=\int\cos^2{t}dt-\arcsin(y)=$$ $$=\frac12\int(1+\cos{2t})dt-\arcsin(y)=\frac12\left[t+\frac12\sin{2t}\right]_{t=\arcsin{y}}-\arcsin(y)+C=$$ $$=\frac12\arcsin{y}+\frac14\sin{(2\arcsin{y})}-\arcsin(y)+C=$$ $$=-\frac12\arcsin{y}+\frac12\sin{(\arcsin{y})}\cos{(\arcsin{y})}+C=$$
$$=-\frac12\arcsin{y}+\frac12y\sqrt{1-y^2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using identity Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using the identity $$\frac{1}{(1-x)(1-x^2)}=\frac{1}{2}\left(\frac{1}{(1-x)^2}\right)+\frac{1}{2}\left(\frac{1}{1-x^2}\right)$$
where $p_k(n)$ is the number of partitions of an integer $n$ into a most $k$ parts. The generating function $P_k(x)$ of $\{p_k(n)\}$ is $$P_k(x) = \frac{1}{\prod_{r=0}^{k}(1-x^k)}$$
Therefore, the generating function $P_2(x)$ of ${p_2(n)}$ is
$$P_2(x) = \frac{1}{(1-x)(1-x^2)}$$
Now I see here that we can use the identity given above, but I am confused of how to apply it to prove the desired statement.
| You have $$P_2(x) = \frac{1}{2}\left(\frac{1}{(1-x)^2}\right)+\frac{1}{2}\left(\frac{1}{1-x^2}\right) = \frac12 (1-x)^{-2} + \frac12 (1-x^2)^{-1}$$
Now, the generalised binomial expansion is $(1+x)^a = 1 + \frac{a}{1!}x + \frac{a(a-1)}{2!}x^2 + \cdots$
Therefore $$\begin{eqnarray}
(1 - x)^{-2} &=& 1 + \frac{(-2)}{1!}(-x) + \frac{(-2)(-3)}{2!}(-x)^2 + \cdots \\
&=& 1 + 2x + 3x^2 + \cdots
\end{eqnarray}$$
and $$\begin{eqnarray}(1-x^2)^{-1} &=& 1 + \frac{(-1)}{1!}(-x^2) + \frac{(-1)(-2)}{2!}(-x^2)^2 + \cdots \\
&=& 1 + x^2 + x^4 + \cdots
\end{eqnarray}$$
Combine and you're done. It may be helpful to consider separately the cases where $n$ is even vs odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the inverse laplace transform of the given problem Find the inverse laplace transform of $ \ Y(s)= \large \frac{\large \frac{82}{ \large s-6}-2s+2}{s^2+6s+10} \ $
Answer:
Let $ \ \mathcal{L}^{-1} \ $ be the inverse laplace operator.
Then,
$ y(t)=\mathcal{L}^{-1} [Y(s);t] \\ \Rightarrow y(t)= \mathcal{L}^{-1} \left[\frac{\large \frac{82}{ \large s-6}-2s+2}{s^2+6s+10} \right] \ = \mathcal{L}^{-1} \left[\frac{ \large -2s^2+12s+70}{ \large (s-6)(s^2+6s+10)} \right] $
Now,
$ \frac{ \large -2s^2+12s+70}{ \large (s-6)(s^2+6s+10)}= \frac{A}{s-6}+\frac{\large Bs+C}{s^2+6s+10} \ $ where $ \ A,B,C \ $ are unknown constants to be determined.
Is this the correct partial fraction?
Help me find the inverse laplace transform.
| You are on the right track, after you work out the constants $A$, $B$ and $C$ you should end up with
\begin{eqnarray}
Y(s) &=& \frac{1}{s - 6} - \frac{3s + 10}{s^2 + 6s + 10} \\
&=& \frac{1}{s - 6} - \frac{3(s + 3) + 1}{(s + 3)^2 + 1} \\
&=& \frac{1}{s - 6} - 3\frac{(s + 3)}{(s + 3)^2 + 1} - \frac{1}{(s + 3)^2 + 1}
\end{eqnarray}
Now use the fact that
$$
\mathcal{L}[e^{at}\sin bt] = \frac{b}{(s-a)^2 + b^2}
$$
and
$$
\mathcal{L}[e^{at}\cos bt] = \frac{s-a}{(s-a)^2 + b^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Which one of the following are correct?
Let, $A= \left[ {\begin{array}{cc} -1 & 2 \\ 0 & -1 \\
\end{array} } \right]$ , and $B = A + A^2 + A^3 +···+ A^{50}$. Then
$(A) B^2 = I $
$(B) B^2 = 0$
$(C) B^2 = A$
$(D) B^2 = B$
Eigenvalues of $A$ are $-1,-1$. So, Eigenvalues of $B$ are $0,0$. So, $\det(B)=0$. So, options (A) and (C) can be eliminated. How do I eliminate further?. By Cayley-Hamilton Theorem $A^2+2A+I=0$. I found $A^3,A^4,...$ using Cayley-Hamilton Theorem. It was time-consuming. Given matrix is not diagonalizable. Can you please help me?
| Since $A-I$ is triangular with nonzero diagonal entries, it follows that $A-I$ has nonzero determinant, so is invertible, hence $I-A$ is also invertible.
Note also that any two poynomials in $A$ commute.
\begin{align*}
\text{Then}\;\;(I-A)B&=(I-A)(A+A^2+A^3+\cdots+A^{50})\\[4pt]
&=(A + A^2 + A^3 +···+ A^{50})-(A^2 + A^2 + A^3 +···+ A^{51})\\[4pt]
&=A-A^{51}\\[4pt]
&=(I-A^{50})A\\[4pt]
&=(I+A^{25})(I-A^{25})A\\[4pt]
&=(I+A)(1-A+A^2-\cdots+A^{24})(I-A^{25})A\\[4pt]
&=(I+A)C,\;\text{where}\;C=(1-A+A^2-\cdots+A^{24})(I-A^{25})A\\[4pt]
\implies\;\bigl((A-I)B\bigr)^2&=\bigl((I+A)C\bigr)^2\\[4pt]
\implies\;(A-I)^2B^2&=(I+A)^2C^2\qquad\text{[since $B,C$ commute with all polynomials in $A$]}\\[4pt]
\implies\;(A-I)^2B^2&=0\qquad\text{[since $(I+A)^2=A^2+2A+I=0$]}\\[4pt]
\implies\;B^2&=0\qquad\text{[since $(I-A)$ is invertible]}\\[4pt]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solve this system of equations for the value of $a,b$ and $c$ such that $a^3+3ab^2+3ac^2-6abc=1$,$b^3+3ba^2+3bc^2-6abc=1$,$c^3+3cb^2+3ca^2-6abc=1$ Solve this system of equations for the value of $a,b$ and $c$.
$$a^3+3ab^2+3ac^2-6abc=1$$
$$b^3+3ba^2+3bc^2-6abc=1$$
$$c^3+3cb^2+3ca^2-6abc=1$$
This is symmetric equation.
My first attempt was,
$a^3+3ab^2+3ac^2=1+6abc$,
$b^3+3ba^2+3bc^2=1+6abc$,
$c^3+3cb^2+3ca^2=1+6abc$,
then, from above three equations,
$a^3+3ab^2+3ac^2=b^3+3ba^2+3bc^2=c^3+3cb^2+3ca^2$
Then, what will I continue?
| Let $$a^3+3ab^2+3ac^2-6abc-1=b^3+3a^2b+3bc^2-6abc-1$$ then we get
$$a^3-b^3-3ab(a-b)+3c^2(a-b)=0$$ Now use that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
Then you will get
$$(a-b)\left((a-b)^2+3c^2)\right)=0$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int e^{-3x} \cos^3x\,dx$
Evaluate $\int e^{-3x}\cos^3x\,dx$
My Attempt
\begin{align}
& \int e^{-3x} \cos^3x\,dx=\cos^3x \cdot \frac{e^{-3x}}{-3}-\int3\cos^2x\sin x \cdot \frac{e^{-3x}}{-3} \, dx \\
= {} &\frac{-1}{3}\cos^3x \cdot e^{-3x}+\int\cos^2x\sin x \cdot e^{-3x} \, dx\\
= {} & \frac{-1}{3}\cos^3x\cdot e^{-3x}+\cos^3x\cdot e^{-3x} - \int\bigg[-2\cos x\sin x\cdot e^{-3x}-3\cos^2x\cdot e^{-3x}\bigg](-\cos x)\,dx\\
= {} & \frac{-1}{3}\cos^3x\cdot e^{-3x}+\cos^3x\cdot e^{-3x}+2\int \cos^2x\sin x \cdot e^{-3x} \, dx-3\int\cos^3x\cdot e^{-3x}\,dx
\end{align}
How do I solve this integral ?
| $$\int e^{-3x}\cos^3x dx=\frac{1}{8}\int e^{-3x}(e^{ix}+e^{-ix})^3dx\\
=\frac{1}{8}\int e^{-3x}(e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix})dx\\
=\frac{1}{8}e^{-3x}\left(\frac{e^{3ix}}{-3+3i}+\frac{e^{-3ix}}{-3-3i}+\frac{3e^{ix}}{-3+i}+\frac{3e^{-ix}}{-3-i}\right)\\
=-\frac{1}{8}e^{-3x}\left(\frac{3(e^{3ix}+e^{-3ix})+3i(e^{3ix}-e^{-3ix})}{18}+\frac{9(e^{3ix}+e^{-3ix})+3i(e^{ix}-e^{-ix})}{10}\right)\\
=-\frac{1}{8}e^{-3x}\left(\frac{\cos3x-\sin3x}{3}+\frac{9\cos x-3\sin x}{5}\right).\\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$
Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$
My Attempt
Using Partial fractions
$$
\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}=\frac{1}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)}+\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}=I_1+I_2\\
I_1=\bigg[\frac{1}{b^2}.\frac{1}{ab}\tan^{-1}\frac{bx}{a}\bigg]^\infty_0=\bigg[\frac{1}{ab^3}\tan^{-1}\frac{bx}{a}\bigg]^\infty_0
$$
$$
I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}=\frac{b^2-a^2}{b^2}\bigg[\frac{x}{(a^2+b^2x^2)^2}-\int\frac{-2.2x.x}{(a^2+b^2x^2)^3}dx\bigg]
$$
How do I evaluate the integral $I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}$ ?
| $$I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}$$
If $x = \frac ab \tan \theta$, then
*
*$(a^2+b^2x^2)=a^2(1+tan^2 \theta)=a^2 sec^2 \theta$
*$dx = \frac ab sec^2 \theta \ d\theta$
*$\displaystyle
\int \frac{dx}{(a^2+b^2x^2)^2}
= \int \frac{\frac ab sec^2 \theta \ d\theta}{a^4 sec^4 \theta}
= \int \frac{1}{a^3b}\cos^2 \theta \ d\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving the DE: $y''+1/(2y^3)=0$. \begin{align}
y''+\frac{1}{2y^3}=0\\
\end{align}
This is a second order differential equation which doesn't contain $x$ explicitly.
Let $y'=p(y)$, so that $y''=p(y)*p'(y)$.
\begin{align}
pp'+\frac{1}{2y^3}=0\\
\int{pdp}=\int{-\frac{1}{2y^3}dy}\\
\frac{p^2}{2}=-\frac{1}{2}*\frac{-1y^{-2}}{2}+C_1\\
p^2=\frac{1}{2y^2}+C_1\\
p=\pm\sqrt{\frac{1+C_1y^2}{2y^2}}\\
\end{align}
$p$ was equal to $y'$
\begin{align}
y'=\pm\sqrt{\frac{1+C_1y^2}{2y^2}}\\
\end{align}
I've tried to continue with trigonometric substitution but that didn't work for me.
Can you please help me out?
Thank you in advance!
Edit: adding final solution
Because of the answers below, I found the solution:
\begin{align}
\frac{\sqrt2ydy}{\sqrt{1+C_1y^2}}=\pm{x}dx\\
\int{\frac{\sqrt2ydy}{\sqrt{1+C_1y^2}}}=\int{\pm{x}dx}\\
\frac{1}{2C_1}\int{\frac{d({1+C_1y^2})}{\sqrt{1+C_1y^2}}}=\pm\frac{1}{\sqrt2}x\\
2\sqrt{1+C_1y^2}=\pm{\frac{2C_1}{\sqrt{2}}}(x+C_2)\\
\sqrt{1+C_1y^2}=\pm{\frac{C_1}{\sqrt{2}}}(x+C_2)\\
2(1+C_1y^2)=C_1^2(x+C_2)\\
\end{align}
| $y' = \frac {\sqrt {Cy^2 + 1}}{\sqrt 2 y}\\
\int \frac {y}{\sqrt {Cy^2 + 1}} \ dy =\int \frac {1}{\sqrt 2} \ dt \\
\frac 1C \sqrt {Cy^2 + 1} = \frac {1}{\sqrt 2} t + D $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
General term of a sequence. So i have the following sequence:
${1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, ...}$
Where the number $i$ appears $i + 1$ times.
I would like to know the $n$-th term of this sequence. I tried to analise certain patterns within the sequence, but wasn´t able to conclude anything so far.
| I would like to put a "number pyramid" like this:
$$1-1$$
$$2-2-2$$
$$3-3-3-3$$
$$4-4-4-4-4$$
$$5-5-5-5-5-5$$
$$\cdots$$
$$k-\text{th floor: } k-k-k-k-k-...-k-k\text{ ($k+1$ times)}$$
The number of numbers appear in all the floors from $1$ to $k-1$:
$$2+3+4+5+\cdots+k=\dfrac{(k+2)(k-1)}{2}$$
The number of numbers appear in all the floors from $1$ to $k$:
$$2+3+4+5+\cdots+k+k+1=\dfrac{(k+3)k}{2}$$
Assume that the $n$-th term of the sequence above is on the $k$-th floor (the bottom floor of the pyramid above), then $n$ may or may not be the last term of the $k$-th floor, so this inequality must hold (we need to find $k\in\mathbb{Z^+}$ given $n\in\mathbb{Z^+}$):
$\dfrac{(k+2)(k-1)}{2}<n\le\dfrac{(k+3)k}{2}$
$\Leftrightarrow k^2+k-2<2n\le k^2+3k$
$\Leftrightarrow \begin{cases}k^2+k-2n-2<0\\k^2+3k-2n\ge 0\end{cases}$
$$k^2+k-2n-2<0$$
$\Leftrightarrow k^2+2\times k\times 0.5+0.25-2n-2.25<0$
$\Leftrightarrow (k+0.5)^2<2n+2.25$
$\Leftrightarrow k+0.5<\sqrt{2n+2.25}$ (because $k,n>0$)
$\Leftrightarrow k<-0.5+\sqrt{2n+2.25}$
$\Leftrightarrow k<\dfrac{-1+\sqrt{8n+9}}{2}$
$$k^2+3k-2n\ge 0$$
$\Leftrightarrow (k+1.5)^2\ge 2n+2.25$ (similar to above)
$\Leftrightarrow k+1.5\ge \sqrt{2n+2.25}$ (because $k,n>0$)
$\Leftrightarrow k\ge \dfrac{-3+\sqrt{8n+9}}{2}$
Combine both equations, we have
$$\dfrac{-3+\sqrt{8n+9}}{2}\le k<\dfrac{-1+\sqrt{8n+9}}{2}$$
Other notes:
*
*The conclusion above is still true for $n\in\mathbb\{1;2\}$ and $k=1$. When $k=1$, the number of numbers appear in all the floors from $1$ to $0$ is zero (no numbers exist), because when $k=1$ we have $\dfrac{(k+2)(k-1)}{2}=0$.
*There is always exactly one positive integer $k$ satisfy the conclusion above (for all $n\in\mathbb{Z^+}$), because the difference between the right hand side and the left hand side is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Simplifying $\frac{\sin\alpha\cos^2\beta}{\sin\beta}+\frac{\sin\beta\cos^2\alpha}{\sin\alpha}$ How to simplify the expression and get a closed form?
$$\frac{\sin\alpha\cos^2\beta}{\sin\beta}+\frac{\sin\beta\cos^2\alpha}{\sin\alpha}$$
where $\alpha := \frac12(m+1)\theta$ and $\beta:=\frac12(m-1)\theta$; and where $m$ is an integer.
I wanna get rid of the denominators or make it a constant without $m$, but don't know how.
| \begin{align}\frac{\sin\frac{m+1}{2}\theta\cos^2\frac{m-1}{2}\theta}{\sin\frac{m-1}{2}\theta}+\frac{\sin\frac{m-1}{2}\theta\cos^2\frac{m+1}{2}\theta}{\sin\frac{m+1}{2}\theta}&=\frac{\left( \sin \frac{m+1}2\theta \cos \frac{m-1}2\theta\right)^2+\left( \sin \frac{m-1}2\theta\cos\frac{m+1}2\theta\right)^2}{\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}\\
&=\frac{\left( \sin \frac{m+1}2\theta \cos \frac{m-1}2\theta\right)^2+\left( \sin \frac{m-1}2\theta\cos\frac{m+1}2\theta\right)^2}{\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}\\
&=\frac{\left( \sin m\theta +\sin \theta\right)^2+\left( \sin m\theta-\sin\theta\right)^2}{4\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}\\
&=\frac{\sin^2 m\theta + \sin^2 \theta}{{2\sin \frac{m-1}2\theta \sin \frac{m+1}2\theta}} \\
&=\frac{\sin^2 m\theta + \sin^2 \theta}{\cos \theta- \cos m\theta}
\end{align}
From this desmos link, the vertical asymptotes is a function of $m$. Hence $m$ is a term in the denominator.
The vertical asymptotes (which due to dision by $0$ in the denominator) appear where $\sin \left( \frac{m-1}2 \theta\right)=0$ or $\sin \left( \frac{m+1}2 \theta\right)=0$, that is when
$$\frac{m \pm 1}2 \theta = k\pi, k \in \mathbb{Z}$$
$$\theta = \frac{2k\pi}{m \pm 1} , k \in \mathbb{Z}$$
which is a function of $m$. The denominator is dependent on $m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $\lim_{n\to\infty}\int_0^{\frac {\pi}{3}}\frac {\sin^nx}{\sin^nx+\cos^nx}dx$
Evaluate $\lim_{n\to\infty}\int_0^{\frac {\pi}{3}}\frac {\sin^nx}{\sin^nx+\cos^nx}dx$
I tried using the substitution $u=\frac {\pi}{2}-x$ and maybe thought this function might be symmetrical in some way and got: $$\lim_{n\to\infty}\int_{\frac {\pi}{6}}^{\frac {\pi}{2}}\frac {\cos^nx}{\sin^nx+\cos^nx}dx$$
but it's not really helping me... any other ideas?
| A heuristic solution. Interchanging the order of integral and limit, we have
$$ \lim_{n\to\infty} \int_{0}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx = \int_{0}^{\frac{\pi}{3}} \lim_{n\to\infty} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx. $$
Now depending on the relative size of $\sin x$ and $\cos x$, we have
$$ \lim_{n\to\infty}\frac{\sin^n x}{\sin^n x+\cos^n x} =
\begin{cases}
0, & \text{if } 0 \leq x < \frac{\pi}{4} \\
\frac{1}{2}, & \text{if } x = \frac{\pi}{4} \\
1, & \text{if } \frac{\pi}{4} < x \leq \frac{\pi}{3}.
\end{cases} \tag{1} $$
So it follows that the limit is $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} dx = \frac{\pi}{12}$.
Remark. Indeed we took a huge leap by assuming that the integral and the limit can be switched. This is indeed possible in our case, although a direct justification required some advance results such as dominated convergence theorem.
A preliminary analysis level solution. Fix any sufficiently small $\epsilon > 0$ and consider $a=\frac{\pi}{4}-\epsilon$ and $b = \frac{\pi}{4}+\epsilon$. If we write $I_n = \int_{0}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx$, then the integrand is monotone increasing in $x$ and hence
\begin{align*}
I_n
&\geq \int_{b}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx \\
&\geq \int_{b}^{\frac{\pi}{3}} \frac{\sin^n b}{\sin^n b+\cos^n b} \, dx \\
&= \left(\frac{\pi}{3}-b\right)\frac{\sin^n b}{\sin^n b+\cos^n b}
\end{align*}
This gives
$$ \liminf_{n\to\infty} I_n \geq \lim_{n\to\infty} \left(\frac{\pi}{3}-b\right)\frac{\sin^n b}{\sin^n b+\cos^n b} = \frac{\pi}{12}-\epsilon.$$
But since the LHS of the above inequality is a constant independent of $\epsilon$, letting $\epsilon \downarrow 0$ proves that $\liminf_{n\to\infty} I_n \geq \frac{\pi}{12}$. Similarly,
\begin{align*}
I_n
&\leq \int_{0}^{a} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx + \int_{a}^{\frac{\pi}{3}} \frac{\sin^n x}{\sin^n x+\cos^n x} \, dx \\
&\geq \int_{0}^{a} \frac{\sin^n a}{\sin^n a+\cos^n a} \, dx + \int_{a}^{\frac{\pi}{3}} dx \\
&= a\frac{\sin^n a}{\sin^n b+\cos^n a} + \left(\frac{\pi}{3}-a\right)
\end{align*}
and hence
$$ \limsup_{n\to\infty} I_n \leq \lim_{n\to\infty} \left[ a\frac{\sin^n a}{\sin^n b+\cos^n a} + \left(\frac{\pi}{3}-a\right) \right] = \frac{\pi}{12}+\epsilon. $$
Letting $\epsilon \downarrow 0$, we obtain $\limsup_{n\to\infty} I_n \leq \frac{\pi}{12}$. These together tell that
$$\liminf_{n\to\infty} I_n = \limsup_{n\to\infty} I_n = \frac{\pi}{12}$$
and therefore the limit of $I_n$ exists and has the value $\frac{\pi}{12}$.
Addendum. The following demonstrates graphs of the integrand for different values of $n$.
$\hspace{5em}$
The graph is already quite close to $\text{(1)}$ when $n=100$, which provides a sanity check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Another way to find the sum $S=1-\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{7}{8}+\cdots+(-1)^{n-1}\cdot\dfrac{2n-1}{2^{n-1}}$ I want to know the sum
$$S=1-\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{7}{8}+\cdots+(-1)^{n-1}\cdot\dfrac{2n-1}{2^{n-1}}.$$ This is my way.
First, we find the sum
$$1 + 3 x + 5x^2 + \cdots + (2n-1) \cdot x^{n-1}=\sum _{k=1}^n (2 k-1) x^{k-1}.$$
We have
$$\sum _{k=1}^n (2 k-1) x^{k-1} = 2\sum _{k=1}^n k\cdot x^{k-1}-\sum _{k=1}^n x^{k-1}. $$
Note that
$$\sum _{k=1}^n k\cdot x^{k-1} = \left (\sum _{k=1}^n x^k\right)' = \left (\dfrac{x\left(x^n-1\right)}{x-1}\right)'=\dfrac{n x^{n+1}-(n+1)x^n+1}{(x-1)^2}.$$
Another way
$$\sum _{k=1}^n x^{k-1} = \dfrac{x^n-1}{x-1}.$$
From the above results, we have
$$1 + 3 x + 5x^2 + \cdots + (2n-1) \cdot x^{n-1}= \dfrac{(2 n-1) x^{n+1}-(2 n+1)x^n+x+1}{(x-1)^2}.$$
With $x=-\dfrac{1}{2}$,we get
$$S=\dfrac{2^n + (-1)^{n+1} \cdot(6n+1)}{9 \cdot 2^{n-1}}.$$
How to find the result with another way?
| $$S+\frac S2=\sum_1^n\frac{(-1)^{k-1}(2k-1)}{2^{k-1}}+\sum_1^n\frac{(-1)^{k-1}(2k-1)}{2^{\color{red}k}}\\
=\sum_1^n\frac{(-1)^{k-1}(2k-1)}{2^{k-1}}+\sum_\color{red}2^{\color{red}{n+1}}\frac{(-1)^{k-2}(2k-3)}{2^{\color{red}{k-1}}}\\
=1+\sum_2^n\frac{(-1)^{k-2}(\color{red}{2k-1-2k+3})}{2^{k-1}}+\frac{(-1)^{n-1}(2n-1)}{2^n}.$$
The rest is well-known.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find smallest $A\in \mathbb{R}$ such that $|f'(0)|\leq A.$ Problem
Find smallest $A\in \mathbb{R}$ such that for 3-degree polynomial satisfying the condition $\forall x\in[0;1],\ |f(x)|\le 1$ holds $|f'(0)|\le A$.
Solution
Assume that the 3-degree polynomial is $$ax^3+bx^2+cx+d,(a \neq 0).$$Let $x=0,\dfrac{1}{3},\dfrac{2}{3},1$ respectively. Then $$\begin{cases}d=f(0),\\\dfrac{a}{27}+\dfrac{b}{9}+\dfrac{c}{3}+d=f\left(\dfrac{1}{3}\right),\\\dfrac{8a}{27}+\dfrac{4b}{9}+\dfrac{2c}{3}+d=f\left(\dfrac{2}{3}\right),\\a+b+c+d=f\left(1\right).\end{cases}$$
By solving the linear equation system, we obtain $$\begin{align*}f'(0)&=c=-\frac{11}{2}f(0)+9f\left(\frac{1}{3}\right)-\frac{9}{2}f\left(\frac{2}{3}\right)+f(1)\\&\leq\frac{11}{2}|f(0)|+9\left|f\left(\frac{1}{3}\right)\right|+\frac{9}{2}\left|f\left(\frac{2}{3}\right)\right|+|f(1)|\\&\leq\frac{11}{2}+9+\frac{9}{2}+1\\&=20.\end{align*}$$
Hence, $$A=20.$$
This is right? I'm not sure for that.
| I think an approach from interval arithmetic could be appropriate. Here the equation
$$
f([0,1])=[0,1]
$$
must be satisfied for your requirements.
$$
f([0, 1]) = a[0,1]^3+b[0, 1]^2+c[0,1]+d
$$
$$
= a[0,1] + b[0,1] +c[0,1] + d
$$
$$
= (a +b + c) [0,1] + d = [0,1]
$$
In order for the last equation to be valid statement $d = [d_1, d_2]$ must be an interval, too. From the the rule of addition of intervals we obtain $d_1=0$. So we can look at only the upper limits of the intervals.
$$
a + b + c + d_2 = 1
$$
Any combination of the coefficients satisfying this equation will fulfill the requirement. Thus you can choose and let $c = A = 0$ leaving
$$
a+b+d = 1,
$$
where $d\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Same integration with 2 different answers? $$\int x(x^2+2)^4\,dx $$
When we do this integration with u substitution we get
$$\frac{(x^2+2)^5}{10}$$
as $u=x^2+2$
$du=2x\,dx$
$$\therefore \int (u+2)^4\,du = \frac{(x^2+2)^5}{10} + C$$
Although when we expand the fraction and then integrate the answer we get is different:
$x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$
$$\int x^9+8x^7+24x^5+32x^3+16x \,dx$$
we get
$$\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2 + C$$
For a better idea of the questions, let's say the questions asks us to find the value of C when y(0)=1
Now,
$x=0$
$$\frac {0^{10}}{10} + 0^8 + 4(0)^6 + 8(0)^4 + 8(0)^2 + C = 1$$
$$\therefore C= 1$$
AND
$$\frac {(0+2)^5}{10} + C= 1$$
$$\therefore \frac {32}{10} + C = 1$$
$$\therefore C = 1 - 3.2 = -2.2$$
| Like mentioned in the comments this is all fixed if you remember your constant of integration.
$$\int x(x^2+2)^4\ dx= \frac{(x^2+2)^5}{10}+C$$
Note if you expand
$$
\begin{split}
\frac{(x^2+2)^5}{10}&=\frac{1}{10}\left(x^{10}+5x^8(2)+10x^6(2^2)+10x^4(2^3)+5x^2(2^4)+2^5\right)\\
&=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\frac{32}{10}
\end{split}
$$
Notice the relation to your other way of computing the integral
$$
\int x(x^2+2)^4\ dx = \frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2 +C
$$
So lets call $F(x)=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2$ and $G(x)=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\frac{32}{10}$
then $F(x)-G(x)=-\frac{32}{10}$ a constant. All antiderivatives of a continuous function only differ by a constant.
Just for fun Let's see another one:
First lets use double angle for sine
$$
\int \cos x\sin x\ dx=\frac{1}{2}\int\sin 2x\ dx=-\frac{1}{4}\cos 2x +C
$$
Then substitutions $u=\sin x$
$$
\int \cos x\sin x\ dx=\int u\ du =\frac{u^2}{2}+C=\frac{\sin^2 x}{2}+C
$$
Then substitutions $u=\cos x$
$$
\int \cos x\sin x\ dx=\int -u\ du =\frac{-u^2}{2}+C=\frac{-\cos^2 x}{2}+C
$$
If you find the constant differences and combine them in the right way you get the half angle formulas:
$$
\sin^2 x=\frac{1-\cos 2x}{2},\quad \cos^2 x=\frac{1+\cos 2x}{2}
$$
Note you can pretty quickly derive some funky trig identities in this way. For instance if you consider $\int \cos^3 x \sin^5 x\ dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\sum_{r=1}^{n}{\cot{}^{ - 1}(3r^2-\frac{5}{12})}$
Find the sum of
$$\sum_{r=1}^{n}{\cot{}^{ - 1}(3r^2-\frac{5}{12})}$$
I tried to convert all into
$\tan {}^{ - 1}(x) $
using
$\cot {}^{ - 1} (x) = { \tan {}^{ - 1} ( \frac{1}{x} ) }$
and then tried to simplify them using property
$ \tan {}^{ - 1} ( \frac{x - y}{1 - xy} ) = \tan {}^{ - 1} (x) - \tan {}^{- 1} (y) $
so that some terms cancel each other.
But I do not manage to simplify the problem. What to do?
| After the fix of sign in front of $\frac{5}{12}$ in the question body, the sum becomes pretty simple.
Let $a_{\pm} = 3(r\pm \frac12)$, notice
$$\frac{a_+ - a_-}{1 + a_+ a_-} = \frac{3}{1 + 9(r^2-\frac14)}
= \frac{12}{36r^2 - 5} = \left(3r^2 - \frac{5}{12}\right)^{-1}$$
We have
$$\begin{align}\cot^{-1}\left(3r^2 - \frac{5}{12}\right)
&= \tan^{-1}\left(\frac{a_+ - a_-}{1 + a_+a_-}\right)
= \tan^{-1}a_+ - \tan^{-1}a_-\\
&= \tan^{-1}(3(r + \frac12)) - \tan^{-1}(3(r - \frac12))
\end{align}
$$
The new sum is a telescoping one and
$$\sum_{r=1}^n
\cot^{-1}\left(3r^2 - \frac{5}{12}\right)
= \tan^{-1}(3(n + \frac12)) - \tan^{-1}(\frac32)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2817270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Value of $\cos\frac{2k\pi}{n}$ We know that $\cos\frac{2\pi}{5} = \frac{-1 + \sqrt{5}}{4}$ and $\cos\frac{4\pi}{5} = \frac{-1 - \sqrt{5}}{4}$. In general is there any formula for
$$\cos\frac{2k\pi}{n}$$
where $k\leq n/2$.
| Claim: for any $n\geq 3$, $\cos\frac{2\pi}{n}$ is an algebraic number over $\mathbb{Q}$ with degree $\frac{1}{2}\varphi(n)$, hence it can be found by solving $p(x)=0$ where $p(x)\in\mathbb{Z}[x]$ is such that $\deg p = \frac{1}{2}\varphi(n)$.
Proof: we first invoke a well-known lemma, i.e. that the cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$. The complex roots of $\Phi_n(x)$ are the primitive $n$-th roots of unity, i.e. the numbers of the form $\exp\left(\frac{2\pi i k}{n}\right)$ with $\gcd(k,n)=1$. In particular $\deg \Phi_n=\varphi(n)$. $\Phi_n$ is a palindromic polynomial, since if $\Phi_n(z)=0$ then $\Phi_n(z^{-1})=0$ too. Thus there is a polynomial $q(x)\in\mathbb{Q}[x]$ such that
$$ \frac{\Phi_n(z)}{z^{\varphi(n)/2}} = q\left(z+\frac{1}{z}\right),\quad \deg q=\frac{\varphi(n)}{2} $$
and a polynomial $p(x)\in\mathbb{Q}[x]$ with $\deg p=\frac{\varphi(n)}{2}$ such that
$$ p\left(\cos\frac{2\pi k}{n}\right)=0,\quad \forall k:\gcd(k,n)=1.$$
This polynomial $p(x)$ is necessarily irreducible over $\mathbb{Q}$, since we have $\left[\mathbb{Q}(e^{2\pi i/n}):\mathbb{Q}(\cos(2\pi/n))\right]=2$. If $p(x)$ were reducible, we would have $[\mathbb{Q}(e^{2\pi i/n}):\mathbb{Q}]<\varphi(n)$, contradicting the irreducibility of $\Phi_n(x)$.
Corollary 1: The regular eptagon and nonagon cannot be constructed with straightedge and compass. Indeed both $\varphi(7)$ and $\varphi(9)$ equal $6$, so both $\cos\frac{2\pi}{7}$ and $\cos\frac{2\pi}{9}$ are algebraic numbers over $\mathbb{Q}$ with degree $3$ and the construction of the associated polygons implies the extraction of a cube root, which is impossible with straightedge and compass. The involved minimal polynomials are
$$ 8x^3+4x^2+4x-1,\quad 8x^3-6x+1. $$
Corollary 2: the regular $5$-agon, $6$-agon, $8$-agon, $10$-agon, $12$-agon, $15$-agon, $16$-agon and $17$-agon are constructible with straightedge and compass. Indeed
$$ \cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4},\quad \cos\frac{2\pi}{6}=\frac{1}{2},\quad \cos\frac{2\pi}{8}=\frac{\sqrt{2}}{2},\quad \cos\frac{2\pi}{10}=\frac{\sqrt{5}+1}{4},$$
$$\cos\frac{2\pi}{12}=\frac{\sqrt{3}}{2},\quad \cos\frac{2\pi}{15}=\frac{1}{8} \left(1+\sqrt{5}+\sqrt{6 \left(5-\sqrt{5}\right)}\right),\quad \cos\frac{2\pi}{16}=\frac{1}{2}\sqrt{2+\sqrt{2}}$$
$$\cos\frac{2\pi}{17}=\frac{1}{16}\left[-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}\right].$$
Corollary 3: the regular $N$-agon is constructible with straightedge and compass iff $N$ is a number of the form $2^\alpha M$, with $M$ being a product of distinct Fermat primes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2817963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding $x-\frac{1}{x}$ if $x^{3}-\frac{1}{x^{3}}=76$. If $x$ is a real number that satisfies $x^{3}-\frac{1}{x^{3}}=76$, determine the value of $x-\frac{1}{x}$.
How can I solve this problem?
| $a^3-b^3=(a-b)(a^2+ab+b^2)$ and note that $(x-\frac{1}{x})^2 =x^2+\frac{1}{x^2}-2$ then if we set $(x-\frac{1}{x})=t$ we have $t(t^2+3)=76$ now find $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of given sequence Find the limit of the sequence {$a_{n}$}, given by$$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=\dfrac {1}{3}(1+a_{n}+a^{3}_{n-1}), \ for \ n \ > \ 1$$
My try:
$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54$ that is the sequence is incresing and each term is positive. Let the limit of the sequence be $x$.
Then $ \lim _{n\rightarrow \infty }a_{n+1}=\lim _{n\rightarrow \infty }a_{n}=x$
$$ \lim _{n\rightarrow \infty }a_{n+1}= \lim _{n\rightarrow \infty }1+a_{n}+a^{3}_{n-1}$$
$\Rightarrow x=\dfrac {1}{3}( 1+x+x^3)$
$\Rightarrow x^3-2x+1=0$
and this equation has three roots $x=\dfrac {-1\pm \sqrt {5}}{2},1$
So the limit of the sequence is $\dfrac {-1 + \sqrt {5}}{2}$.
how can i say that the limit is
$\dfrac {-1 + \sqrt {5}}{2}$?
| Because $f(x)={1\over 3}(1+x+x^3)$ is a strictly increasing function, compute its derivative, show recursively that that $0<a_n$, remark that if $a_{n-1},a_{n-2}$ are strictly inferior to $1$, $a_n\geq f(max(a_{n-1},a_{n-2}))$ and $f(max(a_{n-1},a_{n-2}))<f(1)=1$. _{n-1})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve Differential equation $2t\ y(t)\ y'(t)=t^2+3y(t)^2$ with substitution I am trying to solve this differential equation $2t\ y(t)\ y'(t)=t^2+3y(t)^2$ using substitution, with the initial value $y(1)=0$ This is what I've done so far:
$$
\begin{align*}
y'(t) &= \frac{t^2+3y(t)^2}{2ty(t)} = \frac{t^2}{2ty(t)} + \frac{3y(t)^2}{2ty(t)} \\
&= \frac{t}{2y(t)} + \frac{3y(t)}{2t} \\
&= \frac{1}{\frac{2y(t)}{t}} + \frac{3y(t)}{2t} \\
\\
\text{Substitution:}\\
f\Big(\frac{y(t)}{t}\Big) &= y'(t) \ \ \text{with} \ \ f(z(t)) = \frac{1}{2z(t)} + \frac{3}{2}z(t) \\
z(t)&= y(t) \ \ \text{and} \ \ z_0 = z(t_0) = z(1) = 0 \\
\\
z'(t) &= \frac{1}{t} \Big( f(z(t) - z(t)\Big) = \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{3z(t)}{2} - \frac{2z(t)}{2} \Big) \\
&= \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{z(t)}{2} \Big) = \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{z(t)^2}{2z(t)} \Big) \\
&= \frac{1}{t} \frac{1+z(t)^2}{2z(t)} \\
\\
\text{Seperation of variables:}\\
g(t) &= \frac{1}{t} \ ,\ \ h(z(t)) = \frac{1+z(t)^2}{2z(t)} \\
\\
\int_{t_0=1}^t \frac{1}{t} d\tau &= \Big[\ln(\tau)\Big]_1^t = ln(t) - 0 = ln(t) \\
\\
\int_{z_0=0}^{z(t)} \frac{1+x}{2x} dx &= \int_{0}^{z(t)} \frac{1}{2x} dx + \frac{x}{2x} dx = \int_{0}^{z(t)} \frac{1}{2x} dx + \int_{0}^{z(t)} \frac{1}{2} dx \\
&= \frac{1}{2} \int_{0}^{z(t)} \frac{1}{x} dx + \frac{1}{2} \int_{0}^{z(t)} 1 \ dx = \frac{1}{2} \int_{0}^{z(t)} \frac{1}{x} dx + \frac{z(t)}{2} \\
&= \frac{1}{2} \int_{\alpha}^{z(t)} \frac{1}{x} dx + \frac{z(t)}{2} = \frac{1}{2} \Big[ \ln(x) \Big]_\alpha^{z(t)} + \frac{z(t)}{2} \ \ \ \Big( \alpha \in \big(0,z(t)\big)\Big) \\
&= \frac{1}{2} \ln(z(t)) - \frac{1}{2} \ln(\alpha) + \frac{z(t)}{2} \\
\\
\lim_{\alpha \rightarrow 0} \ &\frac{1}{2} \ln(z(t)) - \frac{1}{2} \ln(\alpha) + \frac{z(t)}{2} = \frac{1}{2} \ln(z(t)) + \frac{z(t)}{2} \\
\\
\text{Now solve for z(t):} \\
ln(t) &= \frac{1}{2} \ln(z(t)) + \frac{z(t)}{2}
\end{align*}
$$
This is the point where I don't know how to continue.
| Or start with setting $v=y^2$,
$$
tv'=t^2+3v
$$
which is linear with integrating factor $1/t^3$ for the normalized equation,
$$
\left(\frac{v}{t^3}\right)'=\frac{tv'-3v}{t^4}=\frac1{t^2}\implies \frac{v}{t^3}=-\frac1t+C,~~v=Ct^3-t^2,\\~\\y=\pm t\sqrt{Ct-1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving $a_{n+1}^2=a_n·a_{n+2}+(-1)^n$ for Fibonacci sequence $\{a_n\}$ Let $a_1,a_2,a_3,...,a_n$ be the sequence of Fibonacci numbers.
Then we are required to prove by induction that $$a_{n+1}^2=a_n \cdot a_{n+2} + (-1)^n$$
My Attempt:
Initially putting a few values like $n=1,2 \: \text{or}\ 3$ makes it certain that we can move ahead.
Following the conventional way we can assume that the proposition holds true for some $k$ s.t.$$a_{k+1}^2=a_k \cdot a_{k+2} + (-1)^k \tag1$$
To prove that the proposition holds for all values of $n$, we need to prove for the same for $n=k+1$ too.
In short we need to prove that:
$$a_{k+2}^2=a_{k+1} \cdot a_{k+3} + (-1)^{k+1} \tag2$$
We know that
$$ a_{k+2}^2=(a_{k+3}-a_{k+1})^2$$
$$\Rightarrow a_{k+2}^2=a_{k+3}^2+a_{k+1}^2-2\cdot a_{k+3}\cdot a_{k+1}$$
$$\Rightarrow a_{k+2}^2=a_{k+3}^2+a_k \cdot a_{k+2} + (-1)^k-2\cdot a_{k+3}\cdot a_{k+1}$$
$$\Rightarrow a_{k+2}^2=a_{k+3}[a_{k+3}- 2\cdot a_{k+1}]+a_k \cdot a_{k+2} + (-1)^k \tag{*}$$
It is easy to see that $$a_{k+3}-a_{k+1}=a_{k+2} \tag3$$
And $$a_{k+2}-a_{k+1}=a_{k} \tag4$$
Adding equation $3$ and $4$, we get$$a_{k+3}-2a_{k+1}=a_{k} \tag5$$
Putting eq$(5)$ in eq$(*)$
$$a_{k+2}^2=a_{k+3}[a_k]+a_k \cdot a_{k+2} + (-1)^k $$
$$\Rightarrow a_{k+2}^2=a_k[a_{k+3}+a_{k+2}]+(-1)^k $$
$$\Rightarrow a_{k+2}^2=a_k(a_{k+4})+(-1)^k$$
This I indeed completely different from something that I wanted to prove. Is there any way to get ahead of this or is there something wrong with what I have done?
| Your original statement is equivalent to
$$a_{n+1}^2=a_n(a_{n+1}+a_n)+(-1)^n,$$
that is
$$a_{n+1}^2-a_na_{n+1}-a_n^2=(-1)^n.\tag{1}$$
To prove $(1)$ all you need to do is to prove it for $n=1$ and verify
$$a_{n+2}^2-a_{n+1}a_{n+2}-a_{n+1}^2=-
(a_{n+1}^2-a_na_{n+1}-a_n^2)^.\tag{2}$$
This is an easy induction.
One can prove $(2)$ just by substituting $a_{n+1}+a_n$
for $a_{n+2}$ on the LHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given $a, b$ so that $0\leqq a, b\leqq 2, a^{-1}+ 2b^{-1}\geqq 2$. Prove that $a^{2}+ b^{2}\leqq 5$ .
Given $a, b$ so that $0\leqq a, b\leqq 2, a^{-1}+ 2b^{-1}\geqq 2$. Prove that $a^{2}+ b^{2}\leqq 5$ (equality: $a= 1, b= 2$).
For $a\leqq 1$, the inequality is clearly true!
For $a\geqq 1, F= a^{-1}+ 2b^{-1}- 2\geqq 0$
$$5- a^{2}- b^{2}= \frac{ab\left ( 2(a- 1)(b+ 1)+ b+ 2 \right )F+ (a- 1)( - 4a^{3}+ 15a- 5)}{(2a- 1)^{2}}\geqq 0$$
But $-4a^{3}+ 15a- 5= (2- a)(4a^{2}+ 8a+ 1)- 7\geqq - 7$, we need to prove $-4a^{3}+ 15a- 5\geqq 0$ .
| Note that if $a=2$ and $b=4/3$ then $0\leq a,\,b\leq 2$, $\displaystyle \frac{1}{a}+ \frac{2}{b}= 2$, but
$$a^{2}+ b^{2}=\frac{52}{9}> 5.$$
P.S. Actually on the compact set
$$K:=\left\{(a,b)\in [0,2]\times[0,2]: \frac{1}{a}+ \frac{2}{b}\geq 2\right\}$$
the continuous function $f(a,b)=a^2+b^2$ attains its maximum value just at $(2,4/3)\in K$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is $x$ if $\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$? I need to find $x$, given that
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
I simplified this to $x^4+14x^3+105x^2+68x-188=0$. According to Symbolab, that is not correct. That's why I'm goint to write out my attempt so you can point out where my mistake is.
My attempt
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
$$\left(\sqrt{x+4+2\sqrt{x+3}}\right)^2=\left(\frac{x+8}{3}\right)^2$$
$$x+4+2\sqrt{x+3}=\frac{x^2+16x+64}{9}$$
$$9x+36+18\sqrt{x+3}=x^2+16x+64$$
$$-x^2-7x-28=-18\sqrt{x+3}$$
$$x^2+7x+28=18\sqrt{x+3}$$
$$(x^2+7x+28)^2=(18\sqrt{x+3})^2$$
$$x^4+7x^3+28x^2+7x^3+49x^2+196x+28x^2+196x+784=324x+972$$
$$x^4+14x^3+105x^2+68x-188=0$$
Where is my mistake? Even if this were true, I still wouldn't be able to solve it without a calculator (I can't use Rational Root Theorem on such a big numbers!).
By the way, the solution should be (again, according to Symbolab) $x \in \{1,-2\}$.
| Your question was how to determine the value of $x$ in the given equation:
$$
\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}
$$
1.)
$
x+4+2 \sqrt{x+3} = \bigg( \frac{x+8}{3} \bigg)^2
$
2.)
$
x+4+2 \sqrt{x+3} = \frac{x^2+8^2}{3^2}
$
3.)
$
9(x+4+2 \sqrt{x+3}) = (x+8)^2
$
4.)
$
9x+36+18 \sqrt{x+3} = (x+8)(x+8)
$
5.)
$
18 \sqrt{x+3}) = (x^2+16x+64)-9x-36
$
6.)
$
\sqrt{x+3}) = \frac{x^2+7x+28}{18}
$
7.)
$
x+3 = \bigg( \frac{(x^2+7x+28)({x^2+7x+28})}{18 \times 18} \bigg)
$
8.)
$
x+3 = \frac{x^4+14x^3+105x^2+392x+784}{324}
$
9.)
$
324x+972 = x^4+14x^3+105x^2+392x+784
$
10.)
$
x^4+14x^3+105x^2+68x=188
$
The outcome of the equation is that it is a $4^ \text{th}$ degree polynomial of the form
$
a_1x^4+a_2x^3+a_3x^2+a_4x+a_4=0
$
$$
x^4+14x^3+105x^2+68x-188=0
$$
The solution given by my calcuator is:
$
x_1=1,
x_2=-2
$
To find out how that comes about, refer to Quartic polynomial solutions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integral $\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x}dx$ By integration by parts and the substitution $x = \sin t$ we can easily calculate the integral $\int_{0}^{1} \ln (x+ \sqrt{1-x^2})dx$ which equals to $\sqrt{2} \ln (\sqrt{2} +1) -1.$
I’ve tried to use the same substitution $x = \sin t$ to calculate the integral $ \int_{0}^{1} \frac {\ln (x+ \sqrt{1-x^2})}{x}dx,$ which becomes
$$ \int_{0}^{\frac {\pi}{2}} \frac {\ln \sin (t+ \frac {\pi}{4})}{\sin t}dt$$
It seems difficult to solve the particular integral. Any help?
| Split the integral at $\frac{1}{\sqrt{2}}$ and use the substitution $x = \sqrt{1-y^2}$ in the second part to obtain
\begin{align}
I &\equiv \int \limits_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x = \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x + \int \limits_{\frac{1}{\sqrt{2}}}^1 \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x \\
&= \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x + \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{y \ln(y+\sqrt{1-y^2})}{1-y^2} \, \mathrm{d} y = \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{ \ln(x+\sqrt{1-x^2})}{x(1-x^2)} \, \mathrm{d} x \, .
\end{align}
Now let $x = \sin (t/2)$ to find
\begin{align}
I &= \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln \left(\sin\left(\frac{t}{2}\right) + \cos\left(\frac{t}{2}\right)\right)}{\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)} \, \mathrm{d} t = \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln \left[\left(\sin\left(\frac{t}{2}\right) + \cos\left(\frac{t}{2}\right)\right)^2\right]}{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)} \, \mathrm{d} t \\
&= \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)\right)}{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)} \, \mathrm{d} t = \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+\sin(t)\right)}{\sin(t)} \, \mathrm{d} t \\
&=\frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+\cos(t)\right)}{\cos(t)} \, \mathrm{d} t \, .
\end{align}
Define (idea from this question)
$$ f(a) \equiv \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+\cos(a)\cos(t)\right)}{\cos(t)} \, \mathrm{d} t $$
for $ a \in [0,\frac{\pi}{2}]$ and observe that $f(0)=I$ and $f(\frac{\pi}{2}) = 0$. Compute (using $\tan(\frac{t}{2}) = s$)
\begin{align}
f'(a) &= - \frac{\sin(a)}{2} \int \limits_0^{\frac{\pi}{2}} \frac{1}{1+\cos(a)\cos(t)} \, \mathrm{d} t = - \sin(a) \int \limits_0^1 \frac{\mathrm{d} s}{1+\cos(a) + (1-\cos(a))s^2} \\
&= - \frac{\sin(a)}{1+\cos(a)} \sqrt{\frac{1+\cos(a)}{1-\cos(a)}} \arctan \left(\sqrt{\frac{1-\cos(a)}{1+\cos(a)}}\right) \\
&= - \frac{\sin(a)}{\sqrt{1-\cos^2 (a)}} \arctan\left(\tan\left(\frac{a}{2}\right)\right) = - \frac{a}{2} \, .
\end{align}
And finally,
$$ I = f(0) = f \left(\frac{\pi}{2}\right) + \int \limits_{\frac{\pi}{2}}^0 f'(a) \, \mathrm{d} a = 0 + \int \limits_0^{\frac{\pi}{2}} \frac{a}{2} \, \mathrm{d} a = \frac{\pi^2}{16} \, .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"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.