Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods???
i only know quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ i tried many algebraic manipulations and i get $(x^2+1)^2=3(x^3+x)$, so can we have solution using only high school methods? i guess i have to do an algebraic substitution to reduce $x^3+x$ to polynomial of degree 2?? i also know $$(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1r_2$$ where $r_1,r_2$ are roots of polynomial of degree 2
| You have $(x^2+1)^2=3x(x^2+1)$ This means that $(x^2+1)^2-3x(x^2+1)=0$ or $(x^2+1)(x^2+1-3x)=0$ Can you solve it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Some exact values of $\cos \theta$ using de Moivre's theorem Presently I am faced with the following question:
By showing that $$\cos5\theta = \cos\theta(16\cos^4\theta - 20\cos^2\theta + 5)$$ and then solving the equation $\cos5\theta = 0$, deduce that
$$\cos^2\left(\frac{\pi}{10}\right) = \frac{5+\sqrt{5}}{8}$$
and hence find the exact values of $\cos^2\left(\frac{3\pi}{10}\right), \cos^2\left(\frac{7\pi}{10}\right), \cos^2\left(\frac{9\pi}{10}\right)$.
I have proved the first part using de Movire's theorem and after solving $\cos5\theta = 0$ reached the following equation:
$$\cos^2 \left(\frac{\pi}{10}\right) = \frac{5\pm\sqrt{5}}{8}$$
Where $\frac{\pi}{10}$ can be replaced with $\frac{3\pi}{10}$, $\frac{7\pi}{10}$, etc. as they are other possible solutions when finding $\theta$ from $\cos5\theta = 0$.
However I am unsure how I am supposed to know whether or not to take the positive or negative root dependent on the angle (i.e. for $\frac{\pi}{10}$, $\frac{7\pi}{10}$, $\frac{3\pi}{10}$, $\frac{9\pi}{10}$), in order to get the exact value.
| Since $\cos^2 \theta$ is strictly decreasing on $\left[0, \frac{\pi}{2}\right]$, we must have that
$$\cos^2 \left(\frac{\pi}{10}\right) > \cos^2 \left(\frac{3 \pi}{10}\right),$$ and so
$$\cos^2 \left(\frac{\pi}{10}\right) = \frac{5 + \sqrt{5}}{8} \qquad \text{and} \qquad \cos^2 \left(\frac{3\pi}{10}\right) = \frac{5 - \sqrt{5}}{8}.$$
The remaining values can be determined by a similar argument, or by using these values together with the easy identity $\cos^2 \theta = \cos^2 (\pi - \theta)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Cannot understand how does he find the minimum value of this funtion My teacher has find the minimum value of $y = x + \frac{2}{x}$ when $x \ge 0$ without using
differential:
$y = x + \frac{2}{x}$
Multiplied both side by $x$ and moved $y$ to other side
$x^2 - yx + 2 = 0$
Then he wrote that having an answer in this equation is conditioned by
$y^2 - 8 \ge 0$ (He used $b^2 - 4ac$ formula)
$y \ge 2\sqrt{2}$
The answer is $2\sqrt{2}$ and it's correct.
Can anyone explain this solution?
| If the discriminant is smaller than zero($b^2-4ac<0$) than $x_{1,2}\in\mathbb{C}$ so clearly $b^2-4ac\geq0$ now if $y\in(-2\sqrt{2},2\sqrt{2})$ it means that there is no solution in reals,and since $x>0$ than $y>0$ so we have that $y\geq2\sqrt{2}$ clearly since we are looking for the minimum we pick the smallest $y$ which is $y=2\sqrt{2}$ And since we have a positive $x$ that satisfies that equation we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Probablity of rolling the same number twice after n times Im trying to figure out what the probability of rolling a dice and getting exactly two of the same numbers after we throw the dice n times is?
The chance to get some number is 1/6 and getting that same number two times in a row must be 1/36, but with this calculation we don't take into account that we throw the dice n times which would increase the probablity that we would get those two numbers exactly two times during n dice throws i'd assume.
What method is used to calculate the probability of such task?
|
I'm trying to figure out what the probability of rolling a dice and getting exactly two of the same numbers after we throw the dice $n$ times is?
Exactly two of a specified number, or exactly two of any number?
Case 1 : The probability of getting exactly two sixes. $n\in \{2,3,....\}$
$$\mathsf P(T_6) = {n\choose 2} 5^{n-1}6^{-n} = \frac{5^{n-2}n(n-1)}{2\cdot 6^n}$$
Similarly, $\mathsf P(T_1)=\mathsf P(T_2) = \cdots = \mathsf P(T_6)$
Case 2 : The probability of getting any number exactly twice. $n\in \{2,3, ...\}$
We need a rather complicated inclusion and exclusion, depending on the size of $n$.
$\begin{align}
\mathsf P(\bigcup_{k=1}^6 T_k) & = \sum_{k=1}^6\mathsf P(T_k) - \sum_{k=1}^5\sum_{h=k+1}^6\mathsf P(T_k\cap T_h) + \sum_{k=1}^4\sum_{h=k+1}^5\sum_{j=h+1}^6 \mathsf P(T_k\cap T_h\cap T_j) - \cdots
\\[4ex] & =
6 {n\choose 2, n-2}\frac{5^{n-2}}{6^n} \operatorname{\bf 1}_{\{2,..,\infty\}}(n)
- 15 {n\choose 2,2,n-4}\frac{4^{n-4}}{6^n} \operatorname{\bf 1}_{\{4,..,\infty\}}(n)
+ 20{n\choose 2,2,2,n-6}\frac{3^{n-6}}{6^n} \operatorname{\bf 1}_{\{6,..,\infty\}}(n)
- 15 {n\choose 2,2,2,2,n-8}\frac{2^{n-8}}{6^n} \operatorname{\bf 1}_{\{8,..,\infty\}}(n)
+ 6{n\choose 2,2,2,2,2,n-10}\frac 1{6^n} \operatorname{\bf 1}_{\{10,..,\infty\}}(n)
- {n\choose 2,2,2,2,2,2}\frac 1{6^n} \operatorname{\bf 1}_{\{12\}}(n)
\\[4ex] & =
\frac{6\cdot 5^{n-2}n(n-1)}{2\cdot 6^n} \operatorname{\bf 1}_{\{2,..,\infty\}}(n)
- \frac{15\cdot 4^{n-4}n(n-1)(n-2)(n-3)}{4\cdot 6^n} \operatorname{\bf 1}_{\{4,..,\infty\}}(n)
+ \frac{20\cdot 3^{n-6}n(n-1)(n-2)\cdots(n-5)}{8\cdot 6^n} \operatorname{\bf 1}_{\{6,..,\infty\}}(n)
- \frac{15\cdot 2^{n-8}n(n-1)\cdots(n-7)}{16\cdot 6^n} \operatorname{\bf 1}_{\{8,..,\infty\}}(n)
+ \frac {6\cdot n(n-1)\cdots(n-9)}{32\cdot 6^n} \operatorname{\bf 1}_{\{10,..,\infty\}}(n)
- \frac {12!}{64\cdot 6^{12}} \operatorname{\bf 1}_{\{12\}}(n)
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1028078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Area of $\triangle ABC$ whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is The maximum area of a triangle whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is
$\bf{My\; Try::}$ Area of $\displaystyle \triangle ABC = \frac{1}{2}ab\sin C = \frac{1}{2}ab\cdot \sqrt{1-\cos^2 C}\;,$ bcz Largest side has largest angle.
Now Using $\displaystyle \cos C = \frac{a^2+b^2-c^2}{2ab}.$ So we get area of $\displaystyle \triangle ABC = \frac{1}{2}ab\cdot \sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}$
Now How can i find Maximum area.
| You would be better off using Heron's formula. But there is an even better way: what is the maximum value that $\frac12ab\sin \theta$ can attain, if $a \le 1$, $b \le 2$, and $\theta$ can be any angle? Is this value attainable given the additional constraint $2 \le c \le 3$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\sum_{n = 1}^{\infty} \frac{2}{2^{n}}$
Evaluate $$\sum_{n = 1}^{\infty} \frac{2}{2^{n}}$$
This is a geometric series and since $a = \dfrac{1}{2}$ Then the infinite sum is jsut $S = \dfrac{1}{1-\frac{1}{2}} = 2$ Then I multiply by $2$ to get $4$ right? But the actual answer should just be $2$. Am I missing something?
| The first term in $\sum_{n=1}^{\infty}\frac{1}{2^n} = \frac{1}{2}+\frac{1}{4}+\dots$ is $\frac{1}{2}$, so the sum of the geometric series is $\frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$.
Another way to do it is to absorb the $2$ into the denominator and then change the index from $n$ to $k = n-1$:
$$\sum_{n=1}^{\infty}\frac{2}{2^n} = \sum_{n=1}^{\infty}\frac{1}{2^{n-1}} = \sum_{k=0}^{\infty}\frac{1}{2^{k}} = 1 + \frac{1}{2}+\frac{1}{4}+\dots= \frac{1}{1-\frac{1}{2}} = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate $\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx$
Evaluate
$$
\int_{0}^{1} \arctan^{2}\left(\, x\,\right)
\ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x
$$
I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got
$$
-\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right) \over \cos^{2}\left(\,\theta\,\right)}\,{\rm d}\theta
$$
After this, I thought of using the Taylor Expansion of
$\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right)$ near zero but that didn't do any good.
Please Help!
| Samurai, this is for the second time you post problems that relate to me. First you posted this question here which is exactly similar with my rated problem on Brilliant.org. I have raised objection to mods but they can do nothing since your post doesn't violate any rules here. Okay, fine. I can accept their reason. Now you post this question which I believe it's taken from one of proposed problems in Brilliant Integration Contest - Season 1 that I held on Brilliant.org. The original problem was proposed by Jatin Yadav as PROBLEM 7 but a day later he deleted this problem and changed to another problem after no-one can solve it included himself. According to him, it's taken from here, on Math S.E. You may want to take a look there.
I tried to solve this problem for hours but no success. Here is my attempt:
Set $x=\tan y$, we get
\begin{align}
I&=\int_0^1\arctan^2x\,\ln\left(\frac{1+x^2}{2x^2}\right)\,dx\\
&=-\int_0^{\pi/4} \frac{y^2\ln\left(2\sin^2y\right)}{\cos^2y}\,dy\\
&=-2\int_0^{\pi/4} \frac{y^2\ln\left(1-\cos2y\right)}{1+\cos2y}\,dy\\
&=-\frac{1}{4}\int_0^{\pi/2} \frac{t^2\ln\left(1-\cos t\right)}{1+\cos t}\,dt\qquad\Rightarrow\qquad t=2y\\
\end{align}
Use integration by parts by taking $u=t^2$ and $dv=\dfrac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt$, then
\begin{align}
v&=\int\frac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt
\end{align}
Use integration by parts by taking $u=\ln\left(1-\cos t\right)$ and $dv=\dfrac{dt}{1+\cos t}$, by Weierstrass substitution: $x=\tan\left(\dfrac{t}{2}\right)$ then
\begin{align}
v=\int\frac{dt}{1+\cos t}=\int \,dx=\tan\left(\frac{t}{2}\right)=\frac{\sin t}{1+\cos t}
\end{align}
Hence
\begin{align}
\int\frac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt
&=\frac{\sin t}{1+\cos t}\ln\left(1-\cos t\right)-\int\frac{\sin t}{1+\cos t}\cdot\frac{\sin t}{1-\cos t}\,dt\\
&=\frac{\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t
\end{align}
and
\begin{align}
I&=-\frac{1}{4}\left[\frac{t^2\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t^3\right]_0^{\pi/2}+\frac{1}{2}\int_0^{\pi/2}\left[\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t^2\right]\,dt\\
&=\frac{\pi^3}{32}+\frac{1}{2}\int_0^{\pi/2}\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)\,dt-\frac{\pi^3}{48}\\
&=\frac{\pi^3}{96}+\frac{1}{2}\int_0^{\pi/2}\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)\,dt
\end{align}
Consider
\begin{equation}
I(a)=\int_0^{\pi/2} \frac{t\sin t}{1+\cos t}\ln\left(1-a\cos t\right)\,dt
\end{equation}
so that $I(0)=0$ and
\begin{align}
I'(a)&=-\int_0^{\pi/2} \frac{t\sin t\cos t}{(1-a\cos t)(1+\cos t)}\,dt\\
&=\frac{1}{1+a}\int_0^{\pi/2} \left(\frac{t\sin t}{1+\cos t}-\frac{t\sin t}{1-a\cos t}\right)\,dt\\
\end{align}
Now consider
\begin{equation}
I(b)=\int_0^{\pi/2} \frac{t\sin t}{1+b\cos t}\,dt
\end{equation}
Use integration by parts by taking $u=t$ and $dv=\dfrac{\sin t}{1+b\cos t}\,dt$, then
\begin{align}
I(b)&= \frac{t\ln(1+b\cos t)}{b}\bigg|_0^{\pi/2}-\frac{1}{b}\int_0^{\pi/2}\ln(1+b\cos t)\,dt\\
&=-\frac{1}{b}\int_0^{\pi/2}\ln(1+b\cos t)\,dt
\end{align}
Consider
\begin{equation}
J(b)=\int_0^{\pi/2}\ln(1+b\cos t)\,dt
\end{equation}
so that $J(0)=0$ and
\begin{align}
J'(b)&=\int_0^{\pi/2} \frac{\cos t}{1+b\cos t}\,dt\\
&=\frac{1}{b}\int_0^{\pi/2} \left(1-\frac{1}{1+b\cos t}\right)\,dt\\
&=\frac{\pi}{2b}-\int_0^{\pi/2} \frac{dt}{1+b\cos t}\qquad\Rightarrow\qquad x=\tan\left(\frac{t}{2}\right)\\
&=\frac{\pi}{2b}-\int_0^{1} \frac{2}{1+b+(1-b)x^2}\,dx\qquad\Rightarrow\qquad x=\sqrt{\frac{1+b}{1-b}}\tan z\\
&=\frac{\pi}{2b}-\frac{2}{\sqrt{1-b^2}}\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\\
J(b)&=\frac{\pi}{2}\ln b-\int\frac{2}{\sqrt{1-b^2}}\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\,db\\
\end{align}
Again we use integration by parts by taking $u=\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)$ and $dv=\dfrac{2}{\sqrt{1-b^2}}$, we have
\begin{align}
J(b)&=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\int\frac{\arcsin b}{1-b}\sqrt{\frac{1-b}{1+b}}\,db\\
&=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\int\frac{\arcsin b}{\sqrt{1-b^2}}\,db\\
&=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\frac{\arcsin^2 b}{2}
\end{align}
Therefore
\begin{align}
I'(a)&=\frac{\pi^2}{8(1+a)}-\frac{1}{a(1+a)}\left[\frac{\pi}{2}\ln (-a)-2\arctan\left(\sqrt{\frac{1+a}{1-a}}\right)\arcsin (-a)-\frac{\arcsin^2(-a)}{2}\right]\\
\end{align}
From this step, I give up. Perhaps someone else want to continue it. Be my guest...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 0
} |
Differentiate $f(x)=\dfrac{\sin^2(3x)}{2x}$ Consider following function
$$ f(x) =
\begin{cases}
\dfrac{\sin^2(3x)}{2x}, & x\neq0 \\
0, & x=0
\end{cases}$$
Evaluate $f'(0)$.
Is this function differentiable at $x=0$ ?
| Take the derivative first, using the quotient rule:
$$\frac{d}{dx}\left(\frac{\sin^2(3x)}{2x}\right)=\frac{2x\cdot2\sin(3x)\cdot\cos(3x)\cdot3-\sin^2(3x)\cdot2}{4x^2}$$
Then we need to take the limit as x approaches 0 of this expression:
$$\lim_{x\rightarrow 0} \frac{2x\cdot2\sin(3x)\cdot\cos(3x)\cdot3-\sin^2(3x)\cdot2}{4x^2}$$
If we directly plug in the numbers, we find a fraction of the form $\frac{0}{0}$, which is a case in which we can use L'Hospital's Rule:$\frac{f}{g}\rightarrow\frac{f'}{g'}$.
So now we differentiate the top and bottom independently:
$$\lim_{x\rightarrow 0} \frac{2x\cdot2\sin(3x)\cdot\cos(3x)\cdot3-\sin^2(3x)\cdot2}{4x^2}=
\lim_{x\rightarrow 0} \frac{12\sin(3x)\cos(3x)+36x\cos^2(3x)-36\sin^2(3x)-12\sin(3x)\cos(3x)}{8x}$$
This form is again $\frac{0}{0}$, so we need to apply L'Hospital's Rule again:
$$\lim_{x\rightarrow 0} \frac{36x\cos^2(3x)-36\sin^2(3x)}{8x}=
\lim_{x\rightarrow 0} \frac{36\cos^2(3x)-72\cos(3x)\sin(3x)3-72\sin(3x)\cos(3x)3}{8}
$$
This final form removes the $0$ in the denominator, so we can now evaluate the left and right-hand sides of the limit.
$$\lim_{x\rightarrow 0} \frac{36\cos^2(3x)-432\cos(3x)\sin(3x)}{8}=
\lim_{x\rightarrow 0} \frac{9\cos^2(3x)-108\cos(3x)\sin(3x)}{2}$$
$$\rightarrow_{x\rightarrow0^+}\frac{9\cos^2(3\cdot0^+)-108\cos(3\cdot0^+)\sin(3\cdot0^+)}{2}=
\frac{9\cos^2(0^+)-108\cos(0^+)\sin(0^+)}{2}=\frac{9\cdot1-108\cdot1\cdot0^+}{2}=\frac{9}{2}
$$
$$\rightarrow_{x\rightarrow0^-}\frac{9\cos^2(3*0^-)-108\cos(3\cdot0^-)\sin(3\cdot0^-)}{2}=
\frac{9\cos^2(0^-)-108\cos(0^-)\sin(0^-)}{2}=\frac{9\cdot1-108\cdot1\cdot0^-}{2}=\frac{9}{2}
$$
Therefore, the limit does exist. It approaches $\frac{9}{2}$ from below on the right, and from above on the left.
Since the derivative at $0$ is $0$, the derivative does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Improper Integral of $\int\frac{dx}{(2x-1)^3}$ Improper Integral of
$$\int_{-\infty}^0\frac{dx}{(2x-1)^3}$$
from Anton Calculus 8th Edition, page 576, question 9. Answer is $-\frac{1}{4}$ but I'm finding $-1$
The integral, substituting $u= (2x-1)$:
$$\frac{1}{2}\cdot\frac{-2}{(2x-1)^2}+C$$
Definite solution to use with limit:
$$ \frac{1}{2}\cdot\frac{-2}{(2x-1)^2)}\Big|^0_a = \frac{1}{2}\cdot\left(-2+\frac{2}{(2a-1)^2}\right)$$
Then solving limit:
$$\lim_{x \to {-\infty}} \frac{1}{2}\cdot\left(-2+\frac{2}{(2a-1)^2}\right) = \frac{1}{2}\cdot(-2+0) = -1$$
Any leads?
Thank you.
| Rather than multiplying by $-2,$ you should be dividing, since $$\int u^{n}\,du=\frac1{n+1}u^{n+1}$$ for all integers $n\ne-1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove reduction formula for $\int \cos^n (x)\sin^m (x) \, dx$ $$\displaystyle\int \:\sin^n\left(x\right)\cos^m\left(x\right)\mathrm dx=\frac{\sin^{n+1}x\cos^{m-1}x}{m+n}+\frac{m-1}{m+n}\int \:\sin^nx\cos^{m-2}x\,\mathrm dx$$
I have been trying to solve for over a week now can someone please help me.
| To start, we rewrite:
$$I=\int\sin^m\left(x\right)\cos^n\left(x\right)dx=\frac{1}{m+1}\int\big[\sin^{m+1}(x)\big]'\cos^{n-1}(x)dx$$
Partial integrating:
\begin{align}
I&=\frac{1}{m+1}\sin^{m+1}(x)\cos^{n-1}(x)-\frac{1}{m+1}\int\sin^{m+1}(x)\big[\cos^{n-1}(x)\big]'dx\\
&=\frac{1}{m+1}\sin^{m+1}(x)\cos^{n-1}(x)+\frac{n-1}{m+1}\int\sin^{m+2}(x)\cos^{n-2}(x)dx\\
\end{align}
Now we use $\sin^2(x)+\cos^2(x)=1$:
\begin{align}
I&=\frac{1}{m+1}\sin^{m+1}(x)\cos^{n-1}(x)+\frac{n-1}{m+1}\int\sin^m(x)\cos^{n-2}(x)dx\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad-\frac{n-1}{m+1}\int\sin^{m}(x)\cos^{n}(x)dx\\
&=\frac{1}{m+1}\sin^{m+1}(x)\cos^{n-1}(x)+\frac{n-1}{m+1}\int\sin^m(x)\cos^{n-2}(x)dx-\frac{n-1}{m+1}I\\
\end{align}
From this we get:
\begin{align}\frac{m+n}{m+1}I=\frac{1}{m+1}\sin^{m+1}(x)\cos^{n-1}(x)+\frac{n-1}{m+1}\int\sin^m(x)\cos^{n-2}(x)dx
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Finding inverse of a function $h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$ I have a function:
$$h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$$
With just pen and paper, how can I determine if there exists an inverse function? Am I supposed to sketch it on paper to see if it can have an invers? Or is there another/simplier way to do it?
How would I go about solving it? This is what I did:
$$h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}} = y$$
$$y(1 + \sqrt{x}) = 1-\sqrt{x}$$
$$y + y\sqrt{x} = 1-\sqrt{x}$$
$$\text{Cancel out roots:}$$
$$y^2 + y^2x = 1+x$$
$$y^2 + y^2x - x = 1$$
$$y^2 + x(y^2 - 1) = 1$$
$$x(y^2 - 1) = 1 - y^2$$
$$x = \frac{1 - y^2}{y^2 - 1}$$
$$\text{Swap x and y and we get the inverse function:}$$
$$f^{-1}(x) = \frac{1 - x^2}{x^2 - 1}$$
$$\text{But the correct answer is supposed to be:}$$
$$f^{-1}(x) = \frac{(1 - x)^2}{(1 + x)^2}, -1< x \le 1$$
What am I doing wrong?
| It is important to remember that
$$(A+B)^2=A^2+\mathbf{2AB}+B^2$$
This what I'd do:
$$y(1+\sqrt x)=1-\sqrt x$$
$$y+y\sqrt x=1-\sqrt x$$
$$(y+1)\sqrt x=1-y$$
$$(y+1)^2x=(1-y)^2$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find Cartesian coordinates of polar curve $r =5\sin(\theta) + 5\cos(\theta)$ Polar equation of the form $r = 5\sin(\theta) + 5\cos(\theta)$
The Cartesian equation is of the form $(x-A)^2+(y-B)^2 = R^2$
Find $A,B$, and $R$.
Guess: Let $x = R\cos(\theta) + A$ and $y = R\sin(\theta)+B$. Plug them in and get on the left hand side:
$$50\sin^2(\theta)\cos^2(\theta)+25\cos^4(\theta)+25\sin^4(\theta)+50\sin(\theta)\cos^3(\theta)+50\sin^3(\theta)\cos(\theta)$$
Plug in the right hand side:
$$R^2 = 25 + 50\sin(\theta)\cos(\theta)$$
Then set the 2 sides equal to each other. It looks as if $R = 5$.
Not sure how to go about this further.
| I would say
$r = 5\sin \theta + 5\cos \theta,\quad x = r\cos \theta, \quad y = r\sin \theta$
$\Rightarrow r^2 = 5x + 5y\Rightarrow x^2+y^2=5x + 5y\Rightarrow (x-\frac{5}{2})^2+(y-\frac{5}{2})^2=25/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Strange double integral What is wrong with this computation of $\int_0^1\int_{-y}^y \sqrt[3]{x} \, dx \, dy$?
I'm considering real functions only.
Since $x^{4/3}$ is an antiderivative of the integrand, we will get $\frac{3}{4}[x^{4/3}]_{-y}^y =\frac{3}{4}(y^{4/3}-(-y)^{4/3})=\frac{3}{4}(y^{4/3}-y^{4/3})=0$. Thus $\int_0^1 \int_{-y}^y \sqrt[3]{x} \, dx \, dy=0$.
However, maple is giving me a complex (nonzero) number as the answer. Why is that? Any hint?
| As others have already stated, Maple is interpreting $\sqrt[3]{x}$ as a complex function. Here's how it's being calculated
$$ \int_0^1\int_{-y}^y\sqrt[3]{x}\ dxdy= \int_0^1\int_{-y}^y x^{\frac{1}{3}}\ dxdy $$
$$ = \int_0^1\left[\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_{-y}^y dy = \int_0^1\left[\frac{x^{\frac{4}{3}}}{\frac{4}{3}}\right]_{-y}^y dy $$
$$
=\frac{3}{4}\int_0^1\left[x^{\frac{4}{3}}\right]_{-y}^y dy=\frac{3}{4}\int_0^1 y^{\frac{4}{3}}-(-y)^{\frac{4}{3}} dy
$$
$$
= \frac{3}{4}\left(\int_0^1 y^{\frac{4}{3}}dy-\int_0^1 (-y)^{\frac{4}{3}} dy\right)
$$
Let $u=-y$, then $-du=dy$. So now we have
$$
\frac{3}{4}\left(\int_0^1 y^{\frac{4}{3}}dy+\int_0^{-1} u^{\frac{4}{3}} du\right)
$$
$$
= \frac{3}{4}\left(\left[\frac{y^{\frac{4}{3}+1}}{\frac{4}{3}+1}\right]_0^1 +\left[\frac{u^{\frac{4}{3}+1}}{\frac{4}{3}+1}\right] _0^{-1} \right)
$$
$$
= \frac{3}{4}\left(\left[\frac{y^{\frac{7}{3}}}{\frac{7}{3}}\right]_0^1 +\left[\frac{u^{\frac{7}{3}}}{\frac{7}{3}}\right] _0^{-1} \right)
$$
$$
= \frac{3^2}{4\cdot 7}\left(\left[y^{\frac{7}{3}}\right]_0^1 +\left[u^{\frac{7}{3}}\right] _0^{-1} \right)
$$
$$
= \frac{9}{28}\left(1-0 +(-1)^{\frac{7}{3}}-0\right)
$$
$$
= \frac{9}{28}\left(1+(-1)^{\frac{7}{3}}\right) = \frac{9}{28}\left(1+(-1)^{\frac{6}{3}+\frac{1}{3}}\right)
$$
$$
=\frac{9}{28}\left(1+(-1)^{\frac{6}{3}}(-1)^{\frac{1}{3}}\right)= \frac{9}{28}\left(1+(-1)^{2}\sqrt[3]{-1}\right)
$$
$$
=\frac{9}{28}\left(1+\sqrt[3]{-1}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $\lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try}::$
$\bullet\; $If $x\in \mathbb{Z}\;,$ and $x\rightarrow \infty\;,$ Then $x^2<x^2+x+1<x^2+2x+1$
So $x<\sqrt{x^2+x+1}<(x+1)\;,$ So $\lfloor x^2+x+1 \rfloor = x\;,$ bcz it lies between two integers.
So $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right) = \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right)\cdot \frac{\left(\sqrt{x^2+x+1}+x\right)}{\left(\sqrt{x^2+x+1}+x\right)}$
So $\displaystyle \lim_{x\rightarrow \infty} \frac{x+1}{\sqrt{x^2+x+1}+x} = \frac{1}{2}$
$\bullet\; $ If $x\notin \mathbb{Z}$ and $x\rightarrow \infty\;,$ Then How can i solve the above limit in that case,
Help me, Thanks
| I would say
$\displaystyle \lim_{x\to \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor=\lim_{x\to \infty} \lfloor \sqrt{x^2+x+1 }\rfloor+\{ \sqrt{x^2+x+1 } \}-\lfloor \sqrt{x^2+x+1 }\rfloor=$
$\displaystyle=\lim_{x\to \infty} \{ \sqrt{x^2+x+1 } \}= (0 \sim 1) \Rightarrow\, $limit not exist
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve $\lfloor \sqrt x \rfloor = \lfloor x/2 \rfloor$ for real $x$ I'm trying to solve $$\lfloor \sqrt x \rfloor = \left\lfloor \frac{x}{2} \right\rfloor$$
for real $x$.
Obviously this can't be true for any negative reals, since the root isn't defined for such.
My approach is the following: Let $x=:n+r$, $n \in \mathbb{N}_0, 0\leq r < 1$.
For the left hand side $\lfloor \sqrt {n+r} \rfloor = \lfloor \sqrt {\lfloor n+r\rfloor} \rfloor = \lfloor \sqrt n \rfloor$ holds (without further proof).
$$\left\lfloor \frac{x}{2} \right\rfloor = \left\lfloor \frac{n+r}{2} \right\rfloor = \left\{ \begin{array}{l l} \frac{n}{2} & \quad \text{for n even} \\ \frac{n-1}{2} & \quad \text{for n odd} \end{array}\right.$$
Now I don't really know if that'd lead me in the right direction, but I'll write my thoughts down anyways.
Let $\sqrt n =: n'+r'$, $n \in \mathbb{N}_0, 0\leq r < 1$.
Therefore $\lfloor n'+r' \rfloor = n'$. And $n = (n'+r')^2 = n'^2 +2n'r' +r'^2$
For which $n',r'$ holds $$ n' < n'^2 + 2n'r' + r'^2.$$
Well and now I'm stuck and don't know how to proceed.
I'd appreciate any help.
| It is completely obvious that $x$ must be a non-negative real number. We decompose it as the following format:
$$x=k^2+n+r$$
in which $k^2$ is the largest integer less than or equal to $x$ in which is a squre of an integer number. So it is completely apparent that
$$n<(k+1)^2-k^2\rightarrow n<2k+1$$
$r$ is a real number such that $0\leq r<1$
We can say that $k\leq\sqrt{x}<k+1$, hence $\lfloor \sqrt{x}\rfloor=k$.
So the problem reduces to $k=\lfloor \frac{k^2+n+r}{2}\rfloor$.
So from the above eqaulity we can deduce
$\lfloor\frac{k^2}{2}\rfloor\leq k$, but for $k\geq 3$, the left hand side of the inequlity grows much faster than the right hand side, hence it is true only for three values for $k$. In other words $k=0,1,2$.
$k=0:$
In this case $n<2k+1=1$, hence we can say that $n=0$, so we have
$$k=\lfloor \frac{k^2+n+r}{2}\rfloor\rightarrow 0=\lfloor\frac{r}{2}\rfloor\rightarrow 0\leq r<2$$
which is a true inequlity for all values of $r$(note that at the first we assumed $0\leq r<1$).
So in this case $x$ reduces to $k^2+n+r=0+0+r=r$ for all values of $r$, so $x\in[0,1)$
$k=1:$
In this case we have
$$k=\lfloor \frac{k^2+n+r}{2}\rfloor\rightarrow 1=\lfloor\frac{1+n+r}{2}\rfloor$$
In addition we have
$$n<2k+1=3\rightarrow n=0,1,2$$
We can say
$$1=\lfloor\frac{1+n+r}{2}\rfloor\rightarrow 1\leq n+r<3$$ in which for $n=0$, this condition is not satisfied, but is satisfied for $n=1,2$. Hence in this case we have:
$$x=k^2+n+r=1+1+r=2+r$$ or $$x=k^2+n+r=1+2+r=3+r$$
So $x\in[2,4)$
$k=2:$
In this case we have $$2=\lfloor\frac{4+n+r}{2}\rfloor\rightarrow 0\leq n+r<2$$
but in this case $n<2k+1=5\rightarrow n=0,1,2,3,4$. But for $n=2,3,4$, the condition of $0\leq n+r<2$ is not satisfied and hence we can say in this case
$$x=k^2+n+r=4+n+r=4+0+r\space\space\space\space or\space\space\space x=4+1+r$$
So in this case $x\in[4,6)$
So the solution for the intended equality is the union of these intevals:
$$x\in[0,1)\cup [2,4)\cup [4,6)\rightarrow x\in[0,1)\cup [2,6)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 0
} |
Number of real solutions of the equation $1+8^x+27^x = 2^x+12^x+9^x$
Find the number of real solutions $x\in\mathbb{R}$ of the equation
$$
1+8^x+27^x = 2^x+12^x+9^x
$$
My Attempt:
Let $2^x=a>0$ and $3^x=b>0$ where $x\in \mathbb{R}$. This allows us to change the equation to
$$
1+a^3+b^3 = a+a^2b+b^2
$$
This can be rewritten as
$$
(a+b)^3-3ab(a+b)+1 = a+ab(a+b)
$$
How can I solve the problem from this point?
| A different method. Define:
$$f(a,b)= 1 + a^3 + b^3 - (a + a^2 b + b^2)$$
Show by solving $f'_a = f'_b=0$ that the minimum of $f$ is at $a=b=1$, and that $f(1,1)=0$, proving there is a single solution, $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\end{align}$$
Both nominator and denominator will never be negative because nominator is powered to two and cosx & sinx are positive when angel is acute.
Is it correct? Is there another way to solve?
| Nice Work
Just don't start by assuming the equality or inequality
Indeed there is another way, There's an elementary way
Since we are only concerned about acute angles
$$\left(a-\frac1a\right)^2\ge0$$
$$a^2-2+\frac1{a^2}\ge0$$
$$a^2+\frac1{a^2}\ge2$$
put $a^2=\tan x$
$$\tan x+\frac1{\tan x}\ge2$$
$$\tan x+\cot x\ge2$$
$\text{Game Over!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
} |
Lagrange Multipliers: Find $\min$ of $f(x,y)=3(x+1) +2(y-1)$ subject to the constraint $x^2+y^2=4$
Find the minumum value of the function $f(x,y)=3(x+1) +2(y-1)$, subject to the constraint that $x^2+y^2=4$.
The problem states to use Lagrange Multipliers. In doing so I obtained the point $(\frac 6{\sqrt{13}},\frac 4{\sqrt{13}})$. But when checking if it is a maximum I found that the result was inconclusive. What did I do wrong?
| $$minimize \hspace{3mm} 3(x+1) + 2(y-1)$$
$$s.t. \hspace{3mm} x^2+y^2=4.$$
The lagragian function is $\mathcal{L} = 3(x+1) + 2(y-1) + \lambda (x^2+y^2-4)$. Thus,
$$\frac{\partial\mathcal{L} }{\partial x} = 3 + 2\lambda x = 0 \Rightarrow \lambda = \frac{-3}{2x},$$ and
$$\frac{\partial\mathcal{L} }{\partial y} = 2 + 2\lambda y = 0 \Rightarrow \lambda = \frac{-2}{2y}.$$ Then, $x = \frac{3y}{2}$.
Substituting in the restriction,
$\frac{9y^2}{4}+ y^2 = 4 \Rightarrow y = \pm \frac{4}{\sqrt{13}},$ and $x = \pm \frac{6}{\sqrt{13}}.$ We have two critical points.
For $(x,y) = (\frac{6}{\sqrt{13}},\frac{4}{\sqrt{13}}), \lambda = -\sqrt{13}/4$.
For $(x,y) = (-\frac{6}{\sqrt{13}},-\frac{4}{\sqrt{13}}) , \lambda = \sqrt{13}/4$.
The Hessian is
\begin{equation}
H = \left [ \begin{array}{cc}
2\lambda & 0 \\
0 & 2\lambda \\
\end{array} \right]
\end{equation} and
For $\lambda = \sqrt{13}/4$, the Hessian is a positive definite matrix, and for $\lambda = -\sqrt{13}/4$, the Hessian is a negative definite matrix (we don't need analyse the Hessian on the tangent space). Thus, we have a minimal and a maximum point.
By substituting both points in cost function $f(x,y)$ we have,
$f(\frac{6}{\sqrt{13}},\frac{4}{\sqrt{13}}) \approx 8.2111$ (global maximum).
$f(-\frac{6}{\sqrt{13}},-\frac{4}{\sqrt{13}}) \approx -6.2111$ (global minimum).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Intersection multiplicity of the curves I want to find the intersection multiplicity of the curves $f(x,y)=x^5+x^4+y^2$ and $g(x,y)=x^6-x^5+y^2$ at the point $P=(0,0)$.
That`s what I have tried:
$f$ and $g$ have a common tangent, the $y=0$.
So $I(P, f\cap g) > m_P(f) \cdot m_P(g)=4$
$$f(x, 0)=x^5+x^4 \Rightarrow s=\deg f(x, 0)=5$$
$$g(x, 0)=x^6-x^5 \Rightarrow r =\deg g(x, 0)=6$$
$$s \leq r$$
So we consider $h(x, y)=g(x, y)-x f(x, y)$
$$h(x, y)=x^6-x^5+y^2-x(x^5+x^4+y^2)=x^6-x^5+y^2-x^6-x^5-xy^2 \\ \Rightarrow h(x, y)=-2x^5+y^2-xy^2$$
$\deg h(x, 0)=5<r$
So $I(P, f\cap g)=I(P,f\cap h)$
$$f(x,0)=x^5+x^4\Rightarrow \deg f(x,0)=5=s$$
$$h(x,0)=-2x^5\Rightarrow \deg h(x,0)=5=p$$
They have a common tangent, $x=0$, so they don`t intersect traverrsally.
We consider the polynomial $$h_1(x,y)=h(x,y)+2f(x,y)=3y^2-xy^2+2x^4$$
$deg h_1(x,0)=4<s,p$
So, $I(P, f\cap h)=I(P,f\cap h_1)$
$f(x,0)=x^5+x^4 \Rightarrow \deg f(x,0)=5=s$
$h_1(x,0)=2x^4\Rightarrow \deg h_1(x,0)=4=t$
They have a common tangent $x=0$,so they don`t intersect traversally.
We consider the polynomial $h_2(x,y)=2f(x,y)-xh_1(x,y)=2x^4+2y^2-3xy^2+x^2y^2$
$\deg h_2(x,0)=4<s$
So $I(P, f\cap h_1)=I(P, h_1\cap h_2)$
$h_1(x,0)=2x^4\Rightarrow \deg h_1(x,0)=4=s$
$h_2(x,0)=2x^4\Rightarrow \deg h_2(x,0)=4=m$
They have a common tangent $x=0$, so they don`t intersect traversally.
We consider the polynomial $h_3(x,y)=h_1(x,y)-h_2(x,y)=y^2(1+2x-x^2)$
$\deg h_3(x,0)=0<s,m$
So $I(P,h_1\cap h_2)=I(P,h_2\cap h_3)$
$h_2(x,0)=2x^4\Rightarrow \deg h_2(x,0)=4=m$
$h_3(x,0)=0\Rightarrow \deg h_3(x,9)=9=n$
So $I(P,h_2\cap h_3)=I(P,h_2\cap y^2)+I(P,h_2\cap (1+2x-x^2))$
$I(P,h_2\cap y^2)=8$
$I(h_2\cap (1+2x-x^2))=0$
Therefore, $I(P, f \cap g)=8$.
Is it right? Do we find that $f$ and $h$ have a common tangent from $f(x,0)$ and $h(x,0)$ ?
| From the system of equations,
$$x^5+x^4+y^2=x^6-x^5+y^2=0,$$
you can eliminate $y$, and
$$-x^6+2x^5+x^4=0.$$
This polynomial has a quadruple root at $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving $\lim_{x\rightarrow 1}x^{\frac{1}{n-x-x^2-x^3-\ldots -x^n}}=\frac{1}{\sqrt[k]{e}}$ Proving
$$\lim_{x\rightarrow 1}x^{\left(\frac{1}{n-x-x^2-x^3\cdots -x^n}\right)}=\frac{1}{\sqrt[k]{e}}$$
when $$k=1+2+3+\ldots+n$$
| Hint: $$x^{\left(\frac{1}{n-x-x^2-x^3\cdots -x^n}\right)}=\exp\left(\ln x^{\left(\frac{1}{n-x-x^2-x^3\cdots -x^n}\right)}\right)=\exp\left(\frac{\ln x}{n-x-x^2-\ldots-x^n}\right)$$ and now due to continuity of $\exp$ it suffices to show with L' Hopitals rule that $$\frac{\ln x}{n-x-x^2-\ldots-x^n} \to -\frac{1}{1+2+3+\ldots+n}=-\frac{1}{k}$$ as $x\to 1$.
Indeed $$\begin{align*}\lim_{x \to 1} \frac{\ln x}{n-x-x^2-\ldots-x^n}&\overset{\frac{0}{0}}=\lim_{x\to 1}\frac{(\ln x)'}{(n-x-x^2-\ldots-x^n)'}\\&=\lim_{x \to 1}\frac{\frac1x}{-1-2x-3x^2-\ldots-nx^{n-1}}\\[0.2cm]&=\frac{1}{-1-2-3-\ldots-n}=-\frac{1}{1+2+3+\ldots+n}=-\frac{1}{k}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove using De Moivre's formula,that $\sum\limits_{k=0}^{n}\sin(kx)=\frac{1}{2}\cot(x/2)-\frac{\cos(nx+(x/2))}{2\sin(x/2)}$ I've been asked to prove that:
$$
\sum\limits_{k=0}^{n}\sin(kx)=\frac{1}{2}\cot(x/2)-\frac{\cos(nx+(x/2))}{2\sin(x/2)}
$$
When $0<x<2\pi$.
I know there are many similar posts on this site, but using $\cos(kx)$ instead, that's why I created this post, I can't get to the $1/2\cot$ in this one.
Thanks!
$$\begin{align}\sum\limits_{k=0}^{n}\sin(kx) &= \Im\left(\frac{e^{i(n+1)x}-1}{e^{ix}-1}\right)\\ &= \Im\left(\frac{e^{\frac{i(n+1)x}2}}{e^{\frac{ix}2}}\frac{\left(e^{\frac{i(n+1)x}2}-e^{-\frac{i(n+1)x}2}\right)}{\left( e^{\frac{ix}2}-e^{-\frac{ix}2}\right)}\right)\\&= \Im\left(e^{\frac{inx}2}\frac{2i\sin\frac{(n+1)x}2}{2i\sin{\frac{x}2}}\right)\end{align}$$ Applying $e^{iy}-e^{-iy}=2i\sin y,$
$$\begin{align}\sum\limits_{k=0}^{n}\sin(kx) &= \Im\left(\left(\cos\frac{nx}2+i\sin\frac{nx}2\right)\frac{\sin\frac{(n+1)x}2}{\sin{\frac{x}2}}\right)\\ &= \sin\frac{nx}2 \frac{\sin\frac{(n+1)x}2}{\sin{\frac{x}2}},\end{align}$$ then I don't know what to do.
| You asked to prove this by DeMovire's formula, which states $$e^{ikx} = (\cos x + \sin x)^k = \cos kx + \sin kx.$$
Everything you did until the final line is correct.
The final move is to do the following, $$\sin \left ( \frac{nx}{2} \right ) \frac{\sin\frac{(n+1)x}2}{\sin{\frac{x}2}} = \frac{\cos(x/2) - \cos(n + 1/2)x}{2\sin \left ( x/2 \right )} = \frac{1}{2}\cot x/2 - \frac{\cos(n + 1/2)x}{2\sin \left ( x/2 \right)}.$$
where we have used the fact that $$\sin u \sin v = \frac{1}{2} \left ( \cos(u - v) - \cos(u + v) \right )$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that $\sin45°+\sin15°=\sin75°$ Steps I took:
1) Finding the value of the left hand side
$$\sin45=\sin\frac { 90 }{ 2 } =\sqrt { \frac { 1-\cos90 }{ 2 } } =\sqrt { \frac { 1 }{ 2 } } =\frac { \sqrt { 2 } }{ 2 } $$
$$\sin15=\sin\frac { 30 }{ 2 } =\sqrt { \frac { 1-\cos30 }{ 2 } } =\sqrt { \frac { 1-\frac { \sqrt { 3 } }{ 2 } }{ 2 } } =\frac { \sqrt { 2-\sqrt { 3 } } }{ 2 } $$
So, $\sin45+\sin15=\frac { \sqrt { 2 } +\sqrt { 2-\sqrt { 3 } } }{ 2 } $
2) Rewriting $\sin75$
$$\sin75=\sin(45+30)=\sin45\cos30+\sin30\cos45$$
$$=\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } \cdot \frac { \sqrt { 2 } }{ 2 } $$
$$=\frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 } $$
So now I have these two expressions that are obviously equal (when I compute them) but how do I show them as being equal to each other. I assume the question is implying for the resulting expressions to look equal.
| This gets easier if you compute $\sin 15^\circ$ in the same way that you computed $\sin 75^\circ$:
$$
\sin 15^\circ = \sin(45^\circ-30^\circ)=\sin 45^\circ \cos 30^\circ - \sin 30^\circ \cos 45^\circ = \frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}-\frac{1}{2}\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4} \, .
$$
For an alternate approach, you could use a sum-to-product formula to note that
$$\sin 15^\circ+\sin 45^\circ = 2 \sin 30^\circ \cos 15^\circ $$
and proceed from there...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Maximum area of a isosceles triangle in a circle with a radius r As said in the title, I'm looking for the maximum area of a isosceles triangle in a circle with a radius $r$.
I've split the isosceles triangle in two, and I solve for the area $A=\frac{bh}{2}$*. I have made my base $x$, and solve for the height by using the Pythagorean theorem of the smaller triangle (seen in picture).
$h=r+\sqrt{r^2-x^2}$
So my the formula, I think, for both triangles should be $A=x(r+\sqrt{r^2-x^2})$
But after I solved for the derivative, when put "$= 0$", and checked on my calculator, I got the maximum to be about $4.3301r$, which differs a lot from my book's answer of $\frac{3\sqrt{3}}{4}r^2$*. Is my formula for the area right? Am I going about this the wrong way, or is it just my derivative that is wrong? Thanks in advance
*
Edited from original post
$A=rx+x\sqrt{r^2-x^2}$
$A'=r+x(\frac{1}{2})(r^2-x^2)^{-\frac{1}{2}}(-2x)+\sqrt{r^2-x^2}$
$r+\sqrt{r^2-x^2}=\frac{x^2}{\sqrt{r^2-x^2}}$
$r\sqrt{r^2-x^2}+(r^2-x^2)=x^2$
$r^2+r\sqrt{r^2-x^2}=2x^2$
$r^2(r^2-x^2)=(2x^2-r^2)^2$
$r^4-r^2x^2=4x^2-4x^2r^2+r^4$
$4x^4=3x^2r^2$
$x=\frac{\sqrt{3}}{2}r$
|
We can also split the triangle into three smaller triangles using $\frac{1}{2} ab \sin C$. Therefore $\Delta ABC$ equals:
$$\frac{1}{2} r^2 \big(\sin \theta + \sin \theta + \sin(360º - 2 \theta ) \big)$$
$$\frac{1}{2} r^2 \big(2 \sin \theta + \sin(- 2 \theta ) \big)$$
Since the area is only dependent on $\theta$, we can take the derivative of:
$$f(\theta) = 2 \sin \theta + \sin(-2 \theta) \implies f' (\theta) = 2 \cos \theta - 2 \cos(-2\theta)$$
and setting $f'(\theta)$ equal to $0$:
$$\cos \theta - \cos(-2 \theta) = 0$$
$$\Rightarrow \cos \theta - (2 \cos^2 \theta - 1) = 0$$
$$\Rightarrow 2 \cos^2 \theta - \cos \theta - 1 = 0$$
$$\Rightarrow \cos \theta = -\frac{1}{2}, \cos \theta = 1$$
and the only solution that makes sense in the range $\theta = (0º, 180º)$ is $\theta = 120º$.
Therefore the maximum possible area is:
$$\frac{1}{2} r^2 \big(2 \sin 120º + \sin(- 2 \times 120º ) \big) = \frac{3 \sqrt3}{4} r^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How can I prove the pattern $\sqrt{1 + 155555…5} = 2 \sqrt{3888…89}?$ How can I prove this
$$\sqrt{1+155}=2\sqrt{39}$$
$$\sqrt{1+1555}=2\sqrt{389}$$
$$\sqrt{1+15555}=2\sqrt{3889}$$
$$\sqrt{1+155555}=2\sqrt{38889}$$
| Square both sides, and you get
$$4\cdot 3\overset{n\ times}{8\cdots8}9 = 4(9+8\sum_{k=1}^n 10^k+3\cdot 10^{n+1})=
36+32\sum_{k=1}^n 10^k +12\cdot 10^{n+1}= $$
$$=6+3\cdot 10+ 2\sum_{k=1}^n 10^k+3\sum_{k=1}^n 10^{k+1}+2\cdot 10^{n+1}+10^{n+2}=
$$
$$=
6+ 2\sum_{k=1}^{n+1} 10^k+ 3\sum_{k=1}^{n+1} 10^k +10^{n+2}=
6+5\sum_{k=1}^{n+1} 10^k+10^{n+2}=1\overset{n+1\ times}{5\cdots 5}6=
1\overset{n+2\ times}{5\cdots 5}+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Prove if $\left\{x_n\right\}$ converges to $2$, then $\left\{\frac{1}{x_n}\right\}$ converges to $\frac{1}{2}$ We know that for all $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\lvert x_n - 2 \rvert < \epsilon$ for all $n \geq N$, and we want to show that for all $\varepsilon' > 0$, there exists an $N' \in \mathbb{N}$ such that $\left| \frac{1}{x_n} - \frac{1}{2} \right| < \epsilon'$ for all $n \geq N'$.
Let $\varepsilon = \varepsilon' - \frac{3}{2}$. Then by the triangle inequality, we have $$\left| \frac{1}{x_n} - \frac{1}{2} \right| \leq \left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right|.$$
Because we proved earlier that $x_n > 1$ for all $n \geq N$, we know that $$\left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right| < \left| x_n\right| + \left| -\frac{1}{2}\right|.$$
By the triangle inequality again, we know that $$\left| x_n - \frac{1}{2}\right| \leq \left| x_n\right| + \left| -\frac{1}{2}\right|.$$
Note that $\left| x_n - \frac{1}{2}\right| = \left| (x_n - 2) + (2 - \frac{1}{2})\right|$, so we get $$\left| (x_n - 2) + (2 - \frac{1}{2})\right| \leq \left| x_n - 2\right| + \left| 2 - \frac{1}{2}\right|.$$
Then we have $$\left| x_n - 2\right| + \left|2 - \frac{1}{2}\right| < \varepsilon + \frac{3}{2} = \left(\varepsilon' - \frac{3}{2}\right) + \frac{3}{2} = \varepsilon',$$ so $\frac{1}{x_n} \rightarrow \frac{1}{2}$.
I just realized that I don't know how to make sure that $\varepsilon' - \frac{3}{2}$ is a positive number. How can I do this?
| Note that
$$\left|\frac{1}{x_n}-\frac{1}{2}\right|= \left|\frac{2-x_n}{2x_n} \right|$$
Choose $N$ large enough that $x_n>1$ and $|x_n-2|<\epsilon$ for all $n>N$. Then
$$ \left|\frac{2-x_n}{2x_n}\right|\leq \epsilon/2< \epsilon $$
for all $n>N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $y'(x)\left(4-3y(x)x^2\right)=4x$
Solve the differential equation
$$y'(x)\left(4-3y(x)x^2\right)=4x$$
I would appreciate some help with this problem.
| $$\frac{dy}{dx}\left(4-3y(x)x^2\right)=4x$$
$$4-3yx^2=4x\frac{dx}{dy}$$
You can use the sustitution $u=x^2$, so $\frac{du}{dy}=2x\frac{dx}{dy}$
In the equation:
$$ 4-3yu=2\frac{du}{dy} $$
Rewriting:
$$ 2\frac{du}{dy}+(3y)u=4 $$
$$ \frac{du}{dy}+\frac{3}{2}yu=2 $$
That is a first order linear ordinary differential equation and its integrating factor is $ z(y)=e^{\int\frac{3y}{2}dy}=e^{3\frac{y^2}{4}} $.
Multyply both sides by z(y):
$$ e^{\frac{3y^2}{4}}\frac{du}{dy}+\frac{1}{2}(3e^{\frac{3y^2}{4}}y)u=2e^{\frac{3y^2}{4}} $$
Substitute $\frac{3}{2}e^{\frac{3y^2}{4}}y=\frac{d}{dy}(e^{\frac{3y^2}{4}})$.
Apply the reverse product rule $g\frac{df}{dy}+f\frac{dg}{dy}=\frac{d(fg)}{dy}$ to the left-hand side:
$$\frac{d}{dy}(e^\frac{3y^2}{4}u)=2e^{\frac{3y^2}{4}}$$
Integrate both sides with respecto to $y$:
$$\int\frac{d}{dy}(e^\frac{3y^2}{4}u)dy=\int2e^\frac{3y^2}{4}dy$$
Evaluate the integrals:
$$e^\frac{3y^2}{4}u=2\sqrt\frac{π}{3}f*e*(\frac{\sqrt3y}{2})+c_1$$
Divide both sides by $z(y)=e^\frac{3y^2}{4}$
$$u=\frac{{2\sqrt\frac{π}{3}}f*e*(\frac{\sqrt3y}{2})+c_1}{{e^\frac{3y^2}{4}}}$$
So:
$$x^2=\frac{{2\sqrt\frac{π}{3}}f*e*(\frac{\sqrt3y}{2})+c_1}{{e^\frac{3y^2}{4}}}$$
Where
$f*e*$: error function
$c_1$ : arbitrary constant
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the determinant of a matrix definition Let $A$ be a matrix that is defined like this:
$$A_{ij}=\begin{cases}
\alpha, & \text{if i=j} \\
\beta , & \text{if i $\ne$ j}
\end{cases}
$$
So I realized this matrix looks somehow like this
$$
\begin{pmatrix}
\alpha & \beta & \beta \\
\beta & \alpha & \beta \\
\beta & \beta & \alpha \\
\end{pmatrix}
$$
I tried to manipulate the rows to get an upper triangular matrix but couldn't succeed, am I in the right direction... some help?:)
| We know that each determinant should consist of $n!$ products of $n$ entries each if we have an $n\times n$ matrix. Considering each $\alpha$ on the diagonal, such an $\alpha$ is multiplied by the determinant of the $(n-1)\times(n-1)$ version of the matrix. So if we take the previous determinant, multiply it by $n\cdot \alpha$, and divide each term by the exponent of $\alpha$ in the given term and add enough times $\beta^n$ to have a total of $n!$ products we should be good:
$$
\begin{align}
n&=2&&&det&=\alpha^2-\beta^2\\
n&=3&&&first&=3\alpha(\alpha^2-\beta^2)\\
&&&&&=3\alpha^3-3\alpha\beta^2\\
&&&&&\sim\alpha^3-3\alpha\beta^2\\
&&&&det&=\alpha^3-3\alpha\beta^2+2\beta^3\\
n&=4&&&first&=4\alpha(\alpha^3-3\alpha\beta^2+2\beta^3)\\
&&&&&=4\alpha^4-12\alpha^2\beta^2+8\alpha\beta^3\\
&&&&&\sim\alpha^4-6\alpha^2\beta^2+8\alpha\beta^3-6\beta^4+3\beta^4\\
&&&&det&=\alpha^4-6\alpha^2\beta^2+8\alpha\beta^3-3\beta^4\\
\end{align}
$$
Note here that in $n=4$ I have subtracted six $\beta^4$ and added three to have 12 negative and 12 positive terms as a total. Remember that we should have $4!=24$ products of $\alpha$'s and $\beta$'s and half of them should be positive, the other half negative. I do not know about a formula for the general $n$ yet, but it may be easy to derive from this ...
The reason for dividing by the exponent of $\alpha$ is that if we have $\alpha^k\beta^{n-k}$ each such product is counted repeated $k$ times when we move through the diagonal and consider which products each $\alpha$ participates in. If $k$ of the $\alpha$'s are in it, it will be counted when consulting each of these $k$ of the $\alpha$'s so $k$ times too often ...
One has to be careful when continuing this. We have to use the $24$-product version of $det(A_4)$ to have the correct number of products in $det(A_5)$:
$$
\begin{align}
first&=5\alpha(\alpha^4-6\alpha^2\beta^2+8\alpha\beta^3-6\beta^4+3\beta^4)\\
&=5\alpha^5-30\alpha^3\beta^2+40\alpha^2\beta^3-30\alpha\beta^4+15\alpha\beta^4\\
&\sim \alpha^5-10\alpha^3\beta^2+20\alpha^2\beta^3-30\alpha\beta^4+15\alpha\beta^4-20\beta^5+24\beta^5\\
det(A_5)&=\alpha^5-10\alpha^3\beta^2+20\alpha^2\beta^3-15\alpha\beta^4-4\beta^5
\end{align}
$$
This result is confirmed by Wolfram Alpha. Note also how the second to last line above has 60 positive and 60 negative contributions. A total of $5!=120$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
A Sum that came up while solving a integral While evaluating $I$, I did the following-
$$\begin{align}I= \int_{0}^{1} \log \left(\dfrac{1+x}{1-x}\right) \dfrac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x &= 2 \int_{0}^{1}\sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{2n+1} \dfrac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x\\ &=2\sum_{n=0}^{\infty} \int_{0}^{1} \dfrac{x^{2n}}{(2n+1)\sqrt{1-x^2}} \ \mathrm{d}x \end{align}$$ Then I used the substitution $x \mapsto \sin \theta $.
$$\begin{align} \therefore I &=2\sum_{n=0}^{\infty} \int_{0}^{\pi/2} \dfrac{\sin^{2n} {\theta}}{2n+1} \ \mathrm{d}\theta\\ &=\pi \sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)} \end{align}$$ The last step is due to Wallis' formula.
However, I couldn't solve the last series. My question is that how do we prove that $$\displaystyle\sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}=\dfrac{\pi}{2} \ ?$$
| Instead of series expansion, you can do this:
$$\int_0^1 \ln\left(\frac{1+x}{1-x}\right)\frac{1}{x\sqrt{1-x^2}}\,dx=\frac{1}{2}\int_{-1}^1 \ln\left(\frac{1+x}{1-x}\right)\frac{1}{x\sqrt{1-x^2}}\,dx$$
$$=\int_{-1}^1 \frac{\ln(1+x)}{x\sqrt{1-x^2}}\,dx=\int_{-\pi/2}^{\pi/2} \frac{\ln(1+\sin\theta)}{\sin\theta}\,d\theta$$
Consider
$$I(a)=\int_{-\pi/2}^{\pi/2} \frac{\ln(1+a\sin\theta)}{\sin\theta}\,d\theta \Rightarrow I'(a)=\int_{-\pi/2}^{\pi/2} \frac{1}{1+a\sin\theta}\,d\theta=\frac{\pi}{\sqrt{1-a^2}}$$
$$\Rightarrow I(a)=\sin^{-1}a+C$$
Since $I(0)=0$, we have $C=0$ and hence,
$$\boxed{I(1)=\dfrac{\pi}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Why is this intuitive method valid? Problem. There are $2$ white and $3$ black balls in the urn. A person randomly picked $2$ balls and put $1$ white ball. What is the probability of the event that the next randomly-picked ball would be white.
To solve this problem formally you have to consider conditional probabilities and so on. But in the childhood I did not know such words so I was solving problems of this type in the following way.
Intuitive solution. Let's grind balls into powder and mix it. We will have $5$ (whatever you want, for instance) kilograms of powder. $\frac{2}{5}$ of each kilogram is white and $\frac{3}{5}$ is black. A person randomly took $2$ balls in my model means that he took $2$ kilogram of powder. Putting $1$ white ball means that he put $1$ kilogram of white powder.
After all actions there are $\left(5 \cdot\frac35-2\cdot \frac35\right)=\frac95$ kilograms of black powder and $\left(5 \cdot\frac25-2\cdot \frac25\right) + 1=\frac95=\frac{11}5$ kilograms of white powder. So the final probability of picking white ball is $\frac{11}{5}/\left(\frac{11}{5}+\frac{9}{5}\right) = \frac{11}{20}$.
You can check that this is the right answer, but the question is why is this intuitive method valid?
| Grinding balls into powder is mathematically the same as calculating expected value.
In fact, the reason why mathematicians often speak of expected value calculations as the sum of the possible values, weighted by their probabilities, is exactly why your intuition of weights of grinded balls is relevant.
The expected number of black balls taken out, $b$, is
$$E(b) = P(b=0)\cdot 0 + P(b=1) \cdot 1 + P(b=2) \cdot 2$$
The $P(b=k)$ for $k=0,1,2$ are
$$P(b=0) = \frac{1}{\binom{5}{2}} = \frac{1}{10}$$
$$P(b=1) = \frac{2 \cdot 3}{\binom{5}{2}} = \frac{3}{5}$$
$$P(b=2) = \frac{\binom{3}{2}}{\binom{5}{2}} = \frac{3}{10}$$
This gives us
$$E(b) = 0 + \frac{3}{5} + 2 \cdot \frac{3}{10} = \frac{6}{5}$$
This means that the expected number of black balls left after two are taken out is
$$3-E(b) = \frac{9}{5}$$
Similarly, the expected number of white balls taken out, $w$, is
$$E(w) = \frac{4}{5}$$
And the expected number of white balls remaining:
$$2-E(w) = \frac{6}{5}$$
You can calculate this directly, or notice that $$(3-E(b)) + (2-E(w)) = 5-E(b+w) = 5-2 = 3$$
After adding one white ball, the new expected number of white balls is
$$1+(2-E(w)) = 1 + \frac{6}{5} = \frac{11}{5}$$
There is now a total of $4$ balls, so the probability of choosing a white ball is
$$\frac{\frac{11}{5}}{4} = \frac{11}{20}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why sometimes we get only one root of quadratic equations? What is logic behind getting (sometimes) only one root of a quadratic equation which satisfies the equation?
| The logic is that every positive number has two square roots, but zero only has one.
The quadratic equation comes from the algebraic process known as completing the square. Given an equation $ax^2 + bx + c = 0$, divide by $a$:
$$
x^2 + \frac{b}{a}x + \frac{c}{a} = 0
$$
Then add $\left(\frac{b}{2a}\right)^2$ and to both sides:
$$
x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = \left(\frac{b}{2a}\right)^2
$$
Then subtract $\frac{c}{a}$:
$$
x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}
$$
The left-hand side is a perfect square (by design). We rearrange the right-hand side at the same time:
$$
\left(x + \frac{b}{2a}\right)^2 = \frac{b^2-4ac}{4a^2}
$$
At this point we would take the square root of both sides. Since the denominator is positive, the number of roots depends only on the numerator. If the numerator is positive, there are two roots. If the number is negative, there are zero (real) roots. If the numerator is zero there is only one.
Now you see why $b^2-4ac$ is called the discriminant of a quadratic equation. It detects the number of roots (or the type of roots if you allow complex ones). To finish up, we have
$$
x+\frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2a}
$$
so
$$
x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
A faster way of calculating this determinant? I'm doing a problem involving Cramer's rule, and one of the determinants I have to work with is as follows:
\begin{vmatrix}
1&1&1\\
a&b&c\\
a^3&b^3&c^3
\end{vmatrix}
So I started off by getting the matrix to a triangular matrix so I can just take the product of the diagonal cells by doing this:
\begin{equation}
R_1 \times-c + R_2
\end{equation}
\begin{equation}
R_3 \times -\frac{1}{c^3} + R_1
\end{equation}
\begin{equation}
R_2 \times(\frac{1-\frac{b^3}{c^3}}{b-c})+R_1
\end{equation}
I then got this matrix
\begin{vmatrix}
((1-\frac{a^3}{c^3})-(a-c)(\frac{1-\frac{b^3}{c^3}}{b-c}))&0&0\\
a-c&b-c&0\\
a^3&b^3&c^3\\
\end{vmatrix}
By summing the diagonal cells, I got this as a final answer:
\begin{equation}
(c-a)(a-b)(b-c)(a+b+c)
\end{equation}
However, it was a lengthy process, and I can't help but thinking this is not the type of calculations I can afford the time in a written exam, especially if I have to repeat this four times solving three linear equations with Cramer's rule.
I'd really appreciate it if you have any thoughts as to speed this process up.
Just as an added detail, the original question was (translated from Chinese):
With regards to the following set of equations
\begin{equation}
x + y + z = 1\\
ax + by + cz = d\\
a^3x + b^3y + c^3z = d^3
\end{equation}
*
*Under what conditions can Cramer's rule be used?
*Please solve the set of equations with Cramer's rule.
For the first question, the immediate thought that pops up is that
\begin{equation}
det(A) ≠ 0
\end{equation}
For the second question, it's simply
\begin{equation}
\frac{\begin{vmatrix}A_1\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}},
\frac{\begin{vmatrix}A_2\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}},
\frac{\begin{vmatrix}A_3\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}}
\end{equation}
which is the part I'm having trouble solving quickly.
Again, appreciate any hints or thoughts on this. Thanks.
| Determinants are not changed by adding multiples of a column to another. So
$$
\begin{vmatrix}1&1&1\\ a&b&c\\ a^3&b^3&c^3\end{vmatrix}=\begin{vmatrix}1&0&0\\ a&b-a&c-a\\ a^3&b^3-a^3&c^3-a^3\end{vmatrix}=(b-a)(c^3-a^3)-(b^3-a^3)(c-a)
$$
Edit: since in the comments to the other answer you were asking how to factor, here it is: you have the identity $x^3-y^3=(x-y)(x^2+xy+y^2)$. Then
\begin{align}
(b-a)(c^3-a^3)-(b^3-a^3)(c-a)&=(b-a)(c-a)(c^2+ac+a^2)-(b-a)(c-a)(b^2+ab+a^2)\\
&=(b-a)(c-a)(c^2+ac+a^2-b^2-ab-a^2)\\
&=(b-a)(c-a)(c^2+ac-b^2-ab)\\
&=(b-a)(c-a)[(c^2-b^2)+a(c-b)]\\
&=(b-a)(c-a)(c-b)(c+b+a).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int\frac{\sin(8x)}{9+\sin^4(4x)}\,\mathrm dx$ I have tried to evaluate $$∫\frac{\sin(8x)}{9+\sin^4(4x)}\,\mathrm d x$$ using the following identity:
$$\frac{d(\sin^{-1}{u})}{du} = \frac{du}{1+u^2}$$
So I then reformed the integral to this:
$$1/9\int\frac{\sin(8x)}{1+\sin^4(4x)/9}\,\mathrm dx = 1/9\int\frac{\sin(8x)}{1+(\sin^2(4x)/3)^2}\,\mathrm dx$$
My $u$ in this case would be $\sin^2(4x)/3$. But I do not know where to go from here because I don't know how to reform $\sin(8x)$ into $d(u)$.
Could anyone please explain how I could turn $\sin(8x)$ into my $d(u)$?
| $\sin^4 (4x) = \dfrac{\left(1- \cos (8x)\right)^2}{4} \Rightarrow u = \cos (8x) \Rightarrow du = -8\sin (8x)dx \Rightarrow \displaystyle \int \dfrac{\sin (8x)}{9+ \sin^4(4x)} dx = -\dfrac{1}{8}\cdot \displaystyle \int \dfrac{1}{9+\dfrac{(1-u)^2}{4}} du = -\dfrac{1}{2}\cdot \displaystyle \int \dfrac{1}{6^2 + (1-u)^2} du = -\dfrac{1}{72}\cdot \displaystyle \int \dfrac{1}{1+\left(\dfrac{1-u}{6}\right)^2}du = \dfrac{1}{12}\cdot \displaystyle \int \dfrac{dv}{1+v^2} = \dfrac{\tan^{-1}v}{12} + C = \dfrac{1}{12}\cdot \tan^{-1}\left(\dfrac{1-\cos (8x)}{6}\right) + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Finding number of integral solutions I am really getting confused in this question.
Number of integral solutions of the equation.
$x_1x_2x_3x_4=770$
options-
*
*$2^{11}$
*$2^{10}$
*$4^4$
*$5^5$
I attemtemted it by saying that $7*11*5*2=770$
so solutions should be $4!$
then we can say $77*5*2*1=770$
so $4!$
and $77*10*1*1=770$
so $4!/2$
and then $770*1*1*1=770$
so $4!/3!$
now multiplying them i get $27648$.
i am sure i have made a big mistage here but i dont know what.
please help
| Case 1: $2\cdot 5\cdot 7\cdot 11 = 770 \to 4! = 24$
Case 2: $10\cdot 7\cdot 11\cdot 1 = 770 \to 4!$
Case 3: $14\cdot 5 \cdot 11\cdot 1 = 770 \to 4!$
Case 4: $22 \cdot 5\cdot 7\cdot 1 = 770 \to 4!$
Case 5: $35 \cdot 2\cdot 11\cdot 1 = 770 \to 4!$
Case 6: $55 \cdot 2\cdot 7\cdot 1 = 770 \to 4!$
Case 7: $77\cdot 2\cdot 5\cdot 1 = 770 \to 4!$
Case 8: $70\cdot 11\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$
Case 9: $110\cdot 7\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$
Case 10: $154\cdot 5\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$
Case 11: $385\cdot 2\cdot 1\cdot 1 = 770 \to 4\cdot 3 = 12$
Case 12: $770\cdot 1\cdot 1\cdot 1 = 770 \to 4$.
Thus the total number of solutions is: $7\cdot 24 + 4\cdot 12 + 4 = 220$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove two identities relating to series Show that:
$$(1)\sum_{n=1}^\infty\ln\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$$
$$(2). \sum_{n=1}^\infty\frac1{2^n}\tan \frac{x}{2^n}=\frac1x-\cot x$$
Thank you in advance.
NOTE:
The origial problem (1) I stated is $\sum_{n=1}^\infty\lim_{x\to \infty}\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$, which is a mistake as noted by the comments below.
| For (1) Consider the product $P_N = \displaystyle\prod_{n = 1}^{N}\cos\dfrac{x}{2^n}$. Multiply by $\sin\dfrac{x}{2^N}$ and use the identity $\cos\theta\sin\theta = \dfrac{1}{2}\sin 2\theta$ repeatedly to get $P_N\sin\dfrac{x}{2^N} = \dfrac{1}{2^N}\sin x$.
Therefore, $P_N = \dfrac{\sin x}{2^N\sin\frac{x}{2^N}} \to \dfrac{\sin x}{x}$ as $N \to \infty$. Hence, $\displaystyle\prod_{n = 1}^{\infty}\cos\dfrac{x}{2^n} = \dfrac{\sin x}{x}$.
Take the natural log of both sides to get $\displaystyle\sum_{n = 1}^{\infty}\ln\left(\cos\dfrac{x}{2^n}\right) = \ln\left(\dfrac{\sin x}{x}\right)$, as desired.
For (2) differentiate both sides of the result from (1) and multiply by $-1$. (Thanks David H for this suggestion).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve these equations for x and y.. equations are
$(x-y)(x+2y)(2x+y) = 20$
and
$x^2+xy+y^2 = 7$
i want the METHOD not the solutions
| Drawing the two associated curves, we see that there are $6$ simple solutions (the maximum according to Bezout theorem). Let $s=x+y,p=xy$. If $(x,y)$ is a solution, then $(-y,-x)$ is also a solution. Thus the solutions in $p$ have multiplicity $2$ and $p$ is any root of a polynomial of degree $3$. In the same way, the solutions in $s$ are pairwise opposite.
We obtain $s^2=p+7$ and $(x-y)(2s^2+p)=2$ or $(x-y)(3p+14)=20$ that implies $(s^2-4p)(3p+14)^2=400$ or $(7-3p)(3p+14)^2=400$. The solutions are $p=-6,-3,2$ ; the corresponding values of $s$ are $\pm 1,\pm 2,\pm 3$. The associated solutions in $(x,y)$ are $(-2,3),(-3,2)$ and $(3,-1),(1,-3)$ and $(-1,-2),(2,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Solving an algebraic inequality For any $a$, $b$, and $c$ prove
$$3a^2+3b^2-2b+2a+1>0$$
I tried the following
$$(a+1)^2+(b-1)^2+2(a^2+b^2)-1>0\\
(a+1)^2-1+(b-1)^2+2(a^2+b^2)>0\\
(a+1-1)(a+1+1)+(b-1)^2+2(a^2+b^2)>0\\
(a^2+2a)+(b-1)^2+2(a^2+b^2)>0\\
$$
$a^2$ and $(b-1)^2$ and $2(a^2+b^2)$ are always positive and greater than zero, when $a>0$ our inequality is proved. However when a is negative I can not say $2a$ is always less than or equal to $a^2$. I think this problem should be solved by algebraic identities alone and not by discussing the different signs of $a$? I appreciate your help
| Hint:
$$(a-b+1)^2=a^2+b^2+1-2ab-2b+2a$$
Solution:
$$3a^2+3b^2-2b+2a+1=2a^2+2b^2+(a-b+1)^2+2ab=a^2+b^2+(a+b)^2+(a-b+1)^2>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_{-2}^{-1}\frac{\text{d}x}{\sqrt{-x^2-6x}}$. Problem statement [from Charlie Marshak's Math GRE Prep Problems]:
Evaluate $\displaystyle \int\limits_{-2}^{-1}\dfrac{\text{d}x}{\sqrt{-x^2-6x}}$.
My work: notice that
$$\begin{align}
-x^2-6x &= -(x^2 - 6x) \\&= -\left[x^2-6x+\left(\dfrac{6}{2}\right)^2-\left(\dfrac{6}{2}\right)^2\right] \\
&= -(x^2-6x+9-9) \\
&= 9 - (x^2 - 6x+9) \\
&= 9 - (x-3)^2\text{.}
\end{align}$$
So,
$$\int\limits_{-2}^{-1}\dfrac{\text{d}x}{\sqrt{-x^2-6x}} = \int\limits_{-2}^{-1}\dfrac{\text{d}x}{3\sqrt{1-\left(\dfrac{x-3}{3}\right)^2}}\text{.}$$
If $u = \dfrac{x-3}{3}$, $\text{d}u = \dfrac{1}{3}\text{ d}x$, so that
$$\int\limits_{-2}^{-1}\dfrac{\text{d}x}{3\sqrt{1-\left(\dfrac{x-3}{3}\right)^2}} = \int\limits_{-5/3}^{-4/3}\dfrac{\text{d}u}{\sqrt{1^2-u^2}} = \sin^{-1}\left(\dfrac{-4}{3}\right) - \sin^{-1}\left(\dfrac{-5}{3}\right)\text{.}$$
The answer is supposed to be $\sin^{-1}(2/3)-\sin^{-1}(1/3)$. Where did I go wrong?
| Where you have $x-3$ you should have $x+3$. The exact point you went wrong is in the first line, where you put $-6x$ rather than $+6x$. There is already a minus sign. This is the error:
$$
-x^2-6x \neq -(x^2 - 6x)
$$
$$
-x^2-6x = -(x^2 + 6x)
$$
Note also that $\sin^{-1}(x)$ is meaningless if $|x|>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What are the intermediate fields of $\mathbb Q(\sqrt[3]2,e^{\frac{2i\pi}{3}})$ (Galois group) The elements of Galois group are
\begin{align*}
\sigma _1:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _2:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{2i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _3:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{4i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _4:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _5:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{2i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _6:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{4i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}}.
\end{align*}
We remark that
\begin{align*}
\sigma _2^3=1\\
\sigma _3^3=1\\
\sigma _4^2=1\\
\sigma _5^2=1\\
\sigma_6^2=1
\end{align*}
But $$\sigma _2\sigma _4(\sqrt[3]2)=\sigma _2(\sqrt[3]2)=\sqrt[3]2e^{\frac{2i\pi}{3}}$$
and
$$\sigma _4\sigma _2(\sqrt[3]2)=\sigma _4(\sqrt[3]2e^{\frac{2i\pi}{3}})=\sqrt[3]2e^{\frac{4i\pi}{3}},$$
therefore $\{\sigma _i\}_{i=1}^6$ is not a commutatif group and thus
$$\text{Gal}(E/\mathbb Q)=\{\sigma _i\}_{i=1}^6\cong \mathfrak S_3.$$
My problem is that it should have a $\sigma _i$ such that $\sigma _i^6=1$ and $\sigma _i^n\neq 1$ for $i\in\{1,2,3,4,5\}$ because one of my intermediate field must be $\mathbb Q$. And if not, is there a mistake in my applications $\sigma _i$?
| As you have found $Aut_{\mathbb{Q}}L \simeq S_3$ then the intermediate fields of $Gal (X^3-2, \mathbb{Q})$ are in correspondence to the subgroups of $Aut_{\mathbb{Q}}L$, according to the Fundamental Theorem of Galois. Now,
$$S_3 = \{Id, \sigma, \sigma^2, \tau, \sigma\tau, \sigma^2\tau\}$$
Notice that there is a element $\sigma$ of order $3$ generating a cyclic subgroup of order $3$, i.e., $\langle\sigma\rangle$ , and 3 elements of order $2$. What is missing is to identify such elements $\sigma$ and $\tau$ among your automorphisms $\sigma_i$ in order to find exactly the intermediate fields of $L = Gal (X^3-2,\mathbb{Q})$ through Galois correspondence.
Hint: $\sigma = \sigma_2$ (need to check)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
solving for sin of the sum of two angles of a triangle In triangle $ABC$,
$3\sin B+4\cos C=6$ and $4\sin C+3\cos B=1$.
Show that $\sin(B+C)=0.5$.
Can we assume $\angle A = 180 - ( B + C)$ and use sum formula.
| Hint: By squaring both sides of the equation:
$$9\sin^2 B + 24 \sin B\cos C + 16 \cos^2 C = 36$$
$$9\cos^2 B + 24 \cos B\sin C + 16 \sin^2 C = 1$$
By adding the equations:
$$9(\sin^2B+\cos^2B)+24(\sin B\cos C+\cos B\sin C)+16(\cos^2 C+\sin^2C)=37$$
Using $\sin^2x+\cos^2x=1$
$$9(1)+24(\sin B\cos C+\cos B\sin C)+16(1)=37\implies$$
$$\sin B\cos C+\cos B\sin C=\frac 12$$
Using $\sin B\cos C+\cos B\sin C=\sin(B+C)$
$${\sin(B+C)=\frac 12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3y \ e^{3\cos x\sin y+4\sin x\sin y}\,dy\,dx$ I am working on this double integral
$$\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3y \ e^{3\cos x\sin y+4\sin x\sin y}\,dy\,dx$$
so far, I don't know how to start.
Can someone give a hint?
Thanks
| You can employ the following formula
$\displaystyle\bbox[#EFF,15px,border:3px solid blue]{\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin^3(y)f\big[\sin\left(x\right)\sin\left(y\right)\big]{\mathrm{d}}x{\mathrm{d}}y=\frac{\pi}{4}\int_{0}^{1}\left(t^{2}+1\right)f\left(t\right){\mathrm{d}}t}$
Then
\begin{align*} &\quad\quad \int_{0}^{2\pi}{\mathrm{d}}x\int_{0}^{\pi}e^{3\cos\left(x\right)\sin\left(y\right)+4\sin\left(x\right)\sin\left(y\right)}\,\,\sin^3\left(y\right){\mathrm{d}}y\\ &=\int_{0}^{2\pi}{\mathrm{d}}x\int_{0}^{\pi}e^{5\sin\left(x+\arctan\frac{3}{4}\right)\sin\left(y\right)}\,\sin^3\left(y\right){\mathrm{d}}y\\ &=\int_{0}^{2\pi}{\mathrm{d}}x\int_{0}^{\pi}e^{5\sin\left(x\right)\sin\left(y\right)}\,\sin^3\left(y\right){\mathrm{d}}y\\ &=2\int_{0}^{2\pi}{\mathrm{d}}x\int_{0}^{\frac{\pi}{2}}e^{5\sin\left(x\right)\sin\left(y\right)}\,\sin^3\left(y\right){\mathrm{d}}y\\ &=2\int_{0}^{\frac{\pi}{2}}\sin^3\left(y\right){\mathrm{d}}y\int_{0}^{2\pi}e^{5\sin\left(x\right)\sin\left(y\right)}\,{\mathrm{d}}x\\ &=2\int_{0}^{\frac{\pi}{2}}\sin^3\left(y\right){\mathrm{d}}y\left[\int_{0}^{\pi}e^{5\sin\left(x\right)\sin\left(y\right)}\,{\mathrm{d}}x+\int_{\pi}^{2\pi}e^{5\sin\left(x\right)\sin\left(y\right)}\,{\mathrm{d}}x\right]\\ &=2\int_{0}^{\frac{\pi}{2}}\sin^3\left(y\right){\mathrm{d}}y\left[\int_{0}^{\pi}e^{5\sin\left(x\right)\sin\left(y\right)}\,{\mathrm{d}}x+\int_{0}^{\pi}e^{-5\sin\left(x\right)\sin\left(y\right)}\,{\mathrm{d}}x\right]\\ &=2\int_{0}^{\frac{\pi}{2}}\sin^3\left(y\right){\mathrm{d}}y\int_{0}^{\pi}\left[e^{5\sin\left(x\right)\sin\left(y\right)}+e^{-5\sin\left(x\right)\sin\left(y\right)}\,\right]\!{\mathrm{d}}x\\ &=4\int_{0}^{\frac{\pi}{2}}\sin^3\left(y\right){\mathrm{d}}y\int_{0}^{\frac{\pi}{2}}\left[e^{5\sin\left(x\right)\sin\left(y\right)}+e^{-5\sin\left(x\right)\sin\left(y\right)}\,\right]\!{\mathrm{d}}x\\ &=4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\left[e^{5\sin\left(x\right)\sin\left(y\right)}+e^{-5\sin\left(x\right)\sin\left(y\right)}\,\right]\sin^3\left(y\right){\mathrm{d}}x{\mathrm{d}}y\\ \end{align*}
Hence
\begin{align*} &\quad\quad \int_{0}^{2\pi}{\mathrm{d}}x\int_{0}^{\pi}e^{3\cos\left(x\right)\sin\left(y\right)+4\sin\left(x\right)\sin\left(y\right)}\,\,\sin^3\left(y\right){\mathrm{d}}y\\ &=4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\left[e^{5\sin\left(x\right)\sin\left(y\right)}+e^{-5\sin\left(x\right)\sin\left(y\right)}\,\right]\sin^3\left(y\right){\mathrm{d}}x{\mathrm{d}}y\\ &=4\cdot\dfrac{\pi}{4}\int_{0}^{1}(t^2+1)\left(e^{5t}+e^{-5t}\right){\mathrm{d}}t\\ &=\frac{2\pi}{125}\left(21e^{5}-\frac{31}{e^{5}}\right) \end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}$ I was just playing around with a calculator, and came to the conclusion that:
$$\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}} \approx 1.29$$
Now I'm curious. Is it possible to evaluate the exact value of the following?
| It is easy to show, assuming the limit exists:
$$\sqrt{a+\sqrt{a+\sqrt{a+...}}}=\frac{1+\sqrt{1+4a}}{2}$$
It is also easy to see the following set of inequalities:
$$\sqrt{\frac{1}{2}+1} < \sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}<\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+...}}}$$
$$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+1}} < \sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}<\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+...}}}$$
$$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{8}+1}}} < \sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}<\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{8}+\sqrt{\frac{1}{8}+...}}}}$$
And so on, getting better and better approximations.
Calculating the nested radicals:
$$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+...}}}=\frac{1+\sqrt{3}}{2}$$
$$\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+...}}}=\frac{1+\sqrt{2}}{2}$$
$$\sqrt{\frac{1}{8}+\sqrt{\frac{1}{8}+\sqrt{\frac{1}{8}+...}}}=\frac{1+\sqrt{1.5}}{2}$$
We get following set of boundaries for the value we need:
$$R=\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}=1.285737\dots$$
$$1.22474<R<1.36603$$
$$1.27202<R<1.30656$$
$$1.28251<R<1.29120$$
And so on. There is no closed form, but it's not hard to evaluate this nested radical with good precision.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 3,
"answer_id": 1
} |
Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$ $$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$
How to integrate?
I tried the substitution $x=\sin\theta$ but didn't work.
| i will try $x = {2t \over 1 + t^2}$
$$dx = 2{ (1+t^2)dt - t \cdot2tdt \over (1+t^2)^2} = {2(1-t^2)dt \over (1+t^2)^2}
\\ 1 - x = {(1-t)^2 \over 1 + t^2}, 1 + x = {(1+t)^2 \over 1 + t^2}, \sqrt{{1-x \over 1 + x}}= {1 - t \over 1 + t} $$
putting all these together,
$$\int{1 \over x} \sqrt{{1-x \over 1 + x}} dx = \int{1 + t^2 \over 2t} {1 - t \over 1 + t} {2(1-t^2)dt \over (1+t^2)^2}
\\= \int {(1-t)^2 \over t(1+t^2)} \ dt = \int {1 + t^2 - 2t \over t(1+t^2)} dt = \int {dt \over t} - 2\int{dt \over 1 + t^2}
\\ = \ln(t) - 2 \tan^{-1}(t) + C $$
you can solve $$x = {2t \over 1 + t^2}$$ for $x$ by writing $$xt^2 - 2t + x =0 $$ as a quadratic equation for $t$. by quadratic formula we get $$t = {1 \pm \sqrt{1 + x^2} \over x} $$
put all these in $$ \int{1 \over x} \sqrt{{1-x \over 1 + x}} dx = \ln(t) - 2 \tan^{-1}(t) + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
} |
Algebra on a Louvre tablet Problem: On a Louvre tablet of about 300 B.C. are four problems concerning rectangles of unit area and given semiperimeter. Let the sides and semiperimeter be $x,y$ and $a$. Then we have
\begin{equation}
xy=1, \qquad x+y=a. \tag{1}
\end{equation}
Solve this system by using the identity
$$
\biggl(\frac{x-y}{2}\biggr)^2 = \biggl(\frac{x+y}{2}\biggr)^2 - xy \tag{2}
$$
Attempted solution:
By elimination: To eliminate $y$, we note that $y = \frac{1}{x}$, and we substitute this $y$-value into the equation $x+y=a$ to obtain
$$
x = \frac{a \pm \sqrt{a^2-4}}{2},
$$
where this solution was obtained by using the quadratic formula.
By using the given identity: Start by letting $y=a-x$. Using this and the fact that $x+y=a$, we have the following:
\begin{align*}
\biggl(\frac{2x-a}{2}\biggr)^2 = \biggl(\frac{a}{2}\biggr)^2 - 1 &\longleftrightarrow 4x^2-4xa+a^2 = a^2-4\\ &\longleftrightarrow x^2-ax+1=0,
\end{align*}
whereby we get that $x = \frac{a \pm \sqrt{a^2-4}}{2}$, as we did above by using elimination.
Main question: My solution using (2) seems to work, but is there a "slicker" approach? That is, it seems like the originally posed problem wants me to do something nifty with (2) rather than using it in such a brutish fashion. Anyone have a cleaner approach in mind in regards to solving (1) by using (2)?
| I think the intended solution was as follows:
$$
\left(\frac{x-y}{2}\right)^2=\left(\frac{a}{2}\right)^2-1.
$$
Thus $x-y$ can be solved for:
$$
x-y=\pm2\sqrt{\left(\frac{a}{2}\right)^2-1}=\pm\sqrt{a^2-4}
$$ Adding and subtracting with $x+y$ gives us:
$$
x=\frac{(x+y)+(x-y)}{2}=\frac{a\pm \sqrt{a^2-4}}{2},
$$
and likewise for $y$.
Note that this is a re-derivation of the quadratic formula; in particular, it defeats the purpose of this approach to use it!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator
It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$?
I did the following:
$$\begin{array}
& 5 \ \text{mod} \ 7=& &5 \\
55 \ \text{mod} \ 7= & &6 \\
555 \ \text{mod} \ 7= & &2 \\
5555 \ \text{mod} \ 7= & &4 \\
55555 \ \text{mod} \ 7= & &3 \\
555555 \ \text{mod} \ 7= & &0 \\
5555555 \ \text{mod} \ 7= & &5 \\
55555555 \ \text{mod} \ 7= & &6 \\
555555555 \ \text{mod} \ 7= & &2 \\
5555555555 \ \text{mod} \ 7= & &4 \\
\end{array}$$
It can be seen that the cycle is: $\{5,6,2,4,3,0\}$.
$$\begin{array}
& 1 \ \text{number =} &5 \\
7 \ \text{numbers =} &5 \\
13 \ \text{numbers =} &5 \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots & \\
985 \ \text{numbers =} &5 \\
991 \ \text{numbers =} &5 \\
997 \ \text{numbers =} &5 \\
998 \ \text{numbers =} &6 \\
999 \ \text{numbers =} &2 \\
\color{red}{1000} \ \color{red}{\text{numbers =}} &\color{red}{4} \\
\end{array}$$
From here, we can conclude that $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7 = 4$.
However, I wasn't allowed to use a calculator and solved this in about 12 minutes. Another problem was that there was a time limit of about 5 minutes. My question is: Is there an easier and faster way to solve this?
Thanks a lot in advance!
| There are many great answers already written up, so I'm not sure if this is going to add anything, but for what it's worth, here's how I did it.
Recognise that $\underbrace{555\cdots555}_{1000\ \text{times}} = 5\cdot \underbrace{111\cdots111}_{1000\ \text{times}} = 5 \cdot 9^{-1} \cdot \underbrace{999\cdots999}_{1000\ \text{times}} = 5 \cdot 9^{-1} \cdot (10^{1000}-1)$
Working modulo $7$,
$10^{1000}-1 \equiv 3^{1000} - 1 = 3^{6\cdot 166 + 4} - 1 {\equiv} 80 \equiv 3 \pmod 7$
The modular inverse of $9$ is $4 \pmod 7$.
So the final residue is $5\cdot 4 \cdot 3 \equiv 4 \pmod 7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 10,
"answer_id": 8
} |
Can one always map a given triangle into a triangle with chosen angles by means of a parallel projection? This is something that seems to be true from experience by playing with shadows from the sun:
If one cuts a paper triangle, he can turn it in a way to make its shadow be a triangle of any given angles (of course, not exceeding internal sum of 180°), for example: If one draws a right triangle with internal angles 90°, 60° and 30°, one can turn it in a way that the shadow triangle can be of angles 60°, 60°, 60° (equilateral) or 120°, 30° and 30° (isosceles), or any other triangle possibilities.
Is it true? How can we prove it? (if possible, showing not only that it's possible, but how to obtain the specific mapping that does it)
| Model of the projection
Let the light come from above, along the $z$-axis.
The shadow of an triangle above the $x-y$-plane is then the projection onto the $x-y$-plane ($z = 0$).
The triangle has a unit normal vector $n$ to describe its orientation ($n^2 = 1$).
The shadow projection $P$ is
$$
P =
\left(
\begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{matrix}
\right)
\quad
P^c =
\left(
\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1
\end{matrix}
\right)
$$
it satisfies $P^2 = P$ and $I = P + P^c$.
Triangle description
We specify the triangle corners by vectors $r_1$, $r_2$, $r_3$ with the origin at $0$.
We define the oriented triangle sides $a_i$ as
$$
a_1 = r_2 - r_1, \quad a_2 = r_3 - r_1, \quad a_3 = r_3 - r_2
$$
The side vectors are linear dependent:
$$
a_1 - a_2 + a_3 = 0
$$
We have the angles $\alpha_i$ at the corner pointed to by $r_i$:
\begin{align}
a_1 \cdot a_2 &= \lVert a_1 \rVert \lVert a_2 \rVert \cos \alpha_1 \\
(-a_1) \cdot a_3 &= \lVert -a_1 \rVert \lVert a_3 \rVert \cos \alpha_2 \quad (*) \\
(-a_2) \cdot (-a_3) &= \lVert -a_2 \rVert \lVert -a_3 \rVert \cos \alpha_3
\end{align}
The equations $(*)$ are still quite general, they just relate the lengths of the sides and the angles, and embody the given orientations of the sides.
Now we get more specific: We use a triangle with these sides and
$$
\lVert a_1 \rVert = 1/2 \quad
\lVert a_2 \rVert = \sqrt{3}/2 \quad
\lVert a_3 \rVert = 1 \quad (**)
$$
and these angles
$$
\alpha_1 = 90^\circ \quad \alpha_2 = 60^\circ \quad \alpha_3 = 30^\circ \\
\cos\alpha_1 = 0 \quad \cos\alpha_2 = 1/2 \quad \cos\alpha_3 = \sqrt{3}/2
$$
This assumes that the scaling of that triangle plays no role, e.g. having all sides twice the length has no qualitative influence on the problem.
To describe the $a_i$ (9 unknowns) in terms of $n$ we have these three equations
$$
a_i \cdot n = 0 \\
$$
plus the three lengths $(**)$ plus the three equations $(*)$ which for this triangle are
\begin{align}
a_1 \cdot a_2 &= 0 \\
a_1 \cdot a_3 &= -1/4 \\
a_2 \cdot a_3 &= 3/4
\end{align}
The Shadow Triangle
The sides of the shadow triangle are $b_i$ and related via the projection:
$$
b_i = P a_i
$$
The projected angles are $\beta_i$ and they should satisfy equations similar to $(*)$:
\begin{align}
b_1 \cdot b_2 &= \lVert b_1 \rVert \lVert b_2 \rVert \cos \beta_1 \\
(-b_1) \cdot b_3 &= \lVert -b_1 \rVert \lVert b_3 \rVert \cos \beta_2 \quad (***) \\
(-b_2) \cdot (-b_3) &= \lVert -b_2 \rVert \lVert -b_3 \rVert \cos \beta_3
\end{align}
We have
$$
b_i \cdot b_j = P a_i \cdot P a_j = a_i \cdot a_j - a_{i3} a_{j3} \\
\lVert b_i \rVert^2 = \lVert P a_i \rVert^2 = \lVert a_i \rVert^2 - a_{i3}^2
$$
Plan: We input our $\beta_i$ into $(***)$ and get three equations for the three unknown $a_{i3}$. From this we try to determine a suitable $n$.
First case $\beta_i = 60^\circ$
The interesting bit is that $\alpha_2 = \beta_2$, which means this angle should stay the same under projection while the other two angles expand or shrink.
Looking at the equation for $\beta_2$:
\begin{align}
(b_1 \cdot b_3)^2
&=
\lVert b_1 \rVert^2 \lVert b_3 \rVert^2 (\cos \beta_2)^2 \iff
\\
\left(\frac{1}{4}+xz\right)^2
&=
\left(\frac{1}{4}-x^2\right)\left(1-z^2\right)\frac{1}{4}
\end{align}
where $x = a_{13}$, $y = a_{23}$ and $z = a_{33}$.
This gives
$$
z = \frac{\sqrt{12} x \sqrt{1- 4 x^2} - 4x}{12x^2 + 1}
$$
plus we know $x - y + z = 0$.
If we have $(x,y,z)$, we can calculate all angles for the projection:
\begin{align}
\beta_1 &= \arccos\frac{-xy}{\sqrt{\frac{1}{4}-x^2}\sqrt{\frac{3}{4}-y^2}} \\
\beta_2 &= \arccos\frac{\frac{1}{4} + x z}{\sqrt{\frac{1}{4}-x^2}\sqrt{1-z^2}} \\
\beta_3 &= \arccos\frac{\frac{3}{4} - y z}{\sqrt{\frac{3}{4}-y^2}\sqrt{1-z^2}}
\end{align}
We get this graph
So while we can keep the projected angle at $\beta_2$ at $60^\circ$, we can not reach $60^\circ$ for $\beta_1$ and $\beta_3$. $\beta_1$ grows from $90^\circ$ and $\beta_2$ shrinks from $30^\circ$.
(At this point I could not resist to rip off a piece from the paper I did the calculations on, and folded a triangle to try it out :-)
(It looked not bad, $\beta_1$ seemed indeed growing)
I am not sure about the other cases, it is easy to make a mistake regarding the solutions, taking the wrong sign etc.
(In progress)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1088066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Evaluating the sum : $\;\frac{1}{3}+\frac{1}{4}.\frac{1}{2!}+\frac{1}{5}.\frac{1}{3!}+\ldots$ How to evaluate this sum?
$$\frac{1}{3}+\frac{1}{4}.\frac{1}{2!}+\frac{1}{5}.\frac{1}{3!}+\ldots$$
Please give some technique. Binomial not working.
| Note
\begin{align*}\sum_{n=1}^{\infty}\dfrac{1}{(n+2)n!}&=\sum_{n=1}^{\infty}\dfrac{n+1}{(n+2)!}=\sum_{n=1}^{\infty}\dfrac{(n+2)-1}{(n+2)!}=\sum_{n=1}^{\infty}\left(\dfrac{1}{(n+1)!}-\dfrac{1}{(n+2)!}\right)\\
&=\dfrac{1}{2}-\lim_{n\to\infty}\dfrac{1}{(n+2)!}\\
&=\dfrac{1}{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Quadratic equation with parameter and conditions on the roots For which values of $m$ the equation
$$(2m+1)x^2-(4m+2)x+m-1=0$$
has two real roots $x_1$ and $x_2$ that satisfy the condition: $x_1<x_2<2$?
I found that $$m\in\left(-\frac{1}{2}; -\frac{1}{3}\right).$$ Is it correct?
| The discriminant is $$\Delta=b^2-4ac=(4m+2)^2-4(2m+1)(m+1)=8m^2+20m+8$$ To get real roots, $m$ must satisfy $$\Delta>0$$
which reduces as $$2m^2+5m+2>0$$
and has for solution $m>-1/2$ or $m<-2$. The roots are then $$x_{2,1}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{4m+2\pm\sqrt{8m^2+20m+8}}{4m+2}$$
with $x_1=1-\frac{\sqrt{8m^2+20m+8}}{4m+2}$ and $x_2=1+\frac{\sqrt{8m^2+20m+8}}{4m+2}$.
If we want $x_1<x_2$, then
$$1-\frac{\sqrt{8m^2+20m+8}}{4m+2}<1+\frac{\sqrt{8m^2+20m+8}}{4m+2}$$
which yields $m>-1/2$, consistent with what we found above.
Now, we also want $x_2<2$, so we solve
$$\frac{\sqrt{8m^2+20m+8}}{4m+2}<1$$
and we find $m>1$ or $m\le-2$.
Combining all these conditions, if we want real roots $x_1$ and $x_2$ that verify $x_1<x_2<2$, we need
$$m\in\left]-\infty;-2\right[\cup\left]1;+\infty\right[$$
Of course, you should always check these results, and also interpret them graphically. For instance, using Mathematica, the command
Manipulate[Plot[(2*m + 1)*x^2 - (4*m + 2)*x + m - 1, {x, -1, 3}], {m, -10, 10}]
creates a nice parametric plot and you can observe the behaviour of the solutions as they approach the boundaries.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
An infinite matrix series Playing around with my CAS, I found that apparently
$$\sum_{n=1}^\infty\begin{bmatrix}1/2 & 1/3\\1/4 & 1/5\end{bmatrix}^n = \begin{bmatrix}29/19 & 20/19\\15/19 & 11/19\end{bmatrix}$$
If that is indeed true, it's pretty cool.
How can I prove this, and what is the general formula if the coefficients are replaced by arbitrary numbers $a,b,c,d<1$?
| Given any $n \times n$ square matrix $A = (a_{ij})$ over a field $K = \mathbb{R}$ or $\mathbb{C}$.
If its operator norm $\|A\|_{op}$:
$$\|A\|_{op} \stackrel{def}{=} \sup\big\{\; \|Av\| : v \in K^n \text{ with } \|v\| = 1 \;\big\}$$
or any sub-multiplicative matrix norm, e.g. the
Frobenius norm
$\|A\|_F$:
$$\|A\|_{F} \stackrel{def}{=} \sqrt{ \sum_{i=1}^n\sum_{j=1}^n |a_{ij}|^2 } = \sqrt{\text{tr}\left(A^{\dagger}A\right)}$$
is smaller than $1$, then the matrix series $\displaystyle\;\sum\limits_{k=1}^n A^n\;$
converge like what you find for a geometric series $\displaystyle\;\sum\limits_{k=1}^\infty z^k = \frac{z}{1-z}\;$ for $z \in K, |z| < 1$. More precisely,
$$\sum_{k=1}^\infty A^k = A\left(I_n - A\right)^{-1}\quad\text{ when }\quad\|A\| < 1 \tag{*1}$$
In particular, when $A = \begin{bmatrix}a & b\\c & d\end{bmatrix}$ and $\|A\|_F^2 = a^2 + b^2 + c^2 + d^2 < 1$, we have
$$\sum_{k=1}^\infty A^k
=
\begin{bmatrix}a & b\\c & d\end{bmatrix}
\begin{bmatrix}1-a & -b\\-c & 1-d\end{bmatrix}^{-1}
=
\frac{\begin{bmatrix}a - (ad-bc) & b\\ c & d - (ad-bc)\end{bmatrix}}{1-(a+d)+(ad-bc)}$$
In your case, $(a,b,c,d) = \left(\frac12,\frac13,\frac14,\frac15\right)$, its Frobenius norm $$\|A\| = \sqrt{\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2}} = \frac{\sqrt{1669}}{60} < 1$$
and the series converges to what you find
$$
\sum_{k=1}^\infty \begin{bmatrix}\frac12 & \frac13 \\ \frac14 & \frac15\end{bmatrix}^k =
\frac{\begin{bmatrix}\frac12 - \frac{1}{60} & \frac13\\ \frac14 & \frac15 - \frac{1}{60}\end{bmatrix}}{1 - \frac{7}{10} + \frac{1}{60}}
= \begin{bmatrix}\frac{29}{19} & \frac{20}{19}\\ \frac{15}{19} & \frac{11}{19}\end{bmatrix}$$
The proof of $(*1)$ is just like how you show $\sum_{k=1}^\infty z^k = \frac{z}{1-z}$ for $z \in K, |z| < 1$. You look at following partials sums and send $N$ to $\infty$
$$\require{cancel}
(1-z)\sum_{k=1}^N z^k = z - \cancelto{0 \text{ when } |z| < 1}{\color{grey}{z^{N+1}}}
\quad\leftrightarrow\quad
(I_n - A)\sum_{k=1}^N A^k = A -
\cancelto{0 \text{ when } \|A\| < 1}{\color{grey}{A^{N+1}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Hint for Estimating an Infinite Series I am currently working through E. V. Shchepin's book Uppsala Lectures on Calculus: On Euler's Footsteps, and I've encountered a problem with estimating an infinite sum.
I need a hint estimating: S = $\sum_{k=1}^\infty{1 \over2^kk}$. This problem appears early in the book in the section on unordered sums.
So far I have bounded S by comparing it to $\sum_{k=1}^\infty{1 \over2^k}$ = 1
and $\sum_{k=1}^\infty{1 \over2^k 2^k}$ = ${1 \over 3}$
I've also tried various groupings of terms such as:
$\sum_{k=1}^\infty{1 \over2^kk}$ = ${1 \over 2}$ ${1 \over 1 }$ + ${1 \over 2^2}$${1 \over 2}$ + ${1 \over 2^3}$${1 \over 3}$ + ${1 \over 2^4}$${1 \over 4}$ +${1 \over 2^5}$${1 \over 5}$ +${1 \over 2^6}$${1 \over 6}$ + ${1 \over 2^7}$${1 \over 7}$ +${1 \over 2^8}$${1 \over 8}$ +$\cdots$
$\ge$ ${1 \over 2}$ ${1 \over 1 }$ + ${1 \over 2^2}$${1 \over 2}$ + (${1 \over 2^3}$${1 \over 4}$ + ${1 \over 2^4}$${1 \over 4}$) + (${1 \over 2^5}$${1 \over 8}$ +${1 \over 2^6}$${1 \over 8}$ + ${1 \over 2^7}$${1 \over 8}$ +${1 \over 2^8}$${1 \over 8}$) +$\cdots$
$=$ ${1 \over 2}$ + ${1 \over 2^3}$+ (${1 \over 2^5}$+ ${1 \over 2^6}$) + (${1 \over 2^8}$ +${1 \over 2^9}$ + ${1 \over 2^{10}}$ +${1 \over 2^{11}}$) +$\cdots$
which is $\sum_{k=1}^\infty{1 \over2^k}$ with some terms missing. However, I can't find any pattern to the missing terms.
Any hints on how to proceed?
| Write
$$
\frac{1}{k} = 1 - \frac{1}{1\cdot 2} -\frac{1}{2\cdot 3} -\frac{1}{3\cdot 4} \cdots \frac{1}{(k-1)k}
$$
Then group terms to get
$$S = 1 \sum \frac{1}{2^k} - \frac{1}{1 \cdot 2} \sum \frac{1}{2^{k+1}} + \cdots
$$
So for example your fifthapproximation will be
$$
1 - \frac{1}{2} \frac{1}{2} - \frac{1}{4} \frac{1}{6} - \frac{1}{8} \frac{1}{12} - \frac{1}{16} \frac{1}{20} = \frac{111}{160}
$$
which is accurate to three decimal digits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Probability of rolling a sum of N with up to infinite rolls of a die I'm trying to figure out if there is some (relatively simple) formula for calculating the probability of rolling a sum of N with as many rolls as needed with a single regular six-sided die.
For example:
$N=1$ is $0.16666 = 1/6 = 1/6$ (1)
$N=2$ is $0.19444 = 7/36 = 1/6 + 1/36$ (2 or 1,1)
$N=3$ is $0.22685 = 49/216 = 1/6 + 2/36 + 1/216$ (3 or 1,2/2,1 or 1,1,1)
$N=4$ is $0.26466 = 343/1296 = 1/6 + 3/36 + 3/216 + 1/1296$ (4 or 2,1/1,2/2,2 or 1,1,2/1,2,1/2,1,1 or 1,1,1,1)
...
Does this series converge to essentially one and is there some good formula?
It seems like pascal's triangle is involved somehow (at least for n=1-6, but I'm not sure how (if even possible) to convert it into a formula. Any help/advice is appreciated.
| Let $q(x)=\frac{1}{6}\left(x^1+x^2+x^3+x^4+x^5+x^6\right)$. Write:
$$\begin{align}G(x)&=\sum_{n=0}^\infty q(x)^n\\&=\frac{1}{1-q(x)}\\&=\frac{6}{6-x-x^2-x^3-x^4-x^5-x^6}
\end{align}$$
Then the coefficient of $x^N$ in $G(x)$ is your probability.
So you need to know something about the roots of the denominator. $1$ is one such root, but you are going to need to deal with complex roots, as well, or at least get an upper bound for the roots.
The formula is:
$$p(N)=a_1\alpha_1^N+a_2\alpha_2^N+\cdots a_6\alpha_6^N$$
For some constants $a_i$ and with the $\alpha_i$ the inverse roots of the denominator.
According to Wolfram Alpha, the roots of the denominator are $1$ and a bunch of values (one real, four complex) with absolute value greater than $1$.
So for large enough $N$, $p(N)\approx a_1$.
We can find $a_1$ by doing the usual partial fractions computation:
$$\lim_{x\to 1} G(x)(1-x)=\frac{2}{7}=\frac{1}{3.5}$$ which is what Arthur conjectured in comments above.
It turns out for $i>1$, $|\alpha_i|<\frac{3}{4}$, so we have:
$$p(N)=\frac{2}{7} + o\left(\left(\frac 34\right)^N\right)$$
We can also get a recursion:
$$P(n+6)=\frac{1}{6}\left(P(n)+P(n+1)+P(n+2)+P(n+3)+P(n+4)+P(n+5)\right)$$
We get $P(28)=.2857\dots.$ For $n\geq 88$ we get:
$$P(n)=0.285714285714\dots$$ which is $12$ digits of accuracy to $\frac{2}{7}=0.\overline{285714}.$
We also get that $$\frac{P(100)-\frac{2}{7}}{\left(\frac34\right)^{100}}\approx -0.016$$
and
$$\frac{P(1000)-\frac{2}{7}}{\left(\frac34\right)^{1000}}\approx 7.34\times 10^{-13}$$
(Computing the left side with python Fraction class to get exact value, then converted to float.)
This gives:
$$P(1000)-\frac{2}{7}\approx 8.45\times 10^{-138}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Is there a closed-form of $ \sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$ Is there a closed-form of $$\sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$$
Thanks for any help
| Let $$f(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+4}}{(2n+1)(2n+2)(2n+3)(2n+4)}.$$
Then taking the derivatives wrt $x$, the factors in the denominator disappear:
$$f''''(x)=\sum_{n=0}^{\infty }(-x^2)^n=\frac1{1+x^2}.$$
Integrate four times from $0$ to $x$ (factors added for convenience).
$$\begin{align}f'''(x)&=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+1}}{2n+1}\\
&=\arctan x.\end{align}$$
$$\begin{align}2f''(x)&=2\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)}\\
&=2x\arctan x-\log(x^2+1).\end{align}$$
$$\begin{align}2f'(x)&=2\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+3}}{(2n+1)(2n+2)(2n+3)}\\
&=(x^2-1)\arctan x-x\log(x^2+1)+x.\end{align}$$
$$\begin{align}12f(x)&=12\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+4}}{(2n+1)(2n+2)(2n+3)(2n+4)}\\
&=2(x^2-3)x\arctan x-(3x^2-1)\log(x^2+1)+5x^2.\end{align}$$
Then
$$f(1)=\dfrac{-\pi-2\log2+5}{12}.$$
For the same "price", you also get the sums with $1$ to $3$ factors: $f'''(1)=\pi/4$, $f''(1)=\pi/4-\log2/2$, $f'(1)=-\log2/2+1/2$.
For efficient evaluation of the integrals, we can proceed as follows using complex numbers.
Notice that $\dfrac1{x^2+1}$ is the imaginary part of $\dfrac1{x-i}$, and use the change of variable $t=x-i$.
Integrate four times (by parts) between $-i$ and $t$:
$$\frac1t,$$
$$\log t,$$
$$t(\log t-1),$$
$$\frac{t^2}4(2\log t-3),$$
$$\frac{t^3}{36}(6\log t-11).$$
Then take the imaginary parts of the differences of the antiderivative values between $1-i$ and $-i$:
$$\Im\left(\log(1-i)-\log(-i)\right)=\frac\pi4,$$
$$\Im\left((1-i)\left(\log(1-i)-1\right)-(-i)\left(\log(-i)-1\right)\right)=-\frac\pi4-\frac{\log2}2,$$
$$\Im\left(\frac{(1-i)^2}4(2\log(1-i)-3)-\frac{(-i)^2}4(2\log(-i)-3)\right)=\frac32-\pi-\frac{\ln2}2$$
$$\Im\left(\frac{(1-i)^3}{36}\left(6\log(1-i)-11\right)-\frac{(-i)^3}{36}\left(6\log(-i)-11\right)\right)=\frac\pi{12}-\frac{\log2}6-\frac{11}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Investigate convergence of the series Investigate convergence of the series:
$$\left( \frac{n^2+3n+10}{n^2+5n+17} \right)^{n^2 (\sqrt{n+1}-\sqrt{n-1})}$$
It should be solvable with simple manipulations with the formula, i guess, but how to do that?
| Let $a_n$ be the above term. Note first that
$$\frac{n^2+3 n+10}{n^2+5 n+17} = \left (1+\frac{3}{n}+\frac{10}{n^2} \right ) \left (1+\frac{5}{n}+\frac{17}{n^2} \right )^{-1} = 1-\frac{2}{n}+\frac{3}{n^2}+O\left (\frac1{n^3}\right )$$
Also,
$$n^2 \left ( \sqrt{n+1}-\sqrt{n-1}\right ) = n^{3/2} \left (1+\frac{3}{8 n^2} + O\left (\frac1{n^3}\right )\right ) $$
Then
$$\begin{align}\log{a_n} &= n^2 \left ( \sqrt{n+1}-\sqrt{n-1}\right ) \log{ \left [ 1-\frac{2}{n}+\frac{3}{n^2}+O\left (\frac1{n^3}\right )\right]} \\ &= n^{3/2} \left (1+\frac{3}{8 n^2} + O\left (\frac1{n^3}\right )\right ) \left (-\frac{2}{n} + \frac1{n^2} + O\left (\frac1{n^3}\right ) \right )\\ &= -2 \sqrt{n} + \frac1{\sqrt{n}} + O \left ( n^{-3/2} \right ) \end{align}$$
which means that the general term in the series behaves as
$$e^{-2 \sqrt{n}} \left ( 1+ \frac1{\sqrt{n}} \right ) $$
for large $n$. Thus the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Matrices, determinants, and applications to identities involving Fibonacci numbers Preamble
It is well known that since:
$$
\begin{pmatrix}
F_{n+1} \\
F_n \\
\end{pmatrix} =
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}
\begin{pmatrix}
F_n & F_{n-1} \\
\end{pmatrix}
$$
it is valid that:
$$
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}^n =
\begin{pmatrix}
F_{n+1} & F_n \\
F_n & F_{n-1} \\
\end{pmatrix}
$$
By calculating determinants, it immediately follows that:
$$F_{n-1}F_{n+1} - {F_n}^2
=\det\left[\begin{matrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{matrix}\right]
=\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]^n
=\left(\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]\right)^n
=(-1)^n$$
or
$$F_{n-1}F_{n+1} - {F_n}^2=(-1)^n$$
This is known as Cassini's identity.
I find both Cassini's identity and this proof exceptionally attractive.
Now, there are other identities that resemble Cassini's identity:
$$F_{n-2}F_{n-1}F_{n+3} - {F_n}^3=(-1)^nF_{n-3}$$
$$F_{n+2}F_{n+1}F_{n-3} - {F_n}^3=(-1)^nF_{n+3}$$
$${F_{n-3}}{F_{n+1}}^2-{F_{n-2}}^2{F_{n+3}}=4 (-1)^n{F_{n}}$$
$${F_{n-1}}^2{F_{n+1}}^2-{F_{n-2}}^2{F_{n+2}}^2=4 (-1)^n{F_{n}}^2$$
$$F_{n-2}F_{n-1}F_{n+1}F_{n+2} - {F_n}^4=-1$$
Question
Is there a proof of five identities above that relies on matrices, the proof that would be attractive and similar to the mentioned proof
of Cassini's identity?
Also, is there any other Fibonacci identity that follows from a suitable correspondant matrix equation?
Trivia
Not directly related to any regular (meaning having the form of an equality) Fibonacci identity, I found also some fairly surprising matrix identities involving Fibonacci numbers, and among them the strangest is:
$$\pmatrix{
3&6&-3&-1\\
1&0&0&0\\
0&1&0&0\\
0&0&1&0
}^n\pmatrix{
2197&512&125&27\\
512&125&27&8\\
125&27&8&1\\
27&8&1&1
}=\pmatrix{
F_{n+7}^3&F_{n+6}^3&F_{n+5}^3&F_{n+4}^3\\
F_{n+6}^3&F_{n+5}^3&F_{n+4}^3&F_{n+3}^3\\
F_{n+5}^3&F_{n+4}^3&F_{n+3}^3&F_{n+2}^3\\
F_{n+4}^3&F_{n+3}^3&F_{n+2}^3&F_{n+1}^3\\
}
$$
Here, one more unusual fact: $8$, $27$, and $125$ are also cubes of a Fibonacci number.
| Let's look at your first identity. Express everything in terms of $F_{n-1}$ and $F_n$.
$$ \eqalign{F_{n-2} F_{n-1} F_{n+3} - F_{n}^3 &= 3 F_{n}^2 F_{n-1} - F_n F_{n-1}^2 - 2 F_{n-1}^3 - F_{n}^3 \cr
&= (F_n - 2 F_{n-1})(-F_n^2 + F_n F_{n-1} + F_{n-1}^2)\cr
(-1)^n F_{n-3} &= -(-1)^n (F_n - 2 F_{n-1})\cr}$$
so the identity reduces to $F_n^2 - F_n F_{n-1} - F_{n-1}^2 = (-1)^n$
which is equivalent to Cassini.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find $ \binom {1}{k} + \binom {2}{k} + \binom{3}{k} + ... + \binom{n}{k} $
Find $$ \binom {1}{k} + \binom{2}{k} + \binom{3}{k} + ... + \binom {n}{k} $$ if $0 \le k \le n$
Any method for solving this problem? I've not achieved anything so far. Thanks in advance!
| For $k=0$, the answer is $n$.
For $k>0$, the answer is ${{n+1}\choose{k+1}}$.
This can be seen by considering how many ways there are to choose $k+1$ integers from the set $\{1,2,\dots,n+1\}$ and conditioning on the greatest integer chosen, because there have to be $k$ choices less than the greatest integer chosen.
For instance we look at ${{n+1}\choose{k+1}}={6\choose 4}$. To choose $4$ elements from $\{1,2,3,4,5,6\}$, the greatest chosen element could be $4$, and then we would have to choose $3$ elements from $\{1,2,3\}$; or the greatest chosen element could be $5$, and then we would have to choose $3$ elements from $\{1,2,3,4\}$; or the greatest chosen element could be $6$, and then we would have to choose $3$ elements from $\{1,2,3,4,5\}$.
So altogether, ${6\choose 4}={3\choose 3}+{4\choose 3}+{5\choose 3}$.
Note: the other terms in the sum, of the form ${i\choose k}$ with $i<k$, are all $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Finding the derivative $f(x)=\sqrt{x^2 -9}$, I need to find the slope at a=5, using the definition for the function $f(x)=\sqrt{x^2 -9}$,
$$f'(x) = \lim_{\Delta x \to 0} {f(x+\Delta x)\over \Delta x}$$
The answer book says the slope is ${1\over 4}$
Here's what I did,
$$f'(x) = \lim_{\Delta x \to 0} {(\sqrt{(x+\Delta x)^2 -9} - \sqrt {x^2 -9} )(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)})\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(1)\\=\lim_{\Delta x \to 0} {(x+\Delta x)^2 -9 -x^2 +9\over \Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(2)\\=\lim_{\Delta x \to 0}{x^2 +2x \Delta x+ \Delta x^2 -x^2\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(3)\\=\lim_{\Delta x \to 0}{2x\Delta x +\Delta x^2\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(4)\\=\lim_{\Delta x \to 0} {2x+\Delta x\over(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(5)\\={2x\over \sqrt{x^2-9+x^2-9}}
(6)\\={2x\over2 \sqrt{x^2 -9}}
(7)\\={x\over \sqrt {x^2 -9}} (8)$$
Now I substitute 5, and I don't get 1/4!!
What have I done wrong??
Thanks
| In step 6, the denominator should be $2 \sqrt{x^2-9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the integral $\int_0^\infty \frac{x (\ln(x))^2}{x^4 + x^2 + 1}\text{ d}x$ What is the value of $\displaystyle\int_0^\infty \frac{x (\ln(x))^2}{x^4 + x^2 + 1}\text{ d}x$?
This is a question I came up with myself. It is not homework.
I constructed this example to make the following technique work:
Integrate $\frac{z (\log(z))^3}{z^4 + z^2 + 1}$ along a "key-hole" contour. The argument can be made rigorous by splitting the contour into two parts, and using two different branch cuts for each part. Warning: This method is time-consuming and not for the faint-hearted
| \begin{align}
\int^\infty_0\frac{x\ln^2{x}}{x^4+x^2+1}dx
&=\frac{1}{8}\int^\infty_0\frac{\ln^2{x}}{x^2+x+1}dx\\
&=\frac{1}{4}\int^1_0\frac{(1-x)\ln^2{x}}{1-x^3}dx\\
&=\frac{1}{4}\sum^\infty_{n=0}\int^1_0\left(x^{3n}-x^{3n+1}\right)\ln^2{x}dx\\
&=\frac{1}{2}\sum^\infty_{n=0}\left(\frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3}\right)\\
&=-\frac{1}{2}\operatorname*{Res}_{z=-1/3}\frac{\pi\cot(\pi z)}{(3z+1)^3}\\
&=-\frac{1}{108}\left(2\pi^3\cot(\pi z)\csc^2(\pi z)\right)\Bigg{|}_{z=-1/3}\\
&=\frac{2\pi^3}{81\sqrt{3}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Writing a piecewise function for $f(x) = \mid x+3\mid -\mid x-1\mid $
I am wanting to write a piecewise function for the following:
$$f(x)= \mid x+3\mid -\mid x-1\mid $$
I know how to write piecewise functions for functions that have a single set of absolute value brackets, but I don't know how to deal with two sets of brackets. A hint would be appreciated.
| The idea is the following:
The absolute value of $y$ is given by
$$|y| = \begin{cases} y&\text{if } y\geq 0\\-y&\text{if } y<0\end{cases}$$
The point is now to check when $x+3$ and $x-1$ are positive, and when they are negative.
Let us look at the case where $x-1\geq 0$ first. Then $x\geq 1$, and it trivially follows that $x+3\geq 0$ as well. So for now we can write
\begin{align}f(x) = |x+3|-|x-1| &= \begin{cases} x+3-(x-1)&\text{if } x\geq 1\\\text{To be determined}&\text{otherwise }\end{cases}\\ &= \begin{cases} 4&\text{if } x\geq 1\\\text{To be determined}&\text{otherwise }\end{cases}.\end{align}
Next let's consider the case where $x+3<0$, i.e. $x<-3$ and it trivially follows that $x-1<0$ as well. We can now write
\begin{align}f(x) = |x+3|-|x-1| &= \begin{cases} 4&\text{if } x\geq 1\\-(x+3)-(-(x-1))&\text{if } x<-3\\\text{To be determined}&\text{otherwise}\end{cases}\\
&= \begin{cases} 4&\text{if } x\geq 1\\-x-3+x-1&\text{if } x<-3\\\text{To be determined}&\text{otherwise}\end{cases}\\
&=\begin{cases} 4&\text{if } x\geq 1\\-4&\text{if } x<-3\\\text{To be determined}&\text{otherwise}\end{cases}.\end{align}
Lastly, we should consider the case where $-3 \leq x<1$. Then $x+3\geq 0$ and $x-1<0$, so finally
\begin{align}f(x) = |x+3|-|x-1| &= \begin{cases} 4&\text{if } x\geq 1\\x+3-(-(x-1))&\text{if } -3\leq x < 1\\-4&\text{if }x<-3\end{cases}\\
&=\begin{cases} 4&\text{if } x\geq 1\\x+3+x-1&\text{if } -3\leq x < 1\\-4&\text{if }x<-3\end{cases}\\
&=\begin{cases} 4&\text{if } x\geq 1\\2x+2&\text{if } -3\leq x < 1\\-4&\text{if }x<-3\end{cases}.
\end{align}
And that is the final result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the sum of $\sum\limits_{n=1}^\infty \bigl[ \frac{(3 - \sqrt 5)^n}{2^nn^3}\bigr]$ Please help me to find the sum of
$\sum\limits_{n=1}^\infty \left[ \frac{\left(\frac{3 - \sqrt 5}{2}\right)^n}{n^3}\right]$
Is there any special technique to solve this one ?
| According to Maple,
$$ \text{polylog}\left(3, \dfrac{3-\sqrt{5}}{2}\right) = \dfrac{4}{5} \zeta(3) + \dfrac{\pi^2}{15} \ln \left(\dfrac{3-\sqrt{5}}{2}\right) - \dfrac{1}{12} \ln\left(\dfrac{3-\sqrt{5}}{2}\right)^3 $$
I don't know where it gets this rather remarkable identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Pythagorean triples So I am given that $65 = 1^2 + 8^2 = 7^2 + 4^2$ , how can I use this observation to find two Pythagorean triangles with hypotenuse of 65.
I know that I need to find integers $a$ and $b$ such that $a^2 + b^2 = 65^2$, but I don't understand how to derive them from that observation.
Here is my attempt.
$65^2 = (8^2+1^2)(7^2+4^2) = 8^27^2 + 1^24^2 + 1^27^2 + 8^24^2 = (8\cdot7)^2 + 4^2 + 7^2 + (8\cdot4)^2$ but now I am stuck here, any suggestions!
| Use the Brahmagupta-Fibonacci Identity
$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2.$$
This identity can be verified by multiplying out each side, and in nicer ways.
From the Identity, we get
$$65^2=(8^2+1^2)(7^2+4^2)=(8\cdot 7-1\cdot 4)^2 +(8\cdot 4+1\cdot 7)^2.$$
We can get another representation of $65^2$ as the sum of two squares by letting $c=4$ and $d=7$.
Remark: The Identity gives the useful result that the product of two numbers, each the sum of two squares, is itself the sum of two squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Example of Stirling Numbers of the First Kind I am trying to calculate the stirling numbers of the first kind. I am not very good in math and there is not a single example somewhere in the internet.
So I would really appreciate it if you could help me there.
The stirling number which I want to calculate is s1(4,2). I know the solution is 11.
How far I came:
| You could use the rising factorial to find $\left[\begin{array}{c} 4 \\ 2 \end{array}\right]$. It goes as follows:
$$(x)^4 = x(x+1)(x+2)(x+3) = x^4 + 6x^3 + 11x^2 + 6x$$
$\left[\begin{array}{c} 4 \\ k \end{array}\right]$ is now the coefficient with $x^k$. Thus $\left[\begin{array}{c} 4 \\ 2 \end{array}\right]$ = 11.
For the recurrence relation with the unsigned Stirling numbers of the first kind, you would have:
$$\begin{align}
\left[\begin{array}{c} 4 \\ 2 \end{array}\right] & = 3\left[\begin{array}{c} 3 \\ 2 \end{array}\right] + \left[\begin{array}{c} 3 \\ 1 \end{array}\right]\\
& = 3\left(2\left[\begin{array}{c} 2 \\ 2 \end{array}\right] + \left[\begin{array}{c} 2 \\ 1 \end{array}\right]\right) + 2\left[\begin{array}{c} 2 \\ 1 \end{array}\right] + \left[\begin{array}{c} 2 \\ 0 \end{array}\right] = 6\left[\begin{array}{c} 2 \\ 2 \end{array}\right] + 5\left[\begin{array}{c} 2 \\ 1 \end{array}\right] + 0 \\
& = 6\left(\left[\begin{array}{c} 1 \\ 2 \end{array}\right] + \left[\begin{array}{c} 1 \\ 1 \end{array}\right]\right) + 5\left(\left[\begin{array}{c} 1 \\ 1 \end{array}\right] + \left[\begin{array}{c} 1 \\ 0 \end{array}\right]\right) = 6\left[\begin{array}{c} 1 \\ 2 \end{array}\right] + 11\left[\begin{array}{c} 1 \\ 1 \end{array}\right] + 0 \\
& = 6\left(0 + \left[\begin{array}{c} 0 \\ 1 \end{array}\right]\right) + 11\left(0\left[\begin{array}{c} 0 \\ 1 \end{array}\right] + \left[\begin{array}{c} 0 \\ 0 \end{array}\right]\right) = 11
\end{align}$$
For the recurrence relation with the signed Stirling numbers of the first kind, you would get:
$$\begin{align}
\mathrm{s}(4,2) & = -3 \mathrm{s}(3,2) + \mathrm{s}(3,1) \\
& = -3[-2\mathrm{s}(2,2) + \mathrm{s}(2,1)] -2\mathrm{s}(2,1) + \mathrm{s}(2,0) = 6\mathrm{s}(2,2) -5\mathrm{s}(2,1) + 0 \\
& = 6[-\mathrm{s}(1,2) + \mathrm{s}(1,1)] -5[-\mathrm{s}(1,1) + \mathrm{s}(1,0)] = -6\mathrm{s}(1,2) + 11\mathrm{s}(1,1) - 0 \\
& = -6[0 + \mathrm{s}(0,1)] + 11[0 + \mathrm{s}(0,0)] = 11
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
System of congruences and Chinese remainder theorem Find all the integers satisfying this system of congruences
$$\begin{cases}
x \equiv 2 \pmod 5\\
x \equiv 1 \pmod {10}\\
x \equiv 0 \pmod 3
\end{cases}
$$
I think you use Chinese remainder theorem but I'm not sure how to.
| Basically the two equivalences
\begin{align}
x &\equiv a \pmod A \\
x &\equiv b \pmod B
\end{align}
have a common solution if and only if $$a \equiv b \pmod{\gcd(A,B)}$$
In the case of \begin{align}
x &\equiv 2 \pmod 5\\
x &\equiv 1 \pmod {10}\\
x &\equiv 0 \pmod 3
\end{align}
We notice that $1 \not \equiv 2 \pmod{\gcd(5,10)}$, so there is no common solution to all three equivalences.
Another method is to reduce each equivalence into prime-power congruences and then remove redundant equivalneces. If there are no contradictory congruences , what you are left with is amenable to the regular CRT.
for your problem, $x \equiv 2 \pmod 5$ and $x \equiv 0 \pmod 3$ are already prime-power congruences. Since $10 = 2 \times 5$ we can break $x \equiv 1 \pmod {10}$ into the two prime-power congruences
\begin{align}
x &\equiv 1 \pmod {2} \\
x &\equiv 1 \pmod {5}
\end{align}
and we notice that $x \equiv 1 \pmod {5}$ and $x \equiv 2 \pmod {5}$ contradict each other. So, again, we know that the system of congruences has no common solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that if $a$ is an integer, then 3 divides $a^3 - a $ Show that if $a$ is an integer, then 3 divides $a^3 - a $
we can write, where $k$ is an integer;
$a^3 - a = 3k $
$a(a^2 - 1) = 3k $
Now if $a = k$ then
$a^2 -1 = 3$ and $a= \pm2 $ so $ a^3 - a = 24 = 3 \times 8$
If $ a $ is not equal to $k$;
then
$a(a^2 - 1) = a(a+1)(a-1) = 3k$
since $a(a+1)(a-1)$ is the product of 3 consecutive integers, the expression is divisible by 3.
Is this ok?, just a check. I'm not really all that good at number theory.
| Dividing $a$ by $3$ allows to write $a=3q+r$. Now
$$
(3q+r)^3=27q^3+9q^2r+3qr^3+r^3=3s+r^3
$$
collecting together all multiples of $3$ in the expansion. Dividing again the result by $3$ we have
$$
(3q+r)^3=3Q+R
$$
where $R$ is the remainder of the division by 3 of $r^3$ (because $3s$ is a multiple of $3$).
But there are only 3 possible remainders $r$, namely:
*
*$r=0$, thus $r^3=0$ and $R=0$,
*$r=1$, thus $r^3=1$ and $R=1$,
*$r=2$, thus $r^3=8$ and $R=2$.
This shows that $r=R$ always, so that $a^3-a$ divided by 3 has remainder $0$ (i.e. is divisible by $3$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Calculate $ S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}. $ Calculate $S =\displaystyle\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}$. This sequence is neither arithmetic nor geometric. How can you solve this. Thanks!
| $$u_k = \dfrac{1}{k(k+1)(k+2)} = \dfrac{1}{2k(k+1)} - \dfrac{1}{2(k+1)(k+2)} =
f_k - f_{k+1}$$
so $$u_1 + u_2 + \cdots u_n = (f_1 - f_2) + (f_2 - f_3) + \cdots + (f_n - f_{n+1}) = f_1 - f_{n+1} =
\dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Check: Find and plot all of the $6^{th}$ roots of unity and the $6^{th}$ roots of $7-3i$ . Find all of the $6^{th}$ roots of unity and the $6^{th}$ roots of $7-3i$ .
For each natural number $n$ there are exactly $n$ $n$-th roots of unity, which can be expressed as: $$z_k=e^{i\frac{2\pi}{n}k},k=0,1,\cdot, n-1$$ In our case $n=6$, hence, $$z_k=e^{i\frac{2\pi}{6}k}=e^{i\frac{\pi}{3}k},k=0,1,2,3,4,5$$
Thus, the $6$-th roots of $1$ are: $$z_0=e^{0i}=1, z_1=e^{i\frac{\pi}{3}}, z_2=e^{i\frac{2\pi}{3}}, z_3=e^{i}=-1, z_4=e^{i\frac{4\pi}{3}}, z_5=e^{i\frac{5\pi}{3}}$$
So the $6^{th}$ roots of $7-3i$ are: $$
z_0=58^\frac{1}{12}e^{i\frac{\pi}{6}}=1,
z_1=58^\frac{1}{12}e^{i\frac{\pi}{2}},
z_2=58^\frac{1}{12}e^{i\frac{5\pi}{6}},
z_3=58^\frac{1}{12}e^{i\frac{7\pi}{6}},
z_4=58^\frac{1}{12}e^{i\frac{3\pi}{2}},
z_5=58^\frac{1}{12}e^{i\frac{11\pi}{6}}$$
| First, start with finding the polar notation of $7 - 3i$ which is :
$$7-3i = \sqrt{58}e^{-i\tan^{-1}{(\frac{3}{7})}}$$
Then go even further to get only exponential :
$$7-3i = e^{\frac{1}{2}\ln{(58)}}e^{-i\tan^{-1}{(\frac{3}{7})}}$$
Now, it's pretty simple : the $n$th roots of $7-3i$ are :
$$e^{\frac{1}{2\cdot 6}\ln{(58)}}e^{-i\cdot \frac{1}{6}(\tan^{-1}{(\frac{3}{7})}+2k\pi)} \text{ for k in 0, ..., 6}$$
$$= \sqrt[\leftroot{-2}\uproot{2}12]{58}e^{-i\cdot \frac{1}{6}(\tan^{-1}{(\frac{3}{7})}+2k\pi)} \text{ for k in 0, ..., 6}.$$
Generally speaking, for any complex $a$, the $n$th roots of unity are :
$$\sqrt[\leftroot{-2}\uproot{2}n]{|a|}e^{i\frac{\arg(a)+2k\pi}{n}}\text{ for k in 0, ..., 6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\sqrt{2}=\prod_{n=1}^{\infty }\left(1+\frac{0.75}{4n^2-1}\right)$ Proving $$\sqrt{2}=\prod_{n=1}^{\infty }\left(1+\frac{0.75}{4n^2-1}\right)$$
By using the numerical calculation I saw that the convergence of product series is slow, so I need the proving.
thanks.
| Use the infinite product representation: $\displaystyle \dfrac{\sin \pi x}{\pi x} = \prod\limits_{n=1}^{\infty} \left(1-\frac{x^2}{n^2}\right)$ to rewrite the product $$\displaystyle \prod_{n=1}^{\infty }\left(1+\frac{3/4}{4n^2-1}\right) = \prod_{n=1}^{\infty }\frac{\left(1-\frac{1}{16n^2}\right)}{\left(1-\frac{1}{4n^2}\right)} = \frac{2\sin \frac{\pi}{4}}{\sin \frac{\pi}{2}} = \sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Help me find the following limit : $\lim_{{n}\to{\infty}} (\frac{2^x+3^x+\cdots+n^x}{n-1})^\frac{1}{x} = ?$ I have no idea where to start.$$\begin{align}\lim_{{n}\to{\infty}} \left(\dfrac{2^x+3^x+\cdots+n^x}{n-1}\right)^{1/x} = ?, n >1\\\end{align}$$
| Another approach:
$$\left(\frac{n-1}{2^x+3^x+\ldots+n^x}\right)^{1/x}\le\left(\frac{n-1}{(n-1)2^x}\right)^{1/x}=\frac12\implies \sum_{n=1}^\infty\left(\frac{n-1}{2^x+3^x+\ldots+n^x}\right)^{1/x}$$
converges, and then
$$\left(\frac{n-1}{2^x+3^x+\ldots+n^x}\right)^{1/x}\xrightarrow[n\to\infty]{}0\implies\left(\frac{2^x+3^x+\ldots+n^x}{n-1}\right)^{1/x}\xrightarrow[n\to\infty]{}\infty$$
since everything positive here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Volume using triple Integrals, cylindrical coordinates I want to calculate the volume of a solid with $z+1\ge x^2+y^2$ and $3\left(z-1\right)\le -\left(x^2+y^2\right)$. After cylindrical coordinates
x=rcosϕ, y=rsinϕ
I got $r^2-1\le z$, and $z\le \frac{\left(3-r^2\right)}{3}$ , so $r^2-1\le z\le \frac{\left(3-r^2\right)}{3}$.
Then $r^2-1\le \frac{\left(3-r^2\right)}{3}$ ==> $0\le r\le 1$, $0\le \phi \le 2\pi $. After calculating the integral I got $\frac{4\pi }{3}$, which doesn't check the answer considered correct by the one who made the exercise, which is $\frac{3\pi }{2}$. I don't understand what I'm doing wrong. Or could that answer be wrong? Thanks.
| Since $r^2-1=1-\frac{r^2}{3}\implies r^2=\frac{3}{2}\implies r=\sqrt{\frac{3}{2}}$,
$V=\displaystyle\int_0^{2\pi}\int_0^{\sqrt{\frac{3}{2}}}\int_{r^2-1}^{1-\frac{r^2}{3}} r\;dz dr d\theta=\int_0^{2\pi}\int_0^{\sqrt{\frac{3}{2}}}\left(2r-\frac{4}{3}r^3\right)\;dr d\theta$
$\displaystyle\;\;=\int_0^{2\pi}\left[r^2-\frac{1}{3}r^4\right]_0^{\sqrt{\frac{3}{2}}}=2\pi\left(\frac{3}{2}-\frac{3}{4}\right)=\frac{3\pi}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that if $xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}}$, then $x+y+z\geq \sqrt{3}$ If $x, y, z$ are positive real numbers such that
$$xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}},$$
then prove that $x+y+z\geq \sqrt{3}$.
|
Sketch indicates the truth of the proposition. Blue boxes : $x+y+z > \sqrt{3}$.
Blue with purple centers $$xy + yz + zx > \frac{1}{\sqrt{x^2+y^2+z^2}}$$
The surfaces indicate where equality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
The formula of the order of multistep methods How can I derive this $$(1+\xi) \left(1+\frac{1}{2}\xi-\frac{1}{12}\xi^{2}\right)+O(\xi^3)$$ from
$$\frac{1+\xi}{1-\frac{1}{2}\xi+\frac{1}{3}\xi^{2}}+O(\xi^3)$$
?
The whole formula is below. This is from "A first course in the numerical analysis of differential equations by Arieh Iserles"
$$\frac{\rho(w)}{\ln w}=\frac{\xi+\xi^{2}}{\xi-\frac{1}{2}\xi^2+\frac{1}{3}\xi^{3}+O(\xi^4)}=
\frac{1+\xi}{1-\frac{1}{2}\xi+\frac{1}{3}\xi^{2}}+O(\xi^3)
$$
$$=(1+\xi) \left(1+\frac{1}{2}\xi-\frac{1}{12}\xi^{2}\right)+O(\xi^3)=1+\frac{3}{2}\xi+\frac{5}{12}\xi^{2}+(\xi^3)$$
| Use the series $\frac1{1-x}=1+x+x^2+O(x^3)$ to get
$$
\begin{align}
\frac1{1-\frac12\xi+\frac13\xi^2}
&=1+\left(\frac12\xi-\frac13\xi^2\right)+\left(\frac12\xi-\frac13\xi^2\right)^2+O(\xi^3)\\
&=1+\frac12\xi-\frac13\xi^2+\frac14\xi^2+O(\xi^3)\\
&=1+\frac12\xi-\frac1{12}\xi^2+O(\xi^3)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Taylor series expansion of the function $f(x)=x \arctan x-0.5 \log(1+x^2)$ about the origin int the region {$|x|\le1$} Find the Taylor series expansion of the function
$\color {green}{f(x)=x \tan^{-1} x-0.5 \log(1+x^2)}$ about the origin int the region {$|x|\le1$}
My effort: I know $\displaystyle \log (1+x)=x-x^2/2+x^3/3-x^4/4...$ now replace $\displaystyle x$ by $\displaystyle x^2$ then we get
$ \displaystyle \frac{1}{2} \log (1+x^2)=x^2/2-x^4/4+x^6/6-x^8/8+....$ (1)
and
$\displaystyle x \arctan x=x\int (1+x^2)^{-1/2}dx= x^2-x^4/6+3x^5/32 -....$ (2),
by (2)-(1) I have the answer which does not match with the answer of the book which is$$ \sum_{0}^{\infty}\frac{(-1)^{n-1}x^{2n}}{2n(2n-1)}.$$
My question is my process wrong? Or there is a miscalculation in my process? Is there any better approach to solve this kind of problem? Please give a answer.
| it is easier to find the maclaurin series for
$$f^\prime(x) = \tan^{-1}(x) = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} + \cdots$$ and integrate this series alsin rthe fact that $f(0) = 0$ which gives you
$$f(x) = x\tan^{-1}x - 0.5 \ln{(1+x^2)} = \frac{x^2}{2} - \dfrac{x^4}{3.4} + \dfrac{x^6}{5.6} + \cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Three mutually-tangent circles have centers at given distances from each other; find each radius, and find the area between the circles Three circles of different radii are tangent to each other externally. The distance between their centers are $9\ cm$, $8\ cm$, and $11\ cm$.
*
*Find the radius of each circle.
*Find the area in between the three circles.
The distance between the centers of two tangent circles should be the sum of their radii. If we call the radii $r_1,r_2,r_3$ then we get a system of equations $$\begin{align}r_1+r_2&=9\\r_2+r_3&=8\\r_3+r_1&=11\end{align}$$ From this we can find $r_1,r_2,r_3$.
Let us call $O_1,O_2,O_3$ the centers of the circles. For the area between the centers we could compute it as the difference of the area of the triangle $O_1O_2O_3$ minus the three circular sectors that the circles cover. We have the area of the triangle because we have its sides $9,8,11$ and we could use Heron's formula with them. For the circular sectors we know their radii but we also need their amplitude. Ah, but the amplitude is the same as the angles of the triangle. I guess we can compute the angles, because we have the sides of the triangle and we can use The law of cosines to solve for the angles.
Is this correct?
Would I have to use the $\arccos(x)$ function to get the final answer or there is a simpler way?
| Solving for the radii, we get $\{3,5,6\}$.
For the area we get
$$
\begin{align}
&\sqrt{14(14-8)(14-9)(14-11)}\\[9pt]
&-\frac{(8+9-11)^2}8\arccos\left(\frac{8^2+9^2-11^2}{2\cdot8\cdot9}\right)\\
&-\frac{(11+8-9)^2}8\arccos\left(\frac{11^2+8^2-9^2}{2\cdot11\cdot8}\right)\\
&-\frac{(9+11-8)^2}8\arccos\left(\frac{9^2+11^2-8^2}{2\cdot9\cdot11}\right)
\end{align}
$$
which is $3.05537320587455$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$ Let a and b be natural numbers so that $2a-1,2b-1$ and $a+b$ are prime numbers. Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$.
I get that $gcd(a,b)=1$.
I haven't got anything special for now but if I do I will update the question.
| Case $a+b\mid a^a+b^b$
Suppose $a$ is odd. Because $a+b\mid a^a+b^a$ we have $a+b\mid b^b-b^a$, hence $a+b\mid b^{|b-a|}-1$ because $\gcd(a,b)=1$, hence $a-b\mid a+b-1$ by Fermat. By symmetry, we get the same if $b$ is odd.
Case $a+b\mid a^b+b^a$
Suppose $a$ is odd. Because $a+b\mid a^a+b^a$ we have $a+b\mid a^b-a^a$. As before we get $a-b\mid a+b-1$. The same if $b$ is odd.
Either way, we have $a-b\mid a+b-1$ from which $a-b\mid a+b-1+(a-b)=2b-1$ and $a-b\mid 2a-1$, hence $a-b=\pm\,1$, that is, $2a-1$ and $2b-1$ are twin primes.
Wlog suppose $b=a+1$. Let $q=a+b=2b-1$. Modulo $q$, $$a^a+b^b\equiv\left(\frac{-1}2\right)^{\frac{q-1}2}+\left(\frac{1}2\right)^{\frac{q+1}2}\equiv\pm\,1\pm\,\frac12\not\equiv0\pmod q$$ and $$a^b+b^a\equiv\left(\frac{-1}2\right)^{\frac{q+1}2}+\left(\frac{1}2\right)^{\frac{q-1}2}\equiv\pm\,\frac12\pm\,1\not\equiv0\pmod q$$ since $q>3$, a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Integrating: $ \;\int \frac{1}{x^2+3x+2} dx $ How can I solve the following integral:
$$ \int \frac{1}{x^2+3x+2} dx $$
Should I proceed by changing the variable (substitution)? or should I use integration by parts? Or another method altogether?
Thank you!
| Just for fun, I tried it another way.
Complete the square
$$
\frac{1}{x^2+3x+2} = \frac{1}{(x+\frac{3}{2})^2-\frac{1}{4}}
$$
Change variables $y=x+\frac{3}{2}$
$$
\int\frac{dx}{x^2+3x+2} = \int\frac{4}{4y^2-1} = {-2\;\mathrm{atanh}(2y)+C}
$$
Substitute back
$$
\int\frac{dx}{x^2+3x+2}=-2\;\mathrm{atanh}(2x+3) + C
$$
Check by differentiating
$$
\frac{d}{dx}\;\big(-2\;\mathrm{atanh}(2x+3)\big)
=\frac{-4}{1-(2x+3)^2} = \frac{1}{x^2+3x+2}
$$
OK
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the sum $x + \frac{x^3}{3} + \frac{x^5}{5} + ... $
Evaluate the sum $$x + \frac{x^3}{3} + \frac{x^5}{5} + ... $$
I was able to notice that: $$ \sum_{n=0}^\infty \frac{x^{2n-1}}{2n-1} = \sum_{n=0}^\infty \int x^{2n-2}dx = \lim_{N\to\infty} \sum_{n=0}^N \int x^{2n-2} dx $$
Where should I take it from here? (assuming I'm not the right way)
EDIT
Following the anwer:
$$\int \sum_{n=0}^\infty x^{2n-1} dx = \int \frac{\sum_{n=0}^\infty (x^2)^n}{x} dx = \int \frac{\frac{1}{1-x^2}}{x} dx= \int \frac{1}{x(1-x^2)}dx = \int \frac{1-x^2+x^2}{x(1-x^2)} = \int \frac{1}{x} dx + \int \frac{x}{1-x^2}dx = \ln(x) + \int \frac{1}{2}\frac{1}{1-t}dt + C = \ln(x) - \frac{\ln(x^2)}{2} + C$$
*
*Is that right?
*How to evaluate $C$?
| Use only your first identity. You should check where the series is absolutelly convergent. In that interval you can change the integral with the sum. That will lead to a known series you shall be able to evaluate. (If you can't solve with the hint I edit this answer later).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Ring with many one-sided zero-divisors Does there exist a ring all of whose elements are left zero-divisors but only one element is a right zero-divisor?
The motivation for asking this question is that if there exists atleast one left zero-divisor there exist atleast one-right zero divisor. Now, it seems a natural question to aks if assuming "many" left-zero divisors will result in "more" right-zero divisors. A specific and extremal version if this idea is the question formulated above.
| It seems the following.
Such a non-trivial ring is unique. Indeed, assume that $R$ is a ring all of whose elements are left zero-divisors but only one element $r\ne 0$ is a right zero-divisor and $R$ contains at least three distinct elements. Let $x\in R$ be an arbitrary non-zero element. Since $x$ is a left zero-divisor, there exists a non-zero element $t\in R$ such that $xt=0$. So, $t$ is a right zero-divisor, so $t=r$. Therefore $xr=0$ for each element $x\in R$. In particular, $rr=0$. Since $x(r+r)=xr+xr=0$ for each element $x\in R$, $r+r$ is a right zero-divisor. Since $r\ne 0$, $r+r\ne r$, so $r+r=0$. Again, let $x\in R$ be an arbitrary non-zero element. Then $rx$ is a right zero-divisor, so $rx=0$ or $rx=r$. But if $rx=0$ then $x$ is a right zero-divisor, therefore $x=r$. So $rx=r$ for each element $x\in R$ distinct from $0$ or $r$. Let $x,y$ be elements of the ring $R$ distinct from $0$ or $r$. Then $r(x+y)=rx+ry=r+r=0$. Therefore $x+y=0$ or $x+y=r$. In particular, $x+x=0$ or $x+x=r$. If $x+x=r$ then $0=xr=x(x+x)=$ $xx+xx=(x+x)x=rx=r$, a contradiction. Therefore $x+x=0$. Hence $y=-x=x$ or $y=r-x=r+x$. Thus the ring $R$ contains exactly four elements: $0$, $r$, $x$ and $x+r$. Since $x$ is not a right zero-divisor, $xx\ne 0$. If $xx=r$ then $0=xr=xxx=rx=r$, a contradiction. If $xx=x+r$ then $xx+r=xx+rx=$ $(x+r)x=xxx=x(x+r)=$ $xx+xr=xx,$ a contradiction. So $xx=x$. Then $(x+r)x=xx+rx=xx+r=x+r$, $x(x+r)=xx+xr=xx=x$, and $(x+r)(x+r)=xx+xr+rx+rr=xx+r=x+r$.
The table of addition $a+b$ for the ring $R$:
$$\begin{array}{c}
a\setminus b & 0 & r & x & x+r\\
0 & 0 & r & x & x+r\\
r & r & 0 & x+r & x\\
x & x & x+r & 0 & r\\
x+r & x+r & x & r & 0
\end{array}$$
The table of multiplication $ab$ for the ring $R$:
\begin{array}{c}
a\setminus b & 0 & r & x & x+r\\
0 & 0 & 0 & 0 & 0\\
r & 0 & 0 & 0 & 0\\
x & 0 & 0 & x & x\\
x+r & 0 & 0 & x+r & x+r
\end{array}
Thus the ring $R$ can be realized as the ring of $2\times 2$ matrices of the form
$\left(
\begin{array}{}
a & 0\\
b & 0
\end{array}
\right)$ over the ring $\Bbb Z_2$, where $0=\left(\begin{array}{}
0 & 0\\
0 & 0
\end{array}\right)$,
$r=\left(\begin{array}{}
0 & 0\\
1 & 0 \end{array}\right)$,
$x=\left(\begin{array}{}
1 & 0\\
0 & 0 \end{array}\right)$, and
$x+r=\left(\begin{array}{}
1 & 0\\
1 & 0 \end{array}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Triangle in parabola I have a problem. In my triangle one vertex is in the vertex of the parabola and two others are in parabola. This is a isosceles triangle and I know one angle in this triangle : 120 grades. The question is: angle 120 grades must be beside of the vertex of parabola?
Thanks a lot for answer, I am beginner in math.
Which option (1) or (2) is possible?
| choose the coordinates so that the equation of the parabola is $y = x^2.$ let the triangle be $OAB$ with $O =(0,0), A = (a, a^2), B= (b, b^2)$ and $\angle AOB = 120^\circ$ by cosine rule, $$AB^2 = OA^2 + OB^2 + OA*OB $$ so that
$$(a-b)^2 + (a^2 - b^2)^2 = a^2 + a^4 + b^2 + b^4 + \sqrt{(a^2 + a^4)(b^2+b^4)} $$
cleaning it up a bit we end up with $$3a^2b^2 + 8ab - a^2 - b^2 + 3 = 0$$
which simplifies to $$-2ab(1+ab) = \sqrt{(a^2 + a^4)(b^2+b^4)} \tag 1$$ from $(1)$ we can see that $a$ and $b$ cannot be of the same sign. using symmetry we can assume that $a > 0, b < 0.$
a special solution is $-b= a = \sqrt{1+\sqrt 2}$ to find the general solution we will square $(1)$ and get a quadratic equation for $b$ $$b^2(3a^2 - 1) + 8ab + (3-a^2) = 0 $$ and the solution is $$ b = \dfrac{-2a \pm \sqrt 3 |a^2 -1|}{3a^2 - 1} \text{ and } b < 0.$$
if you take the negative solution you find that a negative solution exist for $0 \le < \sqrt 3$ and for the positive solution you have $\dfrac{1}{\sqrt 3} < a.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
I need my work checked(properties of operations) Let $x * y = |x + y|.$
$x * y = |x + y| = |y + x| = y * x,$ so $*$ is commutative.
$(x * y) * z = ||x + y| + z| = |x + |x + z|| = x * (y * z),$ so $*$ associative.
$x * e = |x + e| = x,$ so $e = 0.$ Further, $e * x = |0 + x| = x.$ So, $*$ has an identity.
$x * x' = |x + x'| = e,$ so $x' = e - x.$ Further, $x' * x = |e - x + x| = e.$ So, every element has an inverse.
Let $x * y = |xy|.$
$x * y = |xy| = |yx| = y* x,$ so $*$ commutative.
$(x * y) * z = |(xy)z| = |x(yz)| = x * (y * z),$ so $*$ associative.
$x * e = |xe| = x,$ so, $e = 1$. Further, $e * x = |1x| = x$. So, $*$ has no identity.
$x * x' = |xx'| = e, $ so $x' = \frac ex$. Further, $x' * x = |\frac ex x| = e$. So, every element has an inverse.
Let $x * y = \sqrt{x^2 + y^2}.$
$x * y = \sqrt{x^2 + y^2} = \sqrt{y^2 + x^2} = y*x$, so $*$ is commutative.
$(x * y) * z = \sqrt{(\sqrt{x^2 + y^2})^2 + z^2} = \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2 + (\sqrt{y^2 + z^2})^2} = x * (y * z),$ so $*$ is associative.
$x * e = \sqrt{x^2 + e^2} = x,$ so $e = 0.$ Further, $e * x = \sqrt{0 + x^2} = x.$ So, $*$ has an identity.
$x * x' = \sqrt{x^2 + x'^2} = e,$ so $x' = \sqrt{e^2 - x^2} \text { and } x' = -\sqrt{e^2 - x^2}$. Further, $x' * x = \sqrt{(\sqrt{e^2 - x^2})^2 + x^2} = e.$ So, every element has an inverse.
Let $x * y = x - y.$
$x * y = x - y \neq y - x,$ so $*$ is not commutative.
$(x * y) * z = ((x - y) - z) = (x - (y - z),$ so $*$ is not associative.
$x * e = x - e = x,$ so $e = 0$. Further, $e * x = 0 - x = -x.$ So, $*$ has no identity.
$x * x' = x - x' = e,$ so $x' = x - e$. Further, $x' * x = x - e - x = -e.$ So, not every element has an inverse.
Let $x * y = xy + 1.$
$(x * y) * z = xy+ 1 = 1 + xy,$ so $*$ is commutative.
$(x * y) * z = (xyz + z) + 1 = xyz + (z + 1),$ so $*$ is associative.
$x * e = xe + 1 = x, $ so $e = 0$. Further, $e * x = 0x + 1 = 1.$ So, $*$ has no identity.
$x * x' = xx' + 1 = e,$ so $x' = \frac {e - 1}{x}.$ Further, $x' * x = \frac {e - 1}{x} x + 1 = e$, so every element has an inverse.
Let $x * y = max\{x, y\}.$
$(x * y) * z = max\{max\{x, y\}, z\} = max\{max \text { } x, \{y , z\}\},$ so $*$ is associative.
$x * e = max\{x, e\} = x,$ so $e = min\{x, e\}.$ Futher, $e * x = max\{min\{x, e\}, x\} = x,$ so $*$ has an identity.
$x * x' = max\{x, x'\} \neq e,$ so $x'$ is undefined. Not every element has an inverse.
| For number 5:
it is not associate.
you did not check:
$(x∗y)∗z=x∗(y∗z)$
Also, you said "$xe+1=x$, so $e=0$" well if $e=0$ then we get $1=x$ but we need $x=x$ either way, you are right there is no identity but you did not provide the right reason.
This is important: If there is no identity then there is no inverse. So you contradicted yourself in the inverse part.
Hopefully this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to show that $\sqrt[3]{-1+\sqrt{-7}}+\sqrt[3]{-1-\sqrt{-7}}$ is a real number at a time before the invention of complex numbers I have read this PDF from ocw.mit.edu about complex numbers. There is one interesting question: Imagine yourself at the time, when complex numbers had to be invented yet. How to show that $$\sqrt[3]{-1+\sqrt{-7}}+\sqrt[3]{-1-\sqrt{-7}}$$ is a real number.
| I think it's a stretch to claim that this result can be proven without complex numbers, especially because of the presence of $\sqrt{-7}$. Anyway, here's a way to look at it. Let $A = \sqrt[3]{-1 + \sqrt{-7}}$ and $B = \sqrt[3]{-1 - \sqrt{-7}}$. If $t = A + B$, then $$t^3 = A^3 + B^3 + 3AB(A + B) = (-1 + \sqrt{-7}) + (-1 - \sqrt{-7}) + 3t\sqrt[3]{1 - (-7)} = -2 + 6t$$ So $t$ is a root of the cubic $x^3 - 6x + 2$. Let $f(x) = x^3 - 6x + 2$. Then $f(-3) = -7 < 0$, $f(0) = 2 > 0$, $f(1) = -3 < 0$, and $f(3) = 11 > 0$. Hence, by the intermediate value theorem, $f$ has three real roots. Therefore, $t$ is real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Does the limit of this double sequence exist? Consider
$$a_{mn}=\frac{m^2n^2}{m^2+n^2}\left(1-\cos\left(\frac{1}{m}\right)\cos\left(\frac{1}{n}\right)\right)$$
Does $\lim_{m,n\to\infty}a_{mn}$ exist?
It can be seen that $$\lim_{m\to\infty}\left(\lim_{n\to\infty}a_{mn}\right)=\lim_{n\to\infty}\left(\lim_{m\to\infty}a_{mn}\right)=\frac{1}{2}$$
However, I still cannot determine whether the limit exists or not. Any one can help? Thanks!
| We have
\begin{align}a_{mn} &= \frac{m^2n^2}{m^2 + n^2}\left(1 - \left(1 + \frac{1}{2m^2} + O\left(\frac{1}{m^4}\right)\right)\left(1 + \frac{1}{2n^2} + O\left(\frac{1}{n^4}\right)\right)\right)\\
&=\frac{m^2n^2}{m^2 + n^2}\left(1 - \left(1 - \frac{1}{2}\left(\frac{1}{m^2} + \frac{1}{2n^2}\right) + O\left(\frac{1}{m^4n^4}\right)\right)\right)\\
&= \frac{m^2n^2}{m^2 + n^2}\left(\frac{1}{2}\frac{m^2 + n^2}{m^2 n^2} + O\left(\frac{1}{m^4n^4}\right)\right)\\
&= \frac{1}{2} + O\left(\frac{1}{m^2n^2(m^2 + n^2)}\right)\\
&= \frac{1}{2} + O\left(\frac{1}{m^3n^3}\right),\quad \text{as $m,n \to \infty$}
\end{align}
So there is a positive integer $k$ and a constant $C > 0$ such that $|a_{mn} - \frac{1}{2}| < \frac{C}{m^3n^3}$ for all $m, n \ge k$. Given $\varepsilon > 0$, choose $N > \max\{N, (\frac{C}{\varepsilon})^{1/6}\}$. Then $|a_{mn} - \frac{1}{2}| < \varepsilon$ for all $m, n \ge N$. Therefore, $\lim_{m,n\to \infty} a_{mn}$ exists and equals $1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to find $\int_0^{1/4}\frac{1}{x\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)dx$ Let $H_n$ be the harmonic series. I want to find the value of $A=\displaystyle\sum_{n=0}^\infty \binom{2n}{n}\left(\frac{1}{4}\right)^n\frac{H_n}{n} $.
From this paper : https://cs.uwaterloo.ca/journals/JIS/VOL15/Boyadzhiev/boyadzhiev6.pdf
, I found that $$\sum_{n=0}^\infty \binom{2n}{n}H_n x^n=\frac{1}{\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)=f(x)$$
So I get $A=\displaystyle\int_0^{1/4}\frac{f(x)}{x}dx$.
How to find the exact value of
$$\int_0^{1/4}\frac{1}{x\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)dx$$.
Thank in advances.
| Hint :
by $$u=\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}$$
$$\Rightarrow \sqrt{1-4x}=\frac{1}{2u-1}$$
$$\Rightarrow x=\frac{1}{4}(1-\frac{1}{(2u-1)^2})$$
$$dx=\frac{1}{(2u-1)^3}du$$
you get $$I=\int_1^{\infty}\frac{\ln u}{u(u-1)}du=-\sum_{n=0}^{\infty}\int_1^{\infty}u^{n-1}\ln u du=-\sum_{n=0}^{\infty}-\frac{1}{n^2}=\frac{\pi^2}{6}$$
for the last integral use integration by part
notice that $$\frac{1}{u-1}=-\sum_{n=0}^{\infty}u^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the smallest fraction produced by a sum of fractions with bounded denominator? For $x$ a sum of fractions:
$$
x = \sum_{i=1}^{N}\frac{a_i}{b_i}
$$
for all $a_i, b_i \in \mathbb{Z}$ with $ 0 < b_i \leq D$ and $N$ are non-zero positive integers, I know that the denominator of $x$ will not exceed the least common multiple of all positive integers less than or equal to $D$. See OEIS A003418.
So the denominator of $x$ will not grow without bound. My question is, what is the smallest fraction (closest to $0$) that such a sum can produce? Please provide a worked example assuming that $D = 12$. $N$ can be any integer greater than $1$.
Edit
I'd also like to know how to find sequences that produce the smallest value for a given $D$ and how many such sequences there are with minimal $N$ and fractions in lowest terms.
| The LCM of $1, \ldots, 12$ is $27720$. One example with sum $1/27720$ (there are infinitely many others) is
$$ \dfrac{3}{7} + \dfrac{1}{8} + \dfrac{ 2}{9} + \dfrac{3}{10} + \dfrac{1}{11} -\dfrac{14}{12}$$
Hint: if $L(d)$ is the LCM of $1, \ldots, d$, then
$$ \dfrac{1}{L(d)} = \dfrac{\gcd(d,L(d-1))}{d\; L(d-1)}$$
Now use Bezout's identity.
EDIT: There are two sets of $5$ integers $\le 12$ whose LCM is $27720$:
$\{5,7,8,9,11\}$ and $\{7,8,9,10,11\}$. Let's try using five fractions with denominators $5,7,8,9,11$.
From $3 \times 7 - 4\times 5 = 1$ we have $$\dfrac{3}{5} + \dfrac{-4}{7} = \dfrac{1}{35}$$
From $-13 \times 8 + 3 \times 35 = 1$ and this we have
$$ \dfrac{1}{280} = \dfrac{-13}{35} + \dfrac{3}{8} = \dfrac{-39}{5} + \dfrac{52}{7} + \dfrac{3}{8}$$
From $-31 \times 9 + 1 \times 280 = 1$ and this we have
$$ \dfrac{1}{2520} = \dfrac{-31}{280} + \dfrac{1}{9} = \dfrac{1209}{5} + \dfrac{-1612}{7} + \dfrac{ -93}{8} + \dfrac{1}{9}$$
Finally, from $-229 \times 11 + 1 \times 2520 = 1$ and this we have
$$ \dfrac{1}{27720} = \dfrac{-276861}{5} + \dfrac{369148}{7} + \dfrac{21297}{8} + \dfrac{-229}{9} + \dfrac{1}{11}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find $z$ that satisfies $z^2=3+4i$ Super basic question but some reason either I'm not doing this right or something is wrong.
The best route usually with these questions is to transform $3+4i$ to $re^{it}$ representation.
Ok, so $r^2=3^2+4^2 = 25$, so $r=5$. And $\frac{4}{3}=\tan(t)$ so that means $t \approx 0.3217$ and I'm not going to get an exact answer like that.
Another method would be to solve quadratic formula $z^2-3-4i=0$ that means $z_0=\frac{\sqrt{12+16i}}{2}$ and $z_1=\frac{-\sqrt{12+16i}}{2}$
But now I have the same problem, $12+16i$ doesn't have a "pretty" polar representation so its difficult to find $\sqrt{12+16i}$
I want to find an exact solution, not approximate, and it should be easy since the answers are $2+i$ and $-2-i$
Edit:
Also, something else is weird here. I know that if $z_0$ is some root of a polynomial then it's conjugate is also a root.but $2+i$ and $-2-i$ are not conjugates.
| We can use the double angle formula to find the components of $z$:
$$ \cos \theta = \pm \sqrt{\frac{1 + \cos 2\theta}{2}}$$
$$ \sin \theta = \pm \sqrt{\frac{1 - \cos 2\theta}{2}}$$
Here $\theta$ is the argument of $z$ and $2\theta$ the argument of $z^2$.
We know $ \cos 2\theta = \frac{3}{5}$ so $\cos \theta = \pm \sqrt{\frac{4}{5}}$ and $\sin \theta = \pm\sqrt{\frac{1}{5}}$. Since $z^2$ lies on the first quadrant (between $0$ and $\pi/2$), its two square roots must lie on the first and third quadrants, respectively, therefore the final answer is
$$ z_1 = \sqrt{5} \left( \sqrt{\frac{4}{5}} + i\sqrt{\frac{1}{5}} \right) = 2 + i $$
$$ z_2 = -z_1 = -2 - i $$
Alternatively: Since the numbers are relatively simple, we can let $z = x + yi$. Then
$$ z^2 = (x^2 - y^2) + 2xyi = 3 + 4i $$
Now equate the real and imaginary part to get this system of equations:
$$ x^2 - y^2 = 3 $$
$$ 2xy = 4 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Simple algebra involving trigonometry, but confusing How do I get from
$$\frac{\sqrt3}{2} + \frac12 \tan x = 2 \tan x \cdot \frac{\sqrt3}{2}$$
to
$$\frac{\sqrt3}{2 \sqrt3 - 1} = \tan x$$
and then to
$$11 \tan x = 6 + \sqrt3$$
| $$\frac{\sqrt{3}}{2\sqrt{3}-1}=\tan(x)$$
$$\frac{\sqrt{3}}{2\sqrt{3}-1}*\frac{2\sqrt{3}+1}{2\sqrt{3}+1}=\tan(x)$$
$$\frac{6+\sqrt{3}}{11}=\tan(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Let $a,b,c,d>0$ and $a+b+c+d=1$. Prove that $\frac{abc}{1+bc}+\frac{bcd}{1+cd}+\frac{cda}{1+ad}+\frac{dab}{1+ab}\le \frac{1}{17}$ Let $a,b,c,d>0$ and $a+b+c+d=1$. Prove that $\dfrac{abc}{1+bc}+\dfrac{bcd}{1+cd}+\dfrac{cda}{1+ad}+\dfrac{dab}{1+ab}\le \dfrac{1}{17}$
My attempt:
I figured out that if each of the element could be like $\dfrac{abc}{1+bc}\le \dfrac{1}{68}$ then we would be done.
From a little manipulation we get, $\dfrac{1}{bc}+1\le 68a$ , $\dfrac{1}{cd}+1\le 68b$ , $\dfrac{1}{da}+1\le 68c$ , $\dfrac{1}{ab}+1\le 68d$. Summing then we get,
$\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{cd}+\dfrac{1}{da}\le 64$
But, I suppose my first assumption could be wrong, if not then please help me further and if so then please help with the solution. Thank you.
| You could approach it as follows:
$$
\sum_{cyc} \frac{abc}{1+bc}=\sum_{cyc} a\left(1-\frac{1}{1+bc}\right)=\sum_{cyc} \left(a-\frac{a}{1+bc}\right)=1-\sum_{cyc} \frac{a}{1+bc}
$$
So the inequality is equivalent to:
$$
1-\sum_{cyc} \frac{a}{1+bc}\le\frac{1}{17}\iff\frac{16}{17}\le\sum_{cyc} \frac{a}{1+bc}
$$
Using CS, this can be reduced to prove the following:
$$
\left(\sum_{cyc} \frac{a}{1+bc}\right)\cdot\left(\sum_{cyc} a(1+bc)\right)\ge(a+b+c+d)^2=1\iff\sum_{cyc} \frac{a}{1+bc}\ge\frac{1}{\sum_{cyc} a(1+bc)}=\frac{1}{a+b+c+d+abc+bcd+cda+dab}=\frac{1}{1+abc+bcd+cda+dab}
$$
So if
$$
\frac{1}{1+abc+bcd+cda+dab}\ge\frac{16}{17}\iff abc+bcd+cda+dab\le\frac{1}{16}
$$
is true, the original inequality would be true as well.
Edit:
The inequality
$$
abc+bcd+cda+dab\le\frac{1}{16}
$$
is true due to Maclaurin's inequality , which, in a special case, states that:
$$
\left(\frac{abc+bcd+cda+dab}{4}\right)^{\frac13}\le\frac{a+b+c+d}{4}=\frac{1}{4}\iff abc+bcd+cda+dab\le\frac{1}{16}
$$
And your inequality is proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Probability of $(a+b\omega+c\omega^{2})(a+b\omega^{2}+c\omega)=1$
A fair die is thrown three times. If $a$, $b$, $c$ are the numbers obtained on the die, then what is the probability that $$(a+b\omega+c\omega^{2})(a+b\omega^{2}+c\omega)=1$$
(where $\omega$ is a cube root of unity)
My attempt:
On simplification, $a^{2}+b^{2}+c^{2}=1+(ab+bc+ca)$. I can't figure how to find the number of cases where the condition is satisfied.
| Hint: Since $a,b,c\in\mathbb R$ and $\omega$ is the complex cube root of unity, we can write
$$a^3+b^3+c^3-3abc=(a+b\omega+c\omega^2)(a+c\omega+b\omega^2)(a+b+c)$$
$$(a+b\omega+c\omega^2)(a+c\omega+b\omega^2)=\frac{a^3+b^3+c^3-3abc}{a+b+c}$$
Now, we only have to find out how many integer solutions are there for $a^3+b^3+c^3-3abc=a+b+c\forall 1\leq a,b,c\leq 6$.
On further simplification,
$$(a^2+b^2+c^2-ab-bc-ca)(a+b+c)=(a+b+c)$$
$$a^2+b^2+c^2-ab-bc-ca=1$$
The total positive integer solutions less than $6$ to the above equation by the total possible dice rolls ($6^3$) will be your required probability.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove $\lim_{x\to \infty} \frac{4x^2 - 7}{2x^3 - 5} = 0$ using $\epsilon$-$N$ limit definition I am having difficulties manipulating the problem so that I can find a $N$ value to choose.
Suppose $x > N$, then
$$\left|\frac{4x^2 - 7}{2x^3 - 5}\right| \leq \frac{4x^2}{|2x^3 - 5|} + \frac{7}{|2x^3 - 5|} $$
This is basically as far as I gone which I know is correct. I am not sure what would be the next step. Any help/hints would be appreciated!
Edit:
I found an interesting way to approach it, not sure if it is right.
Assume $x > 2$, then
$$\left|\frac{4x^2 - 7}{2x^3 - 5}\right| < \frac{4x^2}{2x^3} = \frac{2}{x} < \frac{2}{N} < \epsilon$$
That means I can choose $N = \max\left\lbrace 2, \frac{2}{\epsilon}\right\rbrace$
Is that correct?
| Not quite, you made your denominator larger there, so that step is not obvious. That is to say, $2x^3-5\le 2x^3\iff {1\over 2x^3-5}\ge {1\over 2x^3}$, so you cannot conclude that ${4x^2-7\over 2x^3-5}\le {4x^2\over 2x^3}$. Instead I recommend the following:
For $x>2$ we have $4x^2>4x^2-7$ and $2x^3-5\ge 2x^3-x^3$, i.e. ${1\over 2x^3-5}\le {1\over 2x^3-x^3}$, since both quantities are positive. So we have
$$\left|{4x^2-7\over 2x^3-5}\right|\le {4x^3\over x^3}={4\over x}$$
From this we can let $N=\max\left\lbrace 2,\left[{4\over\epsilon}\right]+1\right\rbrace$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that the sum is always greater than $1$ Prove that $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+ \frac{1}{n^2} \ge 1$, for all natural numbers $n$.
I tried to use mathematical inducion but failed and I tried to figure out a short formula for the sum but I couldnt find any.
| Using the Arithmetic-Harmonic Inequality:
$$\frac{n^2-n+1}{n+(n+1)+\cdots+n^2}\leq\frac{1}{n^2-n+1}\left(\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{n^2}\right)$$
Now, the denominator of the left-hand fraction is:
$$\sum_{k=0}^{n^2-n}(n+k)=\sum_{k=1}^{n^2-n+1}(n+k-1)=\sum_{k=1}^{n^2-n+1}(n-1)+\sum_{k=1}^{n^2-n+1}k=\frac{n^4+n}{2}$$
Hence, we have that
$$1\leq\frac{2(n^2-n+1)}{n(n+1)}=\frac{2(n^2-n+1)^2}{n(n+1)(n^2-n+1)}=\frac{2(n^2-n+1)^2}{n^4+n}\leq\frac{1}{n}+\cdots+\frac{1}{n^2}$$
We can recover the first inequality here from the fact that $0\leq (n-1)(n-2)$ for all $n\geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How to solve $y′′′+y'=2-\sin(x)$ I have tried to solve this but with no luck.
So far I just get,
$$y_p(x) = A \sin x + B \cos x \\
y'_p(x) = A \cos x - B \sin x \\
y''_p(x) =-A \sin x - B \cos x \\
y'''_p(x) =-A \cos x + B \sin x $$
$$y'''+ y' = -A \cos x + B \sin x + A \cos x - B \sin x \\ \\
= \cos x (-A+A) + \sin x (B-B) = 2 - \sin x $$
I would really appreciate help in solving this
| Rewrite the equation as $$(D^3+D)y=2-\sin(x)$$
Differentiate twice to obtain
$$\begin{align}D^2(D^3+D)y&=\sin(x)\\
\implies (D^2+1)(D^3+D)y&=2\\
\implies D(D^2+1)(D^3+D)y=D^2(D^2+1)^2y&=0\end{align}$$
Which I leave to you to show has the solution
$$y=C_1+C_2x+(C_3+C_4x)\sin(x)+(C_5+C_6x)\cos(x)$$
Plug this back into the original equation to obtain
$$(D^3+D)y=y'''+y'=C_2-2C_4\sin(x)-2C_6\cos(x)=2-\sin(x)$$
Equating coefficients, we have $C_2=2,C_4=1/2,C_6=0$. Shifting the coefficients on $y$, we have that the general solution to the equation is
$$y=C_1+C_2\sin(x)+C_3\cos(x)+2x+\frac12x\sin(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Monkeys Dividing Pile of Bananas Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio 3:2:1, what is the least possible total for the number of bananas?
I tried bashing this problem out using long algebraic expressions, but I kept messing up and was unable to find a good solution. Could I get some hints on one? Thanks!
| Let $A,B,C$ be the banana's removed by monkeys $1,2,3$ respectively and $x$ be the number of bananas monkey $3$ has at the end.
Then:
$\frac{1}{8}A+\frac{3}{8}B+\frac{1}{12}C=x$
$\frac{1}{8}A+\frac{1}{4}B+\frac{11}{24}C=2x$
$\frac{3}{4}A+\frac{3}{8}B+\frac{11}{24}C=3x$
Solving the system we get:
$A=\frac{22}{17}x$
$B=\frac{26}{17}x$
$C=\frac{54}{17}x$.
Now, let $x=17k$.
We need $A=22k$ to be a multiple of $8$ and $4$ (hence $8$)
We need $B=26k$ to be a multiple of $8$ and $4$ (hence $8$)
We need $C=54k$ to be a multiple of $24$.
Hence $k$ must be a multiple of $4$.
So $x$ must be $17\times 4$.
Therefore the number of bananas is $6\times 17\times 4=408$
And monkeys $1,2,3$ take $88,104,216$ bananas each.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What are all the twin primes $p$ and $q = p + 2$ for which $pq - 2$ is also prime? It seems that $p = 3$ and $q = 5$ ($pq - 2 = 13$) are the only solutions. However I'm having a difficult time proving this.
I have that all primes can be represented as $3k + 1$ or $3k + 2$ so if $p = 3k + 2$ then $q = 3k + 4$ and
\begin{align*}
pq - 2 &= (3k + 2)(3k + 4) - 2\\
&= 9k^2 + 18k + 8 - 2\\
&= 9k^2 + 18k + 6\\
&= 3(3k^2 + 6k + 2) \text{ which can't be prime}
\end{align*}
So that seems promising. But when I let $p = 3k + 1$ and $q = 3k + 3$ I get
\begin{align*}
pq -2 &= (3k + 1)(3k + 3) - 2\\
&= 9k^2 + 12k + 3 - 1\\
&= 9k^2 + 12k + 2,
\end{align*}
which could very well be prime. So what do I do? It appears, though, that the only solution is $p = 3$.
| We notice that for $p=3$ and $q=5$ we have a solution.
Let $p \geq 5$
Then p has form $p=6k \pm 1$ where $k \in N$
If $p=6k+1$ then $q=6k+3=3(2k+1)$ so $q$ is not a prime then. That leaves us with $p=6k-1$. Then $q=6k+1$ so $pq-2=(6k-1)(6k+1)-2=36k^2-3=3(12k^2-1)$ which is not a prime.
We conclude that there is no solution for $p \geq 5$ so the only solution is $p,q (3,5)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to prove this floor function equation? How can I prove the following equation?
$$
\lfloor nx \rfloor =
\lfloor x \rfloor +
\Big\lfloor x + \frac{1}{n} \Big\rfloor +
\Big\lfloor x + \frac{2}{n} \Big\rfloor +
\Big\lfloor x + \frac{3}{n} \Big\rfloor +
\Big\lfloor x + \frac{4}{n} \Big\rfloor +
\Big\lfloor x + \frac{5}{n} \Big\rfloor+ \dotsb +
\Big\lfloor x + \frac{n-1}{n} \Big\rfloor
$$
$n∈N$ and $x∈R$
| Try to rewrite $x$ as
$$x=\lfloor x \rfloor + \frac kn + \alpha$$
where $0\le k<n$ and $0\le\alpha<\frac1n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Integration by parts or using u-sub? Integrate: $4(x^3+1)^{\frac{1}{3}}(x^5)$
I tried using u-sub but it is getting me absolutely nowhere. I also tried using integration by parts and eventually was able to evaluate the integral but it does not seem to simplify to the answer. Thank you for your help.
| Maybe this will help: let $u = x^{3} + 1$. Then $u - 1 = x^{3}$, and $du = 3x^{2}\, dx$ (i.e., $\frac{1}{3} \, du = x^{2} \, dx$), so we get:
$\int 4(x^{3} + 1)^{\frac{1}{3}}(x^{3}x^{2}) \,dx = \int \frac{4}{3}(u)^{\frac{1}{3}}(u - 1) \,du = \frac{4}{3} \int u^{\frac{4}{3}} - u^{\frac{1}{3}} \,du = \frac{4}{3}(\frac{3}{7}u^{\frac{7}{3}} - \frac{3}{4}u^{\frac{4}{3}}) + C$
$ = \frac{4}{7}u^{\frac{7}{3}} - u^{\frac{4}{3}} + C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Area of triangle bounded by line and degenerate "crossed lines" conic The question is
Show that the two lines given by
$$(A^2 - 3B^2)x^2 + 8ABxy +(B^2 - 3A^2)y^2=0$$
and the line given by $$Ax+By+C=0$$ determine an equilateral triangle of area $$\frac{C^2}{\sqrt{3}\;(A^2+B^2)}$$
I tried factorizing the pair of straight lines into two straight lines. But that seemed to be not really simple. And I doubt I was proceeding rightly. It would be great if anyone helps.
| The crossed lines defined by
$$( A^2 - 3 B^2 ) x^2 + 8 A B x y + ( B^2 - 3 A^2 ) y^2 = 0 \qquad(\star)$$
pass through the origin, so one of our vertices is $O(0,0)$.
Convenient manipulation allows us to re-write $(\star)$ thusly:
$$x^2 + y^2 = \frac{4( Ax + B y )^2}{3(A^2+B^2)} \qquad(\star\star)$$
If $P(x,y)$ satisfies $(\star)$ (that is, $(\star\star)$) and also $A x + B y + C = 0$, then we have
$$x^2 + y^2 = \frac{4C^2}{3(A^2+B^2)} \quad\text{so that}\quad |\overline{OP}| = \frac{2|C|}{\sqrt{3}\;\sqrt{A^2+B^2}}$$
But $d := |C|/\sqrt{A^2+B^2}$ is precisely the distance from $O$ to the line defined by $Ax+By+C=0$. Consequently, if $P_1$ and $P_2$ are the other two vertices of our triangle, then
$$|\overline{OP_1}| = |\overline{OP_2}| = \frac{2}{\sqrt{3}}d$$
That is, $\triangle OP_1P_2$ is isosceles. Moreover, the ratio of its base altitude ($d$) to its leg makes it equilateral, and we can write
$$|\triangle OP_1P_2| = \frac{1}{2} \;d\;|\overline{OP_1}| = \frac{1}{2} d^2 \frac{2}{\sqrt{3}} = \frac{C^2}{\sqrt{3}\;(A^2+B^2)}$$
as desired. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving that the sum $\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2}=\frac{\pi^2}{8}$ Proving that the sum
$$\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2}=\frac{\pi^2}{8}$$
When $m$=integer even number
I know that the $\frac{\pi^2}{8}$ comes from $\sum_{n=0}^{\infty }\frac{1}{(2n+1)^2}$ and I know how to prove it but I don't know how to prove the above sum especially with the power $m$
| $\sin(n\frac{\pi}{2}) \in \{-1, 0, 1\}$
For even $m$,
$\sin^m(n\frac{\pi}{2}) = \begin{cases}
0 : n \text{ is even} \\
1 : n \text{ is odd}
\end{cases}$
So, we can rewrite this summation as:
$$
\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2} \\
= \sum_{n=1}^{\infty }\frac{1}{n^2} - \sum_{n=1}^{\infty }\frac{1}{(2n)^2} \\
= \sum_{n=1}^{\infty }\frac{1}{n^2} - \frac{1}{4}\sum_{n=1}^{\infty }\frac{1}{n^2} \\
= \frac{\pi^2}{6} - \frac{1}{4}\frac{\pi^2}{6} \\
= \frac{\pi^2}{8}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Prove that $a^2 + b^2 \geq 8$ if $ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $ has at least one real root. If it is known that the equation
$$ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $$
has a (real) root, prove the inequality
$$ a^2 + b^2 \geq 8. $$
I am stuck on this problem, though, it is a very easy problem for my math teacher. Anyway, I can't figure out.
| Let $x$ be a root.
Thus, by C-S $$(x^2+1)^2=-x(ax^2+b^2)\leq\sqrt{x^2(ax^2+b)^2}\leq\sqrt{x^2(x^4+1)(a^2+b^2)}.$$
Id est, it's enough to prove that:
$$\frac{(x^2+1)^4}{x^2(x^4+1)}\geq8$$ or
$$(x^2-1)^4\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Deriving an expression for $\cos^4 x + \sin^4 x$
Derive the identity $\cos^4 x + \sin^4 x=\frac{1}{4} \cos (4x) +\frac{3}{4}$
I know $e^{i4x}=\cos (4x) + i \sin (4x)=(\cos x +i \sin x)^4$. Then I use the binomial theorem to expand this fourth power, and comparing real and imaginary parts, I conclude that $\cos^4 x + \sin^4 x = \cos (4x) + 6 \cos^2 (x) \sin^2 (x)$.
So now I need to show that $\cos (4x) + 6 \cos^2 (x) \sin^2 (x)=\frac{1}{4} \cos (4x) +\frac{3}{4}$, which has stumped me.
| $\cos^2 x \sin^2 x = (1 - \sin^2 x) \sin^ 2 x = \sin^2 x - \sin^ 4 x$ and by the same argument $\cos^2 x \sin^2 x = \cos^2 x - \cos^ 4 x$. Take the average of these two identities to obtain
$$
\cos^2 x \sin^2 x = \frac{1}{2} - \frac{\cos^4x + \sin^4 x}{2} \, .
$$
Now substitute this and simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Real part of Arcsine of a Complex variable How do we show: $\operatorname{Re}({\arcsin{z}}) = \frac{1}{2}(\sqrt{x^2+y^2+2x+1} -\sqrt{x^2+y^2-2x+1})$ ?
None of the things I tried including using the formula of inverse sine seems to work... Please Help
| Solving the equation $(\cosh^2 v)^2-(x^2+y^2+1)\cosh^2 v +x^2=0\,\,$ we have
$$
\cosh ^2v=\frac{1}{2}\left(x^2+y^2+1+\sqrt{(x^2+y^2+1)^2-4x^2}\right)$$
since $\cosh^2 v=1+\sinh^2 v\ge 1$.
Noting $$
\left(\frac{1}{2}\left(\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\right)\right)^2=\frac{1}{2}\left(x^2+y^2+1+\sqrt{(x^2+y^2+1)^2-4x^2}\right),$$
we have$$
\cosh v=\frac{1}{2}\left(\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\right).$$
Therefore $$\sin u=\frac{x}{\cosh v}=\frac{1}{2}\left(\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}\right),
$$which ensures $$u=\operatorname{Re}(\arcsin (x+iy))=\arcsin \left(\frac{1}{2}\left(\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}\right)\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation, hat consists on an arithmetical progression. $$1+x+x^2+x^3+\cdots+x^{99}=0.$$
I said to prove with $0+1+2+3+\cdots+99=0$. How should I proceed?
| another observation $$1+x+x^2+x^3+...+x^{99}=\\(1+x^2+x^4+...+x^{98})+(x+x^3+x^5+...+x^{99})=0\\(1+x^2+x^4+...+x^{98})+x(1+x^2+x^4+...+x^{98})=\\(1+x^2+x^4+...+x^{98})(1+x)=0\\\rightarrow (1+x)=0 \rightarrow x=-1$$note that ! $$1+x^2+x^4+...+x^{98} \neq 0\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Unable to calculate pseudo-inverse $A^TA$ I'm trying to calculate the pseudo-inverse of $A^TA$ as described in this paper:
The SVD is particularly simple to calculate when
the matrix is of the form $A^TA$ because $U=V$ and the rows of $U$
are the eigenvectors of $A^TA$ and the singular values in $D$ are the
eigenvalues of $A^TA$.
Since $U$ and $V$ are orthogonal matrices, the inverse of $M$ is then
$M^{-1}=V^TD^{-1}U$
Since the pseudo-inverse of a non-singular matrix is it's inverse I tried to do this with a non-singular matrix and see if I get the inverse. I tried it with the following:
$A^TA = \begin{pmatrix}
90 & 51 \\
51 & 29
\end{pmatrix}$ where $A = \begin{pmatrix}
3 & 2 \\
9 & 5
\end{pmatrix}$
Eigenvalues and eigenvectors:
$\lambda_1 = 118.9243;$ $\lambda_2=0.0757;$ $v_1 = \begin{pmatrix}
1 \\
0.5671
\end{pmatrix};$ $v_2 = \begin{pmatrix}
1 \\
-1.7632
\end{pmatrix}$
$U = V=\begin{pmatrix}
1 & 0.5671 \\
1 & -1.7632
\end{pmatrix}$
$D = \begin{pmatrix}
118.9243 & 0\\
0 & 0.0757
\end{pmatrix}$
And my result is:
$V^TD^{-1}U= \begin{pmatrix}
13.22 & -23.29 \\
-23.29 & 41.07
\end{pmatrix} \neq (A^TA)^{-1}$
Can someone help me? What am I doing wrong?
| Pick up the thread at the eigenvectors or the product matrix. The ordering was mixed. It should be:
$$
v_{1} =
\left[
\begin{array}{c}
1.76322 \\
1 \\
\end{array}
\right],
\qquad
v_{2} =
\left[
\begin{array}{c}
-0.567144 \\
1 \\
\end{array}
\right].
$$
Normalized, these are the column vectors of the domain matrix:
$$
\mathbf{V} =
\left[
\begin{array}{cc}
0.869844 & -0.493327 \\
0.493327 & 0.869844 \\
\end{array}
\right]
$$
Next, $\mathbf{U} \ne \mathbf{V}!$:
$$
\mathbf{A} = \mathbf{U} \, \Sigma \, \mathbf{V}^{T}
\qquad \Rightarrow \qquad
\mathbf{U} = \mathbf{A} \, \mathbf{V} \, \Sigma^{-1}.
$$
Now perform your cross check:
$$
\begin{align}
%
\mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{T} \\
%
\left[
\begin{array}{cc}
3 & 2 \\
9 & 5 \\
\end{array}
\right]
%
&=
% U
\left[
\begin{array}{cr}
0.329767 & 0.944062 \\
0.944062 & -0.329767 \\
\end{array}
\right]
% Sigma
\left[
\begin{array}{cc}
10.9052 & 0. \\
0. & 0.275097 \\
\end{array}
\right]
% V^T
\left[
\begin{array}{rc}
0.869844 & 0.493327 \\
-0.493327 & 0.869844 \\
\end{array}
\right]
%
\end{align}
$$
All null spaces are trivial and the classic inverse is the pseudoinverse:
$$
\mathbf{A}^{-1} = \mathbf{A}^{\dagger} = \mathbf{V} \, \Sigma^{-1} \, \mathbf{U}^{T} =
\frac{1}{3}
\left[
\begin{array}{rr}
-5 & 2 \\
9 & -3 \\
\end{array}
\right].
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1160429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prime numbers and divisibility Prove that for all prime numbers $p,q>3$ holds
$$48|p^4-q^4$$
My solution:
$$48=2^43$$
$$p^4-q^4=(p-q)(p+q)(p^2+q^2)$$
Since adding or substracting odd numbers, we get a even number, $p^4-p^4$ is divisible by $2^3$. How about another $2$ and $3$?
| For $3$, note that $2^4 \equiv 1^4 \equiv 1 \pmod 3$, so $p^4 - q^4 \equiv 1 - 1 \equiv 0 \pmod 3$.
For $2^4$, you have $(p + q) \equiv (p - q) \equiv 0 \pmod 2$. Since $p, q$ are odd, they are equivalent to $1$ or $3 \pmod 4$. $3^2 \equiv 1^2 \equiv 1 \pmod 4$ so $p^2 - q^2 \equiv 1 - 1 \equiv 0 \pmod 4$.
Putting these together gives the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
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.