Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Summation of infinite series, $\sum((3n+1)^{-1}-(3n+2)^{-1})$ Find the sum of:
$$\sum_{n=0}^\infty \left(\frac{1}{3n+1}- \frac{1}{3n+2}\right) $$
Anwser given was $\dfrac{\pi}{3\sqrt{3}}$.
Thanks in advance.
| \begin{align}
\sum_{n=0}^\infty \left(\frac{1}{3n+1}- \frac{1}{3n+2}\right)
&=\sum_{m=0}^\infty\left(\int^1_0x^{3n}-x^{3n+1}dx\right)\\
&=\int^1_0\frac{1-x}{1-x^3}dx\\
&=\int^1_0\frac{1}{(x+1/2)^2+3/4}dx\\
&=\frac{2}{\sqrt{3}}\left[\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right]^1_0\\
&=\frac{2}{\sqrt{3}}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)\\
&=\frac{\pi}{3\sqrt{3}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrating $\frac 1 {\sqrt{x^2 - a^2}}$ I'm trying to understand how to integrate:
$$\frac 1 {\sqrt{x^2 - a^2}}$$
I tried substituting $t$ as $\sqrt{x^2 - a^2}$, but I can't get an answer that way, and I don't know any other way to integrate other than by parts, which I don't think can be used here. How should I integrate this?
| $$\begin{align}
\text{Integrating} &\int \frac {1\cdot dx}{\sqrt {(x^2-a^2)}}\\\\
\text{Put x} &= a \cdot \cosh (\theta)\\
\Rightarrow \theta &= \cosh^{-1} \left(\frac {x}{a} \right)\\
\text{Therefore}\\ dx &= a \cdot \sinh(\theta) \cdot d\theta\\\\
\text{Now}\\
&\int \frac {a \cdot \sinh (\theta) \cdot d \theta}{\sqrt {a^2\cdot(\cosh^2(\theta)-1)}}\\\\
&\Rightarrow \frac {a \cdot \sinh (\theta) \cdot d \theta}{\sqrt {a^2\cdot \sinh ^2(\theta)}}\\
&\Rightarrow \int \frac{a \cdot \sinh (\theta) \cdot d\theta}{a \cdot \sinh (\theta)}\\
&\Rightarrow \int 1 \cdot d\theta\\
&\Rightarrow \theta + c\\
&\Rightarrow \cosh^{-1} \left(\frac {x}{a} \right) +c
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/888251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
How many different positive integer factors does have? How many different positive integer factors does $(2^7)(3^4)(7^3)(23^5)$ have?
Do we have to do any combinations between the powers here?
| What you are looking for is the number of the positive divisors of $2^7\cdot 3^4\cdot 7^3\cdot 23^5$. Then, the answer is
$$(7+1)(4+1)(3+1)(5+1)=8\cdot 5\cdot 4\cdot 6=960.$$
In general, if
$$N={p_1}^{\color{red}{q_1}}\cdot {p_2}^{\color{red}{q_2}}\cdots {p_k}^{\color{red}{q_k}}$$
where $q_i,k\in\mathbb N$ and $p_1\lt p_2\lt\cdots\lt p_k$ are primes, then the number of the positive divisors of $N$ can be represented as
$$(\color{red}{q_1}+1)(\color{red}{q_2}+1)\cdots (\color{red}{q_k}+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/891005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluating a limit. What makes the equality right? I'm reading a proof of a limit calculation. The limit is:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x}$$
where $a,b>0$.
The aother claims that:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x} =
\exp\left( \lim\limits_{x\to 0}\frac{\frac{a^x+b^x}{2} - 1}{x} \right)$$
How come?
Update:
Of course,
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x} =
\exp\left(\lim\limits_{x\to 0} \frac{\ln\left( \frac{a^x+b^x}{2} \right)}{x} \right)$$
But how to proceed to reach the auther's expression?
| First use simple fact, that $\displaystyle\lim_{y \to 0}\frac{\ln(1+y)}{y}=1$, so:
$$\lim_{x \to 0}\frac{\ln(\frac{a^x+b^x}{2}-1+1)}{y}=\lim_{x \to 0}\frac{\ln(\frac{a^x+b^x}{2}-1+1)}{\frac{a^x+b^x}{2}-1} \cdot \lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{y}=\\=1 \cdot \lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{x}$$
Now $\lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{x}=\lim_{x \to 0}\frac{1}{2}\frac{a^x-1}{x}+\lim_{x \to 0}\frac{1}{2}\frac{b^x-1}{x}$
But $a^x=e^{x \ln a }$, so $\lim_{x \to 0}\frac{1}{2}\frac{a^x-1}{x}=\lim_{x \to 0}\ln a\frac{1}{2}\frac{e^{\ln a x}-1}{x \ln a}=\frac{1}{2}\ln a$. The same with second limit. Finally the result is $\frac{\ln ab}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/893118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Arithmetic Progression. Q. The ratio between the sum of $n$ terms of two A.P's is $3n+8:7n+15$. Find the ratio between their $12$th term.
My method:
Given:
$\frac{S_n}{s_n}=\frac{3n+8}{7n+15}$
$\frac{S_n}{3n+8}=\frac{s_n}{7n+15}=k$
$\frac{T_n}{t_n}=\frac{S_n-S_{n-1}}{s_n-s_{n-1}}=\frac{k\left(\left(3n+8\right)-\left(3\left(n-1\right)+8\right)\right)}{k\left(\left(7n+15\right)-\left(7\left(n-1\right)+15\right)\right)}=\frac{3}{7}$
As this applies for any term:
$\frac{T_{12}}{t_{12}}=\frac{3}{7}$
But this is not the answer. The actual answer is $\frac7{16}$. I know how to obtain that answer.
But why is my solution wrong? Its probably the concept I guess.
| Hint.
We have that
$$
S_n=A_1+\cdots+A_n,
$$
where $A_n=Kn+L$, and hence
$$
S_n=K\frac{n(n+1)}{2}+Ln.
$$
Similarly, $s_n=a_1+\cdots+a_n$, where $a_n=kn+\ell$ and $s_n=k\frac{n(n+1)}{2}+\ell n$,
and
$$
\frac{S_n}{s_n}=\frac{K\frac{n(n+1)}{2}+Ln}{k\frac{n(n+1)}{2}+\ell n}=\frac{K(n+1)+2L}{k(n+1)+2\ell}=\frac{Kn+2L+K}{kn+2\ell+k}=\frac{3n+8}{7n+15},
$$
and
$$
\frac{A_{12}}{B_{12}}=\frac{12K+L}{12k+\ell}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/894079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Question on Factoring I have very basic Question about factoring, we know that,
$$x^2+2xy+y^2 = (x+y)^2$$
$$x^2-2xy+y^2 = (x-y)^2$$
But what will
$$x^2-2xy-y^2 = ??$$
$$x^2+2xy-y^2 = ??$$
| There is no simple factorization of $x^2+2xy-y^2$ nor $x^2-2xy-y^2$, although you can write:
$$\begin{align}x^2+2xy-y^2 &= (x+y)^2-2y^2 \\&= \left(x+y(1+\sqrt{2})\right)\left(x+y(1-\sqrt{2})\right)
\end{align}$$
and similarly:
$$x^2-2xy-y^2 = \left(x-y(1+\sqrt{2})\right)\left(x-y(1-\sqrt{2})\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/894392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
On the decomposition of stochastic matrices as convex combinations of zero-one matrices Let "stochastic" matrix be the matrix whose rows sum to one and deterministic matrix be a stochastic matrix whose all rows consist of a one and zero.
For example $\left [ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0 \end{array} \right] $ is a deterministic matrix.
I am trying to show that any stochastic matrix can be written as a convex combination of deterministic matrices.
| Constructive proof:
Consider some nonzero substochastic matrix $P$ with equal-row-sums. Call $P_{is(i)}$ one of the minimal nonzero entries of row $i$, and consider the deterministic matrix $D$ such that $D_{is(i)}=1$ for every $i$ and $D_{ij}=0$ otherwise. Finally, define $m(P)=\min\{P_{ij}\mid P_{ij}\ne0\}$.
Then $P-m(P)D$ is a substochastic matrix with equal-row-sums and with at least one more zero entry than $P$.
Starting from some stochastic matrix $P$ and iterating the step $P\mapsto P-m(P)D$ at most a finite number $N$ of times, one gets a sequence $(P_k)_{0\leqslant k\leqslant N}$ of substochastic matrices with equal-row-sums and a sequence $(D_k)_{0\leqslant k\leqslant N-1}$ of deterministic matrices such that $P_0=P$, $P_N=0$, and $P_{k+1}=P_k-m(P_k)D_k$ for every $0\leqslant k\leqslant N-1$.
This yields the decomposition of $P$ as a convex combination of deterministic matrices $$P=\sum_{k=0}^{N-1}m(P_k)D_k.$$
Example: Consider the stochastic matrix $$P=\frac1{12}\begin{pmatrix}2&4&6\\ 2&2&8\\ 3&3&6\end{pmatrix},$$ then a sequence of reductions, showing the selected entries on $12\cdot P$ at each step, is
$$
\begin{pmatrix}\underline 2&4&6\\ \underline2&2&8\\ \underline3&3&6\end{pmatrix}\
\begin{pmatrix}0&\underline4&6\\ 0&\underline2&8\\ \underline1&3&6\end{pmatrix}\
\begin{pmatrix}0&\underline3&6\\ 0&\underline1&8\\ 0&\underline3&6\end{pmatrix}\
\begin{pmatrix}0&\underline2&6\\ 0&0&\underline8\\ 0&\underline2&6\end{pmatrix}\
\begin{pmatrix}0&0&\underline6\\ 0&0&\underline6\\ 0&0&\underline6\end{pmatrix}$$ wich yields the decomposition
$$P=\frac16\begin{pmatrix}1&0&0\\ 1&0&0\\ 1&0&0\end{pmatrix}+
\frac1{12}\begin{pmatrix}0&1&0\\ 0&1&0\\ 1&0&0\end{pmatrix}+
\frac1{12}\begin{pmatrix}0&1&0\\ 0&1&0\\ 0&1&0\end{pmatrix}+
\frac16\begin{pmatrix}0&1&0\\ 0&0&1\\ 0&1&0\end{pmatrix}+
\frac12\begin{pmatrix}0&0&1\\ 0&0&1\\ 0&0&1\end{pmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to prove that a diffrensiation of a formula equals to another formula. QUESTION 1) if $y =\dfrac{ \sin x-x\cos x}{x\sin x+\cos x}$ show that $\dfrac{dy}{dx}= \dfrac{x^2}{(x\sin x+\cos x)^2}$
QUESTION 2) if $y = \dfrac{\tan x+1}{\tan x-1}$ show that $\dfrac{dy}{dx}= \dfrac{-2}{1-\sin 2x}$
| For the first, we get
\begin{align}
\frac{d}{dx}\left(\frac{ \sin x-x\cos x}{x\sin x+\cos x}\right)
&= \frac{(x\sin x + \cos x)(\cos x + x\sin x-\cos x)-(\sin x - x\cos x)(\sin x + x\cos x - \sin x)}{(x\sin x + \cos x)^2} \\
&= \frac{x^2\sin^2 x + x\sin x\cos x - (x\sin x\cos x - x^2\cos^2 x)}{(x\sin x + \cos x)^2} \\
&= \frac{x^2(\sin^2 x + \cos^2 x)}{(x\sin x + \cos x)^2} \\
&= \frac{x^2}{(x\sin x + \cos x)^2} .
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can I perform the quadratic formula on polynomial with complex coefficient? 2 weeks ago, we had a Math test on complex number. One of the question was:
Let $z=x+iy$ be a non-zero complex number, where $x,y \in \mathbb{R}$.
Given that $z+\frac{1}{z} = k$, where $k$ is a real number, show that if $y \neq 0$, then $| k | \leq 2$.
My friend had tried solving it this way (as follows), but instead of proving the above, she had gotten a contradiction!
$z+\frac{1}{z} = x+iy + \frac{1}{x+iy} = \frac{x^2 + 2xyi - y^2 + 1}{x+ iy} = k$.
This is equivalent to saying that
$x^2 + 2xyi - y^2 + 1 = k(x + iy) = kx + kyi$
Then, $x^2 + (2yi-k)x + (1-y^2-kyi)=0 $
Since we know that $x \in \mathbb {R}$, then the discriminant of this polynomial will be greater or equal to 0. Hence,
Discriminant $= (2yi-k)^2 - 4(1-y^2-kyi) = -4y^2 - 4kyi + k^2 - 4 + 4y^2+4kyi = k^2 - 4 \geq 0$.
Thus, we can conclude from the above that $k^2 \geq 4, |k| \geq 2$, which is incorrect!
Can anyone help us explain why this is so? (we know the answer and correct solution but are just perplexed with why a contradiction occurs when performing the above. we have asked the class and the teacher but they too couldn't provide a convincing answer...)
| $$x^2-y^2+1+2xy(i)=xk+i(yk)$$
Equating the imaginary parts, $$yk=2xy\implies x=\frac k2\text{ as }y\ne0$$
Equating the real parts, $$x^2-y^2+1=xk\iff\frac{k^2}4-y^2+1=\frac{k^2}2\iff k^2=4(1-y^2)\le4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/897996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Ordinary Differential Equation with trigs! I need help and alternative ways of solving this problem:
$$ y'(x^2 + 1) - 2xy = (x^2 + 1)\arctan x $$
Thank you for all kind of help and hints :)
| $$
\left(x^2+1\right)y' -2xy = \left(x^2+1\right)\arctan x
$$
$$
y' -\frac{2x}{x^2+1}y = \arctan x
$$
$$
\frac{g'}{g} = \frac{2x}{x^2+1} = \frac{d}{dx}\ln g(x)
$$
thus
$$
y\mathrm{e}^{-\ln g} = \frac{y}{g} = \int \frac{1}{g}\arctan x dx = \int \frac{1}{x^2+1}\arctan x dx
$$
$$
\frac{d}{dx}\arctan x = \frac{1}{x^2+1}
$$
therefore
$$
\frac{y}{x^2+1} = \int \left(\frac{d}{dx}\arctan x\right)\arctan x dx = \int \frac{d}{dx}\frac{\arctan^2 x}{2} dx = \frac{1}{2}\arctan^2(x) +C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Summation of general series One of the problems in Donald Knuth's Art of Programming is phrased as follows:
Find and prove a simple formula for the sum $$\sum\limits_{n=0}^k\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}.$$
I have very little experience with summations. My method was to try to find some function such that
$$\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}=f(x)-f(x-1).$$
However, I have no idea how to do that systematically. Fortunately, partial fraction decomposition happens to give a function in that form, with
$$f(x)=\frac{(-1)^x(x+1)}{4(x+1)^2+1}$$
which is the solution in the back of the book. I was wondering how you might go about finding this solution. This specific problem is volume 1, chapter 1.2.1, problem 11.
| By computing the residues of $\frac{x^3}{x^4+4}$ in $x=\pm 1\pm i$ we have:
$$\frac{x^3}{x^4+4}=\frac{1}{2}\left(\frac{x-1}{(x-1)^2+1}+\frac{x+1}{(x+1)^2+1}\right)\tag{1}$$
hence:
$$\begin{eqnarray*}\sum_{n=0}^{k}\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}&=&\frac{1}{2}\sum_{n=0}^{k}(-1)^n\left(\frac{2n}{4n^2+1}+\frac{2(n+1)}{4(n+1)^2+1}\right)\\&=&\frac{(-1)^k\cdot(k+1)}{4(k+1)^2+1}.\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find arc length of curve on the given interval I was asked to find the arc length of the curve of the following curve:
$24xy = x^4 + 48$ from $x = 2$ to $x = 4$
This has turned out to be a very difficult problem, I get stuck using the arc length formula with the derivative I have calculated.
| We can put
$$\begin{cases} x&=t \\y&=\frac{1}{24}t^3+2t^{-1}\end{cases}$$
Arc length formula $$s=\int_{2}^{4}{\sqrt{(x')^2+(y')^2}\text{d}t}$$ give us
\begin{align*}
s&=\int_{2}^{4}{\sqrt{1+\left(\frac{1}{8}t^2-2t^{-2}\right)^2}\text{d}t}\\
&=\int_{2}^{4}{\sqrt{\left(\frac{1}{8}t^2+2t^{-2}\right)^2}\text{d}t}\\
&=\int_{2}^{4}{\left(\frac{1}{8}t^2+2t^{-2}\right)\text{d}t}\\
&=\left[\frac{1}{24}t^3-2t^{-1}\right]_2^4\\
&=\frac{17}{6}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/901249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How can I bring $\sin(x)$ to the following form? What steps do we take for the following?
$$\sin x = \frac{{2\tan\frac{x}{2}}}{1+\tan^2\frac{x}{2}}$$
| $$\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} = 2\sin\frac{x}{2}\cos\frac{x}{2}\frac{\cos\frac{x}{2}}{\cos\frac{x}{2}} = 2\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\cos^2\frac{x}{2} = \frac{2\tan\frac{x}{2}}{\frac{1}{\cos^2\frac{x}{2}}} = \frac{2\tan\frac{x}{2}}{\frac{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}} = \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/902853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Should I throw the dice again if I have rolled 4? My math skills are very basic so it might be a stupid question, I had a discussion with my brother in law and now we have a 'math problem'.
We were playing a game with dices and he threw 4. The challenge was to throw the highest number, you can stop or throw again once, you cant see what the opponent has thrown, you both reveal after finishing. He said if you have 4 you have a 50% change the next time to throw the same or higher.
(4,5,6) vs (1,2,3)
and should throw again.
But I said that I won't throw again on 4 because you already threw 4 and you are not likely to throw 4, 2 times in a row and therefore I would stop at 4. Am I right or is he and do you have a 50 % chance on throwing 4 or higher again?
Game rules
*
*You throw a normal dice 1/6
*You can choose to throw again or keep the current value
*Your opponent cant see your value, you cant see his.
*The one with the highest value wins.
Short version:
If I threw 4, how big is the chance I throw 4 or more the next time I throw the dice, and should i take that chance?.
Thoughts
Average of 1 dice is 3.5, if i throw 4 im above Average and am more likely to throw below average next time.
| If you and your brother-in-law both have the same strategy, then the game is even. So the only question is, what happens if you re-throw on a 4 and your brother-in-law doesn't? Here is a table of the probability of each outcome:
1 2 3 4 5 6
You 1/9 1/9 1/9 1/9 5/18 5/18
Brother-in-law 1/12 1/12 1/12 1/4 1/4 1/4
The probability of a tie is $3 \cdot \dfrac{1}{9} \cdot \dfrac{1}{12} + \dfrac{1}{9} \cdot \dfrac{1}{4} + 2 \cdot \dfrac{5}{18} \cdot \dfrac{1}{4} = \dfrac{7}{36}$
The probability that you win is $\dfrac{1}{12} \cdot \dfrac{8}{9} + \dfrac{1}{12} \cdot \dfrac{7}{9} + \dfrac{1}{12} \cdot \dfrac{6}{9} + \dfrac{1}{4} \cdot \dfrac{5}{9} + \dfrac{1}{4} \cdot \dfrac{5}{18} = \dfrac{29}{72}$
The probability that you lose is $\dfrac{1}{9} \cdot \dfrac{11}{12} + \dfrac{1}{9} \cdot \dfrac{10}{12} + \dfrac{1}{9} \cdot \dfrac{9}{12} + \dfrac{1}{9} \cdot \dfrac{1}{2} + \dfrac{5}{18} \cdot \dfrac{1}{4} = \dfrac{29}{72}$
So it makes no difference!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/905438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Dividing by $\sqrt n$ Why is the following equality true?
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$
| Alternate form:
$$\frac{\frac{\sqrt{n}}{ \sqrt{n}}}{\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}}{\sqrt{n}}}=\frac{1}{\sqrt{\frac{n+\sqrt{n+\sqrt{n}}}{n}}}=\frac{1}{\sqrt{1+ \frac{\sqrt{n+\sqrt{n}}}{n}}}=\frac{1}{\sqrt{1+ \frac{\sqrt{n+\sqrt{n}}}{\sqrt{n} \sqrt{n}}}}=\frac{1}{\sqrt{1+ \frac{1}{\sqrt{n}} \frac{\sqrt{n+\sqrt{n}}}{ \sqrt{n}}}}=\frac{1}{\sqrt{1+\frac{1}{\sqrt{n}} \sqrt{\frac{n+\sqrt{n}}{n}}}}=\frac{1}{\sqrt{1+\frac{1}{\sqrt{n}} \sqrt{1+\frac{1}{\sqrt{n}}}}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\frac{5x}{2x^2+5x+1}=\frac13$ then the value of $x+\frac{1}{2x}$ is If $\frac{5x}{2x^2+5x+1}=\frac13$ then the value of $x+\frac{1}{2x}$ is
note $5x=1$
$2x^2+5x+1=3$
or
$15x=2x^2+5x+1$
so $2x^2-10x+1$
so $\frac{5+\sqrt{23}}2$ or $\frac{5-\sqrt{23}}2$
| Here are the steps
$$
\frac{5x}{2x^{2}+5x+1}=\frac{1}{3}
$$
$$
15x=2x^{2}+5x+1
$$
$$
10x=2x^{2}+1
$$
What would we get if we divide both sides of the last equation by $2x$?
| {
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Catagorising a Differential Equation I have
$$
\frac{d^{2}}{d\epsilon^{2}}g^{\star}+\frac{\left(R^{2}+3\epsilon^{2}\right)}{\epsilon\left(R^{2}-\epsilon^{2}\right)}\frac{d}{d\epsilon}g^{\star}+\frac{\left(5R^{2}+3\epsilon^{2}\right)}{\left(R^{2}-\epsilon^{2}\right)^{2}}g^{\star}=0
$$
and I wonder if anyone could tell me some facts about it.
I know its linear, second order etc, that its Fuchsian (with its three regular singular points) and that it can be transformed (therefore) into the hypergeometric differential equation. Perhaps there's a DE expert who can tell me a bit more about its characterisation. I have (I think) both linearly independent solutions.
| Assume $R\neq0$ for the key case:
Let $r=\epsilon^2$ ,
Then $\dfrac{dg^\star}{d\epsilon}=\dfrac{dg^\star}{dr}\dfrac{dr}{d\epsilon}=2\epsilon\dfrac{dg^\star}{dr}$
$\dfrac{d^2g^\star}{d\epsilon^2}=\dfrac{d}{d\epsilon}\left(2\epsilon\dfrac{dg^\star}{dr}\right)=2\epsilon\dfrac{d}{d\epsilon}\left(\dfrac{dg^\star}{dr}\right)+2\dfrac{dg^\star}{dr}=2\epsilon\dfrac{d}{dr}\left(\dfrac{dg^\star}{dr}\right)\dfrac{dr}{d\epsilon}+2\dfrac{dg^\star}{dr}=2\epsilon\dfrac{d^2g^\star}{dr^2}2\epsilon+2\dfrac{dg^\star}{dr}=4\epsilon^2\dfrac{d^2g^\star}{dr^2}+2\dfrac{dg^\star}{dr}=4r\dfrac{d^2g^\star}{dr^2}+2\dfrac{dg^\star}{dr}$
$\therefore4r\dfrac{d^2g^\star}{dr^2}+2\dfrac{dg^\star}{dr}+\dfrac{2(R^2+3r)}{R^2-r}\dfrac{dg^\star}{dr}+\dfrac{5R^2+3r}{(R^2-r)^2}g^\star=0$
$\dfrac{d^2g^\star}{dr^2}+\left(\dfrac{1}{2r}-\dfrac{3r+R^2}{2r(r-R^2)}\right)\dfrac{dg^\star}{dr}+\dfrac{3r+5R^2}{4r(r-R^2)^2}g^\star=0$
$\dfrac{d^2g^\star}{dr^2}+\left(\dfrac{1}{r}-\dfrac{2}{r-R^2}\right)\dfrac{dg^\star}{dr}+\left(\dfrac{5}{4R^2r}-\dfrac{5}{4R^2(r-R^2)}+\dfrac{2}{(r-R^2)^2}\right)g^\star=0$
Let $g^\star=(r-R^2)u$ ,
Then $\dfrac{dg^\star}{dr}=(r-R^2)\dfrac{du}{dr}+u$
$\dfrac{d^2g^\star}{dr^2}=(r-R^2)\dfrac{d^2u}{dr^2}+\dfrac{du}{dr}+\dfrac{du}{dr}=(r-R^2)\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}$
$\therefore(r-R^2)\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}+\left(\dfrac{1}{r}-\dfrac{2}{r-R^2}\right)\left((r-R^2)\dfrac{du}{dr}+u\right)+\left(\dfrac{5}{4R^2r}-\dfrac{5}{4R^2(r-R^2)}+\dfrac{2}{(r-R^2)^2}\right)(r-R^2)u=0$
$(r-R^2)\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}+\left(\dfrac{r-R^2}{r}-2\right)\dfrac{du}{dr}+\left(\dfrac{1}{r}-\dfrac{2}{r-R^2}\right)u+\left(\dfrac{5(r-R^2)}{4R^2r}-\dfrac{5}{4R^2}+\dfrac{2}{r-R^2}\right)u=0$
$(r-R^2)\dfrac{d^2u}{dr^2}+\dfrac{r-R^2}{r}\dfrac{du}{dr}-\dfrac{u}{4r}=0$
$r(r-R^2)\dfrac{d^2u}{dr^2}+(r-R^2)\dfrac{du}{dr}-\dfrac{u}{4}=0$
Which reduces to Gaussian hypergeometric equation.
| {
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Verification of Laurent Series calculation I tried to calculate the Laurent series of these functions but I have no way to verify my answers.
i)
$$
\begin{align}
f(z)=\frac{e^{z^2 }-1}{z^4}, \mathbb{D}=\mathbb{C} \backslash \{0\}
\end{align}
$$
We know that $ e^z=\sum_{n=0}^\infty(\frac{1}{n!})z^n $ , hence
$$
f(z)=\frac{e^{z^2 }-1}{z^4}=\frac{1}{z^4}(-1+e^{z^2})=\frac{1}{z^4}(-1+\sum_{n=0}^\infty(\frac{1}{n!})(z^2)^n)=\\
=\frac{1}{z^4}(-1+\sum_{n=0}^\infty(\frac{1}{n!})z^{2n})=\frac{1}{z^4}\sum_{n=1}^\infty(\frac{1}{n!})z^{2n}=\\
\sum_{n=1}^\infty(\frac{1}{n!})z^{2n-4}
$$
ii)
$$
\begin{align}
f(z)=\frac{Logz}{(z-1)^2}, \mathbb{D}=\{ z\in \mathbb{C}:0<|z-1|<1\}
\end{align}
$$
We know that $Logz=-\sum_{n=1}^\infty\frac{(-1)^n}{n}(z-1)^n$, for $|z-1|<1$ ,hence
$$
f(z)=\frac{Logz}{(z-1)^2}=\frac{1}{(z-1)^2}\left(- \sum_{n=1}^\infty\frac{(-1)^n}{n}(z-1)^n \right)= \sum_{n=1}^\infty\frac{(-1)^n}{n}(z-1)^{n-2}
$$
iii)
$$
\begin{align}
f(z)=z^6\cos^2{z^{-2}},z_0=0, \mathbb{D}=\mathbb{C} \backslash \{0\}
\end{align}
$$
We know that $$\cos z=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}z^{2n}$$ therefore
$$\cos(\frac{1}{z})=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}(\frac{1}{z})^{2n}$$ and
$$\cos(\frac{1}{z^2})=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}(\frac{1}{z^2})^{2n}$$
We also know that when we a function to the power of 2, its correspondent series will be the derivative of the original series, so:
$$\cos^2(\frac{1}{z^2})=1+\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}2^{2n-1}(\frac{1}{z^2})^{2n}$$ (actually my result here was slightly diferrent, but wolframalfa returned this)
Finally, $$f(z)=z^6+ \left( 1+\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}2^{2n-1}(\frac{1}{z^2})^{2n} \right)$$
iv)
$$f(z)=\frac{1-2z}{z^2-z},\mathbb{D}=\{ z\in \mathbb{C}:|z-1|>1\} $$
$$f(z)=\frac{1-2z}{z^2-z}=\frac{1-2z}{z(z-1)}=-\frac{1}{z}-\frac{1}{z-1}=-\frac{1}{z}+\frac{1}{1-z}==-\frac{1}{z}+\sum_{n=0}^\infty z^n$$
v)
$$f(z)=\frac{1}{1+z^2} , \mathbb{D}=\{ z\in \mathbb{C}:1<|z-2i|<3\} $$
$$f(z)=\frac{1}{1+z^2}=\frac{1}{(i-z)(i+z)}=\frac{1}{(2i)}\frac{1}{(i-z)}+\frac{1}{(2i)}\frac{1}{(i+z)}$$
Let $z=2i+u$ so $$\frac{1}{(2i)}\frac{1}{(i-(2i+u))}+\frac{1}{(2i)}\frac{1}{(i+2i+u)}=\frac{1}{(2i)}\frac{1}{(-i-u)}+\frac{1}{(2i)}\frac{1}{(3i+z)}=\\
\frac{1}{(2i)}\frac{1}{u}\frac{-1}{(1-\frac{1}{iu})}+\frac{1}{(2i)}\frac{1}{3i}\frac{1}{(1-\frac{iu}{3})}=\\
\frac{1}{(2i)}\frac{1}{u}\sum_{n=0}^\infty(\frac{1}{iu})^n-\frac{1}{(6)}\sum_{n=0}^\infty(\frac{iu}{3})^n=\\
\frac{1}{(2i)}\frac{1}{(z-2i)}\sum_{n=0}^\infty(\frac{1}{i(z-2i)})^n-\frac{1}{(6)}\sum_{n=0}^\infty(\frac{i(z-2i)}{3})^n
$$
If anyone will take the time to verify my results and point my errors I will be grateful because I have no other means of checking their validity.
Sorry for the long post.
| You first computation is correct. The second is mostly correct, only at the end you dropped a minus sign. (It might have been better to absorb it into the sign in the sum and write $(-1)^{n-1}$ there rather than $(-1)^n$.)
In the third,
We also know that when we a function to the power of 2, its correspondent series will be the derivative of the original series
is wrong, consider for example polynomials to see that that generally isn't the case.
The square of $\sum\limits_{n=0}^\infty a_n w^n$ is the Cauchy product of the series with itself,
$$\sum_{n=0}^\infty \left(\sum_{k=0}^n a_k a_{n-k}\right) w^n,$$
and $\sum\limits_{k=0}^n a_k a_{n-k}$ rarely is equal to $(n+1)a_{n+1}$. For the cosine, a trigonometric identity makes the computation of the power series for $\cos^2 w$ easy:
$$\cos (2w) = \cos^2 w - \sin^2 w = 2\cos^2 w - 1,$$
and hence $\cos^2 w = \frac{1}{2}(1+\cos (2w))$. Thus
$$\cos^2 \frac{1}{z^2} = \frac{1}{2}\left(1+\cos \frac{2}{z^2}\right) = \frac{1}{2} + \frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \frac{2^{2n}}{z^{4n}} = 1 + \sum_{n=1}^\infty \frac{(-1)^n2^{2n-1}}{(2n)!z^{4n}}.$$
Multiplying that with $z^6$ yields the result.
In part iv), you need to expand the function in powers of $z-1$ since the domain is an annulus with centre $1$.
$$\frac{1-2z}{z^2-z} = \frac{1}{z-1}\left(\frac{1}{z} - 2\right) = (z-1)^{-1}\left(\frac{1}{(z-1)+1}-2\right).$$
Now expand $\frac{1}{(z-1)+1}$ into a geometric series, keeping in mind that $\lvert z-1\rvert > 1$ in the domain.
In the last, you made a sign error in your partial fraction decomposition,
$$\frac{1}{1+z^2} = \frac{1}{(z-i)(z+i)} = \frac{1}{2i}\left(\frac{1}{z-i} - \frac{1}{z+i}\right).$$
From then on, you have once typed $z$ where it ought to have been $u$, but made no real mistake. So to obtain the correct Laurent series, you only need to adjust the signs to match the correct partial fraction decomposition.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving determinants using properties of determinants $$\begin{vmatrix}
1 & a^2+bc & a^3\\
1 & b^2+ca & b^3\\
1 & c^2+ab & c^3
\end{vmatrix}
= (a-b)(b-c)(c-a)(a^2+b^2+c^2)$$
we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calculation to apply so was hoping for an hint.
Edit: just tried the problem and here is how I have done it
$$\begin{vmatrix}
1 & a^2+bc & a^3\\
1 & -(a^2+b^2)+c(a-b) & -(a^3-b^3)\\
1 & c^2-a^2-b(c-a) & c^3-a^3
\end{vmatrix}$$
then
$$\begin{vmatrix}
1 & a^2+bc & a^3\\
0 & -((a-b)(a+b))+c(a-b) & -((a-b)(a^2+ab+b^2))\\
0 & (c-a)(c+a)-b(c-a) & (c-a)(c^2+ca+a^2)
\end{vmatrix}$$
then
$$(a-b)(c-a)\begin{vmatrix}
1 & a^2+bc & a^3\\
0 & a+b+c & -(a^2+ab+b^2)\\
0 & a-b+c & (c^2+ca+a^2)
\end{vmatrix}$$
then
$$(a-b)(c-a)\begin{vmatrix}
a+b+c & -(a^2+ab+b^2)\\
a-b+c & (c^2+ca+a^2)
\end{vmatrix}$$
then
$$(a-b)(c-a) * ( (a+b+c)(c^2+ca+a^2)-(a-b+c)(-(a^2+ab+b^2)) )$$
Here I am Confused on how to multiply them and get the answer
note: typed because I own a very bad handwriting
Thank you every one for your help
| The solution-attempt in the question is excellent (after your edit)!
But unfortunately you made a small mistake. After taking out the factor $(a-b)(c-a)$, the middle element should be $c-a-b$, not $a+b+c$. If you fix that, the remaining determinant can be expanded and easily factorised.
| {
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Taking Calculus in a few days and I still don't know how to factorize quadratics Taking Calculus in a few days and I still don't know how to factorize quadratics with a coefficient in front of the 'x' term. I just don't understand any explanation. My teacher gave up and said just use the formula to find the roots or something like that..
Can someone explain to me simply how I would step by step factorize something like $4x^2 + 16x - 19$ ?
| The term $4x^2+16x$ is almost square of $2x+4$, more precisely
$$
4x^2+16x=(2x+4)^2-4^2.
$$
Therefore, we have
\begin{align}
4x^2+16x-19&=(2x+4)^2-4^2-19\\
&=(2x+4)^2-35\\
&=(2x+4)^2-\left(\sqrt{35}\right)^2.
\end{align}
Knowing that
$$
p^2-q^2=(p-q)(p+q),\tag1
$$
then we have
\begin{align}
4x^2+16x-19
&=\left(2x+4-\sqrt{35}\ \right)\left(2x+4+\sqrt{35}\ \right).\qquad\blacksquare
\end{align}
Addendum :
Here is a general approach to factorize a quadratic equation. Suppose that we want to factorize quadratic equation
$$
ax^2+bx+c=0.\tag2
$$
Now, multiplying $(2)$ by $4a$ yields
$$
4a^2x^2+4abx+4ac=0.\tag3
$$
The term $4a^2x^2+4abx$ is almost square of $2ax+b$, more precisely
$$
4a^2x^2+4abx=(2ax+b)^2-b^2,\tag4
$$
then $(3)$ turns out to be
\begin{align}
4a^2x^2+4abx+4ac&=(2ax+b)^2-b^2+4ac\\
&=(2ax+b)^2-(b^2-4ac)\\
&=(2ax+b)^2-\left(\sqrt{b^2-4ac}\right)^2.
\end{align}
Using $(1)$, we have
$$
4a^2x^2+4abx+4ac=\left(2ax+b+\sqrt{b^2-4ac}\ \right)\left(2ax+b-\sqrt{b^2-4ac}\ \right).\tag5
$$
Final step, dividing $(5)$ by $4a$ yields
$$
ax^2+bx+c=\color{blue}{\frac{\left(2ax+b+\sqrt{b^2-4ac}\ \right)\left(2ax+b-\sqrt{b^2-4ac}\ \right)}{4a}}.
$$
The process might look complicated, but once you understand the logic behind the process, especially for $(4)$, it will not be necessary anymore to memorize every step and your hand will automatically drive you to factorize every quadratic equation.
| {
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"source": "stackexchange",
"question_score": "13",
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Rationalized limit denominator, still undefined (divide by zero), how to solve? I am trying to solve:
$$\lim_{x \to 2}\frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \sqrt{5x-1}}$$
My first step is to multiply by the conjugate to rationalize the denominator.
$$\lim_{x \to 2}\frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \sqrt{5x-1}} \cdot \frac{\sqrt{4x+1} + \sqrt{5x-1}}{\sqrt{4x+1} + \sqrt{5x-1}}$$
Which gives
$$\lim_{x \to 2}\frac
{(\sqrt{x+2} - \sqrt{3x-2}) \cdot (\sqrt{4x+1} + \sqrt{5x-1})}
{({4x+1}) - ({5x-1})} $$
Simplifying denominator
$$\lim_{x \to 2}\frac
{(\sqrt{x+2} - \sqrt{3x-2}) \cdot (\sqrt{4x+1} + \sqrt{5x-1})}
{2-x} $$
Substituting $2$ for $x$ in the denominator and it's zero
$$\lim_{x \to 2}\frac
{(\sqrt{x+2} - \sqrt{3x-2}) \cdot (\sqrt{4x+1} + \sqrt{5x-1})}
{2-2} $$
Did I make a calculation error, if so where, and / or did I use the wrong approach, if so what's a working approach?
| Since we have
$$\begin{align}\sqrt{x+2}-\sqrt{3x-2}&=\frac{(\sqrt{x+2}-\sqrt{3x-2})(\sqrt{x+2}+\sqrt{3x-2})}{\sqrt{x+2}+\sqrt{3x-2}}\\&=\frac{-2(x-2)}{\sqrt{x+2}+\sqrt{3x-2}}\end{align}$$
$$\begin{align}\sqrt{4x+1}-\sqrt{5x-1}&=\frac{(\sqrt{4x+1}-\sqrt{5x-1})(\sqrt{4x+1}+\sqrt{5x-1})}{\sqrt{4x+1}+\sqrt{5x-1}}\\&=\frac{-(x-2)}{\sqrt{4x+1}+\sqrt{5x-1}}\end{align}$$
We have
$$\frac{\sqrt{x+2}-\sqrt{3x+2}}{\sqrt{4x+1}-\sqrt{5x-1}}=\left(\frac{-2(x-2)}{\sqrt{x+2}+\sqrt{3x-2}}\right)\div\left(\frac{-(x-2)}{\sqrt{4x+1}+\sqrt{5x-1}}\right)$$$$=\left(\frac{-2\color{red}{(x-2)}}{\sqrt{x+2}+\sqrt{3x-2}}\right)\times\left(\frac{\sqrt{4x+1}+\sqrt{5x-1}}{-\color{red}{(x-2)}}\right)=2\cdot\frac{\sqrt{4x+1}+\sqrt{5x-1}}{\sqrt{x+2}+\sqrt{3x-2}}$$$$\to 2\cdot \frac{\sqrt 9+\sqrt 9}{\sqrt 4+\sqrt 4}=3\ (x\to 2).$$
| {
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Determine the irrational numbers $x$ such that both $x^2+2x$ and $x^3-6x$ are rational numbers I did not make any progress. The problem is from RMC 2008.
The only idea that I have is:
Try to find sets of irrational numbers such that every number in the set multiplied by another number in the set yields a rational number.
| I add yet another solution hopefully a simple one.
Since $x^2+2x$ is rational so too is $x^2+2x+1=(x+1)^2$.
Let $y=x+1$, so $y^2$ is rational, then $x^3-6x=y(y^2-3)-3y^2+5$ is rational, which means that $y(y^2-3)$ is rational.
Now since $y$ is irrational and $y^2-3$ rational we must have $y^2-3=0$, thus $y=\pm \sqrt{3}$ and $x=-1\pm\sqrt{3}$
| {
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"source": "stackexchange",
"question_score": "5",
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specific magma examples Give an example of a magma $S$ such that $S$ has a zero and $S$ has a left zero divisor that is not a right zero divisor
an example of a magma with an identity such that there is an element with exactly $2$ left inverses but only one right inverse
For the first, I was thinking $\{0,a,b\}$ where $ab=0$, $ba=a$, $b^2=b$, $a^2=a$.
Thanks
| Your answer to the first question may be right (thanks @celtschk for commenting):
$$
\begin{array}{c|ccc}
\cdot & 0 & a & b\\
\hline
0 & 0 & 0 & 0\\
a & 0 & a & 0\\
b & 0 & a & b
\end{array}
$$
Here $0$ denotes an absorbing element, $a$ is only a left zero divisor, while $b$ is only a right zero divisor.
Second question: here they ask for an identity, let us call it $1$. From your comment
@drhab a^2=a, ab=b, ba=1, b^2=a?
I get
$$
\begin{array}{c|ccc}
\bullet & 1 & a & b\\
\hline
1 & 1 & a & b\\
a & a & a & b\\
b & b & 1 & a
\end{array}
$$
but to obtain
an element with exactly 2 left inverses but only one right inverse
we must have, in the Cayley table, $1$ appearing exactly twice in the column labeled by the element and exactly once in the line labeled by the same element, for example:
$$
\begin{array}{c|ccc}
\bullet & 1 & a & b\\
\hline
1 & 1 & a & b\\
a & a & 1 & b\\
b & b & 1 & a
\end{array}
$$
In this last table (I got this from the previous changing only one cell, $a\bullet a$), $1$ is the identity, $a$ and $b$ are both left inverses of $a$, while only $a$ is a right inverse of $a$.
With four or more elements in the underlying set we could get more various examples, like:
$$
\begin{array}{c|cccccc}
* & 1 & 2 & 3 & 4 & 5 & 6\\
\hline
1 & \color{blue}{1} & \color{blue}{2} & \color{blue}{3} & \color{blue}{4} & \color{blue}{5} & \color{blue}{6}\\
2 & \color{blue}{2} & 3 & 2 & 3 & 2 & 3\\
3 & \color{blue}{3} & 3 & 2 & 3 & 2 & \color{red}{1}\\
4 & \color{blue}{4} & 3 & 2 & 3 & 2 & \color{red}{1}\\
5 & \color{blue}{5} & 3 & 2 & 3 & 2 & 3\\
6 & \color{blue}{6} & 3 & 2 & 3 & \color{red}{1} & 3\\
\end{array}
$$
Here, $1$ is the identity (in blue), $3$ and $4$ are left inverses of $6$ and $5$ is right inverse of $6$ (in red).
| {
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Find a vector that bisects the smaller of the two angles formed by vectors <3,4> and <5,-12>. The solution is <8,-1>. I tried finding the angle between the two vectors, but wasn't sure what to do next.
| The cosines of the angles have to be the same and positive (for the smaller angle). Therefore
$$\frac{3 a + 4 b}{\sqrt{3^2 + 4^2} \cdot \sqrt {a^2 + b^2} } = \frac{5 a + (-12) b}{\sqrt{5^2 + (-12)^2} \cdot \sqrt {a^2 + b^2} }>0$$ or
$$\frac{3 a + 4 b}{5 } = \frac{5 a + (-12) b}{13 }>0$$
Check that $(a,b) = (8,-1)$ is a solution. Indeed:
$$\frac{3 \cdot 8 + 4 \cdot (-1)}{5}= 4$$ and
$$\frac{5 \cdot 8 + (-12) \cdot (-1)}{13}=4$$
| {
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Proof by induction that $a_0 = 1, a_1 = 1, a_n=2a_{n-1} + 3a_{n-2}$ satisfies $a_n = \frac12 (3^n) + \frac12 (-1)^n$ The question:
The terms of a sequence are given recursively as
\begin{cases}
a_0 = 1,\\ a_1 = 1 \\a_n=2a_{n-1} + 3a_{n-2} \quad\text{ for } n \geq 2
\end{cases}
prove by mathematical induction
$a_n = \frac12(3^n) +\frac12(-1)^n$ for all $n \geq 0$
I've started by
proving that the formula works e.g
$$a_0= 1\implies a_0 = \frac12(3^0)+\frac12(-1)^0 = 1 \\a_1=1\implies a_1 = \frac12(3^1) + \frac12(-1)^{1} = 1$$
Therefore $p(1)$ and $p(2)$ are true
Now I have to do the hard part.. the inductive step
suppose $p(1), p(2),\ldots, p(k)$, that is $a_{k+1} = 2a_{k} + 3a_{k-1}$
Im not sure where to really go from here.
| We have, by induction hypotheses, that :
$a_k=\frac 12(3^k)+\frac 12(-1)^k$
and :
$a_{k-1}=\frac 12(3^{k-1})+\frac 12(-1)^{k-1}$.
If we "plug them" into :
$a_{k+1}=2a_k+3a_{k-1}$
we get :
$$a_{k+1}=2\left[\frac 12(3^k)+ \frac 12(-1)^k\right]+3\left[\frac 12(3^{k-1})+ \frac 12(-1)^{k-1}\right] =$$
$$= 3^k+(-1)^k+3 \frac 12 3^{k-1}+ \frac 32(-1)^{k-1} =$$
$$=3^k+ \frac 12 3^k+(-1)(-1)^{k-1}+ \frac 32(-1)^{k-1}=\frac 32 3^k+ \frac 12 (-1)^{k-1}=$$
$$=\frac 12 3^{k+1}+ \frac 12 (-1)^{k-1}=\frac 12 3^{k+1}+ \frac 12 (-1)^{k-1}(-1)^2=$$
$$=\frac 12 3^{k+1}+ \frac 12 (-1)^{k+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/927549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Integral $\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}dx$ I have a problem with the following integral:
$$
\int_{0}^{1}\ln\left(\,3 + x \over 3 - x\,\right)\,
{{\rm d}x \over \,\sqrt{\,x\left(\,1 - x\,\right)\,}\,}
$$
The first idea was to use the integration by parts because
$$
\int{{\rm d}x \over \,\sqrt{x\left(\,1 - x\,\right)\,}\,}
=\arcsin\left(\,2x - 1\,\right) + C
$$
but what would be the next step is not clear. Another idea would be expand
$\ln\left(\,\cdot\right)$ into Taylor series but it seems to be even worse option.
So, what are the other options?
| Split the integral into two forms by expanding the logarithm function
$$
\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx=\int_{0}^1\frac{\ln(3+x)}{\sqrt{x(1-x)}}\ dx-\int_{0}^1\frac{\ln(3-x)}{\sqrt{x(1-x)}}\ dx
$$
Let $t=\sqrt{x}\ \rightarrow\ dt=\dfrac{dx}{2\sqrt{x}}$, we have
$$
2\int_{0}^1\frac{\ln(3+t^2)}{\sqrt{1-t^2}}\ dt-2\int_{0}^1\frac{\ln(3-t^2)}{\sqrt{1-t^2}}\ dt
$$
Let $t=\sin\theta\ \rightarrow\ dt=\cos\theta\ d\theta$, we have
$$
2\int_{0}^{\pi/2}\ln(3+\sin^2\theta)\ d\theta-2\int_{0}^{\pi/2}\ln(3-\sin^2\theta)\ d\theta
$$
Using identity $\sin^2\theta=\dfrac12(1-\cos2\theta)$ and setting $y=2\theta$, we have
$$
\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx=\int_{0}^{\pi}\ln\left(7-\cos y\right)\ dy-\int_{0}^{\pi}\ln\left(5+\cos y\right)\ dy
$$
We will use Feynman's way to evaluate integral above. Consider
$$
I(k)=\int_{0}^{\large\pi}\ln\left(k\pm\cos y\right)\ dy
$$
then
\begin{align}
I'(k)&=\int_{0}^{\large\pi}\frac{1}{k\pm\cos y}\ dy
\end{align}
Using formula
$$
\int_0^\pi\frac{1}{a^2+b^2-2ab\cos x}dx=\frac{\pi}{a^2-b^2}
$$
Setting $b=\pm\dfrac{1}{2a}$, we have
$$
\int_0^\pi\frac{1}{a^2+\frac{1}{4a^2}\pm\cos x}dx=\frac{4\pi a^2}{4a^4-1}
$$
Clearly $k=a^2+\dfrac{1}{4a^2}\ \rightarrow\ a^2=\dfrac{k+\sqrt{k^2-1}}{2}$ and $dk=\dfrac{4a^4-1}{2a^3}da$, then
\begin{align}
I(k)&=\int\int_{0}^{\large\pi}\frac{1}{k\pm\cos y}\,dy\,dk\\
&=\int\frac{4\pi a^2}{4a^4-1}\cdot\dfrac{4a^4-1}{2a^3}da\\
&=2\pi\int\frac{1}{a}\,da\\
&=2\pi\ln(a)+C\\
&=\pi\ln(a^2)+C\\
&=\pi\ln\left(\dfrac{k+\sqrt{k^2-1}}{2}\right)+C\\
\end{align}
Finally
\begin{align}
\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx&=I(7)-I(5)\\
&=\pi\ln\left(\dfrac{7+\sqrt{7^2-1}}{5+\sqrt{5^2-1}}\right)\\
&=\pi\ln\left(\dfrac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\\
\end{align}
Yeayyy, I'm done! (>‿◠)✌
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Find solution of equation $(z+1)^5=z^5$ I attempt to solve the equation
$(z+1)^5=z^5$.
My first approach is to expand the left hand side but ı get more complicated equation. So I couldn't go further. Secondly, I write equation as, since $z\neq0$,
$(\frac{z+1}{z})^5=1$, put $\xi=\frac{z+1}{z}$
and attempt to solve equivalent equation $\xi^5=1$. But this time it requires more computation to find solutions $z$. Can anyone suggest a simple way to solve this equation?
Thanks in advance..
| To exploit symmetry, put $z=w-\frac 12$, which gives $z+1=w+\frac 12$.
Solving:
$$\begin{align}(z+1)^5&=z^5\\
\left(w+\frac 12\right)^5&=\left(w-\frac 12\right)^5\\
2\left[5w^4\left(\frac12\right)+10w^2\left(\frac 12\right)^3+\left(\frac 12\right)^5\right]&=0\\
80w^4+40w^2+1&=0\\
w^2&=\frac{-40\pm\sqrt{1600-320}}{160}\\
&=-\frac14 \left(1+\frac25\sqrt5\right)\\
w&=\pm\frac i2\sqrt{1+\frac25\sqrt5}\\
z&=-\frac12 \pm\frac i2\sqrt{1+\frac25\sqrt5}
\end{align}$$
Alternative method using comments on the OQ:
$$(z+1)^5=z^5\\
\left(\frac{z+1}z\right)^5=1\\
\left(1+\frac 1z\right)^5=1\\
1+\frac 1z=e^{i2n\pi/5}\\
z=\frac 1{e^{i2n\pi/5}-1}
$$
where $n\in\Bbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/928735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the integral $\int_{0}^{\frac{\pi}{4}}\frac{\sin{x}\cos{x}}{\sin{x}+\cos{x}}dx$ find the integeral
$$\int_{0}^{\frac{\pi}{4}}\dfrac{\sin{x}\cos{x}}{\sin{x}+\cos{x}}dx$$
I know
$$\dfrac{2\sin{x}\cos{x}}{\sin{x}+\cos{x}}=\dfrac{(\sin{x}+\cos{x})^2-1}{\sin{x}+\cos{x}}=\sin{x}+\cos{x}-\dfrac{1}{\sin{x}+\cos{x}}$$
and
$$\int(\sin{x}+\cos{x})dx=\sin{x}-\cos{x}+C$$
and
$$\int\dfrac{1}{\sin{x}+\cos{x}}dx=\int\dfrac{1}{\sqrt{2}}\csc{(x+\dfrac{\pi}{4})}dx=\dfrac{1}{\sqrt{2}}\cdot(-\ln{(\cot{x}-\csc{x})})+C$$
I think this intgral have other methods.because this is simple form,
so My Question:
can you have other methods?
| Here is a very slightly (and I mean very slightly) different approach compared to those already given.
Recognising
$$\sin x + \cos x = \sqrt{2} \sin \left (x + \frac{\pi}{4} \right ).$$
Now
\begin{align*}
(\sin x + \cos x)^2 &= \sin^2 x + 2 \sin x \cos x + cos^2 x\\
\Rightarrow 2 \sin^2 \left (x + \frac{\pi}{4} \right ) &= 1 + 2 \sin x \cos x\\
\Rightarrow \sin x \cos x &= \sin^2 \left (x + \frac{\pi}{4} \right ) - \frac{1}{2}.
\end{align*}
So our integral becomes
\begin{align*}
\int^{\frac{\pi}{4}}_0 \frac{\sin x \cos x}{\sin x + \cos x} \,dx &= \int^{\frac{\pi}{4}}_0 \frac{\sin^2 \left (x + \frac{\pi}{4} \right ) - \frac{1}{2}}{\sqrt{2} \sin \left (x + \frac{\pi}{4} \right )} \, dx\\
&= \frac{1}{\sqrt{2}} \int^{\frac{\pi}{4}}_0 \sin \left (x + \frac{\pi}{4} \right ) \, dx - \frac{1}{2 \sqrt{2}} \int^{\frac{\pi}{4}}_0 \text{cosec} \left (x + \frac{\pi}{4} \right ) \, dx\\
&= \frac{1}{\sqrt{2}} \left [- \cos \left (x + \frac{\pi}{4} \right ) \right ]^{\pi/4}_0 + \frac{1}{2\sqrt{2}} \ln \left (\text{cosec} \left (x + \frac{\pi}{4} \right ) + \cot \left (x + \frac{\pi}{4} \right ) \right ) \Big{|}^{\pi/4}_0\\
&= \frac{1}{2} - \frac{1}{2\sqrt{2}} \ln (1 + \sqrt{2}).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/928816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
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Find two real numbers x and y such that x, y, x + y, x^2, y^2, x^2 - y^2, x^3, y^3, x^4, y^4 are all irrational but x^2 + y^2 is rational. Find two real numbers $x$ and $y$ such that $x, y, x + y, x^2 , y^2 , x^2 - y^2, x^3, y^3, x^4, y^4$ are all irrational but $x^2 + y^2$ is rational.
What is the approach to solve this problem? It cannot be guess and check, right?
(Shout-outs to CI students. D is for diploma!)
| If we want to avoid transcendentals, use $x=\sqrt{3}+\sqrt{2}$ and $y=\sqrt{3}-\sqrt{2}$.
The price is that we need to work a little harder to prove the irrationality of $x$, $y$, $x+y$, $x^2$, $y^2$, $x^2-y^2$, $x^3$, $y^3$, $x^4$ and $y^4$. But not much harder, there is a lot of symmetry.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$? Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align}$$
I was able to evaluate the limit as follows:
$$\begin{align}
\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ] &\sim \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{\dfrac x2 + \dfrac{x^2} 8}{\dfrac x2 + \dfrac{x^2} 8}} - 1\right ] =\\
&= \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{1 + \frac{2x^2}{4x - x^2}} - 1\right ] \sim\\
&\sim \lim_{x \to 0} \frac{2x^2}{12x^2 - 3x^3} = \frac 16
\end{align}$$
What are other ways to evaluate it? Maybe pure algebraically? I tried to rationalize the denominator, but got stuck at some point...
| Using:
$$
1 -\sqrt{1-x} = \frac{x}{1+\sqrt{1-x}}\qquad \frac{1}{\sqrt{1+x}-1} = \frac{\sqrt{1+x}+1}{x}
$$
we have:
$$
\frac{1 -\sqrt{1-x}}{\sqrt{1+x}-1} = \frac{\sqrt{1+x}+1}{1+\sqrt{1-x}} = 1 + \frac{\sqrt{1+x}-\sqrt{1-x}}{1+\sqrt{1-x}} = 1 + \underbrace{\frac{2x}{\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(1+\sqrt{1-x}\right)}}_{w}
$$
And using
$$
\sqrt[3]{1+w} - 1 = \frac{w}{1 + \sqrt[3]{1+w} + \sqrt[3]{\left(1+w\right)^2} }
$$
we have
$$
\lim_{x \to 0}\frac{1}{x} \left(\sqrt[3]{ \frac{1 -\sqrt{1-x}}{\sqrt{1+x}-1} } - 1\right) = \lim_{x \to 0} \frac{w}{x} \cdot \frac{1}{1 + \sqrt[3]{1+w} + \sqrt[3]{\left(1+w\right)^2} } = \frac{2}{4} \cdot \frac{1}{3} = \frac{1}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Existence of solution in $x,y \in (a,b)$ of $(\frac { a+b}2)^{x+y}=a^xb^y$ Let $a<b$ be positive real numbers , then is it true that there exist $x,y \in (a,b)$ such that
$ \bigg(\dfrac { a+b}2\bigg)^{x+y}=a^xb^y$ ?
| We can solve the equation by
$$(x(t),y(t)) = \left(\ln\left(\frac{2b}{a+b}\right)t,\ln\left(\frac{a+b}{2a}\right)t\right)$$
Now $\frac{a+b}{2}\geq a$ and therefore $\ln\left(\frac{a+b}{2a}\right)\geq 0$
and by $\frac{a+b}{2}\leq b$ we conclude $\ln\left(\frac{2b}{a+b}\right)\geq 0$.
Furthermore $\ln\left(\frac{a+b}{2a}\right) \geq \ln\left(\frac{2b}{a+b}\right)$
Thus for $$t \in \left(\frac{a}{\ln\left(\frac{2b}{a+b}\right)},\frac{b}{\ln\left(\frac{a+b}{2a}\right)}\right)$$
we have $x(t),y(t)\in (a,b)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $p > 5$ is a prime number, then the last digit of $p^4-1$ is $0$. If $p > 5$ is a prime number, then the last digit of $p^4-1$ is $0$ (ex.: $7^4-1=2400$).
How do I prove this?
| By Fermat theorem
$$p^4\equiv1\pmod{5}\\p^4-1\equiv0\pmod{5}\\p^4\equiv1\pmod{2}\\p^4-1\equiv0\pmod{2}\\p^4-1\equiv0\pmod{10}$$
Since $p$ is odd so is $p^4$ clearly it gives $1$ as remainder by $2$.Also since $2$ and $5$ are co-prime we have that $0$ is remainder of $2\cdot 5$ which gives that the last digit is $0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/935417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove $m\cos^2{\theta} + n\sin^2{\theta} < l \implies \sqrt{m}\cos^2{\theta} + \sqrt{n}\sin^2{\theta} < \sqrt{l} $ Prove that $m\cos^2{\theta} + n\sin^2{\theta} < l \implies \sqrt{m}\cos^2{\theta} + \sqrt{n}\sin^2{\theta} < \sqrt{l} $ for every $m, n, l >0$.
| This problem can be found in
LARSON, LOREN C., 1983: Problem-Solving Through Problems.
Springer-Verlag, p. 255, 267.
The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = \sqrt{x} $, is left as an exercise.
I will write here the solution based on the second method:
\begin{align*}
(\sqrt{m}\cos^2{\theta} + \sqrt{n}\sin^2{\theta})^2
&= m \cos^4{\theta} + 2 \sqrt{mn}\cos^2{\theta}\sin^2{\theta}+ n\sin^4{\theta} \\
& \leq m \cos^4{\theta} + (m + n)\cos^2{\theta}\sin^2{\theta}+ n\sin^4{\theta} \\
&= (m\cos^2{\theta} + n\sin^2{\theta})(\cos^2{\theta} + \sin^2{\theta}) \\
&= (m\cos^2{\theta} + n\sin^2{\theta})\\
&<l.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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When $x^2+6xy+y^2$ a square number? Find all natural numbers $x$ and $y$ such that $x^2+6xy+y^2$ is a square number.
For example, $(x,y)=(2,3)$ or $(x,y)=(3,10)$.
Obviously, we can consider $gcd(x,y)=1$.
| For non-trivial cases, $xy\ne0$
Let $x^2+6xy+y^2=(x+ky)^2$ where $k$ is any integer
$\iff y(6x+y)=y(2kx+k^2y)\implies6x+y=2kx+k^2y\iff x(6-2k)=y(k^2-1)$
So, $\dfrac x{k^2-1}=\dfrac y{6-2k}=m$(say an integer)
As $x,y>0$ if $m<0,$ we need $6-2k<0\iff k>3\ \ \ \ (1)$ and $k^2-1<0\iff -1<k<1\ \ \ \ (2)$
There can be no $k$ satisfying both $(1),(2)$
Similarly if $m>0,$ we need $6-2k>0\iff k<3$ and $k^2-1>0\iff k>1$ or $k<-1$
$k<3,k<-1\implies k<-1$
or $k<3,k>1\implies1<k<3\implies k=2$
So, there are infinitely natural values of $x,y$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Another parametric integral relating to hyperbolic function if $0<a\leq1$, then canwe get a closed form of
$$I(a)=\int_0^\infty\frac{x}{\tanh x}\frac{1}{\cosh^2(ax)}dx.$$
In fact,if $a=1$,$I(a=1)=\pi^2/8$.
| If you are interested in particular cases:
$$\begin{align}
I\left(\frac14\right) & = \pi^2 + 2 \\
I\left(\frac13\right) & = -12\operatorname{Li}_2\left( \frac{2}{i\sqrt{3}-1} \right) -12\operatorname{Li}_2\left( -\frac{2}{i\sqrt{3}+1} \right) - \frac{5\pi^2}{8} \\
I\left(\frac12\right) & = \frac{\pi^2}{4} + 1 \\
I\left(\frac34\right) & = -\frac{13\pi^2}{243} + \frac{32\pi\sqrt{3}}{81} + \frac29\\
I\left(1\right) & = \frac{\pi^2}{8} \\
\end{align}$$
There are also some closed-form for $a>1$:
$$\begin{align}
I\left(2\right) & = \frac{3\pi^2}{32} - \frac{\pi}{8} \\
I\left(3\right) & = \frac{49\pi^2}{648} - \frac{2\pi \sqrt{3}}{27}
\end{align}$$
I've got them by using CAS and some manual simplification.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying a proof by contradiction: if $a\equiv 1\bmod 5$, then $a^2\equiv 1\bmod5$ Prove the following either by Direct Proof or by Contraposition:
Suppose $a\in\mathbb{Z}$, if $a\equiv 1\pmod 5$, then $a^2\equiv 1\pmod5$
Suppose $a\equiv 1\pmod 5$
Then $5|\left(a-1\right)$, therefore $a-1=5k$
$a^2-1=\left(a-1\right)\left(a+1\right)=5k\left(a+1\right)$
Thus, $a^2-1=5m$ where $m=k\left(a+1\right)$
Hence $a^2\equiv1\pmod5$
So, that is basically a proof by Contraposition, right?
| If
$a\equiv 1\pmod 5$,
$a = 5n+1$
for some integer $n$.
Therefore
$a^2
= 25n^2+10n+1
= 5(5n+2)+1
$
so
$a^2\equiv 1\pmod 5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $5^2x^3-x^5 = x^3(x-5)(x+5)$ or $-x^3(5-x)(5+x)$ Geogebra's Factor function says that
$5^2x^3-x^5$
is
$-x^3(x-5)(x+5)$
but from what I do, it is positive, $x^3(5+x)(5-x)$
Note the x isnt in the same position
Am I wrong?
| Your answer and Geogebra's answers are equivalent.
$x^3(5+x)(5-x)=-x^3(5+x)(x-5)=-x^3(x+5)(x-5)$.
They just factored out a negative out of $(5-x)$ in order to make it $-(x-5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Exercise in algebra - Express different terms
We have $ a = \dfrac{1}{\sqrt{1-b^2}}, c = \sqrt{\dfrac{1+b}{1-b}}, 0 \leq b < 1 $
Express $b$ in terms of $a$, $b$ in terms of $c$, $c$ in terms of $a$ and $a$ in terms of $c$.
So I want to do this in the quickest and algebraically "cleanest" way. Here's my own attempt:
Part 1:
\begin{align*}
a\sqrt{1-b^2} &= 1\\
\sqrt{1-b^2} &= \dfrac{1}{a}\\
b^2 &= 1- \dfrac{1}{a^2}\\
b &= \sqrt{1 - \dfrac{1}{a^2}}
\end{align*}
Part 2:
\begin{align*}
c^2 &= \dfrac{1+b}{1-b}\\
b+1 &= c^2(1-b)\\
b+ bc^2 &= c^2 -1\\
b(c^2+1) &= c^2 - 1\\
b &= \dfrac{c^2-1}{c^2+1}
\end{align*}
3 and 4. I can't seem to do these. We want the denominator completely rationalized, etc. and I can't do this, I tried but my algebra becomes to hairy.
Are my solutions at 1/2 optimal, and how would we do 3/4 the best way?
| Now your 1 and 2 look fine.
For 3 : $$c=\frac{\sqrt{1+b}}{\sqrt{1-b}}\cdot\frac{\sqrt{1+b}}{\sqrt{1+b}}=\frac{1+b}{\sqrt{1-b^2}}=a(1+b)=a\left(1+\sqrt{1-\frac{1}{a^2}}\right)=a+\sqrt{a^2-1}.$$
For 4 : $$(c-a)^2=a^2-1\Rightarrow c^2-2ac=-1\Rightarrow a=\frac{c^2+1}{2c}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove or disprove $ p^{r+s}\mid q^{ke} - 1 \iff p^s \mid k$. Let $p$ be an odd prime and $q$ be a power of prime. Suppose $e := \min\{\, e \in \mathbb{N} : p \mid q^e - 1 \,\}$ exists. Put $r := \nu_p(q^e - 1)$ (that is, $p^r \mid q^e - 1$ and $p^{r+1} \nmid q^e - 1$). What I want to prove is the following:
$ \forall k \in \mathbb{N},~ \forall s \in \mathbb{N}_0,~p^{r+s} \mid q^{ke} - 1 \iff p^s \mid k.$
I feel that this is true (but perhaps some upper bounds may needed for $k$ and $s$). How can I prove or disprove this statement?
Example:
If $p = 3,~q = 5$ then $e = 2,~r = 1$.
*
*$5^{1 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 0}$
*$5^{2 \cdot 2} - 1 = 2^4 \cdot 3^{1 + 0} \cdot 13$
*$5^{3 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 1} \cdot 7 \cdot 31$
*$5^{4 \cdot 2} - 1 = 2^5 \cdot 3^{1 + 0} \cdot 13 \cdot 313$
*$5^{5 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 0} \cdot 11 \cdot 71 \cdot 521$
My attempt:
In case $(\Leftarrow)$,
$$\begin{align}
q^{ke} - 1 &= ((q^e - 1) + 1)^k - 1 \\
&= (q^e - 1)^k + k(q^e - 1)^{k - 1} + \dotsb + \binom{k}{2}(q^e - 1)^2 + k(q^e - 1).
\end{align}$$
I can see that the last term is divisible by $p^{r + s}$ from the assumption. But is it true for the former terms?
In case $(\Rightarrow)$, we may start by introducing the quotient and the remainder $k = ap^s + b~(0 \le b < p^s)$ and calculate similar expansion as before.
| I write an answer to reorganize the Slade's second answer for oneself.
What I would like to prove is the following proposition.
Proposition. Let $p$ be an odd prime. For every $t, k \in \mathbb{N}$, if $p \mid t - 1$ then
$$\nu_p\left(\frac{t^k - 1}{t - 1}\right) = \nu_p(k).$$
Once this has proved, the substitution $t = q^e$ gives an answer to the original question. Now for the proof.
Proof.
The proof is well-founded induction on $k$ with respect to strict divisibility.
In case $k = 2$. Since $2 \neq p \mid t - 1$ and
$ \frac{t^k - 1}{t - 1} = t + 1 = (t - 1) + 2$,
the desired identity $\nu_p(\frac{t^k - 1}{t - 1}) = 0 = \nu_p(k)$ holds.
In case $k$ is odd prime. Since $p \mid t - 1$, the congruence relation $t^m - 1 \equiv 0 \pmod{p}$ hold for every $m \in \mathbb{N}$. And hence $(t^m - 1)^2 \equiv 0 \pmod{p^2}$, which is equivalent to $t^{2m} + 1 \equiv 2t^m \pmod{p^2}$. Therefore, as $k$ is odd,
$$\begin{align*}
\frac{t^k - 1}{t - 1} &= t^{k - 1} + \dotsb + t + 1 \\
&= (t^{k - 1} + 1) + (t^{k - 2} + t) + \dotsb + (t^{(k + 1)/2} + t^{(k - 3)/2}) + t^{(t - 1)/2} \\
&= (t^{k - 1} + 1) + (t^{k - 3} + 1)t + \dotsb + (t^2 + 1)t^{(k - 3)/2} + t^{(k - 1)/2} \\
&\equiv 2t^{(k - 1)/2} + 2t^{(k - 3)/2}t + \dotsb + 2tt^{(k - 3)/2} + t^{(k - 1)/2} \pmod{p^2}\\
&= kt^{(k - 1)/2}.
\end{align*}$$
From $p \nmid t$, a fact that $k$ is a prime and this congruence relation, we conclude that
$\nu_p\left(\frac{t^k - 1}{t - 1}\right) = \nu_p(k)$.
In case $k$ is composite number. Suppose $k = ab$ for some $a, b \in \mathbb{N} \setminus \{1\}$. Then from induction,
$$\begin{align*}
\nu_p\left(\frac{t^k - 1}{t - 1}\right) &=
\nu_p\left(\frac{t^{ab} - 1}{t - 1}\right) \\ &=
\nu_p\left(\frac{(t^a)^b - 1}{t^a - 1}\right) + \nu_p\left(\frac{t^a - 1}{t - 1}\right) \\ &=
\nu_p(b) + \nu_p(a) = \nu_p(ab) = \nu_p(k).
\end{align*}$$
From induction on $k$, the proposition established. Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving the Volume of an Ellipsoid This is the question:
The solid generated by rotating the region inside the ellipse with equation
$$ \left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1 $$
around the $x$-axis is called an ellipsoid.
(a) Show that the ellipsoid has volume $\displaystyle \frac{4}{3} \pi a b^2.$
(b) What is the volume if the ellipse is rotated around the $y$-axis.
(I want to accomplish this using integrals and basic plane geometry. To give you an idea of how much I know about integrals I'm $4$ weeks into my calculus II course, which is my first exposure to integrals)
I started by drawing an ellipse on a Cartesian plane. The ellipse went from $a$ to $b$, I then rotated this ellipse around the $x$-axis to get an ellipsoid. Now to get the volume I have to find a cross-sectional area, so I noticed that the ellipsoid is really made up of a bunch of circles stacked along the $x$-axis. Where I'm stuck at right now is how I can find the radius of these circles, which would give me my integrand.
| We know that the volume of the solid generated by rotating the curve $y=f(x)\space $ about the x-axis is given as follows
$$V=\int_{x_1}^{x_2}\pi y^2dx$$
Hence, the volume of the solid generated by rotating the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \space$ about the x-axis is obtained by setting $y^2=\frac{b^2}{a^2} (a^2-x^2) \space$ in the above formula & applying the proper limits
$$V=\pi \int_{-a}^{a} \frac{b^2}{a^2} (a^2-x^2)dx= \frac{\pi b^2}{a^2} \int_{-a}^{a} (a^2-x^2)dx$$
$$=\frac{2\pi b^2}{a^2} \int_{0}^{a} (a^2-x^2)dx=\frac{2\pi b^2}{a^2} \left[a^2x-\frac{x^3}{3} \right]_{0}^{a}$$
$$=\frac{2\pi b^2}{a^2} \left[a^3-\frac{a^3}{3} \right]=\frac{2\pi b^2}{a^2} \left[\frac{2a^3}{3} \right]=\frac {4}{3}\pi ab^2$$
We know that the volume of the solid generated by rotating the curve $y=f(x)\space $ about the y-axis is given as follows
$$V=\int_{y_1}^{y_2}\pi x^2dy$$
Hence, the volume of the solid generated by rotating the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \space$ about the y-axis is obtained by setting $x^2=\frac{a^2}{b^2} (b^2-y^2) \space$ in the above formula & applying the proper limits
$$V=\pi \int_{-b}^{b} \frac{a^2}{b^2} (b^2-y^2)dy= \frac{\pi a^2}{b^2} \int_{-b}^{b} (b^2-y^2)dy$$
$$=\frac{2\pi a^2}{b^2} \int_{0}^{b} (b^2-y^2)dy=\frac{2\pi a^2}{b^2} \left[b^2y-\frac{y^3}{3} \right]_{0}^{b}$$
$$=\frac{2\pi a^2}{b^2} \left[b^3-\frac{b^3}{3} \right]=\frac{2\pi a^2}{b^2} \left[\frac{2b^3}{3} \right]=\frac {4}{3}\pi a^2 b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/946198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is the sequence $x_n = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} +\dots + \frac{\sin n}{2^n}$ Cauchy? $$x_{n} = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} + ... + \frac{\sin n}{2^n}$$
I came across this sequence while studying, and while it is convergent, I'm curious as to whether or not it is a Cauchy sequence. If so, what's the proof?
| For $n > m > 0$, we have
$\vert x_n - x_m \vert = \vert \sum_{m + 1}^n \dfrac{\sin i}{2^i} \vert \le \sum_{m + 1}^n \dfrac{\vert \sin i \vert}{2^i} \le \sum_{m + 1}^n \dfrac{1}{2^i}, \tag{1}$
since $\vert \sin i \vert \le 1$. But
$\sum_{m + 1}^n \dfrac{1}{2^i} = \dfrac{1}{2^{m + 1}} \sum_0^{n - m -1}\dfrac{1}{2^i} \le 2\dfrac{1}{2^{m + 1}} = \dfrac{1}{2^m}, \tag{2}$
since $\sum_0^\infty \dfrac{1}{2^i} = 2$. Thus by taking $m$ sufficiently large, we may make $\vert x_n - x_m \vert$ arbitrarily small; the sequence is Cauchy.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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number of matrices of rank 3? Let M be the space of all $4\times 3$ matrices with entries in the finite field of three elements.The number of matrices of rank 3 in M are?
A. $(3^4-3)(3^4-3^2)(3^4-3^3)$
B.$(3^4-1)(3^4-2)(3^4-3)$
C.$(3^4-1)(3^4-3)(3^4-3^2)$
D.$(3^4)(3^4-1)(3^4-2)$
which of the following is the answer??
| The first column can be any non-zero vector in $\mathbb{F}^4$. There are $3^4-1$ of those. Having chosen one of them, call it $c_1$, and then the second column needs to be any vector not in $span(c_1)$. There are $3^4-3$ of those. The third column needs to be outside $span(c_1,c_2)$, so it is one of $3^4-3^2$ options.
The answer is thus C.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the integer solutions of the equation $3\sqrt {x + y} + 2\sqrt {8 - x} + \sqrt {6 - y} = 14$ $
3\sqrt {x + y} + 2\sqrt {8 - x} + \sqrt {6 - y} = 14
$ .
I already solved this using the Cauchy–Schwarz inequality and got $x=4$ and $y=5$. But I'm sure there is a prettier, simpler solution to this and I was wondering if anyone could suggest one.
| The Cauchy-Schwarz inequality goes like this:
$LHS \leq \sqrt{3^2+2^2+1^2}\cdot \sqrt{\left(\sqrt{x+y}\right)^2+\left(\sqrt{8-x}\right)^2+\left(\sqrt{6-y}\right)^2}=14=RHS$.
Thus you have equality when $\dfrac{3}{\sqrt{x+y}}=\dfrac{2}{\sqrt{8-x}}=\dfrac{1}{\sqrt{6-y}} \to x=4,y=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/948688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How do I make this simple proof better (and more correct?) Let $x$ and $y$ be real numbers. If $x\cdot{y}>\frac{1}{2}$, then $x^2+y^2>1$.
Proof: We will prove with the direct method. Let $x$ and $y$ be real numbers. Since
$$ x\cdot{y}>\frac{1}{2} $$
it follows that
$$ 2xy>1,$$
which means
$$x^2+y^2 \geq 2xy.$$
Therefore,
$$ x^2+y^2>1. $$
| There is a much simpler proof.
Since $xy > \frac{1}{2}$, $x$ and $y$ must be non zero numbers, which are both positive or negative.
Suppose $x>0$ and $y>0$.
If $x>1$, then also $x^2>1$, and $x^2 +y^2>x^2>1$ is obvious.
Take $0<x\leq 1$ and $y>0$.
Since $0<x\leq 1$, the function $f$ with $f(x)=\sqrt{1-x^2}$ is well defined. Let $g$ be the function with $g(x)=\frac{1}{2x}, \ 0<x\leq 1$.
It can be easily proved that the function h with $h(x)=-x+\sqrt{2}, \ 0<x\leq 1$
has a graph which is a line tangent to the graphs of $f$ and $g$ at the point
$(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$.
One can also verify easily that $f$ is a concave function and that $g$ is a convex funtion. Therefore the tangent line $h$ is above the graph of $f$ and below the graph of $g$.
So if $0<x\leq 1$,the previous remark implies that $f(x)\leq h(x)\leq g(x)$ and $\sqrt{1-x^2}\leq \frac{1}{2x}$.
Now, the proposition we want to prove is obvious: If $xy > \frac{1}{2}$, then $y>\frac{1}{2x}$,
and $\frac{1}{2x} \geq \sqrt{1-x^2}$ from the previous inequality, hence
$y> \frac{1}{2x} \geq \sqrt{1-x^2}$. Consequently, $y>\sqrt{1-x^2}$, which implies
$y^2>1-x^2$ and therefore $x^2+y^2>1.$
The case $x<0$ and $y<0$ is similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/949661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
integer $n$ for which $n^6+3n^5-5n^4-15n^3+4n^2+12n+3$ is a perfect Square Prove that the no integer $n\;,$ for which $n^6+3n^5-5n^4-15n^3+4n^2+12n+3$ is a perfect
Square.
My Try:: We can write $(n^6+3n^5-5n^4-15n^3+4n^2+12n+3) = (n^3+an^2+bn+c)^2$
Now Here we have to find values of $a,b,c$.
But
this become very complex.
is there is any other way to solve the above question.
Help me
Thanks
| Modulo $4$ we have
$$\eqalign{n^6+3n^5-5n^4-15n^3+4n^2+12n+3
&\equiv n^6-n^5-n^4+n^3+3\cr
&\equiv n^3(n+1)(n-1)^2+3\cr
&\equiv3\ ;\cr}$$
but a square can only be congruent to $0$ or $1$ modulo $4$.
Reason for the last step: if $n$ is even then $n^3$ is a multiple of $4$, while if $n$ is odd then $(n-1)^2$ is a multiple of $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/950523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate this square root $\sqrt{6 + 2\sqrt{5}} + \sqrt{6 - 2\sqrt{5}}$
I have no clue where to begin. I would appreciate a hint, the answer should be
$2\sqrt{5}$
In general, how do you evaluate
$\sqrt{a + b} + \sqrt{a - b}$?
Thanks!
| Notice that $(\sqrt{5}-1)^2 = 6-2\sqrt{5}$ and $(\sqrt{5}+1)^2 = 6+2\sqrt{5}$, hence:
$$ \sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=(\sqrt{5}+1)+(\sqrt{5}-1)=2\sqrt{5}. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $1(0!)+3.(1!)+7(2!)+13(3!) +21(4!) + \cdots $ n terms.... Question( from sequences) :
If $1(0!)+3.(1!)+7(2!)+13(3!) +21(4!) + \cdots $ n terms = $(4000)(4000!)$ Then what is the value of n.
How to proceed in this please suggest , will be of great help to me thanks...
| The summation is
\begin{align}
S_{n} = \sum_{r=0}^{n} \left( r(r+1) + 1\right) \, r!
\end{align}
and by selecting $n$ values it is seen that
\begin{align}
S_{0} &= 1 \\
S_{1} &= 0! + 3 \cdot 1! = 2 \cdot 2! \\
S_{2} &= 0! + 3 \cdot 1! + 7 \cdot 2! = 3 \cdot 3!
\end{align}
which leads to
\begin{align}
S_{n} = \sum_{r=0}^{n} [ r(r+1) + 1] \, r! = (n+1) \cdot (n+1)!.
\end{align}
For the case of $S_{n} = (4000) \cdot (4000)!$ it is seen that $n = 3999$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$.
$$z^{10}+\frac{1}{z^{10}}=?$$
For the solution:
*
*the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
*converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ and $z_2=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}$
*How do I proceed?
| The solutions to $z^2 + z + 1 = 0$ are the complex $z = 1 \angle (\pm 1/3)$.
So
$$\begin{align}
z^{10} + z^{-10} & = \left(1\angle (\pm 1/3)\right)^{10} + \left(1\angle (\pm 1/3)\right)^{-10}
\\& = \left(1\angle (\pm 10/3)\right) + \left(1\angle (\mp 10/3)\right) \
\\ & = (1\angle 10/3) + (1\angle -10/3)
\\ & = (1\angle 1/3) + (1\angle -1/3)
\\ & = -1
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Leibniz test for convergence of non alternating series I am aware that one can use the comparison test and the integral test to show that the series $$\sum_{n=1}^{\infty}\frac{1}{n(n+3)}$$ converges. Is it possible to use the Leibniz test to show that the series converges?
| Your sequence $$\sum\frac1{n(n+3)}=\frac14+\frac1{10}+\frac{1}{18}+\frac1{28}+\frac1{40}+\cdots$$
$$=\sum\left(\frac1n-\frac{n+2}{n(n+3)}\right)=1-\frac{3}{4}+\frac12-\frac4{10}+\frac13-\frac{5}{18}+\frac14-\frac{6}{28}+\frac15-\frac{7}{40}\pm\cdots$$
$$=\sum\left(\frac{n+4}{n(n+3)}-\frac1n\right)=\frac{5}{4}-1+\frac6{10}-\frac12+\frac7{18}-\frac13+\frac{8}{28}+\frac14+\frac{9}{40}-\frac15\pm\cdots$$
Those are the two nice cases of $\sum\left(\frac{n+A+1}{n(n+3)}-\frac{n+A}{n(n+3)}\right)$
Another variation is $=\sum\left(\frac{2n+A+1}{n(n+3)}-\frac{2n+A}{n(n+3)}\right).$ For $A=3$ this gives $$\frac{5}{4}-\frac{4}{4}+\frac{7}{10}-\frac{6}{10}+\frac{9}{18}-\frac{8}{18}+\frac{11}{28}-\frac{10}{28}+\frac{13}{40}-\frac{12}{40}\pm \cdots $$
This has the nice feature that , ignoring signs, each term is close to half way between those on either side.
In some sense one can always do this. Given a convergent positive series $\sum a_i$ we wish to rewrite it as an alternating series $$a_1+a_2+a_3\cdots=b_1-b_2+b_3-b_4+b_5-b_6\pm\cdots$$ with
*
*$ b_{2n-1}-b_{2n}=a_n$
*$b_{2n} \gt b_{2n+1}$
*all the $b_k$ are positive with $\lim b_k=0$
The series would instead be called telescoping if we changed the second condition to $b_{2n}=b_{2n+1}.$ Then we see that everything is forced. $b_{2n-2}=b_{2n-1}=\sum_{k}^{\infty}a_n.$ The catch is that we might not recognize the $b_k$ as anything nice.
For the given series $b_{2n-1}=\frac{11}{18},\frac{13}{36},\frac{47}{180},\frac{37}{180},\cdots$
It took me a long time to realize that this is pretty close to $\frac12,\frac13,\frac14,\frac15\cdots.$ $$
b_{2n-1}=\frac{1}{n+1}+\frac{2}{3n(n+1)(n+2)}=\frac{3n^2+6n+2}{3n(n+1)(n+2)}=\frac{n+1}{n(n+2)}-\frac{1}{3n(n+1)(n+2)}. $$ You can prove that after the fact, but that isn't how I found it.
Evidently for an honest alternating series we must have $b_{2n-1} \gt \sum_n^{\infty}a_k.$ Then $b_{2n}=b_{2n-1}-a_n$ and we will always have a little room to choose the next term with $b_{2n} \gt b_{2n+1} \gt \sum_{n+1}^{\infty}a_k.$
Again it may be difficult to make this nice.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}
\end{align}
I know the final answer is $\frac{9}{2}$. After multiplying by the conjugate, I see where the $9$ in the numerator comes from. I just can't remember how I solved the rest of the problem.
| Setting $\dfrac1x=h,$ the limit reduces to $$\lim_{h\to0^+}\frac{\sqrt{1+2h}-\sqrt{1-7h}}h$$
$$=\lim_{h\to0^+}\frac{1+2h-(1-7h)}{h(\sqrt{1+2h}+\sqrt{1-7h})}$$
$$=\lim_{h\to0^+}\frac9{\sqrt{1+2h}+\sqrt{1-7h}}$$
$$=\frac9{\sqrt{1+2\cdot0}+\sqrt{1-7\cdot0}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/962458",
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"source": "stackexchange",
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Find the number of common tangents to $y^2=2012x$.... Problem :
Find the number of common tangents to $y^2=2012x$ and $xy =(2013)^2$
Solution :
Common tangent will have slope equal to both curves.
therefore, differentiation both the curves we get the slopes .
$\therefore 2y\frac{dy}{dx}=2012 \Rightarrow \frac{dy}{dx} = \frac{2012}{2y} .....(1)$ and $x\frac{dy}{dx}+y=0$ $\Rightarrow \frac{dy}{dx}=-\frac{y}{x}.....(2)$
(1) and (2) represent the slope of the same tangent therefore, 1 =2
$\Rightarrow \frac{2012 }{2y } = \frac{-y}{x} $ But I am not getting anything here,
So, I used the second method : Solving for both the curves we will get point of intersection,
$y^2=2012x \Rightarrow x =\frac{y^2}{2012}.....(1) $ $y =\frac{(2013)^2}{x}.......(2)$
Now putting value of y from (2) in equation (1) we get
$\frac{\frac{(2013)^4}{x^2}}{2012}=x$ But I think this will also not give any solution please suggest thanks.
| Let $(s,t)$ be the point on $y^2=2012x$ and let $(u,v)$ be the point on $xy=(2013)^2$.
Then, we have
$$t^2=2012s,\tag1$$
$$uv=(2013)^2.\tag2$$
Since
$$y^2=2012x\Rightarrow 2y\cdot\frac{dy}{dx}=2012,$$
we know that the $y$-axis is tangent to this curve at the origin. However, since $y$-axis is not tangent to the curve $xy=(2013)^2$, we may assume that $t\not=0$. Then, we have
$$\frac{dy}{dx}=\frac{2012}{2y}.\tag3$$
On the other hand, since $u\not=0$, we have
$$xy=(2013)^2\Rightarrow y+x\cdot\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac yx.\tag4$$
From $(1),(2),(3),(4)$, the tangent line both at $(s,t)$ and at $(u,v)$ can be represented as
$$y-t=\frac{2012}{2t}(x-s)\iff y=\frac{2012}{2t}x+\frac{2t^2-2012s}{2t}=\frac{2012}{2t}x+\frac t2$$
and
$$y-v=-\frac{v}{u}(x-u)\iff y=-\frac{v^2}{(2013)^2}x+2v$$
respectively.
Hence, we have
$$\frac{2012}{2t}=-\frac{v^2}{(2013)^2},\ \ \frac t2=2v\Rightarrow v^3=-\frac{2012(2013)^2}{8},\ t=4v.$$
Thus, we know that there is only one such set $(s,t,u,v)$. Hence, the answer is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/963562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove that if $abc\ne0$ and $ab+bc+ac=0$ then $a+b+c\ne0$ I tried to do proof by contradiction, but problem is how to get from $ab+bc+ac$ to $a+b+c$
Assuming $a+b+c=0$
my approachs:
*
*Adding $ab+ac+bc=0$ and $a+b+c=0$ and try to factor
*Deriving $$a^2+ab+ac=0\\ac+bc+c^2=0\\ab+b^2+bc=0$$ and trying to derive something but nothing useful.
Please just some hint will do the help to me
| If $ab+bc+ac=a+b+c=0$, then $a,b,c$ are the three roots of a cubic equation $X^3 - abc=0$. But any nonzero number has two non-real cube roots, so $abc=0$—assuming that at least two of $a,b,c$ are real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\int _0^1\int _0^{\left(1-x^n\right)^{1/n}}\left(-x^n-y^n+1\right)^{1/n}dydx$ Let $n>0$. How does one integrate
$$\int _0^1\int _0^{\left(1-x^n\right)^{1/n}}\left(-x^n-y^n+1\right)^{1/n}dydx$$
?
This integral represents the volume enclosed by $$x>0,y>0,z>0,x^n+y^n+z^n<1$$.
By substitution of $y=t\left(1-x^n\right)^{1/n}$ we get
$$\int _0^1\left(1-x^n\right)^{1/n}\int _0^1\left(-x^n-\left(t\left(1-x^n\right)^{1/n}\right)^n+1\right)^{1/n}dtdx$$
$$=\int _0^1\left(1-x^n\right)^{1/n}\int _0^1\left((1-x^n)(1-t^n)\right)^{1/n}dtdx$$
$$=\int _0^1\left(1-x^n\right)^{2/n}dx\int _0^1\left(1-t^n\right)^{1/n}dt$$
Some help from Mathematica
$$=\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{\Gamma \left(\frac{n+2}{n}\right)}\int _0^1\left(1-x^n\right)^{2/n}dx$$
I am not sure how to do the rest.
Judging by the pattern, it looks like the final answer is
$$\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{\Gamma \left(\frac{n+2}{n}\right)}
\frac{2 \Gamma \left(1+\frac{1}{n}\right) \Gamma \left(\frac{2}{n}\right)}{3 \Gamma \left(\frac{3}{n}\right)}$$
but I don't see how to actually prove it.
| Final steps thanks to Lucian starting from
$$\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{\Gamma \left(\frac{n+2}{n}\right)}\int _0^1\left(1-x^n\right)^{2/n}dx$$
Let $x=s^{1/n}$ and we get
$$\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{n\Gamma \left(\frac{n+2}{n}\right)}\int _0^1\left(1-s\right)^{2/n}s^{1/n-1}ds$$
$$=\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{n\Gamma \left(\frac{n+2}{n}\right)}\frac{\Gamma \left(\frac{1}{n}\right) \Gamma \left(\frac{n+2}{n}\right)}{\Gamma \left(\frac{n+3}{n}\right)}$$
$$=\frac{\Gamma \left(1+\frac{1}{n}\right)^3}{\Gamma \left(\frac{n+3}{n}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/971320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
What is the value of $\sum_{m=1}^{19} \frac{1}{\zeta^{3m}+\zeta^{2m}+\zeta^{m}+1}$ with $\zeta=e^{2\pi i/19}$? Given that $\zeta=e^{2\pi i/19}$, how to find the value of $$S=\sum_{m=1}^{19} \dfrac{1}{\zeta^{3m}+\zeta^{2m}+\zeta^{m}+1}$$?
All I could think of was to somehow factorize the denominator and apply some sort of a partial fraction method, but I didn't succeed in that too.
Answers and hints in the right direction appreciated.
| Let $X$ be the set $\{\; \zeta, \zeta^2, \zeta^3, \ldots, \zeta^{18}\; \}$.
Since $19$ is a prime number, for any integer $m$ relative prime to $19$, the map
$$X \in x\quad \mapsto \quad x^m \in X$$
is a permutation of $X$. Together with the obvious identity $\sum\limits_{x\in X} x = -1$, we have
$$\begin{align}
&\sum_{m=1}^{19} \frac{1}{\zeta^{3m} + \zeta^{2m}+\zeta^m + 1}\\
=& \frac14 + \sum_{x\in X} \frac{1}{x^3+x^2+x+1}
= \frac14 + \sum_{x\in X} \frac{x-1}{x^4-1}
= \frac14 + \sum_{x\in x} \frac{(x^4)^5 - 1}{x^4 - 1}\\
=& \frac14 + \sum_{y\in X} \frac{y^5-1}{y-1}
= \frac14 + \sum_{y\in X} ( y^4 + y^3 + y^2 + y + 1 )\\
=& \frac14 + \sum_{z\in X} (z + z + z + z + 1)\\
=& \frac14 -4 + 18 = \frac{57}{4}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/971757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Code is not cyclic for any q I have code $C$ over $F_p$ with generator matrix which looks like
$G =
\begin{pmatrix}
0 &0& 0& 1& 0& 1& 1 &1\\
1& 0 &0& 0 &1 &0 &1& 1\\
1& 1& 0& 0& 0& 1& 0& 1\\
1 &1& 1& 0 &0 &0& 1 &0\end{pmatrix}$
I need to show that this code is not cyclic for any $p$.
I constructed vector which looks like
$$(c_2+c_3+c_4, c_3+c_4, c_4, c_1, c_2, c_1+c_3, c_1+c_2+c_4, c_1+c_2+c_3)$$ and its shifts but it didn't help me a lot.
Does somebody have any ideas how it can be proven?
| $C$ must contain every cyclic shift of its codewords. Consider the cyclic shift of the last row of $G$, $x = \left( {01110001} \right)$. If $C$ is a cyclic code, $x$ must be a linear combination of the rows of $G$, i.e.
${x^T} = a\left( {\begin{array}{*{20}{c}}
0\\
0\\
0\\
1\\
0\\
1\\
1\\
1
\end{array}} \right) + b\left( {\begin{array}{*{20}{c}}
1\\
0\\
0\\
0\\
1\\
0\\
1\\
1
\end{array}} \right) + c\left( {\begin{array}{*{20}{c}}
1\\
1\\
0\\
0\\
0\\
1\\
0\\
1
\end{array}} \right) + d\left( {\begin{array}{*{20}{c}}
1\\
1\\
1\\
0\\
0\\
0\\
1\\
0
\end{array}} \right)$
for some $a,b,c,d \in {F_p}$.
Then, it is not hard to see that
$b + c + d = 0\left( p \right)$ and $c + d = 1\left( p \right)$ from the sum of first and second components of the rows, respectively. From the fifth components, $b=0$. So,
$c + d = 1\left( p \right)$ and $c + d = 0\left( p \right)$. This is not possible for all prime $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how to solve $\int\frac{1}{1+x^4}dx$ i want find the answer and method of solve of $\int\frac{1}{1+x^4}dx$.
I know $$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$$,
How I can use this to solve of that integration.
| Hint:
Use the identity
$$1+x^4=(1+\sqrt{2}x+x^2)(1-\sqrt{2}x+x^2)$$
and Partial fractions decomposition.
Edit:
Then
$$\dfrac{1}{1+x^4}=\dfrac{1}{(1+\sqrt{2}x+x^2)(1-\sqrt{2}x+x^2)}\\=
\dfrac{Ax+B}{1+\sqrt{2}x+x^2}+\dfrac{Cx+D}{1+\sqrt{2}x+x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Eigenvectors of $\left( \begin{array}{ccc} a & 0 \\ 0 & -b \end{array} \right)$ I calculated the eigenvalues of the following matrix to be $a$ and $-b$.
$J = \left( \begin{array}{ccc}
a & 0 \\
0 & -b \end{array} \right)$
But when I use the formula $(J - \lambda I)v = 0$ with either $a$ or $-b$ as eigenvalue to calculate the eigenvectors I get an indeterminate system of equations as they cancel with the $a$ and $-b$ in the matrix.
Yet Wolfram Alpha says the eigenvectors are $(1, 0)$ and $(0, 1)$?
| if $a=b=0$
Then you have
$$\begin{pmatrix}a&0\\0&-b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$
This means that every basis $(v_1,v_2)$ of $\Bbb{R}^2$ is a basis of eigenvectors. One can verify this by seeing that the dimension of the kernel of $A$ is $2$ and the fact that multiplying the zero matrix by $v_1$ and $v_2$ we get the vector $(0,0)^T$.
If $a=-b$ Then we have
$$(J-aI)=\begin{pmatrix}0&0\\0&-b-a\end{pmatrix}=\begin{pmatrix}0&0\\0&-b+b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$
You can do the same reasoning as above to conclude that every basis of $\Bbb{R}^2$ is a basis of eigenvectors.
If $a\neq -b$ we have:
$$(J-aI)v_1=0 \iff\begin{pmatrix}0&0\\0&-b-a\end{pmatrix}v_1=0\implies v_1 = k\cdot \begin{pmatrix}1\\0\end{pmatrix},\quad k\in \Bbb{R},$$
because $$\begin{pmatrix}0&0\\0&-b-a\end{pmatrix} \begin{pmatrix}k\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}.$$
Do you get the second eigenvector?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that $1 + 4 + 9 ... + n^2 = (n/6)(n+1)(2n+1)$ I know that it is true but not sure how to write the proof for: $1 + 4 + 9 ... + n^2 = (n/6)(n+1)(2n+1)$. I need help to guide me in the right direction.
Thanks in advance.
edit:
Okay at n=k I have $ 1+4+9 ... + k^2 = (k/6)(k+1)(2k+1)$
and at $n=k+1$ I have $((k+1)/6))((k+1)+1)(2(k+1)+1) = ((k+1)/6)(k+2)(2k+3).$
Does my base step need to begin at $n=0$ or $n=1?$ How do I tell?
| When $n=1$, it is Okay.
Suppose that $n=k$, $$1 + 4 + 9 \cdots + k^2 = (k/6)(k+1)(2k+1).$$
Then $n=k+1$, $$1 + 4 + 9 \cdots + k^2+(k+1)^2 = (k/6)(k+1)(2k+1)+(k+1)^2=((k+1)/6)(k+2)(2k+3).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/974355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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} |
Integration by parts with Legendre Functions I need help deriving
$\int_{-l}^l [P_l^m(x)]^2 = \frac{2}{2l+1} \frac{(l+m)!}{(l-m)!}$
for the associated Legendre functions
I am supposed to use
$P_l^m(x) = (-1)^{-m}\int_{-l}^l \frac{(1-x^2)^{\frac{m}{2}}}{2^ll!}\frac{d^{l+m}}{dx^{l+m}}(x^2-1)^l$
and
$P_l^m(x) = \frac{1}{2^ll!}^{-m}\int_{-l}^l(1-x^2)^{\frac{m}{2}}\frac{d^{l-m}}{dx^{l-m}}(x^2-1)^l$
I tried to multiply them together, ignoring the terms without x and got
$\int_{-l}^l\frac{d^{l+m}}{dx^{l+m}}(x^2-1)^l\frac{d^{l-m}}{dx^{l-m}}(x^2-1)^l$
I'm supposed to integrate by parts m-times to get both derivates to be $\frac{d^{l}}{dx^{l}}$
This is where I think I'm making my mistake, after doing the integration by parts and getting the same number of derivatives for each term, $\frac{d^{l}}{dx^{l}}$
I get
$(-1)^m\int_{-l}^l(\frac{d^{l}}{dx^{l}}(x^2-1)^l)^2$
I guess the (-1)^m will cancel with part of the original equations but I don't see how to get to the final conclusion
$\frac{2}{2l+1} \frac{(l+m)!}{(l-m)!}$
Any guidance or help would be greatly appreciated, Thanks a lot!
| The Rodrigues' formulae you supplied have gone somewhat astray; they should be
$P_n^m(x)=\frac{1}{2^n n!} (x^2-1)^{m⁄2} [(x^2-1)^n ]^{(n+m)}$
and
$P_n^{-m}(x)=\frac{1}{2^n n!} (x^2-1)^{-m⁄2} [(x^2-1)^n ]^{(n-m)}$
Let us derive the value of $‖P_n^m‖^2$ for positive m:
$\newcommand{\partial}[1]{\left[#1\right]}$
$\newcommand{\bracket}[1]{\left(#1\right)}$
\begin{equation}
\begin{split}
‖P_n^m‖^2&=\int_{-1}^{+1} [P_n^m]^2 dx
\\ &=\frac{1}{2^{2n} n!^2} \int_{-1}^{+1}(x^2-1)^m\cdot [(x^2-1)^n ]^{(n+m)}\cdot [(x^2-1)^n ]^{(n+m)}dx
\end{split}
\end{equation}
Making use of
\begin{equation}
\begin{split}
[(x^2-1)^n]^{(n-m)}=\frac{(n-m)!}{(n+m)!} (-1)^m (1-x^2)^m [(x^2-1)^n]^{(n+m)}
\end{split}
\end{equation}
we get
\begin{equation}
\begin{split}
‖P_n^m‖^2&=\frac{1}{2^{2n} n!^2} \int_{-1}^{+1}(-1)^m\frac{(n+m)!}{(n-m)!}[(x^2-1)^n ]^{(n+m)}\cdot [(x^2-1)^n]^{(n-m)}dx
\end{split}
\end{equation}
Partial integration $n-m$ times then gives
\begin{equation}
\begin{split}
‖P_n^m‖^2&=\frac{1}{2^{2n} n!^2}(-1)^m\frac{(n+m)!}{(n-m)!} \cdot (-1)^{n-m}\int_{-1}^{+1}[(x^2-1)^n ]^{(2n)}\cdot [(x^2-1)^n]^{(0)}dx
\\ &=\frac{1}{2^{2n} n!^2}(-1)^n\frac{(n+m)!}{(n-m)!} \int_{-1}^{+1}(2n)!(x^2-1)^ndx
\\ &=\frac{1}{2^{2n} n!^2}(-1)^n\frac{(n+m)!}{(n-m)!}(2n)!(-1)^n\frac{n!^2}{(2n)!}\cdot \frac{2^{2n+1}}{2n+1}
\\ &=\frac{(n+m)!}{(n-m)!} \frac{2}{2n+1}
\end{split}
\end{equation}
And it is easy (using $P_n^{-m}(x) = (-1)^m \frac{(n-m)!}{(n+m)!} P_n^m(x)$) to derive that
\begin{equation}
\begin{split}
‖P_n^{-m}‖^2&=\frac{(n-m)!^2}{(n+m)!^2} ‖P_n^m‖^2
\\ &=\frac{(n-m)!}{(n+m)!} \frac{2}{2n+1}
\end{split}
\end{equation}
Whence
\begin{equation}
\begin{split}
‖P_n^{m}‖^2&=\frac{(n+|m|)!}{(n-|m|)!}\frac{2}{2n+1}
\end{split}
\end{equation}
For negative and positive $m$, $-n \le m \le n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\tan \{\frac{1}{2} \sin^{–1} (2x/ 1 + x^2) + \frac{1}{2} \cos^{–1} (1 – y^2/1 + y^2) \}$ is equal to. $$
\tan \left\{
\frac{1}{2} \arcsin \frac{2x}{1 + x^2} +
\frac{1}{2} \arccos \frac{1 – y^2}{1 + y^2}
\right\}
$$
is equal to.
Note:
i think $\sin a=2x/1+x^2$, $\cos b=(1 – y^2/1 + y^2)$
| Set $\arctan x=A\implies x=\tan A\ \ \ \ (1)$
Using definition of Principal values, $\displaystyle-\frac\pi2\le A\le\frac\pi2\ \ \ \ (2)$
and using Weierstrass substitution formula, $\displaystyle\frac{2x}{1+x^2}=\sin2A$
Now $\displaystyle\arcsin(\sin2A)=\begin{cases} 2A &\mbox{if } -\dfrac\pi2\le2A\le\dfrac\pi2\iff-\dfrac\pi4\le A\le\dfrac\pi4 \\
\pi-2A & \text{otherwise} \end{cases} $
From $(1),(2); \cos A\ge0\implies\cos A=\dfrac1{\sec A}=+\dfrac1{\sqrt{1+\tan^2A}}=\dfrac1{\sqrt{1+x^2}}$
$\sin A=\cos A\cdot\tan A=\dfrac x{\sqrt{1+x^2}}$
Similarly, setting $\arctan y=B,\tan B=y\ \ \ \ (3)$
and like $(2),\displaystyle-\frac\pi2\le B\le\frac\pi2$
$(3)\implies\dfrac{1-y^2}{1+y^2}=\cos2B$
$\displaystyle\arccos(\cos2B)=\begin{cases} 2B &\mbox{if } 0\le2B\le\pi\iff0\le B\le\dfrac\pi2 \\
-2B & \text{otherwise} \end{cases} $
Can you take care of the four possible combinations?
| {
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"source": "stackexchange",
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Show that $n!^{n+1}$ divides $(n^2)!$ My attempt so far is by induction. Let $f(n) = \frac{(n^2)!}{n!^{n+1}}$, I will try showing that $f(n)$ is a positive integer for all $n$.
We have $f(0) = \frac{0!}{0!^{n+1}} = 1$.
Now assume for induction, that $f(n) = k$ for some positive integer k.
Next, I'll examine $f(n+1) = \frac{((n+1)^2)!}{(n+1)!^{n+2}}$.
Note that $(n+1)!^{n+2} = [(n+1)(n)!]^{n+2} = (n+1)^{n+2}(n)!^{n+2} = (n+1)^{n+2}(n)!(n)!^{n+1}$.
And that $(n+1)!^2 = \frac{(n+1)^2!}{n^2!}n^2!$
So $f(n+1) = \frac{\frac{(n+1)^2!}{n^2!}n^2!}{(n+1)^{n+2}(n)!(n)!^{n+1}} = \frac{\frac{(n+1)^2!}{n^2!}}{(n+1)^{n+2}(n)!} \frac{(n^2)!}{n!^{n+1}} = \frac{\frac{(n+1)^2!}{n^2!}}{(n+1)^{n+2}(n)!}k$.
Additionally, I can show that $\frac{(n+1)^2!}{n^2!} = \prod_{i=0}^{2n}{(n+1)^2-i}$.
And this is where I don't how to proceed. Any tips? I'm willing to scrap all this and start from scratch as this feels like a dead end.
| Here is a combinatorial proof:
Arrange $n^2$ person in $n^2$ chairs in a row, we have $n^2!$ ways.
Alternatively, we can pick $n$ persons for the first $n$ chairs, then $n$ persons for chairs $n+1$ to $2n$, then $n$ persons for $2n + 1$ to $3n$, etc, then arrange each group of $n$ persons, we have $\left(\prod_{k=0}^{n-1}{n^2-kn \choose n}\right)(n!)^n$ ways
So $$n^2! = \left(\prod_{k=0}^{n-1}{n^2-kn \choose n}\right)(n!)^n$$
Then we only need to prove $n!$ divides $\prod_{k=0}^{n-1}{n^2-kn \choose n}$.
Denote by $K$the number of ways of dividing $n^2$ persons into non-ordered $n$ groups, each of which contains $n$ persons , then actually we can note
$$\prod_{k=0}^{n-1}{n^2-kn \choose n} = Kn!$$
because $\prod_{k=0}^{n-1}{n^2-kn \choose n}$ is the way of dividing $n^2$ persons into ordered $n$ groups, each of which contains $n$ persons.
Now it's done. In summary, to arrange $n^2$ person in $n^2$ chairs in a row($n^2$! ways), we can divided them into non-ordered $n$ groups of equal size($K$ ways), make a permutation for those groups($n!$ ways), and then make permutations inside each group($(n!)^n$ ways). So we have $$(n^2)! = K (n!)^{n+1}$$
| {
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"url": "https://math.stackexchange.com/questions/979886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Last 3 digits of $7^{12341}$ I know that I need to reduce $7^{12341} \pmod {1000}$
By Euler I have $7^{\phi(1000)}\equiv 7^{400}\equiv1\pmod{1000}$
That leaves me with the monster $7^{341}\pmod{1000}$
Is there a way to reduce this smoothly without working $7^2, 7^4, 7^8$ etc manually ?
| $$7^2=50-1$$
$$7^{20n+1}=7(7^2)^{10n}$$
Now $(7^2)^{10n}=(50-1)^{10n}=(1-50)^{10n}$
Using Binomial Theorem,
$(7^2)^{10n}=1-\binom{10n}150+\binom{10n}2(50)^2\cdots+(50)^{2n}$
$\implies(7^2)^{10n}\equiv1-500n+\dfrac{10n(10n-1)}250^2\pmod{1000}\ \ \ \ (1)$ for $n\ge2$
Again, $\dfrac{10n(10n-1)}250^2=50n^2(50^2)-5n(2500)\equiv-500n\pmod{1000}$ as $2500\equiv100\pmod{200}\implies5(50^2)\equiv5\cdot100\pmod{5\cdot200}\equiv500$
From $(1),$
$(7^2)^{10n}\equiv1-500n-500n\pmod{1000}\equiv1$
Here $n=\dfrac{12341 -1}{20}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit by L'hospital's rule I have to prove that:
$$\lim \limits_{x \to \infty} \frac{{\int_x^{\infty} \exp(-t^2/2)dt}}{\exp(-x^2/2)
(1/x)}=1$$
Should I use L'hospital rule, if yes what are the derivatives?
| First note that
$$
\int e^{-cx^2}dx=\sqrt{\frac{\pi}{4c}}\mathrm{erf}(\sqrt{c}x)
$$
So now we have
$$
\lim_{x \to \infty} \left[\frac{\int_x^{\infty} e^{-\frac{t^2}{2}}dt}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]=\lim_{x \to \infty} \left[\frac{\lim \limits_{b\to\infty} \int_x^b e^{-\frac{t^2}{2}}dt}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]=\lim_{x \to \infty} \left[\frac{\sqrt{\frac{\pi}{2}}\lim \limits_{b\to\infty}\left[\mathrm{erf}\left(\frac{b}{\sqrt{2}}\right)-\mathrm{erf}\left(\frac{x}{\sqrt{2}}\right)\right]}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]
$$
$$
=\lim_{x \to \infty} \left[\frac{\sqrt{\frac{\pi}{2}}\left[1-\mathrm{erf}\left(\frac{x}{\sqrt{2}}\right)\right]}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]=\lim_{x \to \infty} \left[\frac{\sqrt{\frac{\pi}{2}}\mathrm{erfc}\left(\frac{x}{\sqrt{2}}\right)}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]=\sqrt{\frac{\pi}{2}}\lim_{x \to \infty} \left[\frac{\mathrm{erfc}\left(\frac{x}{\sqrt{2}}\right)}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]
$$
$$
=\sqrt{\frac{\pi}{2}}\lim_{x \to \infty} \left[\frac{\frac{d}{dx}\mathrm{erfc}\left(\frac{x}{\sqrt{2}}\right)}{\frac{d}{dx}\left[\frac{1}{x}e^{-\frac{x^2}{2}}\right]}\right]=\sqrt{\frac{\pi}{2}}\lim_{x \to \infty} \left[\frac{-\sqrt{\frac{2}{\pi}}e^{-\frac{x^2}{2}}}{-\frac{e^{-\frac{x^2}{2}}(x^2+1)}{x^2}}\right]=\sqrt{\frac{\pi}{2}}\lim_{x \to \infty} \left[\frac{\sqrt{\frac{2}{\pi}}}{\frac{(x^2+1)}{x^2}}\right]
$$
$$
=\sqrt{\frac{\pi}{2}}\lim_{x \to \infty} \left[\frac{\sqrt{\frac{2}{\pi}}}{1+\frac{1}{x^2}}\right]=\sqrt{\frac{\pi}{2}}\sqrt{\frac{2}{\pi}}=1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that the locus of a point P is a circle I'm struggling with this geometry question:
The fixed points A and B have coordinates $(-3a,0)$ and $(a,0)$ respectively. Find the equation of the locus of a point P which moves in the coordinate plane so that $AP = 3PB$. Show that the locus is a circle, S, which touches the axis of $y$ and has its centre at the point ($\frac{3}{2}a, 0$)
I managed to get this part. First I assumed it was a circle, and used the ratio to find $x$ in terms of $a$. Let the coordinates of P be $(x,y)$:
$$
\sqrt{(x+3a)^2+y^2}=3\sqrt{(a-x)^2+y^2} \\
(x+3a)^2 + y^2 = 9(a-x)^2+y^2\\
x = 3a
$$
So if one of the coordinates of S is $(3a,0)$, and S also touches the y-axis, we know that its centre is $(\frac{3}{2}a,0)$. So the equation of S is:
$$
\left(x-\frac{3}{2}\right)^2+y^2 = \frac{9}{4}
$$
Is that correct? My approach seems quite hokey. The next part of the question is this:
A point Q moves in such a way that the perpendicular distance of Q from the axis of y is equal to the length of a tangent from Q to the circle S. Find the equation of the locus of Q. Show that this locus is also the locus of points which are equidistant from the line $4x+3a = 0$ and the point $(\frac{3}{4},0)$.
I tried finding the distance between $x=0$ and $Q(x,y)$, and that gave a distance of $x$ - but how do I plug that into the equality:
$$
D = x = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}}
$$
| I think you overworked here. The condition is
$$|AP|^2=9|BP|^2\iff (x+3a)^2+y^2=9\left[(x-a)^2+y^2\right]\iff$$
$$\iff 8x^2-24ax+8y^2=0\iff \left(x-\frac{3a}2\right)^2+y^2=\frac{9a^2}4$$
and we get that the points$\;P=(x,y)\;$ are the locus of circle with center $\;\left(\frac{3a}2\,,\,0\right)\;$ and radius $\;\frac{3a}2\;$ , and since radius = absolute value of $\;x$ - coordinate of circle, the circle is tangent to the $\;y$ - axis
Let $\;Q=(x,y)\;$ , then its distance from the $\;y$ - axis is $\;|x|\;$ , and the length of any of its two tangets to the above circle is
$$\sqrt{\left(x-\frac{3a}2\right)^2+y^2\;-\;\frac{9a^2}4}$$
and thus we get the equation:
$$x^2=\left(x-\frac{3a}2\right)^2+y^2\;-\;\frac{9a^2}4\iff$$
$$\iff -3ax+y^2=0\;\;(*)$$
Now, let us check what is the locus of all point whose distance to $\;4x+3a=0\iff x=-\frac{3a}4\;$ equals its distance to $\;\left(\frac{3a}4\,,\,0\right)\;$:
$$\left(x-\frac{3a}4\right)^2+y^2=\frac{(4x+3a)^2}{16}\iff x^2-\frac{3a}2x+\frac{9a^2}{16}+y^2=x^2+\frac{3a}2x+\frac{9a^2}{16}\iff$$
$$\iff -3ax+y^2=0\;\;(*)$$
and thus both conditions are identical
| {
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Solve $\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$
Find the set of integer solutions $(x,y)$ to
$$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$$
I can't seem to multiply both sides by the expression in the denominator.
Nor can I simplify and cancel any terms. How should I it?
| $$\frac{x^2+2xy+y^2}{x^2-y^2}=\frac{(x+y)^2}{(x-y)(x+y)}=\frac{x+y}{x-y} \ge (x+y) \implies \frac{1}{x-y}\ge 1$$
Without loss of generality, assume that $x \ge y$, and fiddle around with this expression.
Edit: We do the standard LHS-RHS trick:
$$\frac{x^2+2xy+y^2}{x^2-y^2}-(x+y)>0 \\ LHS= \frac{(x+y)^2}{(x-y)(x+y))}-(x+y) = \frac{(x+y)^2-(x-y)(x+y)^2}{(x^2-y^2)}$$ Hence, $$\frac{(x+y)^2(1-(x-y))}{x^2-y^2}>0$$
Then I think from here take cases regarding whether $x^2-y^2>0 \text{ or } x^2-y^2<0$
, and solve the inequation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}$ for $0 < x < 1$ I stumbled upon this question while doing practice inequalities questions, and I do not know how to start...
Problem: Prove that
\begin{align*}
\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}
\end{align*}
for $0 < x < 1$.
I thought possibly of having an intermediate equality, for example
\begin{align*}
\sqrt{\frac{2x^2-2x+1}{2}}\ge\text{something}\ge\frac{1}{x+\frac{1}{x}}
\end{align*}
where the "something" is simple, but I could not deduce anything...any help would be appreciated, thanks!
| Hints:
Notice that $\sqrt{2x^2-2x+1}=\sqrt{x^2+(x-1)^2} \ge \sqrt{(\frac{x+1-x}{2})^2}=\frac12$ and
$\frac{1}{x+\frac1x}\le \frac{1}{2x\frac1x}=\frac12$
| {
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Remainder of $3^7/8$ I read here that the remainder of $\frac{ab}{c}$ is equal to the remainder of $\frac{a}{c}\frac{b}{c}$ implying that the remainder of $\frac{a^b}{c}$ is equal to the remainder of $[\frac{a}{c}]^b$. However, when I apply this here, I would get remainder of $[\frac{3}{8}]^7=3^7$ (while the correct answer is $3$). What am I doing wrong here? And if this method is not valid, are there other quick and simple methods of finding remainders like this?
| Remainder of $ \frac {ab}{c}$ = remainder of $ \frac {a}{c} $ $\times$ remainder of $ \frac {b}{c} $ is FALSE in most cases.
When remainder of $ \frac {a}{c} $ $\times$ remainder of $ \frac {b}{c} \lt c$, only then it is valid.
However when remainder of $ \frac {a}{c} $ $\times$ remainder of $ \frac {b}{c} \ge c$, then
Remainder of $ \frac {ab}{c}$ = Remainder of $ \frac {remainder of \frac {a}{c} \times remainder of \frac {b}{c} }{c}$
Since as you wrote: remainder of $ \frac {a}{c} $ $\times$ remainder of $ \frac {b}{c} = 3^7$ and dividing $ 3^7 $ by $8$ is the very problem we started at,
To apply the rule in this case, you could make use of the fact that:
$$ 3^7 = 3^4 \times 3^ 3$$
So assume $a$ to be $3^4$ and $b$ to be $3^3$.
Now, it's quite easy to calculate that $3^4 = 81 $ and $3^3 = 27 $
Remainder of $ \frac {3^7}{8} $ = Remainder of $ \frac {3^4 \times 3^3}{8}$ = Remainder of $ \frac {81 \times 27}{8} $
By the rule, Remainder of $ \frac {81 \times 27}{8} $ = Remainder of $ \frac {81}{8} $ $ \times $ Remainder of $\frac {27}{8} $ = $1 \times 3 $ = $3$
Since $ 3 \lt 8$, this is indeed our answer.
| {
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To calculate side of the Equilateral triangle The figure is an equilateral triangle. 3 line segments , which meet at a(any) point in the triangle , are of the length 5cm, 4cm, and 3 cm as shown in the figure. Find the side of the equilateral triangle.
| Theorem: Let $ABC$ be an equilateral triangle, with a point $P$ inside it such that $PC^2=PA^2+PB^2$, then $\angle APB=150^{\circ}$.
So to do that, we cleverly construct a point $D$ such that $APD=60^{\circ}$ and $AP=PD$. Hence, $APD$ is equilateral and $AP=AD$.
Now, as $60^{\circ}=\angle CAB=\angle DAP$, so, $\angle CAP = \angle BAD$, which is just $\angle PAB$ subtracted from them. We know that $AB=AC$.
So, by SAS criteria $PAC \cong BAD$. Hence, $PC=BD$. This means that $BD^2=PD^2+BP^2$. Thus by converse of Pythagoras Theorem, $\angle BPD = 90^{\circ}$, which means $\angle APB=150^{\circ}$.
${}$
Corollary: If $ABC$ be an equilateral triangle, with a point $P$ inside it such that $PC=5$, $PB=4$ and $PA=3$ then $AB=\sqrt{25+12\sqrt3}$.
Note that $3^2+4^2=5^2$, so $\angle APB=150^{\circ}$. Hence, by the cosine law, $AB^2=3^2+4^2-2\cdot12\cdot \cos 150^{\circ}=25+12\sqrt3$. So: $$AB=\sqrt{25+12\sqrt3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\int_0^\infty \frac{\ln x}{x^n-1}\,dx = \Bigl(\frac{\pi}{n\sin(\frac{\pi}{n})}\Bigr)^2$ This question inspired me to ask the following.
Prove that
$$I_n = \int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2,$$
for $\Re(n)>1$.
For some cases there is a nice specific form of $I_n$. For example
$$\begin{align}
I_2 & = \frac{\pi^2}{4} \\
I_3 & = \frac{4\pi^2}{27} \\
I_4 & = \frac{\pi^2}{8} \\
I_5 & = \frac{2\left(5+\sqrt 5\right)\pi^2}{125} \\
I_6 & = \frac{\pi^2}{9} \\
I_7 & = \frac{2\pi^2}{49\left(1-\sin\left(\frac{3\pi}{14}\right)\right)} \\
I_8 & = \frac{\left(2+\sqrt 2\right)\pi^2}{32}
\end{align}$$
| \begin{align}
\int_0^\infty \frac{\ln x}{x^n-1}\,dx
=& \>\frac1n\int_0^\infty \int_0^\infty \frac{1}{(1+y)(x^n+y)} dy \ dx \\
=&\>\frac{\pi }{n^2}\csc\frac\pi n \int_0^\infty\frac{y^{\frac1n-1}}{1+y}dy
=\frac{\pi^2}{n^2}\csc^2\frac\pi n
\end{align}
| {
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Closed form for $I=\int_{0}^{\infty}\frac{x^n}{x^2+u^2}\tanh(x) \, dx$ solve
$$I=\int_{0}^{\infty}\frac{x^n}{x^2+u^2}\tanh(x) dx:0<n<2$$
I tried for $n=1$ :
$$I(v)=\int_{0}^{\infty}\frac{x}{x^2+u^2}\tanh(vx) dx$$
$$I'(v)=\int_{0}^{\infty}(\frac{1}{\cosh^2(vx)}-\frac{u^2}{(x^2+u^2)\cosh^2(vx)}) dx$$
$$I'(v)=-\sum_{k=1}^{\infty}(-1)^ku^{-2k}\int_{0}^{\infty}(\frac{x^{2k}}{\cosh^2(vx)})dx$$
$$I'(v)=-\frac{1}{v}\sum_{k=1}^{\infty}(-1)^k(uv)^{-2k}\int_{0}^{\infty}(\frac{x^{2k}}{\cosh^2(x)})dx$$
and i solved the last integral in term of zeta function but i failed to find the result of sum .
I interested to find the result of sum.
If you can find for $n=1$ and for general n ?
using two method complex and real :)
| As Lucian stated in the comments above, the integral converges only if $-2 < n <1$.
I'll do the case $n=-1$ using contour integration.
For $u>0$ and not equal to $\pi \left(k + \frac{1}{2} \right)$ (where $k$ is a nonnegative integer), consider the complex function
$$ f(z) = \frac{\tanh z}{u^{2}+z^{2}} \frac{1}{z}.$$
Integrating the function $f(z)$ around a rectangular contour in the upper half-plane of height $K \pi$ and then letting $K \to \infty$ discretely, we find
$$ \begin{align} \int_{0}^{\infty} \frac{\tanh x}{(u^{2}+x^{2})} \, \frac{dx}{x} &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\tanh x}{(u^{2}+x^{2})} \, \frac{dx}{x} \\ &= \pi i \left(\text{Res}[f(z), iu] + \sum_{n=0}^{\infty} \text{Res}\left[f(z), i \pi \left(k+\frac{1}{2} \right)\right] \right) \end{align}$$
where
$$ \text{Res}[f(z), iu] = \lim_{z \to iu} \frac{\tanh z}{z+iu} \frac{1}{z} = \frac{\tanh (iu)}{2iu} \frac{1}{iu} = \frac{\tan u}{2iu^{2}}$$
and
$$ \begin{align} \text{Res}\left[f(z), i \pi \left(k+\frac{1}{2} \right)\right] &= \lim_{z \to i \pi (k+1/2)} \frac{\sinh z}{\frac{d}{dz} (u^{2}+z^{2})z \cosh z } \\ &= \frac{1}{i \pi} \frac{1}{\left( (u^{2}-\pi^{2}(k+ \frac{1}{2})^{2} \right) (k + \frac{1}{2})} \\ &= - \frac{1}{i \pi^{3}} \frac{1}{\left(k- \frac{u}{\pi} + \frac{1}{2} \right)\left(k + \frac{u}{\pi} + \frac{1}{2} \right) \left(k + \frac{1}{2} \right)} . \end{align}$$
In general, $$ \sum_{k=0}^{\infty} \frac{1}{(k+a)(k+b)(k+c)}= \frac{(b-c) \ \psi(a)+ (c-a) \ \psi(b)+(a-b) \ \psi(c)}{(a-b)(b-c)(c-a)}$$
where $a \ne b$, $b \ne c$, $c \ne a$, and $a, b, c \ne 0, -1, -2, \ldots$ .
Here $\psi(z)$ is the digamma function.
This identity can be derived by starting on the right side of the equation and replacing each digamma function with its series representation.
Using this identity, we get $$ \sum_{k=0}^{\infty} \text{Res} \left[f(z), i \pi \left(k+\frac{1}{2} \right)\right] = \frac{1}{i \pi} \frac{\psi \left(\frac{1}{2} - \frac{u}{\pi} \right) + \psi \left(\frac{1}{2} + \frac{u}{\pi} \right) - 2 \psi \left( \frac{1}{2}\right)}{2u^{2}}.$$
Therefore,
$$ \begin{align} \int_{0}^{\infty} \frac{\tanh x}{u^{2}+x^{2}} \frac{dx}{x} &= \frac{1}{2u^{2}} \left[\pi \tan u + \psi \left(\frac{1}{2} - \frac{u}{\pi} \right) + \psi \left( \frac{1}{2} + \frac{u}{\pi}\right) - 2 \,\psi \left( \frac{1}{2}\right) \right] \\ &= \frac{1}{2u^{2}} \left[\pi \tan u + \psi \left(\frac{1}{2} - \frac{u}{\pi} \right) + \psi \left( \frac{1}{2} + \frac{u}{\pi}\right) +2 \gamma + 4 \log 2 \right] .\end{align}$$
Checking with Wolfram Alpha, this result appears to be correct for various values of $u$.
EDIT:
When $u = \pi$, the formula reduces to
$$\int_{0}^{\infty} \frac{\tanh x}{\pi^{2}+x^{2}} \frac{dx}{x} = \frac{2}{\pi^{2}}.$$
| {
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How do I solve this differential equation? $y'=\frac{y}{x+y^3}$ So, hey.
The equation is $y'=\frac{y}{x+y^3}$
So, I get something like this: $y'\left(x+y^3 \right)-y=0$, which I can't actually solve. I must admit I am slightly confused how to attack this one.
What should I do?
| $y'=\frac{y}{x+y^3}\Leftrightarrow(x+y^3)dy=ydx$.
Case 1: $y=0$.
Case 2: $y\neq0$, then $$\frac{1}{y^2}[(x+y^3)dy-ydx]=ydy-xd\frac{1}{y}-\frac{1}{y}dx=d(\frac{1}{2}y^2-\frac{x}{y})$$
$\frac{x}{y}-\frac{1}{2}y^2=C$.
The solution to the ODE is $x=\frac{1}{2}y^3+Cy$.
| {
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$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots+\frac{1}{1331}=\frac{p}{q}$; is $p$ divisible by $1997$? if $p,q\in \mathbb{N}$ and
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots+\frac{1}{1331}=\frac{p}{q}$$
why is $p$ divisible by $1997$?
| Let $S$ be the LHS. Using
$$
1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{2n}=\frac{1}{n+1}+\cdots+\frac{1}{2n}
$$
(which can be proved by induction), we can deduce that
$$
S-\frac{1}{1332}=\frac{1}{667}+\cdots+\frac{1}{1331}+\frac{1}{1332}\\
\implies S=\frac{2}{1332}+\frac{1}{667}+\cdots+\frac{1}{1331}=\frac{1}{666}+\cdots+\frac{1}{1331}.
$$
This implies
$$
S=1997\left(\frac{1}{666\times 1331}+\frac{1}{667\times 1330}+\cdots+\frac{1}{998\times 999}\right).
$$
Let the expression in the parentheses equal to $\frac{a}{b}$ where $a$ and $b$ are positive integers satisfying $a<b$ and $\text{gcd}(a,b)=1$. Then because $1997$ is prime, $b$ isn't divisible by $1997$ (the prime factors of $b$ constitute a subset of the prime factors of $\prod_{j=666}^{1331}j$), so we have $1997a|p$ and $b|q$. The claim follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed form of $\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^4$ Find the closed form of $$\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^4$$
I know the closed form for smaller powers like $2, 3$ exists, but I'm not sure if there is a
closed form for this variant. Is it possible to tackle the question in an elementary way
and find the answer, without using integrals at all?
Then, if this exists, I'd also propose the alternating variant
$$\sum_{n=1}^{\infty} (-1)^{n+1} \left(\frac{H_n}{n}\right)^4$$
Update (by editor): The last sum by OP is:
$$\small \sum _{n=1}^{\infty } (-1)^{n-1} \left(\frac{H_n}{n}\right){}^4=\frac{633}{128} \zeta(6,2)-\frac{1}{6} \pi ^2\zeta(\bar5,1)+6 \zeta(\bar7,1)-4 \zeta(\bar5,1,\bar1,1)+6 \log ^2(2) \zeta(\bar5,1)+12 \log (2) \zeta(\bar5,1,1)-16 \text{Li}_5\left(\frac{1}{2}\right) \zeta (3)-\frac{1}{9} \pi ^4 \text{Li}_4\left(\frac{1}{2}\right)+\frac{\pi ^2 \zeta (3)^2}{8}-\frac{63 \zeta (3) \zeta (5)}{64}+\frac{2}{15} \zeta (3) \log ^5(2)-\frac{2}{9} \pi ^2 \zeta (3) \log ^3(2)-\frac{31}{12} \zeta (5) \log ^3(2)+3 \zeta (3)^2 \log ^2(2)-\frac{16}{45} \pi ^4 \zeta (3) \log (2)-\frac{65}{48} \pi ^2 \zeta (5) \log (2)+\frac{155}{8} \zeta (7) \log (2)+\frac{149141 \pi ^8}{43545600}-\frac{1}{216} \pi ^4 \log ^4(2)+\frac{25 \pi ^6 \log ^2(2)}{1512}$$
| The result is
\begin{align*}
S_{1^4,4}=\sum\limits_{n = 1}^\infty {\frac{{H_n^4}}{{{n^4}}}} = \frac{{13559}}{{144}}\zeta \left( 8 \right) - 92\zeta \left( 3 \right)\zeta \left( 5 \right) - 2\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) + 26{S_{2,6}},
\end{align*}
Detailed process see the paper ``Multiple zeta values and Euler sum" enter link description here
| {
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How do I write out this proof? The following works with any number greater than the number 9.
How do I write the following examples out in a proof?
Take the two digits in your age for example & add then add those two numbers.
So let's say you're 28. 2+8=10 Now subtract that total from your age:
28-10=18 now add those two digits.
1+8= 9
Alright let's try 35
3+5=8
35-8=27
2+7= 9
Let's just try the number 42
4+2=6 So now subtract that number from 42
32-6=36 Now add those two.
3+6= 9
I just decided to try 1,2345
1+2+3+4+5=15
Subtract that from the original number:
12345-15=12330 add that last number up:
1+2+3+3+0= 9
Okay, let's do a harder one.
456789
4+5+6+7+8+9=39
456789-39=456750
now do it again with that last number:
4+5+6+7+5+0=27
and again:
2+7= 9
Now I'm going to type a random number in a calculator.
1563248
1+5+6+3+2+4+8=29 subtract that number from the original.
1563248-29=1563219 Now add up those numbers
1+5+6+3+2+1+9=27
2+7=9
So how do I write this out in a proof?
| the main point here is that the remainder when a number is divided by $9$ is the same as the remainder when the sum of its digits is divided by $9$. in particular this means that if $\sigma(N)$ is the sum of digits (decimal) of $N$ then $N-\sigma(N)$ is divisible by $9$.
the number may be written as:
$$
N = \sum_{k=0}^n d_k10^k
$$
and
$$
\sigma(N)= \sum_{k=0}^n d_k
$$
so
$$
N-\sigma(N) = \sum_{k=1}^n d_k(10^k-1)
$$
now you need the fact that $10^k-1$ is always divisible by $9$. can you prove this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
proving an invloved combinatorial identity How to prove following Identity?
$$\sum_{k=0}^n (-1)^k {n-k \choose k} m^k (m+1)^{n-2k} = \frac {m^{n+1}-1}{m-1}, m \ge 2$$
This seems very hard to me. Any idea about how to prove it combinatorialy?
[P.S: for $k > \lfloor {\frac n 2}\rfloor$ L.H.S evaluates to $0$. ]
| This is not a combinatorial proof but I still find it rather nice. Let us replace $m$ by $x$ to get a more general polynomial identity.
We can notice that
\begin{equation*}
(-1)^{k} \binom{n-k}{k} x^{k}(x+1)^{n-2k}
\end{equation*}
is the coefficient of $y^{k}$ in $(1+x-xy)^{n-k}$. Consequently the left hand of the identity is the coefficient of $y^{n}$ in
\begin{equation*}
\sum_{k=0}^{n} (y+xy-xy^{2})^{n-k}=\sum_{k=0}^{n} (y+xy-xy^{2})^{k}.
\end{equation*}
For $k > n$, $(y+xy-xy^{2})^{k}$ makes no contribution to the coefficient of $y^{n}$, so we can extend the summation to infinity. Hence the left hand side equals
\begin{equation*}
\frac{1}{n!} \frac{\partial}{\partial^{n} y} \frac{1}{1-y-xy+xy^{2}}
\end{equation*}
evaluated in $y=0$. We have
\begin{align}
\enspace &\frac{1}{n!} \frac{\partial}{\partial^{n} y} \frac{1}{1-y-xy+xy^{2}}\\
= \enspace & \frac{1}{n!}\frac{\partial}{\partial^{n} y} \left(\frac{1}{1-y} \frac{1}{1-xy}\right) \\
= \enspace & \frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k}\left(\frac{\partial}{\partial^{k} y}\frac{1}{1-y}\right)\left(\frac{\partial}{\partial^{n-k} y} \frac{1}{1-xy}\right)
\end{align}
But
\begin{align*}
&\frac{\partial}{\partial^{k} y}\frac{1}{1-y}=\frac{k!}{(1-y)^{k+1}} \\
&\frac{\partial}{\partial^{k} y}\frac{1}{1-xy}=\frac{x^{k}k!}{(1-xy)^{k+1}}.
\end{align*}
When $y=0$ the denominators become $1$ and the numerators simplifly with the binomial coefficient, so that indeed the left hand side equals $\displaystyle\sum_{k=0}^{n}x^{k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proving limits with existing results $\lim_{x\to a} \frac{\sin^2 x - \sin^2 a}{x-a} = \sin 2a$ So in my lecture yesterday I learnt how to prove that $\lim_{x\to +\infty} \left(1 + \dfrac{1}{x}\right)^{x} = e$, but I'm lost as to how to apply the results to prove limits.
Any help would be greatly appreciated.
Use the results $\lim_{x\to 0} \left(\dfrac{\sin x}{x}\right) = 1$ and $\lim_{x\to +\infty} \left(1 + \dfrac{1}{x}\right)^{x} = e $ to show that:
$(i) \lim_{x\to a} \dfrac{\sin^2 x - \sin^2 a}{x-a} = \sin 2a;$
$(ii) \lim_{x\to -\infty} \left(1 + \dfrac{1}{x}\right)^{x} = e;$
$(iii) \lim_{x\to 0} (1 + x)^{\frac{1}{x}} = e$
| i) Let $y=x-a$. Then, when $x\to a$, $y\to 0$ and we have
$$
\lim_{y\to 0}\frac{\sin^2(y+a)-\sin^2{a}}{y} = \lim_{y\to 0}\frac{\sin^2 y\cos^2a+2\sin a\cos a\sin y\cos y +\sin^2 a\cos^2 y-\sin^2{a}}{y} = \lim_{y\to 0}\frac{\sin^2 y\cos^2a+2\sin a\cos a\sin y\cos y -\sin^2 a\sin^2 y}{y}=\lim_{y\to 0}\frac{\sin y\cos a\sin(a+y)+\sin a\sin y\cos(a+y)}{y} = \\
\cos a\underbrace{\lim_{y\to 0}\frac{\sin y \sin(a+y)}{y}}_\text{$\lim \sin (a+y) \lim \frac{\sin y}{y}=\sin a$} + \sin a\underbrace{\lim_{y\to 0}\frac{\sin y \cos (a+y)}{y}}_\text{$\lim \cos(a+y)\lim \frac{\sin y}{y}=\cos a$}=2\sin a\cos a=\sin(2a)
$$
In the step where we break the limits, note that given that $\lim_{y\to 0}\sin(a+y)=\sin a$ and $\lim_{y\to 0}\cos(a+y)=\cos a$, then, when multiplying for $\frac{\sin y}{y}$, the limit of the product becomes the product of the limits.
ii) Let $y=-x$. Then, as $x\to -\infty$, $y\to +\infty$ and we have
$$
\lim_{y\to +\infty} (1-\frac{1}{y})^{-y} = \lim_{y\to +\infty}(\frac{y}{y-1})^{y}=\lim_{y\to+\infty}(\frac{y-1}{y-1}+\frac{1}{y-1})^{y} = \lim_{y\to+\infty}(1+\frac{1}{y-1})^{y} = \lim_{z\to \infty}(1+\frac{1}{z})^{z+1} =\underbrace{\lim_{z\to \infty}(1+\frac{1}{z})}_\text{=1}\lim_{z\to \infty}(1+\frac{1}{z})^z = e
$$
In the last step, we took $z=y-1$, but since $y\to \infty$, $z\to \infty$ as well.
iii) You just make the substitution $y=1/x$. Then, when $x\to 0$, either $y\to \infty$ or $y \to -\infty$, but from the previous item and the definition of $e$, in both cases it goes to $e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/996000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to find all irreducible polynomials in Z2 with degree 5? I am totally lost on how to do this one. I am supposed to accomplish the following:
Find all irreducible polynomials in $\mathbb{Z}_2[x]$ with degree $5$.
I may use the fact that x, $x+1$ and $x^2+x+1$ are the irreducibles of degree less than or equal to 2
If someone could provide a step by step explanation of how to do this, that would be amazing! Thanks in advance!
P.S.: This is not a homework, but is a question on a past exam that I couldn't answer and I want to know how to do it.
| The following argument is specific to the question asked. Although I am
by no means a trained professional, please do not attempt this at home with
polynomials of larger degree.
The irreducible polynomial must be of the form
$x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + 1$, that is, the degree-$5$ term
as well as the constant term must be nonzero. Of the $16$ possible
choices of $(a_4,a_3,a_2,a_1)$ we can eliminate the $8$ vectors of
even Hamming weight because the corresponding degree-$5$ polynomials
have $1$ as a root. So, we are down to $8$ polynomials of which
$6$ are necessarily
irreducible because the $30$ elements of $\mathbb F_{32}-\mathbb F_2$
all have order $31$ and come in sets of $5$ conjugates. Starting with
the lexicographically first polynomial $x^5 + x + 1$, we readily
discover via long division that
$$x^5 + x + 1 = (x^2+x+1)(x^3+x^2+1)$$
and so $x^5 + x + 1$ and its reciprocal $x^5+x^4+1 = (x^2+x+1)(x^3+x+1)$
are reducible polynomials. Thus, the irreducible polynomials of
degree $5$ are
$$\begin{align}
x^5+x^2+1 &\quad x^5+x^3+1\\x^5+x^3+x^2+x+1 &\quad x^5+x^4+x^3+x^2+1\\
x^5+x^4+x^2+x+1 &\quad x^5+x^4+x^3 + x + 1
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/998563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Solving 2nd order ODE with Frobenius method - problems with summation symbol I'm trying to solve the ODE:
$$ y''(x) + \frac{2x}{(x-1)(2x-1)} y'(x) - \frac{2}{(x-1)(2x-1)} y(x) = 0 $$
I'm trying to find a solution by the Frobenius method, expanding a power series of the solution around $x = \frac 12$, that is in a series of terms $(x - 1/2)^{n}$.
The indicial equation has two roots, $\alpha = 0$ and $\alpha=2$. For $\alpha= 2$ the solution will be
$$ y(x) = \sum_{k=0}^{\infty} a_k \left( x - \frac 12 \right)^{k+2}$$
If I say that
$$ p(x) =\frac{1}{x-\frac12} \frac{x}{x-1} =\frac{1}{x-\frac12} \left(-1 + \sum_{i = 0}^{\infty} -2^{i+1} \left( x -\frac 12\right)^i \right) $$
and
$$ q(x) = -\frac{1}{\left( x -\frac12 \right)^2} \frac{x-1/2}{x-1} =\frac{1}{\left( x -\frac12 \right)^2} \left( -1 + \sum_{j=0}^{\infty} 2^{j+1} \left( x - \frac 12\right)^{j+1} \right) $$
and plug that and the power series expansion for $y, y', y''$ in the ODE, I get:
$$\sum_{k=0}^{\infty} (k+2)(k+1) a_k \left( x- \frac 12\right)^k + \frac{1}{x-\frac12} \left(-1 + \sum_{i = 0}^{\infty} -2^{i+1} \left( x -\frac 12\right)^i \right) \left(\sum_{k=0}^{\infty} (k+2) a_k \left( x- \frac 12\right)^{k+1}\right) + \frac{1}{\left( x -\frac12 \right)^2} \left( -1 + \sum_{j=0}^{\infty} 2^{j+1} \left( x - \frac 12\right)^{j+1} \right) \left(\sum_{k=0}^{\infty} a_k \left( x - \frac 12 \right)^{k+2} \right)$$
I go through the math and get
$$\sum_{k=0}^{\infty} (k+2)(k+1) a_k \left( x- \frac 12\right)^k + \sum_{k=0}^{\infty} \left( \sum_{i=0}^{k} -a_i (i+2) 2^{k-i+1} \right) \left( x- \frac 12\right)^{k} + \sum_{k=0}^{\infty} (k+2) a_k \left( x - \frac 12 \right)^k + \sum_{k=0}^{\infty} \left( \sum_{j=0}^{k} 2^{k-j} a_j \right) \left(x - \frac 12 \right)^k - \sum_{k=0}^{\infty} a_k \left( x - \frac 12 \right)^k = 0 $$
Now I equate the coefficients ofequal powers to 0:
$$k = 0 , 2 a_0 - 4 a_ + 2 a_0 + a_0 - a_0 = 0 <=> 0 a_0 = 0 $$
$$k = 1 , 6 a_1 - 8 a_0 - 6 a_1 + 3 a_1 + 2 a_0 + a_1 - a_1 = 0 <=> a1 = (6/3) a_0$$
Am I now getting this right?
| Consider the differential equation
\begin{align}
(x-1)(2x-1) y''(x) + 2x y'(x) - 2 y(x) = 0
\end{align}
with solutions expanded around $1/2$. For this it is seen that
\begin{align}
y(x) = \sum_{n=0}^{\infty} a_{n} \left(x - \frac{1}{2} \right)^{n+\sigma}
\end{align}
which leads to
\begin{align}
0 &= 2(x-1)\left(x-\frac{1}{2}\right) \sum_{n=0}^{\infty} a_{n}(n+\sigma)(n+\sigma-1) \left(x - \frac{1}{2} \right)^{n+\sigma-2} + 2x \sum_{n=0}^{\infty} a_{n}(n+\sigma) \left(x - \frac{1}{2} \right)^{n+\sigma-1}
\\
& \hspace{10mm} - 2\sum_{n=0}^{\infty} a_{n} \left(x - \frac{1}{2} \right)^{n+\sigma}\\
&= 2(x-1) \sum_{n=0}^{\infty} a_{n}(n+\sigma)(n+\sigma-1) \left(x - \frac{1}{2} \right)^{n+\sigma-1} + 2 x \sum_{n=0}^{\infty} a_{n}(n+\sigma)^2 \left(x - \frac{1}{2} \right)^{n+\sigma-1} \\
& \hspace{10mm} - 2 \sum_{n=0}^{\infty} a_{n} \left(x - \frac{1}{2} \right)^{n+\sigma} \\
&= 2[(x-1/2) - 1/2] \sum_{n=0}^{\infty} a_{n}(n+\sigma)(n+\sigma-1) \left(x - \frac{1}{2} \right)^{n+\sigma-1} \\
& \hspace{5mm} +2[(x-1/2) +1/2] \sum_{n=0}^{\infty} a_{n}(n+\sigma)^2 \left(x - \frac{1}{2} \right)^{n+\sigma} + \sum_{n=0}^{\infty} a_{n} \left(x - \frac{1}{2} \right)^{n+\sigma} \\
&= -\sum_{n=0}^{\infty} a_{n}(n+\sigma)(n+\sigma-2) \left(x - \frac{1}{2} \right)^{n+\sigma-1} +2 \sum_{n=0}^{\infty} a_{n}[(n+\sigma)^2-1] \left(x - \frac{1}{2} \right)^{n+\sigma} \\
0 &= -a_{0} (\sigma)(\sigma-2) + \sum_{n=0}^{\infty} (n+\sigma+1)(n+\sigma-1) [ (a_{n+1} - 2 a_{n}) \left(x - \frac{1}{2} \right)^{n+\sigma}
\end{align}
From this it is seen that
\begin{align}
0 &= a_{0} (\sigma)(\sigma-2) \\
a_{n+1} &= 2 a_{n}.
\end{align}
From the first equation it is seen that $a_{0} \neq 0$ and $\sigma = 0,2$. The coefficients $a_{n}$ are given by $a_{n} = 2^{n} a_{0}$. The solutions follow from
\begin{align}
y_{\sigma}(x) &=a_{0} \sum_{n=0}^{\infty} 2^{n} \left(x - \frac{1}{2} \right)^{n+\sigma}\\
&= a_{0} \left(x - \frac{1}{2}\right)^{\sigma} \sum_{n=0}^{\infty} (2x-1)^{n} \\
&= \frac{a_{0}}{2(1-x)} \, \left(x - \frac{1}{2} \right)^{\sigma}.
\end{align}
Since $\sigma = 0,2$ then the general solution follows the form
\begin{align}
y(x) &= \frac{a_{2}}{1-x} + a_{3} \, \frac{(1-2x)^{2}}{1-x} = \frac{a_{2}}{1-x} + a_{3} \left( \frac{1}{1-x} - 4 x \right) \\
&= \frac{A}{1-x} + B x.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$ I am trying to calculate the following series:
$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$$
and I managed to reduce it to this term
$$\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$$
And here I am stuck. I tried writing down a few partial sums but I can't see the pattern, $\frac{1}{2}-\frac{1}{2}+\frac{1}{5}+\frac{1}{4}-\frac{1}{3}+\frac{1}{8}+...$ I cant seem to find a closed formula that we can calculate for $S_n$
How should I go about solving this question
| $$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4}=\\\frac{1}{2n}-\frac{2}{2n+2}+\frac{1}{2n+4}=\\(\frac{1}{2n}-\frac{1}{2n+2})+(-\frac{1}{2n+2}+\frac{1}{2n+4})=\\(\frac{1}{2n}-\frac{1}{2n+2})-(\frac{1}{2n+2}-\frac{1}{2n+4})=\\\frac{1}{2}((\frac{1}{n}-\frac{1}{n+1})-(\frac{1}{n+1}-\frac{1}{n+2}))=\\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
How to prove $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2$? Prove the inequality for a triangle with sides $a,b,c$ we have $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} <2$$
Trial: Since $a,b,c$ are sides of a triangle I know $a+b>c,b+c>a,a+c>b$
| \begin{align*}\frac a{b+c}+\frac b{c+a}+\frac c{a+b}<2&\iff 3-2<1-\frac a{b+c}+1-\frac b{c+a}+1-\frac c{a+b}\\&\iff 1<\frac {b+c-a}{b+c}+\frac {c+a-b}{c+a}+\frac {a+b-c}{a+b}\end{align*}
Now
$$\frac {b+c-a}{b+c}+\frac {c+a-b}{c+a}+\frac {a+b-c}{a+b}>\frac {b+c-a}{a+b+c}+\frac {c+a-b}{a+b+c}+\frac {a+b-c}{a+b+c}=1$$
You need the triangle inequality to ensure the fractions on the left side are positive, so you decrease them by increasing the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
A question on cosine integral So I've read a book and found myself stumped in this integral:
$$\int_{0}^{\pi} \frac{\cos(n\theta)}{b^2-a^2\cos(2\theta)}\, d\theta=\begin{cases}
\,\,0 &,\quad\mbox{if}\,\, n\,\,\mbox{is odd}\\[20pt]
\,\,\dfrac{\pi}{\sqrt{b^4-a^4}}\left(\dfrac{\sqrt{b^2-\sqrt{b^4-a^4}}}{a}\right)^n&,\quad\mbox{if}\,\, n\,\,\mbox{is even}\\
\end{cases}$$
where $b>a$.
Anyone knows how to evaluate it? Or knows a reference for helping me to prove formula above?
| If $n=2k+1$, then
$$\int_{\pi/2}^\pi\frac{\cos((2k+1)\theta)}{b^2-a^2\cos(2\theta)}d\theta=\int_0^{\pi/2}\frac{\cos((2k+1)(\pi-\theta))}{b^2-a^2\cos(2(\pi-\theta))}d\theta=\int_0^{\pi/2}\frac{-\cos((2k+1)\theta)}{b^2-a^2\cos(2\theta)}d\theta$$
and the integral is $0$ because the convergence is trivial. So assume $n=2k$, then
$$\int_0^\pi\frac{\cos(2k\theta)}{b^2-a^2\cos(2\theta)}d\theta=\int_0^{2\pi}\frac{\cos(k\theta)}{2b^2-2a^2\cos\theta}d\theta=\frac1{a^2}\int_0^{2\pi}\frac{\cos(k\theta)}{2c-2\cos\theta}d\theta,$$
where $c=b^2/a^2>1$. Now
\begin{align*}\int_0^{2\pi}\frac{\cos(k\theta)}{2c-2\cos\theta}d\theta&=\text{Re}\int_0^{2\pi}\frac{e^{ik\theta}}{2c-(e^{i\theta}+e^{-i\theta})}d\theta=\text{Re}\frac1i\int_0^{2\pi}\frac{-e^{ik\theta}\cdot ie^{i\theta}}{e^{2i\theta}-2ce^{i\theta}+1}d\theta\\&=\text{Re}\frac1i\int_\varphi\frac{-z^k}{z^2-2cz+1}dz,\end{align*}
where $\varphi(\theta)=e^{i\theta}$ on $[0,2\pi]$. The denominator has 2 roots and only $\gamma=c-\sqrt{c^2-1}$ is inside $\text{Int}\,\varphi$, the residue is
$$\frac{-\gamma^k}{2\gamma-2c}=\frac{(c-\sqrt{c^2-1})^k}{2\sqrt{c^2-1}}.$$
Putting all things together yields the answer
$$\frac1{a^2}\text{Re}\frac{2\pi i}i\frac{(c-\sqrt{c^2-1})^k}{2\sqrt{c^2-1}}=\frac{\pi(c-\sqrt{c^2-1})^k}{a^2\sqrt{c^2-1}}=\frac{\pi\Big(b^2-\sqrt{b^4-a^4}\Big)^k}{a^{2k}\sqrt{b^4-a^4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Minimum of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$ What is the minimum of $$f(a,b,c):=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}$$ where $a,b,c$ are positive real numbers?
When $a=b=c$, we have $f(a,b,c)=\dfrac{3}{\sqrt{2}}\approx 2.12$
When $a=1,b=c\rightarrow\infty$, we have $f(a,b,c)\rightarrow 2$. So the minimum is at most $2$.
| Following mookid's hint, we can also avoid the use of Lagrange multiplicators. Normalize so that $a+b+c=1$, and then use the inequality $\sqrt{\dfrac{a}{1-a}}\geq 2a$. This is equivalent to $a(2a-1)^2\geq 0$.
Hence $f(a,b,c)\geq 2(a+b+c)=2$. Equality cannot hold, since $a=b=c=\dfrac{1}{2}$ doesn't satisfy $a+b+c=1$. But $f(a,b,c)$ can arbitrary get close to $2$, as the example in the original question shows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
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A combination integral and series resulting the inverse tangent integral $\def\Ti{{\rm{Ti}}_2}$I have been able to solve an integral problem, now I tried to use the other method to crack the integral and I have to prove the following expression
\begin{equation}
I=\sum_{n=1}^\infty \frac{z^{2n-1}}{2n-1}\int_0^{\Large\frac{\pi}{2}}\sin(2n-1)x\cot x\,dx=-\frac{1}{2}\Ti\left(\frac{1-z^2}{2z}\right)\qquad\quad\quad(\star)
\end{equation}
where $\Ti(\cdot)$ is the inverse tangent integral.
I managed to evaluate the integral and I obtained (if I am not mistaken)
\begin{equation}
\int_0^{\Large\frac{\pi}{2}}\sin(2n-1)x\cot x\,dx=\frac{(-1)^{n-1}}{2n-1}+2\sum_{k=1}^{n-1}\frac{(-1)^{k-1}}{2k-1}
\end{equation}
then
\begin{equation}
I=\sum_{n=1}^\infty \frac{z^{2n-1}}{2n-1}\left[\frac{(-1)^{n-1}}{2n-1}+2\sum_{k=1}^{n-1}\frac{(-1)^{k-1}}{2k-1}\right]=\Ti\left(z\right)+2\sum_{n=1}^\infty \frac{z^{2n-1}}{2n-1}\sum_{k=1}^{n-1}\frac{(-1)^{k-1}}{2k-1}
\end{equation}
I am stuck in the last expression. Could anyone here please help me to prove the expression $(\star)$? Any help would be greatly appreciated. Thank you.
| I stress that this is not a solution, but such a comment was too long to fit above.
I got a different formula for
\begin{equation}
\int_0^{\Large\frac{\pi}{2}}\sin(2n-1)x\cot x\,dx=\frac{(-1)^{n-1}}{2n-1}+2\sum_{k=1}^{n-1}\frac{(-1)^{k-1}}{2k-1}
\end{equation}
than what you posted. But after close inspection, they can be shown to be equivalent by induction. Anyway, I put my work here for clarity in case others are curious.
I'd also like to point out the identity
\begin{equation*}
2\tan^{-1}(z) = \tan^{-1}\left(\frac{2z}{1-z^{2}}\right)
\end{equation*}
which may help in putting the right hand side in the correct form.
Let $w = e^{ikx}$ and note that
\begin{align*}
\sum_{k=1}^{n-1}w^{2k} & = \frac{w^{2n} - w^{2}}{w^{2} - 1}\\
& = \frac{w^{2n-1} - w}{w - w^{-1}}\\
& = \frac{\cos((2n-1)x) + i \sin((2n-1)x) - \cos{x} - i\sin{x}}{2i\sin{x}}.
\end{align*}
Taking the real parts, we have
\begin{equation*}
\sum_{k=1}^{n-1}\cos(2kx) = \frac{\sin((2n-1)x)}{2\sin{x}} - \frac{1}{2}.
\end{equation*}
Multiplying both sides by $\cos{x}$ and rearranging yields
\begin{equation*}
\sin((2n-1)x)\cot(x) = \cos{x} + 2\sum_{k=1}^{n-1}\cos(2kx)\cos{x}
\end{equation*}
Now we use the fact that
\begin{equation*}
2\cos(2kx)\cos{x} = \cos((2k+1)x) + \cos((2k-1)x)
\end{equation*}
to get
\begin{align*}
\int_{0}^{\pi/2} \sin((2n-1)x)\cot{x}\,dx & = \int_{0}^{\pi/2}\cos{x}\,dx + \sum_{k=1}^{n-1}\int_{0}^{\pi/2}\left[\cos((2k+1)x) + \cos((2k-1)x)\right]\,dx\\
& = 1 + \sum_{k=1}^{n-1}\left[ \frac{\sin((2k+1)x)}{2k+1} + \frac{\sin((2k-1)x)}{2k-1} \right]_{x=0}^{\pi/2}\\
& = 1 + \sum_{k=1}^{n-1}\frac{(-1)^{k}}{2k+1} + \frac{(-1)^{k-1}}{2k-1}\\
& = 1 + \sum_{k=1}^{n-1}(-1)^{k}\frac{-2}{(2k+1)(2k-1)}\\
& = 1 + 2\sum_{k=1}^{n-1}\frac{(-1)^{k-1}}{(2k+1)(2k-1)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 1,
"answer_id": 0
} |
Finding the locus of a $z=x+iy$ and its radius if $|z-1|=2|z+2-3i|$
If the point $P$ in the complex plane corresponds to the complex number $z=x+iy$ show that if $|z-1|=2|z+2-3i|$ then the locus of $P$ is a circle centre at $-3+4i$, and find the radius of the circle.
Putting them into cartesian equations, we have:
$$
(x-1)^2+y^2=2[(x+2)^2+(y-3)^2]\\
x^2-2x+1+y^2=2x^2+8x+8+2y^2-12y+18\\
-x^2-10x-y^2+12y=25
$$
So it seems as if the centre must be $(-5,6)$ or $-5+6i$ with a radius of $5$, but that doesn't match the correct answer ($2\sqrt2)$. Am I doing something wrong?
| With thanks to Anurag A:
$$
(x-1)^2+y^2=4[(x+2)^2+(y-3)^2]\\
0=3x^2+18x+3y^2-24y+51\\
-(x+3)^2-(y-4)^2=-8\\
$$
The centre is therefore $(-3,4)$ or $-3+4i$ and the radius is $\sqrt8$ or $2\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Poisson Integral is equal to 1
Show
$$
\int_{-\pi}^{\pi}P(r, \theta)d\theta = 1
$$
Let $\alpha(r) = \frac{r^2 - 1}{2r}$ and $\gamma(r) = -\big(\frac{r^2 + 1}{2r}\big)$.
Then
$$
\frac{1}{2\pi}
\int_{-\pi}^{\pi}\frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}d\theta
= \frac{\alpha}{2\pi}
\int_{-\pi}^{\pi}\frac{1}{\cos(\theta) + \gamma}d\theta
$$
where
$$
\frac{r^2 - 1}{2r}\frac{1}{\cos(\theta) - \frac{1}{2r} - \frac{r^2}{2r}} = \frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}
$$
Next, let $z = e^{i\theta}$. Then $d\theta = \frac{-i}{z}dz$. Since $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$, $\cos(\theta) = \frac{z + z^{-1}}{2}$.
$$
= \frac{-i\alpha}{\pi}\int_C\frac{1}{z^2 + 2z\gamma + 1}dz
$$
and $C$ is the contour oriented counter clockwise with simple poles at $z = -\gamma\pm\sqrt{\gamma^2 - 1}$. Let $f(z) = \frac{1}{z^2 + 2z\gamma + 1}$. Then
$$
= \Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z)
\tag{1}
$$
The only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$.
Then
$$
2\pi i\lim_{z\to z_j}\Bigg[\big(z + \gamma - \sqrt{\gamma^2 - 1}\big)
\frac{1}{\big(z_j + \gamma + \sqrt{\gamma^2 - 1}\big)
\big(z + \gamma - \sqrt{\gamma^2 - 1}\big)}\Bigg] =
\frac{\pi i}{\sqrt{\gamma^2 - 1}}
$$
Now, we can substitute $\frac{\pi i}{\sqrt{\gamma^2 - 1}}$ for $2\pi i\sum\text{Res}$ in equation (1).
\begin{align*}
\Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z)
&= \frac{\alpha}{\sqrt{\gamma^2 - 1}}\\
&= \frac{r^2 - 1}{2r\sqrt{\frac{(r^2 + 1)^2}{4r^2} - 1}}\\
&= \frac{r^2 - 1}{\sqrt{(r^2 + 1)^2 - 4r^2}}\\
&= \frac{r^2 - 1}{\sqrt{r^4 - 2r^2 + 1}}\\
&= \frac{(r - 1)(1 + r)}{(r - 1)(r + 1)}\\
&= 1
\end{align*}
I have been unable to convince myself that the only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$. I know it is the case because if I use the other pole, the integral becomes $-1$
| Let $z_+ = -\gamma + \sqrt{\gamma^2 - 1}$ and $z_- = -\gamma - \sqrt{\gamma^2 - 1}$. Then
\begin{align}
z_+ &= \frac{r^2 + 1}{2r} + \sqrt{\frac{r^4 + 2r + 1 - 4r^2}{4r^2}}\\
&= \frac{r^2 + 1}{2r} + \frac{r^2 - 1}{2r}\\
&= \frac{2r^2}{2r}\\
&= r\\
z_- &= \frac{r^2 + 1}{2r} - \sqrt{\frac{r^4 + 2r + 1 - 4r^2}{4r^2}}\\
&= \frac{r^2 + 1}{2r} - \frac{r^2 - 1}{2r}\\
&= \frac{2}{2r}\\
&= \frac{1}{r}
\end{align}
So for $0<r<1$, only $z_+$ is in the unit circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
For all positive real numbers, is $f(x)=\sqrt{x}+x+2$ one to one? I understand that in order to prove this to be one to one, I need to prove $2$ numbers, $a$ and $b$, in the same set are equal.
This is what I did:
$$\sqrt{a} + a + 2 = \sqrt{b} + b + 2$$
$$\sqrt{a} + a = \sqrt{b} + b$$
$$a + a^2 = b + b^2$$
How would I arrive at $a = b$? Is it possible?
| $f(x)=\sqrt{x}+x+2$ is strictly increasing on $(0,\infty)$, so it's one-one (suppose not, use the strict monotonicity to draw a contradiction).
Or, from your second step,
$$\sqrt{a} + a = \sqrt{b} + b\iff \sqrt{a} - \sqrt{b}+ a -b=0 \iff(\sqrt{a} - \sqrt{b})(1+\sqrt{a}+\sqrt{b})=0 $$
Since $1+\sqrt{a}+\sqrt{b}>0$, we have $\sqrt{a} - \sqrt{b}=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
poisson gamma mixture Let $N$ be Poisson distributed with parameter $10B$, where $B \sim \Gamma(3,1)$ (i.e. $f(b)=\frac{b^2 e^{-b}}{2}$). Find the p.m.f of $N$. How should I manipulate $10B$ in the integration? What is its pdf?
| $\newcommand{\E}{\operatorname{E}}$We are given this conditional distribution:
$$
\Pr(N=n\mid B) = \frac{(10B)^ne^{-10B}}{n!}.
$$
We need to find a marginal distribution:
\begin{align}
\Pr(N=n) & = \E(\Pr(N=n\mid B)) = \E\left( \frac{(10B)^ne^{-10B}}{n!} \right) \\[8pt]
& = \int_0^\infty \frac{(10b)^n e^{-10b}}{n!} \cdot \frac{b^2 e^{-b}}2\,db \\[8pt]
& = \frac{10^n}{n!2} \int_0^\infty b^{n+2} e^{-11b} \, db \\[8pt]
& = \frac{10^n}{n!2}\cdot\frac{1}{11^{n+3}} \int_0^\infty (11b)^{n+2} e^{-11b} (11\, db) \\[8pt]
& = \frac{10^n}{n!2}\cdot\frac{1}{11^{n+3}} \int_0^\infty c^{n+2} e^{-c}\,dc \\[8pt]
& = \frac{10^n}{n!2}\cdot\frac{1}{11^{n+3}} \cdot (n+2)! \\[8pt]
& = \frac 1 {2\cdot11^3}\left(\frac{10}{11}\right)^n (n+2)(n+1) \\[8pt]
& = \binom{n+3-1}{n} \left(\frac{10}{11}\right)^n \left(1-\frac{10}{11}\right)^3.
\end{align}
So this is a negative binomial distribution: the distribution of the number of failures before third success in i.i.d. Bernoulli trials with probability $10/11$ of failure on each trial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $9\sin\theta+40\cos\theta=41$ then prove that $41\cos\theta=40$. I tried it this way:
$$ 40\cosθ+9\sinθ=41 $$
$$ 9\sinθ=41-40\cos\theta $$
Squaring both the sides:
$$81\sin^2\theta=1681+1600\cos^2\theta-2\cdot 40\cdot 41 \cos\theta$$
$$81-81 \cos^2\theta= 1681+1600\cos^2\theta-3280 \cos\theta$$
$$1681\cos^2\theta-3280\cos\theta+1600=0$$
Solving the quadratic equation gives $\cos\theta=\dfrac{40}{41}$. It is not easy to solve the quadratic equation without calculator so there must be some other method, if yes then please explain.
P.S: I've found the other method so I am self-answering the question.
| Using Brahmagupta-Fibonacci Identity,
$$(9\sin\theta+40\cos\theta)^2+(40\sin\theta-9\cos\theta)^2=(40^2+9^2)(\sin^2\theta+\cos^2\theta)=41^2$$ as $41^2=40^2+9^2$
So, $40\sin\theta-9\cos\theta=0$ and $9\sin\theta+40\cos\theta=41$
Can you solve for $\cos\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
factorise the expression $x^3 - 3x^2 -4x + 12$
Factorize the expression $$x^3-3x^2-4x+12$$ Hence calculate the ranges of values of $x$ for which $x^3-3x^2>4x-12$.
I factorised it to obtain $(x-2)(x-3)(x+2)$ but I don't how how to get to the next step.
| $$
{x}^{3}-3\,{x}^{2}-4\,x+12={x}^{3}-4\,x-3\,{x}^{2}+12=x(x^2-4)-(3x^2-12)=
$$
$$
=x(x^2-4)-3(x^2-4)=(x-3)(x^2-4)=(x-3)(x-2)(x+2).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Inequality problem involving QM-AM-GM-HM or Cauchy Schwarz inequality Question: Prove that if $x$, $y$, $z$ are positive real numbers then the following inequality holds: $$\frac{x+y}{x^2+y^2}+\frac{y+z}{y^2+z^2}+\frac{z+x}{z^2+x^2}\leq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$I tried thinking of applying QM-AM, but I didn't know what variables to use. I also thought that this could be related to Cauchy's inequality, but I wasn't sure, again, of what variables to use. Any help would be greatly appreciated, thanks!
| We want to show that:
$$\frac{x+y}{x^2+y^2}+\frac{y+z}{y^2+z^2}+\frac{z+x}{z^2+x^2}\leq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$
Note that, by the QM-AM Inequality:
$${\left(\frac{x+y}{2}\right)}^2 \leq \frac{x^2 + y^2}{2}.$$
Consequently:
$$\frac{x + y}{x^2 + y^2} \leq \frac{2}{x + y}.$$
Therefore, it remains to show that:
$$\frac{2}{x + y} + \frac{2}{y + z} + \frac{2}{x + z} \leq \frac{1}{x} + \frac{1}{y} + \frac{1}{z}.$$
By the AM-HM Inequality:
$$\frac{2}{\frac{1}{x} + \frac{1}{y}} \leq \frac{x + y}{2}.$$
Therefore:
$$\frac{2}{x + y} \leq \frac{\frac{1}{x} + \frac{1}{y}}{2}$$
Similarly:
$$\frac{2}{y + z} \leq \frac{\frac{1}{y} + \frac{1}{z}}{2},$$
and
$$\frac{2}{x + z} \leq \frac{\frac{1}{x} + \frac{1}{z}}{2}.$$
QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
directional derivatives in (0,0) I've this function : $f(x,y)= \dfrac{(1+x^2)x^2y^4}{x^4+2x^2y^4+y^8}$ for $(x,y)\ne (0,0)$ and $0$ for $(x,y)=(0,0)$
It's admits directional derivatives at the origin?
| A function admits directional derivative at a point if its gradient $\nabla{f}$ exists at that point. The gradient of your function is given by,
$$\nabla{f}=\left(\begin{array}{cc} -\frac{2\, x\, y^4\, \left(2\, x^6 + 3\, x^4 - 2\, y^4 + 1\right)}{{\left(x^6 + x^2 + 2\, y^4\right)}^2} & \frac{4\, x^2\, y^3\, \left(x^6 + x^4 + x^2 + 1\right)}{{\left(x^6 + x^2 + 2\, y^4\right)}^2} \end{array}\right)$$
which doesn't exist at $(x,y)=(0,0)$, i.e. the origin, since you have $\frac{0}{0}$ which is undefined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $\sqrt [3]{2}-\sqrt [3]{4}$ is algebraic How do I show, step by step, that $\sqrt [3]{2}-\sqrt [3]{4}$ is a root of $x^3+6x+2$?
Start with $x=\sqrt [3]{2}-\sqrt [3]{4}$ do not use the cubic, the cubic is given for convenience.
( This is example 4.1.3 from Introductory ANT by Alaca/Williams )
| Raising the given number to the third power and using the identity $$(a-b)^3=a^3-3a^2b+3ab^2-b^3=a^3-b^3–3ab(a–b)$$ you obtain $$\left(\sqrt[3]{2}-\sqrt[3]{4}\right)^3=2-4-3\sqrt[3]8\left(\sqrt[3]{2}-\sqrt[3]{4}\right)=-2-6\left(\sqrt[3]{2}-\sqrt[3]{4}\right)$$ or equivalently $$\left(\color{blue}{\sqrt[3]{2}-\sqrt[3]{4}}\right)^3+6\left(\color{blue}{\sqrt[3]{2}-\sqrt[3]{4}}\right)+2=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Cylindrical coordinates on elliptic paraboloids. I want to compute the volume bounded by:
*
*the cylinder $x^2+4y^2=4$.
*the $z=0$ plane.
*the elliptic paraboloid $z = x^2 + 6y^2$.
I would like to use cylindrical coordinates. However I have never dealt with a problem in which $r$ bounds depend on $\theta$.
Here there's a horrible sketch of the solid:
The base is an ellipse. So I take the transformation to cylindrical coordinates:
\begin{align}
x&=r\cos\theta,\\
y&=r\sin\theta, \\
z&=z.
\end{align}
Which implies $0\leq z\leq x^2+6y^2 = r^2\cos^2\theta + 6r^2\sin^2\theta$ and $0\leq\theta\leq2\pi$.
However I'm having difficulties in determining the $r$ bounds. Also I would like to know the equation of the curve of intersection between the cylinder and the paraboloid, but don't know how to do it.
For the $r$ bounds I tried considering the equation for the base:
$$4y^2+x^2=4\Rightarrow 4r^2\sin^2\theta+r^2\cos^2\theta = 4 \Rightarrow r=\frac{2}{\sqrt{4\sin^2\theta+\cos^2\theta}}.$$
I'm not sure if doing that is correct, also the integral looks absolutely grim. I really appreciate your help.
| Consider the change of variables $x=2u\Rightarrow dx = 2\,du$. Now we are working on a circular cylinder: $u^2+y^2=1$. Now let's take cylindric coordinates. Note that the Jacobian is $2r$. And the volume:
\begin{align}
V=\int_0^1\int_0^{2\pi}\int_0^{4r^2\cos^2\theta+6r^2\sin^2\theta} 2r\,dz\,d\theta\,dr = \int_0^1\int_0^{2\pi} 8r^3\cos^2\theta+12r^3\sin^2\theta\,d\theta\,dr\\
=\int_0^1\int_0^{2\pi} \overbrace{ 4r^3+4r^3\cos2\theta+6r^3-6r^3\cos2\theta}^{2\theta = s\;\Rightarrow\; 2\,d\theta=ds} \,d\theta\,dr\\
=\int_0^2\left[8\pi r^3+\int_0^{4\pi}2r^3\cos s\,ds+12\pi r^3 - \int_0^{4\pi}3r^3\cos s\,ds\right]dr\\
=\int_0^1 20\pi r^3\, dr = 5\pi.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find all integers n such that $\;\frac{n^2-9}{n^2-5n+4}$ is an integer.
Find all integers such that $\;\dfrac{n^2-9}{n^2-5n+4}\;$ is an integer.
I am really struggling to figure this out. I can tell that -3,3, and 5 are solutions but I don't know how to show that these are the only solutions or if they even are the only solutions.
| Longer, but using only divisibility considerations:
This factors as $\frac{(n-3)(n+3)}{(n-4)(n-1)}$. Note $n$ cannot be $1$ or $4$.
So $|n-1|\geq1$, and if $p$ is a prime dividing $n-1$, then $p$ has to divide $n-3$ or $n+3$. So mod $p$, either $n\equiv1\equiv3$ or $n\equiv1\equiv-3$. Either way, $p$ must be $2$, and $n-1=\pm2^k$ for some $k\geq0$.
$$\frac{(\pm2^k-2)(\pm2^k+4)}{(\pm2^k-3)(\pm2^k)}$$
Now we see $k$ can't be $4$ or more, or else the denominator is divisible by $16$ and the numerator is not. So there are two ($\pm$) cases for each of $k=0,1,2,3$:
$$\begin{align}
\frac{(\pm1-2)(\pm1+4)}{(\pm1-3)(\pm1)}&&\frac{(\pm2-2)(\pm2+4)}{(\pm2-3)(\pm2)}&&\frac{(\pm4-2)(\pm4+4)}{(\pm4-3)(\pm4)}&&\frac{(\pm8-2)(\pm8+4)}{(\pm8-3)(\pm8)}
\end{align}$$
There are eight options to check.
$$\begin{align}
\frac{5}{2}&&0&&4&&\frac{9}{5}\\
\frac{-9}{4}&&\frac{4}{-5}&&0&&\frac{5}{11}
\end{align}$$
The integers happened for $$n=2^1+1=3$$ $$n=2^2+1=5$$ $$n=-2^2+1=-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How can I calculate $\sum_{j=0}^{\infty}(j+1)(\frac{1}{1.05})^{j+1}$ I have to calculate the sum:
$$\sum_{j=0}^{\infty}(j+1)\cdot\left(\frac{1}{1.05}\right)^{j+1}$$
I know it is convergent from the ratio test.
| Let
$$f(x)=\sum_{k=0}^{\infty}(k+1)(x)^{k+1}=\sum_{k=0}^{\infty}(k+2-1)(x)^{k+1}=\sum_{k=0}^{\infty}(k+2)x^{k+1}-\sum_{k=0}^{\infty}x^{k+1}$$
for $|x|\lt1$
$$g'(x)=\sum_{k=0}^{\infty}(k+2)x^{k+1}=2x+3x^2+4x^3+\cdots$$
$$g(x)=x^2+x^3+x^4+\cdots=\frac{x^2}{1-x}$$
$$g'(x)=\frac{x(2-x)}{(1-x)^2}$$
$$p(x)=\sum_{k=0}^{\infty}x^{k+1}=x+x^2+x^3+\cdots=\frac{1}{1-x}-1$$
$$f(x)=g'(x)-p(x)$$
$$f(x)=\sum_{k=0}^{\infty}(k+1)(x)^{k+1}=\frac{x(2-x)}{(1-x)^2}-\left(\frac{1}{1-x}-1\right)$$
$$f(x)=\frac{x}{(1-x)^2}$$
Now substitute $x=\dfrac{1}{1.05}$ to get
$$\sum_{k=0}^{\infty}(k+1)\left(\frac{1}{1.05}\right)^{k+1}=420$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What is the probability of two out of three events happening? All events are independent.
$$\Pr(A) = \frac{9}{10}$$
$$\Pr(B) = \frac{9}{10}$$
$$\Pr(C) = \frac{6}{10}$$
What is the probability of at least two events happening?
I'd like to use negation, to negate the possibility that event no event happen plus the probability that only one happens.
$$D = \text{at least two events happen}$$
$$\Pr(D) = 1-\Pr(\text{none happens})-\Pr(\text{exactly one happens})$$
$$\Pr(D) = 1 - \left(\frac{1}{10}\cdot \frac{1}{10}\cdot\frac{4}{10}\right) - \left(\frac{1}{10}\cdot\frac{6}{10}\cdot\frac{1}{10} + \frac{9}{10}\cdot\frac{1}{10}\cdot\frac{4}{10} + \frac{9}{10}\cdot\frac{1}{10}\cdot\frac{4}{10}\right) = 0.918$$
The answer seems a little larger, I can't convince myself that I'm right.
| Your approach works, and your answer is correct.
It can also be calculated straight forwards as:
$\begin{align}
\mathsf P(AB\cup AC\cup BC)
& = \mathsf P(AB)+\mathsf P(A^cBC)+\mathsf P(AB^cC)
\\[1ex]
& = \mathsf P(A)\mathsf P(B) + \bigg(\mathsf P(A^c)\mathsf P(B)+\mathsf P(A)\mathsf P(B^c)\bigg)\mathsf P(C)
\\[0ex]
&= \frac{9^2}{10^2} + \bigg(\frac{1\cdot 9}{10^2} + \frac{9\cdot 1}{10^2}\bigg)\frac 6{ 10}
\\[1ex]
& = \frac{918}{1000}
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)} $ Let $x,y,z$ be real numbers in the range of $(0,1)$.
Prove that
$$\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)}.$$
| It's obviously true after following substitution and full expanding.
$x=\frac{a}{a+1}$, $y=\frac{b}{b+1}$ and $z=\frac{c}{c+1}$, where $a$, $b$ and $c$ are positives.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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convergence series geometric test Prove if this converges:
$$\sum_{n=1}^\infty \frac{2^n+3}{3^n-1}$$
pf: using geometric
$$0 < \frac{2^n+3}{3^n-1} \leq \frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}} = \cdots $$ and so on
I know how to do the rest but my question is that where in the world did my teacher get
$$\frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}}$$
| Ratio test:
$$
\frac{\left(\dfrac{2^{n+1}+3}{3^{n+1}-1}\right)}{\left(\dfrac{2^n+3}{3^n-1}\right)} \to \frac 2 3 \text{ as }n\to\infty,
$$
so the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove or disprove: $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$ Is it true that $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$ for any positive integer $n$?
We cannot prove the statement using induction as it is, because the left-hand side is increasing while the right-hand side stays constant. So we need to modify the right-hand side to something like $3-\dfrac{1}{n}$. But it is hard to quantify how much the left-hand side increases when we go from $n$ to $n+1$.
| You can think of the left hand side as the geometric mean of the multi-set with $2^{n-k}$ values $k$ for $k=2,\dots, n$, plus an additional one, for a total of $2^{n-1}$ elements. The AM/GM rule then gives:
$$\frac{1}{2^{n-1}}\left(1+\sum_{k=1}^n k2^{n-k}\right) = \frac{1}{2^{n-1}} + \sum_{k=2}^n\frac{k}{2^{k-1}}$$
as a strict upper bound for your formula.
Since $\sum_{k=1}^{\infty} \frac{k}{2^{k-1}} = \frac{1}{(1-1/2)^2}= 4$, and noting we subtract the term for $k=1$, that gives an upper bound of $3+\frac{1}{2^{n-1}}$.
Since the original sequence is increasing, this means that the values are$\leq 3$ for every $n$. In particular, then, each value has to be less than $3$, since the sequence is strictly increasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How do you find the imaginary roots of a fourth degree polynomial that cannot be simplified? I started out with $f(x)=16x^6-1$, and I got it down to $64x^4+16x^2+4$ by synthetically dividing by roots $0.5$ and $-0.5$ How should I continue in order to find the other roots?
| $16x^6=1$
$(\sqrt[3]{4}x)^6=1$
$u=\sqrt[3]{4}x$
$u^6=1$
These roots are spaced at u=$\pm 1$ and $\pm \frac{1}{2} \pm \frac{i\sqrt{3}}{2}$ ($n^{th}$ roots of unity)
Or $+1,-1, +\frac{1}{2} -\frac{i\sqrt{3}}{2}, +\frac{1}{2} -\frac{i\sqrt{3}}{2} , - \frac{1}{2} + \frac{i\sqrt{3}}{2}, - \frac{1}{2} - \frac{i\sqrt{3}}{2}$
So substituting x back into the solutions... and solving for x, you get the solutions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.