Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Factor $1+x+x^2+x^3+...+x^{14}$ In a previous task, I was asked to factor $1+x+x^2+x^3$ for $x \in \mathbb{R}$, which I accomplished by solving
$1+x+x^2+x^3 = 0 \to $
$1+x(1+x+x^2) = 0 \to $
$x(1+x+x^2) = -1$
which has a solution $x = -1$, and thus I knew $(x+1)$ was a factor. A bit of guesswork gave me $(x+1)(x^2+1)$.
Now I'm asked to factor $1+x+x^2+x^3+...+x^{14}$ for $x \in \mathbb{R}$ and I'm a bit stuck. Again, we have the implication $x(1+x+x^2+...+x^{13}) = -1$, for which $x=-1$ is a solution, so again we have a factor $x+1$. But now I cannot apply guesswork to determining the rest of the factors, so I feel there is some kind of conclusion I can draw about the powers (perhaps their parity) to solve this problem?
| Careful about concluding from $x(1+x+x^2+...+x^{13}) = -1$ that $x=-1$ must be a solution:
$$\textrm{When}\;\; x = -1,\;\; x(1+x+x^2+...+x^{13}) = -1 [7(1) + 7(-1)] = -1(0) = 0\neq -1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/249988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Finding the minimum of $\frac pq + \frac rs$ for distinct integers $p, q, r, s$ from $\{1,2,3,4,5,\ldots,16,17\}$ Here is the question:
Four distinct integers $p$, $q$, $r$ and $s$ are chosen from the set $\{1, 2, 3, 4, 5, \ldots, 16, 17\}$. The minimum possible value of $\frac pq + \frac rs$ can be written as $\frac ab$, where $a$ and $b$ are positive, co-prime integers. What is the value of $a+b$?
I got the answer as $321$. I took the numbers as $\frac pq$ as $\frac 1{16}$ and $\frac rs$ as $\frac 2{17}$ and added them to get $\frac ab = \frac{49}{272}$. Well I first took $\frac pq$ and $\frac rs$ as $\frac 1{17}$ and $\frac 2{16}$ respectively.
So without calculating both the results I wouldn't have got the answer. So can you help me by giving the way to get the answer?
| $\frac{1}{16} + \frac{2}{17} = \frac{1}{16} + \frac{1}{17} + \frac{1}{17}$
$\frac{2}{16} + \frac{1}{17} = \frac{1}{16} + \frac{1}{16} + \frac{1}{17}$
So obviously the first one is smaller.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/250607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$
Possible Duplicate:
How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$?
How can I calculate precise value of that series: $\sum\limits_{i=0}^{n-1} i2^i$ ?
So far, I tried to differentiate the $ \sum\limits_{i=0}^{n} 2^i = 2^{i-1} - 1 $ series, but by result $2^n(n+2)$ isn't correct according to Wolfram ($2^n (n-2)+2$).
| $$x+x^2+x^3+...+x^{n-1}+x^n=\frac {x^{n+1}-x}{x-1}$$
$+$
$$0x+x^2+x^3+...+x^{n-1}+x^n=\frac {x^{n+1}-x^2}{x-1}$$
$+$
$$0x+0x^2+x^3+...+x^{n-1}+x^n=\frac {x^{n+1}-x^3}{x-1}$$
$$.$$
$$.$$
$$.$$
$+$
$$0x+0x^2+0x^3+...+0x^{n-1}+x^n=\frac {x^{n+1}-x^n}{x-1}$$
After adding we get:
$$x+2x^2+...+nx^n=\sum_{i=1}^{n}\frac {x^{n+1}-x^i}{x-1}=\frac{nx^{n+1}}{x-1}-\sum_{i=1}^{n}\frac {x^i}{x-1}=\frac{nx^{n+1}}{x-1}-\frac {x^{n+1}-x}{(x-1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/250746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
How come $2 \times 3^k + 3^k = 3 \times 3^k$ I get something else
$$\begin{align*}
&2 \times 3^k + 3^k=\\
&2 \times 3^k \times 2=\\
&4 \times 3^k
\end{align*}$$
What does $3^k + 3^k$ give exactly?
| It looks like you are confusing $2\cdot 3^k+3^k$ with $2 \cdot (3^k+3^k)$. The first gives $3 \cdot 3^k=3^{k+1}$, while the second gives $2 \cdot (2 \cdot 3^k)=4\cdot 3^k$. Parentheses are important.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/251905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to integrate $(x^{1/2} + 5 x^{1/3})^{-1}$? $$ \int_0^1 \frac{dx}{x^\frac12+5x^\frac13} $$
Let $$u = 5+x^\frac16 $$
I have change x into u, and find $$\int_5^6 \frac{7}{6u(u-5)^3} $$
Then, i don't know how to integrate it.
| $$
\int \frac{1}{u(u-5)^3}du=\int\frac{A}{u}+\frac{B}{(u-5)}+\frac{C}{(u-5)^2}+\frac{D}{(u-5)^3}du.
$$
Courtesy of partial fraction decomposition. :-)
Edit: It would appear your $u$ sub does not work. (Thanks, Gerry.) Nonetheless, I hope this is a helpful illustration of partial fraction decomposition.
Here is why your $u$ sub does not work: If $u=5+x^{\frac{1}{6}}$ and we wish to write $\dfrac{7}{6u(u-5)^3}$, we find the following:
$$
\frac{7}{6u(u-5)^3}=\frac{7}{6}\frac{1}{(5+x^{\frac{1}{6}})(x^{\frac{1}{6}})^3}=\frac{7}{6}\frac{1}{(5+x^{\frac{1}{6}})(x^{\frac{1}{2}})}=\frac{7}{6}\frac{1}{(5x^{\frac{1}{2}}+x^{\frac{2}{3}})}\ne \frac{1}{x^{\frac{1}{2}}+5x^{\frac{1}{3}}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/253014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $5/2 < e < 3$?
Prove that $$\dfrac{5}{2} < e < 3$$
By the definition of $\log$ and $\exp$,
$$1 = \log(e) = \int_1^e \dfrac{1}{t} dt$$
where $e = \exp(1)$.
So given that $e$ is unknown, how could I prove this problem? I think I'm missing some important facts that could somehow help me rewrite $e$ in some form of $3$ and $5/2$. Any idea would be greatly appreciated.
| Method 1:
Using the power series for $\log(1+x)$, we get
$$
\begin{align}
\log\left(1+\frac1n\right)
&=-\log\left(1-\frac1{n+1}\right)\\
&=\frac1{n+1}+\frac1{2(n+1)^2}+\frac1{3(n+1)^3}+\dots\tag{1}
\end{align}
$$
Multiply $(1)$ by $n+1$:
$$
(n+1)\log\left(1+\frac1n\right)=1+\frac1{2(n+1)}+\frac1{3(n+1)^2}+\dots\tag{2}
$$
$(2)$ is obviously decreasing in $n$, therefore $\displaystyle\left(1+\frac1n\right)^{\large n+1}$ is decreasing in $n$.
Multiply $(1)$ by $n$:
$$
\begin{align}
n\log\left(1+\frac1n\right)
&=((n+1)-1)\log\left(1+\frac1n\right)\\
&=1-\frac1{1\cdot2(n+1)}-\frac1{2\cdot3(n+1)^2}-\dots\tag{3}
\end{align}
$$
$(3)$ is obviously increasing in $n$, therefore $\displaystyle\left(1+\frac1n\right)^{\large n}$ is increasing in $n$.
Therefore, since $\displaystyle e=\lim_{n\to\infty}\left(1+\frac1n\right)^{\large n}$, we have
$$
\left(1+\frac1n\right)^{\large n}< e<\left(1+\frac1n\right)^{\large n+1}\tag{4}
$$
Let $n=6$, then
$$
\frac52<\left(\frac76\right)^{\large 6}< e<\left(\frac76\right)^{\large 7}<3\tag{5}
$$
Method 2:
Since $\displaystyle e=\sum_{n=0}^\infty\frac1{n!}\ $, we have
$$
\begin{align}
\frac52=1+1+\frac12< e
&=1+\frac11+\frac1{1\cdot2}+\frac1{1\cdot2\cdot3}+\frac1{1\cdot2\cdot3\cdot4}+\dots\\
&<1+\frac11+\frac1{1\cdot2}+\frac1{1\cdot2\cdot2}+\frac1{1\cdot2\cdot2\cdot2}+\dots\\
&=3\tag{6}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/254335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 3
} |
Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$ I've tried following this way, but I haven't succeeded.
Thank you!
| Calculate the derivative of both sides:
$$(\arctan x)'=\frac{1}{1+x^2}$$
$$\left(\arcsin\frac{x}{\sqrt{1+x^2}}\right)'=\frac{\sqrt{1+x^2}-\frac{x^2}{\sqrt{1+x^2}}}{1+x^2}\cdot\frac{1}{\sqrt{1-\frac{x^2}{1+x^2}}}=$$
$$=\frac{1}{(1+x^2)\sqrt{1+x^2}}\cdot\frac{\sqrt{1+x^2}}{\sqrt 1}=\frac{1}{1+x^2}$$
Since both derivatives are equal the functions are the same up to the sum of a constant:
$$\arctan x=\arcsin\frac{x}{\sqrt{1-x^2}}+C\,\,\,,\,\,C=\,\text{a constant}$$
Finally, to find what $\,C\,$ is you can, for example, input $\,x=0\,$ in the above...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/254561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 0
} |
simplify summation of factorial (random walk) I suspect that the expression
$$\sum_{n=0}^N \frac{(N-2n)^2}{n!(N-n)!}$$
simplifies to
$$\frac{2^N}{(N-1)!}$$
But I cannot find the intermediate steps. Can someone give me a hint how I can deduce this result?
(The expression comes up when calculating the average final position after a random walk of $N$ steps.)
| $$(N-2n)^2 = (N-n)^2 + n^2 - 2n(N-n)$$
Hence,
\begin{align}
S_N(n) & = \dfrac{(N-2n)^2}{n!(N-n)!}\\
& = \dfrac{(N-n)^2 + n^2 - 2n(N-n)}{n!(N-n)!}\\
& = \dfrac{N-n}{n!(N-n-1)!} + \dfrac{n}{(n-1)! (N-n)!} - 2 \dfrac1{(n-1)!(N-n-1)!}\\
& = \dfrac1{n!(N-n-2)!} + \dfrac1{n!(N-n-1)!} + \dfrac1{(n-2)!(N-n)!} + \dfrac1{(n-1)!(N-n)!} - \dfrac2{(n-1)!(N-n-1)!}
\end{align}
Now note that from the binomial expansion of $2^{N-k-m}$, we have that
$$\sum_{n=\min(k,a)}^{\max(N-m,b)} \dfrac1{(n-k)!(N-n-m)!} = \dfrac{2^{N-k-m}}{(N-k-m)!}$$
Hence, $$\sum_{n=0}^{n=N} S_N(n) = \dfrac{2^{N-2}}{(N-2)!} + \dfrac{2^{N-1}}{(N-1)!} + \dfrac{2^{N-2}}{(N-2)!} + \dfrac{2^{N-1}}{(N-1)!} -2 \dfrac{2^{N-2}}{(N-2)!} = \dfrac{2^N}{(N-1)!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/258882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
No natural numbers sucht that $x+3y+5z=2012$ and $x^2+y^2+3z^2=2013$. Prove that there are not exist the natural numbers, $x,y,z$ such that $x+3y+5z=2012$ and $x^2+y^2+3z^2=2013$.
tell me please if my proof is ok ?
$x^2+y^2=3\cdot671-3z^2=3(671-z^2)=3k$ but if we want we can find out $z$. $671 \geq z^2$ so $z=\overline{0\ldots25}.$
But we know that a perfect square has the following form: $4k$ or $4k+1$. For example $x^2=4p+1$ and $y^2=4q$ so $x^2+y^2=4(p+q)+1\neq3k$ for $k\neq 3$ and $p+q \neq 2$.
I cannot find anothers numbers such that $4(p+q)+1=3k$
Is OK?
thanks :)
| Just work $\bmod\ 2$: Your first equation reads $x+y+z=0$ and your second reads $x^2+y^2+z^2=1$, but since $x^2\equiv x\bmod\ 2$, the second is equivalent to $x+y+z=1$ and the two are immediately incompatible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/258962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Do there exist any odd prime powers that can be represented as $n^4+4^n$? Well, I wrote up a solution on it, but according to the place I found the problem, it isn't quite correct. Ah, I'm simply hoping someone will point out where I got wrong.
Now, let, $n^4+4^n = p^k$, where $p$ is an odd prime and $k$ is a positive integer.
Further, $p^k \equiv 1 \pmod 2$. Therefore, $n^4 + 4^n \equiv n^4 \equiv 1 \pmod 2$, and so $n \equiv 1 \pmod 2$. $n$ must be odd.
Okay, now let, $n = 2m +1$. Substituting it in $n^4+4^n$ and using the Sophie Germain inequality, we have,
$$n^4+4\cdot4^{2m} = n^4 + 4(2^m)^4 = (n^2 + 2^n+2^{m+1}\,n)(n^2 + 2^n-2^{m+1}\,n) = p^k$$
Now, as $p^k$ can only be factorized into smaller powers of $p$, let $n^2 + 2^n+2^{m+1}\,n = p^i$, and let $n^2 + 2^n-2^{m+1}\,n = p^j$ where $i+j= k$, obviously, and $i>j$.
Now consider this:
$$\begin{align}
p^i - p^j & \equiv 0\\
2\cdot2^{m+1}\,n = 2^{m+2}\,n &\equiv 0 \pmod p
\end{align}$$
But, as $p$ is odd, $\gcd(p, 2) = 1$, so $n \equiv 0 \pmod p$.
Now look at this:
$$\begin{align}
p^i + p^j &\equiv 0 \\
2(n^2 + 2^n) &\equiv 0 \\
n^2 + 2^n &\equiv 0 \pmod p
\end{align}$$
But we just established that $n \equiv 0 \pmod p$, so $2^n \equiv 0 \pmod p$.
Therefore, let $2^n = jp$ for some integer $j$.
Now, $2^n$ is its own prime factorization, which is unique according to the Fundamental Theorem of Arithmetic and does not include $p$.
Therefore, the above statement is an impossibility!
There exist no such $p$ and $n$, and no odd prime powers can be written as $n^4+4^n$.
Ah, well, that's it.
Sorry for the tediousness of it. I've still no clue how to use $\LaTeX$.
Thank you everybody,
Cheers.
| For odd $n>1$ the expression is never prime, it can be factored into a product where each factor is the sum of two squares:
$$[((2^m)+n)^2 +(2^m)^2][(2^m)^2 +((2^m)^2-n)^2],$$
where $m = (n-1)/2$.
Looks like I didn't read the question properly thanks for the feedback.
I will try and improve my answer.
Let the first factor be $A$ and the second $B$. Then we have $AB = p^k$
with $k >1$, as if $k= 1$ then we have the case $AB =5$.
This means in $A$ and $B$, $n >1$, and so $m$ is at least $1$, (as $n$ is odd).
Then it's easy to see that neither $A$ or $B$ is trivially $1$. So we can set $k = r+s$ and as $A > B$ we can put $A = p^r$, and $B = p^s$, with $r>s>0$.
Now forming the difference of $A$ and $B$, means, we have;
$$A - B = p^r - p^s = n 2^{m+2}. $$
But as $p>s>0$, we have $p$ divides $n$ ($p$ is an odd prime) but this means $p$ divides $4$, as we have assumed $p^k = n^4 + 4^n$, with $k >1$, which is absurd.
Therefore the original expression is only only the power of an odd prime for the case $p = 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/261925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Geodesic of helicoid What is the geodesic of the helicoid?
$M=\{ (x,y,z)\in \mathbb{R}^3: x\sin z - y\cos z =0\}$
Whose tangent at the point $p = (1,0,0)$ in the line $r = \{(x, y, z) \in \mathbb {R} ^ 3: x = 1, y = z\}$.
I tried with the definition, but I don't know the parameterization, and so do not know which is the tangent plane. I can't do $\alpha ''(t) \perp T_{\alpha(t)}M$.
Any help is appreciated.
| The general parametrisation is $X(u,v) = \left( u \cos(v) , u \sin(v) , kv \right)$, where $k$ is a non-zero constant (greater than zero for right, or less than for left), and looking at your definition of $M$ it appears that $k=1$. It is always important to remember that there is usually more than one way to parametrise a surface, so as long as you can argue that it is one-to-one and continuously differentiable then you're all 'g'!
Step 1: compute the first partial derivatives
\begin{align*}
X_u &= \left( \cos(v) , \sin(v) , 0 \right) \\
X_v &= \left( -u \sin(v) , u \cos(v) , k \right)
\end{align*}
Step 2: compute the coefficients of the first fundamental form:
\begin{align*}
E &= X_u \cdot X_u = 1; & F &= X_u \cdot X_v = 0; & G &= X_v \cdot X_v = u^2+k^2.
\end{align*}
Step 3: compute the Christoffel symbols by solving the following system for $\Gamma_{ij}^k$:
\begin{align*}
\Gamma_{11}^1 E + \Gamma_{11}^2 F &= X_{uu} \cdot X_u = \frac{E_u}{2}; \\
\Gamma_{11}^1 F + \Gamma_{11}^2 G &= X_{uu} \cdot X_v = F_u - \frac{E_v}{2}; \\
\Gamma_{12}^1 E + \Gamma_{12}^2 F &= X_{uv} \cdot X_u = \frac{E_v}{2}; \\
\Gamma_{12}^1 F + \Gamma_{12}^2 G &= X_{uv} \cdot X_v = \frac{G_u}{2}; \\
\Gamma_{22}^1 E + \Gamma_{22}^2 F &= X_{vv} \cdot X_u = F_v - \frac{G_u}{2}; \\
\Gamma_{22}^1 F + \Gamma_{22}^2 G &= X_{vv} \cdot X_v = \frac{G_v}{2}.
\end{align*}
Therefore,
\begin{align*}
\Gamma_{11}^1 &= X_{uu} \cdot X_u = \frac{E_u}{2} = 0; \\
\Gamma_{11}^2 \left( u^2+k^2 \right) &= X_{uu} \cdot X_v = F_u - \frac{E_v}{2}=0; \\
\Gamma_{12}^1 &= X_{uv} \cdot X_u = \frac{E_v}{2}=0; \\
\Gamma_{12}^2 \left( u^2+k^2 \right) &= X_{uv} \cdot X_v = \frac{G_u}{2}=u; \\
\Gamma_{22}^1 &= X_{vv} \cdot X_u = F_v - \frac{G_u}{2}=-u; \\
\Gamma_{22}^2 \left( u^2+k^2 \right) &= X_{vv} \cdot X_v = \frac{G_v}{2}=0.
\end{align*}
This yields $\Gamma_{22}^1 = -u$, $\Gamma_{12}^2 = u/\left( u^2+k^2 \right)$, and the rest are equal to zero.
Step 4: compute the geodesic curves $\alpha(t) = (u(t),v(t))$ by solving
\begin{align}
\frac{\mathrm{d}^2 u}{\mathrm{d}t^2} + \Gamma_{11}^1 \left( \frac{\mathrm{d}u}{\mathrm{d}t} \right)^2 + 2 \Gamma_{12}^1 \frac{\mathrm{d}u}{\mathrm{d}t} \frac{\mathrm{d}v}{\mathrm{d}t} + \Gamma_{22}^1 \left( \frac{\mathrm{d}v}{\mathrm{d}t} \right)^2 &= 0; \\
\frac{\mathrm{d}^2 v}{\mathrm{d}t^2} + \Gamma_{11}^2 \left( \frac{\mathrm{d}u}{\mathrm{d}t} \right)^2 + 2 \Gamma_{12}^2 \frac{\mathrm{d}u}{\mathrm{d}t} \frac{\mathrm{d}v}{\mathrm{d}t} + \Gamma_{22}^2 \left( \frac{\mathrm{d}v}{\mathrm{d}t} \right)^2 &= 0.
\end{align}
Therefore,
\begin{align}
\frac{\mathrm{d}^2 u}{\mathrm{d}t^2} - u \left( \frac{\mathrm{d}v}{\mathrm{d}t} \right)^2 &= 0; \\
\frac{\mathrm{d}^2 v}{\mathrm{d}t^2} + 2 \frac{u}{u^2+k^2} \frac{\mathrm{d}u}{\mathrm{d}t} \frac{\mathrm{d}v}{\mathrm{d}t} &= 0.
\end{align}
The last equation above gives (in shorthand notation)
\begin{align}
\frac{v''}{v'} &= -\, \frac{2uu'}{u^2+k^2} \\
\implies \int \; \frac{v''}{v'} \, \mathrm{d}v' &= \int \; - \, \frac{2uu'}{u^2+k^2} \, \mathrm{d}u \\
\implies \ln \left| v' \right| &= - \ln \left| u^2+k^2 \right| + \text{constant} \\
\implies v' &= \frac{\Theta}{u^2+k^2},
\end{align}
where $\Theta$ is just some constant which is leftover from the integration.
Note:
\begin{align}
\left( v' \right)' &= \left( \frac{\Theta}{u^2+k^2} \right)' \\
\implies v'' &= \frac{- 2u'u \Theta}{\left( u^2+k^2 \right)^2} = \frac{-2u'uv'}{u^2+k^2} \\
\implies v'' + 2 \frac{u}{u^2+k^2} u' v' &= 0,
\end{align}
which is exactly what we needed.
Given the above, are you now able to compute the geodesic curves?
| {
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"url": "https://math.stackexchange.com/questions/262393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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Analytical form of an integral I'm trying to find the analytical form of the following integral:
$$\int_0^{2\pi}\left|\frac{(1-i\cos(\phi))}{(1+a-i\cos(\phi))}\right|^2d\phi$$
I've tried in Wolfram Alpha, which ran out of free calculation time, and in Mathematica I get the original expression for the indefinite integral and the program stays in the "Running..." state for several minutes, when evaluating the definite integral, and I stopped it.
However, I am reading a paper, which says there is an analytical expression for this integral, and also to me it seems rather straightforward. Any hints?
| $$
\begin{align}
\int_0^{2\pi}\left|\frac{(1-i\cos(\phi))}{(1+a-i\cos(\phi))}\right|^2\,\mathrm{d}\phi
&=\int_0^{2\pi}\frac{1+\cos^2(\phi)}{(1+a)^2+\cos^2(\phi)}\,\mathrm{d}\phi\\
&=2\pi-\int_0^{2\pi}\frac{(1+a)^2-1}{(1+a)^2+\cos^2(\phi)}\,\mathrm{d}\phi\\
&=2\pi-2\int_0^\pi\frac{(1+a)^2-1}{2(1+a)^2+1+\cos(2\phi)}\,\mathrm{d}(2\phi)\\
&=2\pi-\oint\frac{4(1+a)^2-4}{4(1+a)^2+2+\left(z+\frac1z\right)}\frac{\mathrm{d}z}{iz}\\
&=2\pi+i\oint\frac{4(1+a)^2-4}{z^2+(4(1+a)^2+2)z+1}\,\mathrm{d}z\\
&=2\pi+i(2\pi i)\frac{(1+a)^2-1}{|1+a|\sqrt{(1+a)^2+1}}\\
&=2\pi\left(1-\frac{(1+a)^2-1}{|1+a|\sqrt{(1+a)^2+1}}\right)
\end{align}
$$
Where $z=e^{i2\phi}$, the contour of integration is the unit circle, and the integrand has one singularity inside the unit circle at $z=-2(1+a)^2-1+2|1+a|\sqrt{(1+a)^2+1}$ with residue $\frac{(1+a)^2-1}{|1+a|\sqrt{(1+a)^2+1}}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that an integral is positive For any positive integer $n$, consider
$$\int_{0}^\infty\frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr.$$
I would like to show that it is positive. I try to write it as
$$\int_{0}^\infty\frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr=\int_{0}^\infty\frac{r^{n+1}}{(r^2+1)^{n+2}}dr-2\int_{0}^\infty\frac{r^{n+1}}{(r^2+1)^{n+3}}dr,$$
but I am not sure that it helps.
EDIT: According to sjasonw, the integral may not be positive as I think. I would be happy to see a proof showing that it's not positive.
| If $$\int \frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr=I$$
$$\int \frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr=\int \frac{r^{n+1}}{(r^2+1)^{n+2}}dr-2\int_{0}^\infty\frac{r^{n+1}}{(r^2+1)^{n+3}}dr$$
Now $$\int\frac{r^{n+1}}{(r^2+1)^{n+2}}dr=\frac1{(r^2+1)^{n+2}}\int r^{n+1} dr-\int\left(\frac{d\frac1{(r^2+1)^{n+2}}}{dr}\int r^{n+1} dr\right)dr$$
$$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}-\int\left( \frac{-(n+2)2r}{(r^2+1)^{n+3}}\cdot\frac{r^{n+2}}{n+2} \right)dr$$
$$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}+2\int\frac {r^{n+3}}{(r^2+1)^{n+3}}dr$$ (assuming $n+2\ne 0$)
So, $$I=\int\frac{r^{n+1}}{(r^2+1)^{n+2}}dr$$
$$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}+2\int\frac {r^{n+3}}{(r^2+1)^{n+3}}dr-2\int\frac{r^{n+1}}{(r^2+1)^{n+3}}dr$$
$$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}+2I$$
So, $$I=-\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}$$
Now apply the limit.
| {
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"source": "stackexchange",
"question_score": "6",
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Confusion on Eigenvalues of Similar Matrices
Please help me to identify where I went wrong:
The completely reduced normal form of the real matrix $A=
$\begin{pmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1 \\
\end{pmatrix}
is the following matrix $B=
$\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
Clearly the eigenvalues of $B$ are $1,0,0$ so should the eigenvalues of $A$ since similar matrices have the same characteristic polynomial. But when I'm trying to formally evaluate the eigenvalues of $A$ the roots of $\chi_A$ become $0,0,3$.
$$\begin{vmatrix}
1-x & 1 & 1 \\
1 & 1-x & 1 \\
1 & 1 & 1-x \\
\end{vmatrix}=0$$
$\implies (1-x)[(1-x)^2-1]-1[(1-x)-1]+1[1-(1-x)]=0$
$\implies (1-x)[x^2-2x]-1(-x)+x=0$
$\implies (1-x)[x^2-2x]+2x=0$
$\implies x^2-2x-x^3+2x^2+2x=0$
$\implies 3x^2-x^3=0$
$\implies x^2(3-x)=0$
So the eigenvalues of $A$ are $0,0,3.$
| Your reduced normal form for the matrix A is wrong.
| {
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What is the difference between $\lfloor f \rfloor (x)$ and $\lfloor f(x) \rfloor $? I have a very basic question: What is the difference between $\lfloor f \rfloor (x)$ and $\lfloor{f(x)}\rfloor$? Are $\lfloor f \rfloor (x)$ and $f(\lfloor x \rfloor)$ equivalent?
| You seem to be asking two different questions. $\lfloor f \rfloor (x)$ is defined as $\lfloor f(x) \rfloor$. Hence, it is indeed true that $\lfloor f \rfloor (x) = \lfloor f(x) \rfloor$.
However, $\lfloor f(x) \rfloor$ is different from $f(\lfloor x \rfloor)$. For instance, consider $f(x) = x^2$. Then $\lfloor f(x) \rfloor = \lfloor x^2 \rfloor$, while $f(\lfloor x \rfloor) = \lfloor x \rfloor ^2$. These two are not equal. For instance, choosing $x = 2.5$, we get that
$$\lfloor f(2.5) \rfloor = \lfloor 2.5^2 \rfloor= \lfloor 6.25 \rfloor = 6$$
whereas $$f(\lfloor 2.5 \rfloor) = f(2) = 2^2 = 4$$
Hence, in general
$$\lfloor f(x) \rfloor \neq f(\lfloor x \rfloor)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing that: $(\frac{a}{b+c})^2+(\frac{b}{a+c})^2+(\frac{c}{a+b})^2+\frac{10abc}{(a+b)(b+c)(c+a)}\ge 2$ Let a;b;c>0. Prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{a+c}\right)^2+\left(\frac{c}{a+b}\right)^2+\frac{10abc}{(a+b)(b+c)(c+a)}\geq 2$$
I think
$$\frac{2a}{b+c}=x;\frac{2b}{c+a}=y;\frac{2c}{a+b}=z$$
We have: $xy+yz+zx+xyz=4$
$$(\frac{x}{2})^2+(\frac{y}{2})^2+(\frac{z}{2})^2+\frac{10xyz}{8} \ge 2$$
$\Leftrightarrow x^2+y^2+z^2+5xyz \ge 8$
$\Leftrightarrow x^2+y^2+z^2-5(xy+yz+zx) +12 \ge 0$
deadlock
Can you help me? Thank you very much
| A full expanding gives
$$\sum_{cyc}(a^6+2a^5b+2a^5c-a^4b^2-a^4c^2-4a^3b^3+a^2b^2c^2)\geq0$$ or
$$\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)+2\sum_{cyc}ab(a^2-b^2)^2\geq0,$$
which is true by Schur.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding Eccentricity from the rotating ellipse formula I see that from a normal ellipse formula, we can acquire the eccentricity via this formula here.
However, for this formula (1):
$A(x − h)^2 + B(x − h)(y − k) + C(y − k)^2 = 1$
When parameter $B = 0$, we would have normal ellipse, and the formula from the link above can be used.
But when $B ≠ 0$, we will have a tilting ellipse, and its eccentricity will change as well. In fact, if we are to find the current eccentricity of the given formula (1), what would be the formula for the eccentricity in this case?
| As the translation does not change the eccentricity of a conic, let's us define $x=X+h,y=Y+k$ to get $AX^2+BXY+CY^2=1$
Using Rotation of axes as eccentricity is one of the invariants in rotation,
$$x'^2(A\cos^2t+B\sin t\cos t+C\sin^2t)+x'y'(B\cos2t-(A-C)\sin2t)+y'^2(A\sin^2t-B\sin t\cos t+C\cos^2t)=1$$
To remove the $x'y'$ term, $\tan 2t=\frac{B}{A-C}\implies \frac{\sin2t}{B}=\frac{\cos2t}{A-C}=\pm\frac1{\sqrt{B^2+(A-C)^2}}$
If we take the $'+'$ sign, $\sin2t=\frac B{\sqrt{B^2+(A-C)^2}},\cos2t=\frac {A-C}{\sqrt{B^2+(A-C)^2}}$
Comparing with the standard form we get
$a^2=\frac1{A\cos^2t+B\sin t\cos t+C\sin^2t}$
$=\frac2{A(1+\cos2t)+B\sin2t+C(1-\cos2t)}$
$=\frac{2\sqrt{B^2+(A-C)^2}}{(A+C)\sqrt{B^2+(A-C)^2}+(A-C)^2+B^2}$
$=\frac2{A+C+\sqrt{B^2+(A-C)^2}}$
Similarly, $b^2=\frac2{A+C-\sqrt{B^2+(A-C)^2}}>a^2$
So, $b$ is the semi-major axis, $a$ is the semi-minor axis.
Consequently, the equation of the conic becomes $$\frac{x'^2}{\frac2{A+C+\sqrt{B^2+(A-C)^2}}}+\frac{y'^2}{\frac2{A+C-\sqrt{B^2+(A-C)^2}}}=1$$
$e^2=1-\frac{a^2}{b^2}=1-\frac{A+C-\sqrt{B^2+(A-C)^2}}{A+C+\sqrt{B^2+(A-C)^2}}=\frac{2\sqrt{B^2+(A-C)^2}}{A+C+\sqrt{B^2+(A-C)^2}}$
If $A+C-\sqrt{B^2+(A-C)^2}=0$ i.e., $4AC=B^2,e^2=1,$ the conic becomes a Parabola.
If $A+C-\sqrt{B^2+(A-C)^2}>0$ i.e., $4AC>B^2,e^2<1,$ the conic becomes an ellipse.
If $A+C-\sqrt{B^2+(A-C)^2}<0$ i.e., $4AC<B^2,e^2>1,$ the conic becomes a hyperbola.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/264877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
prove that there are no number such that $\overline{ab\ldots}=a^2+b^2+\ldots$ Prove that a natural number with at least 2 digits cannot be written like a sum with the the power of digits equal $2$.
What I want to say: $$\overline{ab}\neq a^2+b^2.$$
What I have done:
$$10a+b=a^2+b^2 $$ or
$$a(a-10)=b(1-b).$$
$b(1-b)=2k$ so $a(a-10)=2k$ and this is possible only when $a=2q.$
so: $2 \cdot 8 =b(b-1)$ or $4\cdot 6=b(b-1)$ and this is not possible.
final conclusion for the number $\overline{ab}$ is ok, but what can I do for number formatted with $3,4, \ldots$ digits ?
thanks :)
| For more than 3 digits, if there are $n$ digits the sum of squares is at most $81n$ while the number is at least $10^{n-1}$ and you are done. For specifically $3$ digits, the maximum sum of squares would be $3 \cdot 81=243$, but then the leading digit is no more than $2$, so the maximum sum is no more than $4+162=166$, so the leading digit is $1$, so the maximum sum is $163$, so the maximum sum of squares is $1+36+81=118$, so the maximum sum of squares is $1+1+81=83$, failure.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \leq \frac{9}{10}$ if $a+b+c=1$. Let $a,b,c$ are real number such that $a+b+c=1$. Prove that:
$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \leq \frac{9}{10}.$$
| As you have known, the inequality holds when $a,b,c\ge -\frac{3}{4}$. To complete the proof, without loss of generality, we may assume that $a\le -\frac{3}{4}$.
Note that $\frac{b}{b^2+1},\frac{c}{c^2+1}\le \frac{1}{2}$. Since when $-9\le a\le -\frac{3}{4}$, $\frac{a}{a^2+1}<-\frac{1}{10}$, the inequality holds when $a\ge -9$. When $a\le-9$, $b+c\ge 10$. Without loss of generality, we may assume that $b\ge 5$. Then $\frac{b}{b^2+1}\le \frac{5}{26}<\frac{2}{5}$, so the inequality also holds.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the limit of $(x_n)$ defined by $x_{n+1}=c_nu(x_n)$ Find $\displaystyle \lim_{n\rightarrow \infty }x_n$ :
$$\left\{\begin{matrix}x_1=a>0\\ \\ x_{n+1}=\frac{2x_n\cdot \cos\left(\frac{\pi}{2^n+1}\right)}{x_n+1}\end{matrix}\right.$$
I have tried that :
Let $a_n=\dfrac{1}{x_n}$ . So :
$$a_{n+1}=\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}.a_n+\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}$$
So I tried to have geometric series:
By let :
$$a_{n+1}+f(n+1)=\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}(a_n+f(n))$$
So we must find one $f(n)$:
$$\frac{f(n)}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}-f(n+1)=\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}$$
As $$f(n)-f(n+1)\cdot 2\cos\frac{\pi}{2^{n+1}}=1$$
Can you give me the way to find one $f(n)$ sastisfied that ; or anyone has nice way to solve this problem
| An incomplete answer that assumes the existence of the limit:
Since $\cos\frac{\pi}{2^n+1}$ goes to one for $n\to\infty$, the limit $x_\infty=\lim_{n\to\infty}x_n$ will, if it exists, satisfy the fixed-point equation $x_\infty=2x_\infty/(x_\infty+1)$. The only solutions of this equation are $0$ and $1$.
The derivative of $x\mapsto 2x\cos\frac{\pi}{2^n+1}/(x+1)$ at $x=0$ equals $2\cos\frac{\pi}{2^n+1}$ which is greater than one for $n>1$. Fixed-point theory then tells us that $x=0$ can not be the limit of your recursion. The derivative at $x=1$, in contrast, equals $1/2\cos\frac{\pi}{2^n+1}$ which is always positive and less than unity.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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For which values of $x,y$ does $x^2+y^2$ take the minimum?
Find the values of $x,y$ for which $x^2+y^2$ takes the minimum value where $(x+5)^2+(y-12)^2=14^2$.
Trial:$\begin{align} \ (x+5)^2+(y-12)^2=14^2 \\ \implies x^2+y^2+10x-24y=27 \end{align}$.
Here I am stuck. Please help.
| What about Lagrange's Multipliers? Let
$$f(x,y)=x^2+y^2\,\,,\,\,g(x,y)=(x+5)^2+(y-12)^2-14^2\Longrightarrow$$
$$H(x,y,\lambda)=f+\lambda g=x^2+\lambda(x+5)^2+y^2+\lambda(y-12)^2-14^2\Longrightarrow$$
$$H'_x=2\left((x+\lambda(x+5)\right)=0\Longrightarrow x=-\frac{5\lambda}{1+\lambda}$$
$$H'_y=2\left(y+\lambda(y-12)\right)=0\Longrightarrow y=\frac{12\lambda}{1+\lambda}$$
$$H'_\lambda=(x+5)^2+(y-12)^2-14^2\Longrightarrow H'_\lambda=g(x,y)=0$$
Substituting the values for $\,x,y\,$ in the last equation above, we get:
$$\left(-\frac{5\lambda}{1+\lambda}+5\right)^2+\left(\frac{12\lambda}{1+\lambda}\right)^2=14^2\Longrightarrow (25+144)\frac{1}{(1+\lambda)^2}=14^2\Longrightarrow$$
$$(1+\lambda)^2=\frac{13^2}{14^2}\Longrightarrow 1+\lambda =\pm\frac{13}{14}$$
Try now to take it from here.
| {
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Solve $xy=3$ and $4^{x^2}+2^{y^2}=72$ I have a system of equations $xy=3$ and $4^{x^2}+2^{y^2}=72$ whose solution I know is $x=y=\sqrt 3$, but what are the steps to solve it?
| The first attempt to try to solve a system of two (or more variables) is naturally assumed that a possible solution is one in which all variables are equal.
Then set $x=y=t$ and solve de equation
\begin{cases}
xy=t\cdot t=3\\
4^{x^2}+2^{y^2}=\big(2^{t^2}\big)^2+2^{t^2}=72
\end{cases}
By $t^2=3$ we have $t=\pm\sqrt{3}$. Substituting into the another equation we have that two solutions are $(x,y)=(+\sqrt{3},+\sqrt{3})$ and $(x,y)=(-\sqrt{3},-\sqrt{3})$.
Or by $\big(2^{t^2}\big)^2+2^{t^2}=72$ whe have $2^{t^2}=\frac{-1+\sqrt{289}}{2}=8$.
To further investigate other solutions you might assume
$$
x=t+s,
\\
y=t-s.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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proving that $ab$ is a perfect square. let $a,b \ge 2$ be integers, such that for every positive integer $n$, the expression :
$(a^n-1)(b^n-1)$ is a perfect square.
How to prove that $ab$ is a perfect square.
| It is obvious that for $a=b$ the product $(a^n-1)(b^n-1)$ is square for all $n$. We prove that this is the only possibility.
Let $b^n=a^n+h_n$; one has $(a^n-1)(a^n-1+h_n)=x_n^2$ so
$$(a^n-1)^2+h_n(a^n-1)-x_n^2=0\Rightarrow 2(a^n-1)=-h_n\pm\sqrt{h_n^2+4x_n^2}$$ so we have a Pythagorean triple
$$(h_n, 2x_n,z_n)=(r_n^2-s_n^2,\space 2r_ns_n,\space r_n^2+s_n^2)$$ and
$$2(a^n-1)=-(r_n^2-s_n^2)\pm(r_n^2+s_n^2)$$ thus for all $n$ $$a^n=s_n^2+1\\b^n=r_n^2+1$$ For $n=2$ we have
$$a^2-s_2^2=1\\b^2-r_2^2=1$$ We are done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Euclidean norm and matrix questions Let $A$ be an $n×m$ real matrix and define $\vert A \vert_{2}^{2}={\rm tr}(A^tA)$.
1)Show that $|A|_2$ is the Euclidean norm of $A$, when we view $A$ as a vector in $R^{nm}$ by stacking the columns of $A$.
2)Find the cosine of the angle between
$A=\begin{bmatrix}
1 & 1 & 1\\
1& 0 & 1
\end{bmatrix}$, and $B=\begin{bmatrix}
4 & 0 & 0\\
1& 0 & 0
\end{bmatrix}$
(Here what stacking the columns of $A$ means?)
| This: if
$$
A =
\begin{pmatrix}
1 & 3 & 5 \\
2 & 4 & 6
\end{pmatrix}
$$
then, as a vector of $\mathbb{R}^{3\cdot 2}$, we view it like
$$
\begin{pmatrix}
1 \\
2 \\
3 \\
4 \\
5 \\
6
\end{pmatrix}
$$
and the square of its norm is
$$
\vert A \vert_2^2 =
\mathrm{tr}
\left[
\begin{pmatrix}
1 & 2 \\
3 & 4 \\
5 & 6
\end{pmatrix}
\begin{pmatrix}
1 & 3 & 5 \\
2 & 4 & 6
\end{pmatrix}
\right]
=
\mathrm{tr}
\begin{pmatrix}
1+4 & 11 & 17 \\
11 & 9+16 & 39 \\
17 & 39 & 25 + 36
\end{pmatrix}
=
1 + 4 + 9 + 16 + 25 + 36
$$
Which happens to be the same as
$$
(1,2,3,4,5,6)
\begin{pmatrix}
1 \\
2 \\
3 \\
4 \\
5 \\
6
\end{pmatrix}
=
1 + 4 + 9 + 16+ 25 + 36 \ .
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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a question about uniform convergence of a series The problem is to investigate the uniform convergence of $\sum\limits_{n=0}^\infty \frac{x^2}{x^2+n^2}.$ Any help will be appreciated.
Thanks in advance.
| As a bonus, this series not only converges but there is a closed form expression for the sum. Using the method presented here, we study the integral of $$ f(z) = \frac{x^2}{x^2+z^2} \pi \cot(\pi z)$$ along a circle of radius $R$ in the complex plane where $R$ goes to infinity. This integral vanishes in the limit, so the residues at the poles of $f(z)$ must sum to zero. Let $$S = \sum_{n\ge 1} \frac{x^2}{x^2+n^2} .$$ We thus have $$ \operatorname{Res}_{z=-ix} f(z) + \operatorname{Res}_{z=+ix} f(z) + \operatorname{Res}_{z=0} f(z) + 2S = 0,$$ where we have used the fact that $$\operatorname{Res}_{z=m} \pi\cot(\pi z) = 1,$$ with $m$ an integer.
This gives $$\frac{x^2}{-2ix} \pi\cot(-\pi ix) + \frac{x^2}{+2ix} \pi\cot(+\pi ix) + 1 + 2S=0$$ or $$ \frac{x^2}{2ix} \pi\cot(\pi ix) + \frac{x^2}{2ix} \pi\cot(+\pi ix) + 1 + 2S=0$$ or $$ \frac{x}{i} \pi\cot(\pi ix) + 1 + 2S=0$$ and finally $$ S = -\frac{1}{2} + \frac{\pi}{2}x\coth(\pi x).$$ It follows that for the original sum with the term at $n=0$ we have $$\sum_{n\ge 0} \frac{x^2}{x^2+n^2} = \frac{1}{2} + \frac{\pi}{2}x\coth(\pi x).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there any general formula for $S = 1^1 + 2^2 + 3^3 + \dotsb+(n - 1)^{n - 1} + n^n, n \in N$? Is there any general formula to sum following series:
$$S = 1^1 + 2^2 + 3^3 + \dotsb+(n - 1)^{n - 1} + n^n, n \in N$$
I mean for $S = f(n)$, is there a formula to compute $f(n)$?
| I don't know if a closed form exists, but an asymptotic approximation can be given as
$$
\begin{align}
&n^n\left(1+\frac1n\frac{(n-1)^{n-1}}{n^{n-1}}+\frac1{n^2}\frac{(n-2)^{n-2}}{n^{n-2}}+\frac1{n^3}\frac{(n-3)^{n-3}}{n^{n-3}}+O\left(\frac1{n^4}\right)\right)\\
&=\bbox[5px,border:2px solid #C0A000]{n^n\left(1+\frac1{en}+\frac{3+e}{2e^2n^2}+\frac{52+60e+7e^2}{24e^3n^3}+O\left(\frac1{n^4}\right)\right)}
\end{align}
$$
where
$$
\begin{align}
\log\left(\frac{(n-k)^{n-k}}{n^{n-k}}\right)
&=(n-k)\left(-\frac kn-\frac{k^2}{2n^2}-\frac{k^3}{3n^3}+O\left(\frac1{n^4}\right)\right)\\
&=-k+\frac{k^2}{2n}+\frac{k^3}{6n^2}+O\left(\frac1{n^3}\right)
\end{align}
$$
and so
$$
\frac1{n^k}\frac{(n-k)^{n-k}}{n^{n-k}}
=e^{-k}\left(\frac1{n^k}+\frac{k^2}{2n^{k+1}}+\frac{3k^4+4k^3}{24n^{k+2}}+O\left(\frac1{n^{k+3}}\right)\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/278599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Let $a, b$ and $c$ be the lengths of the sides of an arbitrary triangle. Pick out the true statements. Let $a, b$ and $c$ be the lengths of the sides of an arbitrary triangle. Define
$$x =\frac{ab + bc + ca}{a^2 + b^2 + c^2}.$$
Pick out the true statements.
(a) $1/2 ≤ x ≤ 2$.
(b) $1/2 ≤ x ≤ 1$.
(c) $1/2 < x ≤ 1$.
How can I able to solve this problem
| We know in every triangle $ABC$ there are some useful relations called Law of cosines:
$$a^2=b^2+c^2-2bc\cos(A)\\b^2=a^2+c^2-2ac\cos(B)\\c^2=a^2+b^2-2ab\cos(C)$$ By adding them we have: $$a^2+b^2+c^2=2(bc\cos(A)+ac\cos(B)+ab\cos(C))$$ and if we take $A=B=C=60^{~\text{o}}$ then $x=1$($ABC$ is a Equilateral).
Now take $A=90^{~\text{o}},B=45^{~\text{o}},C=45^{~\text{o}}$ and using $b=a\sin(B), c=a\cos(B)$ we have: $$x=\frac{a^2\sin(B)+a^2\sin(B)\cos(B)+a^2\cos(B)}{a^2}\sim 0.9$$ It seems that $x\leq1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$ This question is from [Number Theory George E. Andrews 1-1 #3].
Prove that $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}).$$
This problem is driving me crazy.
$$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\dots +xy^{n-2}+y^{n-1)}$$
$(x^n-y^n)/(x-y) =$ the sum for the first $n$ numbers and then I added $(xy^{(n+1)-2}+y^{(n+1)-1})$ which should equal $(x^{n+1}-y^{n+1})/(x-y)$ but I can't figure it out
This is a similar problem in the book and I tried this method but it wasn't working out
$\quad$Thereom $\bf1$-$\bf2$: $\,\,\,\,$ If $\,x$ is any real number other than $1$, then $$\sum_{j=0}^{n-1}x^j=1+x+x^2+\ldots+x^{n-1}=\dfrac{x^n-1}{x-1}.$$
$\quad$Remark: $\displaystyle\sum_{j=0}^{n-1}A_j$ is shorthand for $A_0+A_1+A_2+\ldots+A_{n-1}.$
$\quad$Proof: Again we proceed by mathematical induction. If $n=1$ then $\displaystyle\sum_{j=0}^{1-1}x^j=x^0=1$ and $(x-1)/(x-1)=1$. Thus the theorem is true for $n=1$. $\quad$ Assuming that $\displaystyle\sum_{j=0}^{k-1}x^j=(x^k-1)/(x-1)$, we find that $$
\eqalign{
\sum^{(k+1)-1}_ {j=0}x^j & = \sum^{k-1}_ {j=0}x^j+x^k=\dfrac{x^k-1}{x-1}+x^k \\
&= \dfrac{x^k-1+x^{k+1}-x^k}{x-1}\\
&= \dfrac{x^{k+1}-1}{x-1}.
}$$
Hence condition $(\rm ii)$ is fulfilled, and we have established the theorem.
$\quad$Corollary $\bf1$-$\bf1$: $\,\,$ If $\,m$ and $n$ are positive integers and if $m>1$, then $n<m^n.$
| Let $u_n$ $=$ $x^n-y^n$. Now note that $u_n$ $=$ $(x+y)u_n$$_-$$_1$ $+$ $xy$ $u_n$$_-$$_2$. Assume that the given expression is true for all numbers from $1$ to some fixed $k(>1)$. Then apply induction to prove the result for $(k+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 3
} |
Limit for a Recurrence Relation How I can find a limit for this recursively defined sequence?
$$a_0>0, a_{n+1}=\frac{a_{n}+2}{3a_{n}+2}$$
I'm particularly interested in answers involving concepts like contractive sequences and fixed points.
Many thanks.
| A little playing shows that the limit "should" be $\frac{2}{3}$. So it is natural to compute $a_{n+1}-\frac{2}{3}$.
We get
$$a_{n+1}-\frac{2}{3}=\frac{a_n+2}{3a_n+2}-\frac{2}{3}=\frac{\frac{2}{3}-a_n}{3a_n+2}.$$
Thus
$$\left|a_{n+1}-\frac{2}{3}\right|=\left|a_n-\frac{2}{3}\right|\frac{1}{3a_n+2}.$$
In particular,
$$\left|a_{n+1}-\frac{2}{3}\right|\lt \frac{1}{2}\left|a_n-\frac{2}{3}\right|.$$
So with each iteration our distance from $\frac{2}{3}$ shrinks by a factor of at least $\frac{1}{2}$. It follows that $\lim_{n\to\infty}a_n=\frac{2}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
| General comment: as soon as one tries to use AM-GM for
$x=a^3+a+1,$ $y=b^3+b+1$ and $z=c^3+c+1$ the inequality becomes wrong, since
$$a^3+a+b^3+b+c^3+c+3\ge 2(a^2+b^2+c^2)+3=9.$$
Using Lagrange multiplayers, one can reduce this problem to the following system:
$$(3a^2+1)(b^3+b+1)(c^3+c+1)=2\lambda a$$
$$(3b^2+1)(a^3+a+1)(c^3+c+1)=2\lambda b$$
$$(3c^2+1)(a^3+a+1)(b^3+b+1)=2\lambda c.$$
In other words, if $\lambda\ne 0,$ for the function
$$f(x)=\frac{x(x^3+x+1)}{(3x^2+1)}$$
we have $f(a)=f(b)=f(c).$ It is easy to see, that $f$ is monotone for $x\ge 0$ so the only option is $a=b=c=1.$ The rest should be clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
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} |
Find the values of the positive constants $k$ and $c$ such that $-37\le k(3\sin\theta + 4\cos\theta) +c\le 43$ for all values of $\theta$ Hi how do i go about solving this?
Find the values of the positive constants $k$ and $c$ such that $$-37\le k(3\sin\theta + 4\cos\theta) +c\le 43 $$for all values of $\theta$ $$\rightarrow-37\le k(5(\sin\theta + 53.1)) +c\le 43 $$
Then what?
Cheers
| By Cauchy-Schwarz
$$(3\sin(x)+4\cos(x))^2 \leq 25 (\sin^2(x)+\cos^2(x))=25$$
and equality is possible.
Then
$$-5 \leq 3\sin(x)+4\cos(x) \leq 5 $$
this shows that
$$-5k+c \leq - k(3\sin\theta + 4\cos\theta) +c\le 5k+c \,.$$
and the lower/upper bounds can be atatined. You can finish it easely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/284611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the nth term of a recursive sequence I have a the following sequence:
$$\begin{gather}
a_1 = 3 \\
a_{n + 1} = 1 + \frac{a_n}{2}
\end{gather}
$$
How can I find the $a_n$ term?
| Put $n = 1$; you get $a_2 = 1 + \frac{1}{2} * 3 = \frac{5}{2}$
Following this continually, you get:
$$a_{n} = \frac{3 + \Sigma {2^{n - 1}}}{2^{n - 1}}$$
where, $n = 1, 2, 3 .. $
Aletnatively
You have $a_2 = 1 + \frac{a_1}{2}$
$$a_3 = 1 + \frac{a_2}{2} = 1 + \frac{1}{2} * (1 + \frac{a_1}{2}) = 1 + \frac{1}{2} + \frac{a_1}{4}$$
Hence, in this manner, you generate:
$$a_{n + 1} = 1 + \frac{1}{2} + ... \text{n terms} + \frac{a_1}{2^n} = \frac{1 * (1 - \frac{1}{2^n}) }{1 - \frac{1}{2}} + \frac{3}{2^n}$$
Therefore, you conclude:
$$a_{n + 1} = 2 + \frac{1}{2^n}$$
where $n \ge 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ for all $n \le x \le n+1$, using the mean value theorem applied to $f(x) = \frac{1}{x}$
Immediately, I can recognize some components of the mean value theorem. $\frac{1}{n^2}$ likely comes from the slope of $f$ at $n$, and the $|\frac{1}{n} - \frac{1}{x}|$ expression likely comes from the secant expression $\frac{\frac{1}{n} - \frac{1}{x}}{x}$. But I cannot figure out how they fit together in the end.
| $$f(x)=\frac{1}{x}\,\,\,\text{on the interval}\,\,\,[n\,,\,x]\,\,,\,n\leq x\leq n+1\Longrightarrow$$
$$\frac{f(x)-f(n)}{x-n}=f'(c)\,\,,\,\,\text{for some}\,\,\,c\in(n\,,\,x)\Longleftrightarrow$$
$$\frac{1}{x}-\frac{1}{n}=-\frac{1}{c^2}(x-n)$$
But
$$\left|\frac{1}{c^2}(x-n)\right|\leq\frac{1}{c^2}\Longleftrightarrow |x-n|\leq 1$$
and the rightmost inequality is true by the given data (why?) , thus we get
$$\left|\frac{1}{x}-\frac{1}{n}\right|\leq\frac{1}{c^2}<\frac{1}{n^2}\,\,,\,\,\text{since}\,\,\, c>n\,$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
log sin and log cos integral, maybe relate to fourier series I try to use the method of differentiation under integral sign for the first one
And integrate it back, but I failed to find the constant $c$ ....
Anyone hav other method?
$$\begin{align}
& \int_{0}^{\frac{\pi }{2}}{{{x}^{2}}{{\ln }^{2}}\left( 2\cos x \right)\text{d}x} \\
& \int_{0}^{\frac{\pi }{3}}{x{{\ln }^{2}}\left( 2\sin \frac{x}{2} \right)}\text{d}x \\
\end{align}$$
| From
\begin{equation}
\int_{0}^{\frac{\pi }{2}}\left( 2\cos \theta \right) ^{x}\cos y\theta
d\theta =\frac{\pi }{2}F(1+\frac{x+y}{2},1+\frac{x-y}{2}), \tag*{(1)}
\end{equation}
where $F(x,y)=\frac{\Gamma (x+y-1)}{\Gamma (x)\Gamma (y)},$ we can get that
\begin{equation}
\begin{array}{c}
\int_{0}^{\frac{\pi }{2}}\theta ^{q}\left( 2\cos \theta \right) ^{x}\cos
\frac{2y\theta +q\pi }{2}\ln ^{p}\left( 2\cos \theta \right) d\theta \\
=\frac{\pi }{2^{p+q+1}}\sum\limits_{k=0}^{q}(-1)^{q-k}C_{q}^{k}\sum
\limits_{j=0}^{p}C_{p}^{j}F_{k+j,q+p-k-j}(1+\frac{x+y}{2},1+\frac{x-y}{2})
\end{array}
\tag*{(2)}
\end{equation}
In (2) let $x,y=0,q$ replaced by $2q$ we have
\begin{equation}
\int_{0}^{\frac{\pi }{2}}\theta ^{2q}\ln ^{p}\left( 2\cos \theta \right)
d\theta =\frac{\pi }{2^{p+q+1}}\sum\limits_{k=0}^{2q}(-1)^{q-k}C_{2q}^{k}
\sum\limits_{j=0}^{p}C_{p}^{j}F_{k+j,2q+p-k-j}(1,1) \tag*{(3)}
\end{equation}
For $F_{p,qj}(1,1)$ there is the following recurrence relations
\begin{equation}
F_{p,q}(1,1)=p!(q-1)!\sum\limits_{k=0}^{p-1}\sum
\limits_{j=0}^{q-1}C_{p+q-1-k-j}^{p-k}\frac{(-1)^{p+q-k-j}\zeta (p+q-k-j)
}{k!j!}F_{k,j}(1,1). \tag*{(4)}
\end{equation}
By (4) we can get that
\begin{equation}
\begin{array}{c}
F_{0,0}(1,1)=1,F_{0,i}(1,1)=0,i=1,2,3,4,F_{1,1}(1,1)=\frac{\pi ^{2}}{6}, \\
F_{1,2}(1,1)=-2\zeta (3),F_{1,3}(1,1)=\frac{\pi ^{4}}{15},F_{1,4}(1,1)=-24
\zeta (5), \\
F_{2,2}(1,1)=\frac{\pi ^{4}}{90},F_{2,3}(1,1)=-2\pi ^{2}\zeta (3)-24\zeta
(5), \\
F_{2,4}(1,1)=\frac{68\pi ^{6}}{315}+24\zeta ^{2}(3),F_{3,3}(1,1)=\frac{
107\pi ^{6}}{420}+36\zeta ^{2}(3), \\
F_{3,4}(1,1)=-6\pi ^{4}\zeta (3)-48\pi ^{2}\zeta (5)-720\zeta (7), \\
F_{4,4}(1,1)=\frac{3701\pi ^{8}}{3150}+96\pi ^{2}\zeta ^{2}(3)-2304\zeta
(3)\zeta (5),
\end{array}
\tag*{(5)}
\end{equation}
By (3) and (5) we have
\begin{equation}
\int_{0}^{\frac{\pi }{2}}\theta ^{2}\ln ^{2}\left( 2\cos \theta \right)
d\theta =\frac{11\pi ^{5}}{1440} \tag*{(6)}
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Help solving summation series of a recursive function Yesterday in class, we were analyzing the Karatsuba multiplication algorithm and how it applies to recurrence equations. Time ran short, and I feel I missed how to solve the final summation.
First, we defined the recurrence equation as
$$T(n) = 3T \left(\frac{n}{2}\right) + 4n$$
and applied a recurrence tree such like
$$T(n) = 3T \left(\frac{n}{2}\right) + 4n \Rightarrow 4n \cdot \left(\frac{3}{2}\right)^0$$
$$T\left(\frac{n}{2}\right) = 3T \left (\frac{n}{4} \right) + 4\left(\frac{n}{2}\right) \Rightarrow 4n \cdot \left(\frac{3}{2}\right)^1$$
$$T\left(\frac{n}{4}\right) = 3T \left (\frac{n}{8} \right) + 4\left(\frac{n}{4}\right) \Rightarrow 4n \cdot \left(\frac{3}{2}\right)^2$$
$$T\left(\frac{n}{8}\right) = 3T \left (\frac{n}{16} \right) + 4\left(\frac{n}{8}\right) \Rightarrow 4n \cdot \left(\frac{3}{2}\right)^3$$
Because the denominiator increases in a logarithmic fashion, we defined the summation as
$$\sum_{x=0}^{log_2n} 4n \cdot \left(\frac{3}{2}\right)^x$$
Time was running short, so several steps were skipped, and the final solution was given as
$$9\cdot 3^{log_2n} = 9n^{log_23} = 9n^{1.58} = O(n^{1.58})$$
based on the properties
$$a^{lg\, b} = b^{lg\, a}\: \text{and}\: log_2 3 \approx 1.58$$
I've tried applying the summation formula
$$\sum_{x=0}^{n}r^x = \frac{r^{n+1}-1}{r-1}$$
with this result, and end up with
$$\sum_{x=0}^{n}r^x = \frac{r^{n+1}-1}{r-1} = \sum_{x=0}^{log_2 n} 4n \cdot \left(\frac{3}{2}\right)^x $$
$$= 4\left(n\cdot \frac{\frac{3}{2}^{lg_2n+1}-1}{\frac{3}{2}-1}\right) = 4\left(n \cdot \frac{\frac{3}{2}^{log_2n+1}-1}{\frac{1}{2}}\right)
= 2\left(n \cdot \frac{3}{2}^{log_2n+1}+1\right) $$
$$=2n \cdot 3^{log_2n+1} + 2$$
which is very different than the solution given. Where did I go wrong?
| The crucial mistake is that you passed somehow from
$$
(\frac32)^{(\log_2 n) +1}
$$
to
$$
3^{(\log_2 n )+ 1}.
$$
Using the rough formula that
$$
\sum_{x=0}^n r^x = O(r^n) \qquad \qquad\text{if $r>1$},
$$
you should get
$$
O(n (\frac32)^{\log_2 n})=O(n e^{(\ln n) \frac{\ln \frac32}{\ln 2}}) = O(n^{1+\frac{\ln \frac32}{\ln 2}}).
$$
Since
$$
1 + \frac{\ln \frac32}{\ln 2} = \frac{\ln 3}{\ln 2} = \log_2 3 = 1.58496...
$$
this is the same as the result you got in class.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/291681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding range of a function Can I get the range of the following function?
$$y=\dfrac{x^2-3x+2}{x^2-5x+6}$$
I cannot isolate $x$ to get $x=f(y)$.
Thanking you in advance.
| The function $ f(x) = \dfrac{x^{2} - 3x + 2}{x^{2} - 5x + 6} $ has domain $ (- \infty,2) \cup (2,3) \cup (3,\infty) $. By eliminating the common factor $ x - 2 $ in the numerator and denominator, we see that it coincides with the function $ g(x) = \dfrac{x - 1}{x - 3} $ on $ \text{Dom}(g) = (- \infty,3) \cup (3,\infty) $. Next, observe that
\begin{align}
\forall x \in (- \infty,3) \cup (3,\infty): \quad g(x)
= \frac{x - 1}{x - 3} &= \frac{(x - 3) + 2}{x - 3} \\
&= 1 + \frac{2}{x - 3},
\end{align}
so we get $ \text{Range}(g) = (- \infty,1) \cup (1,\infty) $, which is easily obtainable if you sketch the graph of $ g $. Then if you want to determine $ \text{Range}(f) $, simply delete $ g(2) = 0 $ from $ \text{Range}(g) $ because
*
*$ 2 \notin \text{Dom}(f) $ and
*$ g $ is a one-to-one function (this is important, because it says that $ g $ does not attain the value $ 0 $ anywhere else other than at $ x = 2 $).
Therefore, $ \text{Range}(f) = (- \infty,0) \cup (0,1) \cup (1,\infty) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/292225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Elementary Set theory question .... it was asked in Exam There are $21$ people.
$9$ eat dish $A$
$10$ eat dish $B$
$7$ eat dish $C$
$5$ eat dish $A , B$ and $C$
How many people eat at least two dishes?
Answer:
$10$ (given in solutions)
$15$ (as per me )
Please tell me which one is correct. Also, tell if youu have different answer.
| There seems to be something wrong here.
Consider this Venn diagram and its labeled parts.
We are told that:
$$a+e+f+g=|A|=9 \qquad b+d+f+g=|B|=10 \qquad c+e+d+g=|C|= 7$$
$$g = 5$$
Thus
$$\begin{align}
a+e+f=4 \\
b+d+f=5 \\
c+e+d=2
\end{align}$$
so that, by adding all three equations,
$$a+b+c+2d+2e+2f=11 \qquad(1)$$
But, assuming each of the 21 people eats at least one dish, we also know that
$$a+b+c+d+e+f+g=21$$
so that (since $g=5$)
$$a+b+c+d+e+f=16 \qquad(2)$$
Consequently, by subtracting equation (2) from equation (1),
$$d+e+f = -5 \text{ (!)}$$
This problematic result is consistent with @Brian's work, which concludes
$$\begin{align}
|A\cap B| + |A\cap C|+|B\cap C| &= (f+g)+(e+g)+(d+g) \\
&= 3g+(d+e+f) \\
&= 3\cdot 5 + (-5) \\
&= 10 \\
\end{align}$$
Note, though, that this value does not answer OP's question, as it over-counts the people eating all three dishes. The number of people eating at least two dishes should be given by $d+e+f+g$ (that is, $|A\cap B|+|A \cap C|+|B\cap C|-2|A\cap B\cap C|$), but the computed value here is zero. Weird.
Edit. Perhaps the assumption "each of the 21 people eats at least one dish" is in error. Let $h$ be the number of people who eat nothing. Then we have
$$a+b+c+d+e+f+g+h=21$$
and
$$d+e+f=h-5$$
so that (barring negative people from the dinner party) $h \ge 5$. Moreover,
$$21 = a+b+c+(h-5)+5+h = a+b+c+2h$$
Therefore (also barring fractional people from the dinner party), $h \le 10$, and we can write
$$5 \le d+e+f+g \le 10$$
I don't see the conditions that force us to accept $10$ for the value of $d+e+f+g$; indeed, I've found scenarios $(a,b,c,d,e,f,g,h)$ that give rise to each possible value of the expression:
$$\begin{align}
(0,0,1,1,0,4,5,10) \quad &\implies \quad d+e+f+g = h = 10 \\
(1,2,0,1,1,2,5,9) \quad &\implies \quad d+e+f+g = h = 9 \\
(1,3,1,0,1,2,5,8) \quad &\implies \quad d+e+f+g = h = 8 \\
(3,4,0,1,1,0,5,7) \quad &\implies \quad d+e+f+g = h = 7 \\
(3,4,2,0,0,1,5,6) \quad &\implies \quad d+e+f+g = h = 6 \\
(4,5,2,0,0,0,5,5) \quad &\implies \quad d+e+f+g = h = 5
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sub determinant of a matrix If I have the matrix $A(x)$ which is $5 x 5$ and I need to find the $t x t$ sub-determinants of $A(x)$ for $t = 1$ to $t = 5$, how do I do this? $A(x)$ = \begin{pmatrix}
(x-1) & 0 & 0 & 0 & 0 \\
-1 & (x-1) & 0 & 0 & 0 \\
0 & 0 & (x-1) & 0 & 0\\
0 & 0 & 0 & (x+1) & 0\\
0 & 0 & 0 & 0 & (x+1)
\end{pmatrix}
| Let's look for the Smith Canonical Form, since it is your goal behind the question.
For $t=1$, we get in particular $x-1$ and $x+1$, whose $gcd$ is $1$, so the gcd of all $1\times 1$ minors is $1$.
For $t=2$, we get in particular $(x-1)^2$ and $(x+1)^2$, whose $gcd$ is $1$, so the gcd of all $2\times 2$ minors is $1$.
For $t=3$, the only nonzero $3\times 3$ minors are $(x-1)^3, (x-1)^2(x+1), (x-1)(x+1)^2$. So the gcd is $(x-1)(x+1)$.
For $t=4$, the only nonzero $4\times 4$ minors are $(x-1)^3(x+1), (x-1)^2(x+1)^2$. So the gcd is $(x-1)^2(x+1)$.
For $t=5$, the determinant is $(x-1)^3(x+1)^2$.
So the Smith Canonical should be the diagonal matrix
$$
\mbox{diag}(1,1,(x-1)(x+1), x-1, (x-1)(x+1)).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/294493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry with multiple angle and exact value of $\tan\pi/5$ By considering the equation $\tan5\theta=0$, show that the exact value of $\tan\pi/5$ is $\sqrt{5-2\sqrt{5}}$.
Do I need to evaluate the multiple angle for $\tan5\theta=0$?
| The idea is to expand $\tan 5\theta$ so as to get a function of $\tan \theta$. Then by letting $\theta = \pi/5$, you will get an equation in $\tan \pi/5$.
In details, put $x = \tan \pi/5$ $$ \tan 5 \theta = \frac{\tan 2\theta + \tan 3\theta}{1 - \tan 2\theta \tan 3\theta}$$
Then you find that $\tan 2\pi/5 + \tan 3\pi/5 = 0$.
But $$\tan 2\pi/5 = \frac{2 x}{1 - x^2}$$
And $$ \tan 3\pi/5 = \frac{\tan(2\pi/5) + x}{1 - x\tan(2\pi/5)} = \frac{3x - x^3}{1 - 3x^2}.$$
Conclusion $$ \frac{2x}{1 -x^2} + \frac{3x - x^3}{1 - 3x^2} = 0.$$
Because $x \neq 0$, we see that $x$ is solution of the polynomial $x^4 - 10x^2 +5 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/294884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Induction on binomial Identity: $0\cdot {n\choose 0} + 2\cdot {n\choose 2} + 4\cdot {n\choose4}+\ldots = n\cdot2^{n-2}$ I am having trouble proving the following identity:
$0\cdot {n\choose 0} + 2\cdot {n\choose 2} + 4\cdot {n\choose4}+\ldots = n\cdot2^{n-2}$
Here is what I have so far:
Proof:
Base: Let $n=0$:
LHS: $0\cdot {0\choose 0} = 0\cdot1 = 0$
RHS: $0\cdot 2^{0-2} = 0$
Step: Let $k\in \mathbb{Z} s.t k \geq 0$ and assume the identity is true for k.
Consider the LHS for $k+1$ where $k$ is even (I leave out the odd case because I think it will turn out the same?):
\begin{align}
=& 0\cdot{k+1\choose 0}+2\cdot{k+1\choose 2}+4\cdot{k+1\choose 4}+... +k\cdot{k+1\choose k}
\\=&0\cdot\left[{k\choose 0}+{k\choose -1}\right] + 2\cdot\left[{k\choose 2}+{k\choose 1}\right]+ 4\cdot\left[{k\choose 4}+{k\choose 3}\right]+\ldots+ k\cdot\left[{k\choose k}+{k\choose k-1}\right]
\\=&\left[0\cdot{k\choose 0}+2\cdot {k\choose 2}+4\cdot{k\choose 4}+\ldots+k\cdot{k\choose k}\right] + \left[0\cdot{k\choose -1}+ 2\cdot {k\choose 1}+4\cdot {k\choose 3}+\ldots+k\cdot{k\choose k-1}\right]
\\=& k\cdot2^{k-2} + \left[0\cdot{k\choose -1}+ 2\cdot {k\choose 1}+4\cdot {k\choose 3}+\ldots+k\cdot{k\choose k-1}\right]
\\
\end{align}
I know I need to end up with something like:
\begin{align}
=&k\cdot2^{k-2}+ \left[k\cdot2^{k-2} + 2^{k-1}\right]
\\=&2k\cdot 2^{k-2} + 2^{k-1}
\\=&k\cdot 2^{k-1}+2^{k-1}
\\=&(k+1)\cdot 2^{k-1}\end{align}
But, how can I get what I need from the combinations above? It may not end up exactly like that, but what is the reasoning behind this?
| Unlike with other proof methods, proofs by induction sometimes become easier if you make the statement to be proved stronger. In the present case, it seems that your approach should be successful if you add a corresponding statement about the sum of binomial coefficients with odd lower arguments to the claim:
$$
\sum_{j=1}^{\lceil n/2\rceil}2j\binom n{2j-1}=2^{n-2}(n+2)\;.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove using combinatorial argument $\binom{n}{0}.\binom{n}{2}+ \binom{n}{1}.\binom{n}{3}+\cdots+ \binom{n}{n-2}.\binom{n}{n}$ How can I prove it using combinatorial argument
$$
\binom n0\binom n2 + \binom n1\binom n3 + \cdots + \binom{n}{n-2}\binom nn = \binom {2n}{n-2}
$$
Sorry friends I have edited it
| Recall that $\binom{n}k=\binom{n}{n-k}$, so your identity can be written
$$\binom{n}0\binom{n}{n-2}+\binom{n}1\binom{n}{n-3}+\ldots+\binom{n}{n-2}\binom{n}0=\binom{2n}{n-2}\;.\tag{1}$$
Imagine that you have $n$ couples, each consisting of a man and wife, and you want to choose $n-2$ of these $2n$ people. Let $k$ be the number of women you choose; $k$ can be anywhere from $0$ through $n-2$, and you must then choose $n-2-k$ men. There are $\binom{n}k$ ways to choose these $k$ women, and there are $\binom{n}{n-2-k}$ ways to choose the $n-2-k$ men, so there are $$\binom{n}k\binom{n}{n-2-k}$$ ways to choose a group of $n-2$ people containing exactly $k$ women.
The lefthand side of $(1)$ simply sums these numbers over all possible values of $k$; the righthand side counts the number of groups of $n-2$ people directly.
Added: I don’t see a nice combinatorial way to answer the question in the comments:
Evaluate $$\sum_{k=0}^{20}(-1)^k\binom{30}k\binom{30}{k+10}\;.\tag{2}$$
However, I can do it with generating functions. First note that $(2)$ is equal to $$\sum_{k=0}^{20}(-1)^k\binom{30}k\binom{30}{20-k}\;,$$ so the problem is an instance of evaluating $$\sum_{k=0}^m(-1)^k\binom{n}k\binom{n}{m-k}\;.$$ From the binomial theorem we have $$(1-x)^n=\sum_{k=0}^n(-1)^k\binom{n}kx^k\quad\text{and}\quad(1+x)^n=\sum_{k=0}^n\binom{n}kx^k\;,$$ so $$\left(1-x^2\right)^n=(1-x)^n(1+x)^n=\left(\sum_{k=0}^n(-1)^k\binom{n}kx^k\right)\left(\sum_{k=0}^n\binom{n}kx^k\right)\;,\tag{3}$$ a polynomial of degree $2n$. Let $c_m$ be the coefficient of $x^m$ in $(3)$. Since $$\left(1-x^2\right)^n=\sum_{k=0}^n(-1)^k\binom{n}kx^{2k}\;,$$ it’s clear that
$$c_m=\begin{cases}0,&\text{if }m\text{ is odd}\\\\\binom{n}{m/2},&\text{if }m\text{ is even}\;.\end{cases}$$
On the other hand, multiplying out the product of polynomials on the righthand side of $(3)$ shows that $$c_m=\sum_{k=0}^m(-1)^k\binom{n}k\binom{n}{m-k}\;,$$ so
$$\sum_{k=0}^m(-1)^k\binom{n}k\binom{n}{m-k}=\begin{cases}0,&\text{if }m\text{ is odd}\\\\\binom{n}{m/2},&\text{if }m\text{ is even}\;.\end{cases}$$
In your problem $n=30$ and $m=20$, so $(2)$ reduces to $\dbinom{30}{10}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that none of $\{11, 111, 1111,\dots \}$ is the perfect square of an integer Please help me with solving this :
prove that none of $\{11, 111, 1111 \ldots \}$ is the square of any $x\in\mathbb{Z}$ (that is, there is no $x\in\mathbb{Z}$ such that $x^2\in\{11, 111, 1111, \ldots\}$).
| Every integer is of one of the forms $\color{green}4k$, $\color{green}4k+\color{red}1$, $\color{green}4k+\color{red}2$ or $\color{green}4k+\color{red}3$ with $k$ integer. The square of an integer has therefore one of the forms
*
*$(\color{green}4k)^2 = 16k^2 = \color{green}4(4k^2) = \color{green}4k_0$,
*$(\color{green}4k+\color{red}1)^2 = 16k^2+8k+1 = \color{green}4(4k^2+2k)+\color{red}1 = \color{green}4k_1+\color{red}{1}$,
*$(\color{green}4k+\color{red}2)^2 = 16k^2+16k+4 = \color{green}4(4k^2+4k+1)=\color{green}4k_2$ or
*$(\color{green}4k+\color{red}3)^2 = 16k^2+24k+9 = \color{green}4(4k^2+6k+2)+\color{red}1 = \color{green}4k_3+\color{red}1$.
That is, a perfect square is equivalent either to $\color{red}0$ or to $\color{red}1$ modulo $\color{green}4$.
On the other hand, $R_n=\underbrace{1\ldots1}_n=(10^n-1)/9$ for $n>1$ has the form
*
*$R_n=100R_{n-2}+11 = \color{green}4(25R_{n-2}+2)+\color{red}3 = \color{green}4k_r+\color{red}3$.
That is, for $n>1$, $R_n$ is equivalent to $\color{red}3$ modulo $\color{green}4$, which as shown above does not happen for perfect squares.
| {
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"url": "https://math.stackexchange.com/questions/298234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
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Induction Proof: $\sum_{i=1}^{n+1} i \cdot 2^i = n \cdot 2^{n+2}+2 $ Prove by Mathematical Induction . . .
$$\sum_{i=1}^{n+1} i \cdot 2^i = n \cdot 2^{n+2}+2 $$
for all $n \geq 0$
I tried solving it, but I got stuck near the end . . .
a. Basis Step:
$1\cdot 2^1 = 0\cdot 2^{0+2}+2$
$2 = 2$
b. Inductive Hypothesis
$$\sum_{i=1}^{k+1} i \cdot 2^i = k \cdot 2^{k+2} +2 $$
for $k \geq 0$
Prove k+1 is true.
$$\sum_{i=1}^{k+2} i \cdot 2^i = (k+1)\cdot 2^{k+3}+2 $$
$\big[RHS\big]$
$k\cdot 2^{k+3}+2^{k+3}+2$
$\big[LHS\big]$
$$\sum_{i=1}^{k+2} {i \cdot 2^{i}} $$
$= \underbrace{\sum_{i=1}^{k+1} i \cdot 2^i} + (k+2)\cdot 2^{k+2}$ (Explicit last step)
$= \underbrace{k\cdot 2^{k+2}+2}+(k+2)\cdot 2^{k+2}$ (Inductive Hypothesis Substitution)
$= k\cdot 2^{k+2}+2+k\cdot 2^{k+2}+2^{k+3}$
$= 2k\cdot 2^{k+2} + 2^{k+3} + 2$
My [LHS] has one too many $2k\cdot 2^{k+2}$ or did it just do it completely wrong?
| You are done. All you need to do is to regroup the terms properly.
$$2k \cdot 2^{k+2} + 2^{k+3} + 2 = k \cdot 2^{k+3} + 2^{k+3} + 2 = (k+1) \cdot 2^{k+3} + 2$$
which is what you want.
As julien points out, your penultimate step must read $$k \cdot 2^{k+2} + 2 + \color{blue}{k \cdot 2^{k+2}} + 2^{k+3}$$
instead of
$$k \cdot 2^{k+2} + 2 + \color{red}{k \cdot 2{k+2}} + 2^{k+3}$$
I assume you made a typo while typesetting this line since you missed the ^ symbol i.e. you have
$$\text{`k \cdot 2{k+2}` instead of `k \cdot 2^{k+2}`}$$
And also as julien points out, your (Explicit last step) should read
$$\color{blue}{\sum_{i=1}^{k+1} i \cdot 2^i} + (k+2) \cdot 2^{k+2}$$
and not
$$\color{red}{(k+1) \cdot 2^{k+1}} + (k+2) \cdot 2^{k+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/299174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integer solution to equation Does $$b =a-23\sqrt 3 \sqrt a+432$$ have infinitely many integer solutions?
$a = 3$ gives one;
$a = 27$ gives one;
$a = 27\times 9$ gives one.
| Let $a = 3n^2,\;$ where $n$ is any integer (and hence $\,a = 3n^2$ must also be an integer.)
So
$$
\begin{align}
b &= a-23\sqrt{3}\sqrt{a}+432 \\ \\
&= a -23\sqrt{3a}+ 432 \\ \\
&= (3n^2)-23\sqrt{3^2n^2}+432\\&=3n^2-23\cdot3n+432\\ \\
\iff b &= 3(n^2 - 23n + 144)
\end{align}
$$
Hence, for $a = 3n^2,\;b = 3(n^2 - 23n + 144)$ is an also an integer for any given any integer $n$. Since $n$ can be any integer, and since there are infinitely many $n$, there are infinitely many solutions $(a, b)$, one solution per $n$, given by $$a = 3n^2;\;\;b = 3(n^2 - 23n + 144).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/299345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Another improper integral Show that :
$$\int_0^1\frac{(\sin ^{-1}x)^2}{x}\text{d}x=\frac{\pi ^2\ln 2}{4}-\frac78\zeta(3)$$
This integral is in "irresistible integrals" on page 122. I can't prove this one.
| This is not an answer, but you might be interested in knowing why this can work. I thought of writing this as a comment, but it's quite big.
*
*Note that $\displaystyle \ln(1+x) = x-\frac{x^2}{2} + \frac{x^3}{3} + \cdots$
*So $\ln(2) = \displaystyle 1-\frac{1}{2} + \frac{1}{3} + \cdots$
Now I tried the standard integration by parts method. Put $x =\sin\theta$. Then the integral becomes $$I =\int_{0}^{\pi/2} \theta^{2} \cdot \cot\theta \ d\theta$$ Using the standard ILATE rule take $dv=\cot\theta \ d\theta$ so that you have $v=\ln(\sin\theta)$. Take $u=x^{2} \implies du = 2\theta \ d\theta$. So we have $$I = \underbrace{\bigl( u\cdot v\bigr)_{0}^{\pi/2}} - 2\int_{0}^{\pi/2} \theta \cdot \ln(\sin\theta)\ d\theta .$$
The underbrace term $\to 0$ so we are left with $$-2\int_{0}^{\pi/2} \theta \cdot \ln(\sin\theta) \ d\theta$$
Now
\begin{align*}
\int_{0}^{\pi/2} \theta \cdot \ln(\sin\theta) \ d\theta &=\int_{0}^{\pi/2} \theta \cdot \ln\bigl(1+ (\sin\theta-1)\bigr)\ d\theta \\ &= \int_{0}^{\pi/2} \theta \cdot \biggl[ (\sin\theta-1) - \frac{(\sin\theta -1)^{2}}{2}+\frac{(\sin\theta-1)^{3} }{3}- \cdots \biggr] \ d\theta
\end{align*}
Now collecting the $\theta$ involving terms we see that its:
\begin{align*}
\int_{0}^{\pi/2} \biggl( -\theta + \frac{\theta}{2} -\frac{\theta}{3} + \cdots \biggr) \ d\theta &=-\biggl(1-\frac{1}{2}+\frac{1}{3} - \cdots\biggr)\cdot \int_{0}^{\pi/2} \theta \ d\theta \\ &= -\ln(2) \cdot \frac{\pi^{2}}{8}
\end{align*}
Note that we have to multiply by $-2$ to complete the solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Finding all solutions to $ \tan^5x - 9\tan{x} = 0 $ I am stuck when it comes to finding the end value of a trig function. I have the following question:
$$ \tan^5x - 9\tan{x} = 0 $$
I worked the problem and got:
$$ \tan x = 0\\
\tan^4x-9 = 0\\
x = 0, \pi, \frac {\pi}{3}, \frac {2\pi}{3}, \frac {4\pi}{3}, \frac {5\pi}{3} $$
My book answer is $x = \frac {\pi k}{3}$ how do you get that? I understand that tan uses $ \pi $ and sin, cos use $ 2\pi $ but I'm not sure how they got to that answer.
| $\tan^2x=3\implies \cos2x=\frac{1-\tan^2x}{1+\tan^2x}=\frac{1-3}{1+3}=-\frac12=\cos\frac{2\pi}3$
$2x=2n\pi\pm \frac{2\pi}3=\frac{2\pi}3(3n\pm1)$ where $n$ is any integer.
$x=\frac{\pi}3(3n\pm1)$
Now, we need $0\le \frac{\pi}3(3n+1)<2\pi\implies 0\le3n+1<6\implies n=0,1$
For $n=0, x=\frac{\pi}3(3n+1)= \frac{\pi}3; n=1\implies x=\frac{\pi}3(3n+1)= \frac{4\pi}3$
Similarly, we need $0\le \frac{\pi}3(3n-1)<2\pi\implies 0\le3n-1<6\implies n=1,2$
For $n=1, x=\frac{\pi}3(3n-1)= \frac{2\pi}3; n=2\implies x=\frac{\pi}3(3n-1)= \frac{5\pi}3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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The remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$ How to find the remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$
| As Ishan has suggested, $(2m+1)^2=4m^2+4m+1=8\frac{m(m+1)}2+1\equiv1\pmod8$
So, $\sum_{0\le r\le n}(2r+1)^2\equiv\sum_{0\le r\le n}1\pmod 8\equiv n+1\pmod8$
For $1013=2\cdot506+1,$ so $n=506\implies \sum_{0\le r\le 506}(2r+1)^2\equiv 507\pmod8\equiv3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$a,b,c\in\mathbb{N}\:\land\:\frac ab+\frac bc+\frac ca\in\mathbb{N}\Rightarrow abc=n^3,\:n\in\mathbb{N}$
If $a$, $b$ and $c$ are natural numbers such that $$\frac ab+\frac bc+\frac ca\in\mathbb{N},$$ prove that $abc$ is a perfect cube.
All I could get was $ab^2+bc^2+ca^2\equiv_{abc}0$.
| Consider polynom $(X-\frac a b)(X- \frac b c)(X - \frac c a)$.
It is equal to $X^3 - u*X^2 + v*X - 1$, where $u = \frac a b + \frac b c + \frac c a$.
and $v = \frac b a + \frac c b + \frac a c$.
for $X = 1$, you get $1 - u + v - 1 = v - u$. So you have $(1-\frac a b)(1- \frac b c)(1- \frac c a) = v-u$.
You can rewrite that: $(b-a)(c-b)(a-c)=c(b^2-a^2)+a(c^2-b^2)+b(a^2-c^2)$
but not sure it leads anywhere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
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How can I find the possible values that $\gcd(a+b,a^2+b^2)$ can take, if $\gcd(a,b)=1$ If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them.
| $$ \gcd(a+b,a^2+b^2) \mid \gcd((a+b)(a-b), a^2+b^2) = \gcd(a^2-b^2, a^2+b^2) \mid \gcd [ ( a^2+b^2)+ (a^2-b^2) , ( a^2+b^2)+ (a^2-b^2) ]=2 \gcd(a^2,b^2)=2$$
Now it is easy to check that both 1 and 2 are possible...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 2
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Find the shortest distance between $9x^2+9y^2-30y+16=0$ and $y^2=x^3$ Find the shortest distance between $9x^2+9y^2-30y+16=0$ and $y^2=x^3$.
I know the shortest distance exists between the curves on the common normal line. Is there any other shorter way to attempt?
| As the first curve is a circle you only have to find its center, and than compute the normal equations of your second curve. the one which goes through the center of the circle will do the job.
$$
\begin{align*}9x^2 + 9y^2 -30y+16 &= 9 \left( x^2+y^2 - \frac{10}{3}y +\frac{16}{9}\right)\\
&=9 \left(x^2+y^2-2\cdot \frac{5}{3} +\left(\frac{5}{3}\right)^2 -\left(\frac{5}{3}\right)^2 +\frac{16}{9}\right)\\
&=9\left(x^2 +\left( y- \frac{5}{3}\right)^2 -1\right)=0
\end{align*}$$
So the center of the circle wil be $$\begin{pmatrix} 0 \\ \frac{5}{3}\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308097",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find all integers such that $\frac{n^3-3}{n^2-7}$ is an integer Find all integers such that $\frac{n^3-3}{n^2-7}$ is an integer.
I have no idea how to approach these types of proofs.
But I tried a few things, did not get me anywhere.
$n^3 -3 = an^2-7a$ then $n^3-an^2 = 3-7a$, and hence $n^2(n-a) = 3-7a$
And then I have no where to go...
Any help is appreciated thanks.
| Hint $\rm\ \ n^2\!-\color{#0A0}7\mid n^3\!-\color{#C00}3\ \Rightarrow\ mod\,\ n^2\!-7\!:\,\ \color{#0A0}7^3\! = n^6 = \color{#C00}3^2\:\Rightarrow\: n^2\!-7\mid 7^3\!-3^2 =\, 2\cdot 167\ $
$\rm\,2,\,167\,$ are prime so, by unique factorization, $\rm\:n^2\!-7\mid\, 2\cdot 167\,$ $\Rightarrow$ $\rm\, n^2\!-7 = \pm\{1, 2,167, 334\},\,$ so $\ldots$
| {
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"url": "https://math.stackexchange.com/questions/309124",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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"answer_id": 0
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All sequences constructed by using the denominator as nominator and the sum of denominator and nominator as denominator converges to $\phi-1$ Assume we are given any number a. Write it in the form $a = \frac{b}{c}$ (if rational, in the usual way, if irrational, use forms like $\frac{a}{1}$).
Construct a sequence $\frac{b}{c},\frac{c}{b+c},\frac{b+c}{b+2 c},\frac{b+2c}{2b+3c}...$
The nth term is $\frac{F_{n-1}*b+F_n*c}{F_n*b+F_{n+1}*c}$ where $F_n$ is the nth Fibonacci number.
I have noticed that no matter what the starting point $a$ is, the sequence always converges to $\phi-1$ , ie. $\frac{(\sqrt{5}+1)}{2}-1$ or $\frac{(\sqrt{5}-1)}{2}$.
I would like to prove this. If $a=1$, then the nth term is $F_{n+1}/F_{n+2}$ and the limit is $\phi-1$. But If $a \not= 1$ this seems harder to show. Especially if $a < 0$.
| Given $\displaystyle\frac{F_{n+1}}{F_n}\to\varphi$ and $\displaystyle\frac{F_{n-1}}{F_n}\to\varphi^{-1}$, your sequence converges as
$$\frac{1/F_n}{1/F_n}\cdot\frac{F_{n-1}b+F_nc}{F_nb+F_{n+1}c}=\frac{b\displaystyle\frac{F_{n-1}}{F_n}+c}{\displaystyle b+c\frac{F_{n+1}}{F_n}}\longrightarrow \frac{b\varphi^{-1}+c}{b+c\varphi}=\varphi^{-1}\frac{b+c\varphi}{b+c\varphi}=\varphi^{-1}.$$
Note that $\varphi^{-1}=\varphi-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the largest possible value for the number of elements If $S$ is a subset from the set {$1,2,3,...,20$} such that the sum of any two elements is not divisible by $5$.How to find the largest possible value for the number of elements for $S$
| If we choose at least one number of the form $5a+1,$ we can not choose any number of the form $5b+4$ and vice versa where $a,b$ are non-negative integers.
If we choose at least one number of the form $5c+2$ we can not choose any number of the form $5d+3$ and vice versa where $c,d$ are non-negative integers.
Clearly, we can choose at most one number of the form $5e$ as $5\mid(5e_1+5e_2)$
Observe that $5\not\mid(5a_1+1+5a_2+1),5\not\mid(5a_1+1+5b_1+2),5\not\mid(5b_1+2+5b_2+2)$ where $a_i,b_i$ are non-negative integers.
As $0<5a+1\le 20\implies 0\le a\le3,$
$0<5b+4\le20\implies 0\le b\le3$
and $0<5e\le 20\implies 1\le e\le 4,$
we can at most choose $4+4+1=9$ elements
| {
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Given coordinates of center, equation of tangent to a circle, what is their point of contact? A circle centre $(-6,3)$ has a tangent with equation $$3X + 2Y = 40$$ What are the coordinates of the point of contact of the tangent with the circle?
| The radius of the circle must be perpendicular to the tangent. Rewriting $3x+2y=40$ gives us $y=-\frac{3}{2}x+20$, so the slope of the tangent line is $-\frac{3}{2}$. The slope of the perpendicular is the negative reciprocal, so the slope of the radius is $\frac{2}{3}$, and therefore the equation of the line the radius lies on is: $y_r=\frac{2}{3}x+b$. We know the center of the circle $(-6,3)$ is on this line, so we have: $3=\frac{2}{3}(-6)+b\Rightarrow b=7$, so $y_r=\frac{2}{3}x+7$. Now solve the equation $\frac{2}{3}x+7=-\frac{3}{2}x+20$, yielding $x=6$. Plugging $6$ in for $x$ in either equation gives $y=11$, so the coordinates of the point of intersection are $(6,11)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$ I need to show that
$$
\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}
$$
I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from. Using a change of variable $ x \mapsto 2x $ :
$$
\int_0^\infty \frac{\sin^2(2x)}{x^2}dx = \pi
$$
Now using the identity $\sin^2(2x) = 4\sin^2x - 4\sin^4x $, we obtain
$$
\int_0^\infty \frac{\sin^2x - \sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\frac{\pi}{2} - \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
But I am now at a loss as to how to make $x^4$ appear at the denominator. Any ideas appreciated.
Important: I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and use the change of variable and identity mentioned above
| HINT: Use the relation
$$\int_0^\infty \left(\frac{\sin x}{x}\right)^n \mathrm{d}x
= \frac{\pi}{2^n (n-1)!}
\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}$$
You may find a proof here A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$.
| {
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Possible average square value suppose the sum of seven Positive number is 21. what is the minimum possible value of the average of the square of these number?
| Let's assume we have a 7-tuple in which some numbers are not equal. Call two numbers that are different $x$ and $y$, then $\frac {x^2+y^2} 2$ is more than $(\frac {x+y} 2 )^2$, which is the average if $x$ and $y$ were each replaced by their average, $(\frac {x+y} 2)$.To show this, first expand $(\frac {x+y} 2 )^2$ to $\frac {x^2+y^2+2xy} 4$. Then, to show that $\frac {x^2+y^2} 2>\frac {x^2+y^2+2xy} 4$ multiply by $4$, you get $2x^2+2y^2>x^2+y^2+2xy$. Subtract $x^2+y^2+2xy$ from both sides and get $x^2+y^2-2xy>0$, which is clearly true if $x \ne y$ because it's equal to $(x-y)^2$. This shows that any such "reduction" will always lower the average of squares.
Now, assume that there is a 7-tuple with a lower average than $(3,..,3)$. You can easily reduce it with my method to a 7-tuple with four numbers equal, and the other three equal as well.(i.e. $x,x,x,x,y,y,y$.) Just apply the reduction method to the first two numbers, the 3rd and 4th,1st and 3rd,2nd and 4th,5th and 6th,4th and 7th,4th and 5th,and 6th and 7th, in that order.(Try it with an actual group of distinct numbers if you don't follow:)
Now this 7-tuple's average of square is lower than our original 7-tuple, and so that average must be lower than 9. However this cannot be for the following reason. The sum of the squares in terms of $x$ is: $4x^2+3((\frac {21-4x} 3)^2)$. This expands to $\frac 7 3 (4x^2-24 x+63)$. For this to be less than 63, $7(4x^2-24 x+63)$=$28x^2-168x+441$ must be less than 189, so $28x^2-168x+252=28(x-3)^2$ must be less than zero. This clearly cannot be for any real x.
Therefore, any solution that all the numbers are not the same cannot have an average square of less than that of $(3,..,3)$, namely 9.
QED
| {
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Limit of series involving ratio of two factorials $$
\sum^{\infty}_{j=0} \frac{(j!)^2}{(2j)!} = \frac{2 \pi \sqrt{3}}{27}+\frac{4}{3}
$$
The above series is in a homework sheet. We're not expected to find the limit, just prove its convergence. That's easy, but since we were given the limit, it got me thinking about how to find such a limit.
If anyone could point me in the right direction, I'd be happy to discover it on my own, but after a few hours of searching, I don't feel much closer.
| Notice that:
$$
\frac{j!^2}{(2j)!} = (2j+1) \operatorname{Beta}(j+1,j+1) = (2j+1) \int_0^1 t^j (1-t)^j \mathrm{d}t
$$
Now, interchanging the summation and the integration:
$$
\sum_{j=0}^\infty \frac{j!^2}{(2j)!} = \int_0^1 \sum_{j=0}^\infty (2j+1)(t(1-t))^j \mathrm{d} t = \int_0^1 \frac{1+t(1-t)}{(1-t(1-t))^2} \mathrm{d} t
$$
The latter integral is evaluated integrating by parts and reducing it to the table integral:
$$
\int_0^1 \frac{1+t(1-t)}{(1-t(1-t))^2} \mathrm{d} t = \left[ \frac{2}{3} \frac{2t-1}{1-t+t^2} + \frac{2}{3 \sqrt{3}} \arctan\left(\frac{2t-1}{\sqrt{3}} \right) \right]_{t=0}^{t=1} =2 \left(\frac{2}{3} + \frac{\pi}{9 \sqrt{3}}\right)
$$
This establishes the result.
Alternatively, you could note that the summand $c_j$ is a hypergeometric term, which means that the ratio of successive terms is a rational function of $j$
$$
\frac{c_{j+1}}{c_j} = \frac{j+1}{4j+2} = \frac{(j+1)(j+1)}{j+\frac{1}{2}} \frac{1}{4} \frac{1}{j+1}
$$
which means that the sum corresponds to the defining series of the Gauss's hypergeometric function:
$$\begin{eqnarray}
\sum_{j=0}^\infty \frac{j! \cdot j!}{(2j)!} &=& \sum_{j=0}^\infty \frac{(1)_j (1)_j}{\left(\frac{1}{2}\right)_j} \frac{(1/4)^j}{j!} = {}_2F_1\left(\left.\begin{array}{cc} 1 & 1\\ & \frac{1}{2} \end{array} \right| \frac{1}{4}\right) \\
&=& \left.\frac{1}{1-z} \left(1 + \sqrt{\frac{z}{1-z}} \arcsin\sqrt{z}\right)\right|_{z=1/4} = \frac{4}{3} + \frac{2 \pi}{9 \sqrt{3}}
\end{eqnarray}
$$
| {
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Solve the following homogeneous differential equation Initial value problem: $\displaystyle \frac{dy}{dx}= \frac{y}{x}+2 \frac{x^2}{y^2}$, $y(1)=1$. Can anyone help
| Again, start with $y = ux$ so that $\frac{dy}{dx} = u + x\frac{du}{dx}$ (applying the simple chain rule).
Substituting this into the original equation, along with
$ \frac{y}{x} = u $ and $ {\frac{x}{y}}^2 = u^{-2} $ results in
$u + x\frac{du}{dx} = u + 2u^{-2}$.
Since the u terms cancel on each side, the equation may be rearranged as
$u^2du = 2x^{-1} dx$.
At $x=x_0 = 1$, $y=y_0=1$ and $u=u_0 = 1$, so we can integrate the left side from 1 to $u$ and integrate the right side from 1 to $x$. This yields $\frac{1}{3}\left(u^3-1\right)=2ln(x)$ -> $u^3=1+6ln(x)$ -> $u={\left(1+6 ln x\right)}^\frac{1}{3}$.
Since u = $\frac{y}{x}$, the final result is $y=x{\left(1+6lnx\right)}^\frac{1}{3}$.
No answer is complete without validating, so let's check the terms:
$$\frac{dy}{dx} = {\left(1+6lnx\right)}^\frac{1}{3} + x\frac{1}{3}{\left(1+6lnx\right)}^\frac{-2}{3}\left(6\right)\frac{1}{x} = {\left(1+6lnx\right)}^\frac{1}{3} + 2 {\left(1+6lnx\right)}^\frac{-2}{3}
$$
$$ \frac{y}{x} = {\left(1+6lnx\right)}^\frac{1}{3}$$
$$ 2\frac{x^2}{y^2} = 2{\left(1+6lnx\right)}^\frac{-2}{3} $$
And at $x = 1$, $y = 1{\left(1+6ln (1)\right)}^\frac{1}{3} = 1$
| {
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Prove that $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+R_8(x)$ where $|R_8(x)|\leq \frac{x^8}{8!}$ Prove that
$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+R_8(x)$$
where $|R_8(x)|\leq \frac{x^8}{8!}$
| Hint: $f(x)=\sum_{i=0}^n\frac{f^ {(i)}(x)x^i}{i! }$
$\text if$ $$|f^{n+1}(x)|\le M_{n+1}$$then $$|R(x)|\le \frac{M_{n+1}|x|^n}{(n+1)!}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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finding the derivative using quotient rule and product rule find dy/dx;
a) $\frac{1-2x}{\sqrt{2+x}}$
b.) $3x(1-x^2)^{1/3}$
My attempt at a)
use the quotient rule:
so $dy/dx = -2 \sqrt{2+x}+ (1-2x)0.5(2+x)^{-1/2}$
but then I get stuck there, cannot simplify it, wolfram gives a nice simplified answer but not sure how to get it.
b.) Product rule:
$dy/dx= 3(1-x^2)^{0.5} + 3 \times 1/3 \times (1-x^2)^{-2/3}$ and again i can't seem to simplify that either to a nice wolfram answer.
| a) $$\left(\frac{1-2x}{\sqrt{2+x}}\right)'=\frac{-2\sqrt{2+x}-(1-2x)\frac{1}{2\sqrt{2+x}}}{2+x}=\frac{-4(2+x)-(1-2x)}{2\sqrt{2+x}(2+x)}$$
$$=\frac{-9-2x}{2\sqrt{2+x}(2+x)}=-\frac{9+2x}{2(2+x)^{3/2}}$$
b.) $$(3x(1-x^2)^{1/3})'=3(1-x^2)^{1/3}-2x^2(1-x^2)^{-2/3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the area of the Rose's petal.
If a Rose leaf is described by the equation $r = \sin 3\theta$, find the area of one petal.
| A sketch is useful here, but the only important observation is that $r=0$ when $\theta=0$, and again at $\frac{\pi}{3}$. These are your limits for one petal.
Since the area of a polar curve between the rays $\theta=a$ and $\theta=b$ is given by $\int_{a}^{b}\frac{1}{2}r^{2}d\theta$, we have
$$A=\int_{0}^{\pi/3}\frac{1}{2}\sin^{2}(3\theta)d\theta=\frac{1}{2}\int_{0}^{\pi/3}\frac{1-\cos(6\theta)}{2}d\theta$$
$$=\frac{1}{4}\left[\theta-\frac{\sin(6\theta)}{2}\right]^{\pi/3}_{0}=\frac{1}{4}\left(\frac{\pi}{3}-\frac{1}{2}\sin\left(\frac{6\pi}{3}\right)\right)=\frac{\pi}{12}$$
| {
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Is there any procedure to compute the root of a matrix equation?? Actually I want to know the procedure, how could we calculate the value of $A$ where
$A^2$= $\begin{pmatrix} x & y\\ z & w\\ \end{pmatrix}$
| $A=\begin{pmatrix} x & y\\ z & w\\ \end{pmatrix}$
You want to find $B=\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}$ so that $B^2=A$
You can just calculate $B^2$ and solve for $a,b,c,d$ but a better way of doing it is to notice that $A$ commutes with $B$: $AB=B^2B=B^3=BB^2=BA$
From that, you can calculate $AB$ and $BA$ and since you know they are equal, it'll simplify $B$.
$AB=\begin{pmatrix} x & y\\ z & w\\ \end{pmatrix}\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}=\begin{pmatrix} xa+yc & xb+yd\\za+wc & zb+wd\\ \end{pmatrix}$
$BA=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\begin{pmatrix} x & y\\ z & w\end{pmatrix}=\begin{pmatrix} xa+zb & ya+wb\\ xc+zd & yc+wd\end{pmatrix}$
$\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}=AB-BA=\begin{pmatrix} xa+yc & xb+yd\\za+wc & zb+wd \end{pmatrix}-\begin{pmatrix} xa+zb & ya+wb\\ xc+zd & yc+wd\end{pmatrix}=\begin{pmatrix} yc-zb & *\\* & zb-yc \end{pmatrix}$
So you get that $zb=yc$ which helps you simplify $B$:
*
*If $z=0=y$ then $A=\begin{pmatrix} x & 0\\ 0 & w \end{pmatrix}$ is diagonal so you can just take $B=\begin{pmatrix}a & 0\\ 0 & d \end{pmatrix}$ where $a^2=x$ and $d^2=2$
*If $y\not= 0$, you get $c=\cfrac{z}{y}b$ so $B=\begin{pmatrix} a & b\\ \cfrac{z}{y}b & d\\ \end{pmatrix}$
*If $z\not= 0$, you get $b=\cfrac{y}{z}c$ so $B=\begin{pmatrix} a & \cfrac{y}{z}c\\ c & d\\ \end{pmatrix}$
Then you just calculate $B^2$ and solve for $a,b,c,d$ (or the ones that remain in the expression of $B$).
By the way there are some matrix that do no have a square root.
$A=\begin{pmatrix} 0 & 1\\ 0 & 0\\ \end{pmatrix}$
$y=1\not= 0$ so we have $B=\begin{pmatrix} a & b\\ \cfrac{z}{y}b & d\\ \end{pmatrix}=\begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix}$
Now $B^2=\begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix}\begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix}=\begin{pmatrix} a^2 & (a+d)b\\ 0 & d^2\\ \end{pmatrix}$
We want $B^2=A$, that is $\begin{pmatrix} a^2 & (a+d)b\\ 0 & d^2\\ \end{pmatrix}=\begin{pmatrix} 0 & 1\\ 0 & 0\\ \end{pmatrix}$
which implies $a=0=d$ and $1=(a+d)b=0b=0$ so you can not find a $B$ so that $B^2=A$
| {
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How to find the $f\left(x_i\right)$ for midpoint Riemann sum
$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline
\mbox{Time} & \mbox{9am} & \mbox{10:30 am} & \mbox{Noon} & \mbox{1:30 pm} & \mbox{3 pm} & \mbox{4:30 pm} & \mbox{6 pm} & \mbox{7:30 pm} & \mbox{9 pm} \\
\hline
t & 0 & 1.5 & 3 & 4.5 & 6 & 7.5 & 9 & 10.5 & 12 \\
\hline
P\left(t\right) & 200 & 728 & 1193 & 1329 & 1583 & 1291 & 804 & 256 & 0 \\
\hline
\end{array}
$
Use a midpoint Riemann sum with four intervals of equal size to estimate the total number of people seeking care during the 12-hour period.
I know the formula for the midpoint Riemann sum is $ \int_a^b f\left(x\right) \; \mathrm{d}x \approx \frac{b-a}{n} \left[ f\left(x_1\right) + f\left(x_2\right) + \ldots + f\left(x_n\right) \right]$
I know that $b = 12$ and $a = 0$ and $n = 4$, but what is my $f\left(x_1\right), \; f\left(x_2\right), \; f\left(x_3\right), \; f\left(x_4\right)$
| The midpoint of $[0,3]$ is $x_1=1.5$, so $f(x_1)=728$. The next interval to consider is $[3,6]$ with midpoint $x_2=4.5$. I think you can continue.
| {
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Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$ Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$
Im not sure should I use primitive root or quadratic residue.
For primitive root, $U_{33} = \{1,2,4,5,7,8,10,13,14,16,17,19,20,23,25,26,28,29,31,32\}$
Thus $\phi_{(3\cdot11)}=20=2^2\times{5}$
But it takes way too much time to test if each of them is primitive root or not...
But Im not quite familiar with the method of Quadratic Residue.
We have $x^8\equiv5\pmod3$ and $x^8\equiv 5 \pmod {11}$
so.... follows the rule of $x^2\equiv q \pmod n$,
we have $(x^4)^2 \equiv 5 \pmod 3$ and $(x^4)^2 \equiv 5 \pmod {11}$
and then how do I continue with it??
| $$x^8 \equiv 5 \pmod{33} \implies x^8 \equiv 5 \pmod 3 \implies (x^4)^2 \equiv 2 \pmod 3 \implies \text{No solution}$$
In general, when you want to solve for $$x^m \equiv a \pmod n \,\,\, (\spadesuit)$$ and if $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$, the idea is to first solve for $$x^m \equiv a \pmod {p_l^{a_l}} \,\,\, (\clubsuit)$$ You have a solution for your original problem $(\spadesuit)$ iff you have a solution for each $l \in \{1,2,\ldots,k\}$ in $(\clubsuit)$. Once you find solution for each $l$ in $(\clubsuit)$, put them together using Chinese Remainder theorem.
| {
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Evaluate a big expression. If:$$\lambda=\int_{0}^{1}\frac{dx}{1+x^3} ;$$
Then evaluate : $$ p=\lim_{n \rightarrow \infty}\left( \frac{\prod_{r=1}^{n}\left(n^3+r^3\right)}{n^{3n}} \right)^{1/n} ;$$
EDIT: I have now ended upto $$lnp=\int_{0}^{1} \ln(1+t^3).dt$$
Now, we have to evaluate this using $\lambda$ , howsoever .
| First, to evaluate $\lambda$, use partial fractions:
$$\frac{1}{1+x^3} = \frac{1}{3} \left ( \frac{1}{1+x} - \frac{x-2}{x^2-x+1} \right )$$
The second piece in the parentheses is more amenable to integration when expressed as follows:
$$\frac{x-2}{x^2-x+1} = \frac{x-1/2}{(x-1/2)^2+3/4} - \frac{3/2}{(x-1/2)^2+3/4}$$
Now, let
$$A = \int_0^1 \frac{dx}{1+x} = \log{2}$$
$$B = \int_0^1 dx \: \frac{x-1/2}{(x-1/2)^2+3/4} = \left[\log{\left[ \left( x-\frac{1}{2}\right)^2+\frac{3}{4}\right]}\right]_0^1 = 0$$
$$C = \int_0^1 dx \: \frac{3/2}{(x-1/2)^2+3/4} = \sqrt{3} \left[\arctan{\left [ \frac{2}{\sqrt{3}} \left( x-\frac{1}{2}\right)\right]}\right]_0^1 = \frac{\pi}{\sqrt{3}}$$
Therefore
$$\lambda = \frac{1}{3} (A - B + C) = \frac{1}{3} \log{2} + \frac{\pi}{3 \sqrt{3}}$$
Now for the limit, which is evaluated by taking logs of both sides:
$$\log{p} = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^n \log{\left(1+\frac{r^3}{n^3}\right)}$$
This is a Riemann sum, in the sense that
$$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^n f \left(\frac{r}{n}\right) = \int_0^1 dx \: f(x)$$
Therefore
$$\begin{align}\log{p} &= \int_0^1 dx \: \log{(1+x^3)}\\ &= [x \log{(1+x^3)}]_0^1 - \int_0^1 dx \: \frac{3 x^3}{1+x^3}\\ &= \log{2} - 3 \int_0^1 dx \: \left ( 1 - \frac{1}{1+x^3}\right)\\ &= \log{2}-3 + 3 \lambda\\&= 2 \log{2}-3 + \frac{\pi}{\sqrt{3}}\\ \therefore p &= 4 e^{-(3 - \frac{\pi}{\sqrt{3}})}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Showing series expansion How would I show $$\frac{\sinh z}{z^2}=\frac{1}{z}+\sum_0^\infty \frac{z^{2n+1}}{(2n+3)!}$$
I have $\sinh z=\sum_0^\infty\frac{z^{2n+1}}{(2n+1)!}$ and if I multiply by $\frac{1}{z^2}$ then $\frac{1}{z^2}\sinh z=\sum_0^\infty \frac{1}{z^2}\frac{z^{2n+1}}{(2n+1)!}$ but I how would I would factor out $\frac{1}{z}$?
| So far you have
$$\frac{\sinh z}{z^2} = \sum_{n = 0}^\infty \frac{1}{z^2} \frac{z^{2n+1}}{(2n+1)!} = \sum_{n = 0}^\infty \frac{z^{2n-1}}{(2n+1)!} = \frac{1}{z} + \sum_{n = 1}^\infty \frac{z^{2n-1}}{(2n+1)!}.$$
If we introduce a new index $k = n - 1$, we have
$$\frac{\sinh z}{z^2} = \frac{1}{z} + \sum_{k = 0}^\infty \frac{z^{2k+1}}{(2k+3)!}$$
| {
"language": "en",
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} |
How to prove Vandermonde's Identity: $\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}$? How can we prove that
$$\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}?$$
(Presumptive) Source: Theoretical Exercise 8, Ch 1, A First Course in Probability, 8th ed by Sheldon Ross.
| Consider the $K\times K$ matrix
$$B= \left[ \begin{array}{ccccccc}
1 & 0 & 0 & \cdots & 0 & 0 & 0 \\
1 & 1 & 0 & \cdots & 0 & 0 & 0 \\
0 & 1 & 1 & \cdots & 0 & 0 & 0 \\
0 & 0 & 1 & \cdots & 0 & 0 & 0 \\
& & & \ddots & & & \\
0 & 0 & 0 & \cdots & 1 & 1 & 0 \\
0 & 0 & 0 & \cdots & 0 & 1 & 1 \\
\end{array}
\right]$$
When it is multiplied by $K\times 1$ vectors, which starts with k-1st row of Pascals's triangle, the result is a $K\times 1$ vector which contains the 1st $K$ elements of kth row of Pascal's triangle. This is because it effectively mimics addition of elements of k-1st row of Pascal's triangle to produce kth row of Pascal's triangle. Except that it only does it for the 1st $K$ elements of a row. In particular,
$$B \times \left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right]
=\left[ \begin{array}{c}
1 \\
1 \\
\vdots \\
0
\end{array}
\right]
$$
$$
B\times B \times
\left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right]
=B\times \left[ \begin{array}{c}
1 \\
1 \\
\vdots \\
0
\end{array}
\right]
=\left[ \begin{array}{c}
1 \\
2 \\
1 \\
\vdots \\
0
\end{array}
\right]
$$
It is an easy proof that all powers of $B$ are symmetric with respect to their antidiagonal (just consider $B$ as a sum of $I$ and $N=B-I$ and then look at the expansion of $(I+N)^m$). If we designate $\left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right]$
as the zeroth row of Pascal's triangle, then the vector containing the 1st $K$ elements of $m$'s row of Pascal's triangle is $$B^m \times \left[ \begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}
\right],$$
which is the 1st column of $B^m$ (and by the symmetry, the last row of $B^m$).
Now set the $K$ (of this answer) equal to $n$ (of the posited question).
Then the left-hand side of the identity can be considered to be the dot product of $nth$ row of $B^R$ and 1st column of $B^M$, which is the $(n,1)$ element of $B^{R+M}$. Which also happens to be the right-hand side of the identity (in the posited question).
| {
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"question_score": "10",
"answer_count": 7,
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Proof: How to prove $n$ is odd if $n^2 + 3$ is even New to the whole proof thing.
Trying to figure out that, for all integers $n$, if $n^2 + 3$ is even, then $n$ is odd.
Thank you for the help.
| The notation "$m\mid n$" means "$m$ divides $n$," or "$n$ is divisible by $m$." In contrast, $m\not\mid n$ means "$m$ does NOT divide $n$," or put differently, "$n$ is not divisible by $m.$"
Assume $n^2 + 3$ is even.
$(1)$ $n^2 + 3$ is even means, by definition, that $2$ divides $n^2 + 3$:
$$2\mid (n^2 + 3) \iff n^2 + 3 = 2k\;\text{ for some}\; k \in \mathbb Z.$$
That is, $n^2 + 3 $ is some integer multiple of of $2$.
Hence $$n^2 = 2k - 3 = 2(k-2) + 1 \implies 2\not\mid 2(k - 2) + 1,\;\implies 2\not\mid n^2$$ since clearly, $\;2 \mid 2(k-2),\;$ but $\;2 \not\mid 1$. In words, $2$ does not divide $n^2$. So $n^2$ is not even, and hence it must be the case that $n^2$ is odd. From $(2)$, it will follow that $n$ is therefore odd.
$(2)$ $n^2$ odd $\iff n$ is odd.
*
*$ \implies$ We prove this by proving the contrapositive:
Assume $n$ is not odd (assume $n$ is even), so $n = 2k$ for some integer $k$. So then $n^2 = (2k)^2 = 4k^2 = 2(2k^2)$, and so $2$ divides $n^2$, and hence $n^2$ is even, ie. if $n$ is not odd, then $n^2$ is not odd, which gives us, contrapositively: $n^2$ is odd $\implies n$ is odd. N.B. (That's really all we need for your proof, but let's go ahead and show that the implication $(2)$ is bi-directional.)
*$\Longleftarrow$ Assume $n$ is odd. Then $n = 2k + 1 $ for some integer $k$. And so $$n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$$ Now $2 \not\mid [2(2k^2 + 2k) + 1]$, so $n^2$ cannot be even: That is, if $n$ is odd, then $n^2$ is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Mersenne numbers congruent to two Prove that $2^{2^n-1} = 2 \pmod{2^n - 1}$ given that $2^n = 2 \pmod n$.
How would I go about proving that? I started by saying let $m = 2^n - 1$
Then, $2^n = 1 \pmod{m}$.
So I need to prove $2^m = 1 \pmod m$. I'm not sure how to proceed.
| Observe that $2^n \equiv 1 \pmod{ 2^n-1} $.
Given that $2^n \equiv 2 \pmod{n}$, we know that $2^n - 1 \equiv 1 \pmod {n}$, or that $2^n -1 = kn + 1 $ for some integer $k$.
Thus, $2^{ 2^n -1 } = 2^{kn + 1} \equiv 2 \pmod{2^n-1}$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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When computing the Legendre symbol $(\frac{3}{p})$, how to combine different congruences together into one statement? I try to find Legendre symbol for $\left(\dfrac{3}{p}\right)$.
This is what I did so far:
case 1: $p=3\mod{4}. $
so $\left(\dfrac{3}{p}\right)=-\left(\dfrac{p}{3}\right)$. now $\left(\dfrac{3}{p}\right)=-1$ iff $p=2 \mod{3}$
case 2: $p=1\mod{4}$
so $\left(\dfrac{3}{p}\right)=\left(\dfrac{p}{3}\right)$. now $\left(\dfrac{3}{p}\right)=1$ iff $p=1 \mod{3}$
Now, I don't know how to combine the cases (each case with itself and together).
| As you've noted, the cases are as follows:
*
*If $p\equiv 3\bmod 4$, then $(\frac{3}{p})=-(\frac{p}{3})$.
*
*If $p\equiv 1\bmod 3$, then $(\frac{p}{3})=1$, so $(\frac{3}{p})=-1$.
*If $p\equiv 2\bmod 3$, then $(\frac{p}{3})=-1$, so $(\frac{3}{p})=1$.
*If $p\equiv 1\bmod 4$, then $(\frac{3}{p})=(\frac{p}{3})$.
*
*If $p\equiv 1\bmod 3$, then $(\frac{p}{3})=1$, so $(\frac{3}{p})=1$.
*If $p\equiv 2\bmod 3$, then $(\frac{p}{3})=-1$, so $(\frac{3}{p})=-1$.
Thus,
$$\left(\frac{3}{p}\right)=\begin{cases}
1 & \text{ if }\begin{cases}p\equiv 3\bmod 4 \text{ and }p\equiv 2\bmod 3,\text{ or }\\
p\equiv 1\bmod 4\text{ and }p\equiv 1\bmod 3, \end{cases}\\[0.1in]
-1 & \text{ if }\begin{cases}p\equiv 3\bmod 4 \text{ and }p\equiv 1\bmod 3,\text{ or }\\
p\equiv 1\bmod 4\text{ and }p\equiv 2\bmod 3, \end{cases}
\end{cases}$$
The Chinese remainder theorem tells you that any statement of the form
$$n\equiv a\bmod 3\quad\text{ and }\quad n\equiv b\bmod 4$$
can be converted, essentially uniquely, into a statement of the form
$$n\equiv c\bmod 12.$$
The best way of solving this sort of thing in general is to find an $x$ and $y$ such that
$$x\equiv 1\bmod 3 \quad\text{ and }\quad x\equiv 0\bmod 4$$
$$y\equiv 0\bmod 3\quad\text{ and }\quad y\equiv 1\bmod 4$$
so that the statement "$n\equiv a\bmod 3$ and $n\equiv b\bmod 4$" is equivalent to
$$n\equiv xa+yb\bmod 3\quad\text{ and }n\equiv xa+yb\bmod 4,$$
hence
$$n\equiv xa+yb\bmod 12.$$
For example, one choice that works is $x=4$ and $y=9$. Thus, to reformulate the statement that $$p\equiv 1\bmod 4\quad\text{ and }\quad p\equiv 2\bmod 3,$$
you have that $9\cdot 1+4\cdot 2=17$, and
$$p\equiv 1\equiv 17\bmod 4\quad\text{ and }\quad p\equiv 2\equiv 17\bmod 3,$$ so that $$p\equiv 17\equiv 5\bmod 12.$$
You can check that this is correct:
$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|}
n \bmod 12 & 0 & 1 & 2 & 3 & 4 & \color{red}{\large \mathbf{5}} & 6 & 7 & 8 & 9 & 10 & 11 \\\hline
n\bmod 3 & 0 & 1 & 2 & 0 & 1 & \color{red}{\large \mathbf{2}} & 0 & 1 & 2 & 0 & 1 & 2 \\\hline
n\bmod 4 & 0 & 1 & 2 & 3 & 0 & \color{red}{\large \mathbf{1}} & 2 & 3 & 0 & 1 & 2 & 3
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/344909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How do we know the rank is 1? Here's matrix A:
$$
\begin{pmatrix}
2 & 1 \\
-4 & -2 \\
-2 & -1 \end{pmatrix}
$$
Apparently, we can determine that matrix A is of rank 1 by noting that.
$$
\det
\begin{pmatrix}
2 & 1 \\
-4 & -2 \end{pmatrix}
=
\det
\begin{pmatrix}
2 & 1 \\
-2 & -1 \end{pmatrix}
=
\det
\begin{pmatrix}
-4 & -2 \\
-2 & -1 \end{pmatrix}
=0
$$
Why does this statement guarantee that A is of rank 1? What happens, for instance, if
$$
\det
\begin{pmatrix}
2 & 1 \\
-4 & -2 \end{pmatrix}
= 1?
$$
Also, is it possible for a matrix to be of rank 0?
| The size of this matrix is $3\times 2$ so $\mathrm{rank}A\leq \min(2,3)=2$ and since the column $C_1=2C_2\not=0$ so
$$\mathrm{rank}A=1.$$
| {
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"source": "stackexchange",
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"answer_count": 6,
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Exponential and logarithmic series: Find the sum of $2^2 + 3^2/ 2!+4^2/3!+...$ to infinity
Find the sum of the following series:
$ 2^2 + 3^2/2! + 4^2/3! + ...$ to infinity
The answer is given as $5e$ but I got it as $5e+1$
$T_n = 1/(n-2)! +3/(n-1)! + 1/(n)! $ for $n \ge 2$
and $T_1+T_2 + ... $ to infinity = $4 + e + 3(e-1) +(e-2)$
$=5e-1$
Can you teell me which is correct and how?
| You (after correction) : $\ \boxed{5\,e-1}$
\begin{align}
S(x):&=\sum_{k=1}^\infty \frac{x^{k+1}}{k!}&=x\,(e^x-1)\\
x\frac d{dx}S(x)&=\sum_{k=1}^\infty \frac{(k+1)\,x^{k+1}}{k!}&=x\,((x+1)e^x-1)\\
x\frac d{dx}\left(x\frac d{dx}S(x)\right)&=\sum_{k=1}^\infty \frac{(k+1)^2\,x^{k+1}}{k!}&=x\,((x^2+3x+1)e^x-1)\\\
\end{align}
Set $x=1$ to conclude.
| {
"language": "en",
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"source": "stackexchange",
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Divide et impera recurrence, why induction does not work? $$
T(n) = T\left(\frac n2\right) + 2^n
$$
where $n \ge 1$ and $T(1) = 1$. If I understand the substitution method and the induction, I can guess that $T(n) = O(2^n)$.
I must prove that $T(n) = O(2^n)$, meaning constants $c$ and $n_0$ exist such that $T(n) \le c2^n$ for all $n \ge n_0$.
Base case
When $n = 1$ then $T(1) = 1$, choosing $c \ge \frac 12$ the inequality is satisfied:
$$
\\
1 \le c2\\
$$
Inductive step
Hypothesis is that $T(k) = O(2^k)$ for all $k \lt n$ (hence $T\left(\frac n2\right) \le c2^{\frac n2}$). Then I show that is true for $n$:
$$
\begin{align}
T(n) &= T\left(\frac n2\right) + 2^n = c2^{\frac n2} + 2^n = (c2^{\frac 12} + 1)2^n \\
&= (c\sqrt 2 +1)2^n \le c2^n
\end{align}
$$
So I ended up with $c \ge c\sqrt 2 +1$ that has no solution! I know that $T(n) = \Theta(2^n)$, so I'm wrong and I'd like to understand why.
| From the definition,
$T(2) = T(1)+2^2 = 5$,
$T(3) = T(1)+2^3 = 9$,
and $T(4) = T(2)+2^4 = 21$.
So, for $1 \le n \le 4$, $T(n) < 2\cdot 2^n$.
Suppose $n > 4$ and $T(m) < c 2^m$ for
$n/2 \le m < n$.
Then, as before,
$T(n) = T(n/2)+2^n
< c 2^{n/2}+2^n
= 2^n(1+c 2^{-n/2})
< c 2^n
$
if
$1+c 2^{-n/2} < c$
or $1 < c(1-2^{-n/2})$
or $c > 1/(1-2^{-n/2})$.
Since $n > 4$,
$2^{-n/2} < 2^{-2} = 1/4$,
so this inequality becomes
$c > 1/(1-1/4) = 4/3$.
This is certainly true for $c = 2$,
so we can substitute this value of $c$
in the original argument like this,
and disguise how we came up with the value for $c$:
Suppose $n > 4$ and $T(m) < 2 \cdot 2^m$ for
$n/2 \le m < n$.
Then, as before,
$T(n) = T(n/2)+2^n
< 2\cdot 2^{n/2}+2^n
= 2^n(1+2 \cdot 2^{-n/2})
< 2^n (1+2\cdot 2^{-2})
< (3/2) 2^n
< 2 \cdot 2^n
$.
This type of analysis is, in my experience, common.
Some calculations are done
which results in getting a value
that allows an induction (or other)
argument to succeed.
Then the initial analysis is thrown away,
and the magic constant is used to prove the result
without any hint of how the
constant was arrived at.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/352568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve a linear equations with 3 unknowns I have this equations
$$x+3y+6z=3$$
$$x+y+z=-2$$
$$-x+y+4z=7$$
my solution is
$$x=0$$
$$y=-5$$
$$z=3$$
this task is 1 of 3 and there is going to be 1 that can be solved, 1 that will have a solution with rational numbers and the last 1 wont have any solution. This last one wont have any solution but I have one.
What have I done wrong?
| As via Cramer's Rule,
$D=\det\begin{pmatrix} 1 & 3 & 6 \\ 1 & 1 & 1 \\ -1 & 1 & 4\end{pmatrix}=0,$
there will be a family of solutions
From the first two equations,
$$x+3y+6z-3=0$$
$$x+y+z+2=0$$
Solving for $x,y$ we get $x=\frac{3z-9}2,y=\frac{5-5z}2$
Putting the values of $x,y$ in the third equation, $$-\frac{3z-9}2+\frac{5-5z}2+4z=7\implies 7=7$$
$$\text{So, } x=\frac{3z-9}2,y=\frac{5-5z}2,z=z$$ will satisfy the given three equations for any finite value of $z$
In your case, $z=3$ and for Jerry, $z=5$
| {
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"url": "https://math.stackexchange.com/questions/352963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I solve this system? Having this system:
$$\begin{cases}
2 x(x^2 + y - 11) + x + y^2 - 7 = 0,\\
x^2 + y - 11 + 2 y (x + y^2 - 7) = 0 .
\end{cases}$$
How can I solve it without using a computer?.
Thank you!
| Let $a = x^2+y-11$ and $b = x+y^2-7$.
The equations we get are as follows:
$$2ax+b=0$$
$$a+2by=0$$
Replacing $a$ in the first equation yields:
$$-4bxy+b = 0$$
which gives us two options: $1-4xy = 0$ ($x = 1/4y$) or $b = 0$ ($x = 7-y^2)$. In both cases, we have expressed $x$ explicitly, we can insert it back the second original equation and solve it.
1.) $(1/4y)^2 + y - 11 + 2y(1/4y + y^2 -7) = 0$
that is (hopefully) equivalent to
$$ 32y^5 -14y^4+16y^3-21\cdot 18y^2+1 = 0$$
which looks painful.
2.) $(7-y)^2 + y - 11 = 0$ which is easily solvable.
| {
"language": "en",
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"question_score": "1",
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Evaluating $\cos(A+B)$, given $\cos A$ and $\sin(B)$ Find the exact value:
Find $\cos(A+B)$ given that $\cos A=1/3$, with $A$ in the first quadrant, and $\sin B = -1/4$, with $B$ in the fourth quadrant.
| Remember that cosine represents adjacent over hypotenuse, and sine represents opposite over hypotenuse.
$\sin (A) = -\dfrac {1}{4} $ is therefore tells us that we have a triangle with a side length of one, and a hypotenuse of length four. Using Pythagorean theorem, we can find the third side, which is equal to $\sqrt {1^2 + 4^2} $, or $\sqrt{15}$; this lets us find the cosine of that same angle.
By definition, $\cos (B)$ then equals $\dfrac{\sqrt{15}}{4}$.
We then do the same with $\cos (A) = \dfrac {2}{3} $ to get that $\sin A = \dfrac{\sqrt{8}}{3}$.
We can then use the identity $\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$ to solve for $\cos(A + B)$ exactly, as we know $\sin(A), \cos(A), \sin(B)$, and $\cos(B)$.
This then leaves us with $\cos(A + B) = \dfrac {\sqrt{15} + \sqrt{8}}{12}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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MacLaurin series of $\ln(1-x^2)$
The MacLaurin series for $\ln(1 + x)$ is obtained from the series for $\frac{1}{1 + x}$
by integration. Use this and appropriate substitutions to obtain the MacLaurin series for $\ln(1-x^2)$.
I did $$\int\frac{2}{x-\frac{1}{x}}$$ but I'm not sure how I'd go about the rest of the substitution in order to find the series.
| I think what the question is asking is to use
$$
\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\dots\tag{1}
$$
then substitute $x\mapsto-x$
$$
\log(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4-\dots\tag{2}
$$
Then using the identity that $\log(ab)=\log(a)+\log(b)$ to get
$$
\begin{align}
\log(1-x^2)
&=-2\frac{x^2}{2}-2\frac{x^4}4-2\frac{x^6}6-\dots\\
&=-x^2-\frac{x^4}2-\frac{x^6}3-\dots\tag{3}
\end{align}
$$
Note than you could also substitute $x\mapsto-x^2$ into $(1)$ to get $(3)$ directly.
| {
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How to solve for $y$ in $x = \frac{\sqrt[3]{9y-5}}{\sqrt[3]{11}}$ I am trying to solve the following for $y$ but am lost. I tried to multiply by $\sqrt[3]{121}/\sqrt[3]{121}$ but don't think that is how to do it.
$$x = \frac{\sqrt[3]{9y-5}}{\sqrt[3]{11}}$$
| $$x = \frac{\sqrt[3]{9y-5}}{\sqrt[3]{11}}$$
Cube both sides.
$$x^3=\frac{9y-5}{11}$$
Multiply both sides by $11$.
$$11x^3=9y-5$$
Add $5$ to both sides.
$$11x^3+5=9y$$
Divide both sides by $9$.
$$y=\frac{11x^3+5}{9}$$
Very easy.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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inverse trigonometry derivation Prove that :
$sin^{-1}x+sin^{-1}y = sin^{-1}[x\sqrt{1-y^2}+y\sqrt{1-x^2}]$
If -1 $\leq x \leq 1; -1 \leq y \leq 1 $ and $x^2+y^2\leq 1$ or if $xy <0 $ and $x^2+y^2 > 1$
solution :
Let $sin^{-1}x = \theta$ and $sin^{-1}y = \alpha$
Since sin($\theta + \alpha) = sin\theta cos\alpha + cos\theta sin\alpha$
= sin$\theta\sqrt{1-sin^2 \alpha}+\sqrt{1-sin^2\theta}sin\alpha$
=(x$\sqrt{1-y^2}+\sqrt{1-x^2}y)$
=$\theta + \alpha = sin^{1}(x\sqrt{1-y^2}+\sqrt{1-x^2}y)$
=$sin^{-1}x+sin^{-1}y = sin^{-1}[x\sqrt{1-y^2}+y\sqrt{1-x^2}]$
I would like to get an idea on the restriction given in this i.e.
If -1 $\leq x \leq 1; -1 \leq y \leq 1 $ and $x^2+y^2\leq 1$ or if $xy <0 $ and $x^2+y^2 > 1$ ...Request you to please guide on this..
| As $x=\sin\theta, \cos^2\theta=1-x^2\implies 1-x^2\ge 0\implies -1\le x\le 1$
Similarly, $-1\le y\le 1$
As the principal value of $\arcsin x$ lies in $\in[-\frac\pi2,\frac\pi2]$
the given relationship will hold iff $-\frac\pi2\le \arcsin x+\arcsin y\le \frac\pi2$
$(1)$ If $x\cdot y<0$ let $x>0,y<0$
So, $0\le \arcsin x\le \frac\pi2$ and $-\frac\pi2\le \arcsin y\le 0 $
$\implies -\frac\pi2\le \arcsin x+\arcsin y\le \frac\pi2$
$(2)$ Else $x\cdot y>0$
$(2a)$ If $x>0,y>0, 0\le \arcsin x, \arcsin y\le \frac\pi2$
$\implies \arcsin x+ \arcsin y\ge 0$
and $\arcsin x+ \arcsin y$ will be $\le \frac\pi2$
if $\arcsin y\le \frac\pi2-\arcsin x $
As $\sin x $ is increasing function in $\in[0,\frac\pi2]$
$ \sin(\arcsin y)\le \sin(\frac\pi2-\arcsin x) $
$\implies y\le \cos(\arcsin x)=\cos(\arccos\sqrt{1-x^2})=\sqrt{1-x^2}$
$\implies y^2\le 1-x^2\iff x^2+y^2\le 1$
$(2b) x,y<0, -\frac\pi2\le \arcsin x, \arcsin y\le 0$
$\implies \arcsin x+ \arcsin y\le 0$
Put $-x=X>0, -y=Y>0$ in case $(2a)$ to prove $X^2+Y^2\le 1\implies x^2+y^2\le 1 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum : $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}$ Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$
I think this is known (see here), I appreciate any hint or link for the solution (or the full solution).
| Here is another way using Fourier analysis:
Let
\begin{equation*}
f(t)=\begin{cases}
t-t^2 & 0<t<1 \\
-f(-t) & -1 < t < 0
\end{cases}
\end{equation*}
be a function with period 2. Then we can express $f$ in a Fourer series:
\begin{equation*}
f(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n \cos n\pi t+b_n \sin n\pi t
\end{equation*}
where
\begin{align*}
&a_n=\frac{1}{1}\int_{0}^{2}f(t)\cos n\pi t \, dt \\
&b_n=\frac{1}{1}\int_{0}^{2}f(t)\sin n\pi t\, dt
\end{align*}
But $f$ is odd, so $a_n=0$.
It follows that
\begin{align*}
b_n&= \int_{0}^{2}f(t)\sin n\pi t\, dt=\{ f(t)\sin n\pi t \text{ even}\}=2\int_{0}^{1}f(t)\sin n\pi t\, dt = \\
&= 2\int_{0}^{1}(t-t^2)\sin n\pi t\, dt=\frac{4-4(-1)^n}{n^3 \pi^3}
\end{align*}
Plugging in $t=\frac{1}{2}$, we get
\begin{equation*}
\frac{1}{4}=4\sum_{n=1}^{\infty}\frac{1-(-1)^n}{\pi^3n^3}\sin \frac{n\pi}{2}
\end{equation*}
But $1-(-1)^n=0$ only if $n$ is even, so
\begin{equation*}
\frac{1}{16}=\sum_{k=0}^{\infty}\frac{2(-1)^k}{(2k+1)^3\pi^3}\iff \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^3}=\frac{\pi^3}{32}
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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Is my textbook wrong? My textbook says (without explaining how it is done):
$$\begin{pmatrix}
1\ 2\ 3\ 4\\
2\ 1\ 4\ 3
\end{pmatrix}\begin{pmatrix}
1\ 2\ 3\ 4\\
2\ 3\ 4\ 1
\end{pmatrix}=\begin{pmatrix}
1\ 2\ 3\ 4\\
3\ 2\ 1\ 4
\end{pmatrix},$$
and as I understand this is obtained by doing calculations from left to right, but as I read here and elsewhere it should be done from right to left, like function composition...
So what i got is:
$\begin{pmatrix}
1\ 2\ 3\ 4\\
2\ 1\ 4\ 3
\end{pmatrix}\begin{pmatrix}
1\ 2\ 3\ 4\\
2\ 3\ 4\ 1
\end{pmatrix}=\begin{pmatrix}
1\ 2\ 3\ 4\\
1\ 4\ 3\ 2
\end{pmatrix}$,
or in cycle notation:
$(1,2)(3,4)(1,2,3,4)=(2,4)$ am I right?
| Permutations are functions, so most people do the evaluation from right to left.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/360305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How prove this $(abc)^4+abc(a^3c^2+b^3a^2+c^3b^2)\le 4$ let $a,b,c>0$,and such that $a+b+c=3$,prove that
$$(abc)^4+abc(a^3c^2+b^3a^2+c^3b^2)\le 4$$
I first consider
$$abc\le\left(\dfrac{a+b+c}{3}\right)^3=1$$
so it suffices to show that
$$a^3c^2+b^3a^2+c^3b^2\le 3$$
But I find this not true.
| Let $\{a,b,c\}=\left\{x,y,z\right\}$, where $x\geq y\geq z$.
Hence, by Rearrangement and AM-GM we obtain:
$$a^2c+b^2a+c^2b+abc=a\cdot ac+b\cdot ba+c\cdot cb+xyz\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$$
$$=y(x+z)^2=4x\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4.$$
Thus, since
$$\sum_{cyc}ab\sum_{cyc}a^2c=\sum_{cyc}(a^3c^2+a^3bc+a^2b^2c),$$
we obtain:
$$a^4b^4c^4+abc\sum_{cyc}a^3c^2= a^4b^4c^4+abc\left(\sum_{cyc}ab\sum_{cyc}a^2c-\sum_{cyc}(a^3bc+a^2b^2c)\right)\leq$$
$$\leq a^4b^4c^4+abc\left((4-abc)\sum_{cyc}ab-\sum_{cyc}(a^3bc+a^2b^2c)\right)=$$
$$=a^4b^4c^4+abc\left(4(ab+ac+bc)-9abc\right).$$
Id est, it remains to prove that:
$$a^4b^4c^4+abc\left(4(ab+ac+bc)-9abc\right)\leq4.$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the last inequality is a linear inequality of $v^2$,
which says that it's enough to prove this inequality for extremal value of $v^2$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$, where $0<a<\frac{3}{2}$.
Hence, we need to prove that
$$(a-1)^2\left(4+8a+12a^2-56a^3+41a^4+6a^5+7a^6+8a^7-72a^8+64a^9-16a^{10}\right)\geq0,$$
which is true for $0<a<\frac{3}{2}$.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 0
} |
convolution computation involving $e^{-x^2}$ In working a problem involving convolution, I have arrived at the following integral, but do not know how to compute it:
$$2\int_0^{\infty}e^{-a(x-y)^2-by^2}dy$$
I thought that this integrand did not have an antiderivative. Based on the way the problem is formulated, however, there must be an actual solution.
Thank you for your help!
| Expand the argument of the exponential:
$$\begin{align}\int_0^{\infty} dy \: e^{-a (x-y)^2} e^{-b y^2} &= e^{-a x^2} \int_0^{\infty} dy \: e^{-(a+b)y^2 + 2 a x y}\end{align}$$
Now complete the square in the exponential:
$$\begin{align}(a+b) y^2 - 2 a x y &= (a+b) \left[y^2 - \frac{2 a x}{a+b} y + \left( \frac{a x}{a+b} \right )^2 \right ] - \frac{a^2 x^2}{a+b} \\ &=(a+b) \left(y - \frac{a x}{a+b} \right)^2- \frac{a^2 x^2}{a+b} \end{align}$$
so that the integral becomes
$$\begin{align}e^{-(a b/(a+b)) x^2} \int_0^{\infty} dy \: e^{-(a+b) \left(y - \frac{a x}{a+b} \right)^2} &= e^{-(a b/(a+b)) x^2} \int_{-\frac{a x}{a+b}}^{\infty} du \: e^{-(a+b) u^2}\\ &= \frac{e^{-(a b/(a+b)) x^2}}{\sqrt{a+b}} \int_{-\frac{a x}{\sqrt{a+b}}}^{\infty} dv \: e^{-v^2}\\ &= \frac{e^{-(a b/(a+b)) x^2}}{\sqrt{a+b}} \frac{\sqrt{\pi}}{2} \left[ 1+ \text{erf}\left (\frac{a x}{\sqrt{a+b}} \right ) \right ]\end{align}$$
where erf is the standard error function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/363304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Expected Number of Coin Tosses to Get Five Consecutive Heads A fair coin is tossed repeatedly until 5 consecutive heads occurs.
What is the expected number of coin tosses?
| Here is a generating function approach.
Consider the following toss strings, probabilities, and terms
$$
\color{#00A000}{
\begin{array}{llc}
T&\frac12&\qquad\frac12x\\
HT&\frac14&\qquad\frac14x^2\\
HHT&\frac18&\qquad\frac18x^3\\
HHHT&\frac1{16}&\qquad\frac1{16}x^4\\
HHHHT&\frac1{32}&\qquad\frac1{32}x^5\\
\color{#C00000}{HHHHH}&\color{#C00000}{\frac1{32}}&\color{#C00000}{\qquad\frac1{32}x^5}
\end{array}
}
$$
Each term has the probability as its coefficient and the length of the string as its exponent.
Possible outcomes are any combination of the green strings followed by the red string. We get the generating function of the probability of ending after $n$ tosses to be
$$
\begin{align}
f(x)&=\sum_{k=0}^\infty\left(\frac12x+\frac14x^2+\frac18x^3+\frac1{16}x^4+\frac1{32}x^5\right)^k\frac1{32}x^5\\
&=\frac{\frac1{32}x^5}{1-\left(\frac12x+\frac14x^2+\frac18x^3+\frac1{16}x^4+\frac1{32}x^5\right)}\\
&=\frac{\frac1{32}x^5}{1-\frac{\frac12x-\frac1{64}x^6}{1-\frac12x}}\\
&=\frac{\frac1{32}x^5-\frac1{64}x^6}{1-x+\frac1{64}x^6}
\end{align}
$$
The average duration is then
$$
\begin{align}
f'(1)
&=\left.\frac{\left(\frac5{32}x^4-\frac6{64}x^5\right)\left(1-x+\frac1{64}x^6\right)-\left(\frac1{32}x^5-\frac1{64}x^6\right)\left(-1+\frac6{64}x^5\right)}{\left(1-x+\frac1{64}x^6\right)^2}\right|_{\large x=1}\\
&=\frac{\frac4{64}\frac1{64}+\frac1{64}\frac{58}{64}}{\left(\frac1{64}\right)^2}\\[12pt]
&=62
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/364038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "102",
"answer_count": 13,
"answer_id": 3
} |
How to solve this determinant: $a_{ij}=|i-j|+1$? I have to solve determinant of the following form:
$$a_{ij}=|i-j|+1$$
It looks like this:
$$
\begin{pmatrix}
1 & 2 & 3 & 4 & \cdots & n \\
2 & 1 & 2 & 3 & \cdots & n-1 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & n-3 & \cdots & 1
\end{pmatrix}
$$
It looks something like Toeplitz matrix, but I haven't found any method of solving it. I would appreciate also a kind of hint that would help.
EDIT: OEIS gives a formula for absolute value: $(n+1)\cdot2^{n-2}$
http://oeis.org/A001792
Thanks in advance!
| What do you get when you substract (i+1)th row - ith row?
$$
\begin{pmatrix}
1 & 2 & 3 & 4 & \cdots & n \\
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 1 & -1 & \cdots & -1 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & 1 & \cdots & -1
\end{pmatrix}
$$
What do you get when you add nth column to all other columns?
$$\begin{pmatrix}
n+1 & n+2 & n+3 & n+4 & \cdots & n \\
0 & -2 & -2 & -2 & \cdots & -1 \\
0 & 0 & -2 & -2 & \cdots & -1 \\
0 & 0 & 0 & -2 & \cdots & -1 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \cdots & -1
\end{pmatrix}$$
What is determinant of the last matrix?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/365774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating $\frac{1}{1+z^3}$ over a wedge to compute $\int_0^\infty \frac{dx}{1+x^3}$.
Compute $\displaystyle\int_0^\infty \frac{dx}{1+x^3}$ by integrating $\dfrac{1}{1+z^3}$ over the contour $\gamma$ (defined below) and letting $R\rightarrow \infty$.
The contour is $\gamma=\gamma_1+\gamma_2+\gamma_3$ where $\gamma_1(t)=t$ for $0\leq t \leq R$, $\gamma_2(t)=Re^{i\frac{2\pi}{3}t}$ for $0\leq t \leq 1$, and $\gamma_3(t)=(1-t)Re^{i\frac{2\pi}{3}}$ for $0\leq t \leq 1$.
So, the contour is a wedge, and by letting $R\rightarrow \infty$ we're integrating over one third of the complex plane. I believe this means we are integrating over the entire complex plane under the substitution $u=x^3$. There are poles at $-\zeta$ for each third root of unity $\zeta$, so there's only one pole in this wedge. I'll just refer to that pole as $-\zeta$.
I guess this means that we can use the residue theorem to say $$\int_{\gamma}\frac{1}{1+z^3}dz=2\pi i\eta(\gamma,-\zeta)\operatorname{Res}\left(\frac{1}{1+z^3},-\zeta\right)=2\pi i \lim_{z\rightarrow -\zeta}\left[(z+\zeta)\frac{1}{1+z^3}\right]$$
I can't evaluate this limit. Also I don't see how it involves $R$, which I'm supposed to be taking a limit of. I suspect I've done something wrong.
What's the problem? How do I proceed?
Also, after I do properly evaluate this integral, I am assuming that its value is supposed to be $\displaystyle\int_0^\infty\frac{dx}{1+x^3}$. Why? (I think I know why conceptually but I need to see how one rigorously writes that out.)
(Note: This is exam review, not homework.)
| $$\frac{1}{x^3+1}=\frac{1}{3(x+1)}-\frac{x-2}{3(x^2-x+1)}$$
But
$$\frac{x-2}{x^2-x+1}=\frac{1}{2}\frac{2x-1}{x^2-x+1}-\frac{\frac{3}{2}}{\frac{3}{4}+\left(x-\frac{1}{2}\right)^2}=\frac{(x^2-x+1)'}{x^2-x+1}-\frac{4}{3}\frac{\frac{3}{2}}{1+\left(\frac{2}{\sqrt3}\left(x-\frac{1}{2}\right)\right)^2}=$$
$$=\frac{(x^2-x+1)'}{x^2-x+1}-\sqrt3\,\frac{\frac{2}{\sqrt3}dx}{1+\left(\frac{2}{\sqrt3}\left(x-\frac{1}{2}\right)\right)^2}$$
Finally:
$$\int\limits_0^\infty\frac{dx}{x^3+1}=\left.\left[\frac{1}{3}\log\frac{\sqrt{x^2-x+1}}{x+1}+\sqrt 3\arctan\frac{2}{\sqrt 3}\left(x-\frac{1}{2}\right)\right]\right|_0^\infty=$$
$$0+\sqrt3\,\left(\frac{\pi}{2}-\arctan\left(-\frac{1}{\sqrt3}\right)\right)=\sqrt3\left(\frac{\pi}{2}+\frac{\pi}{6}\right)=\frac{2\pi}{\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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An infinite fraction Here's a problem I saw on the AoPS twitter. I thought I might as well post it so that it could be discussed and a solution recorded.
What is the value of the following?
$$\cfrac{4}{{1 + \cfrac{1^2}{2 + \cfrac{3^2}{2 + \cfrac{5^2}{2 + \cfrac{7^2}{2 + \cfrac{9^2}{\ddots} }}}}}}$$
As an aside, I find it cool that it can be compressed to the size of a tweet. The actual tweet is just this single line of symbols and nothing more:
$$4 / (1 + (1^2/(2 + 3^2/(2 + 5^2/(2 + 7^2/(2 + 9^2/(2 + ... ) ) ) ) ) ) ).$$
| $$\arctan(x) = \cfrac{x}{1+\cfrac{1^2x^2}{3-x^2+\cfrac{3^2x^2}{5-3x^2+\cfrac{5^2x^2}{7-5x^2 + \cfrac{7^2x^2}{9-7x^2 + \ddots}}}}}$$
Take $x=1$ to get what you want.
EDIT
The proof for the above is the observation that if
$$S_n = a_1 + a_1 a_2 + a_1 a_2a_3 + a_1 a_2 a_3 a_4 + \cdots + (a_1 a_2 \cdots a_n)$$ then
$$S_n = \cfrac{a_1}{1-\cfrac{a_2}{1+a_2 - \cfrac{a_3}{1+a_3 - \cfrac{a_4}{1+a_4 - \ddots \cfrac{a_n}{1+a_n}}}}}$$where the last term ends as $1+a_n$. Hence sending $n \to \infty$, we get that if
$$S = a_1 + a_1 a_2 + a_1 a_2a_3 + a_1 a_2 a_3 a_4 + \cdots + (a_1 a_2 \cdots a_n) + \cdots,$$ then
$$S = \cfrac{a_1}{1-\cfrac{a_2}{1+a_2 - \cfrac{a_3}{1+a_3 - \cfrac{a_4}{1+a_4 - \ddots}}}}$$
Now we know that
$$\arctan(x) = x - \dfrac{x^3}3 + \dfrac{x^5}5 = \cdots$$
Set $a_1 = x$, $a_2 = -\dfrac{x^2}3$, $a_3 = -\dfrac{3x^2}5$ and in general $a_n = - \dfrac{(2n-3)x^2}{2n-1}$ to get what we want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/368505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Congruence modulo a prime. The goal is to show that $x^2+3\equiv 0 \pmod p$ is solvable for every prime $p$, $p\equiv 1 \pmod 3$.
What I know so far is that, since $3\mid (p-1)$, $x^3\equiv 1 \pmod p$ has exactly three solutions. From that you get that $x^3-1\equiv(x-1)(x^2+x+1) \pmod p$ and therefore $x^2+x+1\equiv 0 \pmod p$ must have two solutions since $(x-1)\equiv 0 \pmod p$ has at most one solution.
Is there some step I'm missing here to relate $x^2+x+1$ to $x^2+3$?
| This type of problem is commonly solved with Legendre symbols: $\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=(-1)^{\frac{p-1}{2}}(-1)^{\lfloor \frac{p+1}{6} \rfloor}$, which is $1$ if $p\equiv 1,7\pmod{12}$ and $-1$ if $p\equiv 5,11\pmod{12}$.
A more direct approach is to note that $-3$ is the discriminant of $x^2+x+1$, so taking an $x$ that solves $x^2+x+1\equiv 0$ we set $y=2x+1$, and $y^2\equiv -3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/368695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing the Semimajor and Semiminor axis of an Ellipse I have the equation of the ellipse which is $\frac {x^2}{4r^2}+\frac{y^2}{r^2}=1$ Putting the (4,2) point on the ellipse we get $r^2=8$ so we get the equation $\frac {x^2}{32}+\frac {y^2}8=1$ and the semi-major axis is $\sqrt {32}= 4\sqrt 2$, the semi-minor axis is $\sqrt 8=2\sqrt 2$.
My question is using the value of $r^2=8$ to $\frac {x^2}{4r^2}+\frac{y^2}{r^2}=1$ , how come the semi-major axis is $\sqrt {32}= 4\sqrt 2$ ? and semi-minor axis is $\sqrt 8=2\sqrt 2$?
I am puzzled on how to get that answer.
...
anyway this question is related to my previous question Finding the Width and Height of Ellipse given an a point and angle
| The ellipse is in standard position and orientation, with the axes along the $x$ and $y$ axes. The ellipse meets the $x$-axis at the two points $x=\pm\sqrt{32}$, $y=0$, and meets the $y$-axis at $x=0$, $y=\pm\sqrt{8}$.
The distance from $(-\sqrt{32},0)$ to $(\sqrt{32},0)$ is bigger than the distance between the two $y$-intercepts.
The major axis therefore has length $2\sqrt{32}$, and the semi (half) major axis is half of that.
| {
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"timestamp": "2023-03-29T00:00:00",
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Need help solving $\int x \sqrt{\frac {a^2 - x^2} { a^2 + x^2 }} dx$ I have a complicated integral to solve. Can someone provide a better way to solve it than what i did - dividing by a inside the root, and then putting $ t = x / a $, and then putting $t^2 = \cos \theta $ and so many other substitutions.
$$
\int x \sqrt{\frac {a^2 - x^2} { a^2 + x^2 }} dx
$$
| We can use the algebraic substitution:
$$t=\frac{a^{2}-x^{2}}{a^{2}+x^{2}}.\tag{0} $$
We have that
$$\begin{eqnarray*}
I &=&\int x\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}dx=-\int \sqrt{t}\frac{a^{2}
}{\left( 1+t\right) ^{2}}dt\tag{1} \\
&=&a^{2}\frac{\sqrt{t}}{1+t}-a^{2}\arctan \sqrt{t}+C \\
&=&\frac{a^{2}+x^{2}}{2}\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}-a^{2}\arctan
\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}+C,
\end{eqnarray*}$$
because the integral in $t$ $(1)$ can be evaluated by using another algebraic substitution, $$u^{2}=t,\tag{2}$$ and expanding into partial fractions the resulting integrand
$$
\begin{eqnarray*}
\int \frac{\sqrt{t}}{\left( 1+t\right) ^{2}}dt &=&2\int \frac{u^{2}}{\left(
1+u^{2}\right) ^{2}}\,du \\
&=&2\int -\frac{1}{\left( 1+u^{2}\right) ^{2}}+\frac{1}{1+u^{2}}\,du.\tag{3}
\end{eqnarray*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the values of $p$ such that $\left( \frac{7}{p} \right )= 1$ (Legendre Symbol)
Show that if $p$ is an odd prime coprime to $7$, then $\left( \frac{7}{p} \right) = 1$ if and only if $p \equiv \pm 1, \pm 3,$ or $\pm 9 \pmod{28}$. HINT: If $p$ is an odd prime, determine which values can $p$ take $\mod28$, and consider each of these values in turn. Note that if we know $p \mod 28$ then we know $p \mod 4$, and hence we know whether $\frac{p-1}{2}$ is odd or even.
Here, $\left( \frac{a}{b} \right)$ is the Legendre symbol.
The bit I don't understand in the hint is, what do they mean by consider the values that $p$ can take $\mod 28$. Do they mean the values that would make $p$ a quadratic residue $\mod 28$, i.e all the $x$ values satisfying $x^2 \equiv \mod 28$, because then isn't this just $1,4,9,16,25$?
What do they mean the to "consider each of these values in turn"?
| Here is a solution that avoids the casework suggested by Zev Chonoles. Note that the computation of $\left(\tfrac{-7}{p}\right)$ is much easier. Once we know this, the computation of $\left(\tfrac{7}{p}\right)$ falls to CRT. By the Quadratic Reciprocity law, $$\left(\frac{-7}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{7}{p}\right)=\left(\frac{p}{7}\right)(-1)^{(p-1)/2}(-1)^{(p-1)(7-1)/4}=\left(\frac{p}{7}\right)(-1)^{2(p-1)}=\left(\frac{p}{7}\right)=p^3.$$ Thus, in order for $\left(\tfrac{-7}{p}\right)=1$, we want $p^3\equiv 1\pmod{7}$. A difference of cubes factorization yields $(p-1)(p^2+p+1)\equiv 0\pmod{7}$, so $p\equiv 1\pmod{7}$ or $p^2+p+1\equiv 0\pmod{7}$. In order to make the latter case easy to work with, note that $$p^2+p+1\equiv p^2+p-6=(p+3)(p-2)\equiv 0\pmod{7},$$ so $p\equiv -3,2\pmod{7}$. Summarizing, $\left(\frac{-7}{p}\right)=1\iff p\equiv 1,2,4\pmod{7}$. Now, using that $\left(\frac{7}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{-7}{p}\right)$ along with a straightforward application of CRT should yield the required result.
EDIT: I'm being silly - no need to solve the cubic, just note that the quadratic residues modulo $7$ are $1$, $2$, and $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding generating function for the recurrence $a_0 = 1$, $a_n = {n \choose 2} + 3a_{n - 1}$ I am trying to find generating function for the recurrence:
*
*$a_0 = 1$,
*$a_n = {n \choose 2} + 3a_{n - 1}$ for every $n \ge 1$.
It looks like this:
*
*$a_0 = 1$
*$a_1 = {1 \choose 2} + 3$
*$a_2 = {2 \choose 2} + 3{1 \choose 2} + 9$
*$a_3 = {3 \choose 2} + 3{2 \choose 2} + 9{1 \choose 2} + 27$
*$a_4 = {4 \choose 2} + 3{3 \choose 2} + 9{2 \choose 2} + 27 {1 \choose 2} + 81$
I know what the generating function of the sequence $3 ^n = (1, 3, 9, 27, 81, \dots)$ is, as well as what the generating functions for some sequences of combinatorial numbers are, but how do I split the sequence up into these pieces I know?
(The problem is those combinatorial numbers "move right" every time. If they were growing left-to-right along with their coefficients, it would be much easier. And there is no constant difference between $a_i$ and $a_{i + 1}$.)
| A related problem. Assume $ F(x) = \sum_{n=0}^{\infty}a_n x^n $, then
$$ a_n = {n \choose 2} + 3a_{n - 1} \implies a_{n+1} = {n+1 \choose 2} + 3a_{n}$$
$$ \sum_{n=0}^{\infty} a_{n+1} x^n = \frac{1}{2}\sum_{n=0}^{\infty}n(n+1)x^n + 3\sum_{n=0}^{\infty}a_{n}x^n $$
$$ \implies \sum_{n=1}^{\infty} a_{n} x^{n-1} = \frac{1}{2}\sum_{n=1}^{\infty}nx^{n}+\frac{1}{2}\sum_{n=1}^{\infty}n^2x^{n} +3F(x) $$
$$ \implies \frac{1}{x}F(x)-\frac{a_0}{x}-3F(x) = \frac{1}{2}\sum_{n=1}^{\infty}nx^{n}+\frac{1}{2}\sum_{n=1}^{\infty}n^2x^{n} $$
$$\implies \left(\frac{1}{x}-3 \right)F(x)=\frac{1}{x}+\frac{1}{2}\frac{x}{(x-1)^2}-\frac{1}{2}\frac{x(x+1)}{(x-1)^3} $$
$$ \implies \left(\frac{1}{x}-3 \right)F(x)=\frac{1}{x}-\frac{x}{(x-1)^3} $$
$$ \implies F(x)=\frac{x}{1-3x}\left( \frac{1}{x}-\frac{x}{(x-1)^3} \right). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Geometry: Hyperbolas I was wondering how would I complete the square for this particular hyperbola?
$4x^2 - 5y^2 + 24y = 16$
I tried this technique but to no avail:
$$4x^2 - 5(y^2 + \frac{24}{5}y) = 16$$
$$ \Rightarrow 4x^2 - 5(y + \frac{12}{5})^2 = 16 + \left(\frac{12}{5}\right)^2$$
$$\Rightarrow 4x^2 - 5(y + \frac{12}{5})^2 = \frac{544}{25} .$$
Am I doing something wrong here? On my calculator it says that the equation should be a hyperbola
| You have wrongly added $\left(\frac{12}5\right)^2$ in place of subtraction
$(2x)^2-5\{y^2-2\cdot y\cdot \frac{12}5+ (\frac{12}5)^2\}=16- \left(\frac{12}5\right)^2$
or, $(2x)^2-5\left(y- \frac{12}5\right)^2=\frac{256}{25}=\left(\frac{16}5\right)^2$
and so on
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Recursive formula for variance I'd like to know how I can recursively (iteratively) compute variance, so that I may calculate the standard deviation of a very large dataset in javascript. The input is a sorted array of positive integers.
| Here are the iterative formulas (with derivations) for the population ($N$ normalized) and sample ($N-1$ normalized) standard deviations, which express the $\sigma_{n+1}$ ($s_{n+1}$ for sample) for the $n+1$ value set in terms of $\sigma_{n}$ ($s_{n}$ for sample), $\bar x_{n}$ of the $n$ value set plus the new value $x_{n+1}$ added to the set.
Essentially we need to find:
$$\bar x_{n+1} = f(n, \bar x_n, x_{n+1})$$
and
$$\sigma_{n+1} = g(n, \sigma_n, \bar x_n, x_{n+1})$$
Derivation for the Average
For both cases, the average for $n\geqslant1$ is,
for $n$ values:
$$
\bar x_n=\frac1n\sum_{k=1}^nx_k
$$
for $n+1$ values:
$$
\bar x_{n+1}=\frac1{n+1}\sum_{k=1}^{n+1}x_k = \frac1{n+1}(n\bar x_n + x_{n+1}) \leftarrow f(n, \bar x_n, x_{n+1})
$$
Derivation for the Standard Deviation
The standard deviation formulas for population and sample are:
\begin{aligned}
\sigma_{n} &= \sqrt {\frac1{n} \sum_{k=1}^{n}(x_k - \bar x_{n})^2 } && \textit{for} \textbf{ population } \textit{Standard Deviation}\\
\\
s_{n} &= \sqrt {\frac1{n-1} \sum_{k=1}^{n}(x_k - \bar x_{n})^2 } && \textit{for} \textbf{ sample } \textit{Standard Deviation } \\
\end{aligned}
To consolidate the derivations for both population and sample formulas we'll write the standard deviation using a generic factor $\alpha_{n}$ and replace it at the end to get the population and sample formulas.
with:
\begin{equation}
\alpha_{n} =
\begin{cases}
n & \textit{for} \textbf{ population } \textit{Standard Deviation} \\
n-1 & \textit{for} \textbf{ sample } \textit{Standard Deviation } \\
\end{cases}
\end{equation}
the equation for the standard deviation for the $n$ values can be written as:
\begin{equation}
\tag{1}
\begin{aligned}
\alpha_{n}\sigma^2_{n}
& = \sum_{k=1}^{n}(x_k - \bar x_{n})^2 \\
& = \sum_{k=1}^{n}\big(x_k^2 - 2 x_k \bar x_n + (\bar x_{n})^2\big) \\
& = \sum_{k=1}^{n}x_k^2 - 2 \bar x_n \sum_{k=1}^{n}x_{k} + (\bar x_{n})^2\sum_{k=1}^{n}1 \\
& = \sum_{k=1}^{n}x_k^2 - 2 \bar x_n n \bar x_n + n (\bar x_{n})^2 \\
& = \sum_{k=1}^{n}x_k^2 - n (\bar x_{n})^2
\end{aligned}
\end{equation}
Thus, the same equation for $n+1$ values is:
\begin{equation}
\begin{aligned}
\alpha_{n+1}\sigma^2_{n+1}
& = \sum_{k=1}^{n+1}(x_k-\bar x_{n+1})^2 \\
& = \sum_{k=1}^{n+1}x_k^2 - (n+1)(\bar x_{n+1})^2 \\
& = \sum_{k=1}^{n}x_k^2 + (x_{n+1})^2 - (n+1)(\bar x_{n+1})^2 \\
& = \sum_{k=1}^{n}x_k^2 + (x_{n+1})^2 - (n+1) \big(\frac1{n+1}(n\bar x_{n} + x_{n+1}) \big)^2 \\
& = \sum_{k=1}^{n}x_k^2 + (x_{n+1})^2 - \frac1{n+1} \big(n^2(\bar x_{n})^2 + 2 n \bar x_{n} x_{n+1} + (x_{n+1})^2 \big) \\
\end{aligned}
\end{equation}
from the equation $(1)$ we substitute $\sum_{k=1}^{n}x_k^2$ with $\alpha_{n}\sigma^2_{n} + n (\bar x_{n})^2$ and get:
\begin{equation}
\begin{aligned}
\alpha_{n+1}\sigma^2_{n+1}
& = \alpha_{n}\sigma^2_{n} + n (\bar x_{n})^2 + (x_{n+1})^2 - \frac1{n+1} \big(n^2(\bar x_{n})^2 + 2 n \bar x_{n} x_{n+1} + (x_{n+1})^2 \big) \\
\end{aligned}
\end{equation}
arranging the terms and simplifying we get:
$$
\sigma_{n+1}
= \sqrt { \Big( \sigma^2_{n} + \frac{n}{n+1} \frac1{\alpha_n} (\bar x_n - x_{n+1})^2 \Big) \frac{\alpha_{n}}{\alpha_{n+1}} }
\leftarrow g(n, \sigma_n, \bar x_n, x_{n+1})
$$
Replacing the $\alpha$ values, the specific iterative formulas for population and sample standard deviations are:
\begin{equation}
\begin{aligned}
\sigma_{n+1} &= \sqrt{ \Big( \sigma^2_{n} + \frac{1}{n+1}(\bar x_n - x_{n+1})^2 \Big) \frac{n}{n+1} } &&\textit{population STD} \\
\\
s_{n+1} &= \sqrt{ \Big( s^2_{n} + \frac{n}{n^2-1}(\bar x_n - x_{n+1})^2 \Big) \frac{n-1}{n} } &&\textit{sample STD} \\
\end{aligned}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/374881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 2
} |
Solving an integral equation using the Fourier transform I have to solve the equation
$\int_0^{\infty} f(x) \cos{(\alpha x)}\, dx=\frac{\sin{\alpha }}{\alpha}$
Using fourier transform. I know this is half of the usual fourier cosine transform, and so that I would get back $f(x)$ using $\frac{2}{\pi} \int_0^{\infty} \frac{\sin{\alpha} \cos{\alpha x}}{\alpha}d{\alpha}$
Is this correct? How do I do this integral?
| $$ \begin{align} \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin (\alpha) \cos (x \alpha)}{\alpha} \ d \alpha &= \frac{1}{\pi} \int_{0}^{\infty} \frac{\sin \big((1+x) \alpha \big)+\sin \big( (1-x) \alpha \big)}{\alpha} \ d \alpha \\ &= \frac{1}{\pi} \Big(\text{sgn}(1+x) \frac{\pi}{2}+\text{sgn}(1-x) \frac{\pi}{2} \Big) \\ &= \frac{1}{2} \Big(\text{sgn}(1+x) + \text{sgn}(1-x) \Big) \\ &= \begin{cases} \frac{1}{2}(-1+1) = 0 & \text{if} \ x <-1 \\ \frac{1}{2}(0+1) = \frac{1}{2} & \text{if} \ x = -1 \\ \frac{1}{2} (1+1) = 1 & \text{if} -1 < x <1 \\ \frac{1}{2}(1+0) = \frac{1}{2} & \text{if} \ x = 1 \\ \frac{1}{2} (1-1) = 0 & \text{if} \ x >1\end{cases} \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
is $f$ analytic inside $C?$
If $$f(z_0)=\dfrac{1}{2\pi i}\int_C\dfrac{f(z)}{z-z_0}dz$$ for all point $z_0$ inside $C,$ is $f$ analytic inside $C?~(C:$ simple closed contour$)$
| For simplicity, assume $C:|z-z_0|=r$ be a circle around $z_0$ and $|f(z)|\leqslant M$ for all $z\in C$. Then
\begin{align*}
\Big|\frac{f(z_1)-f(z_0)}{z_1-z_0}\Big|
=&\Big|\frac{1}{z_1-z_0}\Big|\cdot\Big|\dfrac{1}{2\pi i}\int_C\dfrac{f(z)}{z-z_1}dz-\dfrac{1}{2\pi i}\int_C\dfrac{f(z)}{z-z_0}dz\Big|\\
=&\frac{1}{2\pi}\Big|\frac{1}{z_1-z_0}\cdot\int_Cf(z)\Big(\frac{1}{z-z_1}-\frac{1}{z-z_0}\Big)dz\Big|\\
\leqslant&\frac{1}{2\pi}\int_C|f(z)|\cdot\Big|\frac{1}{z_1-z_0}\Big|\cdot\Big|\frac{1}{z-z_0}-\frac{1}{z-z_1}\Big|\cdot|dz|\\
\leqslant&\frac{1}{2\pi}\int_C\frac{M}{|z_-z_0|\cdot|z-z_1|}\cdot|dz|\\
\leqslant&\frac{1}{2\pi}\cdot\frac{M}{r(r-\varepsilon)}\cdot L_C\rightarrow\frac{1}{2\pi}\cdot\frac{M}{r^2}\cdot2\pi r=\frac{M}{r}~~~(z_1\rightarrow z_0)
\end{align*}
Your statement is not true in general, the following is a counterexample:
Let $C:z=e^{i\theta},~\theta\in[0,2\pi)$. Define $f(z)=\frac{1}{\theta}$ for $z\in C\setminus\{1\}$ and $f(1)=\frac{1}{2\pi}$. Consider the differentiation of the point $z_0=r\in(0,1)$. Let $z_1=r+\varepsilon$. We have
\begin{align*}
&\Big|\frac{f(z_1)-f(z_0)}{z_1-z_0}\Big|=\frac{|f(z_1)-f(z_0)|}{\varepsilon}\\
=&\frac{1}{\varepsilon}\Big|\dfrac{1}{2\pi i}\int_C\dfrac{f(z)}{z-z_1}dz-\dfrac{1}{2\pi i}\int_C\dfrac{f(z)}{z-z_0}dz\Big|\\
=&\frac{1}{2\pi\varepsilon}\Big|\int_Cf(z)\Big(\frac{1}{z-z_1}-\frac{1}{z-z_0}\Big)dz\Big|\\
=&\frac{1}{2\pi\varepsilon}\Big|\int_0^{2\pi}\frac{1}{\theta}\cdot\frac{z_0-z_1}{(z-z_1)(z-z_0)}d(e^{i\theta})\Big|\\
=&\frac{1}{2\pi\varepsilon}\Big|\int_0^{2\pi}\frac{1}{\theta}\cdot\frac{(-\varepsilon)ie^{i\theta}}{(e^{i\theta}-r-\varepsilon)(e^{i\theta}-r)}d\theta\Big|\\
=&\frac{1}{2\pi}\Big|\int_0^{2\pi}\frac{1}{\theta}\cdot\frac{e^{i\theta}}{(e^{i\theta}-r-\varepsilon)(e^{i\theta}-r)}d\theta\Big|\rightarrow\frac{K}{2\pi}\int_0^{2\pi}\frac{d\theta}{\theta}\rightarrow\infty~~(\varepsilon\rightarrow0)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/378086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Factorization of the trinomial $x^{2n}+Dx^n+1$? The following trinomials will factor for any $a$,
$$1+a(-3+a^2)x^3+x^6 = (1+ax+x^2)(1-ax-x^2+a^2x^2-ax^3+x^4)\tag{1}$$
and similarly for,
$$1+a(5-5a^2+a^4)x^5+x^{10}\tag{2}$$
$$1+a(-7+14a^2-7a^4+a^6)x^7+x^{14}\tag{3}$$
$$1+a(9-30a^2+27a^4-9a^6+a^8)x^9+x^{18}\tag{4}$$
and so on. (The second one is notable in that for general $a$ it has two factors, but if $a$ is an even-index Lucas number, then there is a third factor.)
Questions:
*
*Given $P(x) = x^{2n}+Dx^n+1$ for all ODD $n>1$, is there a general formula for $D$ as a polynomial in $a$ such that $(1+ax+x^2)$ is a factor of $P(x)$?
*The fourth generally has 3 factors. Is there an odd $n$ such that $P(x)$ generally has four factors?
| If we work in ${\mathbf Z}[x,a]/(x^2+ax+1)$, with $a$ and $x$ as independent indeterminates then we seek a polynomial $D_n$ in ${\mathbf Z}[a]$ such that $x^{2n}+D_nx^n+1 \equiv 0 \bmod x^2+ax+1$ for odd $n$.
Starting from $x^2 \equiv -ax-1 \bmod x^2+ax+1$ we can compute higher powers of $x$ to that modulus by using this relation for $x^2$, for instance $x^3 \equiv (a^2-1)x+a \bmod x^2+ax+1$ and $x^4 \equiv (-a^3+2a)x + (-a^2+1) \bmod x^2+ax+1$. In general, define polynomials $f_n(a)$ and $g_n(a)$ in ${\mathbf Z}[a]$ to fit
$$
x^n \equiv f_n(a)x + g_n(a) \bmod x^2+ax+1.
$$
Then square both sides and use the reduction formula for $x^2 \bmod x^2+ax+1$ to get
$$
x^{2n} \equiv (2f_n(a)g_n(a)-af_n(a)^2)x + (g_n(a)^2-f_n(a)^2) \bmod x^2+ax+1,
$$
so for an unknown $D$ in ${\mathbf Z}[a]$, we have
$$
x^{2n} + Dx^n + 1 \equiv (f_n(a)D - af_n(a)^2 + 2f_n(a)g_n(a))x + (g_n(a)^2-f_n(a)^2+ g_n(a)D+1).
$$
We want this to be $0 \bmod x^2+ax+1$, so we want to find $D$ satisfying the following two equations:
$$
f_n(a)D = af_n(a)^2 - 2f_n(a)g_n(a), \ \ \ g_n(a)D = f_n(a)^2 - g_n(a)^2 -1
$$
in ${\mathbf Z}[a]$. From the first equation, the obvious choice to make is
$$
D = af_n(a) - 2g_n(a).
$$
Testing this with $n = 3, 5, 7, 9$ already recovers the coefficient of $x^n$ for these $n$ in your question, so we're clearly on the right track.
We also need to have this $D$ fit the second condition above, and that is the same as
$$
g_n(a)(af_n(a)-2g_n(a)) \stackrel{?}{=} f_n(a)^2-g_n(a)^2+1,
$$
or equivalently
$$
f_n(a)^2+g_n(a)^2 = af_n(a)g_n(a) - 1.
$$
I leave it as an exercise to check the coefficients of $x^n \bmod x^2+ax+1$ satisfy this constraint.
Edit: This calculation makes no distinction between even $n$ and odd $n$. It works for all $n$. For instance, when $n = 2$ and $n = 4$ the corresponding trinomials would be $x^4 + (-a^2+2)x^2 + 1$ and $x^6 + (a^3-3a)x^3 + 1$. Once I saw those I realized that in fact the coefficient of $x^n$ for all $n$ is a normalized $n$th Chebyshev polynomial up to a sign. Writing $a^n + 1/a^n = T_n(a + 1/a)$, we have $T_1(a) = a$, $T_2(a) = a^2 - 2$, $T_3(a) = a^3 - 3a$, $T_4(a) = a^4 - 4a^2 + 2$, and $T_5(a) = a^5 - 5a^3 + 5a$. Then your examples are simply $x^{2n}+(-1)^{n-1}T_n(a)x^n+1$. So your task amounts to showing $x^{2n}+(-1)^{n-1}T_n(a)x^n+1 \equiv 0 \bmod x^2+ax+1$ for all $n \geq 1$.
In ${\mathbf Z}[a,x]/(x^2+ax+1)$, $x$ is a unit (with inverse ($-x-a$)) and $x + 1/x = -a$. In this ring \begin{eqnarray*}x^{2n} + (-1)^{n-1}T_n(a)x^n + 1 & = & x^n(x^n + 1/x^n + (-1)^{n-1}T_n(a)) \\ & = & x^n(T_n(x+1/x) + (-1)^{n-1}T_n(a)) \\ & = & x^n(T_n(-a) + (-1)^{n-1}T_n(a)).\end{eqnarray*}
Since $T_n(-a) = (-1)^nT_n(a)$, the sum $T_n(-a) + (-1)^{n-1}T_n(a)$ is 0, and that proves $x^{2n} + (-1)^{n-1}T_n(a)x^n + 1 \equiv 0 \bmod x^2 + ax + 1$ for all $n \geq 1$ (including even $n$).
| {
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"url": "https://math.stackexchange.com/questions/382342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a+b+c=1$, find the minimum of $\frac{4+3abc}{ab+bc+ac}$ I came to ask this because I am really stuck at this problem. I have tried everything from arithmetic mean, geometric mean and harmonic mean. Also, I have tried playing with the variables and such, but it got me to nowhere.
If $a+b+c=1$; $a,b,c$ nonnegative, calculate the minimum of
$$\frac{4+3abc}{ab+bc+ac}$$
All I've got so far is:
$$\frac{3abc}{ab+bc+ac} \le \frac{1}{3}$$
But this is obviously on the wrong side of the inequality.
Also, I think that
$$\frac{1}{ab+bc+ac}\ge3$$
But I haven't been able to prove it.
Playing with the most possible and obvious values, one could think that the answer is 37/3, but the excercise is about proving it. Any help and little hints are greatly apprecieated.
| Substitute $a=1-(b+c)$ in $\displaystyle\frac{4+3abc}{ab+bc+ac}$ to get
$$\frac{4+3(1-(b+c))bc}{(1-(b+c))b+bc+(1-(b+c))c} \tag{1}$$
Differentiating $(1)$ with respect to $b$ gives
$$\frac{\partial}{\partial b}\left(\frac{4+3(1-(b+c))bc}{(1-(b+c))b+bc+(1-(b+c))c}\right) =\frac{(3c^3-3c^2+4)(2b-c+1)}{\left(b^2+b(c-1)+c(c-1)\right)^2}$$
Setting this equal to zero yields
$$2b-c+1=0 \tag{2}$$
Differentiating $(1)$ with respect to $c$ gives
$$\frac{\partial}{\partial c}\left(\frac{4+3(1-(b+c))bc}{(1-(b+c))b+bc+(1-(b+c))c}\right) =\frac{(3b^3-3b^2+4)(2c-b+1)}{\left(c^2+c(b-1)+b(b-1)\right)^2}$$
Setting this equal to zero yields
$$2c-b+1=0 \tag{3}$$
Solving for $(2)$ and $(3)$ yields
$$b=c=\frac{1}{3}$$
Plug these values in $(1)$ and you end up with
$$\frac{4+3(1-(b+c))bc}{(1-(b+c))b+bc+(1-(b+c))c}=\frac{37}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/383918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
calculate the roots of $z = 1 + z^{1/2}$ using Lagrange expansion I am trying to find the roots of equation $z = 1 + \sqrt{z}$ using Lagrange's expansion given on this book page no 15.
I expanded $z = 1 + \zeta z^p$ as
$$z = 1 + \zeta + \frac{2p}{2!}\zeta^2 + \frac{3p(3p-1)}{3!}\zeta^3+\frac{4p(4p-1)(4p-2)}{4!}\zeta^4 + \dots$$
putting $\zeta = 1$ and $p = 1/2$ I obtained the series
$$z = 1 + 1 + \frac{1}{2} + \frac18 + \frac{-1}{128} + 0 + \dots$$
The series seems to have formula $\displaystyle 2 + \sum_{n=2}^\infty \frac 1{n!}\prod_{k=0}^{n-2} (n \cdot \frac12 - k)$ which I found it to be $2 + \frac12(1 + \sqrt 5)$ from Mathematica while the root of $z = 1 + \sqrt z$ is $1 + \frac12(1 + \sqrt 5)$. Can anyone help me where it went wrong? Also can anyone help me to evaluate that series with hand?
| You can solve the equation explicitly by squaring both sides then solving the resulting quadratic. Compare the root you find with sum of the first terms of the series you gave.
In more detail.
We have
$$
z - 1 = \sqrt{z} \quad \Longrightarrow \quad z^2 - 2z + 1 = z,
\tag{1}
$$
so that, by the quadratic formula, the two possible roots are
$$
z = \frac{3 \pm \sqrt{5}}{2}.
$$
The smaller candidate root is not a solution to the original equation since
$$
\frac{3 - \sqrt{5}}{2} < \frac{3 - \sqrt{4}}{2} = \frac{1}{2},
$$
which would make the left-hand side negative and the right-hand side positive in the first equation in $(1)$. Thus we conclude that the only possible root is
$$
z = \frac{3 + \sqrt{5}}{2} \approx 2.61803.
$$
The sum of the first terms of the series you found is approximately
$$
1 + 1 + \frac{1}{2} + \frac18 - \frac{1}{128} \approx 2.61719,
$$
so it looks like you've got everything right.
Evaluating the series directly.
If you'd like, you can get the closed form for your series the hard way by noticing that
$$
\prod_{k=0}^{n-2} (np - k) = \frac{(np)!}{(np-n+1)!},
$$
for $n \geq 2$, so that the general term in your sum is
$$
\frac{(np)!}{n!(np-n+1)!} \zeta^n = \frac{1}{n} \binom{np}{n-1} \zeta^n
\tag{2}
$$
Let's take $p=1/2$ now and consider only the odd terms, so that $n = 2m+1$. The expression in $(2)$ becomes
$$
\frac{1}{2m+1} \binom{m+1/2}{2m} \zeta^{2m+1}.
$$
Thanks to Sasha's answer here this is equal to
$$
\frac{1}{4^m} \binom{1/2}{m} \zeta^{2m+1} = \zeta \binom{1/2}{m} \left(\frac{\zeta^2}{4}\right)^m,
$$
so that the sum of the odd terms in your series is
$$
\zeta \sum_{m=0}^{\infty} \binom{1/2}{m} \left(\frac{\zeta^2}{4}\right)^m = \zeta \sqrt{1+\zeta^2/4} = \frac{\zeta}{2}\sqrt{4+\zeta^2}.
\tag{3}
$$
Next we consider the even terms. By taking $n = 2m$ in $(2)$ we see that the general even term is
$$
\frac{1}{2m} \binom{m}{2m-1} \zeta^{2m}.
$$
For all $m > 1$ the quantity $2m-1$ is a negative integer, and in that case $\binom{m}{2m-1} = 0$. Thus the sum of the even terms is just
$$
1 + \frac{1}{2} \binom{1}{1} \zeta^2 = 1 + \frac{1}{2} \zeta^2.
\tag{4}
$$
Summing $(3)$ and $(4)$ we find that
$$
\begin{align}
z &= 1 + \zeta + \frac{2p}{2!}\zeta^2 + \frac{3p(3p-1)}{3!}\zeta^3+\frac{4p(4p-1)(4p-2)}{4!}\zeta^4 + \dots \\
&= 1 + \frac{1}{2} \zeta^2 + \frac{\zeta}{2}\sqrt{4+\zeta^2}.
\end{align}
$$
By taking $\zeta = 1$ we see that this agrees with the result we found earlier, that $z = (3+\sqrt{5})/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/384257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Showing that $2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty}\left( \frac{x^{a-1}}{\sinh x} - x^{a-2} \right) \mathrm dx$ I want to show that $$2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty}\Big( \frac{x^{a-1}}{\sinh x} - x^{a-2}\Big) \, \mathrm dx \, , \quad 0 < \Re (a) <1. \tag{1}$$
I can show that $$2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty} \frac{x^{a-1}}{\sinh x} \, \mathrm dx \, , \quad \Re (a) >1, \tag{2}$$
but I don't know if $(2)$ can be used to prove $(1)$.
EDIT:
Following the approach used in this paper to show that $$\Gamma(s) \zeta(s) = \int_{0}^{\infty} \left(\frac{1}{e^{x}-1}-\frac{1}{x} \right)x^{s-1} \, \mathrm dx \, , \quad0 < \Re(s) < 1, $$
subtract $$\frac{\Gamma(a)}{1-a} b^{1-a} = \int_{0}^{\infty} e^{-b x}x^{a-2} \, \mathrm dx \, , \quad \left( \Re(a) >1, \, b>0 \right),$$ from $(2)$ to get
$$\Gamma(a) \left(2 \zeta(a) \left(1-\frac{1}{2^{a}} \right) - \frac{b^{1-a}}{1-a} \right) = \int_{0}^{\infty} \left(\frac{1}{\sinh x} - \frac{e^{-bx}}{x} \right)x^{a-1} \, \mathrm dx \, , \quad \Re(a) > 0.$$
What this did is compensate for the fact that $\frac{1}{\sinh z}$ has a simple pole at $z=0$ with residue $1$.
Now restrict $a$ to the vertical strip $0 < \Re (a) <1$, and let $b$ tend to zero to get $$ 2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty}\left( \frac{1}{\sinh x} - \frac{1}{x}\right)x^{a-1} \, \mathrm dx \ , \quad 0 < \Re(a) <1.$$
You could then appeal to the identity theorem to argue that $(1)$ actually holds for $-1 < \Re(a) <1$.
|
$$ \color{blue}{\Gamma(s-N)\zeta(s-N)=\int_{0}^{\infty}x^{s-N-2}\left[\frac{x}{e^x-1}-\left(\sum_{n=0}^{N}B_{n}\frac{x^n}{n!}\right)\right]\,dx} $$
$$ {\small \,0\lt\,Re\{s\}\,\lt1 ,\quad N\in\{\,0,\,1,\,2,\,\cdots\,\} ,\quad B_{n}\,\,{Bernoulli\,Number} ,\quad B_{1}=-1/2} $$
$$
\begin{align}
I\,& =\,\int_{0}^{\infty}\left(\frac{x^{s-1}}{\sinh{x}}-x^{s-2}\right)\,dx =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{e^x}{e^{2x}-1}-\frac{1}{2x}\right)\,dx \\[2mm]
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{e^x\color{red}{+1-1}}{e^{2x}-1}-\frac{1}{2x}\color{red}{-\frac{1}{2x}+\frac{1}{2x}}\right)\,dx \\[2mm]
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}-\frac{1}{e^{2x}-1}+\frac{1}{2x}\right)\,dx
\end{align}
$$
For ${\small\color{red}{\,0\le\,Re\{s\}\,\le+1\,}}\colon\quad\Gamma(s)\zeta(s)=\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^x-1}-\frac{1}{x}\right)\,dx\,$; Thus:
$$
\begin{align}
I\,& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}-\frac{1}{e^{2x}-1}+\frac{1}{2x}\right)\,dx \qquad\color{blue}{\small\{\text{both converge}\}} \\[2mm]
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}\right)\,dx\,-\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{2x}-1}-\frac{1}{2x}\right)\,dx \\[2mm]
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}\right)\,dx\,-\,\frac{2}{2^{s}}\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}\right)\,dx \\[2mm]
& =\,\color{red}{\left(2-2^{1-s}\right)\,\Gamma(s)\zeta(s)}
\end{align}
$$
For ${\small\color{red}{\,-1\le\,Re\{s\}\,\le0\,}}\colon\quad\Gamma(s)\zeta(s)=\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^x-1}-\frac{1}{x}+\frac{1}{2}\right)\,dx\,$; Thus:
$$
\begin{align}
I\,& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}\color{red}{+\frac{1}{2}}-\frac{1}{e^{2x}-1}+\frac{1}{2x}\color{red}{-\frac{1}{2}}\right)\,dx \qquad\color{blue}{\small\{\text{both converge}\}} \\[2mm]
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}+\frac{1}{2}\right)\,dx\,-\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{2x}-1}-\frac{1}{2x}+\frac{1}{2}\right)\,dx \\[2mm]
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}+\frac{1}{2}\right)\,dx\,-\,\frac{2}{2^{s}}\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}+\frac{1}{2}\right)\,dx \\[2mm]
& =\,\color{red}{\left(2-2^{1-s}\right)\,\Gamma(s)\zeta(s)}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/385862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
} |
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