Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate: $\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}}$
Evaluate
$$
\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}} \cdotp
$$
My attempt:
$$
I
= \int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}}
= \int \frac{\mathrm{d}x}{x\sqrt{\left(x + \frac{1}{2}\right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}}
$$
I thought completing the square would bring the integrand into some form, but it did not. Please help.
| First use the substitution $x=\frac{1}{t}$ to get
$$\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}} =-\int \frac{\mathrm{d}t}{\sqrt{1+t+t^2}}.$$
Now complete the squares and use a tan substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is there a sharper upper bound of the spectral norm of Hilbert matrix than $3+2\sqrt{2}$?
$A_n$ is a real symmetric $n \times n$ matrix defined by
$$ A_n = \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} & \cdots &
\frac{1}{n} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{3} & \cdots &
\frac{1}{n} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \cdots &
\frac{1}{n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n} \\
\end{bmatrix} $$
Find an upper bound of the eigenvalues of $A_n$ as tight as possible.
Let $A_n = U+U^T-D,$ where $U$ is the upper triangular part of $A_n$ and $D$ is the diagonal of $A_n$.
Suppose $\lambda$ is the largest eigenvalue of $A_n = U^T U$.
For any $X \in \mathbb{R}^n$,
$$X^T A_n X = X^T U^T U X = \left\Vert UX \right\Vert^2 \leq \lambda \left\Vert X \right\Vert^2 \implies \left\Vert UX \right\Vert \leq \sqrt{\lambda} \left\Vert X \right\Vert.$$
Since $U^T U$ is similar to $U U^T$, $\lambda$ is also the largest eigenvalue of $ U U^T $.
Hence we also have
$$ \left\Vert U^T X \right\Vert \leq \sqrt{\lambda} \left\Vert X \right\Vert.$$
Similarly, $$ \left\Vert D X \right\Vert \leq \left\Vert X \right\Vert.$$
Thus
$$ \left\Vert A_n X \right\Vert = \left\Vert U X + U^T X - D X \right\Vert \leq \left\Vert U X \right\Vert + \left\Vert U^T X \right\Vert + \left\Vert D X \right\Vert \leq \left( 2 \sqrt{\lambda} + 1 \right) \left\Vert X \right\Vert.$$
Take $X$ to be the eigenvector of $A_n$ corresponding to $\lambda$, then
$$ \left\Vert A_n X \right\Vert = \lambda \left\Vert X \right\Vert \leq \left( 2 \sqrt{\lambda} + 1 \right) \left\Vert X \right\Vert,$$
we have $\sqrt{\lambda} \leq 1+\sqrt{2} \implies \lambda \leq 3+2\sqrt{2}.$
Is there any tighter upper bound?
| The matrix $A_n$ is semi-famous because its inverse is a tridiagonal matrix: $$A_n^{-1}=2B_n=2\begin{pmatrix}
1 & -1 & 0 & \cdots & 0 \\
-1 & 4 & -3 & 0 & \\
0 & \ddots & \ddots & \ddots & 0 \\
0 & 0 & -t_{n-2} & (n-1)^2 & -t_{n-1} \\
0 & 0 & 0 & -t_{n-1} & n^2/2\end{pmatrix}$$ where $t_n=n(n+1)/2$ are the triangular numbers. It seems easier to find the smallest eigenvalue of $B_n$ than the largest eigenvalues of $A_n$. If $v=(u_1(x),u_2(x),\ldots,u_n(x))$ is an eigenvector of $B_n$ with eigenvalue $x$, then the components satisfy the recurrence relation $$u_{k+1}(x)=\frac{k^2-x}{t_k}u_k(x)-\frac{t_{k-1}}{t_k}u_{k-1}(x),\quad u_1(x)=1,u_2(x)=1-x$$ $$\left(\frac{n^2}{2}-x\right)u_n(x)=t_{n-1}u_{n-1}(x)\tag{1}$$ the last being the characteristic polynomial equation of $B_n$. To find the smallest root of this polynomial one might use more sophisticated techniques, but for brevity let's keep only the constant and $x$ term. The proofs are omitted but it is not hard to see that $$u_n(x)=1-2(H_n-1)x+\cdots$$ where $H_n$ are the harmonic numbers. Substituting these into $(1)$ and neglecting higher order terms gives $x\approx \frac{n}{2 \left(n H_{n-1}+1\right)}$, so the largest eigenvalue of $A_n$ is less than $$\lambda\le H_n$$ A comparison of the actual largest eigenvalue and $H_n$ is below: $$\begin{matrix}n&1&2&3&4&5&\ldots&20\\
\lambda_n&1&1.31&1.49&1.62&1.71&&2.22\\
H_n&1&1.5&1.83&2.08&2.28&&3.66\\
\end{matrix}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I graph $f(x)=\lfloor{x^2}\rfloor$ when the domain is $ℝ^{-}$.
How can I graph $f(x)=\lfloor{x^2}\rfloor$ when the domain is $ℝ^{-}$.
I know that by definition $(\lfloor{x}\rfloor=m) ≡ (m≤x<m+1)$, so it follows that $(\lfloor{x^2}\rfloor=m ≡ (m≤x^2<m+1) ≡ (\sqrt{m}≤x^2<\sqrt{m+1})$; from this I can do
$\lfloor{x^2}\rfloor=
\begin{cases}
...\\
-3, & -3≤x^2<-2 \\
-2, & -2≤x^2<-1 \\
-1, & -1≤x^2<0 \\
0, & 0≤x^2<1 \\
1, & 1≤x^2<2 \\
2, & 2≤x^2<3 \\
3, & 3≤x^2<4 \\
...
\end{cases}
$
=
$\begin{cases}
...\\
-3, & -3≤x^2<-2 \\
-2, & -2≤x^2<-1 \\
-1, & -1≤x^2<0 \\
0, & 0≤x<1 \\
1, & 1≤x<\sqrt{2} \\
2, & \sqrt{2}≤x<\sqrt{3} \\
3, & \sqrt{3}≤x<2 \\
...
\end{cases}
$
Everything is normal until I "solve" for x in the negative intervals of the domain, which results in this
$\lfloor{x^2}\rfloor=
\begin{cases}
...\\
-3, & 3i≤x<2i \\
-2, & 2i≤x<i \\
-1, & i≤x^2<0 \\
0, & 0≤x<1 \\
1, & 1≤x<\sqrt{2} \\
2, & \sqrt{2}≤x<\sqrt{3} \\
3, & \sqrt{3}≤x<2 \\
...
\end{cases}
$
I don't know how to graph between those intervals.
Nonetheless when i see a graph of this function it seems that the function is symmetrical to the y axis.
Could you explain me, please, how can I graph the function in those negative intervals?
Thanks in advance.
| Guide:
*
*There are no real $x$ which satisfies $-1 \le x^2 < 0$.
*$0 \le x^2 < 1$ is equivalent to $0 \le |x|^2 < 1$ and the solution is $|x| < 1$, that is $-1 < x < 1$.
*$1 \le x^2 < 2$ is equivalent to $1 \le |x|^2 < 2$ and the solution is $1 \le |x| < \sqrt{2}$, that is $\{ x: -\sqrt{2} < x \le -1\} \cup \{ x: 1 \le x < \sqrt2\}$.
Note that the function is an even function.
| {
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$2x^2y''+3xy'-\left(x+1\right)y=0$ I am trying to solve $2x^2y''+3xy'-\left(x+1\right)y=0$ for a while but I can't make a progress.
For a solution near $x=0$ I apply method of series expansion but it doesn't show any closed pattern to related $A_n$ to $A_0$ ; and by more 'direct' method (for $x \ne 0$) it can be reduced to $(x^{3/2} y')'=\dfrac{x+1}{2 \sqrt x} y$ but I don't know how to go further: On one hand there is $y$ is RHS and on the other hand, there is $x^{3/2}$ multiplied by $y'$ in LHS which both ways makes it impossible to proceed.
However, there is a closed form (suggested by wolframalpha - without steps) for the solution : $$y(x) = c_1 \frac1x e^{\sqrt {2x}} (\sqrt 2 - 2 \sqrt x) + c_2 \frac1x e^{-\sqrt {2x}} ( 1+ \sqrt {2x}).$$
| $x=0$ is a regular singular point. The indicial equation is $r^2 + r/2 - 1/2 = 0$
which has roots $1/2$ and $-1$, so there are power series solutions of the form $\sum_{k=0}^\infty a_k x^{1/2+k}$ and $\sum_{k=0}^\infty b_k x^{-1+k}$. The recurrences for $a_k$ and $b_k$ are
$$ \eqalign{a_{k+1} &= \frac{a_k}{(2k+5)(k+1)} \cr
b_{k+1} &= \frac{b_k}{(2k-1)(k+1)} \cr}$$
Thus with $a_0 = 1$ the first series is
$$ \eqalign{&x^{1/2} \left(1 + \frac{x}{5 \cdot 1!} + \frac{x^2}{7 \cdot 5 \cdot 2!} + \frac{x^3}{9 \cdot 7 \cdot 5 \cdot 3!} + \ldots\right)\cr&= x^{1/2} \left(1 + \frac{3 \cdot 2 x}{5 \cdot 3!} + \frac{3 (2 x)^2}{7 \cdot 5!} + \frac{3 (2x)^3}{9 \cdot 7!} + \ldots\right)\cr
&= \frac{3 \cosh(\sqrt{2x})}{2 x^{1/2}} - \frac{3 \sqrt{2} \sinh(\sqrt{2x})}{4 x}} $$
Similarly, with $b_0 = 1$ the second series turns out to be
$$ \frac{\cosh(\sqrt{2x})}{x} - \frac{\sqrt{2} \sinh(\sqrt{2x})}{\sqrt{x}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $(1+x)^\frac{1}{x}+(1+\frac{1}{x})^x \leq 4$ Prove that $f(x)=(1+x)^\frac{1}{x}+(1+\frac{1}{x})^x \leq 4$ for all $x>0.$
We have $f(x)=f(\frac{1}{x}), f'(x)=-\frac{1}{x^2}f'(\frac{1}{x}),$ so we only need to prove $f'(x)＞0$ for $0 < x < 1.$
| Just a remark :
On $x\in [1/2,2]$ we have the inequality :
$$f(x)=\left(1+x\right)^{\frac{1}{x}}\leq \frac{4}{x^{-1+\ln\left(4\right)}+1}$$
Adding the two quantities gives :
$$f(x)+f(1/x)\leq g(x)+g(1/x)=4$$
To show it we have the inequalities on $[1,2]$ the hardest part :
$$\left(1+x\right)^{\frac{1}{x}}\leq 2e^{\left(-\frac{5\left(-1+\ln\left(4\right)\right)}{2}\left(x^{\frac{1}{5}}-1\right)\right)}\leq 2e^{\left(1-x^{\frac{\left(-1+\ln\left(4\right)\right)}{2}}\right)}\leq \frac{4}{x^{\left(-1+\ln\left(4\right)\right)}+1}$$
For the fisrt we can substitute $y^5=x$
It's a bit long but fairly easy with logarithmic derivative .We cannot do simpler for the derivative of $f(x)$ .
| {
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"timestamp": "2023-03-29T00:00:00",
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Difficulty Understanding Tom Apostol's Method of Exhaustion Proof In Tom Apostol's Calculus Vol 1 book, Apostol presents the following equation:
$$3[1^2+2^2+...+(n-1)^2]+ 3[1 + 2+ . . . + (n - 1)] + (n - 1)=n^3-1^3$$
He then states that the second sum on the left, which is $[1 + 2+ . . . + (n - 1)]$, is a sum of terms in an arithmetic progression, so that part will simplify to $\frac{n(n-1)}{2}$, thus the final equation becomes this:
$$1^2+2^2+...+(n-1)^2=\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}$$
What I cannot understand is how Apostol simplified $[1 + 2+ . . . + (n - 1)]$ to $\frac{n(n-1)}{2}$. As far as I understand, a sum of terms of an arithmetic progression has the following format: $\frac{n(a+b)}{2}$, with $a$ and $b$ representing the initial and $b$th term of the sequence, respectively. Where does he get $n-1$? Furthermore, how does he simplify the final formula to $1^2+2^2+...+(n-1)^2=\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}$?
| Note that you have $n-1$ terms, not $n$. Thus the sum will be of the form $\frac{(n-1)(a+b)}{2}$. Now it makes sense.
After substituting this in the equation you get:
$$3\sum_{k=1}^{n-1} k^2 = n^3 - 1 - (n-1) - 3\frac{n(n-1)}{2} = n^3 - n - \frac{3n^2}{2} + \frac{3n}{2} = n^3 - \frac{3n^2}{2} + \frac{n}{2}$$
$$\sum_{k=1}^{n-1} k^2 = \frac{n^3}{3} - \frac{n^2}{2} + \frac n6$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Sum of partition's product modulo 5 up to 41 If we define $S(n)$ as
$4=3+1=2+2=2+1+1=1+1+1+1$
$S(4)=3\cdot1+2\cdot2+2\cdot1\cdot1+1\cdot1\cdot1\cdot1=3+4+2+1=10$
$5=4+1=3+2=2+2+1=2+1+1+1=1+1+1+1+1$
$S(5)=4\cdot1+3\cdot2+2\cdot2\cdot1+2\cdot1\cdot1\cdot1+1\cdot1\cdot1\cdot1\cdot1=4+6+4+2+1=20$
$6=5+1=4+2=4+1+1=3+3=3+2+1=3+1+1+1=2+2+2=2+2+1+1=2+1+1+1+1=1+1+1+1+1+1$
$S(6)=5\cdot1+4\cdot2+4\cdot1\cdot1+3\cdot3+3\cdot2\cdot1+ 3\cdot1\cdot1\cdot1+2\cdot2\cdot2+2\cdot2\cdot1\cdot1+2\cdot1\cdot1\cdot1\cdot1+1\cdot1\cdot1\cdot1\cdot1\cdot1=5+8+4+9+6+3+8+4+2+1=50$
so what quantity of $n$ up to $41$ which $S(n)\equiv0\pmod5$?
| Rephrasing the question: let $S(n) = \sum_{\lambda\, \vdash \, n,\, \lambda_1 < n} \prod_i \lambda_i$. Then what is $S(n) \bmod 5$?
As a calculation strategy, define $S(n, k) = \sum_{\lambda\,\vdash\,n,\, \lambda_1 \le k} \prod_i \lambda_i$ to be the corresponding sum over partitions of $n$ into parts no greater than $k$, so that $S(n) = S(n, n-1)$. Then we can consider the largest part in each partition and find the recursion $S(n, k) = \sum_{j=1}^{\min(k,n)} j S(n - j, j)$ with base case $S(0, k) = 1$ as there is one partition of $0$ and it has an empty product.
Using this recursion it is easy to write a computer program to output a table of $S(n, k)$, but neither the table nor the values of $n$ for which $S(n, n-1) \equiv 0 \pmod 5$ are in OEIS, so this may not have been previously studied.
| {
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$\int_{\pi/6}^{\pi/3} (\sin x + \cos x) / \sqrt{\sin(2x)}\, dx$ Question How to solve the following integral?
$$I = \int_{\pi/6}^{\pi/3} \dfrac{\sin x + \cos x}{\sqrt{\sin(2x)}}\,dx$$
Attempt Using the property of definite integrals and putting $\pi/6 + \pi/3 - x$ in place of $x$ did not help. So I tried simplifying the function and I got $[(\tan x)^{1/2} + (\cot x)^{1/2}]/2^{1/2}$, with which I don't know how to proceed?
Any helps appreciated.
| The interval $\left(\frac{\pi}{6},\frac{\pi}{3}\right)$ is centered at $\frac{\pi}{4}$ and $\sin(x)+\cos(x)= \sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$, hence the given integral can be written as
$$ \int_{-\pi/12}^{\pi/12}\frac{\sqrt{2}\sin\left(x+\frac{\pi}{2}\right)}{\sqrt{\sin\left(2x+\frac{\pi}{2}\right)}}\,dx=\sqrt{2}\int_{-\pi/12}^{\pi/12}\frac{\cos(x)}{\sqrt{\cos(2x)}}\,dx=2\sqrt{2}\int_{0}^{\pi/12}\frac{\cos(x)}{\sqrt{\cos(2x)}}\,dx$$
or as
$$ \sqrt{2}\int_{0}^{\pi/6}\frac{\cos(x/2)}{\sqrt{\cos x}}\,dx=\int_{0}^{\pi/6}\sqrt{\frac{1+\cos x}{\cos x}}\,dx=\int_{\sqrt{3}/2}^{1}\sqrt{\frac{1+x}{x}}\frac{dx}{\sqrt{1-x^2}} $$
or as
$$ \int_{\sqrt{3}/2}^{1}\frac{dx}{\sqrt{x(1-x)}}=2\int_{\sqrt{\sqrt{3}/2}}^{1}\frac{du}{\sqrt{1-u^2}}=\pi-2\arcsin\sqrt{\frac{\sqrt{3}}{2}}=\boxed{\arccos(\sqrt{3}-1)}. $$
| {
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Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $ The task is to find out if this series is convergent or divergent.
$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$
The solution uses the ratio test and says:
$ \left.\begin{aligned} \frac { a _ { n + 1 } } { a _ { n } } & = \frac { ( n + 1 ) ! ( n + 1 ) ^ { n + 1 } ( 2 n ) ! } { ( 2 ( n + 1 ) ) ! n ! n ^ { n } } = \frac { ( n + 1 ) n ! ( n + 1 ) ( n + 1 ) ^ { n } ( 2 n ) ! } { ( 2 n + 2 ) ( 2 n + 1 ) ( 2 n ) ! n ! n ^ { n } } \\ & = \frac { ( n + 1 ) ( n + 1 ) } { ( 2 n + 2 ) ( 2 n + 1 ) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } = \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } \\ & \rightarrow \frac { 1 } { 4 } \cdot e < 1 \text { for } n \rightarrow \infty \end{aligned} \right. $
I understand every step until here
$$ \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } $$
How can all n's on the left site become 1/n? And I understand how the left site can become $\frac{1}{4}$, but how can the right site become e in the last step?
| We have that
$$\left( \frac { n + 1 } { n } \right) ^ { n }=\left( 1+\frac { 1 } { n } \right) ^ { n }\to e$$
and from here
$$\frac { ( n + 1 ) ( n + 1 ) } { ( 2 n + 2 ) ( 2 n + 1 ) } $$
just divide both numerator and denominator by $n^2$.
| {
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Prove this inequality for given conditions For all $x,y>0$, $$\frac{1}{(x+1)^2} + \frac{1}{(y+1)^2} \ge \frac{1}{xy+1}$$
I can only think of substituting $x+1$ with $a$ and $y+1$ with $b$. Then the inequality turns into
$$(a^2 + b^2) (ab-a-b +2) \ge a^2b^2$$
I can proceed no further. Please help.
| After your substitutions we get new conditions $a>1$ and $b>1$, which makes the inequality harder.
By the way, it's just $$xy(x-y)^2+(xy-1)^2\geq0.$$
Also, we can use C-S:
$$\sum_{cyc}\frac{1}{(x+1)^2}=\sum_{cyc}\frac{y}{(x\sqrt{y}+\sqrt{y})^2}\geq\sum_{cyc}\frac{y}{(xy+1)(x+y)}=\frac{1}{xy+1}.$$
| {
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How do I find $\sqrt[3]{-i}$? I'm asked to evaluate $$\sqrt[3]{-i}$$
I suppose $\sqrt[3]{-i}=(a+bi)$
$$\implies (a+bi)^3=-i$$
$$\implies \Im \left( (a+bi)^3 \right) =\Im \left( (-i) \right)$$
$$\implies 3a^2b-b^3=-1$$
Now how am I supposed to find $a$,$b$?
Aren't there infinitely of them instead of just three?
| You can solve this as
\begin{equation}
(-i)^{\frac{1}{3}}
=
( e^{i \frac{3\pi}{2} + i2k\pi})^{\frac{1}{3}}
\end{equation}
which gives us three distinct roots $e^{i z_k}$, for $k = 0,1,2$, where
\begin{align}
z_0 &= \frac{1}{3}\frac{3\pi}{2} + \frac{2(0)\pi}{3} = \frac{\pi}{2} \\
z_1 &= \frac{1}{3}\frac{3\pi}{2} + \frac{2(1)\pi}{3} = \frac{7\pi}{6} \\
z_2 &= \frac{1}{3}\frac{3\pi}{2} + \frac{2(2)\pi}{3} = \frac{11\pi}{6} \\
\end{align}
If you insist on solving it your way, then
\begin{equation}
(a+bi)^3 = -i
\end{equation}
means
\begin{equation}
a^3 + 3a^2bi - 3ab^2 - b^3i = -i
\end{equation}
which means
\begin{align}
a^3 - 3ab^2 &= 0 \\
3a^2b - b^3 &= -1
\end{align}
which is
\begin{align}
a^2 - 3b^2 &= 0 \\
3a^2b - b^3 &= -1
\end{align}
or
\begin{align}
(a - \sqrt{3} b)(a + \sqrt{3} b) &= 0 \\
3a^2b - b^3 &= -1
\end{align}
The first equation suggests either
\begin{equation}
a = \pm \sqrt{3} b
\end{equation}
Replacing this in the second equation gives
\begin{equation}
3(\sqrt{3} b)^2b - b^3 = -1
\end{equation}
which is
\begin{equation}
9b^3 - b^3 = -1
\end{equation}
i.e.
\begin{equation}
b = -\frac{1}{2}
\end{equation}
This will give us
\begin{equation}
a = \pm \sqrt{3} (-\frac{1}{2})
\end{equation}
which means we get two solutions
\begin{align}
(a_1,b_1) &= (-\sqrt{3},-\frac{1}{2}) \\
(a_2,b_2) &= (\sqrt{3},-\frac{1}{2}) \\
\end{align}
which are actually what we found before, i.e.
\begin{align}
a_1 + ib_1 &= e^{i z_1} \\
a_2 + ib_2 &= e^{i z_2} \\
\end{align}
Then you've got one more root you need to find (which is the obvious one), where $a_0 = 0$ and $b_0 = 1$, i.e. $i^3 = -i$. Also, you can verify that
\begin{align}
a_0 + ib_0 &= e^{i z_0}
\end{align}
| {
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Closed form of integral over fractional part $\int_0^1 \left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\}\,dx$ Recenly, several interesting questions have been posted asking for closed forms of integrals over the fractional part of certain functions.
For me the story started with Evaluation of $\int_{0}^{1}\int_{0}^{1}\{\frac{1}{\,x}\}\{\frac{1}{x\,y}\}dx\,dy\,$ which after a long and instructive journey I could solve completely. Another example was symmetric double-integral on fractional part. These are examples of double integrals. There are as well many single integrals, and, as we can see below, the field of single integrals is by far not exhausted.
This time my result is given in the beginning and a proof is asked for.
Let $\{z\}$ be the fractional part of $z$.
Prove that:
$$i := \int_0^1 \left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\} = i_{s} $$
where
$$i_{s}=c_{g}-\frac{\gamma }{2}+\frac{3}{4}+\frac{\log (2)}{2} \simeq 0.28000699470709318696$$
Here $\gamma$ is the Euler-Mascheroni constant and
$$c_{g} = \int_0^{\infty } \frac{t-2 I_1(t)}{2 \left(e^t-1\right) t} \, dt \simeq -0.52795876312211303745$$
where
$I_{n}(t)$ is the modified Bessel function of the first kind.
$c_{g}$ is a (probably) new constant which appears in the asymptotic expansion of the sum
$$g(n) = \sum _{k=1}^n \sqrt{k^2-1} $$
| It is helpful to derive the asymptotic expansion of $g$ first. We can use the binomial series to find
\begin{align}
g(n) &= \sum \limits_{k=2}^n k \sqrt{1-k^{-2}} = \sum \limits_{k=2}^n k \sum \limits_{j=0}^\infty {1/2\choose j} (-k^{-2})^j \\
&= \frac{n(n+1)}{2} - 1 - \frac{H_n}{2} + \frac{1}{2} + \sum \limits_{j=2}^\infty {1/2\choose j} (-1)^j \sum \limits_{k=2}^n k^{1-2j}
\end{align}
with the harmonic numbers $H_n$. The monotone convergence theorem now yields the asymptotic equivalence
$$ g(n) \sim \frac{n(n+1)}{2} - \frac{H_n}{2} + c_g + \mathcal{o}(1)$$
as $n \to \infty$ . The constant term can be written as
$$ c_g = - \frac{1}{2} + \sum \limits_{j=2}^\infty {1/2\choose j} (-1)^j [\zeta(2j-1) - 1] = \sum \limits_{k=2}^\infty \left(\sqrt{k^2-1} - k + \frac{1}{2k}\right) \, ,$$
which agrees with the integral representation after using the series expansion of $I_1$.
In order to find $i$ we use the substitution $x = t - \sqrt{t^2-1}$ :
\begin{align}
i &= \int \limits_0^1 \left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\} \, \mathrm{d} x = \int \limits_1^\infty \{t\} \left(\frac{t}{\sqrt{t^2-1}}-1\right) \, \mathrm{d} t \\
&= \sum \limits_{n=1}^\infty \int \limits_n^{n+1} (t-n) \left(\frac{t}{\sqrt{t^2-1}}-1\right) \, \mathrm{d} t \\
&= \frac{1}{2} \sum \limits_{n=1}^\infty \left[\ln\left(\sqrt{(n+1)^2-1}+n+1\right) - \ln\left(\sqrt{n^2-1}+n\right)\right. \\
&\phantom{= \frac{1}{2} \sum \limits_{n=1}^\infty\left[\right.} \left.- (n+1)\sqrt{(n+1)^2-1} + n \sqrt{n^2-1} + 2\sqrt{(n+1)^2 - 1} - 1 \right] \, .
\end{align}
The remaining series is (mostly) telescoping and we obtain
\begin{align}
i &= \frac{1}{2} \lim_{N \to \infty} \left[\ln\left(\sqrt{N^2-1} + N\right) - N \sqrt{N^2-1} + 2 g(N) - N + 1\right] \\
&= \frac{1}{2} \lim_{N \to \infty} \left[\ln\left(1+\sqrt{1-N^{-2}}\right) + \ln(N) - H_N + N \left(N+1 - \sqrt{N^2-1} - 1\right) + 2 c_g + 1\right] \\
&= \frac{1}{2} \left[\ln(2) - \gamma + \frac{1}{2} + 2 c_g + 1\right] \\
&= \frac{3}{4} + \frac{\ln(2)-\gamma}{2} + c_g \, .
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the sum $\sum_{k=1}^rk^2\binom {n-k}{r-k}$ I was stuck while finding the given summation.
$$\sum_{k=1}^rk^2\binom {n-k}{r-k}$$
Since $n$ and $r$ are both constants, so I have first converted the above summation into this:
$$\sum_{k=1}^rk^2\binom {n-k}{n-r}$$
I have no idea how to proceed next. Any help will be appreciated.
| One method is to use $k^2 = (r-k)(r-k-1) - (2r-1)(r-k) + r^2$ and
$$P_{r}^{n} = \sum_{k=1}^{n} \binom{n-k}{r-k} = \frac{n}{n-r+1} \, \binom{n-1}{r-1}.$$
With this one can show that, for $r\geq 2$:
\begin{align}
S_{r}^{n} &= \sum_{k=1}^{n} k^2 \, \binom{n-k}{r-k} \\
&= \frac{(n-r+2)!}{(n-r)!} \, P_{r-2}^{n} - \frac{(2r-1) \, (n-r+1)!}{(n-r)!} \, P_{r-1}^{n} + r^2 \, P_{r}^{n} \\
&= \frac{n! \, (n^2 + (r+2) \, n + r + 1)}{(r-1)! \, (n-r+3)!} \hspace{5mm} \text{after some reduction} \\
&= \binom{n+1}{r-1} + 2 \, \binom{n+1}{r-2} \hspace{5mm} \text{after further reduction}
\end{align}
A similar result occurs when use is made of
$$\sum_{k=1}^{r} \binom{n-k}{r-k} = \frac{n}{n-r+1} \, \binom{n-1}{r-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $ ab^2 \geq k(ab)^\frac{2(k-1)}{k} $ In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page 8 that-
$$ ab^2 \geq k(ab)^\frac{2(k-1)}{k} \cdots (1)$$
How we prove above above inequality? We know that-
$(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k \implies (a+1)(ab^2+1) \geq (ab+1)^2+ k(ab+1)^{2(k-1)/k} \implies ab^2+a \geq 2ab+ k(ab+1)^{2(k-1)/k} $, but how do we get $ ab^2 \geq k(ab)^\frac{2(k-1)}{k}?$
| We know that-
$(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k$
$\implies (a+1)(ab^2+1) \geq (ab+1)^2+ k(ab+1)^{2(k-1)/k}$
[Note,we are using only two terms of binomial expansion of $((ab+1)^\frac{2}{k}+1)^k$]
$\implies ab^2+a \geq 2ab+ k(ab+1)^{2(k-1)/k} $
$\implies ab^2 \geq 2ab-a + k(\sum_{m=0}^{\frac{2(k-1)}{k}}\binom{2(k-1)/k}{ m} (ab)^{m}) $
$\implies ab^2 \geq 2ab-a + k(\binom{2(k-1)/k)}{ 2(k-1)/k)} (ab)^\frac{2(k-1)}{k} + \sum_{m=0}^{\frac{2(k-1)}{k}-1}\binom{(2(k-1)/k)-1}{ m} (ab)^{m}) $
$\implies ab^2 \geq 2ab-a + k (ab)^\frac{2(k-1)}{k} + k( \sum_{m=0}^{\frac{2(k-1)}{k}-1}\binom{(2(k-1)/k)-1}{ m} (ab)^{m}) $
$\implies ab^2 \geq k (ab)^\frac{2(k-1)}{k} $ [Note, $2ab-a>0$]
| {
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$n^2 - n + 11$ is the product of four prime numbers, some of which may be the same. What is the minimum value of $n$? Let $n$ be a positive integer such that $n^2 - n + 11$ is the product of four prime numbers, some of which may be the same. What is the minimum value of $n$?
The problem is taken from here. I know the answer is $132$, but how can I prove it? That would mean the four numbers are $11$, $11$, $11$, and $13$.
Can you give me a hint, please? Thanks!
| First, find the smallest prime that divides $n^2-n+11$. It is not 2:
$$0^2-0+11 \equiv 1 \pmod{2}$$
$$1^2-1+11 \equiv 1 \pmod{2}$$
It is not 3:
$$0^2-0+11\equiv 2\pmod{3}$$
$$1^2-1+11\equiv 2\pmod{3}$$
$$2^2-2+11\equiv 1\pmod{3}$$
Not 5:
$$0^2-0+11\equiv 1\pmod{5}$$
$$1^2-1+11\equiv 1\pmod{5}$$
$$2^2-2+11\equiv 3\pmod{5}$$
$$3^2-3+11\equiv 2\pmod{5}$$
$$4^2-4+11\equiv 3\pmod{5}$$
Not 7:
$$0^2-0+11\equiv 4\pmod{7}$$
$$1^2-1+11\equiv 4\pmod{7}$$
$$2^2-2+11\equiv 6\pmod{7}$$
$$3^2-3+11\equiv 3\pmod{7}$$
$$4^2-4+11\equiv 2\pmod{7}$$
$$5^2-5+11\equiv 3\pmod{7}$$
$$6^2-6+11\equiv 6\pmod{7}$$
This means the smallest prime that divides $n^2-n+11$ is 11. So, you use trial and error:
$$n^2-n+11=11^4$$
has no solutions over the integers
$$n^2-n+11=11^3\cdot 13$$
Here you get 132.
| {
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How to prove that $\sin(\sqrt{x})$ is not periodic? How to prove that $\sin(\sqrt{x})$ is not periodic?
THe definition of a periodic function is $f(x+P)=f(x)$.
So I assume that $\sin(\sqrt{x+P})=\sin(\sqrt{x})$. This is equivalent to $\sin(\sqrt{x+P})-\sin(\sqrt{x})=0$. This implies $2cos(\frac{\sqrt{x+P}+\sqrt{x}}{2})\sin(\frac{\sqrt{x+P}-\sqrt{x}}{2})$. What should I do next?
| If $\sin {\sqrt x}$ is period with period $P$ then so is $\cos {\sqrt x}$ because $\cos {\sqrt x} = \sqrt {1 - \sin^2 \sqrt x}=\sqrt {1 - \sin^2 \sqrt {x+P}}=\cos (\sqrt{x+P})$. Likewis so is the derivative of $\sin {\sqrt x}$ because $ \lim \frac {\sin (\sqrt {x + h}) -\sin(\sqrt{x})}h= \lim \frac {\sin (\sqrt {x +P + h}) -\sin(\sqrt{x+P})}h$.
But the derivative of $\sin (\sqrt{x})$ is $\frac{cos(\sqrt{x})}{2\sqrt{x}}$ and if $\cos (\sqrt{x + P}) = \cos (\sqrt{x})$ then $\frac{cos(\sqrt{x+P})}{2\sqrt{x+P}}=\frac{cos(\sqrt{x})}{2\sqrt{x+P}} \ne \frac{cos(\sqrt{x})}{2\sqrt{x}}$
This is a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find maximum of succession $a_{n+1}=\frac{n^2+n+4}{2n^2+1}a_n$ knowing $a_1=1$ I want to find out what's the maximum of the following sequence:
$$
\left\{
\begin{array}{c}
a_1=1 \\
a_{n+1}=\frac{n^2+n+4}{2n^2+1}a_n \\
\end{array}
\right.
$$
I know that the limit of $a_n$ is equal to $0$. I have no clue about how to find the maximum. I have tried to find out the values of $a_1, a_2, a_3\dots$ but we couldn't find the maximum. Please note that $n\geq 0$. Any hints?
| In general for a recursion of the form:
$$
a_{n+1} = f(n)a_{n}, a_{1} > 0.
$$
if $f(n) - 1$ has only one zero $> 0$ and $lim_{n\to\infty}f(n) < 1$ then you know $a_{n}$ will attain its maximum for the largest $n$ for which you have $f(n)\geq 1$.
in this case:
$f(n)= \frac{n^2+n+4}{2n^2+1} > 1\Rightarrow - n^2 +n +3 > 0 \Rightarrow n < \frac{1+\sqrt{13}}{2} \approx 2.303$. Thus your maximum is $a_{3}$ ...
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving Homogenous First Order ODE Using Substitution Problem
Find the general solution of the following homogeneous equation.
$$ ty' = y + \sqrt { t^2 - y^2} $$
Attempt
I'm following the algorithm provided in section 1.5 of the William Adkins & Mark G. Davidson, Ordinary Differential Equations textbook on page 63, which suggests rearranging the equation into the form:
$$ y' = ... $$
where $ y = tv $ and $ y' = v + tv' $ (v is the substitution variable).
I haven't been able to get too far in the procedure to make the substitution:
$$ y' = \frac{y + \sqrt { t^2 - y^2}}{t} $$
$$ y' = \frac{\frac{1}{t}}{\frac{1}{t}} \frac{y + \sqrt { t^2 - y^2}}{t} $$
$$ y' = \frac{y}{t} + \frac{\sqrt { t^2 - y^2}}{t} $$
Notes
Can someone please help me reduce the RHS of the above equation to the form $ y' = f(\frac{y}{t}) $ (an equation in terms of $ (\frac{y}{t}) $), using the substitution $ y = tv $? For bonus points, please provide a solution to the ODE so myself and other viewers can check their answers!
Thanks in advance!
Solution
With the help of Isham, I was able to get the solution.
$$ ty' = y + \sqrt { t^2 - y^2 } $$
It is implied that $ t^2 - y^2 \geq 0 $, and so $ t^2 \geq y^2 \gt 0 $ or $ \lt 0 $.
$$ ty' - y = \sqrt { t^2 - y^2} $$
$$ \frac {1}{\sqrt {t^2} } (ty' - y) = \frac {1}{\sqrt {t^2} } \sqrt { t^2 - y^2} $$
$$ \frac {1}{t} (ty' - y) = \sqrt { \frac{t^2}{t^2} - \frac{y^2}{t^2}} $$
$$ y' - \frac{y}{t} = \sqrt { 1 - \frac{y^2}{t^2}} $$
$$ y' - \frac{y}{t} = \sqrt { 1 - (\frac{y}{t})^2} $$
$$ y' = \sqrt { 1 - (\frac{y}{t})^2} + \frac{y}{t} $$
Let $ v = \frac{y}{t} $. Then, $ y = tv $ and so $ y' = \frac{dy}{dt} = v + tv' $. Substituting for $ v $ in the above equation we get:
$$ v + tv' = \sqrt {1 - v^2} + v $$
$$ tv' = \sqrt {1 - v^2} $$
Note, since $ t^2 \geq y^2 $, $ \frac{y^2}{t^2} = v^2 \leq 1 $, so there are no equilibrium cases to consider.
$$ t\frac{dv}{dt} = \sqrt {1 - v^2} $$
$$ \frac{1}{\sqrt { 1 - v^2}}dv = \frac{1}{t}dt $$
$$ \int\frac{1}{\sqrt { 1 - v^2}}dv = \int\frac{1}{t}dt $$
$$ arcsin(v) = ln|t| + C_1, C_1 \in \mathbb R $$
$$ \frac {y}{t} = \sin{(ln|t| + C_1)} $$
$$ y = t \sin{(ln|t| + C_1)} $$
| Hint: To reduce, you need to substitute $y=vt$ in RHS and $y'=v+tv'$ in LHS. It would look like this:
$$v+tv'=\frac{tv+t\sqrt{1-v^2}}{t}$$
$$tv'=\sqrt{1-v^2}$$
Where $\displaystyle v'$ is $\displaystyle\frac{\mathrm{d}v}{\mathrm{d}t}$ Can you proceed?
| {
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Prove that $x^2+xy^2+xyz^2 \ge 4xyz-4$ for positive real $x,y,z$
Prove that $x^2+xy^2+xyz^2 \ge 4xyz-4$ for postive real $x,y,z$.
I tried AM-GM but failed. I also can't apply Cauchy-Schwarz. I think we have to change this inequality to apply any well-known inequality. Please help me.
| $$x^2+xy^2+xyz^2+4=x^2+2\cdot\dfrac{xy^2}2+4\cdot\dfrac{xyz^2}4+4$$
Now using AM-GM inequality, $$\dfrac{x^2+2\cdot\dfrac{xy^2}2+4\cdot\dfrac{xyz^2}4+4}{1+2+4+1}\ge\sqrt[1+2+4+1]{x^2\cdot\left(\cdot\dfrac{xy^2}2\right)^2\cdot\left(\dfrac{xyz^2}4\right)^4\cdot4}$$
Here is how I've identified the coefficients:
let $$x^2+xy^2+xyz^2+4=a\cdot\dfrac{x^2}a+b\cdot\dfrac{xy^2}b+c\cdot\dfrac{xyz^2}c+d\cdot\dfrac4d$$
Now by AM-GM, $$\dfrac{a\cdot\dfrac{x^2}a+b\cdot\dfrac{xy^2}b+c\cdot\dfrac{xyz^2}c+d\cdot\dfrac4d}{a+b+c+d}\ge?$$
Compare the exponents of $x,y,z$ to find
$$a=d,b=2a,c=4a$$
Choose $a=1$
| {
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Algebraic manipulation of a complex valued function I have the following function: $$f(z)=\frac{z}{z-1}$$
With complex domain and range, I have to show that the unit circle $e^{i\theta}$ is mapped by the function as a line with real part equal to $\frac{1}{2}$. Moreover i have to show:$$f( e^{i\theta})=\frac{1}{2}-\frac{1}{2}i\cot\frac{\theta}{2} $$
I tried substituting $z=e^{i\theta}$ and manipulating the expression but I keep getting stuck in messy trig expressions...
| You can calculate it directly as follows:
*
*$z = \cos t + i \sin t$ (I use $t$ instead of $\theta$.)
*$\frac{z}{z-1}= 1 +\frac{1}{z-1}$
$$\frac{1}{z-1} = \frac{1}{\cos t-1 + i \sin t}= \frac{\cos t-1 - i \sin t}{\cos^2 t +1 -2 \cos t + \sin ^2 t}= \frac{\cos t-1 - i \sin t}{2(1-\cos t)}= -\frac{1}{2}-\frac{i}{2}\frac{\sin t}{1-\cos t}$$
$$\frac{\sin t}{1-\cos t} = \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{1 - (\cos^2 \frac{t}{2} - \sin^2 \frac{t}{2})} = \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}= \cot \frac{t}{2}$$
So, all together
$$\frac{z}{z-1}= 1 -\frac{1}{2}-\frac{i}{2}\frac{\sin t}{1-\cos t} = \frac{1}{2} -\frac{i}{2}\cot \frac{t}{2}$$
| {
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If $3$ couples are randomly seated, what is they probability that at least $1$ couple sits together? My attempt:
$$A = \text{couple 1 sits together}$$
$$B = \text{couple 2 sits together}$$
$$C = \text{couple 3 sits together}$$
Solving for
$$P(A \cup B \cup C) = P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A \cap B \cap C)$$
$$P(A)=P(B)=P(C) \quad\text{and}\quad P(A \cap B)=P(A \cap C)=P(B \cap C)\text{, so}$$
$$P(A \cup B \cup C) = 3P(A)-3P(A \cap B)+P(A \cap B \cap C)$$
\begin{align*}
P(A) = \frac{5!}{6!}\quad&\text{if 1 couple acts as a unit}\\
P(A \cap B) = \frac{4!}{6!}\quad&\text{if 2 couples act as single units}\\
P(A \cap B \cap C) = \frac{3!}{6!}\quad&\text{if all couples act as single units}
\end{align*}
| You have the right idea, but you have not taken into account the internal arrangements of the couples that sit together. A couple can sit together in $2!$ ways. Hence, you should have
\begin{align*}
\Pr(A) & = \frac{5!2!}{6!}\\
\Pr(A \cap B) & = \frac{4!2!^2}{6!}\\
\Pr(A \cap B \cap C) & = \frac{3!2!^3}{6!}
\end{align*}
$\Pr(A)$: There is a couple that sits together and four other individuals, so we have five objects to arrange, which can be done in $5!$ ways. The couple can be arranged internally in $2!$ ways. Hence, the number of favorable arrangements is $5!2!$.
$\Pr(B)$: There are two couples and two other individuals, so we have four objects to arrange, which can be done in $4!$ ways. Each couple can be arranged internally in $2!$ ways. Hence, there are $4!2!^2$ favorable cases.
$\Pr(C)$: There are three couples to arrange, which can be done in $3!$ ways. Each couple can be arranged internally in $2!$ ways. Hence, there are $3!2!^3$ favorable cases.
With these corrections, we obtain
$$\Pr(A \cup B \cup C) = \binom{3}{1}\frac{5!2!}{6!} - \binom{3}{2}\frac{4!2!^2}{6!} + \binom{3}{3}\frac{3!2!^3}{6!}$$
| {
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Area of ellipse using double integral I am trying to find the area of a quadrant of an ellipse by double integrating polar coordinates but the answer I'm getting is incorrect.
ellipse : $ x^2/a^2 + y^2/b^2 =1 $
Any point on ellipse : $ ( a\cos(\theta), b\sin(\theta)) $
At $ \theta$, taking $ d\theta $ segment, Thus $ r^2 = a^2\cos^2(\theta) + b^2\sin^2(\theta) $ [Using pythagoras theorem]
$$ Area = \int_{0}^{\pi/2} \int_{0}^{\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)}} rdrd\theta $$
$$ = 1/2 \int_{0}^{\pi/2} r^2 \Big|_{0}^{\sqrt{a^2\cos^2(\theta)+b^2\sin^2(\theta)}} d\theta $$
$$ = 1/2 \int_{0}^{\pi/2} (a^2\cos^2(\theta)+b^2\sin^2(\theta)) d\theta $$
$$ = 1/2 \int_{0}^{\pi/2} ((a^2 - b^2)\cos^2(\theta)+b^2) d\theta $$
$$ = 1/4 \int_{0}^{\pi/2} (a^2 - b^2)(1+ \cos(2\theta)) d\theta +2b^2 d\theta $$
I am getting $$ \pi/8 (a^2 + b^2).$$ But the correct answer is $ \pi ab/4 $
| The given ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.To transform it into polar coordinates,substitute $(x,y)=(r\cos\theta,r\sin\theta)$ to get $r=\dfrac{ab}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}$.
Take an elementary area $rdrd\theta$ inside the ellipse.
Then the area of the ellipse in the first quadrant is the sum of all such elementary areas
$\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\int_{r=0}^{\dfrac{ab}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}} r \,dr \, d\theta$
$=\dfrac12\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\dfrac{a^2b^2}{b^2\cos^2\theta + a^2\sin^2\theta} d\theta$
$=\dfrac12\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\dfrac{a^2b^2\sec^2\theta}{b^2+ a^2\tan^2\theta} d\theta$
$=\dfrac{b^2}{2}\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\dfrac{\sec^2\theta}{\dfrac{b^2}{a^2}+ \tan^2\theta} d\theta$
$=\left.\dfrac{b^2}{2}\dfrac{a}{b}\tan^{-1}\left(\dfrac{a\tan\theta}{b}\right)\right|_{0}^{\frac{\pi}{2}}$
$=\boxed{\boxed{\dfrac{\pi ab}{4}}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to change the base cases of a tribonacci sequence solved by matrix exponentation So the base case of a Fibonacci sequence is the following matrix:
\begin{bmatrix}1&1\\1&0\end{bmatrix}
Which I understood as being:
\begin{bmatrix}F_2&F_1\\F_1&F_0\end{bmatrix}
And that seems logical. But I don't really understand what each of the numbers in a tribonacci base case mean:
\begin{bmatrix}1&1&1\\1&0&0\\0&1&0\end{bmatrix}
I don't really understand what each of these numbers represent so if someone could help me, I would appreciate it. Also, I need to compute a tribonacci sequence where the base cases are not t[0]=0, t[1]=0 and t[2]=1 (like in the above matrix), but t[0]=1, t[1]=1 and t[2]=2 So how exactly could I implement that into the matrix that represents the base case? Do I just replace it like the following:
\begin{bmatrix}2&2&2\\2&1&1\\1&2&1\end{bmatrix}
But what if I would want to replace them with distinct numbers, say for example t[0]=3, t[1]=5 and t[2]=8 ? Where would each of these numbers be placed in the matrix?
| Notice that $F_n = F_{n-1} + F_{n-2}, \forall n \ge 2$ means that
$$\begin{bmatrix} F_{n} \\ F_{n-1} \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 1
& 0\end{bmatrix}\begin{bmatrix} F_{n-1} \\ F_{n-2} \end{bmatrix}$$
and the initial conditions are $F_0 = F_1 = 1$ so $\begin{bmatrix} F_{1} \\ F_{0} \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
Therefore iteration gives
$$\begin{bmatrix} F_{n} \\ F_{n-1} \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 1
& 0\end{bmatrix}\begin{bmatrix} F_{n-1} \\ F_{n-2} \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 1
& 0\end{bmatrix}^2\begin{bmatrix} F_{n-2} \\ F_{n-3} \end{bmatrix} = \cdots = \begin{bmatrix} 1 & 1 \\ 1
& 0\end{bmatrix}^{n-1}\begin{bmatrix} F_{1} \\ F_{0} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1
& 0\end{bmatrix}^{n-1}\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$
Similarly the Tribonacci numbers satisfy $T_n = T_{n-1} + T_{n-2} + T_{n-3}, \forall n \ge 3$ so
$$\begin{bmatrix} T_{n} \\ T_{n-1} \\ T_{n-2}\end{bmatrix}= \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0
\end{bmatrix}\begin{bmatrix} T_{n-1} \\ T_{n-2} \\ T_{n-3}\end{bmatrix}$$
Your initial conditions are $T_0 = 1$, $T_1 = 1$ and $T_2 = 2$ so $\begin{bmatrix} T_{2} \\ T_{1} \\ T_0 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 1\end{bmatrix}$.
Therefore as above we get
$$\begin{bmatrix} T_{n} \\ T_{n-1} \\ T_{n-2}\end{bmatrix}= \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0
\end{bmatrix}^{n-2}\begin{bmatrix} T_{2} \\ T_{1} \\ T_{0}\end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0
\end{bmatrix}^{n-2}\begin{bmatrix} 2 \\ 1 \\ 1\end{bmatrix}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Sum of $1+\frac{1\cdot 3}{6}+\frac{1\cdot 3 \cdot 5}{6 \cdot 8}+\cdots \cdots$
Finding sum of
$\displaystyle 1+\frac{1\cdot 3}{6}+\frac{1\cdot 3 \cdot 5}{6\cdot 8}+\frac{1\cdot 3 \cdot 5 \cdot 7}{6 \cdot 8 \cdot 10}+\cdots \cdots$
Try: We can write sum as
$$ \mathcal{S} = 4\bigg[\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3 \cdot 5}{4\cdot 6 \cdot 8}+\cdots \cdots \cdots \bigg]$$
Now Let $$a_{n} = \prod^{n}_{k=1}(2k-1)=2^n\prod^{n}_{k=1}\bigg(k-\frac{1}{2}\bigg)$$
and $$b_{n} = 2^{-1}\prod^{n}_{k=1}(2k)=2^{n-1}\prod^{n}_{k=1}k$$
So $$\frac{a_{n}}{b_{n}} = 2\cdot \Gamma\left(n+\frac{1}{2}\right)\cdot \frac{1}{\Gamma\left(\frac{1}{2}\right)\cdot \Gamma(n+1)}$$
Above I have used $$\Gamma(x+n) = (x+n-1)(x+n-2)\cdot \cdots x \cdots \Gamma(n).$$
So $$\frac{a_{n}}{b_{n}} = 2\cdot \frac{\Gamma\left(n+\frac{1}{2}\right)\cdot \Gamma\left(\frac{1}{2}\right)}{\pi \cdot \Gamma(n+1)}= \frac{2}{\pi}\int^{1}_{0}x^{n-\frac{1}{2}}(1-x)^{-\frac{1}{2}}dx$$
So our sum is $$\mathcal{S} = \frac{8}{\pi} \sum^{\infty}_{n=1}\int^{1}_{0}x^n \cdot \frac{1}{\sqrt{x-x^2}}dx$$
So $$\mathcal{S} = \frac{8}{\pi}\int^{1}_{0}\frac{x}{(1-x)\sqrt{1-x^2}}dx$$
Put $x=\sin^2 \theta$ and $dx = 2 \sin \theta \cos \theta d\theta$ and changing limits
So we have $$\mathcal{S} = \frac{16}{\pi}\int^{\frac{\pi}{2}}_{0}(\sec^2 \theta -1)d \theta = $$
I did not understand where i am wrong.
and answer is $4$
Could some help me to explain it , thanks
| You almost got everything right, and the only problem you have is a minor error when you define $b_n$. You should have
$$b_n=2^n\,(n+1)!=2^n\,\Gamma(n+2)\text{ for }n=1,2,3,\ldots\,.$$
Thus, for each $n=1,2,3,\ldots$,
$$\begin{align}\frac{a_n}{b_n}&=\frac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)\,\Gamma(n+2)}=\frac{2}{\pi}\,\left(\frac{\Gamma\left(n+\frac12\right)\,\Gamma\left(\frac32\right)}{\Gamma(n+2)}\right)\\&=\frac{2}{\pi}\,\text{B}\left(n+\frac12,\frac32\right)\,,\end{align}$$
where $\Gamma$ and $\text{B}$ are the usual gamma and beta functions, respectively. Hence,
$$\frac{a_n}{b_n}=\frac{2}{\pi}\,\int_0^1\,x^{n-\frac12}\,(1-x)^{\frac12}\,\text{d}x\,,$$
so
$$\begin{align}\sum_{n=1}^\infty\,\frac{a_n}{b_n}&=\frac{2}{\pi}\,\int_0^1\,\frac{x^{\frac12}}{1-x}\,(1-x)^{\frac12}\,\text{d}x
\\&=\frac{2}{\pi}\,\int_0^1\,x^{\frac12}\,(1-x)^{-\frac12}\,\text{d}x\,.\end{align}$$
That is, with $u:=x^{\frac12}$, we obtain
$$\begin{align}\sum_{n=1}^\infty\,\frac{a_n}{b_n}&=\frac{4}{\pi}\,\int_0^1\,\frac{u^2}{\sqrt{1-u^2}}\,\text{d}u\\&=\frac{2}{\pi}\,\left(\text{arcsin}(u)-u\,\sqrt{1-u^2}\right)\Big|_{u=0}^{u=1}\,.\end{align}$$
Ergo,
$$\sum_{n=1}^\infty\,\frac{a_n}{b_n}=1\,,$$
whence
$$1+\frac{1\cdot 3}{6}+\frac{1\cdot 3\cdot 5}{6\cdot 8}+\frac{1\cdot3\cdot 5\cdot 7}{6\cdot 8\cdot 10}+\ldots=4\,\sum_{n=1}^\infty\,\frac{a_n}{b_n}=4\,.$$
In fact, one can show that
$$f(z):=\sum_{n=0}^\infty\,\prod_{k=1}^n\,\left(\frac{k-\frac32}{k}\right)\,z^n=(1-z)^{\frac12}$$
for all $z\in\mathbb{C}$ with $|z|\leq 1$. The requested sum satisfies
$$S=8\,\Biggl(1-\frac12-f(1)\Biggr)=4\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why there are exactly 100 distinct (not necessarily primitive) Pythagorean triples $(a,b,c)$ with $c<100$? Why there are exactly 100 distinct (not necessarily primitive) Pythagorean triples $(a,b,c)$ with $c<100$?
Using the fact that all primitive Pythagorean triples can be generated by the following:
$a=2uv, b=u^2-v^2, c=u^2+v^2,$
where $u>v, u$ and $v$ are of different parity (i.e., one is even and the other odd), and $u$ and $v$ are relatively prime (i.e., their greatest common divisor is 1), then
let $a^2+b^2=c^2$ and suppose $a,b$ to both be odd. Then $a^2+b^2=1+1=2,$ and 2 is not a square root in mod 4, thus one, either $a$ or $b$ must be even. Since $gcd(a,b)=1$ then if, say $a$ is even then $b$ must be odd, meaning $c$ is also odd.
So I have solved for the $2uv$ portion but do not know how to get to the $u^2-v^2$ or $u^2+v^2$?
| Let us find all these c<100. The following formula give all Pythagorean triples:
3) $a=m^2-n^2, b=2mn, c=m^2+n^2$
$c=m^2+n^2<100$ ⇒ $m<10, n<10$ ⇒ $(m,n)= (9,2), (9, 4),(8,1), (8,3),(8,5),(7,2),(7,4),(7,6),(6,1),(6,5),(5,2),(5,4),(4,1),(4,3),(3,2),(2,1)$
That is there is 16 primitive triples.The primitives less than 50 are:
*
*$5$ which gives 18 other triples less than 100.
*$13$ which gives 6 other triples less than 100.
*$17$ which gives 4 other triples less than 100.
*$25$ which gives 2 other triples less than 100.
*$29$ which gives 2 other triples less than 100.
*$37$ which gives 1 other triple less than 100.
*$41$ which gives 1 other triple less than 100.
That is we have 34 triples other than primitives, so number on triples less than 100 is:
$$2\times(34+16)=100$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does the fact that $x^2=(x-1)(x+1)+1$ have a name? Just curious about this pattern
$$x^2 = (x-1)(x+1) +1$$
So:
$$\begin{align}
1^2 &= \phantom{1}0\cdot\phantom{1}2+1 = 1 \\
2^2 &= \phantom{1}1\cdot\phantom{1}3+1 = 4 \\
3^2 &= \phantom{1}2\cdot\phantom{1}4+1 = 9 \\
4^2 &= \phantom{1}3\cdot\phantom{1}5+1 = 16 \\
9^2 &= \phantom{1}8\cdot10+1 = 81 \\
15^2 &= 14\cdot16+1 = 225
\end{align}$$
and so on.
Then, to know any number raised to the power of $2$, you can multiply the previous number $(x-1)$ by the next one $(x+1)$, and add $1$.
So, to know the squared root of a number like $64$, you have to substract $1$ ($63$) and look for two numbers multiplied are $63$ and subtracted are $2$:
$$x \cdot y = 63 \qquad x - y = 2$$
Solving the equation you get $9$ and $7$ (or $-7$ and $-9$). The number between these is the square root ($8$).
I don't know if this apply for power of $3$.
Does this fact/theorem/relation has a name or something?
| The equation in your pattern is $x^2 = (x-1)(x+1) + 1$.
Both sides of the equation evaluate to $x^2$. It's just written differently on the right side. Expand the right side (using FOIL) and you get
$$(x-1)(x+1) + 1$$
$$= x^2 - x + x -1 + 1$$
$$= x^2 $$
This is basic algebra. No special theorems are involved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The line $DN$ bisects the line segment $AC$ if $AD=BC$. Consider a circle with diameter $AB$. A point $D$ on the circle is chosen arbitrarily such that $D\ne A,B$. A point $C\in AB$ is also chosen arbitrarily such that $C\ne A,B$. Draw $CH$ perpendicular to $AD$ at $H$. The internal angular bisector of $\angle DAB$ intersects the circle at $E$, and intersects $CH$ at $F$. The line $DF$ intersects circle again at $N$.
Prove that the points $N$, $C$, and $E$ are collinear. If $AD=BC$, then prove that $DN$ intersects $AB$ at the midpoint $I$ of $AC$.
I proved $N,C,E$ are collinear by cyclic quadrilateral and two congruent triangles. But I have no idea for the second problem.
| This is a trigonometric solution, so it is not very nice.
In this part of my answer, I shall prove the first part of the problem.
Let $\theta:=\angle ABD$, $\alpha:=\angle HFD$, and $\beta:=\angle BEC$. Write $d$ for the diameter of the circle (i.e., $d=AB$), and $l$ for the length of $BC$. We see that
$$CH\parallel BD\text{ so that }DH=l\sin(\theta)\,.$$
Now, $\angle DAE=\angle EAB=\dfrac{\pi}{4}-\dfrac{\theta}{2}$. Hence,
$$\angle AFD=\frac{\pi}{4}+\frac{\theta}{2}+\alpha\,.$$
Since $AD=d\sin(\theta)$, we get
$$\frac{DF}{\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\frac{d\sin(\theta)}{\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha\right)}$$
via the Law of Sines on the triangle $ADF$. Thus,
$$\sin(\alpha)=\frac{DH}{DF}=\frac{l\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha\right)}{d\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}\,.\tag{1}$$
Note that $\angle AEB=\dfrac{\pi}{2}$, so
$$BE=d\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\,.$$
Applying the Law of Sines on the triangle $BEC$ yields
$$\frac{l}{\sin(\beta)}=\frac{d\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\beta\right)}\,,\tag{2}$$
noting that $\angle EBD=\angle EAD=\dfrac{\pi}{4}-\dfrac{\theta}{2}$. Consequently,
$$\frac{\sin(\beta)}{\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\beta\right)}=\frac{l}{d\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\frac{\sin(\alpha)}{\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha\right)}\,,$$
using (1) and (2). Using the trigonometric identity $\sin(x)\sin(y)=\dfrac{1}{2}\big(\cos(x-y)-\cos(x+y)\big)$, we obtain
$$\cos\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha-\beta\right)=\cos\left(\frac{\pi}{4}+\frac{\theta}{2}-\alpha+\beta\right)\,.$$
Hence,
$$\sin\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\sin(\alpha-\beta)=0\,.$$
Because $0<\theta<\dfrac{\pi}{2}$, we get $\sin\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right)>0$, making $\sin(\alpha-\beta)=0$. As $\alpha$ and $\beta$ belong to the interval $\left(0,\dfrac{\pi}{2}\right)$, the condition $\sin(\alpha-\beta)=0$ implies that $\alpha=\beta$.
Finally. we see that $\angle AEB=\dfrac{\pi}{2}$, so
$$\angle AEC=\frac{\pi}{2}-\beta=\frac{\pi}{2}-\alpha=\angle ADN=\angle AEN\,.$$
This shows that $E$, $C$, and $N$ are collinear.
I shall now prove the second part of the problem. In fact, I shall prove a stronger result that $DN$ meets $AB$ at the midpoint $I$ of $AC$ if and only if $AD=BC$.
Let $G$ be the point of intersection between $DN$ and $AB$. Using the Law of Sines on the triangle $GFC$ noting that $\angle GFC=\alpha$ and $\angle GCF=\theta$, we get
$$GC=GF\left(\frac{\sin(\alpha)}{\sin(\theta)}\right)\,.$$
We have
$$\frac{GF}{GD}=\frac{AG}{AG+AD}=\frac{AG}{AG+d\sin(\theta)}$$
by the Angle Bisector Theorem. Therefore, $G=I$ if and only if $AG=GC$, which is equivalent to
$$\frac{\sin(\theta)}{\sin(\alpha)}=\frac{GD}{AG+d\sin(\theta)}\,.$$
The Law of Sines on the triangle $AGD$ gives
$$AG=AD\left(\frac{\cos(\alpha)}{\sin(\alpha+\theta)}\right)=\frac{d\sin(\theta)\cos(\alpha)}{\sin(\alpha+\theta)}$$
and
$$GD=AD\left(\frac{\cos(\theta)}{\sin(\alpha+\theta)}\right)=\frac{d\sin(\theta)\cos(\theta)}{\sin(\alpha+\theta)}\,.$$
That is, $G=I$ iff
$$\frac{\sin(\theta)}{\sin(\alpha)}=\frac{\cos(\theta)}{\cos(\alpha)+\sin(\alpha+\theta)}\,.\tag{3}$$
Note that
$$\begin{align}
\cos(\alpha)+\sin(\alpha+\theta)&=\sin\left(\frac{\pi}{2}-\alpha\right)+\sin(\alpha+\theta)
\\&=2\sin\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}-\alpha\right)
\\&=2\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha\right)\,,
\end{align}$$
where we have implemented the identity $$\sin(x)+\sin(y)=2\sin\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x-y}{2}\right)\,.$$
Furthermore, from $\sin(2x)=2\sin(x)\cos(x)$, we have
$$\begin{align}\cos(\theta)&=\sin\left(\frac{\pi}{2}-\theta\right)\\&=2\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\,.\end{align}$$
That is, (3) is equivalent to
$$\sin(\alpha)=\sin(\theta)\left(\frac{\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha\right)}{\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}\right)\,.$$
Combining the last result with (1), we conclude that $G=I$ if and only if $$BC=l=d\sin(\theta)=AD\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A specific homogeneous polar differential equation In an assignment of our school, we are asked to solve
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x + y - 3}{x - y - 1}$$
by turning it into a homogeneous polar differential equation (equation of the form $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = F\left(\frac{y}{x}\right)$) using substitutions $x = X + a$, $y = Y + b$. My solution was:
Firstly, I determined substitutions $x = X + 2$, $y = Y + 1$, such that
$$\frac{\mathrm{d}Y}{\mathrm{d}X} = \frac{X + Y}{X - Y}$$
Then, let $Y = Xv$, thus
$$\begin{aligned}
v + X\frac{\mathrm{d}v}{\mathrm{d}X} &= \frac{X + Xv}{X - Xv} \\
\int \frac{1 - v}{1 + v^2}\ \mathrm{d}v &= \int \frac{\mathrm{d}X}{X}
\end{aligned}$$
The left-hand side, specifically, gives
$$\int \frac{\mathrm{d}v}{1 + v^2} - \int \frac{v}{1 + v^2}\ \mathrm{d}v = \tan^{-1} v - \frac{\ln \left(1 + v^2\right)}{2} + \mathrm{constant}$$
Therefore,
$$\tan^{-1} v - \frac{\ln \left(1 + v^2\right)}{2} = \ln \left|X\right| + \mathrm{constant}$$
i.e.
$$2\tan^{-1} \frac{y - 1}{x - 2} = \ln \left[1 + \frac{\left(y - 1\right)^2}{\left(x - 2\right)^2}\right] + \ln \left(x - 2\right)^2 + \mathrm{constant}$$
However, our assignment didn't come with a standard solution, so I verified my answer with Wolfram Alpha, which gives
$$
2 \tan^{-1}\left(\frac{y(x) + x - 3}{-y(x) + x - 1}\right) = c_1 + \ln\left(\frac{x^2 + y(x)^2 - 2 y(x) - 4 x + 5}{2 \left(x - 2\right)^2}\right) + 2 \ln\left(x - 2\right)$$
which is different from my solution in
*
*the fraction inside function $\tan^{-1}$ is vastly different
*the denominator given by Wolfram Alpha inside the first $\ln$ is twice the denominator I gave
*the $x - 2$ in the last $\ln$ has no absolute value sign around it, but this seems a common problem of Wolfram Alpha solutions, so we can overlook it for the second
May I know whether I'm wrong, or that this is a problem of the Wolfram Alpha solution? (or that the two solutions are actually equivalent, though seemingly very unlikely?)
| One way to solve
$$
\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x+y-3}{x-y-1}
$$
is to let $u=x-2$ and $v=y-1$. Then we get
$$
\frac{\mathrm{d}v}{\mathrm{d}u}=\frac{u+v}{u-v}
$$
which becomes
$$
u\,\mathrm{d}v-v\,\mathrm{d}u=\tfrac12\mathrm{d}\!\left(u^2+v^2\right)
$$
Dividing by $u^2+v^2$ yields
$$
\frac{u\,\mathrm{d}v-v\,\mathrm{d}u}{u^2+v^2}=\frac12\frac{\mathrm{d}\!\left(u^2+v^2\right)}{u^2+v^2}
$$
which is
$$
\mathrm{d}\arctan\left(\tfrac vu\right)=\mathrm{d}\log\sqrt{u^2+v^2}
$$
and thus,
$$
c\,e^{\arctan\left(\frac{y-1}{x-2}\right)}=\sqrt{(x-2)^2+(y-1)^2}
$$
This is the Logarithmic Spiral $r=ce^{\theta}$ centered at $(2,1)$.
A More General Approach
If
$$
\frac{\mathrm{d}y}{\mathrm{d}x}=f\!\left(\frac yx\right)
$$
then
$$
\begin{align}
\frac{\mathrm{d}\frac yx}{\mathrm{d}x}
&=\frac{x\frac{\mathrm{d}y}{\mathrm{d}x}-y}{x^2}\\
&=\frac{f\!\left(\frac yx\right)-\frac yx}{x}
\end{align}
$$
Therefore,
$$
\int\frac{\mathrm{d}\frac yx}{f\!\left(\frac yx\right)-\frac yx}
=\log(x)
$$
Applying this to the question, after $u=x-2$ and $v=y-1$,
$$
\begin{align}
\log(u)
&=\int\frac{\mathrm{d}\frac vu}{\frac{1+\frac vu}{1-\frac vu}-\frac vu}\\
&=\int\frac{\left(1-\frac vu\right)\mathrm{d}\frac vu}{1+\left(\frac vu\right)^2}\\[9pt]
&=\arctan\left(\frac vu\right)-\frac12\log\left(1+\left(\frac vu\right)^2\right)+C
\end{align}
$$
Therefore,
$$
\sqrt{u^2+v^2}=c\,e^{\arctan\left(\frac vu\right)}
$$
| {
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"url": "https://math.stackexchange.com/questions/2926293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find $\lim_{x\rightarrow 0} \frac{5^x-3^x}{3^x-2^x}$ I want to find the limit $$\lim_{x\rightarrow 0} \frac{5^x-3^x}{3^x-2^x}$$
My efforts:
$$\lim_{x\rightarrow 0} \frac{5^x-3^x}{3^x-2^x}=\lim_{x\rightarrow 0}\frac{5^x((3/5)^x-1)}{3^x((2/3)^x-1)}$$
Multiplying and dividing numerator and denominator by $x$ we get, $$\lim_{x\rightarrow 0}\frac{5^x\frac{((3/5)^x-1)}{x-0}}{3^x\frac{((2/3)^x-1)}{x-0}}\tag{1}$$
Let,
$$f(x)=(3/5)^x-1, g(x)=(2/3)^x-1$$
Now $$\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=f'(x)$$
and similarly for $g$, we can rewrite $(1)$ as $$\lim_{x\rightarrow 0} \frac{5^x}{3^x}\times \frac{f'(0)}{g'(0)}\tag{2}$$
We know if $h(x)=a^x,$ then $h'(x)=a^x \log(a)$
Computing and putting everything in piece, we write $(2)$ as
$$\lim_{x\rightarrow 0}\frac{\log(3/5)}{\log(2/3)}$$
So we get limit equal to $\frac{\log(3/5)}{\log(2/3)}$.
Is my computation correct?
| You may directly use the standard limit $\lim\limits _{x\to 0}\dfrac{a^x-1}{x}=\log a$ and write the expression under limit as $$\dfrac{\dfrac {5^x-1}{x}-\dfrac{3^x-1}{x}}{\dfrac{3^x-1}{x}-\dfrac{2^x-1}{x}} $$ and get the limit as $$\frac{\log 5-\log 3}{\log 3-\log 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Proving formula sum of product of binomial coefficients I have to proof the following formula
\begin{align}
\sum_{k=0}^{n/2} {n\choose2k} {2k\choose k} 2^{n-2k} = {2n\choose n}
\end{align}
I tried to use the fact that ${2n\choose n} = \sum_{k=0}^{n} {n\choose k}^2$, but I don't get any conclusion. Any suggestions? Thanks in advance!
| Starting from
$$\sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} {2k\choose k} 2^{n-2k}$$
we write
$${n\choose 2k} {2k\choose k} =
\frac{n!}{(n-2k)! \times k! \times k!}
= {n\choose k} {n-k\choose k}$$
to obtain
$$\sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose k} 2^{n-2k} {n-k\choose n-2k}
\\ = \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose k} 2^{n-2k} [z^{n-2k}] (1+z)^{n-k}
\\ = [z^n] \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose k} z^{2k} (1+z)^{n-k} 2^{n-2k}.$$
Now when $2k\gt n$ there is no contribution to the coefficient
extractor and we may write
$$ [z^n] 2^n (1+z)^n \sum_{k\ge 0}
{n\choose k} z^{2k} (1+z)^{-k} 2^{-2k}
\\ = [z^n] 2^n (1+z)^n (1+z^2/(1+z)/2^2)^n
\\ = \frac{1}{2^n} [z^n] (2^2+2^2z+z^2)^n
= \frac{1}{2^n} [z^n] (z+2)^{2n}
= \frac{1}{2^n} {2n\choose n} 2^n = {2n\choose n}.$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find real $x$ that satisfies $\frac{1}{1+x}>\frac{x}{x-1}$ $\dfrac{1}{1+x}>\dfrac{x}{x-1}$
Multiply both sides with $(x+1)(x-1)$.
$x-1>x(x+1)$
Subtract $(x-1)$ from both sides.
$0>x^2+1$
This seems to have no real answers, although I have been told there should be. What am I doing wrong?
| The direction of an inequality changes if you multiply by something negative, so a common tactic is to multiply by a square. Multiplying by $(x^2-1)^2$ gives $(x-1)(x^2-1)\ge x(x+1)(x^2-1)$, i.e. $(1+x^2)(1-x^2)\ge 0$. (The $>$ has become $\ge$ in case $x^2-1=0$, although in this case the original problem precludes $x=\pm 1$ due to the divisions by $x\mp 1$.) Cancelling a positive factor, $1-x^2\ge 0$ i.e. $-1\le x\le 1$. As already noted, we have to drop the cases $x=\pm 1$, giving $-1<x<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Finding coefficients in polynomials Find $p$ and $q$, if $x^3-2x^2+p+q$ is divided by $x^2+x-2$.
I have found $x =-2$ and $x=1$ however when placed into polynomial I am unable to find a value for $p$ and $q$.
| One can also reduce $x^3 - 2 x^2 + p x + q$ modulo $x^2 + x - 2$ and then demand that the resulting polynomial is identical to zero. This then means that all the coefficients of that polynomial are zero, and that yields the solution. Modulo $q(x) = x^2 + x - 2$ we have:
$$x^2 \bmod q(x)= -x + 2$$
and
$$x^3 \bmod q(x)= \left(-x^2 + 2 x\right) \bmod q(x) = 3 x-2 $$
Therefore:
$\left(x^3 - 2 x^2 + p x + q\right) \bmod q(x) = (5+p) x + q-6$
which implies that $p = -5$ and $q = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
How we can show this: $\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}=\sqrt{6-2\sqrt{6}}$ Only way in my mind to show that : $$\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}=\sqrt{6-2\sqrt{6}}$$ is to multiply $\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}$ by the conjugate factor which is $\sqrt{3+\sqrt{3}}+\sqrt{3-\sqrt{3}}$ such that we can get :$\frac{2\sqrt{3}}{\sqrt{3+\sqrt{3}}+\sqrt{3-\sqrt{3}}}$ then no simplification I can do after that , any way ?
| Note that $3 + \sqrt{3} > 0$ obviously and thus $\sqrt{3+\sqrt{3}} >0$. Again, obviously $\sqrt{3-\sqrt{3}} < \sqrt{3+\sqrt{3}}$ and thus the LHS is positive.
Also, $6-2\sqrt{6} >0 \Leftrightarrow 3 > \sqrt{6}$ which holds, as $3 = \sqrt{9} > \sqrt{6}$ as $f(x) = \sqrt{x}$ is a strictly increasing function.
This is important to show for a rigorous proof, as proceeding by squaring in another case would be wrong.
Now :
$$\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}=\sqrt{6-2\sqrt{6}} \implies \bigg(\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}\bigg)^2=\bigg(\sqrt{6-2\sqrt{6}}\bigg)^2$$
Can you now apply the standard formula $(a+b)^2 = a^2 + 2ab + b^2$ to show that the desired equality formula indeed holds ?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Finding all $z \in \Bbb C$, expressed in the form $z = a + ib$, with $a, b \in \Bbb R$, satisfying equation $z^3 = -i$ I'm trying to find all $z \in \Bbb C$, expressed in the form $z = a + ib$, with $a, b \in \Bbb R$, satisfying equation $z^3 = -i$.
I've figured out that
$$(a + ib)^3 = -i \Longleftrightarrow a + ib = \sqrt[3]{-i} \Longleftrightarrow a + ib = -i$$
and also that
$$(a+ib)^3 = a^3 + 3a^2bi + 3ab^2i^2 +i^3b^3$$
but I'm not sure how to proceed from here.
I think I'm supposed to use synthetic division to find the other roots, but not sure what factor to use. I tried $a + ib + i$ and that didn't work.
Any ideas?
| Let $z=a+ib$. You have
\begin{align*}
(a+ib)^3 & = -i\\
a^3 + 3a^2bi - 3ab^2 -ib^3 & =-i\\
(a^3-3ab^2)+i(3a^2b-b^3) & =-i
\end{align*}
Equate the real and imaginary parts on both sides to get
\begin{align*}
a^3-3ab^2 =a(a^2-3b^2)& =0\\
3a^2b-b^3 =b(3a^2-b^2)& =-1
\end{align*}
From the first of these equations, we get either $a=0$ or $a^2=3b^2$.
With $a=0$, from the second equation, we get $b^3=1$. Since $b \in \mathbb{R}$, so $b=1$. This gives $\color{red}{z=i}$ as a solution.
With $a^2=3b^2$, from the second equation, we get $8b^3=-1$. Since $b \in \mathbb{R}$, so $b=-\frac{1}{2}$. Consequently we get $a=\pm\frac{\sqrt{3}}{2}$ This gives $\color{red}{z=\frac{\sqrt{3}}{2}+\frac{i}{2}}$ and $\color{red}{z=\frac{-\sqrt{3}}{2}+\frac{i}{2}}$ as other two solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Check the convergence of $\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$ and if it converges find the sum Check whether the series converges or not. If yes, find the sum of the series $$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$$
My Efforts
Observe $n^3<n^4+n^2+1$
$\therefore$
$\frac{1}{n^4+n^2+n+1}<\frac{1}{n^3}$ which further implies that $\frac{n}{n^4+n^2+n+1}<\frac{n}{n^3}=\frac{1}{n^2}$
Since $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, we conclude by comparison test that the series converges.
I am not able to find the sum of this series? Any hints or directions will be appreciated.
| We have
$$n^4+n^2+1=n^4+2n^2+1-n^2=(n^2+1)^2-n^2=(n^2+n+1)(n^2-n+1)$$
$$((n+1)^2-(n+1)+1)=(n^2+2n+1-n-1+1)=(n^2+n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to show that $\dfrac{n^3 + 2n}{3}$ is an integer Show that for each natural number $n, \dfrac{n^3 + 2n}{3}$ is an integer
My try:
Let P(n) be the statement that $n^3 + 2n$ is divisible by $3$.
Base step:
When $n = 0$ we have $0^3 + 0 = 0 = 3 \times 0$
So, the base case true.
Inductive hypothesis:
Assume that $P(k)$ is true.
which means $\dfrac{k^3 + 2k}{3}$ is divisible by $3$ and $\dfrac{k^3 + 2k}{3}=p$ for some integer $p$.
Now we need to show that $P(k+1)$ is true.
$(k+1)^3+2(k+1)$ and we will show that this divisible by $3$.
Proof:
$(k+1)^3+2(k1)=k^3+3k^2+3k+1+2k+2$
$.$
$.$
$.$
$.$
$=3(p+k^2+k+1)$
As $p+k^2+k+1$ is an integer we have that $(k+1)^3+2(k+!)$ is divisible by $3$.
Is my above attempt correct?
Did I show that for each natural number $n, \dfrac{n^3 + 2n}{3}$ is an integer?
| Or you could directly let $P'(n)$ be the proposition that "$\frac{n^3+2n}{3}$ is an integer".
Assume $P'(k)$ is true for some $k \in \mathbb N$, then consider $P'(k+1)$,
$$\begin{align*}
\frac{(k+1)^3+2(k+1)}{3} &= \frac{k^3+3k^2+3k+1 + 2k + 2}3\\
&= \frac{k^3+2k}3 + k^2+k+1\\
&\in \mathbb Z
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find the limit of $(\frac{1}{2} \frac{3}{4} ...... \frac{2n -1}{2n})$? Find the limit of $(\frac{1}{2} \frac{3}{4} ...... \frac{2n -1}{2n})$?
I have calculated it through calculating the limit of the general term $a_{n} = \frac{2n -1}{2n}$ and it was equal to 1, am I correct?
| I. The limit of $a_n$
$$
a_n=\frac{2n-1}{2n}\implies a_n=1-\frac{1}{2n}
$$
$$\displaystyle \lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}(1-\frac{1}{2n})=1
$$
II. The limit of your product (an approximation)
$$\frac{1}{2}\cdot \frac{3}{4}\cdot …\cdot\frac{2(n-1) -1}{2(n-1)} \cdot\frac{2n -1}{2n}=(1-\frac{1}{2})(1-\frac{1}{4})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})
$$
The right side equals into:
$$
[1-0-\frac{1}{2}](1-\frac{1}{4})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})
$$
$$
[1-\frac{3}{4}+\frac{1}{2}\cdot\frac{1}{4}](1-\frac{1}{6})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})=
$$
$$[1-\frac{2}{3}-\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{6}](1-\frac{1}{8})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})=
$$
$$[1-\frac{35}{48}+\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{6}\cdot\frac{1}{8}](1-\frac{1}{10})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})=
$$
$$[1-\frac{2887}{3840}-\frac{1}{10!!}](1-\frac{1}{12})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})=
$$
$$...
$$
$$=1-A_n+(-1)^{n} \displaystyle \prod_{j=1}^{n}\frac{1}{2j}
$$
What I can say about $A_n$ at least is
$$A_n\rightarrow 1+\frac{(-1)^{n}}{2n!}
$$
Hence
$$\displaystyle \lim_{n\rightarrow\infty}(1-A_n+(-1)^{n} \displaystyle \prod_{j=1}^{n}\frac{1}{2j})=1-1 +0=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2940617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Conics based on Intercepts $\hspace {2cm}$
We know that the ellipse
$$\frac {x^2}{a^2}+\frac {y^2}{b^2}=1\tag {1}$$
intercepts the axes at $(\pm a, \pm b)$.
It is interesting to note that the parabola
$$\frac {x^2}{a^2}+\frac yb=1\tag {2}$$
intercepts the axes at $(\pm a,0),(0,b)$.
and that the line
$$\frac xa+\frac yb=1\tag{3}$$
intercepts the axes at $(a,0), (0,b)$.
Question
Are equations $(1),(2), (3)$ related to each other, given that they appear very similar in form, i.e. can they be derived from the same equation with a change in parameter, or as one parameter tends to infinity, etc?
For instance, we know that if we fix $b=a$ in $(1)$ then we get a circle.
Can $(1)$ be modified such that it becomes $(2)$ and then $(1)$?
NB Perhaps the Wiki reference on Superellipse may be useful.
| If you view the line as half of a degenerate "crossed-lines hyperbola" then the three curves can be parameterized by a variable second vertex.
Define
$$A = (2a,-b) \qquad B = (0,b) \qquad C = (-2a,-b) \qquad D = \left(-2a \delta, \frac{b}{\varepsilon}\right) \qquad E = \left(2a\delta, \frac{b}{\varepsilon}\right)$$
Note that I've moved things around a little, placing the origin halfway between the fixed vertex and the line containing the two other fixed points. (I've also made the distance between that point and line $2b$ instead of just $b$, and the distance between the two points on the line $4a$ instead of $2a$.) This is so that three special cases are easily described:
*
*$\varepsilon = \phantom{-}1$: Crossed lines through the upper fixed point.
*$\varepsilon = -1$: Parallel lines through the three fixed points.
*$\varepsilon = \phantom{-}0$. Parabola.
The specified ellipse corresponds to $\varepsilon = -1/3$.
These five points determine a conic with equation
$$\begin{align}
0 &= \left|\begin{array}{cccccc}
x^2 & y^2 & x y & x & y & 1 \\
A_x^2 & A_y^2 & A_x A_y & A_x & A_y & 1 \\
B_x^2 & B_y^2 & B_x B_y & B_x & B_y & 1 \\
C_x^2 & C_y^2 & C_x C_y & C_x & C_y & 1 \\
D_x^2 & D_y^2 & D_x D_y & D_x & D_y & 1 \\
E_x^2 & E_y^2 & E_x E_y & E_x & E_y & 1
\end{array}\right| \\[6pt]
&= \frac{32 a^2 b^2 \delta (1 + \varepsilon)}{\varepsilon^3}
\left(\;(1 - \varepsilon ) \left( - b^2 (1+\varepsilon) x^2 + 2 a^2 (y - b) (\varepsilon y - b)\right)
+ 4 a^2 \delta^2\varepsilon^2 \left( b^2 - y^2 \right)
\;\right)
\end{align}$$
Conveniently ignoring the possibility that $\delta$ could be zero (or that $\varepsilon$ could equal $0$ or $-1$), we remove the leading factor. But then we dutifully take $\delta$ to be $0$ anyway, so that points $D$ and $E$ coalesce at a vertex of the conic.
The resulting equation can be written
$$\frac{x^2}{a^2}\;\frac{1+\varepsilon}{2} \;=\; \frac{(y - b) (\varepsilon y - b)}{b^2}\tag{$\star$}$$
Let's review the special cases ...
*
*$\varepsilon = 1$: Crossed lines.
$$\frac{x^2}{a^2} = \frac{(y - b)^2}{b^2} \qquad\to\qquad y = \pm \frac{b}{a} x + b$$
*
*$\varepsilon = -1$: Parallel lines.
$$0 \;=\; (y - b) (y + b) \qquad\to\qquad y = \pm b$$
*
*$\varepsilon = 0$: Parabola.
$$\frac{x^2}{a^2} \;=\; - \frac{y - b}{b} \qquad\to\qquad y - b = -\frac{b}{a^2}x^2$$
*
*$\varepsilon = -1/3$. Ellipse centered the midpoint of $\overline{AC}$.
$$\frac{x^2}{a^2} \;=\; -\frac{(y - b) (y + 3 b)}{b^2} \qquad\to\qquad \frac{x^2}{(2a)^2} + \frac{(y+b)^2}{(2b)^2} = 1$$
| {
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"url": "https://math.stackexchange.com/questions/2946053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Conditional Probability $10$ boxes with coins, $k^{th}$ box has a prob $1/k$ of landing heads Question
There are 10 boxes in front of me, the $k^{th}$ box contains coins that lands heads with probability $\frac{1}{k}$ when flipped.
I pick a random box, then a coin from the box and flip it. The coin lands heads. What is the probability that it was from the first box?
I do it again, and it lands heads again. What is the probability that it was from the first box?
I do it again, and it lands heads again. What is the probability now that it is from the first box?
My Attempt At a Solution
Let $F$ be the event that the coin is from the first box. $P(F)=\frac{1}{10}$.
Let $K$ be the event that the coin is from the $k^{th}$ box. $P(K)=\frac{1}{10}$.
Let $N$ be the event that the coin is from the first box. $P(F)=\frac{4}{5}$.
Using the definition of conditional probability: $$P(A|B)=\frac{P(A \cap B)}{P(B)} $$
$$P(F|H)=\frac{P(F \cap H)}{P(H)} =\begin{cases} \frac{\frac{1}{10}\frac{1}{2}}{\frac{1}{2}}=\frac{1}{10} \quad \text{ for a fair coin} \\ \frac{\frac{1}{10}\frac{1}{1}}{\frac{1}{1}}=\frac{1}{10} \quad \text{ for an unfair coin with probability } \frac{1}{k}; k=1\end{cases} $$
$$P(F|(H\cap H))=\frac{P(F \cap (H\cap H))}{P(H\cap H)} \begin{cases} \frac{\frac{1}{10}\frac{1}{4}}{\frac{1}{4}}=\frac{1}{10} \quad \text{ for a fair coin} \\ \frac{\frac{1}{10}\frac{1}{1}^2}{\frac{1}{1}^2}=\frac{1}{10} \quad \text{ for an unfair coin with probability } \frac{1}{k}; k=1\end{cases} $$
$$P(F|(H\cap H\cap H))=\frac{P(F \cap (H\cap H\cap H))}{P(H\cap H\cap H)} \begin{cases} \frac{\frac{1}{10}\frac{1}{8}}{\frac{1}{8}}=\frac{1}{10} \quad \text{ for a fair coin} \\ \frac{\frac{1}{10}\frac{1}{1}^3}{\frac{1}{1}^3}=\frac{1}{10} \quad \text{ for an unfair coin with probability } \frac{1}{k}; k=1\end{cases} $$
So the probability is always $\frac{1}{5}$?
Thank you for any help/hints and your time.
| You have calculated the denominator incorrectly. We must divide by the total probability that heads was obtained from any of the boxes that could have been selected.
Let's define our variables as follows: Let $B_k$ be the event that the ball is from the $k$th box, $1 \leq k \leq 10$. Let $H$ be the event that the coin lands heads.
Since the box from which the coin is extracted is chosen at random,
$$\Pr(B_k) = \frac{1}{10}, 1 \leq k \leq 10$$
Since the probability that a coin from the $k$th box lands heads is $1/k$,
$$\Pr(H \mid B_k) = \frac{1}{k}, 1 \leq k \leq 10$$
The probability that the coin is from the first box given that it lands heads is
$$Pr(B_1 \mid H) = \frac{Pr(H \cap B_1)}{\Pr(H)}$$
where $\Pr(H)$ is the total probability that heads was obtained from any of the boxes that could have been selected. Therefore,
\begin{align*}
Pr(B_1 \mid H) & = \frac{\Pr(H \cap B_1)}{\Pr(H)}\\
& = \frac{\Pr(B_1)\Pr(H \mid B_1)}{\sum_{k = 1}^{10} \Pr(B_k)\Pr(H \mid B_k)}\\
& = \frac{\frac{1}{10} \cdot 1}{\sum_{k = 1}^{10} \frac{1}{10}\frac{1}{k}}\\
& = \frac{\frac{1}{10}}{\frac{1}{10} \sum_{k = 1}^{10} \frac{1}{k}}\\
& = \frac{1}{\sum_{k = 1}^{10} \frac{1}{k}}
\end{align*}
The probability that the coin is from the first box given that it lands heads on the first two throws is
\begin{align*}
\Pr(B_1 \mid H \cap H) & = \frac{\Pr(B_1 \cap H \cap H)}{\Pr(H \cap H)}\\
& = \frac{\Pr(B_1)\Pr(H \mid B_1)\Pr(H \mid H \cap B_1)}{\sum_{k = 1}^{10} \Pr(B_k)\Pr(H \mid B_k)Pr(H \mid H \cap B_k)}\\
& = \frac{\frac{1}{10} \cdot 1 \cdot 1}{\sum_{k = 1}^{10} \frac{1}{10} \cdot \frac{1}{k} \cdot \frac{1}{k}}\\
& = \frac{\frac{1}{10}}{\frac{1}{10}\sum_{k = 1}^{10} \frac{1}{k^2}}\\
& = \frac{1}{\sum_{k = 1}^{10} \frac{1}{k^2}}
\end{align*}
I leave the calculations and the third part of the problem to you.
| {
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"timestamp": "2023-03-29T00:00:00",
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Logarithmic equation $\log_2(x+4)=\log_{4x+16}8$ So the problem goes:
What is the product of all solutions in the equation
$$\log_2(x+4)=\log_{4x+16}8$$
The solution to this should be $31\over4$, but I got $-14$. This is what I did:
\begin{align*}
\log_2(x+4) & = \log_{4x+16}8\\
\log_2(x+4) & = \frac{1}{\log_8(4x+16)}\\
\log_2(x+4) & = \frac{1}{\frac{1}{3}\log_2 4(x+4)}\\
\log_2(x+4) & = \frac{3}{\log_2 4+ \log_2(x+4)}\\
\log_2(x+4) & = \frac{3}{2 + \log_2(x+4)}\\
(\log_2(x+4))^2+2\log_2(x+4)-3 & =0\\
t^2+2t-3 & = 0 && \text{let $t = \log_2 (x + 4)$}
\end{align*}
Then $t_1=3,\ t_2=-1$. And from here, I get $x_1=4,\ x_2=-{7\over2}$
$x_1 \cdot x_2=-14$
I'm not sure where I messed up, or if there is a solution I didn't find.
| It should be $t_1=1$ and $t_2=-3$ and the rest is true.
I got $x=-2$ or $x=-\frac{31}{8}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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closed formula for determinant Consider the following matrix
$$
\begin{equation}A_{r-1} :=
\begin{bmatrix}
\frac{1}{x_{1}} & -p & \dots & 0 &\dots &0 \\
-q & \frac{1}{x_{2}} & -p &0 & \dots & 0 \\
0 & -q & \frac{1}{x_{3}} &-p & ~... & 0 \\
0 & 0 &-q &\frac{1}{x_{4}} &-p & 0 \\
0 &\vdots & \ddots & -q & \frac{1}{x_{r-2}} & -p \\
0 &0 &0 &\dots &-q &\frac{1}{x_{r-1}}
\end{bmatrix}
\end{equation}
$$
where $x_{i},p,q \in \mathbb{R}$, $x_{i} \neq 0$ for all $i = 1,2,...,r-1$ and $r \in \mathbb{N}$, $r \geq 3$. I want to find a closed formula for $\det(A)$. For $r=3$ we have
$$
\begin{equation}
\det(A) =
\begin{vmatrix}
\frac{1}{x_1} & -p \\
-q & \frac{1}{x_2} \\
\end{vmatrix} = \frac{1 -pq(x_1 x_2)}{x_1
x_2}\end{equation} .
$$
For $r = 4$ we have
$$
\begin{equation}
\det(A) =
\begin{vmatrix}
\frac{1}{x_1} & -p & 0 \\
-q & \frac{1}{x_2} & -p \\
0 &-q &\frac{1}{x_3} \\
\end{vmatrix} = \frac{1 -pq(x_1 x_2 + x_2 x_3)}{x_1 x_2 x_3}
\end{equation} .
$$
Up to this point I think the formula is given by
$$
\det(A) = \frac{1 - pq(x_{r-1}x_{r-2} + x_{r-2}x_{r-3})}{x_{1}x_{2}...x_{r-1}}.
$$
But this is not correct. For $r = 5$ I get
$$
\begin{equation}
\det(A) =
\begin{vmatrix}
\frac{1}{x_1} & -p & 0 &0 \\
-q & \frac{1}{x_2} & -p &0 \\
0 &-q &\frac{1}{x_3} &-p \\
0 & 0 &-q &\frac{1}{x_4 } \\
\end{vmatrix} = \frac{1 -pq(x_1 x_2 + x_2 x_3 + x_3 x_4) + p^2 q^2x_1 x_2 x_3x_4}{x_1 x_2 x_3 x_4}.
\end{equation} .
$$
Has someone has an idea how to find a closed formula? Thanks for any hints!
| Since the matrix is tridiagonal, its determinant satisfies the following recurrence:
$$
\det A_n = \frac{1}{x_n} \det A_{n-1} -pq \det A_{n-2}
$$
Not a closed formula, but probably just as good.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is $\cot x = \tan (π/2 - x) $ true for any angle $x$? Is $$\cot x = \tan \Big(\frac{π}{2} - x\Big)$$ true even when $x$ is not an acute angle ?
| True for all $x$ for which $\cot(x)$ is defined since
$\tan(\frac{\pi}{2}-x)=\frac{\sin(\frac{\pi}{2}-x)}{\cos(\frac{\pi}{2}-x)}=\frac{ \sin (\frac{\pi}{2}) \cos (x) -\cos( \frac{\pi}{2}) \sin (x)}{ \cos( \frac{\pi}{2}) \cos (x)+\sin (\frac{\pi}{2}) \sin (x) } =\frac{ \cos x }{ \sin x}= \cot (x).$
| {
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Prove that $3\cdot 5^{2n+1} +2^{3n+1}$ is divisible by $17$ for all $n ∈ \mathbb{N}$
Use mathematical induction to prove that $3\cdot 5^{2n+1} +2^{3n+1}$
is divisible by $17$ for all $n ∈ \mathbb{N}$.
I've tried to do it as follow.
If $n = 1$ then $392/17 = 23$.
Assume it is true when $n = p$. Therefore $3\cdot 5^{2p+1} +2^{3p+1} = 17k $ where $k ∈ \mathbb{N} $. Consider now $n=p+1$. Then
\begin{align}
&3\cdot 5^{2(p+1)+1} +2^{3(p+1)+1}=\\
&3\cdot 5^{2p+1+2} + 2^{3p+1+3}=\\
&3\cdot5^{2p+1}\cdot 5^{2} + 2^{3p+1}\cdot 2^{3}.
\end{align}
I reached a dead end from here. If someone could help me in the direction of the next step it would be really helpful. Thanks in advance.
| I'll start with the inductive step
$$3\cdot5^{2(n+1)+1}+2^{3(n+1)+1}$$
$$3\cdot5^{2n+1+2}+2^{3n+1+3}$$
$$3\cdot25\cdot5^{2(n+1)}+8\cdot2^{3n+1}$$
We know that $$3\cdot5^{2n+1}+2^{3n+1}=17k$$
Therefore $$3\cdot5^{2n+1}=17k-2^{3n+1}$$
Plug this in:
$$25\cdot(17k-2^{3n+1})+8\cdot2^{3n+1}$$
$$425k\cdot-25\cdot2^{3n+1}+8\cdot2^{3n+1}$$
$$425k\cdot-17\cdot2^{3n+1}$$
$$17(25-2^{3n+1})$$
| {
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"source": "stackexchange",
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"answer_id": 4
} |
Find the coefficient of $x^{10}$ We have been given the following function.
$f(x)$= $x$ +$x^2$ + $x^4$ + $x^8$ + $x^{16}$ + $x^{32}$ + ...upto infinite terms
The question is as follows:
What is the coefficient of $x^{10}$ in $f(f(x))$?
I tried solving it myself and I found the answer too but the method of solving was too much time-consuming.
I had solved it manually by only considering the first four terms of $f(f(x))$. This method took me about 10 minutes.
But the problem is that this question was asked in a competitive exam called JEE which requires solving the question in max. 3-4 minutes.
So, I wanted to know if there was a faster method to solve this problem.
Thanks in advance.
| Note that we may at any point in our calculation ignore any terms of degree $11$ or higher. Therefore we may also ignore any terms whose inclusion will only lead to degree $11$ or higher terms. We have:
$$
f(f(x)) = (x + x^2 + x^4 + x^8 + \cdots) + (x + x^2 + x^4 + x^8 +\cdots)^2 \\+ (x + x^2 + x^4 + \cdots)^4 + (x + x^2 + \cdots)^8 + \cdots
$$
From here we may look at each of the brackets and simply extract the ones which lead to degree $10$, using the multinomial theorem (basically the binomial theorem) for what it's worth:
*
*$x + x^2 + x^4 + x^8 + \cdots$: no terms
*$(x + x^2 + x^4 + x^8 + \cdots)^2$: we get $2x^2\cdot x^8$
*$(x + x^2 + x^4 + \cdots)^4$: we get $6 (x)^2\cdot (x^4)^2 + 4(x^2)^3\cdot x^4$
*$(x + x^2 + \cdots)^8$: we get $28(x)^6\cdot(x^2)^2$
where I've used brackets to clarify which terms I've picked in each case.
| {
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"source": "stackexchange",
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Squares sharing a side have different colors
Let $n\geq 2$. We have $n^2$ unit squares, each square in one of three different colors (some colors may not appear). What is the largest $f(n)$ such that if there are no more than $f(n)$ squares of each color, then the squares can be put into an $n\times n$ square so that any two unit squares that share a side have different colors?
For $n=2$, obviously $f(n)=2$. With $n=3$ the answer is $5$. We can assume that the $n\times n$ square has a checkerboard pattern with $5$ blacks and $4$ whites. If two colors combine to $5$ unit squares, place them on the black squares. This covers the cases $(5,4,0),(5,3,1),(5,2,2),(4,4,1),(4,3,2)$. For $(3,3,3)$ we can find an arrangement (for example each color going down the main diagonal direction). However, if one color has $6$ unit squares, either some row must contain $3$ of these (which is clearly bad), or each row contains $2$ (which is again impossible).
| Let $n\ge 1$ and $0\le r\le g\le b$ with $r+g+b=n^2$ and $b\le\frac 12 n^2$.
Then $b\ge \frac13n^2\ge r$.
Consider a black/white checkerboard pattern with black in the top left corner. Then there are $k:=\lceil \frac{n^2}2\rceil$ black and $w:=\lfloor \frac{n^2}2\rfloor$ white fields. By the given restrictions, $b\le w$.
Label the fields first black from top left to bottom right, then white from top left to bottom right:
$$\begin{matrix}{1}&k+1&2&k+2&\cdots &\begin{cases}\lceil\frac n2\rceil\\ k+\lfloor \frac n2\rfloor\end{cases}\\
k+\lfloor \frac n2\rfloor+1& \lceil \frac n2\rceil+1&k+\lfloor \frac n2\rfloor+2&\lceil \frac n2\rceil+2&\cdots&\begin{cases}k+n\\ n\end{cases}\\
n+1&k+n+1&n+2&k+n+2&\cdots&\begin{cases}\lceil\frac n2+n\rceil\\ k+\lfloor \frac n2\rfloor+n\end{cases}\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
\cdots&\cdots&\cdots&\cdots&\cdots&k
\end{matrix} $$
Note that going two rows down always increases the label by $n$ and that the labels of adjacent squares differ by
$k$, $k-1$, $k+1$, $k+\lfloor \frac n2\rfloor$, or $k-\lceil\frac n2\rceil$, hence at least by $$\tag1k-\left\lceil\frac n2\right\rceil=\frac{n^2-n}2=w-\left\lfloor \frac n2\right\rfloor.$$
For $n\ge 3$, we have $\frac {n^2}3\le\frac{n^2-n}2$ and therefore
$$\tag2 r\le\frac{n^2-n}2.$$
Note that $(2)$ holds also when $n<3$.
Now we place blue squares on the (by label) first $b$ fields, green squares on the last $g$ fields, and red squares on the remaining $r$ fields.
*
*All blue squares are on black fields because $b\le k$, hence no two are adjacent
*All green squares are on white fields because $g\le w$, hence no two are adjacent
*Because of $(2)$, any two red square labels differ by at most $\frac{n^2-n}2-1$. It follows from $(1)$ that they are not adjacent.
We conclude that
$$f(n)\ge \left\lceil\frac12n^2\right\rceil.$$
On the other hand, we partition the $n\times n$ boards into $\lceil \frac {n^2}2\rceil$ regions, namely $\lfloor \frac{n^2}2\rfloor $ dominoes and possibly one single field. Within each of these regions, at most one blue square is allowed. Hence,
$f(n)\le \left\lceil\frac12n^2\right\rceil$ and ultimately
$$f(n)= \left\lceil\frac12n^2\right\rceil.$$
| {
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"source": "stackexchange",
"question_score": "2",
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How to prove $\sqrt{2} + \sqrt[3]{2}$ is an algebraic number I need to show that this is an algebraic number by showing that it is a solution to:
$x^{6} - 9x^{4} - 4x^{3} + 27x^{2} + 36x -23 = 0.$
I'm really struggling with this one, I tried squaring and cubing $\sqrt{2} + \sqrt[3]{2}$, but I only end up with more squares and cubes.
| The minimal polynomial of $\alpha=\sqrt{2}$ is $a^2-2$ and the minimal polynomial of $\beta=\sqrt[3]{2}$ is $b^3-2$.
$\{1,a,b,b^2,ab,ab^2\}$ is a base of $\mathbb{A}=\mathbb{Q}[a,b]/(a^2-2,b^3-2)$, hence $\alpha+\beta$ is a root of a polynomial belonging to $\mathbb{Q}[x]$ with degree $\leq 6$. Indeed in $\mathbb{A}$ the terms $(a+b)^k$ decompose as follows:
$$\begin{array}{|c|c|c|c|c|c|c|}\hline & 1 & a & b & b^2 & ab & ab^2\\
\hline 1 & 1 &&&&&\\
\hline (a+b) & 1 & 1 &&&&\\
\hline (a+b)^2 & 2 & & & 1 & 2 & \\
\hline (a+b)^3 & 2 & 2 & 6 &&& 3\\
\hline (a+b)^4 & 4 & 8 & 2 & 12 & 8 & \\
\hline (a+b)^5 & 40 & 4 & 20 & 2 & 10 & 20\\
\hline (a+b)^6 & 12 & 80 & 60 & 60 & 24 & 12\\
\hline\end{array}$$
hence by Gaussian elimination (we have seven rows and six columns)
$$ (a+b)^6 = 4+24(a+b)-12(a+b)^2+4(a+b)^3+6(a+b)^4 $$
and $\alpha+\beta$ is a root of $x^6-6x^4-4x^3+12x^2-24x-4$. This is the minimal polynomial of $\alpha+\beta$ over $\mathbb{Q}$ since the involved matrix has rank $6$: the determinant of the matrix formed by the first six rows and columns is
$$ 924 = 3\cdot 2^2\cdot\det\left(\begin{smallmatrix}0&1&2&0\\2&0&0&1\\1&6&4&0\\10&1&5&10\end{smallmatrix}\right). $$
Your polynomial cannot vanish at $\sqrt{2}+\sqrt[3]{2}$ since the conjugates of this algebraic number are $\pm\sqrt{2}+\omega^k\sqrt[3]{2}$, hence the norm of $\sqrt{2}+\sqrt[3]{2}$ is $(-4)\color{red}{\neq(-23)}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Perfect square or cube There is series like N= 3! + 4! +.....+ 64!. It is asked whether it is perfect square or cube . How to identify Whether N is perfect square or cube for any big factorial or for sum of factorial ?
| Let us define $a_n$ as
$$a_n=3!+4!+...+n!$$
Rearrange this sum as
$$a_n=3!(1+4+4\cdot 5+4\cdot 5\cdot 6+...+4\cdot5\cdot ...\cdot n)$$
Since $3!=6$ is not a perfect square or divisible by a perfect square, in order for $a_n$ to be a perfect square, the sum $1+4+4\cdot 5+...+4\cdot5\cdot ...\cdot n$ must be divisible by $6$. Note that all terms including and after the $4\cdot 5\cdot 6$ term are divisible by $6$, so the whole sum is divisible by $6$ if and only if $1+4+4\cdot 5$ is divisible by $6$. It is not divisible by $6$, so $a_n$ cannot be a perfect square.
Of course, this makes the assumption that $n\ge 6$, but you can check the cases $n=3,4,5$ for yourself since they are only finitely many.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to simplify this indices question The question:
$$ \mbox{(d)} \ \ \frac{ \frac{ 4^{ n + 1 } }{ 2^{ -n } } - 8^n }{ 2^{ 3n + 2 } } $$
And here are my workings:
$$ 2. \quad
\frac{ \frac{ 4^{ n + 1 } }{ 2^{ -n } } - 8^n }{ 2^{ 3n + 2 } }
= \frac{ 2^{ 2n + 2 } \div 2^{ -n } - 2^{ 3n } }{ 2^{ 3n + 2 } }
= \frac{ 2^{ 2n -(-n) + 2 } - 2^{ 3n } }{ 2^{ 3n + 2 } }
= \frac{ 2^{ 3n + 2 } - 2^{ 3n } }{ 2^{ 3n + 2 } }
\\ = \frac{ 2^{ 3n } \times 2^2 - 2^{ 3n } }{ 2^{ 3n } \times 2^2 }
= \frac{ 4 \times 2^{ 3n } - 2^{ 3n } }{ 4 \times 2^{ 3n } }
$$
I am unable to proceed to the next step can someone please help? Or did i did something incorrect? Btw from the left to right
| You have $\displaystyle \frac{4 \cdot 2^{3n} - 2^{3n}}{4 \cdot 2^{3n}}=\displaystyle \frac{3 \cdot 2^{3n}}{4 \cdot 2^{3n}}$
| {
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What is wrong with how I am approaching this conditional expectation problem
An individual experiences a loss due to property damage and a loss due
to bodily injury. Losses are independent and uniformly distributed on the
interval $[0,3]$. Calculate the expected loss due to bodily injury, given that at
least one of the losses is less than $1$.
This is my reasoning:
Let $X$ be the property damage loss and $Y$ be the bodily injury loss.
The probability that at least one of the losses is less than $1$ is equal to the sum of the following three probabilities:
$$P(X < 1 , Y > 1) + P(X > 1 , Y < 1) + P(X < 1 , Y < 1)$$
From independence
$${1\over 3} \cdot {2\over 3} + {2\over 3} \cdot {1\over 3} + {1\over 3} \cdot {1\over 3} = {5 \over 9}$$
The probability distribution of $Y$ given that at least one of the losses is less than $1$ is the three cases divided by the sum of all the probabilities that at least one of the losses is less than $1$.
$${P(X < 1 , Y > 1)\over {5 \over 9}} , {P(X > 1 , Y < 1)\over {5 \over 9}} ,\ \text{and} \ {P(X < 1 , Y < 1)\over {5 \over 9}}$$
$${{1\over 3} \cdot {2\over 3}\over {5 \over 9}} , {{2\over 3} \cdot {1\over 3}\over {5 \over 9}} ,\ \text{and} \ {{1\over 3} \cdot {1\over 3}\over {5 \over 9}}$$
Finally, the expectation of $Y$ is given by integrating and summing the distributions
$$\int_1^3y{{1\over 3} \cdot {2\over 3}\over {5 \over 9}}dy + \int_0^1y{{2\over 3} \cdot {1\over 3}\over {5 \over 9}}dy + \int_0^1y{{1\over 3} \cdot {1\over 3}\over {5 \over 9}}dy = 1.9$$
However, the correct solution is $1.1$, what is wrong with my approach?
| In your first case the conditional expectation is obviously $2$, in the second $\frac12$ and in the third $\frac12$, making the overall conditional expectation $$2\cdot{{1\over 3} \cdot {2\over 3}\over {5 \over 9}} + \frac12\cdot{{2\over 3} \cdot {1\over 3}\over {5 \over 9}} + \frac12\cdot{{1\over 3} \cdot {1\over 3}\over {5 \over 9}} = 1.1 $$
In more detail, as BGM says in a comment, your error was in not using the conditional density in your integrals. In the first case the conditional distribution is uniform over $[1,3]$ so the density is $\frac{1}{3-1}=\frac12$. In the second and third cases the conditional distribution is uniform over $[0,1]$ so the density is $\frac{1}{1-0}=1$. So your expression should have been $$\int_1^3 y\cdot \frac12 \, dy \cdot {{1\over 3} \cdot {2\over 3}\over {5 \over 9}} + \int_0^1 y\cdot 1\, dy \cdot {{2\over 3} \cdot {1\over 3}\over {5 \over 9}} + \int_0^1 y\cdot 1 \, dy \cdot {{1\over 3} \cdot {1\over 3}\over {5 \over 9}} = 1.1$$ where each integral gives the conditional expectation for that case
| {
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Show that $(1+x)(1+z) \leq (1-(x+z))^-1$, when x,z > 0 and (x+z) < 1 I think my proof is wrong because I don't need the $(x+z) < 1$ in my argumentation.
Can somebody tell me where I made a mistake in my transformations?
$$(1+x)(1+z) \leq (1-(x+z))^{-1}$$ <=>
$$(1-(x+z))(1+x)(1+z) \leq 1$$ <=>
$$(1+x)(1+z) - (x+z)(1+x)(1+z) \leq 1$$ <=>
$$(1+x+z+xz) - (1+x+z+xz)(x+z) \leq 1$$ <=>
$$(1+x+z+xz) - (1+(x+z)+xz)(x+z) \leq 1$$ <=>
$$(1+x+z+xz) - ((1+(x+z)+xz)(x+z)) -1 \leq 0$$ <=>
$$1+(x+z)+xz - ((x+z)(x+z)^2(xz(x+z))) {-1} \leq 0$$ <=>
$$xz - (x+z)^2 - (xz(x+z)) \leq 0$$ <=>
$$xz - x^2 - 2xz - z^2 - (xz(x+z)) \leq 0$$ <=>
$$-xz - x^2 - z^2 - (xz(x+z)) \leq 0$$
This is true because x,z are positive numbers.
| By Expanding we get
$$-x^2z-xz^2-x^2-z^2-xz+1\le 1$$
| {
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Help find the mistake in this problem of finding limit (using L'Hopital)
Evaluate $$\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right).$$
Attempt
\begin{align*}
&\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right)\\
= &\lim_{x \to 0} \left(\frac{1}{x}-\cot{x}\right)\left(\frac{1}{x}+\cot{x}\right)\\
= &\lim_{x \to 0} \left(\frac{\sin{x}+x\cos{x}}{x\sin{x}}\right)\left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right)\\
= &\lim_{x \to 0} \left(\frac{\sin{x}+x\cos{x}}{x\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right).
\end{align*}
Both the terms are in $\frac00$ form. So applying L'Hopital on both the limits we have,
$$= \lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{x\sin{x}}{x\cos{x}+\sin{x}}\right).$$
The second term is in $\frac00$ form. So applying L'Hopital on the second limit we have,
\begin{align*}
= &\lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{x\cos{x}+\sin{x}}{2\cos{x}-x\sin{x}}\right)\\
=& \lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \left(\frac{x\cos{x}+\sin{x}}{2\cos{x}-x\sin{x}}\right)\\
=& 1
\end{align*}
The correct answer is $\dfrac23$ which can be found using series expansion. But I think I'm making a conceptual mistake in one of the above steps. Could you please point out to the specific step where I've committed a mistake in above solution?
|
Both the terms are in 0/0 form. So applying L'Hospital on both the limits we have,
$= \lim_{x \to 0} (\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}) \times \lim_{x \to 0}(\frac{x\sin{x}}{x\cos{x}+\sin{x}})$
Note that the limit $\lim_{x \to 0} (\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}})$ does not exist.
In fact, even before that $\lim_{x \to 0} (\frac{\sin{x}+x\cos{x}}{x\sin{x}})$ does not exist as well.
This is the graph of $\frac{\sin{x}+x\cos{x}}{x\sin{x}}$.
| {
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Finding value of $ \int \frac{1}{x-\sqrt{9x^2+6x+4}}dx$
Finding value of $\displaystyle \int \frac{1}{x-\sqrt{9x^2+6x+4}}dx$
Let $$\displaystyle I = \int\frac{1}{x-\sqrt{9x^2+6x+4}}dx = \int\frac{x+\sqrt{9x^2+6x+4}}{-8x^2-6x-4}dx$$
$$I = -\frac{1}{8}\int \frac{x}{x^2+3x/4+1/2}dx-\int\frac{\sqrt{9x^2+6x+4}}{8x^2+6x+4}dx$$
$$I=-\frac{1}{2}\int\frac{x}{\bigg(x+\frac{3}{8}\bigg)^2+\bigg(\frac{\sqrt{33}}{8}\bigg)^2}$$
Now substitute $\displaystyle x+\frac{3}{8}=t$ and $dx=dt$
I am struck for second part of $I$
Could some help me how to solve second part of integral $I$
| HINT:
An idea to solve the integral would be this line:
$$\int\frac{x}{\bigg(x+\frac{3}{8}\bigg)^2+\bigg(\frac{\sqrt{33}}{8}\bigg)^2}$$
$$\int\frac{x}{\bigg(x+\frac{3}{8}\bigg)^2+ \frac{33}{64}}$$
$$64\int\frac{x}{\bigg(8x + 3\bigg)^2+ 33}$$
Write $x$ as $16 \cdot \frac{1}{128}(8x + 3) −\frac38$ and split:
$$\int \left( \frac{8x + 3}{8\left((8x + 3)^2+33\right)} - \frac{3}{8\left((8x + 3)^2+33\right)}\right) \,dx$$
$$\underbrace{\frac18 \int \frac{8x + 3}{(8x + 3)^2+33}\,dx}_{I_1} - \underbrace{\frac38 \int\frac{1}{(8x + 3)^2+33}}_{I_2}$$
and continue for $I_1 $with
$$u =(8x+3)^2+33 \to dx = \frac1{16(8x+3)}\,du$$
$$I_1=\frac1{16}\int \frac{1}{u}$$
and $I_2$
$$ u=\frac{8x+3}{\sqrt{33}} \to dx=\frac{\sqrt{33}}{8}\,du $$
$$I_2 = \int \frac{\sqrt{33}}{8(33u^2 + 33)} \,du= \frac{1}{8\cdot\sqrt{33}}\int \frac{1}{u^2 + 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help with induction proof I need help with the following induction proof which I am not sure if I am doing correctly.
$$\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$$
I check for $n=1$ (Base case)
$$\frac{1}{1\cdot3}=\frac13$$
$$\frac{1}{2\cdot1+1}=\frac13$$
Now, is this the correct next step in my proof?
$$\frac{k}{2\cdot k+1}+\frac{1}{(2\cdot (k+1)-1)(2\cdot(k+1)+1)}=\frac{k+1}{2\cdot(k+1)+1}$$
I we assume it is correct for $n=k$ then it is also true for $n=k+1$ which means that the RHS must be equal to the LHS.
| For the induction step we need to prove that
$$\frac{1}{1\cdot3}+\frac{1}{1\cdot5}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1} \\\implies \frac{1}{1\cdot3}+\frac{1}{1\cdot5}+...+\frac{1}{(2n+1)(2n+3)}=\frac{n+1}{2(n+1)+1}$$
then we have
$$\frac{1}{1\cdot3}+\frac{1}{1\cdot5}+...+\frac{1}{(2n+1)(2n+3)}\stackrel{Ind. Hyp.}=\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}\stackrel{?}=\frac{n+1}{2(n+1)+1}$$
then all reduces to prove that
$$\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}\stackrel{?}=\frac{n+1}{2(n+1)+1}$$
| {
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"url": "https://math.stackexchange.com/questions/2972591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integer ordered pair of $(x,y)$ in complex algebraic equation
If $x,y \in \mathbb{Z}.$ Then the ordered pair of $(x,y)$ for which
$3x^4-2(19y+8)x^2+361y^2+2(100+y^4)+64=2(190y+2y^2)$
Try: From $$3x^4-2(19y+8)x^2++2y^4+357y^2-380y+264=0$$
For real roots, its discriminant always $\geq 0$
$$4(19y+8)^2-4\cdot 3 \cdot (2y^4+357y^2-380y+264)\geq 0$$
$$361y^2+64+304y-6y^4-1125y^2+1140y-792\geq 0$$
So $$6y^4+764y^2-1444y+728\leq 0$$
I am struck at that point. did not how to solve further
| Completing a few squares shows that your equation can be written as
$$((19y-(10+x^2))^2+2(y^2-1)^2+2(x^2-9)^2=0.$$
As all terms are nonnegative we find that $x=3$ and $y=1$.
At first I tried completing squares in a more roundabout way, yielding the expression
$$2x^4+(x^2-19y-8)^2+2(y^2-1)^2+198=684y.$$
This immediately shows that $y>0$ and that $2(y^2-1)^2+198<684y$, which implies $y<7$. This leaves you with six quartics in $x$ with integer coefficients; you can check whether they have integer roots using the rational root theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given that $f(x) = 1 - 3/(x+2) + 3/(x+2)$ while $x ≠- 2$, show that $f(x) = (x^2 + x + 1)/(x+2)^2$ What type of problem is this and what are the first step(s) needed to tackle it?
Given that:
$$f(x) = 1-\frac{3}{x + 2}+\frac{3}{(x +2)^2}, \quad \text{with }x ≠ -2$$
Show that:
$$f(x) = \frac{x^2 + x +1}{(x + 2)^2}$$
It's fairly basic, clearly, but from my attempts so far it appears that finding common denominators doesn't take you from the first form to the second.
Should something be factorised?
| No need of factorization, just evaluate the sum
$$1-\frac{3}{x + 2}+\frac{3}{(x +2)^2}=\frac{(x +2)^2-3(x +2)+3}{(x +2)^2}.$$
What do you obtain after expanding the numerator?
| {
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How to show that $x_{n+1} = \frac{x_n^4 + 1}{5x_n}$ bounded below and above by $1\over 5$ and $2$
Given a sequence
$$
\begin{cases}
x_{n+1} = \frac{x_n^4 + 1}{5x_n} \\
x_1 = 2 \\
n \in \mathbb N
\end{cases}
$$
Prove it has lower bound at $1\over 5$ and upper bound at $2$
I've tried to find a closed form for the recurrence relation, but couldn't arrive at anything. Also:
$$
x_{n+1} = \frac{x_n^4+1}{5x_n}=\frac{2(x_n^4 +1)}{2\cdot5x_n}=\frac{2}{5x_n}\cdot\frac{x_n^4+1}{2} \ge\frac{2}{5x_n}\sqrt{x_n^4\cdot1} =\frac{2x_n}{5}
$$
So I got:
$$
x_{n+1} \ge \frac{2x_n}{5} \tag1
$$
I have no ideas how to proceed. I'm not even sure it's valid to use AM-GM here.
So my main questions are:
*
*Does this recurrence have a closed form?
*What else should I try to solve the problem?
Please note this is precalculus. I'm not allowed to use calculus.
Update
Using $(x_n^2 - 1)^2 > 0$ i get the same result as in $(1)$. Expanding the terms only shows that the sequence is greater than $0$:
$$
x_{n+1} \ge 2\cdot \left(2\over 5\right)^n
$$
which is tending to $0$ with growing $n$.
Update 2
Consider the following expressions:
$$
x_1 = 2 \\
x_2 = \frac{x_1^3}{5} + \frac{1}{5x_1} \\
\dots \\
x_{n+1} = \frac{x_n^3}{5} + \frac{1}{5x_n} \\
$$
Multiply both sides of each expression by some $z$ in the power of $n$:
$$
z\cdot x_1 = 2\cdot z \\
z^2\cdot x_2 = \left(\frac{x_1^3}{5} + \frac{1}{5x_1}\right)z^2 \\
\dots \\
z^{n+1}\cdot x_{n+1} = \left(\frac{x_n^3}{5} + \frac{1}{5x_n}\right) \cdot z^{n+1} \\
$$
Now sum them up:
$$
\sum_{k=1}^{n+1}x_k\cdot z^k = 2z + {1 \over 5}\left( \sum_{k=2}^{n+1}x_{k-1}^3z^k + \sum_{k=2}^{n+1}{z^k\over 5x_{k-1}} \right) = \\
= 2z + {1\over 5z} \left(\sum_{k=1}^{n}x_k^3z^k + \sum_{k=1}^{n}{z^k\over x_k}\right)
$$
Now define:
$$
G(z) = \sum_{k=1}^{n+1}x_k\cdot z^k
$$
From this point there may be a way to express RHS in terms of $G(z)$ but i couldn't handle that.
Update 3
This goes beyond precalculus level but anyway here is another observation inspired by @amam_Abdallah.
Define $x_{n+1} = f(x_n)$ if this function have fixed points then:
$$
\overline{x} = \frac{\overline{x}^3}{5} + \frac{1}{\overline{x}} \iff \\
\iff \overline{x} = \sqrt[^3]{5\overline{x} - {1\over \overline{x}}}
$$
This equation has two solutions:
$$
\overline{x} = \sqrt{{5\over 2} \pm {\sqrt{21} \over 2}}
$$
Perhaps this will lead someone to ideas on how to use that fact.
Update 4
Some more thoughts on the sequence:
$$
x_{n+1} = \frac{1}{5}x_n^3 + \frac{1}{5x_n} = \\
= \frac{1}{5}\left(\frac{1}{5}x_{n-1}^3 + \frac{1}{5x_{n-1}}\right)^3 + \frac{1}{5x_n} = \\
= \frac{1}{5}\left(\frac{1}{5}\left(\frac{1}{5}x_{n-2}^3 + \frac{1}{5x_{n-2}} \right)^3 + \frac{1}{5x_{n-1}}\right)^3 + \frac{1}{5x_n} = \dots
$$
Can this somehow be wrapped into something in the form of $\prod \dots$ or $\sum \dots$?
| Over the interval $\left[\frac{1}{5},2\right]$ the function $f(x)=\frac{x^4+1}{5x}$ has an absolute minimum occurring at $x=3^{-1/4}$; if we take $I=\left[\frac{4}{5\cdot 3^{3/4}},2\right]$ we have $f(I)\subset I$. $f(2)=\frac{17}{10}$ and $f\circ f$ (but not $f$!) turns out to be a contraction of the metric space $J=\left[\frac{4}{5\cdot 3^{3/4}},\frac{17}{10}\right]$ since $f(f(J))\subset J$ and $\left|\frac{d}{dx}f(f(x))\right|\leq 0.94$ over $J$. By the Banach fixed point theorem, $x_n$ converges to the only fixed point of $f$ in $J$ and $x_n\in J$ for any $n\geq 1$.
| {
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Define a sequence $\{a_n\}_{n \geq 1}$ of real number by $a_1 = 2, a_{n+1} = \frac {a_n^2 +1 } {2} , \text{for}\ n \geq 1 $ Define a sequence $\{a_n\}_{n \geq 1}$ of real number by $$a_1 = 2, a_{n+1} = \frac {a_n^2 +1 } {2} , \text{for}\ n \geq 1 $$ Then prove that $$\displaystyle {\sum_{j=1}^N \frac {1}{a_j +1} < 1 }$$for every natural number N.
I don't know how to solve this problem, please give me some ideas to start in a right way to reach the solution.
This question is from R.M.O 2018 Tamilnadu region.
| I'm pretty sure I have seen this many years ago (back in the days in the discussion group sci.math). In any event, notice
$$\frac{1}{x-1} - \frac{1}{x+1} = \frac{2}{x^2-1} = \frac{1}{\frac{x^2+1}{2}-1}$$
we have
$$\frac{1}{a_j-1} - \frac{1}{a_j+1} = \frac{1}{a_{j+1}+1}$$
This leads to
$$\begin{align}\sum_{j=1}^N \frac{1}{a_j+1}
&= \sum_{j=1}^N \left(\frac{1}{a_j-1} - \frac{1}{a_{j+1}-1}\right)
= \frac{1}{a_1-1} - \frac{1}{a_{N+1}-1}\\
&< \frac{1}{a_1-1} = \frac{1}{2-1} = 1\end{align}$$
because $\displaystyle\;\frac{1}{a_{N+1}-1}$ is clearly positive.
| {
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Find $\sin \left(\frac x2\right)$ and $\cos \left(\frac x2\right)$ given $\cos(x) =\frac{-15}{17}$ and $x$ in quadrant II I'm having trouble fully understanding these two problems. I know I have to use the half angle formula for one, and the double for another. The first question goes as this
Given that $$\cos(x) =\frac{-15}{17}$$ and $x$ in quadrant II find the exact values of:
$$\sin \left(\frac x2\right)$$ and $$\cos \left(\frac x2\right)$$
For this problem the biggest question I do have is what should I input for $x$ exactly? It states $\cos(x)$ is a fraction, so should I simply add that as $x$ in $$\frac x2 $$? Or if I need to find what $$\sin\left(\frac x2\right)$$ is how could I find that?
Secondly, what would the best course of action be to see if it would be a negative or positive since in QII it could be either or.
Second is Given $$\sin(x) = \frac{-7}{25}$$ and $x$ in quadrant III, find the value of $\sin(2x)$,$\cos(2x)$,$\tan(2x)$.
This one I am kinda lost, would I need to find the value of what sin is and just plug it into what x is (For each corresponding formula.)? Or tackle it differently?
This is somewhat a loaded question, and I am sorry for that. But I do appreciate any help anyone could give.
| For your first question, use the half-angle formulas.
$$\sin\frac{x}{2} = \pm\sqrt{\frac{1-\cos x}{2}}$$
$$\cos\frac{x}{2} = \pm\sqrt{\frac{1+\cos x}{2}}$$
$\sin\frac{x}{2}$ and $\cos\frac{x}{2}$ can take positive or negative values depending on which Quadrant $\frac{x}{2}$ is in.
Since $x$ is in Quadrant $II$, then $\frac{\pi}{2} \leq x \leq \pi$. Divide everything by $2$ to get $\frac{\pi}{4} \leq \frac{x}{2} \leq \frac{\pi}{2}$.
Notice that the angle is is between $\frac{\pi}{2}$ and $\frac{\pi}{4}$, so the half-angle must be in Quadrant $I$. Therefore, $\sin \frac{\pi}{2}$ and $\cos \frac{\pi}{2}$ must take positive values. Use $\cos x = -\frac{15}{17}$.
$$\sin\frac{x}{2} = +\sqrt{\frac{1-\big(-\frac{15}{17}\big)}{2}}$$
$$\implies \sin\frac{x}{2} = +\sqrt{\frac{\frac{32}{17}}{2}}$$
$$\implies \sin\frac{x}{2} = +\sqrt{\frac{32}{34}}$$
$$\implies \sin\frac{x}{2} = +\sqrt{\frac{16}{17}}$$
$$\implies \boxed{\sin\frac{x}{2} = +\frac{4}{\sqrt{17}} = +\frac{4\sqrt{17}}{17}}$$
Do the same thing for finding its cosine value.
The second question is very similar. Use the double-angle formulas. (For cosine, any of the given equations can be used.)
$$\sin 2x = 2\sin x\cos x$$
$$\cos 2x = \cos^2 x-\sin^2 x = 2\cos^2 x-1 = \color{blue}{1-2\sin^2 x}$$
$$\tan 2x = \color{blue}{\frac{\sin 2x}{\cos 2x}} = \frac{2\tan x}{1-\tan^2 x}$$
Use $\sin x = -\frac{7}{25}$. Considering you’re given $\sin x$, the blue formulas are the best to use. (Otherwise you’d have to compute $\cos x$ and $\tan x$ first.)
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x,y,z>0$ and $x+y+z=1$ then $\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$ If $x,y,z>0$ and $x+y+z=1$,
then: $$\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$$
$$\frac{xyz}{(1-x)(1-y)(1-z)}=\frac{x}{(1-x)}\frac{y}{(1-y)}\frac{z}{(1-z)}$$
Let $\frac{x}{(1-x)}=a$, $\frac{y}{(1-y)}=b$, $\frac{z}{(1-z)}=c$
I am stuck here.
| Write $8xyz\le (x+y)(y+z)(x+z)$ and use arithmetic geometric inequality.
| {
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Finding $b$ such that $\frac{2+bi}{1-bi}=\frac7{10}+\frac9{10}i$. Where Have I Gone Wrong?
Find $b$ where
$$\frac{2+bi}{1-bi}=\frac7{10}+\frac9{10}i$$
What I did to solve this problem was to rewrite the problem as
$$\frac{2+bi}{1-bi}=\frac{7+9i}{10} \tag{1}$$
Given this, I then cross multiplied
$$(2+bi)\cdot 10=(1-bi)(7+9i) \tag{2}$$
Then to
$$20+10bi=(7+9b)+(9-7b)i \tag{3}$$
Then I got
$$7+9b=20 \quad\text{and}\quad 9-7b=10b \tag{4}$$
The solutions I got from solving these where $13/9$ and $9/17$.
This was not the correct solution, however. Can someone please show me where I have gone wrong? I don't just want an alternative solution. I'm hoping for a detailed explanation on why my method doesn't work. Thank you.
| We have $$\frac{2+bi}{1-bi}=\frac{(2+bi)(1+bi)}{1+b^2}=\frac{2-b^2+3bi}{1+b^2}=\frac{2-b^2}{1+b^2}+\frac{3bi}{1+b^2}$$
and you will get
$$\frac{2-b^2}{1+b^2}=\frac{7}{10}$$ and $$\frac{3b}{1+b^2}=\frac{9}{10}$$
| {
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Find all $2\times 2$ matrices that commute with $AX = XA$? Find all $2\times 2$ matrices that commute with $$ A = \left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$$ where $AX = XA$.
Solution and ask:
$$ X = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$
$$AX = \left( \begin{array}{cc} 1\cdot a + 1\cdot c & 1\cdot b + 1\cdot d \\ 0\cdot a + 0\cdot c & 0\cdot b + 0\cdot d \end{array} \right) = \left( \begin{array}{cc} a+c & b+d \\ 0 & 0 \end{array} \right)$$
$$XA = \left( \begin{array}{cc} a\cdot 1 + b\cdot 0 & a\cdot 1 + b\cdot 0 \\ c\cdot 1+d\cdot 0 & c\cdot 1+d\cdot 0 \end{array} \right) = \left( \begin{array}{cc} a & a \\ c & c \end{array} \right)$$
Since $AX = XA$, we obtain
$$\left( \begin{array}{cc} a+c & b+d \\ 0 & 0 \end{array} \right) = \left( \begin{array}{cc} a & a \\ c & c \end{array} \right)$$
so that $a+c = a$, $b+d = a$, $0 = c$ and $0 = c$.
Is the calculation correct?
What is the value of $a,b,c,d$ ?
$c = 0, a = ?, b = ?, d = ?$
| @Servaes answer is exactly right, but let me add one more thing: the set of all such matrices $X$ forms a subspace of the space of $2 \times 2$ matrices.
One way to see this is to add two of them and see that the resulting matrix has the same form; the other is a little more abstract: suppose that $AX = XA$ and $AY = YA$. Then $A(X+Y) = AX + AY = XA + YA = (X+Y) A$, so the set of such matrices is closed under addition. It's also closed under scalar multiplication (which you can probably show for yourself).
When you come across a subspace like this, it's not a bad idea to ask yourself, "What's the dimension of the subspace? What's a basis for it?"
You can then choose a different matrix $A$, and ask yourself whether the dimension changes when $A$ changes, and if so, what's the relationship?
I know you're not likely to pursue this if this happens to just be a homework problem and you've got others to do as well, but I offer it as a possible general approach as you advance in mathematics.
| {
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Equation of sphere circumscribing the tetrahedron bounded by planes $x+y=0$, $y+z=0$, $z+x=0$, $x+y+z=1$ I found a following question in 1st year undergrad course:
Find the equation of sphere circumscribing the tetrahedron bounded by the planes
$$x+y=0 \qquad y+z=0 \qquad z+x=0 \qquad x+y+z=1$$
Can anyone tell me in detail how to find the equation of sphere ?
I know if i get four vertices of tetrahedron then i can able find out the equation. But how to find the vertices of tetrahedron?
| There are $4$ planes
$$\begin{cases}
P_0 &: x + y + z = 1\\
P_1 &: y + z = 0\\
P_2 &: x + z = 0\\
P_3 &: x + y = 0
\end{cases}$$
To find the vertices, select $3$ planes from $P_0,P_1,P_2,P_3$ and intersect them.
Each vertex corresponds to a different way to select the planes and there are totally
$4$ selections. For $i = 0, \ldots, 3$, let $v_i = (x_i,y_i,z_i)$ be the vertex belongs to $\bigcap\limits_{j=0,\ne i}^3 P_j$.
For $v_0 \in P_1\cap P_2 \cap P_3$, it is clear $v_0 = (0,0,0)$.
For $v_1 \in P_0\cap P_2 \cap P_3$, since it belongs to $P_2 \cap P_3$, we have
$$x_1 + z_1 = 0, x_1 + y_1 = 0 \implies v_1 = (x_1, -x_1, -x_1)$$
Substitute this into the equation of $P_0$, we get $x_1 = -1 \implies v_1 = (-1,1,1)$
By a similar manner, we can determine the coordinates of $v_2, v_3$. The end result is
$$\begin{cases}
v_0 = (0,0,0)\\
v_1 = (-1,1,1)\\
v_2 = (1,-1,1)\\
v_3 = (1,1,-1)
\end{cases}$$
Given coordinates of the four vertices, the circumsphere is given by following equation:
$$\left|\begin{matrix}
1 & x & y & z & x^2 + y^2 + z^2\\
1 & x_0 & y_0 & z_0 & x_0^2 + y_0^2 + z_0^2\\
1 & x_1 & y_1 & z_1 & x_1^2 + y_1^2 + z_1^2\\
1 & x_2 & y_2 & z_2 & x_2^2 + y_2^2 + z_2^2\\
1 & x_3 & y_3 & z_3 & x_3^2 + y_3^2 + z_3^2\\
\end{matrix}\right|
= 0$$
Using the coordinates of $v_i$ derived above, we get
$$
\left|\begin{matrix}
1 & x & y & z & x^2 + y^2 + z^2\\
1 & 0 & 0 & 0 & 0\\
1 & -1 & 1 & 1 & 3\\
1 & 1 & -1 & 1 & 3\\
1 & 1 & 1 & -1 & 3\\
\end{matrix}\right|
= -
\left|\begin{matrix}
x & y & z & x^2 + y^2 + z^2\\
-1 & 1 & 1 & 3\\
1 & -1 & 1 & 3\\
1 & 1 & -1 & 3\\
\end{matrix}\right| = 0$$
This can be further simplified to
$$\bbox[border:1px solid blue;padding: 1em;]{x^2+y^2+z^2 - 3(x+y+z) = 0}$$
If you don't like evaluating a $5 \times 5$ determinant, here is another
way to determine the circumsphere.
Just like circle inversion in the plane, if you perform a sphere inversion with respect to the unit sphere in $\mathbb{R}^3$, any sphere passing through the origin will get mapped to a plane. Since one of the vertex $v_0$ is origin, the circumsphere
we seek will get mapped to a plane.
Under the sphere inversion, $v_1, v_2, v_3$ get mapped to $\frac13 v_1$, $\frac13 v_2$ and $\frac13 v_3$. It is easy to see they are lying on the plane $x + y + z = \frac13$. If we reverse the sphere inversion, we obtain following equation for the circumsphere.
$$\frac{x+y+z}{x^2+y^2+z^2} = \frac13 \iff (x^2+y^2+z^2) - 3(x+y+z) = 0$$
This is the same equation we obtained before using a $5\times 5$ determinant.
| {
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Struggle with finding equality condition in an AM-GM inquality problem Given $0<x,y,z<1$. Prove that
$$\dfrac{1-x}{1+y+z}+\dfrac{1-y}{1+z+x}+\dfrac{1-z}{1+x+y}\geq 3(1-x)(1-y)(1-z).$$
I've thought that the equality holds when $x=y=z$ and I'm thinking ways to use AM-GM inequality. However, it turns out that
$$3\dfrac{1-x}{1+2x}=3(1-x)^3 \Longleftrightarrow (1-2x+x^2)(1+2x)=1\Longleftrightarrow 2x^3-3x^2=0\Longleftrightarrow x=\dfrac{3}{2},$$
which is totally wrong due to $0<x,y,z<1$! Thank you a lot!
| It should be $$3x^2-2x^3\geq0,$$ which is true for $0<x<1.$
Let $x=\frac{a}{1+a},$ $y=\frac{b}{1+b}$ and $z=\frac{c}{1+c},$ where $a$, $b$ and $c$ are positive numbers.
Thus, we need to prove that
$$\sum_{cyc}\frac{(a+1)^2(b+1)^2}{3ab+2a+2b+1}\geq3.$$
Now, by C-S
$$\sum_{cyc}\frac{(a+1)^2(b+1)^2}{3ab+2a+2b+1}\geq\frac{\left(\sum\limits_{cyc}(ab+2a+1)\right)^2}{\sum\limits_{cyc}(3ab+4a+1)}.$$
Thus, it's enough to prove that
$$\left(\sum\limits_{cyc}(ab+2a+1)\right)^2\geq3\sum\limits_{cyc}(3ab+4a+1)$$ or
$$\sum_{cyc}(a^2b^2+2a^2bc+4a^2+5ab+4a^2b+4a^2c+4abc)\geq0,$$ which is obvious.
We can use also another C-S.
Indeed, we need to prove that:
$$\sum_{cyc}\frac{1}{(1-x)(1-y)(1+x+y)}\geq3.$$
Now, $$\sum_{cyc}\frac{1}{(1-x)(1-y)(1+x+y)}\geq\frac{9}{\sum\limits_{cyc}(1+x+y)(1-x)(1-y)}.$$
Id est, it's enough to prove that
$$3\geq\sum_{cyc}(1+x+y)(1-x-y+xy)$$ or
$$\sum_{cyc}(2x^2+xy)\geq\sum_{cyc}(x^2y+x^2z)$$ or
$$\sum_{cyc}(x^2(1-y)+x^2(1-z)+xy)\geq0,$$ which is obvious.
| {
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Solve recurring sequence using a generating function I have the sequence $a_n=3a_{n-1}-3a_{n-2}+a_{n-3}$, $\forall\ n \ge 3$, with $a_0=2$, $a_1=2$, $a_2=4$ being the known terms, and I want to find a non-recursive equation for $a_n$ using a generating function.
What I have done:
$$
\begin{align}
A(x)
& = \sum_{n\ge0}{a_nx^n}
= a_0+a_1x+a_2x^2+\sum_{n\ge 3}\left({3a_{n-1}-3a_{n-2}+a_{n-3}}\right)x^n\\
& = 2+2x+4x^2+3x\sum_{n\ge 3}{a_{n-1}x^{n-1}}
-3x^2\sum_{n\ge 3}{a_{n-2}x^{n-2}}
+x^3\sum_{n\ge 3}{a_{n-3}x^{n-3}}\\
& = 2+2x+4x^2+3xA(x)-3x^2A(x)+x^3A(x)\\
& = \frac{2+2x+4x^2}{1-3x+3x^2-x^3}\\
& = \frac{4x^2+2x+2}{(1-x)^3}
\end{align}
$$
As shown above, I have reached a solution for $A(x)$, but I'm not sure how to use it to find a solution for $a_n$. Any tips pointing me in the right direction would be greatly appreciated.
| $a_n=3a_{n-1}-3a_{n-2}+a_{n-3}$, $\forall\ n \ge 3$, with $a_0=2$, $a_1=2$, $a_2=4$ being the known
There is a theorem that said the following:
Let $q$ be a non zero number then $a_n = q^n$ is a solution of the linear homogeneous recurrence relation $a_n -c_1 a_{n-1} - a_2 a_{n-2} -\cdots - a_k a_{n-k} = 0$ with constant coefficients iff $q$ is a root of the polynomial equation $x^n - c_1 x^{n-1} - \cdots a_k = 0$ From "Introductory Combinatorics, 5th edition by Richard Brualdi". It is similar to differential equations you will find the roots if a root $r$ has a multiplicity say $k$ then the general solution $a_n = c_1 r^n + c_2 n r^n + c_3 n^2 r^n + \cdots + c_k n^k r^n $
Check the link "Roots of the characteristic polynomial"
https://en.wikipedia.org/wiki/Recurrence_relation
You can try this let
$a_n = p^n$
then
$$p^n -3p^{n-1} +3p^{n-2}-p^{n-3}=0$$ Factor $p^{n-3}$
$$p^{n-3} ( p^3 - 2p^2 + 2p - 1) = 0$$
$$p^3 -3p^2 +3p-1=0$$ Factor
$$(p-1)(p^2-2p+1)=(p-1)(p-1)^2 =0$$
$(p-1)^3=0$ hence the solution is $1$ with multiplicity $3$ so
$a_n = c_1 1^n +c_2 1^n n + c_3 1^n n^2$ But we know that $1^n =1 $ then using the given initials to find the constants $a_0 = c_1 = 2 $ etc..
$$a_n = 2 - n + n^2$$ checking $a_3=8, a_4=14$
| {
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if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then...
if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then
$$\frac{1}{1-\alpha_1} + \frac{1}{1-\alpha_2} + \frac{1}{1-\alpha_3}+\ldots+\frac{1}{1-\alpha_n} = ?$$
Now this has the solution too, but I do not understand the last step of the solution so here is the solution from the book.
$1,\, \alpha_1, \, \alpha_2,\, \ldots, \, \alpha_n$ are the $n^\text{th}$ of unity. These are the roots of $x^n-1=0$
Let $y=\frac{1}{1-\alpha}$ where $\alpha = \alpha_1, \, \alpha_2, \, \alpha_3, \, \ldots, \, \alpha_n$
$$1 - \alpha = \frac{1}{y} \Rightarrow \alpha = \frac{y-1}{y}.$$
But $\alpha$ is a root of $x^n-1=0 \therefore \alpha^n=1 \Rightarrow (y-1)^n = y^n$
$$\Rightarrow y^n - _nC_1y^{n-1} + _nC_2y^{n-2}-\ldots+(-1)^n = y^n \\
\Rightarrow - _nC_1y^{n-1} + _nC_2y^{n-2}-\ldots+(-1)^n = 0.$$
Sum of roots
$$\frac{1}{1-\alpha_1}+\frac{1}{1-\alpha_2}+\ldots+\frac{1}{1-\alpha_{n-1}} = \frac{_nC_2}{_nC_1} = \frac{n-1}{2}$$
So this last part from "Sum of roots" I do not understand. I cannot see how this last shape relates to this binomial theorem notation. Can anyone help?
| Since you already obtained some clarification on the textbook's solution, I am introducing a different solution. You can also prove by noting that the roots of $x^n-1=0$ are $1,\xi,\xi^2,\ldots,\xi^{n-1}$, where
$$\xi=e^{\frac{2\pi i}{n}}.$$
Therefore, we may take $a_k$ to be $\xi^k$ for $k=1,2,\ldots,n-1$. Now,
$$\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\xi^{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\frac{\xi^n}{\xi^k}}.$$
Since $\xi^n=1$, we obtain
$$\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\frac{1}{\xi^k}}=\frac{1}{1-\xi^k}+\frac{\xi^k}{\xi^k-1}=1.$$
Therefore,
$$2\sum_{k=1}^{n-1}\frac{1}{1-a_k}=\sum_{k=1}^{n-1}\left(\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}\right)=\sum_{k=1}^{n-1}1=n-1,$$
so
$$\sum_{k=1}^{n-1}\frac{1}{1-a_k}=\frac{n-1}{2}.$$
| {
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Solve $x^3-x-6=0$
Solve for x, $$x^3-x-6=0\tag1$$
let $x=y-2$
$$(y-2)^3-(y-2)-6=0\tag2$$
$$y^3-6y^2+11y-12=0\tag3$$
let $x=y+3$
$$(y+3)^3-(y+3)-6=0\tag4$$
$$y^3+9y^2+26y+18=0\tag5$$
$(5)-(4)$:
$$15y^2+15y+30=0\tag6$$
$$y^2+y+2=0\tag7$$
This approach, I have tried, it is not working.
How can I solve $(1)$?
| Well, you are lucky.
$x^3 - x - 6 = 0\implies$
$x^3 - x = 6\implies$
$x(x^2 - 1) = x(x+1)(x-1)=6$ and it just happens that $6 = 1*2*3$ so if you set in order $\{(x-1),x, (x+1)\} = \{1, 2,3\}$ you get consistent results that $x -1 = 1; x=2; x+1 = 3$ give consistent results and $x = 2$.
However you need to be aware that this will almost NEVER work and you were just lucky.
In general, you can use the rational root theorem that as the leading term is $x^3$ with coefficient $1$ and the constant term is $-6$. Then if there is a rational root it will be $\frac ab$ were $a|6$ and $b|1$.
So that will be $\pm 1, \pm 2, \pm 3; \pm 6$.
If you try them one after another you find
$2^3 - 2 - 6 = 0$ so $x = 2$ is one solution. So we can factor out $(x-2)$ to get
$x^3 - x -6 = x^2(x-2) + 2x^2 - x -6=$
$x^2(x-2) + 2x(x-2) + 4x - x - 6 = x^2(x-2) + 2x(x-2) + 3x - 6=$
$x^2(x-2) + 2x(x-2) + 3(x-2) = (x-2)(x^2 + 2x + 3)$
You can try the rational root thereom on $x^2 + 2x + 3$ and get that if there is a rational root it is $\pm 1, \pm 3$. But seeing as those weren't roots of $x^3 - x -6$ then won't be roots of $x^3 - x -6 = (x-2)(x^2 +2x + 3)$ so there aren't any other rational roots.
So we must find irrational roots with the quadratic equation or by completing the square.
To solve $x^2 + 2x + 3 = 0$ we have
$x^2 + 2x = -3$
$x^2 + 2x + 1 = -3 +1 = -2$
$(x + 1)^2 = -2$. But there are not solutions to that as $(x+1)^2$ can't be negative.
So $x =2$ is the only solutions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\int_{-\infty}^{\infty}dx \frac{x^2 e^x}{(e^x+1)^2}$ My physics textbook makes use of the result:
$$\int_{-\infty}^{\infty}dx \dfrac{x^2 e^x}{(e^x+1)^2} = \dfrac{\pi^2}{3}$$
I'm really curious on how I can derive this but I honestly don't know what to search for. My instinct is to transform to polar coordinates but I would like some guidance. Any help appreciated!
| First off, notice the integrand is even, so we have $$ \int_{-\infty}^\infty \frac{x^2 e^x}{(e^x+1)^2}dx = 2\int_{0}^\infty \frac{x^2 e^x}{(e^x+1)^2}dx.$$ Then we can expand $$ \frac{1}{(1+x)^2} = \sum_{n=0}^\infty (-1)^{n}(n+1) x^n $$ and write $$ 2\int_{0}^\infty \frac{x^2 e^x}{(e^x+1)^2}dx=\\ =2\int_{0}^\infty \frac{x^2 e^{-x}}{(e^{-x}+1)^2}dx \\ = 2 \int_0^\infty x^2e^{-x}\sum_{n=0}^\infty (-1)^n (n+1) e^{-nx}\\=2\sum_{n=0}^\infty (-1)^n(n+1) \int_0^\infty x^2 e^{-(n+1)x}dx\\=4\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^2}.$$ Then we have $$ \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^2} = 1-\frac{1}{2^2} + \frac{1}{3^2}\ldots = (1-\frac{2}{2^2})(1+\frac{1}{2^2} + \frac{1}{3^2}\ldots) = \frac{\pi^2}{12}$$
Edit
Realized this can be simplified somewhat by first doing an integration by parts $$ 2\int_0^\infty \frac{x^2 e^x}{(e^x+1)^2}dx = 4\int_0^\infty \frac{x}{e^x+1}dx$$ followed by a similar series expansion. Additionally, this solution somewhat 'misses the point' relative to contour integration approaches since that's one of the slicker ways to get $\sum_{n}1/n^2=\pi^2/6$ in the first place (and also transformation from the sums to integrals like this are the source of many zeta function identities).
| {
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$a\mid b^2, b\mid c^2, c\mid a^2\Longrightarrow abc\mid (a+b+c)^k$ Is it true that if $a,b,c$ are positive integers such that $a\mid b^2, b\mid c^2, c\mid a^2$, then $abc$ always divides $(a+b+c)^6$?
If not (I couldn’t find a counterexample), then $abc\mid (a+b+c)^7$ is always true? Or in general, find the minimum value of $k$, such that $abc\mid (a+b+c)^k$.
I proved that $a,b,c\mid (a+b+c)^6$, but unfortunetly that is not enough, since $a,b,c$ are not necessarily relative primes.
| The minimum qualifying value of $k$ is $7$.
To show that $k=6$ is not sufficient, let $p$ be a prime.
Then letting $(a,b,c)=(p,p^4,p^2)$, we get
\begin{align*}
abc&=p^7\\[4pt]
(a+b+c)^6&=p^6(1+p^3+p)^6\\[4pt]
\end{align*}
from which it follows that $abc$ does not divide $(a+b+c)^6$.
Next we show that $k=7$ always works . . .
Let $a,b,c$ be positive integers such that
$$a|b^2,\;\;\;b|c^2,\;\;\;c|a^2$$
Claim:$\;abc$ divides $(a+b+c)^7$.
If any of $a,b,c$ equals $1$, then they must all be equal to $1$, in which case the truth of the claim is immediate.
Thus, assume $a,b,c > 1$.
From the given divisibility conditions on $a,b,c$, it follows that $a,b,c$ have the same set of distinct prime factors.
Let $p$ be an arbitrary prime factor of $abc$.
It follows that $p$ is a common prime factor of $a,b,c$.
Letting $i,j,k$ be positive integers such that
$$p^i||a,\;\;\;p^j||b,\;\;\;p^k||c$$
it follows that $p^{i+j+k}{\,||\,}abc$, and $p^{7\min(i,j,k)}{\;\mid\,}(a+b+c)^7$.
Using the given divisibility conditions,
\begin{align*}
a|b^2\implies i\le 2j\\[4pt]
b|c^2\implies j\le 2k\\[4pt]
c|a^2\implies k\le 2i\\[4pt]
\end{align*}
so we get
\begin{align*}
{\small{\bullet}}\;\,&i+j+k\le i+(2k)+k=i+3k\le i+6i=7i\\[4pt]
{\small{\bullet}}\;\,&i+j+k\le i+j+(2i)=3i+j \le 6j+j=7j\\[4pt]
{\small{\bullet}}\;\,&i+j+k\le (2j)+j+k=3j+k\le 6k+k=7k\\[4pt]
\end{align*}
hence $i+j+k\le 7\min(i,j,k)$.
Thus the highest power of $p$ which divides $abc$ is less than or equal to the highest power of $p$ which divides $(a+b+c)^7$.
Since $p$ was an arbitrary prime factor of $abc$, it follows that $abc$ divides $(a+b+c)^7$, as claimed.
| {
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Find a limit of $\sqrt[3]{3} * \sqrt[9]{3} * ... * \sqrt[3^n]{3}$ I have no idea how can i solve this. When i'm trying to transform a multiplication I always get a $0*\infty$ ambiguity.
I found only that
$\lim_{n \to \infty}{(\sqrt[3]{3} * \sqrt[9]{3} * ... * \sqrt[3^n]{3})} = \lim_{n \to \infty}(\sqrt[3]{3 * \sqrt[3]{3 * \sqrt[3]{3*...}}})$
EDIT: Solution
$\lim_{n \to \infty}{(\sqrt[3]{3} * \sqrt[9]{3} * ... * \sqrt[3^n]{3})} = \lim_{n \to \infty}(\sqrt[3]{3 * \sqrt[3]{3 * \sqrt[3]{3*...}}}) = \lim_{n \to \infty}{3^{(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ... + \frac{1}{3^n})}}$
$\sum_{r=1}^n{\frac{1}{3^n}} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}$
So $\lim_{n \to \infty}{3^{(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ... + \frac{1}{3^n})}} = \lim_{n \to \infty}{3^\frac{1}{2}} = \sqrt{3}$
| $$\sqrt[3]{3 * \sqrt[3]{3 * \sqrt[3]{3 * ...}}} = n$$
$$\sqrt[3]{3 * n} = n$$
$$3*n=n^3$$
$$3=n^2$$
$$n=\sqrt{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}=4\sqrt{3}$ I'm trying to calculate the expression: $$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}$$ and show that it is equal $4\sqrt{3}$.
I was trying to group the summands and calculate sums of $$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}} \hspace{0.5cm}\text{and} \hspace{0.5cm} -\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}$$ where we get $$\frac{2\cos\frac{2\pi}{15}+1}{\sin\frac{2\pi}{15}}-\frac{2\cos\frac{4\pi}{15}-1}{\sin\frac{8\pi}{15}}$$ but unfortunately this sum is not simplified.
How to prove this equality?
| Hint:
Using $\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$,find $\alpha$,
$$\dfrac1{\sin12^\circ}+\dfrac1{\sin24^\circ}+\dfrac1{\sin48^\circ}+\dfrac1{\sin96^\circ}-\dfrac2{\sin48^\circ}$$
$$=\cot6^\circ-\cot96^\circ-\dfrac2{\sin48^\circ}$$
$\cot6^\circ-\cot96^\circ=\cot6^\circ+\tan 6^\circ=\dfrac2{\sin12^\circ}$
Now $\dfrac1{\sin12^\circ}-\dfrac1{\sin48^\circ}=\dfrac{\sin48^\circ-\sin12^\circ}{\sin48^\circ\sin12^\circ}=\dfrac{4\sin18^\circ\cos30^\circ}{\cos36^\circ-\cos60^\circ}$
Using Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$, $\cos36^\circ-\cos60^\circ=\sin18^\circ$
| {
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Checking whether a number is prime or not
Is it true that a natural number $n>1$ is prime if and only if $n|\left ( \frac{1+\sqrt{5}}{2} \right )^n+\left ( \frac{1-\sqrt{5}}{2} \right )^n-1$?
We know that $11$ is a prime number, but let us assume that we do not know, and let us also assume that the statement is true;
$$\left ( \frac{1+\sqrt{5}}{2} \right )^{11}+\left ( \frac{1-\sqrt{5}}{2} \right )^{11}-1=198.$$
Clearly, $11|198$. Therefore, as our assumption that the statement is true, the number $11$ is a prime number.
| Note that $\left ( \frac{1+\sqrt{5}}{2} \right )^n+\left ( \frac{1-\sqrt{5}}{2} \right )^n$ is the $n$-th Lucas number.
It is true that if $p$ is a prime then $L_p-1$ is divisible by $p$: $2$ divides $L_2-1=2$ and for any prime $p>2$,
$$L_p-1=\frac{1}{2^{p-1}}\sum_{k=1}^{(p-1)/2}\binom{p}{2k}5^k=0\pmod{p}$$
because $p$ divides each $\binom{p}{2k}$.
On the contrary $705$ is not a prime but it divides $L_{705}-1$.
See Bruckman-Lucas_pseudoprimes and compare with the Wall-Sun-Sun prime.
| {
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Help with a system of inequalities with absolute values I'm trying to solve this system on inequalities
$$
\left\{
\begin{array}{c}
|x-3|<2x \\
|2x+5|>3
\end{array}
\right.
$$
The steps I'm taking are:
Finding the absolute values sings, so for
$x-3 \geq 0$ we have $x \geq 3$ therefore
$$|x-3| =
\left\{
\begin{array}{c}
x-3 & \text{for $x \geq 3$} \\
-x+3 & \text{for $x < 3$}
\end{array}
\right. $$
and
$2x+5 \geq 0$ we have $x \geq \frac{-2}{5}$ therefore
$$|2x+5| =
\left\{
\begin{array}{c}
2x+5 & \text{for $x \geq \frac{-2}{5}$} \\
-2x-5 & \text{for $x<\frac{-2}{5}$} \\
\end{array}
\right. $$
So I build a few systems with the complete inequalities, for the first one we have:
$$
\left\{
\begin{array}{c}
x \geq 3 \\
x-3<2x = x>-3
\end{array}
\right. $$
So the solution here would be $x>3$, then:
$$
\left\{
\begin{array}{c}
x<3 \\
-x+3<2x = x>1
\end{array}
\right. $$
The solution would be $1<x<3$. Then
$$
\left\{
\begin{array}{c}
x \geq \frac{-2}{5} \\
2x+5>3 = x>-1
\end{array}
\right. $$
So the solution of the system is $x>-1$, then
$$
\left\{
\begin{array}{c}
x< \frac{-2}{5} \\
-2x-5>3 = x<-4
\end{array}
\right. $$
And the solution is $x<-4$
Now the solution my book gives is x>1 for the initial system. But I can't find that one. I can't get a solution at all. I tried finding a common point between the 4 solutions I found (as if it was a 4-inequalities system), but there isn't one really. What am I doing wrong?
| We have $$|x-3|<2x\implies 3x^2+6x-9>0\implies (x+3)(x-1)>0$$ and $$|2x+5|>3\implies 4x^2+20x+16>0\implies (x+4)(x+1)>0$$. I think you could proceed now? For a quick method on how to solve further, see here
| {
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What is the limit of $\lim_{n\to \infty} (1 - \frac{1}{4})(1 - \frac{1}{9})(1 - \frac{1}{16}) \cdots (1 - \frac{1}{(n+1)^2})$? What is the evaluation of the following infinite series?
$$\lim_{n\to \infty} \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right) \cdots \left(1- \frac{1}{(n+1)^2}\right)$$
I've tried to simplify each expression which left me with:
$$\lim_{n\to \infty} \frac{3\times8\times15\times24\times\cdots\times((n+1)^2-1)}{4\times9\times16\times25\times\cdots\times(n+1)^2}$$
Is this a good way to approach this problem?
| $$\prod_{k=1}^n\dfrac{f(k)}{f(k+1)}=\dfrac{f(1)}{f(n+1)}$$
Here $f(m)=\dfrac m{m+1}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of a sequence with a difference of cosines Define the following sequence:
$$x_n = \sqrt[3]{n+1}\cos{\sqrt{n+1}} - \sqrt[3]{n}\cos{\sqrt{n}},\forall n\in \mathbb{N} $$
Does the limit
$$\lim_{n \to \infty} x_n$$
exist?
I believe that it does not, mainly because (i think that) $\cos\sqrt{n+1}$ and $\cos{\sqrt{n}}$ have opposite signs an infinite number of times.
More exactly, I tried showing that there exists an infinite sequence $a_n$ of natural numbers such that
$$ \sqrt{a_n} \text{ mod } 2\pi \in \left(\frac{3\pi}{2} + \frac{\pi}{4} ,2\pi\right) \text{ and } \sqrt{a_n + 1} \text{ mod } 2\pi \in \left( 0, \frac{\pi}{4} \right)$$
and a sequence $b_n$ such that
$$\sqrt{b_n} \text{ mod } 2\pi \in \left(\frac{\pi}{2} + \frac{\pi}{4}, \pi \right) \text{ and } \sqrt{b_n + 1} \text{ mod }\in \left(\pi, \pi + \frac{\pi}{4} \right),$$
because then the subsequences $x_{a_n}$ and $x_{b_n}$ would both be bounded and of different signs, showing that $x_n$ is divergent.
Would this approach work? Is the sequence actually somehow convergent?
| By binomial series
$$\sqrt[3]{n+1}=\sqrt[3]{n}\left(1+\frac1n\right)^\frac13=\sqrt[3]{n}+O\left(\frac1{n^\frac23}\right)$$
then
$$\sqrt[3]{n+1}\cos{\sqrt{n+1}} - \sqrt[3]{n}\cos{\sqrt{n}}=$$$$=\sqrt[3]{n}(\cos{\sqrt{n+1}}-\cos{\sqrt{n}})+O\left(\frac{\cos{\sqrt{n+1}}}{n^\frac23}\right)$$
and by sum to product formula
$$\cos{\sqrt{n+1}}-\cos{\sqrt{n}}=-2\sin\left(\frac{\sqrt{n+1}+\sqrt{n}}{2}\right)\sin\left(\frac{\sqrt{n+1}-\sqrt{n}}{2}\right)=$$
$$=-2\sin\left(\frac{\sqrt{n+1}+\sqrt{n}}{2}\right)\sin\left(\frac{1}{2(\sqrt{n+1}+\sqrt{n})}\right)\sim -\frac{\sin(\sqrt n)}{\sqrt n}$$
therefore the given sequence converges to zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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Domain of functions involving arcsine or arccosine I have two rather simple functions. I don't know how to find that the dominator is different than 0. Find the domain of functions:
Function 1:
$f(x) = \frac{4-x}{\arcsin\frac{x}{4}}$
Assumption 1:
$-1\leq \arcsin\frac{x}{4} \leq 1$
$-1\leq \frac{x}{4} \leq 1$
$-4\leq x \leq 4$
$x\in<4;4>$
Assumption 2:
$\arcsin\frac{x}{4} \neq 0$
Here, I have no idea how to proceed further with assumption 2.
Function 2:
$f(x) = \frac{\sqrt{2x-1}}{2+\arccos\frac{x+1}{4}}$
Assumption 1:
$\sqrt{2x-1} \geq 0$
$2x \geq 1$
$x \geq \frac{1}{2}$
Assumption 2:
$ -1 \leq \arccos \frac{x+1}{4} \leq 1 $
$ -1 \leq \frac{x+1}{4} \leq 1 $
$ -4 \leq x + 1 \leq 4 $
$ -5 \leq x \leq 3 $
Assumption 3:
$ 2 + \arccos\frac{x+1}{4} \neq 0 $
Here once again, no idea how to proceed further.
Would anyone be able to help me understand these examples, how to find that the denominator is not equal to $0$, when there's arccos or arcsin in the denominator? Thanks! I do hope my tags are okay...
| For the first function, the fraction is defined when $\arcsin \frac{x}{4} \ne 0 \implies \frac{x}{4} \ne \sin 0=0 \implies x\ne 0$
So the domain of the first function is $[-4,4]-\{0\}$
For the second function $$2+\arccos \frac{x+1}{4} \ne 0$$
$$\implies \arccos \frac{x+1}{4} \ne -2$$
Apply cos both sides,
$$\implies \frac{x+1}{4} \ne \cos(-2)$$
$$\implies x \ne 4\cos(-2) -1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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solve $\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$ I need to find
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$$
Looking at the graph, I know the answer should be $\frac{20}{17}$, but when I tried solving it, I reached $0$.
Here are the two ways I approached this:
WAY I:
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{x^2}(2- \frac{50}{x^2})} {\require{cancel} \cancel{x^2}(2+ \frac{3}{x}-\frac{35}{x^2})}
=\frac{2-2}{\frac {42}{5}}=0
$$
WAY II:
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{2}(x^2- 25)} {\require{cancel} \cancel{2}(x^2+ \frac{3}{2}x-\frac{35}{2})}
=\lim_{x\rightarrow -5} \frac{{\require{cancel} \cancel{(x-5)}}(x+5)}{{\require{cancel} \cancel{(x-5)}}(x+3.5)}= \frac{-5+5}{-5+3.5}=0
$$
What am I doing wrong here?
Thanks!
| For your way $1$, check the computation of your denominator, it should give you $0$ again.
For your way $2$, check your factorization in your denominator as well.
Use L'hopital's rule:
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}= \lim_{x\rightarrow -5} \frac{4x}{4x+3}=\frac{-20}{-17}=\frac{20}{17}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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General solution of $\sin 2x+\cos x=0$
Solve the trigonometric equation $\sin 2x+\cos x=0$
My Attempt
$$
2\sin x\cos x+\cos x=0\implies\cos x=0 \text{ or }\sin x=\frac{-1}{2}=\sin\frac{-\pi}{6}\\
x=(2n+1)\frac{\pi}{2} \text{ or }x=n\pi+(-1)^n\frac{-\pi}{6}\\
x=n\pi+\frac{\pi}{2} \text{ or }x=2n\pi-\frac{\pi}{6}\text{ or }x=2n\pi+\pi+\frac{\pi}{6}
$$
Reference
$$
\cos x=-\sin 2x=\cos\Big(\frac{\pi}{2}+2x\Big)\implies x=2n\pi\pm\Big(\frac{\pi}{2}+2x\Big)\\
-x=2n\pi+\frac{\pi}{2}\text{ or }3x=2n\pi-\frac{\pi}{2}\implies x=2m\pi-\frac{\pi}{2}\text{ or }x=\frac{2m\pi}{3}-\frac{\pi}{6}
$$
But my reference gives the solution $x=2n\pi-\dfrac{\pi}{2}$ or $x=\dfrac{2n\pi}{3}-\dfrac{\pi}{6}$. I understand how it is reached and both represent the same solutions. But, how do I derive the solution in my reference from what I found in my attempt ? i.e.,
How to derive
$$
\bigg[x=n\pi+\frac{\pi}{2} \text{ or }x=2n\pi-\frac{\pi}{6}\text{ or }x=2n\pi+\pi+\frac{\pi}{6}\bigg]\\
\implies \bigg[x=2n\pi-\dfrac{\pi}{2}\text{ or }x=\dfrac{2n\pi}{3}-\dfrac{\pi}{6}\bigg]
$$
| You and your reference (assuming no typos) have omitted many solutions. For instance, $x = \pi/2$, which you are missing, and $-\pi/2$, which your reference is missing.
Assuming throughout that $k$ is any integer.
From $\cos(x) = 0$, $x = \pm \cos^{-1}(0) + 2\pi k$, giving the two solution families $x = \pm \frac{\pi}{2} + 2\pi k$. These can be written as a single family (note the coefficient of $k$): $\frac{\pi}{2} + \pi k$, which is equivalent to $\frac{-\pi}{2} + \pi k$. This equivalence is most likely what is going on between your solution and your reference's.
From $\sin(x) = \frac{-1}{2}$, $x = \sin^{-1}\left(\frac{-1}{2}\right) + 2\pi k = \frac{-\pi}{6} + 2 \pi k$ or $x = \pi - \sin^{-1}\left(\frac{-1}{2}\right) + 2\pi k = \frac{7 \pi}{6} + 2 \pi k$. Taken together, you should recognize these as your solution and one of them you quote from your reference.
Are you sure you have copied your reference's entire answer and correctly copied its $k$ coefficients?
| {
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Prove for all $a, b, c \in \mathbb {R^+}$ ${a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}} \gt \sqrt{a+b+c}$ is true
Prove for all $a, b, c \in \mathbb {R^+}$ $${a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}} \gt \sqrt{a+b+c}$$ is true
I have been studying about inequalities and how to prove different inequalities, and I stumbled over this question which I failed to answer. I failed to see a connection between $\sqrt{a+b+c} \text{ and} {a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}}$. Is there any way I can prove this question and if there is can you please help. Thank you.
| You need only Jensen applied to convex function $f(x)={1 \over \sqrt{x}}$:
$$LHS=a\space f(a+b)+b\space f(b+c)+c\space f(c+a)\ge(a+b+c)\space f(\frac{a(a+b)+b(b+c)+c(c+a)}{a+b+c})$$
$$LHS\ge(a+b+c)\space f(\frac{(a+b+c)^2-ab-bc-ac}{a+b+c})$$
$$LHS\ge(a+b+c)\sqrt{\frac{a+b+c}{(a+b+c)^2-ab-bc-ac}}$$
$$LHS\gt(a+b+c)\sqrt{\frac{a+b+c}{(a+b+c)^2}}$$
$$LHS\gt\sqrt{a+b+c}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the convergence of the sequence $a_{1} = 4$, $a_{n + 1} = \frac{a_{n}}{2} + \frac{2}{a_{n}}$, $n = 1, 2, \ldots$ Prove the convergence of the sequence $a_{1} = 4$, $a_{n + 1} = \frac{a_{n}}{2} + \frac{2}{a_{n}}$, $n = 1, 2, \ldots$
I'm pretty sure the way to do it is to show $a_{n} > 2$ for $n = 2, \ldots$ and then maybe use the Monotone Convergence Theorem to show it converges to $2$, but I think this also might be wrong. Can someone please help me with this problem? I don't know how to prove a bound for it.
| Note, that $\color{blue}{a= 2}$ is a $\color{blue}{\mbox{fixpoint}}$ of the iteration as
$$2 = \frac{a}{2} + \frac{2}{a} = 1+1$$
AM-GM shows that for all members of the sequence we have
$$\frac{a_n}{2} + \frac{2}{a_n} \stackrel{AM-GM}{\geq} 2$$
Now, consider
$$0 \leq \color{blue}{a_{n+1}-2} = \frac{a_n}{2} + \frac{2}{a_n} - 2 = \frac{a_n -2}{2} - \frac{a_n - 2}{a_n} = \left( \frac{1}{2} - \frac{1}{a_n} \right)(a_n - 2) \color{blue}{\stackrel{a_n \geq 2}{\leq} \frac{1}{2} (a_n - 2)}$$
It follows
$$0 \leq a_{n+1}-2 \leq \left( \frac{1}{2}\right)^n(a_1 - 2)\stackrel{n \to \infty}{\longrightarrow} 0 \Rightarrow \color{blue}{(a_n) \mbox{ is convergent and } \lim_{n \to \infty}a_n = 2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Not able to integrate $\displaystyle\int_{0}^{\pi/2}
\frac{\sin\left(x\right)}
{\left[1 + \,\sqrt{\,\sin\left(2x\right)\,}\,\right]^{2}} \,\mathrm{d}x$
i used the property to change reach
$\displaystyle 2I =
\int_0^{\pi/2}\frac{\sin\left(x\right) + \cos\left(x\right)}
{\left[1 + \,\sqrt{\,\sin\left(2x\right)\,}\,\right]^2} \,\mathrm{d}x$
now writing $\displaystyle\sin\left(2x\right) =
1 - \left[\sin\left(x\right) - \cos\left(x\right)\right]^{\, 2}$, and substituting
$\displaystyle\sin\left(x\right) - \cos\left(x\right) = t$,
how to do further ?.
| As in the OP, the substitution $y=\pi/2-x$ gives
$$
\begin{aligned}
J&:=
\int_0^{\pi/2} \frac {\sin x}{(1+ \sqrt{\sin 2x})^2} \;dx
\\
&=\int_0^{\pi/2} \frac {\cos x}{(1+ \sqrt{\sin 2x})^2} \;dx
\text{ ... and thus}
\\
&=
\frac 12
\int_0^{\pi/2} \frac {\sin x+\cos x}{(1+ \sqrt{\sin 2x})^2} \;dx\dots
\\
&\qquad\qquad\text{Now formally set $t=\sin x-\cos x$,}
\\
&\qquad\qquad\text{so $dt=\cos x+\sin x$,
$t^2=1-2\sin x\cos x=1-\sin 2x$...}
\\
&=
\frac 12
\int_{-1}^{1}
\frac {dt}{(1+ \sqrt{1-t^2})^2}
=
\int_0^1
\frac {dt}{(1+ \sqrt{1-t^2})^2}
\\
&\qquad\qquad\text{Now use $t=\sin u$}
\\
&=
\int_0^{\pi/2}
\frac {\cos u\; du}{(1+ \cos u)^2}
\\
&\qquad\qquad\text{Now use $v=\tan(u/2)$}
\\
&=
\int_0^1
\frac {\frac{1-v^2}{1+v^2}\cdot\frac 2{1+v^2}\; dv}
{\left(\frac 2{1+v^2}\right)^2}
=
\frac 12
\int_0^1(1-v^2)\; dv
\\
&=\frac 12\left[v-\frac 13 v^3\right]_0^1
=
\frac 12\cdot \frac 23 =\frac 13\ .
\end{aligned}
$$
Computer check, pari/gp:
? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )
%1 = 0.33333333333333333333333333333333333333
| {
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"timestamp": "2023-03-29T00:00:00",
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Conditional probability (urns and balls) From an urn, containing 6 white and 12 black balls, one takes balls randomly one by one until the second white ball appears.
What is the probability that:
1) second white ball appears on the second step
2) second white ball appears on the third step
3) second white ball appears on the k-th step
My solution is the following.
The probability of the second white ball is
$$Pr(X=2)=\frac{6}{18}\frac{5}{17}$$
The probability of the third white ball is
$$Pr(X=3)=\frac{6}{18}\frac{12}{17}\frac{5}{16}+ \frac{12}{18}\frac{6}{17}\frac{5}{16} = \frac{2⋅6⋅5⋅12}{18⋅17⋅16} $$
Therefore, the probability
$$Pr(X=k)=\frac{(k-1)⋅6⋅5⋅(18-k)!⋅12!}{18!⋅(12-k+2)!}$$
Could somebody, please, check my solution, especially the third part. Thank you.
| I used the hypergeometric distribution for the $(k-1)$-th draw.
The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is
$$\frac{\binom{6}{1}\cdot \binom{12}{k-2}}{\binom{18}{k-1}}=\frac{6\cdot 12!\cdot (k-1)!\cdot (19-k)!}{18!\cdot (k-2)!\cdot (14-k)!}=\frac{6\cdot 12!\cdot (k-1)\cdot (19-k)!}{18!\cdot (14-k)!}$$
At the k-th draw we have to pick a white ball. The probability is $\frac{5}{19-k}$. In total I get
$$P(X=k)=\frac{5\cdot 6\cdot 12!\cdot (k-1)\cdot (18-k)!}{18!\cdot (14-k)!}$$
This is the same result you got.
| {
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Integrate $\sqrt{(1-\cos(x))^3}$ I need help solving the following integration problem:
$$\int \sqrt{(1-\cos(x))^3} dx$$
I know I am supposed to use a trig identity so that I can do a u-sub, but don't know which one or how.
| $$\int \sqrt{\left(1-\cos \left(x\right)\right)^3} = 4\sqrt{2}\int \:\frac{y^3}{\left(1+y^2\right)^{\frac{5}{2}}}dy=2\sqrt{2}\int \:\frac{1}{z^{\frac{3}{2}}}-\frac{1}{z^{\frac{5}{2}}}dz$$
You will get this result by applying these substitutions in sequence
$$y=\tan\left(\frac{x}{2}\right)\qquad\qquad z= 1+y^2$$
$$\int \sqrt{\left(1-\cos \left(x\right)\right)^3}=\color{red}{4\sqrt{2}\left(\frac{1}{3\sec ^3\left(\frac{x}{2}\right)}-\frac{1}{\sec \left(\frac{x}{2}\right)}\right)+C}$$
| {
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Given $\tan\alpha=2$, evaluate $\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$ I need some help with this exercise.
Given that $$\tan\alpha=2$$
calculate the value of:
$$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$$
I've tried everything but I always end up having something irreducible in the end. Maybe this is an easy question, but I'm new at trigonometry. If anyone can help me by providing a step-by-step solution to this, I would be really thankful! :)
| $$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha} = \frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha} \cdot\frac{1/\cos^3\alpha}{1/\cos^3\alpha} = \frac{\tan^3\alpha-2+3\cdot(1/\cos^2\alpha)}{(3\tan\alpha+2)\cdot(1/\cos^2\alpha)}$$
Now, recall that $\frac{1}{\cos^2\alpha}=\sec^2\alpha=1+\tan^2\alpha=5$, so,
$$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha} = \frac{\tan^3\alpha-2+3\cdot(1/\cos^2\alpha)}{(3\tan\alpha+2)\cdot(1/\cos^2\alpha)} = \frac{8-2+15}{(6+2)5}=\frac{21}{40}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A right triangle with integer sides has area equal to twice its perimeter. Find sum of all possible circumradii.
In right triangle $ABC$, the area is twice the perimeter, and all sides have integer lengths. Compute the sum of all possible circumradii of $ABC$.
I only have set up an equation
$$\frac{xy}{2}=2\left(x+y+\sqrt{x^2+y^2}\right)$$ and
$$R=\frac{xy \sqrt{x^2+y^2}}{2xy}=\frac{\sqrt{x^2+y^2}}{2}$$
| $$\frac{xy}{2} = 2\left(x+y +\sqrt{x^2+y^2}\right) \implies (xy - 4(x+y))^2 = 16(x^2+y^2)\\ \iff xy(xy-8(x+y)+32) = 0 \implies (x-8)(y-8) = 32$$
Since $x$ and $y$ are integers and $32 = 2^5$, the total numbers of factors are $(5+1)=6$ which are $1,2,4,8,16,32$ .
So possible cases avoiding any redundancy for we just want circumradii are as follows:
$$\underset{32}{(x-8)}\underset{1}{(y-8)} = 32$$
$$\underset{16}{(x-8)}\underset{2}{(y-8)} = 32$$
$$\underset{8}{(x-8)}\underset{4}{(y-8)} = 32$$
So possible triplets turn out to be $40,9,41$;
$24,10,26$;
$16,12,20$.
So, possible circumradii are $\frac{41}2,\frac{26}2,\frac{20}2$.
| {
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$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\frac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$?
$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\dfrac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$ ?
This expression is found without mentioning the domain of $x,y,z$, but I don't think its true for all $x,y,z$ as the case with the expression for $\tan^{-1}x+\tan^{-1}y$, but I have trouble proving it.
So what is the complete expression for $\tan^{-1}x+\tan^{-1}y+\tan^{-1}z$ ?
\begin{align}
\tan^{-1}x+\tan^{-1}y+\tan^{-1}z&=
\begin{cases}\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy < 1 \\[1.5ex]
\pi + \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy>1,\:\:x,y>0 \\[1.5ex]
-\pi + \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy>1,\:\:x,y<0 \end{cases}\\
&=
\end{align}
|
This follows from the fact the the argument of a product of complex numbers is the sum of the arguments of the factors.
Let $\alpha=\arctan x$, $\beta=\arctan y$ and $\gamma=\arctan z$. These are the arguments of the complex numbers $z_1=1+ix$, $z_2=1+iy$ and $z_3=1+iz$ respectively.
In light of the above fact we see that $\alpha+\beta+\gamma$ is the argument
(up to an integer multiple of $2\pi$) of the product
$$
z_1z_2z_3=(1+ix)(1+iy)(1+iz)=(1-xy-yz-zx)+i(x+y+z-xyz).
$$
But the argument $\phi$ of a complex number $a+ib$ satisfies $\tan\phi=b/a$.
The claim follows from this.
Just be mindful of the lingering uncertainty in the value of the inverse tangent up to an integer multiple of $\pi$. For example, if $x=y=z=1$ we have $\arctan x=\arctan y=\arctan z=\pi/4$ giving $3\pi/4$ on the left hand side. But, $x+y+z-xyz=2$, $1-xy-yz-zx=-2$, so we have $\arctan(-1)=-\pi/4$ on the right hand side.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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sum of $\sum^{n}_{k=0}(-1)^k\cdot \frac{\binom{n}{k}}{\binom{k+3}{k}}$
Finding sum of $\displaystyle \sum^{n}_{k=0}(-1)^k\cdot \frac{\binom{n}{k}}{\binom{k+3}{k}}$
Try: Using
$$\int^{1}_{0}x^m\cdot (1-x)^ndx = \frac{m!\cdot n!}{(m+n+1)!}=\frac{1}{(m+n+1)\binom{m+n}{n}}.$$
So $$ \frac{1}{(k+4)\binom{k+3}{k}}=\int^{1}x^3\cdot (1-x)^kdx$$
So our sum is $$\sum^{n}_{k=0}(-1)^k(k+4)\binom{n}{k}\int^{1}_{0}x^3(1-x)^kdx$$
$$ = \int^{1}_{0}x^3\sum^{n}_{k=0}(-1)^kk\binom{n}{k}(1-x)^kdx+4\int^{1}_{0}x^3\sum^{n}_{k=0}(-1)^k\binom{n}{k}(1-x)^k$$
$$ = -n\int^{1}_{0}x^{n+2}(1-x)dx+4\int^{1}_{0}x^{n+3}dx$$
$$ = -n\bigg[\frac{1}{n+3}-\frac{1}{n+4}\bigg]+\frac{4}{n+4} = -\frac{n}{(n+3)(n+4)}+\frac{4}{n+4} = $$
but answer given in book as $\displaystyle \frac{3}{n+3}$
Could some help me how to solve it, Thanks
| I think
$$\frac{4}{n+4}-\frac{n}{(n+3)(n+4)}=\frac{4(n+3)-n}{(n+3)(n+4)}=\frac{3(n+4)}{(n+3)(n+4)}=\frac{3}{n+3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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For what $p,q,r$ $\lim_{x \to 0} f(x)=\frac{p + q\cos x + r\sin x}{x^2}=1/2$? Let $$f(x)=\frac{p + q\cos x + r\sin x}{x^2}$$. Then for what values of $p$,$q$ and $r$ is the limit $$\lim_{x \to 0} f(x)=1/2.$$ I’ve tried using L’Hospital’s rule but couldn’t get anywhere. Any help would be appreciated.
| We are given that $$\lim_{x\to 0}\frac{p+q\cos x+r\sin x} {x^2}=\frac{1}{2}\tag{1}$$ so that multiplication by $x^2$ gives $$\lim_{x\to 0}(p+q\cos x+r\sin x) =\lim_{x\to 0}\frac{p+q\cos x+r\sin x} {x^2}\cdot x^2=\frac{1}{2}\cdot 0=0$$ and hence $p+q=0$.
Next note that $$p+q\cos x+r\sin x=p+q-q(1-\cos x) +r\sin x=r\sin x-q(1-\cos x) $$ and since $(1-\cos x) /x^2\to 1/2$ it follows from $(1)$ that $$\lim_{x\to 0}\frac{r\sin x} {x^2}=\frac{1+q}{2}$$ and since $(\sin x) /x\to 1$ we have $$\lim_{x\to 0}\frac{r}{x}=\frac{1+q}{2}\tag{2}$$ Multiplying by $x$ gives us $r=0$ and then from the above equation we get $1+q=0$ so that $p=1,q=-1,r=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Series convergence without sigma notation Consider the following series:
$$\frac{1}{1} + \frac{10}{2} + \frac{100}{3} - \frac{37}{4} - \frac{37}{5} - \frac{37}{6} + \frac{1}{7} + \frac{10}{8} + \frac{100}{9} - \frac{37}{10} - \frac{37}{11} - \frac{37}{12} + \dots$$
This series seems to converge but I am not able to prove it. It seems impossible to write thing whole thing into summation notation. I tried to group by modulo 6, but then for example sum of $\frac{1}{6n+1}$ diverges already so that doesn't seem right. Also tried to change the 37s into powers of 10 to maybe use the comparison test but also failed.
Any suggestions?
| First, notice that $$F_n=\frac{1}{6n+1}+\frac{10}{6n+2}+\frac{100}{6n+3}-\frac{37}{6n+4}-\frac{37}{6n+5}-\frac{37}{6n+6} \geq \frac{111}{6n+3}-\frac{111}{6n+4} > 0.$$
Then, notice that $$F_n \leq \frac{111}{6n+1}-\frac{111}{6n+6}=\frac{555}{(6n+1)(6n+6)}.$$
Thus the sum of $F_n$ converges. How can you infer the final result from this?
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int \frac{\cos x}{\sin^3x+\sin x}dx$ Given the function $$g(x)=\frac{\cos x}{\sin^3x+\sin x},$$ by letting $u=\sin x$, show that $$\int g(x) dx=\int\left(\frac{A}{u}+\frac{Bu+C}{u^2+1}\right)du$$ where $A,B$ and $C$ are constants. Hence, find $A,B$ and $C$. Hence, solve $\int g(x)dx$.
My attempt,
$$g(x)=\frac{\cos x}{\sin^2 x(\sin x+1)}$$
$$=\frac{\sqrt{1-u^2}}{u^2(u+1)}$$
I'm stuck here.
| Method$\#1:$
$\sin x=u\implies $
$\displaystyle\int g(x)\ dx=\int\dfrac{du}{u(u^2+1)}$
Now using Partial Fraction Decomposition let $\dfrac1{u(u^2+1)}=\dfrac Au+\dfrac{Bu+C}{u^2+1}$
$\implies1=A(u^2+1)+u(Bu+C)=u^2(A+B)+Cu+A$
Comparing the coefficients of $u,u^2$ and the constants
$\implies C=0,A=1,A+B=0\iff B=-A=?$
Can you take it from here?
Method$\#2:$
$$I=\int\dfrac{\cos x}{\sin x(\sin^2x+1)}dx=\int\dfrac{\cos x\sin x}{\sin^2x(\sin^2x+1)}dx$$
Set $\sin^2x=v\implies dv=2\sin x\cos x\ dx$
$$2I=\int\dfrac{dv}{v^2(v^2+1)}=\int\dfrac{(v^2+1)-v^2}{v^2(v^2+1)}dv=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The distributive law $4\left (x+y \right)=4x+4y $ because $4\left (x+y \right) =\left (x+y \right) +\left (x+y \right) +\left (x+y \right) +\left (x+y \right)$ , but why is $\left (x+y \right) \left (x+y \right) =xx+xy+yx+yy$?
| Because:
$$\left (x+y \right) \left (x+y \right) = \underbrace{(x+y)+(x+y)+...(x+y)}_{x + y\text{ times}}$$
$$=\underbrace{(x+y)+(x+y)+...(x+y)}_{x\text{ times}}+\underbrace{(x+y)+(x+y)+...(x+y)}_{y\text{ times}}$$
$$=\underbrace{x}_{x\text{ times}}+\underbrace{y}_{x\text{ times}}+\underbrace{x}_{y\text{ times}}+\underbrace{y}_{y\text{ times}}$$
$$=xx+xy+yx+yy$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the ratio of the sides of a given triangle given the ratio of areas. Given a triangle $\triangle ABC$, points $M$, $N$, $P$ are drawn on the sides of the triangle in a way that $\frac{|AM|}{|MB|} = \frac{|BN|}{|NC|}= \frac{|PC|}{|PA|}=k$, where $k>0$.
Calculate $k$, given that the area of the triangle $\triangle MNP$ and the area of the triangle $ \triangle ABC$ are in the following ratio: $Area_{\triangle MNP} = \frac{7}{25} \times Area_{\triangle ABC}$.
I have tried Heron formula, bot the calculations seem to be too complicated, I have also tried to simplify the problem and assume that k is equal to 1 and then calculate the ratio of the triangles area but it also doesn't help.
I was also looking for similar triangles.
I would appreciate some hint.
| Draw $BB1$ is perpendicular to $AC$; $MM1//AC$ and perpendicular to $AC$, then we have $\frac{AM}{AB}=\frac{MM1}{BB1}$
So $$\frac{AMP}{ABC}=\frac{1/2 \cdot AP\cdot MM1}{1/2\cdot AC \cdot BB1} =\frac{AP\cdot AM}{AC\cdot AB}$$
$$\Rightarrow \frac{AMP}{ABC} = \frac{CPN}{ABC} = \frac{BMN}{ABC} = \frac{AM}{AB}. \frac{AP}{AC} = \frac{k}{(k+1)^2}$$
Or $$\frac{MNP}{ABC} = 1 - \frac{AMP}{ABC} - \frac{CPN}{ABC} - \frac{BMN}{ABC} = 1 - \frac{3k}{(k+1)^2}=\frac{7}{25}$$
It is not difficult to find $k=\frac{2}{3}$ or $k=\frac{3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043884",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Finding the Taylor series (complex numbers) I have
$$
{\frac{1}{(z+1)(z-2)}}
$$
I did
$$
{\frac{1}{(z+1)(z-2)}} = {\frac{A}{z+1}}+{\frac{B}{z-2}}
$$
and found
$A=-1/3, B=1/3$
So now I have
$$
-{\frac{1}{3}}\times {\frac{1}{z+1}}+{\frac{1}{3}}\times{\frac{1}{z-2}} = -{\frac{1}{3}} \sum(-1)^k z^k + {\frac{1}{3}} \sum(-1)^k (-{\frac{z}{2}})^k
$$
But the answer in the book is different. What am I doing wrong?
| It is indeed true that$$\frac1{(z+1)(z-2)}=-\frac13\times\frac1{z+1}+\frac13\times\frac1{z-2}$$and that$$\frac1{z+1}=\sum_{k=0}^\infty(-1)^kz^k.$$However,$$\frac1{z-2}=-\frac1{2-z}=-\sum_{k=0}^\infty\frac{z^k}{2^{k+1}}.$$
| {
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How do we solve for all values of $x$? Suppose
$$(x^2 - 5)^{8} (x+1)^{-16} = 1$$
How do we solve for all values of $x$?
I think we can equate the bases to $1$ in order to get $$1 \times 1 = 1$$
$$x^2-5 = 1, x^2-5=-1$$
or
$$x + 1 = 1, x + 1 = -1$$
Could you assist me with this?
| $(x^2-5)^8=(x+1)^{16}$
So
$[(x^2-5)^4-(x+1)^8] [(x^2-5)^4+(x+1)^8]=0$
If and only if
$[(x^2-5)^4-(x+1)^8]=0$
If and only if
$[(x^2-5)^2-(x+1)^4]=0$
If and only if
$[(x^2-5)-(x+1)^2] [(x^2-5)+(x+1)^2]=0$
So
$[-2x-6][2x^2+2x-4]=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the integral $\int_{-1}^1 \frac{1}{(2-x)\sqrt{1-x^2}}\,dx$ I started with the substitution $x=\sin(\theta)$ with $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$
And I got the integral $$\int_{-\pi/2}^{\pi/2} \frac{1}{2-\sin(\theta)}\,d\theta$$
which I am again not able to evaluate.
Even integration by parts is not working.
| Try using the substitution $\displaystyle t = \tan \frac{\theta}{2}$, this is a handy substitution to make when there are trigonometric functions that you cannot simplify very easily. In our case we have that,
$$\frac{\mathrm{d}t}{\mathrm{d}\theta} = \frac{1}{2} \sec^2\frac{\theta}{2} = \frac{1}{2} \left(1 + \tan^2 \frac{\theta}{2} \right) = \frac{1}{2}(1 + t^2) \Rightarrow \frac{2\mathrm{d}t}{1+t^2} = \mathrm{d}\theta $$
The usefulness of this substitution is that we can now use properties of a right angled triangle that has angle $\theta /2$ to say that,
$$\sin \theta = 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} = 2 \frac{1}{\sqrt{1+t^2}} \frac{t}{\sqrt{1+t^2}}= \frac{2t}{1+t^2} $$
Putting this into our integral we have,
$$\int_{-\pi/2}^{\pi/2} \frac{\mathrm{d}\theta}{2 - \sin \theta} = \int_{-1}^1 \frac{2}{1+t^2} \cdot \frac{1}{2 - \frac{2t}{1+t^2}} \ \mathrm{d}t = \int_{-1}^1 \frac{1}{t^2 - t+ 1} \ \mathrm{d}t $$
Now we have no more square roots, we have an irreducible quadratic at the bottom so we can complete the square to obtain,
$$\int_{-1}^1 \frac{1}{t^2 - t + 1} \ \mathrm{d} t = \int_{-1}^1 \frac{1}{(t - \frac{1}{2})^2 + \frac{3}{4}} \ \mathrm{d}t = \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2}{\sqrt{3}}\left(t - \frac{1}{2}\right) \right) \Big|_{-1}^1 = \frac{\pi}{\sqrt{3}} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Summation of series and Taylor series are giving different results $A = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots$
$B = 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots$
$A + B = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots$
$2A = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots$
$2A = A + B$
$A - B = 0$
$0 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots$
Using Taylor expansion for $\ln(1+x)$
$\ln(2) = 1 -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots$
Where did I go wrong ?
| What you did wrong was to assume that you can deal with divergent series as if they were convergent. If you say that
$$
A=\frac12+\frac14+\frac16+\cdots
$$
then $A$ is a name of the series, but it is not a number (since the series diverges).
Therefore
$$
1+\frac12+\frac13+\frac14+\cdots
$$
is just another (divergent) series and it is an unfortunate option to call it $2A$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing $\sqrt[4]{28+16 \sqrt 3}$ I want to compute following radical
$$\sqrt[4]{28+16 \sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$
$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
| Alternatively, you can use Michael Rozenberg's identities, where this identity applies to the question:
$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
We have $a = 28$, $b = 16^2 \cdot 3 = 768$. Applying this identity gives:
$$\sqrt{28+\sqrt{768}}=\sqrt{\frac{28+\sqrt{16}}{2}}+\sqrt{\frac{28-\sqrt{16}}{2}}=\sqrt{16}+\sqrt{12}=4+2\sqrt{3}.$$
Applying the identity once more gives, where $a=4, b=2^2 \cdot 3 = 12$ gives:
$$\sqrt{4+\sqrt{12}}=\sqrt{\frac{4+\sqrt{4^2-12}}{2}}+\sqrt{\frac{4-\sqrt{4^2-12}}{2}}=\sqrt{\frac{4+2}{2}}+\sqrt{\frac{4-2}{2}}=\sqrt{3}+1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$ The following integral was proposed by Cornel Ioan Valean and appeared as Problem $12054$ in the American Mathematical Monthly earlier this year.
Prove $$\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$$
I had small tries for it, such as writting:
$$I=\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx\overset{ x\to \tan \frac{x}{2}}=-\frac12 {\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x} dx}$$
And with Feynman's trick we obtain:
$$J(t)=\int_0^\frac{\pi}{2} \frac{x\ln(1-t\sin x)}{\sin x}dx\Rightarrow J'(t)=\int_0^\frac{\pi}{2} \frac{x}{1-t\sin x}dx$$ But I don't see a way to obtain a closed from for the above integral.
Also from here we have the following relation:
$$\int_0^1 \frac{\arctan x \ln(1+x^2)}{x} dx =\frac23 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx$$
Thus we can rewrite the integral as:
$$I=\frac23 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx -2\int_0^1 \frac{\arctan x \ln(1-x)}{x}dx$$
Another option might be to rewrite:
$$\ln\left(\frac{1+x^2}{(1-x)^2}\right)= \ln\left(\frac{1+x}{1-x}\right)+\ln\left(\frac{1+x^2}{1-x^2}\right)$$
$$\Rightarrow I= \int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x}{1-x}\right)dx+\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{1-x^2}\right)dx$$
And now to use the power expansion of the log functions to obtain:
$$\small I=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1 \frac{\arctan x}{x} \, \left(x^{2n+1}+x^{4n+2}\right)dx=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1\int_0^1 \frac{\left(x^{2n+1}+x^{4n+2}\right)}{1+y^2x^2}dydx$$
This seems like an awesome integral and I would like to learn more so I am searching for more approaches.
Would any of you who also already solved it and submitted the answer to the AMM or know how to solve this integral kindly share the solution here?
Edit:
In the meantime I found a nice solution by Roberto Tauraso here and another impressive approach due to Yaghoub Sharifi here.
| Starting with breaking the integral
$\displaystyle I=\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)\ dx=\int_0^1\frac{\arctan x}{x}\ln(1+x^2)dx-2\int_0^1\frac{\arctan x}{x}\ln(1-x)dx$
then using the identity$\ \displaystyle\arctan x\ln(1+x^2)=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}} {2n+1}x^{2n+1}$ for the first integral and series-expanding $\displaystyle\arctan x$ of the second integral, we get
\begin{align*}
I&=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{2n+1}\int_0^1x^{2n}\ dx-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln(1-x)\ dx\\
&=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{(2n+1)^2}-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(-\frac{H_{2n+1}}{2n+1}\right)\\
&=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{(2n+1)^2}-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(-\frac{H_{2n}}{2n+1}-\frac{1}{(2n+1)^2}\right)\\
&=2\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}=2\beta(3)=\frac{\pi^3}{16}
\end{align*}
where $\beta(3)=\frac{\pi^3}{32}$ is the Dirichlet beta function.
Note that we used the classical result $\int_0^1 x^{n-1}\ln(1-x)dx=-\frac{H_n}{n}$ which can be proved as follows:
$$\int_0^1 x^{n-1}\ln(1-x)dx=-\sum_{k=1}^\infty\frac1k\int_0^1 x^{n+k-1}dx=-\sum_{k=1}^\infty\frac{1}{k(n+k)}\\=-\frac1n\sum_{k=1}^\infty\left(\frac1k-\frac1{n+k}\right)=-\frac1n\sum_{k=1}^n\frac1k=-\frac{H_n}{n}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this number theory question from INMO?
Solve the following equation in integers:
$$m(4m^2 + m + 12) = 3(p^n -1)$$
where $m$ and $n$ are integers and $p$ is a prime greater than equal to 5.
This simplifies to :
$$4m^3 + m^2 + 12m + 3 = (4m + 1)(m^2 + 3) = 3p^n$$
$\gcd(4m+1,m^2 +3)$ is greater than 1 because if it were so then we get only 4 possibilities in every case of which we get one of the numbers to be less than 4, which is not possible.
Moreover, $$(4m+1,m^2 + 3) = (4m + 1, 16m^2 + 48) = (4m + 1,49) = 7\text{ or }49$$
This means the prime $p$ is 7. and $4m + 1 = 3*7^k\text{ or }7^k$
If $\gcd(4m+1,49) = 7$ then $k=1$ and $4m+1 = 21$ which doesnt give any solution.Therefore $\gcd(4m+1,m^2 + 3) = 49$. If $7^3$ divides $4m + 1$ then it doesn't divide $m^2 + 3$, and then the solution says that we get $$m^2 + 3 \leq 3*7^2 < 7^3 \leq 4m + 1$$
How do we arrive at the above inequality chain?
| Since $m^2+3 =7^l$ or $m^2+3= 3\cdot 7^l$ we see that $m^2+3 \leq 3\cdot 7^l$
Now we have $7^2\mid m^2+3$ and $7^3\not{\mid} \;m^2+3$ so $l=2$.
And since $7^3\mid 4m+1$ we have $7^3\leq 4m+1$. Clearly $3\cdot 7^2<7^3$.
| {
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power series for $\frac1{x^2+x+1}$ So I am trying to find the sequence $(a_n)_{n\geq0}$ such that
$$\frac1{x^2+x+1}=\sum_{n\geq0}a_nx^n$$
My attempts:
I defined $$A(x)=\frac1{x^2+x+1}$$
Hence $$A(x)=\sum_{n\geq0}a_nx^n$$
And so $$(x^2+x+1)\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_nx^{n+2}+\sum_{n\geq0}a_nx^{n+1}+\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$a_0(x^2+x+1)+\sum_{n\geq1}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$x=0$:
$$a_0=1$$
So
$$x^2+x+1+a_1(x^3+x^2+x)+\sum_{n\geq2}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$\frac{x^2+x+1+a_1(x^3+x^2+x)+\sum_{n\geq2}a_n(x^{n+2}+x^{n+1}+x^n)}x=\frac1x$$
$$x+1+a_1(x^2+x+1)+\sum_{n\geq2}a_n(x^{n+1}+x^{n}+x^{n-1})=0$$
$x=0$:
$$a_1=-1$$
One more time:
$$x^2+x+1-x^3-x^2-x+a_2(x^4+x^3+x^2)+\sum_{n\geq3}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$-x^3+a_2(x^4+x^3+x^2)+\sum_{n\geq3}a_n(x^{n+2}+x^{n+1}+x^n)=0$$
Divide both sides by $x^2$:
$$-x+a_2(x^2+x+1)+\sum_{n\geq3}a_n(x^{n}+x^{n-1}+x^{n-2})=0$$
$x=0$:
$$a_2=0$$
How do I find $a_n$? Am I doing things right so far? Thanks.
| \begin{align}
\frac1{1+x+x^2} &= \frac{1-x}{1-x^3} \\
&= \frac1{1-x^3}-x\left(\frac1{1-x^3} \right)\\
&= \sum_{i=0}^\infty x^{3i}-x\sum_{i=0}^\infty x^{3i}\\
&=\sum_{i=0}^\infty x^{3i}-\sum_{i=0}^\infty x^{3i+1}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{n\to \infty}\sum_{k=1}^{n}(\sin\frac{\pi}{2k}-\cos\frac{\pi}{2k}-\sin\frac{\pi}{2(k+2)}+\cos\frac{\pi}{2(k+2)})$ Find $\lim_{n\to \infty}\sum_{k=1}^{n}\left(\sin\frac{\pi}{2k}-\cos\frac{\pi}{2k}-\sin\frac{\pi}{2(k+2)}+\cos\frac{\pi}{2(k+2)}\right)$
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\sin\frac{\pi}{2k}-\cos\frac{\pi}{2k}-\sin\frac{\pi}{2(k+2)}+\cos\frac{\pi}{2(k+2)}\right)$$
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\sin\frac{\pi}{2k}-\sin\frac{\pi}{2(k+2)}-\cos\frac{\pi}{2k}+\cos\frac{\pi}{2(k+2)}\right)$$
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(2\cos\frac{(\frac{\pi}{2k}+\frac{\pi}{2(k+2)})}{2}\sin\frac{(\frac{\pi}{2k}-\frac{\pi}{2(k+2)})}{2}+2\sin\frac{(\frac{\pi}{2k}+\frac{\pi}{2(k+2)})}{2}\sin\frac{(\frac{\pi}{2k}-\frac{\pi}{2(k+2)})}{2}\right)$$
I am stuck here.
| Hint. One may recall teslescoping sums, by writing
$$
u_{k+2}-u_k=\left(u_{k+2}-u_{k+1} \right)+\left(u_{k+1}-u_{k} \right)
$$
giving
$$
\sum_{k=1}^n\left(u_{k+2}-u_{k} \right)=\left(u_{n+2}-u_2\right)+\left(u_{n+1}-u_1\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.