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solving Differential Equation $y''+x^2 y'+(2x+1)y=0$ I tried to solve this problem with power series method,
but it became so complicated.
like this: $na_{n-2}+n(n+1)a_{n+1}+a_{n-1}=0 $for $n>=2$
And I cannot solve this a_{n}
How can I get this?
$y''+x^2 y'+(2x+1)y=0$
| Plugging in $y = \sum_{n=0}^\infty a_nx^n$ gives
$$ \sum_{n=2}^\infty n(n-1)a_nx^{n-2} + \sum_{n=0}^\infty (n+2)a_nx^{n+1} + \sum_{n=0}^\infty a_nx^n = 0 $$
Notice the middle sum's lowest-order term is $x^1$, so we need to first take out the constant term from the other sums, and shift the remaining indexes to get
$$ (2a_2 + a_0) + \sum_{n=0}^\infty \left[(n+3)(n+2)a_{n+3} + (n+2)a_n + a_{n+1} \right] x^{n+1} = 0 $$
This gives
\begin{cases} a_2 = -\dfrac{a_0}{2} \\ a_{n+3} = -\dfrac{(n+2)a_n + a_{n+1}}{(n+3)(n+2)} \end{cases}
In this arrangement, $a_0$ and $a_1$ define $a_3$, $a_1$ and $a_2$ define $a_4$, and so on. Every following coefficient will depend on $a_0$ and $a_1$, which you're free to set.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof that $\frac{1+x^2}{n^2} \geq 1-e^{-x^2/n^2}$ for all $x,n \in \mathbb{R}$ I'm looking for a simple proof that $\frac{1+x^2}{n^2} \geq 1-e^{-x^2/n^2}$ for all $x,n \in \mathbb{R}$.
My first attempt was to express the exponential as a Taylor series:
$$\frac{1+x^2}{n^2} \geq \frac{x^2}{n^2}-\frac{1}{2!}\frac{x^4}{n^4}+\frac{1}{3!}\frac{x^6}{n^6}- \, ... \, .$$
Obviously
$$\frac{1+x^2}{n^2} \geq \frac{x^2}{n^2},$$
so if I can show
$$-\frac{1}{2!}\frac{x^4}{n^4}+\frac{1}{3!}\frac{x^6}{n^6}- \, ... <0,$$
then I'm done. But I'm stuck here, and also wondering if there's an even simpler way.
| It is $e^t \geq t + 1$ for all $t \in \mathbb R$ (easily proven by elementary calculus), thus :
$$e^{-x^2/n^2} \geq 1 - x^2/n^2$$
But $1/n^2 >0$ for all $n \in \mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :
$$e^{-x^2/n^2} + 1/n^2 \geq 1 - x^2/n^2 \Leftrightarrow x^2/n^2 + 1/n^2 \geq 1-e^{-x^2/n^2} $$
$$\Leftrightarrow$$
$$\boxed{\frac{1+x^2}{n^2} \geq 1 - e^{-\frac{x^2}{n^2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3019372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Values of $x$ satisfying $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$
For what values of $x$ between $0$ and $\pi$ does the inequality $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$ hold?
My Attempt
$$
\sin x\cos x\cdot(\cos^2x-\sin^2x)=\frac{1}{2}\cdot\sin2x\cdot\cos2x=\frac{1}{4}\cdot\sin4x>0\implies\sin4x>0\\
x\in(0,\pi)\implies4x\in(0,4\pi)\\
4x\in(0,\pi)\cup(2\pi,3\pi)\implies x\in\Big(0,\frac{\pi}{4}\Big)\cup\Big(\frac{\pi}{2},\frac{3\pi}{4}\Big)
$$
But, my reference gives the solution, $x\in\Big(0,\dfrac{\pi}{4}\Big)\cup\Big(\dfrac{3\pi}{4},\pi\Big)$, where am I going wrong with my attempt?
| Finishing what you did
$$\sin(4x)>0\iff$$
$$2k\pi <4x<(2k+1)\pi \iff$$
$$\frac{k\pi}{2}<x<\frac{k\pi}{2}+\frac{\pi}{4}$$
$k=0$ gives $0<x<\frac{\pi}{4}$
and
$k=1$ gives $\frac{\pi}{2}<x<\frac{3\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Why is $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$ not correct? $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$
Apparently, the 2nd step is illegal here. Probably because for $x=-\infty$ I'd get $(+\infty-\infty)$ which is not possible. I see why this wouldn't be possible, I'm not sure if it really is the cause which makes that equation illegal though.
But now, apparently, I could do this:
$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x=\lim_{x\to -\infty}\frac{[\sqrt{x^2+5x+3}+x][\sqrt{x^2+5x+3}-x]}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{x^2+5x+3-x^2}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5x+3}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5+3/x}{\sqrt{1+5/x+3/x^2}-1}=-5/2$
which gives me the correct result.
But in the 3th step I used $x^2-x^2=0$, how is that legal?
Also, in the 2nd step I implicitly used:
$-x\sqrt{x^2+5x+3}+x\sqrt{x^2+5x+3}=0$
Which also seems to be fine, but why?
| You may set $x = -\frac{1}{t}$ and consider the limit for $\stackrel{t\rightarrow 0+}{\longrightarrow}$:
$$\begin{eqnarray*} \sqrt{x^2+5x+3}+x
& \stackrel{x = -\frac{1}{t}}{=} & \frac{\sqrt{1-5t +3t^2} - 1}{t} \\
& \stackrel{t\rightarrow 0+}{\longrightarrow} & f'(0) = -\frac{5}{2}\mbox{ for } f(t) = \sqrt{1-5t +3t^2}
\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$
Find the number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$
\begin{align}
2\cos(\pi\sqrt{x-4})&.\cos(\pi\sqrt{x})=2\\\implies\cos\Big[\pi(\sqrt{x-4}+\sqrt{x})\Big]&+\cos\Big[\pi(\sqrt{x-4}-\sqrt{x})\Big]=2\\
\implies\cos\Big[\pi(\sqrt{x-4}+\sqrt{x})\Big]=1\quad&\&\quad\cos\Big[\pi(\sqrt{x-4}-\sqrt{x})\Big]=1\\
\pi(\sqrt{x-4}+\sqrt{x})=2n\pi\quad&\&\quad\pi(\sqrt{x-4}-\sqrt{x})=2m\pi\\
\sqrt{x-4}+\sqrt{x}=2n\quad&\&\quad\sqrt{x-4}-\sqrt{x}=2m\\
2\sqrt{x}=2(n-m)\quad&\&\quad2\sqrt{x-4}=2(n+m)\\
\sqrt{x}=n-m\quad&\&\quad\sqrt{x-4}=n+m\quad\&\quad x\geq4
\end{align}
How do I properly find the solutions ?
Or can I simply say
$$
x=(n-m)^2=(n+m)^2-4nm=x-4-4nm\implies nm=-1\\
\implies x=\bigg[n+\frac{1}{n}\bigg]^2\in\mathbb{Z}\implies n,\frac{1}{n}\in\mathbb{Z}\\
\implies n\neq0\implies n=1,x=4\text{ is the only solution}
$$
| $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt x) = 1$
implies
$\cos(\pi\sqrt{x-4}) = \cos(\pi\sqrt x) = 1$
or
$\cos(\pi\sqrt{x-4}) = \cos(\pi\sqrt x) = -1$
in either case
$\sqrt{x-4}, \sqrt x$ are both integers. In fact, they are both even, or they are both odd. But that isn't entirely relevant.
There is only one pair of integers that are perfect squares and separated by a distance of 4.
$x = 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Positive divisors of n = $2^{14} \cdot 3^9 \cdot 5^8 \cdot 7^{10} \cdot 11^3 \cdot 13^5 \cdot 37^{10}$ How do I find positive divisors of n that are perfect cubes that are multiples of 2^10 * 3^9 * 5^2 * 7^5 * 11^2 * 13^2 * 37^2
The answer is (1)(1)(2)(2)(1)(1)(3) = 12
I don't understand though because I would have done something like:
2: [(14-10)/3]+1 = 2 (taking the floor)
3: [(9-9)/3]+1 = 1
5: [(8-2)/3]+1 = 3
7: [(10-5)/3]+1 = 2
11: [(3-2)/3]+1 = 1
13: [(5-2)/3]+1 = 2
37: [(10-2)/3]+1 =3
2*1*3*2*1*2*3
| Since we are required to be a multiple of $2^{10}\cdot3^9\cdot5^2\cdot7^5\cdot11^2\cdot 13^2\cdot37^2$ and also a perfect cube, we know that whatever our divisor is, it must be divisible by $2^{12}\cdot3^9\cdot5^3\cdot7^6\cdot11^3\cdot 13^3\cdot37^3$. Now when you run your counting argument, the permissible ranges of exponents should be small enough to match the provided answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a difference equation for coin toss sequence probabilities I want to solve the following difference equation:
$$b_n-b_{n-1} = \frac{1}{8}(1-b_{n-3})$$
I tried to solve it similar to the solution of Fibonacci sequence here, but when I try to assume the solution will be of the form $l^n$, I end up with the following polynomial:
$$l^{n-3}(l^3-l^2+\frac{1}{8}) = \frac{1}{8}$$
Unlike in the Fibonacci sequence, we don't get a polynomial equation that can be easily solved and independent of $n$.
Any other strategy to solve this?
The reason I care about this difference equation is that it describes the probability I will get two consecutive heads on the $n$th toss of a fair coin.
Let $a_n$ be the probability that I'll reach two consecutive heads on the $n$th toss.
We know that at the time when I reach my goal on the $n$th toss, the last two tosses I saw would have both been heads.
Also, the third-from-final toss would have had to be a tails (otherwise, I would have won one toss earlier). The probability of this sequence
of THH is $\frac{1}{2}\times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.
Before these three tosses, my probability of winning would have been by definition, $a_{n-3}$. But if I am to win in the $n$th toss,
I need to exclude that event. And similarly $a_{n-4}$ and so on. This means:
$$a_n=\frac{1}{8}\left(1-\sum_{i=1}^{n-3} a_i\right)$$
Now let's define:
$$b_n = \sum_{i=1}^{n} a_i$$
which represents the probability you would have won by the $n$th toss.
Plugging this equation into the previous one we get:
$$b_n-b_{n-1} = \frac{1}{8}(1-b_{n-3})$$
| I have another approach to this I just worked out. Here, we try to use express the recurrence as a system of linear equations. Then, we leverage the eigen values of the matrix associated with the matrix of this linear system.
So far, we have only one equation. This is:
$$b_n -b_{n-1} + \frac{1}{8}b_{n-3} = \frac{1}{8}$$
To form a linear system, we need two more equations. Let's take some dummy equations.
$$b_{n-1}=b_{n-1}$$
$$b_{n-2}=b_{n-2}$$
Now, we can express this system in matrix form.
$$\left( \begin{array}{c}
b_n \\
b_{n-1}\\
b_{n-2}\\
\end{array} \right) = \left( \begin{array}{ccc}
1 & 0 & -\frac{1}{8} & \\
1 & 0 & 0\\
0 & 1 & 0\\
\end{array} \right) \left( \begin{array}{c}
b_{n-1} \\
b_{n-2}\\
b_{n-3}\\
\end{array} \right) + \left( \begin{array}{c}
\frac{1}{8} \\
0\\
0\\
\end{array} \right)$$
Now let:
$\beta_n = \left(\begin{array}{c}
b_n \\
b_{n-1}\\
b_{n-2}\\
\end{array} \right)$, $\gamma = \left(\begin{array}{c}
\frac{1}{8} \\
0\\
0\\
\end{array} \right)$ and $M = \left( \begin{array}{ccc}
1 & 0 & -\frac{1}{8} & \\
1 & 0 & 0\\
0 & 1 & 0\\
\end{array} \right)$.
We then get:
$$\beta_n = M \beta_{n-1} + \gamma$$
$$=M(M \beta_{n-2}+\gamma) + \gamma$$
$$=M^2 \beta_{n-2} + (I+M)\gamma$$
Extending this all the way we get:
$$\beta_n = M^{n-2}\beta_{2} + (I+M+M^2+ \dots + M^{n-3})\gamma$$
Now, assuming $M$ is diagonalizable (which it is) we can say:
$$M=E\Lambda E^{-1}$$
And this would imply:
$$M^n = E \Lambda^n E^{-1}$$
So we get:
$$\beta^n = E\Lambda^{n-2}E^{-1}\beta_2 + E(I+\Lambda+\Lambda^2+\dots+\Lambda^{n-3})E^{-1}\gamma$$
Now, if $\lambda_1$, $\lambda_2$ and $\lambda_3$ happen to be the eigen values of $M$ then,
$$\Lambda = \left( \begin{array}{ccc}
\lambda_1 & 0 & 0 & \\
0 & \lambda_2 & 0\\
0 & 0 & \lambda_3\\
\end{array} \right)$$
and,
$$(I+\Lambda+\Lambda^2 + \dots + \Lambda^{n-3}) = \left( \begin{array}{ccc}
\frac{1-\lambda_1^{n-2}}{1-\lambda_1} & 0 & 0 & \\
0 & \frac{1-\lambda_2^{n-2}}{1-\lambda_2} & 0\\
0 & 0 & \frac{1-\lambda_3^{n-2}}{1-\lambda_3}\\
\end{array} \right)$$
Using these, it is easy to see that:
$$\beta_n = c_0 + c_1 \lambda_1^{n-2} + c_2 \lambda_2^{n-2} + c_3 \lambda_3^{n-2}$$
And the eigen values happen to be: $\lambda_1, \lambda_2, \lambda_3 = \frac{\phi}{2}$, $\frac{1-\phi}{2}$, $0.5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The series $\frac{1}{2}+\frac{2}{5}+\frac{3}{11}+\frac{4}{23}+...$ Consider the expression $\frac{1}{2}+\frac{2}{5}+\frac{3}{11}+\frac{4}{23}+...$
Denote the numerator and the denominator of the $j^\text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1\qquad D_j=
2D_{j-1}+1$$
What is the $50^\text{th}$ term?
Must we evaluate that term-by-term until we reach the $50^\text{th}$ term?
What is the sum of the first $25$ term?
Must we add them one-by-one?
What is the exact value of the sum to $\infty$?
| $D_j=2D_{j-1}+1\\\ \ \ \ =2(2D_{j-2}+1)+1\\\ \ \ \ =4D_{j-2}+1+2\\\ \ \ \ =4(2D_{j-3}+1)+1+2\\\ \ \ \ \ \ \ \ \vdots\\\ \ \ \ =2^kD_{j-k}+2^k-1\\\ \ \ \ =2^{j-1}D_1+2^{j-1}-1\\\ \ \ \ =3\cdot2^{j-1}-1$
$s_n=N_n/D_n=\displaystyle\frac n{3\cdot2^{n-1}-1}, n\ge1$
$s_n<\displaystyle\frac n{3\cdot2^{n-1}-2^{n-1}}=\frac n{2^n}$
$\displaystyle\sum_1^\infty s_n<\sum_1^\infty \frac n{2^n}$ which is an AP-GP series
$\sum_1^\infty \frac n{2^n}=\frac12+\frac24+\frac38...$
$\frac12\sum_1^\infty \frac n{2^n}=0+\frac14+\frac28+\frac3{16}...$
$\sum_1^\infty \frac n{2^n}-\frac12\sum_1^\infty \frac n{2^n}=\frac12\sum_1^\infty \frac n{2^n}=\frac12+\frac14+\frac18...=1$
$\displaystyle\implies0<\sum_1^\infty s_n<2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3032189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Evaluate the limit of the sequence: $\lim_{n_\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$ Evaluate the limit of the sequence:
$$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$$
My try:
Stolz-cesaro: The limit of the sequence is $\frac{\infty}{\infty}$
$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
For our sequence:
$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}=\lim_{n\to\infty}\frac{\sqrt{n!}-\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})\cdot(1+\sqrt{n+1})-(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}=\lim_{n\to\infty}\frac{\sqrt{(n-1)!}\cdot(\sqrt{n-1})}{\left((1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})\right)\cdot(\sqrt{n}+1)}$
Which got me nowhere.
| $(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n}) \ge \sqrt{1}\cdot\sqrt{2}\cdots \sqrt{n}= \sqrt{n!}$, hence
$0 \le \frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})} \le \frac{1}{\sqrt{n}}.$
Can you proceed ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $\lim_{n\to\infty} \cos(\frac{\pi}{4}) \cos(\frac{\pi}{8})\ldots \cos(\frac{\pi}{2^n}) $ I already know that $$ a_n = \cos\left(\frac{\pi}{2^{n+1}}\right) = \overbrace{\frac{\sqrt{2+\sqrt{2+\ldots + \sqrt{2}}}}{2}}^{n\text{ roots}}$$
Also I know that $$\lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^n}\right) = 2
\text{ and if } a_n \xrightarrow {n\to\infty} a \text{ then } \sqrt[n]{a_1 a_2 \ldots a_n} \xrightarrow{n\to\infty} a $$
With that method I only got indeterminate form
$$ \lim_{n\to\infty} \cos\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi}{8}\right)\ldots \cos\left(\frac{\pi}{2^n}\right) = \Big(\frac{\sqrt[n]{a_1 a_2 \ldots a_n}}{2}\Big)^n = 1^\infty $$
Anyone knows a working solution?
| $$\sin(x)=2\sin(\frac{x}{2})\cos(\frac{x}{2})=2\cos(\frac{x}{2})2\sin(\frac{x}{4})\cos(\frac{x}{4})$$
By repeated application of the half angle formula we find
$$\sin(x)=\lim_{n\to\infty}2^n\sin(\frac{x}{2^n})\prod_{k=1}^{n}\cos(\frac{x}{2^{k}})$$
By expanding $\sin(\frac{x}{2^n})$ into its taylor series we can easily derive
$$ \sin(x)=x\prod_{k=1}^{\infty} \cos(\frac{x}{2^k})$$
So $$\prod_{k=2}^{\infty}\cos(\frac{\pi}{2^k})=\lim_{x\to\pi}\frac{\sin(x)}{x\cos(\frac{x}{2})}=\lim_{x\to\pi}\frac{2\sin(\frac{x}{2})}{x}=\frac{2}{\pi}$$
See also Viete's formula
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find all matrices which satisfy $M^2-3M+3I = 0$ I am trying to find all matrices which solve the matrix equation
$$M^2 -3M +3I=0$$
Since this doesn't factor I tried expanding this in terms of the coordinates of the matrix. It also occurs to me to put it into "vertex" form:
$$M^2 - 3M + \frac{9}{4}I+\frac{3}{4}I=0$$
$$(M-\frac{3}{2}I)^2 = -\frac{3}{4}I$$
but this doesn't look much better.
What I found from expanding by coordinates was, if $M=\pmatrix{a & b \\ c & d}$ then
$$\pmatrix{a^2+bc -3a + 3& ab + bd - 3b \\ ac+cd-3c & bc+d^2-3d+3} = \pmatrix{0&0\\0&0}$$
From the off-diagonal entries I get that either
$$a+d-3=0$$
or
$$b=c=0$$
If $a+d-3\not=0$ then $a^2-3a+3=0$ and likewise for $d$. Then we get more cases for $a$ and $d$.
If $a+d-3=0$ the upper-left is unchanged and the lower-right is
$$bc + (3-a)^2-3(3-a)+3 = 0$$
which simplifies to the same thing from the upper-left and so is redundant. In the off-diagonals
$$ac+c(a-3)-3c = 0 \Rightarrow $$
$$2ac-6c = 0$$
We again get cases, and I suppose after chasing cases enough you get the solution set.
However, it just feels like this can't be the intended solution given how tedious and uninformative all of this case-chasing is. Is there some bigger idea I'm missing?
| $m^2 - 3m + 3 = 0\\
\lambda = \frac {3}{2} \pm i\frac {\sqrt {3}}{2}$
You could say that it is all matrices with eigenvalues equal to $\frac {3}{2} + i\frac {\sqrt {3}}{2},\frac {3}{2} - i\frac {\sqrt {3}}{2}$
If we restrict our universe to real $2\times 2$ matrices.
Then it would be all matrices with characteristic equations equal to:
$\lambda^2 - 3\lambda + 3 = 0$
We are looking for matrices with trace equal to 3, and determinant 3.
$\begin{bmatrix} a & b\\ -\frac {a^2 -3a + 3}{b} & 3-a \end {bmatrix}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the limit $\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}$. We could use L'Hospital here, because both numerator as well as denominator tend towards 0, I guess. The derivative of the numerator is $$x^2\cdot \left(-\sin\left(\frac{1}{x}\right)\right) \cdot \left( -\frac{1}{x^2}\right) + 2x \cos\left(\frac{1}{x}\right)=\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right) $$ The derivative of the denominator is $\cos(x)$. So, $$\lim\limits_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)} = \lim\limits_{x\rightarrow 0}\displaystyle\frac{\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right)}{\cos(x)}$$
Is that right so far?
Thanks for the help in advance.
Best Regards,
Ahmed Hossam
| We don’t need l’Hopital, indeed by standard limits
$$\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}=\frac{x \cos\left(\frac{1}{x}\right)}{\frac{\sin(x)}x}\to \frac01=0$$
indeed by squeeze theorem
$$\left|x \cos\left(\frac{1}{x}\right)\right|\le |x|\to0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit Question $\lim_{x\to\infty} \sqrt{x^2+1}-x+1$ I understand the answer is 1 which kind of makes sense intuitively but I can't seem to get there. I would appreciate if someone pointed out which line of my reasoning is wrong, thanks. I tried writing all my steps
\begin{equation}
\lim_{x\to\infty} \sqrt{x^2+1}-x+1
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x^2+1 - x +1}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x^2 - x +2}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x \left( x - 1 +\frac{2}{x}\right)}{x \left( \sqrt{1+\frac{1}{x}}+1-\frac{1}{x} \right)}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x - 1 +\frac{2}{x}}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\infty - 1 + 0}{1+1-0}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\infty - 1}{2} = \infty
\end{equation}
Edit: Added correct steps for completeness, thanks for the quick answers!
\begin{equation}
\lim_{x\to\infty} \sqrt{x^2+1}-x+1
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x^2+1 - x^2+2x -1}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{2x}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x}{x} \frac{2}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}}
\end{equation}
\begin{equation}
\frac{2}{1+1} = 1
\end{equation}
| At line $3$ you should have
$$\frac{x^2+1-(x-1)^2}{\sqrt{x^2+1}+x-1}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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How many sequences can be made with 5 digits so that the difference between any two consecutive digits is $1$? Using the digits $0$, $1$, $2$, $3$, and $4$, how many ten-digit sequences can be written so that the difference between any two consecutive digits is $1$?
I was wondering if my solution is right.
Let $a(n)$ be the number of $n$ digit sequences that end with $0$ or $4$ so that the difference between any two consecutive digits is $1$.
$b(n)$ be the number of n digit sequences that end with $1$ or $3$ so that the difference between any two consecutive digits is $1$.
$c(n)$ be the number of n digit sequences that end with $2$ so
that the difference between any two consecutive digits is $1$.
$x(n)$ be the number of n digit sequences so that the difference between any two consecutive digits is $1$.
$x(n) = a(n) + b(n) + c(n)$
$a(n) = b(n-1)$
$b(n) = a(n-1) + 2c(n-1)$
$c(n) = b(n-1)$
By substituting $a(n-1)$ and $c(n-1)$ in the formula for $b(n)$ we get $b(n) = 3b(n-2)$
We know that $b(1) = 2, b(2) = 4$.
The characteristic equation for this recursion is $x^2-3 = 0$ with have the roots $3^{1/2}$ and $-3^{1/2}$, so $b(n) = A{(3^{1/2})}^{n} + B{(-3^{1/2})}^{n}$ where $A = {(3^{1/2}+2)}/{3}$ and $B = {(2-3^{1/2})}/{3}$. I think this is an integer.
We get $x(n) = 2b(n-1) + 3b(n-2)$ and by substituting we get something.
| Your approach looks good to me but you have to check the final result somehow. You can do that with 15 lines of code:
#include <iostream>
using namespace std;
int count(int startDigit, int length) {
if(startDigit < 0 || startDigit > 4) {
return 0;
}
if(length == 1)
return 1;
return count(startDigit - 1, length - 1) + count(startDigit + 1, length - 1);
}
int main() {
cout << count(1, 10) + count(2, 10) + count(3, 10) + count(4, 10);
}
...and the result is 567.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Distance from meeting the bisectors to the side of the quadrilateral In a quadrilateral $ABCD,\;AC$ and $BD$ are bisectors of $\angle BAD$ and $\angle ADC$. If $AC$ intersects $BD$ at $P,\;AB=6,\;CD=3$ and $\angle APD= 135º$, calculate the distance from $P$ to $AD$.
I even designed it but I did not find the solution;
|
Let $BE = b, EA = a, BA=c$ and the distance from $P$ to the three edges of $\triangle ABE$ be $r$. Then
$$r = \frac{b+a-c}{2}\tag{1}.$$
Since $BC$ is the angle bisector of $\angle EBA$,
$$\frac{a-6}{b} = \frac{6}c,\text{ or } (a-6)c = 6b\tag{2}.$$
Similarly
$$\frac{b-3}{b} = \frac3c,\text{ or }(b-3)c = 3a\tag{3}.$$
Take $2(3)-(2)$ we have
$$c(2b-a) = 6(a-b)\tag{4}.$$
On the other hand,
$$\color{red}{\frac{r}{a-6}} + \color{blue}{\frac{r}{b}} = \color{red}{\frac{BP}{BC}} + \color{blue}{\frac{PC}{BC}} = 1$$
so $$r(a+b-6) = (a-6)b.\tag{5}$$
And similarly
$$r(a+b-3) = (b-3)a.\tag{6}$$
Taking $(6)-(5)$ we obtain
$$ r = 2b - a.\tag{7}$$
Replace $r$ from $(7)$ into $(1)$ we get
$$a+b-c = 2(2b-a),\text{ or } c = 3(a-b).\tag{8}$$
Now replace $(7)$ and $(8)$ into $(4)$ we get $r=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve the system of equations in the set of real numbers. Solve the system of equations in the set of real numbers:
$$\begin{cases}
\frac1x + \frac1{y+z} = \frac13 \\
\frac1y + \frac1{x+z} = \frac15 \\
\frac1z + \frac1{x+y} = \frac17
\end{cases}$$
I got:
$$\begin{cases}
3(x+y+z)=x(y+z) \\
5(x+y+z)=y(x+z) \\
7(x+y+z)=z(x+y)
\end{cases}$$
However, no matter how I continue from here, I always get $x=y=z=0$, which cannot be true; or I get a new system of equations, but still with 3 variables (which I cannot solve).
How can I solve this problem or how should I approach it?
| Let $x+y+z=m$
Adding all the equations, we get,
$xy+yz+zx=\frac{3m+5m+7m}{2}=\frac{15m}{2}$
Subtracting each equation one by one from this, we get,
$xy=\frac{m}{2}$
$yz=\frac{9m}{2}$
$zx=\frac{5m}{2}$
Dividing by $ xyz$, we get, $$\frac{1}{x}:\frac{1}{y}:\frac{1}{z}=9:5:1$$
$$\Longrightarrow x:y:z=\frac{1}{9}:\frac{1}{5}:\frac{1}{1}$$
$$\Longrightarrow x:y:z=5:9:45$$
Now, let $x=5k, y=9k, z=45k$ and get the result.
Hope it is helpful
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
An exercise on the calculation of a function of operator The operator is given by
$$A=\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 4
\end{pmatrix}$$
I have to write down the operator $$B=\tan(\frac{\pi} {4}A)$$
I calculate $$\mathcal{R} (z) =\frac{1}{z\mathbb{1}-A}=\begin{pmatrix}
\frac{1}{z-1} & 0 & 0\\
\frac{1}{(z-1)^2} & \frac{1}{z-1} & 0\\
0 & 0 & \frac{1}{z-4}\end{pmatrix} $$
Now the B operator is given by:
$$B=\begin{pmatrix}
Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{z-1} & 0 & 0\\
Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{(z-1)^2} & Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{z-1} & 0\\
0 & 0 & Res_{z=4}\frac{\tan(\frac{\pi}{4}z)}{z-4}
\end{pmatrix} $$
For me the result should be
$$ B=\begin{pmatrix}
1 & 0 & 0\\
\frac{\pi}{2} & 1 & 0\\
0 & 0 & 0\end{pmatrix}$$
But the exercise gives as solution:
$$ B=\begin{pmatrix}
1 & 0 & 0\\
\frac{\pi}{4} & 1 & 0\\
0 & 0 & 1\end{pmatrix}$$
Where is the error?
Thank you and sorry for bad English
| Can you try using the exponent formula for matrices? It works the same way: $$e^{A} =1+A + \frac{1}{2!} A^2+\dots \in \text{GL}(2, \mathbb{R})$$
The exponent of any matrix is an invertible $3\times 3$ matrix.
if we need a tangent function defined on matrices all of $\tan, \sin, \cos$ can be done with exponent function:
$$ \tan A = \frac{e^{iA}-e^{-iA}}{e^{iA}+e^{-iA}}$$
What remains is to find the matrix exponent of the lower-triangular matrix.
if we can write $A=A_0+N$ with $A_0$ diagonal and $N$ nilpotent (here $N^3=0$), then $e^{A}=e^{A_0}e^{N}$. Indeed
$$A=
\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 4
\end{pmatrix}=
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 4
\end{pmatrix}+
\begin{pmatrix}
0 & 0 & 0\\
1 & 0 & 0\\
0 & 0 & 0
\end{pmatrix}=A_0+N
$$
This gets you started.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to show that $\sum_{n=1}^{\infty}\frac{\phi^{2n}}{n^2{2n \choose n}}=\frac{9}{50}\pi^2$ Given:$$\sum_{n=1}^{\infty}\frac{\phi^{2n}}{n^2{2n \choose n}}=\frac{9}{50}\pi^2$$
Where $\phi=\frac{\sqrt{5}+1}{2}$
How can I we show that the above sum is correct?
I have checked numerically, it seem correct, but i don't how to proves it
| Consider the well-known series expansion of the squared $\arcsin$ function. Namely
$$2\arcsin^2\left(\frac x2\right)=\sum_{n=1}^{\infty}\frac{x^{2n}}{n^2\binom{2n}n}$$
As you can see the RHS equals your given sum for the case that $x=\phi$. Plugging this in yields to
$$\begin{align}
2\arcsin^2\left(\frac \phi2\right)=\sum_{n=1}^{\infty}\frac{\phi^{2n}}{n^2\binom{2n}n}=\frac{9\pi^2}{50}
\end{align}$$
which further reduces the problem to
$$\sin\left(\frac{3\pi}{10}\right)=\frac\phi2=\frac{1+\sqrt{5}}{4}$$
The latter one can be shown by only using the fact that $\cos(\pi/5)=1/4(1+\sqrt 5)$. It is sufficient to show that
$$\begin{align}
\sin\left(\frac{3\pi}{10}\right)&=\cos\left(\frac{\pi}5\right)\\
\sin\left(\frac{3\pi}{10}\right)&=\cos\left(\frac{7\pi}{10}-\frac{\pi}2\right)\\
\sin\left(\frac{3\pi}{10}\right)&=\sin\left(\frac{7\pi}{10}\right)\\
\sin\left(\frac{3\pi}{10}\right)&=\sin\left(-\frac{3\pi}{10}+\pi\right)\\
\sin\left(\frac{3\pi}{10}\right)&=-\sin\left(-\frac{3\pi}{10}\right)
\end{align}$$
and the last line turn out to be true since the sine function is odd.
The assumed fact concerning the value of $\cos(\pi/5)$ can be easily verified by cosidering the fifths roots of unity. To be precise consider
$$\begin{align}
x^4+x^3+x^2+x+1&=0\\
x^2+\frac1{x^2}+x+\frac1x+1&=0\\
\left(x+\frac1x\right)^2+\left(x+\frac1x\right)-1&=0
\end{align}$$
The latter can be solved by the substitution $z=1/x+x$ which boils the problem down to the solving of two quadratic equations. The real part of the first solution $($the one with positive real and imaginary part$)$ equals the cosine of $\pi/5$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Evaluate:$S_{n}=\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{n-3}-\binom{n-3}{n-6}+.......$ If$$S_{n}=\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{n-3}-\binom{n-3}{n-6}+.......$$
Does $S_{n}$ have a closed form.
My Attempt
$$S_{n}=\binom{n}{0}-\binom{n-1}{n-2}+\binom{n-2}{2}-\binom{n-3}{3}+.......$$
$$S_{n}=\binom{n}{n}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+.......$$
$=$coefficient of $x^n$ in $\left\{(1+x)^n-x^2(1+x)^{n-1}+x^4(1+x)^{n-2}-x^6(1+x)^{n-3}+....\right\}$
After this not able to proceed
| If $S_{n} = \sum_{j\geq 0} (-1)^{j}\binom{n-j}{j}$, then
\begin{align}
S_{n+1} &= \sum_{j\geq 0} (-1)^{j} \binom{n+1-j}{j} \\
&= \sum_{j\geq 0} (-1)^{j} \left[\binom{n-j}{j} + \binom{n-j}{j-1}\right] \\
&= \sum_{j\geq 0} (-1)^{j} \binom{n-j}{j} - \sum_{j\geq 1} (-1)^{j-1}\binom{n-1-(j-1)}{j-1} \\
&= S_{n} - S_{n-1}
\end{align}
and $S_{1} = 1, S_{2} = 0$, so $S_{n}$ is
$$
1, 0, -1, -1, 0, 1, 1, 0, -1, -1, 0, 1, \dots
$$
which is a periodic sequence.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How do I get from $x - x^2 = \frac{1}{4}$ to $x =\frac{1}{2}$? I'm working on a text book problem where I need to sketch the graph of $y = 4x^2 - 4x+1$ by finding where the curve meets the $x$ axis.
To start out I set $y = 0$ then tried to isolate $x$ then,
$4x - 4x^2 = 1$
$x - x^2 = \frac{1}{4}$
From here I want to continue algebraically to reach $x = \frac{1}{2}$ which I know is the solution from plotting the curve using an app and I can see that that makes sense.
However I am missing some concepts which allow me to turn $x - x^2 = \frac{1}{4}$ into $x = \frac{1}{2}$ and wanted some help to get unblocked.
| $$x-x^2 = \frac{1}{4} \implies x^2-x+\frac{1}{4} = 0$$
Here, if you notice, you can see we have two perfect squares: $x^2$ and $\frac{1}{4}$. Also, twice the product of their square roots, which is $2\cdot x\cdot \frac{1}{2}$, gives the coefficient of the middle term. You probably know
$$a^2\pm 2ab+b^2 = (a\pm b)^2$$
Factoring the equation above in the same manner, you get
$$x^2-\color{blue}{x}+\frac{1}{4} = x^2-\color{blue}{2\cdot x\cdot\frac{1}{2}}+\left(\frac{1}{2}\right)^2 = \left(x-\frac{1}{2}\right)^2$$
$$\left(x-\frac{1}{2}\right)^2 = 0$$
$$x-\frac{1}{2} = 0$$
$$x = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using the binomial expansion to show $\sum_{k=0}^n {n \choose k} \frac{1}{2^k} = (3/2)^n$ I have to show $$\sum_{k=0}^n {n \choose k} \frac{1}{2^k} = (3/2)^n$$ using the binomial theorem. I haven't had any practice on these types of questions so i'm unsure as to how to proceed. Would anybody be able to what the method is for solving these types of questions, and, is there any general method of what we're trying to do? What's the thought process to solve these types of questions
Much appreciated.
Thank you
| There is a theorem (Binomial theorem, demostrable by induction) that shows you:
\begin{equation}
(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}
\end{equation}
Then if you use the $\displaystyle a=\frac{1}{2}$ and $b=1$, you obtain:
$$\sum_{k=0}^n \binom{n}{k} \frac{1}{2^k} = (3/2)^n$$
Proof:
The demostration of the Binomial Theorem is not complicated, the case $n=1$ is obviously true, then if we assume the case $n$, and we came to prove the case $n+1$ and the theorem is proven. If:
$$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}$$
Then:
\begin{eqnarray*}
(a+b)^{n+1} &=& (a+b)(a+b)^{n} = a(a+b)^{n} + b(a+b)^{n} \\
(a+b)^{n+1} &=& a\sum_{k=0}^n \binom{n}{k} a^k b^{n-k} + b\sum_{k=0}^n \binom{n}{k} a^k b^{n-k} \\
(a+b)^{n+1} &=& \sum_{k=0}^n \binom{n}{k} a^{k+1} b^{n-k} + \sum_{k=0}^n \binom{n}{k} a^k b^{n-k+1} \\
(a+b)^{n+1} &=& \sum_{k=1}^{n+1} \binom{n}{k-1} a^{k} b^{n-k+1} + \sum_{k=0}^n \binom{n}{k} a^k b^{n-k+1} \\
(a+b)^{n+1} &=& \left[a^{n+1} + \sum_{k=1}^{n} \binom{n}{k-1} a^{k} b^{n-k+1} \right] + \left[b^{n+1} +\sum_{k=1}^n \binom{n}{k} a^k b^{n-k+1} \right] \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\left(a^k b^{n-k+1}\left[ \binom{n}{k-1} + \binom{n}{k}\right]\right) \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\left(a^k b^{n-k+1}\left[ \frac{n!}{(k-1)!(n-k+1)!} + \frac{n!}{k!(n-k)!}\right]\right) \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\left(a^k b^{n-k+1}\left[ \frac{n!k}{k!(n-k+1)!} + \frac{n!(n-k+1)}{k!(n-k+1)!}\right]\right) \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\left(a^k b^{n-k+1}\left[ \frac{n!(n+1)}{k!(n-k+1)!}\right]\right) \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\left[a^k b^{n-k+1} \frac{(n+1)!}{k!(n+1-k)!}\right] \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\binom{n+1}{k}a^k b^{n+1-k} \\
(a+b)^{n+1} &=& \sum_{k=0}^{n+1}\binom{n+1}{k}a^k b^{n+1-k} \\
\end{eqnarray*}
And now the theorem is demostrated by induction. Here is a video of the demonstration.
| {
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"question_score": "1",
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System of equations has no solution
If the system of linear equations,
\begin{cases}
x &+ ay &+ z &= 3\\
x &+ 2y &+ 2z &= 6\\
x &+ 5y &+ 3z &= b\\
\end{cases}
has no solution, then:
*
*$a=-1,b=9$
*$a=-1,b \ne 9$
*$a\ne-1,b = 9$
*$a=1,b \ne 9$
I really fail to understand why the answer cannot be option (1).
Here's my method:
For $a = -1$, the coefficient determinant determinant is 0. And if we create $\Delta_z$, the determinant in which coefficients of $z$ are replaced by $3,6,b$, (b= 9), that determinant is non zero.
My book says that if coefficient determinant is $0$ and at least one of $\Delta_y, \Delta_z, \Delta_x$ is non $0$ then the system has no solution.
But the answer given is $(2)$. What's wrong with my method? What's the best way to tackle such problems?
| I suppose the best way to tackle is to row-reduce the matrix and see what happens. You have to eliminate the third and the first columns:
$$
\begin{pmatrix}
1 & a & 1 & 3\\
1 & 2 & 2 & 6 \\
1 & 5 & 3 & b
\end{pmatrix}
\to
\begin{pmatrix}
1 & a & 1 & 3\\
0 & 2-a & 1 & 3 \\
0 & 5-a & 2 & b-3
\end{pmatrix}
\to
\begin{pmatrix}
1 & 2a-2 & 0 & 0\\
0 & 2-a & 1 & 3 \\
0 & 1+a & 0 & b-9
\end{pmatrix}
$$
so if $a=-1$ the last equation reads $0=b-9$, thus having $b=9$ you have $0=0$, which eliminates the equation, and now you have 2 equations in 3 unknowns, hence an infinite number of solutions. Thus (i) cannot be the answer.
The condition for no solutions would require that $a=-1$ (forcing the last equation to read $0=b-9$) and $b \ne 9$ (forcing it to be false).
UPDATE
In retrospect, as you say, if $a=-1$ then $\Delta = 0$, so there will not be a unique solution. However, if $b = 9$, the last column of $\Delta_z$ is $(3,6,9) = 3 \times (1,2,3)$, so exactly $3$ times the last column of $\Delta$, hence, $\Delta_z = 3\Delta = 0$, and the idea in your book is correct.
| {
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In $\Delta ABC$ if $(\sqrt{3}-1)a=2b$, $A=3B$, then find $C$
In $\Delta ABC$ if $(\sqrt{3}-1)a=2b$, $A=3B$, then find $C$
My Attempt
$$
b=\frac{\sqrt{3}-1}{2}a\quad\& \quad \frac{A-B}{2}=B\quad\&\quad\frac{A+B}{2}=2B\\\frac{a-b}{a+b}=\frac{\tan\frac{A-B}{2}}{\tan\frac{A+B}{2}}\implies \frac{3-\sqrt{3}}{\sqrt{3}+1}=\frac{\tan B}{\tan 2B}=\frac{1-\tan^2B}{2}
$$
| In the standard notation by law of sines we obtain:
$$\frac{\sin\beta}{\sin3\beta}=\frac{\sqrt3-1}{2}$$ or
$$\frac{1}{3-4\sin^2\beta}=\frac{\sqrt3-1}{2}$$ or
$$3-4\sin^2\beta=\sqrt3+1$$ or
$$8\sin^2\beta=(\sqrt3-1)^2$$ or
$$\sin\beta=\frac{\sqrt3-1}{2\sqrt2},$$ which gives $$\beta=15^{\circ}.$$
Can you end it now?
| {
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"url": "https://math.stackexchange.com/questions/3062125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\left(\frac{1}{n}-\frac{1}{n+1}\right)\\
&=H_{1}\left(1-\frac{1}{2}\right)+H_{2}\left(\frac{1}{2}-\frac{1}{3}\right)+H_{3}\left(\frac{1}{3}-\frac{1}{4}\right)\\
&=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}+\frac{1}{3}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{9}-\frac{1}{12}
\end{align}$$
I think this method is not quite useful, so I tried another one:
$$H_n=\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
Then,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
| $\newcommand{\Li}{\operatorname{Li}}$Starting with a generating function for the Harmonic Numbers
$$-\frac{\log(1-x)}{1-x}~=~\sum_{n=1}^\infty H_nx^n\tag1$$
we divide both sides by $x$ and integrate afterwards to get
$$\small\begin{align*}
\int-\frac{\log(1-x)}{x(1-x)}\mathrm dx&=\int\sum_{n=1}^\infty H_nx^{n-1}\mathrm dx\\
\Li_2(x)+\frac12\log^2(1-x)+c&=\sum_{n=1}^\infty \frac{H_n}nx^{n}
\end{align*}$$
Plug in $x=0$ to determine the value of $c$ which turns out to be $0$ aswell. Integrating this equation yields to
$$\small\begin{align*}
\int\Li_2(x)+\frac12\log^2(1-x)\mathrm dx&=\int\sum_{n=1}^\infty \frac{H_n}nx^{n}\mathrm dx\\
x\Li_2(x)-x+(x-1)\log(1-x)+\frac12(x-1)(\log^2(1-x)-2\log(1-x)+2)+c&=\sum_{n=1}^\infty\frac{H_n}{n(n+1)}x^{n}
\end{align*}$$
The new constant again equals $0$ due the same argumentation as above. Now we can plug in $x=1$ to compute the value of your given series which overall is given by $\Li_2(1)$ and this is well-known to equal $\zeta(2)$.
$$\therefore~\sum_{n=1}^\infty\frac{H_n}{n(n+1)}~=~\frac{\pi^2}6\tag2$$
| {
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"source": "stackexchange",
"question_score": "6",
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solving $\lim\limits_{x\rightarrow\infty} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}}$ To investigate the convergence of a series I have to solve the folliwing limit:
\begin{equation}
\lim\limits_{x\rightarrow\infty} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}}
\end{equation}
It should be $\frac{1}{2}$ but i can't quite seem to get to that solution. I've tried to factor the square root out of the quotient which was:
\begin{equation}
\lim\limits_{x\rightarrow\infty} \sqrt{\frac{((x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1})^2}{x^2(x + 1)}}
\end{equation}
Then i worked out the square in the numerator which was:
\begin{equation}
(x^2-1)^2 (x + 2)-2x^2(x^2-1)\sqrt{x+1}\sqrt{x+2}+x^4(x+1)
\end{equation}
I could then factor the terms and take and divide the numerator with the (x+1) from the denominator. Then i could expand the terms in the numerator wich became:
\begin{equation}
\lim\limits_{x\rightarrow\infty}\sqrt{\frac{(x^4+x^3-3x^2-x+2)-\sqrt{16x^8+32x^7-12x^6-20x^5+8x^4}+x^4}{x^2}}
\end{equation}
Now i can take the $16x^8$ out of the root and then i then looked at the terms with the highest exponent so i had:
\begin{equation}
\lim\limits_{x\rightarrow\infty}\sqrt{\frac{x^4-4x^4+x^4}{x^2}} =\lim\limits_{x\rightarrow\infty}\sqrt{\frac{-2x^2}{x^2}}
\end{equation}
Which could only be solved with complex numbers, so i should be wrong somewhere in my calculations, since i know that i should get $\frac{1}{2}$. I also checked my sollution with WolframAlpha which also gave $\frac{1}{2}$ so I know that the sollution i have is correct.
Would anyone know where i was wrong or how i could better solve it?
| $$\lim_{x\to\infty}\frac{(x^2-1)\sqrt{x+2}-x^2\sqrt{x+1}}{x\sqrt{x+1}}=\lim_{x\to\infty}\frac{x^2\big[\sqrt{x+2}-\sqrt{x+1}\big]}{x\sqrt{x+1}}-\frac{\sqrt{x+2}}{x\sqrt{x+1}}$$
The latter goes to $0$ since $$\lim_{x\to\infty}\frac{\sqrt{x+2}}{x\sqrt{x+1}}=\lim_{x\to\infty}\frac1x\cdot\sqrt{\frac{x+2}{x+1}}=\lim_{x\to\infty}\frac1x\cdot\sqrt{1+\frac1{x+1}}=0$$
You are left with $$\lim_{x\to\infty}\frac{x\big[\sqrt{x+2}-\sqrt{x+1}\big]}{\sqrt{x+1}}=\lim_{x\to\infty}\frac{x\big[\sqrt{x+2}-\sqrt{x+1}\big]}{\sqrt{x+1}}\cdot\frac{\big[\sqrt{x+2}+\sqrt{x+1}\big]}{\big[\sqrt{x+2}+\sqrt{x+1}\big]}\\=\lim_{x\to\infty}\frac x{\sqrt{x+1}\big[\sqrt{x+2}+\sqrt{x+1}\big]}=\lim_{x\to\infty}\frac 1{\sqrt{1+\frac1x}\Big[\sqrt{1+\frac2x}+\sqrt{1+\frac1x}\Big]}=1/2$$
| {
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Simplify $\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} * \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$ to $\frac{\sqrt{mnc}}{a^9cmn}$ I need to simplify $$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$$
The solution provided is: $\dfrac{\sqrt{mnc}}{a^9cmn}$.
I'm finding this challenging. I was able to make some changes but I don't know if they are on the right step or not:
First, I am able to simplify the left fractions numerator and the right fractions denominator:
$\sqrt{mn^3}=\sqrt{mn^2n^1}=n\sqrt{mn}$
$\sqrt{m^2c^4}=m\sqrt{c^2c^2} = mcc$
So my new expression looks like:
$$\frac{n\sqrt{mn}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{mcc}$$
From this point I'm really at a loss to my next steps. If I multiply them both together I get:
$$\frac{(n\sqrt{mn})(a^{-7}n^{-2})}{(a^2\sqrt{c^{-3}})(mcc)}.$$
Next, I was thinking I could multiply out the radical in the denominator but I feel like I need to simplify what I have before going forwards.
Am I on the right track? How can I simplify my fraction above in baby steps? I'm particularly confused by the negative exponents.
How can I arrive at the solution $\dfrac{\sqrt{mnc}}{a^9cmn}$?
| \begin{align}
\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}} &= \frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{1}{a^7n^{2} mc^2}\\
& = \frac{\sqrt{mn^3}}{a^2} \cdot \frac{\sqrt{c^{3}}}{a^7n^{2}mc^2} \\
& = \frac{\sqrt{m}n\sqrt{n}c\sqrt{c}}{a^9 n^{2}m c^2} \\
& = \frac{\sqrt{nmc}}{a^9 n m c} \\
\end{align}
| {
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"source": "stackexchange",
"question_score": "2",
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Associated elements in $\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$ I got four elements and want to check whether these elements are associated or not. For
$\alpha = \frac{1+\sqrt{-3}}{2}$
my elements are:
$a_1 = 2 - \alpha = \frac{3 - \sqrt{-3}}{2}$
$a_2 = 1 - 2\alpha = -\sqrt{-3}$
$ a_3 = 3 + 2\alpha = 4 + \sqrt{-3}$
$ a_4 = 3 - 2\alpha = 2 - \sqrt{-3}$
I just calculated $\frac{a_1}{a_2} = \frac{1-\sqrt{-3}}{2}$ and $\frac{a_2}{a_1} = \frac{-1+\sqrt{-3}}{2}$. All other pairs $\frac{a_1}{a_3}, \frac{a_1}{a_4},\frac{a_2}{a_3} \frac{a_2}{a_4}, \frac{a_3}{a_4}$ give an odd denominator which is not possble in $\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$ (except the denomitor 1), therefore only $a_1$ and $a_2$ are associated... is this right? Or do I miss something?
Thx for any help on this
| I would just compute the norms first. We have
$$N(a_1)=\frac{3^2+3}{4}=3,$$
$$N(a_2)=3,$$
$$N(a_3)=4^2+3=19,$$
and
$$N(a_4)=2^2+3=7.$$
So the only candidates to be conjugates are $a_1$ and $a_2$, which you have verified to be the case.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute the following floor sum? I want to compute
$$S = \sum_{k=0}^{m} \left\lfloor \frac{n-5k}{2}\right\rfloor.$$
This sum was motivated by the need to compute the number of solutions to
$$x_1+2x_2 +5x_3 = n (*)$$
In particular we observe that the number of solutions to $x_1+2x_2=n$ is $\left\lfloor\frac{n}{2}\right\rfloor+1$ and so the number of solutions for $(*)$ is
$$a(n) =\sum_{k=0}^{\lfloor n/5\rfloor} \left\lfloor \frac{n-5k}{2}\right\rfloor+1.$$
This is because we can write $(*)$ as $x_1+2x_2 = n-5x_3$ and then we can range $0\leq x_3 \leq \lfloor n/5\rfloor$ to get all solutions. We have denoted $m = \lfloor n/5\rfloor$ and so
$$a(n) = S + m+1.$$
I am not sure how to start computing $S$, perhaps someone can give some indications.
| First, suppose $n$ and $m$ are even. Then the sum of the even terms of $S$ is
$$S_2 = \sum\limits_{k=0}^{m/2}\frac{n-10k}{2} = \frac{nm}{4} - \frac{5m(m+2)}{8}.$$
and similarly, if $n$ is even and $m$ is odd, the sum of the odd terms of $S$ is
$$S_1 = \sum\limits_{k=0}^{(m-1)/2}\frac{n - 6 - 10k}{2} = \frac{(n-6)(m-1)}{4} - \frac{5(m-1)(m+1)}{8}.$$
For a general $m$, if $l := \left\lfloor\frac{m}{2}\right\rfloor$, then $$S_2 = \dfrac{nl}{2}-\dfrac{5l(l+1)}{2},$$ and similarly, with $p := \left\lfloor\frac{m-1}{2}\right\rfloor$, we have
$$S_1 = \frac{p(n-6)}{2} - \frac{5p(p+1)}{2}.$$
Now, adding those together, we have
\begin{align*}S = S_1 + S_2 &= \frac{p(n-6)-5p(p+1)+nl-5l(l+1)}{2}
\\&= \frac{(p+l)n-11p-5p^2-5l^2-5l}{2}\end{align*}
But $p+l = m-1$ (if neither had been rounded down, it would sum to $\frac{2m-1}{2}$, but exactly one has been rounded down by exactly $\frac{1}{2}$, so it sums to $\frac{2m-2}{2} = m-1$), so we can do some simplifications:
$$S = \frac{(m-1)n-6p-5m-5(p^2+l^2)}{2}$$
Similarly if there were no rounding, then we would have, $p^2+l^2 = \frac{m^2+(m-1)^2}{4} = \frac{2m^2-2m+1}{4}$, but our rounding reduces this by exactly $\frac{1}{4}$, so in fact $p^2 +l^2 = \frac{m^2-m}{2}$, allowing a further simplification:
$$S = \frac{2n(m-1)-12p-15m-5m^2}{4}.$$
Now, if $m$ is odd, then $12p = 6(m-1)$, while if $m$ is even, then $12p = 6(m-2)$, so if $I_o(m)$ is $1$ when $m$ is odd and $0$ when $m$ is even, then $12p = 6m-12+6I_o(m)$
$$S = \frac{2n(m-1)-21m-5m^2+12-6I_o(m)}{4}$$
which is as nice a format as we're going to get it into.
In the case where $n$ is odd, we see that $S_2$ is reduced by $1$ in each term, so by $l$ in total, and similarly $S_1$ is increased by $1$ in each term, so by $p$ in total, so $S$ changes by $p - l$ in total, and $p - l = I_o(m) - 1$, so our final sum in this case is
$$S = \frac{2n(m-1)-21m-5m^2+8-2I_o(m)}{4}.$$
| {
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How can I calculate $\int\frac{x-2}{-x^2+2x-5}dx$? I'm completely stuck on solving this indefinite integral:
$$\int\frac{x-2}{-x^2+2x-5}dx$$
By completing the square in the denominator and separating the original into two integrals, I get:
$$-\int\frac{x}{x^2-2x+5}dx -\int\frac{2}{(x-1)^2 + 4}dx$$
The second one is trivial, but the first one has me stuck. Whatever substitution I apply or form I put it in, I just can't figure it out. They're meant to be solved without partial integration, by the way.
| The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):
\begin{align}\int\frac{x-2}{-x^2+2x-5}dx &= - \int \frac{x-2}{x^2 - 2x + 5}dx
\\&= -\frac 12 \int \frac{2x-4}{x^2 - 2x + 5}dx. \end{align} The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields
\begin{align}
\int\frac{x-2}{-x^2+2x-5}dx = -\frac 12 \left(\int \frac{2x-2}{x^2-2x+5}dx - \int \frac{2}{x^2-2x+5}dx\right).
\end{align}
The second integral you can solve, can you solve the first?
| {
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"timestamp": "2023-03-29T00:00:00",
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Envelope Question: Five letters addressed to individuals 1-5 are randomly placed in five addressed envelopes, one letter in each envelope. I'm trying to find the probability of:
*
*Exactly three letters go in the correct envelopes.
*Exactly two letters go in the correct envelopes
*No letters go in the correct envelopes
Here is my approach:
So there is clearly a total of 5! distinct ways of arranging the letters.
*
*If exactly three letters go in the correct envelopes, then there are $5 \choose 3$ ways of choosing the positions for the three correct envelopes, and for the remaining two letters, there are 2! ways of organizing them. Thus, probability = $\frac{{5 \choose 3} \cdot 2!}{5!}$.
*If exactly two letters go in the correct envelopes, then there are $5 \choose 2$ ways of choosing the positions for the two correct envelopes, and for the remaining three letters, there are 3! ways of organizing them. Thus, probability = $\frac{{5 \choose 2} \cdot 3!}{5!}$.
*I'm not really sure how to approach this problem.
Any input would be great.
| Note the use of the word exactly. You did not take that into account.
Probability that exactly three letters are placed in the correct envelopes
There are indeed $\binom{5}{3}$ ways to select which three letters are placed in the correct envelopes. That leaves two envelopes in which to place the remaining two letters. Those letters must each be placed in the other envelope, which can only be done in one way. Hence, the probability that exactly three letters go in the correct envelopes is
$$\frac{1}{5!} \cdot \binom{5}{3}$$
Probability that exactly two letters are placed in the correct envelopes
There are $\binom{5}{2}$ ways to select which two letters are placed in the correct envelopes. None of the remaining letters go in the correct envelopes. There are just two ways to do this.
\begin{align*}
&\color{red}{1}, \color{red}{2}, \color{red}{3}\\
&\color{red}{1}, 3, 2\\
&2, 1, \color{red}{3}\\
&2, 3, 1\\
&3, 1, 2\\
&3, \color{red}{2}, 1
\end{align*}
Hence, there are $\binom{5}{2} \cdot 2$ favorable cases.
We can use the Inclusion-Exclusion Principle to see why.
There are $3!$ ways to permute the three letters. From these, we must subtract those cases in which at least one letter is placed in the correct envelope.
There are three ways to select a letter to be placed in the correct envelope and $2!$ ways to place the remaining letters.
There are $\binom{3}{2}$ ways to select two letters to be placed in the correct envelopes and $1!$ ways to place the remaining letter.
There are $\binom{3}{3}$ ways to select three letters to be placed in the correct envelopes and $0!$ ways to place the (nonexistent) remaining letters.
By the Inclusion-Exclusion Principle, the number of ways to place the remaining three letters so that none of them is placed in the correct envelope is
$$3! - \binom{3}{1}2! + \binom{3}{2}1! - \binom{3}{3} = 6 - 6 + 3 - 1 = 2$$
as claimed.
Therefore, the probability that exactly two letters are placed in the correct envelopes is
$$\frac{1}{5!} \cdot 2\binom{5}{2}$$
Probability that none of the letters is placed in the correct envelopes
This is a derangement problem. An Inclusion-Exclusion arguments shows that the number of ways of placing none of the five letters in its correct envelope is
$$5! - \binom{5}{1}4! + \binom{5}{2}3! - \binom{5}{3}2! + \binom{5}{4}1! - \binom{5}{5}0!$$
Dividing that number by $5!$ gives the desired probability.
| {
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Let $a, b \in \mathbb{R^+}$ such that $a-b=10$, find smallest value of constant $K$ for which $\sqrt{(x^2 + ax)}- \sqrt {(x^2 +bx)}0$. Let $a, b$ are positive real numbers such that $a - b = 10$ , then the smallest value of the constant $K$ for which $\sqrt{(x^2 + ax)} - \sqrt {(x^2 +bx)} < K$ for all $x>0$.
My try:
Unable to solve further
| Using $$ \sqrt{a}-\sqrt{b} = {a-b\over \sqrt{a}+\sqrt{b}}$$
so $${(x^2+ax)-(x^2+bx)\over \sqrt{(x^2 + ax)} + \sqrt {(x^2 +bx)}} < K$$
thus $${10x\over \sqrt{(x^2 + ax)} + \sqrt {(x^2 +bx)}} < K$$
taking $x\to \infty$ we get $$\lim _{x\to \infty}{10\over \sqrt{1 + {a\over x}} + \sqrt {1 +{b\over x}}} = 5 \leq K$$
| {
"language": "en",
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Primitive of a function with $\sin \frac{1}{x}$ I have the next integral:
$$\int\biggl({\frac{\sin \frac{1}{x}}{x^2\sqrt[]{(4+3 \sin\frac{2}{x})}}}\biggr)\,dx ,\;x\in \Bigl(0,\infty\Bigr)$$
I used the substitution $u=\frac{1}{x}$ and I got $$-\int\biggl({\frac{\sin u}{\sqrt[]{(4+3 \sin2u)}}}\biggr)\,du$$
Can somebody give me some tips about what should I do next, please?
| The substitution $u=1/x$ yields $dx=-\frac{1}{u^2}\,du$, so the integral becomes
$$
\int\frac{-\sin u}{\sqrt{4+3\sin2u}}\,du=
\int\frac{-\sin u}{\sqrt{4+3\sin2u}}\,du
$$
This can be improved by setting $u=\pi/4-v$, so we get
$$
\frac{1}{\sqrt{2}}\int\frac{\cos v-\sin v}{\sqrt{4+3\cos2v}}\,dv=
\frac{1}{\sqrt{2}}\biggl(
\int\frac{\cos v}{\sqrt{7-6\sin^2v}}\,dv
-\int\frac{\sin v}{\sqrt{6\cos^2v+1}}\,dv
\biggr)
$$
that you should be able to manage.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Which point of the graph of $y=\sqrt{x}$ is closest to the point $(1,0)$? This problem was assigned for an AP Calculus AB class and was not allowed a calculator:
Which point of the graph of $y=\sqrt{x}$ is closest to the point $(1,0)$?
We are not given answers and the teacher will be absent for $2$ weeks. I need to check my answer $(1/2,1/4)$ before she returns.
Here is my work:
$d=\sqrt{(x-1)^2+(y-0)^2}$
$d=\sqrt{(x-1)^2+x}$
$d^2=(x-1)^2+x$
$2dd'=2(x-1)+1$
$d'=(2(x-1)+1)/(2d)$
$0=2(x-1)+1$
$x=1/2$
| As a check:
Partial answer, completing the square.
$x,y\ge 0$.
$d^2= (x-1)^2+x= $
$x^2-2x+1+x= x^2-x+1;$
$d^2=(x-1/2)^2 -1/4+1 =$
$(x-1/2)^2 +3/4\ge 3/4$ (why?).
$d^2_{min} =3/4$, at $x=1/2$, $y=\sqrt{1/2}$.
| {
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Can we have two different polynomials of the same degree $d$ here in the factorisation of $x^{p^n} -x$? In the proposition "The polynomial $x^{p^n} -x$ is precisely the product of all the distinct irreducible polynomials in $\Bbb F_p[x]$ of degree $d$ where $d$ runs through all divisors of $n$."
Can we have two different polynomials of the same degree $d$ here in the factorisation of $x^{p^n} -x$?
| Well, another example is
$$x^{16}-x = x(x+1)(x^2+x+1)(x^4+x+1)(x^4+x^3+1)(x^4+x^3+x^2+x+1),$$
where the polynomials $x^4+x+1$ and $x^4+x^3+1$ are primitive and conjugate and the polynomial $x^4+x^3+x^2+x+1$ is not primitive (the roots are 5th roots of unity).
| {
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Find any local max or min of $x^2+y^2+z^2$ s.t $x+y+z=1$ and $3x+y+z=5$
Find any local max or min of
\begin{align}
f(x,y,z)=x^2+y^2+z^2 && (1)
\end{align}
such that
\begin{align}
x+y+z=1 && (2)\\
3x+y+z=5 && (3)
\end{align}
My attempt. Let
$L(x,y,z,\lambda_1, \lambda_2)= f(x,y,z)+\lambda_2 (x+y+z-1) + \lambda_1 (3x+y+z-5)$
$L_x=2x+ 3 \lambda_1 + \lambda_2 =0$
$L_y=2y+ \lambda_1 + \lambda_2=0$
$L_z=2z+\lambda_1 + \lambda_2=0$
Solve for $x,y,z$ we get:
$x=\frac{-3 \lambda_1 - \lambda_2}{2}$
$z=y=\frac{-\lambda_1 - \lambda_2}{2}$
with the use of $(2)$ and $(3)$ $\implies$
$x=2$
$y=z= \frac{-1}{2}$
so the stationary point is $(x,y,z)=(2, \frac{-1}{2},\frac{-1}{2})$
The Hessian of $L$ gives a postive definite matrix for all $(x,y,z)$ thus
$(x,y,z)=(2, \frac{-1}{2},\frac{-1}{2})$ is the only local minimizer of $f$ and there is no maximizors of $f$.
Is this correct?
| Your result is correct. This is an alternative approach.
Here we have two non-parallel planes which intersect along the line $(x,y,z)=(2,t,-1-t)$ where $t\in \mathbb{R}$.
Hence we reduce the problem to a $1$-variable case,
$$f(x,y,z)=4+t^2+(-1-t)^2=2t^2+2t+5$$
The above quadratic polynomial has just one local/global minimum at $t=-1/2$ that is at $(2,-1/2,-1/2)$. There is no local/global maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Not possible to find non-zero terms of series expansion? I've been asked to compute the first 3 nonzero terms of a power series expansion about x=0 for two linearly independent solutions to the ODE:
$$(1+x^3)y''- 6xy =0 $$
I have tried to solve this many different ways and continue to get the same solution, which does not allow me to find three nonzero terms from two linearly independent solutions
I have used $$y(x) = \sum_{n=0}^\infty a_nx^n $$
$$ y'(x)=\sum_{n=1}^\infty a_nnx^{n-1} $$
$$ y''(x)=\sum_{n=2}^\infty a_nn(n-1)x^{n-2} $$
to obtain the series form
$$ \sum_{n=2}^\infty a_nn(n-1)x^{n-2} + x^3\sum_{n=2}^\infty a_nn(n-1)x^{n-2} -6x\sum_{n=0}^\infty a_nx^n$$
Add in x terms:
$$ \sum_{n=2}^\infty a_nn(n-1)x^{n-2} + \sum_{n=2}^\infty a_nn(n-1)x^{n+1} -\sum_{n=0}^\infty 6a_nx^{n+1}$$
Set all the powers of x equal to n:
$$ \sum_{n=0}^\infty a_{n+2}(n+2)(n+1)x^{n} + \sum_{n=3}^\infty a_{n-1}(n-1)(n-2)x^{n} -\sum_{n=1}^\infty 6a_{n-1}x^{n}$$
Peel off terms to have all series start at n=3:
$$ 2a_2x^0+6a_3x^1-6a_0x^1+12a_4x^2-6a_1x^2$$ $$+$$
$$\sum_{n=3}^\infty [a_{n+2}(n+2)(n+1)+a_{n-1}(n-1)(n-2)-6a_{n-1}]x^n=0$$
From this I have deduced:
$ x^0 : a_2=0 $
$ x^1 : a_3=a_0 $
$ x^2 : a_4=\frac{a_1}{2} $
$ x^n, n\geq3 : $
\begin{align}
& a_{n+2}=-\frac{a_{n-1}(n^2-3n+2-6)}{(n+2)(n+1)}\\
& = -\frac{a_{n-1}(n+1)(n-4)}{(n+1)(n+2)}\\
& = -\frac{a_{n-1}(n-4)}{n+2}\\
\end{align}
Using the recursion equation above I obtained these terms:
$n=3 : $
$$ a_5=\frac{a_2}{5}=0 $$
$n=4 : $
$$ a_6=a_3(4-4)=0 $$
$n=5 : $
$$ a_7=\frac{-a_4}{7}=\frac{-a_1}{14} $$
$n=6 : $
$$ a_8=\frac{-2a_5}{5}=\frac{-a_2}{20}=0$$
$n=7 : $
$$ a_9=\frac{-3a_6}{9}=0$$
$n=8 : $
$$ a_10=\frac{-4a_7}{12}=\frac{a_4}{21}=\frac{a_1}{42} $$
$n=9 : $
$$ a_11=\frac{-5a_8}{11}=\frac{a_2}{44}=0 $$
$n=10 : $
$$ a_12=\frac{-6a_9}{12}=\frac{-a_2}{2}=0 $$
$n=11 : $
$$ a_13=\frac{-7a_10}{13}=\frac{-7a_4}{273}=\frac{-a_1}{78} $$
So, every second and third term equal zer0.
My solution is:
$$ y_1(x)=a_0(1+x^3) $$
$$ y_2(x)=a_1\big(x+\frac{x^4}{2}-\frac{x^7}{14}+\frac{x^{10}}{42} - ...) $$
Can someone please tell me what I am doing wrong here?
I cannot come up with three non zero terms from $y_1$ as there are only two terms, and everything else is zero.
When I try to apply the ratio test to this series solution I get inconclusive results as well which further makes me think my solution is incorrect...
Any help or advice would be very greatly appreciated.
| To verify the correctness of your solution I made the following MATHEMATICA script with the results shown below.
Operator = (1 + x^3) D[#, {x, 2}] - 6 x # &;
n = 20;
Sumb = Sum[Subscript[alpha, k] x^k, {k, 0, n}];
res = Operator[Sumb] /. {Subscript[alpha, 0] -> Subscript[a, 0],
Subscript[alpha, 1] -> Subscript[a, 1]};
coefs = CoefficientList[res, x];
equs = Thread[coefs == 0];
For[k = 1; Alphas = {}; equsk = equs[[1]]; subs = {},
k <= Length[equs] - 1, k++,
solalphak = Solve[equsk, Subscript[alpha, k + 1]];
AppendTo[Alphas, Subscript[alpha, k + 1] /. solalphak][[1]];
AppendTo[subs, solalphak];
equsk = equs[[k + 1]] /. Flatten[subs]
]
Alphas0 = Flatten[Alphas];
series = Subscript[a, 0] + x Subscript[a, 1] +
Sum[Alphas0[[k]] x^(k + 1), {k, 1, Length[Alphas0] - 2}]
$$
y(x) = a_0(1+x^3)+a_1\left(x+\frac{x^4}{2}-\frac{x^7}{14}+\frac{x^{10}}{35}-\frac{x^{13}}{65}+\frac{x^{16}}{104}-\frac{x^{19}}{152}+\cdots +\right)
$$
This is a convergent series for $|x| < 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Definite integration when denominator consists of $x\sin x +1$ $$\int_0^\pi\frac{x^2\cos^2x-x\sin x-\cos x-1}{(1+x\sin x)^2}dx$$
The answer is $0$. I tried and made $(x\sin x +1)^2 $ in numerator and proceed,
but not able to do any further.
| First, let us try and simplify the integrand a bit. As
\begin{align}
\frac{x^2 \cos^2 x - x \sin x - \cos x - 1}{(1 + x \sin x)^2} &= \frac{x^2 - x^2 \sin^2 x - x \sin x - \cos x - 1}{(1 + x \sin x)^2}\\
&= \frac{x^2 + x \sin x - \cos x - (1 + x \sin x)^2}{(1 + x \sin x)^2}\\
&= -1 + \frac{x^2 + x \sin x - \cos x}{(1 + x\sin x)^2},
\end{align}
we have for the integral
$$I = -\pi + \int_0^\pi \frac{x^2 + x \sin x - \cos x}{(1 + x \sin x)^2} = -\pi + J.$$
To find the integral $J$ one can make use of the so-called reverse quotient rule (For another example using this method see here).
Recall that if $u$ and $v$ are differentiable functions, from the quotient rule
$$\left (\frac{u}{v} \right )' = \frac{u' v - v' u}{v^2},$$
and it is immediate that
$$\int \frac{u' v - v' u}{v^2} \, dx = \int \left (\frac{u}{v} \right )' \, dx = \frac{u}{v} + C. \tag1$$
For the integral $J$ we see that $v = 1 + x \sin x$. So $v' = \sin x + x \cos x$. Now for the hard bit. We need to find (conjure up?) a function $u(x)$ such that
$$u' v - v' u = u'(1 + x \sin x) - u (\sin x + x \cos x) = x^2 + x \sin x - \cos x.$$
After a little trial and error we find that if
$$u = -x \cos x,$$
as
$$u' = -\cos x + x \sin x,$$
this gives
$$u' v - v' u = x^2 + x \sin x - \cos x,$$
as required.
Our integral can now be readily found as it can be rewritten in the form given by (1). The result is:
\begin{align}
I &= -\pi + \int_0^\pi \left (\frac{-x \cos x}{1 + x \sin x} \right )' \, dx\\
&= -\pi - \left [\frac{x \cos x}{1 + x \sin x} \right ]_0^\pi\\
&= -\pi + \pi\\
&= 0,
\end{align}
as announced.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove combinatorial $\sum_{k=0}^{n/2} {n\choose2k} = \sum_{k=0}^{n/2 - 1} {n\choose2k+1} = 2^{n-1}$ I have problems solving the following formula for even positive integers $n$:
$$\sum_{k=0}^{n/2} {n\choose 2k} = \sum_{k=0}^{n/2 - 1} {n\choose 2k+1} = 2^{n-1}$$
I tried to prove it by induction but it didn't work. Is there any combinatorial proof?
| For brevity we prove $$\sum_{k=0}^{n} \binom{2n}{2k} = \sum_{k=0}^{n-1} \binom{2n}{2k+1} = 2^{2n-1},\hspace{2cm}(*)$$ for $n\in\mathbb{N}$. It is sufficient to show that $\sum_{k=0}^{n} \binom{2n}{2k}=2^{2n-1}$ since we have
$$\sum_{k=0}^{n} \binom{2n}{2k} +\sum_{k=0}^{n-1} \binom{2n}{2k+1}=\sum_{k=0}^{2n} \binom{2n}{k} = 2^{2n}.\hspace{2cm}(**)$$ Let apply induction for $\sum_{k=0}^{n} \binom{2n}{2k}=2^{2n-1}$.
For $n=1$ equality is true. For $n=m$ assume equality is true, namely $\sum_{k=0}^{m} \binom{2m}{2k}=2^{2m-1}$(Here note that $\sum_{k=0}^{m-1} \binom{2m}{2k+1}=2^{2m-1}$ is also true by (**)). For $n=m+1$,
\begin{eqnarray*}\sum_{k=0}^{m+1} \binom{2m+2}{2k}&=&\sum_{k=0}^{m+1}\left\{ \binom{2m}{2k-1}+\binom{2m}{2k-2}+\binom{2m}{2k}+\binom{2m}{2k-1}\right\}\\
&=&\sum_{k=0}^{m+1} \binom{2m}{2k}+ 2\sum_{k=0}^{m+1} \binom{2m}{2k-1}+\sum_{k=0}^{m+1} \binom{2m}{2k-2}\\
&=&\sum_{k=0}^{m} \binom{2m}{2k}+ 2\sum_{k=0}^{m+1} \binom{2m}{2k-1}+\sum_{k=0}^{m+1} \binom{2m}{2k-2}\\
&=&\sum_{k=0}^{m} \binom{2m}{2k}+2\sum_{k=0}^{m-1} \binom{2m}{2k+1}+\sum_{k=0}^{m} \binom{2m}{2k}\\
&=&4.2^{2m-1}=2^{2m+1}\end{eqnarray*} by use of $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$. Hence equality $\sum_{k=0}^{n} \binom{2n}{2k}=2^{2n-1}$ is valid for $n\in\mathbb{N}$ which implies $(*)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that this matrix is unitary I'm currently working on quantum computing.
In a literature source that I have, it is said that the following matrix $U$ is unitary.
My question is, how can we show that in this matrix? I know that a matrix is unitary if:
$U^{*}U=I$
The matrix is an NxN matrix:
$U=\begin{pmatrix} -1+\frac{2}{N} & \frac{2}{N} & ...& \frac{2}{N} \\
\frac{2}{N} & -1+\frac{2}{N} & ...& \frac{2}{N} \\ \vdots & \vdots & \ddots & \vdots \\
\frac{2}{N} & \frac{2}{N} & ...& -1+\frac{2}{N} \end{pmatrix} $
my first thought would be to pull the 1 out:
$U=\begin{pmatrix} \frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\ \vdots & \vdots & \ddots & \vdots \\
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \end{pmatrix} - I $
Then you would have to show now that the matrix is unitary, but I can not get any further here.
I would be very happy if someone could help me to show that the matrix is unitary. I hope that my question is clear and understandable.
| $ U^{*} = U = \begin{pmatrix} \frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\ \vdots & \vdots & \ddots & \vdots \\
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \end{pmatrix} - I $
(conjugate transpose is just transpose here, as conjugate of a real is the real number itself).
Let us take a new matrix $A$,
$ A= \begin{pmatrix} \frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\ \vdots & \vdots & \ddots & \vdots \\
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \end{pmatrix} $
So $U^{*}U= U^2 = (A-I)*(A-I) = A^2 - 2AI + I = A^2 - 2A + I$
$ A^2= \begin{pmatrix} \frac{4}{N} & \frac{4}{N} & ... & \frac{4}{N} \\
\frac{4}{N} & \frac{4}{N} & ... & \frac{4}{N} \\ \vdots & \vdots & \ddots & \vdots \\
\frac{4}{N} & \frac{4}{N} & ... & \frac{4}{N} \end{pmatrix} = 2A $
So $A^2-2A$ is null matrix (all entries $0$)
And so $U^{*}U= I$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compute $\int_C ze^{\sqrt{x^2+y^2}} \mathrm ds$
Compute $\int_C ze^{\sqrt{x^2+y^2}} \mathrm ds$ where
$$C:x^2+y^2+z^2=a^2, x+ y=0, a \gt 0$$
At first I thought to parametrize this as: $x=a \cos t , y=a \sin t, z =0$, but then the integral will result in $0$ and this might not be true.
| Ok, so lets start with $x + y = 0$ since it has been given to you. This implies that $x^2 = y^2$. From this result we can simplify the following:
\begin{align*}
C: 2x^2 + z^2 = a^2
\end{align*}
As you can see this implies more of an elliptical curve. Your initial parametrization assumed a circular curve. Also, it is never acceptable to just choose a value for a variable like $z = 0$. This would have to be given to you.
Now, we only have two variables, $x$ and $z$. I am assuming that $a$ is just a positive scalar since $x$, $y$ and $z$ cover the three spacial dimensions. Since the curve relation only has these two variables, we can write one of them in terms of the other as follows:
\begin{align*}
z^2 &= a^2 - 2x^2 \\
\end{align*}
Now, I did not write the relation as a function, because I want to keep it in this form for easier calculations. If I was to write it in terms of $z$, by itself, it would require piece-wise defined functions, which can be avoided in this scenario as I will soon demonstrate. Applying implicit differentiation on the above relation, we can find the following useful result:
\begin{align*}
\frac{d}{dx}\left(z^2\right) &= \frac{d}{dx}\left(a^2\right) - 2\frac{d}{dx}\left(x^2\right) \\
2z\frac{dz}{dx} &= -4x\frac{dx}{dx} \\
2z\frac{dz}{dx} &= -4x \\
\frac{dz}{dx} &= -\frac{2x}{z} \\
\\
\left(\frac{dz}{dx}\right)^2 &= \frac{4x^2}{z^2} \\
\left(\frac{dz}{dx}\right)^2 &= \frac{4x^2}{a^2 - x^2} \\
\end{align*}
In general we have the relation, for 3-dimensional space, of:
$$ds = (dx^2 + dy^2 + dz^2)^\frac{1}{2}$$
Since we have $x = -y$, we can deduce that $dx^2 = dy^2$ which allows us to simplify the above relation and reformulate it to a sensible form as follows:
\begin{align*}
ds &= (2dx^2 + dz^2)^\frac{1}{2} \\
ds &= \left(2 + \left(\frac{dz}{dx}\right)^2 \right)^\frac{1}{2}dx \\
ds &= \left(2 + \frac{4x^2}{a^2 - x^2} \right)^\frac{1}{2}dx \\
\end{align*}
Let me know if I should give more information.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int\frac{1}{1+x^3}dx$
Calculate$$\int\frac{1}{1+x^3}dx$$
After calculating the partial fractions I got:
$$\frac{1}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2-x}{x^2-x+1}dx=\frac{1}{3}\ln(x+1)+\frac{1}{3}\int\frac{2-x}{x^2-x+1}dx$$
I have no idea on how to proceed. Am I missing a substitution or something?
| Hint: You can break up your second calculated indefinite integral into two components as so:
$$\frac{2-x}{x^2-x+1} = -\frac{1}{2}(\frac{2x -1}{x^2-x+1}) + \frac{3/2}{(x-\frac{1}{2})^2 + 3/4}$$
Note that $x^2-x+1 = (x - \frac{1}{2})^2 + \frac{3}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $\begin{bmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{bmatrix} = \begin{bmatrix}1&1\\1&0\end{bmatrix}^n$ for all $n ∈ N$. The Fibonacci numbers $F_n$ are recursively defined by
$F_0 = 0, F_1 = 1$
$F_{n+2} = F_{n+1} + F_n, n = 0,1,...$
i) Show that $\begin{bmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{bmatrix} = \begin{bmatrix}1&1\\1&0\end{bmatrix}^n$ for all $n ∈ N$.
ii) Show that $F_0^2 + F_1^2 + ...+F_n^2 = F_n F_{n+1}$ for all $n ∈ N$.
iii) Show that $F_{n-1} F_{n+1} - F_n^2 = (-1)^n$ for all $n ∈ N$.
| i) Diagonalize the RHS matrix:
$$\begin{pmatrix}1&1\\ 1&0\end{pmatrix}^n=
\begin{pmatrix}\psi&\phi\\ 1&1\end{pmatrix}
\begin{pmatrix}\psi^n &0\\ 0&\phi^n\end{pmatrix}
\begin{pmatrix}-\frac{1}{\sqrt{5}}&\frac{\phi}{\sqrt{5}}\\ \frac1{\sqrt{5}}&-\frac{\psi}{\sqrt{5}}\end{pmatrix}=\\
\begin{pmatrix}\psi^{n+1}&\phi^{n+1}\\ \psi^n&\phi^n\end{pmatrix}
\begin{pmatrix}-\frac{1}{\sqrt{5}}&\frac{\phi}{\sqrt{5}}\\ \frac1{\sqrt{5}}&-\frac{\psi}{\sqrt{5}}\end{pmatrix}=\\
\begin{pmatrix}\frac{\phi^{n+1}-\psi^{n+1}}{\sqrt{5}}&\frac{\phi^n-\psi^n}{\sqrt{5}}\\ \frac{\phi^n-\psi^n}{\sqrt{5}}&\frac{\phi^{n-1}-\psi^{n-1}}{\sqrt{5}}\end{pmatrix}=\\
\begin{pmatrix}F_{n+1}&F_n\\ F_n&F_{n-1}\end{pmatrix}.$$
Note: $\phi\cdot \psi=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $6^n=6^{n+5} (mod 100)$ How can we prove that $6^n=6^{n+5} (\text{mod}~ 100)$? I tried by writing $6^{n+5}=7776 \cdot 6^n = 76 \cdot 6^n (\text{mod}~ 100)$ but this approach does not lead to the above result.
| $$6^5-1=5(6^4+6^3+6^2+6+1)$$ and since
$$6^4+6^3+6^2+6+1\equiv(1+1+1+1+1)(\mod5)=5,$$ we obtain $6^6-1$ is divisible by $25$.
Also, $6^n$ is divisible by $4$ for all $n\geq2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$ solve $$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$$
my attempt
$$t = \frac{1}{2}\tan(u)$$
$$dt = \frac{1}{2}\sec^2(u)du\\$$
$$\begin{align}
\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt&=\int_{0}^{1} \frac{\tan^2(u)}{4} \sqrt{1+\tan^2(u)}\frac{1}{2}\sec^2(u)du\\
&=\frac{1}{8}\int_{0}^{1} \tan^2(u)\sec^{3}(u)du\\
&= \frac{1}{8} \int_{0}^{1} (\sec^2(u) - 1)(\sec^{3}(u))du\\
&=\frac{1}{8}\int_{0}^{1} \sec^5(u)du - \frac{1}{8} \int_{0}^{1}\sec^3(u)du
\end{align}$$
what now?
| We substitute $$t=\frac{\tan(u)}{2}$$ then$$dt=\frac{\sec^2(u)}{2}du$$ and
$$\sqrt{4t^2+1}=\sqrt{\tan^2(u)+1}=\sec(u)$$ and our integral will be
$$\frac{1}{2}\int\frac{1}{4}\tan^2(u)\sec^3(u)du$$ and this is
$$\frac{1}{8}\int\sec^3(u)(\sec^2(u)-1)du$$ and then we need the formula
$$\int\sec^m(u)du=\frac{\sin(u)\sec^{m-1}(u)}{m-1}+\frac{m-2}{m-1}\int\sec^{m-2}du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Prove $\sqrt{b} - \sqrt{a} < \sqrt{b-a}$
Prove that if $0 < a < b$ then
$$\sqrt{b} - \sqrt{a} < \sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2\sqrt{ab} < b-a$
subtract $b$ from both sides $a-2\sqrt{ab} < -a$
add $a$ to both sides $2a-2\sqrt{ab} < 0$
than add $2 \sqrt{ab}$ to both sides and get $2a < 2\sqrt{ab}$
divide by $2$ and get $a < \sqrt{ab}$
Thus we know $\sqrt{ab}$ is bigger than $a$ so that $\sqrt{a \cdot a} < \sqrt{ab}$ which means $\sqrt{a} < \sqrt{b}$.
Therefore together with the given $a < b$, we have $\sqrt{b} - \sqrt{a} < \sqrt{b - a}$
I'm confused as to where I went wrong and how to fix this.
| You just need to note that every step you did is reversible. For the conclusion, from $a\le\sqrt{ab}$ you get the equivalent $a^2<ab$ and, dividing by $a$, $a<b$, which is true.
More simply, set $x=\sqrt{a}$ and $y=\sqrt{b}$. You need to show that
$$
y-x<\sqrt{y^2-x^2}
$$
Since $y>x$, this is equivalent to
$$
(y-x)^2<y^2-x^2
$$
that is,
$$
y^2-2xy+x^2<y^2-x^2
$$
which in turn is equivalent, with simple algebraic simplifications, to
$$
x^2<xy
$$
which is true because $x<y$ and so $x^2<xy$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Integral $\int\frac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
Integrate $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$.
| Here's a solution using @MichaelRozenberg's hint. Solution requires some careful observation and without Michael's hint, I wouldn't have noticed it. Still, it provides a quick answer.
Using double-angle formula, we get
$$\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)}=\dfrac{x^2}{(x\cos x-\sin x)(x\sin x+\cos x)}.$$
Dividing the numerator and denominator by $\cos^2x $ gives, $$\dfrac{x^2\sec^2x}{(x-\tan x)(x\tan x+1)}.$$
Now observe that $$\left(\dfrac{x-\tan x}{x\tan x+1}\right)^{'}=\dfrac{-x^2\sec^2x}{(x\tan x+1)^2}.$$
So multiply the numerator and denominator by $x\tan x+1$ and take $\dfrac{x-\tan x}{x\tan x+1}=t. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the three cube roots of $z = -2+2i$ $n$th roots of a complex number:
For a positive integer $n$, the complex number $z=r(\cos(\theta) + i\sin(\theta))$ has exactly $n$ distinct $n$th roots given by
$$\sqrt[n]r\left(\cos\left(\frac{\theta + 2\pi k}n\right) + i\sin\left(\frac{\theta + 2\pi k}n\right)\right)$$
where $k=0, 1, 2, ..., n-1$
Find the three cube roots of $z=-2 + 2i$
$$r = \sqrt{(-2)^2 + 2^2} = \sqrt{8}$$
$$\tan(\theta) = \frac ba = \frac {2}{-2} = -1$$
$\theta = 45^o$, because $z$ is in Quad II, $\theta = 135^o$
The trigonometric form of $z$ is $z = -2 + 2i = \sqrt{8}(cos135^o + isin135^o)$
Using the formula of $n$th roots, the cube roots have the form
$$\sqrt[6]8\left(\cos\left(\frac{135^o + 360^ok}3\right) + i\sin\left(\frac{135^o + 360^ok}3\right)\right)$$
For $k = 0$ we get the root $1 + i$
For $k = 1$ we get the root $-1.3660 + 0.3660i$
For $k = 2$ we get the root $0.3660 - 1.3660i$
My question is, using the formula for $nth$ roots, why is the first part of the formula given by $$\sqrt[6]8$$ and not $$\sqrt[3]8$$
because $n=3$, so I'm confused as to where 6 is coming from.
| You have $r=\sqrt{8}$ and then you have to compute $\sqrt[3]{r}=\sqrt[3]{\sqrt{8}}=\sqrt[6]{8}$. This is the basic property that you have to use: if $a\geq 0$ and $n,m\in \mathbb{N}$, with $n,m\geq 1$, then $\sqrt[n]{\sqrt[m]{a}} = \sqrt[nm]{a}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find rational numbers $\alpha $ and $\beta$ in $\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$ How should we find two rational numbers $\alpha$ and $\beta$ such that
$$\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$$
The answer I got $\alpha = 1 $ and $\beta = 1$. If I'm wrong, please correct me. Thank you
| Apply the detesting formula
$$\sqrt[3]{a+b \sqrt R}=\frac12\sqrt[3]{3bs-a}\left(1+\frac1s \sqrt R\right)\tag1$$
where $s$ is the solution of
$$s^3-\frac{3a}b s^2+3R s-\frac{a}b R =0$$
So, for $\sqrt[3]{7+5\sqrt{2}}$, solve for $s$ in
$$s^3-\frac{21}5s^2+6s-\frac{14}5=\frac15(s-1)(5s^2-16s+14)=0$$
which yields $s=1$. Plug into (1) to obtain the denestation
$$\sqrt[3]{7+5\sqrt{2}}=1+\sqrt{2}$$
or, $\alpha=\beta=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine fair price of a digital option
A digital option pays one dollar at time $t = T$ if the asset price is
above a fixed level (strike) $K$ and is worthless otherwise.
Consider the following
model, with $r = 0$:
\begin{array}{|c|c|c|}
\hline
\omega& S(0) & S(1) & S(2) \\ \hline
\omega_1&6 & 10&12 \\ \hline
\omega_2&6 &10 & 7\\ \hline
\omega_3&6 &4 & 7\\ \hline
\omega_4&6 &4 & 3\\ \hline
\end{array}
Evaluate $\Bbb E_\Bbb Q[X]$ and determine the fair price of the digital option struck at $4$.
What I have done for a part that preceded this was find the risk neutral probabilities $\Bbb Q = (p, \frac{1}{3}-p,\frac{5}{12}-\frac{5}{4}p,\frac{1}{4}+\frac{5}{4}p)$ with a restriction on $0<p<\frac{1}{3}$ using systems of equations.
I am not sure how I can find the expectation based on this and what the random variable $X$ represents exactly. I guess the expected value that I am supposed to find will be the fair price?
| The elementary events $\{\omega_j\}$ correspond to paths on a binomial lattice with transition probabilities $p_1, \, p_2,$ and $p_3$ as shown:
$$\omega_1: \quad S(0) = 6 \underbrace{\to}_{p_1} S_u = 10\underbrace{\to}_{p_2} S_{uu}=12\\\omega_2: \quad S(0) = 6 \underbrace{\to}_{p_1} S_u = 10\underbrace{\to}_{1-p_2} S_{ud}=7\\ \omega_3: \quad S(0) = 6 \underbrace{\to}_{1-p_1} S_d = 4\underbrace{\to}_{p_3} S_{du}=7\\ \omega_4: \quad S(0) = 6 \underbrace{\to}_{1-p_1} S_d = 4\underbrace{\to}_{1-p_3} S_{dd}=3\\$$
The risk-neutral probabilities are found by enforcing expected future prices to be equal to forward prices, which coincide with spot prices as the interest rate is assumed to be $r = 0$. Under the risk-neutral measure the asset price process is a martingale.
Thus,
$$S(0) = \mathbb{E}[S(1)] = p_1S_u + (1-p_1) S_d\\ \implies 6 = 10p_1 + 4(1-p_1) \implies p_1 = \frac{1}{3}$$
$$S_u = \mathbb{E}[S(2)|S(1) = S_u] = p_2S_{uu} + (1-p_2) S_{ud}\\ \implies 10 = 12p_2 + 7(1-p_2) \implies p_2 = \frac{3}{5}$$
$$S_d = \mathbb{E}[S(2)|S(1) = S_d] = p_3S_{du} + (1-p_3) S_{dd}\\ \implies 4 = 7p_3 + 3(1-p_3) \implies p_3 = \frac{1}{4}$$
We can now compute risk-neutral path probabilities as,
$$P(\omega_1) = \frac{1}{3}\frac{3}{5} = \frac{1}{5}, \,\,P(\omega_2) = \frac{1}{3}\frac{2}{5} = \frac{2}{15}, \,\,P(\omega_3) = \frac{2}{3}\frac{1}{4} = \frac{2}{12}, \,\,P(\omega_4) = \frac{2}{3}\frac{3}{4} = \frac{1}{2} \,\,$$
The digital option expiring at time $T=2$ with strike $K= 4$ pays $1$ on paths $1,2,3$ and pays $0$ on path $4$. The fair price is the risk-neutral expected payoff which is
$$\mathbb{E}[X]= P(\omega_1) \cdot 1 + P(\omega_2) \cdot 1 + P(\omega_3) \cdot 1 + P(\omega_4) \cdot 0 = \frac{1}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does $d_1$ equal $\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$? I'm going over the proof of the midpoint formula and the solution in my textbook solves its first distance as follows
$$d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}$$
$$=\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$$
How does $d_1$ equal $\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$ ?
I tried to do it on paper but end up with $d_1$ = $\sqrt{\frac{1}{2}\left(x_2-x_1\right)^2 + \frac{1}{2}\left(y_2-y_1\right)^2} = \sqrt{\frac{1}{2}\left(\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2\right)}$
I'm sure you cant just factor out a multiple right?
| $$d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}
=\sqrt{\frac1{\color{red}4}\left(x_1-x_2\right)^2+ \frac1{\color{red}4}\left(y_1-y_2\right)^2}\\
=\frac12\sqrt{\left(x_1-x_2\right)^2+ \left(y_1-y_2\right)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{n\rightarrow \infty }(a_{0}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$ $a_{0},a_{1},...,a_{k}$ are real numbers and $a_{0}+a_{1}+...+a_{k}=0$
$$L=\lim_{n\rightarrow \infty }(a_{0}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$$
$L=?$
My attempt:
$a_{0}=-a_{1}-a_{2}-...-a_{k}$ so $L=\lim_{n\rightarrow \infty }[(-a_{1}-a_{2}-...-a_{k})\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k}]$
$L=\lim_{n\rightarrow \infty }(-a_{1}\sqrt[3]{n}-a_{2}\sqrt[3]{n}-...-a_{k}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$
Now I group the terms and I have
$L=\lim_{n\rightarrow \infty }[a_{1}(\sqrt[3]{n+1}-\sqrt[3]{n})+a_{2}(\sqrt[3]{n+2}-\sqrt[3]{n})+...+a_{k}(\sqrt[3]{n+k}-\sqrt[3]{n})]$
I split in more limits, the constants $a_{0},a_{1},...,a_{k}$ go ahead the limits and the limits are 0 after some calculations.In the end $L=a_{1}\cdot 0+a_{2}\cdot 0+...+a_{k}\cdot 0=0$
| We may expand each term in the sum using the binomial theorem (or, equivalently, as a Taylor series):
$$\sqrt[3]{n+j}=\sqrt[3]{n}\,\left(1+\frac{j}{n}\right)^{1/3}\!\!=\sqrt[3]{n}\left(1+\frac{j}{3n}-\frac{j^2}{9n^2}+\cdots\right).$$
Therefore
$$S_n \doteq a_0 \sqrt[3]{n} + a_1 \sqrt[3]{n+1} + \cdots + a_k \sqrt[3]{n+k} = \\
\big(a_0 + a_1 + \cdots + a_k\big)\sqrt[3]{n} + \frac{0a_0 + 1a_1 + \cdots + ka_k}{3}n^{-2/3} + \cdots.$$
Because $a_0 + a_1 + \cdots + a_k = 0$, for fixed $k$ we have $S_n \propto n^{-2/3}$, so $L = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101601",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$? I need to find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$. Right now I am getting $\frac{0}{0}$ if I don't factor first, or $\frac{2}{0}$ if I do.
Here are my factoring steps:
$\frac {x^2+x} {x^2-x-2}$
$=\frac{x(x+1)}{(x-2)(x+1)}$
replace $x$ with $-1$
$=\frac{-1(-1+1)}{(-1-2)(-1+1)}$
$=\frac{2}{-3 (0)}$
$=\frac{2}{0}$
How can I solve this problem?
| You can't plug it in since the function $f(x) = {x^2+x\over x^2-x-2}$ is not continuous at $x=-1$.
You have to cancel it first by $x+1$, then you can plug it in.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_0^1 \frac{\arctan x}{x^2-x-1}dx$ After seeing this integral I've decided to give a try to calculate:
$$I=\int_0^1 \frac{\arctan x}{x^2-x-1}dx$$
That is because it's common for many integrals to have a combination of a polynomial in the denominator and a logarithm or an inverse trig function in the numerator.
Mostly I tried standard ways such as integrating by parts, random substitutions, or using:
$$\frac{\arctan x}{x}=\int_0^1 \frac{dy}{1+x^2y^2}$$
But I realised that is not a great idea since it gives some mess after partial fractions, so I decided to prepare the integral a little for a Feynman's trick using probably the only helpful thing with that denominator, it won't change while using $x\mapsto 1-x$.
$$I=\int_0^1 \frac{\arctan x}{x^2-x-1}dx=\int_0^1 \frac{\arctan (1-x)}{x^2-x-1}dx$$
$$\Rightarrow 2I=\frac{\pi}{2 \sqrt 5}\ln\left(\frac{3-\sqrt 5}{3+\sqrt 5}\right)-\int_0^1 \frac{\arctan(x^2-x+1)}{x^2-x-1}dx$$
$$\small J(a)=\int_0^1 \frac{\arctan(a(x^2-x-1)+2)}{x^2-x-1}dx\Rightarrow J'(a)=\int_0^1 \frac{1}{1+(a(x^2-x-1)+2)^2}dx$$
But I am stuck now. I would like to see a method which finds a closed form for this integral, hopefully something decent comes out. I already imagine there will be some special functions.
| Here is my attempt. However, it does not result in a typical "nice, closed form."
Let $I$ be the original integral. Let $x^2-x-1 = 0$. Solving for $x$ using the quadratic formula gives us $x \in \left\{\frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}\right\}$, which means we can rewrite $x^2-x-1$ as $\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)$.
Using partial fraction decomposition on $\frac{1}{\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)}$, Let $A$ and $B$ be some arbitrary constants such that
$$
\frac{1}{\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)} = \frac{A}{\left(x-\frac{1+\sqrt{5}}{2}\right)} + \frac{B}{\left(x-\frac{1-\sqrt{5}}{2}\right)}.
$$
Multiplying both sides by the denominator of the left side gives us
$$
1 = A\left(x-\frac{1-\sqrt{5}}{2}\right) + B\left(x-\frac{1+\sqrt{5}}{2}\right).
$$
We solve for $A$ and $B$. If $x = \frac{1+\sqrt{5}}{2}$, then solving for $A$ yields
\begin{align*}
1 &= A\left(\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\right) + B\left(\frac{1+\sqrt{5}}{2}-\frac{1+\sqrt{5}}{2}\right) \\
&= A\left(\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\right) \\
&= A\sqrt{5} \\
\frac{1}{\sqrt{5}} &= A. \\
\end{align*}
Similarly, if $x = \frac{1-\sqrt{5}}{2}$, then solving for $B$ yields
\begin{align*}
1 &= A\left(\frac{1-\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\right) + B\left(\frac{1-\sqrt{5}}{2}-\frac{1+\sqrt{5}}{2}\right) \\
&= B\left(\frac{1-\sqrt{5}}{2}-\frac{1+\sqrt{5}}{2}\right) \\
&= B\sqrt{5} \\
\frac{1}{\sqrt{5}} &= B. \\
\end{align*}
Therefore, our partial fraction decomposition results in
$$
\frac{1}{\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)} = \frac{1}{\sqrt{5}}\left(\frac{1}{x-\frac{1+\sqrt{5}}{2}}\right) + \frac{1}{\sqrt{5}}\left(\frac{1}{x-\frac{1-\sqrt{5}}{2}}\right)
$$
Thus, our integral becomes
$$
I = \frac{1}{\sqrt{5}}\int_{0}^{1}\frac{\arctan\left(x\right)}{x-\frac{1+\sqrt{5}}{2}}d\theta-\frac{1}{\sqrt{5}}\int_{0}^{1}\frac{\arctan\left(x\right)}{x-\frac{1-\sqrt{5}}{2}}d\theta.
$$
Next, let's define these two integrals and two constants:
$$
I_1 = \int_{0}^{1}\frac{\arctan\left(x\right)}{x-\phi}d\theta \text{ and }
I_2 = \int_{0}^{1}\frac{\arctan\left(x\right)}{x-\theta}d\theta,
$$
such that
$$
\phi = \frac{1+\sqrt{5}}{2} \text{ and } \theta = \frac{1-\sqrt{5}}{2}.
$$
Both $I_1$ and $I_2$ are special functions. Namely, they are Generalized Inverse Tangent Functions $\text{Ti}_2\left(1, -\phi\right)$ and $\text{Ti}_2\left(1, -\theta\right)$, respectively (https://mathworld.wolfram.com/InverseTangentIntegral.html).
Therefore, our integral $I$ is
$$
\int_0^{1}\frac{\arctan{x}}{x^2-x-1}d\theta = \frac{1}{\sqrt{5}}\left(\text{Ti}_2\left(1, -\frac{1+\sqrt{5}}{2}\right) - \text{Ti}_2\left(1, \frac{\sqrt{5}-1}{2}\right)\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Proving convergence of $ \sum_{n=1}^{\infty} \left({n \over n+1}\right)^{n^2} $ $$ \sum_{n=1}^{\infty} \left({n \over n+1}\right)^{n^2} $$
We have :
$$ \left({n \over n+1}\right)^n = \left({1 \over 1+{1\over n}}\right)^n =
{1 \over \left( 1+ {1 \over n}\right)^n}$$
So,
$$ \lim_{n \rightarrow \infty} \left({n \over n+1}\right)^n = {1 \over e} \lt 1 $$
And the series converges by Cauchy criterion.
Is my proof correct?
| By comparing this series with respect to this geometric series we found that, your series converge to the a number less than $1$:
$$\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}=\frac{1}{2}+(\frac{2}{3})^4+(\frac{3}{4})^9+(\frac{4}{5})^{16}...< \frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+(\frac{1}{2})^4+...=1$$
If we could a more restricted upper bound to series, we reach to more accurate number.
To reach a lower bound I use similar way:
$$\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}=\frac{1}{2}+(\frac{2}{3})^4+(\frac{3}{4})^9+(\frac{4}{5})^{16}+...> \frac{1}{2}+(\frac{1}{2})^3+(\frac{1}{2})^5++(\frac{1}{2})^7...=\frac{2}{3}$$
So:
$$\frac{2}{3}<\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107739",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Constrained (and non) extrema of $f(x,y)=2x^2-2xy^3+3y^2$ I need to find the critical points on the boundary,inside $D$,outside $D$, and find the image of this function (constrained on $D$).
$f(x,y)=2x^2-2xy^3+3y^2$
$D=\{2x^2+3y^2\le 9\}$
Critical points non-constrained:
$f_x=4x-2y^3=0$
$f_y=-6xy^2+6y=0$ --> $6y(-xy+1)=0$
$f_y$ is equal to $0$ if $y=0$ or $x=\frac{1}{y}$, plugging it in $f_x$ brings to these critical points : $(0,0),(\pm \frac{1}{2^{\frac{1}{4}}},\pm 2^{\frac{1}{4}})$
using the second derivate test: $(0,0),(\pm \frac{1}{2^{\frac{1}{4}}},\pm 2^{\frac{1}{4}})$ seem to be saddle points (?).
Constrained Critical points (extremas):
$$
\left\{
\begin{array}{c}
4x-2y^3=4x\lambda\\
-6xy^2+6y=6y\lambda\\
2x^2+3y^2-9=0
\end{array}
\right.
$$
$-6xy^2+6y=6y\lambda$ --> $6y(-x+1-\lambda)=0$ from this one I can see that it nullifies when $y=0$ or $x=1-\lambda$.
If $y=0$ the first equation gets to $4x(1-\lambda)=0$ which means that it nullifies when $x=0$ or $\lambda = 1$. Looking at the third equation tells us that $(x,y)=(0,0)$ can't be used;
Plugging $y=0$ in the third equation : possible critical points $(\pm \frac{3}{\sqrt(2)},0)$
What I can say about $(\pm \frac{3}{\sqrt(2)},0)$ ?
I think I missed some points.
| By AM-GM
$$2x^2+3y^2-2xy^3\leq9+|2xy^3|=9+\frac{1}{\sqrt2}\sqrt{2x^2(y^2)^3}\leq$$
$$\leq9+\frac{1}{\sqrt2}\sqrt{\left(\frac{2x^2+3y^2}{4}\right)^4}\leq9+\frac{1}{\sqrt2}\sqrt{\left(\frac{9}{4}\right)^4}=9+\frac{81}{16\sqrt2}.$$
The equality occurs for $2x^2=y^2$, $-xy^3=|xy^3|$ and $2x^2+3y^2=9,$ which gives
$$(x,y)=\left(-\frac{3}{2\sqrt2},\frac{3}{2}\right),$$
which says that we got a maximal value.
The minimal value is $0$, of course.
In your solution it should be $y=0$, which does not give something with $2x^2+3y^2=9,$ or
$$1-xy=\lambda.$$
| {
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How can I find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$ efficiently with combinatorics? To find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$,
I used factorization on $(1+x+\frac{x^2}{2})$ to obtain $\frac{((x+(1+i))(x+(1-i)))}{2}$, then simplified the question to finding the coefficient of $x^6$ in $(x+(1+i))^{10}(x+(1-i))^{10}$, then dividing by $2^{10}$.
Then, we find that the coefficient of $x^6$ would be:
$$\sum_{i=0}^{6} \binom{10}{6-i} \binom{10}{i} (1-i)^{10-(6-i)} (1+i)^{10-i}$$
with the knowledge that $(1-i)(1+i)=2$, I simplified to
$$\binom{10}{6}\binom{10}{0}2^4((1-i)^6+(1+i)^6)+\binom{10}{5}\binom{10}{1}2^5((1-i)^4+(1+i)^4)+\binom{10}{4}\binom{10}{2}2^6((1-i)^2+(1+i)^2)+\binom{10}{3}\binom{10}{3}2^7$$
Note: the formula $(1+i)^x+(1-i)^x$ gives:
$ 2(2^{\frac{x}{2}}) \cos(\frac{x\pi}{4})$
After simplifying and reapplying the division by $2^{10}$, I get $(\frac{0}{1024}) + (-8)(2520)(\frac{32}{1024}) + (\frac{0}{1024}) + (120)(120)(2^7)$, which gives $0-630+0+1800,$ which is 1170, and I checked this over with an expression expansion calculator.
If the original equation was $(1+x+x^2)^{10}$, I would have used binomials to find the answer, however, the $x^2$ was replaced by $\frac{x^2}{2}$.
My question is whether anyone has a combinatorics solution to this question, rather than just algebra. It would be nice if the solution did not require complex numbers.
| The number of ways to partition 6 with just 2, 1, and 0 are
\begin{align*}
6 &= (3, 0, 7)\cdot(2, 1, 0)\\
&= (2, 2, 6)\cdot(2, 1, 0)\\
&= (1, 4, 5)\cdot(2, 1, 0)\\
&= (0, 6, 4)\cdot(2, 1, 0)
\end{align*}
where $\cdot$ indicates the dot product. These partitions represent the choices of $x^2, x^1, x^0$ in the expansion of $(1 + x + x^2)^{10}$ to obtain $x^6$. However, we must account for the $x^2/2$, and so the sums are weighted by $(1/2)^n$, where $n$ is the number of times $x^2$ is chosen. Therefore, we have
\begin{align*}
\left(\frac{1}{2}\right)^3\binom{10}{3, 0, 7} + \left(\frac{1}{2}\right)^2\binom{10}{2, 2, 6} + \left(\frac{1}{2}\right)^1\binom{10}{1, 4, 5} + \left(\frac{1}{2}\right)^0\binom{10}{0, 6, 4} = 1170
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the area of the shaded region of two circle with the radius of $r_1$ and $r_2$ In the given figure , $O$ is the center of the circle and $r_1 =7cm$,$r_2=14cm,$ $\angle AOC =40^{\circ}$. Find the area of the shaded region
My attempt: Area of shaded region $=\pi r^2_2 - \pi r^2_1= \pi( 196-49)= 147\pi$
Is it true ?
| The area of a sector of a circle with angle $\theta$ is $$ \frac{\theta}{360}\pi r^2$$
For your smaller circle, the shaded area is \begin{align}\frac{360-\theta}{360}\pi r_1^2&= \frac{360-40}{360}\pi 7^2\\
&= \frac{320}{360}\times 49\pi\\
&= \frac 89 \times 49\pi\end{align}
For the larger circle we want to calculate the whole sector area and subtract the smaller white sector:
\begin{align}\frac {\theta}{360}\pi r_2^2 - \frac{\theta}{360}\pi r_1^2 &= \frac{40}{360}\pi\times 14^2 - \frac{40}{360}\pi \times 7^2\\
&= \frac19\times 196\pi - \frac 19 \times 49\pi\\
&= \frac \pi 9 (196-49)\\
&= \frac \pi 9 \times 147\end{align}
Therefore the total shaded area is \begin{align}\frac 89 \times 49\pi+\frac \pi 9 \times 147&= \frac \pi 9(8\times 49+147)\\
&=\frac \pi 9 \times 539\\
&= \frac {539}9\pi\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Determinant of a particular matrix. What is the best way to find determinant of the following matrix?
$$A=\left(\begin{matrix}
1&ax&a^2+x^2\\1&ay&a^2+y^2\\ 1&az&a^2+z^2
\end{matrix}\right)$$
I thought it looks like a Vandermonde matrix, but not exactly. I can't use $|A+B|=|A|+|B|$ to form a Vandermonde matrix. Please suggest. Thanks.
| \begin{align}
&|A|\\
&=\det\left(\begin{matrix}
1&ax&a^2+x^2\\1&ay&a^2+y^2\\ 1&az&a^2+z^2
\end{matrix}\right) \\
&=\begin{vmatrix}
1&ax&a^2\\1&ay&a^2\\ 1&az&a^2
\end{vmatrix}+\begin{vmatrix}
1&ax&x^2\\1&ay&y^2\\ 1&az&z^2
\end{vmatrix} \tag{multilinearity on 3rd column} \\
&=0+a\begin{vmatrix}
1&x&x^2\\1&y&y^2\\ 1&z&z^2
\end{vmatrix} \\
&= a (x-y)(y-z)(z-x)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
How to integrate $\cos^3x$ by parts I've converted $\cos^3(x)$ into $\cos^2(x)\cos(x)$ but still have not gotten the answer.
The answer is $\dfrac{\sin(x)(3\cos^2x + 2\sin^2x)}{3}$
My answer was the same except I did not have a $3$ infront of $x$ and my $2\sin^2x$ was not squared.
Help!
| $$
\begin{align}
\int\cos^3{x}\,dx
&=\int\cos^2{x}\cdot\cos{x}\,dx\\
&=\int\cos^2{x}(\sin{x})'\,dx\\
&=\cos^2{x}\sin{x}+2\int\sin^2{x}\cos{x}\,dx\\
&=\cos^2{x}\sin{x}+2\int(1-\cos^2{x})\cos{x}\,dx\\
&=\cos^2{x}\sin{x}+2\int\cos{x}\,dx-2\int\cos^3{x}\,dx\\
&=\cos^2{x}\sin{x}+2\sin{x}-2\int\cos^3{x}\,dx
\end{align}
$$
$$
I=\int\cos^3{x}\,dx
$$
$$
I=\cos^2{x}\sin{x}+2\sin{x}-2I\implies\\
3I=\cos^2{x}\sin{x}+2\sin{x}\implies\\
I=\frac{\cos^2{x}\sin{x}+2\sin{x}}{3}+C.
$$
Check:
$$
\frac{d}{dx}\left[\frac{1}{3}(\cos^2{x}\sin{x}+2\sin{x})+C\right]=\\
\frac{1}{3}(2\cos{x}(-\sin{x})\sin{x}+\cos^2{x}\cos{x}+2\cos{x})=\\
\frac{1}{3}(-2\sin^2{x}\cos{x}+\cos^3{x}+2\cos{x})=\\
\frac{1}{3}(-2(1-\cos^2{x})\cos{x}+\cos^3{x}+2\cos{x})=\\
\frac{1}{3}((-2+2\cos^2{x})\cos{x}+\cos^3{x}+2\cos{x})=\\
\frac{1}{3}(-2\cos{x}+2\cos^3{x}+\cos^3{x}+2\cos{x})=\\
\frac{1}{3}(2\cos^3{x}+\cos^3{x})=\\
\frac{1}{3}(3\cos^3{x})=\\
\cos^3{x}.
$$
The answer you provided is equivalent to mine:
$$
\dfrac{\sin{x}(3\cos^2x + 2\sin^2x)}{3}=
\dfrac{\sin{x}(3\cos^2x + 2(1-\cos^2{x}))}{3}=\\
\dfrac{\sin{x}(3\cos^2x + 2-2\cos^2{x})}{3}=
\dfrac{\sin{x}(\cos^2x + 2)}{3}=\\
\dfrac{\cos^2x\sin{x} + 2\sin{x}}{3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Understanding a proof from the APMO 1998 on inequalities. I was having trouble with proving the following inequality.The question was from the book Secrets to Inequalities by Pham Kim Hung.
$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{xyz} }$
I know that the AM-GM inequality must be applied but I am having a hard time proving the the above inequality. In the book the author was able to show that by applying the trick shown below and then applying AM-GM we can prove the inequality holds.
$3\left (\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \right ) = \left ( \frac{2x}{y} +\frac{y}{z}\right ) + \left ( \frac{2y}{z}+\frac{z}{x} \right ) + \left ( \frac{2z}{x}+\frac{x}{y} \right ) \geq \frac{3x}{\sqrt[3]{xyz}} + \frac{3y}{\sqrt[3]{xyz}} + \frac{3z}{\sqrt[3]{xyz}}$
The problem I had was after applying the AM-GM to the LHS I could not prove the RHS.
| You can get the Pham Kim Hung's proof by the following way.
We'll try to find non-negatives $a$, $b$ and $c$ such that $a+b+c=1$ and
$$a\cdot\frac{x}{y}+b\cdot\frac{y}{z}+c\cdot\frac{z}{x}\geq\frac{x}{\sqrt[3]{xyz}}$$
Now, by AM-GM
$$a\cdot\frac{x}{y}+b\cdot\frac{y}{z}+c\cdot\frac{z}{x}\geq\left(\frac{x}{y}\right)^a\left(\frac{y}{z}\right)^b\left(\frac{z}{x}\right)^c=x^{a-c}y^{b-a}z^{c-b}.$$
Thus, we need $$x^{a-c}y^{b-a}z^{c-b}=\frac{x}{\sqrt[3]{xyz}}$$ or
$$x^{a-c}y^{b-a}z^{c-b}=x^{\frac{2}{3}}y^{-\frac{1}{3}}z^{-\frac{1}{3}},$$ which gives the following system.
$$a-c=\frac{2}{3},$$
$$b-a=-\frac{1}{3}$$ and $$a+b+c=1.$$
After solving of this system we obtain:
$$(a,b,c)=\left(\frac{2}{3},\frac{1}{3},0\right),$$ which gives that by AM-GM
$$\frac{2x}{3y}+\frac{y}{3z}\geq\frac{x}{\sqrt[3]{xyz}}$$ and since
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\sum_{cyc}\frac{x}{y}=\sum_{cyc}\left(\frac{2x}{3y}+\frac{y}{3z}\right)\geq\sum_{cyc}\frac{x}{\sqrt[3]{xyz}}=\frac{x+y+z}{\sqrt[3]{xyz}},$$
we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Show that $\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ=\frac34$ The original exercise is to
Prove that
$$4(\cos^320^\circ+\sin^310^\circ)=3(\cos20^\circ+\sin10^\circ)$$
Dividing both sides by $\cos20^\circ+\sin10^\circ$ leads me to the problem in the question title.
I've tried rewriting the left side in terms of $\sin10^\circ$:
$$4\sin^410^\circ+2\sin^310^\circ-3\sin^210^\circ-\sin10^\circ+1\quad(*)$$
but there doesn't seem to be any immediate way to simplify further. I've considered replacing $x=10^\circ$ to see if there was some observation I could make about the more general polynomial $4x^4-2x^3-3x^2-x+1$ but I don't see anything particularly useful about that. Attempting to rewrite in terms of $\cos20^\circ$ seems like it would complicate things by needlessly(?) introducing square roots.
Is there a clever application of identities to arrive at the value of $\dfrac34$? I have considered
$$\cos20^\circ\sin10^\circ=\frac{\sin30^\circ-\sin10^\circ}2=\frac14-\frac12\sin10^\circ$$
which eliminates the cubic term in $(*)$, and I would have to show that
$$4\sin^410^\circ-3\sin^210^\circ+\frac12\sin10^\circ=0$$
$$4\sin^310^\circ-3\sin10^\circ+\frac12=0$$
| The obvious solution
$\cos3(20^\circ)=4\cos^320^\circ-3\cos20^\circ$
and $\sin3(10^\circ)=3\sin10^\circ-4\sin^310^\circ$
and $\cos3(20^\circ)=\sin3(10^\circ)$
Alternatively, dividing both sides by $\cos20^\circ+\sin10^\circ$ and replacing $\cos20^\circ=1-2\sin^210^\circ$
with $\sin10^\circ=s,3s-4s^3=\sin3(10^\circ)=\dfrac12$
$$4(1-2s^2)^2-4(1-2s^2)s+4s^2=3$$
$$\iff16s^4+8s^3-12s^2-4s+1=0$$
$$\iff-4s\left(3s-4s^3-\dfrac12\right)-2\left(3s-4s^3-\dfrac12\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Disprove or prove using delta-epsilon definition of limit that $\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$ I want to prove if the following limit exists, using epsilon-delta definition, or prove it doesn't exist:$$\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$$
My attempt:
First I proved some directional limits, like for $y=mx$ , and $y=ax^n$, and for all of them I got 0. So I conjectured that this limit exists and it's 0. Then I have to prove:
$$\forall \delta \gt 0 : \exists \epsilon \gt 0 : \|(x,y)\| \lt \epsilon \rightarrow \left| \frac{x^3-y^3}{x^2-y^2}\right| \lt \delta$$
First I noted that $\frac{x^3-y^3}{x^2-y^2} = \frac{(x^2+xy+y^2)(x-y)}{(x+y)(x-y)} = \frac{x^2+xy+y^2}{x+y}$.
Then I did $\left|\frac{x^2+xy+y^2}{x+y}\right| \leq \left|\frac{x(x+y)}{x+y}\right|+\frac{y^2}{\vert x+y\vert} = \vert x \vert + \frac{y^2}{\vert x+y\vert}$
Using $\|(x,y)\| = \vert x\vert + \vert y\vert$ and assuming $\|(x,y)\| \lt \epsilon$
$\vert x\vert + \vert y\vert = \vert x \vert + \frac{ y^2}{\vert y \vert} \geq \vert x \vert+\frac{y^2}{\vert y\vert+\vert x\vert}$
but I can't continue from that since $\vert x + y \vert \leq \vert x \vert + \vert y \vert$
.
I don't know what else to try.
| OK, here is a completely different answer using $(x,y)$ coordinates only and no parameters.
Basically, try a power curve $y=x^n$.
BUT this gives you a path approaching the origin near the $x$ or $y$ axis, which is not a problem. So we need to modify it to place it near the "problem" line $y=-x$.
Consider the curve
$$y=-x+x^2\ .$$
This will lie within the domain of your function as long as $x\ne0,2$. So we can take $x\to0$, which implies $y\to0$, so the curve approaches the origin. Along this curve we have
$$\frac{x^3-y^3}{x^2-y^2}=\frac{2x^3-3x^4+3x^5-x^6}{2x^3-x^4}
=\frac{2-3x+3x^2-x^3}{2-x}$$
which tends to $1$ as $x\to0$. Hence the limit does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Finding out the remainder of $\frac{11^\text{10}-1}{100}$ using modulus
If $11^\text{10}-1$ is divided by $100$, then solve for '$x$' of the below term $$11^\text{10}-1 = x \pmod{100}$$
Whatever I tried:
$11^\text{2} \equiv 21 \pmod{100}$.....(1)
$(11^\text{2})^\text{2} \equiv (21)^\text{2} \pmod{100}$
$11^\text{4} \equiv 441 \pmod{100}$
$11^\text{4} \equiv 41 \pmod{100}$
$(11^\text{4})^\text{2} \equiv (41)^\text{2} \pmod{100}$
$11^\text{8} \equiv 1681 \pmod{100}$
$11^\text{8} \equiv 81 \pmod{100}$
$11^\text{8} × 11^\text{2} \equiv (81×21) \pmod{100}$ ......{from (1)}
$11^\text{10} \equiv 1701 \pmod{100} \implies 11^\text{10} \equiv 1 \pmod{100}$
Hence, $11^\text{10} -1 \equiv (1-1) \pmod{100} \implies 11^\text{10} - 1 \equiv 0 \pmod{100}$ and thus we get the value of $x$ and it is $x = 0$ and $11^\text{10}-1$ is divisible by $100$.
But this approach take a long time for any competitive exam or any math contest without using calculator. Any easier process on how to determine the remainder of the above problem quickly? That will be very much helpful for me. Thanks in advance.
| A really quick way to "see" the answer is with binomial theorem.
$11^{10} = (10+1)^{10} = 10^{10} + k_1\cdot 10^9 + k_2 \cdot 10^8 + ... + 10\cdot 10^1 + 1$
where the $k$'s represent various combinatorial constants. The values are unimportant. What's important is that when we take the whole thing modulo $100$, the expression reduces to $1$. Subtracting one, we get the required result $11^{10} - 1 \equiv 0 \pmod{100}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limiting value of a sequence when n tends to infinity Q) Let, $a_{n} \;=\; \left ( 1-\frac{1}{\sqrt{2}} \right ) ... \left ( 1- \frac{1}{\sqrt{n+1}} \right )$ , $n \geq 1$. Then $\lim_{n\rightarrow \infty } a_{n}$
(A) equals $1$
(B) does not exist
(C) equals $\frac{1}{\sqrt{\pi }}$
(D) equals $0$
My Approach :- I am not getting a particular direction or any procedure to simplify $a_{n}$ and find its value when n tends to infinity.
So, I tried like this simple way to substitute values and trying to find the limiting value :-
$\left ( 1-\frac{1}{\sqrt{1+1}} \right ) * \left ( 1-\frac{1}{\sqrt{2+1}} \right )*\left ( 1-\frac{1}{\sqrt{3+1}} \right )*\left ( 1-\frac{1}{\sqrt{4+1}} \right )*\left ( 1-\frac{1}{\sqrt{5+1}} \right )*\left ( 1-\frac{1}{\sqrt{6+1}} \right )*\left ( 1-\frac{1}{\sqrt{7+1}} \right )*\left ( 1-\frac{1}{\sqrt{8+1}} \right )*.........*\left ( 1-\frac{1}{\sqrt{n+1}} \right )$
=$(0.293)*(0.423)*(0.5)*(0.553)*(0.622)*(0.647)*(0.667)* ....$
=0.009*...
So, here value is tending to zero. I think option $(D)$ is correct.
I have tried like this
$\left ( \frac{\sqrt{2}-1}{\sqrt{2}} \right )*\left ( \frac{\sqrt{3}-1}{\sqrt{3}} \right )*\left ( \frac{\sqrt{4}-1}{\sqrt{4}} \right )*.......\left ( \frac{\sqrt{(n+1)}-1}{\sqrt{n+1}} \right )$
= $\left ( \frac{(\sqrt{2}-1)*(\sqrt{3}-1)*(\sqrt{4}-1)*.......*(\sqrt{n+1}-1)}{{\sqrt{(n+1)!}}} \right )$
Now, again I stuck how to simplify further and find the value for which $a_{n}$ converges when $n$ tends to infinity . Please help if there is any procedure to solve this question.
| The hint:
$$0<a_n=\prod_{k=1}^n\left(1-\frac{1}{\sqrt{k+1}}\right)<\prod_{k=1}^n\left(1-\frac{1}{k+1}\right)=\frac{1}{n+1}\rightarrow0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Formulas for sequences and then how to find the limsup and liminf For each of the following sequences, calculate the limit superior and the limit inferior. If the sequence converges, calculate its limit. Justify your answers.
A) $\ \ \left\{\frac{1}{2},\frac{1}{2},-\frac{1}{2},\frac{1}{3},\frac{2}{3},-\frac{2}{3},\frac{1}{4},\frac{3}{4},-\frac{3}{4},\frac{1}{5},\frac{4}{5},-\frac{4}{5},\ldots\right\}$.
B) $\ \ \left\{\frac{1}{2},1,\frac{1}{4},\frac{1}{3},\frac{1}{6},\frac{1}{5},\frac{1}{8},\frac{1}{7},\ldots\right\}$.
I can calculate the inf and sup of equations but lim inf and lim sup confuse me as well as finding the formulas for these sequences.
| A) Limes superior is $1$, limes inferior is $-1$.
The important subsequence for proving lim sup is
$$a_n =\left\{\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\ldots\right\}$$
as its limit is $1$, and for every its element are two previous elements in the original sequence smaller.
For lim inf the reasoning is similar - with the subsequence
$$b_n =\left\{-\frac{2}{3}, -\frac{3}{4},-\frac{4}{5},-\frac{5}{6},\ldots\right\}$$
B) limit of this sequence is $0$ - it is just a "masked" sequence
$$\left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8},\ldots\right\}$$
with swapped pairs of adjacent members. So lim sup and lim inf are both $0$, too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126678",
"timestamp": "2023-03-29T00:00:00",
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How to prove/show this actually defines a homomoprhism
We define the homomorphism $f: \text{SL}_2(\mathbb Z / 2 \mathbb Z) \to \text{SL}_2(\mathbb Z / 2 \mathbb Z)$ that maps the generators to:
$ \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
$ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \to \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
How do we know this is a homorphism? In a previous exercise I explored that these two elements generate the group, the first has order 3, the second has order 2. Essentially this maps any power of the first matrix to the identity, and any power of the second to a power of the matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ is it as simple as:
$$f\left(\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^{m \bmod 3}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}^{n \bmod 2}\right)=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}^{n \bmod 2}$$
But since the matrices don't commute I find it hard to prove this is a homomorphism ($f(AB)=f(A) f(B))$.
Essentially I want to prove that:
$f$ defined by:
$$\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^2= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^3 =\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \to \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$
is a homomorphism.
| A very concrete way to verify that this is a homomorphism, despite the fact that the generators don't commute, is to find another relation between the generators. This approach turns out to be effective quite often. In this case, setting
$$T:=\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}
\qquad\text{ and }\qquad
S:=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},$$
a quick computation shows that $ST=T^2S$. You already know that $T$ and $S$ have order $3$ and $2$, respectively, so this shows that $ST=T^{-1}S$, so in general for integers $a$ and $b$ this yields
$$S^bT^a=T^{(-1)^ba}S^b.$$
It also shows that for every $M\in\operatorname{SL}_2(\Bbb{Z}/2\Bbb{Z})$ there exists unique $a\in\{0,1,2\}$ and $b\in\{1,2\}$ such that $M=T^aS^b$, and moreover that
$$(T^cS^d)(T^aS^b)=T^c(S^dT^a)S^b=T^c(T^{(-1)^da}S^d)S^b=T^{(-1)^da+c}S^{b+d}.$$
Now it is easy to check that $f$ is a homomorphism, because
$$f((T^cS^d)(T^aS^b))=f(T^{(-1)^da+c}S^{b+d})=I^{(-1)^da+c}S^{b+d}=S^{b+d}$$
$$f(T^cS^d)f(T^aS^b)=(I^cS^d)(I^aS^b)=S^dS^b=S^{b+d}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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About an integral problem (Leibniz integral rule)? I want to evaluate this integral $\int_{0}^\frac{\pi}{2} \ln(1-a^2\sin^2x)dx$, where $|a|<1.$
First, I differentiate with respect to $a.$ Then, it would become a terrible integral. That is $$\int_{0}^\frac{\pi}{2} \frac{-2a\sin^2x}{1-a^2\sin^2x}dx.$$ After that, I think it may be down by cosine since it is symmetric with respect to $\frac{\pi}{4}$. However, it fails. I think that this integral may be done like $\sin^2x$.
Hope that teachers here could give me some hints to crack this done.
| $$F(a)=\int_0^{\pi/2}\log\left[1-a^2\sin(x)^2\right]\mathrm dx\Rightarrow F(0)=0$$
$$F'(a)=\frac2a\int_0^{\pi/2}\frac{-a^2\sin(x)^2}{1-a^2\sin(x)^2}\mathrm dx$$
$$F'(a)=\frac2a\int_0^{\pi/2}\frac{1-a^2\sin(x)^2}{1-a^2\sin(x)^2}\mathrm dx-\frac2a\int_0^{\pi/2}\frac{\mathrm dx}{1-a^2\sin(x)^2}$$
$$F'(a)=\frac\pi a-\frac2a\int_0^{\pi/2}\frac{\mathrm dx}{1-a^2\sin(x)^2}$$
Then we consider $$J(a)=\int_0^{\pi/2}\frac{\mathrm dx}{1-a^2\sin(x)^2}$$
Let $x=t/2$ to get
$$J(a)=\frac12\int_0^{\pi}\frac{\mathrm dt}{1-a^2\left(\frac{1-\cos t}2\right)}$$
$$J(a)=\int_0^{\pi}\frac{\mathrm dt}{a^2\cos t+2-a^2}$$
Then let $x=\tan(t/2)$ to get
$$J(a)=2\int_0^\infty \frac1{a^2\frac{1-x^2}{1+x^2}+2-a^2}\frac{\mathrm dx}{1+x^2}$$
$$J(a)=2\int_0^\infty \frac{\mathrm dx}{a^2(1-x^2)+(2-a^2)(1+x^2)}$$
$$J(a)=\int_0^\infty \frac{\mathrm dx}{(1-a^2)x^2+1}$$
and since $$\int\frac{\mathrm dx}{ax^2+bx+c}=\frac2{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}},\qquad \text{assumng}\ 4ac>b^2$$
We have $$J(a)=\frac\pi{2\sqrt{1-a^2}}$$
So we have
$$F'(a)=\frac\pi a-\frac\pi{a\sqrt{1-a^2}}$$
Hence
$$F(a)=\pi\int_0^a \frac{\sqrt{1-x^2}-1}{x\sqrt{1-x^2}}\mathrm dx$$
Setting $u=\sqrt{1-x^2}$, we have
$$F(a)=\pi\int_1^{\sqrt{1-a^2}}\frac{u-1}{u\sqrt{1-u^2}}\frac{u\mathrm du}{\sqrt{1-u^2}}$$
$$F(a)=\pi\int_1^{\sqrt{1-a^2}}\frac{\mathrm du}{u+1}$$
$$F(a)=\pi\log\frac{1+\sqrt{1-a^2}}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\max\{y-x\}$
If $x+y+z=3, $ and $x^2+y^2+z^2=9$ , find $\max\{y-x\}$.
I tried to do this geometrically, $x+y+z=3$ is a plane in $\Bbb{R}^3$ and $x^2+y^2+z^2=9$ is a ball with radius 3 and center of origin . So the candidate points for $y-x$ are on the intersection of the plane and the ball. But now I am confused how to choose to make $y-x$ maximized.
| By taking the rotation $X=(y-x)/\sqrt{2}$ and $Y=(y+x)/\sqrt{2}$ we have that the equations become
$$\begin{cases}
X^2+Y^2+z^2=9\\
\sqrt{2}Y+z=3\end{cases}$$
Hence $z=3-\sqrt{2}Y$ and
$$X^2=9-(3-\sqrt{2}Y)^2-Y^2=3Y(2\sqrt{2}-Y)\leq 6$$
with equality for $Y=\sqrt{2}$ (and $z=1$).
It follows that
$$X=\frac{y-x}{\sqrt{2}}\leq \sqrt{6}\implies y-x\leq 2\sqrt{3}.$$
Equality is attained for $x=-(\sqrt{3}-1)$ , $y=\sqrt{3}+1$, and $z=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Finding integer solutions out of a a | b Determine all positive integer values of (n) such that
$$ { n \choose 0 } + { n \choose 1 } + { n \choose 2 } + { n \choose 3 } \ \bigg| \ 2 ^ { 2008 } $$
What is the sum of all these values?
CURRENT PROGRESS:
I was able to find out that this is equivalent to $(n+1)(n^2 - n + 6) \,| \,3\times (2)^{2009}$ and opened in 2 cases, $n+1 = 2^a$ and $n+1 = 3\times 2^a$, trying to solve like: $n = 2^a - 1$, then $ n^2 - n + 6 = 2^{2a} - 3\times2^a + 8$, doing the same to the 2nd case but couldn't find solutions. Something that should be useful is that $ n^2 - n + 6 = 2^{2a} - 3\times2^a + 8 | \space\space 3\times2^{2019-a}$, it has a factor 3 in it.
| So $(n+1)(n^2 -n + 6) = 3\cdot 2^{2009}$ so
So $n+1 = 3^t2^s; n^2 -n+6 = 3^r2^w; t+r \le 1; s + w \le 2009$.
Case 1:$t= 0$.
$n = 2^s -1$ and $n^2 - n + 6 = 2^{2s} - 2^{s+1} -2^s + 6$.
If $s = 0$ we have $n=0$ and that's a solution (if we assume ${0\choose k}=0$ for $k \ne 0$). If $s \ge 1$ then
$n^2 -n +6 = 2(2^{2s-1} -2^s - 2^{s-1} + 3)$.
If $s = 1$ we have $n=1$ and that's a solution (if we assume ${n\choose k} = 0$ if $n > k$). If $s \ge 1$ then
$2^{2s-1} - 2^s - 2^{s-1} + 3$ is odd so either is either equal to $3$ or $1$.
So either $2^{2s-1}= 2^s + 2^{s-1}$
$2^{s} = 2 + 1$ which is impossible, or
$2^{2s-1}+2 = 2^s + 2^{s-1}$ so
$2^{2s-2} + 1 = 2^s + 2^{s-2}$ so $s-2 =0$ and $s = 2s -2$ or $s=2$ so $n= 3$ and we have $n+1 = 4$ and $n^2-n+6 = 9$. That's a solution.
So far: $n = 0,n=1, n=3$ are solutions
Case 2: $t = 1$
$n = 3*2^s - 1$ and $n^2 - n + 6 = 9*2^{2s} - 3*2^{s+1} + 1 - 3*2^{s} - 1 + 6 = 3(3*2^{2s} - 2^{s+1} + 2^s + 2) = 2^w$.
That's impossible.
So unless I made a mistake the only solutions are $0, 1,3$ and if we don't consider ${n< k \choose k}$ a legitimate value then the only solution is $n=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate :$\int_{-1}^{1} 2\sqrt{1-x^2} dx $
Evaluate: $$\int_{-1}^{1} 2\sqrt{1-x^2} dx $$
The answer is $\pi$
My attempt
$x = \sin(u), dx = \cos(u)du$
$$\int_{-1}^{1} 2 \sqrt{1-\sin^2(u)}\cos(u)du = \int_{-1}^{1} 2 \cos^2(u)du =\int_{-1}^{1} \frac{1}{2}(1+\cos(2u))du = \bigg(\frac{u}{2} + \frac{1}{2}\sin(2u) \bigg)\Bigg|_{-1}^{1}$$
confused how to proceed ?
| Before evaluating the definite integral, you need to make the following back-substitution:
$$
u=\arcsin{x}
$$
And use this trigonometric identity ($-\frac{\pi}{2}\le u\le \frac{\pi}{2} \implies \cos{u}\ge0$):
$$
\sin{2u}=2\sin{u}\cos{u}=2\sin{(\arcsin{x})}\sqrt{1-\sin^2{(\arcsin{x})}}=2x\sqrt{1-x^2}
$$
You also forgot that you have a $2$ in front of your integral:
$$
2\left[\frac{\arcsin{x}}{2}+x\sqrt{1-x^2}\right]_{-1}^{1}=\\
2\left(\frac{\arcsin{(1)}}{2}+1\cdot\sqrt{1-1^2}-\frac{\arcsin{(-1)}}{2}-1\cdot\sqrt{1-(-1)^2}\right)=\\
2\left(\frac{\pi}{4}+0+\frac{\pi}{4}-0\right)=2\frac{2\pi}{4}=\frac{4\pi}{4}=\pi.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is value of $\prod_{x\geq 2} \frac{x^2+1}{x^2-1}\;?$ For $x \in \mathbb{Z^+}$ what is:
$$
\prod_{x\geq 2} \frac{x^2+1}{x^2-1}\;?
$$
Numerically it seems to be approximately $3.676$.
| Very partial answer:
$$\prod_{x\geq2} \frac{x^2+1}{x^2-1} =
\prod_{x\geq2} \frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}.$$
Then Maple says that both the numerator and denominator converge. It gives
$$\prod_{x\geq2} 1-\frac{1}{x^2} = \frac{1}{2}$$
and
$$\prod_{x\geq2} 1+\frac{1}{x^2} = \frac{\sinh \pi}{2\pi}.$$
So your product is apparently equal to $(\sinh \pi)/\pi.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\sqrt{a}-\sqrt{b}$ is a root of a polynomial with integer coefficients, then so is $\sqrt{a}+\sqrt{b}$.
If $\sqrt{a} - \sqrt{b}$, where $a$ and $b$ are positive integers and non-perfect squares, is a root of a polynomial with integer coefficients, then $\sqrt{a} + \sqrt{b}$ also is.
It seems to hold some relationship with the quadratic formula. However, I have no idea on how to prove it.
| (Let's agree first that $a, b \ge 0$, so that $\sqrt{a}$ and $\sqrt{b}$ are real and sensible)
This statement:
For any polynomial $p(x) \in \Bbb{Z}[x]$, if $p(\sqrt{a} - \sqrt{b}) = 0$ then $p(\sqrt{a} + \sqrt{b}) = 0$
is in general false, with the easiest counter example probably being $a = b = 1,\ p(x) = x$. But that's no fun as an answer, so here is an almost-classification of every situation in which this statement is true:
$1)$ Firstly, if $b = 0$ then the statement holds $\forall a$, because then $p(\sqrt{a} - \sqrt{b}) = p(\sqrt{a} + \sqrt{b}) = p(\sqrt{a})$
$2)$ So now suppose $b \ne 0$, in which case $b \gt 0$. If $b$ is a square (ie: $\sqrt{b} \in \Bbb{Z}$), then the statement never holds, because of the counterexample where $p(x) = \left(x + \sqrt{b}\right)^2 -a$, which is an integer-coefficient polynomial by the assumption that $\sqrt{b} \in \Bbb{Z}$. In this case, $p(\sqrt{a} - \sqrt{b}) = 0$ but $p(\sqrt{a} + \sqrt{b}) = 4 \sqrt{ab} + 4b \ge 4b \gt 0$
$3)$ So now suppose $b \ne 0$ and that $b$ is not a square (ie $\sqrt{b} \not\in \Bbb{Z}$). If $a$ is a square ($\sqrt{a} \in \Bbb{Z}$), then surprisingly the statement always holds. Here is why:
Suppose $p(\sqrt{a} - \sqrt{b}) = 0$ for some $p$ - polynomial with integer coefficients. It is sufficient actually to consider polynomials with rational coefficients, as someone observed in the comments, in which case we can use the division algorithm. Specifically, to divide $p(x)$ by the polynomial $h(x) = \left(x - \sqrt{a}\right)^2 - b$
(Observe that $h(x)$ has both $\sqrt{a} - \sqrt{b}$ and $\sqrt{a} + \sqrt{b}$ as roots). Then we obtain $p(x) = h(x)q(x) + r(x)$ for some rational-coefficient polynomials $q, r$ and where $r$ is a polynomial of degree $1$. Now, $r(\sqrt{a} - \sqrt{b}) = 0$ by staring at the equation $p(x) = h(x)q(x) + r(x)$
Since $r$ is a degree $1$ polynomial we have $m(\sqrt{a} - \sqrt{b}) + n = 0$ for some rational $m, n$. But actually, this is impossible unless $m = n = 0$, because if $m \ne 0$ then $\sqrt{b} = {m\sqrt{a} + n \over m}$, which implies $\sqrt{b}$ is rational (remember in this case we are assuming $\sqrt{a} \in \Bbb{Z}$), and this is impossible because $b$ is not a square, in which case $\sqrt{b}$ must be irrational. And if $m = 0$, then $n = 0$ too by plugging in.
In which case, $p(x) = h(x)q(x) + 0$, and therefore $p(\sqrt{a} + \sqrt{b}) = h(\sqrt{a} + \sqrt{b})q(\sqrt{a} + \sqrt{b}) = 0$.
$4)$ Then the next case is $b \ne 0$, $b$ not a square and $a$ not a square (in particular, $a \ne 0$ as well). This case again has two more subcases: namely if $ab$ is a square or not.
Because if $ab$ is a square, then the statement never holds because of this counterexample: $p(x) = x^2 - a + 2\sqrt{ab} - b$. Then $p(\sqrt{a} - \sqrt{b}) = 0$, but $p(\sqrt{a} + \sqrt{b}) = 4\sqrt{ab} \gt 0$ since $a$ and $b$ are both positive.
$5)$ So now suppose $b \ne 0$, $b, a, ab$ all not squares. You guessed it - two more subcases. If $a = b$ then the statement never holds, by the example $p(x) = x$. This case was necessary to weed out for certain steps in the last case (we need to know that $\sqrt{a} - \sqrt{b} \ne 0$).
$6)$ Which brings me to the final case I was able to stomach before I had to put my pen down: the case where $b \ne 0$, $b, a, ab$ all not squares, and $a \ne b$. I suspect that in this case, the statement will always hold, which would be the complete classification of solutions to when the statement holds, and be the end of the problem.
Here is why I think so: start with $x = \sqrt{a} = \sqrt{b}$. If your goal is to kill $x$ with an integer-coefficient polynomial, then it seems that these are the optimal steps you would want to take:
$\begin{align}
x = \sqrt{a} - \sqrt{b}\\
& \implies x^2 = a - 2\sqrt{ab} + b\\
& \implies x^2 -a -b = 2\sqrt{ab}\\
& \implies (x^2 -a -b)^2 = 4ab\\
& \implies (x^2 -a -b)^2 - 4ab = 0
\end{align}$
Let's call the polynomial that is written in the last step there $H(x)$. The idea is that this has to be the "minimal killing polynomial" of $\sqrt{a} - \sqrt{b}$.
Observe that in the earlier cases - specifically in the construction of $h(x)$ from case $3)$ - $H(x)$ was not minimal - we were able to use other tricks and facts to stop this process early (ie: if $\sqrt{a}, \sqrt{b}, \sqrt{ab}$ were themselves integers, then we could take advantage of that and get a smaller killing polynomial - smaller than this fourth-degree monster, at any rate).
Observe also that not only does $H(x)$ kill $\sqrt{a} - \sqrt{b}$, but it also kills $\sqrt{a} + \sqrt{b}$. This property was also shared by $h(x)$ in case $3)$, which was the crux of that whole argument - in case $3)$, $h(x)$ turned out to be minimal, in the sense that it necessarily had to be a factor of any polynomial that killed $\sqrt{a} - \sqrt{b}$, which implied that the polynomial also killed $\sqrt{a} + \sqrt{b}$
So if $H(x)$ is minimal in this case, case $6)$, and if $H(x)$ must be a factor of any $p(x)$ with $p(\sqrt{a} - \sqrt{b}) = 0$, then we will finally be done.
So, consider an arbitrary killing polynomial $p(x)$ and divide by $H$ - in which case we have $p(x) = H(x)q(x) + r(x)$. This problem is much harder than in case $3)$, because since $H$ is of degree four, then $r(x)$ is some arbitrary cubic polynomial - totally unwieldy. But if you can prove that, under the circumstances of case $6)$, it is impossible for a cubic to kill $\sqrt{a} - \sqrt{b}$, then $r(x)$ will necessarily have to be $0$, in which case we are done.
I proved that in this case too it is impossible for a degree $1$ polynomial to kill $\sqrt{a} - \sqrt{b}$, and I attempted the quadratic case too, but it turned out to be too much for the room I had on my scratch paper. The process involves deriving some sort of contradiction (showing that one of $\sqrt{a}, \sqrt{b}, \sqrt{ab}$ must be rational)
Here is the degree $1$ proof: suppose again that $m(\sqrt{a} - \sqrt{b}) + n = 0$ for some rational $m, n$. Note that if either of $m$ or $n$ are $0$, then in fact $m = n = 0$ (this is where we use the assumption that $a \ne b$ which implies $\sqrt{a} - \sqrt{b} \ne 0$).
Then a nontrivial sol'n would have both $m \ne 0 \ne n$, in which case rewrite the eqn as $m\sqrt{a} + n = m\sqrt{b}$ and square both sides to obtain $m^2 + 2mn\sqrt{a} + n^2 = m^2b$, from which we obtain $\sqrt{a} = {m^2b - m^n - n^2 \over 2mn}$ which is rational, contradicting again that $a$ is not a square and therefore $\sqrt{a}$ is irrational. Thus, we know the case of $r(x)$ being a degree 1 polynomial is impossible.
I won't write what I attempted for the quadratic case, just know that it involves a lot of squaring and algebra. But I suspect that this process can be continued all the way through to $r(x)$ - degree $3$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove, that for every real numbers $ x \ge y \ge z > 0 $, and $x+y+z=\frac{9}{2}, xyz=1$, the following inequality takes place Prove, that for every real numbers $ x \ge y \ge z > 0 $, and $x+y+z=\frac{9}{2}, xyz=1$, the following inequality takes place:
$$ \frac{x}{y^3(1+y^2x)} + \frac{y}{z^3(1+z^2y) } + \frac{z}{x^3(1+x^2z)} > \frac{1}{3}(xy+zx+yz) $$
I've tried using the fact that $(xy+yz+zx)^2 \ge xyz(x+y+z) $ or $xy+yz+zx \le \frac{(x+y+z)^2}{3} $
I've also arrived to the fact that the inequality is equivalent to
$$ \sum_{cyc}{\frac{(xz)^{7/3}}{y^{5/3}(z+y)} > \frac{1}{3}(xy+yz+zx)} $$ which is homogenous.
I can't seem to find a nice way of using the given conditions for the sum and their order, thank you.
| With the two equations $$x+y+z=\frac{9}{2}$$ and $$xyz=1$$ we can express the variables $$x,y$$ by $z$ for instance and your inequality problem reduces to a one variable problem
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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waiting for TT and H? We toss a fair coin until we see one consecutive TT pattern and one H, their order is irrelevant.
For instance, HHTHTHTT, or TTH.
Let X be the number of tossed required. What is E(X)?
Here is my solution: Let
$X_H$= number of tosses required to get H.
$X_{TT}$= number of tosses required to get TT.
First, note that $E(X_{TT})=6$. Then
$E(X)=\frac{1}{2}(2+6)+\frac{1}{4}(2+6)+\frac{1}{4}(2+2)=6.5$
Is it correct?
| Look at this flow chart (each arrow has probability $\frac{1}{2}$ except the last absorbing state has probility $1$ in its self-loop):
Your answer $6.5$ is confirmed by calculating the expected number of steps for this absorbing Markov chain starting from the empty state and reaching the target state $(\text{has seen } H, TT)$.
It has $Q$-matrix
$$Q = \left[\begin{array}*
0 & \frac{1}{2} & 0 &\frac{1}{2} & 0 \\
0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\
0 & 0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
0 & 0 & 0 & \frac{1}{2} & 0 \\
\end{array}
\right]$$
Sum the first row of $(I-Q)^{-1}$ to get the answer:
$$1+\frac{1}{2} +\frac{1}{2}+3+\frac{3}{2} = \frac{13}{2}.$$
PS. I calculated the inverse with Wolfram Alpha with this input:
inverse ( 1/2*[[2,-1,0,-1,0], [0,2,-1,-1,0], [0,0,1,0,0], [0,0,0,1,-1], [0,0,0,-1,2]])
| {
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$ f(\frac{x^2+1}{x^2})=x^2+\frac{1}{x^2}-2 \Rightarrow f(x)=?$
$ f(\frac{x^2+1}{x^2})=x^2+\frac{1}{x^2}-2 \Rightarrow f(x)=?$
I'm getting a wrong answer even though my solution looks valid to me:
Inverse of $(1+\frac{1}{x^2})$ is $(\frac{1}{{\sqrt{x-1}}})$, so I plug this expression in wherever I see $x$'s
$$f(x)=(\frac{1}{{\sqrt{x-1}}})^2+(\sqrt{x-1})^2-2$$
Finally, I get $f(x)=\frac{(x-2)^2}{x-1}$ which is wrong according to the answer key. The answer should have been $x^2-4$
What am I doing wrong?
| In my opinion your answer is correct. Mimicking your computations I got the same result.
Defining $t=\dfrac{x^2+1}{x^2}$ we got that $$x^2=\dfrac{1}{t-1}.$$ So
$$f(t)=f\left(\dfrac{x^2+1}{x^2}\right)=\dfrac{1}{t-1}+t-1-2=\frac{(t-2)^2}{t-1}.$$
This holds for any $t\in (1,\infty)$ assuming $x\in \mathbb R\setminus \{0\}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Differential Equations - Exact ODE's
Hi all, I've attempted the question part b) after rearranging the 1st expression to :
$$v(x,y)\, dx - u(x,y) \, dy = 0.$$
After this I tried using the method of differentiating each expression and trying to calculate if they were exact. It was very lengthy so I am unable to post on here but it seems there is an easier way to go about this. Any help is appreciated.
| $$(-v)dx+(u)dy=\frac{2xy}{(x^2+y^2)^2}dx+\left(1+\frac{y^2-x^2}{(x^2+y^2)^2} \right)dy=0$$
$$\frac{\partial (-v)}{\partial y}= \frac{\partial }{\partial y}\left(\frac{2xy}{(x^2+y^2)^2} \right)= \frac{2x}{(x^2+y^2)^2}-\frac{8xy^2}{(x^2+y^2)^3} =\frac{2x(x^2-3y^2)}{(x^2+y^2)^3}$$
$$\frac{\partial u}{\partial x}= \frac{\partial}{\partial x}\left(1+\frac{y^2-x^2}{(x^2+y^2)^2} \right)=\frac{-2x}{(x^2+y^2)^2}-\frac{4x(y^2-x^2)}{(x^2+y^2)^3} = \frac{2x(x^2-3y^2)}{(x^2+y^2)^3}$$
$$\frac{\partial (-v)}{\partial y}= \frac{\partial u}{\partial x}\quad\text{Thus the ODE is exact}$$
Solving the ODE :
$$\int(-v)dx=\int\frac{2xy}{(x^2+y^2)^2}dx=\frac{-y}{(x^2+y^2)}+f(y)$$
$$\int(u)dx=\int\left(1+\frac{y^2-x^2}{(x^2+y^2)^2}\right)dx=\frac{-y}{(x^2+y^2)}+y+g(x)$$
$$\frac{-y}{(x^2+y^2)}+f(y)=\frac{-y}{(x^2+y^2)}+y+g(x) \quad\implies\quad f(y)=y\text{ and } g(x)=0$$
The ODE is transformed into : $\quad d\left(\frac{-y}{(x^2+y^2)}+y\right)=0$. Integrating leads to the solution of the ODE :
$$\frac{-y}{(x^2+y^2)}+y=C$$
| {
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Why don't we have to use polynomial long division on quadratic irreducible partial fractions in this case? For $(x^2+3x+1)
/(x^2+1)^2 $
Why don't we have to use polynomial long division in this case since the degree of the denominator has the same magnitude of degree as the numerator before doing partial fraction decomposition?
| You can use long division in this particular case, where the denominator is the power of a quadratic. But it is not much help in other cases.
\begin{eqnarray}
\frac{x^2+3x+1}{(x^2+1)^2}&=&\frac{x^2+3x+1}{x^2+1}\cdot\frac{1}{x^2+1}\\
&=&\left(1+\frac{3x}{x^2+1}\right)\cdot\frac{1}{x^2+1}\\
&=&\frac{1}{x^2+1}+\frac{3x}{(x^2+1)^2}
\end{eqnarray}
ADDENDUM: Here is another demonstration of the technique:
Decompose $\dfrac{x^4+1}{(x^2+x+1)^3}$
\begin{eqnarray}
\frac{x^4+1}{(x^2+x+1)^3}&=&\left(\frac{x^4+1}{x^2+x+1}\right)\cdot\frac{1}{(x^2+x+1)^2}\\
&=&\left(x^2-x+\frac{x+1}{x+x+1}\right)\cdot\frac{1}{(x^2+x+1)^2}\\
&=&\left(\frac{x^2-x}{x^2+x+1}+\frac{x+1}{(x^2+x+1)^2}\right)\cdot\frac{1}{x^2+x+1}\\
&=&\left(1-\frac{2x+1}{x^2+x+1}+\frac{x+1}{(x^2+x+1)^2}\right)\cdot\frac{1}{x^2+x+1}\\
&=&\frac{1}{x^2+x+1}-\frac{2x+1}{(x^2+x+1)^2}+\frac{x+1}{(x^2+x+1)^3}
\end{eqnarray}
| {
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About the fact that $\frac{a^2}{b} + \frac{b^2}{a} + 7(a + b) \ge 8\sqrt{2(a^2 + b^2)}$ There's a math competition I participated yesterday (19/3/2019).
In these kinds of competitions, there will always be at least one problem about inequalities.
Now this year's problem about inequality is very easy. I am more interested in last year's problem, which goes by the following:
If $a$ and $b$ are positives then prove that $$ \frac{a^2}{b} + \frac{b^2}{a} + 7(a + b) \ge 8\sqrt{2(a^2 + b^2)}$$
Now I know what you are thinking. It's a simple problem.
By the Cauchy - Schwarz inequality, we have that $\dfrac{a^2}{b} + \dfrac{b^2}{a} \ge \dfrac{(a + b)^2}{a + b} = a + b$. $$\implies \dfrac{a^2}{b} + \dfrac{b^2}{a} + 7(a + b) \ge (a + b) + 7(a + b) = 8(a + b) \le 8\sqrt{2(a^2 + b^2)}$$
And you have fallen into the traps of the people who created the test. Almost everyone in last year's competition did too.
But someone came up with an elegant solution to the problem. He was also the one winning the contest.
You should have 15 minutes to solve the problem. That's what I also did yesterday.
| By the AM-GM inequality, we have that
$$\begin{align*}
\dfrac{a^2}{b} + \dfrac{b^2}{a} + 7(a + b) &= \dfrac{a + b}{ab} \cdot (a^2 + b^2 - ab + 7ab)\\
&= \dfrac{a + b}{ab} \cdot [(a + b)^2 + 4ab]\\
&\ge \dfrac{a + b}{ab} \cdot 4(a + b)\sqrt{ab}\\
&= \dfrac{(a + b)^2}{2\sqrt{2ab \cdot (a^2 + b^2)}} \cdot 8\sqrt{2(a^2 + b^2)}\\
&\ge \dfrac{a^2 + b^2 + 2ab}{a^2 + b^2 + 2ab} \cdot 8\sqrt{2(a^2 + b^2)}\\
&= 8\sqrt{2(a^2 + b^2)}
\end{align*}$$
15 minutes have gone by. Did you do it?
| {
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$\mathbb{Q}\!\left(\sqrt{2}+\sqrt{3}\right)$ definition question I know the definition of $\mathbb{Q}\!\left(\sqrt{2}\right)=\{a+b\sqrt{2}\mid a,b\in\mathbb{Q}\}$. Why can't you similarly say $\mathbb{Q}\!\left(\sqrt{2}+\sqrt{3}\right)=\{a+b\!\left(\sqrt{2}+\sqrt{3}\right)\mid a,b\in\mathbb{Q}\}$? Any intuitive answers without jargon are much appreciated, as I have minimal background.
| Recall polynomials with rational coefficients is the set
$$\mathbb{Q}[x] = \{q_0 + q_1x + q_2x^2 + \dots q_nx^2 : q_i \in \mathbb{Q} \}$$
So if we let $x = \sqrt{2}$, we would get
\begin{align}
\mathbb{Q}[\sqrt{2}] &= \{q_0 + q_1\sqrt{2} + q_2\sqrt{2}^2 + \dots + q_n\sqrt{2}^n \}\\
&\approx\{q_0 + q_1\sqrt{2} \}
\end{align}
But if we let $x = \sqrt{2} + \sqrt{3}$
$$\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \{q_0 + q_1(\sqrt{2} + \sqrt{3})+ q_2(\sqrt{2} + \sqrt{3})^2 + \dots + q_n(\sqrt{2} + \sqrt{3})^n : q_i \in \mathbb{Q}\} $$
You can see there is going to be a problem the further we go. But if you work out the expansion, you can see that
\begin{align}
\mathbb{Q}[\sqrt{2} + \sqrt{3}] &= \{q_0 + q_1(\sqrt{2} + \sqrt{3})+ q_2(\sqrt{2} + \sqrt{3})^2 + \dots + q_n(\sqrt{2} + \sqrt{3})^n \} \\
&\approx \{q_0 + q_1\sqrt{2} + q_2\sqrt{3} + q_3\sqrt{2}\sqrt{3} \} \\
&= \mathbb{Q}[\sqrt{2},\sqrt{3}]
\end{align}
| {
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How to integrate $\frac{1}{\sqrt{x^2+x+1}}$
How to integrate $$\frac{1}{\sqrt{x^2+x+1}}$$
I tried to solve this integral as follows
$\displaystyle
\int \frac{1}{\sqrt{x^2+x+1}} \ dx=
\int \frac{1}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}} \ dx=
\int \frac{1}{\sqrt{(\frac{2x+1}{2})^2+\frac{3}{4}}} \ dx=
\int \frac{1}{\sqrt{\frac{3}{4}((\frac{2x+1}{\sqrt{3}})^2+1)}} \ dx=
\int \frac{1}{\sqrt{(\frac{2x+1}{\sqrt{3}})^2+1}}\frac{2}{\sqrt{3}} \ dx$
Substitution $t=\frac{2x+1}{\sqrt{3}} ;dt=\frac{2}{\sqrt{3}} \ dx$
$\displaystyle\int \frac{1}{\sqrt{(\frac{2x+1}{\sqrt{3}})^2+1}}\frac{2}{\sqrt{3}} \ dx=
\int \frac{1}{\sqrt{t^2+1}} \ dt$
Substitution $\sqrt{u-1}= t;\frac{1}{2\sqrt{u-1}} \ du= dt$
$\displaystyle
\int \frac{1}{\sqrt{t^2+1}} \ dt=
\int \frac{1}{2 \sqrt{u(u-1)}} \ du=
\frac{1}{2}\int \frac{1}{\sqrt{u^2-u}} \ du=
\frac{1}{2}\int \frac{1}{\sqrt{(u-\frac{1}{2})^2-\frac{1}{4}}} \ du=
\int \frac{1}{\sqrt{(2u-1)^2-1}} \ du$
Substitution $g= 2u-1;dg= 2 \ du$
$\displaystyle
\int \frac{1}{\sqrt{(2u-1)^2-1}} \ du=
\frac{1}{2}\int \frac{2}{\sqrt{(2u-1)^2-1}} \ du=
\frac{1}{2}\int \frac{1}{\sqrt{g^2-1}} \ dg=
\frac{1}{2}\arcsin g +C=\frac{1}{2}\arcsin (2u-1) +C= \frac{1}{2}\arcsin (2(t^2+1)-1) +C=\frac{1}{2}\arcsin (2((\frac{2x+1}{\sqrt{3}})^2+1)-1) +C$
However when I tried to graph it using desmos there was no result, and when i used https://www.integral-calculator.com/ on thi problem it got the result $$\ln\left(\left|2\left(\sqrt{x^2+x+1}+x\right)+1\right|\right)$$
where have I made a mistake?
| Here $$\int \frac{1}{\sqrt{g^2-1}} \ dg=...$$
Correct is:
$$\int \frac{1}{\sqrt{1-g^2}} \ dg=
\arcsin g $$
| {
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For which $a>0$ series is convergent?
For which $a>0$ series $$\sum { \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right) \right)^a } $$ $(n \in \mathbb N)$ is convergent?
My try:From Taylor theorem I know that:$$a_{n}={ \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right) \right)^a } = (\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a}$$
Then I have:
$$(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a} \le (\frac{1}{n^{4}}+o(\frac{1}{n^{8}}))^{a}$$At this point, my problem is that if I had:
$$(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a} \le (\frac{1}{n^{4}})^{a}$$I could say that $0 \le a_{n} \le (\frac{1}{n^{4}})^{a}$ so for $a>\frac{1}{4}$ this series is convergent.However in this task I have also $o(\frac{1}{n^{8}}))^{a}$ and I don't know what I can do with it to finish my sollution.Can you help me?
| The answer is indeed as you concluded $a > \frac{1}{4}$.
Note that (from the first line in your attempt)
$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) = \theta\left(\frac{1}{n^4} \right),$$
for $n$ sufficiently large, and that is all you need to conclude
$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right)^{a} = \theta \left(\frac{1}{n^4} \right)^a = \frac{1}{n^{4a}},$$
as $\sum_{n=1}^{\infty} \theta \left(\frac{1}{n^4} \right)^a$ converges iff $a > \frac{1}{4}$.
To elaborate:
$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) = \theta\left(\frac{1}{n^4} \right),$$
which implies for some positive constants $C_1, C_2$
$$ \frac{C_1}{n^4} \le \left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) \le \frac{C_2}{n^4} $$
which implies for positive $a$:
$$ \frac{C_1}{n^{4a}} \le \left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right)^a \le \frac{C_2}{n^4a} $$
However, $\sum_{n=1}^{\infty} \frac{C_1}{n^{4a}}$ diverges for all positive $a \leq \frac{1}{4}$ so if $a$ is positive then $a$ must satisfy $a > \frac{1}{4}$. On the other hand $\frac{C_2}{n^{4a}}$ converges for all $a > \frac{1}{4}$ so it suffices that $a > \frac{1}{4}$.
Can use a similar line of reasoning to show that $a$ cannot be nonnegative for the sum to converge.
| {
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"answer_count": 4,
"answer_id": 1
} |
Integers $n$ satsifying $\frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}}$
If $\displaystyle \frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}},n\in \mathbb{Z}$, then number of $n$ satisfies given equation ,is
What I tried:
Let $\displaystyle \frac{\pi}{n}=x$ and equation is $\sin 5x=\sin 3x$
$\displaystyle \sin (5x)-\sin (3x)=2\sin (4x)\cos (x)=0$
$\displaystyle 4x= m\pi$ and $\displaystyle x= 2m\pi\pm \frac{\pi}{2}$
$\displaystyle \frac{4\pi}{n}=m\pi\Rightarrow n=\frac{4}{m}\in \mathbb{Z}$ for $m=\pm 1,\pm 2\pm 3,\pm 4$
put into $\displaystyle x=2m\pi\pm \frac{\pi}{2}$
How do i solve my sum in some easy way Help me please
| I'm afraid you have used the wrong property: $\sin A-\sin B=2\cos\left(\dfrac{A+B}2\right)\sin\left(\dfrac{A-B}2\right)$. So you get$$2\cos(4x)\sin x=0$$Thus, $x=k\pi\vee4x=(2k+1)\pi/2$ for $k\in\Bbb Z$. For the former, $x=\pi/n=k\pi\implies nk=1$ which is true iff $n=k=1\vee n=k=-1$. In the latter case, $(2k+1)\pi/8=\pi/n\implies(2k+1)n=8$ which is true iff $n=-8,k=-1\vee n=8,k=0$.
Also, remember that $\sin 3x$ and $\sin 5x$ have to be non-zero, because they are in the denominator in the original problem statement. So we have to reject $\pm1$ to get two solutions, $n=\pm8$.
| {
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How to find an exact solution to a certain linear second-order non-homogeneous differential equation The equation below came up in some work I'm doing and I'm at a loss as to how to get a non-numerical solution:
$\ddot x + k_1 \tan( k_1 t) \sec(k_1 t)\dot x+k_2\cos(k_1t) x=k_3\cos(k_1t)$
| Assume $k_1,k_2\neq0$ for the key case:
$\ddot x+k_1\tan(k_1t)\sec(k_1t)\dot x+k_2\cos(k_1t)x=k_3\cos(k_1t)$
$\dfrac{d^2x}{dt^2}+\dfrac{k_1\sin(k_1t)}{\cos^2(k_1t)}\dfrac{dx}{dt}+\cos(k_1t)(k_2x-k_3)=0$
$\cos^2(k_1t)\dfrac{d^2x}{dt^2}+k_1\sin(k_1t)\dfrac{dx}{dt}+\cos^3(k_1t)(k_2x-k_3)=0$
Let $u=x-\dfrac{k_3}{k_2}$ ,
Then $\cos^2(k_1t)\dfrac{d^2u}{dt^2}+k_1\sin(k_1t)\dfrac{du}{dt}+k_2\cos^3(k_1t)u=0$
Let $r=\cos(k_1t)$ ,
Then $\dfrac{du}{dt}=\dfrac{du}{dr}\dfrac{dr}{dt}=-k_1\sin(k_1t)\dfrac{du}{dr}$
$\dfrac{d^2u}{dt^2}=\dfrac{d}{dt}\left(-k_1\sin(k_1t)\dfrac{du}{dr}\right)=-k_1\sin(k_1t)\dfrac{d}{dt}\left(\dfrac{du}{dr}\right)-k_1^2\cos(k_1t)\dfrac{du}{dr}=-k_1\sin(k_1t)\dfrac{d}{dr}\left(\dfrac{du}{dr}\right)\dfrac{dr}{dt}-k_1^2\cos(k_1t)\dfrac{du}{dr}=-k_1\sin(k_1t)\dfrac{d^2u}{dr^2}(-k_1\sin(k_1t))-k_1^2\cos(k_1t)\dfrac{du}{dr}=k_1^2\sin^2(k_1t)\dfrac{d^2u}{dr^2}-k_1^2\cos(k_1t)\dfrac{du}{dr}$
$\therefore\cos^2(k_1t)\left(k_1^2\sin^2(k_1t)\dfrac{d^2u}{dr^2}-k_1^2\cos(k_1t)\dfrac{du}{dr}\right)-k_1^2\sin^2(k_1t)\dfrac{du}{dr}+k_2\cos^3(k_1t)u=0$
$k_1^2\sin^2(k_1t)\cos^2(k_1t)\dfrac{d^2u}{dr^2}-k_1^2(\cos^3(k_1t)+\sin^2(k_1t))\dfrac{du}{dr}+k_2\cos^3(k_1t)u=0$
$k_1^2r^2(1-r^2)\dfrac{d^2u}{dr^2}-k_1^2(r^3-r^2+1)\dfrac{du}{dr}+k_2r^3u=0$
$k_1^2r^2(r^2-1)\dfrac{d^2u}{dr^2}+k_1^2(r^2(r-1)+1)\dfrac{du}{dr}-k_2r^3u=0$
$\dfrac{d^2u}{dr^2}+\left(\dfrac{1}{r+1}+\dfrac{1}{r^2(r+1)(r-1)}\right)\dfrac{du}{dr}-\dfrac{k_2r}{k_1^2(r+1)(r-1)}u=0$
$\dfrac{d^2u}{dr^2}+\left(\dfrac{1}{2(r+1)}+\dfrac{1}{2(r-1)}-\dfrac{1}{r^2}\right)\dfrac{du}{dr}-\dfrac{k_2}{2k_1^2}\left(\dfrac{1}{r+1}+\dfrac{1}{r-1}\right)u=0$
| {
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Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Let $y = \sqrt{x + 3} \implies 3 = y^2 - x$.
$$\large \begin{align}
&13x + 2(3x + 2)\sqrt{x + 3} +42\\
= &14(x + 3) + (6x + 4)y - x\\
= &14y^2 + [6(x + 3) - 14]y - x\\
= &14y(y - 1) - (y^2 - x - 9)y^3 - x\\
= &14y(y - 1) + x(y^3 - y) - y^3(y^2 - 1) + 8y^3\\
= &14y(y - 1) + (xy^2 + xy + x)(y - 1) - (y^4 + y^3)(y - 1) + 8y^3\\
= &(-y^4 + y^3 + xy^2 + xy + x + 14y)(y - 1) + 8y^3\\
\end{align}$$
And I'm stuck here.
| Rearranging and squaring both sides of the original equation gives
$$4(3x+2)^2(x+3)=(13x+42)^2$$
$$36x^3+156x^2+160x+48=169x^2+1092x+1764$$
$$36x^3-13x^2-932x-1716=0$$
$$(x-6)(4x+11)(9x+26)=0$$
$$x=6, -\frac{11}{4}, -\frac{26}{9}$$
But $x=6$ is an extra solution created due to squaring both sides and will not work in the original equation.
| {
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Find number of ways to choose $3n$-subset with repetitions from set $\left\{A,B,C\right\}$ Find number of ways to choose $3n$-subset with repetitions from set $\left\{A,B,C\right\}$ such that:
1. Letter $A$ occur at most $2n$
2. Letter $B$ occur at most $2n$
3. Letter $C$ occur odd times
Approach
I want to use there enumerators. Ok, so a factor responsible for $A$ will be
$$(1+x+x^2+ \cdots + x^{2n}) $$
(We can choose $A$ $0$ times, $1$ time, ... $2n$ times). The same will be for $B$.
Enumerator for $C$ will be
$$(x+x^3+x^5 + \cdots) $$
(We can choose $C$ 1 time, 3 times, etc)
Ok, so I want to find
$$[x^{3n}](1+x+x^2+ \cdots + x^{2n})(1+x+x^2+ \cdots + x^{2n})(x+x^3+x^5 + \cdots) = $$
$$ [x^{3n}] \left(\frac{1-x^{2n+1}}{1-x}\right)^2 \cdot\frac{x}{1-x^2} $$
but... how I can get from there factor at $x^{3n}$?
|
We obtain for $n\geq 1$:
\begin{align*}
[x^{3n}]&\left(\frac{1-x^{2n+1}}{1-x}\right)^2\frac{x}{1-x^2}\tag{1}\\
&=[x^{3n-1}]\frac{1-2x^{2n+1}}{(1-x)^2\left(1-x^2\right)}\tag{2}\\
&=\left([x^{3n-1}]-2[x^{n-2}]\right)\sum_{k=0}^\infty\binom{-2}{k}(-x)^k\sum_{j=0}^\infty x^{2j}\tag{3}\\
\end{align*}
Comment:
*
*In (2) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and we expand the numerator skipping the term $x^{4n+2}$ which does not contribute to $[x^{3n-1}]$.
*In (3) we apply the rule from (2) again and do a geometric and a binomial series expansion.
Next we calculate the coefficient of $x^n$. We obtain from (3)
\begin{align*}
[x^n]&\sum_{j=0}^\infty x^{2j}\sum_{k=0}^\infty\binom{k+1}{1}x^k\tag{4}\\
&=\sum_{j=0}^{\left\lfloor n/2\right\rfloor}[x^{n-2j}]\sum_{k=0}^\infty (k+1)x^k\tag{5}\\
&=\sum_{j=0}^{\left\lfloor n/2\right\rfloor}(n-2j+1)\tag{6}\\
&=(n+1)\sum_{j=0}^{\left\lfloor n/2\right\rfloor}1-2\sum_{j=0}^{\left\lfloor n/2\right\rfloor}j\\
&=(n+1)\left(\left\lfloor\frac{n}{2}+1\right\rfloor\right)-\frac{n}{2}\left(\left\lfloor\frac{n}{2}+1\right\rfloor\right)\\
&=\begin{cases}
(n+1)\left(\frac{n}{2}+1\right)-\frac{n}{2}\left(\frac{n}{2}+1\right)&\qquad\qquad\qquad n\text{ even}\\
(n+1)\left(\frac{n-1}{2}+1\right)-\frac{n-1}{2}\left(\frac{n-1}{2}+1\right)&\qquad\qquad\qquad n\text{ odd}\\
\end{cases}\\
&=\begin{cases}
\frac{1}{4}(n+2)^2&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ n\text{ even}\\
\frac{1}{4}(n+2)^2-\frac{1}{4}&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ n\text{ odd}\tag{7}\\
\end{cases}
\end{align*}
Comment:
*
*In (4) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.
*In (5) we apply again $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and we set the upper limit of the outer sum to $\left\lfloor\frac{n}{2}\right\rfloor$ since the coefficient is non-negative.
*In (6) we select the coefficient of $x^{n-2j}$.
We can now evaluate (3) with the help of (7) and note that if $n$ is even we have odd $3n-1$ and even $n-2$. On the other hand if $n$ is odd we have even $3n-1$ and odd $n-2$.
We obtain from (3) and (7)
\begin{align*}
\color{blue}{[x^{3n}]}&\color{blue}{\left(\frac{1-x^{2n+1}}{1-x}\right)^2\frac{x}{1-x^2}}\\
&=\left([x^{3n-1}]-2[x^{n-2}]\right)\sum_{k=0}^\infty\binom{-2}{k}(-x)^k\sum_{j=0}^\infty x^{2j}\\
&=\begin{cases}
\frac{1}{4}(3n+1)^2-\frac{1}{4}-2\cdot\frac{1}{4}n^2&\qquad\qquad\qquad n\text{ even}\\
\frac{1}{4}(3n+1)^2-2\left(\frac{1}{4}n^2+\frac{1}{4}\right)&\qquad\qquad\qquad n\text{ odd}\\
\end{cases}\\
&\,\,\color{blue}{=\frac{1}{4}\left(7n^2+6n+3[[n\text{ odd}]]\right)}\tag{8}
\end{align*}
In (8) we use Iverson brackets as compact notation for even and odd cases.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $p$ is prime and $a^7-b^3=p^2$ then $\text{gcd}(a,b)=1$
Prove that if $p$ is prime and $a^7-b^3=p^2$ then $\text{gcd}(a,b)=1$.
Any help is appreciated.
| Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....
What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.
For each of these choices what do you get when you look at $\frac {a^7}{hcf(a,b)} - \frac {b^3}{hcf(a,b)} = \frac {p^2}{hcf(a,b)}$?
Which of those choices will allow this to be true?
===
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
So $a^6*\frac {a}{hcf(a,b)} - b^2\frac {b}{hcf(a,b)} = \frac {p^2}{hfc(a,b)}$. So that means $\frac {p^2}{hfc(a,b)}$ is also a multiple of $hcf(a,b)$.
So $a^5 (\frac {a}{hcf(a,b)} )^2 - b(\frac {b}{hcf(a,b)})^2 = \frac {p^2}{hcf^2(a,b)}$. So that means that $\frac {p^2}{hcf^2(a,b)}$ is also a multiple of $hcf(a,b)$.
So $a^5(\frac {a}{hcf(a,b)} )^3- (\frac {b}{hcf(a,b)})^3 = \frac {p^2}{hcf^3(a,b)}$ is an integer.
How can $\frac {p^2}{hcf^3(a,b)}$ possibly be an integer?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding intersection of a sphere and a plane Suppose I have the sphere $\ x^2 + y^2 + z^2 = 4 $ and the plane $\ x-y\sqrt{3} =0. $
How do I find the intersection curve and write it in polar terms?
In polar coordinates the sphere is just $\ r = 2 $
substituting $\ x = \sqrt3 y $ in ball equation:
$$ \ (\sqrt3 y)^2 + y^2 + z^2 = 4 \\ 3r^2\sin^2\theta\sin^2\phi + r^2 \sin^2\theta \sin^2 \phi + r^2 \cos^2\phi =4 \\ r^2 \sin^2\phi \cdot 3 + r^2\cos^2\phi = 4 \\ r^2(3\sin^2\phi+\cos^2\phi) = 4$$
| Find orthogonal vectors each of length (radius) $2$ in the plane given:
$(1, \sqrt{3},0)$ and $(0,0,2)$
Now create a trigonometric (polar) sum of these vectors:
$$\cos \theta (1, \sqrt{3}, 0) + \sin \theta (0,0,2) = (\cos \theta, \sqrt{3} \cos \theta, 2 \sin \theta)$$
In traditional polar coordinates:
$$(r, \theta, \phi) \to (2, 60^\circ, \phi)\ {\rm for}\ 0 < \phi < 2 \pi $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to simplify the the binomial coefficient in the binomial series? Use binomial series to expand the function $\frac{5}{(6+x)^3}$ as a power series.
I understand the process to get the following summation:
$\frac{5}{6^3}\sum_{n=0}^{\infty} {-3 \choose n} (\frac{x}{6})^n $
However, I am stuck on seeing what's going on with ${-3 \choose n}$.
From the Stewart Calculus textbook, it says that ${k \choose n} = \frac{k(k-1)(k-2)...(k-n+1)}{n!}$.
By applying that, I would get:
${-3 \choose n}=\frac {(-3)(-4)(-5)...[-(n+2)]}{n!}$.
I think the next step would be to extract out the negative, so I will get $(-1)^n$.
The summation would then be:
$\frac{5}{6^3}\sum_{n=0}^{\infty} {\frac {(-1)^n(3)(4)(5)...[(n+2)]}{n!}} (\frac{x}{6})^n $
The solution to this problem is $\frac{5}{2}\sum_{n=0}^{\infty} {\frac {(-1)^n(n+1)(n+2)x^n}{6^{n+3}}}$
I am not sure what has happened to the factorial. Was it cancelled out due to the (3)(4)(5)... in the numerator? Where did (n+1) come from?
| We have the following:
\begin{align}
\frac{5}{6^3}\sum_{n=0}^\infty \frac{(-1)^n (3)(4)(5)\ldots(n+2)}{n!}\left(\frac{x}{6}\right)^n &= \frac{5}{6^3}\sum_{n=0}^\infty \frac{(-1)^n (3)(4)(5)\ldots(n)(n+1)(n+2)}{(1)(2)(3)\ldots(n)}\left(\frac{x}{6}\right)^n\\
&= \frac{5}{6^3}\sum_{n=0}^\infty \frac{(-1)^n(n+1)(n+2)}{(1)(2)}\left(\frac{x}{6}\right)^n\\
&= \frac{5}{2}\sum_{n=0}^\infty (-1)^n(n+1)(n+2)\frac{1}{6^3}\frac{x^n}{6^n}\\
&= \frac{5}{2}\sum_{n=0}^\infty \frac{(-1)^n(n+1)(n+2)x^n}{6^{n+3}}\\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\lim_{x\to 1}\frac{(x^n -1)(x^{n-1}-1)\cdots(x^{n-k+1}-1)}{(x-1)(x^2-1)\cdots(x^k-1)}$
Evaluate:$$
\lim_{x\to 1}\frac{(x^n -1)(x^{n-1}-1)\cdots(x^{n-k+1}-1)}{(x-1)(x^2-1)\cdots(x^k-1)}
$$
I'm trying to spot an error in my calculations. It is known that $x^n - 1$ may be factored out as $(x-1)(1+x+x^2+\cdots+x^{n-1})$. Using that fact consecutively for all the brackets one may obtain:
$$
k\ \text{times}\begin{cases}
x^n-1 = (x-1)(1+x+\cdots+x^{n-1})\\
x^{n-1}-1 = (x-1)(1+x+\cdots+x^{n-2})\\
\cdots\\
x^{n-k+1}-1 = (x-1)(1+x+\cdots+x^{n-k})
\end{cases}
$$
For the denominator:
$$
k\ \text{times}\begin{cases}
(x-1) = (x-1)\\
(x^2-1) = (x-1)(1+x)\\
\cdots \\
(x^k-1) = (x-1)(1+x+\cdots+x^{k-1})
\end{cases}
$$
So if we denote the expression under the limit as $f(x)$ we get:
$$
f(x) = \frac{(x-1)^k\prod\sum\cdots}{(x-1)^k\prod \sum\cdots}
$$
Now if we let $x\to1$ we get:
$$
\lim_{x\to1}f(x) = \frac{(n-1)(n-2)\cdots (n-k)}{1\cdot 2\cdot 3\cdots (k-1)} = \frac{(n-1)(n-2)\cdots (n-k)}{(k-1)!}
$$
But this doesn't match the keys section which has $n\choose k$ as an answer. I've checked several times but couldn't spot a mistake. Looks like I'm missing a $+1$ somewhere.
| You are indeed missing a $+1$ on the top. $x^2-1=(x-1)(x+1)$. You divide by $(x-1)$ and set $x=1$ and you get $2$, not $n-1=2-1=1$. In the $1+x+...+x^{n-1}$ there are $n$ terms, since you have $x^0$ as well. Same at the bottom, you go up to $k$, not $k-1$
| {
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Solve for $x$ when $e^{2x}-3e^x=4$
Solve for $x$ when $e^{2x}-3e^x=4$
This is what I've gotten so far:
\begin{array}
he^{2x}-3e^x&=&4\\
\ln(e^{2x}-3e^x)&=&\ln(4) \\
\dfrac{\ln(e^{2x})}{\ln(3e^x)}&=&\ln(4)\\
\dfrac{2x}{\ln(3) + \ln(e^x)}&=&\ln(4) \\
\dfrac{2x}{\ln(3) + x}&=&\ln(4) \\
x= \dfrac{\ln(3) \cdot \ln(4)}{2-\ln(4)}
\end{array}
I clearly made a mistake somewhere, does anyone know where?
| Let, $e^x=y$ then the equation $e^{2x}-3e^x=4$ becomes $y^2-3y=4$
Solving $y^2-3y-4=0$ we get, $$\Delta=b^2-4ac=9+16=25$$ $$y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ $$\implies y=\frac{3\pm\sqrt{25}}{2}$$ $$\therefore y=-1,4$$
Substituting $y=e^x$
$e^x=-1\implies$ No Solution Exist $\forall x\in\mathbb R$ because $e^x>0$
So, $$e^x=4\implies x=\ln{4}$$
$$\therefore x=\ln{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ab+bc+ca \geq 3k^2-1$, prove that: $a^3+b^3+c^3-3abc \geq 9k$.
If $ab+bc+ca \geq 3k^2-1$, prove that: $a^3+b^3+c^3-3abc \geq 9k$.
I recently came across a question in which we had to prove the above inequality using the given condition as mentioned above. Here $a,b,c$ are distinct positive integers and $k$ is also a positive integer. I absolutely have got no idea how to solve it or efficiently use the condition 'positive integers'. Furthermore, although the expression $a^3+b^3+c^3-3abc$ seems a bit familiar but I'm not able to understand how to make the condition useful.
Please help.
| From $\displaystyle a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\bigg[(a-b)^2+(b-c)^2+(c-a)^2\bigg]\geq 3.$
Because $a,b,c$ are distinct integers.
So here we have taken $a,b,c$ as $3$ consecutive integers.
And $\displaystyle (a+b+c)^2-3(ab+bc+ca)\geq 0\Rightarrow (a+b+c)^2\geq 3(ab+bc+ca)=3(3k^2-1)$
So $\displaystyle (a+b+c)\geq \sqrt{9k^2-3}$
So we have taken $(a+b+c)\geq 3k.$
So we have $\displaystyle a^3+b^3+c^3-3abc\geq 3\cdot (3k)=9k.$
Equality hold when $a,b,c$ are $3$ consicutive posotive integers
Namely we have taken $k-1,k,k+1$
| {
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How does this infinite series $1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots$ simplify to an integral $\int_0^1\frac{dx}{1+x^3}$? How does the infinite series below simplify to that integral?
$$1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots=\int_0^1\frac{dx}{1+x^3}$$
I thought of simplifying the series to the sum to infinity of $\frac{1}{6n-5} - \frac{1}{6n-2}$, but this did not help.
| $$\int_{0}^{1}{\frac{dx}{1-(-x)^3}=\int_{0}^{1}{\sum_{n=0}^{\infty}{(-x)}^{3n}}}dx=\sum_{n=0}^{\infty}{(-1)^{3n}\int_{0}^{1}{x^{3n}}dx}$$
$$=\sum_{n=0}^{\infty}{\frac{(-1)^{3n}}{3n+1}}= 1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find Delta, given Epsilon lim x->0 root (x+1) = 1. epsilon = 0.1 Can I solve this using rationalization and then assuming delta <= 1? If so, why is my answer wrong?
we know |√(x+1)-1| < 0.1
rationalization of LHS,
x<0.1*(|√(x+1)+1|)
hence, delta = 0.1*(|√(x+1)+1|)
to find x, consider delta <= 1, x <= 1 hence √(x+1)+1 <= √2 + 1
substiting the value, delta = 0.1*(√2 + 1) = 0.2 (incorrect answer)
| There are two mistakes you are making:
*
*The $\delta$ you are choosing in the first place is not independent of $x$.
*In trying to rectify this, you find an upper bound $c$ for your expression $f(x)$ for $\delta$ that is dependent on $x$ and then choose $\delta$ equal to that upper bound $c$. However, you previously found that $\delta$ must be less than or equal to $f(x)$ for all $x$. But you have proven your chosen $\delta=c$ to be greater than $f(x)$ for $x<1$. In conclusion, this approach cannot work.
Rather, consider the following: We have, as you already noted
$$x<\varepsilon\cdot \left|\sqrt{x+1}+1\right|.$$
Since the expression inside the absolute value is always positive, we can drop the absolute value and obtain
$$\begin{split}
x&<\varepsilon\cdot\left(\sqrt{x+1}+1\right) \\
\Leftrightarrow x-\varepsilon&<\varepsilon\cdot\sqrt{x+1} \\
\Leftrightarrow x^2-2\varepsilon x+\varepsilon^2&<\varepsilon x+\varepsilon \\
\Leftrightarrow x^2-3\varepsilon x+(\varepsilon^2-\varepsilon)&<0 \\
\end{split}$$
We will find the zeroes of the quadratic function on the left. Since its leading term is positive, all $x$ between those zeroes will satisfy the inequality.
$$x_{1,2}=\frac{3}{2}\varepsilon\pm\sqrt{\frac{9}{4}\varepsilon^2-\varepsilon^2+\varepsilon}=\frac{3}{2}\varepsilon\pm\sqrt{\frac{5}{4}\varepsilon^2+\varepsilon}.$$
In the case of $\varepsilon=0.1$, we get $x_1=\frac{3+3\sqrt{5}}{20}$ and $x_2=\frac{3-3\sqrt{5}}{20}$. We have $|x_1|>|x_2|$ and thus we have to take $\delta=|x_2|=\frac{3\sqrt{5}-3}{20}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving polynominals equations (relationship of roots)
The roots of $x^3-4x^2+x+6$ are $\alpha$, $\beta$, and $\omega$.
Find (evaluate):
$$\frac{\alpha+\beta}{\omega}+\frac{\alpha+\omega}{\beta}+\frac{\beta+\omega}{\alpha}$$
So far I have found:
$$\alpha+\beta+\omega=\frac{-b}{a} = 4 \\
\alpha\beta+\beta\omega+\alpha\omega=\frac{c}{a} = 1 \\
\alpha×\beta×\omega=\frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$\frac{\alpha^2\beta+\alpha\beta^2+\alpha^2\omega+\alpha\omega^2+\beta^2\omega+\beta\omega^2}{\alpha\beta\omega}$$
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-\dfrac{11}{3}$
| Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 \Rightarrow (x+1)(x-2)(x-3)=0 \Rightarrow \\
\alpha =-1, \beta =2,\omega=3.$$
Hence:
$$\frac{\alpha + \beta}{\omega} + \frac{\beta + \omega}{\alpha} + \frac{\alpha + \omega}{\beta}=\\
\frac{-1+ 2}{3} + \frac{2 + 3}{-1} + \frac{-1 + 3}{2}=\\
\frac13-5+1=\\
-\frac{11}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$ Find $\frac{a_7}{a_{13}}$ Given $$(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$$ Find Value of $\frac{a_7}{a_{13}}$
My try:
I assumed $A=2x^2$,$B=3x$ and $C=4$
Then we have the following cases to collect coefficient of $x^7$:
Case $1.$ $A^3 \times B^1 \times C^6$
Case $2.$ $A^2 \times B^3 \times C^5$
Case $3.$ $A^1 \times B^5 \times C^4$
case $4.$ $A^0 \times B^7 \times C^3$
Using multinomial theorem we get Coefficient of $x^7$ as:
$$a_7=10! \times \left(\frac{2^33^14^6}{3!6!}+\frac{2^13^54^4}{1!5!4!}+\frac{2^23^34^5}{2!3!5!}+\frac{2^03^74^3}{3!7!}\right)$$
Like wise we need to find $a_{13}$
But is there any better way?
| Given $$(2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}x^k$$
Now using $\displaystyle x\rightarrow \frac{2}{x}$
$$\Rightarrow 2^{10}(2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}2^{k}x^{20-k}$$
$$\Rightarrow (2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}2^{k-10}x^{20-k}$$
Now Comparing Coefficient of $x^{7}$ on both side, we get
$$a_{7}=a_{13}\cdot 2^{3}\Rightarrow \frac{a_{7}}{a_{13}}=8.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimizing $\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2$ While solving a problem I came across this task, minimizing
\begin{align}
\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2.
\end{align}
One can easily do it with calculus to show that the minimum value is $12.5$.
I tried to do it using trigonometric identities and fundamental inequalities (like AM-GM, Cauchy-Schwarz, etc.) but failed. Can someone help me to do it using trig identities and inequalities?
| Here is a solution using AM-GM and the double-angle formulae
*
*$(\star):\sin^2 x = \frac{1-\cos 2x}{2}$, $\cos^2 x = \frac{1+\cos 2x}{2}$
\begin{eqnarray*}\left ( \sin^2 x + \frac{1}{\sin^2 x} \right )^2 + \left ( \cos^2x + \frac{1}{\cos^2x} \right )^2
& \stackrel{AM-GM}{\geq} & 2\left ( \sin^2x + \frac{1}{\sin^2x} \right ) \left ( \cos^2x+ \frac{1}{\cos^2x} \right ) \\
& = & 2\frac{(\sin^4 x + 1)(\cos^4 x + 1)}{\sin^2 x \cdot \cos^2 x}\\
& \stackrel{AM-GM}{\geq} & 8\frac{(\sin^4 x + 1)(\cos^4 x + 1)}{(\sin^2 x + \cos^2 x)^2}\\
& \stackrel{\star}{=} & 8\left(\frac{(1-\cos 2x)^2}{4}+1 \right)\left(\frac{(1+\cos 2x)^2}{4}+1 \right)\\
& = & \frac{1}{2}(5+\cos^2 2x - 2\cos 2x)(5+\cos^2 2x + 2\cos 2x)\\
& = & \frac{1}{2}(25+\cos^4 2x +6 \cos^2 2x)\\
& \geq & \frac{25}{2}
\end{eqnarray*}
Note that equality is attained for $\cos 2x = 0 \Leftrightarrow \cos^2 x = \sin ^2 x$, where the last condition is required to produce equality for AM-GM.
| {
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"url": "https://math.stackexchange.com/questions/3209521",
"timestamp": "2023-03-29T00:00:00",
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$p^2-p+1=P^3$ prove how many $p$, $P$ pairs are there Determine with a proof all prime numbers $p$ such that $p^2-p+1$ is a cube of a prime number.
$19^2-19+1=7^3$
But is it the only $p$?
How should I prove it?
| We will use here: If $a$ and $b$ are positive integers then $$\boxed{a\mid b\implies a\leq b}$$
Rename $q=P$. We can assume that $p\geq 23$ and so $q\geq 8$. We have $$ p(p-1) = (q-1)(q^2+q+1)$$
*
*If $p\mid q-1$ then $q^2+q+1\mid p-1$. So $p\leq q-1$ and $q^2+q+1\leq p-1$ so we have $$q^2+q+1\leq q-2\implies q^2+3\leq 0$$
which is impossible.
*If $p\mid q^2+q+1 $ then $q-1\mid p-1$. Let $r=q-1$ then we have: $p\mid r^2+3r+3$ and $r\mid p-1$ (so $\color{red}{r\leq p-1}$) then $$pr\mid r^2p+3pr+3p-r^2-3r-3$$ and thus $$pr\mid r^2+3r+3-3p$$
Case 1: $r^2+3r+3>3p$ then $$pr\leq r^2+3r+3-3p \implies p\leq r+{3\over r+3}< r+1\leq p$$
A contradiction.
Case 2: $r^2+3r+3<3p$ then $$pr\leq 3p-r^2-3r-3 < 3p \implies r<3$$
A contradiction.
Case 3: $r^2+3r+3=3p =q^2+q+1$ then $$p-1 = 3q-3 = 3r$$ and thus we have $$r^2+3r+3 = 9r+3\implies r=6 \implies q=7$$
A contradiction, since $q>7$.
So the only solution is $P=7$ and $p=19$.
Notice that we didn't use $P$ is prime.
| {
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Check for equality of the integrals I need to show the following identity
$$
\int_a^b\frac{dx}{\sqrt{x-x^2}}=\int_{\tfrac{1-\sqrt{1-a}}{2}}^{\tfrac{1-\sqrt{1-b}}{2}}\frac{dx}{\sqrt{x-x^2}}+\int_{\tfrac{1+\sqrt{1-b}}{2}}^{\tfrac{1+\sqrt{1-a}}{2}}\frac{dx}{\sqrt{x-x^2}}
$$
the sum of the integrals on the right side
$\arcsin \left(-\sqrt{-b+1}\right)-\arcsin \left(-\sqrt{-a+1}\right)+\arcsin \left(\sqrt{-a+1}\right)-\arcsin \left(\sqrt{-b+1}\right)$
and
$$\int_a^b\dfrac{dx}{\sqrt{x-x^2}}=\arcsin \left(2b-1\right)-\arcsin \left(2a-1\right).
$$
Is there any trigonometric identity to show that they are the same? How does this show?
| The integrals in the right-hand side sum to
$$
2\arcsin\sqrt{1-a}-2\arcsin\sqrt{1-b}
$$
If you look at https://math.stackexchange.com/a/3211273/62967, you'll see that
$$
\arcsin(2x-1)=2\arcsin\sqrt{x}-\frac{\pi}{2}
$$
For $x=1-a$, we have $2x-1=2-2a-1=1-2a$.
| {
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What is all N that make $2^N + 1$ divisible by 3? When $N = 1$, $2^N + 1 = 3$ which is divisible by $3$.
When $N = 7$, $2^N + 1 = 129$ which is also divisible by $3$.
But when $N = 2$, $2^N + 1 = 5$ which is not divisible by $3$.
A quick lookout can determine that when $N$ is an odd number, $2^N + 1$ is divisible by $3$. If is that correct, how can it be proven? Otherwise, what is the counterexample and the correct $N$?
Can this be made into a more general form, for example $2^N + C$ or something similar?
| When $N = 1 + 2k$;
\begin{align}2^{1+2k} +1 \pmod 3 & \equiv 2(2^{2k})+1 \pmod 3\\
&\equiv 2(4^{k})+1 \pmod 3\\
&\equiv 2+1 \pmod 3\\
&\equiv \pmod 3
\end{align}
When $N = 2k$
$$2^{2k} +1 \pmod 3\equiv 2 \pmod 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
Here's my working:
Start with $$4+\frac{1}{x}-\frac{1}{x^2}=0$$
Rearranging into standard form:
$$-\frac{1}{x^2}+\frac{1}{x}+4=0$$
Multiply by $-1$ to get a positive leading coefficient $a$:
$$\frac{1}{x^2}-\frac{1}{x}-4=0$$
I'm not sure how to determine my inputs $a,b$ and $c$ with these fractions but I guess $a=\dfrac{1}{x^2}$, $b=\dfrac{1}{x}$ and $c=-4$.
Plugging into quadratic function:
$$x = \frac{-\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
I find this challenging due to the coefficients $a$ and $b$ being fractions.
How can I apply the quadratic formula to $4+\dfrac{1}{x}-\dfrac{1}{x^2}=0$ to arrive at $\dfrac{-1\pm\sqrt{17}}{8}$?
| None of the other answers so far has really addressed the error that you made in your attempt. After rearranging the original equation into $$\frac{1}{x^2}-\frac{1}{x}-4=0,$$ you then decided that $a=1/x^2$, $b=-1/x$ and $c=-4$. Substituting these names for the corresponding values in this equation gives you $$a+b+c=0.$$ This is no longer a quadratic equation—it doesn’t even have an unknown to solve for!
Instead, as other answers explain, you need to either multiply by $x^2$ to eliminate $x$ from all of the denominators, or introduce a new variable such as $y=1/x$. Either approach will give you an equation that looks more like one you’re used to. Remember to reject $x=0$ if it comes up as a solution to the modified equation (it won’t).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
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