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Find the area of the surface generated when the given curve is revolved about the x-axis Find the area of the surface generated when the given curve is revolved about the x-axis
The part of the curve $y=12x-2$ between the points $(\frac{5}{12},3)$ and $(\frac{13}{12},11)$
I understand how to solve this when I am dealing one point, but I am confused then there are two points, I hope someone can guide me through this question.
I used the formula:
$$S=\int_{a}^{b} 2π \left(f(x)\right)\sqrt{1+f'(x)^2} $$
$$S=\int_{\frac{5}{12}}^{\frac{13}{12}} 2π \left(12x-2\right)\sqrt{1+(12)^2} $$
$$S=\int_{\frac{5}{12}}^{\frac{13}{12}} 2π \left(12x-2\right)\sqrt{145} $$
$$[2π( \left(12x-2\right)\sqrt{145})]_{\frac{5}{12}}^{\frac{13}{12}} $$
$$[2π( \left(6x^2-2x\right)\sqrt{145})]_{\frac{5}{12}}^{\frac{13}{12}} $$
$$[2π( \left(6\frac{5}{12}^2-2\frac{5}{12}\right)\sqrt{145})-(6\frac{13}{12}^2-2\frac{13}{12}\sqrt{145})]_{\frac{5}{12}}^{\frac{13}{12}} $$
$$[2π( \left(6\frac{24}{144}-\frac{10}{12}\right)\sqrt{145})-((6\frac{169}{144}-\frac{26}{12})\sqrt{145})] $$
$$[2π( \left(-\frac{48}{3}\right)\sqrt{145})-((-\frac{858}{144})\sqrt{145})] $$
|
Since $\displaystyle S=\int_{\frac{5}{12}}^{\frac{13}{12}} 2π \left(12x-2\right)\sqrt{145} $, letting $u=12x-2$ and $du=12\;dx$ gives
$\displaystyle\frac{2\pi\sqrt{145}}{12}\int_3^{11} u\;du=\frac{\pi\sqrt{145}}{6}\left[\frac{u^2}{2}\right]_3^{11}=\frac{\pi\sqrt{145}}{12}(121-9)=\frac{28\pi\sqrt{145}}{3}$.
Alternatively, using $x=\frac{1}{12}(y+2),\;\; 3\le y\le 11$ gives
$\displaystyle S=\int_3^{11}2\pi y\sqrt{1+\frac{1}{144}}dy=\frac{2\pi\sqrt{145}}{12}\int_3^{11}y\;dy$.
|
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"url": "https://math.stackexchange.com/questions/1162393",
"timestamp": "2023-03-29T00:00:00",
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|
How to factorise $x^4 - 3x^3 + 2$, so as to compute the limit of a quotient? Question:
Find the limit: $$\lim_{x \to 1}\frac{x^4 - 3x^3 + 2}{x^3 -5x^2+3x+1}$$
The denominator can be simplified to: $$(x-1)(x^2+x)$$
However, I am unable to factor the numerator in a proper manner (so that $(x-1)$ will cancel out)
I know upon graphing that the limit is $5\over4$. What should I do here?
Note: To be done without the use of L'Hospital Rule
|
Hint: The numerator can be factorized as $$x^4-3x^3+2=x^4-1-3(x^3-1)=(x-1)((x^2+1)(x+1)-3(x^2+x+1))$$ and the denominator as $$x^4-5x^2+3x+1\\=x^4-1-(5x^2-3x-2)=(x-1)((x^2+1)(x+1)-(5x+2))$$
|
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|
Surface area of ellipsoid given by rotating $\frac{x^2}{2}+y^2=1$ around the x-axis Calculate the surface area of the ellipsoid that is given by rotating
$\frac{x^2}{2}+y^2=1$ around the x-axis.
My idea is that if $f(x)=\sqrt{1-\frac{x^2}{2}}$ rotates around the x-axis we will end up with the same figure. The formula for calculating the surface area is:
$ 2\pi \int_{-\sqrt{2}}^{\sqrt{2}} f(x)\sqrt{1+f'(x)^2}dx $ and with $f'(x)^2 = \frac{x^2}{4-2x^2}$ the integral becomes $$ \int \sqrt{1-\frac{x^2}{2}}\sqrt{1+\frac{x^2}{4-2x^2}} \space dx$$ Now how do I integrate this?
|
Looks like one of those times you'll just have to tangle with the algebra monster. $$\begin{align} \sqrt{1-\frac{x^2}{2}}\sqrt{1+\frac{x^2}{4-2x^2}} = \sqrt{\frac{2-x^2}{2}}\sqrt{\frac{4-x^2}{4-2x^2}} \\ = \sqrt{\frac{2-x^2}{2}}\sqrt{\frac{4-x^2}{2(2-x^2)}} \\ = \frac{1}{2} \sqrt{\frac{(2-x^2)(4-x^2)}{2-x^2}}\end{align} \\ = \frac{1}{2}\sqrt{4-x^2} $$ Can you take it from here?
|
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|
Simplification of $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$ I was trying to simplify $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$. Numerical evaluation suggested that the answer is $\sqrt{2}$ and it checked out when I substituted $\sqrt{2}$ in the equation $x= \sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$.
But I still cannot simplify the initial expression by any obvious means (squaring, multiplying by conjugate etc). Any idea how to do it? Can it be done?
|
The nested radical $\sqrt{16 - 4 \sqrt{7}}$ is equal to $\sqrt{14}-\sqrt{2}$. Simply square both sides to prove that. Using $(a-b)^2=a^2+b^2-2ab$, we get$$(\sqrt{14}-\sqrt{2})^2=14+2-2\sqrt{2}\sqrt{14}=16-2\sqrt{2}\sqrt{2}\sqrt{7}=16-4 \sqrt{7}$$
You can read more about denesting on Wikipedia.
EDIT: Lucian's answer is explicitly solving the denesting equation
$$\sqrt{a-b \sqrt{c}\ } = \sqrt{y}-\sqrt{x}$$
for this particular case in which half the answer ($y=14$) was given to you, that's why he gets an equation in a single unknown. If you can simply guess/know both $x$ and $y$, then you can just verify the solution by squaring both sides as I did.
The general solution formula is also obtained by squaring both sides and doing an identification, yielding the system $a = x+y$ and $b^2c= 4xy$. This is actually exactly the same system that you would get when solving $\sqrt{a+b \sqrt{c}\ } = \sqrt{y}+\sqrt{x}$, the full proof for which is given in the Wikipedia article I linked. So the general solution for $\sqrt{a-b \sqrt{c}\ } = \sqrt{y}-\sqrt{x}$ is also $x= \frac{a-\Delta}{2},$ $y=\frac{a+\Delta}{2}$ where $\Delta=\sqrt{a^2 - b^2c}$ must be a rational number, otherwise there's no solution in rationals for $x$ and $y$. Applying this general method to $a= 16$, $b= 4$ and $c=7$ yields $\Delta=\sqrt{16^2-4^2\cdot7} = \sqrt{256-112}=12$, so $x=\frac{16-12}{2}=2$ and $y=\frac{16+12}{2}=14$.
Also, if you already know $y$ (in your problem $y=14$), then from the equation $a=x+y$, i.e. in your problem $16=x+14$, it follows immediately that the only possible solution is with $x=2$, so you can just check that the second equation $b^2c=4xy$ is verified, i.e. $4^2\cdot7=4\cdot2\cdot14$.
|
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|
How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$
|
\begin{align} x^3-y^3 &=x^3\color{red}{-x^2y+x^2y}-y^3\\
&= x^2(x-y)+y(x^2-y^2)\\
&= x^2(x-y)+y(x-y)(x+y)\\
&=(x-y)\Big( x^2+y(x+y)\Big)\\
&=(x-y)( x^2+xy+y^2)
\end{align}
|
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|
Inequality between real numbers $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$ let $a,b$ and $c$ positive reals. Shows that $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$
How prove this inequality. Thanks
a suggestion please to prove this inequality
|
The inequality $a^ab^bc^c\le(abc)^{\frac{a+b+c}{3}}$ is incorrect. Let $a=b=1$, $c=4$
Then
$$a^ab^bc^c=4^4$$
while
$$(abc)^{\frac{a+b+c}{3}}=4^2$$.
However, the opposite inequality $a^ab^bc^c\ge(abc)^{\frac{a+b+c}{3}}$ is correct as demonstrated in Angel's answer.
|
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|
How do I solve the recurrence relation without manually counting? Given the recurrence relation : $a_{n+1} - a_n = 2n + 3$ , how would I solve this?
I have attempted this question, but I did not get the answer given in the answer key.
First I found the general homogenous solution which is $C(r)^n$ where the root is 1 so we get $C(1)^n$. Then I found the particular non homogenous solution which was $A_1(n) + A_0$.
I then plugged the particular solution into the given recurrence relation and solved for $A_0$ and $A_1$. I got $A_0 = -1$ and $A_1 = 5$. After further steps I got
$a_n = 5n-1 + 2(1)^n$
That answer is however completely off, in the answer key they have
$a_n = (n+1)^2$. Can someone explain to me how to arrive at this answer?
EDIT: Initial condition is that $n\ge 0$ and $a_0 = 1$.
|
Let $$f(x) = \sum_{n=0}^\infty a_n x^n.$$
Multiplying both sides of the given recurrence equation by $x^n$ and summing over $n\geqslant0$, the left-hand side becomes
$$\sum_{n=0}^\infty a_{n+1}x^n-\sum_{n=0}^\infty a_nx^n = \frac1x\sum_{n=0}^\infty a_{n+1}x^{n+1} - f(x) = \frac1x\left(f(x)-1\right)+f(x).$$
The right-hand side becomes
$$\sum_{n=0}^\infty (2n+3)x^n = 2\sum_{n=0}^\infty(n+1)x^n + \sum_{n=0}^\infty x^n=\frac2{(1-x)^2}+\frac1{1-x}. $$
Equating the two we have
$$\frac1x\left(f(x)-1\right)+f(x) = \frac2{(1-x)^2}+\frac1{1-x},$$
and solving for $f(x)$,
$$f(x) = \frac{1+x}{(1-x)^3}. $$
Now
$$\frac1{(1-x)^3}=\sum_{n=0}^\infty\frac12(n+1)(n+2)x^n = 1 + \sum_{n=1}^\infty\frac12(n+1)(n+2)x^n, $$
and
$$\frac x{(1-x)^3}=\sum_{n=0}^\infty\frac12(n+1)(n+2)x^{n+1} = \sum_{n=1}^\infty \frac12n(n+1)x^n. $$
Hence
$$
\begin{align*}
\frac{1+x}{(1-x)^3} &= 1 + \sum_{n=1}^\infty\frac12(n+1)(n+2)x^n + \sum_{n=1}^\infty\frac12n(n+1)x^n\\
&= 1 + \sum_{n=1}^\infty\frac12(n+1)(n+n+2)x^n\\
&= 1 + \sum_{n=1}^\infty(n+1)^2x^n\\
&= \sum_{n=0}(n+1)^2x^n.
\end{align*}
$$
It follows that $a_n=(n+1)^2$ for $n\geqslant0$.
|
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|
Find all primes of the form Find all primes of the form $3\cdot2^{2^t-1}+1$.
For $t=1$ it works.And I was thinking that it should have some logic because the other numbers go really large and is meaningless to take them one by one.
|
For $t>1$, the expression can never be prime.
Modulo 5 we get
$3 \cdot 2^{2^t - 1} + 1 \equiv (-2) \cdot 2^{2^t - 1} + 1 \mod 5$
$\equiv - 2^{2^t} + 1 \mod 5$
Since $t>1$, write $2^t = 4\cdot 2^{t-2}$. So $2^{2^t} = (2^4)^{2^{t-2}}$ and
$- 2^{2^t} + 1 \equiv -(2^4)^{2^{t-2}} + 1 \mod 5$
$\equiv -(1^{2^{t-2}}) + 1 \equiv 0 \mod 5$
Since the expression is obviously greater than 5, it follows that it is never prime.
|
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|
with this inequality condition $xyz(x+y+z)=3$ Let $x,y,z$ be postive real numbers such that $xyz(x+y+z)=3$,show that
$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{54}{(x+y+z)^2}\ge 9$$
I found this problem on math magazine and the solution does not hold,see:
and I want to use Holder's inequality
$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{27}{(x+y+z)^2}$$
so we must show that the following inequality is true
$$\dfrac{81}{(x+y+z)^2}\ge 9$$
In fact, this also does not hold. Because $$xyz(x+y+z)=3\Longrightarrow x+y+z\ge 3$$
Now I suspect if $(1)$ is correct? Although I have not found any counter-example so far.
|
We'll prove a stronger inequality.
Let $a$, $b$ and $c$ be positives such that $ abc(a+b+c)=3$. Prove that:
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{72}{(a+b+c)^2}\ge 11$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $v^2$.
In another hand, $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{72}{(a+b+c)^2}=\frac{9v^4-6uw^3}{w^6}+\frac{8}{u^2}$ is an increasing function of $v^2$.
Thus, by $uvw$ (http://www.artofproblemsolving.com/community/c6h278791) it remains to check one case only: $b=a$ and $c=\sqrt{a^2+\frac{3}{a^2}}-a$, which gives
$\frac{2}{a^2}+\frac{1}{\left(\sqrt{a^2+\frac{3}{a^2}}-a\right)^2}+\frac{72}{\left(\sqrt{a^2+\frac{3}{a^2}}+a\right)^2}\geq11$ or
$146a^8+219a^4-99a^2+18\geq142a^6\sqrt{a^4+3}$.
Let $a^2=x$. Hence, we need to prove that
$(146x^4+219x^2-99x+18)^2\geq142^2x^6(x^2+3)$ or
$(x-1)^2(128x^6+256x^5+768x^4-1932x^3+1281x^2-324x+36)\geq0$, which is obvious.
|
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|
2-adic expansion of (2/3) I have been asked in an assignment to compute the 2-adic expansion of (2/3). It just doesn't seem to work for me though. In our definition of a p-adic expansion we have $x= \sum_{n=0}^{\infty}a_np^n$ with $0 \leq a_0<p$.
So I use the same method as for similar questions which seems to work fine. This involves solving each congruence $\frac{2}{3} = \sum_{n=0}^{\infty}a_np^n \ (mod \ p^N)$ for $N=1,2,3,4,5$ and then we are supposed to guess the pattern and show it by summing the series.
Doing this, I get $\frac{2}{3} = 2 + 2^2 + 2^4 + 2^5 + ... \\ = (2+2^2)(1+2^3+2^6+...) \\ = 6(\frac{1}{1-2^3}) \\ = \frac{6}{-7}.$
Which is just completely wrong?
An example of me solving a congruence:
$\frac{2}{3} \equiv a_0 \ (mod \ 2)$ so $0 \equiv 2 \equiv 3a_0 \ (mod \ 2)$ so $a_0 = 0$.
|
If I am not mistaken you can also do the following:
$\frac{2}{3}=1+\frac{1}{-3}$ where
$\frac{1}{-3}=\frac{1}{1-4}=\frac{1}{1-2^2}=\sum_{n=0}^{\infty}2^{2n}$ (geometric series expansion).
This yields the whole expansion of $\frac{2}{3}$.
|
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|
Problem with a simple convergent power series Create a power series around $x=0$ and find its radius. What is $f(1)$?
$$f(x)=\frac{1}{2+x}$$
Well according to me it is easy to see that the radius is $\left|x\right|<2$ and $f(1)=\frac{1}{3}$...right?
If I try to manipulate it to the geometric serie $f(x)=\frac{1}{1-x}$ I get $f(x)=\frac{1}{1-(-1-x)} =\frac{1}{2+x} $ which would be $\sum (-1-x)^k$ because $f(x)=\frac{1}{1-x}=\sum x^k $ but $\sum (-1-x)^k$ diverge when $x=1$. Why doesnt that work when $f(x)=\frac{1}{2}\frac{1}{1+x/2)}=\frac{1}{2}\sum (\frac{-x}{2})^k$ seems to do fine.$f(1) =1/3$
Do not use maclaurin to prove the series.
I am doing something fundamentally wrong here, please help me out.
|
I start with
$$\frac{1}{t+1}=\sum _{k=0}^{\infty } (-1)^k t^k$$
Now I substitude
$$t=\frac{x}{2}$$
It gives me
$$\frac{1}{\frac{x}{2}+1}=\sum _{k=0}^{\infty } (-1)^k \left(\frac{x}{2}\right)^k$$
Multiply with $\frac{1}{2}$ I get
$$\frac{1}{2 \left(\frac{x}{2}+1\right)}=\frac{1}{2} \sum _{k=0}^{\infty } (-1)^k \left(\frac{x}{2}\right)^k=\sum _{k=0}^{\infty } \frac{(-1)^k x^k}{2^{k+1}}$$
And this is
$$\frac{1}{x+2}=\sum _{k=0}^{\infty } \frac{(-1)^k x^k}{2^{k+1}}$$
|
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|
Solve the following in vector form:
So i did a substitution to solve the system normally, and got
$x=17.67$
$y=9.67$
$z=10.67$
Where I am stuck is how to represent something like this in a vector form, maybe my solution was wrong in the first place.
|
For example
\begin{align}
(1,2,-3)^T \cdot x &= 5 \\
(2,1,-3)^T \cdot x &= 13 \\
(-1,1,0)^T \cdot x &= -8
\end{align}
where $x = (x_1,x_2,x_3)^T$.
But more usual is using a matrix equation
$$
A x = b
$$
with $A$ using the above first argument vectors as row vectors and
$b = (5, 13, -8)^T$.
The solution would be $x = A^{-1} b$, if $A$ is invertible.
In your case it is not. Your solution works, so there must be infinite many solutions.
$$
(A | b) = \\
\left(
\begin{array}{rrr|r}
1 & 2 & -3 & 5 \\
2 & 1 & -3 & 13 \\
-1 & 1 & 0 & -8
\end{array}
\right)
\to \\
\left(
\begin{array}{rrr|r}
1 & 2 & -3 & 5 \\
0 & -3 & 3 & 3\\
0 & 3 & -3 & -3
\end{array}
\right)
\to \\
\left(
\begin{array}{rrr|r}
1 & 0 & -1 & 7 \\
0 & 1 & -1 & -1 \\
0 & 0 & 0 & 0
\end{array}
\right)
$$
This gives the general solution $x = (7 + x_3,-1 + x_3, x_3)^T$ with arbitrary $x_3 \in \mathbb{R}$.
|
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|
order of operation For the following expressions,why we can get the right answer even we do addition /subtraction first ?
*
*$3+4\times 11-5 = (3+4)\times(11-5) = 42$
*$6+4\times7-4$
*$5+2\times13-10$
*$4+7\times16-6$
*$3-2\times1+5$
|
Expressed algebraically, you have:
$$\begin{align}
a+bc+d&=(a+b)(c+d)\\
a+bc+d&=ac+ad+bc+bd\\
a+d-ac-ad-bd&=0\\
a(1-c-d)+d(1-b)&=0\\
\end{align}$$
This is a single simultaneous equation in 4 unknowns, therefore it has an infinite number of solutions - you have just found some of them.
Rewriting to address OP comment
$$a=\frac{d(b-1)}{1-c-d}$$
Pick any integers $c$ and $d$. Choose integer $b$ such that $a$ is an integer.
|
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|
The $n$’th Harmonic number The $n$’th Harmonic number is defined as follows: $$H_n = 1 + \frac12 + \frac13 +\ldots + \frac1n$$ for $n\geq 1$.
a) Show that for all $n≥0$: $H_{2^n} \geq 1+\frac n2$.
b) Show that for all $n≥0$: $H_{2^n} \leq 1+n$.
|
(a). Base case:
for $n = 0,$ we have $H_{2^n}= H_1 = 1,$ and $1 + 0/2 = 1.$
Induction Step:
Let's assume $k \geq 0$ is an arbitrary number, and it holds for $k$:
$H_{2^k} \geq 1+ k/2.$
Then we prove it for $k+1$:
$$\begin{eqnarray}
H_{2^{k+1}} & = & H_{2^k} + \underbrace{\frac{1}{2^k+1} + \ldots +
\frac{1}{2^{k+1}}}_{\text{($2^k$ terms)}} \\
& \geq &1 + k/2 + \underbrace{\frac{1}{2^k+1} + \ldots +
\frac{1}{2^{k+1}}}_{\text{($2^k$ terms)}} \\
& \geq & 1 + k/2 + \underbrace{\frac{1}{2^{k+1}} + \ldots +
\frac{1}{2^{k+1}}}_{\text{($2^k$ terms)}} \\
& = & 1 + k/2 + \frac{2^k}{2^{k+1}} \\
& = & 1 + k/2 +1/2 \\
& = & 1+ (k+1)/2
\end{eqnarray}$$
so, it holds for $k+1.$ Therefore, by induction we proved it is true for all $n \geq 0.$
|
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Prove that there exists a choice of $\pm$ signs Let $x$, $y$ be two real numbers. Prove that there exists a choice of $\pm$ signs so that:
$$ \pm \cos x \pm \cos y \pm \cos (y - x) \leq -1 $$
|
Without loss of generality, we can assume $-\pi\le y\le x\le \pi$.
(If $x<y$, swap $x$ and $y$ and note that $\cos (x-y)=\cos(y-x)$.) Then
$-\frac{\pi}2\le \frac{x}2 \le \frac{\pi}2,\
-\frac{\pi}2\le \frac{y}2 \le \frac{\pi}2,\ \text{and}\ 0\le \frac{x-y}2\le \pi.$
\begin{align*}
A&=\cos x+\cos y+\cos (x-y)+1\\
&= 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}+2\cos^2\frac{x-y}{2}\\
&= 2\cos\frac{x-y}{2}\left(\cos\frac{x+y}{2}+\cos\frac{x-y}{2}\right)\\
&= 4\cos\frac{x-y}{2}\cos\frac{x}{2}\cos\frac{y}{2},\\
B&=-\cos x-\cos y+\cos (x-y)+1\\
&= -2\cos\frac{x+y}{2}\cos\frac{x-y}{2}+2\cos^2\frac{x-y}{2}\\
&= 2\cos\frac{x-y}{2}\left(-\cos\frac{x+y}{2}+\cos\frac{x-y}{2}\right)\\
&= 4\cos\frac{x-y}{2}\sin\frac{x}{2}\sin\frac{y}{2},\\
C&=\cos x-\cos y-\cos (x-y)+1\\
&= 2\sin\frac{x+y}{2}\sin\frac{x-y}{2}+2\sin^2\frac{x-y}{2}\\
&= 2\sin\frac{x-y}{2}\left(\sin\frac{x+y}{2}+\sin\frac{x-y}{2}\right)\\
&= 4\sin\frac{x-y}{2}\sin\frac{x}{2}\cos\frac{y}{2},\\
D&=-\cos x+\cos y-\cos (x-y)+1\\
&= -2\sin\frac{x+y}{2}\sin\frac{x-y}{2}+2\sin^2\frac{x-y}{2}\\
&= 2\sin\frac{x-y}{2}\left(-\sin\frac{x+y}{2}+\sin\frac{x-y}{2}\right)\\
&= -4\sin\frac{x-y}{2}\cos\frac{x}{2}\sin\frac{y}{2}.
\end{align*}
Case 1: If $\frac{\pi}2\le \frac{x-y}2\le \pi$, then $A\le 0$.
Case 2: If $0\le \frac{x-y}2\le \frac{\pi}2$, then either Case2-1, Case2-2 or Case2-3 holds:
Case2-1: if $-\frac{\pi}2\le \frac{x}2 \le 0$, then $C\le 0$.
Case2-2: if $0\le \frac{y}2 \le \frac{\pi}2$, then $D\le 0$.
Case2-3: if $0\le \frac{x}2 \le \frac{\pi}2$ and $-\frac{\pi}2\le \frac{y}2 \le 0$,
then $B\le 0$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
From the set $\{1,2,\ldots,n\}$ two numbers are chosen uniformly, with replacement. Find the probability that the product of the numbers is even. The answer is given, just the reasoning is unclear. Result: $${2\over n}\left[{n\over 2}\right]-\left({\left\lfloor{n \over 2}\right\rfloor\over n}\right)^2,\text{ where $\lfloor\cdot\rfloor$ is the whole part of a number}.$$
The second part of the question is: The probability that the sum of the squares of the numbers is even: again the answer: $$1-{2\over n}\left\lfloor{n\over 2}\right\rfloor+{2\over n^2}\left\lfloor{n \over 2}\right\rfloor^2, \text{where $\lfloor\cdot\rfloor$ is the whole part of the number}.$$
|
First Problem
There are $\left\lfloor\frac{n+1}2\right\rfloor$ odd numbers in $1\dots n$, so the probability that each number chosen is odd is
$$
\frac1n\left\lfloor\frac{n+1}2\right\rfloor
$$
Therefore, the probability that the product is even is
$$
\bbox[5px,border:2px solid #F0A000]{1-\frac1{n^2}\left\lfloor\frac{n+1}2\right\rfloor^2}
$$
There are $n-\left\lfloor\frac{n}2\right\rfloor$ odd numbers in $1\dots n$, so the probability that each number chosen is odd is
$$
\frac1n\left(n-\left\lfloor\frac{n}2\right\rfloor\right)
$$
Therefore, the probability that the product is even is
$$
1-\left(1-\frac1n\left\lfloor\frac{n}2\right\rfloor\right)^2=\bbox[5px,border:2px solid #F0A000]{\frac2n\left\lfloor\frac{n}2\right\rfloor-\frac1{n^2}\left\lfloor\frac{n}2\right\rfloor^2}
$$
Second Problem
There are $\left\lfloor\frac{n+1}2\right\rfloor$ odd numbers and $n-\left\lfloor\frac{n+1}2\right\rfloor$ even numbers in $1\dots n$. Therefore, the number of even-odd draws or odd-even draws is $\left\lfloor\frac{n+1}2\right\rfloor\left(n-\left\lfloor\frac{n+1}2\right\rfloor\right)$. Therefore, the probability of an even sum is
$$
1-\frac2{n^2}\left\lfloor\frac{n+1}2\right\rfloor\left(n-\left\lfloor\frac{n+1}2\right\rfloor\right)
=\bbox[5px,border:2px solid #F0A000]{1-\frac2n\left\lfloor\frac{n+1}2\right\rfloor+\frac2{n^2}\left\lfloor\frac{n+1}2\right\rfloor^2}
$$
There are $n-\left\lfloor\frac{n}2\right\rfloor$ odd numbers and $\left\lfloor\frac{n}2\right\rfloor$ even numbers in $1\dots n$. Therefore, the number of even-odd draws or odd-even draws is $\left\lfloor\frac{n}2\right\rfloor\left(n-\left\lfloor\frac{n}2\right\rfloor\right)$. Therefore, the probability of an even sum is
$$
1-\frac2{n^2}\left\lfloor\frac{n}2\right\rfloor\left(n-\left\lfloor\frac{n}2\right\rfloor\right)
=\bbox[5px,border:2px solid #F0A000]{1-\frac2n\left\lfloor\frac{n}2\right\rfloor+\frac2{n^2}\left\lfloor\frac{n}2\right\rfloor^2}
$$
|
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"url": "https://math.stackexchange.com/questions/1188970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Why does $\binom{n}{2} = \frac{n^2 - n }{2}$? In a proof in Introduction to Algorithms, the book says $\binom{n}{2} \cdot \frac{1}{n^{2}} = \frac{n^2 - n }{2}\cdot \frac{1}{n^{2}}$, which implies $\binom{n}{2} = \frac{n^2 - n }{2}$.
Why are these equivalent? Is it a special case of a more general rule?
|
$\binom{n}{2}$ is the number of subsets of size $2$ from a set of size $n$. There are $n$ ways to choose the first element, and $n-1$ ways to choose the secon element. but sets are not "ordered", so we must divide by two and we have $\frac{n\cdot(n-1)}{2}$ such subsets. Thus $\binom{n}{2}=\frac{n(n-1)}{2}=\frac{n^2-n}{2}$.+
Alternatively use induction:
Base:
$\binom{2}{2}=1=\frac{2^2-2}{2}$
Induction:
$\binom{n}{2}=\binom{n-1}{2}+\binom{n-1}{n}$ by pascal recurrence.
$\binom{n-1}{2}+\binom{n-1}{2}=\frac{(n-1)^2-(n-1)}{2}+(n-1)$ by induction.
$\frac{(n-1)^2-(n-1)}{2}+(n-1)=\frac{(n-1)^2-(n-1)+2(n-1)}{2}=\frac{(n-1)^2+(n-1)}{2}=\frac{n^2-2n+1+n-1}{2}=\frac{n^2-n}{2}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
My Attempt: I start with direct proof.
Let $a,b,c$ be consecutive integers and $a< b < c $, there exists a integer $k$ such that $a=k, b=k+1, c=k+2$. Then $a^3+b^3=k^3+(k+1)^3=k^3+(k^3+3k^2+3k+1)$ and $c^3=(k+2)^3=k^3+6k^2+12k+8=(k^3+3k^2+3k+1)+(3k^2+9k+7)$. Hence, we have $k^3+(k^3+3k^2+3k+1)\neq (k^3+3k^2+3k+1)+(3k^2+9k+7)$ which implies $a^3+b^3\neq c^3$.
Does my proof valid ? And, is use Contradiction easier? If not, can anyone give me a hit or suggestion ?
Thanks
|
You should prove one more thing: $k^3\neq3k^2+9k+7$. Since the equation $k^3-3k^2-9k-7=0$ has no integer solutions, the conclusion follows.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Infinite sequence series. Limit If $0<x<1$ and
$$A_n=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\ldots+\frac{x^{2^n}}{1-x^{2^{n+1}}},$$
then $\lim_{n\to\infty} A_n$ is
$$\text{a) }\ \dfrac{x}{1-x} \qquad\qquad \text{b) }\ \frac{1}{1-x} \qquad\qquad \text{c) }\ \frac{1}{1+x} \qquad\qquad \text{d) }\ \frac{x}{1+x}$$
How to do this? Not able to convert in any standard form.
|
If you expand each of the geometric series in the $A_n$, and combine each of the series together as one sum (you can since these are each absolutely convergent), then you can demonstrate that this is tending to $x(1-x)^{-1}$.
What you need to make sure of though, is that $x^k$ can appear only once between each of the terms $x^{2^n}(1-x^{2^{n+1}})^{-1}$ for each $k \in \mathbb{N}$.
This amounts to showing that each integer $k$ can be expressed uniquely as $2^n+m\cdot 2^{n+1}$. Or in other words that each integer can be written as $2^n(2m+1)$ (a power of 2 times an odd number). This representation is certainly unique.
More explicitly consider, $A_0$, $A_1$ and $A_2$:
$$A_0 = \frac{x}{1-x^2} = x+x^3+x^5+x^7+\cdots$$
Here we have all of the odd numbers.
$$A_1 = A_0 + \frac{x^2}{1-x^4} = (x+x^3+x^5+x^7+\cdots) + (x^2 + x^6 + x^{10} + \cdots)$$
$$=x+x^2+x^3+\underline{ }+x^5+x^6+x^7+\underline{ }+x^9+x^{10}+x^{11}+\underline{ } +\cdots$$
Notice that we are slowly filling in the spaces for the geometric series.
$$A_2 = A_1 + \frac{x^{4}}{1-x^8} = A_1 + (x^4+x^{12}+x^{20}+\cdots)$$
$$=x+x^2+x^3+x^4+x^5+x^6+x^7+\underline{ }+x^9+x^{10}+x^{11}+x^{12} +\cdots$$
With $A_2$ we see that we are still missing $x^8$, but that will appear with $A_3$. The uniqueness at the top of my answer tells us that we will never have $2x^k$ since each $x^k$ will appear only once. Thus $A_n$ tends to the geometric series (times $x$) and $$A_n \to \frac{x}{1-x}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$ How do we compute this integral ?
$$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$
I have tried partial fraction but it is quite hard to factorize the denominator. Any help is appreciated.
|
$$
\begin{align}
\int_0^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\,\mathrm{d}x
&=\int_0^1\frac{(1+x^6)(1+x^2)}{1+x^{10}}\,\mathrm{d}x\tag{1a}\\
&=\int_1^\infty\frac{(1+x^6)(1+x^2)}{1+x^{10}}\,\mathrm{d}x\tag{1b}\\
&=\frac12\int_0^\infty\frac{(1+x^6)(1+x^2)}{1+x^{10}}\,\mathrm{d}x\tag{1c}
\end{align}
$$
Explanation:
$\mathrm{(1a)}$: multiply by $\frac{1+x^2}{1+x^2}$
$\mathrm{(1b)}$: substitute $x\mapsto1/x$ and simplify
$\mathrm{(1c)}$: average $\mathrm{(1a)}$ and $\mathrm{(1b)}$
Depending on the context of the question, there are a few ways to compute $\mathrm{(1c)}$.
One is to use this answer, which gets $\int_0^\infty\frac{x^n}{1+x^{10}}\,\mathrm{d}x=\frac\pi{10}\csc\!\left(\frac{(n+1)\pi}{10}\right)$ by contour integration:
$$
\begin{align}
\int_0^\infty\frac{(1+x^6)(1+x^2)}{1+x^{10}}\,\mathrm{d}x
&=\int_0^\infty\frac1{1+x^{10}}\,\mathrm{d}x
+\int_0^\infty\frac{x^2}{1+x^{10}}\,\mathrm{d}x\\
&+\int_0^\infty\frac{x^6}{1+x^{10}}\,\mathrm{d}x
+\int_0^\infty\frac{x^8}{1+x^{10}}\,\mathrm{d}x\\
&=\frac\pi{10}\left[\csc\left(\frac{\pi}{10}\right)
+\csc\left(\frac{3\pi}{10}\right)
+\csc\left(\frac{7\pi}{10}\right)
+\csc\left(\frac{9\pi}{10}\right)\right]\\
&=\frac\pi5\left[\csc\left(\frac{\pi}{10}\right)+\csc\left(\frac{3\pi}{10}\right)\right]\\
&=\frac{2\pi}{\sqrt5}\tag{2}
\end{align}
$$
Therefore,
$$
\int_0^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\,\mathrm{d}x=\frac\pi{\sqrt5}\tag{3}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1}$$
$$\cos (x)=\frac{1-u^2}{u^2+1}$$
$$dx=\frac{2du}{u^2+1}$$
$$=\int_{0}^{\infty }\frac{2(1-\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+1)}du$$
$$=\int_{0}^{\infty }\frac{2(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{(u+1)^2(u^2+1)}du $$
$$=2\int_{0}^{\infty }(\frac{2}{(u+1)^2}-\frac{1}{u^2+1})du $$
$$=-2\int_{0}^{\infty }\frac{1}{u^2+1}du+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du $$
$$\lim_{b\rightarrow \infty }\left | (-2\tan^{-1}(u)) \right |_{0}^{b}+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=(\lim_{b\rightarrow \infty}-2\tan^{-1}(b))+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=-\pi+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
Let $$s=u+1$$
$$ds=du$$
$$=-\pi+4\int_{1}^{\infty}\frac{1}{s^2}ds$$
$$=-\pi+\lim_{b\rightarrow \infty}\left | (-\frac{4}{s}) \right |_{1}^{b}$$
$$=-\pi+(\lim_{b\rightarrow \infty} -\frac{4}{b}) +4$$
$$=4-\pi$$
$$\approx 0.85841$$
|
First use partial fractions to get rid of the sine in the numerator:
$$ \int_0^{\pi} \frac{1-\sin{x}-1+1}{1+\sin{x}} \, dx = \int_0^{\pi} \left( \frac{2}{1+\sin{x}} - 1 \right) \, dx = -\pi + \int_0^{\pi} \frac{2 \, dx}{1+\sin{x}}. $$
We have the identity
$$ 1+\sin{x} = 2\sin^2{\left( \frac{x}{2} + \frac{\pi}{4} \right)} $$
(from $\cos{2\theta}=1-2\sin^2{\theta}$ and $\sin{\theta}=-\cos{(\theta+\pi/2)}$), so the remaining integral is
$$ \int_0^{\pi} \csc^2{\left( \frac{x}{2} + \frac{\pi}{4} \right)} \, dx = \left[ -2 \cot{\left( \frac{x}{2} + \frac{\pi}{4} \right)} \right]_0^{\pi} = -2(-1-1) = 4 $$
|
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|
Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$ I recently ran into this series:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$
Of course this is just a special case of the Beta Dirichlet Function , for $s=3$.
I had given the following solution:
$$\begin{aligned}
1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\
&\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\
&=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\
&= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\
&=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4} \right ] \\
&=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32}
\end{aligned}$$
where I used polygamma identities and made use of the absolute convergence of the series at $(*)$ in order to re-arrange the terms.
Any other approach using Fourier Series, or contour integration around a square, if that is possible?
|
Here is another way, and it combines integration and probability.
First off, consider the triple integral:
$$I=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{1}{1+x^2y^2z^2}dzdydx.$$
Convert this integrand into a geometric series:
$$\frac{1}{1+x^2y^2z^2}=\sum_{n=0}^{\infty} (-1)^n(xyz)^{2n}.$$
Replace the integrand with this series and integrate term by term to get that: $$I=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}.$$
Now we proceed to evaluate $$I=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{1}{1+x^2y^2z^2}dzdydx.$$ Make the change of variables $$x=\frac{\sin(u)}{\cos(v)},y=\frac{\sin(v)}{\cos(w)},z=\frac{\sin(w)}{\cos(u)}$$ which has a nice Jacobian $\frac{\partial(x,y,z)}{\partial(u,v,w)}=1+x^2y^2z^2$ that cancels with the integrand. The region of integration is the open polytope $U$ described by inequalities $$0<u+v<\frac{\pi}{2},0<v+w<\frac{\pi}{2},0<u+w<\frac{\pi}{2}, u,v,w>0.$$ We need to compute the volume of $U$ to get the value of $I$.
For the purpose of this proof, we consider the scaled polytope $V$ defined by inequalities: $$0<u+v<1,0<v+w<1,0<u+w<1, u,v,w>0.$$ Notice $$\text{Vol}(U)=\left(\frac{\pi}{2}\right)^3 \text{Vol}(V)$$ where $\text{Vol}$ means volume.
We compute $\text{Vol}(V)$ through probability.
Suppose $n=(n_1,n_2,n_3) \in V.$ We first intend to find the probability: $$\text{Pr}\left(n\in V \cap \text{each } n_i <\frac{1}{2}\right)$$ It turns out that:
$$\text{Pr}\left(n\in V \cap \text{each } n_i <\frac{1}{2}\right)=\left(\frac{1}{2}\right)^3=\frac{1}{8}.$$ This is the case because $n$ would lie in the open hypercube $\left(0,\frac{1}{2}\right)^3,$ and one can verify that $\left(0,\frac{1}{2}\right)^3 \subset V.$
Next, we intend to find the probability: $$\text{Pr}\left(n\in V \cap \text{exactly one } n_i \geq \frac{1}{2}\right)$$ It turns out that:
$$\text{Pr}\left(n\in V \cap \text{exactly one } n_i \geq \frac{1}{2}\right)=3\int_{\frac{1}{2}}^{1}\int_{0}^{1-x}\int_{0}^{1-x}=\frac{1}{8}.$$
Here, to get this answer, I computed the probability $\text{Pr}\left(n\in V \cap n_1 \geq \frac{1}{2}\right)$ and multiplied this answer by $3$ to account for all the possible $3$ cases of this event happening.
Notice that we cannot have more than one $n_i \geq \frac{1}{2}$ at the same time, as this will violate the constraints of $V$. This means we add up the computed probabilities to get that : $$\text{Vol}(V)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}.$$ This means that
$$I=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}=\text{Vol}(U)=\left(\frac{\pi}{2}\right)^3 \text{Vol}(V)=\left(\frac{\pi^3}{8}\right)\frac{1}{4}=\frac{\pi^3}{32}.$$
|
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|
Fake induction proofs Question: Can you provide an example of a claim where the base case holds but there is a subtle flaw in the inductive step that leads to a fake proof of a clearly erroneous result? [Note: Please do not answer with the very common all horses are the same color example.]
Comment: Sometimes inductive arguments can lead to controversial conclusions, such as the surprise exam paradox, Richard's paradox and a host of other paradoxes. However, I am interested in examples of a more mathematical nature (as opposed to linguistic) where the inductive argument is subtly flawed and leads to erroneous conclusions.
Note: If you provide an answer, please do so in a way similar to how current answers are displayed (gray out the flaw so people can be challenged to discover it).
|
Claim:
If $a$ is an odd square modulo $m$, then $a$ is a square modulo $2m$.
Proof:
Let $m=2^v k$ where $k$ is $m$'s largest odd factor. Proceed by induction on $v$.
Base case:
$v=0$, so $m=k$ is odd.
$a$ is a square modulo $k$, so, for some $x$, $k\mid x^2-a$. Modulo $2k$,
\begin{align*}
(k+x)^2 &= k^2+2kx+x^2\\
&= k+x^2\tag{$k^2=k$ as $k$ is odd; $2kx=0$}
\end{align*}
so $x^2$ and $(k+x)^2$ are $a$ and $k+a$ in some order. So $a$ is a square modulo $2k$.
Inductive step:
$a$ is a square modulo $2^v k$ so there is an $x$ where $x^2=a\mod 2^v k$.
As the base case is proved, we may suppose that $m$ is even and $v\geqslant1$.
This means that, as $a$ is odd, $x$ is odd.
Knowing that $m$ is even also enables us to refine $x$:
Modulo $m$,
\begin{align*}
(m/2+x)^2 &= (2^{v-1} k+x)^2\\
&= 2^{2v-2} k^2+2\cdot2^{v-1} kx+x^2\\
&=2^{v} kx+x^2\\
&=x^2\\
&=a
\end{align*}
as the other terms are multiples of $m=2^v k$. That is, we may take $0<x<m/2$.
Modulo $2m$,
\begin{align*}
(m/2+x)^2 &= (2^{v-1} k+x)^2\\
&= 2^{2v-2}k^2+2\cdot2^{v-1} kx+x^2\\
&=2^v kx+x^2\\
&=mx+x^2\\
&=m+x^2\tag{as $x$ is odd}
\end{align*}
Thus $x^2$ and $(m/2+x)^2$ are $a$ and $m+a$ in some order. Thus $a$ is a square modulo $2m$. QED.
Flaw:
The base case is valid. The inductive step relies on $2^{v+1}k\mid2^{2v-2}k^2$, which is true only if $2v-2\geqslant v+1$ i.e. $v\geqslant3$ i.e. $8\mid m$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the following limit without L'Hopital I tried to evaluate the following limits but I just couldn't succeed, basically I can't use L'Hopital to solve this...
for the second limit I tried to transform it into $e^{\frac{2n\sqrt{n+3}ln(\frac{3n-1}{2n+3})}{(n+4)\sqrt{n+1}}}$ but still with no success...
$$\lim_{n \to \infty } \frac{2n^2-3}{-n^2+7}\frac{3^n-2^{n-1}}{3^{n+2}+2^n}$$
$$\lim_{n \to \infty } \frac{3n-1}{2n+3}^{\frac{2n\sqrt{n+3}}{(n+4)\sqrt{n+1}}}$$
Any suggestions/help? :)
Thanks
|
For the first limit, it breaks into 2 factors with finite limits.
$$\lim{n \to \infty} \frac{2n^2-3}{7-n^2} = \frac{2n^2}{-n^2} =-2\\
\lim{n \to \infty} \frac{3^n-2^{n-1}}{3^{n+2}+2^n2} = \frac{3^n}{3^{n+2}} = \frac{1}{9}
$$
so the answer is $-\frac{2}{9}$.
For the second, rewrite it as
$$
\left(\frac{(3n-1)(2n-3)}{4n^2-9} \right) ^{\frac{\sqrt{n}2n(1+\frac{3}{2n}+\ldots)}{\sqrt{n}(n+4)(1+\frac{1}{2n}+\ldots)}}
$$
and expand to next-lowest order in $1/n$ to get
$$
\left( \frac{3}{2} \left[ 1-\frac{11}{6n}+\ldots\right] \right)^{2(1+\frac{3}{2n}+\ldots-\frac{9}{2n}+\ldots)}
$$
Since the exponent does not go to infinity we can in fact just use the lowest order terms, getting
$$\left( \frac{3}{2} \right)^2 = \frac{9}{4}$$
|
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|
Jacobi Triple product cases Hello I want some guidance how can I prove these identities of Jacobi Tripple Product
$$\sum_{n=-\infty}^\infty q^{2n^2+n}=\prod_{n=1}^\infty\frac{(1-q^{2n})^2}{(1-q^n)}$$
$$\sum_{n=-\infty}^\infty q^{n^2}=\prod_{n=1}^\infty\frac{(1-(-q)^n)}{(1+(-q)^n)}$$
$$\sum_{n=-\infty}^\infty q^{n(n+1)}=\prod_{n=1}^\infty\frac{(1-q^{4n})}{(1-q^{4n-2})}$$
|
Start with the triple product identity
$$\sum_{n = -\infty}^\infty z^n q^{n^2} = \prod_{n = 1}^\infty (1 - q^{2n})(1 + zq^{2n-1})(1 + z^{-1}q^{2n-1}).$$
*
*Replacing $q$ by $q^2$, then setting $z = q$, we obtain
\begin{align}
\sum_{n = -\infty}^\infty q^{2n^2+n} &= \prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n-1})(1 + q^{4n-3})\\
& = \prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{2n-1})\\
& = \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^{2n})(1 + q^{2n-1})\\
& = \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^n)\\
& = \prod_{n = 1}^\infty \frac{(1 - q^{2n})^2}{1 - q^n}.
\end{align}
*Setting $z = 1$, we get
\begin{align}\sum_{n = -\infty}^\infty q^{n^2} &= \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^{2n-1})(1 + q^{2n-1})\\
&= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n\, \text{odd}}^\infty (1 + q^n)\\
&= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n = 1}^\infty \frac{1 + q^n}{1 + q^{2n}}\\
&= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n = 1}^\infty \frac{1 - q^{2n}}{(1 - q^n)(1 + q^{2n})}\\
&= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n = 1}^\infty \frac{1 - q^{2n}}{(1 - q^{2n})(1 - q^{2n-1})(1 + q^{2n})}\\
&= \prod_{n = 1}^\infty \frac{1 - (-q)^n}{(1 - q^{2n-1})(1 + q^{2n})}\\
&= \prod_{n = 1}^\infty \frac{1 - (-q)^n}{1 + (-q)^n}.
\end{align}
*Setting $z = q$ results in
\begin{align}\sum_{n = -\infty}^\infty q^{n^2 + n} &= \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^{2n})(1 + q^{2n-2})\\
& = \prod_{n = 1}^\infty (1 - q^{4n})\prod_{n = 2}^\infty (1 + q^{2n-2})\\
& = \prod_{n = 1}^\infty (1 - q^{4n}) \prod_{n = 1}^\infty (1 + q^{2n})\\
& = \prod_{n = 1}^\infty (1 - q^{4n}) \prod_{n = 1}^\infty \frac{1 - q^{4n}}{1 - q^{2n}}\\
& = \prod_{n = 1}^\infty (1 - q^{4n}) \prod_{n = 1}^\infty \frac{1 - q^{4n}}{(1 - q^{4n})(1 - q^{4n-2})}\\
&= \prod_{n = 1}^\infty \frac{1 - q^{4n}}{1 - q^{4n-2}}.
\end{align}
|
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|
Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous.
So lets take:
${\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}^2 \, \leqslant \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}{\mid \sqrt{x^2+x} + \sqrt{y^2+y} \mid} \,\, = \,\, {\mid x^2+x-y^2-y \mid} \,\, = \,\, {\mid (x+y)(x-y)+(x-y) \mid} \,\, = \,\, {\mid (x-y)(x+y+1) \mid } < {\epsilon}^2 \Rightarrow \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid} < \epsilon$
So now we know that if we take $\delta = {\epsilon}^2$ the condition for uniform continuity of this function will be met because
${\mid x - y \mid} < (\delta = {\epsilon}^2) \Rightarrow {\mid f(x)-f(y) \mid} < \epsilon$
Is this proof valid? Or I miss something?
|
Hint: Multiply by conjugate:
$$\left|\sqrt{x^2+x}-\sqrt{y^2+y}\right|=\left|\sqrt{x^2+x}-\sqrt{y^2+y}\right|\frac{|\sqrt{x^2+x}+\sqrt{y^2+y}|}{|\sqrt{x^2+x}+\sqrt{y^2+y}|}$$
Note that this part of the proof only works with $[A,\infty)$, where $A > 0$. For the interval $[0,A]$, observe that $[0,A]$ is closed, and $f$ is continuous over that closed interval. There should be a theorem that says if $f$ is continuous over the closed interval then $f$ is uniformly continuous over that interval.
|
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|
Product of repeated cosec.
$$P = \prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$$
I realize that there must be some sort of trick in this.
$$P = \csc^2(1)\csc^2(3).....\csc^2(89) = \frac{1}{\sin^2(1)\sin^2(3)....\sin^2(89)}$$
I noticed that: $\sin(90 + x) = \cos(x)$ hence,
$$\sin(89) = \cos(-1) = \cos(359)$$
$$\sin(1) = \cos(-89) = \cos(271)$$
$$\cdots$$
$$P \cdot P = \frac{\cos^2(-1)\cos^2(-3)....}{\sin^2(1)\sin^2(3)....\sin^2(89)}$$
But that doesnt help?
|
I'm using a non-standard notation inspired by various programming languages in this evaluation because I think it's a bit easier to follow than the traditional big pi product notation. Hopefully, it's self-explanatory. Also, all angles are given in degrees.
This derivation uses the following identities:
$$
\cos x = \sin(90 - x) \\
\sin(180 - x) = \sin x \\
\sin 2x = 2 \sin x \cdot \cos x \\
\text{and hence} \\
\cos x = \frac{\sin 2x}{2 \sin x}\\
$$
$$\begin{align}
\text{Let } P & = prod(\csc^2 x: x = \text{1 to 89 by 2}) \\
1 / P & = prod(\sin x: x = \text{1 to 89 by 2})^2 \\
& = prod(\sin x: x = \text{1 to 89 by 2}) \cdot prod(\cos x: x = \text{1 to 89 by 2}) \\
& = prod(1/2 \sin 2x: x = \text{1 to 89 by 2}) \\
& = 2^{-45} prod(\sin x: x = \text{2 to 178 by 4}) \\
& = 2^{-45} prod(\sin x: x = \text{2 to 86 by 4}) \cdot \sin 90 \cdot prod(\sin x: x = \text{94 to 178 by 4})
\end{align}$$
$$
\text{But } prod(\sin x: x = \text{94 to 178 by 4}) = prod(\sin x: x = \text{2 to 86 by 4}) \\
\text{Let } Q = prod(\sin x: x = \text{2 to 86 by 4}), \text{so } P = \frac{2^{45}}{Q^2}
$$
$$\begin{align}
Q & = prod(cos x: x = \text{4 to 88 by 4}) \\
& = \frac{prod(sin 2x: x = \text{4 to 88 by 4})}{prod(2 sin x: x = \text{4 to 88 by 4})} \\
& = 2^{-22}\frac{prod(sin x: x = \text{8 to 176 by 8})}{prod(sin x: x = \text{4 to 88 by 4})} \\
& = 2^{-22}\frac{prod(sin x: x = \text{8 to 88 by 8}) \cdot prod(sin x: x = \text{96 to 176 by 8})}
{prod(sin x: x = \text{4 to 84 by 8}) \cdot prod(sin x: x = \text{8 to 88 by 8})} \\
\end{align}$$
$\text{But } prod(\sin x: x = \text{96 to 176 by 8}) = prod(\sin x: x = \text{4 to 84 by 8})$
$\text{Hence } Q = 2^{-22} \text{ and } P = 2^{45} / (2^{-22})^2 = 2^{89}$
|
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|
Parabolic sine approximation Problem Find a parabola ($f(x)=ax^2+bx+c$) that approximate the function sine the best on interval [0,$\pi$].
The distance between two solutions is calculated this way (in relation to scalar product): $\langle u,v \rangle=\int_0^\pi fg$.
My (wrong) solution I thought that I would get the solution by calculating the orthogonal projection $w=a+bx+cx^2$ of $v=\sin x$ on subspace $W=\langle u_1,u_2,u_3 \rangle=\langle 1,x,x^2\rangle$ using Gramm matrix. Then I have
$$\begin{pmatrix}\langle u_1, u_1\rangle&\langle u_1, u_2\rangle&\langle u_1, u_3\rangle\\
\langle u_2, u_1\rangle&\langle u_2, u_2\rangle&\langle u_2, u_3\rangle\\
\langle u_3, u_1\rangle&\langle u_3, u_2\rangle& \langle u_3, u_3\rangle\end{pmatrix}
\begin{pmatrix} a\\b\\c\end{pmatrix}=
\begin{pmatrix} \langle u_1, v\rangle\\
\langle u_2,v\rangle\\
\langle u_3, v\rangle \end{pmatrix}$$
So then
$$\begin{pmatrix} \int_0^\pi 1&\int_0^\pi x&\int_0^\pi x^2\\
\int_0^\pi x& \int_0^\pi x^2& \int_0^\pi x^3 \\
\int_0^\pi x^2&\int_0^\pi x^3&\int_0^\pi x^4\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} \int_0^\pi \sin x\\\int_0^\pi x\sin x\\\int_0^\pi x^2 \sin x \end{pmatrix}$$
But solving these equations didn't give me any good answer. So my question is - is my way of solving it completely wrong (if so, can you give me hints how to do it otherwise)?
Thank you
Edit
Then$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
\frac{\pi^2}{2}&\frac{\pi^3}{3}&\frac{\pi^4}{4}\\
\frac{\pi^3}{3}&\frac{\pi^4}{4}&\frac{\pi^5}{5}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\\pi\\\pi^2-4 \end{pmatrix}$$
$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&\frac{\pi^4}{12}&\frac{4\pi^5}{45}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&0&\frac{\pi^5}{180}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
Edit II My way (I have found the mistake I did) will get the result
$$f(x)=\dfrac{60(\pi^2-12)}{\pi^5}x^2-\dfrac{60(\pi^2-12)}{\pi^4}x+\dfrac{12(\pi^2-10)}{\pi^3}$$
which I hope is the right answer with error approx. $0,000936$
|
First,I donnot think the orthogonal projection is proper here. The orthogonal projections are often used to make infinity series to approach function,because the coefficients there is more simple,you can try to use error square or the maximun of error
|
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|
Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational
How can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?
I know that $\sqrt{3}, \sqrt{5}$ and $\sqrt{7}$ are all irrational and that $\sqrt{3}+\sqrt{5}$, $\sqrt{3}+\sqrt{7}$, $\sqrt{5}+\sqrt{7}$ are all irrational, too. But how can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?
|
Assume $$\sqrt{3}+ \sqrt{5}+ \sqrt{7}= \frac{a}{b}$$ for some integers $a,b$. Multiblying both sides by $\sqrt{3}+ \sqrt{5}- \sqrt{7}$ yields $$ (\sqrt{3}+\sqrt{5})^2 - 7 = \frac{a(\sqrt{3}+ \sqrt{5}- \sqrt{7})}{b}=\frac{a\left(a/b-2\sqrt{7}\right)}{b} \\ 2\sqrt{15}+8-7=\frac{a^2/b-2a\sqrt{7}}{b} \\ 2b\sqrt{15} = \frac{a^2}{b}-2a\sqrt{7}-b \\ 60b^2=\frac{a^4}{b^2}+28a^2+b^2-\frac{4a^3\sqrt{7}}{b}+4ab\sqrt{7}-2a^2 \\ 59b^2-26a^2-\frac{a^4}{b^2}=4a\sqrt{7}\left(b-\frac{a^2}{b}\right). $$ We can divide both sides by $4a(b^2−a^2)=4a(b-a)(b+a)$ because $a$ can't be $0$ by definition, $b+a$ is a positive integer and $b−a$ can't be $0$ because that would imply $a/b=1$, which is impossible. Thus we get $$\frac{b\left(59b^4-26a^2b^2-a^4\right)}{4a(b^2-a^2)}=\sqrt{7},$$ contradiction.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to show that $zw=1\implies w = z^{-1}, z = w^{-1}$? I need to show that
$$zw=1\implies w = z^{-1}, z = w^{-1}$$
But I don't know how to start. I've tried to consider $z=a+bi, w = c+di$ then:
$$zw = ac+adi+bci-bd\implies zw = ac-bd + (ad+bc)i = 1\implies \\ac-bd = 1\\ad+bc = 0$$
$$ac = 1+bd\\ad = -bc$$
$$c = \frac{1+bd}{a}\\d = \frac{-bc}{a}$$
$$c = \frac{1+b(\frac{-bc}{a})}{a}=\frac{1-\frac{b^2c}{a}}{a} = \frac{a-b^2c}{a^2}\implies ca^2 = a-b^2c\implies ca^2+b^2c=a\implies c[a^2+b^2] = a\implies c = \frac{a}{a^2+b^2}$$
Using $$d=\frac{-bc}{a}\implies d = \frac{-b\frac{a}{a^2+b^2}}{a} = \frac{-b}{a^2+b^2}$$
LoL I did it :D
|
If you know in advance that the complex numbers form a field then $$zw = 1 \implies z^{-1}zw = z^{-1}1 \implies w = z^{-1}.$$
|
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|
Factoring derivatives Well I have $$r\frac{d^2T}{dr^2} + \frac{dT}{dr} = -\frac{qr}{k}$$
I wonder how to do algebraic manipulation to go from this step to
$$\frac{d}{dr}(r\frac{dT}{dr}) = \frac{-qr}{k}$$
Is it even possible ?
I wonder how to write a second derivative in terms of a first derivative ?
|
Use
\begin{align}
\frac{d}{dx}(f(x) g(x)) = f(x) \, \frac{d g(x)}{dx} + \frac{d f(x)}{dx} \, g(x)
\end{align}
to put
\begin{align}
x \frac{d^{2} f}{dx^{2}} + \frac{d f}{dx} = - \frac{q x}{k}
\end{align}
into the form
\begin{align}
\frac{d}{dx} \left( x \frac{df}{dx} \right) = - \frac{q x}{k}.
\end{align}
For the sake of solution integrating both sides of this last equation leads to
\begin{align}
x \frac{d f}{dx} = - \frac{q x^{2}}{2 k} + c_{1}
\end{align}
and can be continued to be of the form
\begin{align}
f(x) &= \int \left[ - \frac{q x}{ 2 k } + \frac{c_{1}}{x} \right] \, dx \\
&= - \frac{q x^{2}}{4 k} + c_{1} \ln(x) + c_{0}
\end{align}
|
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|
Integration of $\int\log(\sqrt{1-x}+\sqrt{1+x}) \, dx$ Integration of
$$\int\log\left(\sqrt{1-x}+\sqrt{1+x}\right) \, dx$$
Please help to go through this problem as i have started with putting $x$= $cos2y$.
|
Consider the integral
\begin{align}
I = \int\log(\sqrt{1-x}+\sqrt{1+x})dx
\end{align}
It is seen that
\begin{align}
I &= \int \ln\left[ \sqrt{1 -x} \, \left(1 + \sqrt{\frac{1+x}{1-x}} \right) \right] dx \\
&= \frac{1}{2} \int \ln(1-x) \, dx + \int \ln\left(1 + \sqrt{\frac{1+x}{1-x}} \right) dx
\end{align}
Now make the substitution
\begin{align}
u^{2} = \frac{1+x}{1-x}
\end{align}
to obtain
\begin{align}
I &= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \int \frac{4 \ln(1+u) \, u \, du}{(1+u^{2})^{2}}.
\end{align}
Integration by parts leads to
\begin{align}
I &= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \frac{2 \ln(1+u)}{1+u^{2}} - \int \frac{du}{(1+u)(1+u^{2})} \\
&= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \frac{2 \ln(1+u)}{1+u^{2}} + \frac{1}{4} \ln(1 + u^{2}) - \frac{1}{2} \ln(1+u) - \frac{1}{2} \tan^{-1}(u).
\end{align}
Now reverse substitution yields
\begin{align}
I &= \frac{1}{2} [ (x-1) \ln(1-x) - x] + \left(\frac{1}{2}-x\right) \ln\left(1+ \sqrt{\frac{1 + x}{1-x}} \right) + \frac{1}{4} \ln\left(\frac{2}{1-x} \right)
- \frac{1}{2} \tan^{-1}\left( \sqrt{\frac{1 + x}{1-x}} \right) \\
&= \frac{\ln(2)}{4} - \frac{x}{2} + \frac{1}{2} \left( x - \frac{3}{2}\right) \ln(1-x) + \left(\frac{1}{2}-x\right) \ln\left(1+ \sqrt{\frac{1 + x}{1-x}} \right) - \frac{1}{2} \tan^{-1}\left( \sqrt{\frac{1 + x}{1-x}} \right)
\end{align}
|
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|
Number of solution of Frobenius equation Oke I am trying to find all presentable of a number $n$ as sum $ax+(a+1)y$ where $a=0,1,\ldots$ and $x,y\geq0$ are integers.
I find that
$5=1+1+1+1+1=1+1+1+2=1+2+2=2+3=5$ so we have $5$ ways.
$7=1+1+1+1+1+1+1=1+1+1+1+1+2=1+1+1+2+2=1+2+2+2=2+2+3=3+4$ so we have $7$ ways.
I suppose that if $p\geq3$ is a prime number then there are exactly $p$ ways. However, I found that
$11=1+1+1+1+1+1+1+1+1+1+1=1+1+1+1+1+1+1+1+1+2=1+1+1+1+1+1+1+2+2=1+1+1+1+1+2+2+2=1+1+1+2+2+2+2=1+2+2+2+2+2$
and
$=2+2+2+2+3=2+3+3+3=3+4+4 =11$ so there are only $10$ ways.
My question are that: What is the formula for the number ways to preprent a prime $p$ to sum of $x$ numbers $a$ and $y$ numbers $a+1$ where $a,x,y$ are non-negative integers?
|
The problem is slightly misstated; what you're actually counting is the number of different ways to write $n=ax+(a+1)y$ with $a,x\ge1$ and $y\ge0$, not $a,x,y\ge0$. Changing the lower limit on $a$ and $x$ rules out what would otherwise be an infinite number of solutions, $(a,x,y)=(0,x,n)$ with $x=0,1,2,\ldots$, as well as double counting the sum $n=1+1+\cdots+1$ as $(0,1,n)$ and $(1,n,0)$ and the "sum" $n=n$ as $(n-1,0,1)$ and $(n,1,0)$.
With this in mind, note that
$$n=ax+(a+1)y=a(x+y)+y=au+y$$
where $u=x+y$ is strictly greater than $y$ (since $x\ge1$). But this means $y$ is the remainder when you divide $u$ into $n$. So now it's clear what's happening: for each value $u=1,2,3,\ldots$, there is precisely one way to write $n=au+y$ with $0\le y\lt u$. But if $u\gt n$, we have $a=0$. So there are precisely $n$ ways to write $n=au+y$ with $a\ge1$ and $0\le y\lt u$, and these translate into precisely $n$ ways to write $n=a(x+y)+y$ with $a,x\ge1$ and $y\ge0$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Simultaneusly solving $2x \equiv 11 \pmod{15}$ and $3x \equiv 6 \pmod 8$ Find the smallest positive integer $x$ that solves the following simultaneously.
Note: I haven't been taught the Chinese Remainder Theorem, and have had trouble trying to apply it.
$$
\begin{cases}
2x \equiv 11 \pmod{15}\\
3x \equiv 6 \pmod{8}
\end{cases}
$$
I tried solving each congruence individually.
The inverse for the first is 8: $x \equiv 8\cdot11 \equiv 88 \equiv 13 \pmod{15}$.
The inverse for the second is 3: $x \equiv 3\cdot6 \equiv 18 \equiv 2 \pmod{8}$.
But I can't figure out how where to go from here.
|
You must start from Bézout's identity between the moduli $15$ and $8$: $\,2\cdot 8-1\cdot 15=1$. From the solutions to the individual congruences $\color{red} {13} \pmod{15}$ and $\color{red}2 \pmod{8}$, you deduce the solutions to the system:
$$2\cdot 8\cdot\color{red} {13}-1\cdot 15\cdot\color{red}2=178\equiv 58\pmod{120}.$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that the map $f(z)=\frac{1}{z}$ sends any line onto either a line or a circle. Show the cases in which the image is a line and the case in which the image is a circle.
I understand that representing the equation of line, ($ax+by+c=0$ $a,b,c\in\Bbb R$ $a,b\neq0$ at the same time), as $pz+\overline {pz} +2c=0$ where $p$ is a suitable nonzero complex number, should help me achieve my end goal in this proof. However, I'm not sure how to use this fact to do so as well as how to justify why I should think of the line in this way.
|
First of all, I think that the rquation of the line $ax + by + c = 0$, when written in terms of $z = x + iy$, should be
$\bar p z + p \bar z + 2c = 0, \tag{1}$
and not
$pz + \bar p z + 2c = 0; \tag{2}$
here $p = a + bi$. To see the problem with (2), note that it may be written
$(p + \bar p)z + 2c= 0, \tag{3}$
and that
$p + \bar p = (a + bi) + (a - bi) = 2a; \tag{4}$
thus (2) becomes
$2az + 2c = 0. \tag{5}$
Assuming $a \ne 0$, the only solution of (5) is
$z = -\dfrac{c}{a}; \tag{6}$
we see that in this case (5) represents the single point (6), not the line
$ax + by + c = 0, \tag{7}$
and if $a = 0$, then (5) forces $c = 0$ and any $z \in \Bbb C$ solves (5); in neither case does (2) represent a line. (1), however, is easily seen to be equivalent to (7), as follows: with $p$ and $z$ as above, it follows that
$\bar p z = (a - bi)(x + iy) = (ax + by) + i(ay - bz), \tag{8}$
and hence, since
$p \bar z = \overline{\bar p z} \tag{9}$
we have from (8) that
$\bar p z + p \bar z = 2(ax + by), \tag{10}$
whence
$\bar p z + p \bar z + 2c = 2ax + 2by + 2c; \tag{11}$
it is easy to see from (11) that (1) and (7) represent the same line.
Granting that the correct equation for a line, in terms of $z = x + iy$, is (1), we investigate the action of $f(z) = 1/z$. If $c \ne 0$, then clearly (1) does not pass through the origin; $z \ne 0$ for all $z$ satisfying (1). Then $z \bar z \ne 0$ as well, and we may write
$f(z) = \dfrac{1}{z} = \dfrac{\bar z}{z \bar z} \tag{12}$
for any such $z \in \Bbb C$. If we apply $f(z)$ in the form of (12) to (1) we obtain
$\bar p \dfrac{1}{z} + p \dfrac{1}{\bar z} + 2c = 0 \tag{13}$
or
$\bar p \dfrac{\bar z}{z \bar z} + p \dfrac{z}{z \bar z} + 2c = 0; \tag{14}$
multiplying (14) though by $z \bar z = x^2 + y^2 \ne 0$ yields
$\bar p \bar z + p z + 2c z \bar z = 0, \tag{15}$
which I claim is the equation of a circle provided $c \ne 0$. To wit: with $z = x + iy$ and $p = a + bi$,
$pz = (a + bi)(x + iy) = (ax - by) + i(ay + bx), \tag{16}$
and since
$\bar p \bar z = \overline{pz} \tag{17}$
we see that
$\bar p \bar z + p z + 2c z \bar z = 2(ax - by) + 2c(x^2 + y^2), \tag{18}$
and with $c \ne 0$ we may thus infer
$\dfrac{a}{c}x - \dfrac{b}{c}y + x^2 + y^2 = 0; \tag{19}$
we complete the squares in $x$ and $y$ separately by adding $(a^2 + b^2)/4c^2$ to both sides:
$(x + \dfrac{a}{2c})^2 + (y - \dfrac{b}{2c})^2 = \dfrac{a^2 + b^2}{4c^2}. \tag{20}$
Since by hypothesis $a^2 + b^2 \ne 0 \ne c$, (20) describes a circle of non-zero radius $\sqrt{(a^2 + b^2/4c^2}$ centered at $(-a/2c, b/c)$. This covers the case $c \ne 0$.
If $c = 0$, then $z = 0$ is a solution to (1) and we may only use (12) away from $0$; indeed, caution must be taken with the transformation $f(z) = 1/z$ since $0$ must be handled separately. For $z \ne 0$ we may set $c = 0$ in (18) and thus deduce that
$ax - by = 0; \tag{21}$
this is the equation of a line passing through the origin, normal to the vector $(a, - b)$, since (21) may be written
$(a, - b) \begin{pmatrix} x \\ y \end{pmatrix} = 0. \tag{22}$
Apparently, away from $0$, the transformation $1/z$ maps the line $ax + by = 0$ (from (1), (10)) to the line $ax - by = 0$; $0$ maps to the so-called "point at infinity", which maps back to $0$. We summarize:
$f(z) = \dfrac{1}{z}$ maps
Lines not passing through $0$ to circles;
and
Lines passing throug $0$ to other lines passing through $0$.
|
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|
Can flooring and ceiling thirds sum up to more than a whole? I'm working on a web layout which needs to divide an area into three columns, but do so using whole pixel values (due to subpixel rendering issues on some mobile devices). For this purpose I've decided to go with the following approach:
*
*The first two columns are rounded down (floored)
*The last column is rounded up (ceil)
Is there any numeric value for which using the above approach might break the layout? That is, does any integer $x$ exist which satisfies the following condition?
$$ \left \lfloor \frac{x}{3} \right \rfloor + \left \lfloor \frac{x}{3} \right \rfloor + \left \lceil \frac{x}{3} \right \rceil > x\\
x\in \mathbb{Z} $$
How would I approach disproving the existence of such a value?
|
The answer is no:
*
*$x\equiv0\pmod3\implies\left\lfloor\frac{x}{3}\right\rfloor+\left\lfloor\frac{x}{3}\right\rfloor+\left\lceil\frac{x}{3}\right\rceil=\frac{x}{3}+\frac{x}{3}+\frac{x}{3}=\frac{3x}{3}=x$
*$x\equiv1\pmod3\implies\left\lfloor\frac{x}{3}\right\rfloor+\left\lfloor\frac{x}{3}\right\rfloor+\left\lceil\frac{x}{3}\right\rceil=\frac{x-1}{3}+\frac{x-1}{3}+\frac{x+2}{3}=\frac{3x-2+2}{3}=x$
*$x\equiv2\pmod3\implies\left\lfloor\frac{x}{3}\right\rfloor+\left\lfloor\frac{x}{3}\right\rfloor+\left\lceil\frac{x}{3}\right\rceil=\frac{x-2}{3}+\frac{x-2}{3}+\frac{x+1}{3}=\frac{3x-4+2}{3}<x$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If $e_0 = \frac{1}{2} $and $\forall n\in\Bbb{N}:e_n=\frac{(2n-1)^2}{2n(2n+1)}e_{n-1}$, find $\sum_{n\geq 0} e_n$ Consider the following optimal stopping game: The controller is presented with steps in a fair random walk (fair coin flips, $P(h)=P(t) = \frac{1}{2}$) and at each stage of the game, the controller can stop and take a payoff of $\frac{H}{H+T}$ where $H$ is the number of heads encountered up till that point, and $T$ is the number of tails.
A proposed stopping strategy is to stop the first time the payoff exceeds $\frac{1}{2}$. Note that this strategy does eventually stop, with probability 1. (This is probably in fact the optimal strategy, but that is not the point of this question.) What is the expectation value of the payoff if that strategy is used?
I have been able to calculate the probability $g_m$ of stopping at step $2m+1$ as
$$
g_m = \frac{1}{2^{2m+1}}\frac{1}{m+1} \binom{2m}{m}
$$
and the contribution to the expected payoff is from the possibility of stopping at step $2m+1$ is
$$
e_m = \frac{1}{2} g_m \left( 1+\frac{1}{2m+1} \right)
$$
It is then easy to show that $e_0 = \frac{1}{2}$ and for all integer $n>0$
$$
e_n=\frac{(2n-1)^2}{2n(2n+1)}e_{n-1}
$$
To find the expected payoff I need to find
$$\sum_{n=0}^\infty e_n
$$
Thus the question:
If $e_0 = \frac{1}{2} $and $\forall n\in\Bbb{N}:e_n=\frac{(2n-1)^2}{2n(2n+1)}e_{n-1}$, find $\sum_{n\geq 0} e_n$
|
$$\begin{align}e_n &= \frac{(2 n-1)^2}{(2 n+1) (2 n)} e_{n-1} \\ &=\frac{(2 n-1)^2}{(2 n+1) (2 n)} \frac{(2 n-3)^2}{(2 n-1) (2 n-2)}\cdots \frac{1^2}{3 \cdot 2} \cdot\frac12 \\ &= \frac{(2 n)!}{(2 n+1) n!^2} \frac1{2^{2 n+1}} \end{align} $$
Thus,
$$ \sum_{n=0}^{\infty} e_n = \sum_{n=0}^{\infty} \frac{(2 n)!}{(2 n+1) n!^2} \frac1{2^{2 n+1}} $$
This sum is straightforward. Consider
$$f(x) = \sum_{n=0}^{\infty} \frac{(2 n)!}{(2 n+1) n!^2} x^{2 n+1} $$
Then
$$f'(x) = \sum_{n=0}^{\infty} \binom{2 n}{n} x^{2 n} = \left ( 1-4 x^2 \right )^{-1/2}$$
Then
$$f(x) = \frac12 \arcsin{2 x} $$
The sum is then
$$f \left ( \frac12 \right ) = \frac{\pi}{4} $$
|
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"url": "https://math.stackexchange.com/questions/1215997",
"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\left(\cos{\frac{5\pi}{9}}\right)^{11}+\left(\cos{\frac{7\pi}{9}}\right)^{11}+\left(\cos{\frac{11\pi}{9}}\right)^{11}$ How to evaluate $$\left(\cos{\frac{5\pi}{9}}\right)^{11}+\left(\cos{\frac{7\pi}{9}}\right)^{11}+\left(\cos{\frac{11\pi}{9}}\right)^{11}?$$
I found the problem on this page.
|
Using for an odd $n$
$${\cos ^n}(x) = \frac{1}{{{2^{n - 1}}}}\sum\limits_{k = 0}^{\frac{{n - 1}}{2}} {\left( \begin{gathered}
n \\
k \\
\end{gathered} \right)} \cos ((n - 2k)x)$$
gives me:
$${\cos ^{11}}(t) = \frac{{462\cos (t) + 330\cos (3t) + 165\cos (5t) + 55\cos (7t) + 11\cos (9t) + \cos (11t)}}{{1024}}$$
It follows therfore:
$${\cos ^{11}}\left( {\frac{{5\pi }}{9}} \right) = \frac{{154 - 462\sin \left( {\frac{\pi }{{18}}} \right) + 56\cos \left( {\frac{\pi }{9}} \right) - 165\cos \left( {\frac{{2\pi }}{9}} \right)}}{{1024}}$$
$${\cos ^{11}}\left( {\frac{{7\pi }}{9}} \right) = \frac{{154 - 56\sin \left( {\frac{\pi }{{18}}} \right) + 165\cos \left( {\frac{\pi }{9}} \right) - 462\cos \left( {\frac{{2\pi }}{9}} \right)}}{{1024}}$$
and
$${\cos ^{11}}\left( {\frac{{11\pi }}{9}} \right) = \frac{{154 - 56\sin \left( {\frac{\pi }{{18}}} \right) + 165\cos \left( {\frac{\pi }{9}} \right) - 462\cos \left( {\frac{{2\pi }}{9}} \right)}}{{1024}}$$
summing up:
$${\cos ^{11}}\left( {\frac{{5\pi }}{9}} \right) + {\cos ^{11}}\left( {\frac{{7\pi }}{9}} \right) + {\cos ^{11}}\left( {\frac{{11\pi }}{9}} \right) = \frac{{462 - 574\sin \left( {\frac{\pi }{{18}}} \right) + 386\cos \left( {\frac{\pi }{9}} \right) - 1089\cos \left( {\frac{{2\pi }}{9}} \right)}}{{1024}}$$
$$ \approx - 0.10661630948635201594863746012996982756343832484191$$
|
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|
Residue Integration I am attempting to calculate the integral of $\frac{(1+sin(\theta))}{(3+cos(\theta))}$ from $0$ to $2\pi$. I have already changed $sin$ and $cos$ into $\frac{1}{2i(z-z^{-1})}$ and $\frac{1}{2(z+z^{-1})}$. I am really stuck now. Can anyone please guide me?
|
Since the integrand is periodic, then
$$\begin{align}
\int_0^{2\pi} \frac{1+\sin \theta}{3 + \cos \theta} d\theta&=\int_{-\pi}^{\pi} \frac{1+\sin \theta}{3 + \cos \theta} d\theta\\
&=\int_{-\pi}^{\pi} \frac{1}{3 + \cos \theta} d\theta
\end{align}$$
where we exploited the fact that $\frac{\sin \theta}{3+\cos \theta}$ is an odd function.
Next, let $u=\tan (\theta /2)$ so that $du = \frac12 \sec^2(\theta /2)$, and $\cos \theta =\frac{1+u^2}{1-u^2}$. Then, we find the anti-derivative of $\frac{1}{3+\cos \theta}$ is $\frac{\sqrt{2}}{2} \arctan (\sqrt{2}\tan (\theta /2)/2) +C$. Evaluating the anti-derivative between limits of integration reveals
$$\int_{-\pi}^{\pi} \frac{1}{3+\cos \theta} d\theta =\frac{\sqrt{2}\pi}{2}$$
Now, let's use contour integration.
Let $z=e^{i \theta}$ so that $d\theta=dz/(iz)$.
Next note that
$\frac{1+\sin \theta}{3+\cos \theta}=\frac{z^2+2iz-1}{i(z^2+6z+1)}$. The only root of the denominator that lies inside the unit circle is at $z=-3+2\sqrt{2}$.
Now the integral of interest is
$$-\int_C \frac{z^2+2iz-1}{z(z^2+6z+1)} dz$$
The integral has two simple poles, one at $0$ and the other at $-3+2\sqrt{2}$. The evaluation of the integral therefore is, by the Residue Theorem,
$$2\pi i \sum \text{Res} \left(- \left( \frac{z^2+2iz-1}{z(z^2+6z+1)}\right)\right)$$
The residue at $z=0$ is found by evaluating the term $- \frac{z^2+2iz-1}{z^2+6z+1}$ at $z=0$. We find this first residue to be 1.
The residue at $z=-3+2\sqrt{2}$ is found by evaluating the term $-\frac{z^2+2iz-1}{z(z+3+2\sqrt{2})}$ at $z=-3+2\sqrt{2}$. We find this second residue to be $-1-i\sqrt{2}/4$.
Putting it all together recovers the aforementioned result!
|
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|
Extension of the additive Cauchy functional equation Let $f\colon (0,\alpha)\to \def\R{\mathbf R}\R$ satisfy $f(x + y)=f(x)+f(y)$
for all $x,y,x + y \in (0,\alpha)$, where $\alpha$ is a positive real number. Show that there exists an additive function $A \colon \R \to \R$ such that $A(x) = f(x)$ for all $x \in (0, \alpha)$.
Simply I want to define a function A In specific form as an extension of the function f wich is additive functional equation. I tried to define the function A .
|
Let $x > 0$. Choose $n \in \def\N{\mathbf N}\N$ with $\frac xn < \alpha$. Define $A(x) := nf(\frac xn)$. Note that this is well-defnined: If $m\in \N$ is another natural number such that $\frac xm < \alpha$, we have
\begin{align*}
mf\left(\frac xm\right) &= mf\left(\sum_{k=1}^n \frac x{mn}\right)\\
&= m\sum_{k=1}^n f\left(\frac x{mn}\right)\\
&= \sum_{l=1}^m n f\left(\frac x{mn}\right)\\
&= nf\left(\sum_{l=1}^m \frac x{mn}\right)\\
&= nf\left(\frac x{n}\right).
\end{align*}
For $x < 0$ choose $n \in \N$ with $\frac xn > -\alpha$ and define $A(x) := -nf(-\frac xn)$, finally, let $A(0) = 0$. Then $A$ is an extension of $f$, to show that it is additive, let $x,y \in \def\R{\mathbf R}\R$. Choose $n \in \N$ such that $\frac xn, \frac yn, \frac{x+y}n \in (-\alpha, \alpha)$. We have if $x,y \ge 0$:
\begin{align*}
A(x+y) &= nf\left(\frac{x+y}n\right)\\
&= nf\left(\frac xn\right) + nf\left(\frac yn\right)\\
&= A(x) + A(y)
\end{align*}
If both $x,y \le 0$, we argue along the same lines. Now suppose $x \ge 0$, $y \le 0$, $x+y \ge 0$. We have $A(y) = -A(-y)$ be definition of $A$. Hence
\begin{align*}
-A(y) + A(x+y) &= A(-y) + A(x+y)\\
&= A(-y+x+y)\\
&= A(x).
\end{align*}
If $x \ge 0$, $y \le 0$, $x+y \le 0$, we have $-x \le 0$ and
\begin{align*}
-A(x) + A(x+y) &= A(-x) + A(x+y)\\
&= A(y)
\end{align*}
|
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|
how to solve $2(2y-1)^{\frac{1}{3}}=y^3+1$ how can I solve $2(2y-1)^{\frac{1}{3}}=y^3+1$
I got by $x=\frac{y^3+1}{2}$ that $y=\frac{x^3+1}{2}$ but I was told earlier that can't say that $x=y$.
So I can I solve this equation?
Thanks.
|
By cubing you get
$$ y^9+3 y^6+3 y^3 - 16 y + 9 = 0$$
It factors (using Wolfram alpha) into
$$ (y-1) (y^2+y-1) (y^6+2 y^4+2 y^3+4 y^2+2 y+9) = 0$$
According to WA, all the roots of the last polynomial are complex. The second one has $ \frac{(-1-\sqrt{5})}{2}$ and $ \frac{(\sqrt{5}-1)}{2}$ as roots. I don't know if you can get exact forms for the other (complex) roots.
|
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|
Is there a closed-form of $\sum_{n=1}^{\infty} \frac{\sin(n)}{n^4}$ Is there a closed-form summation result for Fourier series:
$$\sum_{n=1}^{\infty}\frac{\sin(n)}{n^4}?\tag{1}$$
I tried using available result of the following (odd) function :
$$\frac{\pi-x}{2}=\sum_{n\geq 1}\frac{\sin(nx)}{n}.\tag{2}$$
However I couldn't get the series to agree because when I integrate the formula to evaluate coefficients, the powers of $n$ become odd..
|
Consider the series
\begin{align}
f(x) = \sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{4}}
\end{align}
for which upon differentiation the following is obtained.
\begin{align}
f'(x) &= \sum_{n=1}^{\infty} \frac{\cos(n x)}{n^{3}} \\
f''(x) &= - \sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{2}} \\
f'''(x) &= - \sum_{n=1}^{\infty} \frac{\cos(n x)}{n} = \frac{1}{2} \left( \ln(1 - e^{i x}) + \ln(1 - e^{- i x}) \right).
\end{align}
By integration the following is obtained
\begin{align}
f''(x) &= \frac{i}{2} \left( Li_{2}(e^{i x}) - Li_{2}(e^{-i x}) \right) + c_{2} \\
f'(x) &= \frac{1}{2} \left( Li_{3}(e^{i x}) - Li_{3}(e^{- i x}) \right) + c_{2} x + c_{1} \\
f(x) &= \frac{-i}{2} \left( Li_{4}(e^{i x}) - Li_{4}(e^{-i x}) \right) + \frac{c_{2} x^{2}}{2} + c_{1} x + c_{0}
\end{align}
where $Li_{m}(x)$ is the polylogarithm function. In order to determine the constants of integration evaluation of the series is required. This is determined by setting $x = 0$.
\begin{align}
f(0) &= 0 = c_{0} \\
f'(0) &= \zeta(3) = c_{1} \\
f''(0) &= 0 = c_{2}.
\end{align}
The result is then
\begin{align}
\sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{4}} &= \frac{-i}{2} \left( Li_{4}(e^{i x}) - Li_{4}(e^{-i x}) \right) + \zeta(3) \, x
\end{align}
|
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|
diophantine equation $x^3+x^2-16=2^y$ Solve in integers: $x^3+x^2-16=2^y$.
my attempt:
of course $y\ge 0$, then $2^y\ge 1$, so $x\ge 1$.
for $y=0,1,2,3$ there is no good $x$.
so $y\ge 4$ and we have equation $x^2(x+1)=16(2^z+1)$, where $z=y-4\ge 0$.
what now?
|
$x^3+x^2=2^y+16$. The RHS is positive, so $x^2(x+1)>0\iff x\ge 1$. Since $2^y$ is an integer, we have $y$ is a positive integer too ($y=0$ won't give a solution).
$x,y$ are positive integers.
$x^3+x^2-16=2^y$. You see a cubic polynomial on the LHS that could easily be strictly bounded between two consecutive cubes (namely $x^3$ and $(x+1)^3$) for most values of $x$, making it impossible for it to be a cube itself. So if $2^y$ is a cube, i.e. $3\mid y$, we're done. And indeed it is a cube.
$2^y\equiv 1,2,4\pmod{7}$ for $y\equiv0,1,2\pmod 3$, respectively.
$x^3+x^2-16\equiv 5,0,3,6,1,1,5\pmod{7}$ for $x\equiv 0,1,2,3,4,5,6\pmod{7}$, respectively.
The only common residue is $1$, so $y\equiv 0\pmod{3}$. This implies $x^3+x^2-16$ is a cube.
But $x^3<x^3+x^2-16<(x+1)^3,\forall x\ge 5$, so $1\le x\le 4$, which only give $(x,y)=(4,6)$.
|
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|
Norm of the sum of two vectors This problem has two parts.
Part a): $x$ and $y$ are vectors. If $||x|| = 7, ||y|| = 11$, what is the smallest value possible for $||x+y||$? (Note: the || || denotes the norm of a vector).
This is what I have tried so far: I put vector $x$ equal to $\begin{pmatrix} a \\ b \end{pmatrix}$ and vector $y$ equal to $\begin{pmatrix} c \\ d \end{pmatrix}$. $||x|| = 7$ would then be, after simplification, $a^2+b^2 = 49$. Similarly, for $||y|| = 11$, after simplification, $c^2+d^2 = 121$.
Then, $||x+y|| = \sqrt{(a+c)^2 + (b+d)^2}$. Expanding gives us $\sqrt{(a^2+b^2) + (c^2+d^2) + 2(ac+bd)} = \sqrt{49+121+2(ac+bd)} = \sqrt{170+2(ac+bd)}$. That is where I was stuck--any hints for the next few steps?
Part b): $x$ and $y$ are vectors (these are not the same vectors as in part a). If $||x|| = 4, ||y|| = 5, ||x+y|| = 7,$ what is $||2x-3y||$?
Using the same approach as in part a), where vector $x$ is equal to $\begin{pmatrix} a \\ b \end{pmatrix}$ and vector $y$ is equal to $\begin{pmatrix} c \\ d \end{pmatrix}$, $a^2+b^2 = 16$ and $c^2+d^2 = 25$. Similarly, for $||x+y||$, after simplification, it equals $41+2(ac+bd) = 49$ -> $ac+bd = 4$. I'm not sure what to do next after this part too. Any hints?
|
for part (a), $$|x+ y| \ge |y|-|x| = 11-7 = 4$$ this is achieved when $x = -y|x|/|y|.$
for part (b), you can find the angle $t$ between the $x,y$ by using the cosine rule in the triangle made up of $x, y, x+y.$ that is $$\cos t = \frac{5^2 + 4^2 - 7^2}{2 \times 5 \times 4} = -\frac 15 \to t =\pi-\cos^{-1}\left(\frac 15\right).$$
$\bf edit:$
we have $$|2x-3y|^2 = 4|x|^2 + 9|y|^2 - 12x.y=64+225+12\times 4\times5\times \frac15=337 $$ that is $$|2x-3y| = \sqrt{337}. $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the 3-volume of the 3-parallelepiped defined by $\left\{\vec{v_1},\vec{v_2},\vec{v_3}\right\}$? We have $\left\{\vec{v_1},\vec{v_2},\vec{v_3}\right\}=\left\{\begin{bmatrix}1\\0\\0\\0\end{bmatrix},\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}1\\2\\3\\4\end{bmatrix}\right\}$
QR-factorization gives $\text{det }(A^TA)=\text{det }(R^TR)=\text{det }(R^2)$ such that $\sqrt{\text{det }(A^TA)}$ yields the volume of our desired parallelepiped. Letting $A=\begin{pmatrix}1&1&1\\0&1&2\\0&1&3\\0&1&4\end{pmatrix}$,
$\sqrt{\text{det }(A^TA)}=\sqrt{\begin{pmatrix}1&0&0&0\\1&1&1&1\\1&2&3&4\end{pmatrix}\begin{pmatrix}1&1&1\\0&1&2\\0&1&3\\0&1&4\end{pmatrix}}=\sqrt{\text{det }\begin{pmatrix}1&1&1\\1&4&10\\1&10&30\end{pmatrix}}$
Expanding down the first column of $\begin{pmatrix}1&1&1\\1&4&10\\1&10&30\end{pmatrix}$ using Laplace yields
$\text{det }\begin{pmatrix}1&1&1\\1&4&10\\1&10&30\end{pmatrix}=6$, such that $\sqrt{\text{det }(A^TA)}=\sqrt{6}$, which is the volume of the desired parallelepiped.
Is there a quicker and more energy-efficient algorithm to solve this? I know that for $3\times3$ matrices we can use $\text{det }A=\vec{u}\cdot(\vec{v}\times\vec{w})$, given $\left\{\vec{u},\vec{v},\vec{w}\right\}$ span the columns of $A$. Is there an analog here?
|
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Since volume of a parallelipiped spanned by a set of vectors is invariant under the operation of adding a scalar multiple of one vector to another, it suffices to compute the volume of the parallelipiped spanned by
$$
\left[\begin{array}{@{}c@{}}
1 \\ 0 \\ 0 \\ 0 \\
\end{array}\right],\qquad
\left[\begin{array}{@{}c@{}}
0 \\ 1 \\ 1 \\ 1 \\
\end{array}\right],\qquad
\left[\begin{array}{@{}c@{}}
0 \\ 0 \\ 1 \\ 2 \\
\end{array}\right].
\tag{1}
$$
The last two vectors (each lying in the coordinate $3$-space $\{x_{1} = 0\}$ orthogonal to the first vector) span a parallelogram $P$ whose area is the magnitude of the cross product
$$
(1, 1, 1) \times (0, 1, 2) = (1, -2, 1),
$$
namely $\sqrt{6}$. Since the first vector has unit length and is orthogonal to $P$, the three vectors in (1) span a prism of volume $\sqrt{6}$.
Naturally, there's "good fortune" in this example: After "column reduction", the first column was a standard basis vector orthogonal to the remaining columns.
Alternatively, there's a "cross product" for $(n - 1)$ vectors in $\Reals^{n}$ whose definition generalizes the usual cross product in $\Reals^{3}$: Assemble your vectors into an $(n - 1) \times n$ matrix, prepend the standard basis vectors into the first row (obtaining a square matrix) and take the determinant. The magnitude of the resulting vector is the volume of the parallelipiped of the original set of vectors. (See below.) Here you'd have
$$
\left\lvert\begin{array}{@{}cccc@{}}
\Basis_{1} & \Basis_{2} & \Basis_{3} & \Basis_{4} \\
1 & 0 & 0 & 0 \\
1 & 1 & 1 & 1 \\
1 & 2 & 3 & 4 \\
\end{array}\right\rvert
= \left\lvert\begin{array}{@{}cccc@{}}
\Basis_{1} & \Basis_{2} & \Basis_{3} & \Basis_{4} \\
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
0 & 0 & 1 & 2 \\
\end{array}\right\rvert
= \Basis_{2} - 2\Basis_{3} + \Basis_{4},\qquad
\|\Basis_{2} - 2\Basis_{3} + \Basis_{4}\| = \sqrt{6}.
$$
(I leave to you the assessment of whether this is more energy-efficient for your purposes, though again I think there's a favorable comparison in this example.)
This "generalized cross product" has the advertised property for reasons analogous to the situation with the ordinary cross product:
*
*If $(v_{j})_{j=1}^{n-1}$ and $w$ are vectors in $\Reals^{n}$ and $v_{1} \times \dots \times v_{n-1}$ denotes the "cross product" described above, then
$$
\left\lvert\begin{array}{@{}c@{}}
w \\ v_{1} \\ \vdots \\ v_{n-1} \\
\end{array}\right\rvert
= (v_{1} \times \dots \times v_{n-1}) \cdot w.
\tag{2}
$$
*By (2), the cross product $v_{1} \times \dots \times v_{n-1}$ is orthogonal to each $v_{j}$, and therefore orthogonal to the parallelipiped spanned by the $(v_{j})$.
*The preceding item and the fact that the determinant in (2) is the $n$-dimensional volume of the parallelipiped spanned by the $(v_{j})$ and $w$ implies that the $(n - 1)$-dimensional volume of the parallelipiped spanned by the $(v_{j})$ is the magnitude of their cross product.
|
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|
How to deduce that $1\cdot 1 + 2\cdot 1 + 2\cdot 2 + 3\cdot 1+3\cdot 2+3\cdot 3 +...+(n\cdot n) = n(n+1)(n+2)(3n+1)/24$ I know how to reason $$1\cdot2 + 2\cdot3 + 3\cdot4 + n(n-1) = \frac{1}{3}n(n-1)(n+1)$$
However, I'm stuck on proving $$1\cdot1 + (2\cdot1 + 2\cdot2) + (3\cdot1+3\cdot2+3\cdot3) + \cdots +(n\cdot 1+...+n\cdot n) = \frac{1}{24}n(n+1)(n+2)(3n+1).$$
Could somebody shed some light on me?
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The combinatorial identity:
$$ \sum_{k=0}^{K}\binom{n+k}{n}=\binom{n+K+1}{n+1}\tag{1}$$
can be easily proved by induction. It gives:
$$\sum_{j=1}^{n-1}j(j+1) = 2\sum_{j=1}^{n-1}\binom{j+1}{2} = 2\binom{n+1}{3} = \frac{(n+1)n(n-1)}{2}\tag{2}$$
as well as:
$$\begin{eqnarray*} \sum_{k=1}^{n}k(1+2+\ldots+k) &=& 2\sum_{k=1}^{n}k\binom{k+1}{2}=2\sum_{k=1}^{n}\binom{k+1}{2}+6\sum_{k=1}^{n}\binom{k+1}{3}\\&=&2\binom{n+2}{3}+6\binom{n+2}{4}.\tag{3}\end{eqnarray*}$$
|
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|
Find the integral $\int \:x^{-\frac{1}{2}}\cdot \left(1+x^{\frac{1}{4}}\right)^{-10} dx$ Help me find the integral. I think we have to somehow replace apply.
$$\int \:x^{-\frac{1}{2}}\cdot \left(1+x^{\frac{1}{4}}\right)^{-10} dx =\int \frac{1}{\sqrt{x} (1+x^{\frac{1}{4}})^{10}} dx $$
|
You may just perform the change of variable $x=u^4$, to get
$$
\begin{align}
\int \:x^{-\frac{1}{2}}\cdot \left(1+x^{\frac{1}{4}}\right)^{-10} dx&=4\int \:u^{-2}\cdot \left(1+u\right)^{-10} u^3du\\\\
&=4\int \frac{1}{(1+u)^9} du-4\int \frac{1}{(1+u)^{10}}du\\\\
&=-\frac{1}{2 (1+u)^8}+\frac{4}{9 (1+u)^9}+C\\\\
&=-\frac{1}{2 \left(1+x^{1/4}\right)^9}+\frac{4}{9 \left(1+x^{1/4}\right)^9}+C.
\end{align}
$$
|
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|
Find the sum of all odd numbers between two polynomials I was asked this question by someone I tutor and was stumped.
Find the sum of all odd numbers between $n^2 - 5n + 6$ and $n^2 + n$ for $n \ge 4.$
I wrote a few cases out and tried to find a pattern, but was unsuccessful.
Call polynomial 1, $p(n) = n^2 - 5n + 6,$ then $p(4)=2.$ Next, call polynomial 2, $q(n)=n^2 + n,$ then $q(4)=20.$ Then adding all the odd numbers between 2 and 20 gives the following sum:
$3+5+7+9+11+13+15+17+19= 99. \\$
$p(5)= 6$ and $q(5)=30.$ Then adding all the odd numbers between 6 and 30 gives the following sum: $7+9+11+13+15+17+19+21+23+25+27+29=216 \\$
$p(6)=12$ and $q(6)= 42.$ Then adding all the odd numbers between 12 and 42 give the following sum: $13+15+17+19+21+23+25+27+29+31+33+35+37+39+41 = 405.$
From here I do not see any apparent patters. This problem was given in a Pre-Calculus course, so clearly only elementary methods are expected by the students.
Any help or advice would be much appreciated. Thank you!!!!!
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Because $n^2+n$ is even we can write the odd numbers between the two polynomials to sum as:
$$n^2+n-1,n^2+n-3,\cdots,n^2+n-(6n-7)$$
and because the sum of odd numbers between $1$ and $2k+1$ is $(k+1)^2$ we have :
$$\sum_{i=0}^{3n-4}(n^2+n-(2i+1))=(3n-3)(n^2+n)-(3n-3)^2 $$
|
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|
Proving $\lim_{x\to c}x^3=c^3$ for any $c\in\mathbb R$ using $\epsilon$-$\delta$ definition
$\lim_{x\to 3}x^3=c^3$ for any $c\in\mathbb R$
Let $\epsilon>0$. Then
$$|x^3-c^3|=|x-c||x^2+xc+c^2|.$$
Let $\delta=\min\{1,\epsilon/x^2+xc+c^2\}$.
Then if $0<|x-c|<\delta$ and therefore since $|x-c|<\epsilon/(x^2+xc+c^2)$,
$$|x^3-c^3|=|x-c||x^2+xc+c^2|<ε.$$
Does this make sense or are the steps done in the right way?
|
Hint: $\delta$ should not be depending of $x$.
Consider $c > 0$ then
$$ c - 1 < x < c+ 1 \implies |x| < c + 1 $$
Then $$\begin{align}|x^3 - c^3| &= |x - c| |x^2 + xc +c^2| \\&\leq |x-c|\bigg(|c+1|^2 + |c+1||c| + + |c|^2\bigg) \\&=|x-c|\bigg[|c+1|(|c + 1| + |c|) + |c|^2\bigg]\end{align} $$
Take $$\delta = \min \Bigg\{1, \frac{\epsilon}{|c+1|(|c + 1| + |c|) + |c|^2}\Bigg\}$$
|
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|
Prove that if $k \in \mathbb{N}$, then $k^4+2k^3+k^2$ is divisble by $4$ I am trying to solve by induction and have established the base case (that the statement holds for $k=1$).
For the inductive step, I tried showing that the statement holds for $k+1$ by expanding $(k+1)^4+2\cdot(k+1)^3+(k+1)^2$, but this equals $16k^4+34k^3+3k^2+16k+4$, and $4$ cannot be factored out.
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First note that $n^4+2n^3+k^2= n^2(n^2+2n+1)=n^2(n+1)^2$. Hence, the problem is to show that for every $n\geq 1$,
$$
S_n : 4\mid n^2(n+1)^2
$$
is true.
Without induction, the direct proof follows from the fact that among $n$ and $n+1$, one is even, and so one of $n^2$ and $(n+1)^2$ has a factor of $4$. But we will do it by induction anyway.
Base step: When $n=1, n^2(n+1)^2=4$, so $S_1$ holds.
Inductive step: Let $k\geq 1$ and let the induction hypothesis be that $S_k$ holds; in particular, suppose that $m$ is such that $k^2(k+1)^2=4m$. It remains to show that
$$
S_{k+1} : 4\mid (k+1)^2(k+2)^2
$$
follows. Indeed
\begin{align}
(k+1)^2(k+2)^2 &= (k^2+4k+4)(k+1)^2\\[0.5em]
&= k^2(k+1)^2+(4k+4)(k+1)^2\\[0.5em]
&= 4m+4(k+1)(k+1)^2\quad\text{(by $S_k$)}
\end{align}
shows that $(k+1)^2(k+2)^2$ is also divisible by $4$, concluding the proof of $S_{k+1}$ and hence the inductive step.
By mathematical induction, it is true that for any positive integer $n$, the number $n^2(n+1)^2$ is divisible by $4$. $\blacksquare$
|
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Maximum value of $ x^2 + y^2 $ given $4 x^4 + 9 y^4 = 64$ It is given that $4 x^4 + 9 y^4 = 64$.
Then what will be the maximum value of $x^2 + y^2$?
I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .
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$x^2=s,\ y^2=t \geq 0$ then $$ (2s)^2+ (3t)^2=8^2 \Rightarrow
2s=4\cos\ \theta,\ 3t=4\sin\ \theta,\ 0\leq\theta \leq
\frac{\pi}{2} $$
Then $$ s+t=2\cos\ \theta + \frac{4}{3}\sin\ \theta=\sqrt{2^2+
\bigg(\frac{4}{3}\bigg)^2} \sin\ (\theta+\alpha),\ 0< \alpha <
\frac{\pi}{2} $$
Let $\theta =\frac{\pi}{2}-\alpha$
|
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|
Find the value of $\sum_{m=1}^\infty \tan ^ {-1}\frac{2m}{m^4+m^2+2}$ How to find value of this sum?
$$\sum\limits_{m=1}^\infty \tan^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$$ I can't understand how to simplify this. Should I use any trigonometric substitution to simplify the fraction? Hints and help needed!
|
\begin{eqnarray}
\frac{2m}{m^4+m^2+2}&=&\frac{2m}{m^4+2m^2-m^2+1+1}\\
&=&\frac{2m}{(m^2+1)^2-m^2+1}\\
&=&\frac{2m}{(m^2+m+1)(m^2-m+1)+1}\\
&=&\frac{(m^2+m+1)-(m^2-m+1)}{(m^2+m+1)(m^2-m+1)+1}
\end{eqnarray}
so almost done!
$$
\arctan(a)-\arctan(b)=\arctan\left(\frac{a-b}{1+ab}\right)
$$
\begin{eqnarray}
\arctan(m^2+m+1)-\arctan(m^2-m+1)&=&\arctan\left(\frac{(m^2+m+1)-(m^2-m+1)}{(m^2+m+1)(m^2-m+1)+1}\right)\\
&=&\arctan\frac{2m}{(m^2+m+1)(m^2-m+1)+1}
\end{eqnarray}
now you have telescopic sumation, because
\begin{eqnarray}
\arctan(m^2+m+1)-\arctan(m^2-m+1)&=&\arctan(m(m+1)+1)-\arctan(m(m-1)+1)\\
&=&f(m)-f(m-1)
\end{eqnarray}
$$
\sum_{m=1}^\infty\tan^{-1}\left(\frac{2m}{m^4+m^2+2}\right)=\\\sum_{m=1}^\infty(f(m)-f(m-1))=\\\lim_{m \to \infty}\tan^{-1}(m^2+m+1)-\tan^{-1}1=\\\frac\pi2-\frac\pi4=\\\frac\pi4.
$$
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Induction for Vandermonde Matrix
Given real numbers $x_1<x_2<\cdots<x_n$, define the Vandermonde
matrix by $V=(V_{ij}) = (x^j_i)$. That is, $$V =
\left(\begin{array}{cccccc} 1 & x_1 & x^2_1 & \cdots &
x^{n-1}_1 & x^n_1 \\
1 & x_2 & x^2_2 & \cdots & x^{n-1}_2 & x^n_2 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
1 & x_{n-1}& x^2_{n-1} & \cdots & x^{n-1}_{n-1} & x^n_{n-1} \\
1 & x_n & x^2_n & \cdots & x^{n-1}_n & x^n_n
\end{array}\right).$$ Prove that $\det(V) = \prod_{1\le i<j\le n} (x_j-x_i)$ by the following inductive steps. Recall first
that adding a multiple of one row to another will not change the
determinant of a matrix. This is also true if you add a multiple of
one column to another. Finally remember that the determinant is
linear in each row when you leave the other ones fixed.
a. Subtract $x_1$ times each column from the column to its right,
starting with the last column. That is, subtract $x_1$ times the
$(n-1)$-st column from the $n$-th column, $x_1$ times the $(n-2)$-nd
column from the $(n-1)$-st column, etc.
b. Then subtract the first row from all of the other rows.
c. Finally observe that each row has a common factor that can be
pulled out of the determinant.
d. After these three steps are done, expand the resulting determinant
in cofactors across the first row.
e. You should see at this point how to apply the induction step.
What I have so far: I have been told that for the induction step I have to first first show that it is true for an $(n-1)$$\times$$(n-1)$ matrix and then show that it is true for an $n$$\times$$n$ matrix (I believe?). I kind of think that my resulting matrix after the steps is incorrect...
|
Add the last row multiplied by -1 to all other rows, and we get,
det$|V|$=det$\left|\begin{array}{}
0 & x_1-x_n & x_1^2-x_n^2 & \cdots & x_1^n-x_n^n \\
0 & x_2-x_n & x_2^2-x_n^2 & \cdots & x_2^n-x_n^n \\
\vdots & \vdots & \vdots & & \vdots \\
0 & x_{n-1}-x_n & x_{n-1}^2-x_n^2 & \cdots & x_{n-1}^n-x_n^n \\
1 & x_n & x_n^2 & \cdots & x_n^n \\
\end{array} \right|
$
=det$\left|\begin{array}{}
x_1-x_n & x_1^2-x_n^2 & \cdots & x_1^n-x_n^n \\
x_2-x_n & x_2^2-x_n^2 & \cdots & x_2^n-x_n^n \\
\vdots & \vdots & & \vdots \\
x_{n-1}-x_n & x_{n-1}^2-x_n^2 & \cdots & x_{n-1}^n-x_n^n \\
\end{array} \right|
$
=$\prod\limits_{k=1}^{n-1}(x_k-x_n)$ det$\left|\begin{array}{}
1 & x_1+x_n & x_1^2+x_1x_n+x_n^2 & \cdots & \sum \limits_{k=0}^{n-1}x_1^{n-k-1}x_n^{k} \\
1 & x_2+x_n & x_2^2+x_2x_n+x_n^2 & \cdots & \sum \limits_{k=0}^{n-1}x_2^{n-k-1}x_n^{k} \\
\vdots & \vdots & \vdots & & \vdots \\
1 & x_{n-1}+x_n & x_{n-1}^2+x_{n-1}x_n+x_n^2 & \cdots & \sum \limits_{k=0}^{n-1}x_{n-1}^{n-k-1}x_n^{k} \\
\end{array} \right|
$
=$\prod\limits_{k=1}^{n-1}(x_k-x_n)$ det$|V_1|$
In $V_1$, first add 1st column multiplied by $-x_n$ to 2nd column, and add 1st column multiplied by $-x_n^2$ to 3nd column, ..., and add 1st column multiplied by $-x_n^{n-1}$ to $(n-1)$'s column, we get
det$|V_1|$ = det$\left|\begin{array}{}
1 & x_1 & x_1^2+x_1x_n & \cdots & \sum \limits_{k=0}^{n-2}x_1^{n-k-1}x_n^{k} \\
1 & x_2 & x_2^2+x_2x_n & \cdots & \sum \limits_{k=0}^{n-2}x_2^{n-k-1}x_n^{k} \\
\vdots & \vdots & \vdots & & \vdots \\
1 & x_{n-1} & x_{n-1}^2+x_{n-1}x_n & \cdots & \sum \limits_{k=0}^{n-2}x_{n-1}^{n-k-1}x_n^{k} \\
\end{array} \right|
$
Then add 2nd column multiplied by $-x_n$ to 3rd column, ..., and add 2nd column multiplied by $-x_n^{n-2}$ to $(n-1)$'s column, we get
det$|V_1|$ = det$\left|\begin{array}{}
1 & x_1 & x_1^2 & \cdots & \sum \limits_{k=0}^{n-3}x_1^{n-k-1}x_n^{k} \\
1 & x_2 & x_2^2 & \cdots & \sum \limits_{k=0}^{n-3}x_2^{n-k-1}x_n^{k} \\
\vdots & \vdots & \vdots & & \vdots \\
1 & x_{n-1} & x_{n-1}^2 & \cdots & \sum \limits_{k=0}^{n-3}x_{n-1}^{n-k-1}x_n^{k} \\
\end{array} \right|
$
Repeat above process, and use induction hypothesis, we have
det$|V_1|$ = det$\left|\begin{array}{}
1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\
1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\
\vdots & \vdots & \vdots & & \vdots \\
1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-1} \\
\end{array} \right|
$ = $\prod_{1\le i<j\le n-1} (x_j-x_i)$
So finally
det$|V|$ =$\prod\limits_{k=1}^{n-1}(x_k-x_n)$ det$|V_1|$ = $\prod_{1\le i<j\le n} (x_j-x_i)$
|
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|
Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction:
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove that given inequality holds for $n+1$ case by using the given $n$ case. Ok, so this is what I've got by now:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{(n+1)^2}\overset{?}{>}1$$
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1$$
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
From the $n$ case we know that:
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}$$
So we basically have to prove that:
$$1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}\frac{1}{n}$$
Since
$$n^2+1<n^2+2<\dots<n^2+2n+1<2n^2+n$$
for $n\geq2$, then also:
$$\frac{1}{n^2+1}>\frac{1}{2n^2+n}$$
$$\frac{1}{n^2+2}>\frac{1}{2n^2+n}$$
.
.
$$\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}$$
so then we have:
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}+\dots+\frac{1}{2n^2+n}=(2n+1)\frac{1}{2n^2+n}=\frac{1}{n}$$
|
Here is an alternate solution. We want to prove
$$
\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1
$$
Or, equivalently, that
$$
\frac{\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}}{n^2 - n + 1}>\frac{1}{n^2 - n + 1}
$$
where the left-hand side is the arithmetic mean of all the fractions. We know that the arithmetic mean is greater than the harmonic mean (equality only when all terms are equal, which isn't the case since $n \geq 2$):
$$
\frac{\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}}{n^2 - n + 1} > \frac{n^2 - n + 1}{n + (n+1) + \cdots + n^2}\\
= \frac{n^2 - n + 1}{\frac{n + n^2}{2}(n^2 - n + 1)} = \frac{2}{n^2 + n}
$$
and the proof is completed by noting that we have
$$
n^2 - 3n + 2 \geq 0\\
2n^2 - 2n + 2 \geq n^2 + n\\
2(n^2 - n + 1) \geq n^2 + n\\
\frac{2}{n^2 + n} \geq \frac1{n^2 - n + 1}
$$
|
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|
Inverse function. A function $h$ is defined by $h:x\rightarrow 2-\frac{a}{x}$, where $x\neq 0$ and $a$ is a constant. Given $\frac{1}{2}h^2(2)+h^{-1}(-1)=-1$, find the possible values of $a$.
Can someone give me some hints? Thanks
|
Firstly, find the inverse function $h^{-1}$. I.e. let $y=2-a/x$ and hence $h^{-1}$ is defined as
\begin{align}x=2-\frac{a}{y} \ \ & \Longrightarrow \ \ \frac{a}{y}=2-x\\
&\Longrightarrow y=\frac{a}{2-x}.
\end{align}
Now, $h(2)=2-a/2$ and $h^{-1}(-1)=a/3$. You should be able to continue from here and get a quadratic equation in $a$.
Edit. Here is some extra help. Your equation can be evaluated with the information above:
\begin{align}
\frac{1}{2}h^2(2)+h^{-1}(-1)&=\frac{1}{2}(h(2))^2+h^{-1}(-1)\\
&=\frac{1}{2}\left(2-\frac{a}{2}\right)^2+\frac{a}{3}\\
\end{align}
|
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|
Find basis so matrix is in Jordan Canonical Form $M = \left(\begin{array}{ccc}0 & -3 & -2 \\1 & 3 & 1 \\1 & 2 & 3\end{array}\right)$
I want to find a basis $B$ such that matrix for $M$ w.r.t $B$ has the form:
$\left(\begin{array}{ccc}2 & 1 & 0 \\0 & 2 & 1 \\0 & 0 & 2\end{array}\right)$
The eigenvalues for $M = 2,2,2$.
$M-2I = \left(\begin{array}{ccc}-2 & -3 & -2 \\1 & 1 & 1 \\1 & 2 & 1\end{array}\right)$.
I tried to find eigenvectors. This is what I came up with.
$(M-2I)v_1 = 0$, then $v_1$ = $(-1,0,1)^t$
And so forth but I either get a wrong answer or it's not invertible.
How do I find the generalized eigenvectors so M is in Jordan Normal Form?
|
let us call $$A- 2I = B = \left(\begin{array}{ccc}-2 & -3 & -2 \\1 & 1 & 1 \\1 & 2 & 1\end{array}\right).$$ row reducing we find that $B \to \pmatrix{1&0&1\\0&1&0\\0&0&0}$ so that the null of $B$ has dimension one and $$u = \pmatrix{1\\0\\-1}, Bu= 0$$ is a basis.
you can also find that null of $B^2$ has dimension $2.$ let $v$ solve $Bv = u.$ we will find $v$ by row reducing the augmented matrix $[B|u].$ we find that $$[B|u] \to \pmatrix{1&0&1&1\\0&1&0&-1\\0&0&0&0}$$ and $$v = \pmatrix{1\\-1\\0}, Bv = u$$
now, we solve $Bw = v$ and find $$[B|v] \to \pmatrix{1&0&1&-2\\0&1&0&1\\0&0&0&0}$$ and $$w = \pmatrix{-2\\1\\0}, Bw = v$$
you can verify that $\{u, v, w\}$ is a basis and with respect to this basis $B, A = B+2I$ have the representations $$ \pmatrix{0&1&0\\0&0&1\\0&0&0}, \pmatrix{2&1&0\\0&2&1\\0&0&2}$$
|
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|
Evaluation of an integral. $$I_1=\int \frac{a^2\sin^2(x)+b^2\cos^2(x)}{a^4\sin^2(x)+b^4\cos^2(x)}dx$$
I tried writing it as $$\int \frac{a^2+\cos^2(x)(b^2-a^2)}{a^4+\cos^2(x)(b^4-a^4)}dx$$
But I don't know how to proceed. What's the procedure to evaluate this integral?
|
$$\frac{a^2u^2+b^2}{a^4u^2+b^4}=\frac{(a^2+b^2)(a^2u^2+b^2)}{(a^2+b^2)(a^4u^2+b^4)}=\cdots =\frac{1}{a^2+b^2}+\frac{a^2(u^2+1)}{(a^2+b^2)(\frac{a^4}{b^2}u^2+b^2)}.$$
So if $u=\tan{x}$ and integrate the previous you get
$$\frac{x}{a^2+b^2}+\frac{a^2}{a^2+b^2}\int \frac{\tan^2{x}+1}{\frac{a^4}{b^2}\tan^2{x}+b^2} dx=$$
$$\frac{x}{a^2+b^2} +\frac{1}{a^2+b^2}\arctan(\frac{a^2}{b^2}\tan{x})+c$$
|
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|
Is this a good proof of the binomial identity? Prove that the binomial identity ${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$ is true using the following expression: $(1+x)^n = (1+x)(1+x)^{n−1}$ and the binomial theorem.
What I have:
We know from the binomial theorem that:
$$(x+1)^n= {n\choose 0} x^0 + {n\choose 1} x^1+\cdots+{n\choose k} x^k+\cdots+ {n\choose n-1} x^{n-1}+ {n\choose n} x^n.$$
By using the property that $(1+x)^{n} = (1+x)(1+x)^{n−1}$, we can take out a factor of $x$ to get:
$$x(x+1)^{n-1}={n-1\choose 0} x^0+\cdots+{n-1\choose k-1} x^{k-1}+\cdots+{n+1\choose n-1} x^{n-1}$$
If we then divide by a factor of $x$ we get:
$$1(x+1)^{n-1}={n-1\choose 0} x^0+\cdots +{n-1\choose k-1} x^k+\cdots+{n-1\choose n-1} x^{n-1}$$
Substituting $(x+1)^{n-1}={n-1\choose 0} x^0 +\cdots+{n-1\choose k-1} x^k+\cdots+{n-1\choose n-1} x^{n-1}$ into the equation $x(x+1)^{n-1}= {n-1\choose 0} x^0+\cdots+{n-1\choose k-1} x^{k-1}+\cdots+ {n+1\choose n-1}x^{n-1}$, the equation can be reduced to:
$${n\choose k}x^k = {n-1\choose k-1} x^{k-1}x+{n-1 \choose k}x^k 1$$
Dividing by $x$ we get: ${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$
Is this a good proof? What can I do to improve it? Is there a better way to solve this problem?
|
First of all, in general
$$x(x+1)^{n-1} \neq \binom{n-1}{0} x^0+\cdots+\binom{n-1}{k - 1}x^{k-1}+\cdots
+ \binom{n+1}{n-1}x^{n-1}.$$
Notice that as $x\to 0$ the left side goes to zero but the right side does not.
Also, in general
$$1(x+1)^{n-1} \neq \binom{n-1}{0}x^0+\cdots+\binom{n-1}{k - 1}x^{k}+\cdots+
\binom{n-1}{n - 1}x^{n-1}.$$
If you change $x^k$ to $x^{k-1}$ on the right hand side, the two sides are equal.
The step where you "reduce" an equation to
$\binom nk x^{k} = \binom{n-1}{k - 1} x^{k-1}x + \binom{n-1}{k}x^{k}1$
has no clear explanation, but it doesn't really matter since the
premises of that statement are already false.
|
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|
Show that a polynomial over $\mathbb{Z}_{2}$ is irreducible Given the polynomial: $p(x)=x^4+x^3+x^2+x+1$ over $\mathbb{Z}_{2}$, to show that it is irreducable, is it enough to show that $p(0)=p(1)=1$?
|
On factorising into quadratics, just try by hand. The coefficient of $x^2$ has to be $1$ in both factors to give $x^4$ and the constant in each factor has to be $1$ too to give a constant term $1$ in the product.
So $$(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+abx^2+(a+b)x+1=x^4+x^3+x^2+x+1$$
We have $a+b=1$ so $a$ and $b$ have different parities, and one is $0$. But then $ab=0$ and the coefficient of $x^2$ is zero. So there is no possible factorisation.
|
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|
Sum of polynomial coefficient
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x)=2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$. Can you tell me the values of $a$ and $c$?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve says, "You’re right. Here is the value of $a$." He writes down a positive integer and asks, "Can you tell me the value of $c$?"
Jon says, "There are still two possible values of $c$."
Find the sum of the two possible values of $c$.
$$P(x) = 2x^3-2ax^2+(a-9)(a+9)x-c$$
Let the roots be $r_1, r_2, r_3$ then:
$$P(x) = 2(x - r_1)(x - r_2)(x - r_3)$$
$$= 2x^3 - 2x^2\overbrace{(r_1 + r_2 + r_3)}^{=a} + x\overbrace{(2r_1r_2 + 2r_1r_3 + 2r_2r_3)}^{= (a-9)(a+9)} - \overbrace{2r_1r_2r_3}^{=c}$$
Yikes!
$$r_1 + r_2 + r_3 = a$$
$$2(r_1r_2 + r_1r_3 + r_2r_3) = (a-9)(a+9)$$
|
We have
$$2(r_1r_2+r_2r_3+r_3r_1) = a^2 - 81 = (r_1+r_2+r_3)^2 - 81$$
This gives us
$$r_1^2 + r_2^2 + r_3^2 = 81$$
Since $81 \equiv 1\pmod8$, we obtain that one of them is odd, say $r_1$ and the other two ($r_2 = 2k_2$ and $r_3 = 2k_3$) are even. Further, $1 \leq r_1,r_2,r_3 \leq 8$, i.e., $1 \leq k_2,k_3 \leq 4$ and $r_1 \in \{1,3,5,7\}$.
*
*$r_1 = 1$. This gives us $k_2^2 + k_3^2 = 20$. The only possible solution is $(4,2)$ and $(2,4)$.
*$r_2 = 3$. This gives us $k_2^2 + k_3^2 = 18$. The only possible solution is $(3,3)$ and $(3,3)$.
*$r_2 = 5$. This gives us $k_2^2 + k_3^2 = 14$. No solution exists.
*$r_2 = 7$. This gives us $k_2^2 + k_3^2 = 8$. This gives $k_2=k_3=2$.
Hence, the possible roots are $(1,4,8)$, $(3,6,6)$ and $(7,4,4)$. Note that the value of $r_1+r_2+r_3$ are $13$, $15$ and $15$ respectively. Further given $a$, there are two possible values of $c$. This means the roots are $(3,6,6)$ or $(7,4,4)$. Hence, the possible values of $c$ are $216$ and $224$.
|
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|
Is $\sqrt{x^2} = (\sqrt x)^2$? Take $x=4$ for example:
$ \sqrt{(4)^2} = \sqrt{16} = \pm4 $
However:
$ (\sqrt{4})^2 = \sqrt{\pm2}$
Case 1: $ (-2)^2 = 4$
Case 2: $ (2)^2 = 4$
Solution : $+4$
How come the $ \sqrt{(4)^2} = \pm4$; but $ (\sqrt{4})^2 = 4 $ ?
What is missing?
|
Disclaimer: In the following we restrict ourselves to real numbers, not taking complex numbers and the like into consideration.
By convention, the square root of a positive number $t$, written as $\sqrt t$, has been defined to be the positive solution to the equation $x^2=t$. This gives meaning to the following way of specifying the two solutions (for positive $t$):
$$
x^2=t\iff x\in\{\sqrt t,-\sqrt t\}
$$
With this your example becomes $\sqrt{4^2}=4$ and $(\sqrt 4)^2=4$, but on the other hand we have $\sqrt{(-4)^2}=4$ whereas $(\sqrt{-4})^2$ is undefined, because $\sqrt{-4}$ is undefined, since $x^2=-4$ has no solutions.
In general, $\sqrt{x^2}$ agrees with $(\sqrt x)^2$ for non-negative input $x$, whereas only the first is defined for negative values of $x$.
|
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|
Write a function as $\sum _{n=0} ^{\infty} a_n x^n$ We have $f(x) = (x+ x^2 + x^3 + x^4 + x^5 + x^6)^4$.
Now I want to write this as $\sum _{n=0} ^{\infty} a_n x^n$.
What I got:
$f(x) = (x+ x^2 + x^3 + x^4 + x^5 + x^6)^4 = x^4 (1+ x + x^2 + x^3 + x^4 + x^5)^4$
We know: $\frac{1}{1-x} = \sum _{n=0} ^{\infty} x^n$.
Note that $\displaystyle \frac{d^3}{dx^3} (\frac{1}{1-x})= \frac{6}{(1-x)^4}$.
Then $\frac{1}{(1-x)^4} = \frac{1}{6} \sum n (n-1) (n-2) x^{n-3}$, what gives
$x^4 \frac{1}{(1-x)^4} = \frac{1}{6} \sum n (n-1) (n-2) x^{n+1} = \frac{1}{6} \sum (n-1) (n-2) (n-3) x^{n}$
If I look now on Wolfram Alpha at the expanded form, this expression is only correct for $4 \leq n \leq 9$. For $n \geq 10$ it gives other values of the coefficients. What goes wrong?
|
We know that for any Taylor series representation:
$$ \sum_{n=0}^{\infty} \frac{ f^{(n)}(a) } {n!} (x-a)^n $$
It looks like you want the series centered at a=0.
So to get the coefficient for terms 1,2,3,4,.... evaluate:
$$b_n=\frac{ f^{(n)}(a)}{n!}$$
|
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|
Number of times $2^k$ appears in factorial
For what $n$ does: $2^n | 19!18!...1!$?
I checked how many times $2^1$ appears:
It appears in, $2!, 3!, 4!... 19!$ meaning, $2^{18}$
I checked how many times $2^2 = 4$ appears:
It appears in, $4!, 5!, 6!, ..., 19!$ meaning, $4^{16} = 2^{32}$
I checked how many times $2^3 = 8$ appears:
It appears in, $8!, 9!, ..., 19!$ meaning, $8^{12} = 2^{36}$
I checked how many times $2^{4} = 16$ appears:
It appears in, $16!, 17!, 18!, 19!$ meaning, $16^{4} = 2^{16}$
In all,
$$2^{18} \cdot 2^{32} \cdot 2^{36} \cdot 2^{16} = 2^{102}$$
But that is the wrong answer, its supposed to be $2^{150}$?
|
How many times $2$ divides the product $\prod_{i=1}^{19}i!$ ?
Let's call each term inside a factorial $i$. That way, $i = 1$ occurs in 19 factorials, $i = 2$ occurs in 18 factorials, and $i = 3$ occurs in 17 factorials etc.
$i = 2$ occurs 18 times. $1 \times 18 = 18$
$i = 4$ occurs 16 times. $2 \times 16 = 32$
$i = 6$ occurs 14 times. $1 \times 14 = 14$
$i = 8$ occurs 12 times. $3 \times 12 = 36$
$i = 10$ occurs 10 times. $1 \times 10 = 10$
$i = 12$ occurs 8 times. $2 \times 8 = 16$
$i = 14$ occurs 6 times. $1 \times 6 = 6$
$i = 16$ occurs 4 times. $4 \times 4 = 16$
$i = 18$ occurs 2 times. $1 \times 2 = 2$
$$18 + 32 + 14 + 36 + 10 + 16 + 6 + 16 + 2 = 150$$
|
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|
Partial fractions - different results when done in steps than not We have: $\frac 1 {(1-x)(1+x)(1-2x)}$
If I do the partial fractions straight: $\frac 1 {(1-x)(1+x)(1-2x)}= \frac a {1-x} + \frac b {1+x} + \frac c {1-2x}$
I get: $a=-\frac 12, b = \frac 1 6, c=\frac 4 3$.
But when I do it in steps, i.e: $\frac 1 {(1-x^2)(1-2x)}=\frac a {1-x^2} + \frac b {1-2x} $
I get: $a=2, b=-1$, then again partial fractions: $\frac 2 {1-x^2}=\frac a {1-x} + \frac b {1+x}$ Get: $a=1, b=-1$
In the first method I got: $\frac {-\frac 12} {1-x} + \frac {\frac 1 6} {1+x} + \frac {\frac 4 3} {1-2x}$
But in steps: $\frac {1} {1-x} - \frac {1} {1+x} - \frac {1} {1-2x}$
How is that possible? Shouldn't they be the same?
|
$\frac 1 {(1-x^2)(1-2x)}\ne\frac a {1-x^2} + \frac b {1-2x}$
$\frac 1 {(1-x^2)(1-2x)}=\frac {a+bx} {1-x^2} + \frac c {1-2x}$
One finds that $a=-1/3$, $b=-2/3$ and $c=4/3$.
|
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|
How do i evaluate the following integral? Hi I was wondering if someone can help me evaluate the following integral.
Show that if $-1 < x < 1$, then
$$\int_{0}^{\pi} \frac{\log{(1+x\cos{y})}}{\cos{y}}dy= \pi \arcsin{x} $$
thank you in advance.
|
Setting
$$
f(x)=\int_0^\pi\frac{\log(1+x\cos y)}{\cos y}\,dy, \quad x\in (-1,1),
$$
we have $f(0)=0$, and
$$
f'(x)=\int_0^\pi\frac{\cos y}{(1+x\cos y)\cos y}\,dy=\int_0^\pi\frac{1}{1+x\cos y}\,dy\quad \forall x\in (-1,1)
$$
Setting
$$
t=\tan\frac{y}{2},
$$
we have
$$
y=2\arctan t,\, \frac{1}{1+x\cos y}=\frac{1}{1+x\frac{1-t^2}{1+t^2}}=\frac{1+t^2}{1+x+(1-x)t^2} \,\mbox{ and } \, dy=2\frac{dt}{1+t^2}.
$$
Now, for every $x\in (-1,1)$ we get:
\begin{eqnarray}
f'(x)&=&\int_0^\infty\frac{1+t^2}{1+x+(1-x)t^2}\cdot\frac{2}{1+t^2}\,dt=\frac{2}{1-x}\int_0^\infty\frac{1}{(1+x)(1-x)^{-1}+t^2}\,dt\\
&=&\frac{2}{\sqrt{1-x^2}}\arctan\left(\frac{t}{\sqrt{(1+x)(1-x)^{-1}}}\right)\Big|_0^\infty=\frac{2}{\sqrt{1-x^2}}\cdot\frac\pi2=\frac{\pi}{\sqrt{1-x^2}}.
\end{eqnarray}
Since $f(0)=0$, we deduce that
$$
\int_0^\pi\frac{\log(1+x\cos y)}{\cos y}\,dy=f(x)=f(0)+\int_0^xf'(\xi)\,d\xi=\int_0^x\frac{\pi}{\sqrt{1-\xi^2}}\,d\xi=\pi\arcsin x.
$$
|
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|
Real Canonical Form Question: Let us consider the quadratic form $q: R^3 -> R$, $$q(x,y,z) = x^2+25y^2+10xy+2xz$$Find the corresponding symmetric bilinear form $f$ and a basis $B$ such that $[f]_B$ has the real canonical form. State the signature and rank of $[f]_B$.
I thought that the best way to start this question was to find an orthogonal basis, C, using the Gram Schmitd process. I got the vectors $(1,0,0),(-5,1,1),(\frac{5}{2},-\frac{1}{2},\frac{1}{2})$ as my orthogonal basis, but I dont know how to use these to find a basis B such that $[f]_B$ has the real canonical form. Any help would be greatly appreciated. Thanks.
|
A way to proceed to get the signature and the rank is to reduce $q$ thanks to the Gauss' algorithm :
$$q\left(x,y,z\right)=x^2+25y^2+10xy+2xz
=\left(x+5y+z\right)^2-10yz-z^2
=\left(x+5y+z\right)^2-\left(z+5y\right)^2+25y^2.$$
Let then $X=\left(x+5y+z\right)$, $Y=\left(z+5y\right)$ and $Z=5y$. Then the reduced equation of $q$ is given by
$$X^2-Y^2+Z^2=0.$$
It is a cone (signature $\left(2,1\right)$, rank=$3$).
Furthermore, we have
$$\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}=\underbrace{\begin{pmatrix}1 & 5 & 1\\
0 & 5 & 1\\
0 & 5 & 0
\end{pmatrix}}_{=:P}\begin{pmatrix}x\\
y\\
z
\end{pmatrix}
$$
and the matrix
$$P^{-1}=\begin{pmatrix}1 & -1 & 0\\
0 & 0 & \frac{1}{5}\\
0 & 1 & -1
\end{pmatrix}$$
is the passage matrix from the canonical basis to a base where $q$ as the above reduced form.
Verification :
The representative matrix in the canonical base of $\mathbb{R}^3$ of the polar bilinear form associated to $q$ is
$$\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}.$$
We have :
$$q\left(x,y,z\right)=\begin{pmatrix}x & y & z\end{pmatrix}\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}\begin{pmatrix}x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}x\\
y\\
z
\end{pmatrix}^{T}\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}\begin{pmatrix}x\\
y\\
z
\end{pmatrix}$$
$$=\left(P^{-1}\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}\right)^{T}\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}\left(P^{-1}\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}\right)=\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}^{T}\left(P^{-1}\right)^{T}\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}P^{-1}\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}$$
$$=\begin{pmatrix}X & Y & Z\end{pmatrix}\left(\left(P^{-1}\right)^{T}\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}P^{-1}\right)\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}
=\begin{pmatrix}X & Y & Z\end{pmatrix}\begin{pmatrix}1 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & 1
\end{pmatrix}\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}=X^{2}-Y^{2}+Z^{2}
$$
hence $q$ is reduced on the basis associated to the matrix $P^{-1}$ (where we read its rank and its signature).
|
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|
Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.
I was thinking of using induction, but wasn't really sure how to do it.
|
In case $n$ is a multiple of $5$ the $n^2$ on the rhs is also.
If $n=2,3\pmod 5$ then $n^2\equiv4\pmod5$ , hence $n^2+1$ would be a multiple of $5$.
In the remaining cases, i.e.,$n=1,4\pmod 5$ then in $n^2-1= (n+1)(n-1)$ one of the term on the right will be a multiple of $5$.
So $n^2(n^2+1)(n^2-1)$ is always a multiple of $5$.
|
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|
Solving a cubic equation involving trigonometric functions Question:
If $$\sin(x) - \cos(x) = \frac{\sqrt{3}}2$$
Then $$\sin^3(x) - \cos^3(x) = ?$$
I have turned first equation into a quadratic so I got $$\sin(x) = \frac{\sqrt{3}\mp\sqrt{5}}4$$ and $$\cos(x) = \frac{-\sqrt{3}\mp\sqrt{5}}4$$ But stuck here. I don't know what should I do, please help
|
$$\sin(x) - \cos(x) = \frac{\sqrt{3}}2$$
$$\implies \sin^2(x) + \cos^2(x)-2\sin x \cos x = \dfrac 34$$
$$\implies \sin x\cos x=\dfrac 18$$ (using $\sin^2x+\cos^2x=1$)
Now, using $a^3-b^3=(a-b)(a^2+ab+b^2)$,$$\sin^3(x)-\cos^3x = (\sin x-\cos x)(\sin^2x+\cos^2x+\sin x\cos x)$$
$$=\dfrac {\sqrt 3}{2}\times \left (1+\dfrac 18\right)$$
$$=\dfrac {9\sqrt 3}{16}$$
|
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|
How can I simplify $\sqrt{3^2 + 3^2\tan^2\theta}$? $$\sqrt{3^2 + 3^2\tan^2\theta}$$
$$ = (3)(3\tan\theta) = 9\tan\theta $$
I've simplified it like this but I'm not sure if that's correct.
|
Hint
$$5x+5y=5(x+y) \text{ distributive property}$$
$$\sqrt{9\cdot 16} = \sqrt{9}\cdot\sqrt{16}\text{ root of a product property}$$
$$\begin{align}
opposite^2 + adjacent^2 &= hypotenuse^2\\
\frac{opposite^2 + adjacent^2}{adjacent^2}&=\frac{hypotenuse^2}{adjacent^2}\\
\bigg(\frac{opposite}{adjacent}\bigg)^2 + \bigg(\frac{adjacent}{adjacent}\bigg)^2&=\bigg(\frac{hypotenuse}{adjacent}\bigg)^2\\
\tan^2 x + 1 &= \sec^2 x
\end{align}$$
|
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|
Find conditions for $\left(\frac{-3}{p}\right)=1$ $\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right)^{\frac{p-1}{2}}$
$=\begin{cases}1,\:p\equiv 1\pmod{4}\text{ or }\begin{cases}p\equiv 3\pmod{4}\\p\equiv 2\pmod {3}\end{cases}\\-1,\:\text{otherwise}\end{cases}$
$p$ is an odd prime. Something is wrong here, since $\left(\frac{-3}{5}\right)=-1$, yet $5\equiv 1\pmod {4}$. What's wrong?
|
$$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right)(-1)^{\frac{p-1}{2}}$$
The problem is in your application of the quadratic-reciprocity law.
|
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|
Algorithm to generate an arbitrary matrix of special linear group $SL(2,\mathbb{Z})$ I have a given $2\times 2$ special linear matrix, for example
\begin{equation}
m=\begin{pmatrix} 55 & 8469 \\ 1 & 154 \end{pmatrix}
\end{equation}
and I would like to get the generating form of it from the s and t matrices which are these:
\begin{equation}
t=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\qquad\qquad\qquad s=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
\end{equation}
I don't know if exist an algoritmh for this problem or any proceedings which could help me.
|
This rather small example can also be done by hand without any algorithms.
Note that
$$
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}
=
\begin{pmatrix}
a & a+b \\
c & c+d
\end{pmatrix}
$$
While $S$ switches the columns and puts a minus sign in.
Then by guesstimating powers, we reduce the size of scalars quite efficiently.
For example:
$$
\begin{pmatrix}
55 & 8469 \\
1 & 154
\end{pmatrix}
\begin{pmatrix}
1 & -153 \\
0 & 1
\end{pmatrix}
=
\begin{pmatrix}
55 & 54 \\
1 & 1
\end{pmatrix}.
$$
Multiply by $S^3=S^{-1}$:
$$
\begin{pmatrix}
55 & 54 \\
1 & 1
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}=
\begin{pmatrix}
-54 & 55 \\
-1 & 1
\end{pmatrix}
$$
This look quite close to a power of $T$. So mulitply it by $T$:
$$
\begin{pmatrix}
-54 & 55 \\
-1 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}=
\begin{pmatrix}
-54 & 1 \\
-1 & 0
\end{pmatrix}
$$
This we recognize as $T^{54}S$. Working backwards, this all means that
$$
MT^{-153}ST=T^{54}S.
$$
Hence $M=T^{54}ST^{-1}S^{-1}T^{153}$. (this might be simplified though)
|
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|
Proof that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$. I proved that $$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$$ is a multiple of $3$ through the use of Little Fermat's theorem but i want to know if there exist other proofs(maybe for induction). How can I demonstrate it?
This my proof:
$$n^3(n^4-1)(n^5+n^3)+n^{13}(n^8-1)$$
Now i know that $$n^4-1\equiv 0\pmod 3 (n\neq 3k)$$ and $$n^8-1\equiv 0\pmod 3 (n\neq 3k).$$
Therefore I proved that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$.
|
$$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}=n^6(n^2-1)(n^2+1)^2+n^{13}(n^8-1)=$$
$$=n^6\left((n^2-1)(n^2+1)^2+n^7(n^2-1)(n^2+1)(n^4+1)\right)=$$
$$=n^6(n^2-1)(n^2+1)\left[n^2+1+n^7(n^4+1)\right]=$$
$$=\color{red}{n(n-1)(n+1)}\cdot n^5(n^2+1)\left[n^2+1+n^7(n^4+1)\right]$$
Can you see why the red part is already a multiple of $\;3\;$ ?
|
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|
Find the sum of values of $x$ such that $|x+2| +|x-3| +|x+4| + |x+5| = 18$ I tried it by finding the different values at the $4$ inflection points of the graph. Then didn't know how to proceed. Am I correct till here?
|
*
*$x \le -5 $ We need to solve $-(x+2)-(x-3)-(x+4)-(x+5)=18$
*$-5 \le x \le -4 $ We need to solve $-(x+2)-(x-3)-(x+4)+(x+5)=18$
*$-4 \le x \le -2$ We need to solve $-(x+2)-(x-3)+(x+4)+(x+5)=18$
*$-2 \le x \le 3$ We need to solve $(x+2)-(x-3)+(x+4)+(x+5)=18$
*$x \ge 3$ We need to solve $(x+2)+(x-3)+(x+4)+(x+5)=18$
Remember solve each of these and see if they are actually s solution to the original equation or even a contradiction with the inequality for that case equation. Hope this helps.
|
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|
Is Vieta the only way out? Let $a,b,c$ are the three roots of the equation $x^3-x-1=0$. Then find the equation whose roots are $\frac{1+a}{1-a}$,$\frac{1+b}{1-b}$,$\frac{1+c}{1-c}$.
The only solution I could think of is by using Vieta's formula repeatedly which is no doubt a very messy solution. Is there any easier and more slick way of doing this ?
|
Since the polynomial $x^3-x-1$ is irreducible over$\def\Q{\Bbb Q}~\Q$ by the rational root test, one approach would be to identify the element $\frac{1+a}{1-a}$ where $a$ is the image of $x$ in the field $K=\Q[x]/(x^3-x-1)$, and to compute its minimum polynomial over$~\Q$; since the Galois group of the splitting field of $x^3-x-1$ permutes its roots transitively, that polynomial will also have as roots the elements obtained by substituting another root ($b$ or $c$) for$~a$.
To compute the inverse of $1-a$ in $K$, find the Bézout coefficients of $\gcd(1-x,x^3-x-1)=1$ which are $-x-x^2$ and $-1$ since $1=(-x-x^2)(1-x) -(-1-x+x^3)$; the first one gives the inverse $-a-a^2$ of $1-a$. Now compute $ q=\frac{1+a}{1-a}=(1+a)(-a-a^2)=-a-2a^2-a^3$, which reduces to $q=-1-2a-2a^2$ using $a^3=a+1$. Using the same relation, one finds the square and cube of $q$ to be respectively $q^2=9+16a+12a^2$ and $q^3=-65-114a-86a^2$. Now simple linear algebra finds the linear dependence of $1,q,q^2,q^3$ to be
$1-q+7q^2+q^3=0$ so the polynomial that was asked for is $1-x+7x^2+x^3$.
|
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|
How to determine if $\sum_{n=1}^{\infty}\left (\frac{n^2-5n+1}{n^2-4n+2}\right)^{n^2}$ converges or diverges $$\sum_{n=1}^{\infty} \left(\frac{n^2-5n+1}{n^2-4n+2}\right)^{n^2}$$
Using root test seems not a efficient way since I got stuck without knowing what to do next
$$\lim_{n\to\infty}\left(\frac{n^2-5n+1}{n^2-4n+2}\right)^{n}$$
Using ratio test is too complicated
$$\left(\frac{(n+1)^2-5(n+1)+1}{(n+1)^2-4(n+1)+2}\right)^{(n+1)^2}=\left(\frac{n^2-3n-3}{n^2-2n-1}\right)^{n^2+2n+1}$$
$$\to\left(\frac{(n+1)^2-5(n+1)+1}{(n+1)^2-4(n+1)+2}*\frac{n^2-4n+2}{n^2-5n+1}\right)^{n^2}*\left(\frac{n^2-3n-3}{n^2-2n-1}\right)^{2n}*\frac{n^2-3n-3}{n^2-2n-1}$$
|
To complete your root test:
$$=\lim_n\left[\left(1+\frac{-5n-1}{n^2-4n+2}\right)^{\frac{n^2-4n+2}{-5n-1}}\right]^{\frac{n(-5n-1)}{n^2-4n+2}}=e^{\lim_n\frac{n(-5n-1)}{n^2-4n+2}}=e^{-5}<1$$
|
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|
Calculate sides of right triangle with hypotenuse and area or perimeter I'm trying to find if it is possible to find the lengths of the base and height of a right triangle with only the hypotenuse and the area (or the perimeter) of the triangle. I would have just figured that it was impossible, but I found an online calculator that could do it, but it didn't provide formulas.
|
For hypotenuse $h$ and area $A$, let the legs be $x$ and $y$. Then $x^2+y^2=h^2$ and $xy=2A$. It follows that
$$(x+y)^2=x^2+y^2+2xy=h^2+4A$$
and
$$(x-y)^2=x^2+y^2-2xy=h^2-4A.$$
Thus $x+y=\sqrt{h^2+4A}$ and $x-y=\pm\sqrt{h^2-4A}$ and now we can solve for $x$ and $y$.
For hypotenuse $h$ and perimeter $p$, we have $x^2+y^2=h^2$ and $x+y+h=p$. Thus $x+y=p-h$.
Note that
$$(x-y)^2=2(x^2+y^2)-(x+y)^2=2h^2-(p-h)^2.$$
Now we know $x-y$ and $x+y$, so as before we can find $x$ and $y$.
|
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|
Cyclic Equation. Prove that: $\small\frac { a^2(b-c)^3 + b^2(c-a)^3 + c^2(a-b)^3 }{ (a-b)(b-c)(c-a) } = ab + bc + ca$? This is how far I got without using polynomial division:
\begin{align}
\tiny
\frac { a^{ 2 }(b-c)^{ 3 }+b^{ 2 }(c-a)^{ 3 }+c^{ 2 }(a-b)^{ 3 } }{ (a-b)(b-c)(c-a) }
&\tiny=\frac { { a }^{ 2 }\{ { b }^{ 3 }-{ c }^{ 3 }-3bc(b-c)\} +{ b }^{ 2 }\{ { c }^{ 3 }-{ a }^{ 3 }-3ca(c-a)\} +c^{ 2 }\{ a^{ 3 }-b^{ 3 }-3ab(a-b)\} }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })-3{ a }^{ 2 }bc(b-c)+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })-3ab^{ 2 }c(c-a)+c^{ 2 }(a^{ 3 }-b^{ 3 })-3abc^{ 2 }(a-b) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })+c^{ 2 }(a^{ 3 }-b^{ 3 })-3{ a }^{ 2 }bc(b-c)-3ab^{ 2 }c(c-a)-3abc^{ 2 }(a-b) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })+c^{ 2 }(a^{ 3 }-b^{ 3 })-3{ a }bc\{ (b-c)+(c-a)+(a-b)\} }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })+c^{ 2 }(a^{ 3 }-b^{ 3 }) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { -a }^{ 3 }(b^{ 2 }-c^{ 2 })+{ a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })-b^{ 2 }c^{ 2 }(b-c) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { -a }^{ 3 }(b+c)+{ a }^{ 2 }({ b }^{ 2 }+{ bc+c }^{ 2 })-b^{ 2 }c^{ 2 } }{ (a-b)(c-a) }
\end{align}
Would it be possible to solve this answer easily without direct polynomial division? By the use of some known identities, perhaps?
|
Set $f(a,b,c) = a^2(b-c)^3+b^2(c-a)^3+c^2(a-b)^3$. Note that $f(a,b,c)$ is cyclic, i.e., $f(a,b,c) = f(c,a,b) = f(b,c,a)$.
Further, we also have $$f(a,a,c) = f(a,b,b) = f(c,b,c) = 0$$
This means $f(a,b,c) = (a-b)(b-c)(c-a)g(a,b,c)$, where $g(a,b,c)$ is also cyclic polynomial of degree $2$.
The most general cyclic polynomial of degree $2$ in $3$ variables is $$g(a,b,c) = k_1\left(a^2+b^2+c^2\right) + k_2\left(ab+bc+ca\right)$$ Setting $c=0$, we obtain
$$f(a,b,0) = a^2b^3-a^3b^2 = a^2b^2(b-a) = (b-a)abg(a,b,0)$$
This gives us $g(a,b,0) = ab = k_1(a^2+b^2)+k_2ab \implies k_1 =0\text{ and }k_2=1$. Hence, we obtain $g(a,b,c) = ab+bc+ca$.
|
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|
Factor $z^4 +1$ into linear factors $z$ is a complex number, how do I factor $z^4 +1$ into linear factors?
Do I write z in terms of $x+yi$ so that $z^4+1=(x+yi)^4+1?$
|
If $z^4+1=0$ then $z^4=-1$ so $z^2=\pm i$. If $z^2=i=\cos90^\circ+i\sin90^\circ$ then $z=\pm(\cos45^\circ+i\sin45^\circ) = \pm\left(\frac1{\sqrt2} + i\frac 1 {\sqrt2}\right)$, and similarly for $-i=\cos(-90^\circ)+i\sin(-90^\circ)$.
So
$$
z^4+1 = \left(z-\frac{1+i}{\sqrt2}\right)\left(z-\frac{-1-i}{\sqrt2}\right)\left(z-\frac{1-i}{\sqrt2}\right)\left(z-\frac{-1+i}{\sqrt2}\right).
$$
Let us note parenthetically that if we group conjugates together, thus
$$
z^4+1 = \underbrace{\left(z-\frac{1+i}{\sqrt2}\right)\left(z-\frac{1-i}{\sqrt2}\right)}_\text{conjugates}\quad \underbrace{\left(z-\frac{-1+i}{\sqrt2}\right)\left(z-\frac{-1-i}{\sqrt2}\right)}_\text{conjugates}
$$
then when conjugates get multiplied, the imaginary parts cancel, so in this case we get:
$$
z^4+1 = \big(z^2 - z\sqrt2 + 1\big)\big(z^2+z\sqrt2+1\big).
$$
|
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|
Let $a,b$ be relative integers such that $2a+3b$ is divisible by $11$. Prove that $a^2-5b^2$ is also divisible by $11$. The divisibility for $11$ of $a^2 - 5b^2$ can be easily verified; in fact: $$a \equiv \frac {-3}{2}b \pmod {11}$$ therefore $$\frac {9}{4}\cdot b^2 - 5b^2 = 11(-\frac{b^2}{4}) \equiv 0 \pmod {11}.$$
The solution doesn't need the use of the rules of modular-arithmetic. How can I demonstrate it?
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We can simplify and rigorize your proof by scaling by $\,4\,$ to clear denominators, yielding
$$2a\equiv -3b\,\overset{\rm square}\Rightarrow\,\color{#c00}{4a^2\equiv 9b^2}\,\Rightarrow\,\color{#c00}4(\color{#c00}{a^2}-5b^2) \equiv \color{#c00}{9b^2}-20b^2 \equiv -11b^2\equiv 0\!\pmod{\!11}\quad $$
Remark $\ $ In fact your fractional arithmetic can be made rigorous in any ring where $\,2\,$ is invertible, say $\,2c = 1,\,$ so $\, c = 1/2.\,$ Then we can rewrite your proof as follows
$\ 2a = -3b\,\Rightarrow\, a = -3b/2\,\Rightarrow\,a^2-5b^2 = (9/4 -5)b^2 = (9-5\cdot 4)b^2/4 = -11b^2/4,\,$
${\rm i.e.}\,\ \ c = 2^{-1},\,\ a = -3bc\,\Rightarrow\ \ a^2-5b^2 =\ (9c^2 -5)b^2 = (9-5\cdot 4)b^2c^2 = -11b^2 c^2$
This is true mod $\,m\,$ for any odd modulus $\, m = 2c-1\,$ since $\,2c\equiv 1\pmod m.\,$ In particular it is true for modulus $\,m = 11,\,$ where $\,-11b^2c^2\equiv 0.\,$
In the same way, arithmetic of fractions whose denominators are all coprime to the modulus can be made rigorous. This universality of fraction arithmetic will become conceptually clearer when one studies localizations (a generalization of the fraction field construction).
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Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$? This isn't a homework question but one I found online.
Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$?
We just started doing field theory in my class and I want extra practice, but I have no idea how to even start this problem. It's no, right? I'm not sure why.
|
On the other hand, $-1$ is the sum of two squares here. What Lam calls Siegel's Theorem is that, in an algebraic number field, the stufe is one of $1,2,4,\infty.$ Either $-1$ is not the sum of any number of squares in the field, or it is itself a square, or it is the sum of two or four squares.
I will see if I can find an explicit expression; others would be quicker than I, though.
There we go, looked at Berrick's expressions:
$$ \left( \frac{1}{2} \sqrt{-2} \right)^2 + \left( \frac{1}{2} \sqrt{-2} \right)^2 = \frac{-1}{2} + \frac{-1}{2} = -1 $$
or
$$ 1^2 + \left( \sqrt{-2} \right)^2 = 1-2 = -1 $$
http://en.wikipedia.org/wiki/Stufe_%28algebra%29#Examples
Extra: found details in a book I don't have by Rajwade; the stufe of a quadratic field, $\mathbb Q(\sqrt {-D})$ with $D>0$ squarefree, is $1$ if $D=1,$ is $2$ if $D \neq 8n+7,$ is $4$ if $D = 8n+7.$ Theorem 3.2 on pages 31-32 in his 1993 book.
$$ \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \sqrt{-7} \right)^2 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{-7}{4} = -1 $$
or
$$ 2^2 + 1^2 + 1^2 + \left( \sqrt{-7} \right)^2 = 4+1+1-7 = -1 $$
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I have a ball that intersects a cylinder and I need the volume. How do I do it? I have an exam coming up and I am stressing out about it really hard. I don't even know how to actually do this. Is it a triple integral?
I have a ball $\{(x,y,z)|x^2 + y^2 + z^2 \leq 9\}$ and a cylinder $\{(x,y,z) | x^2 + y^2 \leq 5, 0 \leq z \leq 3\}$. The volume V is obtained by intersecting the ball and the cylinder. Compute V.
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Let's start with picking convenient coordinate system. Since the $Oz$ axis is an axis of symmetry for both of the bodies, let's consider cylindrical coordinate system $(r, \theta, z)$ with $r = \sqrt{x^2+y^2}$ so $x = r \cos \theta, y = r \sin \theta$.
Now the ball $B$ is described as $r^2 + z^2 \leq 3^2$ and the cylinder as $r^2 \leq 5, 0 \leq z \leq 3$. Note how nicely $\theta$ disappeared.
The volume V is given by this tripple integral:
$$
V = \iiint\limits_{\substack{x^2+y^2+z^2 \leq 3^2\\x^2+y^2\leq 5\\0 \leq z \leq 3}} dxdydz = \iiint\limits_{\substack{r^2+z^2 \leq 3^2\\r^2\leq 5\\0 \leq z \leq 3}} rdrd\theta dz = 2\pi \iint\limits_{\substack{r^2+z^2 \leq 3^2\\r^2\leq 5\\0 \leq z \leq 3}} rdrdz.
$$
Since integral function and region don't depend on $\theta$ the integral became double and $2\pi$ arose from integrating by $\theta$.
$$
V = 2\pi\int\limits_0^3 dz \int\limits_0^\sqrt{\min(5,9-z^2)} rdr =
2\pi\int\limits_0^3 \frac{\min(5,9-z^2)}{2} dz =
2\pi\int\limits_0^2 \frac{5}{2} dz +
2\pi\int\limits_2^3 \frac{9-z^2}{2} dz = \\ =
10\pi + \frac{8\pi}{3} = \frac{38\pi}{3}
$$
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Numbers that can be expressed as the sum of two cubes in exactly two different ways It seems known that there are infinitely many numbers that can be expressed as a sum of two positive cubes in at least two different ways (per the answer to this post: Number Theory Taxicab Number).
We know that
$$1729 = 10^3+9^3 = 12^3 + 1^3,$$
and I am wondering if there are infinitely many numbers like this that can be expressed as the sum of two positive cubes in exactly two ways?
In fact, are there even any other such numbers?
EDIT:
As provided by MJD in the comments section, here are other examples:
$$4104 = 2^3+16^3 = 9^3+15^3,$$
$$13832 = 20^3+18^3=24^3+2^3,$$
$$20683 = 10^3 +27^3 = 19^3 +24^3.$$
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In the paper Characterizing the Sum of Two Cubes, Kevin Broughan gives the relevant theorem,
Theorem: Let $N$ be a positive integer. Then the equation $N = x^3 + y^3$ has a solution in positive integers $x,y$ if and only if the following conditions are satisfied:
*
*There exists a divisor $m|N$ with $N^{1/3}\leq m \leq (4N)^{1/3}.$
*And $\sqrt{m^2-4\frac{m^2-N/m}{3}}$ is an integer.
The sequence of integers $F(n)$,
$$\begin{aligned}
F(n)
&= a^3+b^3 = (2 n + 6 n^2 + 6 n^3 + n^4)^3 + (n + 3 n^2 + 3 n^3 + 2 n^4)^3\\
&= c^3+d^3 = (1 + 4 n + 6 n^2 + 5 n^3 + 2 n^4)^3 + (-1 - 4 n - 6 n^2 - 2 n^3 + n^4)^3
\end{aligned}$$
for integer $n>3$ apparently is expressible as a sum of two positive integer cubes in exactly and only two ways.
$$\begin{aligned}
F(4) &= 744^3+756^3 = 945^3+15^3\\
F(5) &= 1535^3+1705^3 = 2046^3+204^3\\
&\;\vdots\\
F(60) &=14277720^3+26578860^3 = 27021841^3+12506159^3
\end{aligned}$$
Using Broughan's theorem, I have tested $F(n)$ from $n=4-60$ and, per $n$, it has only two solutions $m$, implying in that range it is a sum of two cubes in only two ways. Can somebody with a faster computer and better code test it for a higher range and see when (if ever) the proposed statement breaks down? Incidentally, we have the nice relations,
$$a+b = 3n(n+1)^3$$
$$c+d = 3n^3(n+1)$$
Note: $F(60)$ is already much beyond the range of taxicab $T_3$ which is the smallest number that is the sum of two positive integer cubes in three ways.
$$T_3 \approx 444.01^3 = 167^3+436^3 = 228^3+423^3 = 255^3+414^3$$
(Using the theorem, this yields 3 values for $m$.)
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Not understanding solution to $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz$ computation Not understanding solution to $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz$ computation.
What was shown in class: $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz=\large \int_{|z-2|=2} \frac {5z+7}{(z+3)(z-1)}=\cdots= \int_{|z-2|=2} \frac {2}{z+3}dz+\int_{|z-2|=2} \frac {3}{z-1}dz $
And then it is said that $\large \int_{|z-2|=2} \frac {2}{z+3}dz=0$ while $\large \int_{|z-2|=2} \frac {3}{z-1}dz=3 \cdot 2 \cdot \pi \cdot i$.
Can you please help me understand the last step? What's the difference bewteen the integrals?
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Cauchy's Integral Formula states $$ \oint_C \frac{g(z)}{z-a}dz =g(a)2\pi i.$$ for every $a$ in the interior of $C$.
Here $g(z) = \dfrac {5z+7}{(z+3)}$, and $a= 1$ hence $$ \oint_C \frac{g(z)}{z-1}dz = \dfrac {5(1)+7}{(1+3)}2\pi i = 6\pi i.$$
|
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confusion regarding 'o' function . could one explain me the following steps ?
my books have written ,
$$\sec(x) = \frac{1}{1-x^2/2 + o(x^2)} = 1 + x^2/2 + o(x^2)$$ $$\sec^2(x) = \left(1 + x^2/2 + o(x^2)\right)^2 = 1 + x^2 + o(x^2)$$
the the expression for 'sec' from Taylor series expansion I understand. but how can we write , $$ \frac{1}{1-x^2/2 + o(x^2)} = 1 + x^2/2 + o(x^2)$$
?
And how do we integrate
$$o(x^2) ?$$
in my books they have written integrating $$\sec^2(x)$$ we get $$\tan(x) = x + x^3/3 + o(x^3)$$ how do we do such integration ?
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$\frac{1}{1-x^2/2 + o(x^2)} = 1 + x^2/2 + o(x^2)$
$[1-(x^2/2-o(x^2))]^{-1}=1+x^2/2-o(x^2)+(x^2/2-o(x^2))^2+O(f(x))$
Here for sufficiently small "$x$" we can ignore $2nd$ terms and merge $O(f(x))$ with $o(x^2)$ because it will contain terms at least larger than $x^2$.
Gives $1+x^2/2-o(x^2)+O(f(x))=1+x^2/2+o'(x^2)$
|
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Finding real coefficients of equation given that $a+ib$ is a root Below is the question present in a past examination paper. I'll be giving my attempts and how I thought it through. Do feel free to point out any mistakes I make throughout my working even if unrelated to the question itself.
*(a) Find $a+ ib$ such that $\frac{2+3i}{1-i} = a+ib$. If $a+ib$ is a root of the equation $px^2+qx+r=0$ where $p$ and $q$ are real numbers, find $p$,$q$ and $r$.
Before getting down to the technical side of the problem, I analyzed the question (somewhat). If the complex number $a+ib$ is a root of the given equation, then its conjugate $a-ib$ is also a root of the equation. However, since this is a quadratic equation, does the number of roots exceed the highest degree (in this case, 2)? If not, then are $a+ib$ and $a-ib$ the only roots of the equation?
Attempt to find $a+ib$
$$\frac{2+3i}{1-i} = a + ib$$
$$\frac{2+3i}{1-i} * \frac{1+i}{1+i} = a+ib$$
$$\frac{(2+3i)(1+i)}{2} = \frac{-1+5i}{2} = a+ib$$
$\therefore \frac{-1+5i}{2}$ is the first root and $\frac{-1-5i}{2}$ is the second root.
With these two roots, how would I proceed to obtain the coefficients of the above equation? I know it seems as though I haven't attempted this, I have however I cannot find any resources online that relate to this particular problem (somewhat identically atleast).
Method of solving (results matched with answersheet)
Since I got both roots $\frac{-1+5i}{2} = \alpha$ and $\frac{-1-5i}{2} = \beta$. I was able to structure my quadratic as $(x-\alpha)(x-\beta)$.
From there on all I did was subsitute, expand, simplify.
$$(x-(\frac{-1+5i}{2})(x-(-\frac{-1-5i}{2})$$
$$(x+\frac{1}{2} - \frac{5i}{2})(x+\frac{1}{2}+\frac{5i}{2})$$
$$x^2+x+\frac{13}{2}$$
$$2x^2+2x+13$$
$p=2,q=2, r=13$
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First $px^2 + qx + r = 0$ and $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$ have the same set of solutions. Note that we are sure that $p \neq 0$ (Why??) Think.
We have $\frac{q}{p} = 1$ and $\frac{r}{p} = \frac{1}{2}.$ We get then the equation $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$ as $x^2 + x + \frac{1}{2} = 0,$ so one of several possibilities of $px^2 + qx + r = 0$ is $2x^2 + 2x + 1 = 0.$
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What is the value of $ \lim_{x \to 0} \left [\frac{1}{1 \sin^2 x}+ \frac{1}{2 \sin^2 x} +....+ \frac{1}{n \sin^2 x}\right]^{\sin^2x} $ I took out $\sin^2x$ out of the brackets . Inside the brackets , I think I should use the formula $\frac{ n(n-1)}{2}$ . Am I doing right ? If yes, then what should I do next ? Thanks !
I am sorry for the heading . I know it is not clear and I have no idea why it is not being displayed properly .
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we have $$ \Big( \frac{1}{\sin^2x} \Big)^{\sin^2x} = e^{\sin^2x \ln \frac{1}{\sin^2x}}$$knowing that ${\sin^2x} \rightarrow 0$ as $x \rightarrow 0$, and $ \lim_{x\rightarrow 0} x\ln \frac{1}{x} = 0 $ then the limit is
$$\lim_{x \rightarrow 0} \Big( \frac{1}{\sin^2x} \Big)^{\sin^2x} =1. $$
Similarly, $$\lim_{x \rightarrow 0} \Bigg( 1+\frac{1}{2}+...+\frac{1}{n} \Bigg)^{\sin^2x } = \lim_{x \rightarrow 0} e^{\sin^2x \ln{( 1+\frac{1}{2}+...+\frac{1}{n} )} }=1. $$
SO $$\lim_{x \rightarrow 0} \Bigg( \frac{1}{\sin^2x}+\frac{1}{2\sin^2x}+...+\frac{1}{n\sin^2x} \Bigg)^{\sin^2x }=\lim_{x \rightarrow 0} \Big( \frac{1}{\sin^2x} \Big)^{\sin^2x} \Bigg( 1+\frac{1}{2}+...+\frac{1}{n} \Bigg)^{\sin^2x }=1. $$
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Limit of sequence and Riemann sum problem work verification I have to calculate $$\lim_{n \to \infty}{\sum_{k=1}^{n}{\frac{n}{k^2-4n^2}}}$$ My attempt:
$$\lim_{n \to \infty}{\sum_{k=1}^{n}{\frac{n}{k^2-4n^2}}} = \lim_{n \to \infty}{\frac{1}{n^2}\frac{1}{n}\sum_{k=1}^{n}{\frac{1}{\frac{k^2}{n^2}-4}}} = \left(\lim_{n \to \infty}{\frac{1}{n^2}}\right) \cdot \int_0^1{\frac{1}{x^2-4}} = 0 \cdot \left(-\frac{\log(3)}{4}\right) =0$$
Am I right?
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$$\sum_{k=1}^{n} \frac{n}{k^2-4n^2}=\sum_{k=1}^{n} \frac{n}{n^2\left(\frac{k^2}{n^2}-4\right)}=\frac{1}{n}\sum_{k=1}^{n} \frac{1}{\left(\frac{k^2}{n^2}-4\right)}$$
So
$$\lim_{n \to\infty} \sum_{k=1}^{n} \frac{n}{k^2-4n^2}= \int_0^1 \frac{1}{x^2-4}\mathrm{d}x=\frac{-\ln(3)}{4} $$
I hope I make no mistake !
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$x^2y''+(2x^2+x)y'+(2x^2+x)y=0$ A Bessel equation $$x^2y''+(2x^2+x)y'+(2x^2+x)y=0$$
The solution is $$e^{-x}J_o(x)+e^{-x}Y_o(x)$$
How does one approach a problem like this?
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One method is to recognize that $x^2 y''$ is frequently seen in many differential equations whereas the other terms are not. With this then one can consider a function of the type $y(x) = f(x) g(x)$.
\begin{align}
y(x) &= f g \\
y' &= f g' + f' g \\
y'' &= f g'' + 2 f' g' + f''
\end{align}
Now,
\begin{align}
0 &= x^2y''+(2x^2+x)y'+(2x^2+x)y \\
&= x^2 f g'' + 2 x^2 f' g' + x^2 f'' g + (2 x^2 + x) f g' + (2 x^2 + x) f' g + (2x^2 + x) f g \\
&= x^2 f g'' + (2 x^2 f' + 2 x^2 f + x f) g' +(x^2 f'' +(2 x^2 + x)f' + (2x^2 + x)f )g \\
&= x^2 g'' + \left( 2 x^2 \frac{f'}{f} + 2x^2 + x \right) g' + \left( x^2 \frac{f''}{f} + (2 x^2 + x) \frac{f'}{f} + 2 x^2 + x \right) g.
\end{align}
Let, from the coefficient of $g'$,
\begin{align}
\frac{f'}{f} = -1
\end{align}
which yields $f = e^{-x}$, $f' = - f$, $f'' = f$ and
\begin{align}
0 &= x^2 g'' + x g' + x^2 g
\end{align}
Compare this to the known Bessel function equation which is
\begin{align}
x^2 w'' + x w' + (x^2 + \nu^2) w = 0 \hspace{10mm} w = A J_{\nu}(x) + B Y_{\nu}(x)
\end{align}
to obtain $g = A J_{0}(x) + B Y_{0}(x)$. Placing the components in the proper order it has then been found that
\begin{align}
y(x) = e^{-x} \, \left( A J_{0}(x) + B Y_{0}(x) \right).
\end{align}
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Determining the sign of a term I have a problem in proving the sign of a term. It is as follows:
$$x=\dfrac{1-a}{b_1b_2-a}+1,\qquad y=\dfrac{1-a}{b_1-a}+\dfrac{1-a}{b_2-a},\qquad z=x-y$$
with $0<b_1<1,\quad 0<b_2<1,\quad 0<a<b_1b_2$.
Based on some simulations I have noticed that $z>0$ for several instances of $a$, $b_1$ and $b_2$, but I want to prove it mathematically. What is the simplest method to do this?
Thanks in advance.
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We can write
\begin{align}
x-y &= \frac{1-a}{b_1-a}\cdot \frac{1-a}{b_2-a} - \frac{1-a}{b_1-a} - \frac{1-a}{b_2-a} + 1 + \frac{1-a}{b_1b_2-a} - \frac{1-a}{b_1-a}\cdot \frac{1-a}{b_2-a}\\
&= \biggl(\frac{1-a}{b_1-a} - 1\biggr)\biggl(\frac{1-a}{b_2-a}-1\biggr) + \frac{1-a}{b_1b_2-a} - \frac{1-a}{b_1-a}\cdot \frac{1-a}{b_2-a}.
\end{align}
Now, since $a < b_k < 1$, we have $\frac{1-a}{b_k-a} > 1$, and so the product is strictly positive. Therefore it suffices to see
$$\frac{1-a}{b_1b_2-a} \geqslant \frac{1-a}{b_1-a}\cdot \frac{1-a}{b_2-a}.$$
Cancelling the positive $1-a$ and cross-multiplying, we need to see that
$$(b_1-a)(b_2-a) \geqslant (1-a)(b_1b_2-a).$$
Multiplying both sides out and cancelling the common terms, we need to see that
$$-ab_1 - ab_2 \geqslant -a - ab_1b_2,$$
or
$$a(1-b_1)(1-b_2) \geqslant 0,$$
which is true. Since all transformations we did were equivalences, we have shown that $x-y > 0$, as desired.
|
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Show that $\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$ Question:
Show that:
$$\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$$
then go on to prove the general case that:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k=1}^n \binom{n}{k}x^{k-1}$$
Attempted solution:
It might be doable to first prove the general case and then say that it is true for the specific case, but for the specific case I decided to write out the term and show that they were identical since they were so few.
For the LHS
$$\sum_{k = 0}^{4} (1+x)^k = (1+x)^0 + (1+x)^1 + (1+x)^2 + (1+x)^3 + (1+x)^4$$
$$ = (1) + (1 + x) + (x^2 +2x + 1) + (x^3 + 3x^2 + 3x + 1) + (x^4+4 x^3+6 x^2+4 x+1)$$
$$=5 + 10x + 10x^2 + 5x^3 + x^4$$
For the RHS:
$$\sum_{k=1}^5 \binom{5}{k}x^{k-1} = \binom{5}{1}x^{1-1} + \binom{5}{2}x^{2-1} + \binom{5}{3}x^{3-1} + \binom{5}{4}x^{4-1} + \binom{5}{5}x^{5-1}$$
$$= \frac{5!}{1!4!} x^0 + \frac{5!}{2!3!} x^1 + \frac{5!}{3!2!} x^2 + \frac{5!}{4!1!} x^3 + \frac{5!}{5!0!} x^4$$
$$ = 5 + 10x + 10x^2 + 5x^3 + x^4$$
That completes the first step of the question. So far so good.
For the general case, I started by using the binomial theorem for the binom and then writing out the inner-most sum:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k = 0}^{n-1} \sum_{i = 0}^{k} \binom{k}{i}x^i = \sum_{k = 0}^{n-1} \left( \binom{k}{0}x^0 + \binom{k}{1}x^1 + ... + \binom{k}{k}x^{k}\right)$$
I can imagine that each step in the sum will decide the coefficients for the various powers of x and thus be identical to the RHS of the general case.
However, I run of out steam here and do not at the moment see any obvious way forward. What are some productive approaches for the general case? Am I doing things too complicated?
|
You can also do a direct proof following your first computations. You can write
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k = 0}^{n-1} \sum_{i = 0}^{k} \binom{k}{i}x^i = \sum_{k = 0}^{n-1} \sum_{i = 0}^{n-1} \binom{k}{i}x^i$$
using the fact that $\binom{k}{i}=0$ when $i$ is greater that $k+1$. Then, if you invert the sums, you find that $$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{i = 0}^{n-1} \sum_{k = 0}^{n-1}\binom{k}{i}x^i.$$
Then, a classical result (see just after equation (8) here) says that
$$ \sum_{k = 0}^{n-1} \binom{k}{i} = \binom{n}{i+1}.$$
Therefore, you have eventually
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{i = 0}^{n-1} \binom{n}{i+1} x^i$$ which is exactly what you want.
|
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|
For which values does the Matrix system have a unique solution, infinitely many solutions and no solution? Given the system:
$$\begin{align}
& x+3y-3z=4 \\
& y+2z=a \\
& 2x+5y+(a^2-9)z=9
\end{align}$$
For which values of a (if any) does the system have a unique solution, infinitely many solutions, and no solution?
So I am getting that it has:
infinitely many solutions at: (-1)
No solution at (1)
Unique solution at (-1,1)
AM I RIGHT??
|
Simply by Gauss-Jordan elimination you get:
$$\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
2 & 5 & a^2-9 & 9
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 &-1 & a^2-3 & 1
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 & 0 & a^2-1 & a+1
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 & 0 & (a-1)(a+1) & a+1
\end{array}\right)$$
Now we would like to divide by $(a-1)(a+1)$. But we can do this only if this expression is non-zero. So we have to deal with the cases $a=\pm1$ separately.
If $a=1$, the last row is
$\left(\begin{array}{ccc|c}
0 & 0 & 0 & 2
\end{array}\right)$,
which corresponds to the equation $0x+0y+0z=2$. The equation $0=2$ clearly has no solution.
If $a=-1$ you get the system
$$\left(\begin{array}{ccc|c}
1 & 3 &-3 & 4\\
0 & 1 & 2 & -1\\
0 & 0 & 0 & 0
\end{array}\right)$$
which has infinitely many solutions. (You can also compute them if you want.)
In all other cases (i.e., for $a\ne\pm1$) you can continuo and you get
$$
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 & 0 & (a-1)(a+1) & a+1
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 & 0 & 1 & \frac1{a-1}
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 0 &-9 & 4-3a\\
0 & 1 & 2 & a\\
0 & 0 & 1 & \frac1{a-1}
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 0 & 0 & \frac{-3a^2+7a+5}{a-1}\\
0 & 1 & 0 & \frac{a^2-a-2}{a-1}\\
0 & 0 & 1 & \frac1{a-1}
\end{array}\right)$$
So in these cases the solution is $x=\frac{-3a^2+7a+5}{a-1}$, $y=\frac{a^2-a-2}{a-1}$, $z=\frac{1}{a-1}$.
|
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|
Probability distribution of number of columns that has two even numbers in a chart
We distribute numbers $\{1,2,...,10\}$ in random to the following chart:
Let $X$ be the number of columns that has two even numbers.
What is the distribution of $X$?
My attempt:
$|\Omega|=10!$
$P(x=0)=\frac {(5!)^2\cdot 2 ^5}{10!}=\frac 8 {63}$ Explanation: Arrange the even and odd numbers, choose which one is up 5 times.
I'm having trouble with the rest, for $P(x=1)$: choose two even numbers and a column $5\binom 5 2$, this force a column with odd numbers so choose them and a column $4\binom 5 2$, now we're left with arranging the rest like before: $(3!)^2\cdot 2^3$ so:
$P(x=1)=\frac{5\binom 5 24\binom 5 2 (3!)^2\cdot 2^3}{10!}= \frac {10}{63}$ which is too small to be right, and if I use the same reasoning $P(x=2)$ is too small again so they don't add up to 1.
|
You can first simplify the problem:
Let 1 denote an odd number and 0 and even number.
Case 1: P(X=0)
$11111$
$00000$
The even number can be arranged in 5! ways. The odd numbers can be arranged in 5! ways, too. Thus you have the factor $(5!)^2$ in all cases. In each coloumn 0 can be in the first row or in the second row. This gives an additional factor of $2^5$. You have calculated P(X=0) right.
Case 2: P(X=1)
$00001$
$01111$
The columns can be arranged in $\frac{5!}{3!\cdot 1!\cdot 1!}$ ways. There are 3 columns with different numbers. Thus the additional factor is $2^3$
Case 3: P(X=2)
$00011$
$00111$
The columns can be arranged in $\frac{5!}{2!\cdot 2!\cdot 1!}$ ways. There are 2x2 columns with equal numbers and one column with different numbers. Thus the additional factor is $2^1$
In total you get the distribution of X:
$\begin{array}{|c|c|c|c|} \hline x & 0 & 1 & 2 \\ \hline P(X=x) & \frac{5!\cdot 5!}{10!}\cdot 2^5 & \frac{5!\cdot 5!}{10!}\cdot \frac{5! \cdot 2^3}{3!} & \frac{5! \cdot 5!}{10!}\cdot \frac{2\cdot 5!}{2!\cdot 2!\cdot 1!} \\ \hline \end{array}$
|
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|
Inequality for sides and height of right angle triangle
Someone recently posed the question to me for the above, is c+h or a+b greater, without originally the x and y lengths. I used this method: (mainly pythagorus)
$a^2+b^2=c^2=(x+y)^2=x^2+y^2+2xy$
$a^2=x^2+h^2$ and $b^2=y^2+h^2$
therefore $x^2+h^2+y^2+h^2=x^2+y^2+2xy$
$x^2+y^2+2h^2=x^2+y^2+2xy$
so $2h^2=2xy$ and $$xy=h^2$$
also $Area={ab\over 2}={ch\over 2}$ so $$ab=ch$$
$(a+b)^2=a^2+b^2+2ab$
$(c+h)^2=c^2+h^2+2ch=a^2+b^2+xy+2ab$
therefore $$(a+b)^2+xy=(c+h)^2$$
so $$c+h>a+b$$
I feel like I have made a mistake somewhere, is this incorrectly generalised? And is there an easier way to show the inequality. Also, if this is correct can it be expanded to non right angle triangle, I tried to do this using trig but was pretty much going in circles, thanks!
|
The correct inequality is
$$(a+b)^2=(c+h)^2-xy<(c+h)^2$$
which implies $$a+b<c+h$$
|
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|
How to find differential equation $$\frac{dy}{dx}-8x=2xy^2\quad y=0\,x=1$$
I separated $x$ and $y$.
\begin{align*}
\color{red}{\frac{dy}{y^2}}&=\color{red}{2x+8x dx}\\
\frac{dy}{y^2}&=\color{red}{10x dx}\\
\color{red}{\ln y^2} &= 5x^2\\
y^2&=Ae^{5x^2}
\end{align*}
When I plug $y$ and $x$ in, i get $A=0$.
I think I did it wrong somewhere.
After I fixed it, I get.
$dy/y^2+4=2xdx$
$\ln(y^2+4)=x^2$
$y^2+4=Ae^{x^2}$
After plug in $y$ and $x$
$4=Ae^1$
$A=1.471$
|
You've made a couple of mistakes.
$\frac{dy}{dx}-8x = 2xy^2 \Rightarrow \frac{dy}{dx} = 2x(y^2+4) $
Therefore
$\frac{dy}{(y^2+4)} = 2xdx$
And try to continue from here.
|
{
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|
Finding out $S:=1+\frac12-\frac13-\frac14+\frac15+\frac16-\frac17-\frac18+\cdots$ I was willing to determine the sum of following
$$S:=1+\frac12-\frac13-\frac14+\frac15+\frac16-\frac17-\frac18+\cdots$$
I tried the following
\begin{align*}
S=&\sum\limits_{n=0}^\infty (-1)^n\left(\frac{1}{2n-1}+\frac{1}{2n}\right)\\
=&\sum\limits_{n=0}^\infty (-1)^n\left(\int_0^1 x^{2n-2}dx+\int_0^1 x^{2n-1} dx\right)\\
=&\int_0^1 \sum\limits_{n=0}^\infty (-1)^n\left(x^{2n-2}+x^{2n-1}\right)\\
=&\int_0^1 [x^{-2} \sum\limits_{n=0}^\infty (-1)^nx^{2n}+x^{-1} \sum\limits_{n=0}^\infty (-1)^nx^{2n}]
\end{align*}
and don't know after his what to do . Can you please help me on this regard?
Thanking you in advance
|
Let $$S = \sum_{n = 1}^{\infty} (-1)^n\left(\frac{1}{2n-1}+\frac{1}{2n}\right).$$ Then
$\begin {eqnarray}
S & = & \sum_{k = 1}^{\infty} (-1)^{2k}\left(\frac{1}{4k-1}+\frac{1}{4k}\right) + \sum_{k = 1}^{\infty} (-1)^{2k+1}\left(\frac{1}{4k+1}+\frac{1}{4k+2}\right) \nonumber \\ & = & \sum_{k = 1}^{\infty} \left(\frac {1} {4k - 1} - \frac {1} {4k + 1}\right) + \sum_{k =1}^{\infty} \left(\frac {1} {4k} - \frac {1} {4k + 2}\right),
\end {eqnarray}$
which is easier to proceed now.
|
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|
Determinant of a real skew-symmetric matrix What will be the value of the determinant of a skew-symmetric matrix of even order when a single element is interchanged between first row and first column?
For,
$\left| \begin{array}{cccc}
0 & 1 & 2 & -1 \\
-1 & 0 & 1 & 2 \\
-2 & -1 & 0 & 1 \\
1 & -2 & -1 & 0 \end{array} \right|$ = 16
$\left| \begin{array}{cccc}
0 & 1 & -2 & -1 \\
-1 & 0 & 1 & 2 \\
2 & -1 & 0 & 1 \\
1 & -2 & -1 & 0 \end{array} \right|$ = 16
$\left| \begin{array}{cccc}
0 & 1 & 2 & -1 \\
-1 & 0 & 1 & -2 \\
-2 & -1 & 0 & 1 \\
1 & 2 & -1 & 0 \end{array} \right|$ = 16
$\left| \begin{array}{cccc}
0 & 1 & -2 & -1 \\
-1 & 0 & 1 & -2 \\
2 & -1 & 0 & 1 \\
1 & 2 & -1 & 0 \end{array} \right|$ = 16
but,
$\left| \begin{array}{cccc}
0 & 1 & 2 & 1 \\
-1 & 0 & 1 & -2 \\
-2 & -1 & 0 & 1 \\
-1 & 2 & -1 & 0 \end{array} \right|$ = 36.
In the above example the value of determinant doesn't changed when the elements 2 and -2 are interchanged but the value of determinant changed when the elements 1 and -1 are interchanged.
Kindly any one answer me when does the value of determinant change?
|
I suspect the answer to "when does the value of the determinant change?" is "nearly always". Consider what happens when you interchange the $(1,3)$ and $(3,1)$ elements. Doing the algebra (by computer or by hand) gives
$$\eqalign{\det\pmatrix{0&a&b&c\cr -a&0&d&e\cr-b&-d&0&f\cr -c&-e&-f&0\cr}
-&\det\pmatrix{0&-a&b&c\cr a&0&d&e\cr-b&-d&0&f\cr -c&-e&-f&0\cr}\cr
&=-4be(af+cd)\ .\cr}$$
In your example it happens that $af+cd=0$, so the two determinants are the same. But this will not usually be the case.
|
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|
Uniform convergence: $\sum \frac{x^2}{(1 + x^2)^n}$ Consider the series of functions:
$$\sum_{n \ge 1} \frac{x^2}{(1 + x^2)^n}$$
Q: Where does this series converge uniformly?
We have if $x \neq 0$:
$$\lim_{n \to \infty} \left| \frac{x^2}{(1 + x^2)^{n+1}} / \frac{x^2}{(1 + x^2)^n} \right|= \frac{1}{1 + x^2} < 1$$
And if $x = 0$, then the series converges. Therefore, we have pointwise convergence everywhere.
For uniform convergence:
$$R_n(x) = \sum_{k = n+1}^{\infty} \frac{x^2}{(1 + x^2)^k} \ge \sum_{k = n+1}^{2n} \frac{x^2}{(1 + x^2)^k} \ge \sum_{k = n+1}^{2n} \frac{x^2}{(1 + x^2)^{2n}} = \frac{nx^2}{(1 + x^2)^{2n}}, \ \forall x \in \mathbb {R}$$
Then:
$$R_n(\frac{1}{\sqrt{n}}) \ge \frac{1}{(1 + 1/n)^n)^2} \longrightarrow \frac{1}{e^2} > 0$$
So, we can't have uniform convergence in a neighbourhood of $0$.
It seems that for $a > 0$, we don't have uniform convergence on $[a, \infty)$ or $(-\infty, -a]$ because we can make $x$ as large as we desire. But, I'm not sure about this.
However, on any compact interval $[a,b]$ with $b < -1$ or $a > 1$, we do have uniform convergence because of normal convergence:
$$\sup_{x \in [a,b]} \left( \frac{x^2}{(1 + x^2)^n} \right) \le \frac{b^2}{a^{2n}}$$
Can someone help me find all the sets on which we have uniform convergence?
Thank you.
|
Perhaps I am missing something, but I believe we can write your sequence as:
$$
\frac{x^2}{(1 + x^2)^n} = \frac{x^2 + 1 - 1}{(1 + x^2)^n} = \frac{1 + x^2}{(1 + x^2)^n} - \frac{1}{(1 + x^2)^n} = \frac{1}{(1 + x^2)^{n - 1}} - \frac{1}{(1 + x^2)^n}
$$
This gives two infinite sums:
\begin{align}
\sum_1^\infty \frac{x^2}{(1 + x^2)^n} =& \sum_1^\infty \left(\frac{1}{1 + x^2}\right)^{n - 1} - \sum_1^\infty \left(\frac{1}{1 + x^2}\right)^n \\
=&\sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n - \left(\frac{1}{1 + x^2}\right)\sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n \\
=& \left(1 - \left(\frac{1}{1 + x^2}\right)\right)\sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n \\
=& \frac{x^2}{1 + x^2} \sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n
\end{align}
Since the only sum is now a geometric series with $r = \frac{1}{1 + x^2}$ giving the sum (when $|x| > 0$):
\begin{align}
\sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n = \frac{1}{1 - r} = \frac{1}{1 - \frac{1}{1 + x^2}} = \frac{1 + x^2}{x^2} \\
\sum_1^\infty \frac{x^2}{(1 + x^2)^n} = \frac{x^2}{1 + x^2} \sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n = \frac{x^2}{1 + x^2} \frac{1 + x^2}{x^2} = 1
\end{align}
This series is identically $1$ everywhere except at $x = 0$ when the sum is clearly $0$. Therefore this function $f(x) = \sum_1^\infty \frac{x^2}{(1 + x^2)^n}$ is discontinuous at the value $x = 0$ but continuous and differentiable everywhere else.
Another way to see the above is that the original rearrangement gave a telescopic series:
$$
1 - \frac{1}{1 + x^2} + \frac{1}{1 + x^2} - \frac{1}{(1 + x^2)^2} + \frac{1}{(1 + x^2)^2} ... - \frac{1}{(1 + x^2)^n}
$$
If $|x| > 0$ then $1 + x^2 > 1$ and thus $\lim_{n \rightarrow \infty}\left(1 - \frac{1}{(1 + x^2)^n}\right) = 1$. Even when $x = 0$, for all $n$ we have: $1 - \frac{1}{1^n} = 0$ (which agrees with the observation that the sum is $0$ when $x = 0$).
|
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|
Verifying $\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} - 1}{\cot{\theta} + 1}$ My math teacher gave us an equality involving trigonometric functions and told us to "verify" them. I tried making the two sides equal something simple such as "1 = 1" but kept getting stuck. I would highly appreciate if someone could show me (step by step) how to verify or solve this problem.
$$\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} - 1}{\cot{\theta} + 1}$$
|
Let us look at the RHS only: (It is fine to manipulate both sides but it is usually preferred if you manage to make one into the other.)
$$
\frac{\cot{\theta}-1}{\cot\theta +1} = \frac{\frac{\cos\theta}{\sin\theta}-1}{\frac{\cos\theta}{\sin\theta} +1}
$$
Now we multiply the numerator and denominator by $\sin\theta$ to get,
\begin{equation}
\frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}.
\end{equation}
If we consider the double angle formulas for $\sin$ and $\cos$ i.e.
$$
\cos{2\theta} = \cos^2\theta-\sin^2\theta, \\ \sin 2\theta= 2\sin\theta\cos\theta,
$$
we can see that we can get a $\cos2\theta$ in the numerator of $\frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}$ by multiplying by $\cos\theta+\sin\theta$ as the $\cos2\theta$ formula is a difference of two squares. Proceeding with this idea we get;
$$
\left(\frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}\right)\left(\frac{\cos\theta+\sin\theta}{\cos\theta+\sin\theta}\right) = \frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta+2\sin\theta\cos\theta +\sin^2\theta}.
$$
Now the numerator is equal to $\cos2\theta$ so that is done and we only need to fix the denominator, by a quick inspection we see that we have a term $2\sin\theta\cos\theta$ which gives our $\sin2\theta$ term we desired and using the fact $\cos^2\theta+\sin^2\theta=1$ we find we get
$$
\frac{\cos2\theta}{1+\sin2\theta}
$$
which is what we wanted!
|
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|
Find all 4 digits numbers that $ABCD=(CD)^2$ Please help me to solve following problem:
Find all 4 digits numbers such that $ABCD=(CD)^2$.(any of $A,B,C,D$ is a digit!)
I know one of solutions is $5776=(76)^2$.
|
$ABCD=(CD)^2$ can be translated as $100n + x = x^2$ which implies
$x^2 - x \equiv 0 \pmod{100}$.
$$ \text{$x^2-x \equiv 0 \pmod{100} \iff x(x-1) \equiv 0 \pmod{4}$
and $x(x-1) \equiv 0 \pmod{25}$}$$
$$x^2-x \equiv 0 \pmod{4} \iff x \in \{\bar 0_{4}, \bar1_{4} \}$$
$$x^2-x \equiv 0 \pmod{25} \iff x \in \{\bar 0_{25}, \bar1_{25} \}$$
$$x\equiv a \pmod{4} \ \wedge \ x \equiv b \pmod{25} \iff x \equiv 25a - 24b \pmod{100}$$
\begin{align}
x \equiv 0 \pmod{4} \ \wedge \ x \equiv 0 \pmod{25} \iff x \equiv 0 \pmod{100} \\
x \equiv 0 \pmod{4} \ \wedge \ x \equiv 1 \pmod{25} \iff x \equiv 76 \pmod{100} \\
x \equiv 1 \pmod{4} \ \wedge \ x \equiv 0 \pmod{25} \iff x \equiv 25 \pmod{100} \\
x \equiv 1 \pmod{4} \ \wedge \ x \equiv 1 \pmod{25} \iff x \equiv 1 \pmod{100} \\
\end{align}
So, we get
*
*00^2 = 0000
*01^2 = 0001
*25^2 = 0625
*76^2 = 5776
|
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|
how to parameterize the ellipse $x^2 + xy + 3y^2 = 1$ with $\sin \theta$ and $\cos \theta$ I am trying draw the ellipse $x^2 + xy + 3y^2 = 1$ so I can draw it. Starting from the matrix:
$$ \left[ \begin{array}{cc} 1 & \frac{1}{2} \\ \frac{1}{2} & 3 \end{array}\right]$$
I computed the eigenvalues $2 \pm \frac{1}{2}\sqrt{5}$ and the eigenvectors (not normalized):
$$\left[ \begin{array}{c} x\\ y \end{array}\right]
= \left[ \begin{array}{c} 1\\ 2\pm \sqrt{5} \end{array}\right] $$
So then I tried writing down some combination of the data I generated:
$$ \left[ \begin{array}{c} x(\theta)\\ y(\theta) \end{array}\right]
= \cos \theta \left[ \begin{array}{c} 1\\ 2+ \sqrt{5} \end{array}\right] + \sin \theta \left[ \begin{array}{c} 1\\ 2- \sqrt{5} \end{array}\right]
$$
However, I have a hard time checking the ellipse equation holds true for all $\theta$:
$$x(\theta)^2 + x(\theta)y(\theta) + 3y(\theta)^2 = 1$$
What are the correct functions $x(\theta), y(\theta)$ ?
Following the comments, rescaling the eigenvectors and multiplying the eigenvalues:
$$ \left[ \begin{array}{c} x(\theta)\\ y(\theta) \end{array}\right]
= \frac{2 + \frac{1}{2}\sqrt{5}}{\sqrt{10 + 4 \sqrt{5}}} \left[ \begin{array}{c} 1\\ 2+ \sqrt{5} \end{array}\right]\cos \theta +
\frac{2 - \frac{1}{2}\sqrt{5}}{\sqrt{10 - 4 \sqrt{5}}}\left[ \begin{array}{c} 1\\ 2- \sqrt{5} \end{array}\right]\sin \theta
$$
Is it clear that the ellipse equation is satisfied? I am not sure how to check this.
|
If you need a parametrization, it is best to just complete the square, rather then exploiting the full power of the spectral theorem.
$$ x^2+xy+3y^2 = 1 $$
is equivalent to:
$$ \left(2x+y\right)^2 + 11 y^2 = 4$$
hence $2x+y=2\cos\theta,y=\frac{2}{\sqrt{11}}\sin\theta$ is a valid parametrization, that leads to:
$$ x = \cos\theta-\frac{1}{\sqrt{11}}\,\sin\theta,\quad y=\frac{2}{\sqrt{11}}\,\sin\theta.$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1294442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.