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How to determine equation for $\sum_{k=1}^n k^3$ How do you find an algebraic formula for $\sum_{k=1}^n k^3$? I am able to find one for $\sum_{k=1}^n k^2$, but not $k^3$. Any hints would be appreciated.
|
One classic method is
$$\begin{align} \sum_{k=1}^{n} k^4 &= \left( \sum_{k=1}^n (k+1)^4 \right) + 1 - (n+1)^4
\\&= \left( \sum_{k=1}^n k^4 + 4 k^3 + 6 k^2 + 4 k + 1 \right) + 1 - (n+1)^4
\\&= \left( \sum_{k=1}^n k^4 \right)
+ 4 \left( \sum_{k=1}^n k^3 \right)
+ 6 \left( \sum_{k=1}^n k^2 \right)
+ 4 \left( \sum_{k=1}^n k^1 \right)
+ \left( \sum_{k=1}^n 1 \right)
+ 1 - (n+1)^4
\end{align}$$
The sum of fourth powers cancel out, and you can solve the equation for the sum of curves.
Another approach is to guess that the sum of cubes should be a fourth degree polynomial, then solve for the coefficients of the polynomial using the equations
*
*$f(0) = 0$
*$f(1) = 1$
*$f(2) = 1 + 2^3$
*$f(3) = 1 + 2^3 + 3^3$
*$f(4) = 1 + 2^3 + 3^3 + 4^3$
Five equations in five unknowns. You can cut this down to 2 unknowns if you recognize some easy patterns: the leading coefficient for the sum of $n$-th powers is always $1/n$ and the next is always $1/2$, and the constant term is always 0.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $\int \frac {du}{(a-u^2)^2}$ I am stuck on the following integration problem:
$\int \frac {du}{(a-u^2)^2}; a$ being a constant.
Can someone point me in the right direction? Thanks in advance for your time.
|
Use $u = \sqrt{a} \sin{x}$; we get:
$$
\begin{align}
\int{\dfrac{du}{ (a - u^2)^2}} &= \int{\dfrac{\cos{x} dx}{ (a - a\sin^2{x})^2}} \\
&= \dfrac{1}{a^2}\int{\dfrac{\cos{x} dx}{ \cos^4{x}}} \\
&= \dfrac{1}{a^2}\int{\sec^3{x} dx} \\
&= \dfrac{1}{a^2} \cdot \left( \dfrac{1}{2} \sec{x} \tan{x} + \dfrac{1}{2} \ln{|\sec{x} + \tan{x} |}\right) + C
\end{align}
$$
Where, I used this method to integrate $\sec^3{x}$.
Substitute back the value of $x = \sin^{-1}{ \left(\dfrac{u}{\sqrt{a}} \right)} $
|
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|
Questions regarding p-adic expansion and numbers
As opposed to real number expansions which extend to the right as sums
of ever smaller, increasingly negative powers of the base $p$,
$p$-adic numbers may expand to the left forever, a property that can
often be true for the $p$-adic integers. For example, consider the
$p$-adic expansion of 1/3 in base 5. It can be shown to be
$…1313132_5$, i.e., the limit of a sequence $2_5$, $32_5$, $132_5$,
$3132_5$, $13132_5$, $313132_5$, $1313132_5$, … :
$\dfrac{5^2-1}{3}=\dfrac{44_5}{3} = 13_5; \, \dfrac{5^4-1}{3}=\dfrac{4444_5}{3} = 1313_5$
$\Rightarrow-\dfrac{1}{3}=\dots 1313_5$
$\Rightarrow-\dfrac{2}{3}=\dots 1313_5 \times 2 = \dots 3131_5$
$\Rightarrow\dfrac{1}{3} = -\dfrac{2}{3}+1 = \dots 3132_5.$ (Wikipedia, p-adic number)
I am unable to comprehend this. How does $\dfrac{5^2-1}{3}=\dfrac{44_5}{3} = 13_5; \, \dfrac{5^4-1}{3}=\dfrac{4444_5}{3} = 1313_5$ result in $-\dfrac{1}{3}=\dots 1313_5$?
|
Because is says that if $x=\cdots1313_5$ then $3x+1\equiv 0 \text{ mod } 5^n$ for all $n$, which is precisely what it means to be $0$ in $\mathbb{Q}_5$. Thus, you see that $3x+1=0$ so that $\displaystyle x=\frac{-1}{3}$.
EDIT:
Now that I have more time, let me be less glib about this response.
Whenever possible, we want to turn problems about $\def\Qp{\mathbb{Q}_5}$ $\def\Zp{\mathbb{Z}_5}$ $\Qp$ into problems about $\Zp$ since they are easy to deal with. So, how can we interpret $\displaystyle \frac{-1}{3}\in\Qp$, well since $\Qp$ is $\text{Frac}(\Zp)$ the only clear interpretation is that it is the element $x$ of $\Zp$ which satisfies $3x+1=0$. So, instead let us try to look for a solution $3x+1=0$.
To begin, let us recall how we can think about $\Zp$. Intuitively, $\Zp$ is the set $\{z\}$ of solutions to systems of equations as follows:
$$\begin{cases}z &\equiv a_1 \mod 5\\ z &\equiv a_2 \mod 5^2\\ z &\equiv a_3 \mod 5^3\\ &\vdots\end{cases}$$
where the equations are "consistent" (i.e. $a_i\equiv a_j\mod p^i$ for $i\leqslant j$). So, now if $x$ satisfies $3x+1=0$ then this should translate to mean
$$\begin{cases}3x+1 &\equiv 3a_1+1 \equiv 0 \mod 5\\ 3x+1 &\equiv 3a_2+1 \equiv 0 \mod 5^2\\ 3x+1 &\equiv 3a_3+1\equiv 0 \mod 5^3\\ &\vdots\end{cases}$$
So, we can solve each of these equations piecewise and find that
$$(a_1,a_2,a_3,\ldots)=(3,8,83,\ldots)$$
Ok, so, this tells us that $x=(3,8,83,\ldots)$...this doesn't look right? How can we go from this to the desired $x=\ldots1313_5$? The key is that we have the same element of $\Qp$ expressed in different forms. Indeed, the notation $x=1313_5$ means that
$$x=3+1\cdot 5+3\cdot 5^2 +\cdots$$
To reconcile this ostensible difference, let us write $3+1\cdot 5+3\cdot 5^2+\cdots$ in the same notation that we already have $x$ in. Recall that the correspondence between $\mathbb{Z}$ and these sequences is
$$m\mapsto (m\mod 5,m\mod 5^2,m\mod 5^3,\ldots)$$
Thus, we see that
$$\begin{aligned} 3 & \mapsto (3,3,3,\cdots)\\ 5 &\mapsto (0,5,5,\ldots)\\ 5^2 & \mapsto (0,0,5^2,\ldots)\end{aligned}$$
Thus, we see that
$$ \begin{aligned}3+5+3\cdot 5^2 &=(3,3,3,\ldots)+(0,5,5,\ldots)+(0,0,75,\ldots)\\ &= (3,8,83,\ldots)\end{aligned}$$
and voilà!
|
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|
Definite Integral $\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$ I want to prove that
$$\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$$
|
Note that for $|x| < \frac{\pi}{2}$, we have
$$ \frac{\log\cos x}{\log^2 \cos x + x^2} = \Re \frac{1}{\log\left( \frac{1+e^{2ix}}{2} \right)}. $$
Thus if $I$ denotes the given integral, we have
\begin{align*}
I
&= \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \Re \frac{1}{\log\left( \frac{1+e^{2ix}}{2} \right)} \, dx \\
&= \frac{1}{4} \mathrm{PV}\int_{-\pi}^{\pi} \Re \frac{1}{\log\left( \frac{1+e^{ix}}{2} \right)} \, dx \\
&= \frac{1}{4} \Re \mathrm{PV}\int_{-\pi}^{\pi} \frac{1}{\log\left( \frac{1+e^{ix}}{2} \right)} \, dx.
\end{align*}
Now let $C_{\epsilon}$ be the counter-clockwised contour consisting of the circle of radius 1 centered at the origin, with two semicircular indents $\gamma_{1,\epsilon}$ around $1$ and $\gamma_{2,\epsilon}$ around $-1$ as follows:
By writing
\begin{align*}
I = \frac{1}{4} \Re \lim_{\delta\to0^{+}}\left( \int_{-\pi+\delta}^{-\delta} + \int_{\delta}^{\pi-\delta} \right) \frac{1}{\log\left( \frac{1+e^{ix}}{2} \right)} \, dx
\end{align*}
and plugging the substitution $z = e^{ix}$, we observe that
\begin{align*}
I = \frac{1}{4} \Im \lim_{\epsilon\to 0^{+}}\left(\oint_{C_{\epsilon}} - \int_{\gamma_{1,\epsilon}} - \int_{\gamma_{2,\epsilon}} \right) \frac{dz}{z \log\left(\frac{1+z}{2}\right)}
\end{align*}
Let
\begin{align*}
f(z) = \frac{1}{z \log\left(\frac{1+z}{2}\right)}.
\end{align*}
It is plain from the logarithmic singularity that
\begin{align*}
\lim_{\epsilon \to 0^{+}} \int_{\gamma_{2,\epsilon}} f(z) \, dz = 0.
\end{align*}
Also it follows that
\begin{align*}
\lim_{\epsilon\to 0^{+}} \oint_{C_{\epsilon}} f(z) \, dz
&= 2\pi i \operatorname{Res}_{z=0} f(z)
= -\frac{2\pi i}{\log 2}, \\
\lim_{\epsilon\to 0^{+}} \int_{\gamma_{1,\epsilon}} f(z) \, dz
&= -\pi i \operatorname{Res}_{z=1} f(z)
= -2\pi i.
\end{align*}
Therefore we have
\begin{align*}
I
&= \frac{1}{4} \Im \left( 2\pi i - \frac{2\pi i}{\log 2} \right)
= \frac{\pi}{2} \left( 1 - \frac{1}{\log 2} \right).
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Divisibility criteria for $7,11,13,17,19$ A number is divisible by $2$ if it ends in $0,2,4,6,8$. It is divisible by $3$ if sum of ciphers is divisible by $3$. It is divisible by $5$ if it ends $0$ or $5$. These are simple criteria for divisibility.
I am interested in simple criteria for divisibility by $7,11,13,17,19$.
|
$(1)$
The formulae for $2,3,5,9,11$ can be derived from $\sum_{0\le r\le n}{a_r10^r}$
Observe that $\sum_{0\le r\le n}{a_r10^r}\equiv a_0\pmod 2$
$\sum_{0\le r\le n}{a_r10^r}\equiv a_0\pmod 5$
$\sum_{0\le r\le n}a_r10^r\equiv \sum_{0\le r\le n}a_r\pmod 3$ as $9\mid(10^r-1)$
$\sum_{0\le r\le n}a_r10^r\equiv \sum_{0\le r\le n}(-1)^ra_r\pmod {11}$ as $10^r\equiv(-1)^r\pmod{11}$
$\sum_{0\le r\le n}a_r10^r\equiv(a_0+a_2+a_4+\cdots)-(a_1+a_3+a_5+\cdots)\pmod{11}$
$(2)$
$N=\sum_{0\le r\le n}a_r10^r\equiv \sum_{0\le r\le m-1}a_r10^r\pmod {10^m}\equiv \sum_{0\le r\le m-1}a_r10^r\pmod {2^m}$ as $2^s\mid 10^s$ where integer $s\ge0$
This explains why $2^m\mid N\iff $ the numbers with lower $m$ digits of $N$ is divisible by $2^m$
For example, $2524$ will be divisible by $2^2=4$ as $24$ is, but $2514$ will not be divisible by $2^2=4$ as $14$ is not.
Similarly for $5^m$
$(3)$
For any number $y$ co-prime with $10,$ we can have a reduction formula as follows:
If a number be $10a+b,$ we can find $u,v$ in integers such that $10u+y\cdot v=1$ (using Bézout's Identity)
So, $u(10a+b)+v\cdot y\cdot a=a(10u+y\cdot v)+u\cdot b=a+u\cdot b\implies 10a+b$ will be divisible by $y\iff y\mid(a+u\cdot b)$
For example if $y=7, $ we find $3\cdot7+(-2)10=1\implies u=-2,v=3$
So, $(a+u\cdot b)$ becomes $a-2b$
If $y=19,$ we find $2\cdot10+(-1)19=1\implies u=2\implies a+u\cdot b=a+2b$
We can always use convergent property of continued fractions to find $u,v$.
There is no strong reason why this can not be generalized to any positive integer bases.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Generating Functions: Solving a Second-Order Recurrence I'm self-studying generating functions (using GeneratingFunctionology as a text). I came across this programming problem, which I immediately recognized as a modification of the Fibonacci sequence. I wanted to place my newly found generating function techniques to work, so I tried to solve the recurrence.
The problem basically boils down to solving:
$$a_{n+2} = a_{n+1} + ka_n$$
Where $n \ge 0$, and $k$ is some positive integer.
And, $a_0 = 1$, $a_1 = 1$.
However, my solution (below, but not really simplified, since I'm just plugging it into a computer) yields values that too great by a factor of $k$.
My solution:
$$a_n = \frac{r_2^{n+1} - r_1^{n+1}}{(r_2-r_1)r_1^{n+1}r_2^{n+1}}$$
Where:
$$r_1 = \frac{-1 - \sqrt{1+4k}}{2k}$$
$$r_2 = \frac{-1 + \sqrt{1+4k}}{2k}$$
If someone could help me understand where I went wrong, I'd much appreciate it. (I don't see why I'm too great by a factor of $k$.)
My work is as below:
$$a_{n+2} = a_{n+1} + ka_n$$
$$\sum_{n\ge0}a_{n+2}x^n = \sum_{n\ge0}a_{n+1}x^n + k\sum_{n\ge0}a_{n}x^n$$
Let $A(x) = \sum_{n\ge0}a_nx^n$. Then:
$$\frac{A(x) - a_0 - xa_1}{x^2} = \frac{A(x) - a_0}{x} + kA(x)$$
$$A(x)\left(\frac{1}{x^2} -\frac{1}{x} - k\right) = \frac{1+x}{x^2} - \frac{1}{x}$$
$$A(x)\left(\frac{1 - x - kx^2}{x^2}\right) = \frac{1}{x^2}$$
$$A(x)= \frac{1}{1 - x - kx^2}$$
Now, factoring the denominator yields two roots, $r_1$ and $r_2$ as denoted above. This, in turn, yields:
$$A(x) = \frac{1}{(x-r_1)(x-r_2)}$$
Partial fractions:
$$A(x) = \frac{1}{r_2-r_1}\frac{1}{(r_1 - x)} - \frac{1}{r_2-r_1}\frac{1}{(r_2 - x)}$$
$$A(x) = \frac{1}{(r_2-r_1)(r_1)}\frac{1}{(1 - \frac{x}{r_1})} - \frac{1}{(r_2-r_1)(r_2)}\frac{1}{(1 - \frac{x}{r_2})}$$
$$A(x) = \frac{1}{(r_2-r_1)(r_1)}\sum_{n\ge0}{\left(\frac{x}{r_1}\right)^n} - \frac{1}{(r_2-r_1)(r_2)}\sum_{n\ge0}{\left(\frac{x}{r_2}\right)^n}$$
Combining:
$$A(x) = \sum_{n\ge0} x^n \left(\frac{1}{(r_2-r_1)(r_1^{n+1})} - \frac{1}{(r_2-r_1)(r_2^{n+1})}\right)$$
Therefore:
$$a_n = \frac{r_2^{n+1} - r_1^{n+1}}{(r_2-r_1)r_1^{n+1}r_2^{n+1}}$$
|
Something went wrong between
$$A(x)= \frac{1}{1 - x - kx^2}$$
And
$$A(x) = \frac{1}{(x-r_1)(x-r_2)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Wolframalpha step-by-step of $\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$ I wonder, where the minus sign goes after the first $u$-substitution of integral $\displaystyle\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$?
|
Integrate
$$\int \frac{\cos x}{\sqrt{2+\cos(2x)}}dx$$
Known Identity
$$\cos(2x)=1-2\sin^2(x)$$
Replacing
$$\int \frac{\cos x}{\sqrt{3-2\sin^2(x)}}dx$$
Let $u = \sin(x)$, $du = \cos(x)dx$
Substituting
$$\int \frac{u}{\sqrt{3-2u^2}}du$$
Let $z = \sqrt{3-2u^2}$
$$dz = \frac{1}{2}\frac{-4u}{\sqrt{3-2u^2}}du$$
$$dz = -\frac{2u}{\sqrt{3-2u^2}}du$$
Substituting
$$-\int \frac{1}{2}dz$$
$$=-\frac{z}{2} + C$$
Substituting back
$$=- \frac{1}{2} \sqrt{- 2 u^{2} + 3}+ C$$
Substituting back $u = \sin(x)$
$$=-\frac{1}{2}\sqrt{-2\sin^2(x)+3}+ C$$
Simplifying
$$=-\frac{1}{2}\sqrt{2(1-\sin^2(x))+1}+ C$$
$$=-\frac{1}{2}\sqrt{2\cos^2(x)+1}+ C$$
$$=-\frac{1}{2}\sqrt{1+2\cos^2(x)}+ C$$
|
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|
$ \sum_{n=2}^\infty \frac{1}{n^3(n^3+1)}. $ The series is:
$$
\sum_{n=2}^\infty \frac{1}{n^3(n^3+1)}.
$$
I tried splitting the whole thing into simple fractions but I don't seem to get anywhere.
Any ideas?
|
$$
\begin{align}
\sum_{n=2}^\infty\frac1{n^3(n^3+1)}
&=\sum_{n=2}^\infty\left(\frac1{n^3}-\frac1{n^3+1}\right)\\
&=\sum_{n=2}^\infty\left(\frac1{n^3}-\frac1{n^3(1+1/n^3)}\right)\\
&=\sum_{n=2}^\infty\left(\frac1{n^3}-\frac1{n^3}\left(1-\frac1{n^3}+\frac1{n^6}-\dots\right)\right)\\
&=\sum_{n=2}^\infty\left(\frac1{n^6}-\frac1{n^9}+\frac1{n^{12}}-\dots\right)\\
&=\sum_{n=2}^\infty(-1)^n(\zeta(3n)-1)\\
&\stackrel.=0.01555356082097039944\tag{22 terms}
\end{align}
$$
Added Note: Since $\zeta(3n)-1\sim1/8^n$, each term in the sum gives over $0.9$ digits.
|
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|
Power series involving complex variable. Show that
$1 + \binom{m+1}{1}z + \binom{m+2}{2}z^2 +...+ \binom{m+n}{n}z^n +... = \frac{1}{(1-z)^{m+1}}$
for non-negative integers $m$ and $|z| < 1$.
|
Hint: start with the geometric series:
$$1+z+z^2+\ldots = \frac{1}{1-z}$$
and differentiate. For example,
$$1+2 z+3 z^2+\ldots=\frac{1}{(1-z)^2}$$
$$2+(3)(2)z+(4)(3)z^2+\ldots=\frac{2}{(1-z)^3} \implies 1+\binom{3}{1} z+\binom{4}{2}z^2+\ldots=\frac{1}{(1-z)^3}$$
Differentiate $m$ times to get your result.
|
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|
Factoring $x^8-x^4+1$ over $GF(7)$ Could anyone suggest any good way to do it? (The only way I can think of is by looking for roots (There are none), checking a factorization into the product of a 6 and a 2 polynomial (Many unknowns for the coefficients), checking a factorization of a 5 and 3, and finally one of two 4 degrees. (And each check probably takes lot's of time, since it contains many variables)
Is there any smarter way to do it?
|
First one substitutes $z=x^4$ and arrives at $z^2-z+1=0$. Quadratic polynomials are easy to factor (over any field of characteristic $\neq 2$), here we get $(z-3)(z-5)$. Using a bit of Galois theory (see below) one can find $x^4-3=(x^2+2x+2)(x^2+5x+2)$ and $x^4-5=(x^2+x+4)(x^2+6x+4)$ and these polynomials of degree $2$ are irreducible.
The Galois group of the polynomial is generated by the Frobenius $\alpha \mapsto \alpha^7$, and it acts on the roots. This action is transitive iff the polynomial is irreducible. More precisely, every orbit gives rise to an irreducible factor. Let $\alpha \in \overline{\mathbb{F}_7}$ be a root of $x^4-3$. One checks directly that $\alpha \notin \mathbb{F}_7$, hence $\alpha \neq \alpha^7$. But $\alpha^{7^2}=\alpha \cdot (\alpha^4)^{12}=\alpha \cdot 3^{12}=\alpha$. Therefore, $(x-\alpha)(x-\alpha^7) \in \mathbb{F}_7[x]$ is an irreducible factor of $x^4-3$. The constant term is $\alpha^8=(\alpha^4)^2=3^2=2$. The linear coefficient is $\alpha+\alpha^7$, whose square is $\alpha^2+2 \alpha^8 + \alpha^{14}=\alpha^2 + 2 (\alpha^4)^2+\alpha^2 (\alpha^4)^3=\alpha^2+2 \cdot 2 + \alpha^2 \cdot 6=4$, hence $\alpha+\alpha^7 = \pm 2$. Thus, we find $x^4-3=(x^2+2x+2)(x^2+5x+2)$.
Of course this can also be verified by a direct computation, but using Galois theory (which is quite easy for finite fields) we can derive the irreducible factors. No brute force search or algorithm is necessary.
|
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|
Computing $\int \frac{6x}{\sqrt{x^2+4x+8}} dx$ I am trying to compute an indefinite integral. Thanks!
$$\int \frac{6x}{\sqrt{x^2+4x+8}} dx.$$
|
$$\int \frac{6x}{\sqrt{x^2+4x+8}}dx=\int\frac{6(x+2)-12}{\sqrt{(x+2)^2+2}}dx$$
$$=6\int\frac{x}{\sqrt{x^2+2}}dx-12\int\frac{1}{\sqrt{x^2+2}}dx$$
$$=6\sqrt{x^2+2}-12\sinh^{-1}\left(\frac{x+2}{2}\right)+C.$$
|
{
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|
Balls of 3 colours in a bag - Probability There are $12$ balls in a bag. $3$ of them are red, $4$ of them are green, and $5$ of them are blue. We randomly take out $3$ balls from the bag at the same time. What is the probability that all three balls are of the same colour?
My answer: $(3/12)^3 + (4/12)^3 + (5/12)^3$. Is this correct?
EDIT: My explanation is that since there are the odds of getting a red ball are 3/12, I simply multiplied it 3 times since I consider multiplication to be the "and" operator. Similarly, I consider addition to be the "or" operator. So, the way I think of it is:
(1 red and 1 red and 1 red) or (1 green and 1 green and 1 green) or (1 blue and 1 blue and 1 blue).
|
For $3$ balls of red color $$\mathrm{Prob}=\frac{3}{12}\cdot\frac{2}{11}\cdot\frac{1}{10}$$
For $3$ balls of green color $$\mathrm{Prob}=\frac{4}{12}\cdot\frac{3}{11}\cdot\frac{2}{10}$$
For $3$ balls of blue color $$\mathrm{Prob}=\frac{5}{12}\cdot\frac{4}{11}\cdot\frac{3}{10}$$
Total probability of all ball same color is
$$
\frac{3}{12}\cdot\frac{2}{11}\cdot\frac{1}{10}+\frac{4}{12}\cdot\frac{3}{11}\cdot\frac{2}{10}+\frac{5}{12}\cdot\frac{4}{11}\cdot\frac{3}{10}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof that there exists an integer $n \geqslant2$ such that $n^2$ divides $2^n + 3^n$ Proof that there exists an integer $n \geqslant 2$ such that $n^2$ divides $2^n + 3^n$. I came up with this problem and I don't have a clue how to start, or even if it is not trivial at all.
|
Suppose there is an $n\in\mathbb{N}$ such that $2^n+3^n$ is divisible by $n^2$. Then, $2^n+3^n\equiv 0 \bmod n^2$. In particular, $2^n+3^n\equiv 0 \bmod n$. Suppose for convenience that $n=p\neq 2,3$ is a prime number. Then, by Fermat's little theorem, $2^p\equiv 2 \bmod p$ and $3^p\equiv 3 \bmod p$, so
$$2^p+3^p\equiv 2+3\equiv 5 \bmod p.$$
Hence, if $2^p+3^p\equiv 0 \bmod p$, we must have $p=5$. And indeed, $p=5$ works,
$$2^5+3^5 = 275 = 11\cdot 25.$$
Hence, we have shown that the only prime $p$ such that $p^2$ divides $2^p+3^p$ is $p=5$.
|
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|
Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a + b^3)(4)} \ge 3\sqrt{a} + \sqrt{b^3}$$
$$\sqrt{(3a + b^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2}$$
Also I get:
$$\sqrt{(3b + c^3)} \ge \frac{3\sqrt{b} + \sqrt{c^3}}{2}$$
$$\sqrt{(3c + a^3)} \ge \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
If I add add 3 inequalities I get:
$$\sqrt{(3a + b^3)} + \sqrt{(3b + c^3)} + \sqrt{(3c + a^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2} + \frac{3\sqrt{b} + \sqrt{c^3}}{2} + \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
Now i need to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} + \frac{3\sqrt{b} + \sqrt{b^3}}{2} + \frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge 6 = 2(a+b+c)$$
It's enough now to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} \ge b+c = 3-a$$
$$\frac{3\sqrt{b} + \sqrt{b^3}}{2} \ge a+c = 3-b$$
$$\frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge b+a = 3-c$$
All three inequalities are of the form:
$$\frac{3\sqrt{x} + \sqrt{x^3}}{2} \ge 3-x$$
$$3\sqrt{x} + \sqrt{x^3} \ge 6-2x$$
$$(3\sqrt{x} + \sqrt{x^3})^2 \ge (6-2x)^2$$
$$9x + x^3 + 6x^2 \ge 36 - 24x + 4x^2$$
$$x^3 + 2x^2 + 33x - 36 \ge 0$$
$$(x-1)(x^2 + 3x + 33) \ge 0$$
Case 1:
$$(x-1) \ge 0 \ \ \ \ \text{ for any }\ x \geq 1$$
$$(x^2 + 3x + 33) \ge 0 \ \ \ \ \text{ for any x in R} $$
Case 2:
$$0 \ge (x-1) \ \ \ \ \text{ for any }\ 1 \geq x$$
$$0 \ge (x^2 + 3x + 33) \ \ \ \ \text{there are no solutions in R} $$
This proves that for $$x \geq 1$$ $$(x-1)(x^2 + 3x + 33) \ge 0$$ is true and so it is
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$$, but a, b, c can be every non-negative number. I proved it's true for $$a,b,c \geq 1$$, but i can't for $$a,b,c \geq 0$$
|
I think your first method can still do something, but you probably have to use it in some other way, since using Cauchy-Schwarz directly will fail.
This is a suggestion.
Suppose
$\sqrt{3a+b^3} = u$
$\sqrt{3b+c^3} = v$
$\sqrt{3c+a^3} = w$
Then
$u^2+v^2+w^2 = 3(a+b+c)+a^3+b^3+c^3 = 9 + \sum a^3$
we are going to prove $u+v+w\ge 6$, which is equivalent to prove
$$u^2+v^2+w^2+2uv+2vw+2wu\ge 36$$
That means to prove
$$9 + \sum a^3+2uv+2vw+2wu\ge 36$$
we have proved
$u\ge \dfrac{1}{2}(3\sqrt{a}+\sqrt{b^3})$, thus
$2uv+2vw+2wu \ge \dfrac{1}{2}\sum(9\sqrt{ab}+\sqrt{b^3c^3}+3b^2+3\sqrt{ac^3}) = \\\dfrac{1}{2}\sum( 9\sqrt{ab} +\sqrt{a^3b^3}+3a^2 + 3\sqrt{ab^3})$
We then prove
$$\sum a^3 + \dfrac{1}{2}\sum( 9\sqrt{ab} +\sqrt{a^3b^3}+3a^2 + 3\sqrt{ab^3})\ge 27$$
I checked this with MATLAB, $(1,1,1)$ indeed is a local minimum, but I do not how to deal with the unsymmetric term $\sqrt{ab^3}$, I have tried to use $\sqrt{b^3}\ge 3\sqrt{b}-2$ to make it symmetric, but this will make the inequality fail.
|
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|
Evaluate $\lim\limits_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
*
*Evaluate $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$.
*Examine whether $x^{1/x}$ possesses a maximum or minimum and determine the same.
|
$\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
= $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3+2^3+3^3+\dots +n^3}{n^4}\right\}$ =
$\displaystyle\lim_{n\rightarrow \infty}\left\{\left(\frac{n(n+1)}{2n^2}\right)^2\right\}$
=$\frac{1}{4}\displaystyle\lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2} \right)^2$
=$\frac{1}{4}$
|
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|
Divisibility problem involving the floor function I am trying to prove the following:
$$2^{m+1}\mid\lfloor(1+\sqrt{3})^{2m+1} \rfloor$$ but $2^{m+2}\nmid\lfloor(1+\sqrt{3})^{2m+1} \rfloor$ for every natural number $m$.
I tried to use induction on $m$ but am not sure how to use the induction hypothesis. (The base case is easy to verify). Is there some other way to approach this problem?
Thanks.
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Consider the recurrence $x_0=2, x_1=20$ and $x_n=8x_{n-1}-4x_{n-2}$ for $n \geq 2$. It has characteristic polynomial $x^2-8x+4$, with roots $4 \pm 2\sqrt{3}=(1 \pm \sqrt{3})^2$, so $x_n=a(1+\sqrt{3})^{2n}+b(1-\sqrt{3})^{2n}$ for some contants $a, b$.
Using the values for $x_0, x_1$, we get:
$$2=a+b$$
$$20=a(1+\sqrt{3})^2+b(1-\sqrt{3})^2=4(a+b)+2\sqrt{3}(a-b)=8+2\sqrt{3}(a-b)$$
$$a-b=\frac{12}{2\sqrt{3}}=2\sqrt{3}$$
$$a=\frac{2+2\sqrt{3}}{2}=1+\sqrt{3}, b=\frac{2-2\sqrt{3}}{2}=1-\sqrt{3}$$
Therefore $x_n=(1+\sqrt{3})^{2n+1}+(1-\sqrt{3})^{2n+1}$. Note that $x_n$ is an integer, and $-1<(1-\sqrt{3})^{2n+1}<0$, so $\lfloor(1+\sqrt{3})^{2n+1}\rfloor=\lfloor x_n-(1-\sqrt{3})^{2n+1} \rfloor=x_n$.
The problem is thus equivalent to proving that $2^{n+1} \| x_n$. We proceed by induction.
When $n=0$, clearly $2^1 \|2=x_0$.
When $n=1$, clearly $2^2 \|20=x_1$.
Suppose that the statement holds for $0 \leq n \leq k, k \geq 1$. Then $2^{k} \|x_{k-1}$ and $2^{k+1}\|x_k$. Therefore $x_{k+1}=8x_k-4x_{k-1} \equiv 2^{k+2} \pmod{2^{k+3}}$, so $2^{k+2} \|x_{k+1}$.
We are thus done by induction.
|
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|
Finding the locus of a complex number Find the locus of $\arg\left(\frac{z-3}{z}\right) = \frac{\pi}{4}$ where $z$ represent complex number.
Working: $\arg\left(\frac{z-3}{z}\right) $ can be written as $\arg(z-3)-\arg(z) = \frac{\pi}{4}$, or $\arg\left((x-3)+iy\right) - \arg(x+iy)=\frac{\pi}{4}$.
If we take tangent to both side we get :
$$
\begin{align*}
\tan \left[\arg\left((x-3)+iy\right) -\arg (x+iy)\right] = \tan \frac{\pi}{4},\\
\tan\left[\frac{\arg\left((x-3)+iy\right) - \arg(x+iy)}{1+ \arg\left((x-3)+iy\right) \arg(x+iy)}\right] = 1.
\end{align*}
$$
Please suggest further...
|
$\arg(\dfrac{z-3}{z})=\dfrac{\pi}{4}\implies\Re(\dfrac{z-3}{z})=\Im(\dfrac{z-3}{z})$ and$\Re(\dfrac{z-3}{z})>0,z\neq0 $$\implies\Re(\dfrac{|z|^2-3\bar z}{|z|^2} )=\Im(\dfrac{|z|^2-3\bar z}{|z|^2})$$\implies x^2+y^2-3x=3y,y>0$(As$\Im(\dfrac{z-3}{z})=\dfrac{3y}{x^2+y^2}>0 $ if and only if $y>0$)
|
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|
If $x > 1$, prove that $f(x) = \dfrac{1}{\sqrt{x^2+1}}\log(x+\sqrt{x^2-1})$ $$f(x) = 2\int_{0}^{1}\dfrac{du}{u^2(1-x)+1+x}$$
I have used partial fractions but solves nothing.
|
I don't see a logarithm but an arctangent. Check this:
$$(1-x)u^2+1+x=(1+x)\left(\frac{1-x}{1+x}u^2+1\right)=(1+x)\left(1+\left(\sqrt\frac{1-x}{1+x}\;u\right)^2\right)\Longrightarrow$$
$$2\int\limits_0^1\frac{du}{(1-x)u^2+1+x}=\frac{2}{\sqrt{(1-x)(1+x)}}\;\int\limits_0^1\frac{\sqrt{\frac{1-x}{1+x}}\;du}{\left(1+\left(\sqrt\frac{1-x}{1+x}\;u\right)^2\right)}=$$
$$=\frac{2}{\sqrt{(1-x)(1+x)}}\arctan\left(\sqrt\frac{1-x}{1+x}\;u\right)_0^1=\frac{2}{\sqrt{(1-x)(1+x)}}\arctan\left(\sqrt\frac{1-x}{1+x}\right)$$
|
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|
Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stuck here.
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$$ \frac{4}{x+4} < 2 $$
(We'll do a case-by-case analysis here, which I feel is slightly more illuminating.)
Case 1: $x+4 > 0 \quad$ ($x > -4$)
In this case, we just multiply both sides by $x+4$ to get:
$$ 4 \le 2(x+4) $$
$$ 4 \le 2x + 8 $$
$$ -4 \le 2x $$
$$ -2 \le x $$
So this solution occurs when both $-4 < x$ and $-2 \le x$, which is the same condition as $x \ge -2$. (Draw a number line to see why this is the case.)
Case 2: $x+4 < 0 \quad$ ($x < -4$)
Now, we must flip the inequality when we multiply both sides.
$$ 4 \le 2(x+4) $$
$$ 4 \le 2x + 4 $$
$$ 0 \le 2x $$
$$ 0 \le x $$
$$ x \ge 0 $$
We have a solution whenever $x < -4$ and $x \ge 0$, in other words, $x < -4$.
|
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|
Solving $x+2y+5z=100$ in nonnegative integers I have not done combinatorics since high school, so this is an embarrassingly simple question.
We can solve the diophantine equation $x+y+z=100$ in nonnegative integers using the "bars and boxes" combinatorial method. We have $100$ dots, and we want place 2 partition markers among them, so the answer is ${ 102 \choose 2}$. Is there a way to generalize this (by a change of variable, perhaps) to equations like $x+2y+5z=100$? I know we can handle the case where we need to solve in positive integers by making the substitutions $x\rightarrow x+1$ and so forth, but I can't think of a way to apply a similar technique when the coefficients aren't $1$.
If not, is there a slick way to handle these more general equations?
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You can solve this problem by using the idea of generating functions; specifically, for the example above, let $c_n$ be the number of positive integer solutions to the equation
$$x+2y+5z=n$$
Then the generating function $f(a)$ for the sequence $a_i$ is
$$f(a)=c_0+c_1a+c_2a^2+c_3a^3+...+c_na^n+...$$
We have
$$f(a)=(1+a+a^2+...+a^i+...)(1+a^2+a^4+...+a^{2j}+...)(1+a^5+a^{10}+...+a^{5k}+...)$$
The exponents $i, j, k$ correspond the values of $x, y, z$ respectively in your equation above.
However, expanding that polynomial is still quite inconvenient. As such, we can re-write the formula as follows
$$f(a)=\frac{1}{1-a}\frac{1}{1-a^2}\frac{1}{1-a^5}$$
Using partial fractions, we re-write $f(a)$ as follows
$$f(a)=\frac{-a^4-a^3+a^2+1}{5(1-a^5)}+\frac{13}{40(1-a)}+\frac{1}{8(1+a)}+\frac{1}{4(1-a)^2}+\frac{1}{10(1-a)^3}$$
$$f(a)=\frac{1}{5}(-a^4-a^3+a^2+1)(1+a^5+a^{10}+a^{15}+...)+\frac{13}{40}(1+a+a^2+a^3+...)+\frac{1}{8}(1-a+a^2-a^3+a^4-...)+\frac{1}{4}(1+2a+3a^2+4a^3+...)+\frac{1}{10}(1+3a+6a^2+10a^3+15a^4+...)$$
Therefore, $$c_{100}=\frac{1}{5}+\frac{13}{40}+\frac{1}{8}+\frac{1}{4}\times 101+\frac{1}{10}\times \binom{102}{2}=541$$
|
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Please, I need a more detailed explanation of the particular solution of the problem with vectors Here is the problem and its solution (link to the source if you are interested):
Two different points $A$ and $B$ are given. Find a set of such points $M$, that $\overrightarrow{MA}\cdot\overrightarrow{MB}=k^2$, where $k$ is a given non-zero number.
The solution:
Let coordinates of points be: $A(x_a, y_a), B(x_b, y_b), M(x, y)$.
Then:
$$
\begin{align}
& \overrightarrow{MA}=(x_a - x, y_a - y), \overrightarrow{MB}=(x_b - x, y_b - y), \\
& \overrightarrow{MA}\cdot\overrightarrow{MB}= (x_a - x)(x_b - x)+(y_a - y)(y_b - y)=k^2
\end{align}
$$
Transform the last expression, opening brackets:
$$
x_a x_b - (x_a+x_b)x + x^2 + y_a y_b - (y_a+y_b)y + y^2 = k^2
$$
At this point I ask someone to provide a more detailed explanation, please, cause I failed to understand the next steps. What a transformation has been made here?
$$
\left( x - \frac{x_a + x_b}{2} \right)^2 - \frac{(x_a - x_b)^2}{4} +
\left( y - \frac{y_a + y_b}{2} \right)^2 - \frac{(y_a - y_b)^2}{4} = k^2 \\
\left( x - \frac{x_a + x_b}{2} \right)^2 + \left( y - \frac{y_a + y_b}{2} \right)^2
= k^2 + \frac{(x_a - x_b)^2}{4} + \frac{(y_a - y_b)^2}{4}
$$
Thus, this is a circumference with a center at the middle of segment $AB$ and its radius. (This I also failed to understand)
$$
r = \sqrt{k^2 + \frac{|AB|^2}{4}}
$$
|
Are you familiar with how to "complete the square?" To get from
$$
x_a x_b - (x_a+x_b)x + x^2 + y_a y_b - (y_a+y_b)y + y^2 = k^2
$$
to
$$
\left( x - \frac{x_a + x_b}{2} \right)^2 - \frac{(x_a - x_b)^2}{4} +
\left( y - \frac{y_a + y_b}{2} \right)^2 - \frac{(y_a - y_b)^2}{4} = k^2 \\
$$
consider the following:
$$
\begin{align}
x_ax_b-(x_a+x_b)x+x^2&=x_ax_b-(x_a+x_b)x+x^2+\frac{(x_a - x_b)^2}{4}-\frac{(x_a - x_b)^2}{4}\\
&=\frac{4x_ax_b}{4}-(x_a+x_b)x+x^2+\frac{x_a^2-2x_ax_b+x_b^2}{4}-\frac{(x_a - x_b)^2}{4}\\
&=-(x_a+x_b)x+x^2+\frac{x_a^2+2x_ax_b+x_b^2}{4}-\frac{(x_a - x_b)^2}{4}\\
&=-(x_a+x_b)x+x^2+\frac{(x_a + x_b)^2}{4}-\frac{(x_a - x_b)^2}{4}\\
&=\left( x - \frac{x_a + x_b}{2} \right)^2-\frac{(x_a - x_b)^2}{4}.\\
\end{align}
$$
Similarly,
$$
\begin{align}
y_ay_b-(y_a+y_b)y+y^2&=\left( y - \frac{y_a + y_b}{2} \right)^2-\frac{(y_a - y_b)^2}{4}.\\
\end{align}
$$
As for the second part of your question, the standard equation for a circle with center at $P(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$. Therefore,
$$
\left( x - \frac{x_a + x_b}{2} \right)^2 + \left( y - \frac{y_a + y_b}{2} \right)^2
= k^2 + \frac{(x_a - x_b)^2}{4} + \frac{(y_a - y_b)^2}{4}
$$
is the equation for the circle with center at $$O\left(\frac{x_a+x_b}{2},\frac{y_a+y_b}{2}\right)$$ and radius
$$\begin{align}
r&=\sqrt{k^2 + \frac{(x_a - x_b)^2}{4} + \frac{(y_a - y_b)^2}{4}}\\
&=\sqrt{k^2 + \frac{|AB|^2}{4}}.
\end{align}
$$
Note that the coordinates of $O$ are the coordinates for the midpoint of segment $AB$.
Hope this helps.
|
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|
What goes wrong in this derivative? $$ f(x) = \frac{2}{3} x (x^2-1)^{-2/3} $$
and f'(x) is searched.
So, by applying the product rule $ (uv)' = u'v + uv' $ with $ u=(x^2-1)^{-2/3} $ and $ v=\frac{2}{3} x $, so $ u'=-\frac{4}{3} x (x^2-1)^{-5/3} $ and $ v' = \frac{2}{3} $, I obtain
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3} x^2 + \frac{2}{3} (x^2-1)^{-2/3} $$
whereas according to Wolfram Alpha (see alternate form), the correct result is:
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3} x^2 - \frac{2}{3} (x^2-1)^{-5/3} $$
So apparently, my calculation for $u'v$ is correct, but $uv'$ is wrong. What am I missing here?
|
You're so close, but you have simply multiplied incorrectly.
Note that if $u' = -\frac{4}{3}x(x^2 - 1)^{-5/3}$ and $v = \frac{2}{3}x$, then $$u'v = -\frac{8}{9}x^2(x^2-1)^{-5/3}$$
This will make your answer correct. To match W|A, you just need to combine fractions carefully.
|
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|
Find the floor value of a finite continued surd Given $x=20062007$, and let
$$A=\sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}}.$$
Find the greatest integer not exceeding $A$.
|
$$10x+1<\sqrt{100x^2+39x+\sqrt{3}}<10x+2$$
$$4x+1<\sqrt{16x^2+10x+1}<\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}<\sqrt{16x^2+10x+2}<4x+2$$
$$2x+1=\sqrt{4x^2+4x+1}<\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}<\sqrt{4x^2+4x+2}<2x+2$$
$$x+1=\sqrt{x^2+2x+1}<A<\sqrt{x^2+2x+2}<x+2$$
Thus $\lfloor A \rfloor=x+1=20062008$.
|
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|
How to factor $2b^2c^2 + 2c^2a^2 + 2a^2b^2 -a^4-b^4-c^2$? The term is: $2b^2c^2 + 2c^2a^2 + 2a^2b^2 -a^4-b^4-c^2$
And the answer is : $(a+b+c)(b+c-a)(c+a-b)(a+b-c)$
I have tried a lot, but could't accomplish. Please don't bring up any complex method, it is just a high school math problem. But in vain I just can't do it.
|
$$2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4$$
$$=(2bc)^2-\{(a^2)^2+(b^2)^2+(c^2)^2-2a^2b^2+2b^2c^2-2a^2c^2\}$$
$$=(2bc)^2-(a^2-b^2-c^2)^2$$
$$=(2bc+a^2-b^2-c^2)(2bc-a^2+b^2+c^2)$$
$$=\{a^2-(b-c)^2\}\{(b+c)^2-a^2\}$$
|
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"url": "https://math.stackexchange.com/questions/349260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Residue Integral
Verify the integral with the aid of residues:
$$\int^{\infty}_0\frac{x^2+1}{x^4+1}dx=\frac{\pi}{\sqrt 2}$$
I got:
$f(z)=\frac{z^2+1}{z^4+1}$ and now I must find the residues for $f(z)$ and I got that the poles are: $e^{i\frac{\pi}{4}}$ and $e^{i\frac{3\pi}{4}}$. But I do not know how to finish.
|
This answer, which uses residues, says that
$$
\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x
$$
Using $m=4$ and $n=0$ and $n=2$ yields
$$
\begin{align}
\int_0^\infty\frac{x^2+1}{z^4+1}\,\mathrm{d}x
&=\frac\pi4\csc\left(\frac14\pi\right)+\frac\pi4\csc\left(\frac34\pi\right)\\
&=\frac\pi4\sqrt2+\frac\pi4\sqrt2\\
&=\frac\pi{\sqrt2}
\end{align}
$$
The Long Way
Using the curve $\gamma$ which runs from $-R$ to $+R$ along the real axis then circles counter-clockwise from $+R$ to $-R$ through the upper half-plane, we get
$$
\begin{align}
\int_0^\infty\frac{x^2+1}{x^4+1}\,\mathrm{d}x
&=\frac12\int_{-\infty}^\infty\frac{x^2+1}{x^4+1}\,\mathrm{d}x\\
&=\frac12\int_\gamma\frac{z^2+1}{z^4+1}\,\mathrm{d}z\\
&=\frac{2\pi i}2\left(\frac{e^{2\pi i/4}+1}{4e^{3\pi i/4}}+\frac{e^{6\pi i/4}+1}{4e^{9\pi i/4}}\right)\\
&=\frac{2\pi i}2\left(\frac{e^{\pi i/4}+e^{-\pi i/4}}{4e^{2\pi i/4}}+\frac{e^{\pi i/4}+e^{-\pi i/4}}{4e^{2\pi i/4}}\right)\\
&=\pi\cos\left(\frac\pi4\right)\\
&=\frac\pi{\sqrt2}
\end{align}
$$
Since the residue at the singularities inside the contour $z=e^{\pi i/4}$ and $z=e^{3\pi i/4}$ is $\dfrac{z^2+1}{4z^3}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving Recursions Without Initial Conditions I am trying to solve the following recursion but it does not appear that I can use characteristic equations or generating functions since I do not have initial conditions. Is there another way I am missing or can I assume initial conditions to use the above methods?
$$a_n = 3a_{n-1} -2a_{n-2} + 2^n + n^2$$
For example, if I found roots x = -1 and x = -2 I could then write the general solution as $α(−1)^n+β(−2)^n + P(n)$ for some particular solution? How would I get this particular solution? I am rusty at undetermined coefficients.
|
A general way of solving such, as expounded in Wilf's "generatingfunctionology":
$$
a_{n + 2} = 3 a_{n + 1} - 2 a_n + 4 \cdot 2^n + (n + 2)^2
$$
Define the ordinary generating function:
$$
A(z) = \sum_{n \ge 0} a_n z^n
$$
By the properties of generating functions, with the operator $D = \dfrac{d}{d z}$:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 3 \frac{A(z) - a_0}{z} - 2 A(z) + 4 \frac{1}{1 - 2 z}
+ (z D + 2)^2 \frac{1}{1 - z}
$$
Solving this, written as partial fractions:
$$
A(z) = \frac{5 - a_0 + a_1}{1 - 2 z}
+ \frac{2}{(1 - 2 z)^2}
+ \frac{2 a_0 - a_ 1 - 1}{1 - z}
- \frac{3}{(1 - z)^2}
- \frac{1}{(1 - z)^3}
- \frac{2}{(1 - z)^4}
$$
Expanding each of the terms by the binomial theorem, with $\displaystyle \binom{-k}{n} = (-1)^n \binom{k - 1 + n}{k - 1}$:
$$
\begin{align*}
a_n &= (5 - a_0 + a_1) \cdot 2^n
+ 2 \binom{n + 1}{1} \cdot 2^n
+ (2 a_0 - a_1 - 1)
- 3 \binom{n + 1}{1}
- \binom{n + 2}{2}
- 2 \binom{n + 3}{3} \\
&= \frac{1}{6} ((12 n + 6 a_1 - 6 a_0 + 42) \cdot 2^n
- 2 n^3 - 15 n^2 - 49 n
- 6 a_1 + 12 a_0 - 42)
\end{align*}
$$
Getting the complete solution this way isn't substantially harder than getting the solution by more traditional methods.
The help of maxima with the algebra is gratefully acknowledged.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$k-$ Subsets of $\{1,\cdots,n\}$ with no consecutive integers This is a practice-exam question in discrete mathematics.
Denote this number (as described in the title) with $f(n,k)$ where $f(n,0)= 1$. I figured out that the recurrence $f(n+2,k) = f(n+1,k) + f(n,k-1)$ holds if $k \leq \left \lfloor \frac{n+1}{2}\right \rfloor$. Now I have to show that
$$
\sum_{n=0}^\infty f(n,k)x^n = \frac{x^{2k-1}}{(1-x)^{k+1}}, \quad k\geq1
$$
Can I use induction on $k$ or is this straight-forward ?
I have now shown that $\sum_{k=0}^{\lfloor \frac{n+1}2 \rfloor} f(n,k) = f_n$. Does this help ?
|
OK, get at this with generating functions, à la Wilf's "generatingfunctionology", but in two indices. Define:
$$
F(x, y) = \sum_{n, k \ge 0} f(n, k) x^n y^k
$$
Write the recurrence as:
$$
f(n + 2, k + 1) = f(n + 1, k + 1) + f(n, k)
$$
Clearly $f(n, 0) = 1$ for all $n$, $f(0, k) = [k = 0]$, $f(1, k) = [0 \le k \le 1]$ (here $[\text[{condition}]$ is Iverson's convention, if the $\text{condition}$ is true, it is 1, else 0). We will need:
$$
\begin{align*}
\sum_{n \ge 0} f(n, 0) x^n &= \frac{1}{1 - x} \\
\sum_{k \ge 0} f(0, k) y^k &= 1 \\
\sum_{k \ge 0} f(1, k) x y^k &= x (1 + y)
\end{align*}
$$
Multiplying the recurrence by $x^n y^k$ and adding over $n, k \ge 0$ gives:
$$
\frac{F(x, y) - 1 - x (1 + y) - 1 / (1 - x) + 1 + x}{x^2 y}
= \frac{F(x, y) - 1 - 1 / (1 - x) + 1}{x y} + F(x, y)
$$
The terms added in have been subtracted twice, and must be restored once. We get:
$$
\begin{align*}
F(x, y) &= \frac{1 + x y}{1 - x - x^2 y} \\
&= \sum_{r \ge 0} x^r (1 + x y)^{r + 1} \\
&= \sum_{r \ge 0} \sum_{s \ge 0} \binom{r + 1}{s} x^{r + s} y^s
\end{align*}
$$
The terms with $x^n y^k$ are those with $n = r + s$, $k = s$:
$$
f(n, k) = \binom{n - k + 1}{k}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/355057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
limit with $\arctan$ I have to find the limit
and want ask about a hint:
$$\lim_{n \to \infty} n^{\frac{3}{2}}[\arctan((n+1)^{\frac{1}{2}})- \arctan(n^{\frac{1}{2}})]$$
I dont have idea what to do. Derivatives and L'Hôpital's rule are so hard
|
For $x>0$ we have $$\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}$$
so
\begin{align}(u_n=\arctan^{\frac{1}{2}}(n+1)- \arctan^{\frac{1}{2}}(n)&=\sqrt{\frac{\pi}{2}}\left[(1-\frac{2}{\pi}\arctan\frac{1}{n+1})^{1/2}-(1-\frac{2}{\pi}\arctan\frac{1}{n})^{1/2}\right]\\&=\sqrt{\frac{\pi}{2}}\left(\frac{1}{\pi}\frac{1}{n(n+1)}+O(\frac{1}{n^3})\right)\end{align}
and then we find
$$u_n\underset{n\rightarrow \infty}{\sim}\frac{1}{\sqrt{2\pi}}\frac{1}{n^2}$$
hence we find that your desired limit is $0$ but we find also
$$\lim_{n\to\infty}n^2\left(\arctan^{\frac{1}{2}}(n+1)- \arctan^{\frac{1}{2}}(n)\right)=\frac{1}{\sqrt{2\pi}}$$
Added
We have $\arctan(x)=_0 x-\frac{x^3}{3}+O(x^5)$ so we find
$$\arctan(n+1)^{\frac{1}{2}}- \arctan(n^{\frac{1}{2}})=\arctan(\frac{1}{n^{1/2}})-\arctan(\frac{1}{(n+1)^{1/2}})=\frac{1}{2}n^{-3/2}+O(n^{-5/2})$$
and then we have
$$\lim_{n \to \infty} (n^{\frac{3}{2}}(\arctan((n+1)^{\frac{1}{2}})- \arctan((n)^{\frac{1}{2}}))=\frac{1}{2}$$
|
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|
Asymptotics of sum of binomials How can you compute the asymptotics of
$$S=n + m - \sum_{k=1}^{n} k^{k-1} \binom{n}{k} \frac{(n-k)^{n+m-k}}{n^{n+m-1}}\;?$$
We have that $n \geq m$ and $n,m \geq 1$.
A simple application of Stirling's approximation gives
$$S \approx T = n + m - \frac{n^{3/2-m}}{\sqrt{2\pi}} \sum_{k=1}^n \frac{(n-k)^{m-1/2}}{k^{3/2}}$$
A more accurate approximation is given by
$$n+m- \frac{\left(1+\frac{1}{12 n}\right) n }{\sqrt{2 \pi }} \sum _{k=1}^{n-1} \frac{ (1-\frac{k}{n})^{m-\frac{1}{2}}}{\left(1+\frac{1}{12 k}\right) k^{3/2} \left(1+\frac{1}{12 (n-k)}\right) }$$
Via an indirect and handy wavy argument, my guess is guess for constant $m$ is that the answer is
$$S \sim \sqrt{2n} \frac{\Gamma(m+\frac{1}{2})}{(m-1)!}$$
Update. When $m$ grows almost as quickly as $n$ I think my guess is an underestimate. For example when $n=m$ it seems numerically that $S \sim 1.841 n$ and in fact if $n=m$ then it is suggested that $S \sim n\left(2-\left( -W\left(-\frac{1}{\mathrm{e}^2}\right)\right)\right)$ (see Is $\sum_{k=1}^{n} k^{k-1} (n-k)^{2n-k} \binom{n}{k} \sim\frac{n^{2n}}{2\pi} $?).
Update 2. When $m=1$ then $S$ is precisely the average number of people required to find a pair with the same birthday. This is solved at the wikipedia entry for the Birthday Problem and so $S \sim \sqrt{\frac{\pi n}{2}}$ (which equals my guess above). I would however ideally like to find the asymptotics in terms of $m$ and $n$ without assuming that $m$ is fixed.
Update 3. For the $m=1$ case we can prove that the correct asymptotics is $\sqrt{\frac{\pi n}{2}}$ in two ways.
*
*We will first show the result using the following identity.
$$n + 1 - \sum_{k=1}^{n} k^{k-1} \binom{n}{k} \frac{(n-k)^{n+1-k}}{n^{n+1-1}}=1 + \sum_{k=1}^n \frac{n!}{(n-k)!n^k} = 1+Q(n)$$
The numerator of the sum on the left is $n^{n+1}-Q(n)n^n$ (Q(n) is called Ramanujan's function by Knuth) according to A219706 and A063169. This immediately gives the identity.
We also know that $Q(n) \sim \sqrt{\frac{\pi n}{2}}$ from the wikipedia (is there a better reference?).
*
*The second proof follows from the amazing answer of GEdgar where he shows that
$$\sum_{k=1}^n \binom{n}{k} k^{k-1}(n-k)^{n-k+1} =
n^n\Bigg( n
-\frac{\sqrt{2\pi}}{2} n^{1/2} + \frac{1}{3}
-\frac{\sqrt{2\pi}}{24} n^{-1/2}
+\frac{4}{135}n^{-1}
-\frac{\sqrt{2\pi}}{576}n^{-3/2}
+O\left(n^{-2}\right)\Bigg)
$$
|
Computation for constant $m$ positive integer.
This uses the same method as in Estimate $\sum_{k=1}^{n} k^{k-1} \binom{n}{k} (n-k)^{n+1-k}$ ,
further explanation is there.
Write
\begin{equation*}
u_{-1}(z) = \sum_{n=1}^\infty \frac{n^{n-1}}{n!} z^n
\tag{1}\end{equation*}
Then the unique singularity nearest to the origin is at $z=e^{-1}$,
and we have an expansion there:
\begin{equation*}
u_{-1}(z) = 1 - \sqrt{2}(1-ez)^{1/2}+\frac{2}{3}(1-ez)
+O\left((1-ez)^{3/2}\right)
\tag{2}\end{equation*}
as $z \to e^{-1}$ from the left.
Define recursively $u_m(z) = z u_{m-1}'(z)$ for $m = 0,1,2,\dots$.
Then we have
\begin{equation*}
u_m(z) = \sum_{n=0}^\infty \frac{n^{n+m}}{n!}z^n
\tag{3}\end{equation*}
by induction. To expand these at $e^{-1}$ we will also need
the expansion of $z$:
\begin{equation*}
z = e^{-1} - e^{-1}(1-ez) = e^{-1} +O\left((1-ez)^1\right)
\tag{4}\end{equation*}
Now differentiate (2) and multiply by (4) to get
\begin{align*}
u_0(z) = \frac{1}{\sqrt{2}}(1-ez)^{-1/2}-\frac{2}{3}
+O\left((1-ez)^{1/2}\right)
\tag{5}\end{align*}
Differentiate this and multiply by (4) to get
\begin{align*}
u_1(z) &= \frac{1}{2\sqrt{2}} (1-ez)^{-3/2}
+O\left((1-ez)^{-1/2}\right)
\\ &=
\frac{\Gamma(3/2)}{\sqrt{2\pi}} (1-ez)^{-3/2}
+O\left((1-ez)^{-1/2}\right)
\end{align*}
Continuing, by induction we get
\begin{equation*}
u_m(z) = \frac{\Gamma(m+1/2)}{\sqrt{2\pi}}(1-ez)^{-m-1/2}
+O\left((1-ez)^{-m+1/2}\right)
\tag{6}\end{equation*}
for $m \ge 1$.
Now fix positive integer $m$. (The extra term $-2/3$ in (5)
mean that the formula for $m=0$ is different, but can also be done
by this method.)
Multiply (1) and (3) to get
\begin{equation*}
h(z) := u_{-1}(z)u_m(z)
= \sum_{n=1}^\infty\left(\frac{1}{n!}
\sum_{k=1}^n \binom{n}{k}\frac{k^{k-1}}{k!}\,
\frac{(n-k)^{n-k+m}}{(n-k)!}\right)z^n
=:\sum_{n=1}^\infty c_n z^n
\end{equation*}
Multiply (2) and (6) to get
$$
h(z) = \frac{\Gamma(m+1/2)}{\sqrt{2\pi}}(1-ez)^{-m-1/2}
-\frac{\Gamma(m+1/2)}{\sqrt{\pi}}(1-ez)^{-m}
+O\left((1-ez)^{-m+1/2}\right)
$$
as $z \to e^{-1}$ from the left. So by the Szegö method, we get
an asymptotic series
\begin{align*}
c_n &\approx e^n\left[
\frac{\Gamma(m+1/2)}{\sqrt{2\pi}}\binom{n+m-1/2}{n}
-\frac{\Gamma(m+1/2)}{\sqrt{\pi}}\binom{n+m-1}{n}+\dots
\right]
\\
c_n &= e^n\Bigg[
\frac{\Gamma(m+1/2)}{\sqrt{2\pi}}\left(
\frac{1}{\Gamma(m+1/2)}n^{m-1/2}+O(n^{m-3/2})\right)
\\ &\qquad\qquad
-\frac{\Gamma(m+1/2)}{\sqrt{\pi}}\left(
\frac{1}{(m-1)!}n^{m-1} +O\left(n^{m-2}\right)
\right) +O\left(n^{m-3/2}\right)
\Bigg]
\\ &=
e^n\left[\frac{1}{\sqrt{2\pi}}n^{m-1/2}
-\frac{\Gamma(m+1/2)}{(m-1)!\sqrt{\pi}}n^{m-1}+O\left(n^{m-3/2}\right)\right]
\end{align*}
as $n \to \infty$.
Multiply by Stirling's formula,
$$
n! = e^{-n}n^n\sqrt{2\pi}\left(
n^{1/2}+O\left(n^{-1/2}\right)\right)
$$
to get
\begin{align*}
&\sum_{k=1}^n\binom{n}{k} \frac{k^{k-1}
(n-k)^{n-k+m}}{n^{n+m-1}} = c_n\frac{n!}{n^{n+m-1}}
% \\ &\qquad
= n - \frac{\sqrt{2}\,\Gamma(m+1/2)}{(m-1)!}\,n^{1/2} + O\left(1\right)
\end{align*}
so
\begin{align*}
S &=
n+m-\sum_{k=1}^n\binom{n}{k} \frac{k^{k-1}
(n-k)^{n-k+m}}{n^{n+m-1}}
\\ &=
n+O\left(1\right)
-n + \frac{\sqrt{2}\,\Gamma(m+1/2)}{(m-1)!}\,n^{1/2}
+ O\left(1\right)
\\
&=
\frac{\sqrt{2}\,\Gamma(m+1/2)}{(m-1)!}\,n^{1/2}
+ O\left(1\right)
\end{align*}
as $n \to \infty$, since $m$ is constant.
|
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|
Looking for help understanding the asymptotic expansion of the digamma function I was recently given an example using this asymptotic expansion of the digamma function where:
$$\frac{d}{dx}(\ln\Gamma(x)) = \psi(x) \sim \ln(x) - \frac{1}{2x} - \frac{1}{12x^2}$$
Here's the example:
$$\frac {\psi\left(\frac x4\right)}4 - \frac {\psi\left(\frac x5 + \frac 12\right)}5 - \frac {\psi\left(\frac x{20} + \frac 12\right)}{20}\sim -\frac {\ln(4)}4 + \frac{\ln(5)}5 +\frac {\ln(20)}{20}-\frac 1{2\,x}-\frac {11}{8\,x^2}$$
I'm unclear on the following points:
*
*What happened to each $x$ term?
*Why is the first term negative and the rest of the terms positive? Why isn't the signs of the original terms would be preserved?
I would have expected something like this:
$$\frac{\ln(\frac{x}{4})}{4} - \frac{\ln(\frac{x}{4})}{4} - \frac{\ln(\frac{x}{4})}{4} - \ldots $$
*
*How is $\frac{11}{8x^2}$ being determined? Why does $-\frac{1}{12x^2}$ change but $-\frac{1}{2x}$ stays the same?
Sorry for the elementary questions. The explanation will really help! :-)
Thanks,
-Larry
|
The $\log(x)$ terms cancel because $\dfrac{1}{4} - \dfrac{1}{5} - \dfrac{1}{20} = 0$.
In somewhat more detail,
$$ \eqalign{\frac{\psi(x/4)}{4} &= -\frac{1}{2}\,\ln \left( 2 \right) +\frac{1}{4}\,\ln \left( x \right) -\frac{1}{2\,x}
-\frac{1}{3\, x^2}+O \left( {x}^{-4} \right)\cr
\frac{\psi(x/5+1/2)}{5} &= -\frac{1}{5}\,\ln \left( 5 \right) +\frac{1}{5}\,\ln \left( x \right) +{\frac {5}{24
\,x^2}}+O \left( {x}^{-4} \right) \cr
\frac{\psi(x/20+1/2)}{20} &= -\frac{1}{20}\,\ln \left( 20 \right) +\frac{1}{20}\,\ln \left( x \right) +\frac{5}{6\,x^2}+O \left( {x}^{-4} \right)
}$$
|
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|
Problem related to a clock I faced the following problem:
At what time after 4 o'clock, the hour and the minute hand will lie opposite to each other?
*
*$\quad$ 4-50'-31"
*$\quad$ 4-52'-51"
*$\quad$ 4-53'-23"
*$\quad$ 4-54'-33"
Can someone point me in the right direction?
|
At $4$ o'clock the angle between the hour & the minute hand is $-\frac4{12}\cdot360^\circ=-120^\circ$
To be at $180^\circ,$ the difference of angle to be generated will be $180^\circ-(-120^\circ)=300^\circ$
Now, the hour hand moves in $12$ hours $360^\circ$
So, it moves in $1$ hour $=60$ minutes $\frac{360^\circ}{12}=30^\circ$
Similarly, the minute hand moves in $1$ hour $=60$ minutes $360^\circ$
So, the minutes hand moves $(360-30)^\circ=330^\circ$ faster in $60$ minutes
So, it will move $300^\circ$ faster in $60\cdot\frac{300^\circ}{330^\circ}$ minutes $=\frac{600}{11}$ minutes $=54+\frac6{11}$ minute $=54$ minute $32+\frac8{11}$ second
Alternatively, let's start with $12$ o'clock when the angle between the hands is $0^\circ$
Now, the hands will lie opposite to each other if the angle between them is $n360^\circ+180^\circ=(2n+1)180^\circ$ where $n$ is any integer
As we have seen the difference will be $330^\circ$ in $60$ minutes
So, the difference will be $(2n+1)180^\circ$ in $\frac{(2n+1)180^\circ}{330^\circ}60$ minutes $=\frac{(2n+1)360}{11}$ minutes after $12$ o'clock
Now, as we need the time to after $4$ o'clock i.e., $4$ hours $=4\cdot60$ minutes after $12$ o'clock,
$\frac{(2n+1)360}{11}$ must be $>240\implies n>\frac{19}6$
As $n$ is an integer, $n_{\text{min}}=4$
So, the time will $\frac{(2\cdot4+1)360}{11}$ minutes after $12$ o'clock
i.e., $\frac{(2\cdot4+1)360}{11}-240$ minutes after next $4$ o'clock
$$\text{Now, }\frac{(2\cdot4+1)360}{11}-240=\frac{120}{11}(27-2\cdot11)=\frac{600}{11}$$
|
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|
Can someone check the solution to this recurrence relation? Here's the recurrence relation: $a_n = 4a_{n−1} − 3a_{n−2} + 2^n + n + 3$ with $a_0 = 1$ and $a_1 = 4$
Here's the solution:Write:
$$
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 2^n + n + 3 \quad a_0 = 1, a_1 = 4
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$. If you multiply the recurrence by $z^n$ and sum over $n \ge 0$ you get:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 4 \frac{A(z) - a_0}{z} - 3 A(z)
+ \frac{1}{1 - 2 z} + \frac{z}{(1 - z)^2} + 3 \frac{1}{1 - z}
$$
This gives:
$$
\begin{align*}
A(z) &= \frac{1 - 4 z + 9 z^2 - 12 z^3 + 5 z^4}
{1 - 8 z + 24 z^2 - 34 z^3 + 23 z^4 - 6 z^5} \\
&= \frac{23}{8} \cdot \frac{1}{1 - 3 z}
- \frac{1}{1 - 2 z}
+ \frac{3}{8} \cdot \frac{1}{1 - z}
- \frac{3}{4} \cdot \frac{1}{(1 - z)^2}
- \frac{1}{2} \cdot \frac{1}{(1 - z)^3}
\end{align*}
$$
Expanding the geometric series, and also:
$$
(1 - z)^{-k} = \sum_{n \ge 0} (-1)^n \binom{-k}{n} z^n
= \sum_{n \ge 0} \binom{n + k - 1}{k - 1} z^n
$$
gives:
$$
a_n = \frac{23}{8} \cdot 3^n
- 2^n
+ \frac{3}{8}
- \frac{3}{4} \cdot \binom{n + 1}{1}
- \frac{1}{2} \cdot \binom{n + 2}{2}
= \frac{23}{8} \cdot 3^n - 2^n + \frac{3}{8}
- \frac{1}{6} (n^3 + 6 n^2 + 5 n)
$$
The problem is that when I check this with Wolfram, it has the solution of $a_n = -4(2^n) - n^2 / 4 - 5n / 2 + 1/8 + (39/8)(3^n)$. I just wanted to know if this was an error or what..thanks!
|
You substituted $n+2$ instead of $n$ in your first step, but forgot to change the powers of $2$, etc. In other words, you should have
\begin{align}
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 2^{n+2} + (n+2) + 3 \quad a_0 = 1, a_1 = 4
\end{align}
instead of
\begin{align}
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 2^n + n + 3 \quad a_0 = 1, a_1 = 4
\end{align}
|
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|
How to find the last two digit of $7^{81}$? Could any one tell me how to find the last two digit of $7^{81}$? I have succeeded in finding the last digit only which is $7$.
Any group theoretic approach or any other approach is welcome.
|
An alternative method of determining the value that does not rely on calculating $\phi(100)$ is to perform the calculation in modular arithmetic using efficient exponentiation.
$81=64+16+1$, so here's the process (all lines are mod 100):
$$7 \equiv 7\\7^2 \equiv 49\\7^4 \equiv 49^2 \equiv 1\\7^5 \equiv 1\cdot 7 \equiv 7\\7^{10} \equiv 7^2 \equiv 49\\7^{20} \equiv 49^2 \equiv 1\\7^{40} \equiv 1^2 \equiv 1\\7^{80} \equiv 1^2 \equiv 1\\7^{81} \equiv 1\cdot 7 \equiv 7$$
This approach works for essentially any set of base, exponent, and modulo number.
|
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|
Computing Legendre symbol value using quadratic reciprocity I have an example in my lecture notes that says:
Check
$$\left( \frac{31}{1019} \right) = -1$$
where $\left( \frac{a}{b} \right)$ is the Legendre symbol.
I said, because $1019$ and $31%$ are both primes, we can use quadratic reciprocity and get:
$$ \left( \frac{31}{1019} \right)\left( \frac{1019}{31} \right) = (-1)^{509 \times 15} = -1$$
and so we get
$$\left( \frac{31}{1019} \right) = - \left( \frac{1019}{31} \right) = - \left( \frac{27}{31} \right).$$
Now I'm a little stuck as $27 = 3^3$ and so this isn't $+1$ and so I end up with $-(-1) = +1$ which isn't the answer. Then I thought of doing quadratic reciprocity on
$$\left( \frac{3}{31} \right)^3$$
which gives me
$$\left( \frac{3}{31} \right) \left( \frac{31}{3} \right) = (-1)^{15 \times 1} = -1$$
and so
$$\left( \frac{3}{31} \right) = - \left( \frac{31}{3} \right) = - \left( \frac{1}{3} \right) = -(+1) = -1$$
so now I have
$$- \left(-1 \right)^3 = -(-1) = +1$$
which, again, is the wrong answer.
Where have I gone wrong?
|
The Question seems to be wrong
$$\left(\frac{27}{31}\right)=\left(\frac{-4}{31}\right)=\left(\frac{2^2}{31}\right)\left(\frac{-1}{31}\right)=\left(\frac{-1}{31}\right)\text{ as } (\pm2)^2\equiv 4\pmod {31}$$
Again, we know, $\left(\frac{-1}p\right)=1\iff $ prime $p\equiv1\pmod 4$
|
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|
How to factor $x^4-7x^2-18$ I am not sure how I would factor this. The $x^4$ and $x^2$ are really throwing me off. Can someone explain how I would factor this?
|
Solution 1.
\begin{eqnarray*}
x^4-7x^2-18&=&(x^4+2x^2)-(9x^2+18)\\
&=&x^2(x^2+2)-9(x^2+2)\\
&=&(x^2+2)(x^2-9)\\
&=&(x^2+2)(x-3)(x+3)
\end{eqnarray*}
Solution 2.
\begin{eqnarray*}
x^4-7x^2-18&=&(x^4-9x^2)+(2x^2-18)\\
&=&x^2(x^2-9)+2(x^2-9)\\
&=&(x^2-9)(x^2+2)\\
&=&(x-3)(x+3)(x^2+2)
\end{eqnarray*}
Solution 3.
\begin{eqnarray*}
x^4-7x^2-18&=&(x^4-81)-(7x^2-63)\\
&=&(x^2+9)(x^2-9)-7(x^2-9)\\
&=&(x^2-9)(x^2+9-7)\\
&=&(x-3)(x+3)(x^2+2)
\end{eqnarray*}
|
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|
Evaluating Sums Algebraically or Combinatorially Consider
(1) $$\sum_{k=0}^{n}\binom{n}{k}2^{k-n}$$
(2) $$\sum_{k=0}^{n}\binom{n}{k}\frac{k!}{(n+k+1)!}$$
These sums appear too difficult (in my mind) to evaluate combinatorially. What are some good methods to attack these problems algebraically?
|
The second sum
$$ \sum_{k=0}^{n} \binom{n}{k} \frac{k!}{(n+k+1)!} $$
can be rewritten as
$$\frac{n!}{(2n+1)!} \sum_{k=0}^{n} \binom{2n+1}{n-k} = \frac{2^{2n} n!}{(2n+1)!}$$
expand $\binom{n}{k} = \frac{n!}{(n-k)!k!}$, cancel the $k!$ and multiply and divide by $(2n+1)!$ and,
$\sum_{k=0}^{n} \binom{2n+1}{n-k} = \binom{2n+1}{n} + \binom{2n+1}{n-1} + \dots + \binom{2n+1}{0} = \frac{1}{2} \sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n}$
|
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|
How to find limit of this function- with summation notation i.e Sigma notation Find the value of $\displaystyle\lim_{n \rightarrow \infty} \dfrac 1{n^4}\left[1\left(\sum^n_{k=1}k\right)+2\left(\sum^{n-1}_{k=1}k\right)+3\left(\sum^{n-2}_{k=1}k\right)+\dots+n.1\right]$
Please guide how to proceed in this case .....Thanks..
Here answer is :
The required limit is $\dfrac{1}{24}$
|
$$\displaystyle\lim_{n \rightarrow \infty} \dfrac 1{n^4}\left[1\left(\sum^n_{k=1}k\right)+2\left(\sum^{n-1}_{k=1}k\right)+3\left(\sum^{n-2}_{k=1}k\right)+\dots+n.1\right]=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\left[1\dfrac{n(n+1)}{2}+2\dfrac{(n-1)(n)}{2}+\dots +k\dfrac{(n-k+1)(n-k+2)}{2}+\dots+n\cdot1\right]=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^{n}k\dfrac{(n-k+1)(n-k+2)}{2}=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^{n}\dfrac{k}{2}((n-k)^2+3(n-k)+2)=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\left(\sum\limits_{k=1}^{n}\dfrac{k}{2}\left((n-k)^2+3(n-k)\right)+\dfrac{n(n+1)}{2}\right)=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^\infty\dfrac{k(n-k)^2}{2}+\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\sum\limits_{k=1}^\infty\dfrac{3k(n-k)}{2}+\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\dfrac{n(n+1)}{2}=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1-0}{n}\sum\limits_{k=1}^\infty\dfrac{\frac{k}{n}(1-\frac{k}{n})^2}{2}+\lim\limits_{n\rightarrow\infty}\dfrac{1-0}{n^2}\sum\limits_{k=1}^\infty\dfrac{3\frac{k}{n}(1-\frac{k}{n})}{2}+0=$$
$$\int_{0}^{1}\dfrac{x(1-x)^2}{2}dx+0\cdot \int_{0}^{1}\dfrac{3x(1-x)}{2}dx=\dfrac{1}{24}$$
Note: We used the fact that the limit of Riemann Sums converges to the integral.
|
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|
How to prove that $\lim_{n\to \infty} (n^k/2^n) = 0$? I'm having a hard time trying to prove this statement.
$\lim_{n\to \infty} (n^k/2^n) = 0$
k is a positive number.
Please, help me.
Thanks in advance.
|
Let us prove that $\lim_{n \to \infty}\frac{n^k}{a^n}=0,$ for $a>1$. Let $m \in Z$ and $m>k$, then
$$ 0<\frac{n^k}{a^n}\le \frac{n^m}{a^n}=\left(\frac{n}{\sqrt[m]{a^n}}\right)^m=\left(\frac{n}{b^n} \right)^m $$
where $b=\sqrt[m]{a}>1$.
Now,
\begin{align*} 0<\frac{n}{b^n}&=\frac{n}{(1+(b-1))^n}=\frac{n}{\sum\limits_{k=0}^n {n \choose k}1^{n-k}(b-1)^k}\\
&=\frac{n}{1+n(b-1)+\frac{n(n-1)}{2}(b-1)^2+\cdots+(b-1)^n} \\
&<\frac{n}{\frac{n(n-1)}{2}(b-1)^2} = \frac{2}{(b-1)^2(n-1)} \rightarrow 0 ~ (n \to \infty)
\end{align*}
Since $\frac{n}{b^n} \rightarrow 0 \Rightarrow (\frac{n}{b^n})^m \rightarrow 0~(n \to \infty)$
From $0<\frac{n^k}{a^n}\le \frac{n^m}{a^n}=\left(\frac{n}{\sqrt[m]{a^n}}\right)^m=\left(\frac{n}{b^n} \right)^m$, we have that $$\frac{n^k}{a^n} \rightarrow 0~ (n \to \infty) $$
|
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|
Find expansion around $x_0=0$ into power series and find a radius of convergence My task is as in the topic, I've given function $$f(x)=\frac{1}{1+x+x^2+x^3}$$
My solution is following (when $|x|<1$):$$\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)+(x^2+1)}=\frac{1}{1-(-x)}\cdot\frac{1}{1-(-x^2)}=$$$$=\sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k$$ Now I try to calculate it the following way:
\begin{align}
& {}\qquad \sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k \\[8pt]
& =(-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+\cdots)\cdot(-x^2+x^4-x^6+x^8-x^{10}+\cdots) \\[8pt]
& =x^3-x^4+0 \cdot x^5+0 \cdot x^6 +x^7-x^8+0 \cdot x^9 +0 \cdot x^{10} +x^{11}+\cdots
\end{align}
And now I conclude that it is equal to $\sum_{k=0}^{\infty}(x^{3+4 \cdot k}-x^{4+4 \cdot k})$ ($|x|<1$)
Is it correct? Are there any faster ways to solve that types of tasks? Any hints will be appreciated, thanks in advance.
|
If I recall, if you have two power series, based at the same point, then the radius of convergence of their product is at least the smaller of the two radii of convergence. Generally it will be the smaller of the two radii of convergence. The at least part comes from the possibility of numerators in one cancelling with denominators in another. This isn't the case in this problem.
Note that $1+x+x^2+x^3 \equiv (1+x)(1+x^2)$, and so:
$$\frac{1}{1+x+x^2+x^3} \equiv \frac{1}{1+x} \times \frac{1}{1+x^2}$$
Both factors of the right are got from geometric series with initial terms $1$, and then common ratios of $-x$ and $-x^2$ respectively. The converge for $|x|<1$ and $|x^2|<1$ respectively. These both have radius of convergence $\rho=1$, so the radius of convergence of their product is $\rho=1$.
|
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|
What's the Maclaurin Series of $f(x)=\frac{1}{(1-x)^2}$? This function seemed to be pretty much straight forward, but my solution is incorrect.
I have two questions:
1. Where did I make a mistake?
2. I learned that there are shortcuts for finding a series (substitution / multiplication / division / differentiation / integration of both sides). Is there something that I can apply here?
$$f(x)=\frac{1}{(1-x)^2} \qquad f(0)=1$$
$$f'(x)=\frac{2}{(1-x)^3} \qquad f'(0)=2$$
$$f''(x)=\frac{6}{(1-x)^4} \qquad f''(0)=6$$
$$f'''(x)=\frac{24}{(1-x)^5} \qquad f'''(0)=24$$
$$f''''(x)=\frac{120}{(1-x)^6} \qquad f''''(0)=120$$
The series is then
$$1+\frac{2x}{2!}+\frac{6x^2}{3!}+\frac{24x^3}{4!}+\frac{120x^4}{5!}$$
and when simplified is
$$1+x+x^2+x^3+x^4$$
But the correct answer is
$$1+2x+3x^2+4x^3+5x^4 $$
|
As far as shortcuts go, there are two you might find interesting.
You've probably seen the geometric series several times:
$$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots $$
Now the first idea is that your function $f(x)$ is the square of this one, so:
$$ f(x) = (1 + x + x^2 + x^3 + \ldots )^2 $$
If the indicated multiplication is carried out, one gets:
$$ f(x) = 1 + 2x + 3x^2 + 4x^3 + \ldots $$
The second idea is to differentiate the geometric series formula:
$$ \frac{d}{dx}(\frac{1}{1-x}) = \frac{1}{(1-x)^2} = f(x) $$
Thus the term-by-term differentiation of the geometric series also gets:
$$ f(x) = 1 + 2x + 3x^2 + 4x^3 + \ldots $$
|
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|
mathematical induction ($(1+x)^n\ge1+nx+n(n-1)x^2/2$) Suppose that $x > 0$ and let $n \geq 2$ be a positive integer. Prove that $(1 + x)^n \geq 1 + nx + \frac{n(n-1)}{2}x^2$
So for the base case, I have $x=1$, but that really is not getting me anywhere.
Would it be better to manipulate $n$? Is there a way to know which one to manipulate?
|
HINT:
Let $(1+x)^m\ge \{1+mx+\frac{m(m-1)}2x^2\}$
So, $(1+x)^{m+1}=(1+x)(1+x)^m \ge (1+x)\{1+mx+\frac{m(m-1)}2x^2\}$
$=1+(m+1)x+x^2\{\frac{m(m-1)}2+m\}+x^3\frac{m(m-1)}2\ge \{1+(m+1)x+\frac{m(m+1)}2x^2\}$ as $x>0$ and $m\ge2$
Now, for $m=2,1+2x+x^2-\{1+2x+x^2\}=0\implies (1+x)^2\ge \{1+2x+\frac{2(2-1)}2x^2\}$
|
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|
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? In this recent answer to this question by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$
I would like to know if this result can be generalized to other triples of
natural numbers.
Question. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$
For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$
$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$
and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$
A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.
For $(2)$ the very same idea yields
$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$
and
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$
I tried to solve this system for $a,b$ but since the solution is of the form
$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$
where $x$ satisfies the cubic equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach.
Is this problem solvable, at least partially?
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?
|
There is an infinite family of solutions coming from the idea $(2+\sqrt 3)^3=26+15\sqrt 3$. We can form $(2+\sqrt 3)^{3n}$ and find another solution. The next one is $(2+\sqrt 3)^6=1351+780 \sqrt 3$ and $(1351+780\sqrt 3)^{(1/3)}+(1351-780\sqrt 3)^{(1/3)}=14$ There is a recurrence, if $(a,b)$ is a solution, the next is $(26a+45b,15a+26b)$ and so we get triplets $(1351,780,14),(70226,40545,52),(3650401,2107560,194),(189750626,109552575,724)
9863382151,5694626340,2702)$
and on. I have not shown that these are all the solutions.
|
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|
I cannot find the last factor of this expression? I'm supposed to factor $x^8-y^8$ (the exponents are 8 for both if it is too difficult to see) as completely as possible. It is easy to factor this to $(x+y)(x-y)(x^2+y^2)(x^4+y^4)$. However, the book says "(Hint: there are 5 factors. Note that we sat real coefficients, not just integers)". What is this elusive 5th factor? Thanks!
|
Hint $\rm\,\ x^4 + y^4 = (x^2+y^2)^2 - (\sqrt{2}\, xy)^2 =\, \cdots\ $ (factor the difference of squares)
Remark $\ $ One can perform analogous factorizations in more exotic cases, and these prove very useful for factor factoring integers having such forms. $ $ For example, $ $ Aurifeuille, Le Lasseur and Lucas $ $ discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x)\, =\, C_n(x)^2\! - n\, x\, D_n(x)^2\;$. These play a role in factoring integers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations.
$$\begin{eqnarray}
\rm x^4 + 2^2 &=\,&\rm (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\
\rm \frac{x^6 + 3^2}{x^2 + 3} &=\,&\rm (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\
\rm \frac{x^{10} - 5^5}{x^2 - 5} &=\,&\rm (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\
\rm \frac{x^{12} + 6^6}{x^4 + 36} &=\,&\rm (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\
\end{eqnarray}$$
For more on this and related topics see Sam Wagstaff's introduction to the Cunningham Project.
|
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|
Complex definite integrals/residue I am trying to evaluate this integral:
$$
\int_0^\pi \frac{d\theta}{(a + b \cos \theta)^2}
$$
for $0 < b < a$. I feel like the way I'm trying to solve [using $\cos t=(e^{it}+e^{-it})/2$ and $z=e^{it}$, but it doesn't seem to come out right. Can anyone help me step by step?
|
From Taylor series of $\dfrac1{(a+bx)^2}$, we have
$$\dfrac1{(a+b \cos(t))^2} = \sum_{k=0}^{\infty} \dfrac{(k+1)(-b)^k}{a^{k+2}} \cos^{k}(t)$$
Now $$\int_0^{2\pi} \cos^k(t) dt = 0 \text{ if $k$ is odd}$$
We also have that
$$\color{red}{\int_0^{2\pi}\cos^{2k}(t) dt = \dfrac{(2k-1)!!}{(2k)!!} \times \pi = \pi \dfrac{\dbinom{2k}k}{4^k}}$$
Hence,
$$I=\int_0^{2 \pi}\dfrac{dt}{(a+b \cos(t))^2} = \sum_{k=0}^{\infty} \dfrac{(2k+1)b^{2k}}{a^{2k+2}} \int_0^{2\pi}\cos^{2k}(t) dt = \dfrac{\pi}{a^2} \sum_{k=0}^{\infty}(2k+1) \left(\dfrac{b}{2a}\right)^{2k} \dbinom{2k}k$$
Now from Taylor series, we have
$$\color{blue}{\sum_{k=0}^{\infty}(2k+1) x^{2k} \dbinom{2k}k = (1-4x^2)^{-3/2}}$$
Hence,
$$\color{green}{I = \dfrac{\pi}{a^2} \cdot \sum_{k=0}^{\infty}(2k+1) \left(\dfrac{b}{2a}\right)^{2k} \dbinom{2k}k = \dfrac{\pi}{a^2} \cdot \left(1-\left(\dfrac{b}a\right)^2 \right)^{-3/2} = \dfrac{\pi a}{\left(a^2-b^2 \right)^{3/2}}}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$? Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$
I can't seem to wrap my head around that.
|
We proceed by induction on $n$. The base case for $n = 0$ is $0^2 - \frac{0^2}{2} = 0 =\frac{0^2}{2}$, which follows from the fact that $0$ is the additive identity in $\mathbb{Q}$. Suppose the claim holds for some $n = k$. Now
\begin{aligned}
(k+1)^2 - \frac{(k+1)^2}{2} &= k^2 + 2k + 1 - \frac{k^2}{2} - \frac{2k}{2} - \frac{1}{2} \\
&= \color{blue}{\left ( k^2 - \frac{k^2}{2} \right )} + k + \frac{1}{2} \\
&= \color{blue}{\frac{k^2}{2}} + k + \frac{1}{2} \quad \color{blue}{\text{(hypothesis)}} \\
&= \frac{k^2 + 2k + 1}{2} \\
&= \frac{(k+1)^2}{2}.
\end{aligned}
This concludes the inductive step, so our claim holds for all $n \in \mathbb{N}$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find this
$$\ln{2}<\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\dfrac{\sqrt{2}}{2}$$
following is my some nice methods,
use this inequality
$$\dfrac{x-y}{\ln{x}-\ln{y}}>\sqrt{xy},x>y$$
then we let
$x=2,y=1$
so $$\ln{2}<\dfrac{\sqrt{2}}{2}$$
solution 2:
since
$$\dfrac{1}{n+1}\le\dfrac{1}{2}\cdot\dfrac{3}{4}\cdots\dfrac{2n-1}{2n}$$
then
$$\ln{2}=\sum_{n=0}^{\infty}\dfrac{1}{(n+1)2^{n+1}}<\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^22^{3n+1}}=\dfrac{1}{\sqrt{2}}$$
solution 3
since
$$(1+\sqrt{2})^2(t+1)-(t+1+\sqrt{2})^2=t(1-t)>0$$
so
$$\ln{2}=\int_{0}^{1}\dfrac{1}{t+1}dt<\int_{0}^{1}\left(\dfrac{1+\sqrt{2}}{t+1+\sqrt{2}}\right)^2dt=\dfrac{\sqrt{2}}{2}$$
solution 4:
$$\ln{2}=\dfrac{3}{4}-\dfrac{1}{4}\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(2n+1)}<\dfrac{3}{4}-\dfrac{1}{4}\left(\dfrac{1}{1\times 2\times 3}-\dfrac{1}{2\times 3\times 5}\right)=\dfrac{7}{10}<\dfrac{\sqrt{2}}{2}$$
solution 5
$$\dfrac{1}{\sqrt{2}}-\ln{2}=\sum_{n=1}^{\infty}\dfrac{\sqrt{2}}{(4n^2-1)(17+2\sqrt{2})^n}>0$$
But $$\ln{2}>\left(\dfrac{2}{5}\right)^{\frac{2}{5}}$$ I can't have this nice solution
Thank you everyone can help.
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There is a remarkable series converging to $\ln(2)$ on Wikipedia: $$\ln(2)=\sum_{k=0}^\infty\frac{2}{2k+1}\left(\frac{7}{31^{2k+1}}+\frac{3}{161^{2k+1}}+\frac{5}{49^{2k+1}}\right)$$ although I don't see a source for why this is true.
Summing just the $k=0$ and $k=1$ terms, gives
$$\begin{align}
\ln(2)&>\frac{29558488681560}{42643891494953}\\
\ln(2)&>0.693147\\
\left(\ln(2)\right)^5&>(0.693147)^5\\
\left(\ln(2)\right)^5&>0.160002\\
\left(\ln(2)\right)^5&>\frac{4}{25}\\
\ln(2)&>\left(\frac{2}{5}\right)^{2/5}\\
\end{align}$$
|
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"url": "https://math.stackexchange.com/questions/380302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "95",
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"answer_id": 3
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|
Integral of $\sin x \cdot \cos x$ I've found 3 different solutions of this integral. Where did I make mistakes? In case there is no errors, could you explain why the results are different?
$ \int \sin x \cos x dx $
1) via subsitution $ u = \sin x $
$ u = \sin x; du = \cos x dx \Rightarrow \int udu = \frac12 u^2 \Rightarrow \int \sin x \cos x dx = \frac12 \sin^2 x $
2) via subsitution $ u = \cos x $
$ u = \cos x; du = -\sin x dx \Rightarrow -\int udu = -\frac12 u^2 \Rightarrow \int \sin x \cos x dx = -\frac12 \cos^2 x = -\frac12 (1 - \sin^2 x) = -\frac12 + \frac12 \sin^2 x $
3) using $ \sin 2x = 2 \sin x \cos x $
$ \int \sin x \cos x dx = \frac12 \int \sin 2x = \frac12 (- \frac12 \cos 2x) = - \frac14 \cos 2x = - \frac14 (1 - 2 \sin^2 x) = - \frac14 + \frac12 \sin^2 x $
So, we have:
$$ \frac12 \sin^2 x \neq -\frac12 + \frac12 \sin^2 x \neq - \frac14 + \frac12 \sin^2 x $$
|
Antiderivatives are only unique up to adding a constant ('of integration'). If you were to stick limits in your integrals then you'd always get the same number.
|
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|
Mellin transform of $\sin x$ aka $\int^{\infty}_0 x^{s-1}\sin x dx $ I am trying to find the Mellin transform of $\sin x $, put in other words to solve:
$$\int^{\infty}_0 x^{s-1}\sin x \mathrm{d} x $$
And I know that the answer is:
$$\Gamma(s) \sin \left(\frac{\pi s}{2}\right)$$
From several tables on the internet but I couldn't find any justification.
How can this identity be proven?
|
Alternatively, the Mellin transform for $\sin x$ can be found by employing the following useful property for the Laplace transform:
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$
Noting that
$$\mathcal{L} \{\sin x\}(t) = \frac{1}{1 + t^2},$$
and
$$\mathcal{L}^{-1} \left \{\frac{1}{x^{1-s}} \right \} (t)= \frac{1}{\Gamma (1 - s)} \mathcal{L}^{-1} \left \{\frac{\Gamma (1 - s)}{x^{1-s}} \right \} (t) = \frac{t^{-s}}{\Gamma (1 - s)},$$
then
\begin{align}
\mathcal{M} \{\sin x\} &= \int_0^\infty \sin x \cdot \frac{1}{x^{1 - s}} \, dx\\
&= \int_0^\infty \mathcal{L} \{\sin x\} (t) \cdot \mathcal{L}^{-1} \left \{\frac{1}{x^{1 - s}} \right \} (t) \, dt\\
&= \frac{1}{\Gamma (1 - s)} \int_0^\infty \frac{t^{-s}}{1 + t^2} \, dt.
\end{align}
Setting $u = t^2$, one has
\begin{align}
\mathcal{M} \{\sin x\} &= \frac{1}{2 \Gamma (1 - s)} \int_0^\infty \frac{u^{-\frac{s}{2} - \frac{1}{2}}}{1 + u} \, du\\
&= \frac{1}{2 \Gamma (1 - s)} \operatorname{B} \left (\frac{1}{2} - \frac{s}{2}, \frac{1}{2} + \frac{s}{2} \right ) \tag1\\
&= \frac{1}{2 \Gamma (1 - s)} \Gamma \left (\frac{1}{2} - \frac{s}{2} \right ) \Gamma \left (\frac{1}{2} + \frac{s}{2} \right ) \tag2\\
&= \frac{1}{2 \Gamma (1 - s)} \Gamma \left [1 - \left (\frac{1}{2} + \frac{s}{2} \right ) \right ] \Gamma \left (\frac{1}{2} + \frac{s}{2} \right )\\
&= \frac{1}{2 \Gamma (1 - s)} \frac{\pi}{\sin \left (\frac{\pi}{2} + \frac{\pi s}{2} \right )} \tag3\\
&= \frac{1}{2 \Gamma (1 - s)} \frac{\pi}{\cos \left (\frac{\pi s}{2} \right )}\\
&= \frac{\Gamma (s) \sin (\pi s)}{2 \pi} \cdot \frac{\pi}{\cos \left (\frac{\pi s}{2} \right )} \tag4\\
&=\frac{\Gamma (s) \sin \left (\frac{\pi s}{2} \right ) \cos \left (\frac{\pi s}{2} \right )}{\cos \left (\frac{\pi s}{2} \right )}\\
&= \Gamma (s) \sin \left (\frac{\pi s}{2} \right )
\end{align}
This is valid for $-1 < s < 1$.
Explanation
(1) Using $\operatorname{B} (x,y) = \displaystyle{\int_0^\infty \frac{t^{x - 1}}{(1 + t)^{x + y}} \, dt}$.
(2) Using $\operatorname{B}(x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$.
(3) Using the reflection formula for the gamma function: $\Gamma (1 - z) \Gamma (z) = \dfrac{\pi}{\sin (\pi z)}$.
(4) Again using the reflection formula for the gamma function.
|
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|
if $ab=cd$ then $a+b+c+d $ is composite Let $a,b,c,d$ be natural numbers with $ab=cd$.
Prove that $a+b+c+d$ is composite.
I have my own solution for this (As posted) and i want to see if there is any other good proofs.
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From $ab=cd$ you have $$(a+b)^2-(a-b)^2=(c+d)^2-(c-d)^2\Rightarrow(a+b)^2-(c+d)^2=(a-b)^2-(c-d)^2$$ Hence we have $$(a+b+c+d)(a+b-c-d)=(a-b+c-d)(a-b-c+d)$$ Now note that $|a+b+c+d|>|a-b+c-d|$ and $|a-b-c+d|$. If $(a+b+c+d)$ was prime then it must divide one of $(a-b+c-d)$ or $(a-b-c+d)$, which is not possible.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Integral Of $\int\sqrt{\frac{x}{x+1}}dx$ I want to solve this integral
$$\int\sqrt{\frac{x}{x+1}}dx$$
And think about:
1) $t=\frac{x}{x+1}$
2) $dt = (\frac{1}{x+1} - \frac{x}{(x+1)^2})dx$
Now I need your advice! Thanks!
|
put
$$\sqrt{\frac{x}{x+1}}=t,\quad x=\frac{t^2}{1-t^2},\quad dx=\frac{2t}{(1-t^2)^2}\,\,dt$$
So
\begin{align*}
\int\sqrt{\frac{x}{x+1}}\,\,dx &=\int t\cdot \frac{2t}{(1-t^2)^2}\,\,dt=2\int \frac{ t^2}{(1-t^2)^2}\,\,dt\\
&=2\int \frac{ t^2+1-1}{(1-t)^2(1+t)^2}\,\,dt\\
&=2\int \frac{ (t-1)(t+1)}{(1-t)^2(1+t)^2}\,\,dt-2\int \frac{1}{(1-t)^2(1+t)^2}\,\,dt\\
&=-2\int \frac{1}{ 1-t }\,\,dt-2\int \frac{1}{(1-t)^2(1+t)^2}\,\,dt\\
&=-2\ln|1-t|-2\int \frac{A}{1-t}+\frac{B}{(1-t)^2}+\frac{C}{1+t}+\frac{D}{(1+t)^2}\,\,dt\\
&=-2\ln|1-t|-2A\ln|1-t|-\frac{B}{ 1-t} -C\ln|1+t|+\frac{D}{ 1+t }
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Another trigonometric proof...? ...sigh..another problem how shall I prove the following?
$$ {\cot A\over1- \tan A} + {\tan A \over 1- \cot A} = 1 + \tan A + \cot A$$
so what now? the following's what I've done:
$$\cot A - \cot^2 A + \tan A- \tan^2 A \over 2 - \tan A - \cot A$$
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Let $\tan A=a\implies \cot A=\frac1a$
So, the problem reduces to
$$\frac{\frac1a}{1-a}+\frac a{1-\frac1a}=\frac1{a(1-a)}+\frac{a^2}{a-1}$$
$$=\frac1{a(1-a)}-\frac{a^2}{1-a}=\frac{1-a^3}{a(1-a)}=\frac{1+a+a^2}a=a+\frac1a+1$$
Alternatively,
$$\frac{\tan A}{1-\cot A}=\frac{\tan^2A}{\tan A-1} (\text{ multiplying the numerator & the denominator by }\tan A) $$
$$\implies \frac{\tan A}{1-\cot A}=-\frac{\tan^2A}{1-\tan A}$$
$$\text{So,} {\cot A\over1- \tan A} + {\tan A \over 1- \cot A}={\cot A\over1- \tan A} -\frac{\tan^2A}{1-\tan A}=\frac{\cot A-\tan^2A}{1-\tan A}=\frac{1-\tan^3A}{\tan A(1-\tan A)} (\text{ multiplying the numerator & the denominator by }\tan A) $$
$$=\frac{1+\tan A+\tan^2A}{\tan A}\text{ (assuming }1-\tan A\ne0)$$
$$=\cot A+1+\tan A$$
|
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|
Evaluate a sum with binomial coefficients: $\sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$ $$\text{Find} \ \ \sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$$
I expanded the binomial coefficients within the sum and got $$\binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2$$
What does this equal to? I think this can help me evaluate the original sum.
|
This can also be done using a basic complex variables technique.
Start as in @robjohn's answer. Suppose we seek to evaluate
$$\sum_{k=0}^n (-1)^k k {n\choose k}^2
= \sum_{k=1}^n {n\choose k} (-1)^k k {n\choose k}
\\= \sum_{k=1}^n {n\choose k} (-1)^k k \frac{n}{k} {n-1\choose k-1}
= n \sum_{k=1}^n {n\choose k} (-1)^k {n-1\choose k-1}.$$
Introduce the integral representation
$${n-1\choose k-1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^k} (1+z)^{n-1} \; dz.$$
This gives the following integral for the sum
$$\frac{n}{2\pi i}
\int_{|z|=\epsilon}
\sum_{k=1}^n {n\choose k} (-1)^k \frac{1}{z^k} (1+z)^{n-1} \; dz
\\= \frac{n}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{n-1}
\sum_{k=1}^n {n\choose k} (-1)^k \frac{1}{z^k} \; dz
\\= \frac{n}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{n-1}
\left(-1 + \left(1-\frac{1}{z}\right)^n \right) \; dz$$
We may drop the $-1$ because it participates in a product that is
entire. This leaves
$$\frac{n}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{n-1} \frac{(z-1)^n}{z^n} \; dz
\\ = \frac{(-1)^n n}{2\pi i}
\int_{|z|=\epsilon} (1-z) (1+z)^{n-1} \frac{(1-z)^{n-1}}{z^n} \; dz
\\ = \frac{(-1)^n n}{2\pi i}
\int_{|z|=\epsilon} (1-z) \frac{(1-z^2)^{n-1}}{z^n} \; dz.$$
It follows that the value of the sum is given by
$$(-1)^n n [z^{n-1}] (1-z) (1-z^2)^{n-1}.$$
For $n$ even the $z$ from $1-z$ participates and for $n$ odd the one
participates.
We have for $n$ even the result
$$(-1)^n n \times (-1) (-1)^{(n-2)/2} {n-1\choose (n-2)/2}
= n \times (-1)^{n/2} {n-1\choose n/2-1}
\\= n \times (-1)^{n/2} {n-1\choose n/2}$$
and for $n$ odd
$$(-1)^n n \times (-1)^{(n-1)/2} {n-1\choose (n-1)/2}
= n \times (-1)^{(n+1)/2} {n-1\choose (n-1)/2}.$$
Joining these two terms we obtain
$$n \times (-1)^{\lceil n/2 \rceil} {n-1\choose \lfloor n/2 \rfloor}.$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
|
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Integrating a partial fraction with multiple quadratic denominators When integrating a real rational fraction, you first break it into partial fractions. You then end up with fractions with linear denominators $\frac{A}{(x-b)^n}$, which are easy. You also end up with quadratic denominators $\frac{Ax+B}{(x+ax+b)^n}$. I cannot find a straight forward explanation for solving that second general case anywhere online. I might be missing an obvious google result, but I just cannot find one. There are of course multiple SE questions about this, but they're about specific problems. I'd like to ask about the general case:
What is $$\int \frac{Ax+B}{(x^2+ax+b)^n} dx$$ ?
I can reduce it to the case $\frac{1}{(x^2+ax+b)^n}$ easily. Then I tried completing the square, thus obtaining the form (where a, b, and c are different constants to the ones above):
$$\int \frac{1}{((x+b)^2+c)^n} dx=\frac{1}{\sqrt c}\int \frac{\frac{1}{\sqrt c}}{((\frac{x+b}{\sqrt c})^2+1)^n} dx
=\frac{1}{\sqrt c}\int \frac{1}{\sqrt c}\arctan'(\frac{x+b}{\sqrt c})^n dx$$
$$=\frac{1}{\sqrt c}\int \arctan'(\frac{x+b}{\sqrt c})^n d(\frac{x+b}{\sqrt c})$$
Setting $u=\frac{x+b}{\sqrt c}$ for readability, I experimented with:
$$\frac{1}{\sqrt c}\int \arctan'(u)^n du=\frac{1}{\sqrt c}\int \arctan'(u)^{n-1} d\arctan(u)$$
And that's a dead-end, for me.
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As André Nicolas describes, the end result will come down to generating some sort of reduction formula. With the techniques we learn a little way into integral calculus,
$$\int \frac{Ax+B}{(x^2 + ax + b)^n} \ dx $$
can be "stripped down" using completion of squares to
$$ \frac{A}{2}\int \ \frac{[2x + a] }{( \ [x + \frac{a}{2}]^2 + [b - \frac{a^2}{4}] \ )^n} \ dx \ \ + \ \ \int \ \frac{ B - \frac{Aa}{2} }{( \ [x + \frac{a}{2}]^2 + [b - \frac{a^2}{4}] \ )^n} \ dx \ , $$
so that the real work comes down to integrating the part with the constant in the numerator,
$$ \int \frac{C}{( \ u^2 + \beta^2 )^n} \ du \ \ , \text{with} \ \ u \ = \ x + \frac{a}{2} \ \ , \ \ \beta^2 \ = \ b - \frac{a^2}{4} \ \ . \ \ \ \mathbf{[ 1 ] }$$
This is what suggests using a "chain" of integrations-by-parts with a trig substitution.
EDITS (2/28/14) -- Although this answer was accepted recently, I'd been meaning to return to it at some point, in part to correct an algebra error, and in part to elaborate on the integration chain [and later to fix a different mistake made in the "wee hours"].
The sum of squares in the denominators calls for a "tangent substitution", $ \ \tan \theta = \frac{u}{\beta} \ , $ leading to
$$ \longrightarrow \ \ C \ \int \ \frac{\beta \ \sec^2 \theta \ \ d\theta}{( \ \beta^2 \sec^2 \theta \ )^n} \ \ = \ \ \frac{C}{\beta^{2n-1}} \int \ \cos^{2n-2} \theta \ \ d\theta \ \ . \ \ \ \mathbf{[ 2 ] } $$
To make the integration-by-parts a bit more readable, I am switching the exponent of cosine to $ \ 2m \ $ for the present; the reduction formula is obtained from
$$ \int \ \cos^{2m} \theta \ \ d\theta \ = \ \int \ (\cos^{2m-1} \theta) \ \cdot \ (\cos \theta \ \ d\theta) \ = \ \int \ (\cos^{2m-1} \theta) \ \ d( \sin \theta \ ) $$
$$ = \ \sin \theta \ \cos^{2m-1} \theta \ \ - \ \int \ (\sin \theta) \ \cdot \ (2m-1) \ (\cos^{2m-2} \theta \ [-\sin \theta \ ] \ \ d\theta) $$
$$ = \ \sin \theta \ \cos^{2m-1} \theta \ \ + \ \ (2m-1) \int \ \sin^2 \theta \ \ \cos^{2m-2} \theta \ \ d\theta $$
$$ = \ \sin \theta \ \cos^{2m-1} \theta \ \ + \ \ (2m-1) \int \ \ (1 - \cos^2 \theta) \ \cos^{2m-2} \theta \ \ d\theta $$
$$ = \ \sin \theta \ \cos^{2m-1} \theta \ \ + \ \ (2m-1) \int \ \cos^{2m-2} \theta \ \ d\theta \ \ + \ \ (2m-1)\int \ \cos^{2m} \theta \ \ d\theta $$
$$ \Rightarrow \ \ 2m \ \int \ \cos^{2m} \theta \ \ d\theta \ = \ \sin \theta \ \cos^{2m-1} \theta \ \ + \ \ (2m-1) \int \ \cos^{2m-2} \theta \ \ d\theta $$
$$ \Rightarrow \ \ \int \ \cos^{2m} \theta \ \ d\theta \ = \ \left( \frac{1}{2m} \right) \sin \theta \ \cos^{2m-1} \theta \ \ + \ \left( \frac{2m - 1}{2m} \right) \int \ \cos^{2m-2} \theta \ \ d\theta \ \ \ . $$
Applying this to the integral 2 above produces
$$ \int \frac{C}{( \ u^2 + \beta^2 )^n} \ du $$
$$ = \ \ \frac{C}{\beta^{2n-1}} \ \left[ \ \left( \frac{1}{2n - 2} \right) \sin \theta \ \cos^{2n-3} \theta \ \ + \ \left( \frac{2n - 3}{2n - 2} \right) \int \ \cos^{2n-4} \theta \ \ d\theta \ \right] $$
$$ = \ \ C \ \left[ \ \left( \frac{1}{2n - 2} \right) \frac{u}{\beta^2 \ ( \ u^2 + \beta^2 )^{n-1} } \ + \ \left( \frac{[2n - 3]}{[2n - 2] \ [2n-4]} \right) \frac{u}{\beta^4 \ ( \ u^2 + \beta^2 )^{n-2} } \ + \ \ldots \ + \ \left( \frac{[2n - 3] \cdot \ \ldots \ \cdot 3}{[2n - 2] \ [2n-4] \cdot \ \ldots \ \cdot 4} \right) \frac{u}{\beta^{2n-2} \ ( \ u^2 + \beta^2 ) } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \ \left( \frac{[2n - 3] \cdot \ \ldots \ \cdot 1}{[2n - 2] \ [2n-4] \cdot \ \ldots \ \cdot 2} \right) \arctan(\frac{u}{\beta} ) \ \right] \ + \ K \ \ . $$
$$ $$
Here's an example run on WolframAlpha (with all terms placed over a single denominator, and the arctangent term appearing in the penultimate position):
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"timestamp": "2023-03-29T00:00:00",
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|
How can this $T(n) = T(n-1)+T(n-2)+3n+1$ non homogenous recurrence relation be solved How are can the above recurrence relation be solved?
I've reached here:
$(x^{2}-x-1)(x-3)^2(x-1)$
And then here:
$$a_n = l_1 \cdot (x_1)^n+l_2 \cdot (x_2)^n+l_3 \cdot (x_3)^n+l_4\cdot n \cdot (x_3)^n+l_5\cdot (x_4)^n$$
And we are given that these: $T(0) = 2, T(1) = 3$
So, for $n = 0$ and $n = 1$ we get two equations but we need 3 more, yet we don't have any more constants.
|
Modify $T$ as follows:
$$
\begin{aligned}
T(n) &= T(n-1) + T(n-2) + 3n + 1\\
T(n) + 3n &= T(n-1) + T(n-2) + 6n + 1\\
T(n) + 3n &= T(n-1) + 3n + T(n-2) + 3n + 1\\
T(n) + 3n &= T(n-1) + 3(n-1) + T(n-2) + 3(n-2) + 10\\
T(n) + 3n + 10 &= T(n-1) + [3(n-1) + 10] + T(n-2) + [3(n-2) + 10 ]
\end{aligned}
$$
So, $S(n) = T(n) + 3n + 10$ satisfies the Fibonacci recurrence.
|
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|
Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$ I'm trying to solve this recurrence relation:
$$
a_n = \begin{cases}
0 & \mbox{for } n = 0 \\
5 & \mbox{for } n = 1 \\
6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
\end{cases}
$$
I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - 6x + 5x^2} + \frac{x^3}{(1-x)(1-6x+5x^2)} =
\frac{31x - 24x^2}{(x-1)(x-5)} + \frac{x^3}{(1-x)(x-1)(x-5)}
$$
(I'm not sure if that's right)
and its partial fractions decomposition looks like:
$$
A = \left(\frac{-7}{4} \cdot \frac{1}{x-1} - \frac{445}{4} \cdot \frac{1}{x-5}\right) +
\left( \frac{39}{16} \cdot \frac{1}{x-5} + \frac{3}{4} \cdot \frac{1}{(x-1)^2} - \frac{375}{16} \cdot \frac{1}{x-5} \right)
$$
(again - I'm not sure if it's ok)
I'm stuck here... From solutions I know that I should get:
$$
a_n = \frac{-21}{16} - \frac{1}{4}n + \frac{21}{16}5^n
$$
but I have no idea how it's solved... I hope somebody can help me (I spend more than 3h trying to solve this myself...)
|
Use Wilf's technique from "generatingfunctionology". Define $A(z) = \sum_{n \ge 0} a_n z^n$, and write:
$$
a_{n + 2} = 6 a_{n + 1} - 5 a_n + 1 \qquad a_0 = 0, a_1 = 5
$$
Multiply by $z^n$ and add for $n \ge 0$, which gives:
$$
\frac{A(z) - a_0 - a_1 z}{z^2} = 6 \frac{A(z) - a_0}{z} - 5 A(z) + \frac{1}{1 - z}
$$
Solving for $A(z)$ gives:
$$
A(z) = \frac{5 z - 4 z^2}{1 - 7 z + 11 z^2 - 5 z^3}
= \frac{21}{16} \cdot \frac{1}{1 - 5 z}
- \frac{17}{16} \cdot \frac{1}{1 - z}
- \frac{1}{4} \cdot \frac{1}{(1 - z)^2}
$$
Remember the expansions:
$$
(1 - u)^{-m} = \sum_{k \ge 0} \binom{-m}{k} (-u)^k
= \sum_{k \ge 0} \binom{k + m - 1}{m - 1} u^k
$$
(the binomial coefficient is just an $m-1$-degree polynomial in $k$) and you are all set:
$$
a_n = \frac{21}{16} \cdot 5^n - \frac{17}{16} - \frac{1}{4} \binom{n + 1}{1}
= \frac{21 \cdot 5^n - 4 n - 21}{16}
$$
Maxima's help with the algebra is gratefully acknowledged.
|
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|
Integral of $\int \frac{x^4+2x+4}{x^4-1}dx$ I am trying to solve this integral and I need your suggestions.
$$\int \frac{x^4+2x+4}{x^4-1}dx$$
Thanks
|
welcome to math.stackexchange this question were answered already.
Here is the link
use polynomial division, we get $$\int \frac{x^4+2x+4}{x^4-1} dx = \int 1 + \frac{2x+5}{(x^2 - 1)(x^2 + 1)}dx
= \int 1 + \frac{2x+5}{(x+1)(x-1)(x^2+1)} dx $$
Expressing this as partial fractions, we need only find $A, B, C$
$$= \int \left(1 + \frac{A}{x+1} + \frac B{x-1} +\frac{CX+D}{x^2 + 1}\right)\,dx$$
|
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|
Solve $(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$ I got the square root of 14 and 11 but the answer book states that these answers are wrong. Can someone help me? I used this formula to find the individual roots
$x = -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^2 - q}$
|
Given:
$(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$
Let a = $(x^2+5)$ which gives us a simple quadratic trinomial of the form $ax^2+bx+c=0.$
$a^2-15a+54=0$
Factoring this we get:
$(a-9)(a-6)=0$
$a-9=0$
$a-6=0$
$\implies a=9$ or $a=6$
$x^2+5=9$
$x^2+5=6$
$\implies x = \pm2$ or $\pm 1$
You can check that these solutions are correct by simply plugging values back in to see if you get $0$.
$a^2-15a+54=0 \implies$ $(9)^2 - 15(9) + 54 = 0 \implies 81-135+54=0$
|
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|
What type of Hypergeometric series is this? I am trying to find a closed form for the series
$$ \sum^\infty_{n=0} \frac{1}{n!} \frac{1}{n+1}(-z)^n {}_2F_2\left(m,n+1;\frac{1}{2},n+2; b z\right)$$
$m$ is a nonzero positive integer, and $b$, $z$ are positive real numbers. I to rewrite the sum as
$$ \sum^\infty_{n=0} \sum^\infty_{q=0}
\frac{1}{n!} \frac{1}{q!}
\frac{(m)_q}{(\frac{1}{2})_q}
\frac{(1)_{q+n}}{(2)_{q+n}}
(-z)^n (b z)^q$$
any idea what type of multi-variable hypergeometric function is the last equation?
|
$\sum\limits_{n=0}^\infty\dfrac{1}{n!}\dfrac{1}{n+1}(-z)^n{}_2F_2\left(m,n+1;\dfrac{1}{2},n+2;bz\right)$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(m)_k(n+1)_k(-1)^nb^kz^{n+k}}{\left(\dfrac{1}{2}\right)_k(n+2)_k(n+1)!k!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(m)_k(-1)^nb^kz^{n+k}}{\left(\dfrac{1}{2}\right)_kn!k!(n+k+1)}$
$=\dfrac{1}{z}\int_0^z\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(m)_k(-1)^nb^kz^{n+k}}{\left(\dfrac{1}{2}\right)_kn!k!}~dz$
$=\dfrac{1}{z}\int_0^z\sum\limits_{k=0}^\infty\dfrac{(m)_kb^kz^ke^{-z}}{\left(\dfrac{1}{2}\right)_kk!}~dz$
$=-\dfrac{1}{z}\left[\sum\limits_{k=0}^\infty\sum\limits_{n=0}^k\dfrac{(m)_kb^kz^ne^{-z}}{\left(\dfrac{1}{2}\right)_kn!}\right]_0^z$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$=\dfrac{1}{z}\sum\limits_{k=0}^\infty\dfrac{(m)_kb^k}{\left(\dfrac{1}{2}\right)_k}-\dfrac{1}{z}\sum\limits_{k=0}^\infty\sum\limits_{n=0}^k\dfrac{(m)_kb^kz^ne^{-z}}{\left(\dfrac{1}{2}\right)_kn!}$
$=\dfrac{1}{z}{}_2 F_1\left(1,m;\dfrac{1}{2};b\right)-\dfrac{1}{z}\sum\limits_{n=0}^\infty\sum\limits_{k=n}^\infty\dfrac{(m)_kb^kz^ne^{-z}}{\left(\dfrac{1}{2}\right)_kn!}$
$=\dfrac{1}{z}{}_2 F_1\left(1,m;\dfrac{1}{2};b\right)-\dfrac{1}{z}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(m)_{n+k}b^{n+k}z^ne^{-z}}{\left(\dfrac{1}{2}\right)_{n+k}n!}$
$=\dfrac{1}{z}{}_2 F_1\left(1,m;\dfrac{1}{2};b\right)-\dfrac{e^{-z}}{z}\Phi_1\left(m,1,\dfrac{1}{2};b,bz\right)$ (according to http://en.wikipedia.org/wiki/Humbert_series)
|
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|
Evaluation of a specific determinant.
Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.
$$A_2
\begin{pmatrix} 1& 1\\
1& 2
\end{pmatrix};
A_3 = \begin{pmatrix} 1& 1& 1\\
1& 2& 2\\
1& 2& 3
\end{pmatrix};
A_4 = \begin{pmatrix} 1& 1& 1& 1\\
1& 2& 2& 2\\
1& 2& 3& 3\\
1& 2& 3& 4
\end{pmatrix};
A_5 = \begin{pmatrix} 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2\\
1& 2& 3& 3& 3\\
1& 2& 3& 4& 4\\
1& 2& 3& 4& 5
\end{pmatrix}$$
$$A_6 = \begin{pmatrix} 1& 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2& 2\\
1& 2& 3& 3& 3& 3\\
1& 2& 3& 4& 4& 4\\
1& 2& 3& 4& 5& 5\\
1& 2& 3& 4& 5& 6
\end{pmatrix};
A_7 = \begin{pmatrix} 1& 1& 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2& 2& 2\\
1& 2& 3& 3& 3& 3& 3\\
1& 2& 3& 4& 4& 4& 4\\
1& 2& 3& 4& 5& 5& 5\\
1& 2& 3& 4& 5& 6& 6\\
1& 2& 3& 4& 5& 6& 7
\end{pmatrix}
$$
|
If you expand along the bottom row, then most of the $(n-1)\times(n-1)$ subdeterminants are $0$ because the $(n-1)\times(n-1)$ submatrices have their two right-most columns identical.
So expansion along the bottom row leaves only the last two terms: $$(-1)(n-1)\det(A_{n-1})+n\det(A_{n-1})$$ which is just $\det(A_{n-1})$. So inductively, $\det(A_{n})=\det(A_{n-1})=\cdots=\det(A_{1})=1$
|
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|
Integrate by parts: $\int \ln (2x + 1) \, dx$ $$\eqalign{
& \int \ln (2x + 1) \, dx \cr
& u = \ln (2x + 1) \cr
& v = x \cr
& {du \over dx} = {2 \over 2x + 1} \cr
& {dv \over dx} = 1 \cr
& \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr
& = x\ln (2x + 1) - \int 1 - {1 \over {2x + 1}} \cr
& = x\ln (2x + 1) - (x - {1 \over 2}\ln |2x + 1|) \cr
& = x\ln (2x + 1) + \ln |(2x + 1)^{1 \over 2}| - x + C \cr
& = x\ln (2x + 1)^{3 \over 2} - x + C \cr} $$
The answer $ = {1 \over 2}(2x + 1)\ln (2x + 1) - x + C$
Where did I go wrong?
Thanks!
|
Starting from your second to last line (your integration was fine, minus a few $dx$'s in you integrals):
$$ = x\ln (2x + 1) + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C \tag{1}$$
Good, up to this point... $\uparrow$.
So the error was in your last equality at the very end:
You made an error by ignoring the fact that the first term with $\ln(2x+1)$ as a factor also has $x$ as a factor, so we cannot multiply the arguments of $\ln$ to get $\ln(2x+1)^{3/2}$. What you could have done was first express $x\ln(2x+1) = \ln(2x+1)^x$ and then proceed as you did in your answer, but your result will then agree with your text's solution.
Alternatively, we can factor out like terms.
$$ = x\ln(2x + 1) + \frac 12 \ln(2x + 1) - x + C \tag{1}$$
$$= \color{blue}{\bf \frac 12 }{\cdot \bf 2x} \color{blue}{\bf \ln(2x+1)} + \color{blue}{\bf \frac 12 \ln(2x+1)}\cdot {\bf 1} - x + C$$
Factoring out $\color{blue}{\bf \frac 12 \ln(2x + 1)}$ gives us
$$= \left(\dfrac 12\ln(2x + 1)\right)\cdot \left(2x +1\right) - x + C $$ $$= \frac 12(2x + 1)\ln(2x+1) - x + C$$
|
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|
Find the value of $\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}$ I want to know an objective approach to solve these type of expression in a quick time
Which of the expression equals to
$$\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}$$
a)$1-\tan\theta-\cot\theta$
b)$1+\tan\theta-\cot\theta$
c)$1-\tan\theta+\cot\theta$
d)$1+\tan\theta+\cot\theta$
I've tried it several ways like taking LCM,change whole into $\sin\theta$ and $\cos\theta$.but I've stuck.
|
I'm not sure where you got stuck after writing the expression in terms of $ \sin \theta $ and $ \cos \theta $ but here's how I would do it: $$ \begin{align*}\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} &= \frac{\sin \theta \tan \theta}{\sin \theta - \cos \theta} + \frac{\cos\theta \cot \theta}{\cos \theta - \sin \theta} \\ &= \frac{\sin\theta\tan\theta - \cos\theta \cot \theta}{\sin \theta - \cos \theta} \\&= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin\theta - \cos\theta)} \\&= \frac{\sin^2 \theta + \sin \theta \cos \theta + \cos^2\theta }{\sin \theta \cos \theta}\\&= 1 + \tan \theta + \cot \theta\\&=\boxed{\text{D}} \end{align*} $$
|
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|
Limit $\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}$ I want to compute $$\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}.$$
Is it OK how I want to do?
$$\sin{\sqrt{x+1}}-\sin{\sqrt{x}}=2\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}=2\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}$$
I'm not surem but I want to say that $$|\cos \frac{\sqrt{x+1}-\sqrt{x}}{2}| \leq 1$$ $$\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}} \to 0 \mbox{ when } x \to \infty$$
So the limit is $0$ ?
I thinks is not because if I say that $ \displaystyle|\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}| \leq 1$ I will obtain that another limit than $0$.
Thanks :)
|
Assuming you have the Mean Value Theorem at your dispoeal, that's the easiest way to show that the limit equals $0$: with $g(x) = \sin\sqrt x$, we have
$$
\sin\sqrt{x+1}-\sin\sqrt x = g(x+1)-g(x) = ((x+1)-x)g'(\xi) = \frac{\cos\sqrt\xi}{2\sqrt\xi}
$$
for some $x\le\xi\le x+1$. In particular,
$$
|\sin\sqrt{x+1}-\sin\sqrt x| = \frac{|\cos\sqrt\xi|}{|2\sqrt\xi|} \le \frac1{2\sqrt x},
$$
and so $\lim_{x\to\infty} (\sin\sqrt{x+1}-\sin\sqrt x) = 0$ by the Squeeze Theorem.
|
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|
quadratic equation precalculus from Stewart, Precalculus, 5th, p56, Q. 79
Find all real solutions of the equation
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{x^2-4}$$
my solution
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{(x+2)(x-2)}$$
$$(x+2)(x+5)=5(x-2)+28$$
$$x^2+2x-8=0$$
$$\dfrac{-2\pm\sqrt{4+32}}{2}$$
$$\dfrac{-2\pm6}{2}$$
$$x=-4\text{ or }2$$
official answer at the back of the book has only one real solution of $-4$
where did I go wrong?
|
$$\frac{x+5}{x-2}=\frac{5}{x+2}+\frac{28}{x^2-4}$$
multiply both sides with $(x-2)(x+2)\neq 0$ or $x\neq 2,x\neq -2$
we get
$$(x+5)(x+2)=5(x-2)+28,x\neq 2,-2$$
$$x^2+2x-8=0,x\neq 2,-2$$
$$x^2-2x+4x-8=0,x\neq 2,-2$$
$$x(x-2)+4(x-2)=0,x\neq 2,-2$$
$$(x-2)(x+4)=0,x\neq 2,-2$$
so $x+4=0$ or $x=-4\neq 2,-2$ is unique solution because from $x-2=0$ follow solution $x=-2$ that is prohibited
|
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|
How do I evaluate the following expression? How to evaluate the following expression:
$\displaystyle \frac{1}{\sqrt{2}+1}+ \frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} +\cdots +\frac{1}{\sqrt{9}+\sqrt{8}}$
|
Observe that:
$(\sqrt{2} + 1)(\sqrt{2} - 1) = (\sqrt{2})^2 - 1^2 = 2 - 1 = 1.$
Therefore,
$\dfrac1{\sqrt{2} + 1} = \sqrt{2} - 1.$
Similarly,
$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 2 - 1 = 1.$
And therefore,
$\dfrac1{\sqrt{3} + \sqrt{2}} = \sqrt{3} - \sqrt{2}.$
We keep doing this until we reach,
$\dfrac1{\sqrt{8} + \sqrt{9}} = \sqrt{9} - \sqrt{8}.$
Now we add all the above equations to get:
$\displaystyle \frac{1}{\sqrt{2}+1}+ \frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} +\cdots +\frac{1}{\sqrt{9}+\sqrt{8}}.$
$=\displaystyle (\sqrt{2} - 1) + (\sqrt{3}-\sqrt{2}) + (\sqrt{4} - \sqrt{3}) +\cdots + (\sqrt{9}-\sqrt{8}).$
Notice that $+\sqrt{2}$ cancels with $-\sqrt{2}$. And this happens with all the numbers in the expression, except with $1$ and $\sqrt{9}$.
$= \sqrt{9} - 1 = 3 - 1 = 2.$
|
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|
solution to a root inequality I have the inequality
$$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc} \geq \sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+2ab}.$$I tried to do $u=a^2+b^2+c^2$ and $v=ab+ac+bc$ and $x=a^2+2bc$, $y=b^2+2ac$, $z=c^2+2ab$ ...but I did not find any solution. Any help is appreciated.
|
I will assume that $a, b, c \geq 0$. Change to cylindrical coordinates with axis in the direction of $(1, 1, 1)^T$ as follows:
$$\left(\begin{matrix} a \\ b \\ c \end{matrix}\right) = \frac{z}{\sqrt{3}}\left(\begin{matrix} 1 \\ 1 \\ 1\end{matrix}\right) + \rho\left[\frac{\cos\phi}{\sqrt{6}}\left(\begin{matrix} 1 \\ 1 \\ -2\end{matrix}\right) + \frac{\sin\phi}{\sqrt{2}}\left(\begin{matrix} 1 \\ -1 \\ 0\end{matrix}\right)\right]$$
The inequality then becomes:
$$\sum_{j=1}^3 \sqrt{z^2+\rho^2\cos(2\phi+\frac{2\pi}{3}j)} \leq \sqrt{z^2+\rho^2}+2\sqrt{z^2-\frac{1}{2}\rho^2}$$
The left side depends on $\phi$, while the right side does not, and we have equality when $\phi = \frac{k\pi}{3}$ with $k$ an integer. Evaluating the partial derivative with respect to $\phi$ of the left side shows the existence of critical points at $\phi = \frac{k\pi}{6}$ for integer $k$. Plugging these in to the inequality shows that equality holds when $\phi = \frac{k\pi}{3}$, while strict inequality holds when $\phi = \frac{(2k+1)\pi}{6}$ (unless $\rho = 0$).
We first examine the case when $z^2 \geq \rho^2$. Consider the function
$$f(x_1, x_2, x_3) = \sum_{i=1}^3\sqrt{z^2 + \rho^2x_i}$$
subject to the constraints
$$g(x_1,x_2,x_3)=x_1 + x_2 + x_3 = 0$$
$$h(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2=\frac{3}{2}$$
The method of Lagrange multipliers dictates that local maxima will only be found where $\nabla f + \lambda \nabla g + \mu \nabla h = 0$. Then for each $x_i$ we have
$$\rho^2(z^2+\rho^2x_i)^{-1/2}+\lambda+2\mu x_i=0$$
Let $\psi(x) = \rho^2(z^2+\rho^2x)^{-1/2}+\lambda+2\mu x$. Since $\frac{d^2 \psi}{d x^2} = \frac{3}{4}\rho^6(z^2+\rho^2x)^{-5/2} > 0$, $\psi$ can have at most two roots, so local maxima can only be found when two of the $x_i$ are equal. This corresponds to $\phi = \frac{k\pi}{6}$ for integer $k$. Since we have equality at $\phi = \frac{k\pi}{3}$ and strict inequality at $\phi = \frac{(2k+1)\pi}{6}$, the critical points at $\phi=\frac{k\pi}{3}$ are maxima and the critical points at $\phi = \frac{(2k+1)\pi}{6}$ are local minima. Since the inequality holds at the maxima, it holds for all $\phi$.
Now consider the case when $z^2 < \rho^2$. We know that the term $z^2 - \frac{1}{2}\rho^2$ in the right-hand side is always well-defined, so $\frac{1}{2}\rho^2 \leq z^2 < \rho^2$. The left-hand side is then undefined within a neighborhood of $\phi = \frac{(2k+1)\pi}{6}$. Since $\phi = \frac{k\pi}{3}$ are maxima for $z^2 \geq \rho^2$, by continuity with respect to $z$ and the fact that critical points can only occur at $\frac{k\pi}{6}$, it follows that $\phi = \frac{k\pi}{3}$ are maxima for all $z^2 \geq \frac{1}{2}\rho^2$. Then since the inequality holds at the maxima, it holds for all $\phi$ such that it is well-defined, and thus for all $a, b, c \geq 0$. $\square$
|
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How to resolve this algebra equation? $$f = X^3 - 12X + 8$$ $a $- complex number, $a$ is a root for $f$
$b = a^2/2 - 4 $.
Show that $f(b) = 0$
This is one of my theme exercises ... Some explanations will be appreciated ! Thank you all for your time .
|
As $a$ is root of $f = X^3 - 12X + 8=0, a^3-12a+8=0$
Now, $b=\frac{a^2}2-4\implies a^2=2(b+4)=2b+8$
So, from $a^3-12a+8=0, a(2b+8)-12a+8=0\implies a=-\frac8{2b+4}=-\frac4{b+2}$
So, $$\left(-\frac4{b+2}\right)^3-12\cdot \left(-\frac4{b+2}\right)+8=0$$
On simplification we get , $$b^3-12b+8=0$$
|
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Basis of kernel and image of a linear transformation - verification The transformation matrix I found is: $$\begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 0 & 0\end{pmatrix}$$
Is this how a basis for $\ker$ and $\mathrm{im}$ is calculated?
$$\begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 0 & 0\end{pmatrix} \sim \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{pmatrix}$$
Therefore a basis for $\ker$ is $\{0\}$.
For the image:
$$\begin{pmatrix} 1 & 1 & 0 \\ -1 & 1 & 0 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$$
Therefore a basis for $\mathrm{im}$ is $\left\{ \begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix} \right\}$.
|
This CW answer intends to remove the question from the unanswered queue.
As DonAntonio already noted, your calculation is correct.
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Prove with integration the inequality $e(\frac{n}{e})^n < n! < n \times e(\frac{n}{e})^n$ Prove with integration the inequality, I need some advice about how to start prove it.
I know that if function is Monotonically increasing function so :
$$ f(1)+\int^n_1f(x)dx\leq f(1)+f(2)+....+f(n)\leq f(n)+\int^n_1f(x)dx$$
$$e(\frac{n}{e})^n < n! < n \times e(\frac{n}{e})^n$$
Thanks!
|
Depending on how you introduced $e$, you might be able to use the fact that there are two sequences $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}}$ with
$$\begin{align}
a_n ~~~&:=~~~ \left ( 1 + \frac{1}{n} \right ) ^n \\ ~ \\
b_n ~~~&:=~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n}
\end{align}$$
and
$$\underset{n \rightarrow \infty}{\lim} a_n ~~~=~~~ \underset{n \rightarrow \infty}{\lim} b_n ~~~=~~~ e \\ ~ \\$$
While both sequences converge to the same limit, $a_n$ approaches from the bottom and $b_n$ approaches from the top:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import rcParams
rcParams.update({'figure.autolayout': True})
pts = np.arange(0, 20, 1)
a_n = lambda n: (1+1/n)**n
b_n = lambda n: (1-1/n)**(-n)
plt.errorbar(x = pts, xerr = None, y = a_n(pts), yerr = None, fmt = "bx", markersize = "5", markeredgewidth = "2", label = "$a_n$")
plt.errorbar(x = pts, xerr = None, y = b_n(pts), yerr = None, fmt = "rx", markersize = "5", markeredgewidth = "2", label = "$b_n$")
plt.plot(pts, [np.exp(1)]*len(pts), color = "black", linewidth = 2, label = "$e$")
plt.xlim(1.5, 14.5)
plt.ylim(2.0, 3.5)
plt.legend(loc = "best")
plt.setp(plt.gca().get_legend().get_texts(), fontsize = "22")
plt.show()
So we're going to use the following inequality:
$$\forall n \in \mathbb{N} ~ : ~~~~~ \left ( 1 + \frac{1}{n} \right ) ^n ~~~~<~~~~ e ~~~~<~~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n} \tag*{$\circledast$} \\ ~ \\$$
Thesis
$$\forall n \in \mathbb{N}, ~ n \geq 2 ~ : ~~~~~ e \cdot \left ( \frac{n}{e} \right )^n ~~~~<~~~~ n! ~~~~<~~~~ n \cdot e \cdot \left ( \frac{n}{e} \right )^n \\ ~ \\$$
Proof By Induction
Base Case
We begin with $n = 2$ and get
$$\begin{align}
& ~ && e \cdot \left ( \frac{2}{e} \right )^2 ~~~~&&<~~~~ 2! ~~~~&&<~~~~ 2 \cdot e \cdot \left ( \frac{2}{e} \right )^2 \\ ~ \\
& \Leftrightarrow && e \cdot \frac{4}{e^2} ~~~~&&<~~~~ 1 \cdot 2 ~~~~&&<~~~~ 2 \cdot e \cdot \frac{4}{e^2} \\ ~ \\
& \Leftrightarrow && \frac{4}{e} ~~~~&&<~~~~ 2 ~~~~&&<~~~~ \frac{8}{e} \\ ~ \\
&\Leftrightarrow && 2 ~~~~&&<~~~~ e ~~~~&&<~~~~ 4 ~~~~ \\
\end{align}
$$
Which is a true statement.
Inductive Hypothesis
Therefore the statement holds for some $n$. $\tag*{$\text{I.H.}$}$
Inductive Step
$$\begin{align}
& ~ && e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\
& = && (n+1) \cdot \frac{1}{e} \cdot e \cdot \left ( \frac{n+1}{e} \right )^n\\ ~ \\
& = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( \frac{n+1}{n} \right )^n\\ ~ \\
& = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( 1 + \frac{1}{n} \right )^n\\ ~ \\
& \overset{\circledast}{<} && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot e\\ ~ \\
& \overset{\text{I.H.}}{<} && (n+1) \cdot n!\\ ~ \\
& = && (n+1)!\\ ~ \\
& = && (n+1) \cdot n!\\ ~ \\
& \overset{\text{I.H.}}{<} && (n+1) \cdot n \cdot e \cdot \left ( \frac{n}{e} \right )^n\\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n}{e} \right )^{n+1} \cdot e \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( \frac{n}{n+1} \right )^{n+1} \cdot e \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot e \\ ~ \\
& \overset{\circledast}{<} && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{-(n+1)} \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\
\end{align}
$$
Conclusion
Therefore the statement holds $\forall n \in \mathbb{N}, ~ n \geq 2$.
$$\tag*{$\square$}$$
|
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|
Recurrence Relations with single roots I have the following recurrence: $a_{n+3}=3a_{n+2}-3a_{n+1}+a_n$
with initial values $a_1 = 1, a_2 = 4, a_3 = 9$
I have found the characteristic equation to be $x^3-3x^2+3x-1$ and the root to be 1.
My text book is not helpful in how I should go about solving this when I have a single root and don't have the $a_0$ value given.
Any tips on how I could move forward to solve this?
|
Use Wilf's "generatingfunctionology". Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply your recurrence by $z^n$ and sum over $n \ge 0$ to get:
$$
\frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3}
= 3 \frac{A(z) - a_0 - a_1 z}{z^2} - 3 \frac{A(z) - a_0}{z} + A(z)
$$
Using the recurrence "backwards" gives $a_0 = 0$. Solving for $A(z)$ and expanding into partial fractions:
$$
A(z) = \frac{1}{1 - z} - \frac{3}{(1 - z)^2} + \frac{2}{(1 - z)^3}
$$
By the expansions:
$$
(1 - u)^{-m} = \sum_{k \ge 0} \binom{-m}{k} (-u)^k
= \sum_{k \ge 0} \binom{k + m - 1}{m - 1} u^k
$$
(the binomial coeffiecients are just $m - 1$ degree polynomials in $k$) you get:
$$
a_n = 1 - 3 \binom{n + 1}{1} + 2 \binom{n + 2}{2}
= 1 - 3 \frac{(n + 1)}{1} + 2 \frac{(n + 1) (n + 2)}{2}
= n^2
$$
|
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|
Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$ How can I solve this question with out find a,b,c,d
$$a+b+c+d=2$$
$$a^2+b^2+c^2+d^2=30$$
$$a^3+b^3+c^3+d^3=44$$
$$a^4+b^4+c^4+d^4=354$$
so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$
If the qusetion impossible to solve withot find a,b,c,d then is there any simple way to find a,b,c,d
Is there any help?
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Here's an approach which involves AM-GM inequality(To bound the numbers).
I have considered the positive values of $(a,b,c,d)$. I can consider $a^2,b^2,c^2$ and $d^2$, since they are all strictly positive.
$\dfrac{a^2+b^2+c^2+d^2}{4} \ge \sqrt{abcd}$
$\dfrac{30}{4} \ge \sqrt{abcd} \implies 56.25 \ge abcd$
Since $a^2+b^2+c^2+d^2=30$ and $56 >|abcd|$, One of the $|a|,|b|,|c|,|d|$ is at most $5$.
Considering $a^4+b^4+c^4+d^4$, one of the $|a|,|b|,|c|,|d|$ is atmost $4$, since $5^4=625$ .
Now we have $a^2+b^2+c^2+d^2=30$, the value of $(|a|,|b|,|c|,|d|)$ is from set $(1,2,3,4)$.
And also:
Considering $4th$ degree polynomial such that $a,b,c,d$ are the roots of the equation.
$x^4+px^3+qx^2+rx+s=0$
$a+b+c+d=2=-p$
$(a+b+c+d)^2-a^2+b^2+c^2+d^2=2(ab+bc+cd+ad+ac+bd)=q=-13$
Finding $\sum abc$ and $abcd$ with the help of other inequalities and finding the roots of equation is another way to go. (Not quite sure whether constructing polynomial is a great way to go. )
Edit: I have considered the absolute values of $(a,b,c,d)$ . I didn't assume them to be $+ve$.
|
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|
Finding the correct statements about $(5+2\sqrt{6})^{2n+1} = S + t$ with $S$ integer and $0 \leq t < 1$ Problem:
If $n$ is a positive integer and $(5+2\sqrt{6})^{2n+1} = S + t$, where $S$ is an integer and $0 \leq t < 1$, then
(a) $S$ is an odd integer
(b) $S + 1$ is not divisible by $9$
(c) The integer next above $(5+2\sqrt{6})^{2n+1}$ is divisible by $10$
(d) $S-1 = \frac{t}{t-1}$
Please guide me how to proceed with this question...
|
Everything flows from the hint by Ross Millikan.
Let $(5+2\sqrt{3})^{2n+1}=a_n+b_n\sqrt{3}$. (we can imagine computing $a_n$ and $b_n$ by expanding using the Binomial Theorem). Note that $a_n$ and $b_n$ are integers. Now imagine expanding $(5-2\sqrt{3})^{2n+1}$. We get $a_n-b_n\sqrt{3}$. It follows that
$$(5+2\sqrt{3})^{2n+1}+(5-2\sqrt{3})^{2n+1}=2a_n.\tag{$1$}$$
Define $S_n$ and $t_n$ as in the question, except we have added subscripts to make clear the dependence on $n$. By the hint of Ross Millikan, $S_n=2a_n-1$. We can conclude immediately that $S$ is odd.
Note that
$$\begin{align}a_{n+1}+b_{n+1}\sqrt{3}&=(5+2\sqrt{3})^{2n+3}=(a_n+b_n\sqrt{3})(5+2\sqrt{3})^2\\ &= 37a_n+60b_n+(20a_n+37b_n)\sqrt{3}\end{align}.\tag{$2$}$$
Thus
$$a_{n+1}=37a_n+60b_n\qquad\text{and}\qquad b_{n+1}=20a_n +37b_n.\tag{$3$}$$
Note that $a_0=5$ and $b_0=2$. The integer just above $S_n$ is $2a_n$. This is divisible by $10$ if $n=0$. It follows by induction, using the first recurrence in $(3)$, that $10$ divides $2a_n$, that is, $S_n+1$, for all $n$.
About non-divisibility of $S_n+1=2a_n$ by $9$, we prove that $a_n$ is not divisible by $3$. Certainly $a_0=5$, so $a_0$ is not a multiple of $3$. But then by the recurrence $a_{n+1}=37a_n+60b_n$, we see that $a_1$ is not divisible by $3$, but then again by the recurrence, $a_2$ is not divisible by $3$, and so on.
The computation in (d) is left to you.
|
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Not so easy optimization of variables? What is the maximum value of $x^2+y^2$, where $(x,y)$ are solutions to $2x^2+5xy+3y^2=2$ and $6x^2+8xy+4y^2=3$. (calculus is not allowed). I tried everything I could but whenever I got for example $or$ $x^2+y^2=f(y)$ or $f(x)$ the function $f$ would always be a concave up parabola, so I could not find a maximum for either variable. However, I also don't see how you could solve it if you leave both variables on one side. And by the way I know that you can solve for $x$ and $y$ using the quadratic formula and get $4$ different solutions but I am looking for a much more efficient way than that.
This question came from a math competition from the Math Honor Society, Mu Alpha Theta.
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Since my other entry is already rather long, I thought I'd separately add a couple of other methods that work (at least in principle), but which involve too much effort to be practical "contest math" answers.
The approach I discussed already can also be carried out in polar coordinates, with the equations
$$ 2 \cos^2 \theta \ + \ 5 \sin \theta \cos \theta \ + \ 3 \sin^2 \theta \ = \ \frac{2}{r^2} \ \ , \ \ 6 \cos^2 \theta \ + \ 8 \sin \theta \cos \theta \ + \ 4 \sin^2 \theta \ = \ \frac{3}{r^2} $$
$$ \Rightarrow \ \ 6 \cos^2 \theta \ + \ 15 \sin \theta \cos \theta \ + \ 9 \sin^2 \theta \ = \ 12 \cos^2 \theta \ + \ 16 \sin \theta \cos \theta \ + \ 8 \sin^2 \theta $$
$$ \Rightarrow \ \ 6 \cos^2 \theta \ + \ \sin \theta \cos \theta \ - \ \sin^2 \theta \ = \ 0 \ \ \Rightarrow \ \ 7 \cos^2 \theta \ + \ \sin \theta \cos \theta \ = \ 1 $$
$$ \Rightarrow \ \ 7 \ ( \frac{1}{2} [1 + \cos 2\theta] \ ) \ + \ \frac{1}{2}\sin 2 \theta \ = \ 1 \ \ \Rightarrow \ \ 7 \ \cos 2\theta \ + \ \sin 2 \theta \ = \ -5 $$
$$ \Rightarrow \ \ \frac{7}{\sqrt{50}} \ \cos 2\theta \ + \ \frac{1}{\sqrt{50}} \ \sin 2 \theta \ = \ -\frac{5}{\sqrt{50}} \ \ . $$
The typical method of solving this by the use of an auxiliary angle and an "angle-addition formula" is not particularly enlightening, so we will simply use $ \ x = \cos 2\theta \ $ to write
$$ \frac{7}{\sqrt{50}} \ x \ + \ \frac{1}{\sqrt{50}} \cdot \sqrt{1-x^2} \ = \ -\frac{1}{\sqrt{2}} \ \ \Rightarrow \ \ x^2 \ + \ \frac{7}{5} x \ + \ \frac{12}{25} \ = \ 0 $$
$$ \Rightarrow \ \ \cos 2 \theta \ = \ - \frac{3}{5} \ \ \Rightarrow \ \ \sin 2 \theta \ = \ - \frac{4}{5} \ \ , $$
$$ \Rightarrow \ \ \cos 2 \theta \ = \ - \frac{4}{5} \ \ \Rightarrow \ \ \sin 2 \theta \ = \ \frac{3}{5} \ \ , $$
after removing "spurious" solutions. The "tangent half-angle formula" then yields
$$ \tan \theta \ = \ \frac{1 - \cos 2 \theta}{\sin 2 \theta} $$
$$ \Rightarrow \ \ \cos 2 \theta \ = \ - \frac{3}{5} \ \ \rightarrow \ \ \tan \theta \ = \ -2 \ \ , \ \ \cos 2 \theta \ = \ - \frac{4}{5} \ \ \rightarrow \ \ \tan \theta \ = \ +3 \ \ . $$
We have thus confirmed the slopes of the two lines along which the intersections of the conic sections lie. To answer the original question, however, we only need, in polar coordinates, to find the maximum value of $ \ r^2 \ . $ The trigonometric values we have obtained can now be used in either of the conic section equations to show that the extrema are
$$ \tan \theta \ = \ +3 \ \ \Rightarrow \ \ \sin \theta \ = \ \frac{3}{\sqrt{10}} \ , \ \cos \theta \ = \ \frac{1}{\sqrt{10}} $$
$$ \Rightarrow \ \ r^2 \ = \ \frac{2}{2 \ \left(\frac{1}{10} \right) \ + \ 5 \left(\frac{3}{\sqrt{10}} \right) \left(\frac{1}{\sqrt{10}} \right) \ + \ 3 \left(\frac{9}{10} \right) } \ = \ \frac{5}{11} \ \ , $$
$$ \tan \theta \ = \ -2 \ \ \Rightarrow \ \ \sin \theta \ = \ \frac{2}{\sqrt{5}} \ , \ \cos \theta \ = \ -\frac{1}{\sqrt{5}} $$
$$ \Rightarrow \ \ r^2 \ = \ \frac{2}{2 \ \left(\frac{1}{5} \right) \ + \ 5 \left(\frac{2}{\sqrt{5}} \right) \left(\frac{-1}{\sqrt{5}} \right) \ + \ 3 \left(\frac{4}{5} \right) } \ = \ \frac{5}{2} \ \ . $$
[Inserting the trigonometric values into the other conic section equation leads to the same results, as does using the other set of sine and cosine values associated with each tangent value.]
$$ \ \ $$
Because the problem is looking for extremal values of a function at specific points, looking for "critical points" through the use of calculus is not all that helpful, even if it were permitted. Since we are seeking the maximum value of a function for points that lie on two curves simultaneously, this suggests the use of Lagrange multipliers with two constraints. We then wish to extremize $ \ f(x,y) = x^2 + y^2 \ $ with the constraint functions $ \ g(x,y) = 2x^2 + 5xy + 3y^2 - 2 \ $ and $ \ h(x,y) = 6x^2 + 8xy + 4y^2 - 3 \ . $ The Lagrange equations that emerge, however, do not seem to encourage quick solution:
$$ \nabla f \ = \ \lambda \ \nabla g \ + \ \mu \ \nabla h $$
$$ \Rightarrow \ \ 2x \ = \ \lambda \ (4x + 5y) \ + \ \mu \ (12x + 8y) \ \ , \ \ 2y \ = \ \lambda \ (5x + 6y) \ + \ \mu \ (8x + 8y) \ . $$
[By using the results obtained by other means, we find that the two solutions are
$$ ( \pm \frac{1}{\sqrt{2}} \ , \ \mp \sqrt{2} ) \ \ , \ \ \lambda \ = \ -\frac{8}{5} \ \ , \ \ \mu \ = \ \frac{19}{10} \ \ , $$
$$ ( \pm \frac{1}{\sqrt{22}} \ , \ \pm \frac{3}{\sqrt{22}} ) \ \ , \ \ \lambda \ = \ \frac{38}{55} \ \ , \ \ \mu \ = \ -\frac{17}{55} \ \ . $$ This may serve to provide some idea of how readily this answer could be obtained.]
So the original question would seem to be one more easily answered by the application of less sophisticated techniques than with the tools of calculus.
|
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|
Integrality of $\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6}$ I've been asked to provide a proof for
If $n$ is an integer then
$$\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6}$$
is also an integer.
Any help would be appreciated
|
Hint:$$\frac{n}{6}(n^2+3n+2)=\frac{n}{6}(n+1)(n+2)$$
What can you say about the divisibility of $n,(n+1),(n+2)$ by $2$ and $3$?
|
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Two roots of polynomial If a polynomial with rational coefficients has a root $1 + \cos(2\pi/9) + \cos^2(2\pi/9)$, then the one also has a root $1+\cos(8\pi/9)+\cos^2(8\pi/9).$ How to prove it?
|
Take $\omega=e^{i2\pi/9}$.
Let $\alpha=1 + \cos(2\pi/9) + \cos^2(2\pi/9)$.
Then $4\alpha = 6+2\omega+ \omega^2 + \omega^7 + 2\omega^8$, by using $2\cos(2\pi/9)=\omega+\bar\omega$ and $\omega^9=1$.
Let $\beta = 1+\cos(8\pi/9)+\cos^2(8\pi/9)$. Then $4\beta = 6 + \omega+ + 2\omega^4 + 2\omega^5 + \omega^8$, by using $2\cos(8\pi/9)=\omega^4+\bar\omega^4$.
Now note that the map $\omega \mapsto \omega^4$ on $\mathbb Q(\omega)$ sends $\alpha$ to $\beta$ and the map $\omega \mapsto \omega^5$ sends $\beta$ to $\alpha$. This means that for $p$ a polynomial with rational coefficients, $p(\alpha)=0$ iff $p(\beta)=0$, as required.
Alternatively, you can also conclude that
$\alpha$ and $\beta$ are conjugate algebraic numbers and your result follows because their minimal polynomial is the same.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit with roots I have to evaluate the following limit:
$$ \lim_{x\to 1}\dfrac{\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}} . $$
I rationalized both the numerator and the denominator two times, and still got nowhere. Also I tried change of variable and it didn't work.
Any help is grateful. Thanks.
|
For every $x>1$ we have
\begin{eqnarray}
P(x):&=&\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}=\sqrt{x^2-1}+\frac{x+1-(x^3+1)}{\sqrt{x+1}+\sqrt{x^3+1}}\\
&=&\sqrt{x^2-1}-\frac{x(x^2-1)}{\sqrt{x+1}+\sqrt{x^3+1}}=\sqrt{x^2-1}\left(1-\frac{x\sqrt{x^2-1}}{\sqrt{x+1}+\sqrt{x^3+1}}\right)\\
&=&\sqrt{x-1}p(x),
\end{eqnarray}
with
$$
p(x):=\sqrt{x+1}\left(1-\frac{x\sqrt{x^2-1}}{\sqrt{x+1}+\sqrt{x^3+1}}\right).
$$
Similarly for every $x>1$ we have
\begin{eqnarray}
Q(x):&=&\sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}=\sqrt{x-1}+\frac{x^2+1-(x^4+1)}{\sqrt{x^2+1}+\sqrt{x^4+1}}\\
&=&\sqrt{x-1}+\frac{x^2(1-x^2)}{\sqrt{x^2+1}+\sqrt{x^4+1}}=\sqrt{x-1}\underbrace{\left(1-\frac{x^2(x+1)\sqrt{x-1}}{\sqrt{x^2+1}+\sqrt{x^4+1}}\right)}_{q(x)}.
\end{eqnarray}
It follows that
$$
\lim_{x \to 1^+}\frac{P(x)}{Q(x)}=\lim_{x\to 1^+}\frac{p(x)}{q(x)}=\frac{p(1)}{q(1)}=\sqrt{2}.
$$
|
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|
Diagonalising quadratic form Given the quadratic form $$Q(x) = \alpha\alpha_1\alpha_2 + 2\alpha^2\alpha_1\alpha_3$$ on $\mathbb{R}^2$ where $x = (\alpha_1,\alpha_2,\alpha_3)$ in some basis I want to find the signature of $Q$ dependent on $\alpha$ and the diagonal basis of $Q$.
I was trying to rewrite $Q$ into a sum of squares like this:
$$ Q(x) = (\alpha\alpha_1 + \alpha\alpha_3 + \frac{1}{2}\alpha_2)^2 - \alpha^2\alpha_1^2 - \alpha^2\alpha_3^2 - \frac{1}{4}\alpha_2^2 - \alpha\alpha_3\alpha_2 $$
which yields $$ (\alpha\alpha_1 + \alpha\alpha_3 + \frac{1}{2}\alpha_2)^2 - (\alpha\alpha_1)^2 - (\alpha\alpha_3 + \frac{1}{2}\alpha_2)^2 = Q(x)$$
Since now I have $Q$ represented as a sum of squares, I see the signature being $(0,1,2)$ if $\alpha \neq 0$ and $(1,1,1)$ if $\alpha = 0$. Also the terms in parentheses must be the coordinates of $x$ in the diagonal basis of $Q$:
$$ \beta_1 = \alpha\alpha_1 + \alpha\alpha_3 + \frac{1}{2}\alpha_2, \beta_2 = \alpha\alpha_1, \beta_3 = \alpha\alpha_3 + \frac{1}{2}\alpha_2$$
which corresponds to a matrix
\begin{pmatrix}\alpha & \frac{1}{2} & \alpha\\
\alpha & 0 & 0\\
0 & \frac{1}{2} & \alpha
\end{pmatrix}
To get the diagonal basis, I need to invert this matrix (get alphas represented by betas), but it is clearly singular. What am I doing wrong?
|
As in your question, let
$$Q(x) = \alpha \alpha_1 \alpha_2 + 2 \alpha^2 \alpha_1 \alpha_3$$
be a quadratic form on $\mathbb{R}^3$ in some basis. We know what the signature is $(3,0,0)$ if $\alpha = 0$, so let us assume that $\alpha \neq 0$.
Something went wrong in your diagonalization process because you managed to diagonalize to a non-singular quadratic form. The quadratic form
$$ (\alpha\alpha_1 + \alpha\alpha_3 + \frac{1}{2}\alpha_2)^2 - (\alpha\alpha_1)^2 - (\alpha\alpha_3 + \frac{1}{2}\alpha_2)^2 $$
you got has signature $(0,1,2)$. You may want to try expanding and then comparing the result with $Q(x)$.
The Gram matrix
$$\begin{pmatrix} 0 & \frac{\alpha}{2} & \alpha^2 \\ \frac{\alpha}{2} & 0 & 0 \\ \alpha^2 & 0 & 0 \end{pmatrix}$$
of $Q(x)$ is singular because its second and third rows are linearly dependent. Let us first consider the upper left $2 \times 2$ submatrix which corresponds to the form $\alpha \alpha_1 \alpha_2$. This form represents all real numbers, so it must have signature $(0,1,1)$. Thus the first two entries of the diagonalized matrix must be $1$ and $-1$. Since the original Gram matrix is singular, the final element on the diagonal must be $0$. Putting everything together, we see that the signature of $Q(x)$ is $(1,1,1)$.
If we carry out the diagonalization by hand, we will find that indeed
$$ \begin{pmatrix} 1 & 1 & 0 \\ \frac{1}{a} & -\frac{1}{a} & -2 \\ 0 & 0 & \frac{1}{a} \end{pmatrix}^{\top} \begin{pmatrix} 0 & \frac{\alpha}{2} & \alpha^2 \\ \frac{\alpha}{2} & 0 & 0 \\ \alpha^2 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \\ \frac{1}{a} & -\frac{1}{a} & -2 \\ 0 & 0 & \frac{1}{a} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$
The change-of-basis matrix is non-singular.
|
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|
There are no integers $x,y$ such that $x^2-6y^2=7$ How to show that there is no here are no integers $x,y$ such that $x^2-6y^2=7$?
Help me. I'm clueless.
|
First show that since $a^2 \equiv 0,1,4$
$$7 \vert x^2 + y^2 \implies 7 \vert x \text{ and } 7 \vert y$$
We have
$$x^2 - 6y^2 = 7 \implies (x^2+y^2) - 7y^2 = 7 \implies 7 \vert (x^2+y^2) \implies 7 \vert x \text{ and } 7 \vert y$$
Hence, we have $x = 7x_1$ and $y=7y_1$, which gives us
$$7(x_1^2 - 6y_1^2) = 1$$
which is clearly impossible.
|
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|
How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't help as it leaves $\frac{1+x-2x^2}{1-4x^2}$ any ideas?
|
$$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}} = \frac{\frac{-x}{1-2x}}{\frac{-1}{1-2x}} = \frac{-x}{1-2x} \frac{1-2x}{-1} = \frac{-x(1-2x)}{-1(1-2x)} = \frac{-x}{-1} = x.$$
|
{
"language": "en",
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|
How to find the integral of $\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$ $$\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$$
What is the method to find an integral like this?
|
Note that $\displaystyle \int_{0}^{\pi} \sin^n(x) dx = 2 \int_0^{\pi/2} \sin^n(x) dx$. Let $I_n = \displaystyle \int_{0}^{\pi/2} \sin^n(x) dx$.
$I_n = \displaystyle \int_{0}^{\pi/2} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\pi/2} (n-1) \sin^{n-2}(x) \cos^2(x) dx$
The first expression on the right hand side is zero since $\sin(0) = 0$ and $\cos\left(\pi/2\right) = 0$.
Now rewrite $\cos^2(x) = 1 - \sin^2(x)$ to get
$I_n = (n-1) \left(\displaystyle \int_{0}^{\pi/2} \sin^{n-2}(x) dx - \int_{0}^{\pi/2} \sin^{n}(x) dx \right) = (n-1) I_{n-2} - (n-1) I_n$.
Rearranging we get $n I_n = (n-1) I_{n-2}$, $I_n = \dfrac{n-1}{n}I_{n-2}$.
Using this recurrence we get
$$I_{2k+1} = \dfrac{2k}{2k+1}\dfrac{2k-2}{2k-1} \cdots \dfrac{2}{3} I_1$$
$$I_{2k} = \dfrac{2k-1}{2k}\dfrac{2k-3}{2k-2} \cdots \dfrac{1}{2} I_0$$
$I_1$ and $I_0$ can be directly evaluated to be $1$ and $\dfrac{\pi}{2}$ respectively and hence,
$$I_{2k+1} = \dfrac{2k}{2k+1}\dfrac{2k-2}{2k-1} \cdots \dfrac{2}{3} = \dfrac{4^k (k!)^2}{(2k+1)!}$$
$$I_{2k} = \dfrac{2k-1}{2k}\dfrac{2k-3}{2k-2} \cdots \dfrac{1}{2} \dfrac{\pi}{2} = \dfrac{(2k)!}{4^k (k!)^2} \dfrac{\pi}2$$
Hence,
$$\int_0^{\pi} \sin^n(x) dx = \begin{cases} \dfrac{2^{2k+1} (k!)^2}{(2k+1)!} & \text{if $n$ is odd, i.e., $n=2k+1$}\\
\dfrac{(2k)!}{4^k (k!)^2} \pi & \text{if $n$ is even, i.e., $n=2k$}
\end{cases}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why does the tangent of numbers very close to $\frac{\pi}{2}$ resemble the number of degrees in a radian? Testing with my calculator in degree mode, I have found the following to be true:
$$\tan \left(90 - \frac{1}{10^n}\right) \approx \frac{180}{\pi} \times 10^n, n \in \mathbb{N}$$
Why is this? What is the proof or explanation?
|
The power series for cotangent (for $x$ in radians) is
$$\cot x = \frac{1}{x} - \frac{1}{3}x - \frac{1}{45}x^3 - \frac{2}{945}x^5 - \cdots \approx \frac{1}{x} \; \text{for $x$ small}$$
So,
$$\tan\left( 90^\circ - \frac{1^\circ}{10^n} \right)
= \cot\frac{1^\circ}{10^n}
= \cot\frac{\pi/180}{10^n}
\approx \frac{10^n}{\pi/180} = \frac{180}{\pi}\times 10^n$$
Congratulations for noticing the pattern on your calculator!
|
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|
Show that the following matrix is diagonalizable My question is related to this question discussed in MSE.
$J$ be a $3\times 3$ matrix with all entries $1\,\,$. Then prove that $J$ is
diagonalizable.
Can someone explain it in terms of A.M. and G.M. (algebraic and geometric multiplicity) concept ? Thanks in advance for your time.
|
So we have
$$ J = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} $$
Let's compute the characteristic polynomial $\chi_J(X) = \det(J - X)$:
\begin{align*}
\chi_J(X) &= \det \begin{pmatrix} 1-X & 1 & 1 \\ 1 & 1-X & 1 \\ 1 & 1 & 1-X \end{pmatrix}\\
&= (1-X)^3 + 2 - 3(1-X)\\
&= 1 - 3X + 3X^2 - X^3 + 2 - 3 + 3X\\
&= -X^3 + 3X^2\\
&= -X^2(X-3)
\end{align*}
So 0 (with algebraic multiplicity 2) and 3 (with algebraic multiplicity 1) are the eigenvalues of $J$. To check whether $J$ is diagonaziable we will compute the geometric multiplicity of 0, that is $\dim \ker (J-0) = \dim\ker J$. We do Gaussian elimination: Subtrating the first row from the third and the second gives that
$$ \ker J = \ker \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0&0\\ 0 & 0 & 0 \end{pmatrix}$$
This matrix has rank 1, so its kernel has dimension 2. So $\dim\ker J = 2$ and $J$ is diagonaziable.
|
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|
Find the range of $ f(x)=9^x - 3^x+1$
Problem:-Find range of function $ f(x)=9^x - 3^x+1$, here the domain of $f$ is $\mathbb R$.
Solution: $ f(x)=9^x - 3^x+1$. Let $f(x)=y$. Then
$$ \begin{split}y&=9^x - 3^x+1\\
y&=3^{2x} - 3^x+1
\end{split}$$
Let $3^x= u$. Then $ y=u^2 - u+1$, so
$$ u^2 - u+1-y=0.$$
Am I doing right?
|
After some computation, you will get $$f(x)=(3^x-1/2)^2+3/4.$$
|
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|
Prove this equality $\frac{x}{y^2+5}+\frac{y}{z^2+5}+\frac{z}{x^2+5}\le\frac{1}{2}$ let $x^3+y^3+z^3=3,x,y,z>0$ show that
$$\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2}$$
I have show that
let $x,y,z$ be positive numbers,such that $x+y+z=3$,prove that
$$\dfrac{x}{1+y^3}+\dfrac{y}{1+z^3}+\dfrac{z}{1+x^3}\ge\dfrac{3}{2}$$
pf: use $AM-GM$ we have
$$\dfrac{x}{1+y^3}=x-\dfrac{xy^3}{1+y^3}\ge x-\dfrac{xy^3}{2y^{3/2}}=x-\dfrac{xy^{3/2}}{2}$$
and,similarly
$$\dfrac{y}{1+z^3}\ge y-\dfrac{yz^{3/2}}{2},\dfrac{z}{1+x^3}\ge z-\dfrac{zx^{3/2}}{2}$$
Thus,it suffices to show that
$$xy^{3/2}+yz^{3/2}+zx^{3/2}\le 3$$
and it is known that
$$(a^3b^2+b^3c^2+c^3a^2)^2\le\dfrac{1}{3}(a^2+b^2+c^2)^3$$
seting $x=a^2,y=b^2,z=c^2$,by done!
But for this problem :
$$y^2+5=y^2+1+1+1+1+1\ge 6y^{1/3}$$
and similarly
$$z^2+5\ge 6z^{1/3}, x^2+5\ge 6x^{1/3}$$
it suffices prove that
$$xy^{-1/3}+yz^{-1/3}+zx^{-1/3}\le 3$$
with $x^3+y^3+z^3=3$,I use maple find this is ($xy^{-1/3}+yz^{-1/3}+zx^{-1/3}\le 3$,with $x^3+y^3+z^3=3$) not true!,But after I use maple find this
$\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2},x^3+y^3+z^3=3$ is true!
and my other idea
$$\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2}$$
$$\Longleftrightarrow 2\sum x^2y^3+10\sum x^2y+50\sum x-5\sum x^2y^2-25\sum y^2-x^2y^2z^2\le 95$$
so I think my methods can't prove this problem, can someone use other methods show it? Thank you everyone.
|
By Cauchy-Schwarz inequality
$$\frac{x}{y^2 + 5} + \frac{y}{z^2 + 5} + \frac{z}{x^2+5} \le \sqrt{x^2 + y^2 + z^2}\sqrt{\frac{1}{(x^2+5)^2} + \frac{1}{(y^2 + 5)^2} + \frac{1}{(z^2 + 5)^2}}.$$
By AM-GM the RHS is less than
$$\frac{\frac{x^2 + y^2 + z^2}{6} + \frac{6}{(x^2 +5)^2} + \frac{6}{(y^2+5)^2} + \frac{6}{(z^2+5)^2}}{2} = \sum \left(\frac{x^2}{12} + \frac{3}{(x^2+5)^2}\right).$$
By multiplying everything out and factoring, one can prove that for $x \in [0,\sqrt[3]{3}]$
$$\frac{x^2}{12} + \frac{3}{(x^2+5)^2} \le \frac{2x^3 + 7}{54}.$$
Indeed, the inequality is equivalent with
$$(x-1)^2(4x^5 - x^4 + 34x^3 - 7x^2 + 52x + 26) \ge 0,$$
which is clearly true for $x \ge 0$.
After we have this result, we get that
$$\frac{x}{y^2 + 5} + \frac{y}{z^2 + 5} + \frac{z}{x^2+5} \le \frac{2(x^3+y^3+z^3) + 3\cdot 7}{54} = \frac{1}{2}.$$
|
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|
There are infinitely many $N$ such that $\frac{N}{2}$ is a perfect square, $\frac{N}{3}$ is a perfect cube, and $\frac{N}{5}$ is a perfect fifth power
Show that there are infinitely many $N$ such that $\frac{N}{2}$ is a perfect square, $\frac{N}{3}$ is a perfect cube, and $\frac{N}{5}$ is a perfect fifth power.
A hint is given with this question, which is: $7^{30}$ is a perfect square, a perfect cube and a perfect fifth power.
Can anyone solve this? Thanks!
|
The general form of this kind of number is $2^{15}3^{10}5^{6}x^{30}$, for all values of $x$.
When $x=1$, this number is the product of powers of 2, 3, and 5 just under some integer, eg $32768$, $59049$ and $15625$ are the largest powers under, say $60,000$.
The general working for such a number is to treat the indeces by chinese remainder. So, for example, should we seek something where $N/2$ is square, $N/3$ is cube, and $N/5$ is a fifth power, we would note that, eg it is a product of some powers of 2, 3, and 5, eg $2^a 3^b 5^c$.
We have then that 2 divides $a-1, b, c$, 3 divides each of $a, b-1, c$ and 5 divides each of $a, b, c-1$. We get then that 15 divides a, 10 divides b, and 6 divides c.
*
*We find now a multiple of 15 that 2 divides a-1: result = $a=15$
*We find then that 10 that 3 divides b-1, result $b = 10$
*We find a result where 5 divides c-1: result = $6$.
We then note that every number that is a power of 30, is already a square, a cube, and a fifth power, ie $x^{2 \cdot 3 \cdot 5} = x^{30}$, and thus if $N/2$ is square, so is $(N\cdot x^{30})/2$ is also square, and if $N/3$ is a cube, so is $(N\cdot x^{30})/3$, and if $N/5$ is a fifth power, so is $(N\cdot x^{30})/5$, is true for all $x$.
In the worked example, we include a final row, for all $x$, where the power of $x$ is a multiple of 2, 3, 5, and 7, and therefore satisfies all of the conditions above.
If we wish to further add $N/7$ as a seventh power, then we would need to look for some set where $2 | (a-1, b, c, d) $, $3 | (a, b-1, c, d)$, etc. This means that eg $b, c, d$ are even, $a, c, d$ are multiples of 3, etc. Here, we write the modulus in list form, where the modulus is taken over the corresponding point in the mod function.
We see that in the first row, $a$ must give a remainder of 0, when dividided by 3, 5, 7, so it must be a multiple of the product of $3 \cdot 5 \cdot 7$, that leaves a remainder of $1$, when divided by $2$.
*
*a = 1, 0, 0, 0, mod(2, 3, 5, 7), is a multiple of 105, gives $a=105$
*b = 0, 1, 0, 0, mod(2, 3, 5, 7), is a multiple of 70, gives $b=70$
*c = 0, 0, 1, 0, mod(2, 3, 5, 7), is a multiple of 42, gives $c=126$
*d = 0, 0, 0, 1, mod(2, 3, 5, 7), is a multiple of 30, gives $d=120$
*e = 0, 0, 0, 0, mod(2, 3, 5, 7), is a multiple of 210, gives $e=210$
So we can write this as $2^{105} 3^{70} 5^{126} 7^{120} x^{210}$, for all values of $x$. Note that we don't need to write $x^{210y}$, since the range of $x$ already includes $x^y$.
|
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|
Proving the relation $\det(I + xy^T ) = 1 + x^Ty$
Let $x$ and $y$ denote two length-$n$ column vectors. Prove that $$\det(I + xy^T ) = 1 + x^Ty$$
Is Sylvester's determinant theorem an extension of the problem? Is the approach the same?
|
Hint: Decomposing $$ \begin{pmatrix} 1 & -y^T\\ x & I \end{pmatrix} $$
as lower $\cdot$ upper and upper $\cdot$ lower gives
$$ \begin{pmatrix} 1 & 0\\ x & I\end{pmatrix} \cdot \begin{pmatrix} 1 & -y^T\\ 0 & I + xy^T \end{pmatrix} = \begin{pmatrix} 1 + x^Ty & -y^T\\ 0 & I\end{pmatrix} \cdot \begin{pmatrix} 1 & 0\\ x & I \end{pmatrix}. $$
|
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|
How to show that T is a projection operator For $x ∈ [0, 2π]$ let $G(x) = π^{−1}\cos x$, and define an operator $T$ on $L^2([0, 2π])$ as follows:
$$(Tf)(x) = ∫_0^{2π}G(x − x')f(x') \,dx'. $$
Show that $T$ is a projection operator.
I guess I must show that $T$ is selfadjoint and idempotent, in other words that:
$T=T^*$ and $T^2=T$
I am quite new to this and have a little difficulty to getting started
Sincerely Ingvar
|
To show $T^2 = T$, just compute $T^2$. We have for $f \in L^2([0,2\pi])$:
\begin{align*}
(T^2 f)(x) &= T(Tf)(x)\\
&= \int_0^{2\pi} G(x-x')(Tf)(x')\, dx'\\
&= \int_0^{2\pi} G(x-x')\int_0^{2\pi} G(x'-x'')f(x'')\, dx''\, dx'\\
&= \int_0^{2\pi} \int_0^{2\pi} G(x-x')G(x'-x'')\,dx'\, f(x'')\,dx''
\end{align*}
So we have to compute $\int_0^{2\pi} G(x-x')G(x'-x'')\, dx'$, plugin in the given $G$, we have
\begin{align*}
\int_0^{2\pi} G(x-x')G(x'-x'')\, dx'
&= \frac 1{\pi^2}\int_0^{2\pi} \cos(x-x')\cos(x'-x'')\, dx'\\
&= \frac 1{2\pi^2} \int_0^{2\pi} \bigl(\cos(x-x'') + \cos(x+x''-2x')\bigl)\, dx'\\
&= \frac 1{2\pi^2} \cdot \bigl( 2\pi \cos(x-x'') + 0\bigr)\\
&= \frac 1{\pi}\cos(x-x'')\\
&= G(x-x'').
\end{align*}
Continuing our calculation above, we have
\begin{align*}
(T^2 f)(x)&= \int_0^{2\pi} \int_0^{2\pi} G(x-x')G(x'-x'')\,dx'\, f(x'')\,dx''\\
&= \int_0^{2\pi} G(x-x'')f(x'')\, dx''\\
&= (Tf)(x).
\end{align*}
As $f$ and $x$ were arbitrary, we have $T^2 = T$.
For the adjoint, we will do the same (i. e. computing $T^*$ from its definition), we have for $f,g \in L^2([0,2\pi])$:
\begin{align*}
(T^*f, g) &= (f, Tg)\\
&= \int_0^{2\pi} f(x)(Tg)(x)\, dx\\
&= \int_0^{2\pi} f(x)\int_0^{2\pi} G(x-x')g(x')\, dx'\, dx\\
&= \int_0^{2\pi}\int_0^{2\pi} f(x)G(x-x')g(x')\,dx'\,dx
\end{align*}
Note now, that $G$ is symmetric, that is $G(-y) = G(y)$ for all $y$, hence $G(x-x') = G(x'-x)$, continuing we have
\begin{align*}
(T^*f, g) &= \int_0^{2\pi} \int_0^{2\pi} f(x)G(x-x')g(x')\,dx'\,dx\\
&= \int_0^{2\pi} \int_0^{2\pi} G(x'-x)f(x)\, dx\, dx'\\
&= \int_0^{2\pi} (Tf)(x')g(x')\, dx'\\
&= (Tf, g)
\end{align*}
Hence $T^* = T$.
|
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|
How to find a minimal polynomial (field theory) I was asked to find a minimal polynomial of $$\alpha = \frac{3\sqrt{5} - 2\sqrt{7} + \sqrt{35}}{1 - \sqrt{5} + \sqrt{7}}$$ over Q.
I'm not able to find it without the help of WolframAlpha, which says that the minimal polynomial of $\alpha$ is $$19x^4 - 156x^3 - 280x^2 + 2312x + 3596.$$ (Truely it is - $\alpha$ is a root of the above polynomial and the above polynomial is also irreducible over Q.)
Can anyone help me with this?
Thank you!
|
To begin, clear denominators:
$$(1 - \sqrt{5} + \sqrt{7}) \alpha = 3 \sqrt{5} - 2 \sqrt{7} + \sqrt{35}$$
We need to make the coefficient of $\alpha$ rational, so use a difference-of-squares trick to get rid of the $\sqrt{7}$ on the LHS (i.e. multiply both sides by $1 - \sqrt{5} - \sqrt{7}$),
$$((1 - \sqrt{5})^2 - 7) \alpha = (3 \sqrt{5} - 2 \sqrt{7} + \sqrt{35})(1 - \sqrt{5} - \sqrt{7})$$
and after expanding and collecting like terms:
$$(1 + 2 \sqrt{5}) \alpha = 1 + 4 \sqrt{5} + 7 \sqrt{7}$$
Now do the same again to deal with the $\sqrt{5}$ on the LHS:
$$19 \alpha = 39 - 2 \sqrt{5} - 7 \sqrt{7} + 14 \sqrt{35}$$
Next, we have to deal with the irrational numbers on the RHS. First, we deal with $\sqrt{5}$ (and $\sqrt{35}$): move all the other terms over to the LHS, and square the resulting equation,
$$(19 \alpha - 39 + 7 \sqrt{7})^2 = (-2 + 14 \sqrt{7})^2 \cdot 5$$
which expands to this
$$361 \alpha^2 - 1482 \alpha + 266 \sqrt{7} \alpha + 1864 -546 \sqrt{7} = 6880 - 280 \sqrt{7}$$
To finish off, we deal with $\sqrt{7}$: put all multiples of $\sqrt{7}$ on the RHS and all others on the LHS, and then square the resulting equation:
$$(361 \alpha^2 - 1482 \alpha - 5016)^2 = (- 266 \alpha + 266)^2 \cdot 7$$
Note that $19$ divides all the coefficients, so we can cancel that common factor:
$$(19 \alpha^2 - 78 \alpha - 264)^2 = (-14 \alpha + 14)^2 \cdot 7$$
Finally, we obtain,
$$361 \alpha^4 - 2964 \alpha^3 - 3984 \alpha^2 + 41184 \alpha + 69696 = 1372 \alpha^2 - 2744 \alpha + 1372$$
which simplifies to the desired equation:
$$19 \alpha^4 - 156 \alpha^3 - 280 \alpha^2 + 2312 \alpha + 3596 = 0$$
|
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|
Induction to prove that $1^2+ 3^2 + 5^2+ \cdots +(2n + 1)^2=\frac{(n+1)(2n+1)(2n + 3)}{3}$ Have I started this right? I know I have to add $(k+1)$ but why?
Use mathematical induction to prove that $$1^2+ 3^2 + 5^2+ \cdots +(2n + 1)^2=\frac{(n+1)(2n+1)(2n + 3)}{3}.$$
BASIS STEP: $P(0)$ is true since $$\begin{align}(2(0)+1)^2 &= \frac{(0+1)(2(0)+1)(2(0)+3)}{3}\\1 &= \frac{3}{3}\\1 &= 1.\end{align}$$
INDUCTIVE STEP: Assume $P(k)$ holds for an arbitrary integer $k>0$. $$1^2+ 3^2 + 5^2+ \cdots +(2k + 1)^2=\frac{(k+1)(2k+1)(2k + 3)}{3}$$
|
You've proved that the statement is true for $k=0$ and now supposing that is true to an arbitrary $k$,
$$1^2+2^2+\cdots+(2k+1)^2 = \frac{(k+1)(2k+1)(2k+3)}{3}$$
let's see the case $k+1$. Then we use the hypothesis of induction and we get
$$\begin{align}1^2+2^2+\cdots+(2k+1)^2+(2(k+1)+1)^2 &= \frac{(k+1)(2k+1)(2k+3)}{3}+(2(k+1)+1)^2\\ &= \frac{(k+1)(2k+1)(2k+3)+3(2k+3)^2}{3}\\ &= \frac{[(k+1)(2k+1)+3(2k+3)](2k+3)}{3}\\ &= \frac{[2k^2+3k+1+6k+9](2k+3)}{3}\\ &= \frac{(2k+5)(k+2)(2k+3)}{3}\\ &= \frac{[(k+1)+1][2(k+1)+1][2(k+1)+3]}{3},\end{align}$$
where we used that $(2k+5)(k+2)=2k^2+9k+10.$
|
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|
How can i show, this is a fundamental system for a given differential equation? My Problem is: i have a given differential equation: $$y^{\prime\prime\prime}-\frac{x^2}{x^2-2x+2}\cdot y^{\prime\prime}+\frac{2x}{x^2-2x+2}\cdot y^{\prime}-\frac{2}{x^2-2x+2}\cdot y= 0$$and the given functions :
$$y_{1}=x \quad y_{2}=x^2 \quad y_{3}=e^{x}$$
A fundamental system is defined as a:
$\{y_1,\ldots,y_n\} \quad $ with: $\quad \mathcal{L} := \{y \in C^1([a,b]; \mathbb{R}^n)\ |\ y = \sum_{k=1}^na_ky_k\ ,\ a_1, \ldots, a_n \in \mathbb{R}\}$
How can i show, that the given functions are a fundamental system for the given differential equation?
My Approach was: i managed to show, that the given functions are solutions for the differential equation.
for $y_{1} = x$
$$y^{\prime}=1 \quad y^{\prime\prime}=0 \quad y^{\prime\prime\prime}=0$$ that's why:
$$0 - 0 + \frac{2x}{x^2-2x+2}\cdot 1 -\frac{2}{x^2-2x+2}\cdot x =0$$
$$\frac{2x}{x^2-2x+2} -\frac{2x}{x^2-2x+2}=0$$
$$0=0$$
for $y_{2} = x^2$
$$y^{\prime}=2x \quad y^{\prime\prime}=2 \quad y^{\prime\prime\prime}=0$$ that's why:
$$0 - \frac{x^2}{x^2-2x+2}\cdot 2 +\frac{2x}{x^2-2x+2}\cdot 2x -\frac{2}{x^2-2x+2}\cdot x^2 =0$$
$$0 - \frac{2x^2}{x^2-2x+2} +\frac{4x^2}{x^2-2x+2} -\frac{2x^2}{x^2-2x+2} =0$$
$$0=0$$
for $y_{3} = e^x$
$$y^{\prime}= e^x \quad y^{\prime\prime}= e^x \quad y^{\prime\prime\prime}= e^x$$ that's why:
$$ e^x - \frac{x^2 e^x}{x^2-2x+2} +\frac{2x e^x}{x^2-2x+2} -\frac{2 e^x}{x^2-2x+2} =0$$
$$e^x +\frac{ -e^x(x^2-2x+2)}{x^2-2x+2}=e^x-e^x=0$$
But now i am stuck, i don't think, that this is the proof i am looking for. i dont know how to show it... maybe you can help?
P.S.: edits were made to improve language and latex
|
Your proof is missing two parts:
*
*That every function of the form $\sum a_i y_i$ is also a solution.
*That every solution is of the form $\sum a_i y_i$.
$1$ is satisfied since this is a linear ODE. $2$ is satisfied since the solutions are linearly independent and the order of the ODE is equal to the number of solutions.
|
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|
Evaluate $\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx$ I am trying to find a closed form for
$$\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx = 0.094561677526995723016 \cdots$$
It seems that the answer is
$$\frac{\pi^2}{12}\left( 1-\sqrt{3}\right)+\log(2) \log \left(1+\sqrt{3} \right)$$
Mathematica is unable to give a closed form for the indefinite integral.
How can we prove this result? Please help me.
EDIT
Apart from this result, the following equalities are also known to exist:
$$\begin{align*}
\int_0^1 \frac{\log \left( 1+x^{4+\sqrt{15}}\right)}{1+x}\mathrm dx &=\frac{\pi^2}{12} \left( 2-\sqrt{15}\right)+\log \left( \frac{1+\sqrt{5}}{2}\right)\log \left(2+\sqrt{3} \right) \\ &\quad +\log(2)\log\left( \sqrt{3}+\sqrt{5}\right)
\\ \int_0^1 \frac{\log \left( 1+x^{6+\sqrt{35}}\right)}{1+x}\mathrm dx &= \frac{\pi^2}{12} \left( 3-\sqrt{35}\right)+\log \left(\frac{1+\sqrt{5}}{2} \right)\log \left(8+3\sqrt{7} \right) \\
&\quad +\log(2) \log \left( \sqrt{5}+\sqrt{7}\right)
\end{align*}$$
Please take a look here.
|
Unfortunately, the following generalization works for only positive integer $a$:
$$\int_0^1\frac{\ln(1+x^{2a})}{1+x}dx=\ln^2(2)-\frac{2a^2-1}{8a}\zeta(2)+\frac12\sum_{j=0}^{2a-1}\ln^2\left(2\sin\left(\frac{(2j+1)\pi}{4a}\right)\right)$$
$$-\frac12\sum_{j=0}^{a-1}\ln^2\left(2\sin\left(\frac{(2j+1)\pi}{2a}\right)\right).$$
Proof:
$$\int_0^1\frac{\ln(1+x^{2a})}{1+x}dx=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\int_0^1\frac{x^{2an}}{1+x}dx$$
$$=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(\ln(2)+H_{an}-H_{2an}\right)$$
$$=\ln^2(2)+\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(H_{an}-H_{2an}\right).$$
Omram Kouba provided in his paper page (12-13) the following general result:
$$\sum_{n=1}^\infty(-1)^{n-1}\frac{H_{an}}{n} = \frac{a^2+1}{4a}\zeta(2) - \frac{1}{2} \sum_{j=0}^{a-1} \ln^2\left(2 \sin \frac{(2j+1)\pi}{2a} \right)$$
from which, the proof follows.
By the way, we can find the integral representation of $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{an}}{n}$:
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{an}}{n}=a\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{H_{an}}{an}\right)$$
$$=a\sum_{n=1}^\infty (-1)^{n-1}\left(-\int_0^1 x^{an-1}\ln(1-x)dx\right)$$
$$=a\int_0^1\frac{\ln(1-x)}{x}\left(\sum_{n=1}^\infty(-x^a)^n\right)dx$$
$$=a\int_0^1\frac{\ln(1-x)}{x}\left(\frac{-x^a}{1+x^a}\right)dx$$
$$=a\int_0^1\frac{\ln(1-x)}{x}\left(-1+\frac{1}{1+x^a}\right)dx$$
$$=a\zeta(2)+a\int_0^1\frac{\ln(1-x)}{x(1+x^a)}dx.$$
Substitute the result of $\displaystyle\sum_{n=1}^\infty\frac{(-1)^nH_{an}}{n}$, we also get
$$\int_0^1\frac{\ln(1-x)}{x(1+x^a)}dx=\frac{1-3a^2}{4a^2}\zeta(2)- \frac{1}{2a} \sum_{j=0}^{a-1} \ln^2\left(2 \sin \frac{(2j+1)\pi}{2a} \right).$$
|
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|
Polynomials - Solutions How I can find the exact solutions of this polynomial?
I can not get to the exact roots of the polynomial ... what methods occupy for this "problem"?
$$x^3+3x^2-7x+1=0$$
Thanks for your help.
|
How I can find the exact solutions of this polynomial?
I can not get to the exact roots of the polynomial
You have a real polynomial of the third degree with real coefficients. There are exact formulas to find the roots of any polynomial of this kind.
A general cubic equation of the form
$$
\begin{equation*}
ax^{3}+bx^{2}+cx+d=0,\tag{1}
\end{equation*}
$$
can be transformed by the substitution
$$
x=t-\frac{b}{a}\tag{2}
$$
into the reduced cubic equation
$$
\begin{equation*}
t^{3}+pt+q=0.\tag{3}
\end{equation*}
$$
In the present case, we have
$$
\begin{equation*}
x^{3}+3x^{2}-7x+1=0,\quad a=1,b=3,c=-7,d=1.\tag{$\mathrm{A}$}
\end{equation*}
$$
For $x=t-1$, we get the reduced equation
$$
\begin{equation*}
t^{3}-10t+10=0,\qquad p=-10,q=10.\tag{$\mathrm{B}$}
\end{equation*}
$$
It is known from the classical theory of the cubic equation that when the discriminant
$$
\Delta =q^{2}+\frac{4p^{3}}{27}=10^{2}+\frac{4\left(-10\right) ^{3}}{27}<0,\tag{$\mathrm{C}$}
$$
the three roots $t_k$ of $(3)$, with $k\in\{1,2,3\}$, are real and can be written in the following trigonometric form $^1$
$$
\begin{eqnarray*}
t_{k} &=&2\sqrt{-p/3}\cos \left( \frac{1}{3}\arccos \left( -\frac{q}{2}\sqrt{-27/p^3}\right) +\frac{(k-1)2\pi }{3}\right).
\end{eqnarray*}\tag{4}
$$
The roots of $(1)$ are thus
$$x_k=t_k-\frac{b}{a}.\tag{5}$$
Consequently,
$$
\begin{eqnarray*}
x_{1} &=&2\sqrt{10/3}\cos \left( \frac{1}{3}\arccos \left( -5\sqrt{27/10^3}\right) \right) -1 \approx 1.4236, \\
x_{2} &=&2\sqrt{10/3}\cos \left( \frac{1}{3}\arccos \left( -5\sqrt{27/10^3}\right) +\frac{2\pi }{3}\right) -1 \approx -4.5771, \\
x_{3} &=&2\sqrt{10/3}\cos \left( \frac{1}{3}\arccos \left( -5\sqrt{27/10^3}\right) +\frac{4\pi }{3}\right) -1 \approx 0.15347.
\end{eqnarray*}\tag{$\mathrm{D}$}
$$
--
$^1$ A deduction can be found in this post of mine in Portuguese.
|
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|
Proof of inequality $\sum\limits_{k=0}^{n}\binom n k\frac{5^k}{5^k+1}\ge\frac{2^n\cdot 5^n}{3^n+5^n}$ Show that
$$\sum_{k=0}^{n}\binom n k\frac{5^k}{5^k+1}\ge\frac{2^n\cdot 5^n}{3^n+5^n}$$
where $$\binom n k=\frac{n!}{k!(n-k)!}$$
|
It seems that we can prove the inequality as follows: $$S=\sum_{k=0}^{n}\binom n k\dfrac{5^k}{5^k+1}=$$
$$\frac 12+\sum_{k=1}^{n}\binom n k\dfrac{1}{1+5^{-k}}=$$
$$\frac 12+\sum_{k=1}^{n}\binom n k\sum_{i=0}^\infty (-1)^i(5^{-k})^i=$$
$$\frac 12+\sum_{i=0}^\infty \sum_{k=1}^{n}\binom n k (-1)^i5^{-ki}=$$
$$\frac 12+\sum_{i=0}^\infty \sum_{k=1}^{n}\binom n k (-1)^i5^{-ik}=$$
$$\frac 12+\sum_{i=0}^\infty (-1)^i((1+5^{-i})^n-1).$$
Since the last series is alternating, its sum is greater than the sum of its four first members. Therefore
$$S\ge \frac 12+(2^n-1)-\left(\left(\frac 65\right)^n-1\right)+\left(\left(\frac {26}{25}\right)^n-1\right)-\left(\left(\frac {126}{125}\right)^n-1\right)=$$
$$2^n-\left(\frac 65\right)^n+\left(\frac {26}{25}\right)^n-\left(\frac {126}{125}\right)^n+\frac 12.$$
Check when $$2^n-\left(\frac 65\right)^n+\left(\frac {26}{25}\right)^n-\left(\frac {126}{125}\right)^n\ge \frac{2^n\cdot 5^n}{3^n+5^n}.$$
After reducing both sides to the common denominator and simplifying we obtain the inequality
$$650^n+390^n\ge 630^n+450^n+378^n.$$ But $$650^6>630^6+450^6+378^6,$$ thus the inequality holds for each $n\ge 6$.
It rests to check the initial inequality for $n\le 5$, which can be done straightforward.
|
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|
If $A = \tan6^{\circ} \tan42^{\circ},~~B = \cot 66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$ My trigonometric problem is:
If $A = \tan6^{\circ} \tan42^{\circ}$ B = cot$66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$.
Working :
$$B = \cot 66^{\circ} \cot78^{\circ} = 1- \frac{\tan24^{\circ}+\tan18^{\circ}}{\tan42^{\circ}}$$
$$A= \tan6^{\circ} \tan42^{\circ} = 1- \frac{\tan6^{\circ} +\tan42^{\circ}}{\tan48^{\circ}}$$
but it seems this is the wrong way of doing this...please suggest. Thanks!
|
First, note that $A \approx 0.0946362785$, $\ \ B \approx 0.0946362785$.
Now, we will prove that
$\ \ \ \Large{A=B.}$
a).
$$
\dfrac{A}{B} =
\dfrac
{\sin 6^\circ \sin 42^\circ}
{\cos 6^\circ \cos 42^\circ}
\cdot \dfrac
{\sin 66^\circ \sin 78^\circ}
{\cos 66^\circ \cos 78^\circ}
=
\dfrac
{\bigl( 2 \sin 6^\circ \sin 66^\circ \bigr) \cdot \bigl( 2 \sin 42^\circ \sin 78^\circ \bigr)}
{\bigl( 2 \cos 6^\circ \cos 42^\circ \bigr) \cdot \bigl( 2 \cos 66^\circ \cos 78^\circ \bigr)}.
\tag{1}
$$
b).
Applying formulas
$\ \ \ \ 2\sin\alpha\sin\beta = \cos(\alpha-\beta) - \cos(\alpha+\beta)$, $\ \ \ $
$\ \ \ \ 2\cos\alpha\sin\beta = \cos(\alpha-\beta) + \cos(\alpha+\beta)$,
$ \ \ \ \ \ (1) \implies$
$$
\dfrac{A}{B} =
\dfrac
{\bigl( \cos 60^\circ - \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ - \cos 120^\circ \bigr)}
{\bigl( \cos 60^\circ + \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ + \cos 120^\circ \bigr)}
=
\dfrac
{\bigl( \frac{1}{2} - \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ + \frac{1}{2} \bigr)}
{\bigl( \frac{1}{2} + \cos72^\circ \bigr) \cdot \bigl( \cos 36^\circ - \frac{1}{2} \bigr)}.
\tag{2}
$$
c).
It is known, that $\ \ \ \ \cos72^\circ = \sin 18^\circ = \frac{1}{4}(\sqrt{5}-1)$,
so
$\ \ \ \ \ \ \cos 36^\circ = 1 - 2(\sin 18^\circ)^2 = \frac{8}{8} - \frac{1}{8}(5-2\sqrt{5}+1) = \frac{1}{4}(\sqrt{5}+1)$,
and
$\ \ \ \ \cos 72^\circ \cos 36^\circ =
\frac{1}{16}(\sqrt{5}-1)(\sqrt{5}+1) = \frac{4}{16}=\frac{1}{4}$.
Hence (2) $\implies$
$$
\dfrac{A}{B} =
\dfrac
{\frac{1}{2}\cos 36^\circ - \cos 72^\circ \cos 36^\circ +\frac{1}{4}-\frac{1}{2}\cos 72^\circ}
{\frac{1}{2}\cos 36^\circ +\cos 72^\circ \cos 36^\circ -\frac{1}{4}-\frac{1}{2}\cos 72^\circ }
=
\dfrac
{\frac{1}{2}(\cos 36^\circ - \cos 72^\circ)}
{\frac{1}{2}(\cos 36^\circ - \cos 72^\circ) }
=\Large{1}.
\tag{3}
$$
Proved.
|
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|
Solution of Lagrangian Could you give me advice how to solve the following Lagrangian?
$$L=x^3+y^3 - \lambda (x^2-xy+y^2-5)$$
$$ \left\{ \begin{array}{c}
\frac{\partial L}{\partial x} = 3x^2 - \lambda (2x-y) = 0
\\\frac{\partial L}{\partial y} = 3y^2 - \lambda (-x+2y) = 0
\\ \frac{\partial L}{\partial \lambda} = x^2-xy+y^2=5
\end{array} \right. $$
$$ \left\{ \begin{array}{c}
3x^2 / (2x-y)= \lambda
\\ 3y^2 / (-x+2y) = \lambda
\\ x^2-xy+y^2=5 \Rightarrow x^2 = xy-y^2+5
\end{array} \right. $$
$$\Rightarrow \lambda = 3x^2 / (2x-y)= 3y^2 / (-x+2y) \Rightarrow x^2 = \frac{(2x-y)(y^2)}{2y-x} \Rightarrow \frac{(2x-y)(y^2)}{2y-x} = xy-y^2+5 \Rightarrow (2x-y)(y^2) = (xy-y^2+5) (2y-x) \Rightarrow 2xy^2 - y^3 = 2xy^2 - 2y^3 +10y - x^2y +xy^2 -5x \Rightarrow 0 = - y^3 +10y - x^2y +xy^2 -5x $$
I don't know how to transform the above in order to find x,y.
|
First
$$\lambda = \frac{3x^2}{2x-y} = \frac{3y^2}{-x+2y}\implies -x^3 +2x^2y = 2y^2x - y^3.$$
Using the factoring formula for $x^n -y^n$ when $n$ is an odd number:
$$
x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1}).
$$
Hence we have
$$
x^3 -y^3 + 2y^2x - 2x^2y = 0\implies (x-y)(x^2-xy+y^2) = 0.
$$
I believe you can take it from here.
|
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|
Solve $\frac{dx}{dt} = x^3 + x$ for $x$ This is a seemingly simple first order separable differential equation that I'm getting stuck on. This is what I have so far:
$$\frac{dx}{dt} = x^3+x$$
goes to
$$\frac{dx}{x(1+x^2)} = dt$$
Now using partial fractions to integrate the left-hand side:
$$\frac{1}{x(1+x^2)} = \frac{A}{x} + \frac{Bx+C}{1+x^2}$$
Solving for A, B, C:
$1 = A(1+x^2) + (Bx+C)x$, and using coefficient matching, I get $A=1, B=-1, C=0$.
So the integral yields:
$$\int\frac{1}{x} - \frac{x}{1+x^2}dx = \int dt$$
This yields: $$\ln x -\frac{1}{2}\ln(1+x^2) =t + C$$
So I tried using log rules and such to solve for $x$. I think this is where my source of error is.
My attempt:
$$\ln x - \ln(1+x^2)^\frac{1}{2} = t+C$$
$$\ln\frac{x}{(1+x^2)^{\frac{1}{2}}} = t + C$$
$$\frac{x}{(1+x^2)^{\frac{1}{2}}}= Ce^t$$
But this seems wrong. I apologize in advance if it's a silly mistake that I didn't see.
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You should get
$$ C^2e^{2t}=\frac{x^2}{1+x^2}=1-\frac1{1+x^2},$$
hence
$$ x=\pm\sqrt{\frac{1}{1-C^2e^{2t}}-1}.$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/433966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Transforming trigonometric identities The problem goes like this:
If
$$N=2\sec^4x-3\sec^2x+2=\frac{\cos^2x}{\cos^2y}$$
Calculate the equivalent of
$$M=2\tan^4x+3\tan^2x+2$$
The alternaties I have are:
$$\frac{\tan^2x}{\tan^2y},\mbox{ }\frac{\tan^2y}{\tan^2x},\mbox{ }\frac{\tan^2y}{\sec^2x},\mbox{ }\frac{\sec^2y}{\tan^2x},\mbox{ }\frac{\sec^2x}{\tan^2y}$$
The first thing I tried was to "build" the value of $N$ and then use it for $M$
$$\sec^2x=1+\tan^2x\\
\sec^4x=1+2\tan^2x+\tan^4x\\
2\sec^4x=2+4\tan^2x+2\tan^4x\\
-3\sec^2x=-3-3\tan^2x\\
2\sec^4x-3\sec^2x+2=1+\tan^2x+2\tan^4x\\
\frac{\cos^2x}{\cos^2y}=1+\tan^2x+2\tan^4x\\
\frac{\cos^2x}{\cos^2y}+2\tan^2x+1=2+3\tan^2x+2\tan^4x\\
\frac{\cos^2x}{\cos^2y}+2\tan^2x+1=M$$
But then I can't transform the final equation to one of the alternatives, even after trying a massive substition of $\cos^2y$ it didn't helped too much. Any hints or ideas are greatly appreciated.
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Some ideas:
$$1+\tan^2x=\sec^2x\implies (1+\tan^2x)^2=\sec^4x$$
But
$$(1+\tan^2x)^2=1+2\tan^2x+\tan^4x$$
So
$$2\tan^4x+3\tan^2x+2=2(\tan^4x+2\tan^2x+1)-\tan^2x=2\sec^4x-\tan^2x=$$
$$=2\sec^4x-\sec^2x+1$$
Try now to end the exercise.
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{
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"timestamp": "2023-03-29T00:00:00",
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|
Plotting graphs using numerical/mathematica method From the author's equation 13, 14 We can write by inserting V''(A)=0,
Solving for R we get,
$$R= \frac{6^{D/4} \sqrt{D}}{\sqrt{-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2}}$$
Now inserting the V into the article equation (11)$$E= \left(\frac{\pi }{2}\right)^{D/2} R^D V,$$ we get,
$$E= \left(\frac{\pi }{2}\right)^{D/2} \left(-\left(\frac{2}{3}\right)^{D/2} A^3+2^{\frac{1}{2} (-4-D)} A^4+A^2 \left(1+\frac{D}{2 R^2}\right)\right) R^D$$
Now inserting the value of R, we get,
$$E= \left(-\left(\frac{2}{3}\right)^{D/2} A^3+2^{\frac{1}{2} (-4-D)} A^4+A^2 \left(1+2^{-1-\frac{D}{2}} 3^{-D/2} \left(-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2\right)\right)\right) \left(\frac{6^{D/4} \sqrt{D}}{\sqrt{-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2}}\right)^D \left(\frac{\pi }{2}\right)^{D/2}$$
For $D= 3$
we finally get,
$$E= \frac{27 6^{3/4} \left(-\frac{2}{3} \sqrt{\frac{2}{3}} A^3+\frac{A^4}{8 \sqrt{2}}+A^2 \left(1+\frac{-12 \sqrt{6}+48 A-9 \sqrt{3} A^2}{12 \sqrt{6}}\right)\right) \pi ^{3/2}}{\left(-12 \sqrt{6}+48 A-9 \sqrt{3} A^2\right)^{3/2}} \tag{1}$$
the graph for equation (1) must satisfy the article graph (FIG 2)
My graph:
Plot[(27 6^(3/4) (-(2/3) Sqrt[2/3] A^3 + A^4/(8 Sqrt[2]) +A^2 (1 + (-12 Sqrt[6] + 48 A - 9 Sqrt[3] A^2)/( 12 Sqrt[6]))) \[Pi]^(3/2))/(-12 Sqrt[6] + 48 A - 9 Sqrt[3] A^2)^(3/2), {A, 0.5, 2.5}]
But the author got,
Output :
Am I doing wrong in simulation?
Then The author got like this in Fig 3
`
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Mathematica is more powerful that you give it credit for. You don't have to define any quantities explicitly. It is much more efficient to keep things in symbolic terms and numerically substitute values only when you need them. The below works for $d=3$ and gives results consistent with the paper.
Briefly:
*
*Define the function $V(A)=\frac{A^4}{2^{\frac{d+4}{2}}}-A^3 \left(\frac{2}{3}\right)^{d/2}+A^2 \left(\frac{d}{2 R^2}+1\right);$
*Calculate the solution to $V''(A)=0$, take the second $R$ solution
*Plug that $R$ into $E(A)=\left(\frac{\pi }{2}\right)^{d/2} V(A) R^d$ and plot
*Solve $E'(A)=0$, pick the third solution and call the minimum value $E_\infty$
*Solve $\left(\frac{\pi }{2}\right)^{d/2} V(A) R^d=E_\infty$ for $R$ as a function of $A$ (taking the third solution); this is the locus of points that attain the energy $E_\infty$
To get the right answers you have to simply select the "right" solution that Mathematica spits out when solving the various equations. The code for $d=3$ appears below and recreates the paper results.
(*Change this to 2 or 3*)
d = 3;
(*Define the potential*)
V[A_] := (1 + d/(2 R^2)) A^2 - (2/3)^(d/2) A^3 + A^4/2^((d + 4)/2);
(*Find the energy*)
Energy[A_] :=
Evaluate[(\[Pi]/2)^(d/2) R^d V[A] /. Solve[V''[A] == 0, R][[2]]];
Plot[Energy[A], {A, 0, 2.5}, PlotRange -> {0, 100}]
(*This is the minimum energy, matches article for d=2 and d=3*)
NSolve[D[Energy[A], A] == 0, A][[3]]
Einf = Energy[A] /. NSolve[D[Energy[A], A] == 0, A][[3]]
Plot[R /. NSolve[(\[Pi]/2)^(d/2) R^d V[A] == Einf, R][[3]], {A, 0,
2.5}, PlotRange -> {0, 6}]
And the plots are
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"timestamp": "2023-03-29T00:00:00",
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|
How to calculate $\cos(6^\circ)$? Do you know any method to calculate $\cos(6^\circ)$ ?
I tried lots of trigonometric equations, but not found any suitable one for this problem.
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If you grant the use of $\cos 18^\circ$ = $\dfrac{\sqrt{10+2\sqrt5}}{4}$, you may proceed as follows:
First, calculate $\sin 18^\circ$.
Then, calculate $\cos 36^\circ$ and $\sin 36^\circ$ using $\cos 2\theta = 2\cos^2 \theta - 1$ and $\sin 2\theta = 2\sin \theta \cos \theta$ for $\theta = 18^\circ$.
Finally, use $\cos 6^\circ = \cos (36^\circ-30^\circ) = \cos 36^\circ \cos 30^\circ + \sin 36^\circ \sin 30^\circ$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/438387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
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|
Finding $a + b + c$ given that $\;a + \frac{1}{b+\large\frac 1c} = \frac{37}{16}$ Please help me to find the needed sum:
If $a,b,c$ are positive integers such that $\;a + \dfrac{1}{b+\large \frac 1c} = \dfrac{37}{16},\;$
find the value of $\;(a+b+c)$.
Thanks!
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This is asking for the Continued Fraction expansion of $\frac{37}{16}$. We can use the Euclid-Wallis Algorithm to compute the expansion:
$$
\begin{array}{r}
&&2&3&5\\\hline
1&0&1&-3&16\\
0&1&-2&7&-37\\
37&16&5&1&0\\
\end{array}
$$
The continued fraction is above the horizontal line:
$$
\frac{37}{16}=2+\cfrac1{3+\cfrac1{5}}
$$
Then, $a+b+c=2+3+5=10$.
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{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
The least value of $4x^2-4ax +a^2-2a+2$ on $[0,2]$ is $3$. What is the integer part of $a$? The least value of $4x^2-4ax +a^2-2a+2$ on $[0,2]$ is $3$. What is the integer part of $a$?
We know that minimum value of a quadratic is $-\cfrac{b}{2a}$.
We will get one condition from here and $-\cfrac{b}{2a}$ should be equal to $3$.
But the problem is that this limit is for the whole function, not for an interval, and it might not apply to the interval we have been given.
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Hint: there are three possibilities, given the shape of the graph of a quadratic
Let $f(x)=4x^2-4ax +a^2-2a+2$, then
either $f(x)$ is increasing on $[0,2]$, in which case the minimum value occurs at $x=0$
or $f(x)$ is decreasing on $[0,2]$, in which case the minimum value occurs at $x=2$
or $f(x)$ has minimum value within $[0,2]$
Test each of the three possibilities and see which gives you a consistent solution.
Added later
If $x=0$ we need to solve $a^2-2a+2=3$, so $a=1\pm\sqrt2$. But then we have to check those two values of $a$ to see whether $f$ is decreasing (which is the condition for using $x=0$).
If $x=2$ we have $16-8a+a^2-2a+2=a^2-10a+18=3$ whence $a=5\pm\sqrt{10}$, and we need to test the deacreasing condition.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/442654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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.