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If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then what values of $(a,b)$ would work? Let $a$ and $b$ be distinct integers. If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then which of the following statements are true?
1) $c*(a+c)=-b$
2) $a+b=-1$
3) $a+b+c=0$
4) $c=0$
Update
I just tried to sub $c$ into both of the equations:
$c^2+cb+a=0$ and $c^2+ac+b=0$ which then gives us the equality
$c^2+cb+a=c^2+ac+b$
$ => cb+a=ac+b$
$=> b(c-1)=a(c-1)$
which gives me then a=b which is contradictory because the integers are supposed to be distinct.
Update #2
Ok it looks like 1) is true, 2) is true, 3) is true, and 4) is false .. right?
|
Factor Theorem implies $x-c$ is a factor of $x^2+ax+b-(x^2+bx+a)$, and this is equal to $(a-b)(x-1)$, so $c=1$, let $r$ be the another root of $x^2+ax+b$ then $(x-1)(x-r)=x^2-(1+r)x+r=x^2+ax+b$, so $r=b$ and $a=-(1+r)=-1-b$. Therefore $a+b=-1$.
|
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|
Find the limit: limit x tends to zero [arcsin(x)-arctan(x)]/(x^3) I'm having difficulty in finding the following limit.
$$\lim_{x\to 0}\frac{\arcsin(x)-\arctan(x)}{x^3}$$
I tried manipulating the given limit in standard limit(s) but I got nowhere. I tried L'Hôpital's rule and then realized it would be very lengthy and daunting so I left it. The answer is 1/2.
Thanks for any help in advance.
|
It's worth mentioning that l'Hopital does work here. Your limit becomes
$$\lim_{x \rightarrow 0} {{1 \over \sqrt{1 - x^2}} - {1 \over 1 + x^2} \over 3x^2}$$
$$= \lim_{x \rightarrow 0}{(1 + x^2) - \sqrt{1 - x^2} \over 3x^2(1 + x^2)\sqrt{1 - x^2}}$$
Multiply numerator and denominator by $(1 + x^2) + \sqrt{1 - x^2}$ and you get
$$= \lim_{x \rightarrow 0}{(1 + x^2)^2 - (1 - x^2) \over 3x^2 ((1 + x^2) + \sqrt{1 - x^2})(1 + x^2)\sqrt{1 - x^2}}$$
$$= \lim_{x \rightarrow 0}{3 + x^2 \over 3((1 + x^2) + \sqrt{1 - x^2})(1 + x^2)\sqrt{1 - x^2}}$$
You can now just plug in $x = 0$ into the above and get ${1 \over 2}$.
|
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|
$n$ is some natural number. Let $x$ be the integer part of $\sqrt n$ and $y$ be the decimal part. If $x^2 - y^2 = 1+4y$ what is $y^x$? $n$ is some natural number. Let $x$ be the integer part of $\sqrt n$ and $y$ be the decimal part. If $x^2 - y^2 = 1+4y$ what is $y^x$?
This is some high school problem but I can't solve it. Any help? Thanks.
|
Since $y$ is the decimal part, we know that $0\leq{y}<1$, then from $x^2=1+4y+y^2$ we get $$1\leq{x^2}<6$$. $x$ is an integer, then $x$ can be $1$ or $2$.
For $x=1$, $y=0$, then $y^x=0$;
for $x=2$, $y=\sqrt{7}-2$, and we have $n=7$ which satisfies the requirement. Then $y^x=11-4\sqrt{7}$.
|
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Let $n$ be a positive integer. Show that if $2^n -1$ is a prime number, then $n$ is a prime number. Let $n$ be a positive integer. Show that if $2^n -1$ is a prime number, then $n$ is a prime number.
This is how I started to tackle this question:
Assume that instead of $n$ being a prime number, it is a composite number. Let $n=ab$ where a and b are factors. Thus, we have $2^n-1 = 2^a*2^b-1$ When we try to factor out the $2^b$ out, we will get:
$2^b * (2^a-1/2^b)$ Since $2^b$ is one of the factors, this is a contradiction. Therefore n must be a prime number.
I know that the $-1/2^b$ is not correct as it makes one of the factors a fraction. How do I change this a little bit to make sure that my proof is right?
|
It's probably easier to prove the contrapositive: if $n$ is composite then $2^n - 1$ is composite. So suppose $n$ is composite, i.e. $n=ab$ with $a$ and $b$ integers greater than $1$. Using the factorization
$$
y^a - 1 = (y - 1)(y^{a-1} + y^{a-2} + \cdots + y + 1)
$$
and writing $2^n - 1 = 2^{ab} - 1 = (2^b)^a - 1$, we have
$$
2^n - 1 = (2^b - 1)(2^{b(a-1)} + 2^{b(a-2)} + \cdots + 2^b + 1).
$$
Specifically, I let $y=2^b$ in the factorization I gave earlier.
Then $2^b-1$ divides $2^n-1$ but $2^b - 1$ is strictly less than $2^n-1$, as $b<n$. So $2^n-1$ is composite.
|
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|
Find the value of $(a+b+c)$ when $\cos\theta+\cos^2\theta+\cos^3\theta=1$ and $\sin^6\theta=a+b\sin^2\theta+c\sin^4\theta$ Given: $\cos\theta+\cos^2\theta+\cos^3\theta=1$
and $\sin^6\theta=a+b\sin^2\theta+c\sin^4\theta$
Then find the value of $(a+b+c)$
|
Clearly we need to eliminate $\cos\theta$
We have $\displaystyle\cos\theta(1+\cos^2\theta)=1-\cos^2\theta\iff\cos\theta(1+1-\sin^2\theta)=\sin^2\theta$
Squaring we get $$\cos^2\theta(2-\sin^2\theta)^2=(\sin^2\theta)^2$$
$$\iff(1-\sin^2\theta)(2-\sin^2\theta)^2=\sin^4\theta$$
Rearrange to form a six-degree equation in $\sin\theta$ and compare with the given six degree equation
for the constants and the coefficients of $\displaystyle\sin^2\theta,\sin^4\theta,\sin^6\theta$
|
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|
Solving Non-homogenous System (repeated case) I have the following system. $\vec{x^{'}}(t)=\begin{pmatrix}4&-2\\8&-4 \end{pmatrix} \vec{x}+ \begin{pmatrix}t^{-3}\\-t^{-2}\end{pmatrix}$
I get $\lambda=0$ and the eigenvector of $$4x_1=2x_2 \implies \begin{pmatrix} 1\\2 \end{pmatrix}$$
Next I solved $$\begin{pmatrix}4&-2\\8&-4\end{pmatrix}\begin{pmatrix}\vec{o_1}\\ \vec{o_2} \end{pmatrix}=\begin{pmatrix}1\\2 \end{pmatrix} \implies \vec{o_1}=\dfrac{1+2\vec{o_2}}{4}$$
So can choose a eigenvector of $\begin{pmatrix} 0\\\frac{-1}{2} \end{pmatrix}$
Therefore, since $\lambda=0$ the fundamental matrix will be just the eigenvectors coupled together that is $$\phi(t)=\begin{pmatrix} 1&t\\2&2t-\frac{1}{2} \end{pmatrix}\implies \phi^{-1}(t)=\begin{pmatrix}1-4t&2t\\4&-2\end{pmatrix}$$
I multiply this matrix by $g(t)$ to get $$\begin{pmatrix}t^{-3}-4t^{-2}-2t^{-1}\\4t^{-3}+2t^{-2}\end{pmatrix}$$
Taking the integral gives me $$\begin{pmatrix} \frac{-1}{2}t^{-2}+4t^{-1}-2\ln t\\-2t^{-2}-2t^{-1} \end{pmatrix}$$
I multiply by $\phi(t)$ to get $\begin{pmatrix}1&t\\2&2t-\frac{1}{2}\end{pmatrix}\begin{pmatrix} \frac{-1}{2}t^{-2}+4t^{-1}-2\ln t\\-2t^{-2}-2t^{-1} \end{pmatrix}=\begin{pmatrix}-\frac{1}{2}t^{-2}-2\ln t+2t^{-1} -2\\-4\ln t+5t^{-1}-4 \end{pmatrix}$
Why cant I get these right...
Therefore all that is left is to multiply by $\phi$ and add the homogenous solution which is simply $\begin{pmatrix}1\\2 \end{pmatrix}c_1+c_2\bigg[t\begin{pmatrix}1\\2\end{pmatrix}+\begin{pmatrix}\frac{1}{4}\\0 \end{pmatrix}\bigg]+ x_p$ where $x_p$ is the particular solution. In the book they got something way off that is $$\begin{pmatrix}1\\2 \end{pmatrix}c_1+c_2\bigg[\begin{pmatrix}1\\2\end{pmatrix}t-\frac{1}{2}\begin{pmatrix}0\\1 \end{pmatrix}\bigg]-2\begin{pmatrix}1\\2 \end{pmatrix} \ln t+\begin{pmatrix}2\\5 \end{pmatrix}t^{-1}-\begin{pmatrix}\frac{1}{2}\\0 \end{pmatrix}t^{-2}$$
|
Given:
$$\vec{x^{'}}(t)=\begin{pmatrix}4&-2\\8&-4 \end{pmatrix} \vec{x}+ \begin{pmatrix}t^{-3}\\-t^{-2}\end{pmatrix}$$
We have as eigenvalues and eigenvectors:
$$\lambda_{1,2} = 0, v_1 = (1,2), v_2 = \left(\dfrac{1}{4},0\right)$$
Update 1: Note that eigenvectors are not unique and your book chose $v_2 = \left( 0, -\dfrac{1}{2}\right)$.
From this, we can write the general solution as:
$$x(t) = e^{0t}\left(c_1 \begin{bmatrix} 1\\2 \end{bmatrix} + c_2\left(\begin{bmatrix} 0\\-\frac{1}{2} \end{bmatrix} + t\begin{bmatrix} 1\\2 \end{bmatrix} \right) \right) = \begin{bmatrix} c_1 + c_2 t \\2c_1 + c_2\left(-\frac{1}{2} +2 t \right)\end{bmatrix}$$
This means we have:
$$\phi(t) = \begin{bmatrix} 1 & t \\2 & -\frac{1}{2} + 2 t \end{bmatrix}$$
Now, to find the exponential, just take:
$$e^{At} = \phi(t) \phi^{-1}(0) = \begin{bmatrix} 1 + 4 t & -2 t \\ 8 t & 1 - 4 t \end{bmatrix}$$
Do you see where your $\phi(t)$ went astray? Recall, we have a repeated eigenvalue with a generalized eigenvector!
Update 2
Recall, we have:
$$X(t) = e^{At}X_0 + \int_{t_0}^t e^{A(t-s)}F(s)~ds$$
|
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|
Calculation of $\int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$ Calculation of $\displaystyle \int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$
$\bf{My\; Try::}$ Let $x=\tan \psi\;,$ Then $\displaystyle dx = \sec^2 \psi$
So Integral convert into $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi-1+\sec \psi}{\tan \psi+1+\sec \psi}d\psi = \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi+\sec \psi - 1}{\tan \psi+\sec \psi+1}d\psi$
Now Multiply both Numerator and Denominator by $\left(\tan \psi+\sec \psi-1\right)$
So $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi+\sec \psi-1}{\tan \psi+\sec \psi+1} \times \frac{\tan \psi+\sec \psi-1}{\tan \psi+\sec \psi-1}d\psi = \int_{0}^{\frac{\pi}{4}}\frac{\left(\tan \psi+\sec \psi-1\right)^2}{\left(\tan \psi+\sec \psi\right)^2-1}d\psi$
Now I did not understand how can i solve after that
Help required
Thanks.
|
Let $x=\sqrt{y^2-1}$ to reduce the integral
\begin{align}
&\int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx
=\int_1^{\sqrt2}\frac y{1+y}dy=\sqrt2-1+\ln[2(\sqrt2-1)]
\end{align}
|
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|
If $x\equiv2\pmod{3}$ prove that $3|4x^2+2x+1$ I've tried many different things to get a factor of $k-2$ but keep failing.
If $x\equiv2\pmod{3}$ prove that $3 \mid 4x^2+2x+1$
|
Since $x \equiv 2 \pmod 3$,
$4x^2+2x+1 \equiv (16) + (4) + (1) \equiv 21 \equiv 0 \pmod 3$
Hence, $3\mid4x^2+2x+1$.
|
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Does $\int_0^{2 \pi} \sqrt{1-(a+b \sin\phi)^2} d\phi $ have a closed form in terms of elliptic integrals? Consider the following integral for real $a, b$ such that the square root is real:
\begin{equation}
I=\int_0^{2 \pi} \sqrt{1-(a+b \sin\phi)^2} d\phi
\end{equation}
For $a = 0$, the integral is easily expressed in terms of the complete elliptic function of the second kind. Does the integral have a closed form (in terms of special functions) for arbitrary $a$? Mathematica did not yield a useful result, even for specific exact values of $a,b$.
|
Result:
$$\boxed{\displaystyle \mathcal{I}=4\sqrt{\frac{b}{k}}\,\biggl[\mathbf{E}\left(k^2\right)-\left(1-k^2\right)\mathbf{K}\left(k^2\right)+\left(1-k^2\right)\Pi\left(c^{-2}|k^2\right)\biggr]} \tag{$\heartsuit$}$$
where I follow Mathematica conventions for arguments of $\mathsf{EllipticE}$, $\mathsf{EllipticK}$ and $\mathsf{EllipticPi}$, and the parameters $k$ and $c$ are defined below by (0) and (2), respectively.
Derivation:
$\square$ The underlying elliptic curve $\mathcal{C}$ is described by the equation
$$y^2=\left(1-z^2\right)\left(1-(a+bz)^2\right).$$
Topologically this is a torus realized as a two-sheeted covering of $\mathbb{P}^1$ branched at $4$ points $z_{1\ldots4}=\pm1,\frac{\pm1-a}{b}$. These can be put to $\pm1,\pm k^{-1}$ by a fractional linear transformation. Such transformations preserve the anharmonic ratio, which allows to identify
$$\frac{z_{21}z_{43}}{z_{31}z_{42}}=\frac{4b}{a^2-(1-b)^2}=-\frac{4k}{(1-k)^2}.$$
Choose a solution
$$k=\frac{1+b^2-a^2-\sqrt{(1+b^2-a^2)^2-4b^2}}{2b}.\tag{0}$$
The relevant fractional linear transformation bringing $\mathcal{C}$ to the standard Legendre form is thus
$$z=\frac{a w+(k-b)}{(k-b) w+a}.$$
After these preliminary remarks, we can implement the corresponding change of variables in the integral we are interested in:
\begin{align}
\mathcal{I}=&\int_0^{2\pi}\sqrt{1-(a+b\sin\phi)^2}\,d\phi=\\=& 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{1-(a+b\sin\phi)^2}\,d\phi=\\
=&2\int_{-1}^1 \sqrt{\frac{1-(a+bz)^2}{1-z^2}}dz=\\=&2\left[\frac{a^2}{(b-k)^2}-1\right]\sqrt{\frac{b}{k}}\int_{-1}^1\sqrt{\frac{1-k^2w^2}{1-w^2}}\frac{dw}{\left(w-\frac{a}{b-k}\right)^2}=\\
=&2\left(c^2-1\right)\sqrt{\frac{b}{k}}\int_{-1}^1\frac{1-k^2w^2}{\left(w-c\right)^2}\frac{dw}{\lambda},\tag{1}
\end{align}
with $\lambda=\sqrt{(1-w^2)(1-k^2w^2)}$ and
$$c=\frac{a}{b-k}.\tag{2}$$
It is possible to reduce the last integral to a combination of integrals of three types:
$$\int\frac{dw}{\lambda},\qquad \int\frac{w^2dw}{\lambda},\qquad \int\frac{dw}{(w-c)\lambda},\tag{3}$$
producing elliptic integrals of the 1st, 2nd and 3rd type, respectively.
The idea is to write
\begin{align}
\frac{1-k^2w^2}{\left(w-c\right)^2}\frac{1}{\lambda}=\left[\alpha+\frac{\beta}{w-c}+\frac{\gamma}{(w-c)^2}\right]\frac{1}{\lambda}
\end{align}
The $\alpha$- and $\beta$-term give elliptic integrals of the 1st and 3rd type. To reduce the $\gamma$-term to integrals of type (3), compare it with the derivative
\begin{align}\frac{d}{dw}\frac{\lambda}{w-c}=-\frac{(c^2-1)(c^2k^2-1)}{(w-c)^2\lambda}
-\frac{c(c^2k^2-1)+ck^2(c^2-1)}{(w-c)\lambda}+\frac{k^2(w^2-c^2)}{\lambda}.
\end{align}
Altogether this allows us to write
$$\frac{1-k^2w^2}{\left(w-c\right)^2}\frac{1}{\lambda}=\frac{1}{c^2-1}\left[\frac{d}{dw}\frac{\lambda}{w-c}-\frac{c(1-k^2)}{(w-c)\lambda}+\frac{k^2(1-w^2)}{\lambda}\right],$$
which in turn implies
$$\mathcal{I}=2\sqrt{\frac{b}{k}}\int_{-1}^1\left[-\frac{c(1-k^2)}{(w-c)\lambda}+\frac{k^2(1-w^2)}{\lambda}\right]dw.$$
After some straightforward algebra, this finally gives the above result ($\heartsuit$). $\blacksquare$
|
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How to solve this equation or system of equations? I want to solve the equation
$$(5 x-4) \cdot\sqrt{2 x-3}-(4 x-5)\cdot \sqrt{3 x-2}=2.$$
I tried. Put $a = \sqrt{2 x-3}\geqslant 0$ and $b =\sqrt{3 x-2}\geqslant 0 $.
Suppose
$$5x-4=m(2x-3)+n(3x-2)$$
then $m=\dfrac{2}{5}$ and $n=\dfrac{7}{5}$.
Therefore
$$5x-4=\dfrac{2}{5}(2x-3)+\dfrac{7}{5}(3x-2)=\dfrac{2}{5}a^2 + \dfrac{7}{5}b^2.$$
Similarly, we have
$$4x-5=\dfrac{7}{5}a^2+\dfrac{2}{5}b^2.$$
The given system of equation are written
$$\begin{cases}
\left (\dfrac{2}{5}a^2 + \dfrac{7}{5}b^2\right )a -\left (\dfrac{7}{5}a^2 + \dfrac{2}{5}b^2\right )b=2,\\
3a^2 -2b^2=-5.
\end{cases}$$
Equavalent to
$$\begin{cases}
2(a^3-b^3)-7ab(a-b)=10,\\
3a^2 -2b^2=-5.
\end{cases}$$
Now, I can not solve the last system of equations. How can I solve the equation
$$(5 x-4) \cdot\sqrt{2 x-3}-(4 x-5)\cdot \sqrt{3 x-2}=2$$
or solve the system of equations
$$\begin{cases}
2(a^3-b^3)-7ab(a-b)=10,\\
3a^2 -2b^2=-5.
\end{cases}$$
|
Hint: try integer values...usually in this kind of problem first try to see when you get integer result, so both $\sqrt{3x-2} \text{ and } \sqrt{2x-3}$ should be integers...solution: $x=6$
|
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|
inverse of an infinite matrix How to find inverse of an infinite lower triangular matrix all of whose diagonal entries are 1 and the entries of each column are given by coefficients of some power series rings?
|
Just use gaussian elimination. You have an infinite matrix $(a_{ij})_{i,j \in \mathbb{N}}$ where $a_{ii} = 1$ for all $i$, and $a_{ij} = 0$ if $i < j$. We're looking for a (lower triangular) matrix $(b_{ij})$ such that $$
\begin{pmatrix}
1 & 0 & \ldots \\
a_{21} & 1 & 0 & \ldots \\
a_{31} & a_{32} & 1 & \ddots \\
\vdots & \vdots & \ddots & \ddots
\end{pmatrix}
\cdot
\begin{pmatrix}
1 & 0 & \ldots \\
b_{21} & 1 & 0 & \ldots \\
b_{31} & b_{32} & 1 & \ddots \\
\vdots & \vdots & \ddots & \ddots
\end{pmatrix}
=
\sum_{k=1}^\infty a_{ik} b_{kj} = \sum_{k=1}^i a_{ik} b_{kj} = \delta_{i,j} \text{.}
$$
That yields $$\begin{eqnarray}
a_{21} b_{12} &=& \delta_{2,1} &=& 0 & \quad\Rightarrow b_{21} = \ldots\\
a_{31} + a_{32} b_{21} + b_{31} &=& \delta_{3,1} &=& 0 &\quad\Rightarrow b_{31} = \ldots\\
a_{32} + b_{32} &=& \delta_{3,2} &=& 0 & \quad\Rightarrow b_{32} = \ldots\\
a_{41} + a_{42} b_{21} + a_{43} b_{31} + b_{41} &=& \delta_{4,1} &=& 0 & \quad\Rightarrow b_{41} = \ldots
\end{eqnarray}$$
and so on.
|
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|
Generating function of $\binom{3n}{n}$ Wolfram alpha tells me the ordinary generating function of the sequence $\{\binom{3n}{n}\}$ is given by
$$\sum_{n} \binom{3n}{n} x^n = \frac{2\cos[\frac{1}{3}\sin^{-1}(\frac{3\sqrt{3}\sqrt{x}}{2})]}{\sqrt{4-27x}}$$
How do I prove this?
|
As was already mentioned in the comment section the Lagrange Inversion Formula is a proper method to prove this identity. In the following I use the notation from R. Sprugnolis (etal) paper Lagrange Inversion: when and how.
Let us suppose that a formal power series $w=w(t)$ is implicitely defined by a relation $w=t\Phi(w)$, where $\Phi(t)$ is a formal power series such that $\Phi(0)\ne0$. The Lagrange Inversion Formula (LIF) states that:
$$[t^n]w(t)^k=\frac{k}{n}[t^{n-k}]\Phi(t)^n$$
There are several variations of the LIF stated in the paper. We use in the following $G6$:
Let $F(t)$ be any formal power series and $w=t\Phi(w)$ as before, then the following is valid:
\begin{align*}
[t^n]F(t)\Phi(t)^n=\left[\left.\frac{F(w)}{1-t\Phi'(w)}\right|w=t\Phi(w)\right]\tag{1}
\end{align*}
Note: The notation $[\left.f(w)\right|w=g(t)]$ is a linearization of $\left.f(w)\right|_{w=g(t)}$ and denotes the substitution of $g(t)$ to every occurrence of $w$ in $f(w)$ (that is, $f(g(t))$). In particular, $w=t\Phi(w)$ is to be solved in $w=w(t)$ and $w$ has to be substituted in the expression on the left of the $|$ sign.
We prove the following identity
\begin{align*}
\sum_{n\ge0}\binom{3n}{n}t^n=\frac{2\cos\left[\frac{1}{3}\sin^{-1}\left(\frac{3\sqrt{3}\sqrt{t}}{2}\right)\right]}{\sqrt{4-27t}}\tag{2}
\end{align*}
$$ $$
Let $F(t)=1$ and $\Phi(t)=(1+t)^3$.
Since
$$t\Phi'(w)=3t(1+w)^2=\frac{3t\Phi(w)}{1+w}=\frac{3w}{1+w}$$
we obtain
\begin{align*}
\binom{3n}{n}&=[t^n]F(t)\Phi(t)^n=[t^n](1+t)^{3n}\\
&=[t^n]\left[\left.\frac{1}{1-t\Phi'(w)}\right|w=t\Phi(w)\right]=[t^n]\left[\left.\frac{1+w}{1-2w}\right|w=t\Phi(w)\right]\\
\end{align*}
Let
\begin{align*}
A(t):=\sum_{n\ge0}\binom{3n}{n}t^n=\left.\frac{1+w}{1-2w}\right|_{w=t\Phi(w)}
\end{align*}
Expressing $A(t)=\frac{1+w}{1-2w}$ in terms of $w$, we get
$$w=\frac{A(t)-1}{2A(t)+1}$$
Since $w=t\Phi(w)=t(1+w)^3$, we get
\begin{align*}
\frac{A(t)-1}{2A(t)+1}=t\left(1+\frac{A(t)-1}{2A(t)+1}\right)^3
\end{align*}
which simplifies to:
\begin{align*}
(4-27t)A(t)^3-3A(t)-1=0\tag{3}
\end{align*}
In order to get the RHS of $(2)$ we first analyse the structure of $(3)$ which is
$$f(t)A(t)^3-3A(t)=1$$
with $f(t)$ linear and observe a similarity of this structure with the identity
$$4\cos^3{t}-3\cos{t}=\cos{3t}$$
We use the Ansatz:
$$A(t) := \frac{2\cos\left(g(t)\right)}{\sqrt{4-27t}}$$
and obtain
\begin{align*}
(4-27t)&A(t)^3-3A(t)=\\
&=\frac{8\cos^3\left(g(t)\right)}{\sqrt{4-27t}}-\frac{6\cos\left(g(t)\right)}{\sqrt{4-27t}}=\\
&=\frac{2\cos\left(3g(t)\right)}{\sqrt{4-27t}}\\
&=1
\end{align*}
Since
\begin{align*}
2\cos\left(3g(t)\right)&=\sqrt{4-27t}\\
4\cos^2\left(3g(t)\right)&=4-27t\\
\sin^2\left(3g(t)\right)&=\frac{27}{4}t\\
\end{align*}
we get
\begin{align*}
g(t)&=\frac{1}{3}\sin^{-1}\left(\frac{3\sqrt{3t}}{2}\right)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
and conclude the identity $(2)$ is valid.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Integral $\int_0^{\pi/4} \frac{\ln \tan x}{\cos 2x} dx=-\frac{\pi^2}{8}.$ $$I:=\int_0^{\pi/4} \frac{\ln \tan x}{\cos 2x} dx=-\frac{\pi^2}{8}.$$
I am trying to see nice solutions to this integral. I tried the following
$$
I=\int_0^{\pi/4}\frac{\ln \sin x}{\cos 2x} dx-\int_0^{\pi/4} \frac{\ln \cos x }{\cos 2x}dx
$$
but am not sure how to work with this denominator of $\cos 2x$. If this helps:
$$
\int_0^{\pi/4}\log \sin x \, dx=-\frac{1}{4}\big(2K+\pi \ln 2\big)
$$
where K is the Catalan constant (note I am using Borwein convention not mathematica of using a C to define this constant.) It is given by
$$
K=\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2}=\beta(2)
$$
where $\beta(2)$ is the Dirichlet beta function.
However I cannot solve this integral either. Thanks
|
$$
\begin{aligned}
\int_0^{\frac{\pi}{4}} \frac{\ln (\tan x)}{\cos 2 x} d x=&\int_0^{\frac{\pi}{4}} \frac{\ln \left(\tan \left(\frac{\pi}{4}-x\right)\right)}{\cos 2\left(\frac{\pi}{4}-x\right)} d x \\
=& \int_0^{\frac{\pi}{4}} \frac{\ln \left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)}{\sin 2 x} d x\\=&\frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{\ln \left(\frac{1-\sin 2 x}{1+\sin 2 x}\right)}{\sin 2 x} d x\\=&\frac{1}{4}\left[\int_0^{\frac{\pi}{2}} \frac{\ln (1-\sin x)}{\sin x} d x-\int_0^\frac\pi2 \frac{\ln (1+\sin x)}{\sin x} d x\right]
\end{aligned}
$$
From my post,
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{\ln (1+a \sin x)}{\sin x} d x &=\int_0^{\frac{\pi}{2}} \frac{\ln (1+a \cos x)}{\cos x} d x \\
&=\frac{\pi^2}{8}-\frac{\left(\cos ^{-1} a\right)^2}{2}
\end{aligned}
$$
Now we can conclude that
$$\int_0^{\frac{\pi}{4}} \frac{\ln (\tan x)}{\cos 2 x} d x =\frac{1}{4}\left(-\frac{3 \pi^2}{8^2}-\frac{\pi^2}{8}\right)=-\frac{\pi^2}{8} $$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Exploring 3-cycle points for quadratic iterations How do you factorise $z^8 +4cz^ 6 + (6c^2 +2c)z^4 + (4c^3 +4c^2 )z^2 - z+ (c^4 +2c^3 +c^2 +c)$?
I want to find the 3-cycle points for the quadratic iteration $z \rightarrow z^2 + c$. In order to find it I have to solve the above polynomial. I have factorised into two factors
$(z^2 - z + c) (z^6+z^5+(3c+1)z^4+(2c+1)z^3+(3c^2+3c+1)z^4+((c+1)^2)z +(c^3+2c^2+c+1).$.
Note:
I did missed something in the factors previously. Now it is corrected.
|
The Galois group of such a polynomial of degree $6$ will not be $S_6$.
As soon as you know one of the roots $z$, you know two of the others, by applying $z \mapsto z^2+c$ twice.
The splitting field of $P$ should then generally be of degree $6 \times 3 = 18$.
Let $\zeta$ be a primitive thrid root of unity.
If you call the roots $z_{1,1},z_{1,\zeta},z_{1,\zeta^2},z_{2,1},z_{2,\zeta},z_{2,\zeta^2}$ where $z_{i,j}^2 + c = z_{i,\zeta j}$, any automorphism sending $z_{i,j}$ to $z_{i',j'}$ has to send $z_{i,\zeta j}$ to $z_{i',\zeta j'}$, so to determine an automorphism completely you only need to know the images of $z_{1,1}$ (among $6$ possibilities) and $z_{2,1}$ (among $3$ possibilities).
Let $H$ be the subgroup that does not switch the two $3$-cycles. $H$ is normal of index $2$ in the Galois group $G$, and $H$ is in fact isomorphic to $(\Bbb Z/3\Bbb Z)^2$.
So $G$ is a solvable group.
Let $P_1(z) = (z - z_{1,1})(z - z_{1,\zeta})(z - z_{1,\zeta^2})$. $P_1$ is invariant by $H$, so its coefficients should all belong in the fixed field of $H$, a quadratic extension $K$ of $\Bbb Q(c)$
First we should figure out what is $K$.
The extension ramifies when we have $z_{1,j} = z_{2,j}$, so a necessary condition is that the polynomial has multiple roots.
We can compute the discriminant of $P$. Amazingly, it's simpler than we should expect, as it factors as $\Delta = (4c+7)^3(16c^2+4c+7)^2$.
There are two ways that $P$ can have multiple roots. Either the two $3$-cycles are equal (then we should obtain a multiplicity $3$ root of $\Delta$), either one $3$-cycle collapses into a fixpoint of $z \mapsto z^2+c$ (then we should obtain a multiplicity $2$ root of $\Delta$).
We can confirm that the degree $2$ factors belong in the second case by computing the resultant of $z^2-z+c$ and $P$ (which gives $16c^2+4c+7$). So the only time when the quadratic extension can ramify is when $4c+7 = 0$, hence the extension is $\Bbb Q(c)(\sqrt{q(4c+7)})$ for some squarefree integer $q$.
When we plug $c=0$, $P_0(z)= z^6+z^5+z^4+z^3+z^2+z+1$, of which we know the roots.
The two degree $3$ factors over $K$ must then be $(X - \zeta_7)(X - \zeta_7^2)(X-\zeta_7^4)$ and $(X - \zeta_7^3)(X - \zeta_7^5)(X-\zeta_7^6)$.
Since the only quadratic extension contained in the cyclotomic extension $\Bbb Q \subset \Bbb Q(\zeta_7)$ is $\Bbb Q(\sqrt{-7})$, we must have $K = \Bbb Q(c)(\sqrt{-4c-7})$.
Next we have to find out the factorisation of $P$ over $K$.
Using Gauss' lemma, the coefficients of the factors have to be polynomials in $c$ (to be precise, they have to be in $\Bbb Z[c, \frac{1+\sqrt{-4c-7}}2]$)
By making simple assumptions on the degree of the coefficients, and using the factorisation at $c=0$ ($P_0(X) = (X^3 + \frac{1+\sqrt{-7}}2X^2 + \frac{-1+\sqrt{-7}}2X - 1)
(X^3 + \frac{1-\sqrt{-7}}2X^2 + \frac{-1-\sqrt{-7}}2X - 1)$)
we can quickly find
$$P(X) = \left(X^3 + \frac{1+\sqrt{-4c-7}}2X^2 + \frac{2c-1+\sqrt{-4c-7}}2X + \frac{-c-2+c\sqrt{-4c-7}}2\right) \\
\left(X^3 + \frac{1-\sqrt{-4 c-7}}2X^2 + \frac{2c-1-\sqrt{-4c-7}}2X + \frac{-c-2-c\sqrt{-4c-7}}2\right)
$$
Finally we have to solve each cubic polynomial (which has cyclic Galois group)
To do that, let $w_i = (z_{i,1} + \zeta z_{i,\zeta} + \zeta^2 z_{i,\zeta^2})$. The conjugates of $w_i$ are $\zeta w_i$ and $\zeta^2 w_i$, which means that $w_i^3 \in K(\zeta)$. The same happens with $v_i = (z_{i,1} + \zeta^2 z_{i,\zeta} + \zeta z_{i,\zeta^2})$ and in fact $v_i^3$ is the image of $w_i^3$ after switching $\zeta$ with $\zeta^2$. By summing two appropriate cube roots of those two, you obtain $2z_{1,1} - z_{1,\zeta} - z_{1,\zeta^2}$. Add to it the sum of the three roots (which is a coefficent of $P_1$, in $K$) and divide by $3$ to get $z_{1,1}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $f(t)$ be defined by $ f(t) = \left\{ \begin{array}{c} t, &0 \le t < 1 \\ b - t^2, &1 \le t \le 2 \end{array} \right. $ Let $f(t)$ be defined by
$$
f(t) =
\left\{
\begin{array}{c}
t, &0 \le t < 1 \\
b - t^2, &1 \le t \le 2
\end{array}
\right.
$$
and let $F(x)$ be defined by $F(x) = \int_0^xf(t)dt,\,\,\,\,\, 0 \le x \le 2.$
a) Find F(x)
b) For what value of $b$ in the definition of $f$ is $F(x)$ differentiable for all $x \in [0,2]$?
What I have so far is..
if $0 \le t < 1$, then $\int_0^xt dt = \frac{x^2}{2}$
if $1 \le t \le 2$, then $\int_0^1t dt + \int_1^xb - t^2dt = \frac{-x^3}{3} +xb - b + \frac{5}{6}$
therefore
$
F(x) =
\left\{
\begin{array}{c}
\frac{x^2}{2}, &0 \le t < 1 \\
\frac{-x^3}{3} +xb - b + \frac{5}{6}, &1 \le t \le 2
\end{array}
\right.
$
I'm not sure if I did A correctly and I need help with B.
|
This seems correct to me. Now you must check for which $b$ the right and left derivatives at $x = 1$ are equal. The left derivative is $1$ and the right is $-1+b$ so $1 = -1+b$ iff $b =2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\cot(x) +\cot(\frac{\pi}{3}+x) + \cot(\frac{2\pi}{3}+x) = \frac{3-9\tan^2x}{3\tan x-\tan^3x}$ As the title says. I tried simplifying the LHS but got:
$$\frac{\tan^2x+7\tan x- \sqrt{3}}{\tan^2x-\sqrt{3}}$$
What should I do next?
|
From this, $\displaystyle\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}$
$\displaystyle\implies\cot3x=\frac{1-3\tan^2x}{3\tan x-\tan^3x}\ \ \ \ (1)$
Multiplying the numerator & the denominator by $\cot^3x,$
$\displaystyle\cot3x=\frac{\cot^3x-3\cot x}{3\cot^2x-1}\ \ \ \ (2)$
If $\displaystyle\cot3x=\cot3A\iff\tan3x=\tan3A\implies3x=n\pi+3A$ where $n$ is any integer
$\displaystyle\implies x=\frac{n\pi}3+A$ where $n\equiv0,1,2\pmod3$
Usng
$\displaystyle(2),\frac{\cot^3x-3\cot x}{3\cot^2x-1}=\cot3x=\cot3A$
$\displaystyle\iff\cot^3x-3\cot3A\cot^2x-3\cot x+\cot3A=0$
Using Vieta's formula, $\displaystyle\sum_{n=0}^2\cot\left(\frac{n\pi}3+A\right)=\frac{3\cot3A}1=3\cdot\frac{1-3\tan^2A}{3\tan A-\tan^3A}$ (using $(1)$)
|
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"timestamp": "2023-03-29T00:00:00",
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|
To find relatively prime ordered pairs of positive integers $(a,b)$ such that $ \dfrac ab +\dfrac {14b}{9a}$ is an integer How many ordered pairs $(a,b)$ of positive integers are there such that g.c.d.$(a,b)=1$ , and
$ \dfrac ab +\dfrac {14b}{9a}$ is an integer ?
|
If $(a,b)$ is an ordered pair of positive integers such that $\gcd(a,b)=1$ and
$$\frac{a}{b}+\frac{14b}{9a}=\frac{9a^2+14b^2}{9ab},$$
is an integer, then $9ab$ divides $9a^2+14b^2$. In particular $a$ and $b$ both divide $9a^2+14b^2$, and so $a$ divides $14b^2$ and $b$ divides $9a^2$. Because $\gcd(a,b)=1$ it follows that $a$ divides $14$ and $b$ divides $9$. If $b=9$ then $81a$ divides $9a^2+14\cdot 81$, so $81$ divides $9a^2$ and hence $3$ divides $a$, contradicting the fact that $\gcd(a,b)=1$. Of course $b\neq1$ because $9$ does not divide 14. Hence $b=3$, and we conclude that there are precisely four such pairs.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\int_{0}^{\infty}\frac{x}{x^3+1}dx$ =? So guys I have this improper integral $\int_{0}^{\infty}\frac{x}{x^3+1}dx$. I checked that it converges by $ \int_{0}^{1}\frac{x}{x^3+1}dx + \int_{1}^{\infty}\frac{x}{x^3+1}dx $ and using the $\frac{c}{(x-a)^\lambda} $ and $\frac{1}{x^\lambda} $ criteria. But for finding the value I have trouble.I tried $ \int_{0}^{\infty}\frac{x}{(x+1)(x^2-x+1)}dx$ and $\int_{0}^{\frac{\pi}{2}}\frac{tg\alpha}{tg\alpha^3+1}dx$ but it gets me nowhere. Any ideas on solving it?
|
If $I=\int_{0}^{\infty}\frac{x}{x^3+1}dx$ then $2I=\int_{0}^{\infty}\frac{2x-x^2+x^2}{x^3+1}dx$=$\int_{0}^{\infty}\frac{2x-x^2}{x^3+1}dx$+$\int_{0}^{\infty}\frac{x^2}{x^3+1}dx.$
The secod itegral is easy and can be solved by substituting $u=x^3+1$. For the first write $$\int_{0}^{\infty}\frac{2x-x^2}{x^3+1}dx=\int_{0}^{\infty}\frac{(x+1)-(x^2-x+1)}{(x+1)(x^2-x+1)}dx=\int_{0}^{\infty}\frac{1}{x^2-x+1}dx-\int_{0}^{\infty}\frac{1}{x+1}dx$$
but for the $\int_{0}^{\infty}\frac{1}{x^2-x+1}dx$=$\int_{0}^{\infty}\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}dx$ which is of the form $\int \frac{1}{u^2+a^2}du=\frac{1}{a}\arctan(\frac{u}{a})+c.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $y =\sqrt{5+\sqrt{5-\sqrt{5+ \cdots}}}$, what is the value of $y^2-y$? If $$y =\sqrt{5+\sqrt{5-\sqrt{5 + \cdots}}},$$
what is the value of $y^2-y$ ?
I am unable to get the clue due to these alternative signs of plus and minus please help on this thanks...
|
Consider $z_{-+} = -y = -\sqrt {5 + \sqrt{5 - \ldots}}$.
$z_{-+}$ is clearly a solution to the equation $(z^2 -5)^2-5 = z$,
and the other three solutions of this degree $4$ polynomial equation, should be
$z_{+-} = +\sqrt {5 - \sqrt{5 + \ldots}}, z_{++} = +\sqrt {5 + \sqrt{5 + \ldots}}$, and $z_{--} = -\sqrt {5 - \sqrt{5 - \ldots}}$ .
But $z_{++}$ and $z_{--}$ are in fact solution to a simpler polynomial equation, which is $z^2-5 = z$.
So $z^2-z-5$ has to be a factor of $(z^2-5)^2-z-5$, and doing the division, you end up with $(z^2-5)^2-z-5 = z^4-10z^2-z+20 = (z^2-z-5)(z^2+z-4)$, hence $z_{-+}^2 + z_{-+} - 4 = 0$.
From there you conclude that $y^2-y = z_{-+}^2 + z_{-+} = 4$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the determinant of the following;
Find the determinant of the following matrix, and for which value of $x$ is it invertible;
$$\begin{pmatrix}
x & 1 & 0 & 0 & 0 & \ldots & 0 & 0 \\
0 & x & 1 & 0 & 0 & \ldots & 0 & 0 \\
0 & 0 & x & 1 & 0 & \ldots & 0 & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
0 & 0 & 0 & 0 & 0 & \ldots & x & 1 \\
1 & 0 & 0 & 0 & 0 & \ldots & 0 & x
\end{pmatrix}$$
Now I don't really know how to procees as I get find a suitable row operations that will simplify the process so I thought I would look at cases, maybe see a pattern.
$\mathbf{2 \times 2}$
$\begin{bmatrix}x & 1\\1 & x\end{bmatrix}$
This has determinant; $x^2-1$
$\mathbf{3 \times 3}$
$\begin{bmatrix}x & 1 & 0\\0 & x & 1\\1 & 0 & x\end{bmatrix}$
This has determinant $x^3+1$
So is that the pattern?
determinant is $x^n-1$ if $n$ is even,
determinant is $x^n+1$ if $n$ is odd??
|
Develop the determinant by the first column...
$$\begin{vmatrix}\color{red}x&1&0&0&\ldots&0\\
\color{red}0&x&1&0&\ldots&0\\
\color{red}\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\
\color{red}1&0&0&0&\ldots&x\end{vmatrix}=$$$${}$$
$$x\begin{vmatrix}x&1&0&0&\ldots&0\\
0&x&1&0&\ldots&0\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\
0&0&0&0&\ldots&x\end{vmatrix}+(-1)^{n-1}\begin{vmatrix}1&0&0&0&\ldots&0\\
x&1&0&0&\ldots&0\\
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\
0&0&0&\ldots&x&1\end{vmatrix}$$
Now check that both determinants above are triangular matrices's, so piece of cake...
|
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|
Sum of Digits Question If A is the sum of the digits of $5^{10000}$, B is the sum of the digits of A, and C is the sum of the digits of B, what is C? I know it has something to do with mod 9, but I'm not sure how do use it to solve the problem. I found this question in my Math Challenge II Number Theory Packet, and I got confused.
|
Let $S(n)$ denote the sum of digits of $n$. Then, note that $S(n) \le 9(\lfloor\log_{10}(n)\rfloor + 1)$.
Thus,
$(S(5^{10000}) \le 9(\lfloor\log_{10}(5^{10000})\rfloor + 1) = 9(\lfloor10000\log_{10}(5)\rfloor + 1) = 9(\lfloor1000\log_{10}(5^{10})\rfloor + 1) = 9(\lfloor1000\log_{10}(9765625)\rfloor + 1) \le 9(\lfloor1000\log_{10}(10000000)\rfloor + 1) = 9(\lfloor1000\log_{10}(10^7)\rfloor + 1) = 9(\lfloor1000(7)\rfloor + 1) = 9(7001) = 63009$
Also note that the maximum possible sum of digits of the numbers $\le 63009$ is $59999$, with sum of digits $41$. Now, the number with the maximum sum of digits that is $\le 41$ is $39$, with sum of digits $12$. So, $C \le 12$.
Since $5^{10000} \equiv (5^2)^{5000} \equiv 25^{5000} \equiv (-2)^{5000} \equiv 2^{5000} \equiv (2^3)^{1666} \cdot 4 \equiv 8^{1666} \cdot 4 \equiv (-1)^{1666} \cdot 4 \equiv 4 \pmod{9}$.
Since $C \equiv 5^{10000} \pmod{9}$ and $0 \le C \le 12$, $C = 4$.
|
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|
Points on $(x^2 + y^2)^2 = 2x^2 - 2y^2$ with slope of $1$ Let the curve in the plane defined by the equation: $(x^2 + y^2)^2 = 2x^2 - 2y^2$
How can i graph the curve in the plane and determine the points of the curve where $\frac{dy}{dx} = 1$.
My work:
First i found the roots of this equation with a change of variable $z = y^2$ and get:
and then i tried to graph the point $ x - y$ and $x + y $ but i stuck i can't graph this and find the point where the derivative is 1. Some help please.
|
For determining the graph is better plot it as Lemur says in polar coordinates and see what happen at different angles with $r$.
$\hskip2in$
For the points with derivative equal to one:
$$(x^2 + y^2)^2 = 2x^2 - 2y^2$$
$$x^4+y^4+2x^2y^2=2x^2-2y^2$$
$$4x^3+4y^3\frac{dy}{dx}+4xy^2+4x^2y\frac{dy}{dx}=4x-4y\frac{dy}{dx}$$
Setting $\frac{dy}{dx}=1$
$$4x^3+4y^3+4xy^2+4x^2y=4x-4y$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve for polynomial fractions? I'm self-studying. I have this problem I can't wrap my head around.
$\frac{x}{x+1} + \frac{4}{1-x} + \frac{x^2-5x-8}{x^2-1}$
The answer is $\frac{2\left(x-6\right)}{x-1}$
How do I get that answer? I keep getting $2(x^2-5x-2)$...
This is how I solved it:
$\frac{x}{x+1}-\frac{4}{x+1}+\frac{x^2-5x-8}{x^2-1}$
$=\frac{x-4+x^2-5x-8}{(x+1)(x-1)}$
$=\frac{x(x-1)-4(x-1)+x^2-5x-8}{(x+1)(x-1)}$
= $2(x^2−5x−2)$
I see the problem now!
$\frac{4}{1-x}$ = $\frac{4}{x-1}$ ...not $\frac{4}{x+1}$
because when I multiply the denominator by -1, the result is...
$-(1-x) = -1+x = x-1$
_thus_$\,\, \frac{4}{1-x}$ _becomes_$ \frac{4}{x-1}$
and then I changed the sign of the fraction from plus to minus $-\frac{4}{x-1}$ to keep the fraction equal to the original.
Solved!
|
Find the common denominator, and simplify. (You'll be able to utilize the fact that $x^2-1$ is a difference of squares: $$x^2 - 1= (x+1)(x-1)$$
$$\begin{align} \frac{x}{x+1} + \frac{4}{1-x} + \frac{x^2-5x-8}{x^2-1}& = \dfrac {x}{x+1} -\frac{4}{x-1} + \frac{x^2 - 5x-8}{x^2-1}\\ \\& =\dfrac{x(x-1)-4(x+1) +x^2 - 5x -8}{x^2-1}\\ \\ &= \dfrac{x^2 - x + -4x - 4 + x^2 +x^2 -5x - 8}{x^2-1} \\ \\ &= \dfrac{2x^2 -10x -12}{(x^2-1)}\\ \\ &= \dfrac{2(x^2 - 5x-6)}{(x+1)(x-1)}\\ \\ &= \dfrac{2(x+1)(x-6)}{(x + 1)(x-1)}\\ \\ & = \dfrac{2(x-6)}{x-1}, \quad x\neq -1\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The closed form of $\int_0^{\infty} \frac{\log(\cosh(x))}{x} e^{-x} \ dx$ An integral I discussed last days in a chat, and it looks like a hard nut since
after some manipulations of the initial form we reach an integral where the integrand is expressed in
terms of
digamma function that can be further reduced to some very resistent series.
I conjecture that the integral has a closed form expressed in terms of G-Barnes function
(and newer constants?) Well, I might be wrong, but here we have many integration gurus
that could probably answer that.
$$\int_0^{\infty} \frac{\log(\cosh(x))}{x} e^{-x} \ dx$$
|
The integral to evaluate is
\begin{align}
I(1) = \int_{0}^{\infty} e^{-x} \, \ln(\cosh(x)) \, \frac{dx}{x}.
\end{align}
The evaluation is as follows. Consider
\begin{align}
\int_{1}^{\infty} e^{-\alpha x} d\alpha = \left[ - \frac{1}{x} e^{-\alpha x} \right]_{1}^{\infty} = \frac{e^{-x}}{x}.
\end{align}
Let
\begin{align}
I(\alpha) = \int_{0}^{\infty} e^{-\alpha x} \, \ln(\cosh(x)) \, \frac{dx}{x},
\end{align}
where $I(1)$ is the integral to evaluate and $I(\infty) = 0$, for which differentiation with respect to $\alpha$ leads to
\begin{align}
\partial_{\alpha} I(\alpha) = - \int_{0}^{\infty} e^{- \alpha x} \ln(\cosh(x)) dx.
\end{align}
Using integration by parts it is seen that
\begin{align}
\partial_{\alpha} I(\alpha) &= \left[ \frac{1}{\alpha} \, e^{- \alpha x} \ln(\cosh(x)) \right]_{0}^{\infty} - \frac{1}{\alpha} \int_{0}^{\infty} e^{- \alpha x} \tanh(x) dx \\
&= - \frac{1}{\alpha} \int_{0}^{\infty} e^{- \alpha x} \tanh(x) dx \\
&= - \frac{1}{2 \alpha} \left[ \beta\left( \frac{\alpha}{2} \right) - \beta\left( \frac{\alpha + 2}{2} \right) \right] \\
&= - \frac{1}{4\alpha} \left[ 2 \psi\left( \frac{\alpha + 2}{4} \right) - \psi\left( \frac{\alpha}{4} \right) - \psi\left( \frac{\alpha}{4} + 1 \right) \right] \\
&= \frac{1}{\alpha^{2}} - \frac{1}{\alpha} \beta\left(\frac{\alpha}{2}\right),
\end{align}
where $2 \beta(x) = \psi((x+1)/2) - \psi(x/2)$. Since
\begin{align}
\beta\left(\frac{x}{2}\right) = 2 \partial_{x} \ln\left( \frac{\Gamma((x+2)/4)}{\Gamma(x/4)} \right)
\end{align}
then
\begin{align}
\partial_{\alpha} I(\alpha) &= \frac{1}{\alpha^{2}} - \frac{2}{\alpha} \partial_{\alpha} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right)
\end{align}
and
\begin{align}
\int_{1}^{\infty} \partial_{\alpha} I(\alpha) \, d\alpha &= 1 - 2 \int_{1}^{\infty} \frac{1}{\alpha} \partial_{\alpha} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right) \\
I(\infty) - I(1) &= 1 - 2 \left[ \frac{1}{\alpha} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right) \right]_{1}^{\infty} - 2 \int_{1}^{\infty} \ln\left( \frac{\Gamma((\alpha+2)/4)}{\Gamma(\alpha/4)} \right) \, \frac{d\alpha}{\alpha^{2}} \\
- I(1) &= 1 + 2 \ln\left( \frac{\Gamma(3/4)}{\Gamma(1/4)} \right) - \frac{1}{2} \int_{1/4}^{\infty} \ln\left( \frac{\Gamma(u+1/2)}{\Gamma(u)} \right) \, \frac{du}{u^{2}}.
\end{align}
This is
\begin{align}
I(1) &= \ln\left( \frac{\Gamma^{4}(1/4)}{2 \pi^{2} e} \right) - \frac{1}{2} \int_{1/4}^{\infty} \ln\left( \frac{\Gamma(u)}{\Gamma(u + 1/2)} \right) \, \frac{du}{u^{2}}.
\end{align}
It is of note that
\begin{align}
\ln\left( \frac{\Gamma^{4}(1/4)}{2 \pi^{2} e} \right) < \int_{1/4}^{\infty} \ln\left( \frac{\Gamma(u)}{\Gamma(u + 1/2)} \right) \, \frac{du}{u^{2}} < \ln\left( \frac{\Gamma^{4}(1/4)}{4 \pi^{2} e} \right).
\end{align}
|
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|
prove that if $\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma=2$ then the triangle has a right angle prove that if $\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma=2$ then the triangle has a right angle.
$\alpha,\beta,\gamma$ are the angles of the triangle.
I tried to use all kinds of trigonometric identities but it didn't work for me. it's to complex form me
Thanks.
|
Let $\alpha$, $\beta$ and $\gamma$ be the angles of a triangle; then it holds
$$
\boxed{\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2+2\cos\alpha\cos\beta\cos\gamma}
$$
If this equals $2$, we conclude
$$
\cos\alpha\cos\beta\cos\gamma=0
$$
so one of the angles is a right angle.
Proof of the claim
Let's use that $\gamma=\pi-\alpha-\beta$, so $\cos\gamma=-\cos(\alpha+\beta)$. Then
\begin{align}
\cos^2\alpha+\cos^2\beta+\cos^2\gamma&=
\cos^2\alpha+\cos^2\beta+\cos^2(\alpha+\beta)\\
&=\cos^2\alpha+\cos^2\beta+\cos^2\alpha\cos^2\beta+\sin^2\alpha\sin^2\beta\\
&\qquad-2\sin\alpha\sin\beta\cos\alpha\cos\beta\\
&=\cos^2\alpha+\cos^2\beta+1-\cos^2\alpha-\cos^2\beta+2\cos^2\alpha\cos^2\beta\\
&\qquad-2\sin\alpha\sin\beta\cos\alpha\cos\beta\\
&=1+2\cos\alpha\cos\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)\\
&=1-2\cos\alpha\cos\beta\cos\gamma
\end{align}
giving the final relation
$$
\boxed{\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1-2\cos\alpha\cos\beta\cos\gamma}
$$
Now
\begin{align}
\sin^2\alpha+\sin^2\beta+\sin^2\gamma
&=3-\cos^2\alpha-\cos^2\beta-\cos^2\gamma\\
&=2+2\cos\alpha\cos\beta\cos\gamma
\end{align}
as claimed at the beginning.
|
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|
Finding invertible 3x3 matrix A such that Stab(M)A=AStab(N) for given matrices M,N I'm reading that it is possible to find an invertible $3 \times 3$ matrix $A$ such that for the matrices
$M = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ and
$N = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 2 & 2 \\ 1 & 1 & 2 \end{bmatrix}$
we have $\operatorname{Stab}(M).A = A.\operatorname{Stab}(N)$.
Here $\operatorname{Stab}(M) = \left\{ B \in \operatorname{GL}(3,\mathbb{R}) : BM = M \right\}$
I believe it's possible but I can't find it at all!
I've tried right multiplying the equality by $N$ then I get $B.A.N = A.C.N =A.N$ (for any $B \in \operatorname{Stab}(M)$ and the corresponding $C \in \operatorname{Stab}(N)$) so that $\operatorname{Stab}(M)$ is actually a subset of $\operatorname{Stab}(A.N)$. Also right multiplying by $A^{-1}$ gives $\operatorname{Stab}(M) = A.\operatorname{Stab}(N).A^{-1}$ then right multiplying by $M$ and left multiplying by $A^{-1}$ again gives $A^{-1}.M = \operatorname{Stab}(N).A^{-1}$.
Do you have a solution to this? Am I actually on the right path? Do you have other ideas?
Thank you so much!
|
Your ideas would be good if this was a general property of $M$ and $N$, but I don't think it is. You need to use the specific $M$ and $N$.
The names of these two ideas are “orbit-stabilizer” and “Gauss–Jordan elimination.”
GJ:
Notice that $\begin{bmatrix} 1& 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. N
=\begin{bmatrix} 2 & 0 & 2 \\ 0 & 2 & 2 \\ 1 & 1 & 2 \end{bmatrix}$,
and so
$\begin{bmatrix} \tfrac 12 & 0 & 0 \\ 0 & \tfrac 12 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\cdot \begin{bmatrix} 1& 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot N
= \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix}
$
and so
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & -1 & 1 \end{bmatrix}
\begin{bmatrix} \tfrac 12 & 0 & 0 \\ 0 & \tfrac 12 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\cdot \begin{bmatrix} 1& 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot N
= \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} = M
$
Setting $A =
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & -1 & 1 \end{bmatrix}
\begin{bmatrix} \tfrac 12 & 0 & 0 \\ 0 & \tfrac 12 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\cdot \begin{bmatrix} 1& 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
= \begin{bmatrix} \tfrac12&0&\tfrac12\\ 0&\tfrac12&0\\-\tfrac12&-\tfrac12&\tfrac12\\ \end{bmatrix}
$
We have $M=A.N$.
OS:
$C.A \in \operatorname{Stab}(M).A$ iff $C \in \operatorname{Stab}(M)$ iff $C.M=M$ iff $C.(A.N) = (A.N)$ iff $\left(A^{-1}. C .A\right).N = N$ iff $\left( A^{-1}.C.A \right) \in \operatorname{Stab}(N)$ iff $A.\left(A^{-1}.C.A\right) = C.A \in A.\operatorname{Stab}(N)$.
In other words, $X=C.A$ is in $\operatorname{Stab}(M).A$ if and only if $X$ is in $A.\operatorname{Stab}(N)$. So $\operatorname{Stab}(M).A= A.\operatorname{Stab}(N)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I calculate the new matrix from the new basis? $\bf(1)$ Let $\mathbb{R}^{2\times2}$ denote the vector space of all $2\times2$ matrices with real number entries. Set $A=\pmatrix{1&2\\-2&-4}$. Define a linear transformation $T:\mathbb{R}^{2\times2}\longrightarrow\mathbb{R}^{2\times2}$ by $T(B)=AB$ for any $B\in\mathbb{R}^{2\times2}$. $\quad(a)$ Compute the $\alpha$-matrix $[T]_\alpha$ for $\mathrm T$, where $\alpha$ denotes the following ordered basis for $\mathbb{R}^{2\times2}:$ $$\pmatrix{0&0\\1&1},\pmatrix{1&0\\0&0},\pmatrix{0&0\\-1&1},\pmatrix{0&1\\0&1}.$$ $\qquad\bf Solution:$ $[T]_\alpha=\pmatrix{-5 & -1 & -1 & -\tfrac92 \\
2 & 1 & -2 & 0 \\
-1 & 1 & -5 & -\tfrac92 \\
2 & 0 & 2 & 3}$. $\quad(b)$ Is $\rm T$ an isomorphism? $\bf Solution:$ No $\rm T$ is not an isomorphism. $\rm T$ can not be "onto" since $\operatorname{rank}([T]_\alpha)<4$. Another reason $\rm T$ can not be "onto" is that the $2\times2$ identity matrix $I_2$ is not a value of $\rm T$: if it were a value then we would have that $AB=I_2$ for some $2\times2$ matrix $\rm B$, which would imply that $A$ is invertible - which is not true. $\quad(c)$ Compute the $\alpha$-coordinates of $\pmatrix{\pi & 3 \\ \sqrt{2}&0}$. $\qquad\bf Solution:$ $\pmatrix{\tfrac{\sqrt{2}-3}2 \\ \pi \\ \tfrac{-\sqrt{2}-3}2 \\ 3}$
I'm having trouble with parts A and C.
|
$$T\begin{pmatrix}0&0\\1&1\end{pmatrix}:=\begin{pmatrix}\;\;2&\;\;2\\-4&-4\end{pmatrix}=\color{red}{-5}\begin{pmatrix}0&0\\1&1\end{pmatrix}+\color{red}2\begin{pmatrix}1&0\\0&0\end{pmatrix}+\color{red}{(-1)}\begin{pmatrix}\;\;0&0\\-1&1\end{pmatrix}+\color{red}2\begin{pmatrix}0&1\\0&1\end{pmatrix}$$
and that's how you get the wanted matrix's first column. Continue from here applying $\;T\;$ to the other members of that basis and writing the result as a linear combination of that same basis
For (c): write that vector as a linear combination of the basis:
$$\begin{pmatrix}\pi&3\\\sqrt2&0\end{pmatrix}=a\begin{pmatrix}0&0\\1&1\end{pmatrix}+b\begin{pmatrix}1&0\\0&0\end{pmatrix}+c\begin{pmatrix}\;\;0&0\\-1&1\end{pmatrix}+d\begin{pmatrix}0&1\\0&1\end{pmatrix}$$
For example,. it must clearly be that $\;b=\pi\;$ (why?), etc.
|
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|
Compute $ \int\sin(x^2)\, dx + \int \sqrt{\arcsin t}\, dt$ Compute the sum of two integrals
$$
\int_{\large\sqrt{\frac{\pi}{6}}}^{\large\sqrt{\frac{\pi}{3}}}\sin(x^2)\ \ dx +
\int_{\large\frac{1}{2}}^{\large\frac{\sqrt{3}}{2}} \sqrt{\arcsin t}\ \ dt.
$$
|
Let $x=\sqrt{\arcsin t}$, then
$$
\int\limits_{\sqrt{\frac{\pi}{6}}}^{\sqrt{\frac{\pi}{3}}}\sin(x^2)\, dx
= \int\limits_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} t\,d\sqrt{\arcsin t}
= t\sqrt{\arcsin t}\vert_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}
- \int\limits_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \sqrt{\arcsin t}\, dt.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
To solve $2^x+4^x+2^{\lfloor x \rfloor}+4^{\lfloor x \rfloor}+2^{x- \lfloor x \rfloor}-4^{x-\lfloor x \rfloor}=50+\sqrt{50}$ How to solve for positive real $x$: $$2^x+4^x+2^{\lfloor x \rfloor}+4^{\lfloor x \rfloor}+2^{x- \lfloor x \rfloor}-4^{x-\lfloor x \rfloor}=50+\sqrt{50}$$ ?
|
Notice the function on the left hand side is increasing (for $x>0$). [And for $x<0$ the left hand side is $\leq6$ so we don't care about it]. The only term that pushes down is $4^{x-[x]}$ but this is $\leq4$.
For $x=2$ we get $4+16+4+16+1-1=48<50+\sqrt{50}$.
For $x=3$ we get $8+64+8+64+1-1>50+\sqrt{50}$.
Therefore the solution is in $(2,3)$.
The equation becomes
$$2^x+4^x+4+16+2^{x-2}-4^{x-2}=50+\sqrt{50}$$
Or what is the same
$$a+a^2+4+16+a/4-a^2/16=50+\sqrt{50}$$
where $a=2^x$.
This is a quadratic equation in $a$, so you can solve it.
$a=2^{2+1/2}$ or $a=-4/3-2^{2+1/2}$. The latter cannot be equal to $2^x$.
Then $x=\log_2(2^{2+1/2})=2+1/2$.
|
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|
Solve by using modulo. Or something else An infinite sequence of positive integers $a_1, a_2,\ldots$ has the properties that for $k\geq2$, the $k^\text{th}$ element is equal to $k$ plus the product of the first $k-1$ elements of the sequence. Suppose $a_1 =1$. what is the smallest prime number that does not divide $a_{2010}$?
I have a feeling you have to use mod? But how do you find the twenty tenth number of the sequence?
|
Modular arithmetic is congruent under multiplication, so we can define sequences $a_k^{(2)}$, $a_k^{(3)}$, $a_k^{(5)}$, etc. to represent the sequences modulo 2, 3, and 5 respectively.
Thanks to congruency we still have the recurrence:
$$a_k^{(p)} \equiv \prod_{i=1}^{k-1}a_i^{(p)} + k \mod p$$
Now it remains to look at patterns in sequences $a^{(p)}_k$.
Modulo 2:
Note:
$$a_3^{(2)} \equiv 6 \equiv 0 \mod 2$$
So for any $k>3$, the product term is zero mod 2. Hence we simplify for large $k$ to:
$$a_k^{(2)} \equiv k \mod 2$$
So 2 divides $a_{2010}$ since $2010 \equiv 0 \mod 2$.
Mod 3:
Same approach:
$$a_3^{(3)} \equiv 6 \equiv 0 \mod 3$$
Hence, with the same reasoning,
$$a_{2010}^{(3)} \equiv 2010 \equiv 0 \mod 3$$
So 3 is not the factor we are looking for.
Mod 5:
First notice:
$$a_1^{(5)} a_2^{(5)} \equiv 1 \times 3 \equiv 3 \mod 5$$
Let's list out the next few terms:
$$
\begin{split}
a_3^{(5)} &\equiv 3 + 3 \equiv 1 &\mod 5 \newline
a_4^{(5)} &\equiv 3 \cdot 1 + 4 \equiv 2 &\mod 5 \newline
a_5^{(5)} &\equiv 3 \cdot 1 \cdot 2 + 5 \equiv 1 &\mod 5 \newline
a_6^{(5)} &\equiv 3 \cdot 1 \cdot 2 \cdot 1 + 6 \equiv 2 &\mod 5 \newline
a_7^{(5)} &\equiv 3 \cdot 1 \cdot 2 \cdot 1 \cdot 2 + 7 \equiv 4 &\mod 5
\end{split}
$$
Now let's look at $a_8^{(5)}$.
$$a_8^{(5)} \equiv 3 \cdot 1 \cdot 2 \cdot 1 \cdot 2 \cdot 4 + 8 \equiv 3 + 3 \mod 5$$
Note how the product term has reset itself to 3 mod 5! This means we have entered into a cycle, since the recurrence into $a_9^{(5)}$ will be like the recurrence into $a_4^{(5)}$, and so on. Indeed, $a_{k+5}^{(5)} = a_{k}^{(5)}$ for all sufficiently large $k$. Notice how the remainder never reaches 0, so 5 is our final answer.
|
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How prove $\left(\frac{b+c}{a}+2\right)^2+\left(\frac{c}{b}+2\right)^2+\left(\frac{c}{a+b}-1\right)^2\ge 5$ Let $a,b,c\in R$ and $ab\neq 0,a+b\neq 0$. Find the minimum of:
$$\left(\dfrac{b+c}{a}+2\right)^2+\left(\dfrac{c}{b}+2\right)^2+\left(\dfrac{c}{a+b}-1\right)^2\ge 5$$
if and only if $$a=b=1,c=-2$$
My idea: Since
$$\left(\dfrac{b+c+2a}{a}\right)^2+\left(\dfrac{c+2b}{b}\right)^2+\left(\dfrac{c-a-b}{a+b}\right)^2$$
let
$$x=\dfrac{b+c+2a}{a},y=\dfrac{c+2b}{b},z=\dfrac{c-a-b}{a+b}$$
then I can't work.
Thank you.
|
Remark
$$\displaystyle\sum f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$$
means cyclic sum.
$$\displaystyle\prod f(a,b,c)=f(a,b,c)\cdot f(b,c,a) \cdot f(c,a,b)$$
means cyclic product.
$$
\left( \frac{b+c}{a}+2 \right) ^2+\left( \frac{c}{b}+2 \right) ^2+\left( \frac{c}{a+b}-1 \right) ^2
$$
$$
=\left( \frac{a+b+c}{a}+1 \right) ^2+\left( \frac{b+c}{b}+1 \right) ^2+\left( \frac{c}{a+b}-1 \right) ^2
$$
then we can make the substitution $\left( a+b+c,b+c,c \right) \mapsto \left( x,y,z \right) $ to get a pretty form
$$
=\left( \frac{x}{x-y}+1 \right) ^2+\left( \frac{y}{y-z}+1 \right) ^2+\left( \frac{z}{z-x}+1 \right) ^2
$$
then make a substitution again
$$
\left( \frac{x}{x-y},\frac{y}{y-z},\frac{z}{z-x} \right) \mapsto \left( a,b,c \right)
$$
and try to find an equality
$$
a=\frac{x}{x-y}\Rightarrow \frac{a-1}{1}=\frac{y}{x-y}\Rightarrow \frac{a-1}{\left( a-1 \right) +1}=\frac{y}{\left( x-y \right) +y}\Rightarrow \frac{a-1}{a}=\frac{y}{x}
$$
$$
\prod{\frac{a-1}{a}}=1\Rightarrow \prod{\left( a-1 \right)}=abc\Rightarrow \sum{a}=\sum{ab}+1
$$
then we can get final conclusion
$$
\sum{\left( a+1 \right) ^2}=\sum{a^2}+2\sum{a}+3=\sum{a^2}+2\sum{ab}+6=\left( \sum{a} \right) ^2+5\geq 0
$$
==================================
there is a similar problem with similar tricks
Assuming $\displaystyle \sum x=0$. Find the minimum of $\displaystyle \sum \left(\frac{x}{y}\right)^2$. (Answer : 5)
===============Solution=============
$$
\begin{aligned}
&\sum{\left( \frac{x}{y} \right) ^2}\\
=&\left( \sum{\frac{x}{y}} \right) ^2-2\sum{\left( \frac{x}{y}\cdot \frac{y}{z} \right)}\\
=&\left( \sum{\frac{x}{y}} \right) ^2+2\sum{\frac{-x}{z}}\\
=&\left( \sum{\frac{x}{y}} \right) ^2+2\sum{\frac{y+z}{z}}=\left( \sum{\frac{x}{y}} \right) ^2+2\sum{\frac{x}{y}}+6\\
=&\left( \sum{\frac{x}{y}}+1 \right) ^2+5\\
\geq & 5
\end{aligned}
$$
Of course you can make it into perfect square.
$$\frac{x^2}{y^2}+\left( \frac{x+y}{x} \right) ^2+\left( \frac{y}{x+y} \right) ^2=\frac{\left( x^3+x^2y-2xy^2-y^3 \right) ^2}{x^2y^2\left( x+y \right) ^2}+5\geq 5
$$
|
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|
Evaluation of $\int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx$ The value of $\displaystyle \int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx = $
$\bf{My\; Try::}$ Let $\displaystyle I = \int 4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx$
Using Integration by parts, we get
$\displaystyle I = 4x^3\cdot \int\left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx - 12\int\left\{x^2\cdot \int\left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx\right\}dx$
$\displaystyle I = 4x^3\cdot \frac{d}{dx}(1-x^2)^5-12\int \left\{x^2\cdot \frac{d}{dx}(1-x^2)^5\right\}dx$
$\displaystyle I = 4x^3\cdot 5\cdot (1-x^2)^4\cdot -2x+120\int x^2\cdot (1-x^2)^4\cdot xdx$
Now Let $x^2=t\;,$ Then $2xdx = dt$
$\displaystyle I = -10x^4\cdot (1-x^2)^4+60\int t\cdot (1-t)^4dt=-10x^4\cdot (1-x^2)^4+60\int (1-t)t^4dt$
So $\displaystyle I = -10x^4\cdot (1-x^2)^4+60\left\{\frac{t^5}{5}-\frac{t^6}{6}\right\}+\mathbb{C}$
Is my process is right?,If not then how can we solve it.
Help Required.
Thanks
|
This appeared in my test paper today i.e JEE-Advanced 2014, I am curious as to how you got your hands on this one. :P
As for the solution, apply IBP once to get (I am dropping the factor of 4 for the moment):
$$I=\left(x^3 \frac{d}{dx}(1-x^2)^5 \right|_0^1-\int_0^1 3x^2\frac{d}{dx}(1-x^2)^5\,dx$$
Notice that the first term is zero, hence,
$$I= 15\int_0^1 x^2(1-x^2)^4 (2x)\,dx$$
With the substitution $x^2=t$, you get:
$$I=15\int_0^1 t(1-t)^4\,dt=15 \int_0^1 (1-t)t^4\,dt=15\left(\frac{t^5}{5}-\frac{t^6}{6}\right|_0^1$$
$$\Rightarrow I=\frac{1}{2}$$
Multiplying the result by 4, our final answer is: $\boxed{2}$
|
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|
How to solve this weird inequality? $\frac{x-1}{x+1} < x$
Thanks!
I did the following.
$\frac{x-1}{x+1} - x< 0 /-x$
$\frac{x-1 - x(x+1)}{x+1} < 0$
$\frac{-x^2-1}{x+1} < 0$
What to do next?
|
Start with: $x\le\frac{1}{2}(x^2+1)\Rightarrow x-1\le\frac{1}{2}(x^2-1)$
Thus $\frac{x-1}{x+1}\le\frac{1}{2}(x-1)\le x-1\lt x$
|
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|
Simplifying $(1/x-1/5 )/( 1/x^2-1/25)$ How do I get $\frac{5x}{x+5}$ from simplifying the following?
$$\frac{(\frac{1}{x}-\frac{1}{5} )}{( \frac{1}{x^2}-\frac{1}{25})}$$
My work:
I multiplied the top and bottom by the LCD: $25x^2$ (to get the same denominators).
Then I got: $\frac{25x-5x}{25-x^2}$
Then I got this for my answer: $\frac{20}{-(x-5)(x+5)}$
But the real answer is: $\frac{5x}{x+5}$
So my questions is how to I get from $\frac{20}{-(x-5)(x+5)}$ to $\frac{5x}{x+5}$?
|
You should have gotten to
$$\frac{25x-5x^2}{25-x^2} = \frac{5x(5 - x)}{(5+x)(5-x)}$$
Now cancel the common factor to get the desired result, assuming $x\neq 5$.
|
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|
How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$ I need to prove that $\frac{1+\tan^2\theta}{1+\cot^2\theta}= \tan^2\theta.$
I know that $1+\tan^2\theta=\sec^2\theta$ and that $1+\cot^2\theta=\csc^2\theta$, making it now $$\frac{\sec^2\theta}{\csc^2\theta,}$$ but I don't know how to get it down to $ \tan^2\theta.$
HELP!!!!
I also need help proving that $\tan\theta + \cot\theta = \sec\theta\cdot\csc\theta.$
|
$$\frac{1+\tan^2\theta}{1+\cot^2\theta}=\frac{\sec^2\theta}{\csc^2\theta}=\frac{\frac{1}{\cos^2\theta}}{\frac{1}{\sin^2\theta}}=\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$$
As for your second question that $\tan\theta+\cot\theta=\sec\theta\csc\theta,$ just use the fact that $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\cot\theta=\frac{\cos\theta}{\sin\theta}$ to get $$\tan\theta+\cot\theta=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta}\cdot\frac{1}{\cos\theta}=\sec\theta\cdot\csc\theta$$
|
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|
$4 \sin 72^\circ \sin 36^\circ = \sqrt 5$ How do I establish this and similar values of trigonometric functions?
$$
4 \sin 72^\circ \sin 36^\circ = \sqrt 5
$$
|
Using Werner Formula, $$2\sin36^\circ\sin72^\circ=\cos36^\circ-\cos108^\circ$$
$$\cos108^\circ=\cos(180^\circ-72^\circ)=-\cos72^\circ$$
Now, $\displaystyle\cos36^\circ\cos72^\circ=\frac{2\sin36^\circ\cos36^\circ}{2\sin36^\circ}\cos72^\circ=\frac{2\sin72^\circ\cos72^\circ}{4\sin36^\circ}=\frac{\sin144^\circ}{4\sin(180^\circ-144^\circ)}=\frac14$
From Help needed with trigonometric identity, $\displaystyle\cos36^\circ-\cos72^\circ=\sin54^\circ-\sin18^\circ=\frac12$
$\displaystyle\implies2\sin36^\circ\sin72^\circ=\cos36^\circ+\cos72^\circ=+\sqrt{(\cos36^\circ-\cos72^\circ)^2+4\cos36^\circ\cos72^\circ}$
$\displaystyle=\sqrt{\left(\frac12\right)^2+1}=\frac{\sqrt5}2$
|
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|
Please help to prove $\sum\limits_{n=1}^\infty {1\over a_n-1}<{2\over 7}$,in which $a_n=4n(n+1)$ Please help to prove this.
Suppose $a_n=4n(n+1)$,then
$$\sum_{n=1}^{+\infty}{1\over a_n-1}<{2\over7}.$$
Thanks.
|
$$4n(n+1)-1=(2n+1)^2-2>(2n+1)^2-4=(2n+3)(2n-1)$$So, given sum is less than $$\frac{1}{7}+\sum_{n=2}^{\infty}\frac{1}{(2n+3)(2n-1)}=\frac{1}{7}+\frac{1}{4}\sum_{n\ge 2}\left(\frac{1}{2n-1}-\frac{1}{2n+3}\right)=\frac{1}{7}+\frac{1}{4}\left(\frac{1}{3}+\frac{1}{5}\right)\\=\frac{1}{7}+\frac{2}{15}<\frac{2}{7}$$
|
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|
With how many ways can we choose $9$ balls of a box? With how many ways can we choose $9$ balls of a box that contains $12$ balls, of which $3$ are green, $3$ are white, $3$ are blue and $3$ are red?
$$$$
I have done the following:
$x_1=\# \text{ green balls that we choose }$
$x_2=\# \text{ white balls that we choose }$
$x_3=\# \text{ blue balls that we choose }$
$x_4=\# \text{ red balls that we choose }$
$$x_1+x_2+x_3+x_4=9, \ \ \ \ 0 \leq x_i \leq 3$$
$$\\$$
The number of solutions of the equation: $x_1+x_2+x_3+x_4=9, \ \ \ x_i \geq 0 \ \ \ $ is
$$\binom{9+4-1}{3}=\binom{12}{3}$$
Then I take the cases I don't want, to subtract them from the number above.
$x_1>3 \Rightarrow x_1 \geq 4$
$y_1=x_1-4 \geq 0$
So we become the equation:
$$y_1+4+y_2+y_3+y_4=9, \ \ \ y_i \geq 0 \ \ \ \ \Rightarrow y_1+y_2+y_3+y_4=5, \ \ \ y_i \geq 0$$
And the number of solutions are:
$$\binom{4+5-1}{3}=\binom{8}{3}$$
We do the same for $i=2,3,4$.
$$\\$$
Finally the solution is:
$$\binom{12}{3}- 4\cdot \binom{8}{3}=-4$$
That cannot be right..
What have I done wrong??
|
Assuming we do not care about the order in which the balls are selected, the number of ways of choosing $9$ balls from the box is the same as the number of ways of leaving $3$ balls in the box, which is much easier to calculate.
If we number the balls $1$ to $12$, then there are $12\choose 3$ ways of doing this. But removing balls $1,4,7$ would be the same as removing $2,4,7$, since balls $1$ and $2$ are both red. So we need to break it down to:
*
*Number of ways to leave one ball of each of $3$ colors = ${4 \choose 3} = 4$.
*Number of ways to leave two balls of one color and one ball of another color $ = 4\cdot 3 = 12$.
*Number of ways to leave three balls of one color = $4$.
Total number $= 4 + 12 + 4 = 20$ ways of removing $9$ balls.
|
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|
Math Olympiad Algebraic Question Comprising Square Roots
If $m$ and $n$ are positive real numbers satisfying the equation $$m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n=3$$
find the value of $$\frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}$$
I came across this question in a Math Olympiad Competition and had no idea how to solve it. Can anyone help? Thanks.
|
Given: $m$ and $n$ are positive real numbers satisfying the equation
$$m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n=3$$
Just to get a better feeling, substitute
$\sqrt{m}=x$
$\sqrt{n}=y$
Now your equation becomes
$x^2+4xy-2x-4y+4y^2=3$
Combining first, second and last term of L.H.S. , we get,
$(x+2y)^2-2(x+2y)=3$
Substitute: $(x+2y)=t$ to get,
$t^2-2t-3=0$
$\implies t=3$ or $t=-1$
But since $t=x+2y=\sqrt{m}+2\sqrt{n} \implies t \ge 0$ (Since $\sqrt{}$ gives positive value in its domain)
$\implies \sqrt{m}+2\sqrt{n}=3$
$\implies \dfrac{\sqrt{m} +2\sqrt{n} +2014}{4-(\sqrt{m} +2\sqrt{n})}= \boxed{2017}$
|
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|
Singular Values of negative eigenvalue? Consider a $4 \times 4$ matrix with eigenvalues $0,1,-2,4$. What are the singular values of this matrix?
I am aware that singular values are just square root of the eigenvalues, but what will be the answer for $-2$ eigenvalue? Is it undefined?
|
@Alexander Vigodner provides the punchline
$$
\sigma \left( \mathbf{A} \right) = \sqrt{\lambda \left( \mathbf{A} \right)}
$$
only when $\mathbf{A} = \mathbf{A}^{*}$.
We can provide a bit more insight with a smaller matrix. What are the $2\times 2$ matrices with the eigenvalue spectrum $\left( 1, 0 \right)$?
The characteristic polynomial is
$$
p(\lambda) = \lambda^{2} - \lambda.
$$
Symmetric matrices
The symmetric matrices have the form
$$
\mathbf{A}
=
\left[
\begin{array}{cc}
a & b \\
b & c \\
\end{array}
\right]
$$
with the characteristic equation:
$$
p(\lambda) = \lambda^{2} - \lambda \, \text{tr } \mathbf{A} + \det \mathbf{A}
$$
The matrices can be expressed in terms of one parameter, $b$:
$$
\mathbf{A}(b)
=
\left[
\begin{array}{cc}
\frac{1}{2} \left(1 \pm \sqrt{1-4 b^2} \right) & b \\
b & 1 - \frac{1}{2} \left(1 \pm \sqrt{1-4 b^2} \right) \\
\end{array}
\right]
$$
Example:
$$
\mathbf{A}\left( 1 \right) =
\left[
\begin{array}{cc}
\frac{1}{2} \left(i \sqrt{3}+1\right) & 1 \\
1 & \frac{1}{2} \left(1-i \sqrt{3}\right) \\
\end{array}
\right]
$$
Asymmetric matrices
Relax the symmetry requirement
$$
\mathbf{A}
=
\left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right]
$$
$$
\mathbf{A}(b,c)
=
\left[
\begin{array}{cc}
\frac{1}{2} \left(1 \pm \sqrt{1-4 b c} \right) & b \\
c & 1 - \frac{1}{2} \left(1 \pm \sqrt{1-4 b c} \right) \\
\end{array}
\right]
$$
Example:
$$
\mathbf{A}\left( 1, -1 \right) =
\left[
\begin{array}{cc}
\frac{1}{2} \left(1+\sqrt{5}\right) & 1 \\
-1 & \frac{1}{2} \left(1-\sqrt{5}\right) \\
\end{array}
\right]
$$
|
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|
How to integrate $\int \frac{1}{\sin^4x + \cos^4 x} \,dx$?
How to integrate
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx$$
I tried the following approach:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{\sin^4x + (1-\sin^2x)^2} \,dx = \int \frac{1}{\sin^4x + 1- 2\sin^2x + \sin^4x} \,dx \\
= \frac{1}{2}\int \frac{1}{\sin^4x - \sin^2x + \frac{1}{2}} \,dx = \frac{1}{2}\int \frac{1}{(\sin^2x - \frac{1}{2})^2 + \frac{1}{4}} \,dx$$
The substitution $t = \tan\frac{x}{2}$ yields 4th degree polynomials and a $\sin$ substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach:
$\sin^4x + \cos^4 x = (\sin^2 x + \cos^2x)(\sin^2 x + \cos^2 x) - 2\sin^2 x\cos^2 x = 1 - 2\sin^2 x\cos^2 x = (1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)$
and then I tried substituting: $t = \sin x \cos x$ and got
$$\int\frac{t\,dt}{2(1-2t^2)\sqrt{1-4t^2}}$$
Another way would maybe be to make two integrals:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{(1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)} \,dx = \\ \frac{1}{2}\int \frac{1}{1-\sqrt2\sin x \cos x} \,dx + \frac{1}{2}\int\frac{1}{1+\sqrt2\sin x \cos x} \,dx$$
... and again I tried $t = \tan\frac{x}{2}$ (4th degree polynomial) and $t=\sqrt2 \sin x \cos x$ and I get $\frac{\sqrt 2}{2} \int \frac{\,dt}{(1-t)\sqrt{1-2t^2}}$ for the first one.
Any hints?
|
Another approach:
We have
$$
\frac{1}{\sin^4x+\cos^4x},\tag1
$$
Multiply $(1)$ by $\dfrac{\tan^4x}{\tan^4x}$ we obtain
$$
\frac{\tan^4x}{\sin^4x(1+\tan^4x)}=\frac{\sec^4x}{1+\tan^4x}=\frac{(1+\tan^2x)\sec^2x}{1+\tan^4x}.\tag2
$$
Letting $t=\tan x$, the integral turns out to be
$$\eqalign
{
\int\frac{1+t^2}{1+t^4}\ dt&=\frac12\int\left[\frac{1}{t^2-\sqrt2t+1}+\frac{1}{t^2+\sqrt2t+1}\right]\ dt\\
&=\frac12\int\left[\frac{1}{\left(t-\dfrac1{\sqrt2}\right)^2+\dfrac34}+\frac{1}{\left(t+\dfrac1{\sqrt2}\right)^2+\dfrac34}\right]\ dt.\tag3
}
$$
Using substitution $u=\dfrac{\sqrt3}2\left(t-\dfrac1{\sqrt2}\right)$ and $v=\dfrac{\sqrt3}2\left(t+\dfrac1{\sqrt2}\right)$, the integral in $(3)$ can easily be evaluated.
Addendum :
Another way to evaluate $\displaystyle\int\frac{1+t^2}{1+t^4}\ dt$ is dividing the integrand by $\dfrac{t^2}{t^2}$, we obtain
$$\eqalign
{
\int\frac{1+\dfrac1{t^2}}{t^2+\dfrac1{t^2}}\ dt&=\int\frac{1+\dfrac1{t^2}}{\left(t-\dfrac1{t}\right)^2+2}\ dt.
}
$$
Now let $u=t-\dfrac1{t}\;\Rightarrow\;du=\left(1+\dfrac1{t^2}\right)\ dt$, the integral turns out to be
$$
\int\frac{1}{u^2+2}\ du.\tag4
$$
The evaluation of the integral $(4)$ can follow @achillehui's comment.
|
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|
number of non-negative integer solutions of $ax+by+cz=k$ How to find the number of non-negative integer solutions of $2x+3y+z=21$ ?
Please Help , thanks in advance
|
We can look at the number of $(x,y)$ pairs such that $2x+3y\le 21$.
We have:
$$(0,0),(0,1),\dots,(0,7)$$
$$(1,0),\dots,(1,6)$$
$$(2,0),\dots,(2,5)$$
$$\vdots$$
$$(10,0)$$
The maximum $y$ takes the value $\lfloor \frac{21-2x}{3} \rfloor$, i.e.
$$\begin{array} {c|c}
x&\lfloor \frac{21-2x}{3} \rfloor\\
\hline
0&7\\
1&6\\
2&5\\
3&5\\
4&4\\
5&3\\
6&3\\
7&2\\
8&1\\
9&1\\
10&0\\
\hline
\sum&37
\end{array}
$$
The lists are zero-based, so the final answer is $37+11=48$.
|
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|
Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
|
Note that
$$\sqrt{x^2+1}-x=\frac{1}{\sqrt{x^2+1}+x}\rightarrow 0$$ as $x\rightarrow \infty$
Your expression is equal to
$$\sqrt{(x+1)^2+1}-(x+1)+2 \rightarrow 2$$ as $x\rightarrow \infty$.
|
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|
Divisibility of sum of exponents Consider the sequence
$$r, \ ra, \ ra^2, \ ra^3, ... \ , ra^n \mod M $$
such that:
$$ ra^{n+1} \equiv r \mod M$$
and $a \ne 1$ and $a,r$ are both coprime to $M$
Is it always true then that:
$$ r + ra + \ ra^2 + \ ra^3 ... + \ ra^n \equiv 0 \mod M$$
Consider the case of $M = 7$ just to get a better feel for the problem:
For $ r = 2 \ a = 3 $
$$ 2 + 2*3 + 2*3^2 ... 2*3^5 = 2 \frac{1 - 3^6}{1 -3} \equiv 0 \mod 7 $$
We can also look at
$r = 2 \ a = 4 $
$$ 2 + 2*2 + 2*4 \equiv 2 + 4 + 1 \equiv 0 \mod 7 $$
Why does this appear to be the case? Is there counter-example
|
Your example of $M = 7$ does not make sense unless you mean $ra^{n+1} \equiv r$. Then, since $\gcd(M,r) = 1$, we have $a^{n+1} \equiv 1 \pmod M$. You also probably mean $a$ is not $\equiv 1$.
Since $a$ is not $\equiv 1$, the sum is equal to $r \cdot \frac{a^{n+1}-1}{a-1} \equiv 0 \pmod M$.
|
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|
How to calculate the integral of $\operatorname{sgn}(\sin\pi/x)$ in the interval $(0,1)$? How can I calculate the integral of $\operatorname{sgn}(\sin\pi/x)$ in the interval $(0,1)$?
I need to calculate this integral, thanks
|
Outline: In the interval $(1/2,1)$ our function is $-1$.
In the interval $(1/3,1/2)$, our function is $1$.
In the interval $(1/4,1/3)$, our function is $-1$.
In the interval $(1/5,1/4)$, our function is $1$.
And so on. The intervals have length $\frac{1}{1\cdot 2}$, $\frac{1}{2\cdot 3}$, $\frac{1}{3\cdot 4}$, and so on.
So the integral ought to be
$$-\frac{1}{1\cdot2}+\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}-\frac{1}{5\cdot 6}+\cdots.$$
If we want a closed form, note that
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\cdots.\tag{1}$$
Now calculate $\int_0^{1} \ln(1+x)\,dx$. This is the same as what we obtain when we integrate the series (1) term by term.
|
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|
If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$ If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$
My 1st approach :
$\tan(\alpha +2\alpha +4\alpha) = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $
$\Rightarrow 0 = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $ which doesn't give me any solution.
My IInd approach :
U\sing Euler substitution :
\since $\cos\theta +i\sin\theta = e^{i\theta} $.....(i) and $\cos\theta -i\sin\theta =e^{-i\sin\theta}$....(ii)
Adding (i) and (ii) we get $\cos\theta =\frac{e^{i\theta} +e^{-i\theta}}{2}$ and subtracting (i) and (ii) we get $\sin\theta =\frac{e^{i\theta} -e^{-i\theta}}{2}$
By u\sing this we can write : $$\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$$ as $$\frac{1}{4}\left[ (e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}) (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}}) + (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}})(e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) + (e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) (e^{\frac{i\pi}{7}} -e^{\frac{-i\pi}{7}})\right]$$
$$\large= e^{i\frac{6\pi}{7}}-e^{\frac{i2\pi}{7}}-e^{\frac{-i2\pi}{7}} +e^{\frac{-i6\pi}{7}} +e^{\frac{i3\pi}{7}}-e^{\frac{-i5\pi}{7}}-e^{\frac{i5\pi}{7}} +e^{\frac{-3\pi}{7}} +e^0 -e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}+e^0$$
Can anybody please suggest whether this is my correct approach or not. please guide further... Thanks.
|
(1) Note first that $$\tan x\tan(2x)=\frac{\sin x\sin(2x)}{\cos x\cos(2x)}=\frac{2\sin^2x}{\cos(2x)}=\frac{1}{\cos(2x)}-1
$$
(2) It follows that
$$\eqalign{
S~&\buildrel{\rm def}\over{=}~\tan\alpha\tan{2\alpha}+\tan2\alpha\tan{4\alpha}+\tan4\alpha\tan{\alpha}\cr
&=-3+\frac{1}{\cos\alpha}+\frac{1}{\cos2\alpha}+\frac{1}{\cos4\alpha}\cr
&=-3+\frac{\cos\alpha\cos2\alpha+\cos2\alpha\cos4\alpha+\cos4\alpha\cos\alpha}{\cos\alpha\cos2\alpha\cos4\alpha}\cr
&=-3+\frac{\cos\alpha+\cos3\alpha+\cos6\alpha+\cos2\alpha+\cos5\alpha+\cos 4\alpha}{2\cos\alpha\cos2\alpha\cos4\alpha}\tag{1}\cr
}
$$
(3) If $\xi=e^{i\alpha}$ then we have
$$1+\xi+\xi^2+\xi^3+\xi^4+\xi^5+\xi^6=0$$
Taking real parts we get
$$1+\cos\alpha+\cos2\alpha+\cos3\alpha+\cos4\alpha+\cos5\alpha+\cos6\alpha=0\tag{2}$$
also
$$8\cos\alpha\cos2\alpha\cos4\alpha=\frac{8\sin\alpha\cos\alpha\cos2\alpha\cos4\alpha}{\sin\alpha}=\frac{\sin8\alpha}{\sin\alpha}=1\tag{3}$$
(4) Replacing $(2)$ and $(3)$ in $(1)$ we obtain
$$
S=-3+\frac{-4}{1}=-7.
$$
which is the desired conclusion.
|
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|
Prove the sum $\sum_{n=1}^\infty \frac{\arctan{n}}{n}$ diverges. I must prove, that sum diverges, but...
$$\sum_{n=1}^\infty \frac{\arctan{n}}{n}$$
$$\lim_{n \to \infty} \frac{\arctan{n}}{n} = \frac{\pi/2}{\infty} = 0$$
$$\lim_{n \to \infty} \frac{ \sqrt[n]{\arctan{n}} }{ \sqrt[n]{n} } = \frac{1}{1} = 1$$
Cauchy's convergence test undefined.
There is a $E_0\gt0$:
$$\left|\frac{\arctan{n+1}}{n+1} + \frac{\arctan{n+2}}{n+2} + ... + \frac{\arctan{n+p}}{n+p}\right| \ge \frac{\pi}{4}\left|\frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{n+p}\right| \ge \frac{\pi}{4} \frac{p}{n+p} (Let\, p = n) \ge \frac{\pi}{4} \frac{\bcancel{n}}{2\bcancel{n}} = \frac{\pi}{8} (\sim0.4) \ge E_0 = \frac{1}{8} \gt 0;$$
Now am I correct?
|
For large enough $n$, $\text{arctan}(n)\geqslant \pi/4$ and $\sum \frac{\pi/4}{n} $ diverges.
|
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|
Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?
Prove or disprove that if $t$ is a positive integer, $$f(x,y)=\dfrac{x^2+y^2}{xy-t},$$ then $f(x,y)$ has only finitely many distinct positive integer values with $x,y$ positive integers. In other words, there exist $k\in\mathbb N$ such that if $n\gt k$ then $f(x,y)=n$ has no positive integer solutions.
This problem is a generalization of this famous problem.
*
*Below is the list of the set of $f(x,y)$ with $t\le 10$ (may be incomplete):
{t,{f(x,y)}}=
{1,{5}}
{2,{4,10}}
{3,{3,4,8,13,17}}
{4,{5,26}}
{5,{13,25,37}}
{6,{6,10,50}}
{7,{5,8,9,20,29,41,65}}
{8,{4,10,18,34,82}}
{9,{5,29,61,101}}
{10,{20,122}}
Thanks in advance!
|
October 14, 2015. This is with $$ \frac{x^2 + y^2}{xy - t} = q > 0, $$
which I believe to be the intent of the question.
THEOREM: $$ \color{red}{ q \leq (t+1)^2 + 1 } $$
I got some help from Gerry Myerson on MO to finish the thing.
https://mathoverflow.net/questions/220834/optimal-bound-in-diophantine-representation-question/220844#220844
As far as rapid computer computations, for a fixed $t,$ we can demand $1 \leq x \leq 4 t.$ For each $x,$ we can then demand $1 \leq y \leq x$ along with the very helpful $x y \leq 4 t.$ Having found an integer quotient $q,$ we then keep only those solutions with $2x \leq qy$ and $2y \leq qx.$
In particular, for $t=1$ we find $q=5,$ then for $t=2$ we find $q=4,10.$ In both cases we have $q \leq (t+1)^2 + 1.$ We continue with $t \geq 3.$
With $t \geq 3, $ we also have $t^2 \geq 3t > 3t - 1.$
We are able to demand $xy \leq 4t$ by taking a Hurwitz Grundlösung, that is $2x \leq qy$ and $2y \leq qx.$ Define $k = xy - t \geq 1.$
Now, $xy \leq 4t,$ then $k = xy - t \leq 3t,$ then $k-1 \leq 3t - 1.$ Reverse, $3t-1 \geq k-1.$ Since $t^2 > 3t - 1,$ we reach
$$ t^2 > k-1. $$
Next, $k \geq 1,$ so $(k-1) \geq 0.$ We therefore might get equality in
$$ (k-1)t^2 \geq (k-1)^2, $$
but only when $k=1.$
$$ 0 \geq t^2 - k t^2 + k^2 - 2 k + 1, $$
$$ k t^2 + 2 k \geq t^2 + k^2 + 1. $$ Divide by $k,$
$$ t^2 + 2 \geq \frac{t^2}{k} + k + \frac{1}{k}. $$ Add $2t,$
$$ t^2 +2t + 2 \geq \frac{t^2}{k} + 2 t + k + \frac{1}{k}, $$
with equality only when $k=1.$ Reverse,
$$ \frac{t^2}{k} + 2 t + k + \frac{1}{k} \leq t^2 +2t + 2 $$
with equality only when $k=1.$
Here is Gerry's best bit, this would not have occurred to me. Here we are back to considering all solutions $(x,y)$ and all $k=xy-t.$ Draw the graph of the quarter circle $x^2 + y^2 = k q.$ As $x,y \geq 1,$ there are boundary points at $(1, \sqrt{kq-1})$ and $( \sqrt{kq-1},1).$ The hyperbola $xy = \sqrt{kq-1}$ passes through both points, but in between stays within the quarter circle. It follows by convexity (or Lagrange multipliers again) that, along the circular arc,
$$ \color{blue}{ xy \geq \sqrt{kq-1}}. $$ But, of course, $x^2 + y^2 = k q = qxy - t q$ is equivalent to our original equation $x^2 - q x y + y^2 = -tq.$
We have
$$ -tq = x^2 - q x y + y^2 = (x^2 + y^2 ) - q x y = k q - q x y \leq kq - q \sqrt{kq-1}, $$ or
$$ -tq \leq kq - q \sqrt{kq-1}, $$
$$ -t \leq k - \sqrt{kq-1}, $$
$$ \sqrt{kq-1} \leq t + k, $$
$$ kq -1 \leq t^2 + 2k t + k^2, $$
$$ kq \leq t^2 + 2 kt + k^2 + 1, $$ divide by $k,$
$$ q \leq \frac{t^2}{k} + 2 t + k + \frac{1}{k}. $$
For $t \geq 3$ and a solution with $xy < 4t,$ we showed
$$ \frac{t^2}{k} + 2 t + k + \frac{1}{k} \leq t^2 +2t + 2 $$
with equality only when $k=1.$ For all solutions, Gerry showed
$$ q \leq \frac{t^2}{k} + 2 t + k + \frac{1}{k}. $$
Put these together, we get
$$ q \leq t^2 +2t + 2 $$
with equality only when $k=1,$ that is $xy = t+1.$
ADDENDUM, October 15. Here is another way to get Gerry's main observation, with $k = xy - t,$ that $xy \geq \sqrt{kq-1}.$ We have $x,y \geq 1$ and
$kq =x^2 + y^2 .$ So $kq \geq x^2 + 1$ and $kq -(x^2 + 1) \geq 0.$ We also have $x^2 - 1 \geq 0.$ Multiply,
$$ (x^2 - 1) kq - (x^4 - 1) \geq 0. $$
Next, $y^2 = kq - x^2,$ so $x^2 y^2 = kq x^2 - x^4.$ That is
$$ x^2 y^2 = (kq-1) + (x^2 - 1)kq - (x^4 - 1). $$ However,
$$ (x^2 - 1) kq - (x^4 - 1) \geq 0, $$ so
$$ x^2 y^2 \geq kq - 1, $$
$$ \color{blue}{ xy \geq \sqrt{kq-1}}. $$
|
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|
Drawing Marbles Probability My daughter had a math problem in her textbook that we are having a hard time understanding. You have $3$ white and $4$ black marbles in a bag and $3$ people will draw a marble and not replace it in the bag. The first person to draw the white marble wins. What is the probability that the first person will win, 2nd, 3rd? The answer given was $\frac{18}{35}, \frac{11}{35},$ and $\frac{6}{35}$.
First we were able to get these answers by using 3! and 2! but we are not sure why we would do that.
Second, if these answers are correct can anyone explain why the first person drawing against the total number would have the highest probability of winning? I would believe it to be the opposite.
|
I'll expand a bit on what David said. Firstly, it is logical that the first person has the highest probability of winning, as that person can stop the contest before anyone else has a chance. All the other contestants' chances are dependent on all prior contestants NOT winning prior to their turn.
Round 1:
There are four possibilities:
*
*Person A wins
*Person B wins
*Person C wins
*We go to round 2.
Person A has a $\frac{3}{7}$ probability of winning flat out, as there are 3 white marbles and 7 total. There is a $\frac{4}{7}$ probability that Person B gets a draw. If person B gets a draw, that must mean there are 6 marbles left, 3 black and 3 white. So the probability of person B winning in round 1 is$$
\large \overbrace{\frac{4}{7}}^\textrm{A does not win}\underbrace{\frac{3}{6}}_\textrm{B wins} = \frac{4\cdot 3}{7\cdot 6} = \frac{2}{7}
$$
The same argument shows that the probability that C wins in round 1 is $\frac{6}{35}$ and the probability that round 2 is entered is $\frac{4}{35}$.
Round 2:
At this point there must be 3 white and only 1 black marble left (as to get here, everyone had to pick black marbles in round 1). Given we had a $\frac{4}{35}$ probability to get here, person A has a $\frac{4}{35} \cdot \frac{3}{4}$ chance of winning, and if Person A does not win in round 2, person B must as all there are left are white marbles. Person C cannot get a chance in round 2—there aren't enough black marbles.
Putting this all together we have:
$$
\begin{align}
\textrm{Person A} &= \frac{3}{7} + \frac{4}{35}\cdot\frac{3}{4} = \frac{15 + 3}{35} &= \frac{18}{35}\\
\textrm{Person B} &= \frac{2}{7} + \frac{4}{35}\cdot\frac{1}{4} = \frac{10 + 1}{35} &= \frac{11}{35}\\
\textrm{Person B} &= \frac{6}{35} + 0 &= \frac{6}{35}
\end{align}
$$
|
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|
Evaluate the limit or show that is does not exist $$\lim_{x \to 0}\frac{1-\cos(3x)}{2x^2}$$
So far I have noted
$$\frac{1-\cos(3x)}{2x^2} = \frac{1-\cos(3x)}{2x^2} \cdot \frac{1+\cos(3x)}{1+\cos(3x)}$$
and then used the identity $$\sin^2x = 1-\cos^2x$$ to reduce this to
$$\frac{\sin^2(9x)}{2x^2+\cos(9x)}.$$
Any tips on how to proceed?
|
You had the right idea, but your algebra is suspect. For example, $\cos 3x \cdot \cos 3x \ne \cos^2 9x$
$$\lim_{x\to 0} \frac{1-\cos 3x}{2x} = \frac{3}{2}\cdot\lim_{x\to 0} \frac{1-\cos 3x}{3x}$$
$$= \frac{3}{2}\cdot\lim_{3x\to 0} \frac{1-\cos 3x}{3x}$$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{1-\cos \theta}{\theta}$$
where $\theta = 3x$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{1-\cos \theta}{\theta}\cdot \frac{1+\cos \theta}{1+\cos \theta}$$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{1-\cos^2 \theta}{\theta}\cdot \frac{1}{1+\cos \theta}$$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{\sin^2 \theta}{\theta}\cdot \frac{1}{1+\cos \theta}$$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{\sin \theta}{\theta}\cdot \frac{\sin \theta}{1+\cos \theta}$$
$$= \frac{3}{2}\cdot\lim_{\theta\to 0} \frac{\sin \theta}{\theta}\cdot \lim_{\theta\to 0}\frac{\sin \theta}{1+\cos \theta}$$
$$= \frac{3}{2}\cdot\frac{0}{1+1}\cdot\lim_{\theta\to 0} \frac{\sin \theta}{\theta}$$
$$= 0\cdot\lim_{\theta\to 0} \frac{\sin \theta}{\theta}$$
$$= 0$$
Now all that is required is to prove that $\displaystyle\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1$ (or at the very least, that it exists).
on $(0, \pi/2)$
$$0 \lt \sin \theta \lt \theta$$
$$0/\theta \lt \sin \theta / \theta \lt \theta/\theta$$
$$0 \lt \frac{\sin \theta}{\theta} \lt 1$$
thus $ 0 \le \displaystyle\lim_{\theta\to 0}\frac{\sin \theta}{\theta} \le 1$
and
$$0\cdot 0 \le 0\cdot \displaystyle\lim_{\theta\to 0}\frac{\sin \theta}{\theta} \le 0\cdot1$$
$$0 \le 0\cdot \displaystyle\lim_{\theta\to 0}\frac{\sin \theta}{\theta} \le 0$$
$$\therefore 0\cdot \displaystyle\lim_{\theta\to 0}\frac{\sin \theta}{\theta} = 0$$
|
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|
Interesting line integral using green's theorem Find $$\int_C \frac{2x^3+2xy^2-2y}{ x^2+y^2} \, dx+\frac{2y^3+2x^2y+2x}{ x^2+y^2} \, dy$$
Where $C$ is any simple closed loop which contains the origin.
What I figured out
I cannot use the direct version of Green's theorem.
I know that there is another version of greens theorem which is as follows :
$$\oint_{C+S} P \,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)dA,$$
where $C$ and $S$ are oriented opposite to each other and $D$ is the region enclosed between the two loops.
Using that I found
\begin{align*}
\int_{C+S} \left( 2x-\frac{2y}{ x^2+y^2}\right) \, dx+\left(2y+\frac{2x}{ x^2+y^2} \right) \, dy
&= \iint_D \left[\left(2+\frac{4xy}{(x^2+y^2)^2} \right)-\left(2-\frac{4xy}{(x^2+y^2)^2} \right) \right]dA \\
&=\iint_D \frac{8xy}{(x^2+y^2)^2} dA
\end{align*}
The region inside any closed loop can be defined in polar coordinates as $$\{(r, \theta)\in D \mid 0\leq r\leq r(\theta), \, 0\leq \theta \leq 2\pi \}$$
In polar coordinates the integral becomes $$ =\iint_D \frac{8xy}{(x^2+y^2)^2} dA = \int_0^{2\pi}\!\!\int_0^{r(\theta)} \frac{4\sin2\theta}{r} dr\,d\theta $$
Oops, I am again stuck; it's not zero.
|
You can write your integral as
$$2\int_{C} \left(-\frac{y}{ x^2+y^2}dx+\frac{x}{ x^2+y^2}dy \right) + \int_{C} \left( 2x\ dx + 2y\ dy\right)\ ,$$
the second form is exact, so its contribution is 0, while for the first for you can do the following. Note that $$\left(-\frac{y}{ x^2+y^2}dx+\frac{x}{ x^2+y^2}dy \right) =d\ \arctan\left(\frac{y}{x}\right)\ ,$$
which in polar coordinates can be written as $d\theta$. Your integral is thus just
$$-2\int_{C} d\theta = \mp 4\pi$$
where the sign depend on the orientation of the curve you choose: $-$ if it's counterclockwise, $+$ if clockwise.
|
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|
Proof by induction that $(1^3 + 2^3 + 3^3+\cdots+n^3) = (1 + 2 + \cdots + n)^2$ I'm sitting with the proof in front of me, but I do not understand it.
$$A = \{n \in Z^{++} \mid (1^3 + 2^3 + 3^3+\cdots+n^3) = (1 + 2 + \cdots + n)^2\}$$
The first step of proof by induction is simple enough,to prove that $1 \in A$
$1^3 = 1^2$
The next step is where I get tripped up. So I add $n + 1$ to the right hand side
$$(1 + 2 + \cdots + n + (n + 1))^2 = (1 + 2 + \cdots + n)^2 + 2(1 + 2 + \cdots + n)(n + 1) + (n + 1)^2$$
My algebra is failing me here, because I do not understand how the equation was expanded.
|
For the inductive step:
$$(\underbrace{1+2\cdots+n}_{=A=\frac{n(n+1)}{2}}+(n+1))^2=\underbrace{A^2}_{=1^3+2^3+\cdots+n^3}+\underbrace{(n+1)^2+n(n+1)^2}_{=(n+1)^3}$$
|
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|
Find a generating function for $\sum_{k=0}^{n} k^2$
Find a generating function for $\sum_{k=0}^{n} k^2$
I know my solution is wrong, but why?
My solution:
If $F(x)$ generates $\sum_{k=0}^{n} k^2$ then $F(x)(1-x)$ generates $k^2$.
$\frac{x}{(1-x)^4}: \left\{ 0,1,4,9,16,25... \right\}$
$\frac{x}{(1-x)^3}: \left\{ 0,1,3,5,7,9... \right\}$
$\frac{x}{(1-x)^2}: \left\{ 0,1,2,2,2,2... \right\}$
$\frac{x}{1-x}: \left\{ 0,1,1,1,1,1... \right\}$
$x: \left\{ 0,1,0,0,0,0... \right\}$
So, $F(x)=\frac{x}{(1-x)^5}$ generates $\sum_{k=0}^{n} k^2$
|
According to the example, how to get the generating function for squares you can do the following:
Use $$
\sum_{n=0}^{\infty}\binom{n+k}k x^n= \frac{1}{(1-x)^{k+1}} $$
and set
$$
a\binom{n+3}{3}+ b\binom{n+2}{2}+c\binom{n+1}{1}+d=\frac{n^3}{3}+\frac{n^2}{2}+\frac n2=\sum_{k=1}^n k^2
$$
You'll find $a=2,b=-3,c=1$ and $d=0$. So
$$
\frac{2}{(1-x)^4}-\frac{3}{(1-x)^3}+\frac{1}{(1-x)^2}
$$
is the generating function, which sums up to Daniel's answer...
|
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|
Partial fraction (doubt) I have this partial fraction
$$\displaystyle\frac{1}{(2+x)^2(4+x)^2}$$
I tried to resolve using this method:
$$\displaystyle\frac{A}{2+x}+\displaystyle\frac{B}{(2+x)^2}+\displaystyle\frac{C}{4+x}+\displaystyle\frac{D}{(4+x)^2}$$
$$1=A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2$$
When x=-2
$$1=B(4-2)^2$$
$$B=\displaystyle\frac{1}{4}$$
When x=-4
$$1=D(2-4)^2$$
$$D=\displaystyle\frac{1}{4}$$
When x=0
$$1=A(2)(16)+B(16)+C(4)(4)+D(4)$$
$$1=A(32)+B(16)+C(16)+D(4)$$
Replacing the values for B y D
$$1=A(32)+4+C(16)+1$$
$$1-4-1=A(32)+c(16)$$
$$-4=A(32)+C(16)$$
How I can get the values of $A$ and $D$?
|
take LCM of second expression
you get and sum it
{A(2+x)(4+x)^2 +B(4+x)^2+C(2+x)^2(4+x)+D(2+x)^2}/{(x+2)^2(x+4)^2}= your 1st expression
so you have {A(2+x)(4+x)^2 +B(4+x)^2+C(2+x)^2(4+x)+D(2+x)^2}=1
compare coefficients of x^3 ,x^2,x^1 and x^0 which on right side of aboveequation are equal to 0,0,0 and 1 respectively.
you have
A+C=0
8A+B+4C+D=0
32A+8B + 20C+ 4D=0
32A + 16B + 16C +4D=1
soLVING YOU GET
B=1/12, D=7/12 ,A =-1/6, C=1/6
|
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|
Real solution of the equation $(x^2-2x+2)^2-2(x^2-2x+2)+2 = x$
Calculate all real solutions $x\in\mathbb{R}$ of the equation
$$
\tag1(x^2-2x+2)^2-2(x^2-2x+2)+2 = x
$$
My Attempt:
I used the concept of a composite function. Let $f(x) = x^2-2x+2$. Then equation $(1)$ converts into $f(f(x)) = x$. Both $f(x) = x$ and $f(x) = -x$ satisfy the given composite function.
Case 1: If $f(x) = x$, then
$$
x^2-2x+2=x\\
x^2-3x+2=0\\
x\in\{1,2\}
$$
Note that $1,2\in\mathbb{R}$.
Case 2: If $f(x)=-x$, then
$$
x^2-2x+2=-x\\
x^2-x+2=0\\
x=\frac{1\pm \sqrt{1-8}}{2}\notin \mathbb{R}
$$
So only $x\in\{1,2\}$ are the real solutions of above equation.
Is my process correct? Is there is any other method by which we can solve the above question?
|
Setting $T=x-1$ you have $(T^2+1)^2-2(T^2+1)+1-T=0$ or $T^4-T=0$ that is
$T(T-1)(T^2+T+1)=0$ so for real solutions you have $T=0$ or $T=1$, that is $x=1$ or $x=2$.
|
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|
How to solve this quadratic form equation?
Let $Q(x,y,z)=7x^2+7y^2-2z^2-10xy+8xz+8yz$ be a quadratic form and
$A = \begin{bmatrix}
7 & -5 & 4 \\
-5 & 7 & 4 \\
4 & 4 & -2
\end{bmatrix}$ its matrix. Such that $Q=X^TAX$ for $X=\begin{bmatrix} x \\y \\ z \end{bmatrix}$.
Find an $X$ such that $X^TAX=72$.
I found that the eigenvalues of $A$ are $-6,6\text{ and }12$, so the matrix is not positive definite neither negative definite.
How can I solve this kind of problem? Can you give me a hint? Thanks
|
I found the follow technic for this kind of problem, that is very similar to the Mark Bennet's answer.
If $Q(x,y,z)=7x^2+7y^2-2z^2-10xy+8xz+8yz$ then $A$ is the associated matrix.
Let $X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} \neq0 \text{ and } \lambda \in \mathbb{R}$. If $AX=\lambda X$, then $\lambda$ is an eigenvalue of $A$ and $X$ is an eigenvector of $A$, associated to $\lambda$.
So, $Q(x,y,z)=X^TAX= \lambda X^T X= \lambda (x^2+y^2+z^2)$. One have that $(x^2+y^2+z^2) >0$, so if $\lambda (x^2+y^2+z^2)=72$, then $\lambda>0$.
Between the three eigenvalues of $A$, only $6 \text{ and } 12$ can be applied.
Now by finding the vector space span for the eigenvectors for $6 \text{ and } 12$, one got:
$E_{6}=\langle(1,1,1) \rangle$
$E_{12}=\langle(-1,1,0) \rangle$
About the $X$ that is asked, or $X \in E_{6}$ or $X \in E_{12}$. I found that $X \in E_{12}$:
Let $\begin{bmatrix} -a \\ a \\ 0 \end{bmatrix}$ be a generic vector of $E_{12}$. So , $12 \left((-a)^2+a^2+0^2 \right)=72$. Then
$$12 \left(2a^2 \right)=72$$
$$ 24a^2 =72$$
$$a^2=3$$
$$a=\sqrt3$$
Finaly, a $X$ that satisfy what is asked is $X=\begin{bmatrix} -\sqrt3 \\ \sqrt3 \\ 0 \end{bmatrix}$. You can confirm, by computing $Q(x,y,z)$.
$$Q(-\sqrt3,\sqrt3,0)=7 \cdot (-\sqrt3)^2+7\cdot \sqrt3^2 -10 \cdot (-\sqrt3) \cdot \sqrt3$$
$$Q(-\sqrt3,\sqrt3,0)=7 \cdot 3+7\cdot 3+10 \cdot 3$$
$$Q(-\sqrt3,\sqrt3,0)=21+21+30=72$$
|
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|
$ F(x) = \int_0^2 \sin(x+l)^2\ dl$ Consider the function : $ F(x) = \int_0^2 \sin(x+l)^2\ dl$, calculate $ \frac{dF(x)}{dx}|_{x=0}$ the derivative of $F(x)$ with respect to $x$ in zero.
Let $g(x) = \sin (x)$ and $h(x) = (x+l)^2$ then $F(x) = \int_0^2 g(h(x))\ dl$, so $F´(x) = g(h(x)) h´(x)$ then I can evaluate this in $x=0$ , so $F´(0) = \sin{l^2} 2l $ is that correct? Some help to calculate this please.
|
If you can justify permuting the order of limit and integral then
\begin{align}
\frac{dF(x)}{dx}
&= \lim_{h \to 0} \frac{1}{h}\left( \int_0^2 \sin(x + h+ \ell)^2 d\ell -
\int_0^2 \sin(x + \ell)^2 d\ell \right) \\
&= \lim_{h \to 0} \frac{1}{h} \int_0^2 (\sin(x + h+ \ell)^2 - \sin(x + \ell)^2) d\ell \\
&= \int_0^2 \lim_{h \to 0} \frac{\sin(x + h+ \ell)^2 - \sin(x + \ell)^2}{h} d\ell \\
&= \int_0^2 2(x + \ell) \cos(x + \ell)^2 d\ell \\
&= \int_x^{2+x} 2u \cos u^2 du \\
&= \left. \sin u^2 \right|^{2+x}_x \\
&= \sin(2+x)^2 - \sin x^2.
\end{align}
|
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|
Proof of Cauchy's functional equation for rational arguments We have thesis that for every $c\in\mathbb{Q}$ every additive function has form of $f(x)=cx$. In the proof we're showing that $f(nx)=nf(x)$. Then we're supposed to replace $nx$ by $\frac{1}{n}x$. Why can we do this?
|
It follows from $x = \underbrace{\frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \cdots}_{n\text{ times}}$
$$f(x) = f(\underbrace{\frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \cdots}_{n\text{ times}}) = \underbrace{f(\frac{1}{n}x) + f(\frac{1}{n}x) + f(\frac{1}{n}x) + \cdots}_{n\text{ times}} = nf\left(\frac{1}{n}x\right)$$
|
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|
$\forall \ n \in \mathbb{N}: 6\mid 5n^3+n$ Show $\forall \ n \in \mathbb{N}: 6\mid 5n^3+n$
Indeed,
we've to show $5n^3+n=0\pmod 6 $
note that for $n\in \{0,1,2,3,4,5\}$ we've: $n^{3}=n\pmod 6$ then
$5n^{3}=5n\pmod 6 \implies 5n^{3}+n=6n\pmod 6=0\pmod 6$
AM i right ?
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Use induction
$$
5\Big(n+1\Big)^3 + \Big(n+1\Big) = 5 n^3 + n + 6 \Big( \tfrac{5}{2} n(n+1) +1 \Big)
$$
and as $n(n+1)$ is even we see that
$$
5\Big(n+1\Big)^3 + \Big(n+1\Big) = 5 n^3 + n + 6 k
$$
As it is true for $n=1$, it is true for all $n \ge 1$.
|
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|
$a^{12} \equiv 1 \pmod{35}$,knowing that $(a,35)=1$ Prove that $\forall a \text{ with } (a,35)=1:$
$$a^{12} \equiv 1 \pmod{35}$$
$$35 \mid a^{12}-1 \Leftrightarrow 5 \cdot 7 \mid a^{12}-1 \overset{(5,7=1)}{ \Leftrightarrow} 5 \mid a^{12}-1 \text{ and } 7 \mid a^{12}-1$$
Therefore, $\displaystyle{ a^{12} \equiv 1 \pmod{35} \Leftrightarrow a^{12} \equiv 1 \pmod 5, a^{12} \equiv 1 \pmod 7}$
$$(a,35=1) \Rightarrow (a, 5 \cdot 7)=1 \overset{(5,7)=1}{\Rightarrow } (a,5)=1 \text{ and } (a,7)=1$$
According to Fermat's theorem:
$$a^4 \equiv 1 \pmod 5$$
$$a^{12} \equiv (a^4)^3 \equiv 1 \pmod 5$$
Also:
$$a^6 \equiv 1 \pmod 7$$
$$a^{12} \equiv (a^6)^2 \equiv 1 \pmod 7$$
So,we conclude that:
$$a^{12} \equiv 1 \pmod{35}$$
Could you tell me if it is right?
|
Hint $\ $ If $\,a\,$ is coprime to primes $\,p_1\!\ne p_2\,$ and $\,\color{#0a0}{p_i\!-1\mid n}\,$ then by $\rm\color{#c00}{little\ Fermat}$
$\qquad\qquad\qquad {\rm mod}\ p_i\!:\,\ a^n\equiv (\color{#c00}{a^{\,\large p_i-1}})^{\Large\color{#0a0}{\frac{n}{p_i-1}}\!}\equiv \color{#c00}1^{\Large\color{#0a0}{\frac{n}{p_i-1}}}\equiv 1\,\Rightarrow\, p_i\mid a^n-1\,\Rightarrow\, p_1 p_2\mid a^n-1$
|
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|
Trigonometric equation with complex numbers Let $x$, $y$, and $z$ be real numbers such that $\cos x+\cos y+\cos z=\sin x+\sin y+\sin z=0$.
Prove that $\cos 2x+\cos 2y+\cos 2z=\sin 2x+\sin 2y+\sin 2z=0$.
Starting with the given equation, I got that $i\sin x+i\sin y+i\sin z=0$.
Adding this to the other part of the given equation, I then got $\cos x+\cos y+\cos z+i\sin x+i\sin y+i\sin z=0$, which can also be written as $(\cos x+i\sin x)+(\cos y+i\sin y)+(\cos z+i\sin z)=0$.
Here, I let $a=e^{ix}$, $b=e^{iy}$, and $c=e^{iz}$, which, after substituting in to the above equation gives $a+b+c=0$.
What we want to prove can be written as $a^2+b^2+c^2$, but I am not quite sure how to find this result from what I have.
Some help would be very appreciated. Thanks!
|
In the same way you have: $(\cos x-i \sin x)+(\cos y-i\sin y)+(\cos z-i \sin z)=0$, it's equal: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$, so $ab+ac+bc=0$ and $0=(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2=0$
|
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Optimization of parallelepiped inside an ellipsoid Let $K \in R^3$ the ellipsoid given by the equation $ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $ with $a,b,c > 0$ , let $(x,y,z) \in K$ on the first octant, consider the parallelepiped of vertices $(\pm x,\pm y,\pm z)$ inscribed on $K$ with volume $V = 8xyz$.
How can I find the maximum possible value of $V$?
I stuck with this hard problem for me i tried to find the explicit equation and then get the maximum values : Let $P=(x,y,z)$ be a point on the ellipsoid with $x,y,z\gt 0$.Then i took the eight different points with $P_i (\pm x,\pm y,\pm z)$ the vertices of a parallelepiped with the side length $2x , 2y$ and $2z$.
Then, the volume parallelepiped is $V = 2x\cdot 2y\cdot 2z = 8 xyz$ and i remembered that $V$ is maximum if and only if $V^2$ is maximum .
Some help please.
|
Because
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$
We have
$$\left(\frac{x^2}{a^2}\frac{y^2}{b^2}\frac{z^2}{c^2}\right)^{1/3} \le \frac{1}{3}\left(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}\right) = \frac{1}{3}$$
Thus
$$V=8|xyz| \le \frac{8abc}{\sqrt{27}}$$
|
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|
Integral: $\int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx$ Please help me in proving the following result:
$$\displaystyle \int_0^\infty \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; dx=\frac{\pi}{b}e^{-b\sqrt{a^2+c^2}}\sinh (ab)$$
I found this integral from here: http://integralsandseries.prophpbb.com/post2652.html?sid=d6641d4d4a3726f1b27bbb4b98ca840a and the solution uses contour integration. I am wondering if there is a way to solve it without using contour integration. I tried differentiating wrt $a$ and $c$ but in both cases, the resulting expression was dirty which made me reluctant to proceed further. I am out of ideas for this one.
Any help is appreciated. Thanks!
|
In order to prove the final result I will need to state a lemma that will be used later.
Lemma$\require{autoload-all}$ $1$:
$$\int_0^\infty \! \frac{\cos(bx)}{x^2+\alpha} \mathrm{d}x = \frac{\pi e^{-b\sqrt{\alpha}}}{2b\sqrt{\alpha}}\tag{1}$$
Proof here.
Consider
$$I = \int_0^\infty\!\! \tan^{-1}\left(\frac{2ax}{x^2+c^2} \right)\sin(bx) \; \mathrm{d}x$$
Integrate by parts
$$I = \int_0^\infty \!\!\frac{2 a \left(c^2-x^2\right) \cos (b x)}{x^4 +(4 a^2+2 c^2) x^2+c^4} \; \mathrm{d}x$$
Decompose this function by partial fractions $$\frac{2 a \left(c^2-x^2\right) }{x^4 +(4 a^2+2 c^2) x^2+c^4} = \frac{a_-}{x^2+x_0} + \frac{a_+}{x^2+x_1}$$
It so happens that
$$x_0 = 2 a^2+2 a\sqrt{a^2+c^2}+c^2,\quad x_1 = 2 a^2-2a \sqrt{a^2+ c^2}+c^2$$
$$a_- =\frac{2a(c^2+x_0)}{x_1-x_0}, \quad a_+ = \frac{2a(c^2+x_1)}{x_0-x_1}$$
Note that both $x_0$ and $x_1$ are greater than $0$.
Re-write the integral
$$\begin{align} I &= a_-\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_0} \, \mathrm{d}x + a_+\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_1} \, \mathrm{d}x\\[.3cm] &= \frac{2a(c^2+x_0)}{x_1-x_0}\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_0} \, \mathrm{d}x +\frac{2a(c^2+x_1)}{x_0-x_1}\int_0^{\infty} \!\! \frac{\cos(bx)}{x^2+x_1} \, \mathrm{d}x\end{align}$$
Using $(1)$:
$$\begin{align} I &= \frac{2a(c^2+x_0)}{x_1-x_0}\cdot\frac{\pi e^{-b\sqrt{x_0}}}{2b\sqrt{x_0}} - \frac{2a(c^2+x_1)}{x_1-x_0}\cdot\frac{\pi e^{-b\sqrt{x_1}}}{2b\sqrt{x_1}}\\[.3cm] &= \left(\frac{a\pi}{b(x_1-x_0)}\right)\left(\frac{(c^2+x_0)e^{-\sqrt{x_0}}}{\sqrt{x_0}} - \frac{(c^2+x_1)e^{-\sqrt{x_1}}}{\sqrt{x_1}}\right)\end{align}$$
I will digress here to state (without proof but easily verified) that $$\frac{c^2+x_0}{\sqrt{x_0}}= \frac{c^2+x_1}{\sqrt{x_1}} = \frac{x_1-x_0}{2a}.$$ This allows us a tremendous simplification so that we can write
$$\begin{align} I = \left(\frac{\pi}{2b}\right)\left(e^{-b\sqrt{x_1}}-e^{-b\sqrt{x_0}} \right). \end{align}$$
It can also be shown that
$$\sqrt{x_1} = -a+\sqrt{a^2+c^2}$$
$$\sqrt{x_0} = a+\sqrt{a^2+c^2}$$
Simply square each side to find the desired equality.
We can now complete the proof:
$$\begin{align} I &= \frac{\pi}{2b}\left(e^{-b\sqrt{x_1}}-e^{-b\sqrt{x_0}} \right) \\[.2cm]
&= \frac{\pi}{2b}\left(\exp\left(ab-b\sqrt{a^2+c^2}\right)-\exp\left(-ab-b\sqrt{a^2+c^2}\right) \right) \\[.2cm]
&= \frac{\pi}{b}\exp\left(-b\sqrt{a^2+c^2}\right)\frac12(\exp(ab)-\exp(ab)) \\[.2cm]
&= \dfrac{\pi}{b}\exp\left(-b\sqrt{a^2+c^2}\right)\sinh{ab}\end{align}$$
If you are interested in working through the simplifications that I did not prove, I recommend that you begin by squaring each side after verifying that each side shares the same sign.
|
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When is $(a^2+b)(b^2+a)$ a power of $2$? When is $(a^2+b)(b^2+a)$ a power of $2$?
($a$, $b$ positive integers)
I tried some values on the computer and it seems the only solutions are $a=1$, $b=1$. But I am not sure how to prove it?
Thanks for any help.
|
The product $(a^2+b)(a+b^2)$ is a power of two if and only if both factors are:
$$\begin{array}{rcl}
a^2+b&=&2^r\\
a+b^2&=&2^s
\end{array}$$
If $a=b$, then the equation $a^2+a=a(a+1)=2^r$ has only one solution, namely, $a=1$. So we can WLOG assume that $a>b$. Thus, $r>s$.
Since $2^s=a^2+b\geq2^2+1$, we have that $s>1$.
Since $s>1$, $a$ and $b$ have the same parity. Substracting the equations and factoring we have:
$$(a-b)(a+b-1)=2^s(2^{r-s}-1)$$
Since $a+b-1$ is odd, $2^s$ divides $a-b$, that is, $a=2^sk+b$, $k\geq1$. But $a=2^s-b^2<2^s$, a contradiction.
|
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Sketch the set $\{ z \in \mathbb{C} | \left|\frac{z-i}{z+i}\right|<1 \}$ My question is to sketch the set $\{ z \in \mathbb{C} | \left|\frac{z-i}{z+i}\right|<1\}$ in the complex plane.
I substituted $z$ for $a+bi$, but did not get anywhere:
$\left|\frac{a+(b-1)i}{a+(b+1)i}\right|<1\\
\left|\frac{(a+(b-1)i)(a-(b+1)i)}{a^2+(b+1)^2}\right|<1\\
\left|\frac{a^2-(ab+a)i+(ab-a)i+(b^2-1)}{a^2+b^2+2b+1}\right|<1\\
\left|\frac{a^2+b^2-1-2ai}{a^2+b^2+2b+1}\right|<1\\
\left|\frac{a^2+b^2-1}{a^2+b^2+2b+1}-\frac{2ai}{a^2+b^2+2b+1}\right|<1\\
\left(\frac{a^2+b^2-1}{a^2+b^2+2b+1}\right)^2+\left(\frac{2ai}{a^2+b^2+2b+1}\right)^2<1\\
\frac{a^4+2a^2+2a^2b^2-2b^2+b^4+1}{a^4+2a^2+4a^2b+2a^2b^2+4b+6b^2+4b^3+b^4+1}<1$
|
$|z-i|<|z+i| \implies |x+(y-1)i|<|x+(y+1)i| \implies x^2+y^2-2y+1<x^2+y^2+2y+1\implies y>0$
is just an algebraic way to show what Hans Lundmark said
|
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Find the PDF of Y given Y=X(2-X) and X's PDF Suppose that the continuous random variable $X$ has probability density function
$f_X(x)=\begin{cases}\frac{1}{2}x & \text{if } 0<x<2\\0&\text{otherwise}\end{cases}$
Let $Y=X(2-X)$. Calculate $P(Y>y)$ and hence find the probability density function of $Y$.
So far I have tried expressing $P(Y>y)$ as $P(X(2-X)>y)$ but the quadratic inside has me stuck. Sorry - I am revising for an exam and it has been a long time since I have answered questions like this.
|
We complete the square. Note that $X(2-X)=2X-X^2=1-(X-1)^2$. Suppose that $0\le y\le 1$. Then
$$\Pr(X(2-X)\gt y)=\Pr(1-(X-1)^2\gt y)=\Pr((X-1)^2\lt 1-y)=\Pr\left(1-\sqrt{1-y}\le X\le 1+\sqrt{1-y}\right).$$
But between $0$ and $2$, the random variable $X$ has cdf $\frac{1}{4}x^2$. So
$$\Pr(X(2-X)\gt y)=\frac{1}{4}\left( (1+\sqrt{1-y})^2-(1-\sqrt{1-y})^2 \right).$$
This simplifies to $\sqrt{1-y}$. The cdf of $Y$ is therefore $1-\sqrt{1-y}$ for $y$ between $0$ and $1$. (It is $0$ for $y\lt 0$, and $1$ for $y\gt 1$.)
Remark: Completing the square was not necessary, and may not be the best way, though I go to it by reflex.
The inequality $X(2-X)\gt y$ becomes with some manipulation $X^2-2X+y\lt 0$. That means that $X$ falls between the two roots of the quadratic equation $x^2-2x+y=0$. These two roots are $1-\sqrt{1-y}$ and $1+\sqrt{1-y}$.
|
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|
To find the right most non zero digit When expanded 30! ends in 7 zeroes. Find the first non zero digit from right?
|
We desire the last digit of $n=\dfrac{30!}{10^7}$.
Note that $n\equiv0\pmod2$. Now we just need to find $n\pmod5$. By Wilson's theorem, or direct computation, we know $1(2)(3)(4)\equiv-1\pmod5$. This gives us
$$\dfrac{30!}{5^7}\equiv (-1)\frac{5}{5}(-1)\frac{10}{5}\cdots\frac{25}{5^2}(-1)\frac{30}{5}\equiv(-1)^61(2)(3)(4)(6)\equiv(-1)^7\equiv -1\equiv 4\pmod5$$
However, we also have to take into account the factor of $2^7\equiv2(-1)^3\equiv-2\equiv3\pmod5$. Since the integers modulo $5$ are a field, we can divide $4$ by $3$ in this field to reveal $n\equiv 3\pmod5$
Since $n\equiv0\pmod2$ and $n\equiv3\pmod5$ we get $n\equiv8\pmod{10}.$
|
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Geometric intuition for $\pi /4 = 1 - 1/3 + 1/5 - \cdots$? Following reading this great post : Interesting and unexpected applications of $\pi$, Vadim's answer reminded me of something an analysis professor had told me when I was an undergrad - that no one had ever given him a satisfactory intuitive explaination for why this series definition should be true (the implied assumption being that it is so simple, there should be some way to look at this to make it intuitive).
Now it follows simply from putting $x=1$ in the series expansion $$ \tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots $$
but I don't see anything geometrically intuitive about this formula either!
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By elementary means:
In the figure below, we have drawn a quarter circle inside a unit square and divided the top side in $n$ equal parts. Then we have chosen the height of the blue rectangles in such a way that the area of a rectangle equals the area of the corresponding sector (the aperture angles go decreasing).
If you look at the enhanced triangle, its long sides have length $\sqrt{1+x^2}$, by Pythagoras, so that its short side is $\sqrt{1+x^2}$ times longer than the arc on the circle ($x$ horizontal in $[0,1]$). Then by similarity of triangles, the ratio of the width of a rectangle to this short side is also $\sqrt{1+x^2}$, and we conclude that the equation of the green curve is
$$y(x)=\frac1{2(1+x^2)}.$$
Furthermore, the area under the curve is the same as the area of the eighth of a circle, i.e. $\dfrac\pi8$.
Now the (doubled) area is obtained as the sum of the areas of the rectangles,
$$\frac\pi4=2\sum_{k=1}^n\frac1ny\left(\frac kn\right)=\frac1n\sum_{k=1}^n\frac1{1+\dfrac{k^2}{n^2}}.$$
To evaluate it $^{(1)}$, we use the identity
$$\frac1{1+\dfrac{k^2}{n^2}}=1-\frac{k^2}{n^2}+\frac{k^4}{n^4}-\frac{k^6}{n^5}+\cdots$$
and by the Faulhaber formula $^{(2)}$, we know that the sum of the $n$ first $p^{th}$ powers is $\dfrac{n^{p+1}}{p+1}$.
This gives
$$\frac\pi4=\frac1n\left(n-\frac{n^3}{3n^2}+\frac{n^5}{5n^4}-\frac{n^7}{7n^6}+\cdots\right)$$ which is the claimed formula.
$(1)$ This is "justified" by
$$(1+x)(1-x+x^2-x^3+x^4-\cdots)=1+x-x-x^2+x^2+x^3-x^3-x^4+x^4+x^5\cdots=1\pm x^\infty$$ where the last term vanishes when $x<1$.
$(2)$ This is "justified", using the binomial theorem, by
$$n^p=\sum_{k=1}^n k^p-\sum_{k=1}^{n-1} k^p\approx\frac{n^{p+1}}{p+1}-\frac{(n-1)^{p+1}}{p+1}=\frac{n^{p+1}-(n^{p+1}-(p+1)n^p+\binom{p+1}2n^{p-1}-\binom{p+1}3n^{p-2}+\cdots)}{p+1}=n^p+\cdots.$$
The neglected terms are of lower degree in $n$ and can be ignored for large $n$.
|
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|
Express a complex number in modulus amplitude form Express a complex number in modulus amplitude form
$\displaystyle 1+\sin \alpha +i\cos \alpha $
My Attempt:
$\displaystyle r\cos \theta= 1+\sin \alpha $
$\displaystyle r\sin \theta= \cos \alpha $
Squaring and adding.. $\displaystyle r^2= (1+\sin \alpha)^2+ \cos^2 \alpha$
$\displaystyle r^2= 2(1+\sin \alpha) $
$\displaystyle \tan \theta = \frac{\cos \alpha}{1+\sin \alpha} $
How to break up $\displaystyle 1+\sin \alpha $?
|
It is better to solve it this way:
$$\begin{aligned}
1+\sin\alpha+i\cos \alpha &=1+\cos\left(\frac{\pi}{2}-\alpha\right)+i\sin\left(\frac{\pi}{2}-\alpha\right)\\
& \stackrel{*}{=}2\cos^2\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)+i2\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\\
&=2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\left(\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)+i\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\right)\\
&=2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)e^{i\left(\pi/4-\alpha/2\right)}\\
\end{aligned}$$
$(*)$, In this step I used the following formulas:
$\cos(2x)=2\cos^2x-1$ and $\sin(2x)=2\sin x\cos x$
In your method, I am not seeing how you get that expression for $\tan\theta$. Rather it should be $\tan\theta=\dfrac{\cos\alpha}{1+\sin\alpha}$. Hence,
$$\tan\theta=\frac{\sin\left(\frac{\pi}{2}-\alpha\right)}{1+\cos\left(\frac{\pi}{2}-\alpha\right)}=\frac{2\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}$$
$$\Rightarrow \tan\theta=\tan\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$$
|
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|
I need help finding the derivative of this natural logarithm function. Okay so $$f(x)=\ln[x\ln(x+2)]$$
so $$\ln(x)+\ln(\ln(x+2))$$so $$1.a\frac{dy}{dx}\ln(x)=\frac{1}{x}$$and I thought by chain rule that $$1.b\frac{dy}{dx}\ln(\ln(x+2))=\frac{1}{\ln(x+2)}\cdot\frac{1}{x}$$ so $$2.af'(x)=\frac{1}{x}+\frac{1}{\ln(x+2)}*\frac{1}{x}$$But the site says the following is step 2 $$2.bf'(x)=\frac{1}{x}+\frac{\frac{1}{x+2}}{\ln(x+2)}$$and I'm having problems going from $$\ln(x)+\ln(\ln(x+2))$$ to $$f'(x)=\frac{1}{x}+\frac{\frac{1}{x+2}}{\ln(x+2)}$$.
A step by step explanation would be greatly appreciated. Thanks in advance
|
\begin{align*}
f'(x) &= \frac{d}{dx} f(x) \\
&= \frac{d}{dx} ln[xln(x+2)] \\
&= \frac{d}{dx}(ln(x) + ln(ln(x+2))) \\
&= \frac{d}{dx}ln(x)+\frac{d}{dx}ln(ln(x+2)) \\
&= \frac{\frac{d}{dx}x}{x}+\frac{\frac{d}{dx}ln(x+2)}{ln(x+2)} \;\; \text{(Chain Rule)} \\
&= \frac{1}{x}+\frac{\frac{1}{x+2}}{ln(x+2)}
\end{align*}
|
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|
Find the minimum of $\displaystyle \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$ Is it possible to find the minimum value of $E$ where
$$E = \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$$for any $\triangle ABC$.
I've got the feeling that $\min(E) = 4$ and that the critical value occurs when $ABC$ is equilateral.
|
HINT: Use Lagrange multipliers to find the minimum of the function $f(A,B,C) = \frac{1}{\sin^2(A)} + \frac{1}{\sin^2(B)} +\frac{1}{\sin^2(C)}$ with the constraint $A+B+C = \pi$
|
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|
Problem calculating the argument of a complex variable In Signals & Systems 2nd Ed. written by A. V. Oppenheim,
there is a result of Fourier transformation:
$ \begin{align} H (j \omega) = \frac{1 + (j \omega / \omega_{0})^2 - 2 j \zeta (\omega / \omega_0)}{1 + (j \omega / \omega_{0})^2 + 2 j \zeta (\omega / \omega_0)} \end{align} $
And from this, it states that the angle between the line from the origin to that point and the positive real axis in the complex plane (i.e. argument) is
$ \angle H (j \omega) = -2 \tan ^{-1} \left[ \frac{2 \zeta (\omega / \omega_{0})}{1 - (\omega / \omega_{0})^2} \right] $
where $\omega$ is angular frequency in physics, but it's not important. Just say $\omega$ is a real number. $j^2 = -1$.
I have a trouble deriving this.
Let's simplify these equations a little.
Let $x$ be $\frac{\omega}{\omega_{0}}$ in the previous equations.
Then $ \begin{align} H (x) = \frac{1 + (i x)^2 - 2 i \zeta x}{1 + (i x)^2 + 2 i \zeta x}, \tag{1} \end{align} $
and what I want to derive is $ \begin{align} \angle H (x) = -2 \tan ^{-1} \left[ \frac{2 \zeta x}{1 - x^2} \right]. \tag{2} \end{align} $
Here are what I calculated.
$ \mathcal{Re} \left\{ H (x) \right\} = \frac{(1-x^2)^2 - 4 \zeta^2 x^2}{(1 - x^2)^2 + 4 \zeta^2 x^2} $
$ \mathcal{Im} \left\{ H (x) \right\} = - \frac{4 \zeta x (1 - x^2)}{(1 - x^2)^2 + 4 \zeta^2 x^2} $
$ \begin{align}
\angle H (x) &= \tan ^{-1} \frac{\mathcal{Im} \left\{ H (x) \right\}}{\mathcal{Re} \left\{ H (x) \right\}} \\
&= \tan^{-1} - \frac{4 \zeta x (1 - x^2)}{(1-x^2)^2 - 4 \zeta^2 x^2} \\
&= - \tan^{-1} \frac{4 \zeta x (1 - x^2)}{(1-x^2)^2 - 4 \zeta^2 x^2}
\end{align} $
Using $ \tan^{-1} x = 2 \tan^{-1} \frac{x}{1 + \sqrt{1 + x^2} } $ (from Wikipedia),
$ \begin{align}
\angle H (x) &= -2 \tan^{-1} \frac{x^2 - 1}{2 \zeta x}
\end{align} $
However, in the textbook,
$ \angle H (x) = -2 \tan^{-1} \frac{2 \zeta x}{1 - x^2} $
How come?
Is $ -2 \tan^{-1} \frac{x^2 - 1}{2 \zeta x} = -2 \tan^{-1} \frac{2 \zeta x}{1 - x^2} $ or did I miss something?
|
You have probably a mistake in your last calculation:
$$\alpha = -2 \tan^{-1} \left( \frac{y}{1 + \sqrt{1+y^2}} \right) , \text{with } y = \frac{4 \zeta x(1-x^2)}{(1-x^2)^2 - 4 (\zeta x)^2}$$
I got
$$
\sqrt{1+y^2} = \frac{(1-x^2)^2 + 4(\zeta x)^2}{(1-x^2)^2-4(\zeta x)^2}
$$
and therefore
$$
1+\sqrt{1+y^2} = \frac{2(1-x)^2}{(1-x)^2-4(\zeta x)^2}.
$$
Hence
$$
\alpha = -2 \tan \left( \frac{ \frac{4 \zeta x(1-x^2)}{(1-x^2)^2 - 4 (\zeta x)^2} }{ \frac{2(1-x)^2}{(1-x)^2-4(\zeta x)^2} } \right) = \frac{2 \zeta x}{1-x^2}
$$
which is the angle you were looking for.
|
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|
Limit of $x_n = 0.5(x_{n-1} + x_{n-2})$ - help finishing proof... EDIT: Fixed geometric proof off-by-one error.
I am looking for the limit of $x_n = 0.5(x_{n-1} + x_{n-2})$ with $x_2 > x_1$ arbitrary. I can show $\forall n: |x_{n}-x_{n+1}| = \frac{c}{2^{n-1}}$ (where $c = x_2 - x_1$) and thus existence of the limit (cauchy sequence)(see appended). From there I wanted to proceed:
\begin{eqnarray*}
\lim(x_n) &=& x_1 + \lim( \sum_{k=1}^n (x_{k+1}-x_k)) \\
&=& x_1 + x_2 - x_1+ \lim( \sum_{k=2}^n \frac{(-1)^{k-1} c}{2^{k-1}}) \\
&=& x_2 + \lim(\sum_{k=1}^{n/2} \frac{c}{2^{2k}} - \sum_{k=1}^{n/2} \frac{c}{2^{2k-1}} ) \\
&=& x_2 + c \lim(\sum_{k=1}^{n/2} \frac{1}{4^k} - \sum_{k=1}^{n/2} \frac{1}{2}\frac{1}{4^k}) \\
&=& x_2 + c (\frac{1}{4}\frac{1}{1-1/4}-\frac{1}{4}\frac{1}{2}\frac{1}{1-1/4}) \\
&=& x_2 + \frac{c}{8}\frac{1}{1-1/4} \\
&=& x_2 + \frac{1}{6}c.\end{eqnarray*}
I can tell this is wrong by the example in the book (Bartle Sherbert, 3.5), and I can tell it should be $x_1 + \frac{2}{3}c$ on the penultimate line the same way - but I cannot justify why that would be the case...
Any ideas?
Thanks! Best,
Leon
Let $c = x_2 - x_1$. Then we prove by induction that $\forall n: |x_{n}-x_{n+1}| = \frac{c}{2^{n-1}}$. Base: firstly $|x_1 - x_2| = x_2 - x_1 = c/2^0 = c/2^{1-1}$, and secondly $|x_2 - x_3| = |x_2 - (1/2)(x_1+x_2)| = (1/2)|x_1-x_2| = c/2$. Inductive step:
\begin{eqnarray*}
|x_n - x_{n+1}| &=& | \frac{1}{2}(x_{n-1}+x_{n-2}) - \frac{1}{2}(\frac{1}{2}(x_{n-1}+x_{n-2})+x_{n-1)} | \\
&=& |\frac{1}{2}x_{n-2} - \frac{1}{4}x_{n-1}-\frac{1}{4}x_{n-1} | \\
&=& \frac{1}{4}|x_{n-2}-x_{n-1}| \\
&=& \frac{c}{2^{n-1}}.
\end{eqnarray*}
|
$$\sum_{k=1}^{n/2} \frac1{2^{2k-1}}= \sum_{k=1}^{n/2} \frac{1}{2}\frac{1}{2^{2k-2}}= \sum_{k=1}^{n/2} \frac{1}{2}\frac{1}{4^{k-1}}= \sum_{k=0}^{n/2-1} \frac{1}{2}\frac{1}{4^k}$$
|
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|
Finding $\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt$ I'd like to ask something about the following integral:
$$
\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt
$$
I rewrote and took another variable.
$$
-i\int_0^{2\pi}\frac{e^{is}-e^{-is}+8i}{e^{is}+e^{-is} + \frac{10}3} dt
\quad = \quad
-i\int_{C(0,1)^+}\frac{z-\frac{1}{z}+8i}{z+ \frac1z+ \frac{10}3} \cdot \frac{-i}{z}dt
\quad = \quad
- \int_{C(0,1)^+} \frac{z^2-1+8iz}{z(z^2+1+\frac{10}{3} \cdot z)} dz
$$
The only root of $z(z^2+1+\frac{10}{3} \cdot z)$ inside the unit disk is $0$, so I thought that the integral would equal:
$$
Res_{z=i} \ = \ 2 \pi i \cdot \frac{0-1+8i\cdot 0}{0^2+1+\frac{10}{3}\cdot 0} \quad = \quad 2\pi i
$$
I don't understand why $i$ appears in this value. Can you please tell me what I did wrong?
|
You can do this way. Clearly
$$
\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt=\int_0^{2\pi}\frac{\sin t}{\cos t + \frac{5}3} dt+4\int_0^{2\pi}\frac{1}{\cos t + \frac{5}3} dt
$$
and
$$
\int_0^{2\pi}\frac{\sin t}{\cos t + \frac{5}3} dt=-\int_0^{2\pi}\frac{1}{\cos t + \frac{5}3} d(\cos t + \frac{5}3)=0.
$$
You only need to calculate
$$ \int_0^{2\pi}\frac{1}{\cos t + \frac{5}3} dt $$
and I think you can finish it.
|
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|
Find $x > 0$ for which $\int_{0}^{x} [t]^2 \ dt = 2 (x-1)$. What are all possible $x > 0$ for which the following equation is satisfied?
$$\int_{0}^{x} [t]^2 \ dt = 2 (x-1),$$ where $[.]$ denotes the bracket (or floor) function.
I guess we will have to consider different cases depending on whether $\sqrt{n-1} \leq x \leq \sqrt{n}$ for each positive integer $n$.
For $0 \leq x \leq 1$, the integral on the left-hand side equals $1$; so the solution in this case is clearly $x = 1$ alone.
What are all possible solutions for $x > 0$?
|
Differentiate both sides, we get $[x]^2 = 2$. Le us denote by $\{x\}$ the fractional part of the number $x$, so that $\{x\}=x-[x]$. It is clear that $0\le \{x\} <1$ and $[x]=x-\{x\}$. Thus, the equation becomes $$[x]^2=(x-\{x\})^2=2$$ that is $$x^2-2x\{x\}+\{x\}^2=2.$$ Setting $p=\{x\}$, and write $$x^2-2px+p^2-2=0.$$ Now, using the general law we have
\begin{align}
x_{1,2} &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}
\\
&= \frac{{2p \pm \sqrt {4p^2 - 4\left( 1 \right)\left( {p^2 - 2} \right)} }}{2}
\\
&= \frac{{2p \pm \sqrt 8 }}{2}.
\end{align}
Thus, $x_1 =p+\sqrt{2}$ and $x_2 =p-\sqrt{2}$ however, $a\le p<1$ which means that $x_2<0$ (reject), therefore the required solution is $x=p+\sqrt{2}$, for all $p\in[0,1)$. If you want write as $$\left\{ {x:x = p + \sqrt 2 ,0 \le p < 1} \right\}.$$
|
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|
Imaginary part of $\int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x$ and $\int_{0}^{\pi/2} \frac{\log \cos x}{x^2}\:\mathrm{d}x$ I have found the following new result connecting two rational log-cosine integrals.
Proposition. \begin{align}
\displaystyle & {\Im} \int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x =
\frac{\pi^2}{16} - \frac{\ln 2}{4} + \frac{\pi}{8} \int_{0}^{\pi/2} \frac{\log \cos x}{x^2}\:\mathrm{d}x
\end{align} where $\displaystyle \log (z)$ denotes the principal value of the logarithm defined for $z \neq 0$ by
\begin{align}
\displaystyle \log (z) = \ln |z| + i \: \mathrm{arg}z, \quad -\pi <\mathrm{arg} z \leq \pi. \nonumber
\end{align}
How would you prove it?
|
Numeric confirmation
From Mathematica. Left hand side:
Right hand side:
$$
f(z) = \frac{\ln \cos x}{x^2}
$$
$$
\text{Im }f(z) = \frac{z^2}{z^2+\ln^{2} \left(-2 \cos z \right)}
$$
|
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|
What is the closed form for $\sum_{n=1}^\infty \frac1n - \frac1{n+1/p}$? A while ago, I started to look at expressions of the following form:
$$
S_p:=\sum_{n=1}^\infty \frac1n - \frac1{n+1/p},
$$
where $p$ is prime, because otherwise things get too complicated for me at the moment.
What I found so far is the following:
$$
S_p= p -\log(2p) - \frac12 \pi \cot\left(\frac\pi p\right) + T_p,
$$
where $T_p$ contains $\frac{p-1}2$ terms $t_{p,k}$ of the form:
\begin{cases}
\pm 2 \sin\left(\frac{k\pi}{2p}\right) \log\left(\sin\left(\frac{k\pi}{2p}\right)\right) \\
\pm 2 \cos\left(\frac{k\pi}{2p}\right) \log\left(\sin\left(\frac{k\pi}{2p}\right)\right)\\
\pm 2 \sin\left(\frac{k\pi}{2p}\right) \log\left(\cos\left(\frac{k\pi}{2p}\right)\right)
\\
\pm 2 \cos\left(\frac{k\pi}{2p}\right) \log\left(\cos\left(\frac{k\pi}{2p}\right)\right)
\end{cases}
An example can be found here.
How does the closed form for $\sum_{n=1}^\infty \frac1n - \frac1{a+n}$ look like? (EDIT: where $a=1/p$)
|
Write your expression as the limit at $x \to 1$ of
$$\sum_{n = 1}^{\infty} \frac{p x^{pn}}{pn} - \frac{p x^{pn+1}}{pn+1}.$$
If we differentiate this power series, we get
$$\sum_{n = 1}^{\infty} p x^{pn - 1} - p x^{pn}
=
p x^{p-1} \frac{1-x}{1 - x^p}.
$$
Integrating this (say via a partial fraction decomposition)
and then choosing the constant of integration so that you get
zero when $x = 0$ gives a formula for the original power series.
Now putting $x = 1$ gives the value you want.
E.g. if $p = 2$, we have
$$2 x \frac{1}{1+x} = 2 - \dfrac{2}{1+x},$$
whose desired integral is
$$2x - 2\log (1+x),$$
and so the value of the sum in this case is $2-2\log 2.$
E.g. if $p = 3$, we have
$$3 \frac{x^2}{1+x+x^2} = 3 - 3\frac{1+x}{1+x+x^2} = 3 - \frac{3}{2} \frac{2x+1}{1+x+x^2} - \frac{3}{2}\frac{1}{3/4+ (x+1/2)^2},$$
whose desired integral is
$$3 x - \frac{3}{2}\log(1+x+x^2) - {\sqrt{3}}\arctan\bigl(\frac{2x+1}{\sqrt{3}}\bigr)+\sqrt{3} \frac{\pi}{6},$$
and so the value of the sum in this case is
$$3 - \frac{3\log 3}{2} - \frac{\sqrt{3} \pi}{6}.$$
I didn't try to find the general formula here. One part of mathematics where it is
worked out is the theory of special values (at $s = 1$) of Dirichlet $L$-series.
[Added: it seems that between when I began writing this yesterday, and when I finally got a chance to finish and post it now, other answers with related derivations have appeared. Oh well.]
|
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|
A closed form for $\int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:\mathrm{d}x$ We already know that
\begin{align}
\displaystyle & \int_{0}^{\pi/2} x \ln(2 \cos x)\:\mathrm{d}x = -\frac{7}{16} \zeta(3),
\\\\ & \int_{0}^{\pi/2} x^2 \ln^2(2 \cos x)\:\mathrm{d}x = \frac{11 \pi}{16} \zeta(4). \end{align}
Does the following integral admit a closed form?
\begin{align} \displaystyle & \int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:\mathrm{d}x \end{align}
|
We can try the harmonic analysis path. Since:
$$\log(2\cos x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}\cos(2nx),$$
$$\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n},\tag{1}$$
we have, as an example:
$$\int_{0}^{\pi/2}\log^3(2\sin x)dx=-\frac{3\pi}{4}\zeta(3)$$
since
$$\int_{0}^{\pi/2}\cos(2n_1 x)\cos(2n_2 x)\cos(2n_3 x)\,dx = \frac{\pi}{8}\delta_{2\cdot\max n_i=(n_1+n_2+n_3)}\tag{2}$$
Now:
$$\pi/2-x = \frac{\pi}{4}+\frac{2}{\pi}\sum_{m=0}^{+\infty}\frac{\cos((4m+2)x)}{(2m+1)^2}\tag{3}$$
hence by multiplying $(1)$ and $(3)$ we can write the Fouries cosine series of $(\pi/2-x)\log(2\sin x)$ over $(0,\pi/2)$ and grab from $(2)$ a combinatorial equivalent for
$$\int_{0}^{\pi/2}\left((\pi/2-x)\log(2\sin x)\right)^3\,dx.$$
With the aid of Mathematica I got:
$$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(-(2n+1)\left(\psi'\left(\frac{2n+1}{2}\right)-\psi'\left(-\frac{2n+1}{2}\right)\right)+2\left(2\gamma+\psi\left(\frac{2n+1}{2}\right)+\psi\left(-\frac{2n+1}{2}\right)\right)\right),$$
$$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(2 H_{\frac{2n+1}{2}}-(2n+1)\left(\psi'\left(\frac{2n+1}{2}\right)-\psi'\left(-\frac{2n+1}{2}\right)\right)\right).$$
$$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(-4\log 2+\sum_{j=0}^n\frac{4}{2j+1}+\frac{4}{2n+1}+(2n+1)\sum_{j=0}^{n}\frac{8}{(2j+1)^2}\right).\tag{1}$$
So we have the Fourier cosine series of $(\pi/2-x)\log(2\sin x)$ but the path does not look promising from here. However, if we replace $\pi/2-x$ with a periodic continuation we get the way nicer identity:
$$(\pi/2-x)\log(2\sin x)=-\left(\sum_{n=1}^{+\infty}\frac{\sin(2nx)}{n}\right)\left(\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n}\right)\tag{2}$$
that directly leads to:
$$f(x)=(\pi/2-x)\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{H_{n-1}}{n}\sin(2nx).\tag{3}$$
Now since $\int_{0}^{\pi/2}\sin(2mx)dx=\frac{\mathbb{1}_{m\equiv 1\pmod{2}}}{m}$ and $\int_{0}^{\pi/2}\sin(2ax)\sin(2bx)dx=\frac{\pi}{4}\delta_{a,b}$, the first two identites are easily proven. Now the three-terms integral
$$\int_{0}^{\pi/2}\sin(2ax)\sin(2bx)\sin(2cx)dx$$
is a linear combination of $\frac{1}{a+b+c},\frac{1}{-a+b+c},\frac{1}{a-b+c},\frac{1}{a+b-c}$ depending on the parity of $a,b,c$, so it is quite difficult to find, explicitly, the Fourier cosine series of $f(x)^2$ or the integral $\int_{0}^{\pi/2}f(x)^3\,dx$, but still not impossible. In particular, by this answer we know that the Taylor coefficients of the powers of $\log(1-x)$ depends on the generalized harmonic numbers. In our case,
$$-\log(1-x)=\sum_{n=1}^{+\infty}\frac{1}{n}x^n,$$
$$\log(1-x)^2 = \sum_{n=2}^{+\infty}\frac{2H_{n-1}}{n}x^n,$$
$$-\log(1-x)^3 = \sum_{n=3}^{+\infty}\frac{3H_{n-1}^2-3H_{n-1}^{(2)}}{n}x^n,$$
$$\log(1-x)^4 = \sum_{n=4}^{+\infty}\frac{4H_{n-1}^3+8H_{n-1}^{(3)}-12 H_{n-1}H_{n-1}^{(2)}}{n}x^n\tag{4}$$
hence we can just find a closed form for
$$\int_{0}^{\pi/2}x^3 (1-2\cos x)^n dx$$
and sum everything through the third previous identity. Ugh.
|
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|
Simplify expression $(x\sqrt{y}- y\sqrt{x})/(x\sqrt{y} + y\sqrt{x})$ I'm stuck at the expression: $\displaystyle \frac{x\sqrt{y} -y\sqrt{x}}{x\sqrt{y} + y\sqrt{x}}$.
I need to simplify the expression (by making the denominators rational) and this is what I did:
$$(x\sqrt{y} - y\sqrt{x}) \times (x\sqrt{y} - y\sqrt{x}) = (\sqrt{y} - \sqrt{x})^2$$
Divided by
$$(x\sqrt{y} + y\sqrt{x}) \times (x\sqrt{y} - y\sqrt{x} ) = (x\sqrt{y})^2$$
So I'm left with $\displaystyle \frac{(\sqrt{y} - \sqrt{x})^2}{(x\sqrt{y})^2}$.
This answer is incorrect. Can anyone help me understand what I did wrong? If there is a different approach to solve this it will also be much appreciated. Please explain in steps.
|
The answer can be simplified even further (using the difference of two squares):
$$\frac{x-2\sqrt{xy} + y}{x-y} = \frac{(\sqrt{x} - \sqrt{y})^2}{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})} = \frac{1}{\sqrt{x} - \sqrt{y}} \text{ or } \frac{\sqrt{x} - \sqrt{y}}{x+y}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$ $$\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$
My approaches so far has been using substitution with $\tan x = t$ and $\tan \frac x2 = t$ but the calculations has been harder than I think they should.
I've also tried using ordinary polynom integration to simplify the integral but I'm having problems with factorizing the denominator.
|
$$\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$
$$\int \frac{\tan x(\tan^2(x)+1)}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$
$$\int \frac{\tan x(\sec^2(x))}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$
Frome here, let $u=\tan(x)$, $\dfrac{du}{dx} = \sec^2(x)$., thus,
$$\int \frac{u}{u^3 + 3u^2 + 2u + 6}$$
From there, use partial fractions.
|
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|
How to solve this: $\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left(\frac{2n+3k}{n} \right)^2$ How to solve this:
$$\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left(\frac{2n+3k}{n} \right)^2$$
The answer is supposed to be 39.
My attempt:
$$\frac{3}{n}\sum\left(4+\frac{12}{n}k+\frac{9}{n^{2}}k^{2}\right)=\frac{3}{n}\left(4+\frac{12}{n}\sum k+\frac{9}{n^{2}}\sum k^{2}\right)=\frac{3}{n}\left(4+\frac{12}{n}\frac{n}{2}\left(n+1\right)+\frac{9}{n^{2}}\frac{n}{6}\left(n+1\right)\left(2n+1\right)\right)=3\left(6+3\right)=27\neq39$$
|
$$\lim_{n\to\infty}\frac3n\sum_{k=1}^n\left(2+3\frac kn\right)^2$$
$$=3\int_0^1(2+3x)^2\ dx$$
$$\text{as }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
|
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|
Calculating the area For the two graphs $ \frac{x^3+2x^2-8x+6}{x+4} $ and $ \frac{x^3+x^2-10x+9}{x+4} $, calculate the area which is confined by them;
Attempt to solve: Limits of the integral are $1$ and $-3$, so I took the definite integral of the diffrence between $ \frac{x^3+2x^2-8x+6}{x+4} $ and $ \frac{x^3+x^2-10x+9}{x+4} $ since the latter is a graph of a lower-degree polynomial than the first one, but as a result I got a strange negative area result, which is quite funny since the offical answer is $12 -5 \ln5$...
What am I doing wrong?
Note that in this exercise you need to somehow imagine the two graphs by yourself without the use of any comupters programs or whatever. Hence it is not as trivial to evalutae the position of each graph to the other...
|
@Mary knows that because $$\frac{x^3+2x^2-8x+6}{x+4} - \frac{x^3+x^2-10x+9}{x+4}=\frac{x^2+2x-3}{x+4}=\frac{(x-1)(x+3)}{x+4}$$ and since on $-3<x<1$ so the last fraction in negative. A simple figure could clear the point.
|
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|
find a polynomial whose roots are inverse of squares of roots of $x^3+px+q$ Question is :
Given a polynomial $f(x)=x^3+px+q\in \mathbb{Q}[x]$ find a polynomial whose roots are inverse of sqares of roots of $f(x)$
Supposing $a,b,c$ as roots of $f(x)$ we have :
*
*$a+b+c=0$
*$ab+bc+ca=p$
*$abc=-q$
Now i need to know what
*
*$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$
*$\dfrac{1}{a^2}\cdot\dfrac{1}{b^2}+\dfrac{1}{b^2}\cdot\dfrac{1}{c^2}+\dfrac{1}{c^2}\cdot\dfrac{1}{a^2}$
*$\dfrac{1}{(abc)^2}$
All i have to do is use $(a+b+c)^2$ formula and others and conclude what those sums,products are.. I am fairly comfortable with that...
But then, this question was from a Galois theory course.. So, i some how guess there is a better way to do this...
Can some one suggest something..
|
Note that
$$
0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) = a^2 + b^2 + c^2 + 2p
$$
Thus
$$
\dfrac{1}{a^2b^2} + \dfrac{1}{a^2b^2} + \dfrac{1}{a^2b^2} = \dfrac{a^2 + b^2 + c^2}{a^2b^2c^2} = -\dfrac{2p}{q^2}
$$
|
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|
Is $2^{3^n}+1$ always divisible by $3^{n+1}$? Because it certainly seems to be the case for all positiv $n$ but how can I prove it?
|
Suppose
$$2^{3^n} + 1 \equiv 0 \pmod{3^{n+1}}$$
Then
$$2^{3^n} \equiv -1 + 3^{n+1} a \pmod{3^{n+2}}$$
for some value of $a$. And thus
$$\begin{align}2^{3^{n+1}} &\equiv \left(-1 + 3^{n+1} a\right)^3
\\&\equiv (-1)^3 + 3 \cdot (-1)^2 \left(3^{n+1} a \right) + \ldots
\\&\equiv -1 + 3^{n+2}(\ldots) + \ldots
\\&\equiv -1 \pmod{3^{n+2}} \end{align}$$
In fact, we can show more: that $3^{n+2}$ does not divides $2^{3^n}-1$. Repeating a similar inductive argument, if:
$$ 2^{3^n} \equiv -1 + 3^{n+1} a \pmod{3^{n+3}} $$
for some $a$ such that $a \not\equiv 0 \pmod{3} $, then we ave
$$ 2^{3^{n+1}} \equiv -1 + 3^{n+2} a \pmod{3^{n+3}}$$
and so we know that $3^{n+2}$ divides $2^{3^{n+1}} + 1$, but $3^{n+3}$ does not.
|
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|
A Binet-like integral $\int_{0}^{1} \left(\frac{1}{\ln x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{x^s }{1-x}\mathrm{d}x$ I met this integral
$$
\int_{0}^{1}
\left(\frac{1}{\ln x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{ \mathrm{d}x}{1-x} \qquad (*)
$$
while evaluating this log-cosine integral. I made several attemps before being successful.
Find a closed form for
$$
I(s): = \int_{0}^{1}
\left(\frac{1}{\ln x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{x^s }{1-x}\mathrm{d}x, \quad \Re (s)>-1. \qquad (**)
$$
|
Recall Binet's formula
$$
\log \Gamma(z)= \left( z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) +
\int_0^{\infty} \!
\left(\frac{1}{2} - \frac{1}{x} + \frac{1}{e^{x}-1} \right)\frac{e^{-zx}}{x} \mathrm{d}x,\quad \Re z >0
$$
which, upon making $x = - \log v$, can be written as
$$
\log \Gamma(z)= \! \left( z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) - \!\!
\int_0^{1} \!
\left(\frac{1}{\log v}+\frac{1}{1-v}-\frac{1}{2}\right)\frac{v^{z-1}}{\log v}\mathrm{d}v, \, \Re z >0
$$
and Gauss' formula
$$
-\psi(z)+\log z = \int_0^{1} \left(\frac{1}{\log v}+\frac{1}{1-v}\right)v^{z-1} \mathrm{d}v.
$$
One may check that
\begin{multline}
\displaystyle \left(\frac{1}{\log u} + \frac{1}{1-u} - \frac{1}{2} \right) \frac{u^s}{1-u} = u^{s} \frac{d}{du}\left\{u \left(\frac{1}{\log u} + \frac{1}{1-u}\right)\right\} \, \\
+ \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u} - \frac{1}{2} \right) \frac{u^s}{\log u} - \frac{1}{2} \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u}\right)u^s.
\end{multline}
Hence our integral $I(s)$ is the sum of three integrals.
The first,
\begin{multline}
I_{1}(s) =\displaystyle \left.\left(\frac{1}{\log u} + \frac{1}{1-u} \right)u^{s+1} \right|_{0}^{1}
- s \int_{0}^{1}\!
\displaystyle \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u}\right)u^{s}
\mathrm{d}u,
\end{multline}
using Gauss' formula,
$$
I_{1}(s) = \frac{1}{2} + s \left( \psi(s+1)-\log \left(s+1\right) \right).
$$
The second, applying Binet's formula,
\begin{multline}
I_{2} (s)= \displaystyle \int_{0}^{1}\! \left(\displaystyle \frac{1}{\log u} +
\frac{1}{1-u} -\frac{1}{2} \right) \frac{u^{s}}{\log u} \mathrm{d}u \\
= -\log \Gamma(s+ 1)+ \left(s + 1/2 \right)\log \left(s+ 1 \right) - s - 1 + \frac{1}{2}\ln(2\pi).
\end{multline}
The third, using Gauss' formula once more
\begin{equation}
I_{3}(s) = - \frac{1}{2} \displaystyle \int_{0}^{1}\! \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u}\right)u^{s} \mathrm{d}u =
\frac{1}{2}\left( \psi(s+1)-\log \left(s+1\right) \right).
\end{equation}
Consequently, $ I(s)=I_{1}(s)+I_{2}(s)+I_{3}(s)$ is given by
\begin{multline}
I(s):= \displaystyle \int_0^{1} \left(\frac{1}{\log u}+
\frac{1}{1-u}-\frac{1}{2}\right) \frac{u^s}{1-u} \mathrm{d}u \\
= -\log \Gamma(s+1) + \left( s +\frac{1}{2}\right) \psi(s+1) - s - \frac{1}{2} + \frac{1}{2}\ln(2\pi).
\end{multline}
We have, with $s=0$,
$$
\int_{0}^{1}
\left(\frac{1}{\log x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{ \mathrm{d}x}{1-x} = -\frac{1}{2}+\frac{1}{2} \ln (2 \pi)-\frac{1}{2} \gamma.
$$
|
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|
If $a+b+c+d=4$, then $\sum\frac{1}{a+3}\le \frac{1}{abcd}$ This is somehow related to this problem but I don't have any idea about it.
Let $a$, $b$, $c$ and $d$ be positive reals such that $a+b+c+d=4$. Prove that:
$$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}+\frac{1}{d+3}\le \frac{1}{abcd}$$
Now I also tried to prove the $3$ variable version :
$$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\le \frac{1}{abc}$$ where $a,b,c>0$ with $a+b+c=3$. But I haven't been able to solve it too. Anyone can help?? Thanks a lot.
|
Let $a+b+c+d=4u$ and $ab+ac+ad+bc+bd+cd=6v^2$.
Hence, by AM-GM $abcd\leq v^4$ and by C-S
$$\frac{1}{abcd}-\sum_{cyc}\frac{1}{a+3}=\frac{1}{abcd}-\frac{4}{3}-\sum_{cyc}\left(\frac{1}{a+3}-\frac{1}{3}\right)=$$
$$=\frac{1}{abcd}-\frac{4}{3}+\sum_{cyc}\frac{a}{3(a+3)}\geq\frac{1}{abcd}-\frac{4}{3}+\frac{(a+b+c+d)^2}{3\sum\limits_{cyc}(a^2+3)}=$$
$$=\frac{1}{abcd}-\frac{4}{3}+\frac{16}{3\sum\limits_{cyc}(a^2+3)}=\frac{u^2}{abcd}-\frac{4}{3u^2}+\frac{16}{3(16u^2-12v^2+12u^2)}\geq$$
$$\geq\frac{u^2}{v^4}-\frac{4}{3u^2}+\frac{4}{3(7u^2-3v^2)}\geq0,$$
where the last inequality it's
$$(u^2-v^2)(7u^4+4u^2v^2-4v^4)\geq0,$$
which is true because $u^2\geq v^2$.
Done!
|
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|
Evaluation of $\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx$
Compute the indefinite integral
$$
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx
$$
My Attempt:
$$
\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx &= \int\frac{\cos 2x}{\sin^2 x\sqrt{\cos 2x}}\sin xdx\\
&= \int\frac{2\cos^2 x-1}{(1-\cos^2 x)\sqrt{2\cos^2 x-1} }\sin x \,dx
\end{align}
$$
Let $\cos x = t$, so that $\sin x\,dx = -dt$. This changes the integral to
$$
\begin{align}
\int\frac{(2t^2-1)}{(t^2-1)\sqrt{2t^2-1}}\,dt &= \int\frac{(2t^2-2)+1}{(t^2-1)\sqrt{2t^2-1}}\,dt\\
&= 2\int\frac{dt}{\sqrt{2t^2-1}}+\int \frac{dt}{(t^2-1)\sqrt{2t^2-1}}
\end{align}
$$
How can I solve the integral from here?
|
Put $$\cos 2x =\frac{1 -\tan^2x}{1+\tan^2x}$$
$$\int\frac{\sqrt{1-\tan^2 x}}{\tan x}$$
$$1-\tan^2 x =t^2$$
$$\implies -2\tan x \sec^2 x dx=2tdt$$
$$\int\frac{t}{\tan x}.\frac{-tdt}{\tan x\sec^2 x}=-\int\frac{t^2}{(1-t^2)(2-t^2)}$$
$$=-\int\left[\frac{1}{(1-t^2)} -\frac{2}{(2-t^2)}\right]dt$$
$$=-\frac{1}{2}\log\left|{\frac{1+t}{1-t}}\right| +\frac{1}{\sqrt2}\log\left|{\frac{\sqrt2+t}{\sqrt2-t}}\right|$$
|
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|
How to factorize $x^4+2x^2+4$ to a product of polynomials with real coefficients? How do you factor
$$x^4+2x^2+4 $$
so it can be written as
$$ (x^2+2x+2)(x^2-2x+2) $$
|
Certainly the solution using completing the square (already provided in two answers) is in this case the easiest one.
I will just point out that in general if you have a real polynomial which is simple enough so that we are able to find all complex roots, this might also help you find factorization over $\mathbb R$. This works because for every root $z=a+bi$ also the complex conjugate $\overline z=a-bi$ is a root. And these pairs give you quadratic factors
$$(x-z)(x-\overline z)=(x-a-bi)(x-a+bi)=x^2-2a+b^2+b^2.$$
See also: Complex conjugate root theorem
In this case you can use substitution $t=x^2$ to get a quadratic equation
$$t^2+2t+4=0$$
which has the solutions $t_{1,2}=\frac{-2\pm\sqrt{12}i}2=-1\pm\sqrt3i = 2\left(-\frac12\pm\frac{\sqrt3}2i\right)$.
Now you can solve the equations
$$x^2=-1\pm\sqrt3i$$
which have the solutions
\begin{align*}
x_1&=\sqrt2\left(\frac12+\frac{\sqrt3}2i\right)\\
x_2&=\sqrt2\left(-\frac12\pm\frac{\sqrt3}2i\right)\\
x_3&=\overline{x_1}\\
x_4&=\overline{x_2}
\end{align*}
This will give you the following quadratic factors:
\begin{align*}
(x-x_1)(x-\overline{x_1}) = x^2-\sqrt2x+2\\
(x-x_2)(x-\overline{x_2}) = x^2+\sqrt2x+2
\end{align*}
|
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|
In a triangle ABC, prove that cot(A/2)+cot(B/2)+cot(C/2) =cot(A/2)cot(B/2)cot(C/2) In a triangle ABC, prove that $\cot \left ( \frac{A}{2} \right )+\cot \left ( \frac{B}{2} \right )+\cot \left ( \frac{C}{2} \right )=\cot \left ( \frac{A}{2} \right )\times \cot \left ( \frac{B}{2} \right )\times \cot \left ( \frac{C}{2} \right )$. I tried all identities I know but I have no idea how to proceed.
|
Here’s a somewhat weird proof using inverse trigonometric functions
If $$A+B+C=π$$ then $$\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{π}{2}$$
Let $$\frac{A}{2}=tan^{-1} x, \>\frac{B}{2}=tan^{-1} y ,\> \frac{C}{2}=tan^{-1} z$$
Now,
$$tan^{-1} x+ tan^{-1} y+ tan^{-1} z= π/2$$
$$tan^{-1} x+ tan^{-1} y= cot^{-1} z$$
Because $$tan^{-1} z+ cot^{-1} z= \frac{π}{2}$$
Now,
$$tan^{-1} \frac{x+y}{1-xy}= cot^{-1} z$$
$$cot^{-1} \frac{1-xy}{x+y}= cot^{-1} z$$
$$\frac{1-xy}{x+y}=z$$
$$1=xy+yz+zx$$
$$\frac{1}{xyz}= \frac{1}{z}+ \frac{1}{x}+\frac{1}{y}$$
Now,we have
$$cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}= cot \frac{C}{2}+cot \frac{A}{2}+cot \frac{B}{2}$$
|
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|
How to do integral $\int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$ and $\int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$? I met these two integrals but don't know how to do them:
$$I_1 = \int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t$$
$$I_2 = \int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t$$
where $b>0$, $T>0$.
Please kindly help?
Thanks to hits from Fabien, for the first one, let $t=\frac{T}{u^2+1}$:
$$I_1 = \int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t = \frac{2}{T}e^{-\frac{b^2}{2T}} \int_0^{+\infty} e^{-\frac{b^2}{2T}u^2} \text{d}u = \frac{\sqrt{2\pi}}{b\sqrt{T}} e^{-\frac{b^2}{2T}} $$
seems this tally with Mhenni Benghorbal's result.
For the second one, let $t=\frac{T}{u^2+1}$, $a^2=\frac{b^2}{2T}$:
$$I_2 =2 e^{-\frac{b^2}{2T}} \int_0^{+\infty} \frac{1}{1+u^2} e^{-\frac{b^2}{2T}u^2} \text{d}u =2 e^{-a^2} \int_0^{+\infty} \frac{1}{1+u^2} e^{-a^2u^2} \text{d}u := 2 e^{-a^2} I_3 $$
Then
$$I_3 = \int_0^\infty \frac{1}{1+u^2} e^{-a^2u^2} \text{d}u = \int_0^\infty \text{d}u\, e^{-a^2u^2} \int_0^\infty e^{-x(1+u^2)} \text{d}x = \int_0^\infty \text{d}x\, e^{-x} \int_0^\infty e^{-(a^2+x)u^2} \text{d}u$$
So $$I_3 = \int_0^\infty \text{d}x e^{-x} \frac{1}{2} \sqrt{\frac{\pi}{a^2+x}} = \frac{\sqrt{\pi}}{2} \int_0^\infty \frac{e^{-x}}{\sqrt{a^2+x}} \text{d}x$$
Let $t=\sqrt{a^2+x}$, so $x=t^2-a^2$, $\text{d}x = 2t\text{d}t$,
$$I_3 = \sqrt{\pi} \int_a^\infty e^{-(t^2-a^2)} \text{d}t = \sqrt{\pi} e^{a^2} \frac{\sqrt{\pi}}{2}\, \text{erfc}(a)$$
So $$I_2 = 2e^{-a^2} I_3 = \pi \, \text{erfc} \left(\frac{b}{\sqrt{2T}}\right)$$
Same as Mhenni Benghorbal's answer, $$I_2= \, \pi-
{{\rm erf}\left( {\frac {b}{\sqrt {2T}}}\right)}.$$
YEAH!
|
Hint :
Try the substitution $$t=\cfrac{T}{u^2+1}$$
The first integral has the shape of the gaussian.
The second one leads you to $$I(\beta) = \alpha \int_{\mathbb{R}^+} \cfrac{1}{u^2+1} \exp(-\beta(u^2+1) ) \,du$$
Considering $$\begin{cases} I'(\beta)=\alpha \int_{\mathbb{R}^+} \exp(-\beta(u^2+1) ) \,du=\alpha\exp(-\beta)\cfrac{\sqrt{\pi}}{2\sqrt{\beta}} \\ I(0) = \alpha \int_{\mathbb{R}^+} \cfrac{1}{u^2+1} \,du=\cfrac{\alpha\pi}{2}\end{cases}$$
with the substitution $\gamma^2 = \beta$ ends the problem.
|
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|
Scalar triple product in terms of scalar products only I am trying to express the scalar triple product $\bf a \cdot (b \times c)$ only in terms of scalar products $\bf a \cdot b$, $\bf b \cdot c$, $\bf c \cdot a$ and the lengths of vectors $a$, $b$ and $c$. I start by writing $$\mathbf{a} = B\, \mathbf{b} + C\, \mathbf{c} + D\, \mathbf{b \times c} $$ and then performing the dot products with $\bf a, b, c, b \times c$, I arrive at the following equations $$a^2 = B \, (\mathbf{a \cdot b}) + C \, (\mathbf{c \cdot a}) + D \, (\mathbf{a \cdot (b\times c)})$$ $$\mathbf{a \cdot b} = B\, b^2 + C\, (\mathbf{b \cdot c})$$ $$\mathbf{c \cdot a} = B\, (\mathbf{b \cdot c})+ C\, c^2 $$ $$\mathbf{a \cdot (b\times c)} = D\, \mathbf{(b\times c)}^2$$
Unknowns $B$ and $C$ can be easily eliminated, but when eliminating $D$, I necessarily have to square $\bf a \cdot (b \times c)$ to get the final solution $$[\mathbf{a \cdot (b \times c)}]^2 = a^2b^2c^2 - a^2(\mathbf{b \cdot c}) - b^2(\mathbf{c \cdot a}) - c^2(\mathbf{a \cdot b}) + 2(\mathbf{a \cdot b})(\mathbf{b \cdot c})(\mathbf{c \cdot a})$$
Is there a way of getting a similar expression for $\mathbf{a \cdot (b \times c)}$ directly? I would like to see an expression which clearly demonstrates the cyclic property of the scalar triple product. The square in my expression ruins it...
|
A "similar" expression is the determinant representation
$$\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=\operatorname{det} S,\qquad S=\left(
\begin{array}{ccc}
a_x & a_y & a_z \\
b_x & b_y & b_z \\
c_x & c_y & c_z
\end{array}\right).$$
The expression that you have obtained can be easily deduced from this one:
$$\Bigl(\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\Bigr)^2=\operatorname{det} SS^T=
\operatorname{det}\left(\begin{array}{ccc}
\mathbf{a}\cdot\mathbf{a} & \mathbf{a}\cdot\mathbf{b} & \mathbf{a}\cdot\mathbf{c} \\
\mathbf{b}\cdot\mathbf{a} & \mathbf{b}\cdot\mathbf{b} & \mathbf{b}\cdot\mathbf{c} \\
\mathbf{c}\cdot\mathbf{a} & \mathbf{c}\cdot\mathbf{b} & \mathbf{c}\cdot\mathbf{c}
\end{array}\right).$$
However the non-squared triple product cannot be polynomially expressed in terms of scalar products only.
|
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|
Series for logarithms This is more of a challenge than a question, but I thought I'd share anyway. Prove the following identities, and prove that the pattern continues.
\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)=\ln2
\end{equation*}\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{2}{3n+3}\right)=\ln3
\end{equation*}\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{4n+1}+\frac{1}{4n+2}+\frac{1}{4n+3}-\frac{3}{4n+4}\right)=\ln4
\end{equation*}\begin{equation*}
\mathrm{etc.}
\end{equation*}
By the way, a good notation for discussing this problem can be found here. The problem is to prove that:
$[\overline{1,-1}]=\ln2,\;\;$ $[\overline{1,1,-2}]=\ln3,\;\;$ $[\overline{1,1,1,-3}]=\ln4,\;\;$ $[\overline{1,1,1,1,-4}]=\ln5,\;\;$ etc.
|
So far, there is a proof involving differentiation/integration, and a proof involving nothing more than the knowledge that $\gamma$amma exists. Here is my proof, requiring the Taylor series expansion of the logarithm.
The well known Taylor series for $\ln(x)$ is as follows:
$$-\ln(1-x)=\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\cdots+\frac{x^m}{m}+\cdots$$
which I love, because the exponents correspond to the denominators.
Substituting in $x^m$, we get:
$$-\ln(1-x^m)=\frac{x^m}{1}+\frac{x^{2m}}{2}+\frac{x^{3m}}{3}+\cdots=\frac{m\,x^m}{m}+\frac{m\,x^{2m}}{2m}+\frac{m\,x^{3m}}{3m}+\cdots$$
where I rewrote it slightly so that the exponents still correspond to the denominators.
Subtracting the bottom from the top, we get:
$$\ln\left(\frac{1-x^m}{1-x}\right)=\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\cdots+\frac{-(m-1)x^m}{m}+\\
\frac{x^{m+1}}{m+1}+\cdots+\frac{-(m-1)x^{2m}}{2m}+\cdots$$
(i.e. the coefficients are $-(m-1)$ where the denominator is a multiple of $m$, and $1$ otherwise.)
The LHS can be rewritten as $\ln(1+x+x^2+\cdots+x^m)$. So, plugging in $x=1$:
$$\ln(m)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots-\frac{m-1}{m}+\frac{1}{m+1}+\cdots-\frac{-(m-1)}{2m}+\cdots$$
which is what was meant to be proved.
|
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How can I find $\lim_{x \to 0}\frac{\tan(3x)}{\sin(8x)}$ without L'Hospital's Rule Is there a way to solve $\lim_{x \to 0}\frac{\tan(3x)}{\sin(8x)}$ without using the trig identity $\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-\tan^2(x)}$. I want to know because I had to look up this trig identity in order to solve this limit, but if there is a simpler way, then I would like to know it. I don't want to use L'Hospital's Rule because it hasn't been introduced at this point in the book.
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This problem is simpler and does not require the use of $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$. We have $$\begin{aligned}L &= \lim_{x \to 0}\frac{\tan 3x}{\sin 8x}\\
&= \lim_{x \to 0}\frac{\tan x(3 - \tan^{2}x)}{1 - 3\tan^{2}x}\cdot\frac{1}{2\sin 4x\cos 4x}\\
&= \lim_{x \to 0}\frac{3 - \tan^{2}x}{1 - 3\tan^{2}x}\cdot\frac{\sin x}{\cos x}\cdot\frac{1}{4\sin 2x\cos 2x\cos 4x}\\
&= \lim_{x \to 0}\frac{3 - \tan^{2}x}{1 - 3\tan^{2}x}\cdot\frac{\sin x}{\cos x}\cdot\frac{1}{8\sin x\cos x\cos 2x\cos 4x}\\
&= \lim_{x \to 0}\frac{3 - \tan^{2}x}{1 - 3\tan^{2}x}\cdot\frac{1}{\cos x}\cdot\frac{1}{8\cos x\cos 2x\cos 4x}\\
&= \frac{3}{1}\cdot\frac{1}{1}\cdot\frac{1}{8\cdot 1\cdot 1\cdot 1} = \frac{3}{8}\end{aligned}$$ We have used the limits $$\lim_{x \to 0}\tan x = 0, \lim_{x \to 0}\cos x = 1$$ which are on slightly simpler level compared to $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$.
|
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How do I use Euler's result to find the sum of a series? So I am given:
$$ \zeta(4) = \sum_{n=1}^\infty {1\over n^4}={\pi^4 \over 90} $$
I need to use it to find the sum of the following series using the above information.
$$ \sum_{k=1}^\infty {1\over{(k+2)^4}} $$
So, this is what I have so far:
$$ \sum_{k=1}^\infty {1\over{(k+2)^4}} = \sum_{k=3}^\infty {1\over{k^4}}$$
but that is all I have... How do I get rid of the $k=3$?
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Compare $$\sum_{k=3}^{\infty}\frac{1}{k^4}=\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\cdots$$
with$$\sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{1}{1^4}+\frac{1}{2^4}+\color{red}{\frac{1}{3^4}+\frac{1}{4^4}+\cdots}.$$
Hence we have $$\sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{1}{1^4}+\frac{1}{2^4}+\sum_{k=3}^{\infty}\frac{1}{k^4}\iff \sum_{k=3}^{\infty}\frac{1}{k^4}=\left(\sum_{k=1}^{\infty}\frac{1}{k^4}\right)-\frac{1}{1^4}-\frac{1}{2^4}.$$
|
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|
$a,b,c \geq 0$ and $a+b+c=3$ prove that $\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$ $a,b,c \geq 0$ and $a+b+c=3$ prove that $\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$
can anyone help me solve this problem,i've tried to use C-S and also AM-GM for couple in cycle.Sorry for my bad English that caused me problem to explain my trying.
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$$\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$$
$$\iff \frac{a^2+bc}{3b+3ac} + \frac{b^2+ac}{3c+3ab} + \frac{c^2+ab}{3a+3bc} \geq 1$$
$$\iff \frac{a^2+bc}{(a+b+c)b+3ac} + \frac{b^2+ac}{(a+b+c)c+3ab} + \frac{c^2+ab}{(a+b+c)a+3bc} \geq 1$$
$$\Leftarrow \frac{a^2+bc}{(a+b+c)b+(a^2+ac+c^2)} + \frac{b^2+ac}{(a+b+c)c+(a^2+ab+b^2)} + \frac{c^2+ab}{(a+b+c)a+(b^2+bc+c^2)} = 1$$
|
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|
If $~a^3 + b^3 = c^3~$ has nonzero integer solutions and $~c-b~$ is a cubic number and $~c-b \neq 1$ If $~a^3 + b^3 = c^3~$ has nonzero integer solutions, because:
$c^3 - b ^ 3 = (c - b)((c - b) ^ 2 + 3cb) = a ^ 3,\quad (1)$
if $~c-b~$ is a cubic number and $~c-b \neq 1~$, divide both side of $~(1)~$ by $~c - b~$ get
$(c - b) ^ 2 + 3cb = x^3.\quad (2)$
It seems $~3 \mid \ x~$ and $~3 \mid \ c~$ or $~b,~$ how to prove this?
Edit: add tag number-theory
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Let $~(c−b)^2=y^3,~$ from $~(2)~$ get:
$x^3−y^3=(x−y)((x−y)^2+3xy)=3cb,\quad (3) $
if $~x-y \neq 3,c,b~$ suppose $~(x-y) \mid c,~$ divide both side of $(3)\ $ by $~3$:
$(x−y)^2+3xy=3c_1b,\quad (4) $
form $~(4)~$ we see $~3 \mid (x−y)^2,~$ because $~(x-y) \neq 3,~$then $~3 \mid c_1,~$
form $~(4)~$ get $~3^2k^2+3xy=9c_2b,\quad (5) $
divide both side of $(5)\ $ by $~3$:
$~3k^2+xy=3c_2b,\quad (6) $
form $~(6)~$ we see $~3 \mid x~$ or $~y,~$ if $~3 \mid x,~$ problem solved, if $~3 \mid y,~$ form $~(2)~$ we see $~3 \mid x.$
I dont know is the proof above right and the cases when $~(x-y)=3~$or $~c~$ or $~b.$
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Arithmetic Progression. Q. The ratio between the sum of $n$ terms of two A.P's is $3n+8:7n+15$. Find the ratio between their $12$th term.
My method:
Given:
$\frac{S_n}{s_n}=\frac{3n+8}{7n+15}$
$\frac{S_n}{3n+8}=\frac{s_n}{7n+15}=k$
$\frac{T_n}{t_n}=\frac{S_n-S_{n-1}}{s_n-s_{n-1}}=\frac{k\left(\left(3n+8\right)-\left(3\left(n-1\right)+8\right)\right)}{k\left(\left(7n+15\right)-\left(7\left(n-1\right)+15\right)\right)}=\frac{3}{7}$
As this applies for any term:
$\frac{T_{12}}{t_{12}}=\frac{3}{7}$
But this is not the answer. The actual answer is $\frac7{16}$. I know how to obtain that answer.
But why is my solution wrong? Its probably the concept I guess.
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The trick is to notice that we can safely cancel constant common factors (if any), and stick in the $n$ in both numerator and denominator:
$\dfrac{T_n}{t_n} = \dfrac{S_n - S_{n - 1}}{s_n - s_{n - 1}} = \dfrac{n(3n + 8) - (n - 1)(3(n - 1) + 8)}{n(7n + 15) - (n - 1)(7(n - 1) + 15)} = \dfrac{6n + 5}{14n + 8}$
Plug in $n = 12$ to get the ratio of 12th terms.
|
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Question on Factoring I have very basic Question about factoring, we know that,
$$x^2+2xy+y^2 = (x+y)^2$$
$$x^2-2xy+y^2 = (x-y)^2$$
But what will
$$x^2-2xy-y^2 = ??$$
$$x^2+2xy-y^2 = ??$$
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You can use the quadratic formula:
$$x^2 + (-2y)x + (-y^2) = 0\to \\ x= \frac{2y \pm \sqrt{4y^2 - 4(1)(-y^2)}}{2} \\ =y \pm\sqrt{2}y = (1 \pm \sqrt{2})y.$$
So $x^2 - 2xy - y^2 = [x - (1+\sqrt{2})y][x - (1-\sqrt{2})y].$
For the other case,
$$x^2 + (2y)x + (-y^2) = 0\to \\ x= \frac{-2y \pm \sqrt{4y^2 - 4(1)(-y^2)}}{2} \\ =-y \pm\sqrt{2}y = (1 \pm \sqrt{2})y.$$
So $x^2 + 2xy - y^2 = [x - (1-\sqrt{2})y][x - (1+\sqrt{2})y].$
|
{
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If a fair die is thrown three times, what is the probability that the sum of the faces is 9?
If a fair die is thrown thrice, what is the probability that the sum of the faces is 9?
I did like this.
The total number of cases is $6^3=216$
Now,the number of solutions of the equation $x + y + z = 9$ with each of $x,y,z$ greater than equal to $1$ is ${8 \choose 2}$.
But am not sure about my answer. Please help.
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I get that there are only $25$ ways of writing $9$ as a sum of three integers in $[1,6]$, since:
$$[x^9](x+x^2+x^3+x^4+x^5+x^6)^3 = 25.$$
Hence the probability is $\frac{5^2}{6^3}$.
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How to prove that a diffrensiation of a formula equals to another formula. QUESTION 1) if $y =\dfrac{ \sin x-x\cos x}{x\sin x+\cos x}$ show that $\dfrac{dy}{dx}= \dfrac{x^2}{(x\sin x+\cos x)^2}$
QUESTION 2) if $y = \dfrac{\tan x+1}{\tan x-1}$ show that $\dfrac{dy}{dx}= \dfrac{-2}{1-\sin 2x}$
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for the second, we get
$$\begin{align}
y&=\frac{\tan x+1}{\tan x-1}\\
\frac{dy}{dx}&=\frac{d}{dx}\left(\frac{\tan x+1}{\tan x-1}\right)\\&
=\frac{\frac{d}{dx}(\tan x+1)(\tan x-1)-(\tan x+1)\frac{d}{dx}(\tan x-1)}{(\tan x-1)^2}\\&
=\frac{\sec^2 x(\tan x-1)-(\tan x+1)\sec^2 x}{(\tan x-1)^2}\\&
=\frac{\sec^2x\tan x-\sec^2 x-\sec^2 x\tan x-\sec^2 x}{(\tan x-1)^2}\\&
=\frac{-2\sec^2 x}{(\tan x-1)^2}\\&
=-2\sec^2 x\frac{1}{(\tan x-1)^2}\\&
=-2\frac{1}{\cos^2 x}\frac{1}{\left(\frac{\sin x}{\cos x}-1\right)^2}\\&
=-2\frac{1}{\cos^2 x}\frac{1}{\frac{(\sin x-\cos x)^2}{\cos^2 x}}\\&
=-2\frac{1}{\cos^2 x}\frac{\cos^2 x}{(\sin x-\cos x)^2}\\&
=-2\frac{1}{(\sin x-\cos x)^2}\\&
=-2\frac{1}{\sin^2x-2\sin x\cos x+\cos^2x}\\&
=-2\frac{1}{\color{red}{\sin^2x+\cos^2x}-\color{blue}{2\sin x\cos x}}\\&
=\frac{-2}{\color{red}{1}-\color{blue}{\sin2x}}
\end{align}$$
|
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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.