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determine $\left(\sum\limits_{g=1}^{p-1} g \times ((ag \mod p))\right) \bmod p^2$ Let $p$ be a prime, and let $((x))$ be the least positive residue of $x$ modulo $p$. For $a \in \{1, 2, ..., p-1\}$, consider $f(a) = \sum\limits_{g=1}^{p-1} g \times ((a g))$.
Does the following hold: $f(a)=f(b)$ if and only if $a=b$ or $a=b^{-1} \bmod p$？ Furthermore, can we determine $f(a) \bmod p^2$？
|
First,
$f(a) = \frac{1}{2} \sum_{g=1}^{p-1} [ (g + a(g))^2 - g^2 - a(g)^2 ] = \frac{1}{2} \sum_g (g + a(g))^2 - \sum_g g^2$ .
Next, note that $g + a(g) = c(g)$ or $c(g) + p$ for $a \neq p-1$, where $c = a+1 ($mod $p)$, and $g + a(g) = p$ for $a=p-1$.
So for $a=p-1$,
$f(a) = \frac{p^2(p-1)}{2} - \frac{(p-1)p(2p-1)}{6} = \frac{(p-1)p(p+1)}{6}$.
For $a<p-1$,
$f(a) = \frac{1}{2} \sum_g c(g)^2 + \frac{1}{2} \sum_{c(g)<g} (2c(g) p + p^2) - \frac{(p-1)p(2p-1)}{6} $
$ = \frac{1}{2} \sum_{c(g)<g} (2c(g) p + p^2) - \frac{(p-1)p(2p-1)}{12}$.
Now, I believe that for $c\neq 1$, $c(g)<g$ exactly half the time (as $c(-g) = p-c(g)$). So we have in the latter case
$ f(a) = \sum_{c(g)<g} c(g) p + \frac{p^2(p-1)}{4} - \frac{(p-1)p(2p-1)}{12} = \sum_{c(g)<g} c(g) p + \frac{(p-1)p(p+1)}{12}$.
Edited: That's as far as I got.
|
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|
Need help computing $(f \circ g)(x)$ $$f(x) = \frac{1}{x^2} + 1$$
$$g(x) = \frac{1}{x - 1}$$
Do I just insert the $g(x)$ function into the $f(x)$ function giving me this terrible looking thing?:
$$\frac{1}{\frac{1}{x^2}+1-1}$$
Thanks for any help, I really just need confirmation I'm going in the right direction.
|
Actually, it would be the other way around. $(f \circ g)(x) = f(g(x))$, and $f(g(x)) = \frac{1}{g(x)^2}+1 = \frac{1}{\frac{1}{(x-1)^2}}+1$. This last expression is almost equal to $(x-1)^2+1$. The difference is that in the former, $ f \circ g$ is undefined at $x = 1$, while in the latter, it is. The correct function would be $(f \circ g )(x) = (x-1)^2 + 1, \forall x \neq 1$.
What you did is actually $(g\circ f)(x) = g(f(x)) = \frac{1}{f(x)-1} = \frac{1}{\frac{1}{x^2}+1-1} = \frac{1}{\frac{1}{x^2}}$. The same note about the function not existing at $x = 0$ applies; therefore, $(g \circ f)(x) = x^2, \forall x \neq 0$.
|
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|
Finding the Laurent series of $f(z)=1/((z-1)(z-2))$ Let
$$f(z)=\frac{1}{(z-1)(z-2)}$$
and let
$$R_1=\Bigl\{z\Bigm| 1<|z|<2\Bigr\}\quad\text{ and }\quad R_2=\Bigl\{z\Bigm| |z|>2\Bigr\}.$$
How do you find the Laurent series convergent on $R_1$? Also how do you do it for $R_2$?
I'm having serious trouble with this as I can't see how to expand things into series with n as any integer, not just natural number. Also how to apply Cauchy's integral formula to an annulus. If anyone can explain this to me I will be extremely grateful.
|
The function $f(z)$ can be expanded into two partial fractions
$$
f(z):=\frac{1}{\left( z-1\right) \left( z-2\right) }=\frac{1}{z-2}-\frac{1}{z-1}.
$$
We now expand each partial fraction into a geometric series. On $R_{2}$
these series are
$$
\begin{eqnarray*}
\frac{1}{z-2} &=&\frac{1}{z\left( 1-2/z\right) }=\frac{1}{z}
\sum_{n=0}^{\infty }\left( \frac{2}{z}\right) ^{n}\qquad \left\vert
z\right\vert >2 \\
&=&\frac{1}{z}\sum_{n=0}^{\infty }2^{n}\frac{1}{z^{n}}=\sum_{n=0}^{\infty
}2^{n}\frac{1}{z^{n+1}}
\end{eqnarray*}
$$
and
$$
\begin{eqnarray*}
\frac{1}{z-1} &=&\frac{1}{z\left( 1-1/z\right) } \\
&=&\frac{1}{z}\sum_{n=0}^{\infty }\left( \frac{1}{z}\right)
^{n}=\sum_{n=0}^{\infty }\frac{1}{z^{n+1}}\qquad \left\vert z\right\vert >1.
\end{eqnarray*}
$$
And so, the Laurent series is
$$
\frac{1}{\left( z-1\right) \left( z-2\right) }=\sum_{n=0}^{\infty }\frac{1}{
z^{n+1}}(2^{n}-1)\qquad \left\vert z\right\vert >2>1.
$$
On $R_{1}$, the two geometric series are
$$
\begin{eqnarray*}
\frac{1}{z-2} &=&\frac{-1/2}{1-z/2}=\sum_{n=0}^{\infty }\left( -\frac{1}{2}
\right) \left( \frac{z}{2}\right) ^{n}\qquad \left\vert z\right\vert <2 \\
&=&\sum_{n=0}^{\infty }-\frac{1}{2^{n+1}}z^{n}
\end{eqnarray*}
$$
and
$$
\begin{eqnarray*}
\frac{1}{z-1} &=&\frac{1/z}{1-1/z}=\sum_{n=0}^{\infty }\frac{1}{z}\left(
\frac{1}{z}\right) ^{n}\qquad \left\vert z\right\vert >1 \\
&=&\sum_{n=0}^{\infty }\frac{1}{z^{n+1}}.
\end{eqnarray*}
$$
We thus get the following Laurent series
$$
\frac{1}{\left( z-1\right) \left( z-2\right) }=\sum_{n=0}^{\infty }\left( -
\frac{1}{2^{n+1}}z^{n}-\frac{1}{z^{n+1}}\right) \qquad 1<\left\vert
z\right\vert <2.
$$
|
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|
Proving $f(z)=(e^{3iz}-3e^{iz}+2)/z^3$ has a simple pole at $0$ Let $f(z)=(e^{3iz}-3e^{iz}+2)/z^3$, this clearly has a singularity at $z=0$, how do you show that this is a simple pole? i.e. a pole of order one, every way it definity expands to something with a minimum power of $-3$, so I come to the conclusion there is a pole of order $3$?
And if this is the case is $f(z)+3/z$ holomorphic? Does this term cancel some part of the expansion I have missed that eliminates the pole?
|
For $|z|\ll 1$, we have
$$
\begin{split}
e^{3iz} &= 1 + 3 i z - \frac{9}{2} z^2 + O(z^3) \\
e^{iz} &= 1 + i z - \frac{1}{2} z^2 + O(z^3)
\end{split}
$$
so
$$
\begin{split}
\frac{1}{z^3} (e^{3iz} - 3 e^{iz} +2 )
&= \frac{1}{z^3} \left( 1 - 3 + 2 + (3 - 3) i z + (-\frac{9}{2} + \frac{3}{2}) z^2
+ O(z^3) \right) \\
&= - \frac{3}{z} + O(1)
\end{split}
$$
|
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|
Showing whether two numbers are equal or not $\dfrac{\sin (2x+y)}{\sin (2x)} =\dfrac{\sin (x+2y)}{\sin (2y)}$,where $0<x,y\le\dfrac{\pi}{4}$ .
Can I show that $x=y $ or find two numbers $x,y$ such that $x\not=y$?
|
Start with
$$
\frac{\sin (2x+y)}{\sin (2x)} =\frac{\sin (x+2y)}{\sin (2y)}\tag{1}
$$
Regrouping $(1)$, we get
$$
\frac{\sin\left(\frac{3}{2}(x+y)+\frac{1}{2}(x-y)\right)}{\sin\left((x+y)+(x-y)\right)}=\frac{\sin\left(\frac{3}{2}(x+y)-\frac{1}{2}(x-y)\right)}{\sin\left((x+y)-(x-y)\right)}\tag{2}
$$
Expanding $(2)$, yields
$$
\begin{align}
&\frac{\sin\frac{3}{2}\!\!(x+y)\;\cos\frac{1}{2}\!\!(x-y)+\cos\frac{3}{2}\!\!(x+y)\;\sin\frac{1}{2}\!\!(x-y)}{\sin(x+y)\;\cos(x-y)+\cos(x+y)\;\sin(x-y)}\\
&=\frac{\sin\frac{3}{2}\!\!(x+y)\;\cos\frac{1}{2}\!\!(x-y)-\cos\frac{3}{2}\!\!(x+y)\;\sin\frac{1}{2}\!\!(x-y)}{\sin(x+y)\;\cos(x-y)-\cos(x+y)\;\sin(x-y)}\tag{3}
\end{align}
$$
Since $\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{b}{a}=\frac{d}{c}$, $(3)$ implies
$$
\frac{\tan\frac{1}{2}\!\!(x-y)}{\tan\frac{3}{2}\!\!(x+y)}=\frac{\tan(x-y)}{\tan(x+y)}\tag{4}
$$
Assume $x\not=y$. We can rearrange $(4)$ to get
$$
\frac{\tan\frac{1}{2}\!\!(x-y)}{\tan(x-y)}=\frac{\tan\frac{3}{2}\!\!(x+y)}{\tan(x+y)}\tag{5}
$$
Since $0<x,y\le\frac{\pi}{4}$, the left hand side of $(5)$ is positive, and the denominator of the right hand side is also positive; therefore, the numerator of the right hand side must be positive; that is, $0<\frac{3}{2}\!\!(x+y)\le\frac{\pi}{2}$. Thus, the left hand side is less than $1$ and the right hand side is greater than $1$. Contradiction; therefore, $x=y$.
|
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|
Rolling infinitely many dice For every integer $n>1$, I have a die $D_n$ with $a_n$ sides labeled $1$ to $a_n$.
If $a_n=n^k$, for integer $k>0$, and I roll all the dice at once, what is the probability none of them lands on the side labeled $1$? What is the probability exactly one lands on the side labeled $1$?
|
First, suppose there are total of $q$ dice.
Given that, for fixed $m$, $D_m = 1$, the probability that no other die rolled at one is:
$$
\mathbb{P}(\forall_{n \not= m} \, {D_n \not= 1} ; D_m = 1 ) = \frac{1}{1-\frac{1}{a_m}} \prod_{n=2}^q \left( 1- \frac{1}{a_n} \right)
$$
The probability that exactly one die lands on 1 is:
$$
p_k(q) = \sum_{m=2}^q \frac{1}{a_m} \mathbb{P}(\forall_{n \not= m} \, {D_n \not= 1} ; D_m = 1 ) = \prod_{n=2}^q \left( 1- \frac{1}{a_n} \right) \sum_{m=2}^q \frac{1}{a_m-1} =
\prod_{n=2}^q \left( 1- \frac{1}{n^k} \right) \cdot \sum_{m=2}^q \frac{1}{m^k-1}
$$
For small values of $k$ we have
$$
p_1(q) = \frac{1}{q} H_{q-1} \qquad
p_2(q) = \frac{3}{8} \left(1+\frac{1}{12 q}\right)\left(1-\frac{1}{q} \right)
$$
For higher order the probability is expressible in terms of $\Gamma$ functions and poly-gamma functions.
In the limit of infinitely many dice:
$$
\lim_{q \to \infty} p_1(q) = 0 \qquad
\lim_{q \to \infty} p_2(q) = \frac{3}{8} \qquad
\lim_{q \to \infty} p_3(q) = \frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right)}{9 \pi }
$$
|
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|
Modular exponentiation by hand ($a^b\bmod c$) How do I efficiently compute $a^b\bmod c$:
*
*When $b$ is huge, for instance $5^{844325}\bmod 21$?
*When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, for instance $5^{69}\bmod 101$?
*When $(a,c)\ne1$, for instance $6^{103}\bmod 14$?
Are there any other tricks for evaluating exponents in modular arithmetic?
|
Let's try $5^{844325} \bmod 21$:
$$
\begin{align}
5^0 & & & \equiv 1 \\
5^1 & & &\equiv 5 \\
5^2 & \equiv 25 & & \equiv 4 \\
5^3 & \equiv 4\cdot 5 & & \equiv 20 \\
5^4 & \equiv 20\cdot 5 & & \equiv 16 \\
5^5 & \equiv 16\cdot 5 & & \equiv 17 \\
5^6 & \equiv 17\cdot 5 & & \equiv 1
\end{align}
$$
So multiplying by $5$ six times is the same as multiplying by $1$. We want to multiply by $5$ a large number of times: $844325$. How many times do we multiply by $5$ six times? The number of times $6$ goes into $844325$ is $140720$ with a remainder of $5$. That remainder is what matters. Multiply by $5^6$ exactly $140720$ times and that's the same as multiplying by $1$ that many times. Then multiply by $5$ just $5$ more times, and get $17$.
So $5^{844325} \equiv 17 \bmod 21$.
|
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|
Find x: $2^{2x^2}+2^{x^2+2x+2}=2^{4x+5}$ $$2^{2x^2}+2^{x^2+2x+2}=2^{4x+5}$$
I have to find x in this exponential equation.
I tried to write it in another way, like this:
$2^{2x^2}+2\cdot2^{(x+1)^2}=2\cdot2^{4(x+1)}$
But I don't think this would help.
|
Divide by $2^{2x^2}$ to get $1 + 2^{-(x-1)^2} \cdot 2^3 = 2^{-2 (x-1)^2} \cdot 2^7$. Substitute $y = 2^{-(x-1)^2}$, and solve the quadratics.
|
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|
Proof about an infinite sum: $\sum_{k=1}^\infty \frac1{k^2+3k+1} \ge \frac12$ Hello I have a pretty elementary question but I am a bit confused.
I am trying to prove that $$\sum_{k=1}^\infty \frac1{k^2+3k+1} \ge \frac12$$
thanks,
Thrasyvoulos
|
$$\sum_{k=1}^\infty\frac{1}{k^2+3k+1}\geq\sum_{k=1}^\infty\frac{1}{k^2+3k+2}=\sum_{k=1}^\infty\left(\frac{1}{k+1}-\frac{1}{k+2}\right)=$$
$$\lim_{n\to\infty}\;\;\sum_{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)$$
Note that
$$\sum_{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{2}-\frac{1}{n+2}$$
so
$$\lim_{n\to\infty}\;\;\sum_{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)=\lim_{n\to \infty}\left(\frac{1}{2}-\frac{1}{n+2}\right)=\frac{1}{2}$$
Thus, we have shown
$$\sum_{k=1}^\infty\frac{1}{k^2+3k+1}\geq\frac{1}{2}$$
|
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|
Find the minimum of $x^{2}+5y^{2}+8z^{2}$ if $xy+yz+zx=-1$
If $xy+yz+zx=-1$，find the minimum of $x^{2}+5y^{2}+8z^{2}$.
How to solve it use Elementary mathematics methods？
|
Taking the first variation of $xy+yz+zx=-1$, we get
$$
(y+z)\delta x+(z+x)\delta y+(x+y)\delta z=0\tag{1}
$$
Setting the first variation of $x^2+5y^2+8z^2$ to $0$, we get
$$
2x\delta x+10y\delta y+16z\delta z=0\tag{2}
$$
At an extreme $\{x,y,z\}$, any $\{\delta x,\delta y,\delta z\}$ that satisfies $(1)$, must also satsify $(2)$. Orthogonality dictates that
$$
2\lambda x=y+z\qquad 10\lambda y=z+x\qquad 16\lambda z=x+y\tag{3}
$$
Converting $(3)$ to a matrix yields:
$$
\begin{bmatrix}-\lambda&1/2&1/2\\1/10&-\lambda&1/10\\1/16&1/16&-\lambda\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=0\tag{4}
$$
Thus, $\begin{bmatrix}x&y&z\end{bmatrix}^T$ will be an eigenvector of
$$
\begin{bmatrix}0&1/2&1/2\\1/10&0&1/10\\1/16&1/16&0\end{bmatrix}\tag{5}
$$
which has eigenvectors
$$
v_1=\begin{bmatrix}\frac{7+\sqrt{65}}{4}\\\frac{5+3\sqrt{65}}{20}\\1\end{bmatrix}\qquad v_2=\begin{bmatrix}\frac{7-\sqrt{65}}{4}\\\frac{5-3\sqrt{65}}{20}\\1\end{bmatrix}\qquad v_3=\begin{bmatrix}-6\\2\\1\end{bmatrix}\tag{6}
$$
Computing $xy+yz+zx$ for these vectors, we get
$$
v_1\mapsto\frac{195+29\sqrt{65}}{40}\qquad v_2\mapsto\frac{-416}{195+29\sqrt{65}} \qquad v_3\mapsto-16\tag{7}
$$
Since $xy+yz+zx\ge0$ for $v_1$, there is no multiple of $v_1$ so that $xy+yz+zx=-1$. That leaves $v_2$ and $v_3$ as the extrema.
If we scale $v_2$ so that $xy+yz+zx=-1$, we get $x^2+5y^2+8z^2=5+\sqrt{65}$.
If we scale $v_3$ so that $xy+yz+zx=-1$, we get $x^2+5y^2+8z^2=4$.
Thus, under the constraint that $xy+yz+zx=-1$, $x^2+5y^2+8z^2$ has a minimum of $4$ at $\{x,y,z\}=\left\{-\frac{3}{2},\frac{1}{2},\frac{1}{4}\right\}$ and $\left\{\frac{3}{2},-\frac{1}{2},-\frac{1}{4}\right\}$.
|
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|
How do you divide a polynomial by a binomial of the form $ax^2+b$, where $a$ and $b$ are greater than one? I came across a question that asked me to divide $-2x^3+4x^2-3x+5$ by $4x^2+5$. Can anyone help me?
|
You can do long division with polynomials almost the same way you would for integers. For example, $4x^2+5$ can be multiplied by $-\frac{1}{2}x$ to get a leading term of $-2x^3$, so we might say that $4x^2+5$ goes into $-2x^3 + 4x^2 - 3x + 5$ about $-\frac{1}{2}x$ times. As with long division, we then multiply $4x^2+5$ by $-\frac{1}{2}x$ and subtract the result from $-2x^3 + 4x^2 - 3x + 5$; then repeat the process until we can't anymore. If there's a remainder at the end, we divide it by the divisor and add that to the end of the result (again, similar to long division).
In this case you should verify that the result is $-\frac{1}{2}x + 1 + \frac{-x}{8x^2+10}$.
|
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|
Help with integrating $\int \frac{t^3}{1+t^2} ~dt$ What am I doing wrong on this integration problem?
$$
\begin{align*}
\int\frac{t^3}{1+t^2} &=
\frac14 t^4 (\ln(1+t^2) (t+\frac13 t^3))
\\ &= \frac14 t^4(t \ln(1+t^2)+\frac13 t^3 \ln(1+t^2)
\\ &= \frac14 t^5 \ln(1+t^2)+\frac{1}{3}t^7 \ln(1+t^2)
\end{align*}$$
Answer should be $\frac{1}{2}(t^2-\ln(t^2+1))$. I'm way off
Any help appreciated.
Thanks!
|
$$
\int \frac{t^3}{1+t^2}\, dt = \int \frac{t^2}{1+t^2} \Big( \underbrace{{}\quad t\, dt\quad{}}_{\text{HINT}}\Big) = \int \frac{u}{1+u} \Big(\frac 1 2 \, du \Big) = \int \left( 1 - \frac{1}{1+u} \right) \Big(\frac 1 2 \, du \Big)
$$
|
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|
How to prove :If $p$ is prime greater than $3$ and $\gcd(a,24\cdot p)=1$ then $a^{p-1} \equiv 1 \pmod {24\cdot p}$? I want to prove following statement :
If $p$ is a prime number greater than $3$ and $\gcd(a,24\cdot p)=1$ then :
$a^{p-1} \equiv 1 \pmod {24\cdot p}$
Here is my attempt :
The Euler's totient function can be written in the form :
$n=p_1^{k_1}\cdot p_2^{k_2} \ldots \cdot p_r^{k_r} \Rightarrow \phi(n)=p_1^{k_1}\cdot\left(1-\frac{1}{p_1}\right)\cdot p_2^{k_2}\cdot\left(1-\frac{1}{p_2}\right)\ldots p_r^{k_r}\cdot \left(1-\frac{1}{p_r}\right)$
So,
$\phi(24 \cdot p)=2^3\cdot \left(1-\frac{1}{2}\right)\cdot3^1\cdot\left(1-\frac{1}{3}\right)\cdot p\cdot\left(1-\frac{1}{p}\right)=8\cdot(p-1)$
Euler's totient theorem states that :
if $\gcd(a,n)=1$ then $a^{\phi(n)} \equiv 1 \pmod n$
Therefore we may write :
$a^{\phi(n)}-1 \equiv 0 \pmod n \Rightarrow a^{\phi(24\cdot p)}-1=a^{8\cdot(p-1)}-1 \equiv 0 \pmod{24\cdot p} \Rightarrow$
$\Rightarrow \left(a^{p-1}\right)^8-1=(a^{p-1}-1)\cdot \displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \equiv 0\pmod{24\cdot p}$
So we may conclude :
$(a^{p-1}-1) \equiv 0 \pmod {24\cdot p}$ , or $\displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \equiv 0\pmod{24\cdot p}$
How can I prove that $\displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \not\equiv 0\pmod{24\cdot p}$ ?
|
HINT $\ $ By Carmichael's simple generalization of Euler $\phi$, since prime $\rm\:p\:$ is coprime to $2,3$
$\rm\ \ \lambda(8\cdot3\cdot p) = lcm(\lambda(8),\lambda(3),\lambda(p)) = lcm(\phi(8)/2,\phi(3),\phi(p)) = lcm(2,2,p-1) = p-1\ $
therefore $\rm\quad gcd(n,24\:p) = 1\ \ \Rightarrow\ \ 1\ \equiv\ n^{\:\lambda(24\:p)}\: \equiv \ n^{p-1}$
|
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|
Simplifying Quadratic Equation Quick question, say I'm simplying a solution I got using the quadratic equation and I run into this:
Original version (as posted by OP):
x = -7 +- 3 sqrt(5) over 3
Edited version:
$$
x = \frac{-7\pm 3 \sqrt{5} }{3}
$$
Would the two $3$s cross each out leaving the answer to be $x = -7 \pm \sqrt{5}$, or is that illegal in terms of rules and you have to simplify all the terms, including the $-7$ if you were to simplify correctly.
Thanks!
|
I'm a little late on the scene, but from your most recent comments you still haven't seen the light. Maybe this will help.
$$\frac{-7\pm 3 \sqrt{5} }{3} \;\; =\;\; \left(\frac{1}{3}\right) \left(\frac{-7\pm 3 \sqrt{5}}{1}\right) \;\;= \;\; \left(\frac{1}{3}\right)\left(-7 \; \pm \; 3 \sqrt{5}\right) $$
$$= \;\; \left(\frac{1}{3}\right)(-7) \; \pm \; \left(\frac{1}{3}\right)\left(3 \sqrt{5}\right)
\;\; = \;\; \left(\frac{1}{3}\right)\left(\frac{-7}{1}\right) \; \pm \; \left(\frac{1}{3}\right)\left(\frac{3 \sqrt{5}}{1}\right) \;\; = \;\; -\frac{7}{3}\; \pm \; \frac{3 \sqrt{5}}{3}$$
Usually people do all this in one step (see the first comment under your question, the comment by lhf) and write:
$$\frac{-7\pm 3 \sqrt{5} }{3} \;\; = \;\; -\frac{7}{3}\; \pm \; \frac{3 \sqrt{5}}{3}$$
|
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|
Prime Partition A prime partition of a number is a set of primes that sum to the number. For instance, {2 3 7} is a prime partition of $12$ because $2 + 3 + 7 = 12$. In fact, there are seven prime partitions of $12$: {2 2 2 2 2 2}, {2 2 2 3 3}, {3 3 3 3}, {2 2 3 5}, {2 5 5}, {2 3 7}, and {5 7}. The number of prime partitions of a number is given by A000607.
I want to calculate the number of prime partitions of a number. From reading the text of A000607 it seems to me that the formula $\prod\limits_{p \in \pi(n)} \frac1{1 - n^p}$ should work. But it doesn't. Consider the case of $n=12$. The primes less than $12$ are $2, 3, 5, 7$, and $11$. Thus the formula computes $\frac1{1-12^2} \times \frac1{1-12^3} \times \frac1{1-12^5} \times \frac1{1-12^7} \times \frac1{1-12^{11}}$ = $\frac1{-143} \times \frac1{-1727} \times \frac1{-248831} \times \frac1{-35831807} \times \frac1{-743008370687}$ = $\frac{-1}{1636045119596820253743372240719}$ which is obviously incorrect.
How can I compute the number of prime partitions of a number?
|
You need to learn a bit about generating functions. The text associated with A000607 means the following. For each prime $p$ expand the function $1/(1-x^p)$ as a power series:
$$
\frac1{1-x^p}=1+x^p+x^{2p}+x^{3p}\cdots=\sum_{k=0}^\infty x^{kp}.
$$
Call that series $f_p(x)$.
Then you multiply these power series together, and identify the coefficient of $x^n$. That coefficient is then the desired value of the prime partition function. Let's do this for $n=12$. To that end we can throw away all the terms of degree $>12$. I denote those with three dots. So start with
$$
f_2(x)=1+x^2+x^4+x^6+x^8+x^{10}+x^{12}+\cdots
$$
Multiplying this with $f_3(x)=1+x^3+x^6+x^9+x^{12}+\cdots$ gives
$$
\begin{aligned}
f_2(x)f_3(x)=&f_2(x)+x^3f_2(x)+x^6f_2(x)+x^9f_2(x)+x^{12}f_2(x)+\cdots\\
=&1+x^2+x^3+x^4+x^5+2x^6+x^7+2x^8+2x^9+2x^{10}+2x^{11}+3x^{12}+\cdots
\end{aligned}
$$
At this point you should check that the coefficient of $x^k$ counts the number of ways of
writing $k$ as a sum of twos and threes.
Next we add $p=5$ to the mix, and multiply the above with $f_5(x)=1+x^5+x^{10}+\cdots$ and get
$$
\begin{aligned}
&f_2(x)f_3(x)f_5(x)\\
=&1+x^2+x^3+x^4+2x^5+2x^6+2x^7+3x^8+3x^9+4x^{10}+4x^{11}+5x^{12}+\cdots
\end{aligned}
$$
Next we multiply this with $f_7(x)=1+x^7+\cdots$ and get
$$
\begin{aligned}
&f_2(x)f_3(x)f_5(x)f_7(x)\\
=&1+x^2+x^3+x^4+x^5+2x^6+3x^7+3x^8+4x^9+5x^{10}+5x^{11}+7x^{12}+\cdots\\
\end{aligned}
$$
As a laste step we multiply this with $f_{11}(x)=1+x^{11}+\cdots$ to end with
$$
\begin{aligned}
&f_2(x)f_3(x)f_5(x)f_7(x)f_{11}(x)\\
=&1+x^2+x^3+x^4+x^5+2x^6+3x^7+3x^8+4x^9+5x^{10}+6x^{11}+7x^{12}+\cdots\\
\end{aligned}
$$
Here the term $7x^{12}$ appears. Primes $p>12$ won't affect the term of degree $12$, so at long last that tells us that there are seven prime partitions of $n=12$.
|
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|
Coefficients of $(1+x+\dots+x^n)^3$? Consider the following polynomial:
$$ (1+x+\dots+x^n)^3 $$
The coefficients of the expansion for few values of $n$ ($n=1$ to $5$) are:
$$ 1, 3, 3, 1 $$
$$ 1, 3, 6, 7, 6, 3, 1 $$
$$ 1, 3, 6, 10, 12, 12, 10, 6, 3, 1 $$
$$ 1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1 $$
$$ 1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1 $$
Is there a closed-form formula for the $i$th element of this sequence (for different values of $n$)?
Edit This looks similar to the sequence A109439 on OEIS corresponding to the coefficients of the expansion of: $ \left( \frac{1 - x^n}{1 - x} \right)^3 .$
|
Denote by
$$\sum_{k=0}^{3n}{\binom{3}{k}}_{n+1}x^{k}=(1 + x + ... + x^{n})^3\,$$
the expansion of multinomial of degree $n$
From that relation follow that
$$ \sum_{k=0}^{3n)}{\binom{3}{k}}_{n+1}x^{k}=\left(\frac{1-x^{n+1}}{1-x}\right)^{3}=(1-x^{n+1})^{3}(1-x)^{-3}\,$$
from binomial formula we have
$$(1-x^{n+1})^{3}=\sum_{i=0}^{3}(-1)^{i}\binom{3}{i}x^{(n+1)i}$$
and from Taylor formula
$$(1-x)^{-3}=\sum_{j=0}^{\infty}\binom{2+j}{j}x^{j}$$
after substitutions we get
$$ \sum_{k=0}^{\infty}{\binom{3}{k}}_{n+1}x^{k}=\sum_{i=0}^{3}(-1)^{i}\binom{3}{i}x^{(n+1)i}\sum_{j=0}^{\infty}\binom{2+j}{j}x^{j}=$$
$$=\sum_{j=0}^{\infty}\sum_{i=0}^{3}(-1)^{i}\binom{3}{i}\binom{2+j}{j}x^{(n+1)i+j}=$$
$$ =\sum_{k=0}^{\infty}\sum_{i=0}^{3}(-1)^{i}\binom{3}{i}\binom{2+k-(n+1)i}{2}x^{k}\,$$
then equate the coefficients next to x follow formula
$${\binom{3}{k}}_{(n+1)}=\sum_{i=0}^{3}(-1)^{i}\binom{3}{i}\binom{2+k-(n+1)i}{2}\,$$
formula for computing the coefficients of expansion
|
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|
General method for factorizing matrix determinants I'm learning how to factorize determinants of a square matrix in school, but we haven't learnt a general method to do that, besides 'creating zeros'. So I thought maybe I'll ask here if someone does know a method that generalizes the factorization of a square matrix.
An example
I had this determinant to factorize on a test:
$$\left|
\begin{array}{ccc}
x & y & 1 \\
x^2 & y^2 & 1 \\
x^3 & y^3 & 1
\end{array}
\right| $$
So I started and got the following, quite easy:
$$
\begin{align}
\left|
\begin{array}{ccc}
x & y & 1 \\
x^2 & y^2 & 1 \\
x^3 & y^3 & 1
\end{array}
\right|
&= xy \cdot
\left|
\begin{array}{ccc}
1 & 1 & 1 \\
x & y & 1 \\
x^2 & y^2 & 1 \\
\end{array} \right| & &(1)
\end{align}
$$
At this point I was stuck. I had now idea how to go on from this point. But when I got my test back, it was corrected by the teacher:
$$
\begin{align*}
xy \cdot
\left|
\begin{array}{ccc}
1 & 1 & 1 \\
x & y & 1 \\
x^2 & y^2 & 1 \\
\end{array} \right|
&= xy \cdot
\left|
\begin{array}{ccc}
0 & 0 & 1 \\
x - 1 & y -1 & 1 \\
x^2 -1 & y^2 - 1 & 1 \\
\end{array} \right| \\
&= xy \cdot (x-1)(y-1)
\left|
\begin{array}{ccc}
0 & 0 & 1 \\
1 & 1 & 1 \\
x + 1 & y + 1 & 1
\end{array}
\right| & \text{because } x^2 - 1^2 = (x-1)(x+1)\\
&=xy \cdot (x-1)(y-1) \cdot 1 \cdot(1 \cdot (y+1) - 1 \cdot (x + 1)) & \text{Laplace with row 1}\\
&= xy \cdot (x-1)(x+1)(y-x)
\end{align*}
$$
Alright, so I had to see myself at (1) that I had to subtract row 3 from row 1 and 2. Then I had to see I had to use the formula $a^2 - b^2 = (a-b)(a+b)$. I'm guessing a lot of you didn't see you had to do that. So my question to you: is there an easier way?
PS: I'll accept 'no' for an answer!
|
There is generally no easy way. Concise (closed) formula for the determinants of a general matrix is rare. For your example, it is too special, the determiant is equal to $\left|
\begin{array}{cccc}1 & 1 & 1 & 1\\
x & y & 1 & 0\\
x^2 & y^2 & 1 &0\\
x^3 & y^3 & 1&0
\end{array}
\right|$, where the matrix is a Vandermonde matrix whose determiant has a closed expression.
|
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|
Does iterating $n \to 2n+1$ always eventually produce a prime number? Is it the case that for every non-negative integer $n$, iterating $n \to 2n+1$ eventually produces a prime number? (This is the same as asking whether for every positive integer $n$, there is a non-negative integer $k$ such that $2^k n - 1$ is prime.) If this is not settled by proof, is there some heuristic argument either way?
|
(Following up on the connection to Riesel numbers mentioned in other answers ...)
For an infinite set of counterexamples, here is a proof-sketch (adapted from Baczkowski, et al.) showing that
$$n \equiv 33737173 \pmod{3\cdot 5\cdot 7\cdot 13\cdot 19\cdot 37\cdot 73} \implies \forall k\in \mathbb{N}\ \big(f^k(n-1) \text{ is composite}\big),$$ where $f:\mathbb{N}\to\mathbb{Z}^+:n\mapsto 2n+1$ and $f^k$ is the $k$-fold iteration of $f$.
Note that since $f^k(n-1) = n\cdot 2^k - 1$ for $n\in\mathbb{Z}^+$, the above is equivalent to
$$n \equiv 33737173 \pmod{3\cdot 5\cdot 7\cdot 13\cdot 19\cdot 37\cdot 73} \implies \forall k\in \mathbb{N}\ \big(n\cdot 2^k-1 \text{ is composite}\big),$$
which is to say that all such $n$ are Riesel numbers.
The following congruence classes are sufficient for this purpose:
$$\text{congruence classes for }k\text{ and }n\\
\begin{array}{|c|c|}
k\equiv a_i\pmod{m_i} & n\equiv b_i\pmod{p_i} \\
\hline
0\pmod{2} & 1\pmod{3} \\
0\pmod{3} & 1\pmod{7} \\
1\pmod{4} & 3\pmod{5} \\
11\pmod{12} & 2\pmod{13} \\
7\pmod{36} & 4\pmod{73} \\
19\pmod{36} & 18\pmod{37} \\
31\pmod{36} & 13\pmod{19}
\end{array}
$$
As can be directly verified, the entries in the above table have the following properties:
*
*The congruence classes for $k$ form a covering system for the integers; i.e., every integer $k$ satisfies at least one of the congruences $k\equiv a_i\pmod{m_i}$. (To verify this, it is necessary to check only $k\in \{0,1,2,\dots,35 \}$, because $36$ is the least common multiple of the moduli $\{2,3,4,12,36\}$.)
*The congruences are such that in each row $i\ (1\le i\le 7)$, $b_i\cdot 2^{a_i} \equiv 2^{m_i} \equiv 1\pmod{p_i}$. Consequently, for all $i\ (1\le i\le 7)$
$$k\equiv a_i\pmod{m_i}\quad \& \quad n\equiv b_i\pmod{p_i}\\ \implies\ n\cdot2^k \equiv b_i\cdot 2^{a_i+r\cdot m_i} \equiv (b_i\cdot 2^{a_i})(2^{m_i})^r \equiv 1 \pmod{p_i}.$$
Now, by the Chinese Remainder Theorem, there exist simultaneous solutions to all seven of the congruences for $n$; in particular, the Extended Euclidean Algorithm yields the solutions
$$n \equiv 33737173 \pmod{3\cdot 5\cdot 7\cdot 13\cdot 19\cdot 37\cdot 73}.$$
Therefore, for every such $n$,
$$\forall k\in \mathbb{N}\ \big(n\cdot2^k-1\text{ has at least one divisor in the set }\{3,5,7,13,19,37,73 \}\big).\quad\quad\text{QED}$$
|
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|
What is the result of $\lim\limits_{x \to 0}(1/x - 1/\sin x)$? Find the limit:
$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)$$
I am not able to find it because I don't know how to prove or disprove $0$ is the answer.
|
Simplify to have $$\frac{\sin x-x }{x\sin x}$$ and consider Maclaurin's series for $$\sin x=x-\frac {x^3}{3!}+\frac {x^5}{5!}-...$$
So you have $$\frac{(x-\frac {x^3}{3!}+\frac {x^5}{5!}-...)-x}{x(x-\frac {x^3}{3!}+\frac {x^5}{5!}+...)}=\frac{(-\frac {x}{3!}+\frac {x^3}{5!}-...)}{(1-\frac {x^2}{3!}+\frac {x^4}{5!}-...)}.$$
Finding the limit as $x\rightarrow 0$, we have;
$$\frac{\lim_{x\rightarrow 0}(-\frac {x}{3!}+\frac {x^3}{5!}-...)}{\lim_{x\rightarrow 0}(1-\frac {x^2}{3!}+\frac {x^4}{5!}-...)}=\frac{0}{1}=0.$$
which is the required answer.
|
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|
prove$\frac{ \sin a\vphantom{(}}{\sin b} +\frac{\cos a\vphantom{(}}{\cos b} = \frac{2\sin (a+b)}{\sin 2b}$ I've got this far but don't understand where the $2$ on the numerator comes from:
$$\dfrac{\sin a \cos b + \cos a \sin b}{\sin b \cos b}\overset{?}{=}\dfrac{\sin(a+b)}{\sin 2b}$$
|
$$\dfrac {\sin a}{\sin b}+\dfrac{\cos a}{\cos b}=\dfrac{\sin a\cdot\cos b+\cos a\cdot\sin b}{\sin b\cdot \cos b}$$
After getting this far, you need to observe that, $\boxed{\sin 2b=2 \cdot\sin b \cdot \sin b}$, you'll have to multiply the numerator and denominator by $2$, you'll have the following,
$$\dfrac{2(\sin a\cdot\cos b+\cos a\cdot\sin b)}{2\cdot\sin b\cdot \cos b}=\dfrac{2\sin (a+b)}{\sin 2b}$$
Hope this helps.
|
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|
prove for all $n\geq 0$ that $3 \mid n^3+6n^2+11n+6$ I'm having some trouble with this question and can't really get how to prove this..
I have to prove $n^3+6n^2+11n+6$ is divisible by $3$ for all $n \geq 0$.
I have tried doing $\dfrac{m}{3}=n$ and then did $m=3n$
then I said $3n=n^3+6n^2+11n+6$ but now I am stuck.
|
$n^3+6n^2+11n+6=n^3+n^2+5n^2+5n+6n+6=n^2(n+1)+5n(n+1)+6(n+1)=$
$=(n+1)(n^2+5n+6)=(n+1)(n^2+3n+2n+6)=(n+1)(n+2)(n+3)$
Now , since this last expression represents a product of three consecutive integers it has to be divisible by $3$ .
|
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|
how to work out a closed form of a sequence
Consider the following linear recurrence sequence.
$x_1 = 11$, $x_{n+1} = -0.8x_n + 9,\quad n = 1,2,3, \ldots.$
Find a closed form for this sequence.
|
Use generating functions. Define $X(z) = \sum_{n \ge 0} x_{n + 1} z^n$, multiply the recurrence by $z^n$ and sum over $n \ge 0$ to get:
$$
\frac{X(z) - x_1}{z} = - \frac{4 X(z)}{5} + \frac{9}{1 - z}
$$
Thus:
$$
X(z) = \frac{55 - 10 z}{5 - z - 4 z^2}
= 5 \cdot \frac{1}{1 - z} + 6 \cdot \frac{1}{1 + 4 z / 5}
$$
Two geometric series:
$$
x_{n + 1} = 5 + 6 \cdot (-0.8)^n
$$
|
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|
Simplify $\sin^3{\left(\cot^{-1}{\left(x\right)}\right)}$ How can the following function such that no trigonometric functions are present:
$\sin^3{\left(\cot^{-1}{\left(x\right)}\right)}$
Wolfram|Alpha shows the result as $\frac{1}{{\sqrt{x^2+1}}^3}$.
Thank you for your time.
|
You can show that for $x > 0$
$${\cot ^{ - 1}}x = {\sin ^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}$$
Then
$$\sin {\cot ^{ - 1}}x = \frac{1}{{\sqrt {1 + {x^2}} }}$$
and thus
$${\left( {\sin {{\cot }^{ - 1}}x} \right)^3} = \frac{1}{{1 + {x^2}}}\frac{1}{{\sqrt {1 + {x^2}} }}$$
The proof:
$$x = \cot y$$
$$1+x^2 = \csc^2 y $$
$$\sqrt{1+x^2} = \csc y $$
$$\frac{1}{\sqrt{1+x^2}} = \sin y $$
I guess that should do.
|
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|
Find remainder when dividing $9^{{10}^{{11}^{12}}}-5^{9^{10^{11}}}$ by $13$.
Find remainder when dividing
$$9^{{10}^{{11}^{12}}}-5^{9^{10^{11}}} \hspace{1cm} \text{by} \hspace{1.2cm} 13.$$
I tried transforming these who numbers separately to form $13k+n$ but failed.
|
Write this number as $9^N-5^M$.
Since $3^3=1\pmod{13}$, $9^3=1\pmod{13}$. Since $10=1\pmod{3}$ and $N$ is a power of $10$, $N=1\pmod{3}$. Hence $9^N=9\pmod{13}$.
Since $5^2=-1\pmod{13}$, $5^4=1\pmod{13}$. Since $9=1\pmod{4}$ and $M$ is a power of $9$, $M=1\pmod{4}$. Hence $5^M=5\pmod{13}$.
Finally, $9^N-5^M=9-5=4\pmod{13}$.
|
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|
kronecker product property I'm confused about this concept. Is this right? If not what is the correct version
$$
\begin{bmatrix}\mathbf{I}\otimes\mathbf{x}_{1}^{\prime}\\
\vdots\\
\mathbf{I}\otimes\mathbf{x}_{i}^{\prime}\\
\vdots\\
\mathbf{I}\otimes\mathbf{x}_{n}^{\prime}
\end{bmatrix}=\mathbf{I}\otimes\begin{bmatrix}\mathbf{x}_{1}^{\prime}\\
\vdots\\
\mathbf{x}_{i}^{\prime}\\
\vdots\\
\mathbf{x}_{n}^{\prime}
\end{bmatrix}
$$
Thanks in advance for your help.
|
We have
$$\begin{bmatrix} \mathbf I \otimes \mathbf x_1' \\ \vdots \\ \mathbf I \otimes \mathbf x_n' \end{bmatrix} =
\begin{bmatrix}
\mathbf y_1 \\ \vdots \\ \mathbf y_n
\end{bmatrix}$$
where
$$\mathbf y_i = \mathbf I \otimes \mathbf x_i' =
\begin{bmatrix}
\mathbf x_i' & 0 & 0 & \cdots & 0 \\
0 & \mathbf x_i' & 0 & \cdots & 0 \\
0 & 0 & \mathbf x_i' & \cdots & 0 \\
\vdots & \vdots & & \ddots & \vdots \\
0 & 0 & 0 & \cdots & \mathbf x_i'
\end{bmatrix}$$
and
$$\mathbf I \otimes
\begin{bmatrix} \mathbf x_1' \\ \vdots \\ \mathbf x_n' \end{bmatrix} =
\mathbf I \otimes
\mathbf X =
\begin{bmatrix}
\mathbf X & 0 & 0 & \cdots & 0 \\
0 & \mathbf X & 0 & \cdots & 0 \\
0 & 0 & \ddots & \cdots & \vdots \\
\vdots & \vdots & & \ddots\\
0 & 0 & 0 & \cdots & \mathbf X
\end{bmatrix}
=
\begin{bmatrix}
\mathbf x_1' & 0 & \cdots & 0 \\
\mathbf x_2' & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots \\
\mathbf x_n' & 0 & \cdots & 0 \\
0 & \mathbf x_1' & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots \\
0 & \mathbf x_n' & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots \\
0 & 0 & \cdots & \mathbf x_1' \\
\vdots & \vdots & \vdots & \vdots \\
0 & 0 & \cdots & \mathbf x_n'
\end{bmatrix}
$$
So no, they are not the same thing.
|
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|
Simultaneous Differential Equations How can we solve the simultaneous equations:
$$\frac{d}{dt}\left[\frac{1}{\sqrt{1-x^2-y^2}}\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}}\right]=\frac{x\sqrt{\dot x^2+\dot y^2}}{(1-x^2-y^2)^\frac{3}{2}}$$
$$\frac{d}{dt}\left[\frac{1}{\sqrt{1-x^2-y^2}}\frac{\dot y}{\sqrt{\dot x^2+\dot y^2}}\right]=\frac{y\sqrt{\dot x^2+\dot y^2}}{(1-x^2-y^2)^\frac{3}{2}}$$
I am hoping that the solution is $y=x$, fingers-crossed.
|
Take the difference between the two equations and get
$$\frac{d}{dt}\left[\frac{1}{\sqrt{1-x^2-y^2}}\frac{\dot x-\dot y}{\sqrt{\dot x^2+\dot y^2}}\right]=\frac{(x-y)\sqrt{\dot x^2+\dot y^2}}{(1-x^2-y^2)^\frac{3}{2}}$$
and the result follows straightforwardly.
|
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"url": "https://math.stackexchange.com/questions/112813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Limit of $ u_{n}=\sin(\frac{1}{n+1})+\cdots+\sin(\frac{1}{2n})$ How can I find the limit of
$$ u_n =\sin\left(\frac{1}{n+1}\right)+\cdots+\sin\left(\frac{1}{2n}\right)$$
when $n\rightarrow\infty$?
We have:
$$ \sum_{n=1}^\infty u_{n+1}-u_n =u_\infty -\sin\left(\frac{1}{2}\right)$$
So how can I find $$ \sum_{n=1}^\infty u_{n+1}-u_n =\sum_{n=1}^\infty \sin\left(\frac{1}{2n+2}\right)+\sin\left(\frac{1}{2n+1}\right)-\sin\left(\frac{1}{n+1}\right)\ ?$$
|
We can make use of the fact that $$\sin x = \sum\limits_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
and note that when $x\leq 1$, the absolute value of each term is more than the sum of the absolute values of the later terms. Thus $x-\frac{x^3}{6}<\sin x<x$ for $x\leq 1$, so we have $$\sum\limits_{k=n+1}^{2n}\frac{1}{k}-n\frac{1}{6n^3}<\sum\limits_{k=n+1}^{2n}\frac{1}{k}-\frac{1}{6k^3}<u_n<\sum\limits_{k=n+1}^{2n}\frac{1}{k}$$
and $$\begin{eqnarray}\lim\limits_{n\to\infty}\sum\limits_{k=n+1}^{2n}\frac{1}{k}&=&\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^{2n}\frac{1}{k}-\ln(2n)\right)-\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^{n}\frac{1}{k}-\ln(n)\right)+\lim\limits_{n\to\infty}(\ln(2n)-\ln(n))\\
&=&\lim\limits_{n\to\infty}(\ln(2n)-\ln(n))\\
&=&\lim\limits_{n\to\infty}\ln 2\\
&=&\ln 2\end{eqnarray}$$
while clearly $\lim\limits_{n\to\infty}n\frac{1}{6n^3}=0$, so by squeeze theorem $\lim\limits_{n\to\infty} u_n=\ln 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/113401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Prove that $\varlimsup _{n\rightarrow \infty}(u_n)^{\frac{1}{n}}=1$ where $u_{n+1}=\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}$ and $u_0=1$. Prove that
$$\varlimsup_{n\rightarrow \infty} (u_n)^{\frac{1}{n}}=1,$$
where $u_0=1$ and $$u_{n+1}=\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}\;.$$
|
We can do even better: $\lim\limits_{n\to\infty} (u_n)^{1/n}=1$. Note that for $x>0$ we have
$$\frac{d}{dx}\frac{2x^3+2x^2+x}{2x^2+3x+1}=\frac{4x^4+12x^3+10x^2+4x+1}{(2x^2+3x+1)^2}>0$$
thus since clearly each $u_n>0$, $\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}$ is an increasing function of $u_n$. This allows us to show that $u_n\geq \frac{1}{n+1}$ for all $n\in\mathbb N$, as $u_0=\frac{1}{1}$ and if $u_n\geq \frac{1}{n+1}$ then
$$\begin{eqnarray}
u_{n+1}&=&\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}\\
&\geq& \frac{2/(n+1)^3+2/(n+1)^2+1/(n+1)}{2/(n+1)^2+3/(n+1)+1}\\
&\geq&\frac{1}{n+1}-\frac{1}{(n+1)^2+3(n+1)+2}\\
&\geq& \frac{1}{n+1}-\frac{1}{(n+1)^2+(n+1)}\\
&\geq&\frac{1}{n+2}\end{eqnarray}$$
and so by induction $u_n\geq \frac{1}{n+1}$ for all $n$. Since it is clear that $(u_{n})^{1/n}\leq 1$ for all $n$, we need only show that $\lim\limits_{x\to\infty}\left(\frac{1}{x+1}\right)^{1/x}=1$, or equivalently (since $\ln x$ is continuous) that $$\lim\limits_{x\to\infty}\ln\left(\left(\frac{1}{x+1}\right)^{1/x}\right)=\ln 1=0$$
which we can do using L'Hospital's rule
$$\lim\limits_{x\to\infty}\ln\left(\left(\frac{1}{x+1}\right)^{1/x}\right)=\lim\limits_{x\to\infty}\frac{1}{x}\ln\left(\frac{1}{x+1}\right)=\lim\limits_{x\to\infty}\frac{-1}{x+1}=0$$
and so desired result follows from squeezing $\lim\limits_{n\to\infty}(u_n)^{1/n}$ between $\lim\limits_{n\to\infty} \left(\frac{1}{n+1}\right)^{1/n}$ and $\lim\limits_{n\to\infty} 1$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Simplifying an expression $\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$ if we know $x+y+z=0$ The following expression is given:
$$\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$$
Simplify it, knowing that $x+y+z=0$.
|
First $(x^3 + y^3 +z^3)(x+y+z) = x^4 + y^4 + z^4 + xy^3 + yx^3 + xz^3 +zx^3 +yz^3+y^3z = 0 $ which means
$$ x^4 + y^4 + z^4 = -xy(x^2+y^2) - yz(y^2+z^2) -zx (z^2 + x^2) \ \ \text{(1)}$$
$x+y+z=0$ also implies $x^2+y^2=z^2-2xy$, substitute all the sum of square we have
$ x^4 + y^4 + z^4 = -xy(z^2-2xy) - yz(x^2-2xz)-zx(y^2-2zx)=-xyz(x+y+z)+2(x^2y^2+y^2z^2+z^2x^2) $
So
$$ x^4 + y^4 + z^4=2(x^2y^2+y^2z^2+z^2x^2) \ \ \text{(2)}$$
Next consider $ x^3+y^3-(x+y)^3 = -3xy(x+y) $ (this is a basic identity)
So we have $ x^3 + y^3 +z^3 = 3xyz $ for $x+y+z = 0$
$3xyz(x^4+y^4+z^4)=(x^3 + y^3 +z^3)((x^4 + y^4 +z^4)=x^7+y^7+z^7+ x^4y^3 + y^4x^3 + x^4z^3 +z^4x^3 +y^4z^3+y^3z^4 = x^7+y^7+z^7 +x^3y^3(x+y)+y^3z^3(y+z)+z^3x^3(z+x)$
By substitution with $\text{(1)}$ and $\text{(2)}$,
$$3xyz(x^4+y^4+z^4) =x^7+y^7+z^7-xyz(x^4+y^4+z^4)/2 $$
Therefore the fraction is $\frac{7}{2}$
Wow it's not this long in my thought.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing $\zeta(6)=\sum\limits_{k=1}^\infty \frac1{k^6}$ with Fourier series. Let $ f$ be a function such that $ f\in C_{2\pi}^{0}(\mathbb{R},\mathbb{R}) $ (f is $2\pi$-periodic) such that $ \forall x \in [0,\pi]$: $$f(x)=x(\pi-x)$$
Computing the Fourier series of $f$ and using Parseval's identity, I have computed $\zeta(2)$ and $\zeta(4)$.
How can I compute $ \zeta(6) $ now?
Fourier series of $ f $:
$$ S(f)= \frac{\pi^2}{6}-\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n^2}$$
$$ x=0, \zeta(2)=\pi^2/6$$
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One method is to consider the generating function of $\zeta(2k)$:
$$
\begin{align}
f(x)
&=\sum_{k=1}^\infty\zeta(2k)\,x^{2k}\\
&=\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{x^{2k}}{n^{2k}}\\
&=\sum_{n=1}^\infty\frac{x^2/n^2}{1-x^2/n^2}\\
&=\sum_{n=1}^\infty\frac{x^2}{n^2-x^2}\\
&=-\frac{x}{2}\sum_{n=1}^\infty\left(\frac{1}{x-n}+\frac{1}{x+n}\right)\\
&=-\frac{x}{2}\left(\pi\cot(\pi x)-\frac1x\right)\\
&=\frac12(1-\pi x\cot(\pi x))\tag{1}
\end{align}
$$
In light of equation $(1)$, find the power series of
$$
x\cot(x)=\sum_{k=0}^\infty a_kx^{2k}
$$
$$
\cos(x)=\frac{\sin(x)}{x}\sum_{k=0}^\infty a_kx^{2k}
$$
$$
\begin{align}
\sum_{n=0}^\infty(-1)^n\!\frac{x^{2n}}{(2n)!}
&=\sum_{n=0}^\infty(-1)^n\!\frac{x^{2n}}{(2n+1)!}\;\;\sum_{k=0}^\infty a_kx^{2k}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n(-1)^k\!\frac{a_{n-k}}{(2k+1)!}\right)x^{2n}\tag{2}
\end{align}
$$
Comparing the coefficients of the powers of $x$ in $(2)$ yields
$$
\begin{align}
a_n
&=\frac{(-1)^n}{(2n)!}-\sum_{k=1}^n(-1)^k\!\frac{a_{n-k}}{(2k+1)!}\\
&=\frac{(-1)^n2n}{(2n+1)!}-\sum_{k=1}^{n-1}(-1)^k\!\frac{a_{n-k}}{(2k+1)!}\tag{3}
\end{align}
$$
Since $a_n=-2\dfrac{\zeta(2n)}{\pi^{2n}}$ for positive $n$, $(3)$ becomes
$$
\zeta(2n)=\frac{(-1)^{n-1}\pi^{2n}}{(2n+1)!}n+\sum_{k=1}^{n-1}\!\frac{(-1)^{k-1}\pi^{2k}}{(2k+1)!}\zeta(2n-2k)\tag{4}
$$
Equation $(4)$ gives $\zeta(2n)$ recursively for positive $n$:
$$
\begin{align}
\zeta(2)&=\frac{\pi^2}{3!}=\frac{\pi^2}{6}\\
\zeta(4)&=-\frac{\pi^4}{5!}2+\frac{\pi^2}{3!}\zeta(2)=\frac{\pi^4}{90}\\
\zeta(6)&=\frac{\pi^6}{7!}3-\frac{\pi^4}{5!}\zeta(2)+\frac{\pi^2}{3!}\zeta(4)=\frac{\pi^6}{945}\\
\zeta(8)&=-\frac{\pi^8}{9!}4+\frac{\pi^6}{7!}\zeta(2)-\frac{\pi^4}{5!}\zeta(4)+\frac{\pi^2}{3!}\zeta(6)=\frac{\pi^8}{9450}
\end{align}
$$
|
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"url": "https://math.stackexchange.com/questions/115981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Partial Fraction Decomposition of $\frac{x^4+2}{x^5+6x^3}$ I tried answering the following question but I'm getting it wrong for some reason. I would appreciate any help.
$$\frac{x^4+2}{x^5+6x^3}$$
My answer:
$$\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{Dx+E}{x^3}+\frac{Fx+G}{x^2+6}$$
What am I doing wrong?
|
There are two ways to deal with a repeated factor in the denominator, such as your $x^3$:
*
*You can put a single fraction, with denominator the full repeated factor, and undetermined denominator of degree one less. In your example, since the denominator factors as $x^3(x^2+6)$, you would set up the partial fractions as
$$\frac{Ax^2+Bx+C}{x^3} + \frac{Dx+E}{x^2+6}.$$
*While the above works, for the purposes of integration you will then proceed to split up the first fraction and deal with it as
$$\frac{Ax^2+Bx+C}{x^3} = \frac{Ax^2}{x^3} + \frac{Bx}{x^3} + \frac{C}{x^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3};$$
so the second way of dealing with repeated factors is to simply to that to being with: if you have a repeated factor $(x-a)^n$, you set it up with $n$ fractions, each with constant numerator:
$$\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2}+\cdots+\frac{A_n}{(x-a)^n};$$
with a repeated irreducible quadratic factor, $(x^2+ax+b)^n$, it's
$$\frac{A_1x+B_1}{x^2+ax+b} + \frac{A_2x+B_2}{(x^2+ax+b)^2} + \cdots + \frac{A_nx+B_n}{(x^2+ax+b)^n}.$$
In your case, with $x^3$, you would set up that part as
$$\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3}.$$
Mind you: if you do the algebra right with your set-up, you will get the right answer; you'll just work a lot harder, and end up with the conclusion that $B=D=0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How does one find the inequality $\tau(n) < (\frac{2}{\log 2}\log n)^{2^x}n^{\frac{1}{x}}$? Context (Though I don't see how this might help deriving the final inequality):
Example : For $p$ prime, $x$ greater than $0$ a real number, $n$ greater or equal to $2$ an integer: for $p< 2^{x}$ it holds that $$1+\text{ord}_{p}(n)\le 1+ \frac{\log n}{\log p} \le 1+ \frac{\log n}{\log 2} \le \frac{2}{\log 2}\log n$$ and for $p \ge 2^{x}$ it holds that: $$1+\text{ord}_{p}(n)\le 2^{\text{ord}_{p}(n)}\le p^{\text{ord}_{p}(n)/x}.$$
(no proof supplied from the original author, so I will try to do one myself.)
For $p<2^{x}$:
$$1+\text{ord}_{p}(n)\le 1+ \frac{\log n}{\log p} \tag{1}$$
$$1+ \frac{\log n}{\log p} \le 1+ \frac{\log n}{\log 2} \tag{2}$$
$$1+ \frac{\log n}{\log 2} \le \frac{2}{\log 2}\log n \tag{3}$$
(2) holds because $2$ is the smallest prime. (3) because the derivative of $\log(2x)$ is $\frac{1}{x}$ and $\frac{2}{x}$ (the derivative of $2\log(x)$) is greater for every $x$. I have trouble showing (1) because I don't understand how to handle the $\text{ord}_{p}(n)$ (which also appears in the second inequality). ($\text{ord}_{p}(n)$ is the highest natural exponent $k$ such that $p^{k}$ divides $n$).
For $p>2^{x}$:
$$1+\text{ord}_{p}(n)\le 2^{\text{ord}_{p}(n)} \tag{1}$$
$$2^{\text{ord}_{p}(n)}\le p^{\text{ord}_{p}(n)/x} \tag{2}$$
Right after this example, the author mentions an inequality and this is what I am really interested in:
For $n\ge 2$ an integer, $ x\in \mathbb{R_{>0}}$ , how does one find out that
$$\tau (n) < \left(\frac{2}{\log 2}\log n \right)^{2^x}n^{\frac{1}{x}}\ ?$$ It looks like a popular inequality, does it have a name?
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I think this might work. Given a real $x > 0$ partition the prime divisors of $n$ into two classes, those which are bounded from above by $2^{x}$ and those bounded from below by $2^x$. We have
\begin{align}
\tau(n) = \prod_{2^{x} \geqslant p \mid n}( \text{ord}_{p}(n) + 1) \prod_{2^{x} < p \mid n}( \text{ord}_{p}(n) + 1).
\end{align}
Writing $n = \prod_{p \mid n} p^{\text{ord}_{p}(n)}$,
\begin{align}
\log n = \sum_{p \mid n} \text{ord}_p(n) \log p \quad \Longrightarrow \quad \frac{\log n}{\log p} = \text{ord}_{p}(n) + \sum_{p \neq p^{\prime} \mid n} \text{ord}_{p^{\prime}}(n) \ \log_{p} \, p^{\prime},
\end{align}
so the inequality $\text{ord}_{p}(n) \leqslant \frac{\log n}{\log p}$ holds for any prime divisor of $n$. The bounds which you claim are sufficient to prove the upper bound on the number-of-divisors function. Observe that for $p \leqslant 2^{x}$ (actually, for any prime divisor of $n$), then
\begin{align}
\text{ord}_{p}(n) + 1 \leqslant \frac{\log n}{\log p} + 1 \leqslant \frac{ \log n}{\log 2} + 1 \leqslant \frac{2 \log n}{\log 2},
\end{align}
provided that $n \geqslant 2$ and, if $p > 2^{x}$, one has $\text{ord}_{p}(n) + 1 \leqslant 2^{\text{ord}_p(n)} \leqslant p^{\text{ord}_{p}(n)/x}$. Thus,
\begin{align}
\tau(n) & \leqslant \prod_{2^{x} \geqslant p \mid n} \frac{2 \log n}{\log 2} \, \prod_{2^{x} < p \mid n} p^{\text{ord}_{p}(n)/x} \\ & = \prod_{2^{x} \geqslant p \mid n} \frac{2 \log n}{\log 2} \, \left( \prod_{2^{x} < p \mid n} p^{\text{ord}_{p}(n)} \right)^{1/x}.
\end{align}
The first product has at most $2^{x}$ terms, while the second product has at most $\omega(n)$ terms. Hence,
\begin{align}
\tau(n) & \leqslant \left( \frac{2 \log n}{\log 2} \right)^{2^{x}} \left( \prod_{p \mid n} p^{\text{ord}_{p}(n)} \right)^{1/x} \\ & \leqslant \left( \frac{2 \log n}{\log 2} \right)^{2^{x}} n^{1/x}.
\end{align}
It should now be clear how one might sharpen the bound:
\begin{align}
\tau(n) \leqslant \left( \frac{2 \log n}{\log 2} \right)^{\omega(n)} n^{1/x},
\end{align}
where $\omega$ is the number-of-prime-divisors function.
In any event, taking the original inequality, we see
\begin{align}
\frac{\log \tau(n)}{\log n} \leqslant \frac{2^x \log(\frac{2\log n}{\log 2})+\frac{1}{x} \log n}{\log n} =\frac{2^x \log ( \frac{2 \log n}{\log 2})}{\log n}+\frac{1}{x}.
\end{align}
For any $x > 0$ there is an integer $n_0 = n_{0}(x)$ such that for any $n > n_0$, $\frac{\log \tau(n)}{\log n}$ is bounded from above by $\frac{2}{x}$, and so the limit must vanish.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a+b+c+d=16$, then $(a+\frac{1}{c})^2+(c+\frac{1}{a})^2 + (b+\frac{1}{d})^2 + (d+\frac{1}{b})^2 \geq \frac{289}{4}$
If $a,b,c,d$ are positive integers and $a+b+c+d=16$, prove that
$$\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2 \geq \frac{289}{4}.$$
I know I have to use some inequality, not sure AM GM will work here or Minkowski inequality. But I only need hints, not a complete solution. I want to work on it myself.
Please provide only Hints.
|
First apply the Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality (from http://tinyurl.com/84o57u4)
$$\sqrt{\frac{\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2}{4}} \geq \frac{\left(a+\frac{1}{c}\right)+\left(c+\frac{1}{a}\right)+\left(b+\frac{1}{d}\right)+\left(d+\frac{1}{b}\right)}{4}$$
Square both sides
$$\frac{\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2}{4} \geq \frac{\left( (a+b+c+d)+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}) \right)^2}{16}$$
Since we already know the value of $a+b+c+d=16$, apply Arithmetic Mean, Harmonic Mean inequality, i.e.
$$ \frac{a+b+c+d}{4} \geq \frac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}$$
And take it from there, since you said you only need hints.
|
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|
How can we show that $\pi (x+y) - \pi(y) \le \frac{1}{3} x + C$ using the sieve of eratosthenes?
How do we show that For $x,y \ge 0$ real numbers, there exists a constant C suchthat: $$\pi(x+y)-\pi(y) \le \frac{1}{3}x+C$$ Where $\pi(.)$ denotes thes prime counting function, is true?
the hint is to sieve n with $y< n \le x+y $:
$$\pi (x+y) \le 1+ \sum _{n \le x+y} 1+1-1 - \sum_{2|n}1 - \sum_{3|n}1 + \sum_{6|n}1 + \sum_{n\le x+y} 1 = $$
$$1+ \sum _{n \le x+y} 1+1-1 - \sum_{2|n}1 - \sum_{3|n}1 + \sum_{6|n}1 + [x+y]$$
because: $1\le n = dm \le x+y \Leftrightarrow \frac{1}{d}\le m \le \frac{x+y}{y} $
so: $$\sum_{n\le x+y , d|n}1 = [\frac{x+y}{d}]$$
then that gives: $$\pi (x+y) < 1+ [x+y] - [\frac{x+y}{2}] - [\frac{x+y}{3}] + [\frac{x+y}{6}]$$
so that will give: $\pi (x+y) < \frac{x+y}{3} + 3$ but also we get : $\pi(y) < \frac{y}{3} + 3$ so for any constant $C\ge 0$ it will surely hold that:
$$\pi(x+y) - \pi(y) < \frac{x}{3} \le \frac{x}{3} + C$$
Is this correct?
|
Hint: Which numbers modulo $6$ can be prime? (Certainly not those divisible by $2$ or $3\ldots$)
So on an interval of width $x$ from $y$ to $y + x$, how many primes ($\pi(y+x) - \pi(y)$) do we expect at most on this interval?
|
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|
Prove the equation without solving for X My niece asked me this - If $x=1/(5-x)$ prove $x^3 + \dfrac{1}{x^3}=110$ without solving for x. I said its not possible since solving for x itself gives me two roots for x (one being $\approx4.79$) and substituting for it in the second equation approx gives me 110. So proving algebraically without any assumptions is not possible. Is this right ?
|
From your equation you get $5x-x^2=1$, so $5=\frac{x^2+1}{x}$. Now remind the following cubic fromula $(a+b)^3=a^3+b^3+3ab(a+b)$, then
$$
5^3=\left(\frac{x^2+1}{x}\right)^3=\left(x+x^{-1}\right)^3=x^3+x^{-3}+3xx^{-1}(x+x^{-1})=
$$
$$
x^3+x^{-3}+3\frac{x^2+1}{x}=x^3+x^{-3}+15.
$$
So you get
$$
x^3+x^{-3}=125-15=110
$$
|
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|
Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what method to use.
|
$10^{0} +3⋅4^{0} +5=\color{red}{9}$ and $10^{n+1} +3⋅4^{n+1} +5=10^n +3⋅4^n +5+\color{red}{9}\cdot(10^n+4^n)$ for every $n\geqslant0$.
|
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|
First order differential equation homework help Hi I'm really stuck with some homework:
Find the general solution of the differential equation,
$$\left(x+\dfrac{1}x\right)\dfrac{dy}{dx} + 2y = 2\left(x^2+1\right)^2$$
So far, I've divided both sides by $x+\dfrac{1}x$ and integrated $\dfrac{2y}{x + \frac{1}{x}}$ to get $y \ln\left(x^2+1\right)$ but have no idea where to go from here.
Anyone know what I need to do next?
|
$$
\begin{align*}
(x+(1/x))\frac{dy}{dx} +2y &= 2(x^2+1)^2\\
\left(\frac{x^2+1}{x}\right)\frac{dy}{dx} +2y &= 2(x^2+1)^2\\
\frac{dy}{dx} +\frac{2xy}{x^2+1} &= 2x(x^2+1) \tag{A}\\
\frac{dy}{dx} +P(x)y &= Q(x)\\
\end{align*}
$$
where $\displaystyle{P(x) = \frac{2x}{x^2+1}}$ and $\displaystyle{Q(x) = 2x(x^2+1)}$
$$\displaystyle{\int \frac{2x}{x^2+1} = \ln(x^2+1)}$$
The integrating factor = $\displaystyle{e^{\int P(x) dx}}$ which is
$\displaystyle{e^{\ln(x^2+1)}} = x^2+1$ (why?)
$(A)$ simplifies to
$$\frac{d}{dx}\left( (1+x^2)y \right) = 2x(1+x^2)$$
I could finish it completely, but can you figure the rest (by integrating both sides)?
|
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|
Gaussian proof for the sum of squares? There is a famous proof of the Sum of integers, supposedly put forward by Gauss.
$$S=\sum\limits_{i=1}^{n}i=1+2+3+\cdots+(n-2)+(n-1)+n$$
$$2S=(1+n)+(2+(n-2))+\cdots+(n+1)$$
$$S=\frac{n(1+n)}{2}$$
I was looking for a similar proof for when $S=\sum\limits_{i=1}^{n}i^2$
I've tried the same approach of adding the summation to itself in reverse, and I've found this:
$$2S=(1^2+n^2)+(2^2+n^2+1^2-2n)+(3^2+n^2+2^2-4n)+\cdots+(n^2+n^2+(n-1)^2-2(n-1)n$$
From which I noted I could extract the original sum;
$$2S-S=(1^2+n^2)+(2^2+n^2-2n)+(3^2+n^2-4n)+\cdots+(n^2+n^2-2(n-1)n-n^2$$
Then if I collect all the $n$ terms;
$$2S-S=n\cdot (n-1)^2 +(1^2)+(2^2-2n)+(3^2-4n)+\cdots+(n^2-2(n-1)n$$
But then I realised I still had the original sum in there, and taking that out mean I no longer had a sum term to extract.
Have I made a mistake here? How can I arrive at the answer of $\dfrac{n (n + 1) (2 n + 1)}{6}$ using a method similar to the one I expound on above? I.e following Gauss' line of reasoning?
|
You can use something similar, though it requires work at the end.
If $S_n = 1^2 +2^2 + \cdots + n^2$ then
$$S_{2n}-2S_n = ((2n)^2 - 1^2) + ((2n-1)^2-2^2) +\cdots +((n+1)^2-n^2)$$
$$=(2n+1)(2n-1 + 2n-3 + \cdots +1) = (2n+1)n^2$$ using the Gaussian trick in the middle.
Similarly $$S_{2n+1}-2S_n = (2n+1)(n+1)^2$$
So for example to work out $S_9$, you start
$$S_0=0^2=0$$
$$S_1=1 + 2S_0 = 1$$
$$S_2=3+2S_1=5$$
$$S_4=25+2S_2=30$$
$$S_9 = 225+2S_4 = 285$$
but clearly there are easier ways.
|
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|
How to find the partial sum of a given series? On my last exam there was the question if the series $\sum_{n=2}^{\infty}\frac{1}{(n-1)n(n+1)}$ converges and which limit it has. During the exam and until now, I am not able to solve it. I tried partial fraction decomposition, telescoping sum, etc. But I am not able to find the partial sum formula (Wolfram|Alpha):
$$
\sum_{n=2}^{m}\frac{1}{(n-1)n(n+1)} = \frac{m^2+m-2}{4m(m+1)}.
$$
Could somebody push me in the right direction? Is there any trick or scheme how to find partial sum formulas for given series?
|
So let's try partial fraction decomposition. Writing
$$
\frac 1{(n-1)n(n+1)} = \frac a{n-1} + \frac bn + \frac c{n+1}
$$
we obtain
$$
1 = a(n^2 + n) + b(n^2 - 1) + c(n^2 - n)
$$
and therefore
\begin{align*}
1 &= -b\\
0 &= a - c\\
0 &= a + b + c.
\end{align*}
This gives $b = -1$, $a = c = \frac 12$. Hence
\begin{align*}
\sum_{n=2}^m \frac 1{(n-1)n(n+1)}
&= \sum_{n =2}^m \frac 1{2(n-1)} - \sum_{n=2}^m \frac 1n + \sum_{n=2}^m \frac 1{2(n+1)}\\
&= \frac 12 + \sum_{n=2}^{m-1} \frac 1{2n} - \sum_{n=2}^m \frac 1n
+ \sum_{n=3}^m \frac 1{2n} + \frac 1{2(m+1)}\\
&= \frac 12 + \frac 14 - \frac 12 - \frac 1m + \frac 1{2m} + \frac 1{2m+2}\\
&= \frac 14 + \frac{-2(m+1) + m+1 + m}{2m(m+1)}\\
&= \frac 14 + \frac{-1}{2m(m+1)}\\
&= \frac{m(m+1) - 2}{4m(m+1)}.
\end{align*}
|
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|
Prove that $\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$ All the angles in a triangle $A,B,$ and $C$ are less than $120^{o}$
Prove that $\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$
|
Consider triangle with angles $\small{A_1=120-A, B_1=120-B, C_1=120-C}$
Applying the triangle inequality in this triangle with angles $A_1, B_1,$ and $C_1$,
$$\small{B_1 C_1+C_1 A_1 > A_1 B_1}$$
$$\small{\sin A_1 +\sin B_1 > \sin C_1}$$
$$\small{\sin (120-A) +\sin (120-B) > \sin (120-C)}$$
which by applying $\sin(x-y)$ identities
$$\small{\frac{\sqrt{3}}{2}(\cos A+\cos B-\cos C) + \frac{1}{2}(\sin A+ \sin B-\sin C) > 0} \tag{1}$$
And since $\small{a+b > c , \sin A+\sin B - \sin C > 0}$, and therefore dividing by this is perfectly legitimate
Using this observation and re-writing $(1)$, we obtain
$$ \small{\frac{\sqrt{3}}{2} \frac{\cos A + \cos B -\cos C}{\sin A + \sin B - \sin C} + \frac{1}{2} > 0}$$
$$ \small{\Rightarrow \frac{\cos A + \cos B -\cos C}{\sin A + \sin B - \sin C} > \frac{-\sqrt{3}}{3}}$$
|
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|
Find if the sequence $n!/10^{6n}$ converges Find if the sequence $a_n$ converges, $a_n=\frac{n!}{10^{6n}}=\frac{1}{\frac{10^6n}{n!}}$. Now if we take the limit of numerator and denominator as $n \to \infty$, we get $\frac{1}{0}$($\lim\limits_{n\to\infty} \frac{x^n}{n!}=0 $ for any $x$), but division by zero is not defined, so we are stuck, but the answer is $\infty$. Where am I going wrong?
|
Hint for an alternative route,
$$\rm \frac{a_{\large n+10^6}}{a_{\large 10^6}}\normalsize =\frac{10^6+1}{10^6}\cdot\frac{10^6+2}{10^6}\cdots\frac{10^6+n}{10^6}>\left(1+\frac{1}{10^6}\right)^n \to \infty.$$
|
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|
solution to $ 7^{a}+1 =3^{b}+5^{c} $ for natural $a$,$b$ and $c$ How do I solve $ 7^{a}+1 =3^{b}+5^{c} $ for natural $a$,$b$ and $c$?All I got after some modular arithmetic is that the $a$,$b$ and $c$ are all odd.The problem was posted on Art of Problem Solving(with no responses till now) and is supposedly from India International Mathematical Olympiad Training Camp.I was wondering if someone could please shed some insight as to whether it can be solved.
Thanks.(I am a bit skeptical about this problem although I may be wrong)
|
It looks as if the problem can be conquered by persistent use of modular arithmetic. Throughout, it is assumed that $a,b,c$ are sufficiently large.
*
*Apply $\bmod{3}$ to find that
$$ 2 \equiv 2^c \pmod{3} $$
It follows that $c \equiv 1 \pmod{2}$. Write $c = 2c_1 + 1$.
*Apply $\bmod{8}$ to find that:
$$ (-1)^a + 1 \equiv 3^b + 5 \pmod{8} $$
It follows that $a \equiv b \equiv 1 \pmod{2}$. Write $a = 2a_1 + 1, \ b = 2b_1 + 1$.
*Apply $\bmod{5}$ to find that:
$$ 2 \cdot (-1)^{a_1} +1 \equiv 3 \cdot (-1)^{b_1} \pmod{5} $$
It follows that $a_1 \equiv b_1 \equiv 0 \pmod{2}$. Write $a_1 = 2a_2,\ b_1 = 2b_2$. Th
*Apply $\bmod{16}$ to find that:
$$ 7+1 \equiv 3 + 5 \cdot 9^{c_1} \pmod{16} $$
It follows that $c \equiv 0 \pmod{2}$. Write $c_1 = 2c_2$.
The equation is now:
$$ 7^{4a_2 + 1} + 1 = 3^{4b_2+1} + 5^{4c_2+1} $$
*Apply $\bmod{13}$ [note: the number $13$ comes from the observation that $5^4 \equiv 1 \pmod{13}$ and $\phi(13) = 12$ is divisible by $4$] to find that:
$$ 7^{4a_2 + 1} + 1 \equiv 3^{4b_2+1} + 5 \pmod{13} $$
By Fermat's theorem, only the residues of $4a_2 + 1$ and $4b_2+1$ in arithmetic $\bmod \phi(13)$ play a role in the above congruence, so we have just $\frac{\phi(13)}{\gcd(4,\phi(13))} = 3$ values of $a_2,b_2$ to check. It turns out that $a_2,b_2 \equiv 0 \pmod{3}$ are the only solutions. Write $a_2 = 3 a_3$ and $b_2 = 3b_3$.
The equation is now:
$$ 7^{12a_3 + 1} + 1 = 3^{12b_3+1} + 5^{4c_2+1} $$
*Apply $\bmod{9}$ to find that:
$$ 7 + 1 \equiv 5\cdot4^{c_2} \pmod{9} $$
By checking the $6$ possible values of $c_2$, one finds that $c_2 \equiv 2 \pmod{3}$. Write $c_2 = 3c_3 + 9$.
The equation is now:
$$ 7^{12a_3 + 1} + 1 = 3^{12b_3+1} + 5^{12c_3+9} $$
*Apply $\bmod{37}$. Again, it is a reasonable guess because $\phi(37) = 36$ has many common factors with the $12$ that occurs in the exponentials, so the resulting congruence depends only on $a_3,b_3,c_3 \pmod{3}$. A rather mudane [I advise against attempting to do it by hand] direct check proves that there are no solutions.
*We have shown that there are no solutions with the assumption that all numbers are "sufficiently large". One should now check how large the numbers really have to be, and look for small solutions. This will be significantly easier, since at least one of $a,b,c$ will be fixed at some small value. However, a lot of cases will be involved.
|
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|
Limit when circumference shrinks Let $C_1$ be a fixed circumference with equation $(x-1)^2 + y^2 = 1$ and $C_2$ a
circumference to be shrinked, with center at $(0, 0)$ and radius $r$.
Let $P$ be the point $(0, r)$, $Q$ the upper intersection between $C_1$ and $C_2$
and $R$ the intersection between the line $PQ$ with the $x$ axis.
What happens with $R$ when $C_2$ shrinks (i.e., $r \rightarrow 0^+$) ?
|
The upper intersection point $T=(a,b)$ is on both $C_1$ and $C_2$ and has positive $y$-component. Thus
$$\begin{cases}a^2+b^2=r^2 \\ (a-1)^2+b^2=1. \end{cases}$$
Subtracting the first from the second, we obtain $1-2a=1-r^2$, and thus $a=r^2/2$. We can then solve for $b$ as $b=\sqrt{r^2-(r^2/2)^2}=r\sqrt{1-(r/2)^2}$. Therefore $T=\left(r^2/2,r\sqrt{1-(r/2)^2}\right)$. The line going between $T$ and $(0,r)$ is given by
$$\frac{y-r}{x-0}=\frac{b-r}{a-0}=\frac{r\sqrt{1-(r/2)^2}-r}{r^2/2-0}.$$
The $x$-intercept occurs when we set $y=0$; solving for $x$ gives us
$$x=\frac{-r(r^2/2)}{r\sqrt{1-(r/2)^2}-r} \cdot \frac{1+\sqrt{1-(r/2)^2}}{1+\sqrt{1-(r/2)^2}}=2\left(1+\sqrt{1-(r/2)^2}\right).$$
Clearly as $r\to0$, $x\to 4$.
|
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|
Maximize volume of box in ellipsoid I need to find the dimensions of the box with maximum volume (with faces parallel to the coordinate planes) that can be inscribed in ellipsoid
$$\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1$$
A hint given was:
If vertex of box in first octants is (x,y,z) then volume is 8xyz.
So I first find the partial derivatives
$V_x = 8yz$
$V_y = 8xz$
$V_z = 8xy$
Now the partial derivatives for constraint:
$g_x = \lambda\frac{x}{2}$
$g_y = \lambda\frac{2y}{9}$
$g_z = \lambda\frac{z}{8}$
Then $f_\alpha = \lambda g_\alpha$ so:
$8yz = \lambda\frac{x}{2} \rightarrow \lambda = \frac{16yz}{x}$
$8xz = \lambda\frac{2y}{9} \rightarrow \lambda = \frac{36xz}{y}$
$8xy = \lambda\frac{z}{8} \rightarrow \lambda = \frac{64xy}{z}$
$\frac{16yz}{x} = \frac{36xz}{y} \rightarrow y = \frac{3x}{2}$
$\frac{36xz}{y} = \frac{64xy}{z} \rightarrow y = \frac{3z}{4}$
$\frac{3x}{2} = \frac{3z}{4} \rightarrow x = \frac{z}{2}$
Now how do I continue? I seems to be missing something?
|
Very recently, you asked for advice on a proof of the AM/GM inequality in three variables. This inequality says that for $u$, $v$, and $w$ all $\ge 0$, we have
$$\frac{u+v+w}{3}\ge \sqrt[3]{uvw},$$
with equality iff $u=v=w$.
The result you want is a direct consequence of three variable AM/GM.
Let $u=\frac{x^2}{4}$, $v=\frac{y^2}{9}$, and $w=\frac{z^2}{16}$. We are told that $u+v+w=1$. It follows from AM/GM that
$$\frac{1}{3}\ge \sqrt[3]{uvw}=\sqrt[3]{\frac{x^2y^2z^2}{4\cdot 9\cdot 16}}\tag{$\ast$}$$
with equality if $\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}$.
From the inequality $(\ast)$, you can reach a conclusion about the maximum value of $xyz$ for $x$, $y$, $z$ non-negative. Eight times this maximum value gives the maximum volume, as per the hint.
|
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$3^k$ not congruent to $-1 \pmod {2^e}, e > 2$. $3^k \not\equiv -1 \pmod {2^e}$ for $e > 2, k > 0$. Is this true? I have tried to prove it by expanding $(1 + 2)^k$. [Notation: $(n; m) := n! / (m! (n - m)!)$] E.g., for $e = 3$ I get: $(1+2)^k + 1 = 2 + (k; 1) 2 + (k; 2) 2^2 + (k; e) 2^e + ...$ So, here it's enough to prove that $2^3$ does not divide $2 + (k; 1) 2 + (k; 2) 2^2$. The validity for general e seems very hard to prove.
|
Just a bit more explanative in case the other answers are too compressed.
We do this by induction. The remarkable property, that $\small 3^2=9=8+1$ suggests, to try this modulo 8 (but that could as well be suggested by the problem and it was asked that the factor $\small 2^e $ with $\small e=2 $ should not be exceeded)
*
*we observe for the first even exponent
$\small 3^2+1 = 10 = (8 \cdot 1 + 1) +1 $
*we try the general ansatz
$\small 3^k+1 = (8 m_0 + 1) +1 $
then
$\small \begin{eqnarray}
3^{k+2}+1 &=& \\
9 \cdot 3^k+1 &=& (8+1)(8 m_0 + 1) +1 \\
&=& 8 (8 m_0+1) + (8 m_0 + 1) +1 \\
&=& 8 (9 m_0+1) + 1 +1 \\ &=& (8 m_1+1)+1 \end{eqnarray}$
and see, that the structure persists when we increase the exponent by 2. Because that structure is true for exponent k+2 when it is true for k, and we know, that it is true for k=2, we know it is true for all even exponents.
*Then we look at the odd exponents and observe for the first one
$\small 3^1+1 = 4 = (8 \cdot 0+3)+1 $
*we do the general ansatz
$\small 3^k+1 = (8m+3)+1 $
and then
$\small \begin{eqnarray}
9 \cdot 3^k+1 &=& (8+1)(8m_0+3)+1 \\
&=& 8(8m_0+3)+1 \cdot 8m_0+1 \cdot 3+1 \\
&=& 8(9m_0+3)+4 \\
&=& (8m_1+3)+1 \end{eqnarray} $
In 2) and 4) we conclude, that we have for even exponents k $\small 3^k \equiv 1 \pmod 8$ and for odd exponents $\small 3^k \equiv 3 \pmod 8$ and thus $\small 3^k \equiv -1 \pmod 4$ or $\small 3^k \equiv -1 \pmod 2$ but not $\small \pmod 8$
|
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Integration Problem: $\int \frac{x^2+1}{x^5-1}dx$
$$\int \frac{x^2+1}{x^5-1}dx$$
I am unable to integrate it, nothing works. Yes, I can use partial fraction but who remembers factorization of $x^5-1$, I need a better way of doing this.
|
For these cyclotomic denominators, one can get an answer pretty fast by factoring into roots of unity. Let $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$, where $\theta_k=2\pi k/5$. Then
$$\frac{x^2+1}{x^5-1}=\frac{x^2+1}{\prod_{k=0}^4(x-\omega_k)}=\sum_{k=0}^4\frac{A_k}{x-\omega_k}$$
Obeserving that $\omega_k^{-k}=\omega_k^{5-k}$ and applying L'Hopital's rule a couple of times,
$$\begin{align}\lim_{x\rightarrow\omega_k}\frac{(x^2+1)(x-\omega_k)}{x^5-1} & =\lim_{x\rightarrow\omega_k}\frac{(x^2+1)(1)}{5x^4}=\frac15\left(\omega_k^{-2}+\omega_k\right) \\ & =\lim_{x\rightarrow\omega_k}\sum_{j=0}^4\frac{A_j\left(x-\omega_k\right)}{x-\omega_j}=\sum_{j=0}^4A_j\delta_{kj}=A_k\end{align}$$
Then
$$\begin{align}\int\frac{x^2+1}{x^5-1}dx & =\sum_{k=0}^4A_k\int\frac{dx}{x-\omega_k}=\sum_{k=0}^4\frac15\left(\omega_k^{-2}+\omega_k\right)\ln\left(x-\omega_k\right)+C_1 \\
& =\sum_{k=0}^4\left(\cos2\theta_k+\cos\theta_k-i\sin2\theta_k+i\sin\theta_k\right)\ln\left(x-\cos\theta_k-i\sin\theta_k\right)+C_1 \\
& =\sum_{k=0}^4\left(\cos2\theta_k+\cos\theta_k-i\sin2\theta_k+i\sin\theta_k\right)\left\{\frac12\ln\left(x^2-2x\cos\theta_k+1\right)+i\,\text{atan2}\left(-\sin\theta_k,x-\cos\theta_k\right)\right\}+C_1 \\
& =\sum_{k=0}^4\left(\cos2\theta_k+\cos\theta_k-i\sin2\theta_k+i\sin\theta_k\right)\left\{\frac12\ln\left(x^2-2x\cos\theta_k+1\right)+i\,\text{atan2}\left(x-\cos\theta_k,\sin\theta_k\right)\right\}+C \\
& = \sum_{k=0}^4\left\{\frac12\left(\cos2\theta_k+\cos\theta_k\right)\ln\left(x^2-2x\cos\theta_k+1\right)+\left(\sin2\theta_k-\sin\theta_k\right)\tan^{-1}\left(\frac{x-\cos\theta_k}{\sin\theta_k}\right)\right\}+C\end{align}$$
because the imaginary parts cancel out above. The $k=0$ term is just $\frac25\ln|x-1|$ and the $k=4$ term is a copy of the $k=1$ term, just as the $k=3$ term is a copy of the $k=2$ term. So substituting in the values of the trig functions above and simplifying, we get
$$\begin{align}\int\frac{x^2+1}{x^5-1}dx=\frac25\ln|x-1| & -\frac1{10}\ln\left(x^2+\frac{1-\sqrt5}2x+1\right) \\ & +\frac{\sqrt{10-2\sqrt5}\left(1-\sqrt5\right)}{20}\tan^{-1}\left(\frac{4x-\sqrt5+1}{\sqrt{10+2\sqrt5}}\right) \\& -\frac1{10}\ln\left(x^2+\frac{1+\sqrt5}2x+1\right) \\ & -\frac{\sqrt{10+2\sqrt5}\left(1+\sqrt5\right)}{20}\tan^{-1}\left(\frac{4x+\sqrt5+1}{\sqrt{10-2\sqrt5}}\right)+C\end{align}$$
|
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|
Integrating $\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta$ Can someone please help me integrate $$\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta$$ the question says, as a hint, use $\cos\theta = \frac{z + z^{-1}}{2}$ with $|z|=1$. I'm not really sure where to start.
|
An easy way to evaluate:
$I=\int_0^{2\pi}\frac{1}{1+8\cos^2\theta}d\theta$
$$I=4\int_0^{\frac{\pi}{2}}\frac{1}{8+\frac{1}{\cos^2\theta}}\frac{d\theta}{\cos^2\theta }= 4\int_0^{\frac{\pi}{2}}\frac{1}{9+\tan^2\theta}d\tan\theta= 4\int_0^{\infty}\frac{1}{3^2+t^2}dt= 2\pi/3$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculating the regression equations I have four data points $(1,2), (2,4), (3,5), (5,7)$ and Im looking for the least squares regression line that best fits them.
I use the normal equation
$A^tAx=A^tb$
in this form -
$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \\ 1 & 4 \\ \end{bmatrix}\begin{bmatrix} c \\ m \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 2 \\ 4 \\ 5 \\ 7 \\ \end{bmatrix}$
this gives -
$\begin{bmatrix} 4 & 10 \\ 10 & 30 \end{bmatrix}\begin{bmatrix} c \\ m \end{bmatrix} = \begin{bmatrix} 18 \\ 53 \end{bmatrix}$
I solved this system and got
$m = 1.3, c = 1.25$
so
$y = 1.3x + 1.25$
But if I put "linear fit {1,2},{2,4},{3,5},{4,7}" into wolfram alpha it gives
$1.6x + 0.5$
So have I got it wrong?
|
Everything looks good until
$\begin{bmatrix} 4 & 10 \\ 10 & 30 \end{bmatrix}\begin{bmatrix} c \\ m \end{bmatrix} = \begin{bmatrix} 18 \\ 53 \end{bmatrix}$, but then you have got bad solutions and wolframalpha is right. Also you should check last data point, is it $(4,7)$ or $(5,7)$?
|
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|
Sum the series: $ S = \frac{1}{2} \cdot \sin\alpha + \frac{1\cdot 3}{2 \cdot 4} \sin{2\alpha} + \cdots \ \text{ad inf}$ How do I sum the following series?
$$ S = \frac{1}{2} \cdot \sin\alpha + \frac{1\cdot 3}{2 \cdot 4} \sin{2\alpha} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \sin{3\alpha} + \cdots \ \text{ad inf}$$
|
The coefficients in your series can be written as,
$$\frac{1}{4^n} {2n \choose n}$$
This allows us to write your series a bit more compactly as,
$\sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} \sin(n\alpha)$
Consider the following series instead,
$$\sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} e^{in\alpha} = \sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} \left( e^{\frac{i\alpha}{2}} \right)^{2n}$$
Now,
$$\frac{1}{\sqrt{1 - x^2}} = \sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} x^{2n}.$$
So, our earlier series evaluates to,
$$\sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} \left( e^{\frac{i\alpha}{2}} \right)^{2n} = \frac{1}{\sqrt{1 - e^{i\alpha}}}$$
Therefore,
$$\sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} \sin(n\alpha) = Im\left( \frac{1}{\sqrt{1 - e^{i\alpha}}} \right)$$
|
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|
Generalised Binomial Theorem Intuition It was not until recently (why don't they teach it in secondary school?) that I've come across the Generalised Binomial Theorem, which from what I can tell is basically the same as the regular Binomial Theorem, except that the finite sum is replace by an infinite series:
$$
(x+y)^n=\sum^{n}_{r=0}\binom{n}{r}x^{n-r}y^r=\sum^{\infty}_{r=0}\binom{n}{r}x^{n-r}y^r
$$
Unfortunately, I wasn't able to find any clear explanation of how to get from the regular theorem to the generalised one, the only proof I found being based on some obscure mathematics, while my math book entirely skips the explanation.
Hence, my question is how do you prove the generalised theorem by deriving it from the regular theorem or otherwise?
|
Here is one straightforward approach, mentioned in Lang's Analysis book.
Let $ \alpha \in \mathbb{R} $ and $ x \in (-1,1) $.
We'll try to show $$ (1+x)^\alpha = 1 + \binom{\alpha}{1}x + \binom{\alpha}{2} x^2 + \ldots $$ holds (where for $ k \in \mathbb{Z}_{>0} $, $ \binom{\alpha}{k} := \frac{\alpha (\alpha - 1) \ldots (\alpha - (k-1))}{k!} $)
Writing $ (1+x)^\alpha = 1 + \binom{\alpha}{1} x + \binom{\alpha}{2} x^2 + \ldots + \binom{\alpha}{n-1} x^{n-1} + R_n (x) $, we now need to show $ R_n (x) \to 0 $ as $ n \to \infty $.
Recall integral version of Taylor's theorem
Let $ f : I \to \mathbb{R} $ be a $ C^n $ function on open interval $ I $, and say $ a, b \in I $. Then $$ f(b) = f(a) + f'(a) (b-a) + \ldots + \dfrac{f^{(n-1)}(a)}{(n-1)!} (b-a)^{n-1} + \int_{a}^{b} \dfrac{(b-t)^{n-1}}{(n-1)!} f^{(n)}(t) dt $$
So applying it to $ f : (-1, \infty) \to \mathbb{R} $ given by $ f(t) = (1+t)^\alpha $ and with $ a = 0 $,
$$ \begin{align} R_n (x) &= \int_{0}^{x} \dfrac{(x-t)^{n-1}}{(n-1)!} \alpha (\alpha - 1) \ldots (\alpha -(n-1)) (1+t)^{\alpha - n} dt \\ &= n \binom{\alpha}{n} \int_{0}^{x} (x-t)^{n-1} (1+t)^{\alpha - n} dt \end{align} $$
Estimating $\binom{\alpha}{n} $ term :
If $ \alpha = 0 $, $ \binom{\alpha}{n} $ is trivially $ 0 $. So lets take $ \alpha \neq 0 $.
$\begin{align} \left|\binom{\alpha}{n}\right| &= \left|\dfrac{\alpha (\alpha-1) \ldots (\alpha - (n-1))}{n!} \right| \\ &= \left| \dfrac{1}{n} \alpha \left( \dfrac{\alpha}{1} - 1 \right) \left( \dfrac{\alpha}{2} - 1 \right) \ldots \left( \dfrac{\alpha}{n-1} - 1 \right) \right| \\ &\leq \dfrac{1}{n} |\alpha| \left( \dfrac{|\alpha|}{1} + 1 \right) \left( \dfrac{|\alpha|}{2} + 1 \right) \ldots \left( \dfrac{|\alpha|}{n-1} + 1 \right) \\ \end{align}$
Since each factor in RHS is $ > 0 $, we can write it as
$$ e^{ \ln \left[ \frac{1}{n} |\alpha| \left( \frac{|\alpha|}{1} + 1 \right) \left( \frac{|\alpha|}{2} + 1 \right) \ldots \left( \frac{|\alpha|}{n-1} + 1 \right) \right] }. $$
The exponent here is
$\begin{align} &- \ln n + \ln|\alpha| + \ln\left(\frac{|\alpha|}{1} + 1\right) + \ln\left( \frac{|\alpha|}{2} + 1 \right) + \ldots + \ln\left( \frac{|\alpha|}{n-1} + 1 \right) \\ &\leq - \ln n + \ln |\alpha | + \frac{|\alpha|}{1} + \frac{|\alpha|}{2} + \ldots + \frac{|\alpha|}{n-1} \end{align} $
[because $ \ln (1+x) \leq x $ for $ x \geq 0 $. This in turn is because $ x - \ln(1+x) $ is $ 0 $ at $ x = 0 $, and its derivative $ 1 - \frac{1}{1+x} $ is $ > 0 $ on $ (0, \infty) $]
$\begin{align} &\leq -\ln n + \ln |\alpha| + |\alpha| \left( 1 + \ln (n-1) \right) \end{align} $
[because $ 1 + \frac{1}{2} + \ldots + \frac{1}{n-1} $ $= 1 + \int_{1}^{2} \frac{1}{2} dt + \ldots + \int_{n-2}^{n-1} \frac{1}{n-1} dt $ $ \leq 1 + \int_{1}^{2} \frac{1}{t} dt + \ldots + \int_{n-2}^{n-1} \frac{1}{t} dt $ $ = 1 + \ln (n-1) $. This is also clear visually].
So finally,
$\begin{align} \left| \binom{\alpha}{n} \right| &\leq e^{ \ln \left[ \frac{1}{n} |\alpha| \left( \frac{|\alpha|}{1} + 1 \right) \left( \frac{|\alpha|}{2} + 1 \right) \ldots \left( \frac{|\alpha|}{n-1} + 1 \right) \right] } \\ &\leq e^{-\ln n + \ln |\alpha| + |\alpha| (1 + \ln (n-1) ) } \\ &= \frac{1}{n} |\alpha| e^{|\alpha|} (n-1)^{|\alpha|} ( \leq |\alpha| e^{|\alpha|} n^{|\alpha| - 1} ) \end{align} $
Especially,
$$ \fbox{$\left|\binom{\alpha}{n}\right| = \mathcal{O} (n^{|\alpha|-1}) $ } $$
Estimating $R_n(x)$ :
$ \underline{\text{CASE-1}} \, (x \in [0,1)) $
In the integral as $ t $ varies from $ 0 $ to $ x $, $ (1+t)^{\alpha - n} \leq (1+t)^{\alpha} \leq 2^\alpha $, so
$\begin{align} |R_n (x)| &= n \left| \binom{\alpha}{n}\right| \int_{0}^{x} (x-t)^{n-1}(1+t)^{\alpha - n} dt \\ &\leq n \left| \binom{\alpha}{n} \right| 2^\alpha \int_{0}^{x} (x-t)^{n-1} dt \\ &= \left| \binom{\alpha}{n} \right| 2^\alpha x^n \end{align}$
Now RHS is $ \leq K n^{|\alpha|-1} x^n $, and $ n^{|\alpha|-1} x^n $ goes to $ 0 $ as $ n \to \infty $ [if $ x = 0 $, trivial. Else write $ x $ as $ \frac{1}{1+y} $ with $ y > 0 $ and expand the denominator], as needed.
$\underline{\text{CASE-2}} \, (x \in (-1, 0)) $
$\begin{align} R_n (x) &= n \binom{\alpha}{n} \int_{0}^{x} (x-t)^{n-1} (1+t)^{\alpha - n} dt \\ &= n \binom{\alpha}{n} \int_{0}^{-|x|} (-|x| - t)^{n-1} (1+t)^{\alpha - n} dt \end{align}$
Substituting $ s = (-t) $, the integral becomes
$\begin{align} R_n(x) &= n \binom{\alpha}{n} \int_{0}^{|x|} (-|x| + s)^{n-1} (1-s)^{\alpha - n} (-ds) \\ &= n \binom{\alpha}{n} (-1)^n \int_{0}^{|x|} (|x|-s)^{n-1} (1-s)^{\alpha - n} ds \end{align}$
and now keeping in mind $ |x| \in (0,1) $ and that $ s $ varies from $ 0 $ to $ |x| $ in the integral,
$\begin{align} |R_n (x)| &= n \left| \binom{\alpha}{n} \right| \int_{0}^{|x|} \left(\dfrac{|x|-s}{1-s} \right)^{n-1} (1-s)^{\alpha - 1} ds \\ &\leq n \left| \binom{\alpha}{n} \right| \int_{0}^{|x|} |x|^{n-1} (1-s)^{\alpha -1}ds \\ &\leq L n^{|\alpha|} |x|^{n-1}\end{align}$
and $ n^{|\alpha|} |x|^{n-1} $ goes to $ 0 $ as $ n \to \infty $, as needed.
|
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|
What is awry with this proof? Let $x=5$, $y=7$, $z=6$
$x+y = 2z$
Rearranging, $x-2z = -y$
and $x = -y+2z$
Multiply both sides respectively. $x^2-2xz = y^2-2yz$
$$x^2-2xz+z^2 = y^2-2yz+z^2$$
$$(x-z)^2 = (y-z)^2$$
$$x-z = y-z$$
Hence $x=y$, or $5 = 7$
Well, the conclusion is clearly false, but what went wrong? I think it may be the step in which one square roots both sides because it's taking out one solution?
|
You have $x=−y+2z$ and $x−2z=−y$ then you multiply both sides.
I think you are computing $-xy$ in two ways: for the first $-xy=y^2-2zy$
for the second $-xy=-2zx+x^2$ so you have $y^2-2zy=x^2-2zx$. Adding $z^2$ to both sides
gives $(y-z)^2=(x-z)^2$. But now taking square roots gives $|y-z|=|x-z|$.
If you try to remove the absolute value signs then you get $y-z=x-z$ if both are positive or negative but $y-z=-(x-z)$ if these have opposite signs. In the first case this leads to $y=x$, and in the second case $y=-x+2z$. For your example, the second case applies
$7=-5+2\cdot6$.
|
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|
Advice on using the Karamata Inequality Is it possible to use the Karamata inequality to prove the following inequalities to be true ?
$$x^{4}+y^{4}+z^{4}\geq x^{2}yz + xy^{2}z + xyz^{2}$$
$$x^{5}+y^{5}\geq x^{3}y^{2} + x^{2}y^{3}$$
$$x^{4}y + xy^{4}\geq x^{3}y^{2} + x^{2}y^{3}$$
for all $x,y,z \in \mathbb{R}^{+}$
|
As Byron observed all your inequalities follow immediately from AM-GM inequality.
To prove the first one, you can do the following:
$$\frac{x^4+x^4+y^4+z^4}{4} \geq x^2yz \,$$
$$\frac{x^4+y^4+y^4+z^4}{4} \geq xy^2z \,$$
$$\frac{x^4+y^4+z^4+z^4}{4} \geq xyz^2 \,$$
and add them together.
The second one follows from
$$\frac{x^5+x^5+x^5+y^5+y^5}{5} \geq x^3y^2$$
$$\frac{x^5+x^5+y^5+y^5+y^5}{5} \geq x^2y^3$$
while the last one
$$\frac{x^4y+x^4y+xy^4}{3} \geq x^3y^2$$
$$\frac{x^4y+xy^4+xy^4}{3} \geq x^2y^3$$
Actually, all your inequalities are a particular case of the Muirhead's Inequality. Muirhead is similar in idea to Karamata inequality, but I think it is not a consequence of it.... I actually doubt that you can use Karamata inequality here, since you have two sets of variables: $x,y,z$ and the powers.
As for Muirhead, your first inequality is just Muirhead for $[4,0,0] \succeq [2,1,1]$, your second inequality is Muirhead $[5,0] \succeq [3,2]$ and the last is Murihead $[4,1] \succeq [3,2]$.
P.S. Here is a better link for Muirhead Inequality.
Muirhead Explained
P.P.S. The solution above is just the standard "Prove this particular case of Muirhead by using AM-GM" approach...
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|
Showing $\lim _{n\rightarrow \infty } \frac {S_{n}}{n^{2}} = \frac{1}{6}$ I am trying to show that if the arithmetic mean of the products of all distinct pairs of positive integers whose sum is $n$ is denoted by $S_{n}$ then $$\lim _{n\rightarrow \infty } \dfrac {S_{n}}{n^{2}} = \dfrac{1}{6}$$
Solution attempt
If $n$ was even then $$S_{n} = \sum _{i=1}^{i=\dfrac{n} {2}}\dfrac {\left( n-i\right) \left( n-\left( n-i\right) \right) } {\dfrac {n } {2}} = \sum _{i=1}^{i=\dfrac{n} {2}}2i\left( 1-\dfrac {i} {n}\right) $$
so
$$\lim _{n\rightarrow \infty } \dfrac {S_{n}}{n^{2}} = \lim _{n\rightarrow \infty } \sum _{i=1}^{i=\dfrac{n} {2}}\dfrac{2i}{n^{2}}\left( 1-\dfrac {i} {n}\right)$$
Similarly If $n$ was odd then
$$S_{n} = \sum _{i=1}^{i=\dfrac{\left(n+1\right)} {2}}\dfrac {\left( n-i\right) \left( n-\left( n-i\right) \right) } {\dfrac {\left(n+1\right) } {2}} = \sum _{i=1}^{i=\dfrac{\left(n+1\right)} {2}}2i\left(\dfrac {n-i} {n+1}\right)$$
so
$$\lim _{n\rightarrow \infty } \dfrac {S_{n}}{n^{2}} = \lim _{n\rightarrow \infty } \sum _{i=1}^{i=\dfrac{\left(n+1\right)} {2}}\dfrac{2i}{n^{2}}\left(\dfrac {n-i} {n+1}\right)$$
I am unsure how to proceed from here to show the result. Any help would be much appreciated.
|
If $n$ is odd, then
$$\sum_{i=1}^{(n+1)/2} i (n-i) = \frac{(n-1)(n+1)(n+3)}{12}$$
Hence, you get that $$\frac{S_n}{n^2} = \frac{2}{n^2(n+1)} \times \frac{(n-1)(n+1)(n+3)}{12} = \frac{(1-1/n)(1+3/n)}{6}$$
If $n$ is even, then
$$\sum_{i=1}^{n/2} i (n-i) = \frac{n(n+2)(2n-1)}{24}$$
Hence, you get that $$\frac{S_n}{n^2} = \frac{2}{n^2(n+1)} \times\frac{n(n+2)(2n-1)}{24} = \frac{\left(1+\frac1{n+1} \right) \left(1-\frac1{2n} \right)}{6}$$
|
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|
Determine matrix of a set in a certain base I have a set
$S = \{ x^2 + 1, x + 1, 1 - x, x^3 \}$
in a polynomial vector space.
How do I write a vector matrix of $S$ in the base $B = \{ 1, x, x^2, x^3 \}$?
I attempted this using the formula: $M(S) = B^{-1} * T * B$,
where $T$ is the matrix of the set.
I got the following result:
$$
T=\left(\begin{matrix}
1& 0& 0& 0\\
0& 1& 0& 0\\
0& 0& 1& 0\\
0& 0& 0& 1
\end{matrix}\right)
$$
Is this correct or have I missed something?
I don't have the solutions so I cannot confirm it.
|
The purpose of this question is to write the elements of $S$ as linear combinations of elements of the basis $B$, then to give a matrix representation to the result.
Firstly let us call $(1,x,x^2,x^3) = (b_0,b_1,b_2,b_3).$
Then we have
$$
x^2+1 = 1(1) + 0(x) + 1(x^2) + 0(x^3) = b_0+b_2\\\
x+1 = 1(1) + 1(x) + 0(x^2)+0(x^3) = b_0+b_1\\\
1-x = 1(1) -1(x) + 0(x^2)+0(x^3) = b_0-b_1\\\
x^3 = 0(1) + 0(x) + 0(x^2) + 1(x^3) =b_3
$$
so that each element of $S$ is expressed as a linear combination of the basis elements in $B$.
Then let us write each element of $S$ as a column vector in the basis $B$. We have
$$
x^2+1 = b_0+b_2 = \left[\begin{array}{c}1\\0\\1\\0 \end{array} \right]
$$
$$
x+1 = b_0+b_1 = \left[\begin{array}{c}1\\1\\0\\0\end{array} \right]
$$
$$
1-x = b_0-b_1 = \left[\begin{array}{c}1\\-1\\0\\0 \end{array} \right]
$$
$$
x^3 = b_3 = \left[\begin{array}{c}0\\0\\0\\1 \end{array} \right]
$$
and we then put all of these side by side in one matrix
$$
M=\left[\begin{array}{cccc}1&1&1&0\\0&1&-1&0\\1&0&0&0\\0&0&0&1 \end{array}\right]
$$
where $M$ is a matrix representation of $S$ in the basis $B$.
|
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|
Proof of identity $\ln \left|\frac{\sin x}{\cos x - 1}\right| = \ln \left|\frac{\cos x + 1}{\sin x}\right|$ How do you prove this identity:
$$\ln \left|\frac{\sin x}{\cos x - 1}\right| = \ln \left|\frac{\cos x + 1}{\sin x}\right|$$
Mathematica says it's true, but if I try to simplify both sides, I wind up with
$$ \sin^2 x = \cos^2 x - 1$$
which ain't right.
|
We have
$$\left|\frac{\sin x}{\cos x-1}\right|=\left|\frac{\sin x}{\cos x-1}\cdot \frac{\cos x+1}{\cos x+1}\right|=\left|\frac{\sin x(\cos x+1))}{(\cos x)^2-1}\right|.$$
Now, using $(\cos x)^2+(\sin x)^2=1$, we get
$$\left|\frac{\sin x}{\cos x-1}\right|=\left|\frac{\sin x(\cos x+1)}{-(\sin x)^2}\right|=\left|\frac{\cos x+1}{-\sin x}\right|=\left|\frac{\cos x+1}{\sin x}\right|.$$
|
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|
Multiset Combination in Combinatorics I want to buy a $k$-combination of doughnuts, where $k$ is any amount less than or equal to the total doughnuts available. At the bakery there are $n$ different types of doughnuts but there are restricted amount left for each type of doughnut.
For example, In this case, the total doughnuts available is
4 + 2 + 3 + 2 + 7 + 2 + 8 = 28
and n = 7
Doughnut type Amount left
1 4
2 2
3 3
4 2
5 7
6 2
7 8
=====
28
Let's assume I want to buy a combination of $10$ doughnuts but I cannot have more than, for example $7$ of type $5$ doughnut and I can't have more than $3$ of type $3$ doughnuts.
How many combination of $10$ doughnuts can I have altogether?
|
Let $a_i$ be the number of donuts you buy of type $i$. Then you have $$a_1+a_2+\cdots+a_7=10$$ subject to the constraints, $$0\le a_1\le4,0\le a_2\le2,\dots,0\le a_7\le 8$$ If you think about how you would multiply out the following product, you'll see you want the coefficient of $x^{10}$ in $$P(x)=(1+x+x^2+x^3+x^4)(1+x+x^2)\cdots(1+x+x^2+\cdots+x^8)$$ Now all those brackets contain geometric series, so you can rewrite as $$P(x)=(1-x^5)(1-x^3)\cdots(1-x^9)(1-x)^{-7}$$ and you can use the binomial theorem to get $$(1-x)^{-7}=1+{7\choose6}x+{8\choose6}x^2+{9\choose6}x^3+\cdots$$ So all you have to do is pick out the coefficient of $x^{10}$ in $$(1-x^5)(1-x^3)(1-x^4)(1-x^3)(1-x^8)(1-x^3)(1-x^9)\left(1+{7\choose6}x+{8\choose6}x^2+\cdots\right)$$ Now if you multiply out the first 7 terms, you get $$1-3x^3-x^4-x^5+3x^6+3x^7+2x^8-x^9-3x^{10}+{\rm\ higher\ terms}$$ (although you should check my work on that one), where we can ignore the higher terms because they can't contribute to the coefficient of $x^{10}$ that we are looking for. So, finally, you just have to pick out the coefficient of $x^{10}$ in $$(1-3x^3-x^4-x^5+3x^6+3x^7+2x^8-x^9-3x^{10})\left(1+{7\choose6}x+{8\choose6}x^2+\cdots\right)$$
|
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|
System of equations: $x^2+y=7, y^2+x=11$
Possible Duplicate:
Steps to solve this system of equations
During the flight from Moscow to Yerevan my neighbor gave me the following problem:
Solve the system:
$$\left\{\begin{array}{c}x^2+y=7 \\ y^2+x=11. \end{array}\right.$$
It is easy to find 1 of the 4 solutions. Is there a beautiful way to find the other three?
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Every solution of the given system
$$\left\{
\begin{array}{c}
x^{2}+y=7 \\
y^{2}+x=11
\end{array}
\right. \tag{0}$$
is a solution of
$$\left\{
\begin{array}{c}
\left( y-3\right) \left( y^{3}+3y^{2}-13y-38\right) =0 \\
x^{2}=121-22y^{2}+y^{4}.
\end{array}\tag{1}
\right. $$
The same applies to the system
$$\left\{
\begin{array}{c}
y^{2}=49-14x^{2}+x^{4} \\
\left( x-2\right) \left( x^{3}+2x^{2}-10x-19\right) =0.
\end{array}\tag{2}
\right. $$
The integral solution of $(0)$ is $\left( x_{0},y_{0}\right) =\left( 2,3\right) $. Simple ways to find the remaining solutions are only possible in particular cases, as far as I know. The standard way to solve a cubic equation such as
$$y^{3}+3y^{2}-13y-38=0\tag{3}$$
is to make the change of variables $$y=s-\dfrac{3}{3\cdot 1}=s-1\tag{3a}$$ to get the reduced cubic equation
$$s^{3}-16s-23=0.\tag{4}$$
If the discriminant $q^{2}+\frac{4p^{3}}{27}$ of an equation of the form $s^3+px+q=0$ is negative, its three solutions are real numbers. In this case we have $q^{2}+\frac{4p^{3}}{27}=23^{2}-\frac{4\times 16^{3}}{27}<0$ and the solutions of $(4)$ can be written in the trigonometric form$^{1}$
$$s_{k}=2\sqrt{\frac{16}{3}}\cos \left( \frac{1}{3}\arccos \left( \frac{23}{2}\sqrt{\frac{27}{16^{3}}}\right) +\frac{2\left( k-1\right) \pi }{3}\right),\tag{5}$$
with $k=1,2,3$. So $$y_{k}=s_{k}-1\tag{6}$$ and
$$x_{k}=11-y_{k}^2.\tag{7}$$
For $k=1$, we get $\left( x_{1},y_{1}\right) \approx \left(
-1.8479,3.5844\right) $. And similarly for $k=2$ and $k=3$.
--
$^{1}$A deduction can be found in this Portuguese post of mine.
Added. If $\Delta =q^{2}+\frac{4p^{3}}{27}<0$ the three real solutions of the following reduced cubic equation
$$t^{3}+pt+q=0\tag{A}$$
are given by
$$t_{k}=2\sqrt{-\frac{p}{3}}\cos \left( \frac{1}{3}\arccos \left( -\frac{q}{2}\sqrt{-\frac{27}{p^{3}}}\right) +\frac{2\left( k-1\right) \pi }{3}\right) \tag{B},$$
with $k=1,2,3$.
PS. I do not find trigonometric functions nor radicals ugly. But this is just an opinion.
|
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|
find an $A$ so that for all $n \ge A (2^n+3^n)/(4^n+5^n)/(1/1000)$ So $1000(2^n+3^n)\le 4^n+5^n$.
I take $1000\cdot2^n\le 4^n$ and $1000\cdot3^n\le5^n$ so that adding both gives the inequality, theorem in the ordered fields.
So $1000\cdot2^n\le2^2n$. This leads me to an $A$ for $n = 10$ as $1000\cdot2^n < 2^n\cdot2^n$ so $1000\leq2^n$ for $n=10$.
But this does not count for $1000\cdot3^n<5^n$. I think to see that $1000\cdot3^n<3^n\left(\frac{5}{3}\right)^n$ so $1000<\left(\frac{5}{3}\right)^n$. so $n = \log_{\frac{5}{3}} (1000)$.
This leads me to an $A = 14$ which is my final answer.
Any other suggestions.
A similar problem is $\displaystyle\frac{(1.01)^n}{n^{147}} < 100$. This is tougher. Any hints
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For the similar problem, the inequality is false. If we take the base $10$ log of each side, we get $n \log 1.01 - 147 \log n \lt 2$ or $ .00432n - 147 \log n \lt 2$ Now we just need to take $n$ large enough. $n=10^6$ is easily enough.
|
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|
Evaluating $\int_{-20}^{20}\sqrt{2+t^2}\,dt$ I have this integral:
$$\int_{-20}^{20}\sqrt{2+t^2}\,dt$$
I tried solving it many times but without success.
The end result is this:
$$2\left( 10\sqrt{402}+\mathop{\mathrm{arcsinh}}(10\sqrt{2})\right).$$
I can't seem to get this end result. I got a few wrong ones but cant find this one. Perhaps it's wrong? Could anyone confirm it?
I tried Sage, and it calculates it correctly, but not with steps.
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While I was writting this answer, Artem posted his. My answer is essentially his 1st one.
Let $f(t)=\sqrt{2+t^{2}}$. Since $f(-t)=$ $f(t)$, we have
$$\begin{equation*}
I=\int_{-20}^{20}\sqrt{2+t^{2}}dt=2\int_{0}^{20}\sqrt{2+t^{2}}dt
\end{equation*}$$
As the integrand $f(t)$ is a quadratic irrational we can also use the
following Euler substitution
$$\begin{eqnarray*}
x &=&\sqrt{2+t^{2}}-t\Leftrightarrow t=\frac{2-x^{2}}{2x}\Leftrightarrow
\sqrt{2+t^{2}}=\frac{2+x^{2}}{2x} \\
dt &=&-\frac{2+x^{2}}{2x^{2}}dx,
\end{eqnarray*}$$
which transforms the integrand into a rational fraction in $x$
$$\begin{eqnarray*}
\int \sqrt{2+t^{2}}dt &=&\int \frac{2+x^{2}}{2x}\left( -\frac{2+x^{2}}{2x^{2}
}\right) dx \\
&=&-\int \frac{1}{x}+\frac{1}{x^{3}}+\frac{x}{4}dx \\
&=&-\left( \ln \left\vert x\right\vert -\frac{1}{2x^{2}}+\frac{x^{2}}{8}
\right) +C.
\end{eqnarray*}$$
The limits of integration of $I/2$ are
$$\begin{eqnarray*}
t &=&0,x=\sqrt{2} \\
t &=&20,x=\sqrt{2+20^{2}}-20=\sqrt{402}-20.
\end{eqnarray*}$$
Thus
$$\begin{eqnarray*}
I &=&\left. -2\left( \ln \left\vert x\right\vert -\frac{1}{2x^{2}}+\frac{
x^{2}}{8}\right) \right\vert _{\sqrt{2}}^{\sqrt{402}-20} \\
&=&2\ln \left( \frac{\sqrt{2}}{\sqrt{402}-20}\right) +\frac{1}{\left( \sqrt{
402}-20\right) ^{2}}-\frac{\left( \sqrt{402}-20\right) ^{2}}{4}+\frac{1}{2}.
\end{eqnarray*}$$
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|
Matrix transformation mapped onto itself Question:
$$A= \begin{pmatrix} k & -2 \\ 1-k & k \end{pmatrix}\text{, where k is a constant}$$
$$\text {A transformation } T : \mathbb{R}^2 \rightarrow \mathbb{R}^2 \text{ is represented by the matrix A.}$$
$$\text {Find the value of k for which the line } y = 2x \text{ is mapped onto itself under T.}$$
Working:
$$\begin{pmatrix} k & -2 \\ 1-k & k \end{pmatrix}\cdot\begin{pmatrix} x \\ 2x \end{pmatrix}=\begin{pmatrix} x \\ 2x \end{pmatrix}$$
$$\begin{pmatrix} x(k-4) \\ x(1+k) \end{pmatrix}=\begin{pmatrix} x \\ 2x \end{pmatrix}$$
$$x(k-4)=x$$
$$x(1+k)=2x$$
Leaving me with $k=5$ and $k=1$, However the answer is $k = 9$ why?
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On the same line (!) of thought: the line $\,l:y=2x\,$ is the same as the vector space $\,\operatorname{Span}\{(1,2)\}\leq\mathbb{R}^2$ , or if you prefer: $\,l:\{(r,2r)\,/\,r\in\mathbb{R}\}\,$ , and then what we really want to happen is $$\begin{pmatrix}k&-2\\1-k&k\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}r\\2r\end{pmatrix}\Longrightarrow \begin{array}\\k =\,\,\,r+4\\k=2r-1\end{array}$$so $\,r+4=2r-1\Longrightarrow r=5\,$ and thus $\,k=9\,$
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|
Finding square roots of $\sqrt 3 +3i$ I was reading an example, where it is calculating the square roots of $\sqrt 3 +3i$.
$w=\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)\\=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$
Let $z^2=w \Rightarrow r^2(\cos(2\theta)+i\sin(2\theta))=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$.
But how did they get from $\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$?
And can one just 'let $z^2=w$' as above?
Edit:
$w=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})=z^2\\ \Rightarrow z=\sqrt{2\sqrt 3}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})\\ \Rightarrow \sqrt{2\sqrt 3}\frac{\sqrt 3}{2} +i \sqrt{2\sqrt 3} \frac{1}{2}$
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Denesting the radical $\:\sqrt{(1+\sqrt{-3})/2}\:$ can be tackled by employing an easy radical denesting formula that I discovered as a teenager.
Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $
and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$
Here $\ (1+\sqrt{-3})/2\:$ has norm $= 1.\:$ $\rm\ \ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 1\ $ yields $\ \ (-1+\sqrt{-3})/2\:$
and this has $\rm\:\sqrt{trace}\: =\: \sqrt{-1},\ \ thus\ \ \color{brown}{dividing\ it\ out}\ $ of this gives the sqrt: $\:(\sqrt{3}+\sqrt{-1})/2$
Checking we have
$$\smash[t]{\displaystyle \left(\frac{\sqrt{3}+\sqrt{-1}}{2}\right)^2 =\ \frac{3-1 + 2\sqrt{-3}}{4}\ =\ \frac{1+\sqrt{-3}}2}$$
Therefore $\rm\ \sqrt{\sqrt{3}+3{\it i}}\ =\ \sqrt{2\sqrt 3\left(\dfrac{1+\sqrt{-3}}{2}\right)}\ =\ \sqrt{2}\ 3^{1/4}\:\!\dfrac{(\sqrt{3}+\sqrt{-1})}{2}$
See this answer for general radical denesting algorithms.
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|
Factorize the polynomial $P(z) = z^4 - 2z^3-z^2+2z+10$, into linear and/or quadratic factors with real coefficients $2+i$ is given to be one of the roots of the polynomial.
I am doing this as a practice for exam prep.
Since $2+i$, is a root, then $(z-2-i)$ is a factor?
So I have:
$(z-2-i)(z^3-Az^2-Bz+C) = z^4-2z^3-z^2+2z+10$
Then, I group the $z$ terms of the same degrees after I multiply out?
Am I on the right track? Because it seems like this will take quite a while in exam conditions? Thanks for any direction!
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Since the question has the "calculus" tag, letus use a slightly different method. We start with $$(z^2-4z+5)(z^2+bz+c)= z^4-2z^3-z^2+2z+10.$$ as smanoos does. When we plug zero int both sides, we get $5c=10$, so $c=2$. This gives us the equation,
$$(z^2-4z+5)(z^2+bz+2)= z^4-2z^3-z^2+2z+10.$$. Now letus take the derivative of both sides to get,
$$(2z-4)(z^2+bz+2)+(z^2-4z+5)(2z+b)=4z^3-6z^2-2z+2$$. Setting $z=0$ gives,
$$-8+5b=2$$, thus $b=2$.
So we factored the polynomial as $(z^2-4z+5)(z^2+2z+2)$, but then we know how $(z^2+2z+2)$ factors, and we must factor $(z^2+2z+2)$, but for that we will refer to the earlier answers.
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|
Integral of $\int \sqrt{1-4x^2}$ I know I am messing up something with the substitutions but I am not sure what.
$$\int \sqrt{1-4x^2}$$
$$u = 4x, du = 4 \,dx$$
$$\frac{1}{4}\int \sqrt{1-u^2}$$
$u = \sin \theta$
$$\frac{1}{4}\int \sqrt{1-\sin^2 \theta} = \frac{1}{4}\int \sqrt{ \cos^2 \theta} = \frac{1}{4}\int \cos \theta = \frac{\sin \theta}{4}$$
Replace $\theta$ with $u$
$$\frac{\sin (\arcsin u)}{4} = \frac{u}{4} = \frac{4x}{4} = x$$
This is wrong and I have no idea why.
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Some things to watch out for:
*
*As already mentioned, if you substitute $u = 4x$, then you get $u^2 = (4x)^2 = 16x^2$ instead of $u^2 = 4x^2$. If you use $u = 2x$, then $u^2 = (2x)^2 = 4x^2$.
*An annoying but important part of practicing good mathematics is being precise.
*
*In your integrals, you forgot the $dx$ and $du$ several times, which also lead you to forget to express $du$ in terms of $d\theta$. Perhaps you have heard of Riemann integrals; these may help you understand why the "$dx$" is there.
*You forgot the fact that if $F'(x) = f(x)$, then $\int f(x) dx = F(x) + C$. After all, when taking derivatives, all constants disappear, so the derivatives of $F(x)$ and $F(x) + 5$ are exactly the same. Thus, when calculating indefinite integrals, you do not know the constant term of the resulting function. So we generally write $\int f(x) dx = F(x) + C$ for some unknown constant $C$.
Here's a step-by-step approach of how I would do it. First, substitute $u = 2x$, so that $u^2 = (2x)^2 = 4x^2$, and apply $du = 2dx$:
$$\int \sqrt{1 - \color{red}{4x^2}} \color{blue}{dx} = \int \sqrt{1 - \color{red}{u^2}} \color{blue}{\frac{du}{2}} = \frac{1}{2} \int \sqrt{1 - u^2} du.$$
Then substitute $u = \sin \theta$ and $du = \cos \theta d\theta$ to get:
$$\frac{1}{2} \int \sqrt{1 - \color{red}{u^2}} \color{blue}{du} = \frac{1}{2} \int \sqrt{1 - \color{red}{\sin^2 \theta}} \color{blue}{\cos \theta d\theta} = \frac{1}{2} \int \sqrt{\cos^2 \theta} \cos \theta d\theta = \frac{1}{2} \int \cos^2 \theta d\theta.$$
Next, apply the double angle formula for the cosine, $\cos^2 \theta = \frac{1}{2} (1 + \cos 2\theta)$, to get
$$\frac{1}{2} \int \color{red}{\cos^2 \theta} d\theta = \frac{1}{2} \int \color{red}{\frac{1 + \cos 2\theta}{2}} d\theta = \frac{1}{4} \int (1 + \cos 2\theta) d\theta.$$
Now split in two integrals and calculate those separately:
$$\frac{1}{4} \int (\color{red}{1} + \color{blue}{\cos 2\theta}) d\theta = \frac{1}{4} \left(\int \color{red}{1} d\theta + \int \color{blue}{\cos 2\theta} d\theta\right) = \frac{1}{4}\left(\color{red}{\theta + C_1} + \color{blue}{\frac{\sin 2\theta}{2} + C_2}\right) \\ = \frac{\theta}{4} + \frac{\sin 2\theta}{8} + C.$$
Applying a double angle formula for the sine, $\sin 2\theta = 2 \sin \theta \cos \theta$, we get
$$\frac{\theta}{4} + \frac{\color{red}{\sin 2\theta}}{8} + C = \frac{\theta}{4} + \frac{\color{red}{2 \sin \theta \cos \theta}}{8} + C = \frac{\theta}{4} + \frac{\sin \theta \cos \theta}{4} + C.$$
Substituting back $\theta = \arcsin u$, and using $\cos (\arcsin u) = \sqrt{1 - u^2}$, we get
$$\frac{\color{red}{\theta}}{4} + \frac{\sin \color{blue}{\theta} \cos \color{green}{\theta}}{4} + C = \frac{\color{red}{\arcsin u}}{4} + \frac{\sin(\color{blue}{\arcsin u}) \cos(\color{green}{\arcsin u})}{4} + C = \frac{\arcsin u}{4} + \frac{u \sqrt{1 - u^2}}{4} + C.$$
Finally substituting $u = 2x$, you get
$$\frac{\arcsin \color{red}{u}}{4} + \frac{\color{blue}{u} \sqrt{1 - \color{green}{u}^2}}{4} + C = \frac{\arcsin \color{red}{2x}}{4} + \frac{\color{blue}{2x} \sqrt{1 - \color{green}{(2x)}^2}}{4} + C = \boxed{\displaystyle\frac{\arcsin 2x}{4} + \frac{x \sqrt{1 - 4x^2}}{2} + C}.$$
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A problem about rational number for right triangles. If a right triangle has hypotenuse of length 5, and the remaining two sides $a, b$ have rational lengths, what can we say about $a$ and $b$?
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We want to generate the rational solutions of $u^2+v^2=25$. Equivalently, we
generate the solutions of the equation $x^2+y^2=25z^2$ in integers, with $z\ne 0$, and set $u=x/z$, $v=y/z$. Ultimately the procedure we get is a close analogue of the standard way of generating primitive Pythagorean triples. The case where $5$ divides one (and therefore both) of $x$ and $y$ is not interesting, so assume that neither $x$ nor $y$ is divisible by $5$.
The square $3+4i$ of the Gaussian prime $2+i$ must divide one of $x+iy$ or $x-iy$. By changing the sign of $y$ if necessary, suppose that $3+4i$ divides $x+iy$.
So $x+iy=(s+it)(3+4i)=(3s-4t)+(4s+3t)i$, and therefore $x=3s-4t$, $y=4s+3t$.
Squaring, we find that $x^2+y^2=25(s^2+t^2)=25z^2$, so we want $s^2+t^2=z^2$.
It is enough to consider primitive solutions of this equation. These are given by
(i) $(a^2-b^2,2ab,a^2+b^2)$, and (ii) $(2ab, a^2-b^2, a^2+b^2)$, where $a$ and $b$ are relatively prime and of opposite parity. We have to consider both possibilities because of the lack of symmetry between $s$ and $t$ in $x=3s-4t$, $y=4s+3t$.
Case (i) gives us the solutions $x=3(a^2-b^2)-8ab$, $y=4(a^2-b^2)+6ab$, $z=a^2+b^2$, and therefore the rational solutions
$$u=\frac{3(a^2-b^2)-8ab}{a^2+b^2},\qquad v=\frac{4(a^2-b^2)+6ab}{a^2+b^2}.$$
Case (ii) yields something similar.
Comment: The above procedure, with possibly changes of sign, generates all non-trivial rational solutions. There is undoubtedly a more attractive way of doing this. Furthermore, it seems likely that even though we are taking $a$ and $b$ relatively prime and of opposite parity, some rational solutions are generated more than once.
For a more elementary way of getting at a parametric solution, rewrite the equation $x^2+y^2=25z^2$ as $x^2-9z^2=16z^2-y^2$, and imitate the analysis of Pythagorean triples. It is somewhat messy.
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What is wrong with my solution? $\int \cos^2 x \tan^3x dx$ I am trying to do this problem completely on my own but I can not get a proper answer for some reason
$$\begin{align}
\int \cos^2 x \tan^3x dx
&=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \sin^3 x}{ \cos x}dx\\
&=\int \frac{ \sin^2 x \sin x}{ \cos x}dx\\
&=\int \frac{ (1 -\cos^2 x) \sin x}{ \cos x}dx\\
&=\int \frac{ (\sin x -\cos^2 x \sin x) }{ \cos x}dx\\
&=\int \frac{ \sin x -\cos^2 x \sin x }{ \cos x}dx\\
&=\int \frac{ \sin x }{ \cos x}dx - \int \cos x \sin x dx\\
&=\int \tan x dx - \frac{1}{2}\int 2 \cos x \sin x dx\\
&=\ln|\sec x| - \frac{1}{2}\int \sin 2x dx\\
&=\ln|\sec x| + \frac{\cos 2x}{4} + C
\end{align}$$
This is the wrong answer, I have went through and back and it all seems correct to me.
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A very simple way to check if the answer to an indefinite integral is correct is to differentiate the answer. If you get the original function, your answer is correct, and is equal, up to a constant, with any other solutions.
We have
$$\begin{align*}
\frac{d}{dx}\left(\ln|\sec x| + \frac{\cos 2x}{4} + C\right) &= \frac{1}{\sec x}(\sec x)' + \frac{1}{4}(-\sin(2x))(2x)' + 0\\
&= \frac{\sec x\tan x}{\sec x} - \frac{1}{2}\sin(2x)\\
&= \tan x - \frac{1}{2}\left(2\sin x\cos x\right)\\
&= \frac{\sin x}{\cos x} - \sin x\cos x\\
&= \frac{ \sin x - \sin x\cos^2 x}{\cos x}\\
&= \frac{\sin x(1 - \cos^2 x)}{\cos x}\\
&= \frac{\sin^3 x}{\cos x}\\
&= \frac{\sin ^3 x \cos^2 x}{\cos^3x}\\
&= \frac{\sin^3 x}{\cos^3 x}\cos^2 x\\
&= \tan^3 x \cos^2 x.
\end{align*}$$
So your answer is right.
This happens a lot with integrals of trigonometric identities, because there are a lot of very different-looking expressions that are equal "up to a constant". So the answer to $$\int\sin x\cos x\,dx$$
can be either $\sin^2 x + C$ or $-\cos^2x + C$; both correct, though they look different, because they differ by a constant: $-\cos^2x + C = 1-\cos^2x + (C-1) = \sin^2x + (C-1)$.
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Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
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You can use
$$x^4+1=(x^2+1)^2-2x^2=(x^2+1+\sqrt 2x)(x^2+1-\sqrt 2x).$$
Then, find $A,B,C,D$ such that
$$\frac{1}{x^4+1}=\frac{Ax+B}{x^2+1+\sqrt 2x}+\frac{Cx+D}{x^2+1-\sqrt 2x}.$$
You'll find
$$\frac{1}{x^4+1}=\frac{1}{2\sqrt 2}\left\{ \frac{x+\sqrt 2}{x^2+\sqrt2 x+1}+\frac{-x+\sqrt 2}{x^2-\sqrt 2x+1}\right\}.$$
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|
Discriminant for $x^n+bx+c$ The ratio of the unsigned coefficients for the discriminants of $x^n+bx+c$ for $n=2$ to $5$ follow a simple pattern:
$$\left (\frac{2^2}{1^1},\frac{3^3}{2^2},\frac{4^4}{3^3},\frac{5^5}{4^4} \right )=\left ( \frac{4}{1},\frac{27}{4},\frac{256}{27},\frac{3125}{256} \right )$$
corresponding to the discriminants
$$(b^2-4c, -4b^3-27c^2,-27b^4+256c^3,256b^5+3125c^4).$$
Does the pattern for the ratios extend to higher orders? (An online reference would be appreciated.)
|
Yes. Sketch: $b$ is a symmetric polynomial of degree $n-1$ in the roots and $c$ is a symmetric polynomial of degree $n$, whereas the entire discriminant is a symmetric polynomial of degree $n(n-1)$. It follows that the discriminant is a linear combination of $b^n$ and $c^{n-1}$, and the coefficients can be determined by setting $b = 0, c = -1$ and then $b = -1, c = 0$ and reducing to the computation of the discriminant of $x^n - 1$.
|
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|
If $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $, then either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ For $a, b = 1, 2, 3, \cdots$, let $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $. Then prove that either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ holds.
|
Either $\frac{1}{2a}\leq \frac{1}{2b}$, in which case
$$\frac{1}{b}=\frac{1}{2b}+\frac{1}{2b}\geq\underbrace{\frac{1}{2a}+\frac{1}{2b}}_{\atop \dfrac{1}{c}}\geq\frac{1}{2a}+\frac{1}{2a}=\frac{1}{a},$$
or $\frac{1}{2b}\leq \frac{1}{2a}$, in which case
$$\frac{1}{a}=\frac{1}{2a}+\frac{1}{2a}\geq\underbrace{\frac{1}{2a}+\frac{1}{2b}}_{\atop \dfrac{1}{c}}\geq\frac{1}{2b}+\frac{1}{2b}=\frac{1}{b}.$$
Now take reciprocals.
|
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|
Proof an inequality I'm trying to prove that
$$ \frac{3-2\sqrt{1-15 m^2}}{1+12 m^2}\geq 1+3 m^2$$
I have obtained in a CAS software the Taylor expansion in $m=0$
One posibility to prove the inequality is showing coeficients in Taylor expansion are non-negative, by I don't find how.
Really I want only to obtain inequality. Some idea?
EDIT
$m$ must be between $0<m<\frac{1}{\sqrt 15}$
|
$$ \frac{3-2\sqrt{1-15m^2}}{1+12m^2} \geq 1+3m^2 \iff $$
$$ 3-2\sqrt{1-15m^2} \geq (1+3m^2)(1+12m^2) \iff $$
$$ 2\sqrt{1-15m^2} \leq 3-(1+3m^2)(1+12m^2) \iff $$
$$ \sqrt{1-15m^2} \leq \frac{3-(1+3m^2)(1+12m^2)}{2} \iff $$
$$ \sqrt{1-15m^2} \leq \frac{2-15m^2-36m^4}{2} $$
Note that on the interval you're concerned about, the right hand side is always positive. Proof: it's obviously decreasing on $\left(0,\frac{1}{\sqrt{15}}\right)$, and is equal to $\frac{21}{50}$ at the right endpoint. Therefore squaring both sides is legal here with an $\iff$ statement.
$$ \sqrt{1-15m^2} \leq \frac{2-15m^2-36m^4}{2} \iff $$
$$ 1-15m^2 \leq \left(\frac{2-15m^2-36m^4}{2}\right)^2 \iff $$
$$ 1-15m^2 \leq 324m^8 + 270m^6 + \frac{81}{4}m^4 - 15m^2 + 1 \iff $$
$$ 0 \leq 324m^8 + 270m^6 + \frac{81}{4}m^4 $$
This last statement is clearly true.
|
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|
Simplest method to find $5^{20}$ modulo $61$ What is the simplest method to go about finding the remainder of $5^{20}$ divided by $61$?
|
Note that $5^3=125 = 3\pmod{61}$, so $5^{20}=5^{18}.25=(5^3)^6.25=3^6.25\pmod {61}$.Now $3^5=243\pmod {61}=-1\pmod {61}$ $\implies$ $3^6.25\pmod {61}= -3.25\pmod{61}=-75\pmod{61}=47\pmod {61}$.
|
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|
Factoring $x^4z-2z^2-4x^6+x^2z$ We want to factor $8x^4y^4-2y^8-4x^6+x^2y^4 = -2y^8 + (8x^4+x^2)y^4 -4x^6$. We substitute $x^4$ with $z$:
Now we want to compute this $8x^4z-2z^2-4x^6+x^2z = -(x^2-2z)(4x^4-z)$ by hand.
Therefore we transform it into $-(2z^2-(8x^4+x^4)z+4x^6z^0)$ and use the quadratic formula on the (inner) polynomial in $z$. The result is
$$z = \frac 1 2 x^2 \lor z = 4x^4.$$
So the result should be $(z-\frac 1 2 x^2)(z-4x^4)$. But the first factor appears only half? What is the reason for that? Is this a way to factor $8x^4y^4-2y^8-4x^6+x^2y^4$? The exercise is to use the recursive form $-2y^8 + (8x^4+x^2)y^4 -4x^6 \in \mathbb{Q}[y][x]$ (or alternatively $\mathbb{Q}[x][y]$).
|
If I understand your question correctly, the reason is because when you factor a polynomial using its roots or zeroes, you have to keep the main coefficient as a factor also. More explicitly, if you have a polynomial
$$F(x) = a_n x^n + \cdots + a_1 x + a_0$$
with zeroes $c_1, \dots , c_n$ then
$$F(x) = a_n (x - c_1)\cdots (x - c_n)$$
In your case, what happens is that you're using the quadratic formula to find the roots of the polynomial $F(z) = 2z^2 + \cdots$ so when you use the factorization with the roots you found, you're forgetting the $2$ that corresponds to the principal coefficient and when you multiply the term $(z - \frac{1}{2}x^2)$ by $2$ you get precisely $2z - x^2$ which is what you expected.
|
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How to prove: $S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$ If $$m_a, m_b, m_c$$ are the medians of a triangle and let $$m=\frac{m_a+ m_b+ m_c}{2}$$
then Area $S$ of triangle is given by
$$S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$$ This looks very similar to Heron's formula. How to prove this formula?
|
$$S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$$
$$ =\frac{4}{3}\sqrt{ \frac{(m_a+m_b+m_c)}{2}\frac{(-m_a+m_b+m_c)}{2}\frac{(m_a-m_b+m_c)}{2}\frac{(m_a+m_b-m_c)}{2}}$$
$$ =\frac{1}{3}\sqrt{(m_a+m_b+m_c)(-m_a+m_b+m_c)(m_a-m_b+m_c)(m_a+m_b-m_c)}$$
$$ =\frac{1}{3}\sqrt{[(m_a+m_b)^2-m_c^2] [m_c^2-(m_b-m_a)^2] } $$
$$ =\frac{1}{3}\sqrt{-[(m_a+m_b)(m_b-m_a)]^2+m_c^2(m_a^2+m_b^2)+m_c^2(m_b^2-m_a^2)-m_c^4}$$
$$ =\frac{1}{3}\sqrt{-[m_b^2-m_a^2]^2+m_c^2(m_a+m_b)^2+m_c^2(m_b-m_a)^2-m_c^2}$$
$$ =\frac{1}{3}\sqrt{2(m_a^2m_b^2+m_b^2m_c^2+m_c^2m_a^2)-(m_a^4+m_b^4+m_c^4)}$$
Now replacing $m_a$, $m_b$, $m_c$ respectively with $\frac{1}{2}\sqrt{2c^2+2b^2-a^2}$, $\frac{1}{2}\sqrt{2a^2+2c^2-b^2}$, $\frac{1}{2}\sqrt{2a^2+2b^2-c^2}$
You'll arrive at the Heron's Formula:
$$S = \sqrt{s(s-a)(s-b)(s-c)}$$
EDIT
Proof for $m_a = \frac{1}{2}\sqrt{2c^2+2b^2-a^2}$:
Assumptions: $BC = a$, $AC = b$, $AD = DE = m_a$ and $AB = c$
From cosine theorem, we have:
$$a^2 = b^2 + c^2 - 2bc*cos A$$
Now, in triangle ABE(since ACD is congruent to EBD);
$$4m_a^2 = AE^2 = b^2 + c^2 - 2bc*cos(\pi - A)$$
$$a^2 + m_a^2 = 2(b^2 + c^2) - 2[bc*cos A + bc*cos(\pi - A)]$$
or,
$$2m_a^2 = 2b^2 + 2c^2 - a^2$$
Hence,
$$m_a = \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2} $$
|
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|
Some method to solve $\int \frac{1}{\left(1+x^2\right)^{2}} dx$ and some doubts. First approach.
$\int \frac{1}{1+x^2} dx=\frac{x}{1+x^2}+2\int \frac{x^2}{\left(1+x^2\right)^2} dx=\frac{x}{1+x^2}+2\int \frac{1}{1+x^2}dx-2\int \frac{1}{\left(1+x^2\right)^2}dx$
From this relationship, I get:
$2\int \frac{1}{\left(1+x^2\right)^2}dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}dx$ Then:
$\int \frac{1}{\left(1+x^2\right)^2}dx=\frac{1}{2}\left[\frac{x}{1+x^2}+\arctan x\right]+C$ This is a recursive solution.
Second approach.
$x=\tan t$ in $t\in (- \pi/2, \pi/2)$, i.e. $t=\arctan x$, then $dx=(1+x^2) dt$.
$\int \frac{1}{\left(1+x^2\right)^2}dx=\int \frac{1}{1+x^2}dt=\int \frac{\cos^2t}{\sin^2t+\cos^2t}dt=\int \cos^2t dt=\frac{1}{2}\int \left(1+\cos 2t \right) dt=\frac{t}{2}+\frac{1}{4}\sin 2t$
This result can be rewritten (using trigonometric formulas):
$\frac{t}{2}+\frac{1}{4}\sin 2t=\frac{t}{2}+\frac{1}{2}\sin t \cos t$
From $\cos^2 t=\frac{1}{1+x^2}$, I have:
$|\cos t|=\sqrt{\frac{1}{1+x^2}}$ but in $t\in (- \pi/2, \pi/2)$, $|\cos t|=\cos t$. So:
$\cos t=\sqrt{\frac{1}{1+x^2}}$. Now I have a problem: $|\sin t|=\sqrt{\frac{1}{1+x^2}}$, but $|\sin t|\neq \sin t$ for $t\in (- \pi/2, \pi/2)$. Any suggestions, please? This integral can be solved in other ways?
Thanks.
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If we use this identity $\sin(2t)=\frac{2\tan(t)}{1+\tan^2(t)}$, the confusion should not arise.
|
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|
Verify trigonometry equation $\tan A - \csc A \sec A (1-2\cos^2 A)= \cot A$ How would I verify the following trigonometry identity?
$$\tan A - \csc A \sec A (1-2\cos^2 A)= \cot A$$
My work so far is
$$\frac{\sin A}{\cos A}-\frac{1}{\sin A}\frac{1}{\cos A}(1- \cos^2 A- \cos^2 A)$$
|
I just want to comment on how you could take your work so far a little further.
$$\frac{\sin A}{\cos A}-\frac{1}{\sin A}\frac{1}{\cos A}(1- \cos^2 A- \cos^2 A)$$
What you need to do is to get a common denominator
$$\frac{\sin A}{\cos A}\cdot\frac{\sin A}{\sin A}-\frac{1}{\sin A\cos A}(1- \cos^2 A- \cos^2 A)$$
$$=\frac{\sin^2 A}{\sin A\cos A}-\frac{1- \cos^2 A- \cos^2 A}{\sin A\cos A}$$
$$=\frac{\sin^2 A}{\sin A\cos A}-\frac{\sin^2 A- \cos^2 A}{\sin A\cos A}$$
$$=\frac{\cos^2 A}{\sin A\cos A}$$
$$=\frac{\cos A}{\sin A}$$
$$=\cot A$$
But in general I do think thatit would be nice to use more weapons than just breaking everything into sines and cosines.
As a starting point we have
$$\cos^2\theta + \sin^2\theta = 1$$
$$\tan^2\theta + 1 = \sec^2\theta$$
$$\cot^2\theta + 1 = \csc^2\theta$$
$$\tan\theta = \frac{\sin\theta}{\cos\theta}, \cot\theta = \frac{\cos\theta}{\sin\theta}$$
$$\csc\theta = \frac{1}{\sin\theta}, \sec\theta = \frac{1}{\cos\theta}, \cot\theta = \frac{1}{\tan\theta}$$
But we could use the above identities to come up with these identities
$$\cot\theta\tan\theta = \sin\theta\csc\theta = \cos\theta\sec\theta = 1$$
$$\tan\theta = \frac{\sec\theta}{\csc\theta}, \cot\theta = \frac{\csc\theta}{\sec\theta}$$
So lets try the same problem again, but this time deploying these new weapons.
$$\tan A - \csc A \sec A (1-2\cos^2 A)$$
$$=\cot A \tan A(\tan A - \csc A \sec A (1-2\cos^2 A))$$
$$=\cot A( \tan^2 A - \frac{\sec A}{\csc A}\csc A \sec A (1-\frac{2}{\sec^2 A})))$$
$$=\cot A( \tan^2 A - \sec^2 A (1-\frac{2}{\sec^2 A})))$$
$$=\cot A( \tan^2 A - \sec^2 A + 2)$$
$$=\cot A( (\sec^2 A - 1) - \sec^2 A + 2)$$
$$=\cot A$$
The point that I'm trying to make is that sometimes it's more convenient to express everything in terms of sins and coss, but at other times it's better to express everything in terms of tans and secs (or cots and cscs). These other identities will make your trigonometric simplifications all the more powerful.
|
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Prove $\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}$ How would I simplify this difficult trigonometric identity:
$$\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}.$$
I am not exactly sure what to do.
I simplified the right side to
$$\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos^2 A}{\sin^2 A}}$$
But how would I proceed.
|
$$RHS = \frac{\tan A}{1-\tan^2 A} = \frac{\frac{\sin A}{\cos A}}{1-\frac{\sin^2 A}{\cos^2 A}}$$
$$=\frac{\frac{\sin A}{\cos A}}{\frac{\cos^2 A - \sin^2 A}{\cos^2 A}}\cdot \frac{\cos^2 A}{\cos^2 A}$$
$$=\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = LHS$$
|
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Sum of the series : $1 + 2+ 4 + 7 + 11 +\cdots$ I got a question which says
$$ 1 + \frac {2}{7} + \frac{4}{7^2} + \frac{7}{7^3} + \frac{11}{7^4} + \cdots$$
I got the solution by dividing by $7$ and subtracting it from original sum. Repeated for two times.(Suggest me if any other better way of doing this).
However now i am interested in understanding the series 1,2,4,7,11,..... In which the difference of the numbers are consecutive natural numbers.
How to find the sum of
$1+2+4+7+11+\cdots nterms$
This is my first question in MSE. If there are some guidelines i need to follow, which i am not, please let me know.
|
Let T = 1 + 2/7 + 4/7^2 + 7/7^3 + ... ----> (1)
Then
T/7 = 1/7 + 2/7^2 + 4/7^3 + ... ----> (2)
Now,
(2)-(1) => 6T/7 = 1 + 1/7 + 2/7^2 + 3/7^3 ----> (3)
Hence, (1/7)(6T/7) = 1/7 + 1/7^2 + 2/7^3 ----> (4)
So, (3)-(4)=> (1 - 1/7)(6T/7) = 1 + 0 + 1/7^2 + 1/7^3 + ...
=> (6/7)(6T/7) = 1 + (1/7^2 + 1/7^3 + .... infinity)
=> 36T/49 = 1 + (1/7^2)/(1 - 1/7) [Formula = a/(1-r)]
=> 36T/49 = 1 + (1/49)/(6/7)
=> T = (49/36)(1 + (1/49)(7/6))
=> T = 49/36 + (49/36)(1/49)(7/6)
=> T = 49/36 + 7/216
=> T = (294 + 7)/216
=> T = 301/216
|
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Value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$, $P(1)=10$, $P(2)=20$, $P(3)=30$
What will be the value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$
provided that $P(1)=10$, $P(2)=20$, $P(3)=30$?
I put these values and got three simultaneous equations in $a, b, c, d$. What is the smarter way to approach these problems?
|
Other way doing this:
We try to find reals $e,f,g$ such that $P(12)+P(-8)=eP(1)+fP(2)+gP(3)$. So, if we try to equal "$x^k$ evaluated", we gain a system of equations:
$$
\left\{\begin{array}{ccc}
1^ke+2^kf+3^kg&=&12^k+(-8)^k
\end{array}\right.,\quad k=0,\cdots,4
$$
In particular,
$$
\left\{\begin{array}{ccc}
e+f+g&=&1+1\\
e+2f+3g&=&12+(-8)\\
e+2^2f+3^2g&=&12^2+(-8)^2
\end{array}\right.
$$
and we obtain $e=100,f=-198,g=100$. Verifying the others values for $k$:
$$
\left\{\begin{array}{ccc}
100+2^3(-198)+3^3\cdot100&=&12^3+(-8)^3\\
100+2^4\cdot(-198)+3^4\cdot100&=&12^4+(-8)^4 + 19800
\end{array}\right.
$$
So, $P(12)+P(-8)=100P(1)-198P(2)+100P(3)+1980=19840$
|
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|
Spivak 's Calculus (Chapter I, Problem 16c) In problem 16(c) of chapter 1 of Calculus, Spivak asks the reader to determine the conditions under which the expression $(x + y)^4$ equals $x^4 + y^4$. Clearly,
$$ (x + y)^4 = x^4 + y^4 \Leftrightarrow x = 0 \vee y = 0 \vee 4x^2 + 6xy + 4y^2 = 0 $$
From the preceding problem, we know that $0 \leq 4x^2 + 6xy + 4y^2$. If $x = 0$ and $y = 0$ then $4x^2 + 6xy + 4y^2 = 0$. If either $x = 0$ and $y \neq 0$ or $x \neq 0$ and $y = 0$ then $0 < 4x^2 + 6xy + 4y^2$. I want to show that if $x \neq 0$ and $y \neq 0$ then $0 < 4x^2 + 6xy + 4y^2$, which is intuitively true. In order to show that $0 < 4x^2 + 6xy + 4y^2$, it suffices to show that $6xy < 4x^2 + 4y^2$, but I'm not sure how to demonstrate that this inequality is true. I would presumably derive it from the ordered field axioms in conjunction with the local assumptions of the problem.
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You have $(x+y)^4 - (x^4 + y^4) = 2 x y (2 y^2 + 3xy + 2 y^2)$.
To find the zeros of $2 y^2 + 3xy + 2 y^2$, use the quadratic formula to get $ y = \frac{x}{4} ( -3 \pm i \sqrt{7})$, hence the only zero (in $\mathbb{R}^2$) is $(0,0)$.
It follows that the zeros are $x=0$ or $y=0$.
|
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Prove trigonometry identity for $\cos A+\cos B+\cos C$ I humbly ask for help in the following problem.
If
\begin{equation}
A+B+C=180
\end{equation}
Then prove
\begin{equation}
\cos A+\cos B+\cos C=1+4\sin(A/2)\sin(B/2)\sin(C/2)
\end{equation}
How would I begin the problem I mean I think $\cos C $ can be $\cos(180-A+B)$. But I am unsure what to do next.
|
Your observation that $C=180^\circ-(A+B)$ is a good one. Recall also the following trigonometric identities: $$\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y$$ $$\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y$$ By LHS, I'll denote the expression on the left-hand side of the desired identity; by RHS, the expression on the right-hand side.
Now, the RHS is messier, so we'll start there. Let's first apply your observation to $\sin(C/2)$, along with the angle difference and sum formulas for sine, and the angle sum formula for cosine, to see that
$\begin{eqnarray*}
\sin(C/2) & = & \sin\bigl(90^\circ-(A+B)/2\bigr)\\
& = & \sin 90^\circ\cos\bigl((A+B)/2\bigr)-\cos 90^\circ\sin\bigl((A+B)/2\bigr)\\
& = & \cos(A/2+B/2)\\
& = & \cos(A/2)\cos(B/2)-\sin(A/2)\sin(B/2),
\end{eqnarray*}$
since $\sin 90^\circ=1$ and $\cos 90^\circ=0$. Thus, we see that
$$\mathrm{RHS} = 1+4\sin(A/2)\cos(A/2)\sin(B/2)\cos(B/2)-4\sin^2(A/2)\sin^2(B/2)\tag{1}$$
Recall also the double angle formulas for sine (a special case of angle sum with $x=y$): $$\sin(2x)=2\sin x\cos x.$$
Also, the Pythogorean identity and the angle sum formula for cosine (with $x=y$) gives us the following double angle formula for cosine: $$\cos(2x)=\cos^2x-\sin^2x=1-2\sin^2x,$$ from which we derive the identity $$2\sin^2x=1-\cos(2x).$$ Applying this identity, along with the double angle and angle sum formulas for sine, to $(1)$ gives us
$\begin{eqnarray*}
\mathrm{RHS} & = & 1+\bigl(2\sin(A/2)\cos(A/2)\bigr)\bigl(2\sin(B/2)\cos(B/2)\bigr)-\bigl(2\sin^2(A/2)\bigr)(2\sin^2(B/2)\bigr)\\
& = & 1+\sin A\sin B-(1-\cos A)(1-\cos B)\\
& = & \cos A + \cos B - \cos A\cos B+\sin A\sin B\\
& = & \cos A + \cos B - \cos(A+B).
\end{eqnarray*}$
At this point, we can apply your observation again, along with the angle difference formula for cosine, to see that
$\begin{eqnarray*}
\mathrm{LHS} & = & \cos A + \cos B + \cos 180^\circ\cos(A+B)-\sin 180^\circ\sin(A+B)\\
& = & \cos A + \cos B - \cos(A+B),
\end{eqnarray*}$
since $\cos 180^\circ=-1$ and $\sin 180^\circ=0$. Thus, LHS = RHS, as desired.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why isn't this square root $+$ or $-$? I was tasked with proving the identity $\tan(\frac x 2) = \dfrac {\sin(x)}{1+\cos(x)}$
I used the quotient identity for tangent and the half angle identities for sine and cosine to get $ \pm \dfrac {\sqrt{\dfrac {1-\cos(x)}{2}}}{\sqrt{\dfrac {1-\cos(x)}{2}}}$
which I reduced to $\pm \sqrt{\dfrac {1-\cos(x)}{1+\cos(x)}}$
I multiplied the fraction (within the square root) by $ \dfrac {1+ \cos(x)}{1+\cos(x)}$
Resulting in $\pm \sqrt{\dfrac {1-\cos^{2}(x)}{(1+\cos(x))^2}}$
Using the Pythagorean identity, I get $\pm\sqrt{\dfrac {\sin^{2}(x)}{(1+\cos(x))^2}}$
Taking the square root of the numerator and denominator I further reduced to $\pm \dfrac {\sin(x)}{1+\cos(x)}$
I thought I was done but when I checked my work in the answer book, it showed $ \left|\dfrac {\sin(x)}{1+\cos(x)}\right|$
Where do they get the absolute value from?
|
Avoid unnecessary squaring wherever possible.
If we use the following approach, no such confusion arises.
$$\tan\frac{x}{2}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}$$
Now multiply numerator & denominator by $2\cos\frac{x}{2}$
$$\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\frac{2\cos\frac{x}{2}\sin\frac{x}{2}}{2\cos^2\frac{x}{2}}.$$
Now $\sin2A=2\sin A\cos A$ and $\cos2A=2\cos^2A-1$
So, $\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}$
If the multiplier is $2\sin\frac{x}{2}$, $\tan\frac{x}{2}$ will be $\frac{1-\cos x}{\sin x}$ which is same as $\frac{\sin x}{1+\cos x}$
If we follow your approach also, just observe that $1+\cos x$ can not be negative as $-1≤\cos A≤1$ and $\frac{\tan\frac{x}{2}}{\sin x}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}2\cos\frac{x}{2}\sin\frac{x}{2}}=\frac{1}{2\cos^2\frac{x}{2}}$ which also $>0$.
So the sign of $\frac{\tan\frac{x}{2}}{\frac{\sin x}{1+\cos x}}$ is positive.
So the sign of $\tan\frac{x}{2}$ and $\frac{\sin x}{1+\cos x}$ are same.
|
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|
Consider $x = (2+\sqrt[]{3})^6$, $x=[x]+t$, where $[x]$ is the integer part of $x$, and $t$ is the 'non integer' part of $x$. find $x(1-t)$ consider $x = (2+\sqrt[]{3})^6$, $x=[x]+t$, where $[x]$ is the integer part of $x$, and $t$ is the 'non integer' part of $x$. find the value of $x(1-t)$
|
Note that $(2+\sqrt{3})^6+(2-\sqrt{3})^6$ is an integer, indeed an even integer. For imagine expanding each term, using the Binomial Theorem. The terms involving odd powers of $\sqrt{3}$ cancel.
We have $2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$. So $(2-\sqrt{3})^6=\frac{1}{x}$, and
$$(2+\sqrt{3})^6+(2-\sqrt{3})^6=x+\frac{1}{x}.$$
Since $2-\sqrt{3}$ is between $0$ and $0.3$, the number $(2-\sqrt{3})^6$ is positive but close to $0$. So $x$ is close to but below the integer $x+\frac{1}{x}$, and therefore
$$\lfloor x\rfloor=x+\frac{1}{x}-1.$$
Also, $t=x-(x+\frac{1}{x}-1)=1-\frac{1}{x}$, so $1-t=\frac{1}{x}$. It follows that $x(1-t)=(x)(1/x)=1$.
Remark: The result is obviously structural. In particular, the exponent $6$ is irrelevant. The same argument works with, for example, $(3+2\sqrt{2})^{99}$, and in many analogous situations.
A crucial role was played by the conjugate $2-\sqrt{3}$ of the number $2+\sqrt{3}$. This sort of thing happens very frequently.
The fact that medium sized powers of $2+\sqrt{3}$ are almost integers (but just a little bit smaller than an integer) is at first a little startling. It may be enlightening to use the calculator to compute a few powers. For similar but not identical behaviour, compute some powers of $2+\sqrt{5}$.
|
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|
Gre Question Complex Number (plug and chug) This seems like it should be easy, but I can't seem to simplify it: If $z=e^{i\frac{2\pi}{5}}$, then what is $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9$. The choices are $0, 4e^{i\frac{3\pi}{5}}, 5e^{i\frac{4\pi}{5}}, -4e^{i\frac{-2\pi}{5}}, -5e^{i\frac{3\pi}{5}},$ with the answer being $-5e^{i\frac{3\pi}{5}}.$ I can plug in the given $z$ into the equation and get $5+10e^{-i\frac{2\pi}{5}}+5e^{i\frac{2\pi}{5}}+5e^{i\frac{-4\pi}{5}}+5e^{i\frac{4\pi}{5}}$, but have been unsuccessful in simplifying it so far.
|
Note that $z^5=1$, as $z$ is a fifth root of unity, so the expression simplifies to
$$
\begin{align}
1+z+z^2+z^3+5z^4+4+4z+4z^2+4z^3+5z^4 &=5+5z+5z^2+5z^3+10z^4\\
&=5(1+z+z^2+z^3+z^4)+5z^4
\end{align}
$$
However, either by using the formula for the geometric series, or the fact that $1+X+X^2+X^3+X^4$ is the fifth cyclotomic polynomial, it follows that $1+z+z^2+z^3+z^4=0$. So the final answer is
$$
5z^4=5\exp(i8\pi/5)=-5\exp(i3\pi/5).
$$
|
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|
Prove that $\frac{x^5-x^2}{x^5+y^2+z^2}+\frac{y^5-y^2}{x^2+y^5+z^2}+\frac{z^5-z^2}{x^2+y^2+z^5}≥0 $. Given $x, y, z $ are 3 positive reals such that $xyz≥1$. Prove that
$$\frac{x^5-x^2}{x^5+y^2+z^2}+\frac{y^5-y^2}{x^2+y^5+z^2}+\frac{z^5-z^2}{x^2+y^2+z^5}≥0.$$
This question is so complicated. I failed many times to get the proof. Can anyone help me please? Thank you.
|
This is problem №3 from IMO 2005. Here you can find its solution.
|
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|
Proving $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
Possible Duplicate:
Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Prove trigonometry identity?
If $A$, $B$, and $C$ are to be taken as the angles of a triangle, then I beg someone to help me the proof of
$$\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}.$$
Thanks!
|
As Salech pointed out in the comment above, we have $A=180^\circ-(B+C)$, so applying angle sum and difference formulas for cosine and sine, we have
$\begin{eqnarray*}
\cos\frac A2 & = & \cos\left(90^\circ-\frac{B+C}2\right)\\
& = & \cos 90^\circ\cos\left(\frac B2+\frac C2\right)+\sin 90^\circ\sin\left(\frac B2+\frac C2\right)\\
& = & \sin\left(\frac B2+\frac C2\right)\\
& = & \sin\frac B2\cos\frac C2+\cos\frac B2\sin\frac C2,
\end{eqnarray*}$
since $\cos 90^\circ=0$ and $\sin 90^\circ=1$.
Hence, $$4\cos\frac A2\cos\frac B2\cos\frac C2 = 4\sin\frac B2\cos\frac B2\cos^2\frac C2+4\cos^2\frac B2\sin\frac C2\cos\frac C2.$$
Now, we've got the double angle identities $\sin(2x)=2\sin x\cos x$ and $\cos(2x)=2\cos^2x-1$, the latter of which gives us $2\cos^2x=1+\cos(2x)$. Hence, by substitution and application of angle addition formula from sine,
$\begin{eqnarray*}
4\cos\frac A2\cos\frac B2\cos\frac C2 & = & \left(2\sin\frac B2\cos\frac B2\right)\left(2\cos^2\frac C2\right)+\left(2\cos^2\frac B2\right)\left(2\sin\frac C2\cos\frac C2\right)\\
& = & \sin B(1+\cos C)+(1+\cos B)\sin C\\
& = & \sin B\cos C+\cos B\sin C+\sin B+\sin C\\
& = & \sin(B+C)+\sin B+\sin C.
\end{eqnarray*}$
Finally, observe that $\sin(180^\circ-x)=\sin x$, so since $A=180^\circ-(B+C)$, then $\sin A=\sin(B+C)$, and therefore, $$4\cos\frac A2\cos\frac B2\cos\frac C2=\sin A+\sin B+\sin C,$$ as desired.
|
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|
$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$ Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$
prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$$
|
Another way.
Let $a=\frac{y}{x}$ and $b=\frac{z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and by C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{a}{\sqrt{a+b}}=\sum_{cyc}\frac{\frac{y}{x}}{\sqrt{\frac{y}{x}+\frac{z}{y}}}=\sum_{cyc}\frac{\sqrt{y^3}}{\sqrt{x(y^2+xz)}}=\sum_{cyc}\frac{y^2}{\sqrt{xy(y^2+xz)}}\geq$$
$$\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}\sqrt{xy(y^2+xz)}}=\frac{2\sqrt2(x+y+z)^2}{2\sum\limits_{cyc}\sqrt{2xy\cdot(y^2+xz)}}\geq\frac{2\sqrt2(x+y+z)^2}{\sum\limits_{cyc}(2xy+y^2+xz)}=$$
$$=\frac{2\sqrt2(x+y+z)^2}{\sum\limits_{cyc}(x^2+3xy)}\geq\frac{2\sqrt2(x+y+z)^2}{\sum\limits_{cyc}\left(x^2+\frac{1}{3}x^2+\frac{8}{3}xy\right)}=\frac{2\sqrt2(x+y+z)^2}{\frac{4}{3}(x+y+z)^2}=\frac{3}{\sqrt2}.$$
|
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|
Prove that $\left (\frac{a^2 + b^2 +c^2}{a+b+c} \right) ^ {(a+b+c)} > a^a b^b c^c$ Prove that $\left (\dfrac{a^2 + b^2 +c^2}{a+b+c} \right) ^ {(a+b+c)} > a^a b^b c^c$ if $a$, $b$ and $c$ are distinct natural numbers. Is it possible using induction?
|
Rewrite it as:
$$
\left( a \frac{a}{a+b+c} + b \frac{b}{a+b+c} + c \frac{c}{a+b+c} \right) > a^\frac{a}{a+b+c} \cdot b^\frac{b}{a+b+c} \cdot c^\frac{c}{a+b+c}
$$
This is Jensen's inequality:
$$
\log\left(\mathsf{E}\left(X\right)\right) > \mathsf{E}\left(\log\left(X\right)\right) \quad \text{or}\quad \mathsf{E}\left(X\right) > \exp \left( \mathsf{E}\left(\log\left(X\right)\right) \right)
$$
where $X$ is the random variable which can assume one of three possible values $\{a,b,c\}$ with respective probabilities $\{ \frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c} \}$.
|
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|
how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$
Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$
how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$
we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$
now, if $x=\sqrt[]{2+\sqrt{2}}$, then, $(x-\sqrt{2})(x+\sqrt{2})=\sqrt{2}$
|
put $x = \sqrt{2+x}
\implies x^2 = 2+x \implies x^2 -x-2=0$
Now just solve the quadratic equation.
|
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|
Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$ I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \frac{x(2x - 13) - 15(2x+3)}{x(2x + 3)} \lt 0$$
$$\iff \frac{2x^2 - 13x - 30x - 45}{2x(x + \frac{3}{2})} \lt 0 \iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 $$
$$\iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 \iff \frac{(x - \frac{43}{4})^2 - \frac{1939}{4}}{x(x+\frac{3}{2})} \lt 0$$
I don't know how to get any further, and I'm starting to get too high values to handle.
The next step as I can see would be to find an $x$ that makes $(x - \frac{43}{4})^2 = \frac{1939}{4}$. This, alongside the obvious ones for $x$ and $x+\frac{3}{2}$ (creating division by $0$), would help me find the possible values for $x$.
But how do I get the last step? Or am I already dead wrong?
|
After simplifying you have $\frac{2x^2-43x-45}{x(2x+3)}=\frac{(2x-45)(x+1)}{x(2x+3)}<0$. So we consider the function on the intervals $x<-\frac{3}{2}$, $x \in (-3/2,-1)$, $x \in (-1,0)$, $x \in (0,\frac{45}{2})$, $x>\frac{45}{2}$. At $x=-2$ the function is positive and the sign alternates in each successive region.
Thus the function is negative for $x \in (-3/2,-1)$ and $x \in (0,\frac{45}{2})$.
|
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|
Prove $\cos((3\pi/4)+x)-\cos((3\pi/4)-x)=(-\sqrt{2})\sin(x)$ Please help me to solve the below trigonometric function as i am trying it from the last hour.
$$\cos((3\pi/4)+x)-\cos((3\pi/4)-x)=(-\sqrt{2})\sin(x)$$
|
Apply, $\cos(A+B)-\cos(A-B)=-2\sin A \sin B$
So, $\cos(3\frac{\pi}{4}+x)-\cos(3\frac{\pi}{4}-x)=-2\sin x\sin(3\frac{\pi}{4}) $
$=-2\sin x\sin(\pi-\frac{\pi}{4})=-2\sin x\sin(\frac{\pi}{4})$
as $\sin(\pi-y)=\sin y$
So, $\cos(3\frac{\pi}{4}+x)-\cos(3\frac{\pi}{4}-x)=-\sqrt2\sin x$
or apply $\cos 2C- \cos 2D=-2sin(C+D)sin(C-D)$
Put $2C=3\frac{\pi}{4}+x, 2D=3\frac{\pi}{4}-x=>C+D=3\frac{\pi}{4}$ and $C-D=x$
|
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|
If $A>0$ and $B>0$ and $A^{-1}If $A>0$ and $B>0$ and $A^{-1}<B$, then can we claim that $x^T(AB)x\geq x^Tx$?
($A$ and $B$ are real symmetric matrices).
I know these facts (from Matrix Mathematics book) (It seemed to me these might help but I haven't been able to use them in my advantage!),
Let $A,B\in\mathbb{F}^{n\times n}$ (real or complex matrix), and assume that $A$ and $B$ are positive semi-definite. Then, $0\leq A<B$ if and only if $\rho(AB^{-1})<1$.
Let $A,B\in\mathbb{N}^{n\times n}$ (positive semi-definite matrix). Then AB is semi simple, and every eigenvalue of $AB$ is nonnegative. If in addition $A$ and $B$ are positive definite, then every eigenvalue of $AB$ is positive.
|
Since $B=A^{-1}+C$ with $C$ symmetric and positive, one asks whether $x^TACx\geqslant0$ for every vector $x$ and every symmetric positive matrices $A$ and $C$.
Consider $A=\begin{pmatrix}5 & 2 \\ 2 & 1\end{pmatrix}$ and $C=\begin{pmatrix}1 & -2 \\ -2 & 10\end{pmatrix}$. Then $A$ and $C$ are symmetric positive, $A^{-1}=\begin{pmatrix}1 & -2 \\ -2 & 5\end{pmatrix}$, $B=\begin{pmatrix}2 & -4 \\ -4 & 15\end{pmatrix}$ and $AC=\begin{pmatrix}1 & 10 \\ 0 & 6\end{pmatrix}$. Hence $x^TACx\lt0$ for $x^T=(1,-1)$.
Thus, the result above does not hold in full generality.
|
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|
Radius of the spherical image of a circle This is question 5 on page 20 of the book Complex Analysis by Lars Ahlfors. I have no idea how to answer that problem:
Find the radius of the spherical image of the circle in the plane whose center is $a$ and radius is $R$.
Here spherical image means: the image of a subset of complex numbers under the identification of the complex plane with the sphere $\Bbb S^2$ (the Riemann sphere) by stereographic projection:
Thanks.
|
I think I managed to write it properly.
Let's denote the stereographic projection from $\Bbb C$ to the sphere by $\varphi$. Let's denote $C(a,R)$ the circle with center $a$ and radius $R$.
Assume first that $a\in[0,\infty[$. Then since $\varphi$ maps circles into circles, the diameter of the circle image is
$$D = \sup\{d(\varphi(z),\varphi(a+R)):z\in C(a,R)\}.$$
The approach is to find a bound for
$$\{d(\varphi(z),\varphi(a+R)):z\in C(a,R)\}$$
and then see that the bound is reached at a point in the circle so it has to be the sup.
So, observe that
\begin{align}
d(\varphi(z),\varphi(a+R))
&= \frac{2\lvert z - (a+R)\rvert}{\sqrt{(1+\lvert z\rvert^2)(1+\lvert a + R\rvert^2)}} \\
&\leq \frac{2(\lvert z-a\rvert + R)}{\sqrt{(1+\lvert z\rvert^2)(1+\lvert a + R\rvert^2)}} \\
&= \frac{4R}{\sqrt{(1+\lvert z\rvert^2)(1+\lvert a + R\rvert^2)}} \tag{1}
\end{align}
In the other hand, if $z = x + iy$ is in $C(a,R)$ then
$$ R^2 = (x-a)^2 + y^2,$$
so
\begin{align}
\lvert z \rvert^2 &= x^2 + R^2 - (x-a)^2 \\
&= x^2 + R^2 - x^2 + 2xa - a^2 \\
&= R^2 + 2xa - a^2\tag{2}
\end{align}
but since $z$ is in the circle $\Re z = x \in [a-R,a+R]$. So by (2)
\begin{align}
\lvert z \rvert^2 &= R^2 + 2xa - a^2 \\
&\geq R^2 + 2(a-R)a - a^2 \\
&= (a-R)^2.
\end{align}
So
\begin{align}
1 + \lvert z\rvert^2 &\geq 1 + (a-R)^2 \\
\frac{1}{\sqrt{1 + \lvert z\rvert^2}} &\leq \frac{1}{\sqrt{1 + (a-R)^2}}.
\end{align}
By (1)
$$
d(z,a+R) \leq \frac{4R}{\sqrt{(1+\lvert a - R\rvert^2)(1+\lvert a+R\rvert^2)}}
$$
and the bound is reached at $z=a-R$, so the radius is
$$\frac{2R}{\sqrt{(1+\lvert a - R\rvert^2)(1+\lvert a+R\rvert^2)}}.$$
Now, notice that the formula
$$d(\varphi(z),\varphi(z'))=\frac{2|z-z'|}{\sqrt{(1+|z|^2)(1+|z'|^2)}}$$
is invariant under rotations.
So if $a\in\Bbb C \setminus [0,\infty[$, in particular it is not $0$, so rotate by $\frac{\bar a}{\lvert a \rvert}$ and we are done.
|
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|
Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities Please help to solve the following inequality using rearrangement inequalities.
Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that
\begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\end{equation}
Thanks.
|
Since $ab=1$, we have $b = \frac{1}{a}$, and thus your inequality is equivalent to $\displaystyle\frac{a}{a^2+3}+\frac{a}{1+3a^2}\leq\frac{1}{2}$.
You can simply define $f(a) = \displaystyle\frac{a}{a^2+3}+\frac{a}{1+3a^2}$ and maximize $f$.
|
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|
Solution to a second order recurrence relation with non constant coefficient I have the following equation:
$(aq^n+b+s)C_n(s)=aq^nC_{n+1}(s)+bC_{n-1}(s)$ , for $n\geq1$.
$C_0(s)=1$ , and for all $s\geq 0$ we have $0\leq C_n(s)\leq1$.
$a>0$, $b>0$, $0\leq q\leq1$.
In fact, $C_n(s)$ is a Laplace Transform of a non-negative RV, for all n.
Any thoughts on how to solve this equation?
Thanks!
|
As
$$
G_n(s) = \frac{1}{a q^{n-1}}\left((a q^{n-1}+b+s)G_{n-1}(s)-b G_{n-2}(s)\right)
$$
with
$$
G_0(s) = 1, \ \ \ G_1(s) = \frac{c_1}{s+b_1}
$$
the recurrence formula which follows, in MATHEMATICA, generates all the instances for $G_n(s)$
G[s, 0] = 1;
G[s, 1] = Subscript[c, 1] /(s + b_0);
G[s_, n_] := ((a q^(n - 1) + b + s) G[s, n - 1] - b G[s, n - 2]) q^(1 - n)/a
G[s,4] // Expand
$$
G_4(s) = \frac{1}{s+b_1}\left(\frac{b^3 c_1}{a^3 q^6}-\frac{b^3 s}{a^3 q^6}-\frac{b_1 b^3}{a^3 q^6}+\frac{3 b^2 c_1 s}{a^3 q^6}-\frac{2 b^2 s^2}{a^3 q^6}-\frac{2
b_1 b^2 s}{a^3 q^6}+\frac{3 b c_1 s^2}{a^3 q^6}-\frac{b s^3}{a^3 q^6}-\frac{b_1 b s^2}{a^3 q^6}+\frac{c_1 s^3}{a^3 q^6}+\frac{b^2
c_1}{a^2 q^3}-\frac{b^2 s}{a^2 q^3}-\frac{b_1 b^2}{a^2 q^3}+\frac{b c_1 s}{a^2 q^5}+\frac{b c_1 s}{a^2 q^4}+\frac{2 b c_1 s}{a^2
q^3}-\frac{b s^2}{a^2 q^4}-\frac{b_1 b s}{a^2 q^4}-\frac{b s^2}{a^2 q^3}-\frac{b_1 b s}{a^2 q^3}+\frac{c_1 s^2}{a^2 q^5}+\frac{c_1
s^2}{a^2 q^4}+\frac{c_1 s^2}{a^2 q^3}+\frac{b c_1}{a q}-\frac{b s}{a q}-\frac{b_1 b}{a q}+\frac{c_1 s}{a q^3}+\frac{c_1 s}{a
q^2}+\frac{c_1 s}{a q}+c_1\right)
$$
So $G_n(s)$ is non causal for $n \ge 2$.
|
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|
Automorphic numbers
Problem. We say that the $n$-digit number $x$ is automorphic iff $x^2\equiv x \mod(10^n)$. Prove that if $x$ is $n$-digit automorphic number then $(3x^2-2x^3)\mod(10^{2n})$ is $2n$-digit automorphic number. Hint: use Chinese reminder theorem to find the necessary and sufficient condition for number to be automorphic.
So from Chinese reminder theorem we have that $x$ is automorphic iff: $$ \begin{cases} x(x-1)\equiv 0 \mod(2^n)\\ x(x-1)\equiv 0 \mod(5^n)\end{cases} $$ which gives us four systems of equations:
$$ \begin{cases} x\equiv 0 \mod(2^n)\\ x\equiv 0 \mod(5^n)\end{cases} $$
$$ \begin{cases} x\equiv 1 \mod(2^n)\\ x\equiv 1 \mod(5^n)\end{cases} $$
$$ \begin{cases} x\equiv 0 \mod(2^n)\\ x\equiv 1 \mod(5^n)\end{cases} $$
$$ \begin{cases} x\equiv 1 \mod(2^n)\\ x\equiv 0 \mod(5^n)\end{cases} $$
and it's easy to check that thesis is true for the first two cases, just by simple operations. But how to check the last two?
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(3)If $x≡0\pmod{2^n}=a2^n$ for some integer $a$
$3x^2-2x^3=3(a2^n)^2-2(a2^n)^3$ is clearly divisible by $2^{2n}$
Here $x≡1\pmod{5^n}=(b5^n+1)$ for some integer $b$
$3x^2-2x^3=3(b5^n+1)^2-2(b5^n+1)^3=(b5^n+1)^2(3-2(b5^n+1))$
$≡(1+2b5^n+b^25^{2n})(1-2b5^n)≡(1+2b5^n)(1-2b5^n)=1-4b^25^{2n}≡1\pmod{5^{2n}}$
(4)If $x≡0\pmod{5^n}=c5^n$ for some integer $c$
$3x^2-2x^3=3(c5^n)^2-2(c5^n)^3$ is clearly divisible by $5^{2n}$
Here $x≡1\pmod{2^n}=(d2^n+1)$ for some integer $d$
$3x^2-2x^3=3(d2^n+1)^2-2(d2^n+1)^3=(d2^n+1)^2(3-2(d2^n+1))$
$≡(1+2d2^n)(1-2d2^n)=1-4b^22^{2n}≡1\pmod{2^{2n}}$
Alternatively, if $x=ma+b$ where $0 ≤b<m$ So,$x≡b\pmod {m}≡b\pmod {m^2}$,
$3x^2-2x^3=3(ma+b)^2-2(ma+b)^3≡3b^2-2b^3-6mab(b-1)\pmod {m^2}$
If $3b^2-2b^3-6mab(1-b)≡b\pmod {m^2}$
$m^2\mid 2b^3-3b^2+b+6mab(b-1)$
$m^2\mid b(b-1)(2b-1+6ma)$
Clearly two of the three solutions are $b=0,1\pmod {m^2}$
$\implies$
if $x=ma,m^2\mid(3x^2-2x^3)$
if $x=ma+1≡1\pmod {m}, 3x^2-2x^3≡1\pmod {m^2}$
|
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"url": "https://math.stackexchange.com/questions/192720",
"timestamp": "2023-03-29T00:00:00",
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|
Inequality with unusual constraint $a,b,c\in (0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$ Suppose as in the title that $a,b,c$ are three real positive numbers in $(0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$. Then I was asked to prove that $$a+b+c\leq \frac 32.$$ I am not very good at inequalities, so can anybody help me with this task? Especially the first condition seems very unusual to me. Thanks.
|
As $a,b,c \in (0,1)$, let $a=\cos A, b=\cos B,c=\cos C$, so that $0<A,B,C<\frac{\pi}{2}$
So, $\cos^2A+\cos A(2\cos B\cos C)+\cos^2B+\cos^2C-1=0$
$$\implies \cos A=\frac{-2\cos B\cos C±\sqrt{(2\cos B\cos C)^2-4\cdot 1\cdot (\cos^2B+\cos^2C-1)}}{2}$$
$$\implies \cos A=-\cos B\cos C±\sin B\sin C$$ as
$(2\cos B\cos C)^2-4(\cos^2B+\cos^2C-1)=4(1-\cos^2B)(1-\cos^2C)=4\sin^2B\sin^2C$
Taking the '+' sign, $ \cos A=-\cos B\cos C+\sin B\sin C=-\cos(B+C)$
$=\cos(\pi±(B+C))$ or $A+B+C=\pi$ as $0<A,B,C<\frac{\pi}{2}$
If $f(x)=\cos x,f'(x)=-\sin x, f''(x)=-\cos x<0$ as $0<x<\frac{\pi}{2}$
So, $\cos x$ is concave function in $(0,\frac{\pi}{2})$.
So using Jensen's inequality, $$\sum \cos A≤3\cos\left(\frac{A+B+C}{3}\right)=3\cos\frac{\pi}{3}=\frac{3}{2}$$
Taking the '-' sign, $ \sin A=-\sin B\sin C-\cos B\cos C$
$=-\cos(B-C) =\cos(\pi±(B-C))$
$\implies A-B+C=\pi$ or $A+B-C=\pi$ which is impossible as $0<A,B,C<\frac{\pi}{2}$.
Alternatively, we can put $a=\sin A,$ etc., where $0<A,B,C<\frac{\pi}{2}$.
Then $\sin A = \cos(B+C)=\sin(\frac{\pi}{2}±(B+C))\implies A+B+C=\frac{\pi}{2}$, as $A-B-C=\frac{\pi}{2}$ is not allowed as $0<A,B,C<\frac{\pi}{2}$.
Again, $\sin x $ is concave function in $(0,\frac{\pi}{2})$.
So using Jensen's inequality , $$\sum \sin A≤3\sin\left(\frac{A+B+C}{3}\right)=3\sin\frac{\pi}{6}=\frac{3}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/194143",
"timestamp": "2023-03-29T00:00:00",
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|
Showing a sequence is convergent using the $\epsilon$-$N$ definition I need to prove that the following sequence converges:
$\lim_{n\rightarrow \infty} \frac{2n^2+3n+1}{n^2+n+1}=2$
So for the proof/solution I have the following:
Let $\epsilon >0$. Then let $N=\frac{1}{\epsilon}$. Then for all $n\geq N$,
$|\frac{2n^2+3n+1}{n^2+n+1}-2| = |\frac{n-1}{n^2+n+1}| < \frac{1}{n} < \frac{1}{N} = \epsilon$
Thus the sequence converges to 2.
Is this the correct way of going about this? Thanks in advance.
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There is a small error with symbols, it should be like this:
Choose $\epsilon\gt 0$
Let $N=\lceil\frac{1}{\epsilon}\rceil$. Then for all $n\gt N$,
$|\frac{2n^2+3n+1}{n^2+n+1}-2| = |\frac{n-1}{n^2+n+1}| < \frac{1}{n} < \frac{1}{N} = \epsilon$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/195030",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int\frac{dx}{\cos^6 x}$ Please for the substition method solving $\int\frac{dx}{\cos^6 x}$
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$\int\frac{dx}{\cos^6 x}$=$\int\frac{1}{\cos^4 x}\cdot\frac{dx}{\cos^2 x}$=$\int\frac{1}{(\frac{1}{\sqrt{1+\tan^2 x}})^4}\cdot\frac{dx}{\cos^2 x}$=$\int(1+\tan^2 x)\cdot\frac{dx}{\cos^2x}$=$|\tan x=t\Rightarrow\frac{dx}{\cos^2 x}=dt|$=$\int(1+t^2)^2dt$=$\int(1+2t^2+t^4)dt$=$t+\frac{2}{3}t^3+\frac{1}{5}t^5$=$\tan x+\frac{2}{3}\tan^3 x+\frac{1}{5}\tan^5 x+C$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/195088",
"timestamp": "2023-03-29T00:00:00",
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|
Inequality: $7a+5b+12ab\le9$ If we assume that $a,b$ are real numbers such that $9a^2+8ab+7b^2\le 6$, how to prove that :
$$7a+5b+12ab\le9$$
|
We have
$$2(a-b)^2+7\left(a-\frac{1}{2}\right)^2 + 5\left(b-\frac{1}{2}\right)^2 \geq 0$$
which is equivalent to
$$7a+5b+12ab\leq 9a^2+7b^2+8ab+3 \leq 6+3=9$$
The motivation here is to search for equality case by solving the system of equation in real values $a,b$
\begin{equation*}
\begin{cases}
7a+5b+12ab=9 \\
9a^2+7b^2+8ab=6
\end{cases}
\end{equation*}
which yields $a=b=\frac{1}{2}$. Thus the factors $\left(a-\frac{1}{2}\right)^2$, $\left(b-\frac{1}{2}\right)^2$ and $(a-b)^2$ are in order.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/196128",
"timestamp": "2023-03-29T00:00:00",
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|
Find the inverse a matrix with trigonometic entries What is the inverse of
\[
\begin{pmatrix}
1&0&0\\0&\cos x &\sin x\\ 0 &\sin x &-\cos x \end{pmatrix}
\]
Please help me to solve the above problem.
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Implement the formula $\def\adj{\operatorname{adj}}A^{-1}=\frac{1}{\det A}\cdot \adj A$
Find $\det A$
$\det A=\begin{vmatrix} 1&0&0\\0&\cos x &\sin x\\ 0 &\sin x &-\cos x \end{vmatrix}=-1$
Find $\adj A$
$A_{11}=(-1)^{1+1}\left\lvert \begin{array}{cc} \cos x & \sin x \\ \sin x & -\cos x \end{array} \right\rvert=-1$
$A_{12}=(-1)^{1+2}\left\lvert \begin{array}{cc} 0 & \sin x \\ 0 & -\cos x \end{array} \right\rvert=0$
$A_{13}=(-1)^{1+3}\left\lvert \begin{array}{cc} 0 & \cos x \\ 0 & \sin x \end{array} \right\rvert=0$
$A_{21}=(-1)^{2+1}\left\lvert \begin{array}{cc} 0 & 0 \\ \sin x & -\cos x \end{array} \right\rvert=0$
$A_{22}=(-1)^{2+2}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & -\cos x \end{array} \right\rvert=-\cos x$
$A_{23}=(-1)^{2+3}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & \sin x \end{array} \right\rvert=-\sin x$
$A_{31}=(-1)^{3+1}\left\lvert \begin{array}{cc} 0 & 0 \\ \cos x & \sin x \end{array} \right\rvert=0$
$A_{32}=(-1)^{3+2}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & \sin x \end{array} \right\rvert=-\sin x$
$A_{33}=(-1)^{3+3}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & \cos x \end{array} \right\rvert=\cos x$
$A^{-1}=\left\lvert \begin{array}{ccc} 1&0&0\\ 0 &\cos x & \sin x \\ 0 & \sin x & -\cos x \end{array} \right\rvert$
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{
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|
Show $8\mid n^2-1$ if $n$ is an odd positive integer. Show that $n^2-1$ is divisible by $8$, if $n$ is an odd positive integer.
Please help me to prove whether this statement is true or false.
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We know that any odd positive integer is of the form $4q+1$ or, $4q+3$ for some integer $q$.
So, we have the following cases:
Case 1 when $n=4q+1$: In this case, we have
$n^2 -1=(4q+1)^2 -1=16q^2 +8q+1-1=16q^2 +8q=8q(2q+1)$
$∴ n^2 -1$ is divisible by 8 [$∵ 8q(2q+1)$ is divisible by 8]
Case 2 when $n=4q+3$: In this case, we have
$n^2 -1=(4q+3)^2 -1= 16q^2 +24q+9-1=16q^2 +24q+8$
⇒ $n^2 -1=8(2q^2 +3q+1)=8(2q+1)(q+1)$
$∴ n^2 -1$ is divisible by 8 [$∵ 8(2q+1)(q+1)$ is divisible by 8]
Hence, $n^2 -1$ is divisible by 8.
|
{
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"timestamp": "2023-03-29T00:00:00",
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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.