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Generating function - technical issue. Which sequence is generated by $\frac{{5x - 3{x^2}}}{{{{(1 - x)}^3}}}$?
We know that:
$$\frac{1}{{{{(1 - x)}^3}}} = \sum\limits_{j = 0}^\infty {\left( {\begin{array}{*{20}{c}}
{j + 2} \\
2 \\
\end{array}} \right){x^j}} $$
So we have:
$$(5x - 3{x^2}) \cdot \sum\limits_{j = 0}^\infty {\left( {\begin{array}{*{20}{c}}
{j + 2} \\
2 \\
\end{array}} \right){x^j}} $$
Now it's easy to see that for $x^k$, $a_k$ can be defined by:
$$5\left( {\begin{array}{*{20}{c}}
{k + 1} \\
2 \\
\end{array}} \right) - 3\left( {\begin{array}{*{20}{c}}
k \\
2 \\
\end{array}} \right)$$
Because we "used" $x^{k-1}$ and $x^{k-2}$ there may be a problem for $k=0,1$ and I've been told I need to check it directly. Can you help me with that?
|
I think you're overthinking things. First of all,
\begin{align*}
5 {k + 1 \choose 2} - 3{k \choose 2}
&= 5\frac{(k+1)k}{2} - 3\frac{k(k-1)}{2} \\
&= \frac{5}{2} k^2 + \frac52 k - \frac32k^2 + \frac32 k \\
&= 4k + k^2 \\
&= k(k + 4) \\
\end{align*}
So your sequence is $a_k = k(k+4)$. Second of all, they probably want you to plug in $k = 0$ and $k = 1$, giving
$$
a_0 = 0, a_1 = 5
$$
Now check that these particular cases line up with the series expression you had before
$$(5x - 3{x^2}) \cdot \sum\limits_{j = 0}^\infty {\left( {\begin{array}{*{20}{c}}
{j + 2} \\
2 \\
\end{array}} \right){x^j}}. $$
That is, check that the series above starts out with $0 + 5x^1 + \cdots$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
The normal line intersects a curve at two points. What is the other point? The line that is normal to the curve $\displaystyle x^2 + xy - 2y^2 = 0 $ at $\displaystyle (4,4)$ intersects the curve at what other point?
I can not find an example of how to do this equation. Can someone help me out?
|
Given the curve $\mathcal{C}$ as
$$
x^2 + x y - 2 y^2 = 0. \tag 1
$$
Solve the curve $\mathcal{C}$
$$
x^2 + x y - 2 y^2 = 0.\\
\Downarrow\\
4 x^2 + 4 x y - 8 y^2 = 0.\\
\Downarrow\\
\Big( 2 x + y \Big)^2 - 9 y^2 = 0.\\
\Downarrow\\
\Big( 2 x + y \Big)^2 = \Big( 9 y \Big)^2.\\
\Downarrow\\
2 x + y = \pm 3 y.\\
\Downarrow\\
2 x = \Big( \pm 3 - 1 \Big) y.\\
\Downarrow\\
y = \frac{2}{ \pm 3 - 1 } x.\\
\Downarrow\\
y = \frac{2}{2} x \vee y = \frac{2}{-4} x.\\
$$
So the curve $\mathcal{C}$ are two crossing lines,
given by
$$
y = x \vee y = - \frac{1}{2} x. \tag{2}
$$
The normal through point $(4,4)$
The normal through point $(4,4)$ is given by
$$
y = 8 - x. \tag 3
$$
Intersection
We need to find the intersection between
$$
y = - \frac{1}{2} x
$$
and
$$
y = 8 - x.
$$
So we get
$$
y = - \frac{1}{2} x = 8 - x.\tag 4
$$
Solving intersection
As
$$
- \frac{1}{2} x = 8 - x,
$$
we get
$$
\frac{1}{2} x = 8,
$$
so
$$
x = 16.
$$
And we have
$$
y = - \frac{1}{2} x,
$$
so
$$
y = - 8.
$$
Solution
The solution is given by
$$
\bbox[16px,border:2px solid #800000] { (x,y) = (16,-8). } \tag 5
$$
|
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"url": "https://math.stackexchange.com/questions/707716",
"timestamp": "2023-03-29T00:00:00",
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|
Definite integration with natural logarithm $$\int_0^a \ln(x)\ln(a-x)\,dx$$
How to do this? I couldn't proceed at all.
($\ln$ is natural logarithm.)
|
$a>0$
\begin{align}J&=\int_0^a \ln(x)\ln(a-x)\,dx\\
&=\frac{1}{2}\int_0^a \ln^2(x)dx+\frac{1}{2}\int_0^a \ln^2(a-t)dt-\frac{1}{2}\int_0^a \ln^2\left(\frac{u}{a-u}\right)du\\
&\overset{x=a-t,x=\frac{u}{2-u}}=\int_0^a \ln^2(x)dx-\frac{a}{2}\int_0^\infty \frac{\ln^2 x}{(1+x)^2}dx\\
&=\int_0^a \ln^2(x)dx-\frac{a}{2}\int_0^1 \frac{\ln^2 x}{(1+x)^2}dx-\frac{a}{2}\int_1^\infty \frac{\ln^2 u}{(1+u)^2}du\\
&\overset{x=\frac{1}{y}}=\int_0^a \ln^2(x)dx-a\int_0^1 \frac{\ln^2 x}{(1+x)^2}dx\\
&\overset{\text{IBP}}=\int_0^a \ln^2(x)dx+a\left[\left(\frac{1}{1+x}-1\right)\ln^2 x\right]_0^1-2a\int_0^1 \left(\frac{1}{1+x}-1\right)\frac{\ln x}{x}dx\\
&=\int_0^a \ln^2(x)dx+2a\int_0^1 \frac{\ln x}{1+x}dx\\
&=\int_0^a \ln^2(x)dx+2a\int_0^1 \frac{\ln x}{1-x}dx-2a\int_0^1 \frac{2u\ln u}{1-u^2}du\\
&\overset{x=u^2}=\int_0^a \ln^2(x)dx+2a\int_0^1 \frac{\ln x}{1-x}dx-a\int_0^1 \frac{\ln x}{1-x}dx\\
&=\int_0^a \ln^2(x)dx+a\int_0^1 \frac{\ln x}{1-x}dx\\
&=\Big[x\left(\ln^2 x-2\ln x+2\right)\Big]_0^a+a\times -\frac{\pi^2}{6}\\
&=\boxed{a\left(\ln^2 a-2\ln a+2\right)-\frac{a\pi^2}{6}}
\end{align}
NB: i assume that: $\displaystyle \int_0^1 \frac{\ln x}{1-x}dx=-\zeta(2)=-\frac{\pi^2}{6}$
|
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"url": "https://math.stackexchange.com/questions/708072",
"timestamp": "2023-03-29T00:00:00",
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|
Inverse trigonometric Problem For any $x \in [-1,0) \cup (0,1]$, how can I prove that:
$$\sin^{-1}(2x\sqrt{1-x^2})=2\cos^{-1}x$$
Also, can someone explain to me how to understand the graphs of $sin$ and $cos$ functions?
|
Sufficient care has to be taken while dealing with Inverse trigonometric functions
as the principal values of $\cos^{-1}x$ lies in $\left[0,\pi\right]$ whereas it lies in $\left[-\frac\pi2,\frac\pi2\right]$ for $\sin^{-1}y$
$\displaystyle\iff0\le2\cos^{-1}x\le2\pi$
Case $\#1:\displaystyle2\cos^{-1}x$ will be $\displaystyle\sin^{-1}2x\sqrt{1-x^2}$
iff $\displaystyle-\frac\pi2\le2\cos^{-1}x\le\frac\pi2\iff -\frac\pi4\le\cos^{-1}x\le\frac\pi4$
$\displaystyle\implies0\le\cos^{-1}x\le\frac\pi4\iff1\ge x\ge\frac1{\sqrt2}$
Case $\#2:$
If $\displaystyle\frac\pi4<\cos^{-1}x\le\frac\pi2\iff\frac1{\sqrt2}> x\ge0$
$\displaystyle\iff\frac\pi2<2\cos^{-1}x\le\pi,\sin^{-1}2x\sqrt{1-x^2}=\pi-2\cos^{-1}x$
Similarly, Case $\#3:$
$\displaystyle\iff\frac\pi2<\cos^{-1}x\le\frac{3\pi}4,\sin^{-1}2x\sqrt{1-x^2}=\pi-2\cos^{-1}x$
and Case $\#4:$
$\displaystyle\iff\frac{3\pi}4<\cos^{-1}x\le\pi,\sin^{-1}2x\sqrt{1-x^2}=2\pi-2\cos^{-1}x$
For example, if $\displaystyle x=-1,\cos^{-1}x=\cos^{-1}(-1)=\pi$
But, $\displaystyle\sin^{-1}\{2x\sqrt{1-x^2}\}=\sin^{-1}0=0$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is wrong with this method for a rotated and shifted parabola? $(x+2y)^2=4(x-y)$
Disecting the above parabola is the question. (vertex, axis,tangent at vertex,etc).
So at first what I thought of was making its equations at LHS and RHS perpendicular. I thought since it might be tedious (which wasnt at all and got me the correct answer), I thought of something else. I thought that a parabola is symmetric bout its axis. Consider $y^2=4ax$. I get two values of y for every +ve x and that is why its symmetric. But what if I put something in x such that y has no option except one i.e. the terms containing y make a perfect square. So I expanded the above and wrote it in form of a quadratic of second degree in y and put $b^2=4ac$. Ofcourse, a,b and c were in terms of x, which yielded a value which I thought was the answer but the answer came by introducing a k in the square term in LHS and making slope 2 in RHS. But what is wrong in my method?
|
This is actually the
duplicate of an answer
to the following question:
Proving that for each two parabolas, there exists a transformation taking one to the other .
According to the theory presented there, it is advantageous to write the OP's equation as follows:
$$
2(x-y) = \frac{1}{2} (x+2y)^2 \qquad \mbox{or} \qquad \eta = \frac{1}{2} \xi^2
$$
Where it is noted that $\;\xi\;$ physically corresponds with "time" . Anyway, it follows that:
$$
\left\{ \begin{array}{l} \xi = x+2y \\ \eta = 2(x-y) \end{array}\right.
\qquad \Longleftrightarrow \qquad
\left\{ \begin{array}{l}
x = \frac{1}{3}\xi + \frac{1}{3}\eta \\
y = \frac{1}{3}\xi - \frac{1}{6}\eta
\end{array}\right.
$$
Giving the "velocities", "accelerations" and "origin" in the parameter equations of the parabola.
$$
v_x = v_y = \frac{1}{3} \quad ; \quad
a_x = \frac{1}{3} \quad ; \quad
a_y = - \frac{1}{6} \quad ; \quad
s_x = s_y = 0
$$
In matrix form:
$$
\left[ \begin{array}{c} x \\ y \end{array} \right] =
\left[ \begin{array}{cc} v_x & a_x \\ v_y & a_y \end{array} \right]
\left[ \begin{array}{c} \xi \\ \xi^2/2 \end{array} \right] =
\left[ \begin{array}{cc} +\frac{1}{3} & +\frac{1}{3} \\ +\frac{1}{3} & -\frac{1}{6} \end{array} \right]
\left[ \begin{array}{c} \xi \\ \eta \end{array} \right]
$$
If you believe that a linear mapping transforms a parabola into a parabola then we are done.Otherwise.
An orthogonal coordinate transformation, still according to
the same theory, is obtained by a parameter shift $\;\tau$ :
$$
\tau = \frac{v_x a_x + v_y a_y}{a^2_x + a^2_y} = \frac{1/3 \cdot 1/3 - 1/3 \cdot 1/6}{1/9 + 1/36} = \frac{2}{5}
$$
Resulting in:
$$
x = (v_x-a_x\tau)\xi + a_x\eta + (\frac{1}{2}a_x\tau^2-v_x\tau+s_x)
= \frac{1}{5}\xi + \frac{1}{3}\eta - \frac{8}{75} \\
y = (v_y-a_y\tau)\xi + a_y\eta + (\frac{1}{2}a_y\tau^2-v_y\tau+s_y)
= \frac{2}{5}\xi - \frac{1}{6}\eta - \frac{11}{75}
$$
In matrix form:
$$
\left[ \begin{array}{c} x \\ y \end{array} \right] =
\left[ \begin{array}{cc} +\frac{1}{5} & +\frac{1}{3} \\
+\frac{2}{5} & - \frac{1}{6} \end{array} \right]
\left[ \begin{array}{c} \xi \\ \eta \end{array} \right] -
\left[ \begin{array}{c} \frac{8}{75} \\ \frac{11}{75} \end{array} \right] =
\left[ \begin{array}{cc} \frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\
\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{array} \right]
\left[ \begin{array}{c} \frac{1}{\sqrt{5}}\xi \\ -\frac{\sqrt{5}}{6}\eta \end{array} \right] -
\left[ \begin{array}{c} \frac{8}{75} \\ \frac{11}{75} \end{array} \right]
$$
The meaning of this is transforming $\;\eta = \xi^2/2\;$ into the original
$\;(x+2y)^2=4(x-y)\;$ by scaling the $\xi$ coordinate with a factor $1/\sqrt{5}$ ,
scaling the $\eta$ coordinate with a factor $-\sqrt{5}/6$ (i.e. scaling and mirroring),
followed by a rotation over an angle $\;\arctan(2)\;$ and a translation over $\;-(8,11)/75$ . Nobody will deny that $\;\eta = \xi^2/2\;$ represents a "standard" parabola and these transformations don't change that fact.Viewports:
xmin := -5; xmax := 5; xmin := -1/3; xmax := 1.1;
ymin := -5; ymax := 5; ymin := -1/3; ymax := 1.1;
Colors:
Blue: standard parabola $y = x^2/2$
Red: original parabola $(x+2y)^2=4(x-y)$
Grey: global Cartesian coordinate system / unit vectors of standard parabola.
Lime: non-orthogonal coordinate system / unit vectors if "we are done".
Green: orthogonal coordinate system if "Otherwise". Note that the system is left handed and that the axis corresponding with "acceleration" does not change / is parallel to the Lime one.
Black: translation vector $(s_x,s_y)$ corresponding with "time" $\tau$
needed to travel along the (red) parabola from the origin to that position.
A. Einstein: Things should be made as simple as possible. But not simpler.
|
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|
Prove by induction that $\sum_{i=0}^n \left(\frac 3 2 \right)^i = 2\left(\frac 3 2 \right)^{n+1} -2$
Prove, disprove, or give a counterexample:
$$\sum_{i=0}^n \left(\frac 3 2 \right)^i = 2\left(\frac 3 2 \right)^{n+1} -2.$$
I went about this as a proof by induction. I did the base case and got the LHS = RHS. When I went to show $P(k) \implies P(k+1)$ I could not get the LHS to equal the RHS. Is this because it isn't a proof by induction? Or, that it cannot be proved?
Any help would be greatly appreciated.
|
The inductive step:
$$\begin{align}\sum_{i=0}^{k+1}\left(\frac{3}{2}\right)^{i} &= \left(\frac{3}{2}\right)^{k+1} + \sum_{i=0}^{k}\left(\frac{3}{2}\right)^{i}\\
&=\left(\frac{3}{2}\right)^{k+1} + 2\cdot\left(\frac{3}{2}\right)^{k+1} - 2\\
&= 3\cdot\left(\frac{3}{2}\right)^{k+1} - 2\\
&= 3\cdot\frac{2}{3}\left(\frac{3}{2}\right)^{k+2} - 2\\
&= 2\cdot\left(\frac{3}{2}\right)^{(k+1) + 1} - 2\end{align}$$
As an alternative (and arguably, faster) method, we notice that LHS is actually a a geometric series: Let $S$ be defined as
$$S = \left(\frac{3}{2}\right)^0 + \left(\frac{3}{2}\right)^1 + \dots \left(\frac{3}{2}\right)^n$$
Then,
$$\frac{3}{2}S = \left(\frac{3}{2}\right)^1 + \left(\frac{3}{2}\right)^2 + \dots \left(\frac{3}{2}\right)^{n+1}$$
Substract $S$ from $\frac{3}{2}S$ to get
$$\begin{align}\frac{3}{2}S - S &= \left(\frac{3}{2}\right)^{n+1} - \left(\frac{3}{2}\right)^0\\
\frac{1}{2}S &= \left(\frac{3}{2}\right)^{n+1} - 1\\
S &= 2\left(\frac{3}{2}\right)^{n+1} - 2\end{align}$$
|
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|
Length of median extended to the circumcircle A triangle has side length $13,14,15$, and its circumcircle is constructed. The median is then drawn with its base having a length of $14$, and is extended to the circle. Find its length.
|
Here, $AB=13, AC=15, BC=14$. $AD$ is the median on $BC$ and we have to find the length of $AE$. So, we first take out the length of $m=AD$. Noting by law of cosines that:
$$m^2=c^2-(\frac{a}2)^2+am\cos \alpha$$
$$m^2=b^2-(\frac{a}2)^2+am\cos \beta=b^2-(\frac{a}2)^2-am\cos \alpha$$
And adding them up, to get the Apollonius formula:
$$2m^2=c^2+b^2-\frac{a^2}2$$
Thus we solve for $m$, and then substitute the values:
$$m=\frac{\sqrt{2c^2+2b^2-a^2}}2=\frac{\sqrt{2\cdot13^2+2\cdot15^2-14^2}}2=2\sqrt{37}$$
Now, we will use the fact that:
$$AD\cdot DE=BD\cdot DC$$
$$2\sqrt{37}\cdot DE=49$$
Thus:
$$AE=AD+DE=\frac{49}{2\sqrt{37}}+2\sqrt{37}=\frac{197}{2\sqrt{37}}\approx 16.193$$
|
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|
Finding the common ratio from adjacent elements in a geometric sum Is there a general solution for this problem?
Given $S_N$ and $S_{N+1}$ are the sums of geometric series, can we find the common ratio(s). (Assuming $a_0 = 1$).
To be more explicit, given
$$S_N = 1 + r + \ldots + r^{N-1} = \frac{1-r^N}{1-r}$$
and
$$S_{N+1} = 1 + r + \ldots + r^{N} = \frac{1-r^{N+1}}{1-r}\ ,$$
find $r$.
For example, given $S_N = 3$ and $S_{N+1} = 7$, find $r$.
In this example, $r=2$ is a solution, since for $r=2$ and $N=2$, $S_N = 1 + 2 = 3$ and $S_{N+1} = 1 + 2 + 4 = 7$.
|
We have
$$S_{N+1}-1=r+r^2+\cdots+r^N=rS_N\ .$$
So if $S_N$ and $S_{N+1}$ are given you can easily find
$$r=\frac{S_{N+1}-1}{S_N}\ .$$
|
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|
Trigonometric limit $\lim_{x\to0} \frac{\tan^2{(3x)}+\sin{(11x^2)}}{x\sin{(5x)}}$ How to solve this limit:
$$\lim_{x \rightarrow 0}\frac{\tan^2{(3x)}+\sin{(11x^2)}}{x\sin{(5x)}}$$
|
write
$$\frac{\tan^2(3x)+\sin(11x^2)}{x\sin 5x}=\frac{3}{1}\cdot\frac{3}{1} \cdot \frac{1}{5}\cdot \frac{\tan 3x}{3x}\cdot \frac{\tan 3x}{3x}\cdot \frac{5x}{\sin 5x}+ \frac{\sin (11x^2)}{11x^2}\cdot \frac{5x}{\sin 5x}\cdot \frac{1}{5}\cdot \frac{11}{1}$$
Making the passage to the limit when $x\rightarrow 0$ each fraction involving trigonometric numbers tends to $1$ and then we obtain
$$\lim_{x\rightarrow 0}\frac{\tan^2(3x)+\sin(11x^2)}{x\sin 5x}=\frac{9}{ 5}+\frac{11}{5}=\frac{20}{5}=4. $$
|
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|
What is the smallest positive integer n such that $2^5 · 3 · 5^2 · 7^3 ·$ $n$ is a perfect square Is there some kind of rule that I need to solve this? Can someone give me some clue how to solve this? Thanks
My teacher gave the solution as $42$.
Can someone explain why?
|
I will use the rule that:
$$a^2b^2=(ab)^2$$
We want all exponents to be even. Why? Let's use an example.
(Assume all variables are integers, and are not perfect squares themselves).
$a^2b^4$ is a perfect square because it equals $(ab^2)^2$. But $a^2b^3$ is not a perfect square because there is no way to convert it into something of the form $(xy)^2$. The best we can do is $(ab)^2\times b$, but $b$ is not in the squared term, therefore it is not a perfect square (remember, we assumed $b$ is not a perfect square).
We want to find the value of $n$ in $2^5\times 3\times 5^2\times 7^3\times n$ so that the latter expression will be a perfect square.
First step: Attempt to make as many exponents even as possible. We can rewrite our expression as:
$$2^4\times 5^2\times 7^2\times 2\times 3\times 7\times n$$
We can now group the first three terms into a perfect square.
$$(2^2\times 5\times 7)^2\times 2\times 3\times 7\times n$$
$$=140^2\times 2\times 3\times 7\times n$$
We need to find the lowest value of $n$ such that $2\times 3\times 7\times n$ is a perfect square. It is easily seen that the value of $n$ is $2\times 3\times 7$, which is $42$. Therefore:
$$\color{green}{\boxed{n=42}}$$
Hope I helped!
P.S. If you want to find out what happens when $n=2\times 3\times 7$, read on.
When $n=2\times 3\times 7$:
$$140^2\times 2\times 3\times 7\times n=140^2\times 2^2\times 3^2\times 7^2$$
$$=140^2\times (2\times 3\times 7)^2$$
$$=140^2\times 42^2$$
$$=(140\times 42)^2$$
$$=5880^2$$
$$=34574400$$
|
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|
Evaluate limit $\lim_{n \to \infty } {1 \over n^{k + 1}}\left( {k! + {(k + 1)! \over 1!} + \cdots + {(k + n)! \over n!}} \right),k \in \mathbb{N}$ Evaluate the limit:
$$\lim_{n \to \infty } {1 \over n^{k + 1}}\left( {k! + {(k + 1)! \over 1!} + \cdots + {(k + n)! \over n!}} \right),k \in \mathbb{N}$$
It looks like a classic Cesaro-Stolz problem, but applying it didn't bring me any useful result.
I've been told the following equality might be helpful: $(1 - q)(1 + q + \cdots + q^N) = (1 - q^{N + 1})$
|
Hint: It can be shown that
$$\sum_{m = 0}^n \frac{(k + m)!}{m!} = \frac{(n + 1)(k + n + 1)!}{(k + 1)(n + 1)!}.$$
You may find this formula useful.
If you then divide by $n^{k + 1}$ and take the limit as $n \to \infty$, you should get
$$\frac{1}{1 + k}.$$
Edit: Using Stirling's asymptotic formula $N! \sim N^N e^{-N}\sqrt{2\pi N}$, where $\sim$ denotes asymptotic equality, we have
$$\frac{(n + 1)(k + n + 1)!}{(n + 1)!} = \frac{(k + n + 1)!}{k^{k + 1}n!} \sim \frac{1}{n^{k + 1}} \frac{(k + n + 1)^{k + n + 1}e^{-(k + n + 1)}\sqrt{2\pi (k + n + 1)}}{n^n e^{-n}\sqrt{2\pi n}}.$$
If we simplify the right-hand side, we find that
\begin{align*}
\frac{1}{n^{k + n + 3/2}(k + n + 1)^{k + n + 3/2}e^{k + 1}} &= \left(1 + \frac{k + 1}{n}\right)^{-(k + n + 3/2)} e^{-(k + 1)}\\
&= \left(1 - \frac{-(k + 1)}{n}\right)^{-(k + n + 3/2)} e^{-(k + 1)}\\
&= \left[\left(1 - \frac{-(k + 1)}{n}\right)^{(k + n + 3/2)}\right]^{-1} e^{-(k + 1)}\\
&\to e^{k + 1}e^{-(k + 1)} \quad (\text{as } n \to \infty)\\
&= 1
\end{align*}
because
$$e^{-x} = \lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^n$$
and $k + n + 3/2 \sim n$ as $n \to \infty$ since $k \in \mathbb{N}$ is fixed.
|
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|
Compute the square root of a complex number This is a follow up to a previous question. I solved the equation $z^4 - 6z^2 + 25 = 0$ and I found four answer to be $z = \pm\sqrt{3 \pm 4i}$.
However someone in the comment said that the answer is going to be $2+i$, $2-i$, $-2+i$, $-2-i$. I cannot understand how we can find these roots from the answer that I found. How are we supposed to compute the square root of a complex number?
|
Suppose $(x+iy)^2 = a+ib$ with $a,b$ real and you want to find real values for $x$ and $y$.
Then $(x^2-y^2) + i(2xy) = a+ib$. Since $a,b,x,y$ are real, this is equivalent to $x^2-y^2 = a$ and $2xy = b$
$(x^2+y^2)^2 = (x^2-y^2)^2 + 4x^2y^2 = a^2+b^2$, so $x^2+y^2 = \sqrt {a^2+b^2}$. You always pick the positive root because $x^2+y^2$ is positive.
Then, $x^2 = \frac {\sqrt {a^2+b^2}+a}2$, so that $x = \pm \sqrt \frac {\sqrt {a^2+b^2}+a}2$, and finally $y = \frac b{2x}$ if $x \neq 0$.
If $x = 0$ you have to use $y = \pm \sqrt \frac {\sqrt {a^2+b^2}-a}2$ instead.
Here, for $a=3, b=4$, $a^2+b^2 = 9+16 = 25 = 5^2$ so that $x = \pm \sqrt {\frac{5+3}2} = \pm 2$, and $y = \frac 4 {2x} = \pm 1$ (with the same sign as $x$). This gives the two square roots of $3+4i$ .
|
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"timestamp": "2023-03-29T00:00:00",
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|
Specify finite group satisfying two conditions Let $G = \left\{ 0, 1, 2, 3, 4, 5, 6, 7\right\}$ be a group with the operation $\circ$, which satisfies the following conditions:
\begin{align}
a \circ b \leq a + b \quad & \forall a,b \in G & (a) \\
a \circ a = 0 \quad & \forall a \in G & (b)
\end{align}
I have to create the operator table for this Group.
Since $a \circ a = 0$ for all $a \in G$, $0$ is the identity element and each element is self-inverse. Using this information, I get the operator table below. Each free space $\circ(a,b)$ must satisfy $0 < \circ(a,b) \leq a + b$ (because the inverse element is unique in a group and because of the above condition (a)). Also, associativity must be satisfied (since $(G,\circ)$ is a group).
\begin{array}[t]{l|ccccccc}
\circ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
1 & 1 & 0 & & & & & & \\
2 & 2 & & 0 & & & & & \\
3 & 3 & & & 0 & & & & \\
4 & 4 & & & & 0 & & & \\
5 & 5 & & & & & 0 & & \\
6 & 6 & & & & & & 0 & \\
7 & 7 & & & & & & & 0 \\
\end{array}
I wrote a minizinc (a CSP solver) program, which fills out the operation table. There is one unique solution (symmetric among both diagonals):
\begin{array}[t]{l|rr}
\circ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
1 & 1 & 0 & 3 & 2 & 5 & 4 & 7 & 6 \\
2 & 2 & 3 & 0 & 1 & 6 & 7 & 4 & 5 \\
3 & 3 & 2 & 1 & 0 & 7 & 6 & 5 & 4 \\
4 & 4 & 5 & 6 & 7 & 0 & 1 & 2 & 3 \\
5 & 5 & 4 & 7 & 6 & 1 & 0 & 3 & 2 \\
6 & 6 & 7 & 4 & 5 & 2 & 3 & 0 & 1 \\
7 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\
\end{array}
My question: Is there an elegant way of solving this problem rather than trying out all possible combinations? If yes, whats the mapping rule of $\circ?$
Thanks in advance for any help!
|
There is a shorter way to solve the problem, although I wouldn't call it elegant.
But before that, the more important thing: there is also a very short way to describe how this operation $\circ$ that you found operates. To calculate $a \circ b$, you just write $a$ and $b$ in binary, then apply the bitwise XOR, and then convert the result back from binary to decimal.
For instance, look at this identity that I took from your operation table: $3 \circ 5 = 6$. Written in binary we have $3 = 011$ and $5 = 101$. Their bitwise XOR is $011 \oplus 101 = 110$, which is the binary notation for $6$. You can see for yourself that this rule describes your operation table completely.
Now, moving on to the computer-free solution. First of all, from the equality $0 \circ 0 = 0$ we see that $0$ must be the neutral element. Next, from the equality $x \circ x = 0$ for all $x$ it follows that $G$ is an elementary abelian 2-group, so it must be isomorphic to $\mathbb{Z}_2^3$. This also makes $G$ a vector space over the field $\mathbb{Z}_2$.
Now we just need to establish an isomorphism $f: G \to \mathbb{Z}_2^3$. Then we will have our multiplication rule as $x \circ y = f^{-1}(f(x) + f(y))$.
Choosing an isomorphism $f$ is essentially the same as choosing a basis in $G$. To choose a basis in $G$, we first note that $1 \circ 2 = 3$ (we know that $1 \circ 2 \leq 1 + 2$, and it cannot be $0$, $1$ or $2$). Then for our basis we can pick (say) $1$, $2$ and anything else other than $0$ and $3$. I choose to pick the basis $(1,2,4)$.
Now using the rule $x \circ y \leq x + y$ it's not hard to find all the linear combinations of the basis elements. Here they are:
$$
\begin{align*}
1 \circ 2 &= 3 \\
1 \circ 4 &= 5 \\
2 \circ 4 &= 6 \\
1 \circ 2 \circ 4 &= 7
\end{align*}
$$
Now we build the isomorphism $f$ by sending $1$ to $(0,0,1)$, $2$ to $(0,1,0)$, $4$ to $(1,0,0)$. Images of other elements are determined uniquely from linearity of $f$. Here they are:
$$
\begin{align*}
f(0) &= (0,0,0) \\
f(1) &= (0,0,1) \\
f(2) &= (0,1,0) \\
f(3) &= f(1 \circ 2) = f(1) + f(2) = (0,1,1) \\
f(4) &= (1,0,0) \\
f(5) &= f(1 \circ 4) = f(1) + f(4) = (1,0,1) \\
f(6) &= f(2 \circ 4) = f(2) + f(4) = (1,1,0) \\
f(7) &= f(1 \circ 2 \circ 4) = (1,1,1)
\end{align*}
$$
As you see, the image of each number turns out to be that number written in binary code. Now, if we recall that $f$ is an isomorphism, i.e. $f(x \circ y) = f(x) + f(y)$, we see that $\circ$ must operate exactly by XORing the binary codes of its operands.
|
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|
Question Regarding Telescopic Sums How do I find a formula for the partial sum $s_n$ for the series $$\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+2})$$
|
For example: $$\left(\dfrac{1}{1} - \dfrac{1}{3}\right) + \left(\dfrac{1}{2} - \dfrac{1}{4}\right) + \left(\dfrac{1}{3} - \dfrac{1}{5}\right) =
\dfrac{1}{1} + \dfrac{1}{2} - \dfrac{1}{4} - \dfrac{1}{5}$$
Try a few more: do you see the pattern?
|
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|
How to prove for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$? I'm new to induction so please bear with me. How can I prove using induction that, for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$?
I think $9$ can be an example since the sum of the first $9$ positive odd numbers is $1,3,5,7,9,11,13,15,17 = 81 = 9^2$, but where do I go from here.
|
Let $P(n)$ be the statement that the sum of the first $n$ odd positive integers is $n^2$. $1=1^2$, so we have $P(1)$.
Suppose we have $P(k)$ for some positive integer $k$. Then the sum of the first $k$ odd positive integers is $k^2$. Now the $(k+1)$th odd positive integer is $2k+1$, so the sum of the first $k+1$ odd positive integers is $k^2+2k+1=(k+1)^2$. Hence we have $P(k+1)$.
It follows that we have $P(n)$ for all positive integers $n$.
|
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|
One Diophantine equation I wonder now that the following Diophantine equation:
$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$
have only this formula describing his decision?
$a=-(k^2+2(p+s)k+p^2+ps+s^2)$
$b=2k^2+4(p+s)k+3p^2+3ps+2s^2$
$c=3k^2+4(p+s)k+2p^2+ps+2s^2$
$d=2k^2+4(p+s)k+2p^2+3ps+3s^2$
$k,p,s$ - what some integers.
By your question, I mean what that formula looks like this. Of course I know about the procedure of finding a solution, but I think that the formula would be better.
|
The above equation which is mentioned below has another parametrisation,
$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$
$a=(5k+2)^2$
$b=(3k-4)^2$
$c=(2k+6)^2$
$d=4(19k^2+10k+28)$
For $k=0$ we get $(a,b,c,d)=(1,4,9,28)$
|
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|
Analyzing a fourth degree polynomial Let $a,b$ and $c$ be real numbers. Then prove that the fourth degree polynomial in $x$
$acx^4+b(a+c)x^3+(a^2+b^2+c^2)x^2+b(a+c)x+ac$ has either 4 real roots or 4 complex roots.
I have never solved a fourth degree polynomial and don't know the conditions for it to have real/complex roots. How do we approach this problem?
|
Notice that the given expression,
$$acx^4+b(a+c)x^3+(a^2+b^2+c^2)x^2+b(a+c)x+ac = 0$$
can be factorized into (see below for derivation):
$$(ax^2 + bx + c)(cx^2 + bx + a) = 0$$
The discriminant for each is the same, $b^2 - 4ac$.
If this common discriminant is zero or more, then the roots for both $ax^2 + bx + c = 0$ and $cx^2 + bx + a = 0$ are all real (possible with multiplicity $2$). If not, they are all imaginary. Hence the roots are either all real or all imaginary.
Here's how I did the factorization:
$$acx^4+b(a+c)x^3+(a^2+b^2+c^2)x^2+b(a+c)x+ac$$
$$=acx^4 + abx^3 + bcx^3 + a^2x^2 + b^2x^2 + c^2x^2 + abx + bcx + ac$$
$$=(acx^4 + abx^3 + a^2x^2) + (bcx^3 + b^2x^2 + abx) + (c^2x^2 + bcx + ac)$$
$$=ax^2(cx^2 + bx + a) + bx(cx^2 + bx + a) + c(cx^2 + bx + a)$$
$$=(ax^2 + bx + c)(cx^2 + bx + a)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $I=\int_0^\infty \frac{\ln(1+x)\ln(1+x^{-2})}{x} dx$ Hi I am stuck on showing that
$$
\int_0^\infty \frac{\ln(1+x)\ln(1+x^{-2})}{x} dx=\pi G-\frac{3\zeta(3)}{8}
$$
where G is the Catalan constant and $\zeta(3)$ is the Riemann zeta function. Explictly they are given by
$$
G=\beta(2)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}, \ \zeta(3)=\sum_{n=1}^\infty \frac{1}{n^3}.
$$
I have tried using
$$
\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n},
$$
but didn't get very far.
|
The infinite sum in Chen Wang's answer, that is, $ \displaystyle \sum_{n=1}^{\infty} \frac{H_{4n}}{n^{2}}$, can be evaluated using contour integration by considering the function $$f(z) = \frac{\pi \cot(\pi z) [\gamma + \psi(-4z)]}{z^{2}}, $$
where $\psi(z)$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant.
The function $f(z)$ has poles of order $2$ at the positive integers, simple poles at the negative integers, simple poles at the positive quarter-integers, and a pole of order $4$ at the origin.
The function $\psi(-4z)$ does have simple poles at the positive half-integers, but they are cancelled by the zeros of $\cot( \pi z)$.
Now consider a square on the complex plane (call it $C_{N}$) with vertices at $\pm (N + \frac{1}{2}) \pm i (N +\frac{1}{2})$.
On the sides of the square, $\cot (\pi z)$ is uniformly bounded.
And when $z$ is large in magnitude and not on the positive real axis, $\psi(-4z) \sim \ln(-4z)$.
So $ \displaystyle \int_{C_{N}} f(z) \ dz $ vanishes as $N \to \infty$ through the positive integers.
Therefore,
$$\sum_{n=1}^{\infty} \text{Res} [f(z), n] + \sum_{n=1}^{\infty} \text{Res}[f(z),-n] + \text{Res}[f(z),0] + \sum_{n=0}^{\infty} \text{Res}\Big[f(z), \frac{2n+1}{4} \Big] =0 .$$
To determine the residues, we need the following Laurent expansions.
At the positive integers,
$$ \gamma + \psi (-4z) = \frac{1}{4} \frac{1}{z-n} + H_{4n} + \mathcal{O}(z-n) $$
and
$$ \pi \cot (\pi z) = \frac{1}{z-n} + \mathcal{O}(z-n) .$$
At the origin,
$$ \gamma+ \psi(-4z) = \frac{1}{4z} -4 \zeta(2) z -16 \zeta(3) z^{2} + \mathcal{O}(z^{3})$$
and
$$ \pi \cot (\pi z) = \frac{1}{z} - 2 \zeta(2) z + \mathcal{O}(z^{3}) .$$
And at the positive quarter-integers,
$$ \gamma + \psi(-4z) = \frac{1}{4} \frac{1}{z-\frac{2n+1}{4}} + \mathcal{O}(1)$$
and
$$ \pi \cot (\pi z) = (-1)^{n} \pi + \mathcal{O}\Big(z- \frac{2n+1}{4} \Big) .$$
Then at the positive integers,
$$f(z) = \frac{1}{z^{2}} \Big( \frac{1}{4} \frac{1}{(z-n)^{2}} + \frac{H_{4n}}{z-n} + \mathcal{O}(1) \Big), $$
which implies
$$\begin{align} \text{Res} [f(z),n] &= \text{Res} \Big[ \frac{1}{4z^{2}} \frac{1}{(z-n)^{2}} , n \Big] + \text{Res} \Big[ \frac{1}{z^{2}} \frac{H_{4n}}{z-n}, n \Big] \\ &= - \frac{1}{2n^{3}} + \frac{H_{4n}}{n^{2}} .\end{align}$$
At the negative integers,
$$ \text{Res}[f(z),-n] = \frac{\gamma + \psi(4n)}{n^{2}} = \frac{H_{4n-1}}{n^{2}} = \frac{H_{4n}}{n^{2}} - \frac{1}{4n^{3}} . $$
At the origin,
$$ f(z) = \frac{1}{z^{2}} \Big( \frac{1}{4z^{2}} - \frac{\zeta(2)}{2} - 4 \zeta(2) - 16 \zeta(3) z + \mathcal{O}(z^{2}) \Big),$$
which implies
$$\text{Res}[f(z),0] = -16 \zeta(3) .$$
And at the positive quarter-integers,
$$ f(z) = \frac{\pi}{4z^{2}} \frac{(-1)^{n}}{z- \frac{2n+1}{4}} + \mathcal{O}(1),$$
which implies
$$ \begin{align} \text{Res} \Big[ f(z),\frac{2n+1}{4} \Big] &= \text{Res} \Big[\frac{\pi}{4z^{2}} \frac{(-1)^n}{z- \frac{2n+1}{4}}, \frac{2n+1}{4} \Big] \\ &= 4 \pi \ \frac{(-1)^{n}}{(2n+1)^{2}} . \end{align} $$
Putting everything together, we have
$$ - \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{3}} + 2 \sum_{n=1}^{\infty} \frac{H_{4n}}{n^{2}} - \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^{3}} - 16 \zeta(3) + 4 \pi \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} $$
$$ = - \frac{1}{2} \zeta(3) + 2 \sum_{n=1}^{\infty} \frac{H_{4n}}{n^{2}} - \frac{1}{4} \zeta(3) - 16 \zeta(3) + 4 \pi G = 0 .$$
Therefore,
$$ \sum_{n=1}^{\infty} \frac{H_{4n}}{n^{2}} = \frac{67}{8} \zeta(3) - 2 \pi G .$$
EDIT:
I found the Laurent expansion of $\psi(-4z)$ at the positive integers by using the functional equation of the digamma function to express $\psi(4z)$ as
$$ \psi(4z) = \psi(4z+4n+1) - \frac{1}{4z+4n} - \frac{1}{4z+4n-1} - \ldots - \frac{1}{4z} .$$
Then I evaluated the limit $$\lim_{z \to -n} (z+n) \psi(4z) = - \frac{1}{4}$$ and the limit $$\lim_{z \to -n} \Big(\psi(4z) + \frac{1}{4} \frac{1}{z+n} \Big) = - \gamma +H_{4n} .$$
This leads to the expansion $$\gamma + \psi (-4z) = \frac{1}{4} \frac{1}{z-n} + H_{4n} + \mathcal{O}(z-n) .$$
I did something similar to find the expansion at the positive quarter-integers.
|
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|
How do I integrate $\frac{1}{x^6+1}$ My technique so far was substitution with the intent of getting to a sum of three fractions with squares in their denominators.
$t = x^2 \\
\frac{1}{x^6 + 1} = \frac{1}{t^3+1} = \frac{1}{(t+1)(t^2-t+1)}$
Then I try to reduce this fraction into a sum of two fractions
$\frac{A}{t+1} + \frac{B}{t^2-t+1} = \frac{(At^2-(A-B)t) + A + B}{(t+1)(t^2-t+1)}$
And this is where I reach a dead-end
$\begin{cases}
At^2-(A-B)t &= 0 \\
A + B &= 1
\end{cases}$
Any techniques I'm overlooking?
|
$$\begin{align}x^6+1=(x^2)^3+1^3&=(x^2+1)(x^4-x^2+1)=(x^2+1)((x^2+1)^2-3x^2)\\&=(x^2+1)(x^2-\sqrt3\,x+1)(x^2+\sqrt3\,x+1)\end{align}$$
Now you can use partial fractions:
$$\frac1{(x^2+1)(x^2-\sqrt3\,x+1)(x^2+\sqrt3\,x+1)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2-\sqrt3\,+1}+\frac{Ex+F}{x^2+\sqrt3\,+1}$$
|
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|
How to integrate $\int_{1}^{3} {\frac{x^2 - 1}{x^4 + 1}}\, dx$ $$\int_{1}^{3} {\frac{x^2 - 1}{x^4 + 1}}\, dx$$
Well, I can simplify the numerator:
$$\int_{1}^{3} {\frac{(x - 1)(x+1)}{x^4 + 1}}\, dx$$
But I have no idea how to simplify the denumenator:
$$\ x^4+1=? \ $$
How to solve this integral?
|
$$
\begin{align}
\hspace{-1cm}\int_1^3\frac{x^2-1}{x^4+1}\mathrm{d}x
&=\int_1^3\frac{x^2-1}{(x^2+1)^2-2x^2}\mathrm{d}x\tag{1}\\
&=\int_1^3\left[\frac{-1/2+x/\sqrt2}{x^2-\sqrt2x+1}
+\frac{-1/2-x/\sqrt2}{x^2+\sqrt2x+1}\right]\mathrm{d}x\tag{2}\\
&={\Large\int}_1^3\left[\frac{\frac1{\sqrt2}\left(x-\frac1{\sqrt2}\right)}{\left(x-\frac1{\sqrt2}\right)^2+\frac12}
-\frac{\frac1{\sqrt2}\left(x+\frac1{\sqrt2}\right)}{\left(x+\frac1{\sqrt2}\right)^2+\frac12}\right]\mathrm{d}x\tag{3}\\
&=\left[\frac1{2\sqrt2}\log\left(\left(x-\tfrac1{\sqrt2}\right)^2+\tfrac12\right)
-\frac1{2\sqrt2}\log\left(\left(x+\tfrac1{\sqrt2}\right)^2+\tfrac12\right)\right]_1^3\tag{4}\\
&=\left[\frac1{2\sqrt2}\log\left(\frac{x^2-\sqrt2x+1}{x^2+\sqrt2x+1}\right)
\right]_1^3\tag{5}\\
&=\left[\frac1{2\sqrt2}\log\left(\frac{(x^2-\sqrt2x+1)^2}{x^4+1}\right)
\right]_1^3\tag{6}\\
&=\frac1{2\sqrt2}\log\left(\frac{2(10-3\sqrt2)^2}{82(2-\sqrt2)^2}\right)\tag{7}\\
&=\frac1{2\sqrt2}\log\left(\frac{57+28\sqrt2}{41}\right)\tag{8}\\
\end{align}
$$
Explanation:
$(1)$: break the denominator into a difference of squares
$(2)$: using the factorization from $(1)$, do partial fractions
$(3)$: complete the squares in the denominators and adjust the numerators
$(4)$: integrate
$(5)$: expand the squares and combine the logarithms
$(6)$: neaten up the denominator
$(7)$: plug in the limits
$(8)$: simplify the argument to $\log$
|
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|
What converges this series? What exactly converges the series?
$\sum _{k=3}^{\infty \:}\frac{2}{k^2+2k}$
I tried taking out the constant
$=2\sum _{k=3}^{\infty \:}\frac{1}{k^2}$
then $p=2,\:\quad \:p>1\quad \Rightarrow \sum _{k=3}^{\infty \:}\frac{1}{k^2}$
but I really don't know what i'm doing
|
Use partial fractions: Suppose
$$f(n)=\sum\limits_{k=3}^n\dfrac{2}{k^2+2k}=\sum\limits_{k=3}^n(\dfrac{1}{k}-\dfrac{1}{k+2})$$
$$=(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})+\dots+(\frac{1}{n-2}-\frac{1}{n})+(\frac{1}{n-1}-\frac{1}{n+1})+(\frac{1}{n}-\frac{1}{n+2})$$
All the terms cancel out except four:
$$f(n)=\displaystyle \frac{1}{3}+\frac{1}{4}-\frac{1}{n+1}-\frac{1}{n+2}=\frac{7}{12}-\frac{2n+3}{(n+1)(n+2)}$$
Obviously the last fraction converges to zero and you 're left with $\frac{7}{12}$.
|
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|
How to prove $\frac{1}{x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2\sqrt{\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}}$ Question:
Let $a,b,c>0$ are give numbers and $x>0$, such that
$$
\sqrt{\dfrac{a+b+c}{x}}=\sqrt{\dfrac{b+c+x}{a}}+\sqrt{\dfrac{c+a+x}{b}}+\sqrt{\dfrac{a+b+x}{c}}
$$
show that
$$
\dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+2\sqrt{\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}}
$$
I found this problem is very nice and here is my attempt:
Since
$$
\dfrac{a+b+c}{x}=\dfrac{b+c+x}{a}+\dfrac{c+a+x}{b}+\dfrac{a+b+x}{c}+2\sum_{cyc}\sqrt{\dfrac{(b+c+x)(c+a+x)}{ab}}
$$
it follows that
$$
\dfrac{a+b+c}{x}+3=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)(a+b+c+x)+2\sum_{cyc}\sqrt{\dfrac{(b+c+x)(c+a+x)}{ab}}.
$$
I feel it's very ugly and I can't continue it.
Thank you.
|
You can prove this equation using geometry approach. Note that this equation is exactly similar with Descartes 4 Circles Theorem. Descartes' theorem says: If four circles are tangent to each other at six distinct points and the circles have curvatures $k_i$ (for $i = 1,\cdots, 4$), then $k_i$ satisfies the following relation:
$$
(k_1+k_2+k_3+k_4)^2=2(k_1^2+k_2^2+k_3^2+k_4^2),
$$
where $k_i=\pm\dfrac{1}{r_i}$, $r_i$ is the radius of circle. The equation can also be written as:
$$
k_4=k_1+k_2+k_3\pm2\sqrt{k_1k_2+k_2k_3+k_1k_3},
$$
or
$$
\frac{1}{r_4}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\pm2\sqrt{\frac{1}{r_1r_2}+\frac{1}{r_2r_3}+\frac{1}{r_1r_3}}.
$$
The generalization to $n$ dimensions or variables is referred to as the Soddy–Gosset theorem.
$$
\left(\sum_{i=1}^{n+2}k_i\right)^2=n\sum_{i=1}^{n+2}k_i^2.
$$
For detail explanation and complete proof of Descartes' theorem (also to answer your question), you may refer to these sites: 1, 2, or download this journal.
$$\\$$
$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
|
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|
What is the value of $\int x^x~dx$? I am struggling with this puzzle.
Question 1. Is it possible to determine the value of the indefinite integral $\int x^x~dx$ explicitly?
By "explicit" I mean without power series.
Question 2. What are known theorems in the following form?
"$\int x^x~dx$ is not expressible by an expression given by functions in a family $\mathfrak{F}$" e.g. $\mathfrak{F}=\{\sin, \cos, \text{polynomials}\}$
|
$\int x^x~dx=\int e^{x\ln x}~dx=\int\sum\limits_{n=0}^\infty\dfrac{x^n(\ln x)^n}{n!}dx$
For $\int x^n(\ln x)^n~dx$ , where $n$ is any non-negative integers,
$\int x^n(\ln x)^n~dx$
$=\int(\ln x)^n~d\left(\dfrac{x^{n+1}}{n+1}\right)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{x^{n+1}}{n+1}d((\ln x)^n)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)x}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^n(\ln x)^{n-1}}{n+1}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{n(\ln x)^{n-1}}{n+1}d\left(\dfrac{x^{n+1}}{n+1}\right)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{x^{n+1}}{n+1}d\left(\dfrac{n(\ln x)^{n-1}}{n+1}\right)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^{n+1}(\ln x)^{n-2}}{(n+1)^2x}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^n(\ln x)^{n-2}}{(n+1)^2}dx$
$=\cdots\cdots$
$\vdots$
$\vdots$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}-\int\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times1)x^n}{(n+1)^n}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}+\dfrac{(-1)^n(n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+1}}+C$
$=\dfrac{(n+1)x^{n+1}(\ln x)^n}{(n+1)^2}-\dfrac{(n+1)nx^{n+1}(\ln x)^{n-1}}{(n+1)^3}+\cdots\cdots+\dfrac{(-1)^{n-1}((n+1)n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^{n+1}}+\dfrac{(-1)^n((n+1)n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+2}}+C$
$=\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}(n+1)!x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+2}}+C$
$=\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$
$\therefore\int\sum\limits_{k=0}^n\dfrac{(x\ln x)}{n!}dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$
Hence $\int x^x~dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$
|
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|
Integrals Clarification Needed
Missed a couple days of class, can't seem to figure out how to get the numbers marked with the blue arrows. Other than that I understand the concept.
|
Alright, I've done this problem on ms paint. This is how the answer is $-136$
We have the definite integral of
$\int^4_2 (10+4x-3x^3)\,dx$
Before we can evaluate the definite integral, we need to take the antiderivative of $10+4x-3x^3$. To take the antiderivative we need to add one to the exponent and divide by the new exponent number. Since we only have $10$, the antiderivative is $10x$.
How did we get $10x?$
Well, when we had $10$, there was no exponent, so when we add the exponent $x$ that's $x^1$ but we don't write it that often, and divide by the new exponent number, so we have $\frac{10x^1}{1} \rightarrow 10x$
For $4x$, we add one to the exponent so that becomes $4x^{1+1}$. Since we have $4x^2$, we need to divide by the new exponent number which is 2. $\frac{4x^2}{2}$ and that becomes $2x^2$.
For $3x^3$ the new exponent number is $4$ and we divide by $4$. There is nothing to reduce, so we're done.
$3x^{3+1} \rightarrow 3x^4 \rightarrow \frac{3x^4}{4}$
So now we have $10x+2x^2 -\frac{3x^4}{4}$
You can check to see if it's correct by taking the derivative, and it will be $10 +4x-3x^3$
$\int^4_2 (10x+2x^2-\frac{3x^4}{4})\,dx$
Now we evaluate using $F(b) -F(a)$ where $b=4 $ and $ a=2$.
For $F(4)-F(2)$, we have
$10(4)+2(4)^2-\frac{3(4)^4}{4}-[10(2)+2(2)^2-\frac{3(2)^4}{4}]$
$40+2(16)-\frac{256(3)}{4}-[20+8-\frac{48}{4}]$
$40+32-\frac{768}{4}-[20+8-12]$
$72-192-[28-12]$
$72-192-28+12$
$-148+12$
$-136$
|
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|
Why is the derivative of circle area the circumference? Why is the derivative of the volume of a sphere the surface area? And why is the derivative of the area of a circle the circumference? Too much of a coincidence, there has to be a reason!
Also, why is the sum of the cubes the square of the sum of the integers?
|
I'll answer the last question first because that's the easiest one in my opinion.
Prove that
$$1^3 + 2^3 + 3^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
We will prove this using induction, by first confirming the identity when $n = 1$ and then proving the inductive step.
Base Case: $n = 1$
$$1^3 = \left(\frac{1(1+1)}{2}\right)^2 = \left(\frac{2}{2}\right)^2 = 1^2 = 1$$
Inductive Step:
Suppose the equality is true for some $x$. That is, $$1^3 + 2^3 + 3^3 + \cdots + x^3 = \left(\frac{x(x+1)}{2}\right)^2$$
Now we add $(x+1)^3$ to both sides and get
$$1^3 + 2^3 + 3^3 + \cdots + x^3 + (x+1)^3 = \left(\frac{x(x+1)}{2}\right)^2 + (x+1)^3 = (x+1)^2 \left(\frac{x^2}{4} + x + 1\right) = \frac{(x+1)^2}{4} (x^2 + 4x + 4) = \frac{(x+1)^2 (x+2)^2}{4} = \left(\frac{(x+1)(x+2)}{2}\right)^2$$
Therefore, if the equation is true for some $x$, then it is also true for $x+1$. And because it is true for $1$, it will be true for $2$, for $3$, for $4$, etc.
Prove that the derivative of the area of the circle is the circumference.
We note that the area of the bigger circle is $\pi (r+dr)^2$ and that of the smaller circle is $\pi r^2$. The difference in area, therefore, is $\pi (r+dr)^2 - \pi r^2 = \pi (2rdr + dr^2)$
We can then imagine turning the small "outline" (the area between the large circle and the small circle) into a thin, narrow strip. Assuming that $dr$ is small enough, the strip will be almost a rectangle with area $\pi (2rdr + dr^2)$. Because the height of the rectangle is $dr$, we divide the area by the height to get the width, which is $$\frac{\pi (2rdr + dr^2)}{dr} = \pi (2r + dr)$$
And finally, because $dr$ approaches $0$ as the strip gets thinner, we see that the width approaches $2\pi r$, the circumference.
Similar logic applies to the sphere and the surface area.
|
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|
Let $u_{n+3} = u_n + 2u_{n+1}$ . Show that $p$ divides $u_p$ for all $p$ prime number.
Let $(u_n)$ a sequence such that $u_0 = 3$, $u_1 = 0$, $u_2 = 4$ and $u_{n+3} = u_n + 2u_{n+1}$
Show that $p$ divides $u_p$ for all $p$ prime number.
I'm really stuck on this exercise,
Does anyone can give me a good HINT to start ?
Thank you in advance.
|
The characteristic equation for the recurrence relations $u_{n+3} - 2u_{n+1} - u_n = 0$ is
given by
$$\lambda^3 - 2\lambda - 1
= (\lambda-1)\left(\lambda-\frac{1+\sqrt{5}}{2}\right)\left(\lambda-\frac{1-\sqrt{5}}{2}\right)$$
Since the roots are all simple, the general solution for $u_n$ has the form
$$u_n = \alpha (-1)^n + \beta \left(\frac{1+\sqrt{5}}{2}\right)^n
+ \gamma \left(\frac{1-\sqrt{5}}{2}\right)^n$$
for suitably chosen constants $\alpha, \beta, \gamma$. With a little bit of algebra,
the initial conditions $u_0 = 3, u_1 = 0, u_2 = 4$ leads to $\alpha = \beta = \gamma = 1$.
Since $2 \mid u_2$, we just need to figure out what happens to $u_p$ when $p$ is an odd prime. For such an odd prime $p$,
$$\begin{align}
2^{p-1} u_p
&= -2^{p-1} + \frac12\bigg[ (1 + \sqrt{5})^p + (1-\sqrt{5})^p \bigg]\\
&= -2^{p-1} + \sum_{k=0, k\text{ even}}^p \binom{p}{k} \sqrt{5}^k\\
&= - ( 2^{p-1} - 1 ) + \sum_{\ell=1}^{\lfloor p/2\rfloor} \binom{p}{2\ell} 5^\ell
\tag{*1}
\end{align}$$
By Fermat little theorem, $p \mid 2^{p-1} -1$. Together with the fact $p \mid \binom{p}{k}$ for $1 \le k \le p-1$, we get
$$p \mid \text{RHS(*1)}\quad\implies\quad p \mid 2^{p-1} u_p\quad\implies\quad p \mid u_p$$
|
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|
Probability of eight dice showing sum of 9, 10 or 11 Suppose we roll eight fair dice. What is the probability that:
1) The sum of the faces is $9$
2) The sum of the faces is $10$
3) The sum of the faces is $11$
I'm thinking that we start with $8$ dice, each showing $1$. Then we think of the problem as assigning one $1$, two $1$'s or three $1$'s to the $8$ dice.
So that would give us:
1) $P(\sum = 9)$ = $8(\frac{1}{6})^8$
2) $P(\sum = 10)$ = $8^2(\frac{1}{6})^8$
3) $P(\sum = 11)$ = $8^3(\frac{1}{6})^8$
Is this right?
|
In all possible orders:
1) $21111111$. Probability: $\frac{8!}{1!7!}\times6^{-8}$
2) $31111111$ or $22111111$. Probability: $\left(\frac{8!}{1!7!}+\frac{8!}{2!6!}\right)\times6^{-8}$
3) $41111111$or $32111111$ or $22211111$. Probability: $\left(\frac{8!}{1!7!}+\frac{8!}{1!1!6!}+\frac{8!}{3!5!}\right)\times6^{-8}$
|
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|
Prove that positive $x,y,z$ satisfy $\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge \sqrt{2x}+\sqrt{2y}+\sqrt{2z}$
Prove that positive $x,y,z$ satisfy $$\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge \sqrt{2x}+\sqrt{2y}+\sqrt{2z}$$
Actually, this is a part of my solution to another problem, which is:
If $a,b,c$ are sides of a triangle, prove that $$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \sqrt{a+b-c}+\sqrt{a-b+c}+\sqrt{-a+b+c}$$
I substituted $a=x+y$, $b=y+z$, $c=z+x$. It's often called "Ravi substitution".
Here's a similar Math.SE question.
|
\begin{align}
& 2(\sqrt{x+y}+\sqrt{x+z}+\sqrt{y+z}) \geq 2(\sqrt{2x}+\sqrt{2y}+\sqrt{2z}) \\
& \Leftrightarrow (2\sqrt{x+y}-\sqrt{2x}-\sqrt{2y})+(2\sqrt{x+z}-\sqrt{2x}-\sqrt{2z})+(2\sqrt{y+z}-\sqrt{2y}-\sqrt{2z}) \geq 0
\end{align}
Note that
\begin{align}
& (2\sqrt{x+y}-\sqrt{2x}-\sqrt{2y}) \geq 0 \\
& \Leftrightarrow 2\sqrt{x+y} \geq \sqrt{2x}+\sqrt{2y} \\
& \Leftrightarrow 4(x+y) \geq (\sqrt{2x}+\sqrt{2y})^2=2(x+y)+4\sqrt{xy} \\
& \Leftrightarrow 2(\sqrt{x}-\sqrt{y})^2 \geq 0
\end{align}
Thus $(2\sqrt{x+y}-\sqrt{2x}-\sqrt{2y}) \geq 0$. Similarly $(2\sqrt{x+z}-\sqrt{2x}-\sqrt{2z}) \geq 0$ and $(2\sqrt{y+z}-\sqrt{2y}-\sqrt{2z}) \geq 0$ so we are done. Equality holds iff $x=y=z$.
|
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|
The general term formula $a_{n+1}=\frac{1+a_n^2}2$ Let $\ a_1=\dfrac 12\ $ and $\ a_{n+1}=\dfrac{a_n^2+1}2$,
*
*Could we find the general term formula $a_n$?
*If the answer to the question $1$ is "NO", for $\left|a_n-\dfrac{n-1}{n+1}\right|$ and $\left|a_n-\dfrac{n}{n+1}\right|$, which one is small in comparison with the other?
The general term formula is difficult, as for the question $2$, $\left|a_n-\dfrac{n-1}{n+1}\right|> \left|a_n-\dfrac{n}{n+1}\right|$ ? I am not sure. I'd be grateful for any help you are able to provide.
|
(1) As it was noted in the comments above, the sequence $(a_n/2)_{n\geq1}$ is defined by the iteration of the function $f(x)=x^2+c$ with $c=1/4$ and consequently it does not have a closed form.
(2) By a simple induction we see that $\frac{1}{2}\leq a_n<a_{n+1}<1$ for each $n$, This allows us to conclude that $(a_n)$ converges to the unique solution of the equation $x=(x^2+1)/2$ that is $1$.
Let us consider $b_n=1-a_n$. The sequence $(b_n)$ satisfies
$$ b_1=\frac{1}{2},\quad b_{n+1}=b_n\left(1-\frac{b_n}{2}\right)$$
This recursion has a more convenient form:
$$
\frac{1}{b_{n+1}}-\frac{1}{b_n}=\frac{1}{2-b_n}\tag{1}
$$
In particular, adding the inequalities $b_{k+1}^{-1}-b_{k}^{-1}\geq1/2$ for $k=1,\ldots,n-1$, we get
$b_n^{-1}\geq2+\frac{n-1}{2}$,
or equivalently
$$
b_n\leq \frac{2}{n+3}\tag{2}
$$
Using this again in $(1)$ we get $~b_{k+1}^{-1}-b_{k}^{-1}\leq\dfrac{1}{2}+\dfrac{1}{2(k+2)}$, adding these inequalities for $k=1,\ldots,n-1$, and rearranging we obtain
$b_n^{-1}\leq 2+\frac{n-1}{2}+\frac{1}{2}(H_{n+1}-3/2)$, with $H_n=\sum_{i=1}^n1/i$ (the well-known harmonic number.) Thus
$$
b_n\geq \frac{2}{n+3/2+H_{n+1}}\tag{3}
$$
Since clearly we have $H_{n+1}\leq n+\frac{1}{2}$ for $n\geq1$, and $n+3\geq n+1$ we conclude from $(1)$ and $(2)$ that
$$\frac{1}{n+1}\leq \frac{2}{n+3/2+H_{n+1}}\leq b_n\leq\frac{2}{n+3}\leq\frac{2}{n+1}$$
Now, note that
$$\eqalign{\left\vert a_n-\frac{n}{n+1}\right\vert-\left\vert a_n-\frac{n-1}{n+1}\right\vert&=
\left(b_n-\frac{1}{n+1}\right)-\left(\frac{2}{n+1}-b_n\right)
\cr&=2b_n-\frac{3}{n+1}\cr
&\geq \frac{4}{n+3/2+H_{n+1}}-\frac{3}{n+1}\cr
&\geq\frac{2n-1-6H_{n+1}}{2(n+1)(n+3/2+H_{n+1})}
}
$$
Now, the sequence $(2n-1-6H_{n+1})$ is monotonous non-decreasing and the first positive term corresponds to $n=10$, (I checked this with Mathematica.) Thus, we have proved that
$$
\forall\,n\geq 10,\quad
\left\vert a_n-\frac{n}{n+1}\right\vert>\left\vert a_n-\frac{n-1}{n+1}\right\vert
$$
For the first nine terms, we have to check the difference by "hand". Doing this yields the following conclusion:
$$
\forall\,n\geq 9,\quad
\left\vert a_n-\frac{n}{n+1}\right\vert>\left\vert a_n-\frac{n-1}{n+1}\right\vert
$$
and the inequality is reversed for $n\in\{1,2,\ldots,8\}$.
|
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|
How to simplify this trignometric expression: $4( 3 \sin \theta)( 3 \cos \theta)$? I was given a circle with a radius of $3$ and in it was a rectangle and an angle $\theta$ extending from the $x$ axis to up with coordinates of $(3 \cos \theta, 3 \sin \theta)$ and the question asks me to show that the area of the triangle represented by $A$ is equal to $18 \sin 2 \theta$. I figured that the the rectangle has $8$ triangles so using the angle and the coordinates I would find the area of one triangle and multiple it by $8$. So I ended up with $$(8)(.5) b h= (8)(.5)(3 \cos \theta) (3 \sin\theta) = 4( 3 \sin \theta)( 3 \cos \theta).$$ How do you represent this as $18 \sin 2 \theta$?
Thanks and sorry for the question being too long.
|
$\text{Area}= \frac 1 2 3\sin\theta 3\cos\theta$
Use $\sin\theta \cos\theta = \frac 1 2\sin 2\theta$
$=\frac 9 4 \sin 2\theta$
Multiply by $8$ to get $18\sin 2\theta$
|
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|
How to prove that $a+b$ is a multiple of $24$? Let $x$ be an integer one less than a multiple of $24$. Prove that if $a$ and $b$ are positive integers such that $ab=x$, then $a+b$ is a multiple of $24$.
|
So $ab \equiv -1 \pmod {24} \implies \gcd(a,24) = \gcd(b,24) = 1$. Then let $ab = 24k-1$ so $b = \frac{24k-1}a$ and $a+b = \frac{24k-1+a^2}a$. Since $\gcd(a,24) = 1$, it follows that $a^2 \equiv 1 \pmod {24}$ and we are done.
|
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How do I solve $(x^3-4x^2+5x-6)/(x^2-x-6)=4$ algebraically? How do I solve $\frac {(x^3-4x^2+5x-6)}{(x^2-x-6)=4}$ algebraically?
I tried:
$4(x^2-x-6)=x^3-4x^2+5x-6$
$4x^2-4x-24=x^3-4x^2+5x-6$
$x^3-8x^2+9x+18=0$
I don't know how to solve this algebraically.
|
$$x^3-4x^2+5x-6 = (x-3)(x^2-x+2)$$
$$x^2-x-6 = (x-3)(x+2)$$
so the equation becomes: $$\frac{x^2-x+2}{x+2} =4 $$
$$x^2-x+2=4x+8$$
$$x^2-5x-6=0$$
$$(x-6)(x+1)=0$$
$x=6$ or $-1$
|
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|
Cayley-Hamilton Block Matrix In Q4 d) I understand how to get the expression for A (using Q3), but I don't understand how you can then say that C is of the form described.
|
Well, because $C$ is a block diagonal matrix, we have
$$ C^² + C + I_4 = \begin{pmatrix}
A^² & 0 \\
0 & A^² \\
\end{pmatrix} + \begin{pmatrix}
A & 0 \\
0 & A \\
\end{pmatrix} + \begin{pmatrix}
I_2 & 0 \\
0 & I_2 \\
\end{pmatrix} $$
which simplifies to
$$
C^² + C + I_4 = \begin{pmatrix}
A^² + A + I_2 & 0 \\
0 & A^² + A + I_2 \\
\end{pmatrix} = \begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}
$$
|
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|
Estimation of integral Suppose the function $f(x)$ has a Taylor series expansion. Then
$$\int_a^bf(x)dx=\int_a^b(f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2+\cdots)dx=\\ \frac{f(a)}{1!}(b-a)+\frac{f'(a)}{2!}(b-a)^2+\frac{f''(a)}{3!}(b-a)^3+\cdots$$
and
$$\int_a^bf(x)dx=\int_a^b(f(b)+f'(b)(x-b)+\frac{1}{2}f''(b)(x-b)^2+\cdots)dx=\\ \frac{f(b)}{1!}(b-a)-\frac{f'(b)}{2!}(b-a)^2+\frac{f''(b)}{3!}(b-a)^3+\cdots$$
Therefore
$$\int_a^bf(x)dx=\frac{1}{1!}\frac{f(a)+f(b)}{2}(b-a)+\frac{1}{2!}\frac{f'(a)-f'(b)}{2}(b-a)^2+\frac{1}{3!}\frac{f''(a)+f''(b)}{2}(b-a)^3+\cdots$$
However, one can also consider
$$\int_a^bf(x)dx=\int_a^df(x)dx+\int_d^bf(x)dx$$
where $d=\frac{a+b}{2}$
Then
$$\int_a^df(x)dx=\int_a^d(f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2+\cdots)dx=\\
\frac{1}{1!}\frac{f(a)}{2}(b-a)+\frac{1}{2!}\frac{f'(a)}{2^2}(b-a)^2+\frac{1}{3!}\frac{f''(a)}{2^3}(b-a)^3+\cdots$$
$$\int_d^bf(x)dx=\int_d^b(f(b)+f'(b)(x-b)+\frac{1}{2}f''(b)(x-b)^2+\cdots)dx=\\
\frac{1}{1!}\frac{f(b)}{2}(b-a)-\frac{1}{2!}\frac{f'(b)}{2^2}(b-a)^2+\frac{1}{3!}\frac{f''(b)}{2^3}(b-a)^3+\cdots$$
and so
$$\int_a^bf(x)dx=\frac{1}{1!}\frac{f(a)+f(b)}{2}(b-a)+\frac{1}{2!}\frac{f'(a)-f'(b)}{2^2}(b-a)^2+\frac{1}{3!}\frac{f''(a)+f''(b)}{2^3}(b-a)^3+\cdots$$
My question is, the two estimations are different. Which one is correct?
|
Remarkably, they appear to both be correct. I have checked the following integrals using both formulas: $$\int_0^{\pi/2}\sin x\,\mathrm{d}x=1,\;\;\int_0^{\pi/2}\cosh x\,\mathrm{d}x\approx 2.301,\;\;\int_a^{b}x^2\,\mathrm{d}x=\frac{b^3-a^3}{3},\;\;\int_a^{b}x^3\,\mathrm{d}x=\frac{b^4-a^4}{4}$$
See for example
Here and
Here
(scroll down to "Alternate forms")
The first two were performed by a program in Python.
There does not appear to be any mistake in your argument, so I can only conclude that in fact they are equal to each other. Note that the second series converged much more rapidly in my approximations than did the first. Hopefully someone can provide a proof.
|
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|
If $d=\gcd(a+b,a^2+b^2)$, with $\gcd(a,b)=1$, then $d=1$ or $2$ Suppose $\gcd(a,b)=1$. Let $d=\gcd(a+b,a^2+b^2)$. I want to prove that $d$ equals $1$ or $2$.
I get that $d\mid2ab$ but I can't find a linear combination that will give me some help to use the fact that $\gcd(a,b)=1$.
|
Bezout based approach
Since $(a,b)=1$, there are $x,y$ so that $ax+by=1$. Then
$$
\begin{align}
1
&=(ax+by)^3\\
&=a^2\color{#C00000}{(ax^3+3bx^2y)}+b^2\color{#00A000}{(3axy^2+by^3)}
\end{align}
$$
Then, since
$$
(a^2+b^2)+(a+b)(a-b)=2a^2
$$
and
$$
(a^2+b^2)-(a+b)(a-b)=2b^2
$$
we have
$$
\begin{align}
&(a^2+b^2)\left[\color{#C00000}{(ax^3+3bx^2y)}+\color{#00A000}{(3axy^2+by^3)}\right]\\
+&(a+b)(a-b)\left[\color{#C00000}{(ax^3+3bx^2y)}-\color{#00A000}{(3axy^2+by^3)}\right]\\
=&2a^2\color{#C00000}{(ax^3+3bx^2y)}+2b^2\color{#00A000}{(3axy^2+by^3)}\\[6pt]
=&2
\end{align}
$$
Therefore, $(a+b,a^2+b^2)\mid2$.
|
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|
Proving that function with domain (-1,1) is injective. Function $g\colon (-1,1) \rightarrow \mathbb R$ is defined by $g(x)=\dfrac{x}{1-x^2}$. Prove that $g(x)$ is injective.
Work:
I shifted the equation so that it ends up like $\dfrac{x}{y}=\dfrac{1-x^2}{1-y^2}$. From here, I thought about using cases such as $x>y$ or $y>x$ to prove it, but it does not seem to work. I tried to manipulate the equation into various forms, but I cannot end up with $x=y$.
|
Rewrite your function as follows $f(x) = \frac{x}{1-x^2} = \frac{1}{\frac{1}{x} - x}$
Suppose $f(x) = f(y)$. That is
$$ \frac{1}{\frac{1}{x} - x} = \frac{1}{\frac{1}{y} - y} \iff \frac{\frac{1}{y}-y}{\frac{1}{x} - x} = 1$$
We want to show $x = y$. Suppose not. Then we either have $x > y$ or $y > x$.
Suppose $y > x$. Then $-y < -x $ and $\frac{1}{y} < \frac{1}{x} $. Hence
$$ 1 = \frac{\frac{1}{y}-y}{\frac{1}{x} - x} < \frac{\frac{1}{x}-x}{\frac{1}{x} - x} = 1$$
Contradiction!
Similarly, I let you do it the case $x > y$ which will lead you to a contradiction.
Therefore $x = y$ which implies $f$ is injective
|
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|
Prove for all $n \in N$, $\gcd(2n+1,9n+4)=1$ Question: Prove for all $n \in N$, $\gcd(2n+1,9n+4)=1$
Attempt: I want to use Euclid's Algorithm because it seemed to be easier than what my book was doing which was manually finding the linear combination.
Euclid's Algorithm states that we let $a,b \in N $. By applying the Division Algorithm repeatedly...then $\gcd(a,b) = r_j$ will be the last non-zero remainder. By the Well Ordering Principle, there is a smallest element of the set of natural numbers less than or equal to $r_j$.
I have used long division, but since I can't get it to show up here, I will type what I've done.
Starting at $\gcd(2n+1,9n+4)$,
$\frac{2n+1}{9n+4}$
I can multiply $2n+1$ four times and I would have a remainder of $n$ because $(9n+4)-(8n+4) = 9n+4-8n-4=n$
so we have $4 \cdot (2n+1)+n$ if I apply Euclid's Algorithm.
For $\gcd(2n+1, n)$,
$\frac{2n+1}{n}$
I can multiply $n$ 2 times and I will have only 1 as the remainder because $(2n+1)-(2n+0) = 2n+1-2n+0=1$
Therefore, we have $2 \cdot (n) +1$
and $\gcd(n,1)$ which is $1$
Since the end result is $1, \gcd(2n+1,9n+4)=1$
I followed an example from this link
http://cms.math.ca/crux/v33/n5/public_page274-276.pdf
Am I doing this correctly?
|
There is also the systematic matrix approach:
$$
\begin{pmatrix} a \\ b \end{pmatrix}
=
\begin{pmatrix} 2 & 1 \\ 9 & 4 \end{pmatrix}
\begin{pmatrix} n \\ 1 \end{pmatrix}
\implies
\begin{pmatrix} n \\ 1 \end{pmatrix}
=
\begin{pmatrix} -4 & \hphantom-1 \\ \hphantom-9 & -2 \end{pmatrix}
\begin{pmatrix} a \\ b \end{pmatrix}
$$
In particular, $1=9a-2b$, which implies $\gcd(a,b)=1$.
The key point here is that the first matrix has determinant $-1$ and so is invertible over the integers.
|
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|
A tough inequality problem with condition $a+b+c+abc=4$ If, $a+b+c+abc=4$, with $a,b,c$ being positive reals, then prove or disprove the following inequality:
$$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}}+\frac{c}{\sqrt{a+b}}\geq\frac{a+b+c}{\sqrt2}$$
I couldn't do anything, please help. However, from the some values I have tried, the inequality does seem to be true.
|
We have by Holder's inequality:
$$\left(\sum_{cyc} \frac{a}{\sqrt{b+c}}\right)^2\left(\sum_{cyc}a(b+c) \right) \ge (a+b+c)^3$$
So it is sufficient to show that
$$2(a+b+c) \ge \sum_{cyc} a(b+c) \text{ or equivalently, } a+b+c \ge ab + bc + ca$$
Suppose $a+b+c < ab + bc + ca$. Then by Schur's inequality we have
$$\begin{align}
\frac{9abc}{a+b+c} &\ge 4(ab+bc+ca)- (a+b+c)^2 \\
&> (a+b+c)\left( 4- (a+b+c)\right) \\
&= (a+b+c) \cdot abc \\
\end{align}$$
This gives $a+b+c< 3$, further we have from $4= a+b+c+ abc \ge 4\sqrt{abc} \implies abc \le 1$, so $4 = a+b+c+abc < 3+1$, a contradiction. Hence proved.
|
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|
Finding the Asymptotic Curves of a Given Surface I have to find the asymptotic curves of the surface given by $$z = a \left( \frac{x}{y} + \frac{y}{x} \right),$$ for constant $a \neq 0$.
I guess that what was meant by that statement is that surface $S$ can be locally parametrized by $$X(u,v) = \left( u, v, a \left( \frac{u}{v} + \frac{u}{v} \right) \right).$$ Do you think that my parametrization is correct (meaning that I read the description of the surface correctly), and do you know of a more convenient parametrization?
Assuming that parametrization, I derived the following ($E$, $F$, $G$, are the coefficients of the first fundamental form; $e$, $f$, $g$ are coefficients of the second fundamental form; $N$ is the normal vector to surface $S$ at a point; these quantities are all functions of local coordinates $(u,v)$):
$$E = 1 + a^2 \left( \frac{1}{v} - \frac{v}{u^2} \right)^2,$$
$$F = -\frac{a^2 (u^2 - v^2)^2}{u^3 v^3},$$
$$G = 1 + a^2 \left( \frac{1}{u} - \frac{u}{v^2} \right)^2.$$
$$N = \frac{1}{\sqrt{E G - F^2}} \left( a \left( \frac{v}{u^2}-\frac{1}{v} \right), a \left( \frac{u}{v^2}-\frac{1}{u} \right), 1 \right).$$
$$X_{u,u} = \left( 0,0, \frac{2 a v}{u^3} \right), X_{u,v} = \left( 0,0, -a \left( \frac{1}{u^2} + \frac{1}{v^2} \right) \right), X_{v,v} = \left( 0, 0, \frac{2 a u}{v^3} \right).$$
$$e = \frac{2 a v}{u^3 \sqrt{E G - F^2}},$$
$$f = - \frac{a (\frac{1}{u^2} + \frac{1}{v^2})}{\sqrt{E G - F^2}},$$
$$g = \frac{2 a u}{v3 \sqrt{E G - F^2}}.$$
Thus, the Gaussian curvature (from these calculations) is:
$$K = -\frac{a^2 u^4 v^4 (u^2 - v^2)^2}{(u^4 v^4 +
a^2 (u^2 - v^2)^2 (u^2 + v^2))^2}.$$
And the mean curvature would be:
$$H = \frac{a u^3 v^3 (u^4 + v^4)}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2
}}.$$
So, the principal curvatures are:
$$k_{\pm} = H \pm \sqrt{H^2 - K} = a u^2 v^2 \frac{u v (u^4 + v^4) \pm \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2}}.$$
In order to find the asymptotic curves, but trying to avoid the differential equation, I was hoping to find the angles $\theta (u,v)$ such that the normal curvature would always be $0$. In other words I was trying:
$0 = k_n = k_{+} \cos{(\theta)}^2 + k_{-} \sin{(\theta)}^2$, and solving for $\theta$.
Assuming sufficient niceness, this calculate would result in: $$(u v (u^4 + v^4) + \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \cos{(\theta)}^2 + (u v (u^4 + v^4) - \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \sin{(\theta)}^2$$
First of all, is this approach (solving for $\theta$ rather than solving the differential equation) valid? If it is, after I find that angle $\theta$, determined by location $(u,v)$ on $S$, what more work do I have to do? How do I find the equations for the asymptotic curves based on this angle?
If this whole method was for naught, how does one solve the differential equation. in this case, of: $$e (u')^2 + 2f u' v' + g (v')^2 = 2a v^4 (u')^2 - 2a u^3 v^3 \left( \frac{1}{u^2} + \frac{1}{v^2} \right)u' v' + 2a u^4 (v')^2 = 0?$$ (Again, assuming sufficient niceness.)
(See: https://math.stackexchange.com/questions/762195/differential-equation-for-the-asymptotic-directions-of-a-given-surface)
Thank you!
|
Asymptotic lines for a Monge patch X[x,y] has a simple form. Primes on arc.
$ r\, x' ^2 + 2 \,s\, x'y' + t \, y' ^2 = 0 $ where $(r,t,s)$ are second partial derivatives.
|
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|
Prove the matrix satisfies the equation $A^2 -4A-5I=0$ How to prove that
$$
A=\begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{bmatrix}
$$
satisfies the equation $A^2 -4A-5I=0$?
|
$$A^2 = \left(
\begin{array}{ccc}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9 \\
\end{array}
\right)$$
$$-4A = \left(
\begin{array}{ccc}
-4 & -8 & -8 \\
-8 & -4 & -8 \\
-8 & -8 & -4 \\
\end{array}
\right)$$
$$A^2 - 4A = \left(
\begin{array}{ccc}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5 \\
\end{array}
\right)$$
This is just as easy as any other way for this problem.
|
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|
What mistake am I making trying to calculate the line integral $\oint_C3xy^2dx+8x^3dy$.
Evaluate the line integral
$$\oint_C3xy^2dx+8x^3dy$$
where $C$ is the boundary of the region between the circles $x^2+y^2=1$ and $x^2+y^2=64$ having positive orientation.
I actually used Green's theorem to find this. I know $r$ ranges from $1$ to $8$ and $\theta$ ranges from $0$ to $\pi$.
Okay, so I found $\dfrac{d}{dx}8x^3=24x^2$, and $\dfrac{d}{dy}3xy^2=6xy$.
Now I'm integrating $24x^2-6xy$. I converted it to polar coordinates and took the integral and I got $0$. However this is incorrect. What am I doing wrong?
|
A direct integration around the two circles involves (by convention) following the larger circle counter-clockwise ("positive" direction) and then the smaller circle clockwise ("negative" direction). Applying polar coordinates, the integral becomes
$$ \int_0^{2 \pi} \ 3 \ (r \cos \theta) \ (r \sin \theta)^2 \ \ d(r \cos \theta) \ \ + \ \ 8 \ (r \cos \theta)^3 \ \ d(r \sin \theta) $$
with fixed values for $ \ r \ $ . We would trace the two circles by
$$ 8^4 \ \int_0^{2 \pi} \ -3 \ \sin^3 \theta \ \cos \theta \ + \ 8 \ \cos^4 \theta \ \ d \theta $$
$$ - \ \ 1^4 \ \int_0^{2 \pi} \ -3 \ \sin^3 \theta \ \cos \theta \ + \ 8 \ \cos^4 \theta \ \ d \theta $$
$$ = \ \ ( \ 8^4 - 1 \ ) \ \left( \ [ \ 3 \theta \ + \ 2 \ \sin \ 2 \theta \ + \ \frac{1}{4} \sin \ 4 \theta \ ] \ + \ [ \ \frac{3}{4} \sin^4 \theta \ ] \ \right) \ \vert_0^{2 \pi} \ \ . $$
Integrating through one full period, all of the terms except the first produce zeroes. Thus, the value of the integral is
$$ ( \ 8^4 - 1 \ ) \ \cdot \ 3 \cdot \ 2 \pi \ = \ 6 \pi \ ( \ 8^4 - 1 \ ) \ \ \text{or} \ \ 24570 \pi \ \ . $$
This confirms the result found by Jeb , applying Green's Theorem over the annulus bounded by the two circles.
|
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|
Is $\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}}$ convergent? Is $\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}}$ convergent ?
I use it to compare with $1/n^2$, and then I used LHôpitals rule multiple times. Finally , I can solve it. However,I think we have other methods! somebody can help me?
|
Note that if $$k\le\sqrt{n}< k+1$$ then, since the function $x\mapsto 3^x$ is increasing, we have $$3^k\le3^{\sqrt{n}}<3^{k+1},\;\;\text{and}$$ $$\frac{1}{3^k}\ge\frac{1}{3^{\sqrt{n}}}>\frac{1}{3^{k+1}}$$for every $n\in [k^2,(k+1)^2[$ then
\begin{align}
\sum_{n=1}^{N^2-1}{\frac{1}{3^{\sqrt{n}}}} & = \frac{1}{3^1}+\frac{1}{3^{\sqrt{2}}}+\frac{1}{3^{\sqrt{3}}}+\frac{1}{3^2}+\ldots+\underbrace{\frac{1}{3^{\sqrt{(N-1)^2}}}+\frac{1}{3^{\sqrt{(N-1)^2+1}}}+\ldots+\frac{1}{3^{\sqrt{N^2-1}}}}_{2N-1\,\text{terms}}\\
& \le \frac{1}{3^1}+\frac{1}{3^1}+\frac{1}{3^1}+\frac{1}{3^2}+\ldots+\underbrace{\frac{1}{3^{N-1}}+\ldots+\frac{1}{3^{N-1}}}_{2N-1\;\text{terms}}\\
& = \sum_{k=1}^{N-1}{\frac{2k+1}{3^k}}
\end{align}
Then, for comparison test, $\sum_{n=1}^{\infty}{\frac{1}{3^{\sqrt{n}}}}$ converges if $\sum_{k=1}^{\infty}{\frac{2k+1}{3^k}}$ does, and this happens due the ratio test.
|
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|
Logarithmic problem with 2 variables help How on earth do I solve? Any help will be much appreciated.
The value of $M$ is given by $M = a \log_{10}S + b$.
Note: Seismic moment measure the energy of the earthquake.
Using the following information, determine the values of $a$ and $b$ and hence find the seismic moment $(S)$ of an earthquake which has a magnetic moment $(M)$ of $7.9$.
Seismic Moment $(S) = 4.47 \times 10^{25}$, Magnetic Moment $(M) = 7$
Seismic Moment $(S) = 2 \times 10^{27}$, Magnetic Moment $(M) = 7.5$
|
You have
\begin{align}
M=a\log_{10}S+b
\end{align}
Then, from the given values you have
\begin{align}
7=a\log_{10}(4.47\times10^{25})+b\tag1
\end{align}
and
\begin{align}
7.5=a\log_{10}(2\times10^{27})+b\tag2
\end{align}
Subtract $(1)$ from $(2)$ yield
\begin{align}
7.5-7&=a\log_{10}(2\times10^{27})+b-(a\log_{10}(4.47\times10^{25})+b)\\
0.5&=a\log_{10}(2\times10^{27})-a\log_{10}(4.47\times10^{25})\\
0.5&=a(\log_{10}(2\times10^{27})-\log_{10}(4.47\times10^{25}))\\
0.5&=a\log_{10}\left(\frac{2\times10^{27}}{4.47\times10^{25}}\right)\\
a&=\frac{0.5}{\log_{10}\left(\frac{2\times10^{27}}{4.47\times10^{25}}\right)}
\end{align}
The value of $b$ can be obtained by substituting $a$ to $(1)$ or $(2)$.
\begin{align}
7&=\frac{0.5}{\log_{10}\left(\frac{2\times10^{27}}{4.47\times10^{25}}\right)}\log_{10}(4.47\times10^{25})+b\\
b&=7-\frac{0.5\log_{10}(4.47\times10^{25})}{\log_{10}\left(\frac{2\times10^{27}}{4.47\times10^{25}}\right)}
\end{align}
|
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|
$ a+b+c+d=6 , a^2+b^2+c^2+d^2=12$ $\implies$ $ 36 \leq 4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4) \leq48 $ Let $a , b, c, d$ be real numbers such that $$ a+b+c+d=6 \\ a^2+b^2+c^2+d^2=12$$ How do we prove that $$ 36 \space \leq\space 4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4) \space\leq48\space$$ ?
|
lemma1 : $$0\le a,b,c,d\le 3$$
proof: Use Cauchy-Schwarz inequality,we have
$$b^2+c^2+d^2\ge\dfrac{1}{3}(b+c+d)^2\Longrightarrow 12-a^2\ge \dfrac{(6-a)^2}{3}$$
so
$$\Longrightarrow 0\le a\le 3$$
lemma 2:
$$4a^3-a^4\ge 2a^2+4a-3,0\le a\le 3$$
proof:
$$\Longleftrightarrow a^4-4a^3+2a^2+4a-3\le 0$$
since
$$a^4-4a^3+2a^2+4a-3=(a-1)^2(a+1)(a-3)\le 0$$
**lemm3:**$$4a^3-a^4\le 4a^2$$
proof: since
$$4a^3-a^4=4a^2-(a^2-2a)^2\le 4a^2$$
USe lemma 2: we have
$$4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4)=\sum_{cyc}(4a^3-a^4)\ge\sum_{cyc}(2a^2+4a-3)=36$$
Use lemma3: we have
$$4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4)=\sum_{cyc}(4a^3-a^4)\le \sum_{cyc}4a^2=48$$
By Done
|
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How can I get the following recursive relation that explained? if $b(n)$ is the number of words created by the alphabet ${a,b,c}$ with $n$ length that each word has at least one $a$ character and after each $a$ there is no $c$ character write a recursive relation for $b(n)$.
I have tried calculating the words by mind and getting the sequence of $b(n)$ numbers and then guessing the relation but calculating the words with more alphabet is difficult.
|
This is how I derived the recurrence.
Imagine the count $b(n)$ of good words to be split into two disjoint counts:
*
*$\color{blue}{x(n)}$: count of good words which end with $a$.
*$\color{green}{y(n)}$: count of good words which end with $b$ or $c$.
where a good word is defined as a word that meets our requirement - that is, it is generated from the alphabet $\{a, b, c\}$ such that it contains at least one $a$ and no $a$ is followed by $c$. If a word does not satisfy both the properties, we call it bad.
We have $b(n) = \color{blue}{x(n)} + \color{green}{y(n)}\tag{1}$
Next, we see how good words of length $n+1$ are created from words of length $n$. (You may want to spend some time thinking about this).
$$\begin{array}{rc} \text{length} & 1 & 2 & 3 & \cdots & n & n+1\\\hline\\ \color{blue}{x(n)}: & - & - & - & \cdots & \color{blue}{a}& \begin{cases}\color{blue}{a} \\\color{green}{b}\end{cases}\\\hline\\ \color{green}{y(n)}: & - & - & - & \cdots & \color{green}{b|c} & \begin{cases}\color{blue}{a}\\\color{green}{b}\\\color{green}{c}\end{cases}\\\hline\\ \color{red}{z(n)}: & \color{red}{b|c} & \color{red}{b|c} & \color{red}{b|c} & \cdots & \color{red}{b|c} & \begin{cases}\color{blue}{a}\end{cases}\end{array}$$
where $\color{red}{z(n)}$ is the count of bad words that do not contain $a$ (it is equal to $\underbrace{\color{red}{2\cdot2\cdots2}}_{n \text{ times}} = \color{red}{2^n}$).
This gives us the recurrences
$\begin{align}\color{blue}{x(n+1)} &= \color{blue}{x(n)} + \color{green}{y(n)} + \color{red}{z(n)} = \color{blue}{x(n)} + \color{green}{y(n)} + \color{red}{2^n} \\\color{green}{y(n+1)} &= \color{blue}{x(n)} + \color{green}{y(n)} + \color{green}{y(n)} = \color{blue}{x(n)} + \color{green}{2y(n)} \end{align}\tag{2}$
From $(1)$ and $(2)$ we get
$\begin{align}b(n+1) &= \color{blue}{2x(n)} + \color{green}{3y(n)} + \color{red}{2^n}\\b(n+2) &= \color{blue}{5x(n)} + \color{green}{8y(n)} + \color{red}{4\cdot2^n}\end{align}\tag{3}$
If we write $\color{blue}{x(n)}$ and $\color{green}{y(n)}$ in terms of $b(n+1)$ and $b(n+2)$ above and substitute in $(1)$, we get the needed recurrence
$b(n+2) = 3b(n+1) - b(n) + 2^n \text{ for } n \ge 0\tag{4}$
where $b(0) = 0$ and $b(1) = 1$ are the initial terms.
You can also play with this Perl program. It spits out the first few terms using both the recurrence relation and brute force.
EDIT: The OP told me that the second property of a good word is actually this:
*
*$c$ is not allowed in the substring that follows any $a$.
Below is an answer for this version.
Every good word is of the form
$$\begin{array}{rc} \text{length} & 1 & 2 & \cdots &k-1&k &k+1&\cdots & n\\\hline\\ b(n): & b|c & b|c & \cdots &b|c &a &a|b& \cdots & a|b\end{array}$$
*
*It contains at least one $a$ (so $1 \le k \le n$).
*There is no $c$ after the first $a$ (so no substring after any $a$ contains $c$).
Good words of length $n+1$ are created from words of length $n$ in the following two ways.
$$\begin{array}{rc} \text{length} & 1 & 2 & \cdots &k-1&k &k+1&\cdots & n &n+1\\\hline\\ b(n): & b|c & b|c & \cdots &b|c &a &a|b& \cdots & a|b & \begin{cases}a\\b\end{cases}\\\hline\\ \color{red}{z(n)}: & \color{red}{b|c} & \color{red}{b|c} & \color{red}{\cdots} & \color{red}{b|c} & \color{red}{b|c} &\color{red}{b|c} & \cdots & \color{red}{b|c} & \begin{cases}a\end{cases}\end{array}$$
where $\color{red}{z(n) = 2^n}$ is the count of bad words that do not contain $a$.
This gives the recurrence
$$b(n+1) = 2 \cdot b(n) +\color{red}{z(n)} = 2 \cdot b(n) + 2^n \text{ where } n \ge 0$$
and $b(0) = 0$ is the initial term.
|
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|
Whats the limit: $\lim\limits_{x \to 0} \frac{1-\sqrt[3]{\cos x}}{x(1-\cos\sqrt{x})}$ I am stuck at this one, i want to find it without using L'Hôpital's Rule.
$$\lim_{x \to 0} \frac{1-\sqrt[3]{\cos x}}{x(1-\cos\sqrt{x})}$$
|
$1-\cos \alpha=2\sin^2\frac{\alpha}2$, $1-\cos \alpha\sim \frac {\alpha^2} 2(\alpha\to0)$, so
\begin{equation}
\begin{split}\lim_{x \to 0} \frac{1-\sqrt[3]{\cos x}}{x(1-\cos\sqrt{x})} &=\lim_{x \to 0} \frac{1-\cos x}{x(1-\cos\sqrt{x})(1+\sqrt[3]{\cos x}+\sqrt[3]{\cos^2 x})}\\&=\frac13 \lim_{x \to 0} \frac{1-\cos x}{x(1-\cos\sqrt{x})}\\
&=\frac13 \lim_{x \to 0} \frac{\frac {x^2} 2}{x\cdot\frac x2}\\
&=\frac13
\end{split}
\end{equation}
|
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|
How would I solve this system of differential equations? Find a general solution of
$$
x' = \begin{pmatrix}
11 & -25 \\
4 & -9
\end{pmatrix} x + \begin{pmatrix}
e^{-t} \\
0
\end{pmatrix} $$
I found the general solution of the homogeneous system to be
$$ c_1e^t \begin{pmatrix}
5 \\
2
\end{pmatrix} + c_2 \left[te^t\begin{pmatrix}
5 \\
2
\end{pmatrix} +e^t \begin{pmatrix}
3 \\
1
\end{pmatrix} \right]
$$
I am not sure how to go about solving the nonhomogeneous form. The only method I know is variation of parameters but that seems very impractical in this situation. Any help would be greatly appreciated.
|
I agree with the general solution you found.
See my post here for details of this process, resolve an non-homogeneous differential system.
We have:
$$\phi(t) = e^t \left(
\begin{array}{cc}
5 & 5 t+3 \\
2 & 2 t+1 \\
\end{array}
\right)$$
$$\phi^{-1}(t) = \left(
\begin{array}{cc}
-e^{-t} (2 t+1) & e^{-t} (5 t+3) \\
2 e^{-t} & -5 e^{-t} \\
\end{array}
\right)$$
$$\phi^{-1}(t).f(t) = \left(
\begin{array}{cc}
-e^{-t} (2 t+1) & e^{-t} (5 t+3) \\
2 e^{-t} & -5 e^{-t} \\
\end{array}
\right).\left(
\begin{array}{c}
e^{-t} \\
0 \\
\end{array}
\right) = \left(
\begin{array}{c}
-e^{-2 t} (2 t+1) \\
2 e^{-2 t} \\
\end{array}
\right)$$
Integrating this previous result yields:
$$\left(
\begin{array}{c}
-e^{-2 t} (-t-1) \\
-e^{-2 t} \\
\end{array}
\right)$$
We now multiply the previous result with $\phi(t)$, yielding a particular solution of:
$$\left(
\begin{array}{c}
2 e^{-t} \\
e^{-t} \\
\end{array}
\right)$$
Finally, our solution is:
$$x(t) = x_h(t) + x_p(t) = c_1e^t \begin{pmatrix}
5 \\
2
\end{pmatrix} + c_2 \left[te^t\begin{pmatrix}
5 \\
2
\end{pmatrix} +e^t \begin{pmatrix}
3 \\
1
\end{pmatrix} \right]
+ \left(
\begin{array}{c}
2 e^{-t} \\
e^{-t} \\
\end{array}
\right)$$
|
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|
How Do I Integrate? $\int \frac{-2x^{2}+6x+8}{x^{2}(x+2)}$ How do I integrate this one? $$\int \frac{-2x^{2}+6x+8}{x^{2}(x+2)}\,dx$$
Is my answer correct: $$-3\ln\left \| x+2 \right \|+\ln\left \| x \right \|+\frac{4}{x}+C$$
|
Partial fraction decomposition.
$$\dfrac{-2x^2 + 6x +8}{x^2(x + 2)} = \dfrac{A}{x}+ \dfrac{B}{x^2} + \dfrac{C}{x+2}$$
Solve for $A, B, C$.
Your almost there with your answer, but we need $-\frac 4x$, and $\ln |x| - 3\ln|x+2| = \ln\left|\dfrac{x}{(x+2)^3}\right|$, if you want to simplify further.
|
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|
Testing for convergence of a series Test the following series for convergence
$$\sum_{k=1}^\infty \sin k \sin \frac{1}{k}$$.
I am just wondering if the following method is ok:
$$\sum_{k=1}^\infty \sin k \sin \frac{1}{k} = \sum_{k=1}^\infty \sin (\frac{1}{k} - (\frac{1}{k} -k)) \sin \frac{1}{k} \\ \leq \sum_{k=1}^\infty (\sin \frac{1}{k} \cos(\frac{1}{k} -k) - \cos \frac{1}{k}\sin(\frac{1}{k}-1)) \sin \frac{1}{k} \\ \leq
\sum_{k=1}^\infty \sin \frac{1}{k} \cos(\frac{1}{k} -k)\sin\frac{1}{k} - \cos \frac{1}{k}\sin(\frac{1}{k}-1) \sin \frac{1}{k}
\\\sum_{k=1}^\infty \sin \frac{1}{k} \cos(\frac{1}{k} -k)\sin\frac{1}{k}
\\\sum_{k=1}^\infty \sin \frac{1}{k} \sin\frac{1}{k}
\\\sum_{k=1}^\infty \frac{1}{k^2}$$ which converges.
Is there any more elegant proof? Help appreciated thanks.
|
Use the fact that $\sum \frac{ \sin (n)}{n} $ converges. Hence, by limit comparison test,
$$ \frac{ \sin (n) \sin ( \frac{1}{n} ) }{\frac{ \sin (n)}{n}} = \frac{ \sin( \frac{1}{n} )}{\frac{1}{n}} \to 1 $$
|
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|
Power Series Solution to Differential Equation The equation is
$$y'' - xy' + y = 0$$
So far I have the recurrence relation -
$$a_{n+2} = \dfrac{(n-1)a_n}{(n+1)(n+2)} $$
From this -
$a_2 = \dfrac{-a_0}{2!}$
$a_3 = 0$
$a_4 = \dfrac{-a_0}{4!}$
$a_5 = 0$
$a_6 = \dfrac{-3a_0}{6!}$
and so on..
The question asks for the first five non-zero terms of a general series solution of the d.e, seperating out for $a_0$ and $a_1$
How do I compute this?
Thanks
|
The differential equation
\begin{align}
y^{''} - x y^{'} + y = 0
\end{align}
can be solved via a power series of the form
\begin{align}
y(x) = \sum_{k=0}^{\infty} a_{n} x^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \cdots .
\end{align}
It is fairly evident that
\begin{align}
\sum_{k=0}^{\infty} k(k-1) a_{k} x^{k} = \sum_{k=0}^{\infty} (k-1) a_{k} x^{k}
\end{align}
which yields the equation for the coefficients
\begin{align}
a_{k+2} = \frac{ (k-1) a_{k} }{ (k+1) (k+2) }.
\end{align}
It is discovered that $a_{3} = 0 \cdot a_{1}$. Since, for $k$ being odd, say $k \rightarrow 2k+1$,
\begin{align}
a_{2k+3} = \frac{k a_{2k+1} }{(k+1)(2k+3)}
\end{align}
it is clear that all the odd coefficients depend of $a_{3}$ for $k \geq 1$ and leads to $a_{2k+1} = 0$ for $k \geq 1$. The even $k$ values are
\begin{align}
a_{2} &= - \frac{a_{0}}{2!} \\
a_{4} &= - \frac{a_{0}}{4!} \\
a_{6} &= - \frac{(1 \cdot 3) a_{0}}{6!} \\
a_{8} &= - \frac{(1\cdot 3 \cdot 5)a_{0}}{8!}
\end{align}
which has the general form
\begin{align}
a_{2k} = - \frac{a_{0}}{2^{k} k! (2k-1)}.
\end{align}
The series for $y(x)$ now be seen in the form
\begin{align}
y(x) = a_{0} + a_{1} x - a_{0} \sum_{k=1}^{\infty} \frac{ x^{2n} }{2^{k} k! (2k-1)}.
\end{align}
The power series discovered can be evaluated as follows. Consider
\begin{align}
\partial_{x} \left( \sum_{k=1}^{\infty} \frac{ x^{2n-1} }{2^{k} k! (2k-1)} \right) &= \sum_{k=1}^{\infty} \frac{ x^{2n-2} }{2^{k} k!} = \frac{1}{x^{2}}( e^{x^{2}/2} -1).
\end{align}
Integrating both sides
\begin{align}
\sum_{k=1}^{\infty} \frac{ x^{2n} }{2^{k} k! (2k-1)} &= x \int^{x} \frac{e^{u^{2}/2} -1}{u^{2}} du = x \left[ \sqrt{\frac{\pi}{2} } erfi\left( \frac{x}{\sqrt{2}} \right) - \frac{e^{x^{2}/2}}{x} + \frac{1}{x} \right] \\
&= \sqrt{\frac{\pi}{2} } \cdot x \cdot erfi\left( \frac{x}{\sqrt{2}} \right) - e^{x^{2}/2} + 1.
\end{align}
With this series the general solution of $y(x)$ can be sen by
\begin{align}
y(x) &= a_{0} + a_{1} x - \sum_{k=1}^{\infty} \frac{ x^{2n} }{2^{k} k! (2k-1)} \\
&= a_{1} x - a_{0} \left[ \sqrt{\frac{\pi}{2} } \cdot x \cdot erfi\left( \frac{x}{\sqrt{2}} \right) - e^{x^{2}/2} \right],
\end{align}
where $erfi(x)$ is the imaginary error function (erfi(x) = -i erf(ix)).
|
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|
what are the integral sides possible? Let a triangle have sides $a$,$b$,$c$ and $c$ is the greatest side , triangle ABC is obtuse in nature having integral sides. Find the smallest perimeter possible
Given A=2B where A and B are angles of respective vertices
|
Basic conditions for enumeration
A brute force approach could enumerate triangles with integral edge lengths in order of increasing perimeter. In other words, examine partitions of increasingly larger numbers into three summands, taking the triangle inequality into account. What we'd need is some easy approach to detect this $A=2B$ condition. From the law of cosines, we get
$$\cos A=\frac{-a^2+b^2+c^2}{2bc}\qquad \cos B=\frac{a^2-b^2+c^2}{2ac}$$
We also know that $\cos A=\cos(2B)=2\cos^2B-1$ so we obtain the equation
$$
\frac{-2a^2+b^2+c^2}{2bc}=2\left(\frac{a^2-b^2+c^2}{2ac}\right)^2-1 \\
-a^4c + a^2b^2c + a^2c^3 = a^4b - 2a^2b^3 + b^5 - 2b^3c^2 + bc^4 \\
a^4b - 2a^2b^3 + b^5 + a^4c - a^2b^2c - 2b^3c^2 - a^2c^3 + bc^4 = 0 \\
(b + c)(-a - b + c)(a - b + c)(-a^2 + b^2 + bc) = 0
$$
The first factor would cannot be satisfied for positive edge lengths. The second translates to $c=a+b$ which would be a degenerate triangle. Likewise the third, $b=a+c$. So the only condition which remains is
$$a^2-b^2-bc=0\tag{1}$$
Now do a bit of enumeration, in order of increasing perimeter, and you find the following combination which satisfies both this and the triangle inequality:
$$a=15\qquad b=9\qquad c=16$$
However, the resulting triangle is not obtuse. So let's add a condition for that:
$$a^2+b^2<c^2$$
Now you get the solution
$$a=28\qquad b=16\qquad c=33$$
Avoiding the enumeration
If you want to avoid the brute force enumeration, start with this reformulation of $\text{(1)}$:
$$c=\frac{a^2-b^2}{b}\in\mathbb N$$
So $a^2$ must be a multiple of $b$. On the other hand, $a$ itself cannot be a multiple of $b$ because otherwise, $bc$ would also be a multiple of $b^2$ so $c$ would be a multiple of $b$, and you could divide all lengths by $b$ to obtain a smaller solution. So for some $s,t,u\in\mathbb N$ you have
$$a=stu\qquad b=s^2u\qquad c=\frac{s^2t^2u^2-s^4u^2}{s^2u}=(t^2-s^2)u$$
You can see that $u=1$ is required for minimality, otherwise you could again divide all lengths by $u$. So $b=s^2=\gcd(a,b)^2$ is a square number.
The triangle inequality states
\begin{gather*}
c=t^2-s^2<st+s^2=a+b \\
t^2-st-2s^2<0 \\
-s<t<2s \tag2
\end{gather*}
An obtuse angle at $c$ turns into
\begin{gather*}
a^2+b^2=s^2t^2+s^4<t^4-2t^2s^2+s^4=(t^2-s^2)^2=c^2\\
3s^2t^2<t^4 \\
3s^2<t^2 \\
\sqrt3s<t \tag3
\end{gather*}
Taking $\text{(2)}$ and $\text{(3)}$ together, you get $\sqrt3s<t<2s$ which you can turn into the condition $\sqrt3s<2s-1$ since $t$, being an integer, must be at least one smaller than $s$. This in turn becomes $s>\frac1{2-\sqrt3}\approx3.7$ so $s=4$ is the first case where both conditions can be satisfied. You get
$$6.9\approx4\sqrt3<t<8$$
This can be satisfied by $t=7$, leading the result above:
$$a=st=4\cdot7=28\qquad b=s^2=4^2=16\qquad c=t^2-s^2=7^2-4^2=33$$
Larger $s$ lead to larger minimal values for $t$ which together lead to larger perimeter. Therefore the above solution is indeed minimal.
|
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|
Solve $\cos x+8\sin x-7=0$
Solve $\cos x+8\sin x-7=0$
My attempt:
\begin{align}
&8\sin x=7-\cos x\\
&\implies 8\cdot \left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)=7-\cos x\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=7-1+2\sin ^2\frac{x}{2}\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=6+2\sin^2 \frac{x}{2}\\
&\implies 8\sin \frac{x}{2}\cos \frac{x}{2}=3+\sin^2 \frac{x}{2}\\
&\implies 0=\sin^2 \frac{x}{2}-8\sin \frac{x}{2}\cos \frac{x}{2}+3\\
&\implies 0=\sin \frac{x}{2}\left(\sin \frac{x}{2}-8\cos \frac{x}{2}\right)+3
\end{align}
I'm not sure how to proceed from here (if this process is even right at all?) . Any help would be appreciated. Thanks.
|
Divide either sides by $\cos^2\frac x2$ $$\sin^2\frac x2-8\sin\frac x2\cos\frac x2+3=0$$ to get $$\tan^2\frac x2-8\tan\frac x2+3\left(1+\tan^2\frac x2\right)=0$$ which on rearrangement is a Quadratic Equation in $\displaystyle\tan\frac x2$
|
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|
Integral $\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx=2-\frac{4}{\pi}$ Hi I am trying to show$$
I:=\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx=2-\frac{4}{\pi}.
$$
Thank you.
What a desirable thing to want to prove! It is a work of art this one. I wish to prove this in as many ways as we can find.
Note I tried writing
$$
I=\int_0^\infty \log(1+x^2)\coth \frac{\pi x}{2} \sinh^{-2} \frac{\pi x}{2}\mathrm dx
$$
but this didn't help me much. We can also try introducing a parameter as follows
$$
I(\alpha)=\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\alpha \pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx,
$$
But this is where I got stuck. How can we calculate I? Thanks.
|
I will solve the general form
\begin{align}\int^\infty_0\frac{x}{x^2+1}\frac{dx}{\sinh(ax)}&=\int^\infty_0\frac{x}{x^2+a^2}\frac{dx}{\sinh(x)}
\\&=\int^\infty_0 \int^\infty_0e^{-at} \frac{\sin(xt)}{\sinh(x)}\,dt \, dx\\&=\int^\infty_0 e^{-at}\int^\infty_0 \frac{\sin(xt)}{\sinh(x)} \, dx\,dt
\\&=\frac{\pi}{2} \int^\infty_0 e^{-at}\tanh\left(\frac{\pi}{2} t\right)\,dt\\&=\int^\infty_0 e^{-zx}\tanh(x)\,dt \,\,\,\,; z=\frac{2}{\pi}a\\&=\int^\infty_0\frac{e^{-zx}(1-e^{-2x})}{e^{-2x}+1}\,dx\end{align}
By splitting the integral we have
\begin{align} \int^\infty_0\frac{e^{-zx}}{e^{-2x}+1}\,dx &= \sum_{n\geq 0}\int^\infty_0e^{-x(2n+z)}\,dx\\&=\sum_{n\geq 0}\frac{(-1)^n}{2n+z}\\&=\frac{1}{4}\left (\psi \left(\frac{1}{2}+\frac{z}{4}\right)-\psi \left(\frac{z}{4} \right) \right)\end{align}
\begin{align}-\int^\infty_0\frac{-e^{-x(z+2)}}{e^{-2x}+1}\,dx&=-\sum_{n\geq0}\frac{(-1)^n}{z+2+2n}\\&=-\frac{1}{4}\left(-\psi \left(\frac{1}{2}+\frac{z}{4}\right)+\psi\left(1+\frac{z}{4} \right) \right)\end{align}
Hence we have
\begin{align}\int^\infty_0\frac{x}{x^2+1}\frac{dx}{\sinh(ax)}&=\frac{1}{4}\left (2\psi \left(\frac{1}{2}+\frac{z}{4}\right)-\psi \left(1+\frac{z}{4}\right)-\psi \left(\frac{z}{4} \right) \right)\\&=\frac{1}{2}\psi \left(\frac{1}{2}+\frac{z}{4}\right)-\frac{1}{2}\psi \left(\frac{z}{4} \right)-z \\&=\frac{1}{2}\psi \left(\frac{1}{2}+\frac{a}{2\pi}\right)-\frac{1}{2}\psi \left(\frac{a}{2\pi} \right)-\frac{2}{\pi}a \end{align}
Let $a=\pi/2$
\begin{align}\int^\infty_0\frac{x}{x^2+1}\frac{dx}{\sinh(\frac{\pi}{2}x)}&=\frac{1}{2}\psi \left(\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{2}\psi \left(\frac{1}{4} \right)-1\\&=\frac{\pi}{2} \cot(\pi/4)-1\\&=\frac{\pi}{2}-1\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Elegant proof of $\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$? Let $a, b > 0$ satisfy $a^2-4b^2 \geq 0$. Then:
$$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$$
One way to calculate this is by computing the residues at the poles in the upper half-plane and integrating around the standard semicircle. However, the sum of the two residues becomes a complicated expression involving nested square roots, which magically simplifies to the concise expression above.
Sometimes such 'magical' cancellations indicate that there is a faster, more elegant method to reach the same result.
Is there a faster or more insightful way to compute the above integral?
|
It's time to return the favor from here Ruben. (>‿◠)✌
First, we will prove
$$\int_{-\infty}^\infty \frac{dx}{\beta^4x^4+2\beta^2\cosh(2\alpha)\,x^2+1}=\frac{\pi}{2\beta\cosh\alpha}$$
Note that
$$\frac{1}{\beta^4x^4+2\beta^2\cosh(2\alpha)\,x^2+1}=\frac{1}{e^{2\alpha}-e^{-2\alpha}}\left[\frac{1}{\beta^2 x^2+e^{-2\alpha}}-\frac{1}{\beta^2 x^2+e^{2\alpha}}\right]$$
Hence
\begin{align}
\int_{-\infty}^\infty \frac{dx}{\beta^4x^4+2\beta^2\cosh(2\alpha)\,x^2+1}&=\frac{1}{e^{2\alpha}-e^{-2\alpha}}\left[\int_0^\infty\frac{dx}{\beta^2 x^2+e^{-2\alpha}}-\int_0^\infty\frac{dx}{\beta^2 x^2+e^{2\alpha}}\right]\\
&=\frac{1}{e^{2\alpha}-e^{-2\alpha}}\left[\frac{e^{\alpha}}{\beta}\arctan\left(e^{\alpha}x\right)-\frac{e^{-\alpha}}{\beta}\arctan\left(e^{-\alpha}x\right)\right]_{x=-\infty}^\infty\\
&=\frac{\pi}{\beta}\left[\frac{e^{\alpha}-e^{-\alpha}}{\left(e^{\alpha}-e^{-\alpha}\right)\left(e^{\alpha}+e^{-\alpha}\right)}\right]\\
&=\frac{\pi}{2\beta\cosh\alpha}\qquad\qquad\square
\end{align}
Now
$$\int_{-\infty}^\infty \frac{dx}{x^4+\frac{2\cosh(2\alpha)}{\beta^2}\,x^2+\frac{1}{\beta^4}}=\frac{\beta^3\pi}{2\cosh\alpha}$$
Setting $a=\frac{2\cosh(2\alpha)}{\beta^2}$ and $b^2=\frac{1}{\beta^4}$, then using $\cosh\alpha=\sqrt{\frac{\cosh(2\alpha)+1}{2}}$ will give
$$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}\qquad\qquad\square$$
|
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|
jacobian times its inverse - should be identity Here's an easy one.
A Jacobian is $\frac{dx^i}{dy^j}$.
The inverse is $\frac{dy^j}{dx^k}$. So, in tensor notation,
$\frac{dx^i}{dy^j} \frac{dy^j}{dx^k} = \frac{dx^i}{dx^k} = \delta^i_k$
Now I'll try to this as in matrix form, in two dimensions:
$\left[
\begin{array}{cc}
\frac{dx^1}{dy^1} & \frac{dx^1}{dy^2} \\
\frac{dx^2}{dy^1} & \frac{dx^2}{dy^2} \\
\end{array}
\right]
\cdot
\left[
\begin{array}{cc}
\frac{dy^1}{dx^1} & \frac{dy^1}{dx^2} \\
\frac{dy^2}{dx^1} & \frac{dy^2}{dx^2} \\
\end{array}
\right]
$
The 1,2 element of this product is
$\frac{dx^1}{dy^1}\frac{dy^1}{dx^2} + \frac{dx^1}{dy^2}\frac{dy^2}{dx^2} = \frac{dx^1}{dx^2}+\frac{dx^1}{dx^2} = 0
$
as required.
But looking at the 1,1 element,
$
\frac{dx^1}{dy^1}\frac{dy^1}{dx^1} + \frac{dx^1}{dy^2}\frac{dy^2}{dx^1} = 2 \frac{dx^1}{dx^1} = 2
$
which is wrong… but why?
|
In two dimensions, we have functions $f_1$ and $f_2$ which are both functions of $x_1$ and $x_2$. The Jacobian matrix is given by,
$$J=\left(
\begin{matrix}
\partial_1 f_1 & \partial_2 f_1 \\
\partial_1 f_2 & \partial_2 f_2
\end{matrix}
\right)$$
where $\partial_n := \partial/\partial x_n$. Computing the inverse of the Jacobian using the standard formula yields,
$$J^{-1}=\frac{1}{\partial_1f_1 \partial_2f_2-\partial_2f_1\partial_1f_2}\left(
\begin{matrix}
\partial_2 f_2 & -\partial_2 f_1 \\
-\partial_1 f_2 & \partial_1 f_1
\end{matrix}
\right)$$
Multiplying both matrices yields the desired result,
$$JJ^{-1}=J^{-1}J=\left(
\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}
\right)$$
You just used the wrong expression for the inverse.
|
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|
A question on egyptian fractions An Egyptian fraction is the sum of distinct unit fractions. Are there any 2000 egyptian fractions that their sum is 1?
|
Yes!
Because for any $k\geqslant 3$ we can write $1$ as sum of $k$ different fractions $\dfrac{1}{n}$. $\dfrac{1}{n}=\dfrac{1}{n+1}+\dfrac{1}{n(n+1)}$.
Apllying this identity we get: $1=\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{42}=\dots$
|
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|
To find relatively prime ordered pairs of positive integers $(a,b)$ such that $ \dfrac ab +\dfrac {14b}{9a}$ is an integer How many ordered pairs $(a,b)$ of positive integers are there such that g.c.d.$(a,b)=1$ , and
$ \dfrac ab +\dfrac {14b}{9a}$ is an integer ?
|
I first find out what the values of a/b are which make the expression an integer, then check which of the solutions are rational numbers. Let t = a/b, then t + 14/(9t) should be an integer.
Solve t + 14/(9t) = n: t^2 + 14/9 = nt, t^2 - nt + n^2/4 = n^2/4 - 14/9.
t - n/2 = +/- sqrt (n^2/4 - 14/9).
t should be rational, so n^2/4 - 14/9 should be the square of a rational number. Multiply by 36, and 9n^2 - 56 should be the square of a rational number, which means it must be the square of an integer.
When n >= 10, (3n-1)^2 < 9n^2 - 56 < (3n)^2, so 9n^2 - 56 cannot be the square of an integer. n = 1 to n = 9 can be checked manually, and we find that n = 3 makes 9n^2 - 56 = 25 = 5^2, n = 3 makes 9n^2 - 56 = 169 = 13^2.
n = 3: t - 3/2 = +/- sqrt (9/4 - 14/9) = +/- sqrt (25/36) = +/- 5/6; t = 14/6 or t = 4/6.
n = 5: t - 5/2 = +/- sqrt (25/4 - 14/9) = +/- sqrt (169/36) = +/- 13/6, t = 28/6 or t = 2/6.
Together: a/b is one of 7/3, 2/3, 14/3 or 1/3
Checking:
7/3 + (14*3 / 9*7) = 7/3 + 2/3 = 3
2/3 + (14*3 / 9*2) = 2/3 + 7/3 = 3
14/3 + (14*3 / 9*14) = 14/3 + 1/3 = 5
1/3 + (14*3 / 9*1) = 1/3 + 14/3 = 5.
|
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|
Are there nice ways to solve $(2+x)^{0.25}-(2-x)^{0.25}=1$ Are there nice and elegant ways to solve this equation?
$(2+x)^{0.25}-(2-x)^{0.25}=1$
Thanks.
|
Let $A=\sqrt{2+x}$ and $B=\sqrt{2-x}$. See if you can show the following two equations hold:
$$A^2+B^2=4$$
$$2A+2AB+2B=5$$
Adding these two equations together, and using the quadratic formula you should be able to show that $A+B=\sqrt{10}-1$. Then, plugging this into the second equation, and using the fact that $AB=\sqrt{4-x^2}$, you will obtain the equation
$$7-2\sqrt{10}=2\sqrt{4-x^2}$$
Solving for $x$ gives
$$x=\frac{1}{2}\sqrt{28\sqrt{10}-73}$$
|
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|
Remarkable integral: $\int_0^{\infty} x \left(1 - \frac{\sinh x}{\cosh x-\sqrt 3/2} \right) \mathrm dx= -\frac{13 \pi ^2}{72}$? Numerical evidence suggests that
$$\int_0^{\infty} x \left(1 - \frac{\sinh x}{\cosh x-\sqrt 3/2} \right) \mathrm dx= -\frac{13 \pi ^2}{72}$$
How can we prove this? I could not find a nice contour in the complex plane to integrate around. Integration by parts also didn't help.
Mathematica finds a very complicated antiderivative in terms of special functions, but this was a contest problem so there must be a 'human' way to calculate it.
As O.L. helpfully pointed out, I had the sign wrong. It is corrected now.
|
The following is another approach that also involves differentiating under the integral sign.
Let $$I(\theta) = \int_{0}^{\infty} x \left(1- \frac{\sinh x}{\cosh x - \cos \theta} \right) \, \mathrm dx \, , \quad 0 < \theta < \pi. $$
Then $$I'(\theta) = \sin \theta \int_{0}^{\infty} \frac{x \sinh x}{\left(\cosh x - \cos \theta \right)^{2}} \, \mathrm dx.$$
I tried evaluating $I'(\theta)$ using contour integration, but calculating the residues became too tedious.
Instead let $a$ be a positive parameter, and let $$J(\alpha) = \int_{0}^{\infty}\frac{\mathrm d x}{\cosh (\alpha x)- \cos \theta} = \frac{1}{\alpha} \int_{0}^{\infty} \frac{du}{\cosh u - \cos \theta}.$$
Then $I'(\theta) = -\sin (\theta) \, J'(1) $.
To evaluate $J(\alpha)$, let's integrate the function $f(z) = \frac{z}{\cosh z - \cos \theta}$ around a rectangle contour in the upper half of the complex plane of height $2 \pi i $.
We get $$- 2 \pi i \int_{-\infty}^{\infty} \frac{dt}{\cosh t - \cos \theta} = 2 \pi i \, \left(\operatorname{Res} \left[f(z), i \theta \right]+ \operatorname{Res} \left[f(z), i \left(2 \pi - \theta\right)\right] \right)= 4 \pi i \, \frac{\left(\theta - \pi\right)}{\sin \theta}.$$
Therefore, $J(\alpha) = \frac{1}{\alpha}\frac{\pi - \theta}{\sin \theta} $, which means $I'(\theta) =\pi - \theta $.
Integrating with respect to $\theta$, we get $$I(\theta) = \pi \theta -\frac{\theta^{2}}{2} + C. $$
To determine the constant of integration, we can evaluate $I(\theta)$ at $\theta = \frac{\pi}{2}$.
$$\begin{align} I(\pi/2) &= \int_{0}^{\infty} x \left(1- \tanh x \right) \mathrm dx \\ &= (1- \tanh x) \frac{x^{2}}{2}\Bigg|^{\infty}_{0} + \frac{1}{2}\int_{0}^{\infty}\frac{x^{2}}{\cosh^{2} x} \mathrm dx \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{x^{2}}{\cosh^{2} x} \mathrm dx \\ &= 2 \int_{0}^{\infty}x^{2} \, \frac{e^{-2x}}{(1+e^{-2x})^{2}} \, \mathrm dx \\ &= 2 \int_{0}^{\infty}x^{2} \, \sum_{n=1}^{\infty} (-1)^{n-1} n e^{-2nx} \mathrm dx \\ &= 2 \sum_{n=1}^{\infty} (-1)^{n-1} n \int_{0}^{\infty} x^{2} e^{-2nx} \, \mathrm dx \\ &= \frac{1}{2} \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^{2}} \\ &= \frac{\pi^{2}}{24}. \end{align}$$
So $C= \frac{\pi^{2}}{24} -\frac{\pi^2}{2} + \frac{\pi^{2}}{8} = -\frac{\pi^2}{3}$, and, therefore, $$I(\theta) = \pi \theta -\frac{\theta^{2}}{2} - \frac{\pi^{2}}{3}. $$
|
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|
Evaluate $\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\cdots$ Evaluate
$$
\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\frac{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}{4}\cdots .
$$
First, it is clear that terms tend to $1$.
It seems that the infinity product is not 0. This is related to the post Sequence $x_{n+1}=\sqrt{x_n+a(a+1)}$.
|
If we set $a_0=1/2$ and define
$$
a_k=\frac{1+\sqrt{a_{k-1}}}{2}\tag{1}
$$
then the product sought is
$$
\prod_{k=0}^\infty\ a_k\tag{2}
$$
Since
$$
\cos(2x)=2\cos^2(x)-1\tag{3}
$$
we have that
$$
a_k=\cos^2(2^{-k}x_0)\tag{4}
$$
satisfies $(1)$ with $x_0=\frac\pi4$. Then
$$
\sin(2x)=2\sin(x)\cos(x)\tag{5}
$$
implies by telescoping product that
$$
\begin{align}
\prod_{k=0}^\infty\ a_k
&=\prod_{k=0}^\infty\frac{\sin^2(2^{-k+1}x_0)}{4\sin^2(2^{-k}x_0)}\\
&=\lim_{n\to\infty}\left(\prod_{k=0}^n\sin^2(2^{-k+1}x_0)\middle/\prod_{k=1}^{n+1}4\sin^2(2^{-k+1}x_0)\right)\\
&=\lim_{n\to\infty}\left(\frac{\sin^2(2x_0)}{4^{n+1}\sin^2(2^{-n}x_0)}\right)\\
&=\frac{\sin^2(2x_0)}{4x_0^2}\\
&=\frac4{\pi^2}\tag{6}
\end{align}
$$
|
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|
Proving a tough geometrical inequality, with equality in equilateral triangles. For any triangle with sides-lengths $a$, $b$ and $c$ prove or disprove (1) and (2) :
*
*$$\sum_\mathrm{cyc} \frac{1}{\frac{(a+b)^2-c^2}{a^2}+1}\ge \frac34$$
*Equality in (1) holds if and only if the triangle is equilateral.
Playing with GeoGebra tells that they are correct, however, the proof eludes me.
Please help :)
|
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, By C-S we obtain:
$$\sum_\mathrm{cyc} \frac{1}{\frac{(a+b)^2-c^2}{a^2}+1}=\sum_{cyc}\frac{a^2}{(a+b)^2-c^2+a^2}=$$
$$=\sum_{cyc}\frac{(y+z)^2}{y^2+5z^2+6yz+4zx}=\sum_{cyc}\frac{(y+z)^2(y+x)^2}{(y+x)^2(y^2+5z^2+6yz+4zx)}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(y+z)(y+x)\right)^2}{\sum\limits_{cyc}(y+x)^2(y^2+5z^2+6yz+4zx)}=\frac{\sum\limits_{cyc}(x^4+6x^3y+6x^3z+11x^2y^2+24x^2yz)}{\sum\limits_{cyc}(x^4+6x^3y+6x^3z+11x^2y^2+40x^2yz)}.$$
Thus, it remains to prove that
$$4\sum\limits_{cyc}(x^4+6x^3y+6x^3z+11x^2y^2+24x^2yz)\geq3\sum\limits_{cyc}(x^4+6x^3y+6x^3z+11x^2y^2+40x^2yz)$$ or
$$\sum\limits_{cyc}(x^4+6x^3y+6x^3z+11x^2y^2-24x^2yz)\geq0,$$
which is obviously true.
Done!
|
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|
Find the value of the integral Find the value of the following integral conatining a term with natural logarithm$$\int_0^1 (1-y) \ln\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)\, dy.$$
|
Rewrite $\displaystyle\int_0^1 (1-y)\ln\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)\ dy$ as
$$
\int_0^1 (1-y) \ln\left(2+\sqrt{1-y}\right)\ dy-\int_0^1 (1-y) \ln\left(2-\sqrt{1-y}\right)\ dy
$$
then let $x=2+\sqrt{1-y}$ and $z=2-\sqrt{1-y}$. The integral becomes
$$
\int_0^1 (1-y) \ln\left(2+\sqrt{1-y}\right)\ dy=2\int_2^3 (x-2)^3\ \ln x\ dx
$$
and
$$
\int_0^1 (1-y) \ln\left(2-\sqrt{1-y}\right)\ dy=-2\int_1^2 (z-2)^3\ \ln z\ dz,
$$
then
$$
\int_0^1 (1-y)\ln\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)\ dy=2\int_1^3 (x-2)^3\ \ln x\ dx
$$
The last part can be solved using IBP by letting $u=\ln x$ and $dv=(x-2)^3$.
|
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|
How prove this limit $\lim_{n\to\infty}\sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k}=\ln{3}$
show that: this limit
$$\lim_{n\to\infty}\sum_{k=n}^{9n-1}\dfrac{1}{\sqrt{k^2+1}+k}=\ln{3}$$
my idea: since
$$\dfrac{1}{\sqrt{k^2+1}+k}=\sqrt{k^2+1}-k$$
then I can't.Thank you
|
Notice $\frac{1}{\sqrt{k^2+1}+k}$ is monotonic decreasing, we have
$$\sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k} \ge
\int_n^{9n}\frac{dx}{\sqrt{x^2+1}+x} \ge \sum_{k=n+1}^{9n}\frac{1}{\sqrt{k^2+1}+k}\\
$$
This implies
$$\left| \sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k} -
\int_n^{9n}\frac{dx}{\sqrt{x^2+1}+x} \right|
\le \frac{1}{\sqrt{n^2+1}+n} - \frac{1}{\sqrt{(9n)^2+1}+9n} < \frac{1}{2n}
$$
and hence
$$\lim_{n\to\infty} \sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k}
= \lim_{n\to\infty} \int_n^{9n}\frac{dx}{\sqrt{x^2+1}+x}\tag{*1}$$
Notice in the indefinite integral,
$$\int \frac{dx}{\sqrt{x^2+1}+x} = \int\left(\sqrt{x^2+1}-x\right) dx
= \frac12\left( x(\sqrt{x^2+1} -x) + \log(\sqrt{x^2+1}+x) \right)
= \frac12\left[ \color{blue}{\frac{x}{\sqrt{x^2+1}+x} + \log(\sqrt{1+x^{-2}}+1)} + \log x\right]
$$
The first two term (in blue) converges to a constant $\frac12 + \log 2$ as $x\to\infty$.
they won't contribute to the limit in RHS of $(*1)$. As a result,
$$\lim_{n\to\infty} \sum_{k=n}^{9n-1}\frac{1}{\sqrt{k^2+1}+k}
= \lim_{n\to\infty} \frac12\left(\log(9n)-\log(n)\right) = \log 3.$$
|
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|
Stuck with integral Having this:
$\int x\sqrt{1-x^2}dx$
Substitution:
$t = 1-x^2$
$dt = -2xdx => dx=\frac{-2x}{dt}$
So:
$$\int x\sqrt{1-x^2}dx = -\int x t^\frac{1}{2}\frac{2x}{dt} = -\int \frac{2x^2 t^\frac{1}{2}}{dt} = $$
...but here I stuck... I've tried $-4x\frac{t^\frac{-1}{2}}{\frac{-1}{2}} + c$ but it doesn't match correct result... why did I screw here?
|
Use the fact that $\frac{d}{dx}((1-x^2)^\frac{3}{2})=-3x(1-x^2)^\frac{1}{2}$
(This is just using the rule $\frac{d}{dx}f(x)^n=f'(x)f(x)^{n-1}$)
And we have:
$\int x\sqrt{1-x^2}dx=(-1/3)\int 3x\sqrt{1-x^2}dx=(-1/3)\int \frac{d}{dx}((1-x^2)^\frac{3}{2})dx$
$=\int 1d((1-x^2)^\frac{3}{2})=-(1/3)(1-x^2)^\frac{3}{2}+c$
|
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"url": "https://math.stackexchange.com/questions/791992",
"timestamp": "2023-03-29T00:00:00",
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|
Find the linear-to-linear function whose graph passes through the given three points Find the linear-to-linear function whose graph passes through the points
$(1, 1)$, $(4, 2)$ and $(30, 3)$.
So by using the
$$f(x)=\frac{ax +b}{x+d}$$
I got my final answer to be
$$f(x)=\frac{\frac{75}{23}x + \frac{64}{23}}{x + \frac{12}{23}}$$
and it is wrong.
I have to do a bunch of problems like this and I can't seem to figure it out. I am off somewhere because I did
$f(4)=(4a+b)/(4+d)= 2$
$f(30)= (30a+b)/(30+d) = 3$
and $f(1)= (1a+b)/(1 +d) = 1$
then went on the cancel out $d$ and get $a$ and $b$ then find $c$.
|
So we know that $f(1) = 1$, $f(4) = 2$, and $f(30) = 3$.
Since $f(x)$ is a linear to linear function, we know that:
$f(x) = (ax + b)/(x + c)$
Substituting, we have:
$f(1) = (a + b)/(1 + c)$ = 1, or $a + b = 1 + c$
$f(4) = (4a + b)/(4 + c)$ = 2, or $4a + b = 8 +2c$
$f(30) = (30a + b)/(30 + c)$ = 3, or $30a + b = 90 +3c$
Writing these equations so the variables are on the left gives us:
$a + b - c = 1$
$4a + b - 2c =8$
$30a + b - 3c = 90$
You can solve this system however you need to, I used this site.
Solving gives us $a = 75/23$, $b = 12/23$, and $c = 64/23$.
Thus the final result is:
$f(x) = (75/23x + 12/23)/(x + 64/23)$
It could be that you accidentally switched around some of the variables while doing calculations.
|
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|
Why does $\frac{4}{2} = \frac{2}{1}$? I take for granted that $\frac{4}{2} = \frac{2}{1}$.
Today, I thought about why it must be the case. My best answers amounted to $\frac{4}{2}=2$ and $\frac{2}{1}=2$; therefore $\frac{4}{2}=\frac{2}{1}$. However, that explanation seems circular:
*
*one can express $2$ as $\frac{2}{1}$.
*As such, to say $\frac{4}{2}$ equals $\frac{2}{1}$ because both equal $2$, is nearly saying $\frac{4}{2} = \frac{2}{1}$ (the question) and $\frac{2}{1}=\frac{2}{1}$ (trivial, at best).
So why does $\frac{4}{2} = \frac{2}{1}$?
|
The set $\mathbb{Q}$ is a group then for all $x\in \mathbb{Q}$, there is a only Inverse element such that $x+(-x)=0$. In this case note that
$$\frac{4}{2}-\frac{2}{1}=0$$
then you conclude that $$\frac{4}{2}=\frac{2}{1}$$
|
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|
Finding generators for an ideal of $\Bbb{Z}[x]$ We know that $\Bbb{Z}$ is Noetherian. Hence, we can conclude that $\Bbb{Z}[x]$ is Noetherian, too. Consider the ideal generated by $\langle 2x^2+2,3x^3+3,5x^5+5,…,px^p+p,…\rangle$ for all prime natural numbers $p$. How can I determine a finite number of elements which generate this ideal? Clearly the Euclidean algorithm is not valid in $\Bbb{Z}[x]$.
Thank you
|
Hint Use the Euclidian Algorithm in $\mathbb{Q}[X]$, then multiply whatever what you get by the common denominator. If needed.
Edit By the extended EA we have
$$3x^3+3=\frac{3}{2}x (2x^2+2)+3-3x$$
$$2x^2+2=-\frac{2}{3}(-3x+3)+4$$
Therefore
$$4=2x^2+2+\frac{2}{3}(3x^3+3-\frac{3}{2}x (2x^2+2))\\
=(2x^2+2)(x-1)+\frac{2}{3}(3x^3+3)$$
This shows that $12 \in I$.
Now, for each $p \neq 2,3$ we can write $1=12m+pn$, and hence
$$x^p+1=12(x^p+1)m+n(px^p+1) \in I$$
Using again the Euclidian algorithm in $Q[x]$ we get
$$x^7+1=x^2(x^5+1)+(-x^2+1)$$
$$x^5+1=(-x^2+1)(-x^3-x)+x+1$$
$$-x^2+1=(x+1)(-x+1)$$
Thus, $x+1$ is a linear combination of $x^5+1$ and $x^7+1$ with all the coefficients in $Z[X]$. Moreover, as it divides $px^p+p$ for all $p \neq 2$, it is trivial to show
$$I=(2x^2+2, x+1) \,.$$
Now, use again the Euclidean algorithm, for those two polynomials to get
$$4=2x^2+2-2(x-1)(x+1)$$
|
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|
Integral $\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}$ I am trying to prove this interesting integral
$$
I:=\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}.
$$
I tried using $y=1+x^3$ but that didn't help.
We can possibly try
$$
I=\int_0^\infty \frac{\log(1+x^3) x}{1+x^3} \,dx-\int_0^\infty \frac{\log(x^3) x}{1+x^3}\,dx.
$$ These integrals would be much easier had the bounds been from $0 $\ to $\infty$, however they are not. Perhaps partial integration will work but I didn't find the way if we try
$$
dv=\frac{x}{1+x^3}, \quad u= \log(1+x^3)
$$
but I ran into a divergent integral. Thanks how can we prove I?
|
Define
$$
I(a)=\int_0^\infty \log \frac{a+x^3}{x^3} \frac{x \,dx}{1+x^3}.$$
Then $I(0)=0$ and
\begin{eqnarray}
I'(a)&=&\int_0^\infty\frac{x}{(a+x^3)(1+x^3)}dx\\
&=&\frac13\int_0^\infty\frac{1}{x^{1/3}(a+x)(1+x)}dx\\
&=&\frac{1}{3(1-a)}\left(\int_0^\infty\frac{1}{x^{1/3}(a+x)}dx-\int_0^\infty\frac{1}{x^{1/3}(1+x)}dx\right)\\
&=&\frac{1}{3(1-a)}\frac{2\pi}{\sqrt3}\left(\frac{1}{a^{1/3}}-1\right)\\
&=&\frac{2\pi}{3\sqrt3}\frac{1}{a+a^{2/3}+a^{1/3}}
\end{eqnarray}
Here we use
$$ \int_0^\infty\frac{1}{x^{1/3}(a+x)}dx=\frac{2\pi}{\sqrt3 a^{1/3}}. $$
Thus
\begin{eqnarray}
I(1)&=&\frac{2\pi}{\sqrt3}\int_0^1 \frac{1}{a+a^{2/3}+a^{1/3}}da\\
&=&\frac{2\pi}{3\sqrt3}\int_0^1 \frac{b}{b^2+b+1}db\\
&=&-\frac{\pi^2}{9}+\frac{\pi}{\sqrt3}\ln 3.
\end{eqnarray}
|
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|
How to derive $\frac{d}{dx}\left(x+1\right)^{\sin\left(x\right)}$ I need help to find derivative of: $\frac{d}{dx}(x+1)^{\sin x}$
i tried to do something like this..
$$(x+1)^{\sin x}\cdot \ln\left(x+1\right)=\sin x(x+1)^{\sin\left(x\right)-1}\cdot \ln(x+1)-(x+1)^{\sin(x)}\cdot \frac{1}{x+1}\:=\sin x(x+1)^{\sin\left(x\right)-1} \ln(x+1)-\frac{(x+1)^{\sin x}}{x+1}$$
I tried another way:
$\left(x+1\right)^{sinx}\:=\:ln\left(x+1\right)^{sinx}\:=\:sin\left(x\right)ln\left(x+1\right)\:=\:cos\:\cdot \:ln\left(x+1\right)+sin\left(x\right)\cdot \frac{1}{x+1}\:=\:cos\left(x\right)ln\left(x+1\right)\:+\:\frac{sin\left(x\right)}{x+1}$
ok i got the solution! I put it there, maybe this will help someone!
i used this rule : $e^a=e^{a\cdot ln\cdot e}$
$\left(x+1\right)^{sinx}=\:e^{sinx\cdot ln\left(x+1\right)}=\:e^{sinx\cdot ln\left(x+1\right)}\cdot cosx\:\cdot \:ln\left(x+1\right)\:+\:sinx\cdot \frac{1}{x+1}\:=\:\left(x+1\right)\left(cosxln\left(x+1\right)\:+\:\frac{sinx}{x+1}\right)$
|
You have:
$$f(x) = (x+1)^{\sin{x}},$$ so $\ln{f} = \sin{x} \, \ln{(x+1)}$. Now we have:
$$ \frac{d}{dx} \ln {f} = \frac{1}{f} \frac{df}{dx} = \frac{d}{dx}\left[ \sin{x} \, \ln{(x+1)} \right] . $$
You can now solve for $f'$ once you expand the LHS of the equation.
Cheers!
|
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|
Does this sequence of polynomials have a closed form? Consider the following sequence of polynomials in the variable $d$, which I encountered during a calculation:
$$
1 + 3 d^2
\\
1 + 10 d^2 + 5 d^4
\\
1 + 21 d^2 + 35 d^4 + 7 d^6
\\
1 + 36 d^2 + 126 d^4 + 84 d^6 + 9 d^8
\\
1 + 55 d^2 + 330 d^4 + 462 d^6 + 165 d^8 + 11 d^{10}
\\
1 + 78 d^2 + 715 d^4 + 1716 d^6 + 1287 d^8 + 286 d^{10} + 13 d^{12}
\\
1 + 105 d^2 + 1365 d^4 + 5005 d^6 + 6435 d^8 + 3003 d^{10} + 455 d^{12} +
15 d^{14}
\\
1 + 136 d^2 + 2380 d^4 + 12376 d^6 + 24310 d^8 + 19448 d^{10} +
6188 d^{12} + 680 d^{14} + 17 d^{16}
\\
...
$$
These polynomials came from a relatively simple calculation, so there might be a simple way to generate them directly. There are some patterns in the prime factorizations of the coefficients, but I can't exactly point out what.
Is there a closed form or something similar for these polynomials, perhaps involving special functions or binomial coefficients?
|
It looks like $\dfrac{(1+d)^{2k+1}+(1-d)^{2k+1}}{2}$.
|
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|
Is my proof correct regarding the non primality of $2\cdot 17^a +1$? Today I need your help to know if the proof I have provided below is correct or not. I want to prove that there is no prime of the form $2\cdot 17^a+1$ where $a\in \mathbb N$.
Now, first of all, I tried to get some initial information about $a$ and used MAPLE 14. I came to know that if $a=47$ then the above number is indeed prime but otherwise for any $a\in \mathbb N\backslash \{47\}$ with $a\leq 3000$ so far, computationally it has been shown that $2\cdot 17^a+1$ is not prime.
So we can take the advantage of it and conjecture: ** No number of the form $2\cdot 17^a+1$ is prime if $a\geq 3000$**
Proof:
Note that for $a\geq 3000$, we have
\begin{align*}
&2\cdot 17^a+1\\
\equiv &(-1)(-1)^a+1[3]\\
\equiv &(-1)^{a+1}+1[3]\\
\equiv &0[3]
\end{align*}
provided $a\equiv 0[2]$.
Next we check the case when $a$ is odd. Assume that $a=2a_1+1, a_1\in \mathbb N$. (when $a=1$ then $2\cdot 17^1+1=35=5\cdot 7$ hence not prime. So we see for $a=3, 5, 7 $ etc i.e. $a=2a_1+1$ form.)
This time we see that \begin{align*}
&2\cdot 17^a+1\\
= &2\cdot 17^{2a_1+1}+1\\
\equiv &2(2)^{2a_1+1}+1[5]\\
\equiv &2(2^2)^{a_1}2+1[5]\\
\equiv &(-1)(-1)^{a_1}+1[5]\\
\equiv &(-1)^{a_1+1}+1[5]\\
\equiv &0[5]
\end{align*}
provided $a_1+1$ is odd viz $a_1$ is even.
So next we have to check what happens if we let $a_1=2a_2-1$ form. In this case the number will become as $$2\cdot 17^{2a_1-1}+1=2\cdot 17^{4a_2-3}+1.$$
And then taking modulo 5, we obtain as \begin{align*}
&2\cdot 17^{4a_2-3}+1\\
\equiv & 2(2)^{4a_2-3}+1[5]\\
\equiv & 2(2)^{4a_2}2^{-3}+1[5]\\
\equiv & 2(1)(2^3)^{-1}+1[5]\\
\equiv & 2(3)^{-1}+1[5]\\
\equiv & 2\cdot 2+1[5]\\
\equiv & 0[5]
\end{align*}
Thus we conclude that $2\cdot 17^a+1$ when $a\geq 3000$, is composite number.
Please tell if I have made any mistake on proving this.
Thanking to all of you in advance.
|
No, your second computation and third computation rule out exactly the same candidates, all the $a$ congruent to $1$ mod $4$. You have never adressed the case where $a \equiv 3 \pmod 4$
There is absolutely no hope of doing a similar computation ruling out all the $a$ congruent to $3$ mod $4$, because the case $a = 47$ gives out a prime (notice how $47 \equiv 3 \pmod 4$).
This means that a proof only using congruences would have to work modulo some $M \ge 47$ and would have to show that $a \equiv 47 \pmod M \implies f(a) \equiv 0 \pmod {f(47)}$ (this is the only way to rule out the case $47$ mod $M$). Seeing how $f(47)$ is huge, $M$ also has to be huge (it has to be a multiple of the order of $17$ modulo $f(47)$ ; nothing like your $4$), and then you have to find primes to rule out the other $M-1$ cases (which would be nothing short of a miracle)
|
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|
Calculating limits using l'Hôpital's rule. After a long page of solving limits using l'Hôpital's rule
only those 2 left that i cant manage to solve
$$\lim\limits_{x\to0}{\sqrt {\cos x} - \sqrt[3]{\cos x}\over \sin^2 x }$$
$$\lim\limits_{x\to\ {pi\over 2}}{\tan 3x - 3\over \tan x - 3 }$$
Thanks in advance for any help :)
i edit the second one in mistake i entered $0$ insted of $\pi\over 2$
|
$$\lim\limits_{x\to0}{\sqrt {\cos x} - \sqrt[3]{\cos x}\over \sin^2 x }=\lim\limits_{x\to0}{-\frac{1}{2}\sin x(\cos x)^{-\frac{1}{2}} + \frac{1}{3}\sin x (\cos x)^{-\frac{1}{3}}\over 2\sin x \cos x }=\lim\limits_{x\to0}{-\frac{1}{2}(\cos x)^{-\frac{1}{2}} + \frac{1}{3} (\cos x)^{-\frac{1}{3}}\over 2 \cos x }=-\frac{1}{12}$$
For the second one, you can evaluate the limit directly by substituting in $x=0$.
|
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|
How find this sum $ \frac{1}{1999}\binom{1999}{0}-\frac{1}{1998}\binom{1998}{1}+\cdots-\frac{1}{1000}\binom{1000}{999}$
prove or disprove :
$$S=\dfrac{1}{1999}\binom{1999}{0}-\dfrac{1}{1998}\binom{1998}{1}+\dfrac{1}{1997}\binom{1997}{2}-\dfrac{1}{1996}\binom{1996}{3}+\cdots-\dfrac{1}{1000}\binom{1000}{999}=\dfrac{1}{1999}\left(w^{1999}_{1}+w^{1999}_{2}\right)$$
where
$$w_{1}=\dfrac{1+\sqrt{3}i}{2},w_{2}=\dfrac{1-\sqrt{3}i}{2}$$
My idea: since
$$\dfrac{1}{1999-k}\binom{1999-k}{k}=\dfrac{1}{1999-k}\dfrac{(1999-k)!}{k!(1999-2k)!}=\dfrac{(1998-k)!}{k!(1999-2k)!}$$
Then I can't,maybe can use integral deal it
Thank you
|
The Chebyshev polynomials of the first kind are given by
\begin{align}
T_{n}(x) &= \frac{n}{2} \sum_{k=0}^{[n/2]} \frac{(-1)^{k}}{n-k} \binom{n-k}{k} (2x)^{n-2k} \\
&= \frac{1}{2} \left[ (x - \sqrt{x^{2}-1})^{n} + (x - \sqrt{x^{2}-1})^{n} \right].
\end{align}
When $x=1/2$ it is seen that
\begin{align}
\sum_{k=0}^{[n/2]} \frac{(-1)^{k}}{n-k} \binom{n-k}{k} = \frac{1}{n} \left( a^{n} + b^{n} \right)
\end{align}
where $2a = 1+\sqrt{3} i$ and $2b=1-\sqrt{3} i$. When $n=1999$ this becomes
\begin{align}
\sum_{k=0}^{999} \frac{(-1)^{k}}{1999-k} \binom{1999-k}{k} = \frac{1}{1999} \left( a^{n} + b^{n} \right).
\end{align}
Thus the relationship is shown to be true.
If the sum is changed to all positive terms it can easily be seen as a Lucas number, namely,
\begin{align}
\sum_{k=0}^{999} \frac{1}{1999-k} \binom{1999-k}{k} = \frac{L_{1999}}{1999}.
\end{align}
|
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|
How prove $\left(\frac{b+c}{a}+2\right)^2+\left(\frac{c}{b}+2\right)^2+\left(\frac{c}{a+b}-1\right)^2\ge 5$ Let $a,b,c\in R$ and $ab\neq 0,a+b\neq 0$. Find the minimum of:
$$\left(\dfrac{b+c}{a}+2\right)^2+\left(\dfrac{c}{b}+2\right)^2+\left(\dfrac{c}{a+b}-1\right)^2\ge 5$$
if and only if $$a=b=1,c=-2$$
My idea: Since
$$\left(\dfrac{b+c+2a}{a}\right)^2+\left(\dfrac{c+2b}{b}\right)^2+\left(\dfrac{c-a-b}{a+b}\right)^2$$
let
$$x=\dfrac{b+c+2a}{a},y=\dfrac{c+2b}{b},z=\dfrac{c-a-b}{a+b}$$
then I can't work.
Thank you.
|
Thank you ,@Calvin Lin,
I have solve this problem ,let $$F(a,b,c)=\left(\dfrac{b+c}{a}+2\right)^2+\left(\dfrac{c}{b}+2\right)^2+\left(\dfrac{c}{a+b}-1\right)^2$$
note
$$F(a,b,c)-5=\dfrac{\left((a^2+ab+b^2)c+b(a+b)(2a+b)\right)^2}{a^2b^2(a+b)^2}\ge 0$$
|
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|
Non-linear Second Order Differential equation I want to ask for a hint in solving the following ODE,
$y'' (1+x^2) + y' *(x) = C$,
where C is a constant. I've tried a couple of ways to manipulate this ODE; such as letting v = y'
$v' (1+x^2) + v *(x) = C$
but I can't see a way to solve the equation for y
|
$$
\left(1+x^2\right)v' + xv = C
$$
we can obtain
$$
v\sqrt{1+x^2} = C\int\frac{1}{\sqrt{1+x^2}}dx + \lambda_{1}
$$
or
$$
\frac{dy}{dx} = \frac{C}{\sqrt{1+x^2}}\int\frac{1}{\sqrt{1+x^2}}dx + \frac{\lambda_{1}}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+x^2}}\left[C\sinh^{-1}(x) + \lambda_1\right]
$$
so
$$
y = \lambda_1\int \frac{1}{\sqrt{1+x^2}} + C\int \frac{1}{\sqrt{1+x^2}}\sinh^{-1}(x)
$$
the first integral evaluates to $\sinh^{-1}x$ the second can be evaluated as follows
$$
\int \frac{1}{\sqrt{1+x^2}}\sinh^{-1}(x) = \int \sinh^{-1}(x) \frac{d}{dx}\sinh^{-1}(x) dx = \frac{1}{2}\left(\sinh^{-1}x\right)^{2} + \lambda_2
$$
$$
y(x) = \lambda_1 \sinh^{-1}(x) + \frac{C}{2}\left(\sinh^{-1}x\right)^{2} + \lambda_3
$$
|
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|
How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$ I need to prove that $\frac{1+\tan^2\theta}{1+\cot^2\theta}= \tan^2\theta.$
I know that $1+\tan^2\theta=\sec^2\theta$ and that $1+\cot^2\theta=\csc^2\theta$, making it now $$\frac{\sec^2\theta}{\csc^2\theta,}$$ but I don't know how to get it down to $ \tan^2\theta.$
HELP!!!!
I also need help proving that $\tan\theta + \cot\theta = \sec\theta\cdot\csc\theta.$
|
Draw a right triangle A and label the sides appropriately as OPP, ADJ, and HYP with angle $\theta$ opposite side OPP, and HYP opposite the right angle. Next generate similar triangle B by multiplying each side of triangle A by $\displaystyle\frac{HYP}{OPP\times ADJ }$.
Triangle B has a hypotenuse $\displaystyle HYP\cdot \frac{HYP}{OPP\times ADJ}=\frac{HYP}{OPP}\cdot\frac{HYP}{ADJ}=\csc\theta\sec\theta$.
Its opposite side is $\displaystyle OPP\cdot\frac{HYP}{OPP\times ADJ}=\frac{HYP}{ADJ}=\sec\theta$.
Its adjacent side is $\displaystyle ADJ\cdot\frac{HYP}{OPP\times ADJ}=\frac{HYP}{OPP}=\csc\theta$.
This triangle is shown in the diagram below.
From the definition of $\tan\theta$ we have
$$\color{green}{\tan\theta}=\frac{opposite}{adjacent}=\color{green}{\frac{\sec\theta}{\csc\theta}}$$
We also see that
$$\sin\theta=\frac{opposite}{hypotenuse}=\frac{\sec\theta}{\sec\theta\csc\theta}=\frac{1}{\csc\theta}$$
And
$$\cos\theta=\frac{adjacent}{hypotenuse}=\frac{\csc\theta}{\sec\theta\csc\theta}=\frac{1}{\sec\theta}$$
Additionally, using the Pythagorean theorem we have the following identity
$$\color{blue}{\csc^2\theta+\sec^2=\sec^2\theta\csc^2\theta}$$
Now, using these new tools your first problem becomes
$$\frac{1+\tan^2\theta}{1+\cot^2\theta}=\frac{\sec^2\theta}{\csc^2\theta}=\bigg(\color{green}{\frac{\sec\theta}{\csc\theta}}\bigg)^2=(\color{green}{\tan\theta})^2=\tan^2\theta$$
And your second problem becomes
$$\begin{array}{lll}
\tan\theta+\cot\theta&=&\frac{\sec\theta}{\csc\theta}+\frac{\csc\theta}{\sec\theta}\\
&=&\frac{\sec^2\theta}{\sec\theta\csc\theta}+\frac{\csc^2\theta}{\sec\theta\csc\theta}\\
&=&\frac{\color{blue}{\sec^2\theta+\csc^2\theta}}{\sec\theta\csc\theta}\\
&=&\frac{\color{blue}{\sec^2\theta\csc^2\theta}}{\sec\theta\csc\theta}\\
&=&\sec\theta\csc\theta
\end{array}$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
"answer_count": 9,
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|
proving $\tan^{-1} \left( \frac{x}{1-x^2} \right) = \tan^{-1}x + \tan^{-1}(x^3)$ Two related questions, one easy, one just a bit harder:
1) Prove the identity
$$
\tan^{-1} \left( \frac{x}{1-x^2} \right) =
\tan^{-1}x + \tan^{-1}(x^3)
$$
2) Now try to find a geometric or trigonometric proof of that same geometry, without resorting to calculus.
I'll post an answer to both questions in a couple of days if nobody has one yet.
|
The proof can be established starting from the angle addition identity $$\sin(x+y) = \sin x \cos y + \cos x \sin y.$$ Then with $x = \frac{\pi}{2} - a$, $y = -b$, we get $$\cos(a+b) = \sin(\tfrac{\pi}{2} - a - b) = \cos a \cos b - \sin a \sin b.$$ We then find $$\tan(x+y) = \frac{\sin(x+y)}{\cos(x+y)} = \frac{(\sin x \cos y + \cos x \sin y)/(\cos x \cos y)}{(\cos x \cos y - \sin x \sin y)/(\cos x \cos y)} = \frac{\tan x + \tan y}{1 - \tan x \tan y}.$$ Now letting $u = \tan x$, $v = \tan y$, we get $$\tan(\tan^{-1} u + \tan^{-1}v) = \frac{u+v}{1-uv},$$ or $$\tan^{-1} u + \tan^{-1} v = \tan^{-1} \frac{u+v}{1-uv}.$$ Now letting $u = x$, $v= x^3$, we find $$\frac{u+v}{1-uv} = \frac{x+x^3}{1-x^4} = \frac{x(1+x^2)}{(1-x^2)(1+x^2)} = \frac{x}{1-x^2},$$ and the result immediately follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/812123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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|
Prove that $2002^{2001}$ and $2002^{2001}+2^{2001}$ have the same number of digits. I have an arithmetic exercise as follows: Prove that $2002^{2001}$ and $2002^{2001}+2^{2001}$ have a same number of digits. It seems easy but I don't know how to do. I can do it if $2002$ is replaced by $2000$. Any hints are appreciated.
Thanks a lot!
|
$\ln (a + b) = \ln (a(1 + \dfrac{b}{a})) = \ln a + \ln (1 + \dfrac{b}{a})$.
The number of digits in the base-10 numeral for integer $n > 0$ is $\lceil \log_{10}n \rceil$. $\log_{10} n = \dfrac{\ln n}{\ln 10}$
Let $a = 2002^{2001}$ and $b = 2^{2001}$
$\lceil \log_{10} a \rceil = \left\lceil \dfrac{\ln a}{\ln 10} \right\rceil = \left\lceil \dfrac{\ln 2002^{2001}}{\ln 10} \right\rceil=\left\lceil \dfrac{2001 \ln 2002}{\ln 10} \right\rceil = \lceil 6606.2296\dots\rceil = 6697$
$$\begin{align}
\\ \lceil \log_{10} (a + b)\rceil &= \left\lceil \dfrac{\ln (a+ b)}{\ln 10} \right\rceil
\\ & = \left\lceil \dfrac{\ln a + \ln (1 + \dfrac{b}{a})}{\ln 10} \right\rceil
\\ & = \left\lceil \dfrac{\ln a + \ln \left(1 + \left(\dfrac{2}{2002}\right)^{2001}\right)}{\ln 10} \right\rceil
\\ & = \left\lceil \dfrac{\ln a + O(10^{-6003})}{\ln 10} \right\rceil
\\ & = \left\lceil \dfrac{\ln a}{\ln 10} + O(10^{-6003}) \right\rceil
\\ & = \left\lceil 6606.2296\dots + O(10^{-6003}) \right\rceil
\\ &= 6697
\end{align}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Applications of the identity $ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$ I am reading Euclid's elements
I found the algebraic identity
$ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$
I ponder on usage of this identity for $2$ hours.
but I can't click anything.
$a^2 + b^2 = c^2$ can be used when you want to know the direction between $2$ coordinates.
Any example involving this identity?
|
This identity can be rewritten as $$(a-b)^2=(a+b)^2-4ab$$ and, is the secret behind the discriminant of a quadratic equation. Let me explain it. Suppose $Ax^2+Bx+C=0$ is a quadratic equation with real coefficients, then its roots must occur as complex conjugate pairs. If we have two complex roots $a, b$ for this equation, then there are real $\alpha, \beta$ such that $a,b=\alpha\pm i\beta.$ Clearly, $$a-b=2i\beta,\qquad a+b=2\alpha=-\dfrac{B}{A},\qquad ab=\alpha^2+\beta^2=\dfrac{C}{A}.$$ Hence we have $$(a-b)^2=-4\beta^2=\dfrac{B^2-4AC}{A^2}.$$ Since $\beta^2\gt 0$ we have $$\Delta=A^2(a-b)^2=B^2-4AC\lt0.$$ Other two cases follows similarly.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/816588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding all possible values of $x^4+y^4+z^4$ Given real numbers $x,y,z$ satisfying $x+y+z=0$ and
$$ \frac{x^4}{2x^2+yz}+\frac{y^4}{2y^2+zx}+\frac{z^4}{2z^2+xy}=1$$
Find all possible values of $x^4+y^4+z^4$ with proof.
My attempt :
Putting $x=-y-z$ and doing subsequent calculations we get something like $$\displaystyle \sum_{cyc}\frac{x^4}{(z-x)(y-x)}=1$$
But nothing further. Can someone help me with this? Thanks in advance.
|
this is a simple approach for $x^2+xy+y^2=1$, The op has a good start anyway, from his last step:
note:$\dfrac{x^4}{(z-x)(y-x)}=\dfrac{x^4}{y-z}\left(\dfrac{1}{z-x}-\dfrac{1}{y-x}\right)$
$\sum_{cyc}\dfrac{x^4}{(z-x)(y-x)}=\dfrac{x^4}{(z-x)(y-z)}-\dfrac{x^4}{(y-z)(y-x)}+\dfrac{y^4}{(z-y)(x-y)}+\dfrac{z^4}{(x-z)(y-z)}=\dfrac{x^4-z^4}{(z-x)(y-z)}+\dfrac{y^4-x^4}{(y-z)(y-x)}=\dfrac{-(x+z)(x^2+z^2)+(y+x)(y^2+x^2)}{y-z}=\dfrac{y(x^2+z^2)-z(y^2+x^2)}{y-z}=\dfrac{x^2(y-z)-yz(y-z)}{y-z}=x^2-yz=x^2+xy+y^2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/816702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Solve a system of linear equations $\newcommand{\Sp}{\phantom{0}}$There is a system of linear equations:
\begin{alignat*}{4}
&x - &&y - 2&&z = &&1, \\
2&x + 3&&y - &&z =-&&2.
\end{alignat*}
I create the matrix of the system:
$$
\left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
2 & 3 & -1 & -2
\end{array}\right]
$$
then with GEM,
$$
\left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
0 & 5 & 3 & -4
\end{array}\right]
$$
I don't know how to proceed after that?
I have found the correction of this exercise but I still don't understand the way to solve it. Can someone help me please?
|
I would apply two more row operations and present a solution with a free variable:
$$
\begin{align*}
& \left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
2 & 3 & -1 & -2
\end{array}\right] R_2+(-2)R_1 \rightarrow R_2 \\
\equiv & \left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
0 & 5 & 3 & -4
\end{array}\right] \frac{1}{5}R_2 \rightarrow R_2 \\
\equiv & \left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
0 & 1 & \frac{3}{5} & -\frac{4}{5}
\end{array}\right] R_1+R_2 \rightarrow R_1 \\
\equiv & \left[\begin{array}{rrr|r}
1 & 0 & -\frac{7}{5} & \frac{1}{5} \\
0 & 1 & \frac{3}{5} & -\frac{4}{5}
\end{array}\right].
\end{align*}
$$
Read the solution straight from the matrix as
$$
\begin{align*}
x &=\frac{1}{5}+\frac{7}{5}r \\
y&=-\frac{4}{5}-\frac{3}{5}r \\
z&=r,
\end{align*}
$$
with $r\in \mathbb{R}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/817183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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|
Equation of a tangent line to a curve Equation of the curve is $y=(x+9)/(x+5)$, we are looking for the tangent line to that curve that also goes through $O(0,0)$. Answer given is $x+25y=0$ which I found to be true for $A(-15, 3/5)$ being part of the line and the curve. Question is, how we got to that answer. After differentiating the curve equation, we get $y'=-4/(x+5)^2$ which afterwards gives the equation $4x+25y=0$. There is a $4$ that shouldn't be there, where is my mistake? Thanks in advance.
|
This doesn't always work, but when it does, it means that you can find the equation of the tangent line without using calculus.
A line through the origin must have the form $y_{_L} = mx$. When does such a line intersect the graph of the equation $y_{_H}=\dfrac{x+9}{x+5}$?
\begin{align}
y_{_L} &= y_{_H} \\
mx &= \dfrac{x+9}{x+5} \\
mx^2 + 5mx &= x+9 \\
mx^2 +(5m-1)x - 9 &= 0 \\
x &= \dfrac{1-5m \pm \sqrt{(5m-1)^2+36m}}{2m} \\
\end{align}
If the line $y_{_L}=mx$ were a tangent line, there would only be, locally, one solution for $x$. This will happen when the descriminant is equal to $0$.
\begin{align}
(5m-1)^2+36m &= 0 \\
25m^2 +26m + 1 &= 0 \\
(25m+1)(m+1) &= 0 \\
m &\in \left\{-\dfrac{1}{25}, -1 \right\}
\end{align}
So we find two lines: $y_{_\ell} = -\dfrac{1}{25}x$ and $y_{_u} = -x$.
We still need to justify that these are, indeed, tangent lines. The equation
$y_{_H}=\dfrac{x+9}{x+5}$ describes a "tilted" hyperbola and it appears that the line
$y_{_\ell}=-x$ is tangent to the upper branch while the line $y_{_u} = -\dfrac{1}{25}x$ is tangent to the lower branch of the hyperbola.
In the first case, we find
$y_{_H} - y_{_\ell} =
\dfrac{x+9}{x+5} + x =
\dfrac{(x+3)^2}{x+5}$
We see that, near $x=-3$, this is approximately the parabola $y =\dfrac 12(x+3)^2$, which is tangent to the $x-$axis, from above, at $x=-3$. In other words, the two curves are tangent to each other at $x=-3$
In the other case, we find
$y_{_u} - y_{_H} =
-\dfrac{x}{25} - \dfrac{x+9}{x+5} =
-\dfrac{(x+15)^2}{25(x+5)}$
We see that, near $x=-15$, this is approximately the parabola
$y =-\dfrac 52(x+15)^2$, which is tangent to the $x-$axis, from below, at $x=-15$. In other words, the two curves are tangent to each other at $x=-15$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/818181",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Calculate the number of different words, where $0$ appears an even number of times. Let the set of words with an even length $n$ from the alphabet: $\{ 0,1,2\}$.
Calculate the number of different words, where $0$ appears an even number of times.
For example, for $n=6$ , the words $121212,001212,000000$ are allowed,but the word $100011$ is not allowed.
My idea is:
$$\sum_{k=0}^{\frac{n-2}{2}} \binom{n}{2k} \cdot \binom{n-2k}{2}$$
Is this correct?
|
Another way is by using exponential generating function:
Since $0$ appears even number of times and $1,2$ has no restriction, the egf is:
\begin{align*}
G(x) &= \left(1+\frac{x^2}{2}+\frac{x^4}{4!}+\cdots\right)\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)^2 \\
&= \left(\frac{e^x+e^{-x}}{2}\right)e^{2x} \\
&= \frac{e^{3x}+e^{x}}{2}
\end{align*}
and $$\left[\frac{x^n}{n!}\right]G(x) = \frac{3^n+1}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/818385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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|
Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$ Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$
My approach :
I used $\sin A +\sin B = 2\sin(A+B)/2\times\cos(A-B)/2 $
$\Rightarrow \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(8\pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(\pi + \pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)-\sin(\pi/7) $
I am not getting any clue how to proceed further or whether it is correct or not. Please help thanks..
|
Here's my approach, it only use basic trigonometry identities (i.e, it requires no complex numbers), so it's quite lengthy. I would be glad if you guys can help me shorten it a bit.
*
*Use Product to Sum Formula to prove that:
$\displaystyle \begin{align*}& \ \cos\left(\frac{2\pi}{7} \right) \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{2\pi}{7} \right)\cos\left(\frac{8\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right)\cos\left(\frac{8\pi}{7} \right)\\
= & \ \cos\left(\frac{2\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{6\pi}{7} \right) \\
= & \ \cos\left(\frac{2\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{8\pi}{7} \right)\end{align*}$
*
*Let $\displaystyle A = \cos\left(\frac{2\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{8\pi}{7} \right)$. Prove that $A < 0$, then square $A$ and using (1) along with Power Reduction Formula to prove that $A$ is a solution to the equation: $\displaystyle X^2 - \frac{5}{2}X - \frac{3}{2} = 0$. Hence $\displaystyle A = -\frac{1}{2}$.
*Again use the Product to Sum Formula to prove that $\displaystyle \sin\left(\frac{2\pi}{7} \right) \sin\left(\frac{4\pi}{7} \right) + \sin\left(\frac{2\pi}{7} \right)\sin\left(\frac{8\pi}{7} \right) + \sin\left(\frac{4\pi}{7} \right)\sin\left(\frac{8\pi}{7} \right) = 0$.
*Let $\displaystyle B = \sin\left(\frac{2\pi}{7} \right) + \sin\left(\frac{4\pi}{7} \right) + \sin\left(\frac{8\pi}{7} \right)$. Prove that $B > 0$. Using the result in (3), we have $\displaystyle B^2 = \sin^2\left(\frac{2\pi}{7} \right) + \sin^2\left(\frac{4\pi}{7} \right) + \sin^2\left(\frac{8\pi}{7} \right)$, using Power Reduction Formula, and (2), you'll get $\displaystyle B^2 = \frac{3}{2} - \frac{1}{2} A = \frac{7}{4}$. From the fact that $B > 0$, we'll have $\displaystyle B = \frac{\sqrt{7}}{2}$.
The above are just kind of hints, and you'll have to apply some trig formulae (identities) to get the result.
|
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"url": "https://math.stackexchange.com/questions/818749",
"timestamp": "2023-03-29T00:00:00",
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|
The 3 Integral $\int_0^\infty {x\,{\rm d}x\over \sqrt[3]{\,\left(e^{3x}-1\right)^2\,}}=\frac{\pi}{3\sqrt 3}\big(\log 3-\frac{\pi}{3\sqrt 3} \big)$ Hi I am trying evaluate this integral and obtain the closed form:$$
I:=\int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}=\frac{\pi}{3\sqrt 3}\left(\log 3-\frac{\pi}{3\sqrt 3} \right).
$$
The integral and result has all 3's everywhere. I am not sure how to approach this on. The denominator seems to be a problem.
If $\displaystyle x=\frac{2in\pi}{3}$ we have a singularity but I am not sure how to use complex methods. We will have a branch cut because of the root function singularity.
Differentiating under the integral sign did not help either. I tried partial integration with $v=(e^{3x}-1)^{\frac{2}{3}}$ but this did not simplify since I get a power $x^n, \ (n>1)$ in the new integral.
Thanks, how can we evaluate the integral I?
|
Substitute $e^{-x}=t$, then the integral can be written as:
\begin{align*}
\int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}\, dx &= -\int_{0}^{1} \, \frac{t\, \log{t}}{\left(1-t^3\right)^{2/3}}\, dt\tag 1
\end{align*}
Consider:
\begin{align*}
I(a) &= \int_{0}^{1} \, \frac{t^{a+1}}{\left(1-t^3\right)^{2/3}}\, dt \\
&= -\frac{1}{3} \, {\rm B}\left(\frac{1}{3}, \frac{a+2}{3}\right) \\
I'(0) &= \int_{0}^{1} \, \frac{t\, \log{t}}{\left(1-t^3\right)^{2/3}}\, dt \\
&= \frac{1}{9} \, {\left(\gamma + \psi\left(\frac{2}{3}\right)\right)} {\rm B}\left(\frac{1}{3}, \frac{2}{3}\right) \tag{2}
\end{align*}
Simplifying $(2)$ by using Gauss's digamma theorem for $m<k$
$\displaystyle \psi\left(\frac{m}{k}\right) = -\gamma -\ln(2k) -\frac{\pi}{2}\cot\left(\frac{m\pi}{k}\right) +2\sum_{n=1}^{\lfloor \frac{k-1}{2} \rfloor} \cos\left(\frac{2\pi nm}{k} \right) \ln\left(\sin\left(\frac{n\pi}{k}\right)\right)$
and Euler's reflection formula for gamma functions,
$\displaystyle {\rm B}(1-z, z)= \Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin(\pi z)} $
and from $(1)$,
\begin{align*}
\int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}\, dx &= \frac{\pi}{3\sqrt 3}\left(\log 3-\frac{\pi}{3\sqrt 3}\right)
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to integrate $\int \frac{1}{\sin^4x + \cos^4 x} \,dx$?
How to integrate
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx$$
I tried the following approach:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{\sin^4x + (1-\sin^2x)^2} \,dx = \int \frac{1}{\sin^4x + 1- 2\sin^2x + \sin^4x} \,dx \\
= \frac{1}{2}\int \frac{1}{\sin^4x - \sin^2x + \frac{1}{2}} \,dx = \frac{1}{2}\int \frac{1}{(\sin^2x - \frac{1}{2})^2 + \frac{1}{4}} \,dx$$
The substitution $t = \tan\frac{x}{2}$ yields 4th degree polynomials and a $\sin$ substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach:
$\sin^4x + \cos^4 x = (\sin^2 x + \cos^2x)(\sin^2 x + \cos^2 x) - 2\sin^2 x\cos^2 x = 1 - 2\sin^2 x\cos^2 x = (1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)$
and then I tried substituting: $t = \sin x \cos x$ and got
$$\int\frac{t\,dt}{2(1-2t^2)\sqrt{1-4t^2}}$$
Another way would maybe be to make two integrals:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{(1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)} \,dx = \\ \frac{1}{2}\int \frac{1}{1-\sqrt2\sin x \cos x} \,dx + \frac{1}{2}\int\frac{1}{1+\sqrt2\sin x \cos x} \,dx$$
... and again I tried $t = \tan\frac{x}{2}$ (4th degree polynomial) and $t=\sqrt2 \sin x \cos x$ and I get $\frac{\sqrt 2}{2} \int \frac{\,dt}{(1-t)\sqrt{1-2t^2}}$ for the first one.
Any hints?
|
Simplify the denominator in the following way:
$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^2(2x)}{2}=\frac{1+\cos^2(2x)}{2}=\frac{2+\tan^2(2x)}{2\sec^2(2x)}$$
Hence, the integral you are dealing with is:
$$\int \frac{2\sec^2(2x)}{2+\tan^2(2x)}\,dx$$
I guess the next step is pretty obvious now. ;)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inverting $a+b\sqrt{2}$ in the field $\Bbb Q(\sqrt{2})$ I have been reading through my notes and I came across this example and I found it hard to understand so I need some help in explaining how the inverse of this is found.
The set $\mathbb Q(\sqrt 2)=\{a+b\sqrt2$: $a,b$ in $\mathbb Q \}$ is a field. The inverse of an element $a+b\sqrt 2$ in $ \mathbb Q(\sqrt 2$) is
$$\frac{1}{a+b\sqrt{2}}=\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2} $$
Can anyone explain to how the inverse was found? Appreciate your help.
|
Consider an arbitrary element $a + b \sqrt{2} \in \mathbb{Q}[\sqrt{2}]$. From our experience with real numbers, we know that its inverse will be $\displaystyle \frac{1}{a + b\sqrt{2}}$. However, this is not in the "standard" form for elements in $\mathbb{Q}[\sqrt{2}] = \{x + y\sqrt{2} \ | \ x, y \in \mathbb{Q}\}$.
To get this into the desired form, multiply the purported inverse by $\displaystyle \frac{a - b\sqrt{2}}{a - b\sqrt{2}}$. Doing so, we get:
$$\frac{1}{a + b\sqrt{2}} \cdot \Bigg(\frac{a - b\sqrt{2}}{a - b\sqrt{2}}\Bigg) \ = \ \frac{a-b\sqrt{2}}{a^2 - 2b^2} \ = \ \frac{a}{a^2 - 2b^2} + \Bigg(\frac{-b}{a^2-2b^2}\Bigg)\sqrt{2}$$
And this is indeed what we want since $\displaystyle \frac{a}{a^2 -2b^2}$ and $\displaystyle \frac{-b}{a^2 - 2b^2}$ are rational numbers.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate : $\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}dx$
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$
My approach :
Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then
$$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$
But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.
|
$\bf{My\; Solution::\; }$Let $\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)\cdot (x\cos x-\sin x)}dx$
Let $x=\tan \theta\;,$ Then $dx = \sec^2 \theta d\theta.\;\; $ Then $$I = \displaystyle \int \frac{\tan^2 \theta\cdot \sec^2 \theta }{(\tan \theta\cdot \sin(\tan \theta)+\cos(\tan \theta) )\times (\tan \theta\cdot \cos(\tan \theta)-\sin(\tan \theta) )}d\theta$$
So $$\displaystyle I = \int\frac{\tan^2 \theta\cdot \sec^2 \theta\cdot \cos^2 \theta }{(\sin \theta\cdot \sin(\tan \theta)+\cos(\tan \theta)\cdot \cos \theta )\times (\sin \theta\cdot \cos(\tan \theta)-\sin(\tan \theta)\cdot \cos \theta )}d\theta$$
So $$\displaystyle I = \int\frac{2\tan^2 \theta}{2\cos(\theta-\tan \theta)\cdot \sin(\theta-\tan \theta)}d\theta = \int\frac{2\tan^2 \theta}{\sin (2\theta-2\tan \theta)}d\theta$$
Now Let $\displaystyle (2\theta-2\tan \theta) = u\;\;,$ Then $(2-2\sec^2 \theta)d\theta = du\Rightarrow 2\tan^2\theta d\theta = -du$
So $$\displaystyle I = -\int\frac{1}{\sin u}du = -\int \csc u du = -\ln \tan \left(\frac{u}{2}\right)+\mathbb{C}=-\ln \tan \left(\theta -\tan \theta\right)+\mathbb{C}$$
So $$\displaystyle I = \ln\left|\frac{\cos (\theta-\tan \theta)}{\sin (\theta-\tan \theta)}\right|+\mathbb{C} = \ln\left|\frac{\cos \theta \cdot \cos (\tan \theta)+\sin \theta \cdot \sin(\tan \theta)}{\sin \theta \cdot \cos (\tan \theta)-\cos \theta \cdot \sin(\tan \theta)}\right|+\mathbb{C}$$
So $$\displaystyle I = \ln\left|\frac{\sin (\tan \theta)\cdot \tan \theta+\cos(\tan \theta)}{\cos (\tan \theta)\cdot \tan \theta-\sin (\tan \theta)}\right|+\mathbb{C} = \ln \left|\frac{x\sin x+\cos x}{x\cos x-\sin x}\right|+\mathbb{C}$$
So $$\int\frac{x^2}{(x\sin x+\cos x)\cdot (x\cos x-\sin x)}dx=\ln \left|\frac{x\sin x+\cos x}{x\cos x-\sin x}\right|+\mathbb{C}$$
|
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|
Prove that if $2^{p}-1$ is prime then $n=2^{p-1}(2^p-1)$ is a perfect number Prove that if $2^{p}-1$ is prime then $$n=2^{p-1}(2^p-1)$$ is a perfect number
here is what i did:
We need to prove the $\sigma(n)=n$
so $\sigma(n)=\sigma(2^{p-1})\sigma(2^p-1)$
since $2^{p}-1$ is a prime thus $\sigma(2^p-1)=2^p$
since $2$ is prime we have $\sigma(2^{p-1})=\frac{2^p-1}{2-1}=2^p-1$
so we have $\sigma(n)= 2^p(2^p-1)\neq n$
someone please help where did i go wrong?
|
Let $p \geq 2$ and that $d=2^p-1$ is a prime.
Then,the divisors of $2^{p-1} d$ are these:
$$1,2, \dots, 2^{p-1},d,2d, \dots , 2^{p-1} d$$
Therefore,
$$\sigma(n)=1+2+ \dots+ 2^{p-1}+d+2d+ \dots+ 2^{p-1} d=(1+2+ \dots+ 2^{p-1})(1+d) \\ =(2^p-1)2^p=2n$$
So,we conclude that $n$ is a perfect number.
|
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|
Expand $(x-7)^2 + x^2 = (x+2)^2$ algebraically $(x-7)^2 + x^2 = (x+2)^2$
$(x-7)(x-7) + x^2 = (x+2)(x+2)$
$x^2 -7x -7x + 49 + x^2 = x^2 + 4x + 4$
$x^2 + 18x - 45 = x^2 + x^2$
From that point on, everything I do is incorrect. I don't know what to do with the three $x^2$.
|
Expand the lhs and you get $lhs=2x^2-14x+49$. Expand the rhs and get $rhs=x^2+4x+4$.So $$lha-rhs=(2x^2-14x+49)-(x^2+4x+4)=x^2-18x+45$$ What you wrote is correct but you did not finish. You ended with $$x^2 + 18x - 45 = x^2 + x^2$$ so substract the last lhs from the last rhs of your post and you are done.
|
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|
If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$ If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$
My 1st approach :
$\tan(\alpha +2\alpha +4\alpha) = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $
$\Rightarrow 0 = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $ which doesn't give me any solution.
My IInd approach :
U\sing Euler substitution :
\since $\cos\theta +i\sin\theta = e^{i\theta} $.....(i) and $\cos\theta -i\sin\theta =e^{-i\sin\theta}$....(ii)
Adding (i) and (ii) we get $\cos\theta =\frac{e^{i\theta} +e^{-i\theta}}{2}$ and subtracting (i) and (ii) we get $\sin\theta =\frac{e^{i\theta} -e^{-i\theta}}{2}$
By u\sing this we can write : $$\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$$ as $$\frac{1}{4}\left[ (e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}) (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}}) + (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}})(e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) + (e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) (e^{\frac{i\pi}{7}} -e^{\frac{-i\pi}{7}})\right]$$
$$\large= e^{i\frac{6\pi}{7}}-e^{\frac{i2\pi}{7}}-e^{\frac{-i2\pi}{7}} +e^{\frac{-i6\pi}{7}} +e^{\frac{i3\pi}{7}}-e^{\frac{-i5\pi}{7}}-e^{\frac{i5\pi}{7}} +e^{\frac{-3\pi}{7}} +e^0 -e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}+e^0$$
Can anybody please suggest whether this is my correct approach or not. please guide further... Thanks.
|
We can usually solve these kind of problems by building an equation for which the the given trigonometric values are the roots. (See S.L.Loney's book)
It's as much time consuming as doing it by expanding and simplifying the angle sums, but we get several other answer's easily once the equation is in place.
Here is an overview of the method:
First, use (can be derived from Euler's formula):
\begin{align*}
\cos{(7\, \alpha)} &= 64\cos(\alpha)^7-112\cos(\alpha)^5+56\cos(\alpha)^3-7\cos(\alpha)
\end{align*}
Substituting for $\alpha$ and factoring gives:
\begin{align*}
8c^3+4c^2-4c-1 &= 0 \tag 1
\end{align*}
which is the equation whose roots are $\cos\left(\frac{2\pi}{7}\right),\cos\left(\frac{4\pi}{7}\right)$ and $\cos\left(\frac{2\pi}{7}\right)$
Substituting $c=\frac{1}{c}$ in $(1)$ gives:
\begin{align*}
c^3+4c^2-4c-8 &= 0 \tag 2
\end{align*}
which has the roots $\sec \left(\frac{2\pi}{7}\right),\sec\left(\frac{4\pi}{7}\right)$ and $\sec\left(\frac{2\pi}{7}\right)$
Substituting $c=\sqrt{1+c}$ in $(2)$ and rationalizing gives:
\begin{align*}
c^3-21c^2+35c-7 &= 0 \tag 3
\end{align*}
which has the roots $\tan \left(\frac{2\pi}{7}\right)^2 ,\tan\left(\frac{4\pi}{7}\right)^2$ and $\tan\left(\frac{2\pi}{7}\right)^2$
Getting further equations where roots will be square of the current roots is easy, but going back to find square roots is a bit ambiguous for me, I'd like to know if there
is a better way. I did it like this:
Consider an equation,
\begin{align*}
x^3-a\, x^2+b\, x - c &= 0 \tag 4
\end{align*}
Substitute $x=\sqrt{x}$ and rationalize:
\begin{align*}
x^3-(a^2 - 2\, b)x^2+(b^2 - 2ac)\, x - c^2 &= 0 \tag 5
\end{align*}
Compare with the equation $(2)$ (subst. $c=x$ there)
\begin{align*}
x^3-21x^2+35x-7 &= 0 \tag 6
\end{align*}
and solve for $a,b,c$ and choose by inspection:
$(4)$ would be:
\begin{align*}
x^3+\sqrt{7}x^2-7x+\sqrt{7} &= 0 \tag 7
\end{align*}
which is the equation having $\tan \left(\frac{2\pi}{7}\right) ,\tan\left(\frac{4\pi}{7}\right)$ and $\tan\left(\frac{2\pi}{7}\right)$ as the roots.
Hence, the required sum:
\begin{align*}
\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha &= -7
\end{align*}
From this equation, we can obtain several other identites, such as:
\begin{align*}
\tan\alpha +\tan2\alpha + \tan4\alpha &= -\sqrt{7} \\
\tan\alpha \cdot \tan2\alpha \cdot \tan4\alpha &= -\sqrt{7} \\
\end{align*}
And from $(6)$,
\begin{align*} \\
\tan\left(\alpha\right)^2 +\tan\left(2\alpha\right)^2 + \tan\left(4\alpha\right)^2 &= 21 \\
\left(\tan\alpha \tan2\alpha\right)^2 +\left(\tan2\alpha \tan4\alpha\right)^2 +\left(\tan4\alpha \tan\alpha\right)^2 &= 35 \\
\tan\left(\alpha\right)^2 \cdot \tan\left(2\alpha\right)^2 \cdot \tan\left(4\alpha\right)^2 &= 7 \\
\end{align*}
and so on for powers of $4, 8$ etc.
|
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|
What is the sum of this? $ 1 + \frac12 + \frac13 + \frac14 + \frac16 + \frac18 + \frac19 + \frac1{12} +\cdots$ I'm in trouble with this homework.
Find the sum of the series $ 1 + \frac12 + \frac13 + \frac14 + \frac16 + \frac18 + \frac19 + \frac1{12} +\cdots$, where the terms are the inverse of the positive integers whose only prime factors are 2 and 3.
Hint: write the series as a product of two geometric series.
Ok, I found some patterns in these numbers, but I can't find the two series.
|
$$\left(1+\frac{1}{2}+\frac{1}{4}+\cdots \right)\left(1+\frac{1}{3}+\frac{1}{9}+\cdots \right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\cdots$$
both of the first series are geometric and can be summed
$$1+\frac{1}{2}+\frac{1}{4}+\cdots =2$$
$$1+\frac{1}{3}+\frac{1}{9}+\cdots =\frac{3}{2}$$ so
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\cdots=3$$
|
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|
Prove the sum $\sum_{n=1}^\infty \frac{\arctan{n}}{n}$ diverges. I must prove, that sum diverges, but...
$$\sum_{n=1}^\infty \frac{\arctan{n}}{n}$$
$$\lim_{n \to \infty} \frac{\arctan{n}}{n} = \frac{\pi/2}{\infty} = 0$$
$$\lim_{n \to \infty} \frac{ \sqrt[n]{\arctan{n}} }{ \sqrt[n]{n} } = \frac{1}{1} = 1$$
Cauchy's convergence test undefined.
There is a $E_0\gt0$:
$$\left|\frac{\arctan{n+1}}{n+1} + \frac{\arctan{n+2}}{n+2} + ... + \frac{\arctan{n+p}}{n+p}\right| \ge \frac{\pi}{4}\left|\frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{n+p}\right| \ge \frac{\pi}{4} \frac{p}{n+p} (Let\, p = n) \ge \frac{\pi}{4} \frac{\bcancel{n}}{2\bcancel{n}} = \frac{\pi}{8} (\sim0.4) \ge E_0 = \frac{1}{8} \gt 0;$$
Now am I correct?
|
Hint: $\displaystyle\lim_{N\to\infty}\sqrt[^N]{\dfrac\pi2}=1\neq0$. The same holds true if you replace $\dfrac\pi2$ with any other strictly positive
finite quantity.
|
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|
$f(n)$ and $f(2^n)$ are co prime for all natural numbers $n$. Find all such polynomials. Find all polynomials $f(x)$ with integer coefficients such that $f(n)$ and $f(2^n)$ are co prime for all natural numbers $n$.
|
Here is a proof of non-existence of non-constant polynomials of this type: Suppose $f(x) = a_0 + a_1x + a_2x^2 + … + a_mx^m$ is such a polynomial (where $m>0$). We reach a contradiction.
If $a_0$ is even then $f(2)$ and $f(2^2)$ share a common factor of $2$. So $a_0$ must be odd. Thus, for every positive integer $k$, the number $f(2^k)$ must be odd. Since $f(2^k)$ is odd and diverges to $\infty$ or $-\infty$ as $k$ gets large, there must be a positive integer $K$ such that $f(2^K)$ is divisible by some odd prime $p$. That is, $f(2^K) \equiv 0$ (mod $p$).
Claim: We can construct an integer $Z$ such that $f(Z)$ and $f(2^Z)$ are both divisible by $p$ (which yields the contradiction).
Proof: Since $p$ is an odd prime, the sequence $\{mod(2^i, p)\}_{i=1}^{\infty}$ is never $0$, and periodically cycles over a subset of integers in $\{1, \ldots, p-1\}$. Let $n$ be a positive integer that satisfies $2^{n+2^K} \equiv 2^K$ (mod $p$). Define $Z = np+2^K$. By Fermat's Little Theorem we have $2^{np+2^K} \equiv 2^{n+2^K}$ (mod $p$). Thus:
$2^Z = 2^{np+2^K} \equiv 2^{n+2^K} \equiv 2^K$ (mod $p$).
Hence, $f(2^Z) \equiv f(2^K) \equiv 0$ (mod $p$).
However, $Z = np + 2^K \equiv 2^K$ (mod $p$), and so $f(Z) \equiv f(2^K) \equiv 0$ (mod $p$).
Thus, $f(Z)$ and $f(2^Z)$ are both divisible by $p$.
|
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|
Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?
Prove or disprove that if $t$ is a positive integer, $$f(x,y)=\dfrac{x^2+y^2}{xy-t},$$ then $f(x,y)$ has only finitely many distinct positive integer values with $x,y$ positive integers. In other words, there exist $k\in\mathbb N$ such that if $n\gt k$ then $f(x,y)=n$ has no positive integer solutions.
This problem is a generalization of this famous problem.
*
*Below is the list of the set of $f(x,y)$ with $t\le 10$ (may be incomplete):
{t,{f(x,y)}}=
{1,{5}}
{2,{4,10}}
{3,{3,4,8,13,17}}
{4,{5,26}}
{5,{13,25,37}}
{6,{6,10,50}}
{7,{5,8,9,20,29,41,65}}
{8,{4,10,18,34,82}}
{9,{5,29,61,101}}
{10,{20,122}}
Thanks in advance!
|
October 7, 2015. This is with $$ \frac{x^2 + y^2}{xy - t} > 0, $$
which I believe to be the intent of the question.
I proved finiteness, with an explicit bound that is not that bad.
This works. Note that the original question requires $xy> t.$ Otherwise we could have listed $x=1,y=1,t=2$ to get
$(x^2 + y^2)/ (xy-t) = -2.$ This was not done. So we are keeping $xy>t>0,$ in
$$ \frac{x^2 + y^2}{xy-t} = q. $$
We have the arc of the hyperbola
$$ x^2 - q x y + y^2 = -tq $$ in the first quadrant $x,y > 0$ that lies in the sector of the first quadrant defined by
$$ 2 x \leq q y $$ and
$$ 2 y \leq q x. $$ Note that the points of intersection of the two boundary lines with the hyperbola branch give the two points with the minimum values of $x$ and of $y.$
As noted in the other answer, if there are any integer solutions $(x,y)$ with $q$ also an integer, then there is at least one solution between the indicated Hurwitz lines.
Next, we always have $q \geq 3.$ In $ x^2 - q x y + y^2 = -tq ,$ if $q=1$ the quadratic form on the left hand side is positive definite and can never equal the right hand side, which is negative. If $q=2$ the quadratic form on the left hand side is positive semi-definite ($(x-y)^2$) and can never equal the right hand side, which is negative.
The key to finiteness was simply the size of $xy/t.$ We know already that $xy > t,$ that is $xy/t > 1.$ By Lagrange multipliers, the smallest value occurs when $x = y,$ at which point
$$ \frac{xy}{t} = \frac{q}{q-2} = 1 + \frac{2}{q-2}. $$
Once again by Lagrange multipliers, the largest value of $xy/t$ within the Hurwitz region occurs at the boundary point where one of the lines meets the hyperbola. One of them is at
$$ y = \left( \frac{2}{q} \right) x. $$ Plugging this into $ x^2 - q x y + y^2 = -tq $ gives a nice value for $x^2,$ then $ y^2 = \left( \frac{4}{q^2} \right) x^2 $ gives a nice value for $y^2.$ These turn out to be
$$ x^2 = \frac{q^3 t}{q^2 - 4}, \; \; \; y^2 = \frac{4 q t}{q^2 - 4}. $$
Together
$$ x^2 y^2 = \frac{4 q^4 t^2}{(q^2 - 4)^2}, $$ and
$$ x y = \frac{2 q^2 t}{q^2 - 4}, $$ or
$$ \frac{x y}{t} = \frac{2 q^2 }{q^2 - 4} = \frac{2 q^2 - 8 }{q^2 - 4} + \frac{8 }{q^2 - 4} = 2 + \frac{8 }{q^2 - 4} . $$
This gives the maximum. Since $q \geq 3,$
$$ \frac{x y}{t} \leq 2 + \frac{8 }{3^2 - 4} = \frac{18}{5} = 3.6 . $$
Here we finally return to integers. We have $x \geq 1,$ which tells us that a Hurwitz fundamental solution always has
$$ y \leq \frac{18}{5} t. $$ Once again, Lagrange multipliers tell us that $x^2 + y^2$ is maximized at the boundary point $x=1$ on the curve $xy= 18t/5,$ so
$$ x^2 + y^2 \leq 1 + \frac{324}{25} t^2. $$ However, $xy - t \geq 1,$ meaning $q \leq x^2 + y^2.$ We then get finiteness from
$$ q \leq 1 + \frac{324}{25} t^2. $$
Computations, as above, suggest the stronger $q \leq t^2 + 2 t + 2.$
As far as rapid computer computations, for a fixed $t,$ we can demand $1 \leq x \leq 4 t$ because $18/5 < 4.$ For each $x,$ we can then demand $1 \leq y \leq x$ along with the very helpful $x y \leq 4 t.$ Having found an integer quotient $q,$ we then keep only those solutions with $2x \leq qy$ and $2y \leq qx.$
To repeat the good part: if there is any solution $(x,y)$ then there is at least one fundamental solution, that is with
$$ \color{blue}{ 2x \leq qy}$$ and $$ \color{blue}{ 2y \leq qx}.$$
For such a fundamental solution, we have
$$ \color{blue}{ 1 + \frac{2}{q-2} \leq \frac{xy}{t} \leq 2 + \frac{8}{q^2-4} }. $$
Since $x^2 - qxy + y^2$ is positive (semi)-definite when $q = 1,2,$ we know that $q \geq 3$ always. Thus $q^2 - 4 \geq 5.$ As $\frac{8}{5} \leq 2,$ we get
$$ \color{blue}{xy \leq 4t}. $$
Here is a graph for $t=1, q=5,$ showing the region where fundamental solutions must lie:
|
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|
Prove by induction that $(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}$ Prove by induction that $$(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}.$$
I got up to:
$n=1$ is true, and assuming $n=k$ prove for $n=k+1$.
Prove...
$$\frac{k(2k+1)(7k+1)+6(2(k+1))^2}{6} = \frac{(k+1)(2k+3)(7k+8)}{6}$$
I keep trying to expand to $6(2k+2)^2$ and factorising but I end up being short on one factor, e.g., I end up with $\frac{(k+1)(2k+3)(7k+6)}{6}$.
|
Alternative route (also with induction)
You can start proving inductively that: $$\sum_{k=1}^{n}k^{2}=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)$$
Then have a look at: $$\sum_{k=n+1}^{2n}k^{2}=\sum_{k=1}^{2n}k^{2}-\sum_{k=1}^{n}k^{2}$$
|
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|
Partial fraction (doubt) I have this partial fraction
$$\displaystyle\frac{1}{(2+x)^2(4+x)^2}$$
I tried to resolve using this method:
$$\displaystyle\frac{A}{2+x}+\displaystyle\frac{B}{(2+x)^2}+\displaystyle\frac{C}{4+x}+\displaystyle\frac{D}{(4+x)^2}$$
$$1=A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2$$
When x=-2
$$1=B(4-2)^2$$
$$B=\displaystyle\frac{1}{4}$$
When x=-4
$$1=D(2-4)^2$$
$$D=\displaystyle\frac{1}{4}$$
When x=0
$$1=A(2)(16)+B(16)+C(4)(4)+D(4)$$
$$1=A(32)+B(16)+C(16)+D(4)$$
Replacing the values for B y D
$$1=A(32)+4+C(16)+1$$
$$1-4-1=A(32)+c(16)$$
$$-4=A(32)+C(16)$$
How I can get the values of $A$ and $D$?
|
Assign different values to $x$, like $-1, 0, +1, +2$ and solve the linear system of 4 equations in 4 unknowns.
$$\left[\begin{array}{cccc}
1/1& 1/1& 1/3& 1/9\\
1/2& 1/4& 1/4& 1/16\\
1/3& 1/9& 1/5& 1/25\\
1/4& 1/16& 1/6& 1/36
\end{array}\right]
\left[\begin{array}{c}
A\\
B\\
C\\
D\end{array}\right]=
\left[\begin{array}{c}
1/9\\
1/64\\
1/225\\
1/576\end{array}\right]$$
$$\left[\begin{array}{c}
A\\
B\\
C\\
D\end{array}\right]=
\left[\begin{array}{c}
-1/4\\
1/4\\
1/4\\
1/4\end{array}\right]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/833170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Factor the Expression completely$ (a+b)^2 - (a-b)^2$ I don't understand this question. The answer in the book is $4ab$, but how is that term a factor?
I was thinking along the line that this was a difference of squares example. $a^2-b^2 = (a+b)(a-b)$
My answer is $[(a+b)-(a-b)][(a+b)+(a-b)]$
What do I not understand?
|
$$(a + b)^2 - (a-b)^2 = (a+b)(a+b) - (a+b)(a-b) $$ $$= (a+b)((a+b) - (a - b)) $$ $$ = (a + b)(a + b - a + b) $$ $$= (a+b)(2b)$$
POST EDIT: $$(a+b)^2 - (a-b)^2 = a^2 +2ab + b^2 -(a^2 - 2ab + b^2) = 2ab - (-2ab) = 4ab$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/834111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Shears using Matrix Methods Determine the equation of the image of the graph:
$$y=(x-1)^3 -2$$
after a shear of factor $1$ away from the $y$-axis, relative to the line $y=1$.
|
You can parameterize the graph $y=(x-1)^3-2$ with the variable $t$ like so:
$$
\left(\begin{array}{c}
t\\
(t-1)^3-2
\end{array}\right)
$$
You can then transform this vector by translating it downwards one unit, performing the shear, then translating it back up. Because this involves doing a translation, we add an extra $1$ on the end of the previous vector so that it is $3$-dimensional, and perform a shear transformation in the three dimensions.
$$
\left [ \begin{array}{a}
1&0&0\\
0&1&0
\end{array} \right ]
\left [ \begin{array}{a}
1&0&0\\
0&1&1\\
0&0&1
\end{array} \right ]
\left [ \begin{array}{a}
1&1&0\\
0&1&0\\
0&0&1
\end{array} \right ]
\left [ \begin{array}{a}
1&0&0\\
0&1&-1\\
0&0&1
\end{array} \right ]
\left[\begin{array}{c}
t&\\
(t-1)^3-2\\
1
\end{array}\right]
$$
$$=\left[\begin{array}{c}
(t-1)^3+t-3\\
(t-1)^3-2
\end{array}\right]$$
You now have a parametric equation for the resulting line in terms of a variable $t$ over the domain $\mathbb{R}$.
$$\left[\begin{array}{c}
x(t)\\
y(t)
\end{array}\right]=\left[\begin{array}{c}
(t-1)^3+t-3\\
(t-1)^3-2
\end{array}\right]$$
Note that in general the method above works, in this particular circumstance, you could probably skip doing the three linear transformations by observing that a shear of factor $1$ about $y=1$ is the same as doing a shear about $y=0$ and subtracting $1$ from $x$ afterwards.
Thus:
$$
\left [ \begin{array}{a}
1&0&0\\
0&1&0
\end{array} \right ]
\left [ \begin{array}{a}
1&1&-1\\
0&1&0\\
0&0&1
\end{array} \right ]
\left[\begin{array}{c}
t&\\
(t-1)^3-2\\
1
\end{array}\right]
$$
Gives you the same result. Since we're only dealing with one transformation now and we're ignoring the third component of the solution anyway, we can just do:
$$
\left [ \begin{array}{a}
1&1&-1\\
0&1&0
\end{array} \right ]
\left[\begin{array}{c}
t&\\
(t-1)^3-2\\
1
\end{array}\right]
$$
Suppose we have some arbitrary function $y(x)$ sheared by a factor $k$ about the line $y=c$. Then we would do:
$$
\left [ \begin{array}{a}
1&k&-kc\\
0&1&0
\end{array} \right ]
\left[\begin{array}{c}
t\\
y(t)\\
1
\end{array}\right]
$$
We'd then have a parametric equation for the resulting $\left(\begin{smallmatrix}x\\y\end{smallmatrix}\right)$ graph in terms of $t$.
In short, if you want to compose a translation with your linear transformations, invoke an extra dimension and use that. In this case, you may be able to tell immediately that this final matrix is the only transformation necessary. Pretend you're doing a normal shear, but compensate for the fact that it's actually off-axis by subtracting the appropriate amount from $x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/835824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Show $(x+y)^a > x^a + y^a$ for $x,y>0$ and $a>1$ This is a pretty straightforward question. I want to show $(x+y)^a > x^a + y^a$ for $x,y>0$ and $a>1$.
One way would be this. WLOG, suppose $x \leq y$. Then:
$(1+\frac{x}{y})^a >1+\frac{x}{y} \geq 1+(\frac{x}{y})^a$.
Noting that $\frac{1}{y^a}t_1>\frac{1}{y^a}t_2 \implies t_1 > t_2$, it follows that $(x+y)^a > x^a + y^a$.
But what other ways are there to show this inequality?
|
Write
$$ f(a) = \left( \frac{x}{x+y} \right)^a + \left( \frac{y}{x+y} \right)^a $$
It is clear that
$$ f(1) = \frac{x}{x+y} + \frac{y}{x+y} = 1 $$
It is also clear that
$$ f'(a) = \ln\left( \frac{x}{x+y} \right) \left( \frac{x}{x+y} \right)^{a-1} + \ln\left( \frac{y}{x+y} \right) \left( \frac{y}{x+y} \right)^{a-1} $$
whence
$$f'(a) < 0 $$
and therefore
$$ a > 1 \Rightarrow f(a) < 1$$
Thus
$$ \left( \frac{x}{x+y} \right)^a + \left( \frac{y}{x+y} \right)^a < 1 $$
or
$$\Big( x + y \Big)^a > x^a + y^a $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/836337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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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.