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Integration using different substitutions resulting in different answers? I'm going through the MIT online calc course, and I'm trying to find the following from an Integration Problem set:
$$\int \frac{x^2+1}{x^2+2x+2}dx$$
I could use the substitution u = x+1, which results in the correct answer:
$$\int \frac{(u-1)^2+1}{u^2+1}du$$
$$\int 1 + \frac{1}{u^2+1} - \frac{2u}{u^2+1}du$$
$$u + arctan \; u - ln(u^2+1)$$
$$x+1+arctan \; (x+1) - ln(x^2+2x+2)$$
But originally I tried using completing the square and the substitution $(tan \theta = x + 1)$. I get:
$$\int \frac{x^2+1}{(x+1)^2+1}dx$$
$$\int \frac{tan^2 \theta - 2tan \theta + 1 + 1}{sec^2\theta}sec^2\theta d\theta$$
$$\int sec^2\theta - 2tan\theta+1 d\theta$$
$$tan \theta + 2ln(cos \theta) + \theta$$
$$x + 1 - 2ln(x^2+2x+2) + arctan(x+1)$$
There's an extra 2 coefficient there. I don't think there's a math error, so I'm thinking the ambitious substitution is the cause. Any ideas?
|
The mistake is in thinking $\cos\theta=x^2+2x+2$.
Remark: I would probably want to divide first, getting $1-\dfrac{2x+1}{(x+1)^2+1}$.
Then write $\dfrac{2x+1}{(x+1)^2+1}$ as $\dfrac{2x+2}{(x+1)^2+1}-\dfrac{1}{(x+1)^2+1}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Implicit differentiation question Given that $x^n + y^n = 1$, show that $$\frac{d^2y}{dx^2} = -\frac{(n-1)x^{n-2}}{y^{2n-1}}.$$
I found that $\displaystyle nx^{n-1}+ny^{n-1}\frac{dy}{dx} = 0$ so that $\displaystyle y'=\frac{-x^{n-1}}{y^{n-1}}$.
Then $$n(n-1)x^{n-2}+n(n-1)y^{n-2}\left(\frac{dy}{dx}\right)^2 + \frac{d^2y}{dx^2}ny^{n-1} = 0.$$
Therefore $$y'' = \displaystyle \frac{-n(n-1)x^{n-2}-n(n-1)y^{n-2}(y')^2}{ny^{n-1}}.$$
Substituting the first derivative:
$$y'' = \displaystyle\frac{-n(n-1)x^{n-2}-n(n-1)y^{n-2}\left(\dfrac{-x^{n-1}}{y^{n-1}}\right)^2}{ny^{n-1}}.$$
I've been trying tons of different steps and can't establish which way to eliminate the $y$ terms from the numerator. Could someone offer a hint on how to proceed. Thanks.
|
Differentiating $y^{n-1}y_1+x^{n-1}=0$ wrt $x,$
$\displaystyle y^{n-1}y_2+(n-1)y^{n-2}y_1^2+(n-1)x^{n-2}=0$
$\implies \displaystyle -\frac{y^{n-1}y_2}{n-1}=x^{n-2}+y^{n-2}\left(-\frac{x^{n-1}}{y^{n-1}}\right)^2$ as $y_1=-\frac{x^{n-1}}{y^{n-1}}$
$$\implies -\frac{y^{n-1}y_2}{n-1}=x^{n-2}+\frac{x^{2n-2}}{y^n}=\frac{x^{n-2}(y^n+x^n)}{y^n}=\frac{x^{n-2}}{y^n}\text{ as } y^n+x^n=1$$
|
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|
Give $3$ coterminal angles
Give three angle measures in radians which are coterminal with each of the following. Include at least one positive and one negative angle measure.
*
*$$\pi/4\quad \text{rad}$$
*$$5\pi/3\quad \text{rad}$$
|
Firstly let us have a look at the definition of coterminal:
Two angles are coterminal if they are drawn in the standard position and both have their terminal sides in the same location.
The angles can be anything as long as the lines are aligned. Here are some examples:
$\angle ABC$ and $\angle DBC$ are coterminal because they are both $50^\circ$.
$\angle ABC$ and $\angle DBC$ are coterminal because $410^\circ\equiv50^\circ\pmod{360^\circ}$
$\angle ABC$ and $\angle DBC$ are coterminal because $-312^\circ\equiv48^\circ\pmod{360^\circ}$
This works for any angle system and in your case this is radians. To find coterminal angles you have to find angular solutions to this equivalence
$$\angle DBC \equiv \angle ABC \pmod{\angle S}$$
$\angle S$ is $2\pi$ for radians and $360^\circ$ for degrees.
So finally all you have to do is add or subtract $2\pi$.
1)
$$\frac\pi4+2\pi=\frac\pi4+\frac{8\pi}4=\frac{9\pi}4$$
$$\frac\pi4+4\pi=\frac\pi4+\frac{16\pi}4=\frac{17\pi}4$$
$$\frac\pi4-2\pi=\frac\pi4-\frac{8\pi}4=-\frac{7\pi}4$$
2)
$$\frac{5\pi}3+2\pi=\frac{5\pi}3+\frac{6\pi}3=\frac{11\pi}3$$
$$\frac{5\pi}3+4\pi=\frac{5\pi}3+\frac{12\pi}3=\frac{17\pi}3$$
$$\frac{5\pi}3-2\pi=\frac{5\pi}3-\frac{6\pi}3=-\frac{\pi}3$$
|
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|
Effective method to solve $ \frac{x}{3x-5}\leq \frac{2}{x-1}$ I need to solve this inequality. How can I do so effectively?
$$ \frac{x}{3x-5}\leq \frac{2}{x-1}$$
|
Make $0$ one side of the inequality:
$\displaystyle\frac{x}{3x-5}-\frac{2}{x-1}\le 0$
Write the expression in $x$ as a single fraction:
$\displaystyle\frac{x(x-1)-2(3x-5)}{(3x-5)(x-1)}\le 0$
Expand the numerator:
$\displaystyle\frac{x^2-7x+10}{(3x-5)(x-1)}\le 0$
Factorise the numerator:
$\displaystyle\frac{(x-2)(x-5)}{(3x-5)(x-1)}\le 0$
Multiply throughout by the square of the denominator which is positive, thus preserving the inequality, while noting the values $x$ cannot take for the initial fractions to be defined:
$\displaystyle(x-1)(x-\frac{5}{3})(x-2)(x-5)\le 0$ and $x\ne 1$ and $x\ne \frac{5}{3}$
By considering regions of positivity and negativity, we have:
$\displaystyle 1<x<\frac{5}{3}$ or $2\le x\le 5$
|
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|
Solutions of $x^2 + 7y^2 = 2^n$ where $x$ and $y$ are odd numbers Is it true that for any $n\geq 2$ the equation $x^2 + 7y^2 = 2^n$ has a solution with $x$ and $y$ odd ??
|
The result is true for $n=3$, since $2^3=x^2+7y^2$ with $x=y=1$. Moreover, $x\equiv y\pmod{4}$.
We show that if there exist $x$ and $y$, odd, with $x\equiv y\pmod{4}$, such that $x^2+7y^2=1$, then there exist $x_1,y_1$, odd, with $x_1\equiv y_1\pmod{4}$, such that $x_1^2+7y_1^2=2^{n+1}$. Let
$$x_1=\frac{x+7y}{2}\quad\text{and}\qquad y_1=\frac{x+y}{2}.$$
First note that $\frac{x+y}{2}$ and $\frac{x-7y}{2}$ are odd. For $\frac{x+y}{2}$, we use the fact that $x$ and $y$ are odd and $x\equiv y\pmod{4}$ to conclude $\frac{x+y}{2}$ is odd. The argument for $\frac{x-7y}{2}$ is similar. It is easy to verify that $x_1\equiv y_1\pmod{4}$.
To show that $x_1,y_1$ work, we compute
$$\frac{(x-7y)^2}{4}+7\frac{(x+y)^2}{4}.$$
This simplifies to $2x^2+14y^2$, which is $2^{n+1}$. (We used a version of the Brahmagupta Identity.)
Remark: It is an artefact of the proof that the solutions involve negative integers. We can get positive solutions by taking the absolute value.
|
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|
How find the value $\lim_{x\to 0}\dfrac{1}{x^2}\left(1-\cos{x}\cdot\sqrt{\cos{(2x)}}\cdots\sqrt[n]{\cos{(nx)}}\right)$ find the value
$$I_{n}=\lim_{x\to 0}\dfrac{1}{x^2}\left(1-\cos{x}\cdot\sqrt{\cos{(2x)}}\cdots\sqrt[n]{\cos{(nx)}}\right)$$
This is my methods:
\begin{align*}I_{n+1}-I_{n}&=\lim_{x\to 0}\dfrac{1}{x^2}\left(\cos{x}\cdot\sqrt{\cos{(2x)}}\cdots\sqrt[n]{\cos{(nx)}}\left(1-\sqrt[n+1]{\cos{(n+1)x}}\right)\right)\\
&=\lim_{x\to 0}\dfrac{1-\sqrt[n+1]{\cos{(n+1)x}}}{x^2}\\
&=\dfrac{n+1}{2}
\end{align*}
so
$$I_{n}=\dfrac{n(n+1)}{4}$$
Have you other nice methods? Thank you
|
$$
\cos(nx) = 1 - \frac{n^2 x^2}{2} + O(x^4)
$$
as $x \to 0$ so
$$
\sqrt[n]{\cos(nx)} = 1 - \frac{n x^2}{2} + O(x^4)
$$
by the binomial theorem, thus
$$
\begin{align}
\cos(x)\sqrt{\cos(2x)}\cdots\sqrt[n]{\cos(nx)} &= 1 - \frac{x^2}{2} \sum_{k=1}^{n} k + O(x^4) \\
&= 1 - \frac{n(n+1)}{4} x^2 + O(x^4).
\end{align}
$$
|
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|
Evaluation Of Maximum Value Without Calculus $x$,$y$,$z$ are non-negative numbers.
$x+y+z=3$
Find the maximum value of $~$ $x^{2}y+y^{2}z+z^{2}x$ $~$ without calculus.
|
Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Hence, by Rearrangement and AM-GM we obtain:
$$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$
$$=b(a^2+ac+c^2)\leq b(a^2+2ac+c^2)=b(a+c)^2=4b\left(\frac{a+c}{2}\right)^2\leq$$
$$\leq4\left(\frac{b+\frac{a+c}{2}+\frac{a+c}{2}}{3}\right)^3=4.$$
The equality occurs for $z=0$, $x=2$ and $y=1$, which says that the answer is $4$.
|
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|
In a numerical system of base $r$, the polynomial $x^2 − 11x + 22 = 0$ has the solutions $3$ and $6$. What is the base r of the system? From Algebra, the statement is equivalent to say that $(x^2− 11x +
22)_{r}$ = $(x − 3)_{r} \cdot (x − 6)_{r}$. Doing operations we arrive at $3 + 6 = 11_{r} = r + 1$, and $(3)(6) = 22_{r} = 2 \cdot 11_{r}$. In any case, $r = 8$.
This is the solution to the problem, but how do I yield to the conclusion that $3 + 6 = 11_{r} = r + 1$?
|
If the roots are $3$ and $6$, the equation is $(x-3)(x-6)=x^2-(3+6)x+3\cdot 6=0$ So $11_r=3+6$ and $11$ in base $r$ is $r+1$ because the leading digit is $r$. Similarly we have $3\cdot 6 = 22_r=2r+2$ Each of these gives us $r=8$
|
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|
Summing random numbers with different domain This should be simple, but my math skills are rusty.
Given the random integers $X_0,\ldots, X_i,\ldots, X_n$, each of which is uniformly distributed on a domain $0,\ldots,J_i,\ldots,J_n$, what is the probability that the sum of the numbers will be greater than or equal to some number $Y$?
I've found a formula that works when the domain is constant (e.g. using die rolls, every integer can be only 1, 2, 3, 4, 5, or 6), but what about when the domain is different for each (e.g. one is determined by $D_6$ die and the other by a $D_{10}$ die, so $1, \ldots, 6$ and $1, \ldots, 10$)?
|
I will show you how to do the two case example, and hopefully you can then do it for a three case scenario. So, suppose that $X$ models the first dice throw, and therefore has a discrete uniform distribution on $[1,6]$, while $Y$ models the second dice throw, thus being discrete uniform on $[1,10]$. We are interested in describing $Z=X+Y$. The first observation is that the support of $Z$ is $[1+1, 6+10] = [2,16]$.
The second observation (not that obvious) is that, because of the different supports, the distribution of $Z$ will look like a trapezoid. Why? Look at the discrete convolution formula, which gives the distribution of the $Z$, assuming that $X$ and $Y$ are independent (which they are, since we look at dice rolls). The formula states that
$$
P(Z=z) = \sum_{k=0}^z P(X=k) P(Y=z-k)
$$
since when $X=k$, clearly $Y=z-k$. So:
$$P(Z=2) = P(X=1)P(Y=2-1) = \frac{1}{6}\frac{1}{10} = \frac{1}{60}\\
P(Z=3) = P(X=1)P(Y=3-1) + P(X=2)P(Y=3-2) = \frac{2}{60} \\
\vdots\\
P(Z=7) = P(X=1)P(Y=7-1) + \dots +P(X=6)P(Y=7-6) = \frac{6}{60}.
$$
We conclude that for $2\le n \le 7, P(Z) = \frac{n-1}{60}$. Now,
$$
P(Z=8) = P(X=1)P(Y=8-1) + \dots + P(X=6)P(Y=8-6) = \frac{6}{60}\\
P(Z=9) = P(X=1)P(Y=9-1) + \dots + P(X=6)P(Y=9-6) = \frac{6}{60}\\
\vdots\\
P(Z=11) = P(X=1)P(Y=11-1) + \dots + P(X=6)P(Y=11-6) = \frac{6}{60}.
$$
At this point, since $Y$ is capped at a maximum of 10, we can't form $Z=12$ when $X=1$ and therefore conclude that for $ 8\le n \le 11, P(Z) = \frac{6}{60}$. Finally,
$$
P(Z=12) = P(X=2)P(Y=12-2) + \dots + P(X=6)P(Y=12-6) = \frac{5}{60}\\
P(Z=13) = P(X=3)P(Y=13-2) + \dots + P(X=6)P(Y=13-6) = \frac{4}{60}\\
\vdots\\
P(Z=16) = P(X=6)P(Y=16-6) = \frac{1}{60}.
$$
Putting everything together we get that:
$$
P(Z=z) =
\begin{cases}
\dfrac{z-1}{60} & 2 \le z \le 7 \\
\dfrac{6}{60} & 8 \le z \le 11 \\
\dfrac{17-z}{60} & 12 \le z \le 16 \\
\end{cases}
$$
To do your 3 case example, pick $Z$ as described above, and pick $U$ to have any distribution you want. Then see if you can write $Z+U$ using the same type of argument.
|
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|
compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$ Knowing that $1 - \frac 12 + \frac 13 - \cdots = \ln 2$ and $1 - \frac 13 + \frac 15 - \cdots = \frac{\pi}{4}$, compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$. I programmed and found that $\frac{\ln 2}{4}+{\frac{\pi}{4}}/2$ is precise to the eighth digit.
|
$$
{\cal F}\left(x\right)
\equiv
\sum_{n = 0}^{\infty}{x^{4n + 2} \over \left(4n +1\right)\left(4n + 2\right)}
$$
\begin{align}
{\cal F}''\left(x\right)
&\equiv
\sum_{n = 0}^{\infty}x^{4n}
=
{1 \over 1 - x^{4}}
=
{1/2 \over 1 - x^{2}} + {1/2 \over 1 + x^{2}}
=
{1 \over 4}\sum_{\sigma = \pm}{1 \over 1 + \sigma\,x} + {1/2 \over 1 + x^{2}}
\\[3mm]
{\cal F}'\left(x\right)
&\equiv
{1 \over 4}\sum_{\sigma = \pm}\sigma\,\ln\left(1 + \sigma\,x\right)
+
{1 \over 2}\arctan\left(x\right)
\\[3mm]
{\cal F}\left(x\right)
&\equiv
{1 \over 4}\sum_{\sigma = \pm}\sigma\left\lbrack%
x\ln\left(1 + \sigma\,x\right)
-
\int_{0}^{x}x'\,{\sigma \over 1 + \sigma\,x'}\,{\rm d}x'
\right\rbrack
+
{1 \over 2}\,x\,\arctan\left(x\right)
-
{1 \over 2}\int_{0}^{x}{x' \over x'^{2} + 1}\,{\rm d}x'
\\[3mm]&=
{1 \over 4}\sum_{\sigma = \pm}\sigma\left\lbrack%
x\ln\left(1 + \sigma\,x\right)
-
x
+
\sigma\ln\left(1 + \sigma\,x\right)
\right\rbrack
+
{1 \over 2}\,x\arctan\left(x\right)
-
{1 \over 4}\,\ln\left(1 + x^{2}\right)
\\[5mm]&
\end{align}
$$
{\cal F}\left(x\right)
=
{1 \over 4}\left\lbrack%
\left(-x + 1\right)\ln\left(1 - x\right) + \left(x + 1\right)\ln\left(1 + x\right)
\right\rbrack
+
{1 \over 2}\,x\arctan\left(x\right)
-
{1 \over 4}\,\ln\left(1 + x^{2}\right)
$$
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\lim_{x \to 1^{-}}{\cal F}\left(x\right)
=
\sum_{n = 0}^{\infty}{1 \over \left(4n +1\right)\left(4n + 2\right)}
=
{1 \over 8}\,\pi + {1 \over 4}\,\ln\left(2\right)\quad}
\\ \\ \hline
\end{array}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
what is the largest number here? what is the largest number here? how to find these with out inspection but a proper mathematical rule?
$$\Large 2^{3^4},2^{4^3},3^{2^4},3^{4^2},4^{2^3},4^{3^2}$$
Thank you
|
As $\displaystyle a^{m^n}=a^{(m^n)}$ and $\displaystyle (a^m)^n=a^{m\cdot n}$
$2^{3^4}=2^{81}$
$2^{4^3}=2^{64}$
$3^{2^4}=3^{16}$
$3^{4^2}=3^{16}$
$4^{2^3}=(2^2)^{(2^3)}=(2^2)^8=2^{16}$
$4^{3^2}=(2^2)^{(3^2)}=(2^2)^9=2^{18}$
As we are interested only in finding the largest, we can safely consider $2^{81}$ and $3^{16}$ (why?)
$2^{81}=2\cdot(2^5)^{16}>(32)^{16}>3^{16}$
Also, $2^4>3, (2^4)^{16}>3^{16}\implies 2^{64}>3^{16}$
|
{
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|
Minimum and maximum value of function $f(x,y,z) = x^3 + y^3 + z^3$ Find minimum and maximum value of function $$f(x,y,z) = x^3 + y^3 + z^3$$ on set $$ \left\{ (x,y,z): x^2 + y^2 + z^2 = 1 \wedge x+y+z = \sqrt{3} \right\} $$
I don't know what is this set. We have sphere and plane so I suppose that it may be circle or point. How find it?
|
note
$$(x^2+y^2+z^2)(1+1+1)\ge (x+y+z)^2$$
so if
$$x^2+y^2+z^2=1,x+y+z=\sqrt{3}$$
then we have
$$3(x^2+y^2+z^2)=(x+y+z)^2$$
$$\Longrightarrow x^2+y^2+z^2-xy-yz-xz=0$$
so
$x=y=z$
|
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|
What is the sum of $4\sqrt{28}$ and $3\sqrt{7}$ ? As far as I can simplify it -
$$4\sqrt{7*4} + 3\sqrt{7}
= 8\sqrt{7} + 3 \sqrt{7}
= \sqrt{7} * 11$$
However , The options for the correct answer are -
A) $ 8/3$
B) $ 16/3$
C) $ 18/3$
D) $24/3$
I am a ninth grader so please try to explain in simple terms .
|
I think there was a misprint , most probably they meant - $$\dfrac{4\sqrt{28}} {3\sqrt{7}}
= \dfrac{8\sqrt{7}} {3\sqrt{7}}
= \dfrac83 $$
In this case , the answer would be A) 8/3
|
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|
Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$
Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.
I have tried it by substituting $x = \exp(2i\pi/5)$
but it is getting complicated.
|
$x^4+x^3+x^2+x+1=0$. Divide by $x^2$, and substitute $z=x+x^{-1}$, resulting in $z^2+z-1=0$. This has two solutions: $z_1=2cos72^\circ=\dfrac{-1+\sqrt5}{2}$, and $z_2=2cos144^\circ=\dfrac{-1-\sqrt5}{2}$.
|
{
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|
Solving a system of equation modul0 5 Consider the system of linear equations $$\begin{pmatrix} 6 & -3\\ 2 & 6 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}=\begin{pmatrix} 3\\ 1 \end{pmatrix} $$
a) Solve the system in $\mathbb{F}_5$
I just want to make sure my solution is correct:
We have: $$A=\begin{pmatrix} 6 & -3\\ 2 & 6 \end{pmatrix}\\ \Rightarrow A^{-1}= \frac{1}{42}\begin{pmatrix} 6 & 3\\ -2 & 6 \end{pmatrix}\\ \equiv \begin{pmatrix} 3 & \frac{3}{2}\\ -1 & 3 \end{pmatrix} \mbox{ } [5]$$
Therefore: $\begin{pmatrix} x_1\\ x_2 \end{pmatrix}=\begin{pmatrix} 3 & \frac{3}{2}\\ -1 & 3 \end{pmatrix}\begin{pmatrix} 3\\ 1 \end{pmatrix}= \begin{pmatrix} \frac{21}{2}\\ 0 \end{pmatrix} [5] $
|
This method is legitimate since, if a solution exists, the determinant is non-zero mod $5$, and thus is invertible mod $5$.
To continue, $21 \equiv 1 \pmod 5$ and $\frac{1}{2}=2^{-1} \equiv 3 \pmod 5$.
So the solution becomes $$\begin{bmatrix} 3 \\ 0 \\ \end{bmatrix}.$$
|
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|
Trying to calculate the probability that one RV exceeds another RV I am running into a silly mistake when trying to calculate the probability that a random variable, $U$ is less than another random variable, $V$. I am hoping that someone can help me spot my mistake.
My setup is as follows:
*
*$U$ ~ $U(0,1)$
*$W = \frac{1}{2}\min U_i$ where $U_i$~$U(0,1)$ for $i = 1 \ldots n$
Given this setup, I believe that the right approach is to condition on the value of $W$ and then integrate across all possible values of $W$. That is,
$P(U<W) = \int_{w=0}^{1/2} P(U < W ~|~ W = w) f(w)dw$
In order to evaluate this expression, we need to first obtain $f(w)$.
Note that:
$$ \begin{align} F(w) &= P(W<w) \\
&= P( \frac{1}{2}\min U_i < w)\\
&= P( \min U_i < 2w)\\
&= 1-P( \min U_i \geq 2w) \\
&= 1-\prod_{i=1}^n{P(U_i \geq 2w)} \\
&= 1-(1-2w)^n
\end{align}$$
Thus,
$$f(w) = 2n(1-2w)^{n-1} $$
Using this information, I get
$$\begin{align} P(U<W) &= \int_{0}^{1/2} P(U < W ~|~ W = w) f(w)dw \\
&\int_{0}^{1/2} wf(w)dw \\
&\int_{0}^{1/2} 2nw(1-2w)^{n-1} dw \\
\end{align}$$
Unfortunately, however, this integral does not evaluate to a meaningful result.
|
It looks like the integral does evaluate to a meaningful result. I just had a silly algebra mistake.
We can proceed by substitution. Define $x = 1-2w$, so that $w = \frac{1-x}{2}$ and $dx = \frac{dw}{-2}$. Then:
$$\begin{align} P(U<W) &= \int_{0}^{1/2} 2nw(1-2w)^{n-1} dw \\
&= \int 2n \frac{1-x}{2} x^{n-1} \frac{dx}{-2} \\
&= \frac{n}{2} \int x^n - x^{n-1} dx \\
&= \frac{n}{2} \Bigg(\frac{x^{n+1}}{n+1} - \frac{x^{n}}{n} \Bigg) \\
&= \frac{nx^{n}}{2} \Bigg( \frac{x}{n+1} - \frac{1}{n} \Bigg) \\
&= \frac{n(1-2w)^{n}}{2} \Bigg( \frac{(1-2w)}{n+1} - \frac{1}{n} \Bigg) \\
\end{align}$$
Plugging in $w=1/2$ and $w=0$ we get:
$$\begin{align} P(U<W) &= \frac{n(1-2w)^{n}}{2} \Bigg( \frac{(1-2w)}{n+1} - \frac{1}{n} \Bigg) \Bigg|_{0}^{1/2}\\
&= 0 - \frac{n}{2} \Bigg( \frac{1}{n+1} - \frac{1}{n} \Bigg)\\
&= \frac{n}{2} \Bigg( \frac{1}{n} - \frac{1}{n+1} \Bigg)\\
&= \frac{1}{2(n+1)}
\end{align}$$
|
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|
How to solve the following equation $\sqrt[3]{x+3}+\sqrt[3]{x}=\sqrt[3]{3+8x}$ I am trying to solve this equation:
$$\sqrt[3]{x+3}+\sqrt[3]{x}=\sqrt[3]{3+8x}$$
I would like to get some advice, how to solve it.
Thanks.
|
W'll use $$a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\sum_{cyc}(a-b)^2.$$
Since $x+3=x=-3-8x$ is impossible, our equation is equivalent to
$$x+3+x-3-8x+3\sqrt[3]{x(x+3)(8x+3)}=0$$ or
$$\sqrt[3]{x(x+3)(8x+3)}=2x$$ or
$$x((x+3)(8x+3)-8x^2)=0,$$
which gives $x=0$ or $x=-\frac{1}{3}$.
|
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|
Show that the odd prime divisors of the integer $n^2+n+1$ which are different from $3$ are of the form $6k+1$. 1)The odd prime divisors of the integer $n^4+1$ are of the form $8k+1$.
My attempt: Let $p$ be odd divisor of $n^4+1$.Then $n^4+1 \equiv 0$ (mod p) $\Rightarrow n^4 \equiv -1$ (mod p) $\Rightarrow n^8 \equiv 1$ (mod p). Hence, the order of $n$ modulo $p$ is 8, which implies that $8 | \phi(p)=p-1 \Rightarrow p=8k+1$
The part i not sure is the order of $n$ modulo $p$. How do I show the order of $n$ modulo $p$ is 8?
2)Show that the odd prime divisors of the integer $n^2+n+1$ which are different from $3$ are of the form $6k+1$.
My attempt: Let $p$ be odd divisor of $n^2+n+1$. Then $n^2+n+1 \equiv 0 $(mod p). I try to mimick the method above but i fail.
Can anyone guide me?
|
Hint: $$4(n^2+n+1)=(2n+1)^2+3$$
Alternatively:
$$(n-1)(n^2+n+1)=n^3-1$$
|
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|
Integrating $\sec^2 x$ from first principles Is it possible to solve $\int \sec^2 x ~dx$ without knowing that $\frac{d}{dx}\tan x = \sec^2 x$?
|
Yes, this can be done, but the method is far longer than if you know the derivative of $\tan x$ already. The point is to convert this trigonometric integral into the integral of a rational function, integrate that with partial fractions, and then convert back. We use the substitution $u = \tan(x/2)$, for which
$$
\sin x = \frac{2u}{1+u^2}, \ \ \ \cos x = \frac{1-u^2}{1+u^2}, \ \ \ dx = \frac{2\,du}{1+u^2}.
$$
(This is called the $\tan(x/2)$-substitution, and is discovered naturally by comparing the trigonometric parametrization of the unit circle and a rational parametrization of the unit circle. See http://en.wikipedia.org/wiki/Tangent_half-angle_substitution.)
[EDIT: Let me show how to derive those transformation formulas above, especially the one for $dx$ since there is a comment on my answer that it depends on knowing the derivative of $\tan x$ is $\sec^2 x$ already, but that's not really the case. We do need to know that $1 + \tan^2 t = \sec^2 t = 1/\cos^2 t$. Then $\cos^2(x/2) = 1/(1+\tan^2(x/2)) = 1/(1+u^2)$, so from the double-angle formulas for sine and cosine we get
$$
\cos x = 2\cos^2(x/2) - 1 = 2\sec^2(x/2) - 1 = \frac{2}{1+u^2} - 1 = \frac{1-u^2}{1+u^2}
$$
and
$$
\sin x = 2\sin(x/2)\cos(x/2) = 2\tan(x/2)\cos^2(x/2) = \frac{2u}{1+u^2}.
$$
Then differentiate both sides of the equation for $\sin x$ using the quotient rule on the right:
$$
\cos x\,dx = \frac{2(1+u^2)-(2u)(2u)}{(1+u^2)^2}\,du = \frac{2(1-u^2)}{(1+u^2)^2}\,du \Rightarrow dx = \frac{2\,du}{1+u^2}.
$$
]
From the formula for $\cos x$ in terms of $u$ we have $\sec^2 x = 1/\cos^2 x = (1+u^2)^2/(1-u^2)^2$, so
$$
\int \sec^2 x \,dx = \int \frac{(1+u^2)^2}{(1-u^2)^2}\frac{2}{1+u^2}\,du = \int \frac{2(1+u^2)}{(1-u^2)^2}\,du.
$$
The partial fraction decomposition of that last integrand is
$$
\frac{2(1+u^2)}{(1-u^2)^2} = \frac{2(1+u^2)}{(1-u)^2(1+u)^2} = \frac{1}{(1-u)^2} + \frac{1}{(1+u)^2},
$$
so
\begin{eqnarray*}
\int \sec^2 x\,dx & = & \int\left(\frac{1}{(1-u)^2} + \frac{1}{(1+u)^2}\right)\,du \\
& = & \frac{1}{1-u} - \frac{1}{1+u} + C \\
& = & \frac{2u}{1-u^2} + C \\
& = & \frac{2\tan(x/2)}{1 - \tan^2(x/2)} + C.
\end{eqnarray*}
Recall now the addition formula for the tangent function:
$$\tan(a+b) = \frac{\tan a + \tan b}{1-(\tan a)(\tan b)}.$$ Thus the double-angle formula is $\tan(2a) = 2(\tan a)/(1 - \tan^2 a)$. Therefore, using $a = x/2$, we get
$$
\int \sec^2 x\,dx = \tan\left(2\frac{x}{2}\right) + C = \tan x + C.
$$
|
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|
Number theory proof from AoPS http://www.artofproblemsolving.com/Resources/articles.php?page=htw.readers
In the above link, he gives a problem, namely
Let $S(n)$ be the sum of the digits of $n$. Find
$S(S(S(4444^{4444}))).$
Then in the lemma he presents is where I get confused. He states
Every integer $n$, written in decimal notation, is congruent to the sum of its digits modulo $9$.
And he goes on to present the proof. But by the same argument he uses I can also show that it works for mod $8$ as well. Also, how would I represent $18$? $19$?
|
Let's define a $k$-digit number $n$ to have the decimal (base-10) digits $d_0, d_1, \ldots d_{k-1}$. We can then write $n$ as:
$$n = 10^0d_0 + 10^1d_1 + \cdots + 10^{k-1}d_{k-1}$$
Note that we can re-arrange this as:
$$\begin{align}
n &= d_0 + 10^1d_1 + \cdots + 10^{k-1}d_{k-1}\\
&= d_0 + (9d_1+d_1) + \cdots + (10^{k-1}-1)d_{k-1} + d_{k-1}\\
&= (d_0 + d_1 + \cdots + d_{k-1}) + \underbrace{(9d_1 + 99d_2 + \cdots (10^{k-1}-1)d_{k-1})}_{\text{all this is a multiple of 9}}
\end{align}$$
Because anything that is a multiple of $9$ is congruent to $0 \pmod{9}$, we have:
$$\begin{align}
n &\equiv (d_0 + d_1 + \cdots + d_{k-1}) + 0 &\pmod{9}\\
&\equiv d_0 + d_1 + \cdots + d_{k-1} &\pmod{9}
\end{align}$$
An example may help. Let's try $d_0=1, d_1=6,d_2=7,d_3=3,d_4=4$, which makes our $5$-digit number $16734$. Then, we have:
$$\begin{align}
n &= 1 + 6\cdot10 + 7\cdot10^2 + 3\cdot10^3 + 4\cdot10^4\\
&= 1 + (6 + 6\cdot9) + (7 + 7\cdot99)+(3+3\cdot999) + (4+4\cdot9999)\\
&= (1+6+7+3+4) + (6\cdot9 + 7\cdot99 +3\cdot999+4\cdot9999)\\
&\equiv (1+6+7+3+4) + 0 &\pmod{9}\\
&\equiv 11 &\pmod{9}
\end{align}$$
|
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|
Trigonometric Equation $\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$ How can I solve this trigonometric equation?
$$\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$$
|
Ok, there might be nicer ways to do this but here it goes. Let $\theta=\pi/15$ so you have
$$
\tan(\theta)\tan(4\theta)\tan(\frac{9}{2}\theta)\tan(6\theta)
$$
then you use $\tan()=\frac{\sin()}{\cos()}$, and then you use the equivalences
$2\sin(\alpha)\sin(\beta)=\cos(\alpha-\beta)-\cos(\alpha+\beta),$
and
$2\cos(\alpha)\cos(\beta)=\cos(\alpha-\beta)+\cos(\alpha+\beta)$.
You should get
$\frac{\cos(3\theta)\cos(\frac{3}{2}\theta)-\cos(3\theta)\cos(\frac{21}{2}\theta)-\cos(5\theta)\cos(\frac{3}{2}\theta)+\cos(5\theta)\cos(\frac{21}{2}\theta)}{\cos(3\theta)\cos(\frac{3}{2}\theta)+\cos(3\theta)\cos(\frac{21}{2}\theta)+\cos(5\theta)\cos(\frac{3}{2}\theta)+\cos(5\theta)\cos(\frac{21}{2}\theta)} $
** I also used the fact that $\cos(-\alpha)=\cos(\alpha)$. Now, note that you have something of the form
$\frac{a-b-c+d}{a+b+c+d},$
which is equal to $1+\frac{-2b-2c}{a+b+c+d}$. Now, we shall show that $-2b-2c=0$. So we have
$
-2b-2c=-2\cos(3\theta)\cos(\frac{21}{2}\theta)-2\cos(5\theta)\cos(\frac{3}{2}\theta)
$
using the formula above for the product of cosines we get
$-2\cos(3\theta)\cos(\frac{21}{2}\theta)-2\cos(5\theta)\cos(\frac{3}{2}\theta)=-\cos(\frac{15}{2}\theta)-\cos(\frac{27}{2}\theta)-\cos(\frac{7}{2}\theta)-\cos(\frac{13}{2}\theta).$
Now plug in the value of $\theta$, you get
$
-\cos(\pi/2)-\cos(\frac{27}{30}\pi)-\cos(\frac{7}{30}\pi)-\cos(\frac{13}{30}\pi)
$
naturally the first element is $0$. Now, using the fact that $\cos(\alpha\pm\pi/2)=\mp\sin(\alpha)$ you can write
$
-\cos(\frac{27}{30}\pi)-\cos(\frac{7}{30}\pi)-\cos(\frac{13}{30}\pi)=\sin(6\pi/15)-\sin(\frac{4}{15}\pi)-\sin(\pi/15)
$
Now you use $\sin(\alpha)\cos(\beta)=\frac{\sin(\alpha+\beta)+\cos(\alpha-\beta)}{2}$ looking for two numbers such that the equality holds. You should get then
$
\sin(6\pi/15)-\sin(\frac{4}{15}\pi)-\sin(\pi/15)=\sin(6\pi/15)-2\sin(\pi/6)\cos(\pi/10)=\sin(6\pi/15)-\cos(\pi/10)
$
which it is easily seen to be zero.
Therefore
$
\tan(\theta)\tan(4\theta)\tan(\frac{9}{2}\theta)\tan(6\theta)=1, \qquad \theta=\pi/15
$
|
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|
Combinatorial interpretation of an alternating binomial sum Let $n$ be a fixed natural number. I have reason to believe that $$\sum_{i=k}^n (-1)^{i-k} \binom{i}{k} \binom{n+1}{i+1}=1$$ for all $0\leq k \leq n.$ However I can not prove this. Any method to prove this will be appreciated but a combinatorial solution is greatly preferred. Thanks for your help.
|
Suppose we seek to verify that
$$\sum_{q=k}^n (-1)^{q-k} {q\choose k} {n+1\choose q+1} = 1$$
where $n\ge k.$
We first treat the case when $k\gt 0$ and introduce
$${q\choose k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} (1+z)^q \; dz.$$
Observe that this is zero when $0\le q\lt k$ so that we may extend the
limit in the sum to zero, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}}
\sum_{q=0}^n (-1)^{q-k} {n+1\choose q+1} (1+z)^q
\; dz
\\ = (-1)^{k+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} \frac{1}{1+z}
\sum_{q=0}^n (-1)^{q+1} {n+1\choose q+1} (1+z)^{q+1}
\; dz
\\ = (-1)^{k+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} \frac{1}{1+z}
\sum_{q=1}^{n+1} (-1)^{q} {n+1\choose q} (1+z)^{q}
\; dz
\\ = (-1)^{k+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} \frac{1}{1+z}
(-1+(1-(1+z))^{n+1})
\; dz
\\ = (-1)^{k+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} \frac{1}{1+z}
(-1 + (-1)^{n+1} z^{n+1})
\; dz.$$
Now since $n\ge k$ this simplifies to
$$(-1)^{k} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} \frac{1}{1+z} \; dz
= (-1)^k (-1)^k = 1.$$
The second case when $k=0$ yields
$$\sum_{q=0}^n (-1)^{q} {n+1\choose q+1}
= - \sum_{q=1}^{n+1} (-1)^{q} {n+1\choose q}
= - ((1-1)^{n+1}-1) = 1.$$
|
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|
Simplest or nicest proof that $1+x \le e^x$ The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is accessible to students with limited calculus experience, and does not involve too much analysis of different cases.
|
The shortest proof I could think of:
$$1 + x \leq 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = e^x.$$
However, it is not completely obvious for negative $x$.
Using derivatives:
Take $f(x) = e^x - 1 - x$. Then $f'(x) = e^x - 1$ with $f'(x) = 0$ if and only if $x = 0$. But this is a minimum (global in this case) since $f''(0) = 1 > 0$ (the second derivative test). So $f(x) \geq 0$ for all real $x$, and the result follows.
Another fairly simple proof (but it uses Newton's generalization of the Binomial Theorem which is often covered in precalculus):
We proceed by contradiction. Suppose the inequality does not hold, i.e., $e^x < 1 + x$ for some $x$. Then $e^{kx} < (1 + x)^k$. Now set $x = 1/k$ so that
\begin{align*}
e &< \left( 1 + \frac{1}{k} \right)^k\\
&= 1 + \frac{k}{1}\left( \frac{1}{k} \right)^1 + \frac{k(k - 1)}{1 \cdot 2}\left( \frac{1}{k} \right)^2 + \frac{k(k - 1)(k - 2)}{1 \cdot 2 \cdot 3}\left( \frac{1}{k} \right)^3 + \cdots\\
&< 1 + \frac{k}{1}\left( \frac{1}{k} \right)^1 + \frac{k^2}{1 \cdot 2}\left( \frac{1}{k} \right)^2 + \frac{k^3}{1 \cdot 2 \cdot 3}\left( \frac{1}{k} \right)^3 + \cdots\\
&= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots\\
&= e,
\end{align*}
which is absurd. Therefore $1 + x \leq e^x$ for all real $x$.
By the way, this is where
$$e = \lim_{k \to \infty}\left( 1 + \frac{1}{k} \right)^k$$
comes from because
$$\lim_{k \to \infty}\frac{k(k - 1)}{k^2} = \lim_{k \to \infty}\frac{k(k - 1)(k - 2)}{k^3} = \cdots = \lim_{k \to \infty}\frac{k(k - 1)(k - 2) \cdots (k - n)}{k^{n + 1}} = \cdots = 1.$$
|
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|
Integrate $\int x \sqrt{2 - \sqrt{1-x^2}}dx $ it seems that integration by parts with some relation to substitution...
$$
\int x \sqrt{2-\sqrt{1-x^2}}
= \frac{2}{5} \sqrt{2-\sqrt{1-x^2}} \cdot \sqrt{1-x^2}+\frac{8}{15}\sqrt{2-\sqrt{1-x^2}}+c
$$
How can I get that?
|
Let $x=\sqrt{1-y^2}$, then then integral becomes
$$-\int dy \, y \, \sqrt{2-y}$$
Integrate by parts:
$$\frac{2}{3}y (2-y)^{3/2} + \frac{2}{3} \int dy \, (2-y)^{3/2}$$
which is
$$\frac{2}{3}y (2-y)^{3/2} - \frac{4}{15} (2-y)^{5/2} + C$$
Therfore
$$\int dx \, x \, \sqrt{2-\sqrt{1-x^2}} = \frac{2}{3} \sqrt{1-x^2} \left ( 2-\sqrt{1-x^2}\right)^{3/2} - \frac{4}{15}\left ( 2-\sqrt{1-x^2}\right)^{5/2}+C $$
|
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|
A definite integral $\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx$ I need to find a value of this definite integral:
$$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx.$$
Its numeric value is approximately $0.7875720991394284$, and lookups in Inverse Symbolic Calculator Plus and WolframAlpha did not return a plausible closed-form candidate.
Do you have any ideas how I can approach this problem?
|
There is an alternate way to compute this integral w/o a summation over $n$ in the middle steps.
Notice
$$\frac{2-\cos z}{5 - 4\cos z} = \frac12 \left[\frac{(2-e^{iz})+(2-e^{-iz})}{(2-e^{iz})(2-e^{-iz})}\right]
= \frac12\left[\frac{1}{2-e^{iz}} + \frac{1}{2-e^{-iz}}\right]
$$
and $\displaystyle\;\frac{1}{1+z^4}$ is an even function, we have
$$\mathcal{I} \stackrel{def}{=} \int_0^\infty \frac{2-\cos x}{(1+x^4)(5-4\cos x)} dx =
\frac12\int_{-\infty}^\infty \frac{1}{(1+z^4)(2 - e^{iz})}dz$$
Since $\displaystyle\;\frac{1}{2-e^{iz}}$ is entire on the upper half plane $\Im z \ge 0$ and $\displaystyle\;\left|\frac{1}{2-e^{iz}}\right| \le \frac{1}{2-1} = 1$ there, we can complete the contour in the upper half-plane and
$$\mathcal{I} = \lim_{R\to\infty}\frac12 \oint_{C_R} \frac{1}{(1+z^4)(2 - e^{iz})}dz
\quad\text{ where }\quad
C_R = [-R, R ] \cup \big\{\; Re^{i\theta} : \theta \in [0,\pi]\big\}
$$
$\displaystyle\;\frac{1}{1+z^4}$ has $4$ poles $\omega_k = e^{\frac{(2k+1)i}{4\pi}}, k = 0..3$ over $\mathbb{C}$. Two of them $\omega_0 = \frac{1+i}{\sqrt{2}}$ and $\omega_1 = \frac{-1+i}{\sqrt{2}}$ belongs to the upper half-plane. Since
$$\frac{1}{1+z^4} = \sum_{k=0}^3 \frac{1}{(z-\omega_k) 4\omega_k^3}
= -\frac14 \sum_{k=0}^3\frac{\omega_k}{z-\omega_k}$$
The residue of the integrand at $\omega_k$ are $\displaystyle\;-\frac14 \frac{\omega_k}{2 - e^{i\omega_k}}$ for $k = 0, 1$. This leads to
$$\begin{align}
\mathcal{I}
&= \frac12\left[ -\frac{2\pi i}{4}\left(\frac{\omega_0}{2 - e^{i\omega_0}} + \frac{\omega_1}{2 - e^{i\omega_1}} \right)\right]\\
&= \frac{-\pi i}{4\sqrt{2}}\left(
\frac{1+i}{2 - e^{-1/\sqrt{2}} e^{i/\sqrt{2}}} +
\frac{-1+i}{2 - e^{-1/\sqrt{2}} e^{-i/\sqrt{2}}}\right)\\
&= \frac{-\pi i}{4\sqrt{2}}
\left[
\frac{
(1+i)(2 - e^{-1/\sqrt{2}} e^{-i/\sqrt{2}})
+ (-1+i)(2 -e^{-1/\sqrt{2}} e^{ i/\sqrt{2}})
}{
4 - 4 e^{-1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + e^{-\sqrt{2}}
}
\right]\\
&=
\frac{\pi}{2\sqrt{2}}
\left[
\frac{2 - e^{-1/\sqrt{2}}\left(
\cos\left(\frac{1}{\sqrt{2}}\right)
- \sin\left(\frac{1}{\sqrt{2}}\right)
\right)
}{
4 - 4 e^{-1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + e^{-\sqrt{2}}
}
\right]\\
&=
\frac{\pi e^{1/\sqrt{2}} }{2\sqrt{2}}
\left[
\frac{2 e^{1/\sqrt{2}} - \left(
\cos\left(\frac{1}{\sqrt{2}}\right)
- \sin\left(\frac{1}{\sqrt{2}}\right)
\right)
}{
4 e^{\sqrt{2}} - 4 e^{1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + 1
}
\right]
\end{align}
$$
Reproducing what Vladimir get in his answer.
|
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|
How prove this $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1 $ Show that
$$\sum_{k=0}^{n}\dfrac{\binom{2n-k}{n}}{2^{2n-k}}=1$$
I think this problem can be solved with nice methods, such as algebraic ones. Or can I use probability methods? Thank you
|
Suppose we seek to show that
$$\sum_{k=0}^{n} {2n-k\choose n} 2^k = 2^{2n}.$$
Introduce
$${2n-k\choose n}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-k}}{z^{n+1}} \; dz.$$
This gives for the sum that
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}}
\sum_{k=0}^n \frac{2^k}{(1+z)^k}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}}
\frac{1-2^{n+1}/(1+z)^{n+1}}{1-2/(1+z)} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}}
\frac{1+z-2^{n+1}/(1+z)^{n}}{1+z-2} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}}
\frac{1+z-2^{n+1}/(1+z)^{n}}{z-1} \; dz.$$
The difference here yields two pieces, the first is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}}
\frac{2^{n+1}/(1+z)^{n}}{1-z} \; dz
\\ = \frac{2^{n+1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}}
\frac{1}{1-z} \; dz.$$
This yields
$$2^{n+1} \sum_{q=0}^n {n\choose q} = 2^{2n+1}.$$
The second piece is
$$- \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}}
\frac{1+z}{1-z} \; dz
\\ = - \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{n+1}}
\frac{1}{1-z} \; dz.$$
This is
$$-\sum_{q=0}^{n} {2n+1\choose q}
= -\frac{1}{2} 2^{2n+1} = - 2^{2n}.$$
Combining the two pieces we have
$$2^{2n+1} - 2^{2n} = 2^{2n}$$
as claimed.
|
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|
Form of a harmonic function Given that $\phi(x^2+y^2)$ is harmonic, where $\phi: (0, \infty)\to \mathbb{R}$, find the form of $\phi$.
I do not know what they mean by form nor could I find anything online... My book says that for the complex function $f(z)= (y^3-3x^2y)+i(-3xy^2+x^3+C)$ the form $f(z) = i(z^3+c)$ is easily verified and is suggested by taking $y=0$.
Thanks for any help, I will keep searching.
EDIT From the help I got below:
Taking $\psi(x,y) = \phi(x^2+y^2)$ then
\begin{align*}
\psi_x =& \phi_x(x^2+y^2)2x \\
\psi_xx =& 2\phi_x(x^2+y^2)+4x^2\phi_{xx}(x^2+y^2) \\
\psi_y =& \phi_y(x^2+y^2)2y \\
\psi_yy =& 2\phi_y(x^2+y^2) + 4y^2\phi_{yy}(x^2+y^2)
\end{align*}
I can add $\psi_{xx} + \psi_{yy} = 0$ but where does that get me? Where should I go from here?
EDIT2: Using Polar coordinates we know $x^2+y^2 = r^2$ so we have:
\begin{align*}
\phi_r =& \phi_r(r^2)2r \\
\phi_{rr} =& 2\phi_r(r^2) + 4r^2\phi_{rr}(r^2) \\
\phi_\theta =& 0
\end{align*}
|
Hint: Note that $\phi$ is a function of one variable, hence $\psi_{xx} + \psi_{yy} = 0$ implies
$$4\phi'(x^2+y^2)+4(x^2+y^2)\phi''(x^2+y^2) = 0,$$
i.e.
$$\phi'(t)+t\phi''(t) = 0.$$
|
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|
Approximate $\int_0^{\pi /2} \frac{ds}{\sqrt{1-x\sin^2s}}$ I am trying to approximate the following integral
$$K(x)=\int\limits_0^{\pi /2} \frac{ds}{\sqrt{1-x\sin^2s}}$$
with $0<x<1$.
I need to show that for x close to one that $K(x)\sim -\frac{1}{2}\ln(1-x)$.
My first attempt was to Taylor expand the function $f(x)= (1-x\sin^2s)^{-1/2}$ about $x=1$. I did this using $f(x)\approx f(1)+f'(1)(x-1)$. I integrated the result and found that $K(x)\sim \left[\frac{5-x}{4}\ln|\sec s + \tan s|\right]_0^{\pi/2} - \left[\frac{1-x}{4}\sec s \tan s \right]_0^{\pi/2}$, which goes to infinity when $s=\pi/2$.
I can't think of another way to approach this, so any advice would be helpful.
|
The integral as written is the
complete elliptic integral of the first kind and it can be expressed in term of hypergeometric function:
$$K(x) = \frac{\pi}{2}\,_2F_1(\frac12,\frac12;1;x)\tag{*1}$$
We know when $\gamma = \alpha + \beta$, $\alpha$ and $\beta \ne 0, -1, -2, \ldots$, $|\arg(1-x)| < \pi$, $|1-x| < 1$, the hypergeometric function has following
expansion near $x = 1$:
$$\begin{align}
\,_2F_1(\alpha,\beta;\gamma;x) =
& \frac{-\Gamma(\gamma)}{\Gamma(\alpha)\Gamma(\beta)}
\sum_{n=0}^{\infty}\frac{(\alpha)_n(\beta)_n}{n!^2}(1-x)^n \times\\
& \left\{
\psi(\alpha+n)+\psi(\beta+n) -2\psi(1+n) + \log(1-x)
\right\}
\end{align}$$
where $(t)_n$ is the rising Pochhammer symbol and $\psi(t)$ is the
digamma function.
Substitute this back into $(*1)$ for $x$ near $1$, we have up to $O((1-x)\log(1-x))$,
$$K(x) \sim -\frac12 \left\{ 2\psi(\frac{1}{2}) - 2\psi(1) + \log(1-x)\right\}
= \log 4 - \frac12\log(1-x)$$
Actually, we can derive this leading term by elementary means.
Let $\delta = \sqrt{\frac{1-x}{x}}$, $u = \cos s = \delta \sinh\theta$, we have
$$\begin{align}
K(x)
= & \frac{1}{\sqrt{x}}\int_0^{\frac{\pi}{2}} \frac{ds}{\sqrt{\delta^2 + \cos^2\!s}}\\
= & \frac{1}{\sqrt{x}}\int_0^1 \frac{du}{\sqrt{\delta^2 + u^2}\sqrt{1-u^2}}\\
= & \frac{1}{\sqrt{x}}\left\{
\int_0^1 \frac{du}{\sqrt{\delta^2 + u^2}} +
\int_0^1 \frac{1-\sqrt{1-u^2}}{\sqrt{\delta^2+u^2}} \frac{du}{\sqrt{1 - u^2}}
\right\}\\
= & \frac{1}{\sqrt{x}}\left\{
\int_0^1 \frac{du}{\sqrt{\delta^2 + u^2}} +
\int_0^{\frac{\pi}{2}}\frac{1-\sin s}{\sqrt{\delta^2 + \cos^2\!s}} ds
\right\}\\
= & \frac{1}{\sqrt{x}}\left\{
\int_0^{\sinh^{-1}\frac{1}{\delta}} d\theta +
\int_0^{\frac{\pi}{2}}\frac{1-\sin s}{\sqrt{\delta^2 + \cos^2\!s}} ds
\right\}\\
\end{align}$$
The $2^{nd}$ integral in the RHS has a finite limit as $\delta \to 0$.
If we replace it by its limit, we will introduce an error that won't affect
our determination of the leading term. Up to $o(1)$$\color{blue}{^{[1]}}$, we find:
$$\begin{align}
K(x)
\sim & \frac{1}{\sqrt{x}}\left\{\sinh^{-1}\frac{1}{\delta} + \int_0^{\frac{\pi}{2}}\frac{1-\sin s}{\cos s} ds
\right\}\\
= & \frac{1}{\sqrt{x}}\left\{ \log\left(\frac{1+\sqrt{1+\delta^2}}{\delta}\right) + \log 2\right\}\\
= & \frac{1}{\sqrt{x}}\left\{ \log\left(\frac{\sqrt{x}+1}{\sqrt{1-x}}\right) + \log 2\right\}\\
\sim & (1 + O(1-x))
\left\{\log\left(\frac{1 + O(1-x) + 1}{\sqrt{1-x}}\right) + \log 2\right\}\\
\sim & \log 4 - \frac12\log(1-x) + O((1-x)\log(1-x))
\end{align}$$
recovering the leading term as expected.
Notes
*
*$\color{blue}{[1]}$ A more careful analysis shows the error introduced in this step is of the order $O(\delta^2\log\delta) = O((1-x)\log(1-x))$.
|
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|
How do I decompose this partial fraction case? Decompose $$\dfrac{2x}{1+x} $$
Looking at this case, it looks like any simple partial fraction. But it is trickly. This is how I attempted:
$$\dfrac{2x}{1+x} = \dfrac{A}{1+x}$$
multiply by LCD $(1+x)$ to get $2x = A$
How to I reduce this to give me the value of $A$?
What if I do like this: $2x^1=Ax^0$ and conclude that $A = 2$?
by using long division: I am getting:
$2+\dfrac{2}{x+1}$
$$================$$
@Ron Gordon et al: Ok I get what you mean.
Finally, how then do I use the equation $1+\dfrac{1}{1+x}$ to come up with the partial fractions?
In my own thoughts I decided to eliminating the fraction in $1+\dfrac{1}{1+x}$
by multiply with LCD. I get:
$ \dfrac{1}{1}+\dfrac{1}{x+1}$ which gives $\dfrac{(x+1)+1}{(x+1)}$
$ \therefore$ our new equation to decompose is $$\dfrac{x+2}{(x+1)}$$
$\dfrac{(x+2)}{(x+1)} = $ .....is this equation now correct?
if so I proceed as below:
$\dfrac{x+2}{x+1} =\dfrac{A}{x+1}$
multiply both sides by LCD we get
$x+2=A$
to eliminate $x$ and it coefficient, let $x=0$
$$\therefore A = 2$$
$\therefore$ my solution is $ \dfrac{x+2}{x+1}=\dfrac{2}{x+1}$
Either am totally confused or the instructions here are not helping me understand this important concept?
|
$$\frac{x}{1+x} = \frac{1+x}{1+x} - \frac{1}{1+x} = 1-\frac{1}{1+x}$$
|
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|
Extended euclidean algorithm So I am trying to figure this out.
And for one of the problem the question is x*41= 1 (mod 99)
And the answer lists
x | 41x mod 99
0 99
1 41
-2 17
5 7
-12 3
29 1
And conclude x=29
How did they get this value(can someone explain)? To better put, how do you calculate this?
|
A better table would probably be the following
$$\begin{matrix}
99 & 41\\
\hline
1 & 0 & | & 99\\
0 & 1 & | & 41\\
1 & -2 & | & 17\\
-2 & 5 & | & 7\\
5 & -12 & | & 3\\
-12 & 29 & | & 1
\end{matrix}$$
where each line
$$\begin{matrix}
a & b & | & r\\
\end{matrix}$$
means
$$
99 \cdot a + 41 \cdot b = r,
$$
and you go from two consecutive lines
$$\begin{matrix}
a_{1} & b_{1} & | & r_{1}\\
a_{2} & b_{2} & | & r_{2}\\
\end{matrix}$$
to the next by doing Euclidean division of $r_{1}$ by $r_{2}$,
$r_{1} = r_{2} q + r$, with $0 \le r < r_{2}$, and then taking as the next line
$$\begin{matrix}
a_{1} - a_{2} q & b_{1} - b_{2} q & | & r,\\
\end{matrix}$$
which satisfies indeed
$$
99 \cdot (a_{1} - a_{2} q) + 41 \cdot (b_{1} - b_{2} q)
=
99 a_{1} + 41 \ b_{1}
-
(99 \cdot a_{2} + 41 b_{2}) q
=
r_{1} - r_{2} q = r.
$$
So the last column yields the remainders of the Euclidean algorithm. In your table, the first column is omitted, since the required inverse is the last element in the second column. I have left it in, because the full scheme provides the extra Bézout information
$$
99 \cdot (-12) + 41 \cdot 29 = 1,
$$
from which you get
$$
41 \cdot 29 \equiv 1 \pmod{99}.
$$
|
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|
Prove by induction that $1^3 + \dots + n^3 = (1 + \dots + n)^2$ I'm suppose to prove by induction:
$1^3 + \dots + n^3 = (1 + \dots + n)^2$
This is my attempt; I'm stuck on the problem of factoring dots.
|
Hint:
$$
(n+1)^3 =(n+1)(n+1+n(n+1))=(n+1)(n+1+2(1+2+\dots+n))\\
$$
Then
$$
(1 + \dots + n)^2+(n+1)^3=(n+1)^2+2(n+1)(1+2+\dots+n)+(1+2+\dots+n)^2\\
=(1+2+\dots+n+(n+1))^2
$$
|
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|
Solve for $x: 2/\sqrt{2-x^2} = 4- 2x^2?$ How do you expand and solve for $x$?
It is $2 = 4\sqrt{2-x^2} - 2x^2\sqrt{2-x^2}$
Thank you!
How would I solve for x?
|
\begin{align*}
\frac{2}{\sqrt{2-x^2}} &= 4 - 2x^2\\
2 &= 2(2-x^2)^{3/2}\\
1 &= (2-x^2)^{3/2}\\
1 &= 2-x^2\\
1 &= x^2\\
\pm1 &= x
\end{align*}
|
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|
How is it, that $\sqrt{x^2}$ is not $ x$, but $|x|$? As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.
But then I remember that $\sqrt{x}$ is the same as $x^{1/2}$ and thus, $\sqrt{x^2}$ is the same as $x^{1}$.
So, as far as I get it, $\sqrt{x^2} = x^{2/2}$.
But then, I can cancel $\frac{2}{2}$ to $1$. So: $\sqrt{x^2} = x^1 = x$.
Where does the absolute value in this derivation of the calculation come from?
I get it why its there: Because $(-x)^2 = x^2$, so the radicand is always positive. But from $\sqrt{x^n} = x^{n/2}$ I dont understand how it gets there.
|
It is true that $\sqrt{x} = x^{\frac{1}{2}}$ and $(\sqrt{x})^2 = (x^{\frac{1}{2}})^2 = x$. This is true for all $x$ in the domain of $\sqrt{x}$, namely $x \in [0, \infty)$.
The absolute value comes from composing the square root and square in the opposite order; that is $\sqrt{x^2} = (x^2)^{\frac{1}{2}}$. For any $x \in \mathbb{R}$, $x^2 \geq 0$ so $x^2$ is in the domain of the square root function. The square root function however returns only non-negative values, so if $x < 0$, $(x^2)^{\frac{1}{2}} \neq x$ (but if $x \geq 0$, $(x^2)^{\frac{1}{2}} = x$). In fact, as you have noted, if $x < 0$ write $x = -a$ for some $a \in (0, \infty)$, then $x^2 = (-a)^2 = a^2$, so $(x^2)^{\frac{1}{2}} = (a^2)^{\frac{1}{2}}$. Now $a > 0$ so we obtain $(a^2)^{\frac{1}{2}} = a$, but $a$ is just the absolute value of $x$, i.e. $a = |x|$.
|
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|
Given that $gcd(a,b)=1$, prove that $gcd(a+b,a^2-ab+b^2)=1$ or $3$, also when will it equal $1$? It is an exercise on the lecture that i am unable to prove.
Given that $gcd(a,b)=1$, prove that $gcd(a+b,a^2-ab+b^2)=1$ or $3$, also when will it equal $1$?
|
Let $d|a+b \quad(1)$ and $d|a^2-ab+b^2 (2)$
$ d|a^2-ab+b^2 \quad (2)\implies d|(a+b)^2 - 3ab \qquad \qquad(3)$
And as $d|a+b$ then, $(3) \land (1) \implies d|-3ab \implies d|3ab \implies d|3 \vee d|a \vee d|b$
On the other hand, as $d|a+b$, if $d|a \implies d|b$, and vice verse. And as $gcd(a,b)=1$, then $d=1$.
The other possibility was $d|3$, and it's clear that $d|3 \Longleftrightarrow (d=1 \vee d=3)$
|
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|
How to have this definite integral?
Suppose that a function $f$ of $x$ and $y$ be defined as follows:$$f(x,y) =
\begin{cases}
\frac{21}{4}x^2y & \text{for $x^2 \leq y\leq 1$,} \\
0 & \text{otherwise. } \\
\end{cases}$$
I have to determine the value of integral for which $y\leq x$ also holds.
The answer is $\frac{3}{20}$ and I also get it using figure for $x$ and $y$, but don't know how to get it with calculations.
|
The given condition $y\le x$and $x^2\le y\le1$ implies $x^2\le y\le x$.
Since $x^2\le x$, $0\le x\le 1$.
$$\int_{0}^1\int_{x^2}^x\frac{21}{4}x^2ydydx=\int_{0}^1\left[\frac{21}{8}x^2y^2\right]^x_{x^2}dx=\int_{0}^1\frac{21}{8}x^2(x^2-x^4)=\frac{21}{8}\left[\frac{x^5}{5}-\frac{x^7}{7}\right]^1_{0}=\frac{3}{20}$$
|
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|
Evaluating $\int_{-2}^{2} 4-x^2 dx$ with a Riemann sum I'm having problems with a Riemann sum ... I need to find the integral:$$\int_{-2}^2 (4-x^2)\;dx$$Clearly we have $$\int_{-2}^{2}(4-x^2)\;dx=4x-\frac{x^3}{3}\mid_{-2}^{2}=(4\cdot2-\frac{2^3}{3})-(4\cdot(-2)-\frac{(-2)^3}{3})=\frac{32}{3}$$OK.
On the other hand, we have
$$\Delta x=\frac{b-a}{n}=\frac{4}{n}$$
and
$$\xi_1=-2+\frac{4}{n};\;\;\xi_2=-2+2\frac{4}{n};\;\;\ldots\;\;;\xi_n=-2+n\frac{4}{n}$$
then
$$\xi_i=-2+\frac{4i}{n}=\frac{4i-2n}{n},$$
so that
$$ \begin{align} \int_{-2}^2 {4-x^2}\;dx=&\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{4i-2n}{n}\right)^2\right)\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{4i-2n}{n}\right)^2\right)\cdot\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{16i^2-16ni+4n^2}{n^2}\right)\right)\cdot\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n \left(\frac{4n^2-16i^2+16ni-4n^2}{n^2} \right)\cdot\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n \frac{-16i^2+16ni}{n^2}\cdot\frac{4}{n} \\
=&\lim_{n\to+\infty}\sum_{i=1}^{n} \frac{-64i^2+64ni}{n^3}\\
=&\lim_{n\to+\infty} \frac{64}{n^3}\left(-\sum_{i=1}^{n} i^2+n\sum_{i=1}^{n} i\right) \\
=&\lim_{n\to+\infty} \frac{64}{n^3} \left(-\frac{n(n+1)(2n+1)}{6}+n\cdot \frac{n(n+1)}{2} \right)\\
=&\lim_{n\to+\infty} \frac{64}{n^3} \left(-\frac{\color{#ff0000}{2}n^3+3n^2+n}{6}+\frac{n^3+n^2}{2} \right) \\
=&\lim_{n\to+\infty} \frac{64}{n^3} \left(\frac{-\color{#ff0000}{2}n^3-3n^2-n+3n^3+3n^2}{6}\right)\\
=&\lim_{n\to+\infty}\frac{32}{3} \left(\frac{\color{#ff0000}{1\times}n^3-n}{n^3}\right) \\
=&\lim_{n\to+\infty}\frac{64}{\color{#ff0000}{2\times}3}-\frac{32}{3n^2}=\frac{64}{\color{#ff0000}{2\times}3} \color{#ff0000}{=}\frac{32}{3}=\int_{-2}^{2} {4-x^2}\;dx
\end{align}$$
Where is the mistake?
|
Check the step where you expand $n \left( n+1 \right) \left( 2\,n+1 \right)$. I believe the mistake is in that step.
|
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|
Prove that $x^3 \equiv x \bmod 6$ for all integers $x$
Prove that $x^3 \equiv x \bmod 6$ for all integers $x$
I think I got it, but is this proof correct?
We can write any integer x in the form: $x = 6k, x = 6k + 1, x = 6k + 2, x = 6k + 3, x = 6k + 4$, and $x = 6k + 5$.
If $x = 6k$, then $x^3 = 216k^3$. Then $x^3 - x = 216k^3 - 6k = 6(36k^3 - k)$. Thus, $6 | (x^3 - x)$. Thus, $x^3 \equiv x \pmod 6$.
If $x = 6k + 1$. Then $x^3 - x = (216k^3 + 108k^2 + 18k + 1) - (6k + 1) = 6(36k^3 + 18k^2 + 2k)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$.
If $x = 6k + 2$, then $x^3 - x = (216k^3 + 216k^2 + 72k + 8) - (6k + 2) = 6(36k^3 + 36k^2 + 11k + 1)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$
If $x = 6k + 3$, then $x^3 - x = (216k^3 + 324k^2 + 162k + 27) - (6k + 3) = 6(36k^3 + 54k^2 + 26 + 4)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6 $
If $x = 6k + 4$, then $x^3 - x = (216k^3 + 432k^2 + 288k + 64) - (6k + 4) = 6(36k^3 + 72k + 47k + 10)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$.
If $x = 6k + 5$, then $x^3 - x = (216k^3 + 540k^2 + 450k + 125) - (6k + 5) = 6(36k^3 + 90k + 74k + 20)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$.
In all cases, we have shown that $x^3 \equiv x \pmod 6$. QED.
|
One has
$$x^3-x=x(x+1)(x-1)\qquad\forall x\ .$$
When $x$ is an integer then at least one factor on the right is even, and exactly one factor on the right is divisible by $3$. It follows that for any $x\in{\mathbb Z}$ the right hand side is divisible by $6$, and so is the left hand side. That is to say: $x^3=x$ mod $6$ for all integers $x$.
|
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|
Prove the inequality $xyz^3 \leq 27(\frac{x+y+z}{5})^5$. For $x,y,z>0$, prove the inequality $xyz^3 \leq 27(\frac{x+y+z}{5})^5$.
Any ideas? I am stuck.
|
Write $f(x,y,z) = 27 \left( \frac{x+y+z}{5} \right)^5 - x y z^3$. Then
$$\frac{\partial f}{\partial x}(x,y,z) = 27 \left(\frac{x+y+z}{5}\right)^4 - y z^3$$
$$\frac{\partial f}{\partial y}(x,y,z) = 27 \left(\frac{x+y+z}{5}\right)^4 - x z^3$$
$$\frac{\partial f}{\partial z}(x,y,z) = 27 \left(\frac{x+y+z}{5}\right)^4 - 3 x y z^2$$
At a minimum these all should equal $0$, so $yz^3 = xz^3 = 3x y z^2$, which gives us $x=y$ and $z=3x$. Plugging these in the inequality shows that the equality is attained at these points. One also has to check the boundary, i.e. when one or two or three of the variables is $0$, but this is trivial.
|
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|
How to solve the inequality $\frac {5x+1}{4x-1}\geq1$ Please help me solve the following inequality.
\begin{eqnarray}
\\\frac {5x+1}{4x-1}\geq1\\
\end{eqnarray}
I have tried the following method but it is wrong. Why?
\begin{eqnarray}
\\\frac {5x+1}{4x-1}&\geq&1\\
\\5x+1&\geq& 4x-1\\
\\x &\geq& -2
\end{eqnarray}
Thank you for your attention
|
$$
\frac{5x+1}{4x-1}\geq 1\Leftrightarrow \frac{5x+1}{4x-1}-1\geq 0\Leftrightarrow\frac{5x+1-(4x-1)}{4x-1}\geq 0 \Leftrightarrow \frac{5x+1-4x+1}{4x-1}\geq 0
$$
$$
\Leftrightarrow \frac{x+2}{4x-1}\geq 0
$$
Now $x+2=0 \Rightarrow x=-2$, and $4x-1=0\Rightarrow x=\frac{1}{4}$
For $x\in (-\infty, -2]$, and $x\in(\frac{1}{4}, +\infty)$, $\frac{x+2}{4x-1}\geq 0$
|
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|
Finding $\int \frac{x^2}{(a^2-x^2)^{\frac{3}{2}}}dx$ using trigonometric substitution. Where did I go wrong? Evaluate the following integral using trigonometric substitution
$$\int \frac{x^2}{(a^2-x^2)^{\frac{3}{2}}}dx$$
I used the substitution $x=a \sin(u)$, then $dx = a \cos(u) du$. The integral then becomes:
$$\int \frac{a^2 \sin^2(u) a \cos(u)}{(a^2-a^2 \cos^2(u))^{\frac{3}{2}}}du = \int \frac{a^3 \sin^2(u) \cos(u)}{(a^2 \sin^2(u))^{\frac{3}{2}}}du =\int \frac{\cos(u)}{\sin(u)}du = \ln | \sin(u) | + C$$
The last equality comes from the substitution $v=\sin(u)$. Now from the first substitution we have $x=a\sin(u)$ and thus $\sin(u) = \frac{x}{a}$. This gives us
$$\int \frac{x^2}{(a^2-x^2)^{\frac{3}{2}}}dx = \ln \left| \frac{x}{a}\right| + C = \ln|x| + C'$$
Where $C' = C - \ln|a|$. This however is of course not correct (unless I am missing something...). Can anyone tell me where I went wrong on this one? Thanks a lot!
|
The bottom should be $a^3\cos^3 u$, so you are essentially integrating $\tan^2 u$, that is, $\sec^2 u-1$.
|
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|
Integral of quartic function in denominator I'm sorry, I've really tried to use MathJaX but I can't get integrals to work properly.
indefinite integral
$$\int {x\over x^4 +x^2 +1}$$
I set it up to equal
$$x\int {x\over x^4 +x^2 +1} - \int {x\over x^4 +x^2 +1}$$
$$\text{so } (x-1)\int {1\over x^4 +x^2 +1}$$
OKAY, now I set the denominator to $(X^2 +.5)^2 + \frac 34$
So I multiplied the top and bottom by $\frac 43$
then I absorbed it into the squared quantity by dividing it (within the parenthesis) by $\sqrt 3\over 2$
so
$${1\over \left({X^2 +.5\over {\sqrt 3\over 2}}\right)^2 + 1}$$
so $\arctan\left({X^2 +.5\over{\sqrt 3\over 2}}\right)$
FINAL ANSWER: $(x-1)\arctan\left({X^2 +.5\over{\sqrt 3\over 2}}\right) + C$
Thanks for reading, I have no way of checking this work... tutors are always have too many people wanting help.
|
HINT:
As $\displaystyle x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)$ and $\displaystyle (x^2+x+1)-(x^2-x+1)=2x$
$$\frac x{x^4+x^2+1}=\frac12\frac{(x^2+x+1)-(x^2-x+1)}{x^4+x+1}=\frac12\left(\frac1{x^2-x+1}-\frac1{x^2+x+1}\right)$$
Now as, $\displaystyle x^2+x+1=\frac{(2x+1)^2+(\sqrt3)^2}4$ put $2x+1=\sqrt3\tan\theta$
and similarly for $\displaystyle\int\frac{dx}{x^2-x+1}$
|
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|
Showing that a system of Diophantine equations will have irrational solutions as well as integers Solve $\begin{cases} 3xy-2y^2=-2\\ 9x^2+4y^2=10 \end{cases}$
Rearranging the 2nd equation to $x^2=\dfrac{10-4y^2}{9} \Longrightarrow 0\leq x^2 \leq 1$ if $x^2=1$ than $y=\pm\dfrac{1}{2}$ and $x=\pm1$ but how do I show there exists two more solutions to this equation by using number theory vs college algebra. Does it have to do with the relationship that $0\leq y^2 \leq 10$ ? So does this say anything about solutions being irrational?
|
If you add five times the first equation plus the second, you get $9 x^2 + 15 x y - 6 y^2 = 0,$ or
$$ 3 (3x - y)(x+ 2 y) = 0. $$
So the choices lie along either the line $$ y = 3x $$ or the line $$ y = \frac{-x}{2}. $$
From the ellipse equation $9 x^2 + 4 y^2 = 10,$ we get either $45 x^2 = 10$ and $9 x^2 = 2$ and $(3x)^2 = 2,$ or $10 x^2 = 10$ and $x^2 = 1.$
|
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|
How prove this $x+y=0$ if $\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$ Question:
let $x,y$ are real numbers,and such
$$\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$$
show that
$$x+y=0\tag{1}$$
before I have solve following problem:
if
$$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1,\Longrightarrow x+y=0\tag{2}$$
solution:we have
$$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\tag{3}$$
$$x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\tag{4}$$
$(3)+(4)$
then $x+y=0$
But for $(1)$ we can't use this methods to solve it.maybe can have other nice methods, Thank you
|
Rather than use $x$ directly, we will use $z=-x$.
Suppose that $y>0$. Let
$$
f_1(z)=z+\sqrt{y^2+z^3} \ (\text{for} \ z\geq{-y^{\frac{2}{3}}}), \ \
f_2(z)=\frac{y^3}{-y+\sqrt{y^3+z^2}}
\ (\text{for} \ z\in{\mathbb R}), \
$$
We must then show that $f_1(z) \neq f_2(z)$ when $z\neq y$.
It is easy to see that $f_1$ is increasing on its domain and
that $f_2$ is increasing on $(-\infty,0)$ and decreasing on
$(0,\infty)$. Let $c=f_1(y)=f_2(y)$.
When $z > y$, we have $f_2(z)<c<f_1(z)$ by the variations
of $f_1$ and $f_2$ so $f_1(z) \neq f_2(z)$.
When $0 < z < y$, we have
$f_1(z)<c<f_2(z)$ by the variations of $f_1$ and $f_2$ so $f_1(z) \neq f_2(z)$.
When $z <0$, for $f_1(z)$ to be defined we need $z\geq \mu$ where
$\mu=-y^{\frac{2}{3}}$. Now $\mu^2=y^{\frac{4}{3}} \leq y^2$, so
$f_2(\mu) \geq f_2(y)=c$. As $f_1(z) < c$ by the variations of $f_1$, we still
have $f_1(z)<c<f_2(z)$ so $f_1(z) \neq f_2(z)$.
The case $y < 0$ is similar.
|
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|
Find the natural numbers $a$ and $b$ so that $a\cdot b$ has the largest possible value but $a + b = x$ must hold. Is there a way to find the natural numbers $a$ and $b$ so that $a\cdot b$ has the largest possible value but $a + b = x$ must hold. It's easy small numbers but is there any way, through calculus or otherwise, to solve the general case?
|
Take $(a-b)^2$. Any square is positive (except $0^2=0$) so you can conclude that
\begin{align}
0&\le(a-b)^2 &&\text{expand square binom:}\\
0&\le a^2-2ab+b^2 &&\text{add $4ab$ at both sides:}\\
4ab&\le a^2+2ab+b^2 &&\text{factorize square binom:}\\
4ab&\le(a+b)^2 &&\text{replace $x$:}\\
4ab&\le x^2 &&\text{divide by $4$:}\\
a\cdot b&\le\frac14x^2
\end{align}
So we have a max boundary for $a\cdot b$. Now we must find that $a\cdot b=\frac14x^2$. If $x$ is odd it does not as $x^2$ is odd and $4$ does not divide $x^2$.
In the formula above, we can replace all $\le$ with $=$ if we make $a-b=0$, this is when $a=b=\frac12x$ which work for $x$ even.
For $x$ odd we cannot have $a=b$ so we will try the closest alternative: $a=b+1$ which means $a=\frac{x+1}2$ and $b=\frac{x-1}2$. We hav (following the formulas)
\begin{align}
a-b &= 1\\
(a-b)^2 &= 1\\
a^2-2ab+b^2 &= 1\\
a^2+2ab+b^2 &= 4ab+1 \\
(a+b)^2 &= 4ab+1 \\
x^2 &= 4ab+1 \\
x^2-1 &= 4ab \\
ab &= \frac14(x^2-1)
\end{align}
Note that $\frac14(x^2-1)$ is the closest integer bellow $x^2$ for $x$ odd.
So the maximum value for $a\cdot b$ is:
$$
a\cdot b=\left\lfloor\frac14x^2\right\rfloor,
\qquad a=\left\lceil\frac12x\right\rceil,
\qquad b=\left\lfloor\frac12x\right\rfloor.
$$
(And no calculus was required)
|
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|
How to prove this inequality $\sin{\left(\frac{\pi}{2}ab\right)}\le\sin{\left(\frac{\pi}{2}a\right )}\sin{\left(\frac{\pi}{2}b\right)}$? Let $$0\le a\le 1,0\le b\le 1$$
Prove or disprove
$$\sin{\left(\dfrac{\pi}{2}ab\right)}\le\sin{\left(\dfrac{\pi}{2}a\right )}\sin{\left(\dfrac{\pi}{2}b\right)}$$
My try:
Since
$$\sin{x}\sin{y}=\dfrac{1}{2}[\cos{(x-y)}-\cos{(x+y)}]$$
So
$$\sin{\left(\dfrac{\pi}{2}a\right)}\sin{\left(\dfrac{\pi}{2}b\right)}=\dfrac{1}{2}[\cos{\dfrac{\pi}{2}(a-b)}-\cos{\dfrac{\pi}{2}(a+b)}]$$
Then I can't,Thank you very much!
|
The statement is true.
Proof: Given $a\in[0,1]$, define
$$f(x)=\frac{\sin( ax)}{\sin x},\quad x\in (0,\frac{\pi}{2}],\quad\text{and}\quad g(x)=a\tan x-\tan(ax),\quad x\in[0,\frac{\pi}{2}).$$
From $g(0)=0$ and
$$g'(x)=a \big((\sec x)^2-(\sec (ax))^2\big)\ge 0,\quad\forall x\in (0,\frac{\pi}{2}),$$
we know that $g(x)\ge 0$ on $(0,\frac{\pi}{2})$. Therefore,
$$f'(x)=\frac{a\sin x\cdot \cos(ax)-\cos x\cdot\sin(ax)}{(\sin x)^2}=\frac{\cos x\cdot\cos(ax)}{(\sin x)^2}\cdot g(x)\ge 0,\quad\forall x\in (0,\frac{\pi}{2}).$$
It follows that $f(x)\le f(\frac{\pi}{2})$ on $(0,\frac{\pi}{2}]$.
Now we can complete the proof. When $b=0$, the inequality is trivial; when $b\in(0,1]$,
$$f(\frac{\pi b}{2})\le f(\frac{\pi}{2}) \Longleftrightarrow \sin\frac{\pi ab}{2}\le \sin\frac{\pi a}{2}\sin\frac{\pi b}{2}.$$
|
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|
Factorize the polynomial $x^3+y^3+z^3-3xyz$ I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Using Mathematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?
|
$$\underbrace{x^3+y^3}+z^3-3xyz = \underbrace{(x+y)^3-3xy(x+y)}+z^3-3xyz$$
$$=\underbrace{(x+y)^3+z^3}-\underbrace{3xy(x+y)-3xyz} $$
$$=\underbrace{\{(x+y)+z\}}\{(x+y)^2-(x+y)z+z^2\}-3xy\underbrace{\{(x+y)+z\}} \left(\text{ using } a^3+b^3=(a+b)(a^2-ab+b^2)\text{ for the first two terms }\right)
=(x+y+z)\{(x+y)^2-(x+y)z+z^2-3xy\}$$
Now $\displaystyle (x+y)^2-(x+y)z+z^2-3xy=x^2+y^2+z^2-xy-yz-zx$
|
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|
Proper method for solving quadratic equations with exponents $(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2}$ + $(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}$ = $2^{(x+4)/4}$
I have found out, by trial and error method, that $x=0$ and $x=4$ satisfy this equation. But is there a proper way to solve this equation and get the solutions?
|
Let's start by putting $$y_1=\left(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4}\right)^{x/4}$$ and $$y_2=\left(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4}\right)^{x/4}.$$ Now, observe that $$\begin{align}2 &= (x^2-5x+6)-(x^2-5x+4)\\ &= \left(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4}\right)\left(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4}\right),\end{align}$$ and so $$2^{(x+4)/4}=2\cdot 2^{x/4}=2y_1y_2.$$ Thus, the given equation becomes equivalent to solving $$y_1^2+y_2^2=2y_1y_2,$$ or $$(y_1-y_2)^2=0.$$ Hence, we need only solve the equation $$y_1=y_2,$$ so since both $y_1,y_2>0$ (why?), then we face the equivalent task of solving $$y_1^2=y_2^2.$$ As you've seen, $x=0$ readily does the trick, but if $x\ne 0,$ then the only way that we can have $a^x=b^x$ for $a,b>0$ is if $a=b.$ Hence, it remains only to solve $$\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4}=\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4},$$ which I leave to you.
|
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|
Product of two complementary error functions (erfc) I believe that (i.e., it would be convenient if, and visually appears that) the product of the two complementary error functions:
$$\operatorname{erfc}\left[\frac{a-x}{b}\right]\operatorname{erfc}\left[\frac{a+x}{b}\right]$$
will have a solution, or can be approximated with a solution, of a Gaussian form (i.e., $c\operatorname{exp}\left[-\frac{x^2}{2d^2}\right]$, where $c$ and $d$ are functions of $a$ and $b$) when $a>0$ and $c>0$, however I cannot find a proof of this. Any help?
|
In this paper, the authors provide the following simple approximation of the error function:
$$\operatorname{erf}(z) = 1- \exp\{-c_1z-c_2z^2\},\; z\ge 0 $$
with
$$ c_1 = 1.09500814703333,\;\; c_2 = 0.75651138383854$$
Set $\frac{a-x}{b} \equiv z_1$, and $\frac{a+x}{b} =\equiv z_2$. Given also that $\operatorname{erfc} = 1- \operatorname{erf}$, and under the implied restrictions so that $\frac{a-x}{b}\ge 0,\;\frac{a+x}{b}\ge 0$, (for example, for $b>0,\; a>0,\; x\le |a|$) we have, using the approximation,
$$\operatorname{erfc}(z_1)\operatorname{erfc}(z_2) = [1-\operatorname{erf}(z_1)][1-\operatorname{erf}(z_1)] $$
$$\approx [1-1+ \exp\{-c_1z_1-c_2z_2^2\}][1-1+ \exp\{-c_1z_2-c_2z_2^2\}]$$
$$=\exp\{-c_1(z_1+z_2)-c_2(z_1^2+z_2^2)\} \\= \exp\left\{-c_1\left(\frac{a-x}{b}+\frac{a+x}{b}\right)-c_2\left[\left(\frac{a-x}{b}\right)^2+\left(\frac{a+x}{b}\right)^2\right]\right\}$$
$$ =\exp\left\{-\frac{2a}{b}c_1-c_2\left[\left(\frac{a-x}{b}+\frac{a+x}{b}\right)^2-2\frac{a-x}{b}\frac{a+x}{b}\right]\right\} $$
$$=\exp\left\{-\frac{2a}{b}c_1-c_2\left[4\left(\frac{a}{b}\right)^2-2\frac{a^2-x^2}{b^2}\right]\right\} $$
$$=\exp\left\{-\frac{2a}{b}c_1-c_2\left[2\left(\frac{a}{b}\right)^2+2\frac{x^2}{b^2}\right]\right\} $$
$$ = \exp\left\{-\frac{2a}{b}c_1-2c_2\left(\frac{a}{b}\right)^2\right\}\cdot\exp\left\{-2c_2\frac{x^2}{b^2}\right\}$$
$$=C\cdot\exp\left\{-\frac{x^2}{2d^2}\right\} $$
with
$$ C\equiv \exp\left\{-\frac{2a}{b}c_1-2c_2\left(\frac{a}{b}\right)^2\right\},\;\; d^2 = \frac {b^2}{4c_2}$$
|
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|
Solving trigonometry equation Please help me understand how to solve this for $0\leq x\leq360 $
I seem to have a problem with equations with powers.
$$3\sin^2 x-3\cos^2x+\cos x-1=0 $$
thinking that I would start by simplifying:
$$3 (\sin^2 x- \cos^2x)+\cos x - 1=0 $$
How I wish the equation in the bracket was in form $\sin^2 x+ \cos^2x$ which is equal to 1.
I also tried to substitute $\sin^2 x=1- \cos^2x$
|
$3\sin^2 x-3\cos^2 x+ \cos x - 1= 0$
$6\cos^2 x-\cos x - 2 = 0$
By the quadratic formula
$cos x = \dfrac{1 \pm \sqrt{1-4(6)(-2)}}{12} =\dfrac{-1\pm 7}{12}$
So $cos x = \frac{-1}{2}$ and $cos x = \frac{2}{3}$
|
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|
What is the minimum value of $a+b+\frac{1}{ab}$ if $a^2 + b^2 = 1$? For the case when $a,b>0,$ I used AM-GM Inequality as follows that:
$\frac{(a+b+\frac{1}{ab})}{3} \geq (ab\frac{1}{ab})^\frac{1}{3}$
This implies that $(a+b+\frac{1}{ab})\geq 3$. Hence, the minimum value of $(a+b+\frac{1}{ab})$ is 3
But the answer is $2+\sqrt{2}$ ... how is it ?
|
Let the minimum value of $a+b+\frac{1}{ab} = k$. Both curves must be tangent to each other, which means they must have the same gradient. Thus $a^2 + b^2 = 1 \implies \frac{db}{da} = -\frac{a}{b}$, and:
$$a+b+\frac{1}{ab} = k \implies 1 + \frac{db}{da} - \frac{1 + db/da}{(ab)^2} = 0$$
$$\implies (ab)^2 \left(1 - \frac{a}{b} \right) - 1 - -\frac{a}{b} = 0$$
$$\implies a^2b^2 - a^3b - 1 + \frac{a}{b} =0$$
$$\implies a^2b^3 - a^3b^2 - b + a = 0$$
$$\implies a^2b^2(b-a) - (b-a) =0$$
$$\implies (ab+1)(ab-1)(b-a) = 0$$
but only $b-a = 0$ intersects with $a^2 + b^2 = 1$. This gives $a = b = \frac{1}{\sqrt2}$, and hence $k = \frac{1}{\sqrt2} + \frac{1}{\sqrt2} + \frac{1}{1/2} = 2 + \sqrt{2}$.
|
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|
How to prove this inequality $\frac{x^y}{y^x}+\frac{y^z}{z^y}+\frac{z^x}{x^z}\ge 3$
let $x,y,z$ be positive numbers, and such $x+y+z=1$
show that
$$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}\ge 3$$
My try:
let
$$a=\ln{\dfrac{x^y}{y^x}},b=\ln{\dfrac{y^z}{z^y}},c=\ln{\dfrac{z^x}{x^z}}$$
so
$$a=y\ln{x}-x\ln{y},b=z\ln{y}-y\ln{z},c=x\ln{z}-z\ln{x}$$
and we note
$$az+bx+yc=(y\ln{x}-x\ln{y})z+(z\ln{y}-y\ln{z})x+(x\ln{z}-z\ln{x})y=0$$
so
$$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}=e^a+e^b+e^c$$
so
$$\Longleftrightarrow e^a+e^b+e^c\ge 3$$
But then I can't prove it.
If this problem is to prove
$$ze^a+xe^b+ye^c\ge 3,$$
I can prove it,because
$$ze^a+xe^b+ye^c\ge=\dfrac{z}{x+y+z}e^a+\dfrac{x}{x+y+z}e^b+\dfrac{y}{x+y+z}e^c$$
so
use Jensen's inequality,we have
$$ze^a+xe^b+ye^c\ge e^{\dfrac{az+bx+yc}{x+y+z}}=3$$
This problem comes from How prove this $\dfrac{x^y}{y^x}\ge (1+\ln{3})x-(1+\ln{3})y+1$?
Thank you very much!
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This is to prove $a^{b-1}b^{1-a}\ge 1$ as needed in Ron Ford's answer above. Let $f(b)=(1-b)(-\log a)-(1-a)(-\log b)$, $b\in [a,1]$. $f(a)=f(1)=0$. $f$ is concave, as $f''(b)=-\frac{1-a}{b^2}<0$. So $f(b)>0,\,\forall b\in(a,1)$ and the result follows.
|
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Denesting Phi, Denesting Cube Roots I have been looking into denesting square roots but I have found that $\sqrt[3]{2+\sqrt{5}}$ equals $(1+\sqrt{5})/2$. The same is true for $\sqrt[3]{2-\sqrt{5}}$ and $(1-\sqrt{5})/2$. I cannot figure how this is true. I proved this by setting both equal to x and forming polynomial equations. $y= x^2-x-1$ and $y=x^6-4x^3-1$. I do not know how to solve a degree 6 polynomial by hand like that. My objective is to denest the original $\sqrt[3]{2+\sqrt{5}}$ without knowledge of the simplified phi.
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Hint : As a general rule, when dealing with nested radicals of the form $\sqrt[n]{A+B\sqrt[m]C}$ , you write $A+B\sqrt[m]C=(a+b\sqrt[m]C)^n$, and then employ Newton's binomial theorem. In our case, we have
$$\left(a+b\sqrt5\ \right)^3=a^3+3a^2(b\sqrt5)+3a(b\sqrt5)^2+(b\sqrt5)^3=\underbrace{(a^3+15ab^2)}_2+\underbrace{(3a^2b+5b^3)}_1\sqrt5$$ $$\iff a^3+15ab^2=2(3a^2b+5b^3)\iff a^3+15ab^2-6a^2b-10b^3=0\quad|:b^3\iff$$ $$\iff\left(\frac ab\right)^3-6\left(\frac ab\right)^2+15\left(\frac ab\right)-10=0\iff x^3-6x^2+15x-10=0\iff x=1$$ $$\iff\frac ab=1\iff a=b\iff a^3+15a^3=2\iff16a^3=2\iff a=\sqrt[3]\frac18=\frac12$$ or $3b^3+5b^3=1\iff b=\sqrt[3]\frac18=\frac12$ . Similarly for when the second term is $-1$.
|
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|
write formula to predict nth term of sequence $1, 1\cdot3, 1\cdot3\cdot5, 1\cdot3\cdot5\cdot(2n-1)$ How can I write a formula for a sequence with the following behavior:
{$1, 1\cdot3, 1\cdot3\cdot5, 1\cdot3\cdot5\cdot7, 1\cdot3\cdot5\cdot7\cdot9$}
1st term is $1$
2nd term is $1 \cdot 3 = 3$
3rd term is $1 \cdot 3 \cdot 5 = 15$
4th term is $1 \cdot 3 \cdot 5 \cdot 7 = 105$
Please show steps on how to arrive at the answer:
Ted
|
The classic trick is to multiply and divide by the even terms:
$$
a_n=1\cdot3\cdot\dots\cdot(2n-1)=\frac{1\cdot2\cdot3\cdot \dots\cdot 2n}{2\cdot 4\cdot (2n)}
$$
The numerator is $(2n)!$. As for the denominator, we can factor each of the $n$ terms by $2$, so the denominator is $2^nn!$.
Finally, $a_n=\frac{(2n)!}{2^nn!}$
|
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|
Non-Homogeneous System [Problem] "Find a general solution of the system and use that solution to find a general solution of the associated homogeneous system and a particular solution of the given system."
$\begin{bmatrix}3 & 4 & 1 & 2 \\ 6 & 8 & 2 & 5\\9 & 12 & 3 & 10 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix} = \begin{bmatrix}3 \\ 7 \\ 13\end{bmatrix}$
So we solve it like its a homogenous system. I end up getting to $\begin{bmatrix} 3 & 4& 1 & 2\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{bmatrix}$.So would $x_2$ and $x_3$ be free variables? I can set them to $x_2 = \alpha$ and $x_3 = t$, $x_4 = 0$ then we set up:
$\begin{cases}3x_1 + 4x_2 + x_3 + 2x_4 = 0\\0+0+0+x_4 = 0 \\0+0+0+0 = 0\end{cases}$
So, $\underline{v} = \begin{pmatrix}3 \\ 7 \\ 13\end{pmatrix} + \begin{pmatrix}\frac{1}{3}(-4\alpha -t - 2_4) \end{pmatrix}$
I don't think this is right. What would $x_4$ be? I'm confused
|
For the augmented RREF matrix, I get:
$$\begin{bmatrix}1 & \dfrac{4}{3} & \dfrac{1}{3} & 0 & \dfrac{1}{3} \\ 0 & 0 & 0 & 1 & 1\\0 & 0 & 0 & 0 & 0\end{bmatrix}$$
This means we can write:
$$x_4 = 1$$
$$x_1 = \dfrac{1}{3}-\dfrac{4}{3}x_2 - \dfrac{1}{3}x_3$$
So, we have two free variables, $x_2$ and $x_3$.
Can you now write this in a general form?
|
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How do you find the smallest possible value of aan equation with two unknowns? I'm solving a list of problems where I'm given an equation and I find the smallest possible value by comparing the equation to a quadratic equation and completing the square, however the next one involves another unknown $y$:
$$ x^2 - 3x + 2y^2 + 4y + 2. $$
I've been thinking about maybe somehow making $c$ equal $2y^2 + 4y + 2$?
The answer from the answerbook is: $$ x^2 - 3x + 2y^2 + 4y + 2 = \left( x - \dfrac {3}{2} \right)^2 + 2 \left( y + 1 \right)^2 - \dfrac {9}{4}. $$
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copper.hat is right that the easiest way to solve this is to view it as composed from two functions and take calculate the derivate of each. Then set $f'(x)=0$ and solve for $x$ for both functions.
But this problem is from Spivak's Calculus (1.717(b), p. 17), before Spivak has introduced differentiation. I think Spivak's intention here is to pump intuitions on completing the square, leading the student towards deriving the quadratic equation from the exercises.
With that in mind the way to solve problems of this sort is to rewrite the formula such that it is the sum of two quadratics, one for the $x$ terms and one for the $y$ terms:
\begin{equation*}
x^2 - 3x + 2y^2 + 4y + 2 = x^2 - 3x + 1 + 2y^2 + 4y + 1
\end{equation*}
This gives us two separate quadratics:
\begin{equation*}
f(x) = x^2 - 3x + 1
\end{equation*}
\begin{equation*}
g(y) = 2y^2 + 4y + 1
\end{equation*}
Now we just have to find the smallest possible value of each quadratic separately, then add them together, which we can do by "completing the square":
\begin{equation*}
f(x) = x^2 - 3x + 1 = (x - \frac{3}{2})^2 - \frac{5}{4}
\end{equation*}
So the smallest possible value of $f(x)$ is the value when $(x
-\frac{3}{2})^2 = 0$, so $x = \frac{3}{2}$ and the minimum of $f(x) = -\frac{5}{4}$. And,
\begin{align}
g(y) &= 2y^2 + 4y + 1\\
&= 2(y^2 + 2y + 1/2)\\
&= 2((y+1)^2 - 1/2)\\
&= 2(y+1)^2 - 1
\end{align}
Here the smallest possible value of $g(y)$ is when $y + 1 = 0$ or $g(-1) = -1$.
Adding them together we get
\begin{equation*}
f(\frac{3}{2}) + g(-1) = -\frac{5}{4} - 1 = -\frac{9}{4}
\end{equation*}
(You'll note that the values for $x$ and $y$ here are roots for the squares within the quadratics)
|
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Integral $\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$$
|
Define the function $I(s)$ for $s > 0$ by
$$ I(s) = \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - e^{s}\sqrt{x+1} )}{x+1} \, dx. $$
By observing that $I(\infty) = 0$, we have
$$ I(s) = -\int_{s}^{\infty} I'(t) \, dt. $$
Applying Leibniz's integral rule,
$$ I'(s) = \int_{0}^{\infty} \frac{e^{s}\sqrt{x+1}}{1 + (\sqrt{x} - e^{s}\sqrt{x+1} )^{2}} \, \frac{dx}{x+1}. $$
Now with the substitution $x = \tan^{2}\theta$,
\begin{align*}
I'(s)
&= \int_{0}^{\frac{\pi}{2}} \frac{\sin\theta}{\cosh s - \sin\theta}
= \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos\theta}{\cosh s - \cos\theta}.
\end{align*}
Then with the substitution $z = e^{i\theta}$, it follows that
\begin{align*}
I'(s)
&= \frac{i}{2} \int_{\Gamma} \frac{z^{2} + 1}{z^{2} - 2z \cosh s + 1} \frac{dz}{z},
\end{align*}
where the contour $\Gamma$ denotes the counter-clockwise semicircular arc joining from $-i$ to $i$. Note that we have $z^{2} - 2z \cosh s + 1 = 0$ if and only if $\cosh s = \cos \theta$ if and only if $z = e^{i\theta} = e^{\pm s}$. Thus deforming the contour to the straight line joining from $-i$ to $i$ with infinitesimal indent at the origin, we obtain
\begin{align*}
I'(s)
&= \frac{i}{2} \left\{ \operatorname{PV}\! \int_{-i}^{i} f(z) \, dz + 2\pi i \operatorname{Res} (f, e^{-s}) + \pi i \operatorname{Res} (f, 0) \right\},
\end{align*}
where $f$ denotes the integrand
$$ f(z) = \frac{z^{2} + 1}{z (z^{2} - 2z \cosh s + 1)}. $$
Proceeding the calculation,
\begin{align*}
I'(s)
&= \frac{i}{2} \left\{ i \operatorname{PV} \! \int_{-1}^{1} f(ix) \, dx + \pi i - 2 \pi i \coth s \right\} \\
&= - \frac{1}{2} \int_{0}^{1} \{ f(ix) + f(-ix) \} \, dx - \frac{\pi}{2} + \pi \coth s \\
&= \cosh s \int_{0}^{1} \frac{\frac{1}{2} (1 - x^{-2}) }{\{ \frac{1}{2}(x + x^{-1}) \}^{2} + \sinh^{2} s} \, dx - \frac{\pi}{2} + \pi \coth s.
\end{align*}
Now with the substitution $u = \frac{1}{2} \{ x + x^{-1} \}$, it follows that
\begin{align*}
I'(s)
&= - \cosh s \int_{1}^{\infty} \frac{du}{u^{2} + \sinh^{2} s} - \frac{\pi}{2} + \pi \coth s \\
&= - \arctan(\sinh s) \coth s - \frac{\pi}{2} + \pi \coth s.
\end{align*}
Finally, with the substitution $x = \sinh t$,
\begin{align*}
I(s)
= - \int_{s}^{\infty} I'(t) \, dt
&= \int_{s}^{\infty} \left\{ \arctan(\sinh t) \coth t + \frac{\pi}{2} - \pi \coth t \right\} \, dt \\
&= \int_{\sinh s}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx.
\end{align*}
Our problem corresponds to $s = \log 2$, or equivalently, $a := \sinh s = \frac{3}{4}$.
\begin{align*}
&\int_{a}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx \\
&= - \int_{a}^{\infty} \frac{\arctan (1/x)}{x} \, dx + \frac{\pi}{2} \int_{a}^{\infty} \left( \frac{1}{\sqrt{x^{2} + 1}} - \frac{1}{x} \right) \, dx \\
&= - \int_{0}^{1/a} \frac{\arctan x}{x} \, dx + \frac{\pi}{2} \lim_{R\to\infty} \left[ \log\left( \frac{x + \sqrt{x^{2} + 1}}{x} \right) \right]_{a}^{R} \\
&= - \operatorname{Ti}\left(\frac{1}{a}\right) - \frac{\pi}{2} \log\left( \frac{a + \sqrt{a^{2} + 1}}{2a} \right),
\end{align*}
where the function
$$ \operatorname{Ti}(x) = \frac{\operatorname{Li}_{2}(ix) - \operatorname{Li}_{2}(-ix)}{2i} $$
is the inverse tangent integral. Using the following simple identity
$$ \operatorname{Ti}(x) = \operatorname{Ti}\left(\frac{1}{x}\right) + \frac{\pi}{2}\log x, $$
it follows that
\begin{align*}
\int_{a}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx
= - \operatorname{Ti}(a) - \frac{\pi}{2} \log\left( \frac{a + \sqrt{a^{2} + 1}}{2a^{2}} \right).
\end{align*}
Therefore, plugging $a = \frac{3}{4}$ gives
$$ \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - 2\sqrt{x+1} )}{x+1} \, dx
= \pi \log(3/4) - \operatorname{Ti}(3/4) $$
as Cleo pointed out without explanation. More generally, for $k > 1$
$$ \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - k \sqrt{x+1} )}{x+1} \, dx
= \frac{\pi}{2} \log \left( \frac{(k^{2} - 1)^{2}}{2k^{3}} \right) - \operatorname{Ti}\left( \frac{k^{2} - 1}{2k} \right). $$
|
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|
which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$ Which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$?
Squaring both sides will give me something but I could not go any further.
|
$$\frac{f(5)+f(7)}{2} \lt f(\frac{5+7}{2}) $$
for a concave function $f$, and so for $f(x)=\sqrt x$.
But even if you proceed on the way you have started you will get the result.
$$\begin{eqnarray}
\sqrt{7}-\sqrt{6} &\lt& \sqrt{6}-\sqrt{5} &\mid& ^2 \\
13-2\,\sqrt{6}\,\sqrt{7} &\lt& 11-2\,\sqrt{5}\,\sqrt{6} &\mid& -11+2\sqrt{6}\sqrt{7} \mid \div 2 \\
1 &\lt& \sqrt{6}\,\sqrt{7}-\sqrt{5}\,\sqrt{6} & \mid& ^2 \\
1&\lt& 72-12\,\sqrt{5}\,\sqrt{7} & \mid& +12\sqrt{5}\sqrt{7} \\
12\,\sqrt{5}\,\sqrt{7} &\lt& 71 & \mid& ^2 \\
12^2 \cdot 35 &\lt& 71^2
\end{eqnarray}$$
All squaring operations are reversible and don't change the inequality relation because the LHS and the RHS of the inequalities are positive.
But $$12^2 \cdot 35 =12 \cdot 6 \cdot 2 \cdot 35=72 \cdot 70 = (71+1)(71-1)=71^2-1$$
So the last and therefore all inequalities are true.
|
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|
how is a factorial fraction equal to the product notation How is the $\prod_{k=2}^n(2k-3)={(2n-3)!\over 2^{n-2}(n-2)!}$, where $n \geq 2$
Note:
I know that the $(2n-3)!$ is equal to the product of $2k-3$ from $k=2$ to $n$, but I can't figure out the bottom half of how diving by $2^{n-2}(n-2)!$ equals the product notation of $2k-3$.
|
Your equation $\prod_{k=2}^n(2k-3)$ can be expanded to $(2n-3)(2n-5)\cdots(3)(1)$. Clearly, we have every other term for (2n-3)! since we only have the odd values.
Hence,$(2n-3)! = (2n-3)(2n-4)\cdots(2)(1) = (2n-4)(2n-6)\cdots(4)(2) \prod_{k=2}^n(2k-3)$.
Taking the addition terms on the RHS, $(2n-4)(2n-6)\cdots(4)(2) = 2^{n-2}[(n-2)(n-3)\cdots(2)(1)] = 2^{n-2}(n-2)!$
Thus, $(2n-3)! = \prod_{k=2}^n(2k-3) 2^{n-2}(n-2)!$, which you simplify to get the desired result.
|
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|
Is there a closed form for this sum? While generalizing the previous result, I conjectured that the series expansion of
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \arctan \left( \frac{2x \sin\theta}{1-x^{2}} \right) \arctan \left( \frac{2y \sin\theta}{1-y^{2}} \right) \arctan \left( \frac{2z \sin\theta}{1-z^{2}} \right) \arctan \left( \frac{2w \sin\theta}{1-w^{2}} \right) \, d\theta
\end{align*}
is equal to
$$ \frac{\pi}{2} \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \sum_{l=0}^{\infty} (-1)^{i+j+k+l} d(i,j,k,l) \frac{x^{2i+1}}{2i+1} \frac{y^{2j+1}}{2j+1} \frac{z^{2k+1}}{2k+1} \frac{w^{2l+1}}{2l+1}, \tag{1} $$
where $d(i,j,k,l)$ denotes the number of choices of signatures so that
$$ \pm(2i+1) \pm(2j+1) \pm (2k+1) \pm(2l+1) = 0. $$
I checked that the formula $\text{(1)}$ is correct up to degree 40. Compared to the egregious look of the original integral, this series representation is quite neat and tantalizing.
Assuming that $i \leq j \leq k \leq l$, we have the following algorithm:
$$ (-1)^{i+j+k+l} d(i,j,k,l) = \begin{cases}
6 & \text{if } i = j = k = l, \\
4 & \text{else if } i = j < k = l, \\
2 & \text{else if } i + l = j + k, \\
-2 & \text{else if } i + j + k = l-1, \\
0 & \text{else}.
\end{cases} $$
If we only sum the terms of $\text{(1)}$ corresponding to $d = 6$ or $4$, we get a combination of Legendre chi functions. But I have no idea how to simplify further.
|
The OP already gave the series expansion for $\arctan \left( \frac{2x \sin\theta}{1-x^{2}} \right) $ in here. Using that expansion, we have
\begin{align*}
&\int_{0}^{\frac{\pi}{2}} \arctan \left( \frac{2x \sin\theta}{1-x^{2}} \right) \arctan \left( \frac{2y \sin\theta}{1-y^{2}} \right) \arctan \left( \frac{2z \sin\theta}{1-z^{2}} \right) \arctan \left( \frac{2w \sin\theta}{1-w^{2}} \right) \, d\theta \\
&\quad = 16 \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \sum_{p=0}^{\infty} \sum_{q=0}^{\infty} \frac{x^{2m+1}y^{2n+1}z^{2p+1}w^{2q+1}}{(2m+1)(2n+1)(2p+1)(2q+1)} \times \\
& \qquad \times \int_{0}^{\frac{\pi}{2}} \sin(2m+1)\theta \sin(2n+1)\theta \sin(2p+1)\theta \sin(2q+1)\theta \, d\theta \quad \quad \rm{\bf (1)}
\end{align*}
Now we use the following somewhat lengthy trigonometric equality for general $x,y,z,w$ (which can be shown by Euler formulae, or more readily, by combining triple and double product formulae as listed in Wikipedia here)
\begin{align*}
& \quad 8 \sin(w) \sin(x) \sin(y) \sin(z) = \\
& - \cos(w+x+y-z) - \cos(w+y+z-x) - \cos(w+z+x-y) + \cos(w+x+y+z)+\\
&\cos(-w+x+y-z) + \cos(-w+y+z-x) + \cos(-w+z+x-y) - \cos(-w+x+y+z)
\end{align*}
Integrals arise, where 4, 3, or 2 arguments have a positive sign. Let's look at these type of integrals. In the following, $\delta_n = 1$ for $n=0$, $\delta_n =0$ else.
a) 4 positive signs:
$$
\int_{0}^{\frac{\pi}{2}} \cos(2m+2n+2p+2q+4)\theta \, d\theta = \frac{\pi}{2} \delta_{m+n+p+q +2}
$$
Since $m,n,p,q \geq 0$, this term will never be met in the sum.
b) 3 positive signs:
$$
\int_{0}^{\frac{\pi}{2}} \cos(2m+2n+2p-2q +2)\theta \, d\theta = \frac{\pi}{2} \delta_{m+n+p-q +1}
$$
c) 2 positive signs:
$$
\int_{0}^{\frac{\pi}{2}} \cos(2m+2n-2p-2q)\theta \, d\theta = \frac{\pi}{2} \delta_{m+n-p-q}
$$
So the full integral (1) becomes
\begin{align*}
& \frac{\pi}{16}\Big[ \delta_{m+n+p+q +2} - \delta_{-m+n+p+q +1}- \delta_{m-n+p+q +1}- \delta_{m+n-p+q +1}- \delta_{m+n+p-q +1} +\\
& \delta_{m+n-p-q} +\delta_{-m+n-p+q} +\delta_{m-n-p+q} \Big]
\end{align*}
So the conjecture is proved if we establish
\begin{align*}
& \frac12 (-1)^{m+n+p+q} d(m,n,p,q) = \quad \quad \rm{\bf (2)}\\
& \delta_{m+n+p+q +2} - \delta_{-m+n+p+q +1}- \delta_{m-n+p+q +1}- \delta_{m+n-p+q +1}- \delta_{m+n+p-q +1} +\\
& \delta_{m+n-p-q} +\delta_{-m+n-p+q} +\delta_{m-n-p+q}
\end{align*}
Since $d(m,n,p,q)$ denotes the number of choices of signatures so that
$$
\pm(2m+1) \pm(2n+1) \pm (2p+1) \pm(2q+1) = 0
$$
we can convince ourselves of the various cases. This is best seen with the case description given by the OP. (Note my comment above, there appears to be a typo in the one but last case.) Here is a list of terms (with corresponding signs) which apply. Note that we need half as many terms as given by the OP, due to the factor $\frac12$ in (2):
$$
\begin{cases}
\delta_{m+n-p-q} +\delta_{-m+n-p+q} +\delta_{m-n-p+q} & \text{if } m = n = p = q, \\
\delta_{-m+n-p+q} +\delta_{m-n-p+q} & \text{else if } m = n < p = q, \\
\delta_{m+n-p-q} & \text{else if } m + n = p + q, \\
- \delta_{m+n+p-q +1} & \text{else if } m + n + p = q-1, \\
0 & \text{else}.
\end{cases}
$$
This holds as well for all permutations of the indices. Done. $\quad \quad \Box$
|
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|
Use partial fractions to find the integral. Find the integral using partial factions.
$$\int\frac{(2x^2+5x+3)}{(x-1)(x^2+4)}\,dx$$
So do I do...$$\frac{2x^2+5x+3}{(x-1)(x^2+4)}=\frac{A}{x-1} + \frac{Bx+C}{x^2+4},$$
then get
\begin{align*}
2x^2+5x+3 &= A(x^2+4)+(Bx+C)(x-1) \\
2x^2+5x+3 &= Ax^2+4A+Bx^2-Bx+Cx-C?
\end{align*}
|
It seems that you are stuck at the actual partial fraction decomposition, rather than at the integration. So let's pick up at $$2x^2+5x+3= A(x^2+4) + (Bx+C)(x-1)$$
While this must hold true for all values of $x$, certain values of $x$ will lead us to the values of $A,B,C$ more quickly...
$x = 1$ is a natural choice, as that will make the $(Bx+C)$ go away. We then have $$2(1)^2 + 3(1) + 5 = A((1)^2 + 4) + 0$$
$$2 + 3 + 5 = 5A$$
$$A=2$$
Then you could use complex x-values ($\pm 2i$ for example) and equate real and imaginary parts, but you could just pick a few other small (real) x-values for ease of computation.
So $x=2$ gives us $21 = 16 + 2B + C$, or $$2B +C = 5$$
Then $x = 3$ gives us $36 = 26 + (3B+C)(2)$, or $$3B+C = 5$$
Solving these last two equations simultaneously gives $B = 0, C = 5$.
|
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|
How find this inequality minimum $\sum\limits_{cyc}\sqrt{a^2+b^2+ab-2a-b+1}$
Let $0<a,b,c<1$, find this follow minimum
$$\sqrt{(a+b)^2-(a+1)(b+2)+3}+\sqrt{(b+c)^2-(b+1)(c+2)+3}+\sqrt{(c+a)^2-(c+1)(a+2)+3}$$
My try:
since
$$(a+b)^2-(a+1)(b+2)+3=a^2+b^2+2ab-ab-2a-b-2+3=a^2+b^2+ab-2a-b+1$$
so we only find this follow minimum
$$\sum_{cyc}\sqrt{a^2+b^2+ab-2a-b+1}$$
But I can't.Thank you
|
$a^2+b^2+ab-2a-b+1=(b+\dfrac{a-1}{2})^2+\dfrac{3}{4}(a-1)^2$
$\sum\limits_{cyc}\sqrt{a^2+b^2+ab-2a-b+1} \ge \sqrt{(\sum\limits_{cyc} (b+\dfrac{a-1}{2})^2+\dfrac{3}{4}(\sum\limits_{cyc}(a-1))^2}=\sqrt{(\dfrac{3}{2}(\sum\limits_{cyc} a-1)^2+\dfrac{3}{4}(\sum\limits_{cyc} a-3)^2}=\sqrt{3((\sum\limits_{cyc} a)^2-3\sum\limits_{cyc} a+3)}\ge \dfrac{3}{2}$
when $\sum\limits_{cyc} a=\dfrac{3}{2}$ and $ a=b=c=\dfrac{1}{2}$ get min.
|
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|
Finding derivative using product and chain rule I need to find first derivative of $x\sqrt{2-x^2}$. My approach
*
*Using product rule: $(2-x^2)^{1/2} + x\frac{\operatorname{d}(2-x^2)^{1/2}}{\operatorname{d}x}$
*Using chain rule: $(2-x^2)^{1/2} + x\left[\frac{1}{2}(2-x^2)^{-1/2} (-2x)\right]$
*Result: $(2-x^2)^{1/2} - x^2 (2-x^2)^{-1/2}$
Is that correct?
|
It is, indeed, correct. A way to see that the answers are the same is to note that $$\begin{align}(2-x^2)^\frac12-x^2(2-x^2)^{-\frac12} &= (2-x^2)^{1+-\frac12}-x^2(2-x^2)^{-\frac12}\\ &= (2-x^2)(2-x^2)^{-\frac12}-x^2(2-x^2)^{-\frac12}\\ &= \bigl((2-x^2)-x^2\bigr)(2-x^2)^{-\frac12}\\ &= \frac{2-2x^2}{(2-x^2)^\frac12}\\ &= \frac{2(1-x^2)}{\sqrt{2-x^2}}\\ &= -\frac{2(x^2-1)}{\sqrt{2-x^2}}.\end{align}$$
|
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|
Showing $m+n\sqrt 2$ is a unit in set $R$ Let $R=\{m+n\sqrt 2|m,n\in\mathbb{Z}\}$. Show that $m+n\sqrt2$ is a unit in $R$ iff $m^2-2n^2=\pm 1$. Hint: Show that if $(m+n\sqrt2)(x+y\sqrt 2)=1$ then $(m-n\sqrt 2)(x-y\sqrt2)=1$ I needed help with the foward direction
This is what I have so far: ($\rightarrow$). Suppose that $m+n\sqrt2$ is a unit. I want to show that $m^2-2n^2=(m+n\sqrt2)(m-n\sqrt2)=\pm 1$. Since $m+n\sqrt2$ is a unit there exists a $p\in R$ such that $p*(m+n\sqrt2)=1$ where $p=(x+y\sqrt 2)$. But from here i'm stuck showing $(m-n\sqrt 2)(x-y\sqrt2)=1$. Any hints or help would be great.Thanks
Edit:
Using what I got as a hint below I got since $(m-n\sqrt 2)(x-y\sqrt2)=(m^2-2n^2)(x^2-2y^2)=1\implies (m^2-2n^2)=\frac{1}{x^2-2y^2}$. But since $m^2-2n^2$ is an integer it follows that $(x^2-2y^2)=\pm 1$ and $(m^2-2n^2)=\pm 1$ since $1*1=1$ and $-1*-1=1$.
|
Let $(x,y)$ denote $x+y\sqrt 2$. Then, we have multiplication defined as $(x,y)(x',y')=(xx'+2yy',xy'+x'y)$. If we define $N(x,y)=x^2-2y^2$, we have that $$\begin{align}N(x,y)N(x',y')&=(x^2-2y^2)(x'^2-2y'^2)\\&=x^2x'^2-2(x'^2y^2+x^2y'^2)+4y'^2y^2\\&=(xx')^2+(2y'y)^2-2((x'y)^2+(xy')^2)\\&=(xx')^2+\color{red}{2(xx')(2yy')}+(2y'y)^2-2((x'y)^2+\color{red}{2(x'y)(xy')}+(xy')^2)\\&=(xx'+2yy')^2-2(xy'+x'y)^2\\&=N((x,y)(x',y'))\end{align}$$
Now, suppose $z=(x,y)$ and $w=(x',y')$ are such that $zw=1$. Then $N(wz)=N(w)N(z)=1$, so?
|
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|
Double Integral $\int\limits_0^b\int\limits_0^x\sqrt{a^2-x^2-y^2}\,dy\,dx$ What is the best method for evaluating the following double integral?
$$
\int_{0}^{b}\int_{0}^{x}\,\sqrt{\,a^{2} - x^{2} - y^{2}\,}\,\,{\rm d}y\,{\rm d}x\,,
\qquad a > \sqrt{\,2\,}\,\,b
$$
Is there exist an easy method?
My try:
$$\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}dy\,dx=\int_0^{\frac{\pi}{4}}\int_0^{b\sec(\theta)}r\sqrt{a^2-r^2}dr\,d\theta$$
$$=\int_0^{\frac{\pi}{4}}\frac{-1}{3}\left[(a^2-r^2)\sqrt{a^2-r^2}\right]_0^{b\sec(\theta)}d\theta$$
$$=\frac{1}{3}\int_0^{\frac{\pi}{4}}\left[a^3-(a^2-b^2\sec^2(\theta))\sqrt{a^2-b^2\sec^2(\theta)}\right]d\theta$$
but evaluating above integral is very difficult and antiderivative is very complexity! see here.
|
Suppose
$$I=\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}dy\,dx$$
Let $y=\sqrt{a^2-x^2}\sin \theta$ then $dy=\sqrt{a^2-x^2}\cos \theta\,d\theta$, so
$$I=\int_0^b\int_0^{\arcsin\frac{x}{\sqrt{a^2-x^2}}}(a^2-x^2)\cos^2 \theta\,d\theta\,dx$$
$$=\frac{1}{2}\int_0^b(a^2-x^2)\left[\theta+\frac{1}{2}\sin 2\theta\right]_0^{\arcsin\frac{x}{\sqrt{a^2-x^2}}}\,dx$$
and now note that $\sin \theta=\frac{x}{\sqrt{a^2-x^2}}$, so $\cos \theta=\sqrt{\frac{a^2-2x^2}{a^2-x^2}}$ and $\frac{1}{2}\sin 2\theta=\sin \theta\cos \theta=\frac{x\sqrt{a^2-2x^2}}{a^2-x^2}$. therefore
$$I=\frac{1}{2}\left(\int_0^b(a^2-x^2)\arcsin\frac{x}{\sqrt{a^2-x^2}}\,dx+\int_0^bx\sqrt{a^2-2x^2}\,dx\right)=\frac{1}{2}(I_1+I_2).$$
for evaluating $I_1$ use integrating by parts. If you let $u=\arcsin\frac{x}{\sqrt{a^2-x^2}}$ and $dv=(a^2-x^2)dx$, then
$$du=\frac{a^2}{(a^2-x^2)\sqrt{a^2-2x^2}}dx,v=a^2x-\frac{x^3}{3}$$
therefore
$$I_1=\frac{3a^2b-b^3}{3}\arcsin\frac{b}{\sqrt{a^2-b^2}}+I_3$$
and
$$I_3=-\int_0^b\frac{a^2(a^2-\frac{x^2}{3})x}{(a^2-x^2)\sqrt{a^2-2x^2}}\,dx$$
now let $u=\sqrt{a^2-2x^2}$, then $du=\frac{-2x}{\sqrt{a^2-2x^2}}dx$ and $x^2=\frac{a^2-u^2}{2}$. so
$$I_3=\frac{a^2}{2}\int_a^{\sqrt{a^2-2b^2}}\frac{a^2-\frac{a^2-u^2}{6}}{a^2-\frac{a^2-u^2}{2}}\,du=\frac{a^2}{6}\int_a^{\sqrt{a^2-2b^2}}\frac{u^2+5a^2}{u^2+a^2}\,du$$
$$=\frac{a^2}{6}\int_a^{\sqrt{a^2-2b^2}}\left(1+\frac{a^2}{u^2+a^2}\right)\,du=\frac{a^2}{6}(\sqrt{a^2-2b^2}-a)+\frac{2a^3}{3}\left(\arctan\frac{\sqrt{a^2-2b^2}}{a}-\frac{\pi}{4}\right)$$
for $I_2$ we have
$$I_2=\int_0^bx\sqrt{a^2-2x^2}\,dx=\frac{-1}{6}\left[(a^2-2x^2)\sqrt{a^2-2x^2}\right]_0^b=\frac{1}{6}(a^3-(a^2-2b^2)\sqrt{a^2-2b^2})$$
Hence,
$$I=\frac{3a^2b-b^3}{6}\arcsin\frac{b}{\sqrt{a^2-b^2}}+\frac{a^3}{3}\arctan\frac{\sqrt{a^2-2b^2}-a}{\sqrt{a^2-2b^2}+a}+\frac{b^2}{6}\sqrt{a^2-2b^2}.$$
|
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|
Expressing $\sum_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$ in closed form I want to express $$\sum_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$$ in closed form.
I know that $$\pi z\cot(\pi z)=1+2z^2\sum_{n=1}^\infty\dfrac{1}{z^2-n^2}$$ which looks close, but I don’t know how to use it.
|
Here is a closed form by Maple
$$ \frac{2}{3}\,{\frac {\pi \, \left( -\cos \left( 2\,\pi \,z \right) +\sqrt {3}
\sin \left( \pi \,z \right) \sinh \left( \pi \,z\sqrt {3} \right) +
\cos \left( \pi \,z \right) \cosh \left( \pi \,z\sqrt {3} \right)
\right) }{{z}^{2} \left( 2\,\sin \left( \pi \,z \right) \cosh \left(
\pi \,z\sqrt {3} \right) -\sin \left( 2\,\pi \,z \right) \right) }}.$$
|
{
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|
Boolean Algebra simplify minterms I have this equation
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C$$
and need to simplify it. I have got as far as I can and spent a good 2 hours at it. I've realized I now need to use De Morgan's law to continue however I am baffled as to which rule to use. If someone could send me in the right direction that would be great!
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C$$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot\bar{C} + B\cdot\bar{C} + B\cdot C$$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot C + B(\bar{C} + C)) $$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot C + B) $$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(C + B) $$
Thanks
|
The $\mathrm{ExOR}$ function can be denoted by: $\oplus$ ; $X \oplus Y=\overline {X}\cdot Y +X \cdot \overline{Y} $.
Also $\overline{X \oplus Y}=\overline {X}\cdot \overline{Y} +X \cdot {Y} $
Hence $$\begin{align}
\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C &=\left(\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C \right)+ \left(A\cdot B\cdot \bar{C} + A \cdot B\cdot C \right) \\
&=\bar{B}\left(\bar{A}\cdot\bar{C} + A\cdot C \right)+A\cdot B\left(\bar{C} + C \right)\\
&=\bar{B}\left(\overline{A \oplus C}\right)+A\cdot B\\
\end{align}$$
|
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|
Euler numbers grow $2\left(\frac{2}{ \pi }\right)^{2 n+1}$-times slower than the factorial? Stirling's approximation of the factorial for even numbers is given by
$$ (2n)! \sim \left(\frac{2n}{e}\right)^{2n}\sqrt{4 \pi n}. \tag{1} $$
Further, the Euler numbers grow quite rapidly for large indices as they have the following lower bound
$$ |E_{2 n}| > 8 \sqrt { \frac{n}{\pi} } \left(\frac{4 n}{ \pi e}\right)^{2 n}=
2\left(\frac{2}{ \pi }\right)^{2 n+1} \sqrt { 4n\pi} \left(\frac{2 n}{ e}\right)^{2 n}. \tag{2}$$
Why do these formulas look so similar? Or is $(2)$ just a way to say that Euler numbers grow $2\left(\frac{2}{ \pi }\right)^{2 n+1}$-times slower than the factorial?
|
There's more to it, we have
$$\frac{1}{\cos z} = \sum_{n=0}^\infty (-1)^n \frac{E_{2n}}{(2n)!}z^{2n}.\tag{1}$$
Since $\dfrac{1}{\cos z}$ has poles in $(k+\frac12)\pi$ for $k\in\mathbb{Z}$ and is holomorphic everywhere else, the series $(1)$ has a radius of convergence of $\dfrac{\pi}{2}$. The Cauchy-Hadamard formula now says
$$\limsup_{n\to\infty} \sqrt[n]{\frac{\lvert E_{2n}\rvert}{(2n)!}} = \frac{2}{\pi}.$$
More precisely, from the partial fraction development of $\dfrac{\pi}{\cos \pi z}$ (see below) we obtain
$$\frac{\pi}{\cos \pi z} = \sum_{n=0}^\infty \left(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}\right)2^{2n+2}z^{2n}.\tag{2}$$
Replacing $z$ with $\pi z$ in $(1)$ and multiplying with $\pi$ yields
$$\frac{\pi}{\cos \pi z} = \sum_{n=0}^\infty (-1)^n\frac{\pi^{2n+1}E_{2n}}{(2n)!}z^{2n},\tag{3}$$
and comparing coefficients with $(2)$ yields
$$(-1)^n\frac{\pi^{2n+1}E_{2n}}{(2n)!} = 2^{2n+2}\underbrace{\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}}_{s(2n+1)},$$
or
$$E_{2n} = (-1)^n 2 \left(\frac{2}{\pi}\right)^{2n+1}s(2n+1)\cdot(2n)!$$
Since $s(2n+1) \to 1$ for $n\to \infty$, we have
$$\lvert E_{2n}\rvert \sim 2\left(\frac{2}{\pi}\right)^{2n+1}\cdot(2n)!$$
$\cos \pi z$ has zeros in $a_k = k + \frac12,\: k \in \mathbb{Z}$ and nowhere else. All zeros are simple. The residue of
$$f(z) := \frac{\pi}{\cos \pi z}$$
in $a_k$ is therefore
$$\operatorname{Res}\left(f;a_k\right) = \frac{\pi}{\frac{d}{dz}(\cos \pi z)\bigl\lvert_{a_k}} = \frac{1}{-\sin \pi a_k} = (-1)^{k+1}.$$
The series
$$g(z) = \sum_{k\in \mathbb{Z}} (-1)^{k+1}\left(\frac{1}{z-a_k} + \frac{1}{a_k}\right)$$
is absolutely and locally uniformly convergent, and therefore $g$ is an entire meromorphic function with the same principal parts as $f$, thus $f-g$ is an entire holomorphic function.
$$g'(z) = \sum_{k\in\mathbb{Z}} \frac{(-1)^k}{(z-a_k)^2}$$
is easily seen to be an entire meromorphic function with period $2$ and the property that $\lim\limits_{\lvert \operatorname{Im} z\rvert\to \infty} \lvert g'(z)\rvert = 0$ uniformly in $\{ z : \lvert \operatorname{Re} z\rvert \leqslant 1\}$. The same holds for
$$f'(z) = \frac{\pi^2\sin \pi z}{\cos^2\pi z},$$
and furthermore $f'$ and $g'$ have the same poles and principal parts. Hence $f' - g'$ is an entire holomorphic function with period $2$ and $\lim\limits_{\lvert \operatorname{Im} z\rvert\to \infty} \lvert f'(z) - g'(z)\rvert = 0$ uniformly in $\{ z : \lvert \operatorname{Re} z\rvert \leqslant 1\}$, thus bounded, ergo constant, and the vanishing of $f'-g'$ for growing imaginary part determines that constant as $0$.
Therefore $f-g$ is constant, and since $g(0) = 0$, we have
$$\frac{\pi}{\cos \pi z} = \pi + \sum_{k\in\mathbb{Z}} (-1)^{k+1}\left(\frac{1}{z-a_k} + \frac{1}{a_k}\right).$$
Grouping, for $k \geqslant 0$, the terms for the two indices $k$ and $-(k+1)$, noting $a_{-(k+1)} = -a_k$, we obtain for $\lvert z\rvert < \frac12$
$$\begin{align}
\frac{\pi}{\cos\pi z} &= \pi + \sum_{k=0}^\infty (-1)^k \left(\frac{1}{a_k-z} + \frac{1}{a_k+z} - \frac{2}{a_k}\right)\\
&= \pi + \sum_{k=0}^\infty \frac{(-1)^k2}{a_k}\left(\frac{a_k^2}{a_k^2-z^2}-1\right)\\
&= \pi + \sum_{k=0}^\infty \frac{(-1)^k2}{a_k}\sum_{n=1}^\infty \left(\frac{z}{a_k}\right)^{2n}\\
&= \pi + \sum_{n=1}^\infty \sum_{k=0}^\infty \frac{(-1)^k2}{a_k^{2n+1}}z^{2n}\\
&= \pi + \sum_{n=1}^\infty \left(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}\right)2^{2n+2}z^{2n}.
\end{align}$$
Writing $\pi$ as the Leibniz series $\frac{\pi}{4} = 1 - \frac{1}{3} +\frac{1}{5} - \dotsb$ then yields $(2)$.
|
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|
How many ways are there for $2$ teams to win a best of $7$ series? Case $1$: $4$ games: Team A wins first $4$ games, team B wins none = $\binom{4}{4}\binom{4}{0}$
Case $2$: $5$ games: Team A wins $4$ games, team B wins one = $\binom{5}{4}\binom{5}{1}-1$...minus $1$ for the possibility of team A winning the first four.
Case $3$: $6$ games: Team A wins $4$ games, team B wins $2$ = $\binom{6}{4}\binom{6}{2}-2$...minus $2$ for the possibility of team A winning the first four games; and the middle four (games $2,3,4,5$), in which case there would be no game $6$.
Case $4$: $7$ games: Team A wins $4$ games, team B wins $3$ = $\binom{7}{4}\binom{7}{3}-3$...minus $3$ for the possibility of team A winning the first four games; games $2,3,4,5$; and games $3,4,5,6$.
Total = sum of the $4$ cases multiplied by $2$ since the question is asking for $2$ teams.
Is this correct?
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For another approach, if the series ends before the seventh game, extend it to seven games by having the losing team win the rest. The series will now be four games to three. There are $2$ ways to choose the losing team, and ${7 \choose 3}$ ways to choose which game the losing team wins. The total is then $2{7 \choose 3}=70$
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|
Why is this binomial coefficient bounded thus? Source: Miklos Bona, A Walk Through Combinatorics.
$$ \forall k\geq 2,\binom{2k-2}{k-1}\leq4^{k-1}.$$
The RHS is the upper bound of the Ramsey number $R(k,k)$.
How can I prove the inequality without using mathematical induction? I've merely expanded the LHS to obtain $\frac{(2k-2)!}{(k-1)!(k-1)!}$.
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Let $n=k-1$.
We have $\frac {2n!}{n!n!}$.
Then $2n\cdot (2n-1)\cdot(2n-2)\cdot(2n-3)\dotsm 3\cdot 2\cdot 1$.
Taking even terms $2n\cdot(2n-2)\cdot(2n-4)\dotsm 8\cdot 6\cdot 4\cdot 2$
we have 2 common in all term, so it is $\underbrace {2\cdot 2\dotsm2}_{\text{$n$ times}} \cdot n\cdot (n-1)\cdot(n-2)\dotsm 3\cdot2\cdot1 = 2^n \cdot n!$
After canceling with $n!$ we have left $2^n \cdot (2n-1)\cdot(2n-3)\cdot(2n-5)\dotsm 5\cdot 3\cdot 1/n!$
We can see that $2n > 2n-1$, $2n-2> 2n-3$, etc.,
so the product of odd numbers is less than product of even numbers
therefore $(2n-1)\cdot(2n-3)\cdot(2n-5)\dotsm 5\cdot 3\cdot 1/n! \le 2^n$,
so $\binom{2n}n \le 2^n \cdot 2^n = 4^n$
|
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|
How to integrate $\frac{4x+4}{x^4+x^3+2x^2}$? Please could anyone help me to integrate
$\quad\displaystyle{4x + 4 \over x^4 + x^3 + 2x^2}.\quad$
I know how to use partial fraction and I did this:
$$
x^{4} + x^{3} + 2x^{2} = x^{2}\left(x^{2} + x + 2\right)
$$
And then ?.$\quad$ Thanks all.
|
Hint:
$$
\begin{align}
\frac{4x+4}{x^4+x^3+2x}
&=\frac{4x+4}{x^2(x^2+x+2)}\\
&=\frac1x+\frac2{x^2}-\frac{x+1/2}{x^2+x+2}-\frac{5/2}{x^2+x+2}
\end{align}
$$
|
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|
Trigonometric Limit: $\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$ I cannot figure out how to solve this trigonometric limit:
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$
I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I always go back to the indeterminate $\infty-\infty$
Has someone a different approach to solve this limit?
|
Use the series for cotangent
$$\cot x = \frac{1}{x}-\frac{1}{3}x-\frac{1}{45}x^3\pm\dots$$
$$\cot^2 x = \frac{1}{x^2}-\frac{2}{3}+\frac{1}{15}x^2\pm\dots $$
$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)=
\lim_{x\to 0}\left(\frac{1}{x^2}-\cot^2 x\right)=
\lim_{x\to 0}\left(\frac{2}{3}-\frac{1}{15}x^2 \pm \dots\right) = \frac{2}{3}
$$
|
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|
Fibonacci sequence proof Prove the following:
$$f_3+f_6+...f_{3n}= \frac 12(f_{3n+2}-1) \\ $$
For $n \ge 2$
Well I got the basis out of the way, so now I need to use induction: So that $P(k) \rightarrow P(k+1)$ for some integer $k \ge 2$
So, here are my first steps:
$$ \begin{align} & \frac 12(f_{3k+2}-1) + f_{3k+3} = \\
& = \frac 12(f_{3k+2}-1) + f_{3k+1} + f_{3k} + f_{3k+1} \\
& = \frac 12(f_{3k+2}-1) + f_{3k+1} + \frac 12(f_{3k+2}-1) + f_{3k+1} \\
& = f_{3k+2}-1 + 2 \cdot f_{3k+1}
\end{align} $$
And the fun stops around here. I don't see how to get to the conclusion: $\frac 12(f_{3k+5}-1) \\ $. Any help from this point would be great.
|
A combinatorial argument:
The fibonacci number $f_n$ represents the number of paths from 0 to $n - 1$ by taking steps of 1 or 2.
Now for any path for $f_{3n + 2}$,
*
*it's all ones
*for $0 \leq k < 3n$, it starts with k ones, followed by a 2, followed by one of the paths for $f_{3n - k}$
so
\begin{align*}
f_{3n + 2} &= 1 + \sum_{k = 1}^{3n} f_{k} \\
&= 1 + \sum_{k = 1}^n \left(f_{3k} + f_{3k - 1} + f_{3k - 2} \right) \\
&= 1 + 2\sum_{k = 1}^n f_{3k} \tag{Since $f_i = f_{i - 1} + f_{i - 2}$} \\
f_3 + f_6 + \dots f_{3n} &= \frac{1}{2}(f_{3n + 2} - 1)
\end{align*}
|
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|
Calculate the determinant of the matrix. Calculate the determinant.
\begin{bmatrix} C_{n}^{p+n} & C_{n}^{p+n+1} & \dots & C_{n}^{p+2n} \\
C_{n}^{p+n+1} & C_{n}^{p+n+2} & \dots & C_{n}^{p+2n+1} \\
\vdots & \vdots & \dots & \vdots \\
C_{n}^{p+2n} & C_{n}^{p+2n+1} & \dots & C_{n}^{p+3n} \end{bmatrix}
|
For $n=0$, we have
$$\det(C(p,0)) = 1$$
For $n=1$, we have
$$\det\left(\begin{bmatrix} p+1 & p+2\\ p+2 & p+3\end{bmatrix} \right) = -1$$
For $n=2$, we have
$$\det\left(\begin{bmatrix} \dfrac{(p+1)(p+2)}2 & \dfrac{(p+2)(p+3)}2 & \dfrac{(p+3)(p+4)}2\\ \dfrac{(p+2)(p+3)}2 & \dfrac{(p+3)(p+4)}2 & \dfrac{(p+4)(p+5)}2\\ \dfrac{(p+3)(p+4)}2 & \dfrac{(p+4)(p+5)}2 & \dfrac{(p+5)(p+6)}2 \end{bmatrix} \right) = -1$$
From a small MATLAB script I wrote, it looks like:
$$D(n,p) = \begin{cases} +1 & \text{if }n \equiv 0,3\pmod4\\ -1 & \text{if } n \equiv 1,2 \pmod4\end{cases}$$
|
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|
How prove this congruence equation has four zeros solution Question:
let congruence equation
$$\begin{cases}
\left(\overline{a_{1}a_{2}\cdots a_{m}}\right)^2\equiv \overline{a_{m}}(\mod 10)\\
\left(\overline{a_{1}a_{2}\cdots a_{m}}\right)^2\equiv \overline{a_{m-1}a_{m}}(\mod 10^2)\\
\cdots\cdots\cdots\cdots\\
\left(\overline{a_{1}a_{2}\cdots a_{m}}\right)^2\equiv \overline{a_{1}a_{2}\cdots a_{m}}(\mod 10^m)
\end{cases}$$
show that: this congruence equation has $four$ zeros solution.
where $m\ge 2,a_{i}\in\{0,1,2,\cdots,9\}$
My try:
for example
when $m=2$ then
$25$and $76$ ,$0,1$this four solution such
first $0$ and $1$ is such it
and
$$25^2=625\equiv 5(\mod 10),76^2=5776\equiv 6(\mod 10)$$
$$25^2=625\equiv25(\mod 10^2),76^2=5776\equiv 76(\mod 10^2)$$
$$25^2=625\equiv 625(\mod 10^3),76^2=5776\equiv 776(\mod 10^3)$$
But My problem I can't prove it,Thank you very much!
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As @Hurkyl pointed out, if we let $x=\overline{a_1a_2\ldots a_m}$, it suffices to prove that $x^2 \equiv x \pmod{10^m}$ has exactly four solutions $\pmod{10^m}$.
This is straightforward; we get $2^m(5^m)=10^m \mid x(x-1)$. Note that $\gcd(x, x-1)=1$, so we have four cases:
\begin{align}
x \equiv 0 \pmod{2^m}, x \equiv 0 \pmod{5^m} \\
x \equiv 0 \pmod{2^m}, x \equiv 1 \pmod{5^m} \\
x \equiv 1 \pmod{2^m}, x \equiv 0 \pmod{5^m} \\
x \equiv 1 \pmod{2^m}, x \equiv 1 \pmod{5^m}
\end{align}
For each case, Chinese Remainder Theorem ensures the existence of a unique solution for $x \pmod{10^m}$. We thus get exactly four solutions.
|
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|
Find all triples of positive integers (x,y,z) such that Find all triples of positive integers (x,y,z) such that
$x^{z+1} \ - \ y^{z+1}=2^{100}$
The RHS is even, then x and y must be odd and $x^{z+1}>y^{z+1}$, but how to find out them all ?
|
$x \ge 0 \land y \ge 0 \rightarrow x \ge y$.
$$x^{z+1} - y^{z+1} = \left(x - y\right)\left(x^z + x^{z-1}y + x^{z-2}y^2 + \dots +y^z\right) = 2^{k+1} \tag{T1}$$
So we can conclude $x - y = 2^{r_1}$, and that $x$ and $y$ have the same parity.
If $z$ is even, rewrite $(T1)$:
$$x^z + y^z + xy\underbrace{\left(x^{z-2} + x^{z-3}y + \dots + y^{z-2}\right)}_{\text{An odd number of terms}} = 2^{k} \tag{T2}$$
Since $x$ and $y$ have the same parity then $x$ and $y$ must both be even to satisfy the parity of $(T2)$. Let $2a = x$ and $2b = y$.
$$a^{z+1} - b^{z+1} = 2^{k - z}$$
which is a smaller instance of the original problem with the added assumption that $z$ is even. We can repeat the proof to get either $\exists c\, d\, : 2c = a \land 2d = b$ or that $2^{k - \dots}$ is no longer an integer, ad nauseam.
If $z$ is odd,
$$x^{z+1} - y^{z+1} = \left(x + y\right)\left(x^z - x^{z-1}y + x^{z-2}y^2 - \dots -y^z\right) = 2^{k}$$
Conclude odd $z$ leaves a system of equations
$$x-y = 2^{r_1}$$
$$x+y = 2^{r_2}$$
$x = 2^{r_2 -1} + 2^{r_1 - 1}, y = 2^{r_2 -1} - 2^{r_1 - 1}$
Expanding $x^{z+1} - y^{z+1} = 2^{k+1}$ for $z=1$ gives a set of solutions:
$$2^{r_1 + r_2} = 2^{k+1}$$
$$(x,y,z) \in (2^{n - 1} + 2^{k - n},2^{n - 1} - 2^{k - n},1)\, 1 \le n \le k$$
Expanding for $z = 3$ gives another set:
$$2^{3r_2 + r_1 - 1} + 2^{3r_1 + r_2 - 1} = 2^{k+1}$$
with $(x,y,z)$ being found by the $r_1$ and $r_2$ values that satisfy the above.
At this point I'm unsure how to continue. All further expansions $z > 3$ seem to involve only positive terms, some of the form $A\cdot 2^{B(r_1, r_2)}\, ,\, A \ne 2$, which comes from the binomial expansion. It might be provable by induction. If so, then there are no more solutions.
|
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|
Find a limit $\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$ I am to find the limit of
$$\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$$
so I used:
$$\lim_{x \to -\infty} = \lim_{x \to \infty}f(-x)$$
but I just can't solve it to the end...
Please show me all steps, or at least most of them, so I'll know how to solve it. Thank you.
This question was posted on: Find $\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$, and got 3 answers, but I still don't know how should I solve it, because when I try to solve it (with help of those 3 answers) I get :
$$0−12/0$$ every time and that goes to minus infinity...
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$$\left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$$
$$\left(\frac{4^{x}*16- 2\cdot3^{-x}}{4^{-x}+2\cdot3\cdot3^{x}}\right)$$
$$\left(\frac{4^{x}*16- 2\cdot\frac{1}{3^x}}{\frac{1}{4^x}+2\cdot3\cdot3^{x}}\right)$$
$$\left(\frac{16\cdot4^{x}\cdot3^x- 2 }{3^x}\right)\cdot\left(\frac{4^x}{1+6\cdot3^x\cdot4^x}\right)$$
$$\left(\frac{16\cdot4^{x}\cdot3^x- 2 }{1+6\cdot3^x\cdot4^x}\right)\cdot\left(\frac{4^x}{3^x}\right)$$
now when you put $x \rightarrow -\infty$ the $4^x/3^x$ tends to $0$ and the limit tends to $0$
As for $$\left(\frac{16\cdot4^{x}\cdot3^x- 2 }{1+6\cdot3^x\cdot4^x}\right)$$ the limit is finite value which is ultimately gonna multipled to $0$
|
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|
Integer $a$ , If $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots If the equation $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots. Then the integer value of $a$ is
$\bf{My\; Try}::$ Let $\alpha,\beta\in \mathbb{Z}$ be the roots of the equation . Then $\alpha+\beta = (6-a)$ and $\alpha\cdot \beta = a$
Now $D = (a-6)^2-4a = $ perfect square. So $a^2+36-12a-4a = k^2$. where $k\in \mathbb{Z}$
$a^2-16a+36=k^2\Rightarrow (a-8)^2-28=k^2\Rightarrow (a-8)^2-k^2 = 28$
Now How can i solve after that
Help Required
Thanks
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From the first line of your solution, $\alpha.\beta+\alpha+\beta=6-a+a=6$, so $(\alpha+1)(\beta+1)=7$
What are the factors of 7?
|
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|
How to find the minimum of $a+b+\sqrt{a^2+b^2}$ let $a,b>0$, and such
$$\dfrac{2}{a}+\dfrac{1}{b}=1$$
Find this minimum
$$a+b+\sqrt{a^2+b^2}$$
My try: since
$$2b+a=ab$$
so
$$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$
then I can't
maybe this problem can use AM-GM or Cauchy-Schwarz inequality solve it.Thank you very much
|
Lagrange multipliers seems like a good approach. Solve the system of equations $$\displaystyle \frac{\partial }{\partial a}\left(a+b+\sqrt{a^2+b^2}-\lambda \left(\frac{2}{a}+\frac{1}{b}-1\right)\right)=0\textrm,$$,
$$\frac{\partial }{\partial b}\left(a+b+\sqrt{a^2+b^2}-\lambda \left(\frac{2}{a}+\frac{1}{b}-1\right)\right)=0\textrm,$$,
$$\frac{\partial }{\partial \lambda }\left(a+b+\sqrt{a^2+b^2}-\lambda \left(\frac{2}{a}+\frac{1}{b}-1\right)\right)=0\textrm.$$
Mathematica gives the solution as $\left(\lambda = -10, a= \frac{10}{3},b= \frac{5}{2}\right)$, which gives the minimum as $$a+b+\sqrt{a^2+b^2}= \frac{10}{3}+\frac{5}{2}+\sqrt{\left(\frac{10}{3}\right)^2+\left(\frac{5}{2}\right)^2} = 10\textrm.$$
|
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|
Prove that if $p>2$ is prime then $\left(\prod_{k=1}^{p-1}k^k\right)^2\equiv (-1)^{\frac{p+1}2}\pmod p.$ Prove that if $p>2$ is prime then
$$\left(\prod_{k=1}^{p-1}k^k\right)^2\equiv (-1)^{\frac{p+1}2}\pmod p.$$
I find this by computer but cannot prove it, thank you!
|
Let's deal with the stuff inside the square first. Since $p-1$ is even we have that:
$\prod_{k=1}^{p-1} k^k \equiv \left(\prod_{k=1}^{\frac{p-1}{2}} k^k\right) \left(\prod_{k=1}^{\frac{p-1}{2}} (-k)^{p-k}\right) = (-1)^{\frac{p-1}{2}} \prod_{k=1}^{\frac{p-1}{2}}k^p \equiv (-1)^{\frac{p-1}{2}}\prod_{k=1}^{\frac{p-1}{2}}k \quad\bmod p$.
So $\left(\prod_{k=1}^{p-1} k^k\right)^2 \equiv \left(\left(\frac{p-1}{2}\right)!\right)^2 \bmod p$.
Now if $p\equiv 1 \bmod 4$ then:
$\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv (p-1)! \equiv -1 \equiv (-1)^{\frac{p+1}{2}} \bmod p$.
Whereas if $p \equiv 3 \bmod 4$ then:
$\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv -(p-1)! \equiv 1 \equiv (-1)^{\frac{p+1}{2}} \bmod p$.
|
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|
How prove this $1\times 3\times 5\times 7\cdots\times 2009+2\times4\times6\cdots 2010\equiv 0 (\mod2011)$ Question:
show that
$$1\times 3\times 5\times 7\cdots\times 2009+2\times4\times6\cdots 2010\equiv 0 (\mod2011)$$
my try: since
\begin{align*}&1\times 3\times 5\times 7\cdots\times 2009+2\times4\times6\cdots \times2010\\
&=(2009)!!+2^{1005}\cdot1005!!
\end{align*}
then I can't.Thank you for you help
|
Ok so for the first term:
$1\times 3 \times 5 \times ... \times 2009$
$\equiv 1 \times 3 \times 5 \times ... \times 1005 \times (-1004) \times... \times (-4) \times (-2)$
$\equiv (1 \times 2 \times ... \times 1005)$
$\equiv 1005! \bmod 2011.$
The second term is similar:
$2\times 4\times 6 \times ... \times 2010$
$\equiv 2 \times 4 \times 6 \times ... \times 1004 \times (-1005) \times ... \times (-3) \times (-1)$
$\equiv -(1 \times 2 \times ... \times 1005)$
$\equiv -1005! \bmod 2011.$
So the sum is $0 \bmod 11$.
|
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sums of squares of integers We have to prove that there exists infinitely many integers $a,b,c$ such that $a^2 + b^2 = c^2 + 3$ .
This looked like a very straight-forward question . I did some algebraic manipulations but couldn't reach the conclusion . Help me here
|
You need $c^2 - a^2 = b^2 - 3 \iff (c-a)(c+a) = b^2- 3$
Let $b=3k$, then we need $(c-a)(c+a)= 3(3k^2-1)$, so let us try $c-a = 3, c+a = 3k^2-1$
Adding and subtracting, we get $c = \frac{3k^2+2}{2}, a = \frac{3k^2-4}{2}$, both would be integer if $k$ is even.
Hence one set of solutions is $a = 6k^2-2, b = 6k, c = 6k^2+1$, for integer $k$.
|
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Expanding brackets of the form $(a+b)^n$ If we have the equation
$
(a+b)^n = (a+b)\times(a+b)\times(a+b)\times \ldots \times (a+b)
$
expanding the right hand side of the above, is the sum of terms in the form $a^n$, $a^{n-1}b$,$a^{n-2}b^2$, $\ldots$, $b^n$. That is $a^{n-1}b^k$ for $k=0,1,\ldots,n$
Now, my textbook states the following
Each term of the form $a^{n-k}b^k$ arises by choosing the variable $b$
from $k$ of the pairs of brackets on the right hand side of the
equation and the variable $a$ from the remaining $n-k$ pairs of
brackets.
My question is now, why? There is no proof given for this, and I am struggling to understand why the above holds.For a very simple expansion, where $k=1$ and $n=2$, we have 2 terms of the form $ab$, here its pretty easy to see from
$
(a+b)(a+b) = a(a+b) + b(a+b) = a^2 + b^2 + 2ab
$
why the statement in the book holds for this case. But, for bigger ones, such as $(a+b)^3$ i can't see it. I can see that there is 3 ways of choosing one $b$ and two $a$, for the term $a^2b$ when looking at $(a+b)(a+b)(a+b)$, but I cant relate it to the expansion of them, if that makes sense?
|
To simplify talking about the statement a bit, let $C_i = (a + b)$ for $1\le i \le b$, so
$$(a + b)^n = C_1 \times \dots \times C_n.$$
For the case where $n = 3$ and the term $a^2b$:
If we choose $b$ from $C_1$, then we choose $a$ from $C_2$ and $C_3$.
If we choose $b$ from $C_2$, then we choose $a$ from $C_1$ and $C_3$.
If we choose $b$ from $C_3$, then we choose $a$ from $C_1$ and $C_2$.
Thus the coefficient of $a^2b$ is $3 = \binom{3}{1}$.
More generally you should be able to see that a term of $a^{n - k}b^k$ must have taken either $a$ or a $b$ from any term $C_i$. Thus simply follows from
$$(a + b)^n = C_1 \times\dots\times C_n = (a + b)[C_1 \times\dots C_{i - 1} \times C_{i + 1} \times C_n]$$
Since each $C_i$ contributes exactly one $a$ or $b$, we end up with exactly $k$ of the choices being $b$ and exactly $n - k$ of the choices being $a$. Since there are $\binom{n}{k}$ ways to choose to either take $b$ or not to take $b$ from each $C_i$, we end up with $\binom{n}{k}$ terms of $a^{n-k}b^k$.
|
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Induction: $n^{n+1} > (n+1)^n$ and $(n!)^2 \leq \left(\frac{(n + 1)(2n + 1)}{6}\right)^n$ How do I prove this by induction:
$$\displaystyle n^{n+1} > (n+1)^n,\; \mbox{ for } n\geq 3$$
Thanks.
What I'm doing is bunch of these induction problems for my first year math studies.
I tried using Bernoulli's inequality at some point, but no success. Also, tried $(n+1)^{n+2}=(n+1)^{n+1}(n+1)$, then expanding $(n+1)^{n+1}$ by binomial formula to get the $n^{n+1}$ member to apply the induction hypothesis, still no success.
Here's another one I've been struggling with:
$$(n!)^2 \leq \left(\frac{(n + 1)(2n + 1)}{6}\right)^n$$
EDIT: Finally solved the second one!
What I needed was the AM-GM inequality.
Therefore,
$$\frac{(n + 1)(2n + 1)}{6} = \frac{1}{n} \sum_{i=1}^{n} i^2 \geq \sqrt[n]{1^2 \cdot 2^2 \cdots n^2}$$
Thus,
$$\left(\frac{(n + 1)(2n + 1)}{6}\right)^n = \left(\frac{1}{n} \sum_{i=1}^{n} i^2\right)^n \geq 1^2 \cdot 2^2 \cdots n^2 = (n!)^2$$
Done.
|
Your inequality is the same as
$$
n\left(\frac{n}{n+1}\right)^n\gt1
$$
Notice that
$$
\begin{align}
(n+1)\left(\frac{n+1}{n+2}\right)^{n+1}
&=\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+1}n\left(\frac{n}{n+1}\right)^n\\
&=\left(1+\frac1{n(n+2)}\right)^{n+1}n\left(\frac{n}{n+1}\right)^n\\[9pt]
&\gt1\cdot1
\end{align}
$$
|
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|
finding the value of $f(\frac{1}{7})$ $f$ is a function mapping positive reals between $0$ and $1$ to reals. Let $f$ be given by, $f( \frac{x+y}{2} ) = (1-a)f(x)+af(y)$ where $y > x$ and $a$ being a constant. Also,$f(0) = 0$ and $f(1) = 1$. Find $f( \frac{1}{7} )$.
I have tried some things but it is not working out. How to find the $f( \frac{1}{7} )$ .
|
For each $n$ with $1\leq n\leq 6$ we have the equation $f(\frac{n}{7}) = (1-a)f(\frac{n-1}{7}) + af(\frac{n+1}{7})$. Since we know $f(0)$ and $f(1)$, this gives us a system of $6$ linear equations in $6$ unknowns.
|
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|
Prove the following trigonometric identity. $$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\sin x +1}{\cos x}$$
I tried substituting $\sin^2x+\cos^2x = 1$ but I cannot solve it.
|
The above method is really verifying and always quick. Another method to arrive at the answer is by rationalising denominator (mainly when the answer [or RHS] is not known or one is asked to work out only from LHS to RHS):
$$\frac{\sin x - \cos x + 1 }{\sin x + \cos x - 1 }\cdot \frac{\sin x + \cos x + 1}{\sin x + \cos x + 1}$$
$$\frac{ (\sin x + 1)^2 - \cos^2 x }{ 2 \sin x \cos x } $$
$$ \frac{ \sin^2 x + 2 \sin x + 1 - \cos^2 x }{ 2 \sin x \cos x } $$
$$ \frac{ \sin^2 x + 2 \sin x + \sin^2 x + \cos^2 x - \cos^2 x } {2 \sin x \cos x } $$
and the answer follows i.e. $$ \frac{\sin x + 1}{\cos x}. $$
Hope it was helpful.
|
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|
Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^22^n}$ How can I prove that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2).$$
Can anyone help me please?
|
\begin{align*}
\operatorname{Li}_3\left(\frac{1}{2}\right)&=\int _0^{\frac{1}{2}}\frac{\operatorname{Li}_2\left(x\right)}{x}\:dx=-\operatorname{Li}_2\left(\frac{1}{2}\right)\ln \left(2\right)+\int _{\frac{1}{2}}^{1}\frac{\ln \left(x\right)\ln \left(1-x\right)}{1-x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{2}\ln ^3\left(2\right)+\int _0^1\frac{\ln \left(1-x\right)\ln \left(x\right)}{x}\:dx-\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln \left(1-x\right)}{1-x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{2}\ln ^3\left(2\right)+\sum _{k=1}^{\infty }\frac{1}{k^3}-\frac{1}{2}\ln ^3\left(2\right)-\frac{1}{2}\int _0^{\frac{1}{2}}\frac{\ln ^2\left(1-x\right)}{x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)-\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)-\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+\operatorname{Li}_3\left(\frac{1}{2}\right)
\end{align*}
And magically we find the value for that sum
\begin{align*}
\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}=-\frac{1}{2}\ln \left(2\right)\zeta \left(2\right)+\zeta \left(3\right)
\end{align*}
|
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|
Fibonacci proof question: $f_{n+1}f_{n-1}-f_n^2=(-1)^n$ Show that
$$f_{n+1}f_{n-1}-f_n^2=(-1)^n$$
when $n$ is a positive integer and $f_n$ is the $n$th Fibonacci number.
|
If this is a homework exercise then this is likely not how you're meant to prove it, since you'll first need to prove that the $n$-th Fibonacci number can be written as $$f_n=\frac{\varphi^n-(-\frac{1}{\varphi})^{n}}{\sqrt{5}}$$ where $\varphi=\frac{1+\sqrt{5}}{2}$. But if you have that then
$$\begin{align}f_{n-1}f_{n+1}-f_{n}^{2} &= \left(\frac{\varphi^{n-1}-(-\frac{1}{\varphi})^{-\left(n-1\right)}}{\sqrt{5}}\right)\left(\frac{\varphi^{n+1}-(-\frac{1}{\varphi})^{n+1}}{\sqrt{5}}\right)-\frac{\varphi^{2n}-2\varphi^{n}(-\frac{1}{\varphi})^{-n}+\left(-\frac{1}{\varphi}\right)^{2n}}{5} \\
&= \frac{\varphi^{2n}+\left(-\frac{1}{\varphi}\right)^{2n}-\varphi^{n-1}\left(-\frac{1}{\varphi}\right)^{n+1}-\left(\frac{1}{\varphi}\right)^{n-1}\varphi^{n+1}}{5}-\frac{\varphi^{2n}-2(-1)^{n}+\psi^{2n}}{5} \\
&= \frac{\varphi^{2n}+\left(-\frac{1}{\varphi}\right)^{2n}-\left(\frac{1}{\varphi}\right)^{2}\left(-1\right)^{n+1}+\varphi^{2}\left(-1\right)^{n-1}}{5}-\frac{\varphi^{2n}-2(-1)^{n}+\left(-\frac{1}{\varphi}\right)^{2n}}{5} \\
&= \frac{\left(-1\right)^{n}\left[\left(\frac{1}{\varphi}\right)^{2}+\varphi^{2}\right]+2(-1)^{n}}{5}\end{align}$$
and, miraculously, $$\left(\frac{1}{\varphi}\right)^{2}+\varphi^{2} = \left(\frac{\sqrt{5}-1}{2}\right)^{2}+\left(\frac{1+\sqrt{5}}{2}\right)^{2}
= \frac{5-2\sqrt{5}+1+1+2\sqrt{5}+1}{4}
= 3
$$
so finally $$\frac{\left(-1\right)^{n}\left[\left(\frac{1}{\varphi}\right)^{2}+\varphi^{2}\right]+2(-1)^{n}}{5}=\frac{3\left(-1\right)^{n}+2(-1)^{n}}{5}=(-1)^{n}$$
|
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|
Ellipse problem : Find the slope of a common tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and a concentric circle of radius r. Problem :
Find the slope of a common tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and a concentric circle of radius r.
Few concepts about Ellipse :
Equation of Tangent to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at point $(x_1,y_1)$ is given by $T = \frac{xx_1}{a^2} +\frac{yy_1}{b^2}-1$
Also point of contact where line $y =mx +c $ touched the ellipse .
The line is y =mx +c touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ when $c =\pm \sqrt{a^2m^2+b^2}$
$(x_1,y_1) =(\pm \frac{a^2m}{\sqrt{a^2m^2 +b^2}}, \pm \frac{b^2}{\sqrt{a^2m^2+b^2}})$
Now, we know that equation of any tangent to the given ellipse is $y =mx \pm \sqrt{a^2m^2+b^2}$
Now if it touches $x^2+y^2=r^2$
Then
$ \sqrt{a^2m^2+b^2} = r\sqrt{1+m^2}$ I am unable to understand this condition ..please guide for this particular condition only... thanks.
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If we use parametric forms of the ellipse $(a\cos\phi,b\sin\phi)$ and of the circle $(r\cos\psi,r\sin\psi)$
we get the tangents to be $\displaystyle x\frac{\cos\phi}a+y\frac{\sin\phi}b=1\iff x b\cos\phi+y a\sin\phi=ab$
and for the circle $x\cos\psi+y\sin\psi=r$ with slope $\displaystyle-\frac{\cos\psi}{\sin\psi}=-\cot\psi\ \ \ \ (1)$
These two equations have to be identical implies $$\frac{b\cos\phi}{\cos\psi}=\frac{a\sin\phi}{\sin\psi}=\frac{ab}r$$
So, $\displaystyle\cos\phi=\frac{a\cos\psi}r,\sin\phi=\frac{b\sin\psi}r$
Squaring & adding we get, $\displaystyle(a\cos\psi)^2+(b\sin\psi)^2=r^2$
Divide either sides by $\sin^2\psi,$
$\displaystyle(a\cot\psi)^2+b^2=r^2(1+\cot^2\psi)\implies \cot^2\psi=\frac{r^2-b^2}{a^2-r^2}$ which must be $\ge0$
$\displaystyle\implies(r^2-b^2)(a^2-r^2)\ge0\iff (r^2-a^2)(r^2-b^2)\le0$
$\displaystyle\implies $ min$(a,b)\le r\le$max$(a,b)$
The value of $r$ will dictate the value of the slope in $(1)$
|
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How to calculate the size of any rectangle to fit into an ellipse? It's been a long time I do not review my math knowledge, please help me out.
I have an ellipse image with fix size, let's say it has the bounding rect with width=w1, height=h1, and I have any random image with random size let's say width=w2, height=h2, I need to make a thumbnail of this image so that it fits totally inside the ellipse.
The center of the two images is the same, and it's in the center position.
Any thoughts on how to calculate the new w2' and h2'?
Thanks in advance,
|
For your ellipse centred on (0,0) the equation is:
$$
\frac{x^2}{w1^2}+\frac{y^2}{h1^2} = 1 \Rightarrow y^2 = h1^2 \cdot \left( 1 - \frac{x^2}{w1^2} \right)
$$
Now imagine a box centred on (0,0) of width $w2'$ and height $h2'$ where $\frac{h2'}{w2'} = \frac{h2}{w2}$.
Finally imagine a straight line passing through (0,0) and the top right and bottom left corners of the box. The equation of this line will be :
$$y = \frac{h2}{w2} \cdot x$$
We can find $x$ such that $y$ is the same for both equations the note that $x = \frac{w2'}{2}$
Thus squaring our equation for the line and setting it equal to the ellipse we have
$$\begin{align}
\frac{h2^2}{w2^2} \cdot x^2 & = h1^2 \cdot \left( 1 - \frac{x^2}{w1^2} \right) \\
\frac{h2^2}{w2^2} \cdot x^2 & = h1^2 - \frac{h1^2 \cdot x^2}{w1^2} \\
x^2 \left( \frac{h2^2}{w2^2} + \frac{h1^2}{w1^2} \right) & = h1^2\\
x & = \sqrt{ \frac{h1^2}{\left( \frac{h2^2}{w2^2} + \frac{h1^2}{w1^2} \right) }}
\end{align}$$
Thus
$$
w2' = 2 \cdot \sqrt{ \frac{h1^2}{\left( \frac{h2^2}{w2^2} + \frac{h1^2}{w1^2} \right) }}
$$
and
$$
h2' = w2' \cdot \frac{h2}{w2}
$$
|
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What is $\lim_{x\to0} \frac{(\cos x + \cos 2x + \dots+ \cos nx - n)}{\sin x^2}$? What is the limit of $$\lim_{x\to0} \frac{\cos x + \cos 2x + \dots+ \cos nx - n}{\sin x^2}$$
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We have by taylor series
$$\cos(kx)\sim_0 1-\frac{k^2}{2}x^2$$
and
$$\sin x^2\sim_0 x^2$$
hence
$$\lim_{x\to0} \frac{(\cos x + \cos 2x + ...+ \cos nx - n)}{\sin x^2}= -\frac{1}{2}\sum_{k=1}^n k^2=-\frac{n(n+1)(2n+1)}{12}$$
|
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How to prove continued fraction convergents of a number Let $x=1+\sqrt{3}$. Prove that in pairs the continued fraction convergents of $x$ are $a_n$/$b_n$ < x < $c_n$/$d_n$ where $a_1$ = 2, $b_1$ = 1, $c_1$ = 3, $d_1$ = 1, $a_{n+1}$ = 2$c_n$ + $a_n$, $b_{n+1}$ = 2$d_n$ + $b_n$, $c_{n+1}$ = 3$c_n$ + $a_n$, $d_{n+1}$ = 3$d_n$ + $b_n$.
I am not sure where to start with this one. Help please? :)
EDIT:
I started in on an induction proof. I'm kinda stuck on the final portion.
Let $x=1+\sqrt{3}$. The continued fraction expansion is <2, 1, 2, 1, 2, 1, 2, 1, ...> and the first few convergents are 2, 3, 8/3, 11/4, 30/11, 41/15, 112/41. So we have 2 = 2/1 = $a_1$/$b_1$ because 2 < $1+\sqrt{3}$ and 3 = 3/1 = $c_1$/$d_1$ because 2 > $1+\sqrt{3}$, so we have $a_1$ = 2, $b_1$ = 1, $c_1$ = 3, $d_1$ = 1.
Base case: Let n=1. Then $a_2$ = 2$c_1$ + $a_1$ = 2(3) + 2 = 8, etc. and we get $a_2$/$b_2$ = 8/3 and $c_2$/$d_2$ = 11/4 which is true.
Assume true for n=k (the formulas for $a_n$, $b_n$, etc.)
Prove for n=k+1: $a_{n+1}$ = 2$c_n$ + $a_n$ so $a_{k+2}$ = 2$c_{k+1}$ + $a_{k+1}$ = 8$c_k$ + 3$a_k$
There is where I get stuck
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Let $x = [a_0; a_1 ,a_2 , \ldots]$. Find the continued fraction using the common algorithm. You will get $x = [2;\overline{1,2}]$.
Sidenote: If you are confused about the inequalities $a_n/b_n < x < c_n/d_n$, remember that $C_{2k} < C_{2k+1}$, and $C_{2k} < C_{2(k+1)}$ so that $C_0 < C_2 < \ldots < C_n < C_{n-1} < \ldots < C_5 < C_3 < C_1$. This is proven by simply considering $C_i - C_{i-2}$.
Let $C_i = p_i /q_i$ where $$p_0 = a_0, p_1 = a_0 a_1 + 1, p_i = a_i p_{i-1} + p_{i-2}\\q_0 = 1, q_1 = a_1, q_i = a_i q_{i-1} + q_{i-2}.$$ Hence $$p_0 = 2, p_1 = 2 \cdot 1 + 1 = 3\\ q_0 = 1, q_1 = 1.$$ So $$C_0 = \frac{p_0}{q_0} = 2,\\ C_1 = \frac{p_1}{q_1} = 3.$$ This confirms that $a_1 = 2, b_1=1, c_1=3,d_1=1$ in your notation.
$$p_2 = a_2 p_1 + p_0 = 2 \cdot 3 + 2 = 8, q_2 = a_2 q_1 + q_0 = 2 \cdot 1 + 1 = 3 \\ \implies C_2 = 8/3.$$
$$p_3 = a_3 p_2 + p_1 = 1 \cdot 8 + 3 = 11, q_3 = a_3 q_2 + q_1 = 1 \cdot 3 + 1 = 4 \\ \implies C_3 = 11/4.$$
It is easy to check that these results are in line with $n=1$ given by the recursive sequences. Hence the base case of the induction holds. Try to prove the induction.
Induction: $a_n/b_n$ will always correspond to even $n$ in the $C_n$ notation. Hence we need only prove that $a_{n+1} = 2p_n + p_{n-1}$ and $b_{n+1} = 2q_n + q_{n-1}$ for the left-hand side of the inequality $a_n/b_n < x$. But this follows immediately from the recursive equations for $p$ and $q$, as $a_{n+1}=2$ (where this $a$ is the $a$ in the continued fraction). We use these to prove $c_{n+1}$ and $d_{n+1}$. We have $$c_{n+1}=p_{n+2} = a_{n+2}p_{n+1} + p_n = a_{n+2} (a_{n+1} p_n + p_{n-1}) + p_n = p_n(a_{n+2} a_{n+1}+ 1) + a_{n+2} p_{n-1} = 3 c_n + a_n$$ where the $a_n$ after the last equality corresponds to the $a_n/b_n <x$ and the rest to the $a_i$'s in the continued fraction. $d_{n+1}$ is proven in a similar manner.
I must admit I find it terribly confusing dividing up the convergents on either side of $x$, as they are closely linked. Splitting them up into 4 separate sequences seems silly and unnecessary. $C_n$ notation saves a lot of trouble.
|
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All solutions of $a+b+c=abc$ in natural numbers I was observing some nice examples of equalities containing the numbers $1,2,3$ like $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$ and $\log 1+\log 2+ \log 3=\log (1+2+3)$. I found out this only happens because $1+2+3=1*2*3=6$. I wanted to find other examples in small numbers, but I failed. How can we find all of the solutions of $a+b+c=abc$ in natural numbers?The question seemed easy, but it seems difficult to find. I would prefer an elementary way to find them! What I did: We know if $a+b+c=abc$, $a|a+b+c$ so $a|b+c$. Similarly, $b|a+c$ and $c|a+b$. Other than that, if we multiply both sides by $b$, we get $b^2+1=(bc-1)(ab-1)$. If we also divide both sides by $abc$, we get $\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1$. I don't know how to go further using any of these, but I think they are a good start. I would appreciate any help.
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leaving aside the solution in which $a=b=c=0$ order the numbers so $a \le b \le c$
as OP shows, $c|(a+b)$ so we have:
$$
\frac{a+b}{c} \le 2
$$
thus only the values $1$ and $2$ are possible for $\frac{a+b}c$. these give $a+b=c$ and $a+b=2c$ respectively. however the latter is only possible if all three numbers are equal, as $c$ cannot be the arithmetic mean of two smaller numbers.
if all three numbers are equal they must each be $\sqrt{3}$, not a natural number solution.
plugging $a+b=c$ back into the equation $a+b+c=abc$ gives $ab=2$ hence $a=1, b=2, c=3$
|
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How to simplify $\tan{\arcsin{\frac{y}{R}}/2}$? I have verified with Mathematica that, for $R>0, y \in \mathbb{R}$:
$$ \tan{\frac{\arcsin{\frac{y}{R}}}{2}} = \frac{R - \sqrt{R^2 -y ^2}}{y}$$
using
Assuming[Element[y, Reals] && R > 0,
FullSimplify[TrigToExp[Tan[ArcSin[y/R]/2]]]]
How can I prove this withouse messing with complex exponentials and logarithms?
|
Let $\displaystyle\arcsin\frac yr=\phi\implies \sin\phi=\frac yR\ \ \ \ (1)$
and based on the definition of principal value, $\displaystyle-\frac\pi2\le\phi\le\frac\pi2\ \ \ \ (2)$
So, we need to find $\displaystyle\tan\frac\phi2$
Using Weierstrass substitution, $\displaystyle\sin\phi=\frac{2\tan\frac\phi2}{1+\tan^2\frac\phi2}$
So, using $\displaystyle(1),\frac{2\tan\frac\phi2}{1+\tan^2\frac\phi2}=\frac yR\implies y\tan^2\frac\phi2-2R\tan\frac\phi2+y=0\ \ \ \ (3)$
Solving the Quadratic Equation for $\displaystyle\tan\frac\phi2=\frac{R\pm\sqrt{R^2-y^2}}y$
Now using $\displaystyle(2), -\frac\pi4\le\frac\phi2\le\frac\pi4\implies -1\le\tan\frac\phi2\le1$
Observe that for real $\displaystyle\phi, R^2\ge y^2\implies R\ge |y|$ as $R>0$
If $\displaystyle R=|y|,\tan\frac\phi2=\pm1$ where both roots are same
Else $\displaystyle\frac{R+\sqrt{R^2-y^2}}{|y|}> \frac R{|y|}>1,$ hence should be discarded.
Observe that if $\displaystyle\tan\frac{\phi_1}2,\tan\frac{\phi_2}2$ are the roots of $(3),$
using Vieta's formula, $\displaystyle\tan\frac{\phi_1}2\tan\frac{\phi_2}2=1$
$\displaystyle\implies\tan\frac{\phi_1}2=\frac1{\tan\frac{\phi_2}2}=\cot\frac{\phi_2}2=\tan\left(\frac\pi2-\frac{\phi_2}2\right)=\tan\frac{\pi-\phi_2}2$
$\displaystyle\implies\frac{\phi_1}2=\frac{\pi-\phi_2}2\iff\phi_1=\pi-\phi_2 $
$\displaystyle\implies\sin\phi_1=\sin\left(\pi-\phi_2\right)=\sin\phi_2 $
Also, $\displaystyle\sin\phi=\frac{2\tan\frac\phi2}{1+\tan^2\frac\phi2}=\frac{2\cot\frac\phi2}{1+\cot^2\frac\phi2}$ (multiplying the numerator & the denominator by $\displaystyle\cot^2\frac\phi2$)
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|
Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
My approach :
If I divide numerator and denominator by $n^2$ I get :
$$\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0$$
but the answer is $\dfrac{1}{2}$ please suggest how to solve this.. thanks.
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$$ S_n=\sum_{k=1}^n \frac{k}{k+n^2} = \sum_{k=1}^n \left( 1 -\frac{n^2}{k+n^2} \right) \\
= n - \sum_{k=1}^n \left(\frac{n^2}{k+n^2} \right)
$$
but
$$\sum_{k=1}^n \left(\frac{n^2}{k+n^2} \right) = \sum_{k=1}^n \frac1{1+\frac{k}{n^2}}$$
$$ = \sum_{k=1}^n \sum_{j=0}^{\infty} \left(\frac{-k}{n^2} \right)^j \\
= n -\frac1{n^2}\frac{n(n+1)}{2} +O\left(\frac1{n}\right) = n - \frac12 +O\left(\frac1{n}\right) $$
so
$$ \lim_{n \rightarrow \infty} S_n = n - (n - \frac12) = \frac12 $$
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|
Need help simplifiying a rational expression There's a math question on an online test which asks the following
Multiply the following expression, and simplify:
$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$
But no matter how I try I keep getting the answer incorrect with a message telling me to simplify my answer. I can't seem to figure out how to simplify it enough to get it right.
$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$
equals, $\frac{x^4 + 4x^3y - 16x^2y^2 - 64xy^3}{x^2-4xy}$. I then factored x out of the numerator and denominator to get $\frac{x(x^3 + 4x^2y - 16xy^2 - 64y^3)}{x(x-4y)}$ and cancelled out the factored x's to get $\frac{x^3 + 4x^2y - 16xy^2 - 64y^3}{x-4y}$. I don't know what to do from here though.
I've managed to get enough marks to be able to pass it but since it's a readiness test I want to understand all of the material going in.
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$$\frac{x^2+16y^2}{x} \cdot \frac{x^2+4xy}{x-4y}=\frac{x^2+16y^2}{x} \cdot \frac{x(x+4y)}{x-4y}=(x^2+16y^2)\cdot \frac{(x+4y)}{x-4y}$$
$$=(x+i4y)(x-i4y) \frac{x+4y}{x-4y}$$
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|
How to get the simplest form of this radical expression: $3\sqrt[3]{2a} - 6\sqrt[3]{2a}$. How to get the simplest form of this radical expression:
$$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$
Here is my work:
$$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$
Since the radicands are the same, we just add the coefficients.
$$-3\sqrt[3]{2a} \sqrt[3]{2a}$$
Since everything is under the same index it becomes:
$$-3\sqrt[3]{2} \sqrt[3]{a}$$
Did I do this correctly, if not can anyone tell me what I should do?
Thanks :-).
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Your middle step is incorrect, it should be $-3\sqrt[3]{2a}$ not $-3\sqrt[3]{2a}\sqrt[3]{2a}$. It should be $$3\sqrt[3]{2a} - 6\sqrt[3]{2a} = 3\times\sqrt[3]{2a} - 6\times\sqrt[3]{2a} = (3 - 6)\times\sqrt[3]{2a} = -3\times\sqrt[3]{2a} = -3\sqrt[3]{2a}.$$ I don't think $-3\sqrt[3]{2}\sqrt[3]{a}$ is any simpler than $-3\sqrt[3]{2a}$, but that's just my opinion.
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"timestamp": "2023-03-29T00:00:00",
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|
Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$ I was solving an exercise, so I realized that the one easiest way to do it is using a "weird", but nice identity below. I've tried to found out it on internet but I've founded nothingness, and I wondering how to show easily this identity ? As a matter of the fact, it's a beautiful identity.
$$-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}$$
The one way I think about this I'll let here :
$$\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=\frac{(a-b)(b+c)(c+a)+(b-c)(c+a)(a+b)+(c-a)(c+b)(b+a)}{(a+b)(b+c)(c+a)}$$
Now, I'll take one of the term of RHS
$(a −b)(b +c)(c + a)\\ =(a −b)(b−c + 2c)(a −c + 2c)\\ =(a −b)(b −c)(a −c)+ 2c(a −b)(b−c + a −c + 2c) \\=(a −b)(b −c)(a −c)+ 2c(a −b)(a +b)$
Similarly, We'll do that with the remainder
$(b−c)(a+b)(c + a)+(c − a)(a +b)(b+ c)\\=(a +b)[(b −c)(c + a)+(c −a)(b +c)]\\=(a +b)(bc +ba −c^{2} −ca +cb+c^{2} −ab −ac)\\= (a +b)(2bc − 2ac)\\= −2c(a −b)(a +b)$
and we get :
$$\boxed{\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}}$$
So it's too many work to show this identity. I just want to know if there's a simple way to show that or I don't know. I'm questing that because I didn't find anything on internet.
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Observe:
This equation is a cubic(degree 3) in "a".So at maximum it can have 3 roots if it is not an identity. But if you show that it has 4 roots , then it becomes an identity.
Roots to try :
b,c(easy ones),0(also easy),-a(think it over).
Now you have shown that it has 4 roots. Thus it becomes an identity.
If b,c are also variables same procedure can be applied for them also.
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What is the shortest way to compute the last 3 digits of $17^{256}$? What is the shortest way to compute the last 3 digits of $17^{256}$ ?
My solution:
\begin{align}
17^{256} &=289^{128} \\
&=(290 - 1)^{128}\\
&=\binom{128}{0}290^{128} - ... +\binom{128}{126}290^2 - \binom{128}{127}290 + \binom{128}{128}
\end{align}
Computed the last 3 terms whose last 3 digits gave the last 3 digits of $17^{256}$.
Is there any shorter method to do this(which requires much less computation) ?
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Use Euler's Totient Theorem and the Chinese Remainder Theorem.
We have that $ 17 \equiv 1 \mod 8 $ and hence $ 17^{256} \equiv 1 \mod 8 $ as well.
Because $ 17 $ is coprime to $ 125 $, we know that $ 17^{100} \equiv 1 \mod 125 $.
We are left to calculate $ 17^{56} \mod 125 $, which can be quickly be done by hand via the method of doubling: $$\begin{align*} 17^{2} &\equiv 39 &\mod 125 \\ 17^4 &\equiv 21 &\mod 125\\ 17^8 &\equiv 66 &\mod 125\\ 17^{16} &\equiv 106 &\mod 125\\ 17^{32} &\equiv 111 &\mod 125 \end{align*}$$
Hence, $ 17^{32 + 16 + 8} \equiv 111 \cdot 106 \cdot 66 \equiv 56 \mod 125 $.
Using the Chinese Remainder Theorem, we have that the list three digits are $ 681 $.
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|
Indefinite integral $\int(\sin x)^2/(x\cdot\cos x-\sin x)^2\,dx$ How to find the antiderivative of $\dfrac{(\sin x)^2}{(x\cdot\cos x-\sin x)^2}$? I know that series theory can solve the problem. But I wish to solve the problem by the antiderivative way. The numerator and denominator divide the $x^2$ at the same time, don't they?
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As $\displaystyle \frac{d(x\cos x-\sin x)}{dx}=-x\sin x,\int\frac{-x\sin x}{(x\cos x-\sin x)^2}dx=-\frac1{x\cos x-\sin x}$
Integrating by parts,
$$\int\frac{\sin^2x}{(x\cos x-\sin x)^2}dx=\int \frac{(-x\sin x)}{(x\cos x-\sin x)^2}\cdot\frac{-\sin x}x dx$$
$$=\frac{-\sin x}x\int\frac{(-x\sin x)}{(x\cos x-\sin x)^2}dx-\int\left(\frac{d\left(\frac{-\sin x}x\right)}{dx}\cdot\int \frac{(-x\sin x)}{(x\cos x-\sin x)^2}dx\right)dx$$
$$=\frac{\sin x}x\cdot\frac1{x\cos x-\sin x}+\int\left(\frac{x\cos x-\sin x}{x^2}\cdot\frac1{x\cos x-\sin x}\right)dx$$
$$=\frac{\sin x}x\frac1{x\cos x-\sin x}+\int\frac{dx}{x^2}$$
Can you take it from here?
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Inequality in three variables $ \frac{3-(xy+yz+zx)}{2} \geq \sum\limits_{\text{cyc}}\frac{1-x^2y^2}{2+x^2+y^2}$ If $x, y, z$ are positive real numbers with the property $ xy, yz, zx \leq 1 $, then prove that $$ \frac{3-(xy+yz+zx)}{2} \geq \sum_{\text{cyc}}\frac{1-x^2y^2}{2+x^2+y^2}.$$
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Use AM-GM to get:
$$RHS = \sum_{\text{cyc}}\frac{1-x^2y^2}{2+x^2+y^2} \le \sum_{cyc}\frac{1-x^2y^2}{2+2xy}= \frac12\sum_{cyc}(1-xy) = LHS $$
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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.