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Minimal value of the product $(x+y)(y+z)$
Find the min. value of the product $\left(x + y\right)\left(y + z\right)$ given that $xyz\left(x + y + z\right) = 1$ and $x,y,z > 0$.
I want to solve this problem using geometry of a triangle:
*
*Let $a,b,c$ be the sides of the triangle and we can substitute
$a = x + y\,,\,\, b = y + z\,,\,\, c = z + x$.
*$A^{2} = xyz\left(x + y + z\right)$ is just a squared area of a triangle and $\left(x + y\right)\left(y + z\right)$ is a part of another formula of triangle $$
\mbox{i.e.}\quad A = {1 \over 2}\,ab\sin\left(\theta\right)\quad \mbox{or}\quad
A = {1 \over 2}\left(x + y\right)\left(y + z\right)\sin\left(\theta\right)
$$
I' ve tried the same with other formulas but I don't know how to "extract" inequality.
Before the solution I would appreciate that you give a hint, and then a solution. Any help appreciated.
|
Let $a=x+y, \, b=y+z ,\, c = z+x, \, p = \frac{a+b+c}{2},$ then
$$(x+y)(y+z) = ab,$$
and
$$1 = \sqrt{xyz(x+y+z)} = \sqrt{p(p-a)(p-b)(p-c)} = S = \frac{1}{2} ab \sin C.$$
But $\sin C \leqslant 1,$ so
$$2 = ab \sin C \leqslant ab.$$
Therefore $ab \geqslant 2.$ Equality hold for $x = 1, \, y = \sqrt 2 -1, \, z = 1.$
P/s. Using the AM-GM inequality, we have
$$(x+y)(y+z) = xz+y(x+y+z) \geqslant 2\sqrt{xyz(x+y+z)} = 2.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A sequence $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and find $x_{2020}$ The sequence is given by the formula $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and it is known that $x_{2017} + x_{2023} = 990$, then what is $x_{2020}$ = ?
My little approch:
It is given that $x_{2017} + x_{2023} = 990$ ----- (1) and $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$
So,
$x_{2017} = x_{2016+1} = 3 x_{2016} + \sqrt{8x^2_{2016} + 2}$ -------(2)
$x_{2023} = x_{2022+1} = 3 x_{2022} + \sqrt{8x^2_{2022} + 2}$ -------(3)
Then from (1),(2),(3) =>
$3 x_{2016} + \sqrt{8x^2_{2016} + 2} + 3 x_{2022} + \sqrt{8x^2_{2022} + 2} = 990$
$3 (x_{2016}+x_{2022} ) + \sqrt{8x^2_{2016} + 2} + \sqrt{8x^2_{2022} + 2} = 990$
$3 (x_{2016}+x_{2020} + x_{2021}) + \sqrt{8x^2_{2016} + 2} + \sqrt{8x^2_{2020} + 2}+ \sqrt{8x^2_{2021} + 2}+ \sqrt{8x^2_{2022} + 2} = 990$
I stuck here and can't go further. Please help me with it.
|
$$x_{n+1} = 3x_{n} + \sqrt{8x_{n}^2+2}$$
$$x_{n+1}-3x_{n} = \sqrt{8x_{n}^2+2}$$
By squaring both sides,
$$x_{n+1}^2+x_{n}^2-6x_{n}x_{n+1}=2$$
Since this is a general expression, we can write -
$$x_{n+1}^2+x_{n}^2-6x_{n}x_{n+1}=x_{n}^2+x_{n-1}^2-6x_{n-1}x_{n}$$
Simplifying this we get,
$$x_{n+1}+x_{n-1}=6x_{n}$$
Now our task is to find $x_n$ for $n=2020$ when $x_{n+3}+x_{n-3}$ is given.
First we will find $x_{n+2}+x_{n-2}$.
Let $$x_{n+2}+x_{n-2}=y$$.
By adding $2x_{n}$ to both sides of the equation,
$$6(x_{n+1}+x_{n-1})=y+2x_{n}$$
So,
$$y=34x_{n}$$
Now let,
$$x_{n+3}+x_{n-3}=z$$
Add $x_{n+1}+x_{n-1}$ to both the sides,
$$6(x_{n+2}+x_{n-2})=z+6x_{n}$$
$$z=198x_{n}$$
So, we get -
$$x_{2020}=5$$
|
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|
Alternative approaches to prove the following inequality For $a,b,c \in \mathbb{R^+},$ prove that
$$\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \left(\dfrac{2b}{c+a}\right)^{\frac{2}{3}} + \left(\dfrac{2c}{a+b}\right)^{\frac{2}{3}} \geq 3.$$
I managed to prove this problem using the technique of isolated fudging. In particular, one can prove that $\dfrac{1}{3}\left(\frac{2a}{b+c} \right)^{\frac{2}{3}} \geq \dfrac{a}{a+b+c} $ (motivation is explained below) using AM-GM, as follows:
\begin{align*}
\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} \geq \dfrac{a}{a+b+c} & \iff \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} \geq \dfrac{3a}{a+b+c} \\
& \iff \left(\dfrac{2a}{b+c} \right)^{2} \geq \left(\dfrac{3a}{a+b+c}\right)^3 \\
& \iff 4a^2(a+b+c)^3 \geq 27a^3(b+c)^2.
\end{align*}
The preceding inequality is homogeneous, hence W.L.O.G. set $a+b+c=3$. It thus suffices for us to prove that $4 \geq a(3-a)^2 \iff 8 \geq 2a(3-a)^2$. But this is obvious from AM-GM: $2a(3-a)^2 \leq \left(\dfrac{2a+(3-a)+(3-a)}{3}\right)^3=8.$ Similarly, we have $\dfrac{1}{3}\left(\dfrac{2b}{a+c} \right)^{\frac{2}{3}} \geq \dfrac{b}{a+b+c} $ and $\dfrac{1}{3}\left(\dfrac{2c}{a+b} \right)^{\frac{2}{3}} \geq \dfrac{c}{a+b+c} $. Thus, summing cyclically, we obtain:
$$\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \dfrac{1}{3}\left(\dfrac{2b}{a+c} \right)^{\frac{2}{3}} + \dfrac{1}{3}\left(\dfrac{2c}{a+b} \right)^{\frac{2}{3}} \geq 1 \Rightarrow \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \left(\dfrac{2b}{c+a}\right)^{\frac{2}{3}} + \left(\dfrac{2c}{a+b}\right)^{\frac{2}{3}} \geq 3 .$$
And we are done.
Some motivation:
Let $f(a,b,c)=\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} - \dfrac{a^r}{a^r+b^r+c^r}$, for some $r \in \mathbb{R}$. Ideally, we want $f(a,b,c) \geq 0$ for all values of $a,b,c$. Note that $f(1,1,1)=0$, which suggests that we should set $(1,1,1)$ as a local minimum of $f$. Hence, take the partial derivative of $f$ with respect to $a$, and set it to zero at $(1,1,1)$. By solving the resulting equation, we find a corresponding value of $r$:
$$\dfrac{\partial f}{\partial a}= \dfrac{1}{3} \cdot \sqrt[^3]{4} \cdot \left(\dfrac{1}{b+c} \right)^{\frac{2}{3}} \cdot \dfrac{2}{3} \cdot a^{\frac{-1}{3}} - \dfrac{ra^{r-1}(a^r+b^r+c^r)-a^r(ra^{r-1})}{(a^r+b^r+c^r)^2}$$
Hence,
$$\frac{\partial f}{\partial a }\Bigr|_{(1,1,1)} =0 \Rightarrow \dfrac{2}{9}- \dfrac{2r}{9} =0 \Rightarrow r=1.$$
My question is, is there any other way to solve this inequality besides isolated fudging? I.e. would methods such as Cauchy / Holder / Jensen also work here? I would love to see any other alternative approach.
|
We can assume that $a+b+c = 1$ and so with $$ f(x) = \left(\frac{2x}{1-x}\right)^{2/3}$$ the inequality becomes $$ f(a) + f(b) + f(c) \geq 3.$$ Now observe that $f(x) \geq 3x$ for $x \in [0,1]$.
|
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|
Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$ $a$ and $b$ are both positive Real Numbers .
I saw this in a math olympiad, and within two days i couldn't solve it.
There is a simple case where:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
We can solve it like this:
$$(a-1)^{2}\geq 0$$
$$\Leftrightarrow a^{2}-2a+1 \geq0$$
$$\Leftrightarrow a^{2}+1 \geq2a$$
$$\Leftrightarrow \frac{a^{2}+1}{a} \geq2$$
With the same method:
$$\frac{b^{2}+1}{b} \geq2 $$
And when we add the two inequalities:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
But in case where:
$$\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$$
I can’t solve it,
Thank you for your help.
|
I would start with noticing that $a^{2} + 1 \geq 2\sqrt{a^{2}} = 2a$.
Similarly, we have that $b^{2} + 1 \geq 2b$.
Then we have that
\begin{align*}
\frac{a^{2} + 1}{b} + \frac{b^{2} + 1}{a} \geq 2\left(\frac{a}{b} + \frac{b}{a}\right) \geq 4\sqrt{\frac{a}{b}\times\frac{b}{a}} = 4
\end{align*}
Hopefully this helps!
|
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|
$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$ without using L'Hospital or Taylor's series Find limit without using L'Hospital or Taylor's series:
$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$$
$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = \displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} \cdot \frac{\sin(2x^5)}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = $$
$$= \displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} \cdot \frac{\sin(2x^5)}{x^5} \cdot \frac{x^2}{5^{x^{2}} - 1} \cdot \frac{ x^3}{\tan^3(3x)}$$
*
*$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} = 1$
*$\displaystyle \lim_{x \to 0} \frac{\sin(2x^5)}{x^5} = 2$
*$\displaystyle \lim_{x \to 0} \frac{x^2}{5^{x^{2}} - 1} = \frac{1}{\ln5}$
*$\displaystyle \lim_{x \to 0} \frac{ x^3}{\tan^3(3x)} = \displaystyle \lim_{x \to 0} \cos^3(3x) \cdot \displaystyle \lim_{x \to 0} \frac{x^3}{\sin^3(3x)} = 1 \cdot \left( \frac{1}{3} \right)^3 = \frac{1}{27}$
We get:
$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = 1 * 2 * \frac{1}{\ln5} * \frac{1}{27} = \frac{2}{27 \ln5}$$
All credits to @Bernard. Thank you all, question closed.
|
You can do it using equivalents and some standard limits: rewrite the fraction as
$$\frac{\ln(1+\sin(2x^5))}{\sin(2x^5)}\frac{\sin(2x^5)}{\tan^3(3x)} \cdot \frac{x^2}{(5^{x^{2}} - 1)}\cdot\frac1{x^2}.$$
By substitution, the first factor tends to $1$. Also by substitution, $\:\dfrac{5^{x^2}-1}{x^2}$ tends to the derivative of $5^x$ at $0$, i.e. $\ln 5$.
Last, $\sin(2x^5)\sim_0 2x^5$ and $\tan 3x)\sim_03x$, so $\tan^3(3x)\sim_0(3x)^3=27x^3$, and ultimately we obtain the equivalence
$$\frac{\ln(1+\sin(2x^5))}{\sin(2x^5)}\frac{\sin(2x^5)}{\tan^3(3x)} \cdot \frac{x^2}{(5^{x^{2}} - 1)}\cdot\frac1{x^2}\sim_0\frac{2\not x^5}{27\not x^3\cdot\ln5\cdot \not x^2}=\frac 2{27\ln 5}.$$
|
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|
check if a function is holomorphic Let $f:\mathbb{C} → \mathbb{C}$ be defined as
$$f(z)= \begin{cases}\frac{x^3y(y − ix)}{x^6 +y^2}&\text{ if }z = x + iy= 0\\ 0 &\text{ if }z=x+iy=0\end{cases}$$
And holomorph at $z = 0$
I tried to apply the definition of holomorphic application, which is as follows:
$\displaystyle \lim_{(a, b) \to (0,0)} \frac{f (z_0 + h) - f (z_0)}{ h}$
Defining:
$f(x, y) = (xy^2 /x^4 -1, -x^2 y / x^4 -1)$
$f(x + a, y + b) = ((x + a) (y + b)^2 / (x + a)^4 -1, - (x + a)^2. (y + b) / (x + a)^4 -1)$
$f (z_0) = f (x, y) = (xy^2 / x^4 -1, -x^2 y / x^4 -1)$
$ h = (a, b)$
My question is how to solve this limit ...
My question is to define the limit of homologous functions or Verify if it is differentiable from a real point of view and use the cauchy rieman equations.
Thank you!
|
Set $f=u + iv$, so $u=\frac{xy^2}{x^4-1}$ and $v=\frac{-x^2y}{x^4-1}$.
calculating the partial derivatives we have:
$$\frac{\partial u}{\partial x} = \frac{y^2(x^4-1)-xy^2(4x^3)}{(x^4-1)^2}$$
And
$$\frac{\partial v}{\partial y}= \frac{-2xy(x^4-1)-(-x^2y)(4x^3)}{(x^4-1)^2}$$
as the first equation is already different, it follows that f is not holomofa at 0.
|
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|
How do we prove the proposed expression is nonnegative? So the expression which I am interested in is given by
\begin{align*}
f(b,\theta) = \frac{2b-1}{b\sqrt{(1-\theta)(2b-1)^{2} + \theta}} + \frac{\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1}{2b^{2}(\theta - 1)}
\end{align*}
where $b\in(0,1]$, $\theta\in\mathbb{R}_{>0}$ and $\theta\neq 1$.
It is part of my research project and I would like to know a good way to approach it.
Any hint is appreciated.
EDIT
According to WA, the solution is given by $0 < b \leq 1$ when $\theta > 0$.
|
Denote $A = (1-\theta)(2b-1)^2 + \theta$.
Since $(2b-1)^2 \le 1$, we have $A \ge (1-\theta)(2b-1)^2 + \theta (2b-1)^2 = (2b-1)^2$.
We have
\begin{align}
f(b, \theta) &= \frac{2b-1}{b\sqrt{A}} + \frac{\sqrt{A} - 1}{2b^2(\theta - 1)} \\
&= \frac{2b-1}{b\sqrt{A}} + \frac{A - 1}{2b^2(\theta - 1)(\sqrt{A} + 1)}\\
&= \frac{2b-1}{b\sqrt{A}} + \frac{2(1-b)}{b(\sqrt{A} + 1)}\\
&= \frac{2b - 1 + \sqrt{A}}{b\sqrt{A}(\sqrt{A} + 1)}\\
&\ge \frac{2b - 1 + \sqrt{(2b-1)^2}}{b\sqrt{A}(\sqrt{A} + 1)}\\
&\ge 0.
\end{align}
We are done.
|
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|
Evaluate $\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$
How to evaluate:
$$\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$$
Attempt:
$$\int_{0}^{1}\frac{\ln(1 + x + x^2 + \ldots + x^n)}{x} \mathrm dx
= \int_{0}^{1}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\mathrm d x$$
Any hints would be appreciated.
Edit: Testing it with different values of $n$, it seems like the integral evaluates to be $\frac{n \pi^2}{6(n+1)}$
|
I derive under the integral
$$f(n)=\int_{0}^{1}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\,dx$$
$$f'(n)=\int_{0}^{1}\frac{\partial}{\partial n}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\,dx=$$
$$=\int_0^1\frac{x^n \log x}{x^{n+1}-1}\,dx=\frac{\pi ^2}{6 (n+1)^2}$$
$$f(n)=\int\frac{\pi ^2}{6 (n+1)^2}\,dn=C-\frac{\pi ^2}{6 (n+1)}$$
As $$f(1)=\int_0^1 \frac{\log (x)}{x-1} \, dx=\frac{\pi ^2}{6}$$
then $C=\frac{\pi ^2}{6}$ and finally we have
$$f(n)=\frac{\pi ^2}{6}-\frac{\pi ^2}{6 (n+1)}=\frac{\pi ^2 n}{6 (n+1)}$$
|
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|
Calculate integral $\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx$.
Calculate integral $$\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx.$$
My direction: Since this integral can't calculate normally, I tried to use the property following:$$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx.$$
Then, I have
$$I=\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx = \int_0^{\pi/2} \frac{\sin^3x}{\cos^2x + \sin^3x}dx.$$
Therefore
$$2I = \int_0^{\pi/2} \left(\frac{\cos^3x}{\sin^2x + \cos^3x} + \frac{\sin^3x}{\cos^2x + \sin^3x}\right) dx.$$
I stucked here.
|
Rewrite the integral with $t=\tan\frac x2$
$$I=\frac\pi2-\int_0^{\pi/2} \frac{\sin^2x}{\sin^2x + \cos^3x}dx
=\frac\pi2 +8 \int_0^1 \frac{t^2}{t^6-7t^4-t^2-1} dt$$
Note that $t^6-7t^4-t^2-1$ is cubic in $t^2$, with one real root $r=7.159$ (analytically solvable with the Cadano’s formula). Then, factorize
$$ t^6-7t^4-t^2-1 =(t^2-r)[t^4+(r-7)t^2+1/r]$$
and partially-fractionalize the integrand to proceed
\begin{align}
I &= \frac\pi2+ \frac8{2r^3-7r^2+1}\int_0^1 \left( \frac{r^2}{t^2-r}
+\frac{1-r^2t^2}{ t^4+(r-7)t^2+1/r}\right) dt\\
&= \frac\pi2 +\frac8{2r^3-7r^2+1} \left(- r^{3/2}\coth^{-1} \sqrt{r}
-\frac{r^2-\sqrt r}{2\sqrt{ \frac2{\sqrt r}+r-7 }} \cot^{-1} \frac{\frac1{\sqrt r} -1}{\sqrt{ \frac2{\sqrt r}+r-7 }}\\
+ \frac{r^2+\sqrt r}{ 2\sqrt{ \frac2{\sqrt r}-r+7 }}\coth^{-1} \frac{\frac1{\sqrt r} +1}{\sqrt{ \frac2{\sqrt r}-r+7 }}\right)\\
\end{align}
where, as mentioned above
$$r= \frac13\left( 7 + \sqrt[3]{388 +12\sqrt{69}}+ \sqrt[3]{388 -12\sqrt{69}}\right)
$$
|
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|
Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$ Calculate:
$$\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$
The problem with that case is that the roots are in different powers so multiplication in nominator and denominator by conjugate is not an option (at least I think it's not).
|
HINT.-Use the identity $A^3-B^3=(A-B)(A^2+AB+B^2)$ with $A=\sqrt[3]{11-x}$ and $B=x-1$ and multiplying numerator and denominator by $(A^2+AB+B^2)$ you' ll get again the undeterminated form $\dfrac{(4-4)(4+4+4)}{8-8}$. But the new denominator being equal to $11-x-(x-1)^3$ you have its derivative is equal to $-13\ne0$ so the undetermination disappear after apply Hospital's rule.
|
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|
write five digits numbers using $0,0,0,1,1,2,2,2,2,3,3,3,4,4$. It is given that we want to write five digits numbers using $0,0,0,1,1,2,2,2,2,3,3,3,4,4$.
How many ways are there to write it $?$
My thought In the first place, it seems cumbersome question a little bit , because i think that we should write them separately in terms of the leading term such that firstly i check over $1$ as leading term , then $2$ so on. However , this process is exhaustive and i hope to find elegant approach for this question.
|
Here is a method which still requires a bit of work, but which is still more efficient then naive brute force.
Start with a simpler question:
For each $k=1,2,\dots,5$, how many $k$ digit numbers can you make using $3, 3, 4, 4, 4$?
With a little work, the answers are
k
1
2
3
4
5
# numberss with k digits using 3s and 4s
2
4
7
10
10
Next....
For each $k=1,2,\dots,5$, how many $k$ digit numbers can you make using $2,2,2,2,3,3,4,4,4$?
The key is that we can leverage or work in the previous step to make this one easier. Take the summation of all ways to place the $2$'s, times the number of ways to fill in the rest of the empty spaces with $3$'s and $4$'s, the latter of which can be looked up in the previous table. For example,
\begin{align}
\text{# $5$-digit nums with $2$'s, $3$'s and $4$'s}
&= \binom{5}0\cdot \hspace{-1cm}
\overbrace{10}^\text{# 5 digit nums with 3,4}\hspace{-.1cm}
+ \binom{5}1\cdot \hspace{-1cm}\overbrace{10}^\text{# 4 digit nums with 3,4}\hspace{-1cm}
\\
&+ \binom{5}2\cdot \hspace{-1cm}\underbrace{7}_{\substack{\;\\\\\text{# 3 digit nums with 3,4}}}\hspace{-.1cm}
+ \binom{5}3\cdot \hspace{-1cm}\underbrace{4}_\text{# 2 digit nums with 3,4}\hspace{-.1cm}
\\&+ \binom{5}4\cdot \hspace{-1cm}\underbrace{2}_{\substack{\;\\\\\text{# 1 digit nums with 3,4}}}\hspace{-1cm}
\hspace{1.5cm}=\boxed{180}
\end{align}
You then have to do the same for the $1,2,\dots,4$ digit numbers, giving another row to the table. Then, use that row to calculate the number of numbers where you can use digits $1$ to $4$.
k
0
1
2
3
4
5
# numbers with k digits using 3s and 4s
1
2
4
7
10
10
# numbers with k digits using 2, 3, and 4
1
3
9
26
71
180
# numbers with k digits using 1, 2, 3 and 4
1
4
16
62
229
795
Finally, the number of five digit numbers using all available digits is found by summing over all ways to place the zeroes, times the number of ways to fill what remains with $1$ to $4$, which is in the last row of the table:
$$
\binom{4}0\cdot 795+\binom{4}1\cdot 229+\binom{4}2\cdot 62+\binom{4}3\cdot 16=\boxed{2147}
$$
|
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|
Binomial distribution closed by linear operations? I understand that normal distributions are closed by linear operations, but is binomial distribution as well?
In other words, if $X$ is a binomial random variable, does $aX + b$ follow a binomial distribution for any real number $a$ and $b$?
|
Is Y=aX+b binomial for all real a,b? No, if a=2 then the support of Y is 0,2,4,...,2n which is not binomial.
Do the values of Y follow "binomial probabilities"? Yes. If $X\sim Bin(n,p), Y=aX+b$, then the support of Y is $aX+b, X=0,1,...,n$ and the pmf is $f(Y)={n\choose \frac{y-b}{a}}p^{\frac{y-b}{a}}(1-p)^{1-\frac{y-b}{a}}$.
Example
$X\sim Bin(3,.4)$, $Y=2X+1$
$X\begin{cases}0&.6^3\\
1&{3\choose 1}.4.6^2\\
2&{3\choose 2}.4^2.6\\
3&.4^3\end{cases}$
$Y\begin{cases}1&.6^3\\
3&{3\choose 1}.4.6^2\\
5&{3\choose 2}.4^2.6\\
7&.4^3\end{cases}$
|
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|
show $(1+\frac{x}{n})^n \leq e^x$ I want to show that $(1+\frac{x}{n})^n \leq e^x$ for all $x \geq -n$, $n\in \mathbb{N}$.
I already showed that positive case ($x>0$) and zero case $(x=0)$.
Not sure how to approach the negative case, because in the case $x>0$ I used that $x^n$ is an increasing function on $(0, \infty)$. But I can't do the same when $x$ is negative.
|
Assuming you know that
\begin{align}
\lim_{n\rightarrow \infty}\left( 1+ \frac{x}{n}\right)^n = e^x,
\end{align}
then it suffices to prove that
\begin{align}
\left( 1+ \frac{x}{n}\right)^n \le \left( 1+ \frac{x}{n+1}\right)^{n+1}
\end{align}
for all $n \in \mathbb{N}$.
By Bernoulli's inequality, we have that
\begin{align}
\frac{\left( 1+ \frac{x}{n+1}\right)^{n+1}}{\left( 1+ \frac{x}{n}\right)^{n+1}} =&\ \left( 1+\frac{-\frac{x}{n(n+1)}}{1+\frac{x}{n}}\right)^{n+1} = \left( 1-\frac{x}{n(n+1)+(n+1)x}\right)^{n+1} \\
\ge&\ 1-(n+1)\left(\frac{x}{n(n+1)+(n+1)x} \right) = 1-\frac{x}{n+x} = \frac{1}{1+\frac{x}{n}}.
\end{align}
Multiplying both side by $(1+x/n)^{n+1}$ yields the desired inequality.
Remark: This is a standard proof given in a first course of analysis without assuming any knowledge of derivatives.
|
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|
How to compute $1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$ How to compute $S = 1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$
Was thinking $\frac{S}{24} = {4\choose 4} + {6\choose 4} + {8\choose 4} + ... + {100\choose 4}$ but how do I sum this up?
|
General term is:
$$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$$
Now we give values 1, 3,... n :
$1\times2\times3\times4=1^4+6\times1^3+11\times 1^2+6\times 1$
$3\times4\times5\times6=3^4+6\times3^3+11\times 3^2+6\times 3$
.
.
.
$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$
Now we sum up:
$A=S_4+6S_3+11S_2+6S_1$
We use well known formulas for $S_n$ for odd numbers, for example:
$S_3=1^3+3^3+ . . . +n^3=n^2(2n^2-1)$
Here $n= 96$, we plug this in and get the sum.
|
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|
Find the total length of two line segments of two overlapping right triangles
Find the length of $AE+EB$ ?
(A) $\frac{128}{7}$
(B) $\frac{112}{7}$
(C) $\frac{100}{7}$
(D) $\frac{96}{7}$
(E) $\frac{56}{7}$
My solution:
For $\Delta AEB$ :
$\angle BAC = sin^{-1}(\frac{5}{13}) = 22.62^{\circ}$
$\angle ABD = sin^{-1}(\frac{9}{15}) = 36.87^{\circ}$
$\angle AEB = 180 - \angle BAC - \angle ABD = 120.52^{\circ}$
Use Law of Sines:
$\frac{AE}{sin(\angle ABD)} = \frac{AB}{sin(\angle AEB)}$
$AE = \frac{AB}{sin(\angle AEB)} \times sin(\angle ABD)$
$AE = \frac{12 \times \frac{9}{15}}{sin 120.51} = 8,372$
$\frac{EB}{sin(\angle BAC)} = \frac{AB}{sin(\angle AEB)}$
$EB = \frac{AB}{sin(\angle AEB)} \times sin(\angle BAC)$
$EB = \frac{12 \times \frac{5}{13}}{sin 120.51} = 5.367$
$AE+EB = 8.372 + 5.367 = 13,7387 \approx \frac{96}{7}$ Answer: (D)
My question:
Is there a solution without having to compute both arcsin $\angle BAC$ & $\angle ABD$? The reason I'm asking is because the choices all have 7 as denominators. So I'm guessing there may be a solution that contains only integers.
|
Yes there is, drop a perpendicular down from $E$ to $AB$ and denote the point as $X$. Let $EX = x$ and $AX = 12-y$ and $BX = y$. Since $\triangle EBX $ ~ $\triangle DBA$, we have that $\frac9x = \frac{12}y$. Also, since we have $\triangle CBA$ ~ $\triangle EXA$, we have that $\frac5x = \frac{12}{12-y}$. Now, solving for $x,y$ using our equations, we get $x=\frac{45}{14}, y = \frac{30}{7}$. Plugging in these values, we get $AX = \frac{54}{7}$ and $BX = \frac{30}{7}$ and $EX = \frac{45}{14}$. Now using the pythagorean theorem, we get $AE = \frac{117}{14}$ and $BE=\frac{75}{14}$. Now, adding these lengths, we get that the sum is $\frac{96}{7}$
|
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How to simply this radical expression $\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$ $$\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$$
I could not multiply by the conjugate since it is a cube root. Can you show me a way to simplify it?
Thanks!
|
Cubing both sides, we wish to prove:
$$\sqrt[3]{2}-1 = \bigg(\frac{\sqrt[3]{3}}{1 + \sqrt[3]{2}}\bigg)^{3}$$
We have:
$$\bigg(\frac{\sqrt[3]{3}}{1 + \sqrt[3]{2}}\bigg)^{3} = \frac{3}{1 + 3\sqrt[3]{2} + 3\sqrt[3]{2}^{2} + 2}=\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\frac{1}{1+1\cdot\sqrt[3]{2} + \sqrt[3]{2}^{2}}\cdot\frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}$$
Because $(a - b)(a^{2} + ab + b^{2}) = a^{3} - b^{3}$, this simplifies to:
$$=\frac{\sqrt[3]{2}-1}{\sqrt[3]{2}^{3} - 1} = \frac{\sqrt[3]{2}-1}{2-1}=\sqrt[3]{2}-1$$
Thus, this identity is true. $\blacksquare$
|
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|
Evaluate $\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$ Evaluate
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$$
My attempt : I put $t= \sqrt{1-r^2}$ now $dt/dr= \frac{-r}{2\sqrt {1-r^2}}$ $$\implies dr=\frac{2\sqrt {1-r^2}}{r}dt$$
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}\frac{2\sqrt {1-r^2}}{r}dt$$
$$= 2\int_{0}^{1} r^2 dt$$
$t= \sqrt{1-r^2}\implies r^2= t^2-1$
$$2\int_{0}^{1} t^2-1 dt= 2[\frac{t^3}{3} -t]_{0}^{1}==-4/3$$
|
$$I=\int_{0}^{1} \frac{r^3 dr}{\sqrt{1-r^2}}$$
Let $r=\sin t \implies dr= \cos t dt$, then
$$I=\int_{0}^{\pi/2} \sin^3 t dt=\int_{0}^{\pi/2} \sin t(1-\cos^2 t) dt$$
$$I=\int_{0}^{\pi/2} (\sin t -\cos^2 t \sin t) dt$$
$$ I=-\cos t+\frac{\cos^3 t}{3}|_{0}^{\pi/2}=-(-1+1/3)=2/3.$$
|
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|
Step by step method for $\int\frac{x^2+1}{x^2-1}\,dx$ $$\int\frac{x^2+1}{x^2-1}\,dx$$
What is the step-by-step way of solving this integral problem?
I tried using substitution which was $x^2-1=t^2$, but end up with an even more complicated equation.
Substituting trigonometric functions for example :$x^2=\sec^2(a)$ so that $x^2-1$ part becomes $\tan^2(a)$ did not help either.
|
This can be evaluated using partial fraction decomposition. Observe how
$$\begin{align*}\frac {x^2+1}{x^2-1} & =\frac {x^2-1+2}{x^2-1}\\ & =1+\frac 2{x^2-1}\end{align*}$$
The final fraction can be broken down by assuming the partial fractions follow the form
$$\frac 2{(x+1)(x-1)}=\frac A{x+1}+\frac B{x-1}$$
Multiplying both sides by $(x+1)(x-1)$ and setting $x=\pm1$ gives $A=-B=-1$. Or, in other words,
$$\frac {x^2+1}{x^2-1}=1+\frac 1{x-1}-\frac 1{x+1}$$
Integrate piece by piece to get
$$\int\mathrm dx\,\frac {x^2+1}{x^2-1}\color{blue}{=x+\log(x-1)-\log(x+1)+C}$$
|
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|
given $a+b+c=3$ prove that $abc(a^2+b^2+c^2) \leq 3$ I am preparing for inmo and I came accross this problem while solving a worksheet, but couldn't solve it, pl help me...
Problem-
Prove that if a,b,c are non negative real numbers such that a+b+c=3, then $abc(a^2+b^2+c^2)\leq3$ ... (0)
Developments-
Firstly applying am-gm on $a+b+c$ to get $abc\leq1$
Then applying quadratic mean arthematic mean to get
$a^2+b^2+c^2 \ge 3$
Then writing the expansion of $(a+b+c)^2$ we get
$a^2+b^2+c^2 = 9-2(ab+bc+ca)$ ...(I)
Thn applying am-hm on $ab+bc+ca$ to get
$ab+bc+ca \ge 3abc$
Then substituting in (I) we get
$a^2+b^2+c^2 \leq 9-6(abc)$
Then we substituting in (0)
$LHS \leq abc(9-6abc)$
Whose max value is $27/8$ but is not less than three, so I am stuck for this point onwards
|
Assume $a=\min\{a,b,c\} \rightarrow 0\le a \le 1.$
By AM-GM, we have $$abc\left(a^2+b^2+c^2\right) =\dfrac{1}{3} a\cdot
3bc\cdot\left(a^2+b^2+c^2\right)$$
$$\le \dfrac{1}{12} a \left[a^2+bc+\left(b+c\right)^2\right]^2\le \dfrac{1}{12} a \left[a^2+\dfrac{\left(3-a\right)^2}{4}+\left(3-a\right)^2\right]^2\le 3,$$
so it suffices to prove
$$\left( 9\,{a}^{2}-33\,a+64 \right) \left(1-a \right) ^{3}\ge 0,$$
which is obvious since $0\le a\le 1.$
Done!
|
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Simplification - Pre Algebra I have $$\frac{x^4+2x}{x^4}$$ which becomes $$\frac{x^3+2}{x^3}$$ after pulling out the common factor. I wanted to know why you you don't simplify $\frac{x^4+2x}{x^4}$ to $2x$ and the two $x^4$ cancel each other out, or is this because you must apply the denominator fraction to all numerator parts? So $\frac{x^4}{x^4}$ and $\frac{2x}{x^4}$ which doesn't work.
|
You can either do
$$\frac{x^4+2x}{x^4} = \frac{x(x^3+2)}{x(x^3)} = \frac{x^3+2}{x^3}$$
and then split the fraction as
$$\frac{x^3+2}{x^3} = \frac{x^3}{x^3} + \frac{2}{x^3} = 1 + \frac{2}{x^3}$$
Alternatively, split up first, and cancel common factors:
$$\frac{x^4+2x}{x^4} = \frac{x^4}{x^4} + \frac{2x}{x^4} = 1 + \frac{2}{x^3}$$
Either way works, though the second is quicker and a bit more straightforward.
|
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How do I find a number $\frac{a^4}{4}$ to be added to an odd integer $N$ to make it a Perfect Square? Finding the least number to be added to an integer $N$ to make it a Perfect Square is simple:
https://www.geeksforgeeks.org/least-number-to-be-added-to-or-subtracted-from-n-to-make-it-a-perfect-square/
But how do we proceed if we require the number to be added must be a fourth power divided by four $\frac{a^4}{4}$?
Assume $N=p_1\cdot p_2$ is a product of two primes.
How do I find a number $a$ such that $\frac{a^4}{4}+N$ is a perfect square?
An example $N=p\cdot q=7\cdot3=21$ and we find $a=2$ which produces a perfect square $21+\frac{2^4}{4}=25$ too.
Another example $N=p\cdot q=7\cdot11=77$ and we find $a=2$ which produces a Perfect Square $77+\frac{2^4}{4}=81$.
An example including larger numbers is $N=p\cdot q=5\cdot13=65$ and we find $a=8$ which produces a perfect square $65+\frac{8^4}{4}=1089$.
For a given $N$, is there a way to find such an integer $a$ (without brute force)?
|
Given $n=pq$ with $p$ and $q$ prime and $p\leq q$, if $a$ is an integer such that
$$n+\frac{a^4}{4}=c^2,$$
for some positive integer $c$, then $\tfrac{a^4}{4}$ is an integer and so $a$ is even, say $a=2b$. Then also
$$pq=n=c^2-\tfrac{a^4}{4}=c^2-4b^4=(c-2b^2)(c+2b^2).$$
Then either $c-2b^2=1$ and $c+2b^2=pq$, or $c-2b^2=p$ and $c+2b^2=q$, and correspondingly
$$(2b)^2=pq-1\qquad\text{ or }\qquad (2b)^2=q-p,$$
where of course $q-p$ is the smaller of the two. So
*
*If $q-p$ is a perfect square, then $a=\tfrac12\sqrt{q-p}$.
*Else, if $pq-1$ is a perfect quare, then $a=\tfrac12\sqrt{pq-1}$.
*Else there is no such $a$.
|
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Find all $m,n \in \mathbb N$ Such that $3(2^m)+4=n^2$.
Find All $n,m\in \mathbb N$ Such that $3(2^m)+4=n^2$
I’ve tried to plug some values of $m$, and it turns up that the only valid values is $$m\in \{2,5,6\}.$$
So once i saw that i tried to prove that it doesn’t exist a solution to the equation $\forall m\geq7$. But i have no idea how to prove this.
Something i’m not sure whether it’s true:
$$3(2^m)+4=n^2 \iff3(2^m+1)=n^2-1 \\ \iff 3(2^m+1)=(n+1)(n-1)$$
Thus $3\mid (n+1)(n-1) $, But since $3$ is a prime and $\gcd(n+1,n-1)=1$ or $2 \implies$
$3\mid n+1$ or $3\mid n-1$ What’s next?
|
In respect of the stated comment, I add our second case (and so the whole solution) I left for OP to solve.
First, check $m=1$.
If $m≥2, m-2=k$ and $n=2n_1$, then
$$3×2^{k}+1=n_1^2, n_1≥2$$
Case $-1.$
$n_1=3a+1$, where $a≥1.$
$$3×2^k=9a^2+6a$$
$$2^k=3a^2+2a$$
$$a≥1 \Longrightarrow \begin{cases} a=2b \\ k≥3 \end{cases}$$
$$2^k=12b^2+4b$$
$$2^{k-2}=3b^2+b=b(3b+1)$$
$$b=2^x, 3b+1=2^y$$
If $x≥1$, then we get $3b+1≥7$ and $3b+1$ is an odd number, which gives a contradiction.
So, we deduce that $x=0.$
Then, we can check $n_1=3a+2$ by the same way.
Of course, $n_1=3a$ is impossible.
The method of looking at our second case is the same as the first one, as I mentioned earlier.
Case $-2.$
$n_1=3a+2$, where $a≥0.$
First, check $a=0$. If, $a≥1$ then
$$3×2^k+1=9a^2+12a+4$$
$$2^k=3a^2+4a+1, a≥1$$
$$a≥1 \Longrightarrow \begin{cases} a=2b-1 \\ k≥3 \end{cases}$$
We have,
$$2^k=3(2b-1)^2+4(2b-1)+1, b≥1$$
$$2^k=12b^2-4b$$
$$2^{k-2}=3b^2-b=b(3b-1)$$
$$b=2^x, 3b-1=2^y$$
If $x≥1$, then we get $3b-1≥5$ and $3b-1$ is an odd number, which gives a contradiction.
So, we deduce that $x=0.$
Finally, backward all the steps and you will find all required integer values of $m$ and $n.$
|
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For real $a$, $b$, $c$ all greater than $1$, show $\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a} \;\ge\; \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ An inequality question from an olympiad book:
If $a,b,c$ are real numbers satisfying: $a>1$, $b>1$ and $c>1$, then prove that:
$$\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a} \;\ge\; \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$
I tried taking $LCM$, but it doesn't seem to work.
I don't know how to proceed at all. Just a little hint would be very helpful.
|
Note that $$0=\ln\dfrac{a^a}{b^b}+\ln\dfrac{b^b}{c^c}+\ln\dfrac{c^c}{a^a}=\ln\dfrac{a}{b}+\ln\dfrac{b}{c}+\ln\dfrac{c}{a}$$
Let (WLOG) $\;\ln\dfrac{a}{b}\ge\ln\dfrac{b}{c}\ge\ln\dfrac{c}{a}$
We have $\;\dfrac{a}{b}\ge\dfrac{b}{c}\ge\dfrac{c}{a}$
Thus, $\;ac\ge b^2\;$, $a^2\ge bc\;$ and $\;ab\ge c^2\;$
Combining these $3$ inequalities gives us $a\ge b$ and $a\ge c$
Observe that (easy to show) $$\ln\dfrac{a^a}{b^b}\ge \ln\dfrac{a}{b}$$
$$\ln\dfrac{a^a}{b^b}+\ln\dfrac{b^b}{c^c}\ge \ln\dfrac{a}{b}+\ln\dfrac{b}{c}$$
Then we have
$$\left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right) \prec \left(\ln\dfrac{a^a}{b^b}, \ln\dfrac{b^b}{c^c}, \ln\dfrac{c^c}{a^a}\right)$$
or (actually only former majorization holds but showing it requires more work, so, including the possibility is better for the sake of shortness)
$$\left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right) \prec \left(\ln\dfrac{b^b}{c^c}, \ln\dfrac{a^a}{b^b}, \ln\dfrac{c^c}{a^a}\right)$$
Karamata(Majorization) Inequality with using $f(x)=e^x$ ($f''(x)=e^x>0$) gives the desired inequality
$$\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a} \;\ge\; \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$
|
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help solving a geometric sequence after manipulation Evaluate the sum. $\sum_{n=2}^\infty \frac{n(-3)^n}{4^{n+1}}$
I have manipulated it to be in a more common form I thought I could solve but I'm still having trouble getting the correct sum. I'm using wolfram to check against. It says the sum evaluates to $\frac{99}{784}=0.12628$
Here is what I have so far.
$\sum_{n=2}^{\infty}\frac{n(-3)^{n}}{4^{n+1}}$
$=\sum_{n=2}^{\infty}\frac{n(-3)^{n}}{4^{n}\cdot4}$
$=\sum_{n=2}^{\infty}\frac{n(-3)^{n}}{4^{n}}\ \cdot\frac{1}{4}$
$=\sum_{n=2}^{\infty}n\left(-\frac{3}{4}\right)^{n}\ \cdot\frac{1}{4}$
$=\frac{1}{4}\sum_{n=2}^{\infty}n\left(-\frac{3}{4}\right)^{n}$
was trying to get the common difference to solve that way but it's not coming out right. How do I properly evaluate the remaining sum?
|
$1+x+x^2+x^3+...=\frac{1}{1-x} \iff \sum_{n\geq0}x^n=\frac{1}{1-x}\Rightarrow \sum_{n\geq1}nx^{n-1}=\frac{1}{(1-x)^2}$
$\sum_{n\geq2}n\Big(-\frac{3}{4}\Big)^{n}=-\frac{3}{4}\sum_{n\geq2}n\Big(-\frac{3}{4}\Big)^{n-1}=-\frac{3}{4} \Big( \frac{1}{(1+\frac{3}{4})^2}-1\Big)$
|
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|
Prove by induction: $9\mid (4^{3n} + 8)$ for all integers $n \geq 0$. I cannot seem to figure this one out.
My though process so far is the following:
$\textit{Proof}$: For the base case n = 0, we have $9 \mid (4^0 + 8)$. This is true.
$\textit{Inductive step:}$ Suppose for $k \geq 0$ we have that $9\mid (4^{3k} + 8)$. We will show that $9 \mid (4^{3(k + 1)} + 8)$. The statement $9 \mid (4^{3k} + 8)$ can be expressed as $4^{3k} + 8 = 9a$ for integer $a$. Observe that
$$\begin{equation}\begin{split} 4^{3(k + 1)} + 8 &= 4^{3k}4^3 + 8 \\ &=
\dots \\ &\vdots \\ &= 9 (z)\end{split}\end{equation}$$
Where $z$ is some integer. I feel I am forgetting some algebraic properties or a manipulation of the equation above of some sort to help in this prove.
Any help is great! Thanks!
|
Well, we need to use $4^{3k} +8$ is divisible by $9$.
SO we have somehow isolate $4^{3k} + 8$ from $4^{3k}4^3 + 8$.
Only way I see to do that is $4^{3k}4^3 + 8= (4^3 - 1)4^{3k} + \color{blue}{(4^{3k} + 8)}$
Now we know $9$ divides $\color{blue}{(4^{3k} + 8)}$ so it suffices to show $9$ divides $(4^3-1)4^{3k}$ as well.
As the only prime factors of $4^{3k}$ is $2$ at suffices to show $9$ divides $4^3-1$.
And that's a single constant value.
Oh... come on! Did you really need to peek at the spoiler to figure out how to prove $9$ divides into $4^3 -1$?
|
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|
Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$ Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
I attempted to solve this question as follows:
$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
$a^2(b+c)+b^2(c+a)+c^2(a+b)-(a^3+b^3+c^3)\le 3abc$
$(a+b)(b+c)(c+a)-2abc-3abc\le 3abc$
$(a+b)(b+c)(c+a)-5abc\le 3abc$
$(a+b)(b+c)(c+a)\le 8abc$
And I got stuck here. Could you please explain to me how to finish it off?
|
$a=y+z,b=x+z,c=x+y$ then we have to prove $$\sum 2z{(y+z)}^2\le 3(x+y)(y+z)(z+x) \tag 1$$ Now note that $$\sum (2x{(z+y)}^2-8xyz)=\sum 2x{(y-z)}^2$$ and $$(x+y)(y+z)(x+z)-8zxy=\sum z{(x-y)}^2$$ therfore after subtracting $24xyz$ from both sides of $(1)$ it remains to rpove $$\sum z{(x-y)}^2\ge 0$$ which is true
Note that one could actually multiply out inequality $(1)$ to get $$x^2(y+z)+y^2(z+x)+z^2(x+y)\ge 6xyz$$ which is true by AM_GM but i felt writing it as an SOS is more nice
|
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|
Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $ How to determine the range of the function
$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$
It it is straightforward to verify that $f(x)$ is even and
$$f(0)= \frac13, \>\>\>\>\> \lim_{x\to\pm \infty} f(x) \to -\infty$$
which implies $f(x) \in (-\infty,\frac13]$, i.e.
$$\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le \frac13
$$
and is visually confirmed below
However, it is not obvious algebraically that $f(x)$ monotonically decreases away from $x=0$. The standard derivative tests are not viable due to their rather complicated functional forms.
So, the question is how to prove the inequality $f(x) \le \frac13$ with rigor.
Note that it is equivalent to proving
$$\cot \left(\frac{\pi(1+x)}{3+x^2}\right)
\cot \left(\frac{\pi(1-x)}{3+x^2}\right)\le \frac13
$$
|
$$g(x)=\frac{(1+x)^2}{3+x^2}=1+2\frac{x-1}{3+x^2}$$
$$\text{$g$ is continuous for all real numbers and has a derivative} \\ \text{therefore we can find the max and min} \\ g'=2\frac{3-x^2+2x}{(3+x^2)^2}; \quad g'=0\Rightarrow (x-3)(x+1)=0 \\ \text{After that it's obvious that the range of $g$ is } [f(-1),f(3)]=[0,\frac{4}{3}] $$
$$\text{At $g(2)=1$ the $\tan$ function inside $f(g)$ for $\frac{\pi}{2}$ isn't defined}$$
$$f(g)=\tan\Big(g(x)\frac{\pi}{2}\Big)\cdot \tan\Big(g(-x)\frac{\pi}{2}\Big)$$
$$g\cdot\frac{\pi}{2} :\mathbb{R} \rightarrow[0,\frac{2}{3}\pi]$$
$$\text{And also, at }x\geq0,\;g(x)\cdot \frac{\pi}{2}\in[\frac{\pi}{6},\frac{2}{3}\pi] \\ \hspace{135px} g(-x)\cdot\frac{\pi}{2}\in[0,\frac{\pi}{2})$$
$$\text{And since when both $tan$ parts of the function when $g(x \geq 0)\cdot \frac{\pi}{2}\in (\frac{\pi}{2}, \frac{2}{3}\pi]$} \\ \text{are multiplied, we get a negative result,} \\ \text{Therefore we should check for $\forall x\;\;g(x\geq 0)\cdot \frac{\pi}{2}\in[\frac{\pi}{6}, \frac{\pi}{2})$}$$
$$\text{From here it should be simpler to continue, what you should end up} \\ \text{with is that at $\frac{\pi}{6}$ the $\tan$ functions take their maximal value when} \\ \text{$g(x)\frac{\pi}{2} = \frac{\pi}{6} \;$and multiplied together, the result will be $\frac{1}{3}$}$$
|
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|
Factoring $x^4 + 12x^3 + 46x^2 + 59x + 18$
How do I factorize the following?
$$x^4 + 12x^3 + 46x^2 + 59x + 18$$
I've tried looking for a root by trial and error to no avail.
The answer is
$$(x^2 + 5x + 2)(x^2 + 7x + 9)$$
|
The first step is check for linear factors. By the rational roots theorem, the only possible rational roots are $\pm 1, \pm 3, \pm 9, \pm 2, \pm 6, \pm 18$. Once you have eliminated linear factors, the only possible factorization is as two quadratics.
With that out of the way, a good general technique to look for factorizations is to reduce mod n for various n. If $f(x)=p(x)q(x)$, then reducing mod n yields a factorization for $f(x) \pmod n$. If $n$ is bigger than any of the coefficients of $p, q,$ or $f$, then we can read off the actual factorization from the modular factorization.
To figure out what primes we might attempt to reduce modulo, let us compute $f(k)$ for small values of $k$, as a factor here will yield that $k$ is a root mod that factor.
Working mod 2, we have $f(x)=x^4+x=x(x+1)(x^2+x+1)$, the last factor being irreducible mod 2. Working mod 3, $f(x)=x^4+x^2-x=x(x+1)(x^2-x-1)$, the last term also being irreducible mod 3. Computing $f(k)$ for small values of $k$ and factoring to find primes such that $f(k)$ will have at least one linear factor, we see that $(x+8)$ will be a factor mod 13 and 17, and $(x-3)$ will b a factor mod 13. We get $f(x)\equiv (x-3)^3(x+8) \pmod{13}$ and $f(x)\equiv (x+3)(x-1)(x-6)^2 \pmod{17}$.
Thus, either $p(x)\equiv (x-3)^2 \pmod{13}$ and $q(x)\equiv (x-3)(x+8) \pmod{13}$ or vice versa. Similarly, $p(x)\equiv (x-6)^2,(x-6)(x-1),(x-6)(x+3)$, or $(x-1)(x+3)\pmod{17}.$
Just using the mod 13 and 17 information is enough to find the factors mod 221, which is enough to specify the factors entirely. The mod 2 and 3 information serves as a good double-check. There are still a few combinations to check, but it makes the problem tractable.
The advantage of this method is that when a factorization does NOT exist, this can potentially show that too, by narrowing down what the shape of the factorization would have to be, and getting contradictory information. For example, if you had $f(x)=x^5+15x^2+3$, then mod 3, this would either factor as $x(x^4)$ or $x^2(x^3)$, and since the constant term is not divisible by $9$, neither of these can happen. This particular case can be extended to Eisenstein's criterion, but the general idea is useful more broadly.
|
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|
Proof for The Limit $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4},\space n\in\Bbb{N}$ I am revising my knowledge on the topic of real analysis by attempting some simple proofs. The question requires for the proof of the Limit for $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4}$ using the definition of convergence: $$\forall\epsilon>0\space\space\exists N\in\Bbb{R}\space s.t\space\space \forall n>N,|s_n-L|<\epsilon$$ Below is my working: $$|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\epsilon$$ $$|\frac{3n^2-n}{4n^2+1}-\frac{3n^2+\frac{3}{4}}{4n^2+1}|=|\frac{-n-\frac{3}{4}}{4n^2+1}|=\frac{|-n-\frac{3}{4}|}{4n^2+1}<\frac{4n}{4n^2}=\frac{1}{n}<\epsilon $$ $$\frac{1}{n}<\epsilon\rightarrow\frac{1}{\epsilon}<n$$ $$\forall n>N=\frac{1}{\epsilon},\space|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\frac{1}{n}<\epsilon$$ By the definition of convergence,$\space\frac{3n^2-n}{4n^2+1}\rightarrow\frac{3}{4}$ as $n\rightarrow\infty.$
Is this correct? Any feedback would be greatly appeciated.
|
Check:
With $n=\dfrac1{\epsilon}$,
$$\left|\frac{\dfrac3{\epsilon^2}-\dfrac1{\epsilon}}{\dfrac4{\epsilon^2}+1}-\frac{3}{4}\right|=\left|\frac{3-\epsilon}{4+\epsilon^2}-\frac{3}{4}\right|=\left|\frac{4+3\epsilon}{4(4+\epsilon^2)}\right|\epsilon<\left|\frac{4+3\epsilon}{16}\right|\epsilon<\epsilon.$$
|
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|
Does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists? Let $k>2$, is a natural number, does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists? If $k \to \infty$, then I think no, because $\ln k$ will go to infinite. And $\sum \frac{1}{n}$ diverge too, and so it's $\infty - \infty$. But someone say the limit exists? and it's positive. How?
|
Note that
$$
- \log k = - \log \left( {\frac{2}{1} \cdots \frac{{k - 1}}{{k - 2}}\frac{k}{{k - 1}}} \right) = - \sum\limits_{n = 2}^k {\log \left( {\frac{n}{{n - 1}}} \right)} = \sum\limits_{n = 2}^k {\log \left( {1 - \frac{1}{n}} \right)} .
$$
Thus,
$$
\sum\limits_{n = 1}^k {\frac{1}{n}} - \log k = 1+\sum\limits_{n = 2}^k {\left( {\frac{1}{n} + \log \left( {1 - \frac{1}{n}} \right)} \right)} .
$$
Now if $0<x\leq \frac{1}{2}$, then
$$
\left| {x + \log (1 - x)} \right| = \left| { - \int_0^x {\frac{t}{{1 - t}}dt} } \right| \le 2\int_0^x {tdt} = x^2 .
$$
Therefore,
$$
\left| {\sum\limits_{n = 2}^k {\left( {\frac{1}{n} - \log \left( {1 - \frac{1}{n}} \right)} \right)} } \right| \le \sum\limits_{n = 2}^k {\left| {\frac{1}{n} - \log \left( {1 - \frac{1}{n}} \right)} \right|} \le \sum\limits_{n = 2}^k {\frac{1}{{n^2 }}} < \sum\limits_{n = 2}^\infty {\frac{1}{{n^2 }}} < \infty .
$$
Consequently, the sequence
$$
1 + \sum\limits_{n = 2}^k {\left( {\frac{1}{n} - \log \left( {1 - \frac{1}{n}} \right)} \right)} = \sum\limits_{n = 1}^k {\frac{1}{n}} - \log k
$$
is convergent.
|
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|
Evaluating $\int_0^{\pi/2} \log(1 - x \cot x) \, dx\;$ (Is there a closed form?) I am interested in knowing if it is possible to find a closed-form solution to the following challenging log-cotangent integral
$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx$$
I very much doubt a closed form in terms of known mathematical constants can be found and have no reason to suspect one exists.
In the absence of a simple closed-form value being found, maybe the integral can be evaluated in terms of an infinite series. In this direction, since $0 < x \cot x < 1$ for all $x \in (0,\frac{\pi}{2})$ the log term appearing in the integrand can be expanded. Doing so produces
$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx = -\sum_{n = 1}^\infty \frac{1}{n} \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx = -\sum_{n = 1}^\infty \frac{a_n}{n},$$
where
$$a_n = \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx. \tag1$$
Noting that
\begin{align*}
a_1 &= \int_0^{\frac{\pi}{2}} x \cot x \, dx = \frac{\pi}{2} \log (2)\\
a_2 &= \int_0^{\frac{\pi}{2}} (x \cot x)^2 \, dx = \pi \log (2) - \frac{\pi^3}{24}\\
a_3 &= \int_0^{\frac{\pi}{2}} (x \cot x)^3 \, dx = \frac{9\pi}{16} \zeta (3) - \frac{\pi^3}{16} + \frac{3\pi}{2} \log (2) - \frac{\pi^3}{8} \log (2),
\end{align*}
perhaps it is possible to find a general expression for $a_n$. Indeed, an attempt at finding (1) can be found here.
Any other approaches or suggestions to this challenging integral would be welcome.
|
Approximating the integral using rational approximants
Using integration by parts, we easily get
$$I = \int_0^{\pi/2} \ln(1 - x\cot x) \mathrm{d} x
= - \frac{\pi^3}{24} + \frac{\pi}{2}\ln 2 - \int_0^{\pi/2} \frac{x^3}{\tan x - x} \mathrm{d} x.$$
$I \approx -3.353337262889472...$
We focus on
$$I_1 = \int_0^{\pi/2} \frac{x^3}{\tan x - x} \mathrm{d} x.$$
We may approximate $I_1$ by rational approximants to $\tan x$ at $0$.
*
*Pade $(5, 4)$ approximant $\tan x \approx \frac{x(x^4 - 105x^2 + 945)}{15x^4 - 420x^2 + 945}
\triangleq g(x)$
We can bound the error by
$$0 \le \frac{x^3}{g(x) - x} - \frac{x^3}{\tan x - x} \le g(\pi/2) \approx 0.000045.$$
We have
$$I_1 \approx \int_0^{\pi/2} \frac{x^3}{g(x) - x} \mathrm{d} x =
- \frac{5}{112}\pi^3 + \frac{165}{56}\pi - \frac{243}{56}\sqrt{10}
\operatorname{arctanh} \frac{\pi\sqrt{10}}{30}.$$
Then we have $I \approx -3.353344689$ with relative error $< 0.000003$.
|
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|
Evaluate this limit $ \lim_{x\to\infty}\left (\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x}$ Please help to evaluate this limit
$$ \lim_{x\to\infty} \left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x},$$
where $0 \leq a$ and $a \not= 1$.
I tried to logarithm from both sides, and apply taylor series but so far without success.
|
Case 1: If $a>1$ then from $\text{exp}^{\large{\log{u(x)}}}=u(x)$ we deduce
\begin{align}
\left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x} &=
\exp\left(\frac{\log\left(\frac{1}{x}\frac{a^x - 1}{a - 1}\right)}x\right)\\&=
\exp\left(\frac{\log\left(\frac 1x\right)+\log\left(\frac{a^x - 1}{a - 1}\right)}x\right)\\&=
\exp\left(\frac{-\log(x)+\log(a^x-1)-\log(a-1)}x\right)\\&=
\exp\left(\frac{-\log(x)+\log\big[a^x\left(1-a^{-x}\right)\big]-\log(a-1)}x\right)\\&=
\exp\left(\frac{-\log (x)+\log(1-a^{-x})-\log(a-1)}x+\log(a)\right)
\end{align}
Sending $x\to \infty$ will make the middle terms vanish since
$$\lim_{x\to\infty}\frac{\log(x)}{x}=0$$
Therefore
$$\lim_{x\to+\infty}\left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x}=\exp\bigl(\log(a)\bigr)=a.$$
Case 2: If $0< a <1$ then we first write
$$\frac{a^x - 1}{a - 1} = \frac{1-a^x}{1-a}$$
where
\begin{align}
\log\left(\frac{1}{x}\frac{1-a^x}{1-a}\right)&=
\log\left(\frac{1}{x}\right)+\log\left(\frac{1- a^x}{1-a}\right)\\&=
-\log(x)+\log\left(1- a^x\right)-\log\left({1-a}\right)
\end{align}
and
$$\lim_{x\to\infty}\frac{-\log(x)+\log\left(1- a^x\right)-\log\left({1-a}\right)}{x}=0$$
thus
$$\lim_{x\to+\infty}\left(\frac{1}{x}\frac{1-a^x}{1-a} \right)^\frac{1}{x}=\exp\bigl(0)=1.$$
|
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|
Algebraic manipulation of equations and determinant $$a_{11}^2 + a_{12}^2 + a_{13}^2 = 1$$
$$\dfrac{a_{11}}{a_{33}a_{22}-a_{23}a_{32}} = \dfrac{a_{12}}{a_{21}a_{33}-a_{31}a_{23}} = \dfrac{a_{13}}{a_{22}a_{31}-a_{21}a_{32}} = k$$
$$k(a_{11}(a_{33}a_{22}-a_{23}a_{32})+a_{12}(a_{21}a_{33}-a_{31}a_{23})+a_{13}(a_{22}a_{31}-a_{21}a_{32})) = k \det
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix} = 1. $$
Does someone know how to obtain the last equation combining the first and second ones?
|
The trick is to express $a_{1i}^2$ as $a_{1i}.k(...)$
$\dfrac{a_{11}}{a_{33}a_{22}-a_{23}a_{32}} = \dfrac{a_{12}}{a_{21}a_{33}-a_{31}a_{23}} = \dfrac{a_{13}}{a_{22}a_{31}-a_{21}a_{32}} = k$
$a_{11}^2=a_{11}k({a_{33}a_{22}-a_{23}a_{32}})$
$a_{12}^2=a_{12}k({a_{21}a_{33}-a_{31}a_{23}})$
$a_{13^2}=a_{13}k({a_{22}a_{31}-a_{21}a_{32}})$
$a_{11}^2 + a_{12}^2 + a_{13}^2 = 1$
$a_{11}^2 + a_{12}^2 + a_{13}^2$
$ = k(a_{11}({a_{33}a_{22}-a_{23}a_{32}})$$+a_{12}({a_{21}a_{33}-a_{31}a_{23}})+a_{13}({a_{22}a_{31}-a_{21}a_{32}}))$
=k $\det\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}$=1
|
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|
Finding the volume of tetrahedron $ABCD$ if $AB=AC=AD=5$, $BC=3$, $CD = 4$, and $BD = 5$ I am trying to solve this question with out using calculus:
Calculate the volume of a tetrahedron $ABCD$ where $AB = AC = AD = 5$ and $BC = 3, CD = 4$, and $BD = 5$.
I managed to find the area of the base triangle which came up to be $3/4$ ($\sqrt{91}$), I am stuck on finding the height of the tetrahedron...
|
I have another analytical geometry explanation for the height of the tetrahedron:
Consider that $C(0,0,0), B(3,0,0), D(0,4,0).$
Let $A(x,y,z)$. The constraints are:
$$\begin{cases}AB^2&=&25\\AC^2&=&25\\AD^2&=&25\end{cases} \ \ \iff \ \ \begin{cases}(x-3)^2+y^2+z^2&=&25& \ \ (1) \\
x^2+(y-4)^2+z^2&=&25& \ \ (2) \\
x^2+y^2+z^2&=&25& \ \ (3)
\end{cases}$$
Subtracting (3) from (1) gives $-6x+9=0$, i.e., $x=\frac32$;
Subtracting (3) from (2) gives $-8y+16=0$, i.e., $y=2$.
Plugging these two values of $x$ and $y$ in (3) gives :
$$\frac{9}{4}+4+z^2=25 \ \ \ \implies \ \ \ z=\frac52 \sqrt{3} \ \tag{4}$$
which is the height of the tetrahedron.
Remark: we could have taken the negative value for $z$ in (4): the result would have been a tetrahedron with apex below the base instead of being above it...
|
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|
$\frac{(x-a)^2}{(x-2a)^2}=\left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert$ leading to contradictory solutions I have the following equation
$$\frac{(x-a)^2}{(x-2a)^2}=\left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert$$
And to find the critical values for the absolute value function I carried out polynomial long division and ended up with
$$1+ \frac{a}{x-2a}, x>2a$$
$$-(1+ \frac{a}{x-2a}), x<2a$$
I first attempted to solve this using the original form of both equations
For the case where $x>2a$:
$$\frac{(x-a)^2}{(x-2a)^2} =\frac{x-a}{x-2a} $$
$$\frac{x-a}{x-2a} =1$$
$$x-a=x-2a$$
However this lead me to the answer
$$-a=-2a$$
Which is self contradictory, I am extremely confused on where I messed up, have I made a problem cross multiplying or is it something else?
|
For the case you describe, you didn't mess anything up: what you found tells you there is no solution for $ \ x > 2a \ $ . (However, your characterization of the rational function after polynomial division is incomplete.)
In your original equation,
$$\frac{(x-a)^2}{(x-2a)^2} \ = \ \left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert \ \ , $$
the rational function within the absolute-value brackets has two "special points": $ \ x = 2a \ $ , where the function is undefined, and $ \ x = a \ $ , where the function equals zero.
The real numbers are divided into three intervals then, $ \ x < a \ , \ a < x < 2a \ , \ \text{and} \ x > 2a \ . $ In the first interval, the numerator and denominator are both negative, while in the third interval, both are positive.
So for these cases, $ \ \left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert \ = \ \frac{x-a}{x-2a} \ , $ making the original equation
$$ \left( \frac{x-a}{x-2a} \right)^2 \ = \ \frac{x-a}{x-2a} \ \ , $$
which can only be true when both sides equal zero or $ \ 1 \ $ . The former is the case, as remarked above, for $ \ x = a \ $ ; in the latter case, the numerator and denominator cannot be equal for any real number. Hence, there are no solutions to the original equation in the intervals $ \ x < a \ \ \text{or} \ \ x > 2a \ . $
In the interval $ \ a < x < 2a \ , $ however, the numerator is positive while the denominator is negative, so we have
$$ \left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert \ = \ -\frac{x-a}{x-2a} \ = \ \frac{x-a}{2a-x} \ \ \rightarrow \ \ \frac{(x-a)^2}{(x-2a)^2} \ = \ \frac{x-a}{2a-x} \ \ . $$
Since $ \ x \ne a \ , \ x \ne \ 2a \ , $ it is "safe" to cross-multiply and divide out factors, leading to
$$ (x-a)^2 · (2a-x) \ = \ (x-a) · (x-2a)^2 \ \ \Rightarrow \ \ (x-a) · ( -1) \ = \ (x-2a) $$
$$ \Rightarrow \ \ (x-2a) \ + \ (x-a) \ = \ 0 \ \ \Rightarrow \ \ 2x - 3a \ = \ 0 \ \ \Rightarrow \ \ x \ = \ \frac32 a \ \ . $$
The original equation thus has two solutions, $ \ x = a \ $ and $ \ x \ = \ \frac32 a \ \ . $
|
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|
How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8,9$ which are divisible by $3$ and $5$? How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8$ and $9$ which are divisible by $3$ and $5$, without any of the digits repeating?
For the number to be divisible by $5$ it must end with $0$ (or $5$, but we don't have it in the given problem). So we have $V_6^4$ numbers which are divisible by $5$ but how can we exclude those which aren't divisible by $3$?
$120$
|
To be divisible by $5$, the five-digit number must end in $0$. There are $360=6\cdot5\cdot4\cdot3$ integers that can be made by arranging four of the available non-zero digits before the zero. These $360$ five-digit numbers might have remainder $0$, $1$, or $2$ when divided by $3$, and since the six available digits are equally distributed modulo $3$ (two of each possibility mod $3$), each of the three possible remainders is equally likely among the $360$ possible five-digit numbers. Therefore exactly one third of them, or $120$, are divisible by $3$.
|
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In which of the intervals is $\sqrt{12}$ In which of the intervals is $\sqrt{12}:$
a) $(2.5;3);$
b) $(3;3.5);$
c) $(3.5;4);$
d)
$(4;4.5)$?
We can use a calculator and find that $\sqrt{12}\approx3.46$ so the correct answer is actually b. How can we think about the problem without a calculator (if on exam for example)? I was able to conclude that $$\sqrt{9}=3<\sqrt{12}<\sqrt{16}=4,\\3<\sqrt{12}<4,\\\sqrt{12}\in\left(3;4\right).$$ How can I further constrict the interval? Thank you in advance!
|
$$\sqrt{12} = 3 + (\sqrt{12} - 3) = 3 + \frac{(\sqrt{12})^2 - 3^2}{\sqrt{12} + 3} = 3+\frac{3}{\sqrt{12}+3}$$
and since $3 < \sqrt{12} < 4$, $\frac{3}{3 + 4} < \frac{3}{\sqrt{12} + 3} < \frac{3}{3 + 3}$. Thus $\sqrt{12} < 3 + \frac{1}{2}$ so $b$ is correct.
|
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|
False proof that $\frac{13}{6}=0$
At start, the length of a line segment is $a_0=0$. When $3$ hours have elapsed since start, its length is $a_3$. When $1$ hour has elapsed since start, its length increased by $\frac{a_3}{2}$ with respect to $a_0$ (call the new length $a_1$). When $2$ hours have elapsed since start, its length increased by $\frac{a_3}{3}$ (call the new length $a_2$) with respect to $a_1$. When $3$ hours have elapsed since start, its length increased by $\frac{a_3}{4}$ with respect to $a_2$. What is the value of $\frac{a_3}{a_1}$?
At $t=0$, the length is $a_0=0$. At $t=1$ hour, the length is $a_1=\frac{a_3}{2}$. At $t=2$ hours, the length is $a_2=\frac{a_3}{2}+\frac{a_3}{3}$. At $t=3$ hours, the length is $\color{red}{a_3=\frac{a_3}{2}+\frac{a_3}{3}+\frac{a_3}{4}=\frac{13}{12}a_3}$. So
$$\frac{a_3}{a_1}=\frac{\frac{13}{12}a_3}{\frac{a_3}{2}}=\frac{13}{6}$$
is the answer. But notice that the red equation enables solving for $a_3$ which gives $a_3=0$. Therefore $\frac{a_3}{a_1}=0$ and, by transitivity, we have:
Conclusion: $\frac{13}{6}=0$
Where is the mistake?
|
The red equation gives $a_3 = 0$. So dividing by $a_1 = \dfrac{a_3}{2}$ is the same as dividing by $0$.
|
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Simplifying $\frac{2\sqrt n + \frac{1}{\sqrt n}-3}{2\sqrt n -1}$ Consider the term $$\dfrac{2\sqrt n + \dfrac{1}{\sqrt n}-3}{2\sqrt n -1} \tag1$$
I simplified it to
$$\dfrac{4 n-4 \sqrt{n}+\dfrac{1}{\sqrt{n}}-1}{4 n-1} \tag2$$
or also
$$\dfrac{2 n+1-3 \sqrt n}{2 n-\sqrt{n}} \tag3$$
but apparently one can simplify this to
$$\dfrac{n-\sqrt n}{n} \tag4$$
but I don't really get to this expression.
How can one arrive at $(4)$?
|
As you have $\frac{1}{\sqrt n}$ in the numerator, first multiply numerator and denominator by $\sqrt n$ and check if it gets simplified.
$$\frac{2\sqrt n +\frac1{\sqrt n} -3}{2\sqrt n -1}=\frac{2n + 1 -3\sqrt n}{\sqrt n(2\sqrt n - 1)}$$
If you let $\sqrt n = x$, the numerator becomes, $2x^2 - 3x +1 = 2x^2-2x-x+1 = (2x-1)(x-1) = \boxed{(2\sqrt n-1)(\sqrt n -1)}$
So,
$$\frac{2n + 1 -3\sqrt n}{\sqrt n(2\sqrt n - 1)} = \frac{(2\sqrt n-1)(\sqrt n -1)}{\sqrt n(2\sqrt n - 1)}=\frac{\sqrt n-1}{\sqrt n} = \frac{n-\sqrt n}{n}$$
|
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Show $(y/2)/\sqrt{1-x^y}$ is bounded by one for $x\in(0,3/4)$ and $y\in(0,1)$ I would like to show that for all $x\in(0,3/4)$ and $y\in(0,1)$ that $f(x,y) = \frac{y}{2\sqrt{1-x^y}}$ is bounded above by one.
My thought is that for any $x$ I can try to show it an increasing function of $y$ on $(0,1)$. Then it would be bounded by $f(x,1) \leq 1$. To do this I computed the derivative wrt $y$ $\frac{df}{dy} (x,y) = \frac{2(1-x^y)+x^y\log(x^y)}{4 (1-x^y)^{3/2}}$, but I'm not sure how to proceed. I feel like there must be some bound for logs which I can apply here to show the derivative is non-negative.
|
I think I might have figured it out:
Consider the function
\begin{align*}
g(x,y) = \frac{y}{2\sqrt{1-x^y}}.
\end{align*}
This function is clearly increasing (thanks @Eric) in $x$ for all $y\in(0,1)$, so it suffices to set $x=3/4$.
Thus, define
\begin{align*}
f(y) = \log \left( \frac{y}{2\sqrt{1-(3/4)^y}} \right)
\end{align*}
which has derivative
\begin{align*}
f'(y) = \frac{1}{y} - \frac{\log(4/3)}{2((4/3)^y-1)}.
\end{align*}
We aim to show $f'(y)$ is non-negative.
Note that $(4/3)^y-1 > \log(4/3)y$ for all $y$ so
\begin{align*}
\frac{\log(4/3)}{2((4/3)^y-1)}
\leq \frac{\log(4/3)}{2 \log(4/3)y}
= \frac{1}{2y}.
\end{align*}
Therefore $f'(y) \geq 1/(2y) \geq 0$ so $f(y)$ is monotonic. This implies that $g(3/4,y)$ is also monotonic, so it is bounded by $g(3/4,1) = 1$
|
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|
Interpreting Riemann sums as integrals Can you interpret any Riemann sum as an infinite number of definite integrals, all giving you the same value when you plug in the limits? For example, this limit of a sum:
$$\lim_{n \to \infty} \frac{3}{n} \left[ 1+ \sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{n+3(n-1)}}\right]$$
Was evaluated in a question as $$\int_0^1(1+3x)^{\frac{-1}{2}}\text{d}x$$
However, by taking out a $\sqrt{3}$ from the denominator of the general expression, i simplified it into $$\sqrt{3}\int_\frac{1}{3}^{\frac{1}{3}+1}\frac{\text{d}x}{\sqrt{x}}$$ and the value is the same, ie., 2.
Is this chance or are multiple possible integrals possible?
|
The integral that you have written does not equal the limit when you use Reimann Sums, I believe your book is incorrect?
The middle integral is:
$$
\int_0^1 \frac{1}{\sqrt{1+3x}}\text{d}x
$$
which means you can't factor out a $\sqrt{3}$. So I'll answer the question using the correct Reimann Sum.
Using the Left point method:
$$
A_\mathrm{left} = \Delta x \left[f\left(a\right) + f\left(a + \Delta x\right) + f\left(a+ 2*\Delta x\right) + \cdots + f\left(b - \Delta x\right)\right]
$$
$n=\#\ \text{steps}$, $\Delta x = \frac{b-a}{n} = \frac{1}{n}$.
We want to write it in terms of steps rather than step size, so:
$$
\begin{align}
A_\mathrm{left} &= \frac{1}{n} \left[f\left(0\right) + f\left(\frac{1}{n}\right) + f\left(2*\frac{1}{n}\right) + \cdots + f\left(1 - \frac{1}{n}\right)\right] \newline
&= \frac{1}{n} \left[1 + \frac{1}{\sqrt{1+3\left(\frac{1}{n}\right)}} + \frac{1}{\sqrt{1+3\left(\frac{2}{n}\right)}} + \cdots + \frac{1}{\sqrt{1+3\left(\frac{n-1}{n}\right)}}\right]
\end{align}
$$
Therefore:
$$
\int_0^1 \frac{1}{\sqrt{1+3x}}\text{d}x = \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \frac{1}{\sqrt{1+3\left(\frac{1}{n}\right)}} + \frac{1}{\sqrt{1+3\left(\frac{2}{n}\right)}} + \cdots + \frac{1}{\sqrt{1+3\left(\frac{n-1}{n}\right)}}\right]
$$
Which is not really in the nicest form, but I hope this helps!
UPDATE:
Ok so I think I misunderstood the question. So the first limit and last integral are equal to $2$, but the middle integral evaluates to $\frac{2}{3}$ which is not the same. I did the working for the middle integral thinking you were trying to determine the Reimann sum for that, but your initial question had a mistake so rather than deleting everything I will leave the working here in case anyone else finds it helpful, and answer the rest of your question.
Middle integral:
$$
\begin{align}
\int_0^1 \frac{1}{\sqrt{1+3x}}\text{d}x &= \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \frac{1}{\sqrt{1+3\left(\frac{1}{n}\right)}} + \frac{1}{\sqrt{1+3\left(\frac{2}{n}\right)}} + \cdots + \frac{1}{\sqrt{1+3\left(\frac{n-1}{n}\right)}}\right] \newline
&= \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \sum_{k=1}^{n-1}\frac{1}{\sqrt{1+3\left(\frac{k}{n}\right)}}\right] \newline
&= \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \sum_{k=1}^{n-1}\frac{n}{\sqrt{n^2+3kn}}\right] \newline
&= \lim_{n \rightarrow \infty} \frac{1}{n} + \lim_{n \rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n-1}\frac{n}{\sqrt{n^2+3kn}} \newline
&= \lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\sum_{k=1}^{n-1}\frac{1}{\sqrt{n+3k}} \newline
&=\frac{2}{3}
\end{align}
$$
First limit:
$$
\lim_{n \rightarrow \infty}\frac{3}{n} \sum_{k=0}^{n-1}\sqrt{\frac{n}{n+3k}} = 2
$$
Last integral:
$$
\sqrt{3}\int_{\frac{1}{3}}^{\frac{1}{3}+1}\frac{1}{\sqrt{x}}\text{d}{x} = 2
$$
So its clear that the middle and last integral are not the same $2\neq\frac{2}{3}$. But back to your question: Yes certain Reimann Sums may come from multiple integrals.
E.g. If the middle integrals integrand was: $3*f\left(x\right)$, then we would have:
$$
\boxed{3}*\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\sum_{k=1}^{n-1}\frac{1}{\sqrt{n+3k}}
$$
Now we use algebra to see if we can get this to the same form as the first limit.
First let the summation index start at $k=0$ (and subtract the $k_0$th term so we don't change the expression):
$$
\begin{align}
&=\lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\left[\left(\sum_{\boxed{k=0}}^{n-1}\frac{1}{\sqrt{n+3k}}\right)-\frac{1}{\sqrt{n}}\right] \newline
&= \lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\sum_{k=0}^{n-1}\frac{1}{\sqrt{n+3k}}-\lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\frac{1}{\sqrt{n}} \newline
&= \lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\sum_{k=0}^{n-1}\frac{1}{\sqrt{n+3k}} \newline
&= \lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\boxed{\frac{\sqrt{n}}{\sqrt{n}}}\sum_{k=0}^{n-1}\frac{1}{\sqrt{n+3k}} \newline
&= \lim_{n \rightarrow \infty} \frac{3}{n}\sum_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n+3k}} \newline
&= \lim_{n \rightarrow \infty} \frac{3}{n}\sum_{k=0}^{n-1}\sqrt{\frac{n}{n+3k}} \newline
&= 2
\end{align}
$$
As you can see just by modifying the middle integral I was able to reach the first limit!
So I'm going to leave this here and hopefully you can see by this example that you can indeed have multiple integrals linked to a Reiman Sum.
|
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How to find reduction formula for $I_n=\int\frac{1}{x^n \sqrt{x^2-1}}dx$ I have tried manipulating the integral but couldn't come up with an answer that matched textbook answer.
I just need a hint. Thanks in advance.
$\int\frac{dx}{x^n \sqrt{x^2-1}}$
This is what I tried:
Let $I_{n-2}=\int\frac{dx}{x^{n-2} \sqrt{x^2-1}}$
Then I integrated by parts.
The answer given in textbook is
$$(n-1)I=\frac{\sqrt{x^2-1}}{x^{n-1}} +(n-2)I_{n-2}$$
|
Reduce the integral as follows
\begin{align}
I_n= &\int\frac{1}{x^n \sqrt{x^2-1}}dx \\
=& \int\frac{(1-x^2)+x^2}{x^n \sqrt{x^2-1}}dx
=-\int \frac{\sqrt{x^2-1}}{x^n }dx + I_{n-2} \\
=&\ \frac1{n-1}\int \sqrt{x^2-1} \>d\left(\frac{1}{x^{n-1} }\right)+ I_{n-2} \\
\overset{ibp}=&\ \frac1{n-1}\frac{\sqrt{x^2-1}}{x^{n-1} }+\frac{n-2}{n-1}I_{n-2}
\end{align}
|
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|
Weird "hidden answer" in $2\tan(2x)=3\text{cot}(x)$ The question is
Find the solutions to the equation $$2\tan(2x)=3\cot(x) , \space 0<x<180$$
I started by applying the tan double angle formula and recipricoal identity for cot
$$2* \frac{2\tan(x)}{1-\tan^2(x)}=\frac{3}{\tan(x)}$$
$$\implies 7\tan^2(x)=3 \therefore x=\tan^{-1}\left(-\sqrt\frac{3}{7} \right)$$
$$x=-33.2,33.2$$
Then by using the quadrants
I was lead to the final solution that $x=33.2,146.8$ however the answer in the book has an additional solution of $x=90$, I understand the reasoning that $\tan(180)=0$ and $\cot(x)$ tends to zero as x tends to 90 however how was this solution found?
Is there a process for consistently finding these "hidden answers"?
|
Factorize the equation as follows
\begin{align}
2\tan(2x)-3\cot(x)
=& \frac{2\sin2x}{\cos 2x} - \frac{3\cos x}{\sin x}\\
=& \frac{2\sin2x\sin x-3 \cos x\cos2x }{ \sin x\cos 2x}\\
=& \frac{\cos x(10\sin^2x-3 )}{ \sin x\cos 2x}\\
\end{align}
where the factor $\cos x =0$ captures the solution $x=\frac\pi2$ and $10\sin^2x -3=0$ yields $x= \sin^{-1}\sqrt{\frac3{10}}, \>\pi - \sin^{-1}\sqrt{\frac3{10}}$.
|
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A biquadratic equation to have at least two real roots Which of the equations have at least two real roots?
\begin{aligned}
x^4-5x^2-36 & = 0 & (1) \\
x^4-13x^2+36 & = 0 & (2) \\
4x^4-10x^2+25 & = 0 & (3)
\end{aligned}
I wasn't able to notice something clever, so I solved each of the equations. The first one has $2$ real roots, the second one $4$ real roots and the last one does not have real roots. I am pretty sure that the idea behind the problem wasn't solving each of the equations. What can we note to help us solve it faster?
|
There is a test that can be made for biquadratics without the need for square-roots that is akin to a test for the existence of real zeroes of a quadratic polynomial (there is an evident connection with the Rule of Signs that Gregory discusses). There, "completing the square" produces
$$ u^2 \ + \ Bu \ + \ C \ \ = \ \ \left(u \ + \ \frac{B}{2} \right)^2 \ + \ \left(C \ - \ \frac{B^2}{4} \right) \ \ = \ \ 0 \ \ \Rightarrow \ \ \left(u \ + \ \frac{B}{2} \right)^2 \ \ = \ \ \frac{B^2}{4} \ - \ C \ \ , $$
which tells us that there are no real zeroes for $ \ u \ $ if $ \ \frac{B^2}{4} - C \ < 0 \ \Rightarrow \ B^2 \ < \ 4C \ \ , $ one if $ \ B^2 \ = \ 4C \ \ , $ and two if $ \ B^2 \ > \ 4C \ \ . $ This is equivalently to saying that a parabola with its vertex at $ \ x \ = \ h \ = \ -\frac{B}{2} \ , \ y \ = \ k \ = \ C \ - \ \frac{B^2}{4} \ $ does not intersect the $ \ x-$axis in this first case, is just tangent to it in the second, and has two $ \ x-$intercepts in the third case.
With the biquadratic $ \ x^4 \ + \ Bx^2 \ + \ C \ \ , $ which substitutes $ \ u = x^2 \ \ , $ now requires $ \ \left(x^2 \ + \ \frac{B}{2} \right)^2 \ > \ 0 \ $ for $ \ B > 0 \ $ and the polynomial cannot have a value less than $ \ \left( \frac{B}{2} \right)^2 + \frac{B^2}{4} - C \ = \ C \ . $ From this, we see that the biquadratic has no real zeroes for $ \ \mathbf{B > 0 \ , \ C > 0} \ \ $ and two real zeroes for $ \ \mathbf{B > 0 \ , \ C < 0} \ \ . $ In terms of factorization, the first of these corresponds to
$$ (x^2 + \alpha^2)·(x^2 + \beta^2) \ \ = \ \ x^4 \ + \ (\alpha^2 + \beta^2)·x^2 \ + \ \alpha^2·\beta^2 \ \ , $$
while the second is one case for
$$ (x^2 + \alpha^2)·(x^2 - \beta^2) \ \ = \ \ x^4 \ + \ (\alpha^2 - \beta^2)·x^2 \ - \ \alpha^2·\beta^2 \ \ , $$
with $ \ |\alpha| > |\beta| \ \ , \ \ \alpha \ $ being imaginary.
Coming to the cases pertinent to your equations, we have $ \ B < 0 \ \ , $ which permits $ \ \left(x^2 \ + \ \frac{B}{2} \right)^2 \ \ge \ 0 \ \ . $ The significance of this change is that the biquadratic curve has just one "turning point" for $ \ B > 0 \ \ , $ but three for $ \ B < 0 \ \ . $ The function $ \ f(x) \ = \ x^4 \ + \ Bx^2 \ + \ C \ \ $ has even symmetry and can be thought of as "fusing" the portion of the parabola $ \ y \ = \ x^2 \ + \ Bx \ + \ C \ $ for $ \ x \ge 0 \ $ with its "reflection" about the $ \ y-$axis. For $ \ B > 0 \ \ , $ the vertex of the parabola is at $ \ x \ = \ -\frac{|B|}{2} \ < \ 0 \ \ , $ so the biquadratic curve only has a turning point at its $ \ y-$intercept $ \ (0 \ , \ C ) \ \ . $ However, for $ \ B < 0 \ \ , $ the vertex lies at $ \ x \ = \ \frac{|B|}{2} \ > \ 0 \ \ , $ leading to the turning point at $ \ (0 \ , \ C ) \ \ $ and two others at $ \ x^2 \ = \ \frac{|B|}{2} \ \Rightarrow \ x \ = \ \pm \sqrt{ \frac{|B|}{2}} \ \ . $ The $ \ y-$coordinate for these points is then
$$ \left(\frac{|B|}{2} \right)^2 \ + \ B·\left(\frac{|B|}{2} \right) \ + \ C \ \ = \ \ \frac{ B^2 }{4} \ - \ |B|·\left(\frac{|B|}{2} \right) \ + \ C $$ $$ = \ \ \frac{ B^2 }{4} \ - \ \frac{ B^2}{2} \ + \ C \ \ = \ \ C \ - \ \frac{ B^2 }{4} \ \ . $$
[The foregoing can be found more directly by using calculus.]
With this established, we can now sort the cases for $ \ B < 0 \ \ : $
• with the "high" turning point "below" the $ \ x-$axis $ \ [ \ \mathbf{C < 0} \ ] \ \ , $ the biquadratic curve has two $ \ x-$intercepts [the factorization is $ (x^2 + \alpha^2)·(x^2 - \beta^2) \ \ = \ \ x^4 \ - \ (\beta^2 - \alpha^2)·x^2 \ - \ \alpha^2·\beta^2 \ \ , $
with $ \ |\alpha| < |\beta| \ \ , \ \ \alpha \ $ imaginary]
;
• with the other turning points "above" the $ \ x-$axis, for which $ \ C \ - \ \frac{ B^2 }{4} \ > \ 0 \ \Rightarrow \ \mathbf{B^2 \ < \ 4C} \ \ , $ the curve does not intersect the $ \ x-$axis;
• for $ \ C \ - \ \frac{ B^2 }{4} \ < \ 0 \ < \ C \ \Rightarrow \ \mathbf{0 \ < \ 4C \ < \ B^2} \ \ , $ there are four $ \ x-$intercepts "surrounding" the three turning points [the corresponding factorization is $ \ (x^2 - \alpha^2)·(x^2 - \beta^2) \ \ = \ \ x^4 \ - \ (\alpha^2 + \beta^2)·x^2 \ + \ \alpha^2·\beta^2 \ \ . \ ] $
We see then that for your equation (1), $ \ B = -5 \ , \ C = -36 \ $ tells us that the equation has two real zeroes; its factorization is $ \ (x^2 - 9)·(x^2 + 4) \ \ . $ Equation (2) has $ \ B = -13 \ , \ C = 36 \ $ $\Rightarrow \ B^2 = 169 \ > \ 4C = 144 \ > \ 0 \ \ , $ hence there are four real zeroes; the expression factors as $ \ (x^2 - 9)·(x^2 - 4) \ \ . $ Finally, with equation (3), we "factor out" a $ \ 4 \ \ $ (which has no effect on the solutions) to obtain $ \ B = -\frac52 \ , \ C = \frac{25}{4} \ \Rightarrow \ B^2 = \frac{25}{4} \ < \ 4C = 25 \ \ , $ indicating that there are no real zeroes. (The factorization here is not very tidy.)
A table of the conditions for the number of real zeroes is presented below.
|
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|
Number of possible values of $4x-z$ if $x+y+z=20$ Given that $x, y,z$ are non negative integers such that $x+y+z=20$. If $S$ is the set of all possible values of $4x-z$.Find $n(S)$.
My try:
By stars and bars number of non negative integer solutions of $x+y+z=20$ is $\binom{22}{2}=210$
Among these $210$ ordered triplets of $(x,y,z)$ we need to find number of possible values taken by $4x-z$.
Obviously the least value taken by $4x-z$ is $-20$ when $x=0,z=20$.
The next value taken by $4x-z$ is $-19$ when $x=0,z=19$ and so on.
So once we fix $x=0$ ,since $z$ varies from $[0,20]$ number of values taken by $4x-z$ is $21$.
But when we fix $x=1$, we get some overlaps. For example when $x=1,z=18$ we have $4x-z=-14$ and this value has already been counted in the previous case when $x=0,z=14$.
So any way to count total number of values taken by $4x-z$ excluding overlaps?
|
For a given value of $x$, $z$ can range from $0$ up to $20-x$. This means that
$$D_a := \{4a-z \mid a + y + z = 20\} = \{5a-20, 5a-19, \ldots, 4a\}$$
(note that $5a-20 \leq 4a$ when $0 \leq x \leq 20$) where $a$ is fixed. Notice also that $4a \geq 5(a+1)-20$ for $a \leq 15$. This means that $D_a$ and $D_{a+1}$ overlap for $a \leq 15$. As a result, we find that
$$D_0 \cup D_1 \cup D_2 \cup \cdots \cup D_{16} = \{5 \cdot 0 - 20, \ldots, 4 \cdot 16\} = \{-20, \ldots, 64\}$$
For $a \geq 17$, $D_a$ is disjoint from $D_{a-1}$ and $D_{a+1}$. This means that
\begin{align}
n(\{4x-z \mid x + y + z = 20\}) &= n(D_0 \cup D_1 \cup \cdots \cup D_{20}) \\
&= n(D_0 \cup \cdots \cup D_{16}) + n(D_{17}) + n(D_{18}) + n(D_{19}) + n(D_{20}) \\
&= 85 + 4 + 3 + 2 + 1 = \boxed{95}
\end{align}
|
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|
Derive reduction formula for $\int \ln^n(x+\sqrt{x^2+1}) \, dx $ I wrote something like $$I=\int(x)' \ln^n {\left(x+\sqrt{x^2+1} \right)} \, dx$$ and $$I=x\ln(x+\sqrt{x^2+1})-n \displaystyle \int x \frac{\ln^{n-1} {\left(x+\sqrt{x^2+1} \right)}}{\sqrt{x^2+1}}dx.$$
I noticed that ${\left[\ln{{\left(x+\sqrt{x^2+1} \right)}}\right]}'=\dfrac{1}{1+x^2},$ but I don't know what I should do next.
|
Let's use integration by parts.
$\int u_1u_2 \,dx=u_1 \int u_2\,dx-\int u_1' (\int u_2\,dx)\,dx$
Sometimes choosing $u_1,u_2$ in certain ways makes integration by parts simpler. In this case, I'll take logarithmic function as $u_1$ and $1$ as $u_2$ to start with.
Let
$\begin{align} I_n=&\int \ln^n(x+\sqrt{x^2+1}) \times 1 \, dx \\=&\ln^n(x+\sqrt{x^2+1})x-n\int\ln^{n-1}(x+\sqrt{x^2+1})\frac{1}{x+\sqrt{x^2+1}}(1+\frac x{\sqrt{x^2+1}})x \, dx\\=&\ln^n(x+\sqrt{x^2+1})x-n\int \ln^{n-1}(x+\sqrt{x^2+1})\frac x{\sqrt{x^2+1}} \, dx\end{align}$ $\tag 1$
Second integral is
$
\int \ln^{n-1}(x+\sqrt{x^2+1})\frac x{\sqrt{x^2+1}} \, dx
=\ln^{n-1}(x+\sqrt{x^2+1})\sqrt{x^2+1}-\\(n-1)\int\ln^{n-2}(x+\sqrt{x^2+1})
\frac 1{x+\sqrt{x^2+1}}(1+\frac x{\sqrt{x^2+1}})\sqrt{x^2+1} \,dx
\\=\ln^{n-1}(x+\sqrt{x^2+1})\sqrt{x^2+1}-(n-1)I_{n-2}$
So $(1)$ becomes: $I_n=\ln^n(x+\sqrt{x^2+1})x-n\ln^{n-1}(x+\sqrt{x^2+1})\sqrt{x^2+1}+n(n-1)I_{n-2}$, where $n\ge 2$
|
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|
Let $a_1=\sqrt{6},a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to \infty} (a_n-3)6^n$.
Let $a_1=\sqrt{6}$, $a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to
\infty} (a_n-3)6^n$.
First, we may obtain $\lim\limits_{n\to \infty}a_n=3$. Hence, $\lim\limits_{n \to \infty}(a_n-3)b^n$ belongs to a type of limit with the form $0 \cdot \infty$.
Moreover, we obtained a similar result here, which is related to the form $\lim\limits_{n \to \infty} (a_n-3)9^n$.
How should I proceed?
|
Ths approach shows that the results in question emerge quite naturally from the standard fixed-points-stability analysis of the evolution equation. Also the dependence of the limit of $b_n = (a_n -3)6^n$ from the inital value $a_1$ is discussed.
Following the common procedure to analyse the evolution equation
$$a_{n+1} = \sqrt{6+a_{n}}\tag{1}$$
we fist look for fixed points. These follow from the equation $0 = a^2-a-6$ to be $a = 3$ and $a=-2$.
Ruling out the negative solution we study the stabilty letting
$$a_{n} = 3 +\delta_{n}\tag{2}$$
where $|\delta_{n}| << 3$.
Inserting into $(1)$ gives
$$\begin{align}
& 3 +\delta_{n+1} = \sqrt{6+3+\delta_{n}}=\sqrt{9+\delta_{n}}\\&=3\left( \sqrt{1+\frac{\delta_{n}}{9}}\right)\simeq 3\left( 1+\frac{1}{2}\frac{\delta_{n}}{9}\right)=3 +\frac{1}{6} \delta_{n}\end{align} $$
whence follows
$$\delta_{n+1} \simeq \frac{1}{6} \delta_{n}\tag{3}$$
The solution $\delta_{n}=\frac{c}{6^n}$ of $(3)$ with some constant $c$ shows that the fixed point $a_{\infty}=3$ is stable and from $(2)$ we find
$$a_{n} \simeq 3 + \frac{c}{6^n}\tag{4}$$
This in turn means that
$$b_{n} := \left(a_{n} - 3 \right)6^n \simeq c\tag{5}$$
So that the limit in question is the constant $c$. For $0 \lt a_{1} \lt 3$ the constant $c$ is negative, for $a_{1} \gt 3$ it is positive. In the actual case we have $a_{1} = \sqrt{6} \lt 3$, and the limit is numerically
$$b_{\infty} \simeq -3.36571\tag{6}$$.
Limit as a function of the initial value
Here is a plot showing the dependence of the limit $b_{\infty}$ on the initial value $a_{1}$ for $-6 \le a_{1} <10$
Some special values of the limit are: $b_{\infty}(-6) = -121.164$, $b_{\infty}(0) = -20.1939$, and $b_{\infty}(3) = 0$
The limit function is well approximated by
$$b_{\infty}(a_1)=45.1116 (a_1+6)^{0.44966}-121.164\tag{7}$$
|
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|
Is it possible to write any permutation as the product of transpositions $(1 \text{ } 2),(2 \text{ } 3),...(n-1 \text{ } n)$ I understand any permutation $(a_1 \text{ } a_2 ... a_k)$ in $S_n$ where $k \leq n$ can be written as a product of $(a_1 \text{ } a_k)(a_1a_{k-1})...(a_2 \text{ } a_1)$. Similarly, $(a_1a_2...a_k)$ can be written as $(1 \text{ } a_1)(1 \text{ } a_k)...(1 \text{ } a_2)(1 \text{ } a_1)$ (if the permutation has $1$ in it, it can be rearranged so $1$ is the first integer appearing on the left and then the first and last transpositions can be removed) which shows any permutation can be written as the products of $(1 \text{ } 2), (1 \text{ } 3),...,(1 \text{ } n)$.
Is it possible to write a permutation as a product of $(1 \text{ } 2),(2 \text{ } 3),...,(n-1 \text{ } n)$? If the permutation consists of consecutive numbers in increasing order then $(a_1 \text{ } a_2)...(a_{k-1} \text{ } a_k)$ works fine. But I cannot generalize it for any sequence of integers. Is this actually possible?
Edit
It turns out it is possible. I am hoping to generalize this claim so for any permutation, so I can write it as a product of $(1 2), (2 3), ...,(n-1 \text{ } n)$, the way I have shown it works for $(1 2), (1 3),...,(1 n)$. For instance if $\sigma =(352)$ then $\sigma=(1 3)(1 2)(1 5)(13)$. How can the same cycle be written as a product of $(1 2),(2 3),...(n-1 \text{ } n)?$
|
Solution 1:
It is obvious that we can write any permutation as a product of transpositions.
So, we need only to show that any transposition is a product of special transpositions $(1 \text{ } 2),(2 \text{ } 3),...(n-1 \text{ } n)$.
Let $1\leq i<j\leq n$.
$(i\text{ }j)=(i\text{ }i+1)\cdots(j-2\text{ }j-1)(j-1\text{ }j)\cdots(i+1\text{ }i+2)(i\text{ }i+1)$.
For example, $(3\text{ }7)=(3\text{ }4)(4\text{ }5)(5\text{ }6)(6\text{ }7)(5\text{ }6)(4\text{ }5)(3\text{ }4)$.
Solution 2:
Let $\sigma$ be any permutation in $S_n$.
Let $a:=\sigma(1)$.
If $a=1$, let $\tau_1:=\sigma$.
If $a\neq 1$, then, $\tau_1:=(2\text{ }1)\cdots(a-1\text{ }a-2)(a\text{ }a-1)\sigma$.
Then $\tau_1(1)=1$.
Let $b:=\tau_1(2)\in\{2,\dots,n\}$.
If $b=2$, let $\tau_2:=\tau_1$.
If $b\neq 2$, let $\tau_2:=(3\text{ }2)\cdots(b-1\text{ }b-2)(b\text{ }b-1)\tau_1$.
Then $\tau_2(1)=1, \tau_2(2)=2$.
Let $c:=\tau_2(3)\in\{3,\dots,n\}$.
If $c=3$, let $\tau_3:=\tau_2$.
If $c\neq 3$, let $\tau_3:=(4\text{ }3)\cdots(c-1\text{ }c-2)(c\text{ }c-1)\tau_2$.
Then $\tau_3(1)=1, \tau_3(2)=2,\tau_3(3)=3$.
$\vdots$
Let $y:=\tau_{n-2}(n-1)\in\{n-1,n\}$.
If $y=n-1$, let $\tau_{n-1}:=\tau_{n-2}$.
If $y\neq n-1$, let $\tau_{n-1}:=(n\text{ }n-1)\tau_{n-2}$.
Then $\tau_{n-1}(1)=1,\tau_{n-1}(2)=2,\dots,\tau_{n-1}(n-2)=n-2,\tau_{n-1}(n-1)=n-1$.
Let $z:=\tau_{n-1}(n)\in\{n\}$.
Then, $z=n$.
So, $\tau_{n-1}$ is the identity.
Therefore, the inverse of $\sigma$ is a product of special transpositions $(1 \text{ } 2),(2 \text{ } 3),...(n-1 \text{ } n)$.
Since the inverse of the inverse of $\sigma$ is also a product of special transpositions $(1 \text{ } 2),(2 \text{ } 3),...(n-1 \text{ } n)$ and $\sigma$ is equal to the inverse of the inverse of $\sigma$, $\sigma$ is a product of special transpositions $(1 \text{ } 2),(2 \text{ } 3),...(n-1 \text{ } n)$.
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|
Get $f(x)=u_x\frac{x}{u}$ from ODE for $u$ Consider the Cauchy-Euler ODE
\begin{align*}
\frac{1}{2}x^2u_{xx}+xu_x-u=0.
\end{align*}
Guessing $u(x)=Cx^n$ gives
\begin{align*}
\frac{1}{2}n(n-1)Cx^n+nCx^n-Cx^n=0,
\end{align*}
which we can solve to get
\begin{align*}
n_1&=-2,\\
n_2&=1.
\end{align*}
Given initial conditions and boundary behavior, we can pin down a unique solution.
I'm really interested in the function $$f(x)=u_x\frac{x}{u}.$$ Given the solution for $u$, we can compute $f$ to be $f(x)=n$, which is constant!
Q: I wonder whether I can find $f$, without solving the ODE first?
Set $f=u_x\frac{x}{u}$. Then, $u_x=\frac{fu}{x}$ and $u_{xx}=\frac{f_xu}{x}+\frac{fu_x}{x}-\frac{fu}{x^2}$. The ODE then turns into
\begin{align*}
\frac{1}{2}\left(xf_xu+xfu_x-fu\right)+fu-u=0.
\end{align*}
Dividing this ODE by $u$ gives
\begin{align*}
\frac{1}{2}\left(xf_x+f^2-f\right)+f-1=0.
\end{align*}
If I assume $f_x=0$, then I get
\begin{align*}
\frac{1}{2}f^2+\frac{1}{2}f-1=0,
\end{align*}
which is a normal quadratic equation with solutions
\begin{align*}
f_1&=-2,\\
f_2&=1.
\end{align*}
These are precisely the solutions I expected from the previous calculations.
However, I had to assume $f_x=0$. I only knew this because I already knew the solution.
Why doesn't the second approach work? Is there a way to compute $f(x)$ without first solving the ODE for $u(x)$?
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The general solution to your ODE is
$$
u(x)=Ax+Bx^{-2}
$$
and so
$$
f(x)=1-\frac{3 B}{A x^3+B}
$$
which is not in general constant. That is why you do not find $f_x=0$ directly from the ODE.
|
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Determine all ordered pair (x,y) of positive integers such that $y^{2}−(x+ 2)2^x=1$. This is what I have come up so far,
*
*y has to be odd. Since $(x+2) 2^{x}$ is always even, hence $(x+2)2^x+1$ is always odd. Thus, y must be an odd number.
*$$y^2 \equiv1(mod 4)$$Since all square numbers either congruent to 0 or 1 (mod 4), but $y^2$ is odd, it has to be congruent to 1 (mod 4).
After re-writing $y=2k+1$, I am able to use quadratic formula to get k in term of x $$k=\frac{-4\pm\sqrt{16+(4)(4)(x+2)(2^x)}}{8}$$ So far, I only know that $x=5,y=15$ is a solution. I would greatly appreciate if anyone can help.
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$y^2 - 1 = (x+2)2^x$
$(y-1)(y+1)= (x+2)2^x$.
Now $y- 1, y+2$ must both be even and $\gcd(y-1,y+1) = 2$ so
So one of $y-1$ or $y+1$ must be divisible by $2$ but not any higher power of $2$. And the other must be divisible by a power of $2$ that is at least $2^{x-1}$.
The very least the term divisible by $2^{x-1}$ can be is $2^{x-1}$. And the very most the other term, that is only divisible by $2$ and no higher power, can be is $2(x+2)$. But for significantly large $x$ we will have $2^{x-1} > 2(x+2)$. However the difference between the terms, $y+1$ and $y-1$ is fixed at $2$.
So we can find an upper bound for the possible value of $x$.
$2^{x-1} - 2(x+2)$ can not be more than $2$.
A bit of mucking $2^{x-1} - 2(x+2) \le 2$ means $2^{x-1} \le 2x + 6$ means $2^{x-2} \le x + 3$ and for $x =5$ we have $2^{5-2} = 5 + 3$ but $2^{6-2} > 6+3$ and by induction we can just say that $2^{x-2}$ will double as $x$ increases by $1$ whereas $x+3$ will only increase by $1$.
So we must have $x \le 5$.
So just try them.
What we are looking for is that $(y-1)(y+1) = (x+2)2^x = [2m]\cdot [N]$ where $m$ is odd and $N$ is divisible by at least $2^{x-1}$ and the difference between $N$ and $2m$ is exactly $2$.
$x= 5$ gives us a perfect $(x+2)2^x = 7\cdot 32 = (2\cdot 7)\cdot 2^4$ where $2^4 - 2\cdot 7 = $ exactly $2$ so $y-1=14; y+1 = 16$ and $y = 15$ is a solution.
No others work.
$x =4$ gives us $(x+2)2^x= 6\cdot 16=\begin{cases}(2\cdot 3)\cdot 16\\(2\cdot 1)\cdot 48\end{cases}$ neither gives us a difference of $2$.
$x= 3$ gives us $(x+2)2^x = 5\cdot 8= (10)\cdot 4=2\cdot 20$ which fails. As does $x=2$ ($(x+2)2^x=16 =2\cdot 8$) and $x=1$ ($(x+2)2^x= 6=6\cdot 1$... this really fails as when $y$ is odd $(y-1)(y+1)$ is always divisible by $8$).
$x= 5; y = 15$ is the only solution.
|
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If $xy = ax + by$, prove the following: $x^ny^n = \sum_{k=1}^{n} {{2n-1-k} \choose {n-1}}(a^nb^{n-k}x^k + a^{n-k}b^ny^k),n>0$ If $xy = ax + by$, prove the following: $$x^ny^n = \sum_{k=1}^{n} {{2n-1-k} \choose {n-1}}(a^nb^{n-k}x^k + a^{n-k}b^ny^k) = S_n$$ for all $n>0$
We'll use induction on $n$ to prove this.
My approach is to use this formula: $$ \frac{k}{r} {{r} \choose {k}} = {{r-1} \choose {k-1}}$$
I'd like to show: $$S_{n} = xy.S_{n-1}$$. Or:
$$\sum_{k=1}^{n} {{2n-1-k} \choose {n-1}}(a^{n}b^{n-k}x^k + a^{n-k}b^{n}y^k) = (ax+by)\sum_{k=1}^{n-1} {{2n-3-k} \choose {n-2}}(a^{n-1}b^{n-k-1}x^k + a^{n-1-k}b^{n-1}y^k)$$
We have:
$$ (ax+by)\sum_{k=1}^{n-1} {{2n-3-k} \choose {n-2}}(a^{n-1}b^{n-k-1}x^k + a^{n-1-k}b^{n-1}y^k)
= \sum_{k=1}^{n-1} {{2n-3-k} \choose {n-2}}(a^{n}b^{n-k-1}x^{k+1} + a^{n-k}b^{n-1}xy^k + a^{n-1}b^{n-k}x^ky + a^{n-1-k}b^ny^{k+1})
= \sum_{k=2}^{n} {{2n-2-k} \choose {n-2}}(\pmb{a^{n}b^{n-k}x^{k}} + a^{n-k+1}b^{n-1}xy^{k-1} + a^{n-1}b^{n-k+1}x^{k-1}y + \pmb{a^{n-k}b^ny^{k}})
= \sum_{k=2}^{n} \frac{n-1}{2n-1-k} {{2n-1-k} \choose {n-1}} [...]
$$
Now we can almost extract the intended term($S^{'}_{n}$):
$$
\sum_{k=2}^{n} {{2n-1-k} \choose {n-1}} (a^{n}b^{n-k}x^{k} + a^{n-k}b^ny^{k}) + \sum_{k=2}^{n} {{2n-1-k} \choose {n-1}} (a^{n-k+1}b^{n-1}xy^{k-1} + a^{n-1}b^{n-k+1}x^{k-1}y) + \sum_{k=2}^{n} (\frac{n-1}{2n-1-k}-1) {{2n-1-k} \choose {n-1}} [...]
$$
There is further derivation but not seems very promising.
The idea of this theorem is really interesting. I'm asking for a simpler approach or how i should countinue my proof. Thank you in advance!
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The approach by induction is fine. We just need one additional twist (one more induction).
We want to show for $n\geq 1$
\begin{align*}
\color{blue}{(xy)^n=\sum_{k=1}^n\binom{2n-1-k}{n-1}\left(a^nb^{n-k}x^k+a^{n-k}b^ny^k\right)}\tag{1}
\end{align*}
provided
\begin{align*}
\color{blue}{xy=ax+by}
\end{align*}
The base step $n=1$ is easily shown.
Induction hypothesis: $n=N-1$:
We assume the claim (1) is valid for $n=N-1$, i.e. we have
\begin{align*}
(xy)^{N-1}=\sum_{k=1}^{N-1}\binom{2N-3-k}{N-2}\left(a^{N-1}b^{N-1-k}x^k+a^{N-1-k}b^{N-1}y^k\right)\tag{2}
\end{align*}
Induction Step: $n=N$:
Now we want the show the validity of (1) for $n=N$. We consider $(xy)^N=(xy)^{N-1}(xy)$. Multiplication of the right hand side of (2) with $xy$ gives us terms $Ax^ky$ and $Bxy^k$ with constants $A$ and $B$. We obtain
\begin{align*}
xy&=ax+by\\
x^2y&=ax^2+abx+b^2y\\
x^3y&=ax^3+abx^2+ab^2x+b^3y\\
&\vdots\\
x^ky&=\sum_{j=1}^k ab^{k-j}x^j+b^ky\qquad\qquad xy^k=a^kx+\sum_{j=1}^ka^{k-j}by^j\tag{3}
\end{align*}
which can be easily shown by induction.
Putting (3) in (2) we obtain
\begin{align*}
\color{blue}{(xy)^n}&=\left(xy\right)^{n-1}(xy)\\
&=\sum_{k=1}^{N-1}\binom{2N-3-k}{N-2}\left(a^{N-1}b^{N-1-k}x^k+a^{N-1-k}b^{N-1}y^k\right)(xy)\\
&=\sum_{k=2}^{N}\binom{2N-2-k}{N-2}\left(a^{N-1}b^{N-k}x^ky+a^{N-k}b^{N-1}xy^k\right)\tag{4}\\
&=\sum_{k=2}^{N}\binom{2N-2-k}{N-2}\left(a^{N-1}b^{N-k}\left(\sum_{j=1}^kab^{k-j}x^j+b^ky\right)\right.\\
&\qquad\qquad\qquad\qquad\qquad\left.+a^{N-k}b^{N-1}\left(a^kx+\sum_{j=1}^ka^{k-j}by^j\right)\right)\tag{5}\\
&\,\,\color{blue}{=\sum_{k=2}^{N}\binom{2N-2-k}{N-2}\left(\sum_{j=1}^ka^Nb^{N-j}x^j+a^{N-1}b^Ny\right.}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\color{blue}{\left.+a^Nb^{N-1}x+\sum_{j=1}^ka^{N-j}b^Ny^j\right)}\tag{6}
\end{align*}
Comment:
*
*In (4) we shift the index and start with $k=2$.
*In (5) we use the identities from (3).
In order to show that (1) and (6) are equal we show equality of coefficients of the polynomials in $x$ and $y$. We denote with $[x^t]$ the coefficient of $x^t$ and obtain from (1) for $1\leq t\leq N$:
\begin{align*}
[x^t](xy)^N=\binom{2N-1-t}{N-1}a^Nb^{N-t}\tag{7}
\end{align*}
We obtain from (6)
\begin{align*}
[x^t](xy)^N&=\sum_{k=t}^N\binom{2N-2-k}{N-2}a^Nb^{N-t}\\
&\qquad+\sum_{k=2}^N\binom{2N-2-k}{N-2}a^Nb^{N-1}[[t=1]]\tag{8}
\end{align*}
with $[[t=1]]$ denoting Iverson brackets which is $1$ iff $t=1$ and zero otherwise.
Equating (7) and (8) we finally have to show that for $1\leq t\leq N$ we have
\begin{align*}
\color{blue}{\sum_{k=t}^N\binom{2N-2-k}{N-2}+\sum_{k=2}^N\binom{2N-2-k}{N-2}[[t=1]]
=\binom{2N-1-t}{N-1}}\tag{9}
\end{align*}
We obtain for $t\geq 2$:
\begin{align*}
\color{blue}{\sum_{k=t}^N}&\color{blue}{\binom{2N-2-k}{N-2}}\\
&=\sum_{k=0}^{N-t}\binom{2N-2-t-k}{N-2}\\
&=\sum_{k=0}^{N-t}\binom{N-2+k}{N-2}\tag{$k\to N-t-k$}\\
&=\sum_{k=0}^{N-t}[z^{N-2}](1+z)^{N-2+k}\\
&=[z^{N-2}](1+z)^{N-2}\sum_{k=0}^{N-t}(1+z)^{k}\\
&=[z^{N-2}](1+z)^{N-2}\frac{(1+z)^{N-t+1}-1}{(1+z)-1}\\
&=[z^{N-1}]\left((1+z)^{2N-t-1}-(1+z)^{N-2}\right)\\
&=[z^{N-1}](1+z)^{2N-t-1}\\
&\,\,\color{blue}{=\binom{2N-t-1}{N-1}}
\end{align*}
showing the claim (9) is valid for $t\geq 2$.
Since the above result is also valid for $t=1$ we use it to show the case $t=1$:
\begin{align*}
\color{blue}{2\sum_{k=2}^N}&\color{blue}{\binom{2N-2-k}{N-2}}\\
&=2\sum_{k=1}^N\binom{2N-2-k}{N-2}-2\binom{2N-3}{N-2}\\
&=2\binom{2N-2}{N-1}-2\,\frac{N-1}{2N-2}\binom{2N-2}{N-1}\\
&\,\,\color{blue}{=\binom{2N-2}{N-1}}
\end{align*}
showing the claim (9) is valid for $t=1$.
Showing the validity of $[y^t](xy)^N$, $1\leq t\leq N$ follows the same lines.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4106937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
If $x^{x^{x+1}}=\sqrt{2}$, then evaluate $x^{x^{p}}$, where $p = 2x^{x+1}+x+1$ I can't figure out how to give a proper form to this expression to use the root of two.
If
$$x^{x^{x+1}}=\sqrt{2}$$
find the value of $W$ if
$$W=x^{x^{p}} \quad\text{where}\; p = 2x^{x+1}+x+1$$
EDIT: This is an algebraic manipulation problem with the exponents. There was an error in the previous version (see the Edit History) that I have corrected.
|
$x^{x^x} = \sqrt{2} \implies x^x = log_x\sqrt2$
Now:
$W = x^{x^{2x^{x+1}+x+1}} = x^{\left(x^{2x^{x+1}}x^xx\right)} = x^A$
Let the exponent be A for the time being:
$A = x^{2x^{x+1}}x^xx = \left(x^{x^{x+1}}\right)^2x^xx = \left(x^{x^xx}\right)^2x^xx$
Since $x^x = log_x\sqrt2$, we have $A = \left(x^{xlog_x\sqrt{2}}\right)^2xlog_x\sqrt{2} = (\sqrt{2}^x)^2xlog_x\sqrt{2}$
Let $\sqrt{2}^x = B \implies A = B^2log_x(B) = log_x(B^{B^2})$
Remember $W = x^A = B^{B^2} = (\sqrt{2}^x)^{2^x}$
What to do from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4107777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Find $S = a + b$ such that for $\forall m \in \left[a\sqrt{\frac{15}{7}} + b\sqrt{\frac{7}{15}}; 2\right)$ then $2x^2 + 2x - mf(x) + 5 = 0$ has root
Let $f(x)$ be continuos on $\mathbb R$ satisfy $f(0) = 2\sqrt{2}$ and $f(x) > 0, \forall x \in \mathbb R$ and $f(x) f'(x) = (2x+1)\sqrt{1+f^2(x)}$.
For all $m \in \left[a\sqrt{\dfrac{15}{7}} + b\sqrt{\dfrac{7}{15}}; 2\right)$, then $2x^2 + 2x - mf(x) + 5 = 0$ has at least a root. Find $S = a + b$
*
*Here is what I did so far
$$f(x) f'(x) = (2x+1)\sqrt{1+f^2(x)}$$
$$\implies \dfrac{f(x) f'(x)}{\sqrt{1+f^2(x)}} = 2x+1$$
$$\implies \int \dfrac{f(x) f'(x)}{\sqrt{1+f^2(x)}} \,dx = \int2x+1 \,dx = x^2 + x + C$$
Let $t = f^2(x) + 1 \implies dt = 2f'(x)f(x) dx$. Therefore:
$$\int \dfrac{1}{2\sqrt{t}} \,dx = x^2 + x + C$$
$$\implies \sqrt{t} = x^2 + x + C$$
$$\implies \sqrt{1 + f^2(x)} = x^2 + x + C \, (2)$$
Since $f(0) = 2\sqrt{2}$, plug in $(2)$ we get $C = 3 \implies f(x) = \sqrt{(x^2+x+3)^2 - 1}$
I can't proceed further from here
|
COMMENT.-Once find out $f(x)=\sqrt{(x^2+x+3)^2-1}$ the function $g(x)=2x^2+2x-mf(x)+5$ has derivative $g'(x)=(2x+1)\left(2-\dfrac{m(x^2+x+3)}{f(x)}\right)$ so take an extremum at $x=-\dfrac12=-0.5$ and $g(-0.5)=4.5-m\dfrac{\sqrt{105}}{16}$
It follows that $g(x)$ has at least a (real) root in a little interval for $m$ whose first element is given by $g(-0.5)=4.5-m\dfrac{\sqrt{105}}{16}=0\iff m=\dfrac{18}{\sqrt{105}}\approx1.756620$ and it is bounded by $2$.
(for example for $m=1.99$ one has $g(9.31)=0$ but for $m\ge2$ there are not (real) root).
Then we have $\dfrac{18}{\sqrt{105}}=a\sqrt{\dfrac{15}{7}}+b\sqrt{\dfrac{7}{15}}=\dfrac{15a+7b}{\sqrt{105}}\Rightarrow \boxed {15a+7b=18}$
Can someone finish?
|
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|
what is $a^2+9=b^2+16=1+(a+b)^2$ solve for $a,b$ This is for a geometry question, and through a construction arrived at this equation. I could not solve it and after plugging it into wolfram got the correct answer but can anyone show a method for finding a,b
they are $\frac{5}{\sqrt3}$ and $\frac{2}{\sqrt3}$ respectively (note since this is a geometry question the lengths of a side cannot be negative so the negative solution sets do not matter)
|
Maybe try this way:
$b^2+16=1+(a+b)^2\Rightarrow 15=a(a+2b)$ and $a^2+9=1+(a+b)^2\Rightarrow 8=b(2a+b)$
$\frac{8}{15}=\frac{b(2a+b)}{a(a+2b)}=\frac{x(2+x)}{1+2x}$, where $x=\frac{b}{a}$ (Assuming $a$ and $b$ are positive).
On solving, we get $x=\frac{2}{5}=\frac{b}{a}$.
Thus, $5b=2a$.
As $a^2+9=b^2+16=(\frac{2a}{5})^2+16$. This gives $a=\frac{5}{\sqrt{3}}$. Thus, $b=\frac{2}{\sqrt{3}}$. (Considering only positive numbers)
|
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|
How am I supposed to expand $\sin^2 A + \sin^4 A = 1$ into $1 + \sin^2A = \tan^2A$? My question is how can i expand
$$\sin^2 A + \sin^4 A = 1$$
into:
$$1 + \sin^2A = \tan^2A$$
I tried quite a few ways I know but all of them kinda felt random. i am not sure how to share my trials here. I am quite beginner in trigonometry. it is one of the extra test question from my textbook. I don't need it but cant control curiosity. so pls help me.
Thanks in advance!
EDIT:
found the solution, dropping it here,
\begin{align}
\sin^2 A + \sin^4 A & = 1 \\
\sin^4 A & = 1 - sin^2 A \\
\sin^2 A . \sin^2 A & = cos^2 A \\
\sin^2 A . (1 - \cos^2 A) & = cos^2 A \\
\sin^2 A - \sin^2 A.\cos^2 A & = cos^2 A \\
\sin^2 A & = cos^2 A + \sin^2 A.\cos^2 A \\
\sin^2 A & = \cos^2 A(1 + \sin^2 A) \\
1 + \sin^2 A & = \cfrac{\sin^2 A}{\cos^2 A} \\
1 + \sin^2 A & = \tan^2 A \\
\end{align}
|
Hint
If $\sin^2A + \sin^4A = 1 \implies \sin^4A = \cos^2 A$
Can you proceed from here?
Additional Info
$$\sin^2A(1-\cos^2A) = \cos^2A$$
$$\sin^2A = \cos^2A + \sin^2A\cos^2A$$
Divide by $\cos^2 A$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A} \leq \frac{A}{B}+\frac{B}{A}$ for acute angles, $A$ and $B$.
Prove that $\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A} \leq \frac{A}{B}+\frac{B}{A}$ for acute angles, $A$ and $B$.
I'm confused about how to do this since we can't say $\frac{\sin A}{\sin B}\leq \frac{A}{B}$. So I simplified and got $$\frac{\sin^2 A+ \sin^2 B}{\sin A \sin B} \leq \frac{A^2+B^2}{AB}$$
Using $\sin x \leq x$ we can say $\sin^2 A+ \sin^2 B \le A^2+B^2$ but since we cannot divide, this doesn't work either.
|
Here is another solution: let $f(t) = t + 1/t$, and observe that it is increasing on $[1,\infty)$, and decreasing on $(0,1]$.
Since $f(1/t) = f(t)$ for any $t > 0$, we can assume w.l.o.g. that $B \leq A$.
Now, observe that the function $g(t)=\frac{\sin(t)}{t}$ is decreasing on $(0,\frac{\pi}{2})$, and thus:
$$
g(A) \leq g(B) \implies 1\leq\frac{\sin(A)}{\sin(B)}\leq \frac{A}{B}.
$$
Finally
$$
f(\frac{\sin(A)}{\sin(B)})\leq f(\frac{A}{B}) \implies \frac{\sin(A)}{\sin(B)} + \frac{\sin(B)}{\sin(A)}\leq \frac{A}{B} + \frac{B}{A}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Finding limit of term involving binomial coefficient I am required to find the limit of
$$
\frac{\sqrt{n}}{2^n}\cdot \binom{n}{\frac{n}{2} + \sqrt{n}}
$$
I am given the following hints:
*
*Stirling's approximation
$$
n! \approx \sqrt{2\pi n}\cdot\left(\frac{n}{e}\right)^n
$$
*For $k \rightarrow \infty$, with $x$ being a constant.
$$
\left(1 + \frac{x}{k} \right)^k \rightarrow e^x
$$
I've tried doing a direct substitution using Stirling's approximation, but I am stuck with how to further simplify the terms. I am also unsure of how to use (2).
This is the expression I am currently stuck at and have no idea how to proceed:
$$
\frac{n^{n+1}}{2^n \cdot \sqrt{2\pi}}\frac{1}{\sqrt{(\frac{n}{2})^2-n}\cdot(\frac{n}{2}-\sqrt{n})^{\frac{n}{2}-\sqrt{n}}\cdot(\frac{n}{2}+\sqrt{n})^{\frac{n}{2}+\sqrt{n}}}
$$
Any hints is greatly appreciated.
|
Ignoring the $\sqrt{2\pi}$ factor, you can collect terms as follows: $$2 \cdot\left( \frac{n}{2\sqrt{n^2/4 -n}} \right)^{n+1} \cdot \left( \frac{n/2-\sqrt n}{n/2 +\sqrt n} \right)^{\sqrt n} $$ Now, the first factor is further simplified to $$\left( \frac{\sqrt n}{\sqrt {n-4}} \right)^{n+1} \\ = \frac{1}{ (1-\frac 4n)^{\frac{n+1}{2}}}\\ = \frac{1}{((1-\frac 4n)^n)^{1/2} \cdot (1-\frac 4n)^{1/2}} $$
$$ \to \frac{1}{(e^{-4})^{1/2}} \cdot \frac{1}{1}\\ = e^2 $$ and the second factor is $$\left(1-\frac{2\sqrt n}{n/2 +\sqrt n} \right)^{\sqrt n} \\ =\left( 1-\frac{2}{\sqrt n/2 +1} \right)^{\sqrt n} \\= \left( \left( 1-\frac{2}{\sqrt n/2 +1} \right)^{\sqrt n /2 +1} \right)^2 \cdot \left( 1-\frac{2}{\sqrt n/2 +1} \right)^{-2} \\ \to (e^{-2})^2 \cdot 1 \\ =e^{-4} $$
So, all in all the answer should be $$\frac{1}{\sqrt{2\pi}} \times 2 \times e^2 \times e^{-4}=\color{red}{ \sqrt{\frac{2}{\pi}} e^{-2}} $$
|
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|
Maximizing $x^2y$ given $x^2+y^2=100$, without using the AM-GM inequality and calculus tools Problem says:
Let $x^2+y^2=100$, where $x,y>0$. For which ratio of $x$ to $y$, the value of $x^2y$ will be maximum?
I know these possible tools:
*
*AM-GM inequality
*Calculus tools
Here, I want to escape from all of the tools I mentioned above.
I will try to explain my attempts in the simplest sentences. (my english is not enough, unfortunately). I will not prove any strong theorem and also I'm not sure what I'm doing exactly matches the math, rigorously.
Solution I made:
First, it is not necessary to make these substitutions. I'm just doing this to work with smaller numbers.
Let, $x=10m, ~y=10n$,
where $0<m<1,~ 0<n<1$, then we have
$$ x^2+y^2=100 \iff m^2+n^2=1$$
$$x^2y=1000m^2n$$
This means,
$$\max\left\{x^2y\right\}=10^3\max\left\{m^2n\right\}$$
$$m^2n=n(1-n^2)=n-n^3$$
Then suppose that,
$$\begin{align}\max\left\{n-n^3 \mid 0<n<1\right\}&=a, a>0&\end{align}$$
This implies
$$n-n^3-a≤0,~ \forall n\in\mathbb (0,1)$$
$$n^3-n+a≥0,~\forall n\in\mathbb (0,1)$$
Then, we observe that
$$\begin{align}n^3-n+a≥0, \forall n\in (0,1) ~ \text{and} ~ \forall n≥1\end{align}$$
This follows
$$ n^3-n+a≥0, ~ \forall n>0.$$
Using the last conclusion, I assume that there exist $u,v>0$, such that
$$n^3-n+a=(n-u)^2(n+v)≥0.$$
If $n>0$, then the equality occurs, if and only if
$$n=u>0$$
Based on these, we have:
$$\begin{align}n^3-n+a= (n-u)^2(n+v)≥0 \end{align}$$
$$\begin{align}n^3-n+a = & n^3 - n^2(2u-v)+ n(u^2 - 2 u v ) + u^2v & \end{align}$$
$$\begin{align} \begin{cases} 2u-v=0 \\ u^2-2uv=-1 \\u^2v=a \\u,v>0 \end{cases} &\implies \begin{cases} v=2u \\ u^2-4u^2=-1 \\ 2u^3=a \\ u,v>0 \end{cases}\\
&\implies \begin{cases} u=\frac{\sqrt 3}{3} \\ v=\frac{2\sqrt 3}{3}\\ a=2\left(\frac{\sqrt 3}{3} \right)^3=\frac{2\sqrt 3}{9} \end{cases} \end{align}$$
$$\begin{align}n^3-n+\frac{2\sqrt 3}{9} &=\left(n-\frac{\sqrt 3}{3} \right)^2\left(n+\frac{2\sqrt 3}{3}\right)≥0.&\end{align}$$
As a result, we deduce that
$$\begin{align}n-n^3-\frac{2\sqrt 3}{9} &=-\left(n-\frac{\sqrt 3}{3} \right)^2\left(n+\frac{2\sqrt 3}{3}\right)≤0, &\forall n\in (0,1).&\end{align}$$
$$\begin{align}\max\left\{n-n^3 \mid 0<n<1\right\}&=\frac{2\sqrt 3}{9}, ~ \text{at }~ n=\frac{\sqrt 3}{3}&\end{align}$$
Finally, we obtain
$$m=\sqrt{1-n^2}=\sqrt{1-\frac 13}=\frac{\sqrt 6}{3}$$
$$\frac xy=\frac mn=\sqrt 2.$$
Question:
*
*How much of the things I've done here are correct?
|
An alternative approach without calculus. Considering that $x^2y = \lambda$ and $x^2+y^2=10^2$ are analytic, at it's maximum, $x^2y$ should be tangent to $x^2+y^2=10^2$ then
$$
\frac{\lambda}{y}+y^2-10^2=0
$$
so, at tangency $y^3-10^2y+\lambda = (y-r_1)^2(y-r_2)$
Equating coefficients we have
$$
\cases{
r_1^2r_2 +\lambda = 0\\
r_1^2+2r_1r_2+10^2=0\\
2r_1+r_2 = 0
}
$$
now solving, we have
$$
\cases{
r_1 = \pm\frac{10}{\sqrt{3}}\\
r_2 = \mp\frac{20}{\sqrt{3}}\\
\lambda = \frac{2000}{3\sqrt{3}}
}
$$
so the maximum is $\frac{2000}{3\sqrt{3}}$ and choosing $y^* = \frac{10}{\sqrt{3}}$ we have $x^* = 10\sqrt{\frac 23}$
Attached a plot showing in red $x^2+y^2=10^2$ and in black the level curves of $x^2y$
|
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|
How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involving the natural logarithm, that is similar to the identity that:
$$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)} = \frac{\psi(a) - \psi(b)}{a-b}$$
Also potentially related, the Lüroth analogue of Khintchine’s constant can be defined as the following:
$$\sum_{n=1}^{\infty} \frac{\ln (n)}{n(n+1)}$$
as mentioned here.
After some work, the following can be shown:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{5\ln(2) + 4\ln(3)}{16} + \frac{1}{2} \sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right)$$
and furthermore:
$$\sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right) = \int_{0}^{2} \left( \frac{\left(1-\pi x \cot(\pi x) \right)}{2x^2} + \frac{1}{x^2 - 1} + \frac{1}{x^2 -4} \right) \, dx$$
EDIT
I have derived yet another form for my sum of interest, however, I found this one interesting as it seems like it could potentially be solvable?
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \int_{0}^{\infty} \left( \frac{\psi^{(0)} (s+3) + \gamma}{(s+2)(s-2)} - \frac{25}{16 (s-2)(s+1)} \right) \, ds$$
From this, it is possible to obtain the following:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4} i + \frac{25}{48} (\ln (2) - i \pi) - \frac{1}{8} + \frac{1}{16} i \pi + \frac{1}{4} \int_{0}^{i \pi} \psi^{(0)} \left( \frac{4}{1+ e^{u}} \right) \, du$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4}i+\frac{25}{48} (\ln (2)-i \pi )+\frac{7 i \pi }{48}-\frac{1}{8}-\frac{\ln (2)}{3} -2 \int_0^{\infty } \frac{t \ln (\Gamma (1-i t))}{\left(t^2+4\right)^2} \, dt$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = -\frac{1}{8}-\frac{i \pi }{4}+\frac{i \gamma \pi }{4}-\frac{\ln (2)}{16} - 2 \int_{0}^{\infty} \frac{t \ln (\Gamma (-i t)) }{(4+t^2)^2} \, dt$$
$$\implies \sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} =\frac{25}{48} \ln (2) -\frac{1}{8} + \int_{1}^{\infty} \frac{\ln (v-1) \text{li} (v^2)}{v^5} \, dv$$
Where $\text{li}$ is the logarithmic integral function.
$$\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} = \frac{3 \ln (2)}{16} - \frac{\pi^2+1}{8} - \frac{\pi}{2} \int_{0}^{\infty} \sin(4\pi x) (\psi (x) - \ln (x)) \, dx$$
|
Partial Answer
\begin{align}
\sum_{k=3}^\infty\frac{\ln k}{k^2-4}&=\sum_{k=3}^\infty\frac{\ln k}{k^2\left(1-\frac4{k^2}\right)}\\
&=\sum_{k=3}^\infty\frac{\ln k}{k^2}\sum_{l=0}^\infty\frac{2^{2l}}{k^{2l}}\\
&=\sum_{l=0}^\infty2^{2l}\sum_{k=3}^\infty\frac{\ln k}{k^{2l+2}}&&(\text{By Fubini's theorem})\\
&=\sum_{l=0}^\infty2^{2l}\left(-\zeta'(2l+2)-\frac{\ln2}{2^{2l+2}}\right)\\
&=-\sum_{l=0}^\infty\left(\frac{\ln2}{4}+\zeta'(2l+2)4^l\right)
\end{align}
And I couldn't proceed further. But as a side note, the convergence of this sum proves that
\begin{align}
&\lim_{l\to\infty}\frac{\zeta'(2l+2)2^{2l+2}}4=-\frac{\ln2}{4}\\
\implies &\lim_{x\to\infty}\zeta'(x)2^x=-\ln2
\end{align}
where the second step follows by the fact that Riemann zeta function is analytic.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "41",
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|
How to solve $x^4-2x^3-x^2+2x+1=0$? How to solve $x^4-2x^3-x^2+2x+1=0$?
Answer given is: $$\frac{1+\sqrt5}{2}$$
I tried solving it by taking common factors:
$$x^3(x-2)-x(x-2)+1=0 $$ $$x(x-2)(x^2-1)+1=0 $$ $$(x+1)(x)(x-1)(x-2)+1=0$$
But it's not leading me anywhere.
|
Another way to notice the factorization
$$x^4-2x^3-x^2+2x+1=0$$
Since $x=0$ is not the root of the equation, divide by $x^2$ to get
$$x^2 -2x-1 + \frac{2}{x} + \frac{1}{x^2} = 0$$
Rewrite it as
$$x^2+\frac{1}{x^2} - 2\left(x-\frac{1}{x}\right) - 1 = 0$$
or
$$\left(x-\frac1x\right)^2 + 2 - 2\left(x-\frac{1}{x}\right) - 1 = 0$$
Substitute $t = x - 1/x$
$$t^2 + 2 - 2t - 1 = 0\\
t^2 - 2t + 1 = 0 \\ (t-1)^2 = 0$$
Substitute back to get the final result
$$\left(x - \frac{1}{x} - 1\right)^2 = 0$$
which says
$$(x^2-x-1)^2 = 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4126617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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|
Find set of vectors orthogonal to $\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$ $\mathbf{Question:}$ Find set of vectors orthogonal to $\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}$
$\mathbf{My\ attempt:}$
The vector is in $R^3$ so we can let vector $\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix} $ represent orthogonal vectors.
$$\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}
.
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0 \\
0 \\
\end{bmatrix}
$$
$$
x_1 + x_2 + x_3 = 0
$$
$$
x_1 = -x_2 - x_3
$$
$$
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
-x_2 - x_3 \\
x_2 \\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
-1 \\
1 \\
0 \\
\end{bmatrix}x_2
+
\begin{bmatrix}
-1 \\
0 \\
1 \\
\end{bmatrix}x_3
$$
Therefore, the vectors in subspace $W = span(\begin{bmatrix}
-1 \\
1 \\
0 \\
\end{bmatrix} , \begin{bmatrix}
-1 \\
0 \\
1 \\
\end{bmatrix} )$ are orthogonal to vector $\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}$.
|
The space of vectors orthogonal to $\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$ can be defined by $span(\begin{bmatrix}
-1 \\
1 \\
0 \\
\end{bmatrix} , \begin{bmatrix}
-1 \\
0 \\
1 \\
\end{bmatrix} )$ where the dimension of the space is 2.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4129312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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|
Are the following systems of inequalities the same? Suppose $x, y \in \mathbb{R}$, and $\mathcal{S_1}$ is a system of inequalities:
\begin{align*}
\mathcal{S_1} &= \begin{Bmatrix}
x - y \geq 1\\
-x + 2y \geq 1\\
3x - 5y \geq 2
\end{Bmatrix}\\
&= \begin{Bmatrix}
x \geq 1 + y\\
x \leq 2y-1\\
x \geq \frac{2+5y}{3}
\end{Bmatrix}
\end{align*}
I eliminate $x$ from $S_1$ to obtain $S_2$:
\begin{align*}
\mathcal{S_2} &= \begin{Bmatrix}
1+y \leq 2y -1 \\
\frac{2+5y}{3} \leq 2y-1
\end{Bmatrix}\\
&= \begin{Bmatrix}
2 \leq y \\
2+5y \leq 6y-3
\end{Bmatrix}\\
&= \begin{Bmatrix}
2 \leq y \\
5 \leq y
\end{Bmatrix}
\end{align*}
Therefore, I know that $5 \leq y < \infty$. Let $S_3$ denote the following system of inequalities
\begin{align*}
\mathcal{S_3} &= \begin{Bmatrix}
x - y \geq 1\\
-x + 2y \geq 1\\
3x - 5y \geq 2\\
5 \leq y < \infty
\end{Bmatrix}
\end{align*}
My question is, is $S_3 = S_1$?
|
Yes, $S_3=S_1$. We need only to show that
$$5≤y<\infty$$
Therefore, we have,
$$3(-x+2y)+3x-5y≥3\times 1+2 $$
$$\iff y≥5.$$
But, this notation is problematic.
$$5≤y≤\infty$$
The correct one can be written as follows:
$$5≤y<\infty$$
Or,
$$y\in[5,+\infty)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Differentiate $x^{a^x}$ without logarithmic differentiation? Problem. Compute $\frac{d}{dx} x^{a^x}$.
Method 1: Logarithmic Differentiation (Correct):
\begin{align*}
y&= x^{a^x} \\
\ln(y)&=a^x \cdot \ln(x) \\
\frac{1}{y} \frac{dy}{dx} &=a^x \frac{1}{x} + \ln(x) \ln(a) a^x \\
\frac{dy}{dx} &= y\left(\frac{1}{x} a^x + \ln(a) \ln(x) a^x\right) \\
\frac{dy}{dx} &= x^{a^x}\left(\frac{1}{x} a^x + \ln(a) \ln(x) a^x\right) \\
\frac{dy}{dx} &= a^x x^{a^x}\left(\frac{1}{x} + \ln(a) \ln(x)\right) \\
\frac{dy}{dx} &= a^x x^{a^x-1}\left(1 + x\ln(a) \ln(x) \right) \\
\end{align*}
Method 2: Chain Rule (Incorrect):
\begin{align*}
y&= x^{a^x} \\
\frac{dy}{dx} &= a^x x^{a^x-1} \cdot \left[\ln(a) \cdot a^x\right]
\end{align*}
Method 2 is incorrect because it $x^{a^x}$ is not a power function, so we cannot apply power rule (thanks @Alann_Rosas and @Parcly_Taxel).
My Question: Can Ninad Munshi's answer here be adopted to correct method 2 without the use of logarithmic differentiation? What is the name (and/or proof) of this generalized version of chain rule?
Thank you!
|
It is not a misapplication of the chain rule, but of the power rule. The power rule does not work for non-constant exponents; by that logic the derivative of $x^x$ would be $x\cdot x^{x-1}=x^x$, which is flat-out wrong.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Prove $\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8}$ As the title says, prove $$\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8},$$ for $n>1$. This inequality is from Erdős, "Problems and results on the theory of interpolation". I, Lemma 3.
My attempt: since $H_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k} > \int_{1}^n\frac{1}{t}dt = \log(n),$ we then have
$$
\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} = \sum_{k=1}^{n-1}\frac{n^2 - 2kn + k^2}{2k} = \frac{n^2}{2}H_{n-1} - n(n-1) + \frac{n(n-1)}{4} > \frac{n^2\log(n)}{2} - \frac{3n(n-1)}{4}.
$$
Am I missing something here?
|
Using the stronger inequality, for $n>1$ that:
$$H_{n-1}>\frac{1}{2}\left(1-\frac1n\right) +\log n$$
(see proof below) we get:
$$\frac{n^2H_{n-1}}{2}-\frac{3n(n-1)}4>\frac{n^2\log n}2 +\frac{n(n-1)}4-\frac{3n(n-1)}4\\
=\frac{n^2\log n}2-\frac{n(n-1)}2$$
So you need:
$$\frac{3n^2\log n}8>\frac{n(n-1)}2$$
or $$\frac{\log n^3}4> 1-1/n.$$
But for $n\geq 4,$ we have $\log n^3>4$ and $1-1/n<1.$ For $n=2,$ you have $\log 2^3>2$ so $$\frac{\log 2^3}{4}>\frac12=1-\frac 12$$
For $n=3,$ you need $\log 27 >3> \frac 83.$
If you don’t want to use a calculator to show $\log(4^3)>4,$ use the estimate:
$$3\log(4)/4 >3(1/2+1/3+1/4)/4=\frac{13}{16}>1-1/4.$$
For $n= 5,$ $$3\log(n)/4>3(1/2+1/3+1/4+1/5)/4=\frac{77}{80}>\frac45.$$
It’s easier to prove $e^2<8,$ rather than $\log(8)>2.$
Just show $$e^2<1+2+\frac{2^2}2+\frac{2^3}6\cdot\frac1{1-2/4}=\frac{23}3<8.$$
Theorem: $$H_{n-1} >\frac12\left(1-\frac 1n\right)+\log n$$
Proof: The key is that $1/x$ is convex. For any convex function, $f$, the trapezoid estimate of the integral overestimates the integral (part of the Hermite-Hadamard inequality):
$$\frac{b-a}2(f(a)+f(b))\geq \int_a^b f(x)\,dx$$ with equality only if $f$ is linear on $[a,b].$ so for $f(x)=1/x,$ this means: $$\frac12\left(\frac 1{k-1}+\frac1k\right)>\int_{k-1}^k \frac{dx}x$$
That can be rewritten:
$$\frac{1}{k-1}>\frac12\left(\frac1{k-1}-\frac1k\right)+\int_{k-1}^k \frac{dx}x$$
Summing $k=2,\dots,n$, you get: $$H_{n-1}>\frac12\left(1-\frac1n\right) +\log n$$
|
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|
If $\frac{r^2}{AP^2}$ can stated in the form $\frac{a}{b}$ where $\gcd(a, b) = 1$. Determine the value of $a + 10b.$
Given a circle $ω$ with center $O$ and length of radius $r$. There is
a point $A$, $B$ on the circle and point $C$ outside the circle so
that $BC$ tangent to the circle $ω$. Then $AC$ intersects the circle
$ω$ at $D$. Then $P$ is the midpoint of $AC$ and $AD = BP = BC$. If $\frac{r^2}{AP^2}$ can stated in the form $\frac{a}{b}$ where $gcd$ $(a, b) = 1$. Determine the value of $a + 10b.$
Suppose $AP = y$, $DP = x$
with the power of point theorem, i get :
$BC^2=CD.AC\Longrightarrow (x+y)^2=2y(y-x)\Longrightarrow x^2+4xy-y^2=0\Longrightarrow y=x(2+\sqrt{5})$
i need an equation containing $x$ and $r$,but I didn't find it, if someone can help, that's great!!!
There are the original question, Page 4, Number 5.
|
The ratio ${\large{\frac{r^2}{AP^2}}}$ can be computed by brute force as outlined below.
*
*Assume $y=1$.
Since the ratio ${\large{\frac{r^2}{AP^2}}}$ is scale invariant, we can assume $y=1$.
*Compute $x$.
By power of a point we get $(x+1)^2=(1-x)(2)$, so
$$x=\sqrt{5}-2$$
*Compute $\cos\bigl(\angle{BCP}\bigr)$.
By the law of cosines in triangle $BCP$ we get
$$\cos\bigl(\angle{BCP}\bigr)=\frac{1}{2(x+1)}$$
*Compute $BD$.
By the law of cosines in triangle $BCD$ we get
$$BD=\sqrt{3-7x}$$
*Compute $\cos\bigl(\angle{CBD}\bigr)$.
By the law of cosines in triangle $CBD$ we get
$$\cos\bigl(\angle{CBD}\bigr)=\frac{3(1-x)}{2(1+x)\sqrt{3-7x}}$$
*Compute $\cos\bigl(\angle{OBD}\bigr)$.
Since angles $OBD$ and $CBD$ are complementary we get
$$
\cos\bigl(\angle{OBD}\bigr)
=
\sin\bigl(\angle{CBD}\bigr)
=
\sqrt{1-\cos^2\bigl(\angle{CBD}\bigr)}
=
\frac{\sqrt{748-198x}}{44}
$$
*Compute $r$.
By the law of cosines in triangle $OBD$ we get
$$r=\frac{22\sqrt{3-7x}}{\sqrt{748-198x}}$$
*Compute ${\large{\frac{r^2}{AP^2}}}$.
Since $AP=y=1$ and $x=\sqrt{5}-2$ we get
$$
\frac{r^2}{AP^2}
=
r^2
=
\frac{22(3-7x)}{34-9x}
=
\frac{1138-422\sqrt{5}}{209}
$$
Thus we've computed ${\large{\frac{r^2}{AP^2}}}$, but contrary to the claim of the problem statement, it's not a rational number.
|
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|
Modified exponential summation How do we prove that
$$1+\frac{2^3}{2!}+\frac{3^3}{3!}+\frac{4^3}{4!}+\cdots=5e$$
I'll post the answer to this question as a knowledge share.
|
Let $S$ denote the sum in question.
$$S=\sum\limits_{r=1}^{\infty}\left(\frac{r^3}{r!}\right)$$
Here, the general term
$$\begin{equation}\begin{aligned}
t_r &= \frac{r^3}{r!}\\
&=\frac{r^2}{(r-1)!}\\
&=\frac{(r-1+1)^2}{(r-1)!}\\
&=\frac{(r-1)^2}{(r-1)!}+\frac{1}{(r-1)!}+\frac{2(r-1)}{(r-1)!}\\
&=\frac{r-1}{(r-2)!}+\frac{1}{(r-1)!}+\frac{2}{(r-2)!}\\
&=\frac{r-2+1}{(r-2)!}+\frac{1}{(r-1)!}+\frac{2}{(r-2)!}\\
&=\frac{r-2}{(r-2)!}+\frac{1}{(r-2)!}+\frac{1}{(r-1)!}+\frac{2}{(r-2)!}\\
&=\frac{1}{(r-3)!}+\frac{3}{(r-2)!}+\frac{1}{(r-1)!}\\
\end{aligned}\end{equation}$$
Splitting $S$ to my convenience, I’ll write:
$$\begin{equation}\begin{aligned}
S&=1+\frac{2^3}{2!}+\sum\limits_{r=3}^{\infty}\left(\frac{r^3}{r!}\right)\\
&=1+\frac{2^3}{2!}+\sum\limits_{r=3}^{\infty}\left\{\frac{1}{(r-3)!}+\frac{3}{(r-2)!}+\frac{1}{(r-1)!}\right\}\\
\end{aligned}\end{equation}$$
From the exponential series, we have
$$e^x=\sum\limits_{r=0}^{\infty}\left(\frac{x^r}{r!}\right)$$
$$\implies e=\sum\limits_{r=0}^{\infty}\left(\frac{1}{r!}\right)$$
Utilizing this result, we can write,
$$
S=\left(1+\frac{2^3}{2!}\right)+e+(3e-3)+(e-2)=5e
$$
|
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|
Proof of $\int_{0}^{1}\dfrac{\log(x)\log(1 + 2x)}{x(1+x)} = -\dfrac{7}{6}\zeta(3)$ From quite sometime I've been struggling on proving
$$\int_{0}^{1}\dfrac{\log(x)\log(1 + 2x)}{x(1+x)}\,dx = -\dfrac{7}{6}\zeta(3)$$
I have tried partial fraction decomposition which produces a simple integral and a difficult one. I'm quite wondering if we can evaluate the integral without using the Linear combination of polylogarithms and logarithms. Any approach including complex analysis is most welcomed.
Any help would be appreciated. Thanks for reading.
|
Assuming $0 \leq x\leq 1$ a CAS gives for the antiderivative
$$\text{Li}_3\left(1+\frac{1}{2
x}\right)-\text{Li}_3\left(2+\frac{1}{x}\right)+\text{Li}_3(-2 x-1)+\text{Li}_3(-2
x)+\text{Li}_3(-x)+\left(\text{Li}_2\left(2+\frac{1}{x}\right)-\text{Li}_2\left(1+
\frac{1}{2 x}\right)\right) \log \left(\frac{1}{x}+2\right)-\text{Li}_2(-2 x-1)
\log (2 x+1)-(\text{Li}_2(-2 x)+\text{Li}_2(-x)) \log (x)+\frac{1}{2} \log (2)
\left(\log ^2(x)-\pi ^2\right)-\log (x) \log (2 (x+1)) \log (2 x+1)$$ I must confess that using the bounds I have a quite ugly expression
$$\text{Li}_3(-3)+\text{Li}_3(-2)+\text{Li}_3\left(\frac{3}{2}\right)-\text{Li}_3(3)-
\text{Li}_2(-3) \log (3)+2 \text{Li}_2(3) \log (3)-\frac{\log ^3(3)}{2}+(\log (2)+i
\pi ) \log ^2(3)+\frac{1}{6} \left(-\log ^3(2)+3 i \pi \log ^2(2)+2\pi ^2 \log
(2)\right)+\frac{1}{2} \left(\log \left(\frac{3}{2}\right)+i \pi \right)^2 \log
(3)$$ which numerically is the rhs.
If we use partial fraction decomposition,
$$\int \frac{\log (x) \log (2 x+1)}{x (x+1)}\,dx=\int \frac{\log (x) \log (2 x+1)}{x }\,dx-\int \frac{\log (x) \log (2 x+1)}{ (x+1)}\,dx$$
we have for the simplest one
$$int \frac{\log (x) \log (2 x+1)}{x }\,dx=\text{Li}_3(-2 x)-\text{Li}_2(-2 x) \log (x)$$ but, as you wrote, the second is not pleasant
$$\int \frac{\log (x) \log (2 x+1)}{ (x+1)}\,dx=-\text{Li}_3\left(1+\frac{1}{2
x}\right)+\text{Li}_3\left(2+\frac{1}{x}\right)-\text{Li}_3(-2
x-1)-\text{Li}_3(-x)+\left(\text{Li}_2\left(1+\frac{1}{2
x}\right)-\text{Li}_2\left(2+\frac{1}{x}\right)\right) \log
\left(\frac{1}{x}+2\right)+\text{Li}_2(-2 x-1) \log (2 x+1)+\text{Li}_2(-x) \log
(x)+\frac{1}{2} \log (2) \left(\pi ^2-\log ^2(x)\right)+\log (x) \log (2 (x+1))
\log (2 x+1)$$ and still the same problem when using the bounds.
|
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|
Can we use Vieta's formula in solving Trigonometric equations? **The value of $$\sec\frac{\pi}{11}-\sec\frac{2\pi}{11}+\sec\frac{3\pi}{11}-\sec\frac{4\pi}{11}+\sec\frac{5\pi}{11}$$
is ...
My Approach
I used the fact that $$\sec (\pi-x)=-\sec x$$ to simplify the equation to $$\sec\frac{\pi}{11}+\sec\frac{3\pi}{11}+\sec\frac{5\pi}{11}+\sec\frac{7\pi}{11}+\sec\frac{9\pi}{11}$$
Now I tried to devise an equation whose roots are $$\sec\frac{\pi}{11}, \sec\frac{3\pi}{11}, \sec\frac{5\pi}{11}, \sec\frac{7\pi}{11}, \sec\frac{9\pi}{11}$$
Afterwards, I found that the equation $$\cos \frac{11x}{2}=0 $$ satisfy the condition. But the equation has infinite number of roots, so my plan to use Vieta's formula to calculate the required sum did not work.
Please suggest how to proceed in this problem or share any other method.
|
Observe that $\cos(2n+1)\pi=-1$ for any integer $n$
If $11x=(2n+1)\pi,\cos6x=-\cos5x$
With $\cos x=c,$
$$\cos6x=2\cos^23x-1=2(4c^3-3c)^2-1=?$$
Using Prosthaphaeresis Formulas,
$$\cos5x+\cos x=2\cos3x\cos2x=2(4c^3-3c)(2c^2-1)$$ to find
$$0=\cos6x+\cos5x=32c^6+16c^5-48c^4-20c^3+18c^2+5c-1$$
So, the roots of $$32c^6+16c^5-48c^4-20c^3+18c^2+5c-1=0$$ are $c_n=\cos\dfrac{(2n+1)\pi}{11},0\le n\le 5$
But $c_5=-1$
So, the roots of $$32c^5-16c^4-32c^3+12c^2+6c-1=0$$ are $c_n=\cos\dfrac{(2n+1)\pi}{11},0\le n\le 4$
Set $\dfrac1c=s$ to find the roots of $$s^5-6s^4+\cdots-32=0$$ to be $s_n=\sec\dfrac{(2n+1)\pi}{11},0\le n\le 4$
Can you take it from here?
|
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|
Finding the global minimum of the multivariable function using only algebraic tools Problem says:
Find the global minimum of
$$\begin{align}f(x,y): &= x^2 + y^2 + \alpha xy + x + 2y\end{align}$$
where, $\alpha\in\mathbb R$.
The things I have done:
Let,
$$f(x,y):=x^2+x(\alpha y+1)+2y+y^2$$
*
*Algebraic tool I will use along the way:
$$ax^2+bx+c=a(x-m)^2+n$$
$$m=-\frac{b}{2a},~n=-\frac{\Delta}{4a}$$
In this case, we have
$$\begin{align}m:&=-\frac{\alpha y+1}{2} \\
n:&=\frac 14 \left(y^2(4-\alpha ^2) + 2y(4-
\alpha)-1\right)&\end{align}$$
Then,
$$\begin{align}f(x,y):=\left(x+\frac{\alpha y+1}{2}\right)^2+\frac 14 \left(y^2(4-\alpha ^2) + 2y(4-\alpha)-1\right)\end{align}$$
If $\alpha =±2$ , putting
$$x=-\frac{\alpha y+1}{2}$$
we get
$$f(x,y):=\frac{y(4-\alpha)}{2}-\frac 14$$
Since $4-\alpha>0$ for $\alpha=±2$, we observe that the lower bound of $\frac{y(4-\alpha)}{2}-\frac 14$ doesn't exist (applying $y\to -\infty$).
If $|\alpha|>2$, then applying the same method, we have
$$\begin{align}\frac 14 \left(y^2(4-\alpha ^2) + 2y(4-\alpha)-1\right)=y^2\left(1-\frac{\alpha ^2}{4}\right)+y\left(2-\frac{\alpha}{2}\right)-\frac 14=\left(1-\frac{\alpha^2}{4}\right)\left(y+\frac{4-\alpha}{4-\alpha ^2}\right)^2+\frac{2 \alpha - 5}{4 - \alpha^2}\end{align}$$
So, we obtain
$$\begin{align}f(x,y):=\left(x+\frac{\alpha y+1}{2}\right)^2+\left(1-\frac{\alpha^2}{4}\right)\left(y+\frac{4-\alpha}{4-\alpha ^2}\right)^2+\frac{2 \alpha - 5}{4 - \alpha^2}\end{align}$$
Likewise, if $|\alpha|>2$ then putting $$x=-\frac{\alpha y+1}{2}$$ we get,
$$\begin{align}f(x,y):=\left(1-\frac{\alpha^2}{4}\right)\left(y+\frac{4-\alpha}{4-\alpha ^2}\right)^2+\frac{2 \alpha - 5}{4 - \alpha^2}\end{align}$$
Observing that,
$$1-\frac{\alpha^2}{4}<0$$ where $|\alpha|>2$. This means, if $|\alpha|>2$, then the lower bound of
$$\begin{align}\left(1-\frac{\alpha^2}{4}\right)\left(y+\frac{4-\alpha}{4-\alpha ^2}\right)^2+\frac{2 \alpha - 5}{4 - \alpha^2}\end{align}$$
doesn't exist. (applying $y\to +\infty$)
Finally, if $|\alpha|<2\iff -2<\alpha<2$, then $1-\frac{\alpha^2}{4}>0$ and putting
$$x=-\frac{\alpha y+1}{2}, ~~y=\frac{\alpha-4}{4- \alpha^2}$$
in the original function
$$\begin{align}f(x,y):=\left(x+\frac{\alpha y+1}{2}\right)^2+\left(1-\frac{\alpha^2}{4}\right)\left(y+\frac{4-\alpha}{4-\alpha ^2}\right)^2+\frac{2 \alpha - 5}{4 - \alpha^2}\end{align}$$
we conclude
$$\begin{align}\min \left\{x^2 + y^2 + \alpha xy + x + 2y {\large{\mid}} -2<\alpha<2\right\}=\frac{2\alpha-5}{4-\alpha^2} ~\text {at}~ x=\frac{2-2\alpha}{\alpha^2-4} ~\text{and}~ y=\frac{\alpha-4}{4- \alpha^2}. \end{align}$$
Questions:
*
*Is the algebraic method I use rigorous enough? Are there still non-rigorous points in my steps?
*What I'm doing is just finding the minimum of the given function. How can I show that the minimum I found is a global minimum?
Thank you for reviewing.
|
Hint.
Another approach is to include $f(x,y)$ in the quadratic class as
$$
f(x,y) = (x-x_0,y-y_0)'M(x-x_0,y-y_0)+c
$$
Choosing $M = \left(
\begin{array}{cc}
1 & \frac{\alpha }{2} \\
\frac{\alpha }{2} & 1 \\
\end{array}
\right)$ and equating coefficients we arrive at
$$
\cases{
x_0 = -\frac{2 (\alpha -1)}{\alpha ^2-4}\\
y_0 = \frac{4-\alpha }{\alpha ^2-4}\\
c = \frac{5-2 \alpha }{\alpha ^2-4}
}
$$
and now follow the considerations about $M(\alpha)$ avoiding the case with $\alpha = \pm 2$
|
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"url": "https://math.stackexchange.com/questions/4139459",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Filling in details for calculation of the limit $\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right)$ I want to evaluate the following limit using asymptotics
\begin{equation}
\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) \tag{1}
\end{equation}
This is an example problem and this was the solution:
\begin{gather}
\frac{x^{3} +x}{1+x^{3}} =\left( 1+\frac{1}{x^{2}}\right)\left( 1+\frac{1}{x^{3}}\right)^{-1} =\left( 1+\frac{1}{x^{2}}\right)\left( 1-\frac{1}{x^{3}} +\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) \tag{2}\\
=1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{3}
\end{gather}
And then
\begin{gather}
\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} =\left( 1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\right)^{\frac{1}{7}} \tag{4}\\
=1+\frac{1}{7} .\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{5}
\end{gather}
\begin{equation}
\cos\frac{1}{x} =1-\frac{1}{2x^{2}} +\mathcal{O}\left(\frac{1}{x^{4}}\right) \tag{6}
\end{equation}
From above, we obtain:
\begin{equation}
\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{7}
\end{equation}
Hence the required limit is:
\begin{equation}
\lim _{x\rightarrow \infty }\left(\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\right) =\frac{9}{14} \tag{8}
\end{equation}
I tried to fill in the details for the steps involved in the above steps.
I supplied the details for $\displaystyle ( 3)$ as below:
\begin{gather}
\left( 1+\frac{1}{x^{2}}\right)\left( 1-\frac{1}{x^{3}} +\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) =1+\frac{1}{x^{2}} +\frac{1}{x^{3}}\left( -1-\frac{1}{x^{2}} +\left( x^{3} +x\right)\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) \tag{9}\\
=1+\frac{1}{x^{2}} +\frac{1}{x^{3}}\left( -1-\frac{1}{x^{2}} +\frac{\left( x^{3} +x\right)}{x^{6}}\mathcal{O}( 1)\right) =1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{10}
\end{gather}
where in (9) and (10), I have used the standard result: $\displaystyle \frac{\mathcal{O}( f( x))}{g( x)} =\mathcal{O}\left(\frac{f( x)}{g( x)}\right)$, if $\displaystyle g( x) \neq 0$.
I tried to get (5) from (4) but failed.
Then I supplied details for $\displaystyle ( 7)$ as below:
\begin{gather*}
\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) -\mathcal{O}\left(\frac{1}{x^{4}}\right) =\frac{9}{14x^{2}} +\frac{1}{x^{3}}\left(\mathcal{O}( 1) -\mathcal{O}\left(\frac{1}{x}\right)\right)\\
=\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\\
\Longrightarrow \lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\lim _{x\rightarrow \infty }\left(\frac{9}{14} +\mathcal{O}\left(\frac{1}{x}\right)\right) =\frac{9}{14}
\end{gather*}
Any help in getting (5) from (4) is much appreciated. Thanks.
\begin{equation*}
\end{equation*}
\begin{equation*}
\end{equation*}
|
You can make life easier letting $x=\frac 1y$. This makes the expression
$$A=\frac 1{y^2} \left(\frac{\sqrt[7]{1+y^2} }{\sqrt[7]{1+y^3} } - \cos(y)\right)$$ Now, using the binomial theorem or Taylor series for the surds and Taylor series for the cosine, you have
$$A=\frac 1{y^2} \left(\frac{1+\frac{y^2}{7}-\frac{3 y^4}{49}+O\left(y^6\right)}{1+\frac{y^3}{7}-\frac{3 y^6}{49}+O\left(y^9\right) } -\left(1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^{6}\right)
\right)\right)$$ Long division and simplification lead to
$$A=\frac{9}{14}-\frac{y}{7}-\frac{121 y^2}{1176}+O\left(y^3\right)$$ which shows more than the limit.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4142617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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|
Hermite interpolation of even degree Using Hermite interpolation determine the fourth degree polynomial $p(x)$ for which
\begin{equation}
p(0) = p'(0) = 0, \quad p(1) = p'(1) = 1, \quad p(2) = 1
\end{equation}
I can only seem to find content on how to apply Hermite interpolation to data points with odd degree polynomial. Some guidance would be greatly appreciated.
|
Let $P_0(x) =x^2, P_1(x)=(x-1)^2, P_2(x)=x-2$.
Let $A_0(x) = 0, A_1(x) = 1+(x-1)=x, A_2(x)=1$.
We want to find $p(x)$ such that $$p(x) \equiv A_i(x) \pmod{P_i(x)}$$
We let $$Q(x)=x^2(x-1)^2(x-2)$$
Denote $Q_i(x) = \frac{Q(x)}{P_i(x)}.$ We have
$$Q_0(x) =(x-1)^2(x-2)$$
$$Q_1(x) = x^2(x-2)$$
$$Q_2(x)=x^2(x-1)^2$$
Let $$\sum_{i=0}^2S_i(x)Q_i(x) = 1$$
where $degree(S_i)<degree(P_i)$
$$S_0(x)(x-1)^2(x-2)+S_1(x)x^2(x-2)+S_2(x)x^2(x-1)^2=1\tag{1}$$
Let $x=2$, we conclude that $S_2(x) = \frac14$.
Let $x=1$, we have $S_1(1)=-1$. We let $S_1(x)=ax+b$, we have $a+b=-1$.
Differentiating $(1)$ once,
$$\frac{d}{dx}[S_0(x)(x-1)^2(x-2)+S_2(x)x^2(x-1)^2]+\frac{d}{dx}[S_1(x)(x^3-2x^2)]=0$$
$$\frac{d}{dx}[S_0(x)(x-1)^2(x-2)+S_2(x)x^2(x-1)^2]+a(x^3-2x^2)+(ax+b)(3x^2-4x)=0$$
Let $x=1$, we have $-a-(a+b)=0$, that is $b=-2a$. That is $a=1$ and $b=-2$.
$S_1(x)=x-2$.
Now let's compute $A_1(x)S_1(x) \pmod{P_1(x)}$ to reduce the degree:
\begin{align}
A_1(x) S_1(x) &\equiv x(x-2) \pmod{(x-1)^2} \\
&\equiv x^2-2x \pmod{(x-1)^2} \\
&\equiv x^2-2x+1-1 \pmod{(x-1)^2} \\
&\equiv -1 \pmod{(x-1)^2}.
\end{align}
Let $B_i \equiv A_iS_i \pmod{P_i}$,
The desired polynomial is $\sum_{i=0}^2 B_i(x)Q_i(x)$, that is
\begin{align}-Q_1(x)+ \frac14Q_2(x)&=-x^2(x-2)+\frac14x^2(x-1)^2 \\
&=x^2 \left[ -(x-2) + \frac14(x-1)^2\right]\end{align}
Relevant reading from wikipedia
|
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|
Need help with this! $\lvert x-y \rvert\lt\epsilon^2 \Rightarrow \lvert\,\lvert\sqrt x \rvert - \lvert \sqrt y \rvert \,\rvert \lt \epsilon$ Given $\Bbb R$, $ \epsilon \gt 0$. Prove that any positive $\Bbb R$
$$\lvert x-y \rvert\lt\epsilon^2 \Rightarrow \bigl\lvert\lvert \sqrt x \rvert - \lvert \sqrt y \rvert \bigr\vert \lt \epsilon$$
I'm trying to get this done all and I get is an dead end, such as squaring everyone up
$\left( \ \vert \ x-y \ \vert \ \right)^2\lt\left( \ \epsilon^2 \ \right)^2 \Rightarrow \\$
$\left( \ x-y \ \right)^2\lt\ \epsilon^4 \Rightarrow \\$
$x^2 - 2xy + y^2\lt\ \epsilon^4$
and that's pretty much it, from here can't go anywhere close to the objective.
Then again tried another aproach which led me to another dead end.
$$\vert \ x-y \ \vert \lt\epsilon^2$$
$$\Rightarrow u = x-y$$
$$\vert \ u \ \vert \lt \epsilon^2$$
$$\Rightarrow -\left( \epsilon^2 \right) \lt u \lt \epsilon^2$$
$$\Rightarrow \epsilon \gt 0$$
$$u \lt \epsilon^2$$
$$x-y\lt\epsilon^2$$
$$\sqrt {x-y}\lt\epsilon$$
$$\left(x-y\right)^{1/2}\lt\epsilon$$
After this I coudn't go any further and I ran out of any ideas, would anyone be kind to show show a proper solution?
|
Whenever I see something like $a^2 - b^2$ in these kind of problems, I often try to see if factoring it into $(a-b)(a+b)$ is fruitful.
In this problem, we do not have $x^2 - y^2$, but we do have $|x - y|$ and also $|\sqrt{x} - \sqrt{y}|$, which suggested to me that using $|x - y| = |(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})|$ might be fruitful.
In this problem, if $x$ and $y$ are both nonnegative, then:
\begin{align}
||\sqrt{x}| - |\sqrt{y}|| &= |\sqrt{x} - \sqrt{y}| \ge \epsilon \\
& \implies \max(\sqrt{x}, \sqrt{y}) \ge \epsilon \label{max}\\
& \implies \sqrt{x} + \sqrt{y} \ge \epsilon \\
& \implies |\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| \ge \epsilon^2 \\
&\implies |x - y| \ge \epsilon^2
\end{align}
Therefore $|x - y| < \epsilon^2 \implies |\sqrt{x} - \sqrt{y}| < \epsilon$
Edit to add more details, as requested:
Since $\sqrt{x}$ and $\sqrt{y}$ are both greater than or equal to zero, and the distance between them is at least $\epsilon$, then we know that $\max(\sqrt{x}, \sqrt{y}) \ge \epsilon $. Otherwise, they would both be in the interval $[0, \epsilon)$, which would make it impossible for the distance between them to be at least $\epsilon$.
Then, since they are both nonnegative, we know that $\sqrt{x} + \sqrt{y} \ge \max(\sqrt{x}, \sqrt{y}) $, hence $\sqrt{x} + \sqrt{y} \ge \epsilon$.
So we started with assumption that $|\sqrt{x} - \sqrt{y}| \ge \epsilon$, and deduced (using the nonnegativity of $x$ and $y$) that $\sqrt{x} + \sqrt{y} \ge \epsilon$, which combine to give:
$$|\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| \ge \epsilon^2$$
Since $|\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| = |x-y|$, that means that we have shown that:
$$|\sqrt{x} - \sqrt{y}| \ge \epsilon \implies |x - y| \ge \epsilon^2$$
Therefore, we have also shown the contrapositive:
$$|x - y| < \epsilon^2 \implies |\sqrt{x} - \sqrt{y}| < \epsilon $$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4143557",
"timestamp": "2023-03-29T00:00:00",
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|
$f(a)=b,f(b)=c,f(c)=a$, find $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Let $f(x)=x^2-x-2$. It is given that $a, b,c \in \mathbb{R}$ such that
$$f(a)=b,f(b)=c,f(c)=a$$ Then find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.
My attempt:
We have:
$$\begin{array}{l}
a^{2}-a-2=b \\
b^{2}-b-2=c \\
c^{2}-c-2=a
\end{array}$$
Subtracting pair wise we get:
$$(a-b)(a+b)=a-c$$
$$(b-c)(b+c)=b-a$$
$$(c-a)(c+a)=c-b$$
Trivially $a=b=c$ satisfies. So we get $a=b=c=\sqrt{3}+1, 1-\sqrt{3}$
Thus we get
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3\sqrt{3}+3}{2}, \frac{-3-3\sqrt{3}}{2}$$
Now if $a\ne b \ne c$
Then we have:
$$(a+b)(b+c)(c+a)=-1$$
Letting $p=a+b+c$ we have
$$(p-a)(p-b)(p-c)=-1$$ $\implies$
$$p(ab+bc+ca)-abc+1=0$$
Any hint here?
|
From the condition, we need $f\circ f\circ f(a)=a$. Now
$$f(x) = x^2-x-2=(x-2)(x+1)$$
$$f\circ f(x) = (x^2-x-4)(x^2-x-1)=x^4-2x^3-4x^2+5x+4$$
$$\implies f\circ f\circ f(x)=(x^4-2x^3-4x^2+4x+2)(x^4-2x^3-4x^2+4x+5)$$
$$\implies f\circ f\circ f(a)-a=a^8-4a^7-4a^6+26a^5+3a^4-54a^3-3a^2+34a+10 \\=(a^2-2a-2)(a^3-3a-1)(a^3-2a^2-3a+5)$$
It is easy to note all roots are real (using Descartes and evaluation at $a=-1, 0, 1$), so we have $8$ possibilities for $a$, which fall into two cases.
Case 1: The first/quadratic factor gives $a=1\pm\sqrt3$, each of which is also a solution to $f(a)=a$, so as you have already noted, we have two values for $\frac1a+\frac1b+\frac1c=\frac3a$ from these, viz. $\frac12(\pm3\sqrt3-3)$.
Case 2: Now we have no more roots of $f(a)=a$, hence the rest of the roots (six roots from the two cubic factors), must contribute to two possible sets of $\{a, b, c\}$.
Taking the first cubic factor,
$\alpha^3-3\alpha-1=0 \implies f(\alpha)^3-3f(\alpha)-1 = (\alpha^2-\alpha-2)^3-3(\alpha^2-\alpha-2)-1 \\=(\alpha^3-3\alpha-1)(\alpha^3-3\alpha^2+3)=0$
i.e. if $\alpha$ is a root of the first cubic, so is $f(\alpha)$. Thus the roots of that cubic contribute one set of $\{a, b, c\}$. Now using Vieta, sum of the reciprocal of roots is $$\frac1a+\frac1b+\frac1c=\frac{ab+bc+ca}{abc}=-3$$
Similarly the roots of the last cubic form the last set of possible $a, b, c$, hence have a sum or reciprocals $=\frac35$.
Summarising the value of $\frac1a+\frac1b+\frac1c \in \{\frac12(\pm3\sqrt3-3), -3, \frac35\}$, four possibilities.
|
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|
Solve Integral -$ \int \sqrt{3x^2 + 2x}\ dx$ I am trying to work out the integral without just skipping straight to the formula:
$$\int \sqrt{u^2-a^2} du = \frac{u}{2}\sqrt{u^2-a^2} - \frac{a^2}{2} \ln\left|u+\sqrt{u^2-a^2}\right|+C $$
I did all the substitutions, etc.
$$u=x+1/3,\,a=1/3,\,a\sec\theta=u,\,du=\sec\theta\tan\theta d\theta$$
and get
$$\sqrt3 \left [ \frac{u}{2}\sqrt{u^2-a^2} - \frac{a^2}{2} \ln\left|\frac{u}{a}+\frac{\sqrt{u^2-a^2}}{a}\right| \right] +C $$
then
$$\sqrt3 \left [ \frac{1}{2}\left(x+\frac{1}{3}\right)\sqrt{x^2+\frac{2}{3}x} - \frac{1}{18} \ln\left|\left(3\left(x+\frac{1}{3}\right)+3\sqrt{x^2+\frac{2}{3}}\right)\right| \right] +C $$
I have 2 questions -
*
*is my answer correct?
*And what happened to the $a$ under $\ln\left|\frac{u}{a}+\frac{\sqrt{u^2-a^2}}{a}\right|$ in the original general formula assuming $a\ne1$.
Thanks.
|
Yes your answer is correct. Here $a=\frac 13$, so $\frac 1a=3$, as you have written. Note that integration by parts can also be used to solve the problem.
|
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|
Triple integral between $z^2 = {(x - 1)}^2 + y^2$ and the unit sphere. I have to compute $\int_D f$, where $D$ is the region in ${(0 , \infty)}^3$ between the cone $z^2 = {(x - 1)}^2 + y^2$ and the sphere $x^2 + y^2 + z^2 = 1$, and $f : D \to \mathbb{R}$ is given by $f(x , y , z) = z \sqrt{x^2 + y^2}$.
My attempt: If I use spherical coordinates $(\rho , \varphi , \theta) \in (0 , \infty) \times (0 , 2 \pi) \times (0 , \pi)$ given by
$$
\left\{
\begin{array}
xx = \rho \sin \varphi \cos \theta \\
y = \rho \sin \varphi \sin \theta \\
z = \rho \cos \varphi
\end{array}
\right.
$$
then we can start to deduce $\rho \leq 1$ from the equation of the sphere, and $\varphi < \frac{\pi}2$ is deduced from $z > 0$. We can use this to see furthermore that $\theta < \frac{\pi}2$ because $x , y > 0$. The equation of the cone says
$$
{\rho}^2 {\cos}^2 \varphi = {\rho}^2 {\sin}^2 \varphi - 2 \rho \sin \varphi \cos \theta + 1.
$$
This helps appearently nothing. I am wondering if spherical coordinates are helpful to fix this integral.
|
Cylindrical coordinates is easier.
Equation of the surface of the cone, $z^2 = (x - 1)^2 + y^2$
Equation of the surface of the sphere, $x^2 + y^2 + z^2 = 1$
At intersection,
$z^2 = (x - 1)^2 + y^2 = 1 - x^2 - y^2 \implies x^2 - x + y^2 = 0$
This is the projection in xy plane, of the intersection of cone with the sphere. It is circle of radius $\frac{1}{2}$ at $(\frac{1}{2}, 0)$ in xy-plane.
Now in polar coordinates, $x = r \cos\theta, y = r\sin\theta$ so the equation of the circle becomes,
$r^2 - r \cos\theta = 0 \implies r = \cos\theta, 0 \leq \theta \leq \frac{\pi}{2}$ as we are only supposed to consider region in first octant.
Bounds of $z$ simply come from the equations of cone and the sphere. It is bound below by the cone and above by the sphere. So,
$ \sqrt {1 + r^2 - 2 r\cos\theta} \leq z \leq \sqrt{1-r^2}$
So integral becomes,
$\displaystyle \int_0^{\pi/2} \int_0^{\cos\theta} \int_{\sqrt {1 + r^2 - 2 r\cos\theta}}^{\sqrt{1-r^2}} r^2 z \ dz \ dr \ d\theta$
Given the inner most integral will return $z^2$, the integral becomes simpler to evaluate.
|
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"timestamp": "2023-03-29T00:00:00",
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|
System of two quadratics equation, $P(x)$ and $Q(x)$ If $P(x) = ax^2 + bx + c$ and $Q(x) = – ax^2 + dx + c$, $ac \ne 0$, then the equation $P(x) . Q(x) = 0$ has
(A) Exactly two real roots
(B) At least two real roots
(C) Exactly four real roots
(D) No real roots
My approach is as follow Let $T(x)=P(x).Q(x)$
$T\left( x \right) = - a^2{x^4} + a\left( {d - b} \right){x^3} + \left( {bd} \right){x^2} + c\left( {d + b} \right)x + {c^2}$
Not able to approach from here
|
Set $PQ=\left( ax^2+bx+c\right) \left( -ax^2+dx+c\right)$ to zero, to get $$\left( ax^2+bx+c\right) \left( -ax^2+dx+c\right)=0.$$
Check the discriminant for both $P$ and $Q$.
For $P$, you get $D_P=b^2-4ac$. For $Q$, you get $D_Q=d^2+4ac$.
Suppose $P$ has no real roots, such that $D_P=b^2-4ac < 0 \implies4ac>b^2$. In that case, $D_Q=d^2+4ac>0$ and $Q$ thus has two roots, so your polynomial $PQ$ has at least two roots.
Next, suppose $P$ has 1 real root, in which case $4ac=b^2$ and so $D_1 = d^2+b^2$ which has 2 roots, so $PQ$ has 3.
Lastly, suppose $P$ has 2 real roots, and thus $b^2>4ac$, in which case $Q$ may have zero, one, or two real roots.
Therefore, $PQ$ has at least two real roots.
|
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|
Simplifying $\left({\sqrt{x} + \frac{1}{\sqrt{x}}}\right)^2 - \left({\sqrt{x} - \frac{1}{\sqrt{x}}}\right)^2 $ Hi can someone help me please simplify the following showing the working out step by step?
$$
\left({\sqrt{x} + \frac{1}{\sqrt{x}}}\right)^2 - \left({\sqrt{x} - \frac{1}{\sqrt{x}}}\right)^2
$$
I can't get the answer matching the text book but I'd also like to get an idea of the most idiomatic way to solve it in terms of steps.
What I attempt to do based on what I've learned so far is to:
*
*try and simplify the contents of the parens using conjugate and then LCM
*then square
*then handle the subtraction.
But my answer ends up incorrect.
So even steps just to simplify say the left hand term (without the squaring step) would be helpful.
My simplifying the left hand looks like:
Use Conjugate:
$$
\left(\sqrt{x} + \frac{(1)(\sqrt{x})}{(\sqrt{x})(\sqrt{x})}\right)^2
$$
$$
\left(\sqrt{x} + \frac{(\sqrt{x})}{x}\right)^2
$$
Use LCM:
$$
\left(\frac{(\sqrt{x})(x)}{x} + \frac{(\sqrt{x})}{x}\right)^2
$$
$$
\left(\frac{(x)(\sqrt{x}) + \sqrt{x}}{x}\right)^2
$$
Then I'm not sure next best step.
|
Just do it:
$$
\left({\sqrt{x} + \frac{1}{\sqrt{x}}}\right)^2 - \left({\sqrt{x} - \frac{1}{\sqrt{x}}}\right)^2
$$
$$=\left({ {x} + 2+\frac{1}{ {x}}}\right) - \left(x-2+\frac1x\right)$$
$$=4$$
using $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$ with $a=\sqrt x$ and $b=\dfrac1{\sqrt x}$.
ADDENDUM to answer question in OP's comment
To start from where you left off,
$$
\left(\frac{x\sqrt{x} + \sqrt{x}}{x}\right)^2=\frac {(x\sqrt x+\sqrt x)^2}{x^2}
=\frac{x^3+2x^2+x}{x^2}=x+2+\dfrac1x.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations and worked out examples convince me that this inequality indeed holds true. I have tried to solve this problem by induction. Clearly, for $N=1$ we have,
$$\mathcal{P}(1) = \left(\frac{2}{3}\right)^{\tfrac{3}{4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{4}} \approx 0.8774 < 1.$$
Now assume the inequality holds for $N$, then for $N+1$ we have,
\begin{align}
\mathcal{P}(N+1) &=\left(\frac{2N+2}{2N+3}\right)^{\tfrac{2N+3}{2N+4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{2N+4}} \\
&= \left(\left(\frac{2N}{2N+1}\right)\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)\right)^{\left(\frac{2N+1}{2N+2}\right)\left(\frac{2N+2}{2N+1}\cdot\frac{2N+3}{2N+4}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\frac{2N+2}{2N+4}}\\[1em]
&= \left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)\left(1 + \frac{1}{2(N+1/2)(N+2)}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\left(1 - \frac{1}{N+2}\right)}\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \small\left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}}\cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \mathcal{P}(N) \cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \cdot\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}
\end{align}
Now from here we know that the first three terms are all smaller than 1 ($\mathcal{P}(N) < 1$ by induction hypothesis). However the last term is larger than one. For the proof by induction to work out, we need that this last term cancels against,
$$\left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}}.$$ But I do not see how it does. Any help is greatly appreciated.
|
Written in a more compact form
$$P_n=2 \left(\frac{n}{2 n+1}\right)^{1-\frac{1}{2 (n+1)}}$$
Take logarithms and compose Taylor series
$$\log(P_n)=\frac{\log (2)-1}{2 n}+\frac{3-4 \log (2)}{8
n^2}+O\left(\frac{1}{n^3}\right)$$
Continue with Taylor
$$P_n=e^{\log(P_n)}=1-\frac{1-\log (2)}{2 n}+\frac{4+(\log (2)-6) \log (2)}{8
n^2}+O\left(\frac{1}{n^3}\right)$$
Even for $n=1$, the above truncated expansion is not bad. In fact, the relative error is less than $0.01$% as soon as $n \geq 5$.
|
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|
Let $c$ be a positive real number for which the equation $x^4-x^3+x^2-(c+1)x-(c^2+c)=0$ has a real root $\alpha$. Prove that $c=\alpha ^2 - \alpha$ Let $c$ be a positive real number for which the equation
$x^4-x^3+x^2-(c+1)x-(c^2+c)=0$
has a real root $\alpha$. Prove that $c=\alpha ^2 - \alpha$
I tried to to solve using relation between roots and coefficients but unable to progress much. Please help. Thanks in advance.
|
Direct substitution with $\alpha=x$ and $c=x^2-x$ leads to.
$x^4-x^3+x^2-x(x^2-x+1)-((x^2-x)^2+x^2-x)$
$=(1-1)x^4+(2-2)x^3+(2-2)x^2+(1-1)x=0$
|
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|
A First Order Homogenous Differential Equation The following problem is from Schaum's book on Differential Equations.
Problem:
Solve the following differential equation.
$$ y' = \frac{ x^4 + 3x^2y^2 + y^4}{x^3 y} $$
Answer:
We rewrite this equation as:
$$ \frac{dy}{dx} =
\frac{ 1 + 3 \left( \dfrac{y^2}{x^2}\right) + \dfrac{y^4}{x^4} }{ \dfrac{y}{x} } $$
\begin{align*}
\text{Let }y &= xv \\
\frac{dy}{dx} &= x \frac{dv}{dx} + v \\
x \frac{dv}{dx} + v &= \frac{ 1 + 3 v^2 + v^4 }{ v} \\
x \frac{dv}{dx} &= \frac{ v^4 + 2v^2 + 1 } { v } \\
x \frac{dv}{dx} &= \frac{ (v^2 + 1)^2 } { v } \\
\frac{v \,\, dv}{(v^2+1)^2} &= \frac{dx}{x}
\end{align*}
Now we need to perform the following integration:
$$ \int \frac{v}{(v^2+1)^2} \,\, dv $$
We use the substitution $u = v^2 + 1$.
\begin{align*}
du &= 2v \, dv \\
\int \frac{v}{(v^2+1)^2} \,\, dv &= \int \left( \frac{1}{2} \right) u^{-2} \, \, du \\
\int \frac{v}{(v^2+1)^2} \,\, dv &= - \left( \frac{1}{2} \right) u^{-1} \\
\int \frac{v}{(v^2+1)^2} \,\, dv &= - \frac{1}{2(v^2+1)} \\
\end{align*}
\begin{align*}
- \frac{1}{2(v^2+1)} &= \ln{|x|} + C_1 = \ln{|x|} + \ln( C_2 ) \\
- \frac{1}{2 \left( \left( \dfrac{y}{x} \right) ^2 + 1\right) } &= \ln{|C_2 x|} \\
- \frac{1}{2 \left( y^2 + x^2 \right) } &= \ln{|C_2 x|} x^2 \\
- \frac{1}{\left( y^2 + x^2 \right) } &= 2 \ln{|C_2 x|} x^2 = \ln{|C_2 x^2|} x^2 \\
\frac{1}{\left( y^2 + x^2 \right) } &= - \ln{|C_2 x^2|} x^2 \\
\end{align*}
However, the book's answer is:
$$ y^2 = -x^2 \left( 1 + \dfrac{1}{ \ln{|kx^2|} } \right) $$
Where did I go wrong?
Here is my updated and correct solution to the problem.
We rewrite this equation as:
$$ \frac{dy}{dx} =
\frac{ 1 + 3 \left( \dfrac{y^2}{x^2}\right) + \dfrac{y^4}{x^4} }{ \dfrac{y}{x} } $$
\begin{align*}
\text{Let }y &= xv \\
\frac{dy}{dx} &= x \frac{dv}{dx} + v \\
x \frac{dv}{dx} + v &= \frac{ 1 + 3 v^2 + v^4 }{ v} \\
x \frac{dv}{dx} &= \frac{ v^4 + 2v^2 + 1 } { v } \\
x \frac{dv}{dx} &= \frac{ (v^2 + 1)^2 } { v } \\
\frac{v \,\, dv}{(v^2+1)^2} &= \frac{dx}{x}
\end{align*}
Now we need to perform the following integration:
$$ \int \frac{v}{(v^2+1)^2} \,\, dv $$
We use the substitution $u = v^2 + 1$.
\begin{align*}
du &= 2v \, dv \\
\int \frac{v}{(v^2+1)^2} \,\, dv &= \int \left( \frac{1}{2} \right) u^{-2} \, \, du \\
\int \frac{v}{(v^2+1)^2} \,\, dv &= - \left( \frac{1}{2} \right) u^{-1} \\
\int \frac{v}{(v^2+1)^2} \,\, dv &= - \frac{1}{2(v^2+1)} \\
\end{align*}
\begin{align*}
- \frac{1}{2(v^2+1)} &= \ln{|x|} + C_1 = \ln{|x|} + \ln( C_2 ) \\
- \frac{1}{2 \left( \left( \dfrac{y}{x} \right) ^2 + 1\right) } &= \ln{|C_2 x|} \\
- \frac{1}{2 \left( y^2 + x^2 \right) } &= \frac{ \ln{|C_2 x|} } { x^2 } \\
- \frac{1}{\left( y^2 + x^2 \right) } &= \frac{ 2 \ln{|C_2 x|} } { x^2 } \\
- \frac{1}{\left( y^2 + x^2 \right) } &= \frac{ \ln{|C_2 x^2|} } { x^2 } \\
\frac{1}{\left( y^2 + x^2 \right) } &= \frac{ - \ln{|C_2 x^2|} } { x^2 } \\
\end{align*}
Now, I need to simplify my answer. I do this by cross multiplying.
\begin{align*}
x^2 &= -(y^2 + x^2)\ln{|C_2 x^2|} \\
x^2 + x^2 \ln{|C_2 x^2|} &= -y^2 \ln{|C_2 x^2|} \\
\frac{x^2}{\ln{|C_2 x^2|}} + x^2 &= -y^2 \\
x^2 \left( 1 + \dfrac{1 } {\ln{|C_2 x^2|}} \right) &= -y^2 \\
y^2 &= -x^2 \left( 1 + \dfrac{1}{ \ln{|kx^2|} } \right)
\end{align*}
|
Check the step after $$-\frac {1}{2((\frac yx)^2+1)}=\ln|C_2x|$$
You should get $$-\frac {1}{2(x^2+y^2)}=\frac {\ln|C_2x|}{x^2}$$
That is not what you wrote.
|
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|
The solutions of the equation $|p^3-q^2|=2^n$ Let's consider the following equation
$$|p^3-q^2|=2^n$$
where $p$, $q$ are distinct primes and $n$ a non-negative integer. The equation admits some trivial solutions:
$$|2^3-3^2|=2^0$$
$$|3^3-5^2|=2^1$$
$$|5^3-11^2|=2^2$$
$$|17^3-71^2|=2^7$$
Clearly, not all values of $n$ admit a solution (for example, if $3|n$ the equation has no solution).
Are there other solutions besides those mentioned above?
Many thanks.
|
We consider the elliptic curves $y^2 = x^3\pm2^n$ with parameter $n$ where $x$ and $y$ are prime number .
Searching the integer points $(x,y)$ with $\boldsymbol{n \lt 100}$ and $\boldsymbol{x \lt 10^{10}}$, we found the integer points $(x,y,n)=(3, 5, 1)$ and $(5, 11, 2)$ for $y^2 = x^3-2^n.$
Similarly, we found the integer point $(x,y,n)= (17, 71, 7))$ for $y^2 = x^3+2^n.$
There seems to be no other solution.
|
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|
How to solve $\frac{1}{2\sin50^\circ}+2\sin10^\circ$ I want to solve the expression $\frac{1}{2\sin50^\circ}+2\sin10^\circ$ to get a much simpler and neater result. I have tried to manipulate this expression such as using sum/difference formulas, but it didn't help (and made the expression even more messy).
Here is what I did:
\begin{align}
\frac{1}{2\sin50^\circ}+2\sin10^\circ&=\frac{1}{2\sin(60-10)^\circ}+2\sin10^\circ \\
&= \frac{1}{2\left(\frac{\sqrt{3}}{2}\cdot\cos10^\circ-\frac{1}{2}\cdot\sin10^\circ\right)}+2\sin10^\circ \\
&= \frac{1}{\sqrt{3}\cdot\cos10^\circ-\sin10^\circ}+2\sin10^\circ \\
&= \frac{\sqrt{3}\cdot\cos10^\circ+\sin10^\circ}{\left(\sqrt{3}\cdot\cos10^\circ-\sin10^\circ\right)\left(\sqrt{3}\cdot\cos10^\circ+\sin10^\circ\right)}+2\sin10^\circ \\
&= \frac{\sqrt{3}\cdot\cos10^\circ+\sin10^\circ}{3\cos^210^\circ-\sin^210^\circ}+2\sin10^\circ
\end{align}
But I don't know how to continue at this point. Multiplying in $2\sin10^\circ$ into the fraction is clearly unrealistic as it would result in trignometry of third power. Any help or hint would be appreciated. According to a calculator, the result of this expression should come to a nicely $1$, but I just want to know how to algebraically manipulate this expression to show that it is equal to $1$.
|
$$\frac{1}{2\sin{50^{\circ}}} + 2 \sin{10^{\circ}}= \frac{1}{2\sin{50^{\circ}}} + 2 \sin{(60^{\circ}-50^{\circ})}=\frac{1}{2\sin{50^{\circ}}} + \sqrt{3} \cos{50^{\circ}} -\sin{50^{\circ}} =$$
$$=\frac{1-2 \sin^2{50^{\circ}}}{2\sin{50^{\circ}}}+\frac{2\sqrt{3} \cos{50^{\circ}}\sin{50^{\circ}}}{2\sin{50^{\circ}}}=\frac{\cos{100^{\circ}}}{2\sin{50^{\circ}}}+\frac{\sqrt{3}\sin{100^{\circ}}}{2\sin{50^{\circ}}} =$$
$$=\frac{\frac{1}{2}\cos{100^{\circ}} + \frac{\sqrt{3}}{2} \sin{100^{\circ}}}{\sin{50^{\circ}}} = \frac{\sin{(30^{\circ}+100^{\circ})}}{\sin{50^{\circ}}}=1$$
|
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|
Find continuous function $f$ such that $\int_0^{\infty} f(x) \left(\frac{x^2}{1+x^2} \right)^n dx =\infty$ for all integers $n \geq1$. Let $f(x)$ be a continuous function on $[0,\infty)$.
I want to find $f$ such that $\int_0^{\infty} f(x) \left(\frac{x^2}{1+x^2} \right)^n dx =\infty$ for all integers $n \geq1$.
Let $g_n(x) = \left(\frac{x^2}{1+x^2} \right)^n$, then I know $g_n(x) \rightarrow 0$ and $g_n(x)=
\left(\frac{x^2}{1+x^2}\right)^n = \left( \frac{1}{1+\frac{1}{x^2}}\right)^n \leq 1$ is bounded.
but since $f(x)$ is bounded $\int_0^{\infty} f(x) g_n(x) dx \leq \int_0^{\infty} f(x) dx$ and since $f$ is continuous it is integrable, so it seems $<\infty$. Contradiction...
Is $f$ be continuous condition that should be relaxed? Or am I doing something wrong?
|
Note that
$$
\left( {\frac{{x^2 }}{{1 + x^2 }}} \right)^n \ge \left( {\frac{{n^2 }}{{1 + n^2 }}} \right)^n \ge \frac{1}{2}
$$
for $n\geq 1$ and $x\geq n$. Taking, for example $f(x)=\frac{1}{x+1}$ (which is continuous for $x\geq 0$ and is bounded)
\begin{align*}
\int_0^{ + \infty } {f(x)\left( {\frac{{x^2 }}{{1 + x^2 }}} \right)^n dx} & \ge \int_n^{ + \infty } {f(x)\left( {\frac{{x^2 }}{{1 + x^2 }}} \right)^n dx} \ge \frac{1}{2}\int_n^{ + \infty } {f(x)dx} \\ & = \frac{1}{2}\left(\mathop {\lim }\limits_{X \to + \infty } \log (X + 1)\right) - \frac{1}{2}\log (n + 1) = + \infty .
\end{align*}
|
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|
Solve the differential equation: $(y^\prime)^2 - \frac{2y}{x}y^\prime+1=0$ Solve the differential equation: $(y^\prime)^2 - \frac{2y}{x}y^\prime+1=0$
My first step was to solve for $y^\prime$ to get:
$$y^\prime=\frac{y}{x}\pm\sqrt{(\frac{y}{x})^2-1}$$
Solving each individually yields (using substitution $u = \frac{y}{x}$):
$$y^\prime = \frac{y}{x} + \sqrt{(\frac{y}{x})^2-1} \rightarrow u'x + u = u + \sqrt{u^2-1}$$
$$\int \frac{1}{\sqrt{u^2-1}}du = \int \frac{1}{x}dx \rightarrow \ln|u+\sqrt{u^2-1}|=\ln|x|+C$$
Raising to $e$ and subing back and multiplying each equation yields the following integral equation:
$$(y-Cx^2 + \sqrt{y^2-x^2})(y-Cx+\sqrt{y^2-x^2})=0$$
However, I am suppose to yield $(x^2C^2 + 1 - 2Cy)\times(x^2+C^2-2Cy)=0$ ... could anyone point out the issues in my approach?
|
$$(y^\prime)^2 - \frac{2y}{x}y^\prime+1=0$$
This is D'Alembert 's differential equation:
$$y=xf(y')$$
where $$f(y')=\dfrac {1+y'^2}{2y'}$$
From your attempt you should get
$$ u+\sqrt {u^2-1}=cx$$
$$u^2-1=(cx-u)^2$$
$$c^2x^2-2cxu+1=0$$
$$c^2x^2-2cy+1=0$$
$$ \implies y= \dfrac {c^2x^2+1}{2c}$$
There is also a singular solution:
$$y=\pm x$$
|
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|
Find the range of possible values of $\sqrt{a^2+a+1}-\sqrt{a^2-a+1}$
Let $x=\sqrt{a^2+a+1}-\sqrt{a^2-a+1},x\in \mathbb R$. Find range of possible values of $x$.
I tried drawing the graph and obtained this:
Through which the answer came out to be $(-1,1)$.
What should be the procedure through algebra?
|
Multiply your expression by
$$\frac{ \sqrt{a^2+a+1}+\sqrt{a^2-a+1} } {\sqrt{a^2+a+1}+\sqrt{a^2-a+1}}$$
To get
$$\frac{2a}{ \sqrt{a^2+a+1}+\sqrt{a^2-a+1}}.$$
For $a$ positive, divide top and bottom by $a$ and push
the $1/a$ inside the square roots:
$$\frac{2}{ \sqrt{ 1+ \frac{1}{a} +\frac{1}{a^2} } +
\sqrt{ 1 - \frac{1}{a} +\frac{1}{a^2} }}.$$
Now you can see that as $a$ gets very large, the above
fraction gets close to
$$\frac{2}{1+1} = 1.$$
For $a$ negative, do the same thing, but when you push the $1/a$ inside the square roots, you have to leave the negative outside. The expression tends to $-1$.
Also, looking at the last expression, you can work out that the denominator is larger than the absolute value of the numerator, so the range is $(-1,1)$.
|
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|
Solving ODE-systems with Matrix exponential is wrong? Originally I've learned that the solution of a systems of coupled ODE:
$$\underbrace{\left[\begin{array}{cc}{y_1}'(x)\\ \vdots \\{y_n}'(x)\end{array}\right]}_{y'(x)}=
\underbrace{\left[\begin{array}{cccc}&a_{1\,1} &\cdots &a_{1\,n}
\\ &\vdots \quad &&\vdots \\
&a_{n\,1}&\cdots&a_{n\,n}\end{array}\right]}_{A}\,
\underbrace{\left[\begin{array}{cc}{y_1}(x)\\ \vdots \\{y_n}(x)\end{array}\right]}_{y(x)}$$
is determined by: $$y(x) = \exp(A\,x)\,C$$ where $C$ is a vector with constants $\left[\begin{array}{cc}C_1\\ \vdots \\C_n\end{array}\right]$ and $\exp(A\,x)$ the matrix exponential, that can be at best calculated by: $$\exp(A\,x) = V^{-1}\,\exp(\Lambda\,x)\,V$$ where $V$ is a vector full of Eigenvectors: $\left[\begin{array}{cc}v_1&\cdots&v_n\end{array}\right]$
and $\Lambda$ a matrix full of Eigenvalues on its main diagonal: $\left[\begin{array}{ccc}\lambda_1&\\ &\ddots\\&&v_n\end{array}\right]$
Now apparently this leads to another solution compared to: $$y(x) = c_1\,v_1\,\exp(\lambda_1\,x)+\cdots+c_n\,v_n\,\exp(\lambda_n\,x)$$
Even if one told me both solutions were to solve a system of ODE
For example consider the system:
$$\left(\begin{array}{cc}{y_1}'(x) \\ {y_2}'(x)\end{array}\right) = \left(\begin{array}{cc} 4 & 10\\ 8 & 2 \end{array}\right)\,\left(\begin{array}{cc}{y_1}(x) \\ {y_2}(x)\end{array}\right)$$
with Eigenvalues $\lambda_1 = 12, \lambda_2 = -6$ and Eigenvectors: $v_1 = \left(\begin{array}{cc}1 \\ 8/10\end{array}\right), v_2 = \left(\begin{array}{cc}1 \\ -1\end{array}\right)$
According to the second plain solution process I'd obtain:
$$\left(\begin{array}{cc}{y_1}(x) \\ {y_2}(x)\end{array}\right) = \left(\begin{array}{cc}1 \\ 8/10\end{array}\right)\,\exp(12\,x)+\left(\begin{array}{cc}1 \\ -1\end{array}\right)\,\exp(-6\,x)$$
However the matrix exponential spits:
$$\left(\begin{array}{cc}{y_1}(x) \\ {y_2}(x)\end{array}\right) = C\,\left(\begin{array}{cc} \frac{4\,{\mathrm{e}}^{-6\,x}}{9}+\frac{5\,{\mathrm{e}}^{12\,x}}{9} & \frac{5\,{\mathrm{e}}^{12\,x}}{9}-\frac{5\,{\mathrm{e}}^{-6\,x}}{9}\\ \frac{4\,{\mathrm{e}}^{12\,x}}{9}-\frac{4\,{\mathrm{e}}^{-6\,x}}{9} & \frac{5\,{\mathrm{e}}^{-6\,x}}{9}+\frac{4\,{\mathrm{e}}^{12\,x}}{9} \end{array}\right)\,$$
Probably those two are inconvenient, because the constants are set differently. In fact the second approach is independent of constants somehow. So how's that all in relation with each other?
|
We have
$$\left(\begin{array}{cc}{y_1}'(x) \\ {y_2}'(x)\end{array}\right) = \left(\begin{array}{cc} 4 & 10\\ 8 & 2 \end{array}\right)\,\left(\begin{array}{cc}{y_1}(x) \\ {y_2}(x)\end{array}\right)$$
The eigenvalues are
$$\lambda_1 = 12, ~~\lambda_2 = -6$$
The eigenvectors are
$$v_1 = \begin{pmatrix} 5 \\ 4 \end{pmatrix}, ~~v_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$$
The solution can written as
$$y(x) = c_1 e^{\lambda_1 x} v_1 + c_2 e^{\lambda_2 x} v_2 = c_1 e^{12 x}\begin{pmatrix} 5 \\ 4 \end{pmatrix} + c_2 e^{-6x}\begin{pmatrix} -1 \\ 1 \end{pmatrix}$$
We can also use the matrix exponential
$$e^{A x} = P e^{Dx} P^{-1} = \begin{pmatrix} 5 & -1 \\ 4 & 1 \end{pmatrix}\begin{pmatrix} e^{12x} &0 \\ 0 & e^{-6x} \end{pmatrix}
\begin{pmatrix}
\dfrac{1}{9} & \dfrac{1}{9} \\
-\dfrac{4}{9} & \dfrac{5}{9} \\
\end{pmatrix} = \begin{pmatrix}
\dfrac{4 e^{-6 x}}{9}+\dfrac{5 e^{12 x}}{9} & \dfrac{-5}{9} e^{-6 x}+\dfrac{5 e^{12 x}}{9} \\
\dfrac{-4}{9} e^{-6 x}+\dfrac{4 e^{12 x}}{9} & \dfrac{5 e^{-6 x}}{9}+\dfrac{4 e^{12 x}}{9} \\
\end{pmatrix}$$
The solution using the matrix exponential is given by
$$y(x) = e^{Ax} c = \left(
\begin{array}{c}
c_1 \left(\dfrac{4 e^{-6 x}}{9}+\dfrac{5 e^{12 x}}{9}\right)+c_2 \left(\dfrac{-5}{9}e^{-6 x}+\dfrac{5 e^{12 x}}{9}\right) \\
c_1 \left(\dfrac{-4}{9} e^{-6 x}+\dfrac{4 e^{12 x}}{9}\right)+c_2 \left(\dfrac{5 e^{-6 x}}{9}+\dfrac{4 e^{12 x}}{9}\right) \\
\end{array}
\right)$$
Compare the two results while noting that you can combine constants because they are arbitrary.
Also, if you choose an initial condition, both methods produce exactly the same result.
|
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|
Series sum of $\sum_{n = 1}^ {\infty} (\frac{1}{2^n3^n} + \frac{1}{2^{n+1}3^n}$) What is the sum of following series $\sum_{n = 1}^ {\infty} (\frac{1}{2^n3^n} + \frac{1}{2^{n+1}3^n}$)
*
*$\frac{3}{8}$
*$\frac{3}{10}$
*$\frac{3}{14}$
*$\frac{3}{16}$
My Attempt:
$\sum_{n = 1}^ {\infty} (\frac{1}{2^n3^n} + \frac{1}{2^{n+1}3^n}$) = $\sum_{n = 1}^ {\infty}\frac{1}{2^n 3^n} (1 + \frac{1}{2}$) = $\frac{3}{2}\sum_{n = 1}^ {\infty}\frac{1}{2^n 3^n}$ = $\frac{3}{2}\sum_{n = 1}^ {\infty}\frac{1}{6^n}$ = $\frac{3}{2}\cdot\frac{1}{1 - {1\over6}}$ = $\frac{3}{2}\cdot \frac{6}{5}$ = $\frac{9}{5}$ which is different from all the above options. Please point out my mistake or solve it different and easiest approach. Thanks.
|
Your error is that the geometric series starts with $n=1$, not $n=0$, so it should be $\sum_{n=1}^\infty \frac{1}{6^n} = \frac{1}{6} \frac{1}{1-\frac{1}{6}} = \frac{1}{5}$, not $\frac{6}{5}$. Making this correction will give you one of the answers in the multiple choice.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrating $\int{\left( x+2 \right)\sqrt{x-1}dx}$ I’m stuck on this relatively simple indefinite integral but, unfortunately, cannot figure out why my answer is wrong.
Here it is:
\begin{align}
\int{\left( x+2 \right)\sqrt{x-1}dx}&=\int{\left( \left( x-1 \right)+3 \right)\frac{3}{2}\frac{1}{{{\left( \sqrt{x-1} \right)}^{2}}}\left( \frac{2}{3}{{\left( \sqrt{x-1} \right)}^{3}} \right)dx}\\&=
\int{\left( \left( x-1 \right)+3 \right)\frac{3}{2}\frac{1}{{{\left( \sqrt{x-1} \right)}^{2}}}{{\left( \sqrt{x-1} \right)}_{x}}'dx}\\&=\frac{3}{2}\int{\left( \left( x-1 \right)+3 \right)\frac{1}{x-1}d\sqrt{x-1}}\\&=
\frac{3}{2}\int{\left( 1+\frac{3}{x-1} \right)d\sqrt{x-1}}\\&=\frac{3}{2}\int{1d\sqrt{x-1}}+\frac{9}{2}\int{\frac{1}{{{\left( \sqrt{x-1} \right)}^{2}}}d\sqrt{x-1}}\\&=\frac{3}{2}\sqrt{x-1}-\frac{9}{2\sqrt{x-1}}
\end{align}
I'm fully aware there’re more compact methods to solve this but I’m interested to find the error in the solution given above.
|
If you differentiate $\sqrt{x-1}$, what you get is $\dfrac1{2\sqrt{x-1}}$ rather than $\displaystyle\frac23\sqrt{x-1}^3$. If you want to get $\displaystyle\frac23\sqrt{x-1}^3$, then you should differentiate $\dfrac4{15}\sqrt{x-1}^5$.
|
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|
Induction Squares Proof Prove that for every positive integer $n$ there exist positive integers $$a_{11}, a_{21}, a_{22}, a_{31}, a_{32}, a_{33}, \dots ,a_{n1}, a_{n2},\dots,a_{nn}$$
such that
$$
a_{11}^2 = a_{21}^2 + a_{22}^2 = a_{31}^2 + a_{32}^2 + a_{33}^2 = a_{n1}^2 + a_{n2}^2 + \cdots + a_{nn}^2.
$$
We're doing a chapter on proofs by induction so I'm pretty sure that would be the way to go. My general thought is to somehow prove that a square exists that can be the sum of any number of squares but I'm not too sure. Thank you for the help!
|
We prove by induction.
The statement holds true for $n = 2$. Suppose it holds true for $n$, there exists the $\{a_{ij}\}_{1\le j \le i\le n}$ such that
$$
\begin{align}
a_{11}^2 & = a_{21}^2 + a_{22}^2\\
& = a_{31}^2 + a_{32}^2 + a_{33}^2\\
& = \cdots\\
& =a_{n1}^2 + a_{n2}^2 + \cdots + a_{nn}^2 \\
\end{align}
$$
For $n+1$, take $b_{22} = \color{red}{2}a_{11}$, then choose $b_{21} = a_{11}^2-1$ and $b_{11} = a_{11}^2+1$ then
$$b_{11}^2 = b_{21}^2 +b_{22}^2$$
Remark: How I find $(b_{22},b_{21},b_{11})$? The idea here is to apply the Euclid's formula, there exists $(m,n) = (a_{11},1)$ such that
$$
\begin{cases}
b_{22} = \color{red}{2} \times a_{11} \times 1 = 2mn\\
b_{21} = m^2 -n^2\\
b_{11} = m^2+n^2
\end{cases}
$$
Then, take
$$
\begin{cases}
b_{k1} = b_{21} = a_{11}^2 +1 \quad \text{for } 2\le k \le n\\
b_{ij} = \color{red}{2}a_{(i-1),(j-1)} \quad \text{for } 2 \le j \le i \le n+1\\
\end{cases}
$$
So, there exists $\{b_{ij}\}_{1\le j \le i\le n+1}$ such that
$$
\begin{align}
b_{11}^2 & = b_{21}^2 + a_{11}^2= b_{21}^2 + b_{22}^2\\
& = b_{31}^2 + a_{11}^2 = b_{31}^2 + b_{32}^2 + b_{33}^2\\
& = \cdots\\
& = b_{(n+1),1}^2 + a_{11}^2 =b_{(n+1),1}^2 + b_{(n+2),2}^2 + \cdots + b_{(n+1),(n+1)}^2
\end{align}
$$
Q.E.D
|
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|
For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$. For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$.
My attempt: We have
$$2(a^2+b^2)\geq (a+b)^2$$
so $$-2\leq a+b \leq 2$$
In other hand $$ab=\frac{(a+b)^2-2}{2}=(a+b)^2-1$$
|
Let $a = \sqrt{2} \cos u, b = \sqrt{2} \sin u$ which immediately satisfies the condition. Using Simon's Favorite Factoring Trick, $9 + 3a + 3b + ab = (a + 3)(b + 3)$, so by C-S:
$$(\sqrt{2} \cos u + 3)(\sqrt{2} \sin u + 3) ≥(\sqrt{\sqrt{2} \cos u} \sqrt{\sqrt{2} \sin u} + \sqrt{3} \sqrt{3})^2$$
$$= (\sqrt{2 \cos u \sin u} + 3)^2 = (\sqrt{\sin(2u)} + 3)^2$$
There are two values which have $\sqrt{2} \cos u$, $\sqrt{2} \sin u$ as their square. Choosing the negative value for the minimum, this is greater or equal to $(-1 + 3)^2 = 4$, hence $3a + 3b + ab ≥ 4 - 9 = -5$.
|
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|
Finding $n,m\in N$ such that $|\sqrt{e} - \frac{n}{m}| < \frac{1}{100}$ Find $n,m\in N$ such that $|\sqrt{e} - \frac{n}{m}| < \frac{1}{100}$.
I wrote this proof:
Let $f(x)=e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$
$|\sqrt{e}-P_N(\frac{1}{2})|=|R_N(\frac{1}{2})| < \frac{1}{100}$
From the Taylor theorem we get that there exists $0<c<\frac{1}{2}$ such that:
$|R_N(\frac{1}{2})| = |\frac{f^{(N+1)}(c)}{(N+1)!2^N}| = \frac{e^c}{(N+1)!2^N} \leq \frac{e}{(N+1)!2^N}\leq \frac{3}{(N+1)!2^N}\leq \frac{3}{(N+1)!}<10^{-2}$
hence for $N=5$ we get $720>300$ and the inequality holds.
Therefore:
$P_5(\frac{1}{2})=1+\frac{1}{2}+\frac{1}{8}+\frac{1}{6\cdot8}+\frac{1}{16\cdot 24}+\frac{1}{32\cdot 120} = \frac{32\cdot120+16\cdot 120+4\cdot 120+4\cdot 20+2\cdot 5}{32\cdot 120} = \frac{n}{m}$
Did I get it right? Was there a simpler method?
|
You can just compute the continued fraction getting $[1;1,1,1,5,1,1,9,1,1,13,\ldots]$. If you stop after the $5$ this evaluates to $\frac {28}{17}$, which is off by less than $0.0017$
|
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|
Prove $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $ without expansion It's easy to prove that if $a,b,c \neq 0 $: $$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $$
as $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff \frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\iff (ab+bc+ca)(a+b+c)=abc \iff (a+b)(b+c)(c+a)=0 $
I'm curious to prove this some other ways. I tried to use function and inequality but still no progress.
|
not an answer :
If we assume : $$a+b+c=abc$$
Then we obtain :
$$ab+bc+ca=1$$
Now substitute :
$$\frac{1}{a}=u$$
$$\frac{1}{b}=v$$
$$\frac{1}{c}=w$$
We obtain:
$$\frac{1}{uv}+\frac{1}{vw}+\frac{1}{wu}=1$$
Or :
$$uvw=u+v+w$$
So I recall the tangent formula :
$$\tan(a)+\tan(b)+\tan(c)=\tan(a)\tan(b)\tan(c)$$
So there is a geometrical interpretation behind it .
|
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|
Computation of $\int_{0}^{\infty}\frac{x}{\left(1+y^2x^2\right)\left(1+a^2x^2 \right)}dx$ The following integral is simple to compute
$$I=\int_{0}^{\infty}\frac{\arctan(x)}{1+x^2}dx \,\,\tag{1}$$
letting $$u=\arctan x \Rightarrow du=\frac{dx}{1+x^2}$$
$$\int\frac{\arctan(x)}{1+x^2}dx=\int u \, du = \frac{u^2}{2}$$
Therefore
$$\int_{0}^{\infty}\frac{\arctan(x)}{1+x^2}dx=\frac{\arctan^2 (x)}{2}\Big|_{0}^{\infty}=\frac{\pi^2}{8}$$
But if we introduce a parameter $a$ in (1) such that the integral becomes
$$I(a)=\int_{0}^{\infty}\frac{\arctan(ax)}{1+x^2}dx \,\,\tag{2}$$
the same technique above does not apply anymore.
My attempt for this integral is differentiating (2) w.r. to a
$$I^{\prime}(a)=\int_{0}^{\infty}\frac{x}{\left(1+x^2\right)\left(1+(ax)^2 \right)}dx \,\,$$
From this point I got stuck, altough I tried one more step using the fact that
$$\int_{0}^{\infty}e^{-xt}\cos(t)dt=\frac{x}{1+x^2}$$
$$I^{\prime}(a)=\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-xt}\cos(t)}{\left(1+(ax)^2 \right)}\,dt\,dx $$
$$I^{\prime}(a)=\int_{0}^{\infty}\cos(t)\int_{0}^{\infty}\frac{e^{-xt}}{\left(1+(ax)^2 \right)}\,dx\,dt $$
But it seems too hard.
Alternatively, integrating by parts we obtain
$$I(a)=\int_{0}^{\infty}\frac{\arctan(ax)}{1+x^2}dx=-a\int_{0}^{\infty}\frac{\arctan(x)}{1+(ax)^2}dx$$
now using the following integral representation
$$\arctan (x)=\int_{0}^{1}\frac{x}{1+x^2y^2}dy$$
$$I(a)=-a\int_{0}^{\infty}\int_{0}^{1}\frac{x}{\left(1+y^2x^2\right)\left(1+a^2x^2 \right)}dy \,dx$$
$$I(a)=-a\int_{0}^{1}\int_{0}^{\infty}\frac{x}{\left(1+y^2x^2\right)\left(1+a^2x^2 \right)}dx \, dy$$
I dont know the solution of this integral, however I saw these results below and I wanted to proof them
$$\int_{0}^{\infty} \frac{\arctan \frac{x}{\phi}}{1+x^{2}} d x=\frac{\pi^{2}}{12}+\frac{3 \ln ^{2} \phi}{4} $$
$$\int_{0}^{\infty} \frac{\arctan \phi x}{1+x^{2}} d x =\frac{\pi^{2}}{6}-\frac{3 \ln ^{2} \phi}{4}$$
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I've left some ideas below. However I'm having difficulty confirming them numerically.
I'll consider the integral in your title, namely
$$F(x,y)=\int_0^\infty \frac{t}{(1+x^2t^2)(1+y^2t^2)}\mathrm{d}t$$
What we do first is use a substitution $s=t^2\implies \mathrm ds/2=t~\mathrm dt$. So,
$$F(x,y)=\frac{1}{2}\int_0^\infty\frac{\mathrm ds}{(1+x^2s)(1+y^2s)}$$
Though it's tempting to use partial fractions to split the integral up here, it fails, since the individual pieces will fail to converge. So we need to get a little more creative. What we do is expand the denominator, then complete the square.
$$(1+x^2s)(1+y^2s)=x^2y^2\left(\frac{1}{x^2y^2}+\frac{x^2+y^2}{x^2y^2}s+s^2\right)=x^2y^2\left[\left(s+\frac{x^2+y^2}{2x^2y^2}\right)^2+\left(\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}\right)\right]$$
Which leads us to make the substitution
$$z=s+\frac{x^2+y^2}{2x^2y^2}$$
Hence our integral is now
$$F(x,y)=\frac{1}{2x^2y^2}\int\limits_{\frac{x^2+y^2}{2x^2y^2}}^\infty \frac{\mathrm dz}{z^2+\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}$$
Finally now this is a well known integral:
$$F(x,y)=\frac{1}{2x^2y^2\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\arctan\left(\frac{z}{\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\right)~\Bigg|^{z=\infty}_{z=\frac{x^2+y^2}{2x^2y^2}}$$
So
$$F(x,y)=\frac{1}{2x^2y^2\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\left[\frac{\pi}{2}-\arctan\left(\frac{(x^2+y^2)/2x^2y^2}{\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\right)\right]$$
|
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|
Finding the remainder of a polynomial divided by $x^4+x^2+1$ if remainders when dividing by $x^2+x+1$, $x^2-x+1$ are $-x+1$, $3x+5$.
Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.
My attempt was to write $f(x)=(x^4+x^2+1)q(x)+Ax^3+Bx^2+Cx+D=(x^2+x+1)(x^2-x+1)q(x)+Ax^3+Bx^2+Cx+D$ and then factor out $x^2+x+1$ to get $f(x)=(x^2+x+1)[(x^2-x+1)q(x)+B]+Ax^3+(C-B)x+(D-B)$, then do the same for $x^2-x+1$ and then use that along with the known remainder of $f$ divided by $x^2+x+1$ and $x^2-x+1$ to obtain $A$, $B$, $C$, $D$. However, $Ax^3$ is in the way so I don't know how to proceed nor do I have any other ideas to start with.
|
Using the Extended Euclidean Algorithm as described in this answer and adapted to polynomials yields
$$
\begin{array}{r|r|r|r}
\bbox[5px,border:2px solid #C00]{x^2+x+1}&\bbox[5px,border:2px solid #C00]{1}&0\\
\bbox[5px,border:2px solid #090]{x^2-x+1}&0&\bbox[5px,border:2px solid #090]{1}\\
2x\phantom{{}+0}&1&-1&1\\
1&\bbox[5px,border:2px solid #C00]{-\frac12x+\frac12}&\bbox[5px,border:2px solid #090]{\frac12x+\frac12}&\frac12x-\frac12\\
0&x^2\phantom{\frac12}-x+\,1&-x^2\phantom{\frac12}-x-\,1&2x\phantom{{}+\frac12}
\end{array}\tag1
$$
which says that
$$
\overbrace{\color{#C00}{\left(x^2+x+1\right)\left(-\tfrac12x+\tfrac12\right)}}^{\large\frac12-\frac12x^3}+\overbrace{\color{#090}{\left(x^2-x+1\right)\left(\tfrac12x+\tfrac12\right)}}^{\large\frac12+\frac12x^3}=1\tag2
$$
Which, in turn says that
$$
\frac12-\frac12x^3\equiv\left\{\begin{array}{}
0&\bmod x^2+x+1\\
1&\bmod x^2-x+1
\end{array}\right.\tag3
$$
and
$$
\frac12+\frac12x^3\equiv\left\{\begin{array}{}
1&\bmod x^2+x+1\\
0&\bmod x^2-x+1
\end{array}\right.\tag4
$$
Therefore, the desired polynomial is
$$
\begin{align}
\scriptsize(-x+1)\overbrace{\left(\tfrac12+\tfrac12x^3\right)}^{\substack{1\bmod x^2+x+1\\0\bmod x^2-x+1}}+(3x+5)\overbrace{\left(\tfrac12-\tfrac12x^3\right)}^{\substack{0\bmod x^2+x+1\\1\bmod x^2-x+1}}
&=3+x-2x^3-2x^4\\
&\equiv\bbox[5px,border:2px solid #CA0]{5+x+2x^2-2x^3}\bmod x^4+x^2+1\tag5
\end{align}
$$
|
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|
Proof the equality $\prod_{r=1}^{mn}\left(x+mn -r\right)=\prod_{k=1}^{m}\prod_{l=1}^{n}\left(x+mn-(1+ml-k) \right)$
I came across a proof of Gauss multiplication formula for the Gamma function which relies on the following indentity (without a proof)
$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=m^{mn}\prod_{k=1}^{m}\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)} \, \tag{1}$$
I am trying to prove it.
I started expanding the left hand side first. From the recurrence relation of the Gamma function we have that
$$\Gamma \left(x+mn \right)= \left(x+mn -1\right)\Gamma \left(x+mn-1 \right)$$
$$\Gamma \left(x+mn \right)= \left(x+mn -1\right) \left(x+mn -2\right)\Gamma \left(x+mn-2 \right)$$
$$\cdots$$
$$\Gamma \left(x+mn \right)= \left(x+mn -1\right) \left(x+mn -2\right) \cdots \left(x+mn -mn\right)\Gamma \left(x+mn-mn \right)$$
Therefore we can rewrite the L.H.S of $(1)$ as
$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\left(x+mn -1\right) \left(x+mn -2\right) \cdots \left(x\right)$$
$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=\prod_{r=1}^{mn}\left(x+mn -r\right) \, \tag{2}$$
Similarly, for the R.H.S. of $(1)$ we obtain
$$\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\left(\frac{x+k-1}{m}+n-1 \right)\left(\frac{x+k-1}{m}+n-2 \right) \cdots
\left(\frac{x+k-1}{m}+n-n \right)$$
$$\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\prod_{l=1}^{n}\left(\frac{x+k-1}{m}+n-l \right)$$
$$\frac{\Gamma \left(\frac{x+k-1}{m}+n \right)}{\Gamma \left(\frac{x+k-1}{m} \right)}=\prod_{l=1}^{n}\frac{1}{m}\left(x+k-1+mn-ml \right) \, \tag{3}$$
Plugging $(2)$ and $(3)$ in $(1)$ we obtain the following equality.
$$\prod_{r=1}^{mn}\left(x+mn -r\right)=\prod_{k=1}^{m}\prod_{l=1}^{n}\left(x+k-1+mn-ml \right) \, \tag{4} $$
$$\prod_{r=1}^{mn}\left(x+mn -r\right)=\prod_{k=1}^{m}\prod_{l=1}^{n}\left(x+mn-(1+ml-k) \right)$$
suposse $m=2 \,\, \text{and}\,\,n=2$, the right hand side becomes
$$\prod_{k=1}^{2}\prod_{l=1}^{2}\left(x+k-1+4-2l \right)=\prod_{k=1}^{2}\left(x+k-1+4-2 \right)\left(x+k-1+4-2\times2 \right)$$
$$=\prod_{k=1}^{2}\left(x+k+1 \right)\left(x+k-1 \right)$$
$$=\left(x+1+1 \right) \cdot\left(x+1-1 \right)\cdot\left(x+2+1 \right)\cdot\left(x+2-1 \right)$$
$$=\left(x+2 \right)\cdot x \cdot \left(x+3 \right) \cdot\left(x+1 \right)$$
$$= x \cdot\left(x+1 \right)\cdot \left(x+2 \right)\cdot \left(x+3 \right) \, \tag{5}$$
And the left hand side becomes
$$\prod_{r=1}^{4}\left(x+4 -r\right)=\left(x+4 -1\right)\left(x+4 -2\right)\left(x+4 -3\right)\left(x+4 -4\right)$$
$$= x \cdot\left(x+1 \right)\cdot \left(x+2 \right)\cdot \left(x+3 \right)$$
Which equals exactly $(5)$. So intuitively I see that the equality $(1)$ holds.
My question is: How can I go from this heuristic intuitive proof to a formal proof, may be proved by induction?
|
Just notice that any natural number $\le nm $ can be uniquely written in the form $ml - k’$, where $1\le l \le n$, $0 \le k’ < m$ are integers (this is just a division by $m$ with remainder). Changing $k’$ on $k - 1$, we can see, that each factor of the left product in (4) appears exactly one time in the right product. But they both have $mn$ factors. Hence the products are equal.
|
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|
Solving an equation in quaternions How to solve equation $x^4=1$ in quaternions?
I know how to solve that equation in complex numbers but I have no idea how to do it in quaternions.
I have also a question about general way to solve equations in quaternions. And is there similar theorem to fundamental algebra theorem which connects degree of polynomial equation with number of quaternion roots of that equation?
EDIT:
My task was to find roots $x^2+1=0$
EDIT2:
How to prove that $x^2+1=0$ has infinite number of roots?
I represented x as $a+bi+cj+dk$ and get equations: $a^2−b^2−c^2−d^2=−1$ and $ab+cd=0$, $ac−bd=0$, $ad+cb=0$. Is it worth to solve that or is there a smarter way?
|
Let me consolidate my two comments into an answer that may be a bit clearer to the OP.
You want to find all the roots of the polynomial $x^4 - 1$ in the quaternions $\Bbb{H}$. Note that $x^4 - 1 = (x^2 - 1)(x^2 + 1)$, just as in $\Bbb{R}$ or $\Bbb{C}$ and note that $\Bbb{H}$ has no non-trivial zero divisors: if $X, Y \in \Bbb{H}$, then $XY = 0$ implies $X = 0$ or $Y = 0$. So $X^4 -1 = 0$ iff either $X^2 - 1 = 0$ or $X^2 + 1 = 0$.
For a general element $X = a + bi + cj+ dk$ of $\Bbb{H}$, where $a, b, c, d \in \Bbb{R}$, we have the identity:
$$
X^2 = (a + bi + cj+ dk)^2 = a^2 - b^2 - c^2 - d^2 + 2a(bi + cj + dk)
$$
which you can verify by multiplying out using the formulas $ij = k$, $jk = i$, $ji = -k$ etc.
Thus, comparing cooefficients, $X^ 2 - 1 = 0$ iff:
$$
a^2 - b^2 - c^2 - d^2 = 1 \\
ab = ac = ad = 0
$$
From the first of those equations, we must have $a \neq 0$, but then from the other equations we must have $b = c = d = 0$, implying that $a^2= 1$. So the only solutions to $x^2 - 1 = 0$ are the quaternions $X = \pm 1$.
Likewise, if $X^2 + 1 = 0$, we must have:
$$
a^2 - b^2 - c^2 - d^2 = -1 \\
ab = ac = ad = 0
$$
And then the first equation tells us that at least one of $b$, $c$ or $d$ is non-zero, from which one of the the other equations will tell us that $a = 0$. So the solutions of $x^2 + 1 = 0$, are the quaternions $X = bi + cj + dk$ where $b^2 + c^2 + d^2 = 1$.
Putting that all together gives us all the solutions of $x^4 + 1 = 0$: they are the quaternions $X$ such that $X = \pm 1$ or $X = bi + cj + dk$ where $b^2 + c^2 + d^2 = 1$.
|
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|
Solution set to non-linear equation of $3$ variables I have the following trigonometric equation of $3$ variables:
$$f(\theta,\lambda,\phi)=3 \cos (\theta ) \cos (\lambda ) \cos (\phi
)-(\cos (\theta )+3) \sin (\lambda ) \sin
(\phi )$$
$$-\sin (\theta ) \cos (\lambda )+\sin
(\theta ) \cos (\phi )+3 \cos (\theta )+\cos
(\lambda ) \cos (\phi )-7$$
I want to prove that the solution set to the equation $f(\theta,\lambda,\phi)=0$ is
$$S_f=\{(2\pi k,\lambda+2\pi m,-\lambda+2\pi n):k,m,n\in\mathbb{Z},\lambda\in\mathbb{R}\}$$
Graphically, it is easy to see that this is the case but everything I have tried so far has failed to prove the conjecture. Perhaps my best attempt was extrapolating this equation into an unwieldy polynomial of $3$ variables
$$P(x,y,z)=9 x^4 y^2+16 x^4-18 x^3 y^2-192 x^3+9 x^2
y^4+z^4 \left(\left(x^2-1\right)
y^2+1\right) \left(16 \left(x^2-1\right)
y^2+(3 x+5)^2\right)-64 x^2 y^2+z^2
\left((x-1) (x+1) \left(9 x^2+30 x+41\right)
y^4+2 x (x (48 (x-5) x+187)+234) y^2+x (x (9
(x-2) x-64)-78)-74 y^2+151\right)+736 x^2+2
y z^3 \left(9 x^4+12 x^3-20 x^2+(x-1) (x+1)
(x (31 x-78)-33) y^2-96 x-33\right)+2 y z
\left(31 x^4-270 x^3+560 x^2+(x (x (3 x (3
x+4)-20)-96)-33) y^2-210 x-111\right)+30 x
y^4-78 x y^2-960 x+25 y^4+151 y^2+400$$
over the domain $(x,y,z)\in [-1,1]^3$. If I can prove that the solution set to $P(x,y,z)=0$ over this domain is
$$S_P=\{(1,y,y):y\in[-1,1]\}$$
then the original conjecture would be solved. The motivation behind this is actually proving a certain type of quantum error detection encoding exists. It's a little difficult to explain (although if anyone wants details I am more than happy to provide them) but suffice to say that after a lot of work I managed to whittle my existence proof down to the conjecture above.
|
We prove $f(\theta,\lambda,\phi) \le 0$ for each $\theta,\lambda,\phi$ and that equality holds exactly when you conjecture it does.
Write $$f(\theta,\lambda,\phi) = \cos(\theta)a+\sin(\theta)b+\cos(\lambda)\cos(\phi)-3\sin(\lambda)\sin(\phi)-7$$ for $a = 3\cos(\lambda)\cos(\phi)-\sin(\lambda)\sin(\phi)+3$ and $b = \cos(\phi)-\cos(\lambda)$. To maximize $f$, we of course to choose $\theta$ so that $\cos(\theta)a \ge 0$ and $\sin(\theta)b \ge 0$. So, for ease, let's just pretend $a,b \ge 0$ so that $\cos(\theta),\sin(\theta) \ge 0$. To maximize $xa+\sqrt{1-x^2}b$ for $x \in [0,1]$, one takes $x = \frac{a}{\sqrt{a^2+b^2}}$ (easy to prove by basic calculus), yielding a maximum of $\sqrt{a^2+b^2}$; and note that any other $x$ yields a strictly smaller value. Therefore, we wish to show $$\sqrt{(3\cos(\lambda)\cos(\phi)-\sin(\lambda)\sin(\phi)+3)^2+(\cos(\phi)-\cos(\lambda))^2}+\cos(\lambda)\cos(\phi)-3\sin(\lambda)\sin(\phi)-7 \le 0$$ with equality if and only if $\lambda = -\phi+2\pi m$ for some $m \in \mathbb{Z}$ (since then $\cos(\theta)$ will be $1$). For ease, let $x = \cos(\lambda)\cos(\phi)$ and $y = \sin(\lambda)\sin(\phi)$. We can rewrite the above as $$\sqrt{(3x-y+3)^2+(\cos(\phi)-\cos(\lambda))^2}+x-3y-7 \le 0.$$ We will show $$\sqrt{(3x-y+3)^2+(\cos(\phi)-\cos(\lambda))^2+(\sin(\phi)+\sin(\lambda))^2}+x-3y-7 \le 0$$ with equality if and only if $\lambda = -\phi+2\pi m$, which clearly suffices. The point of introducing the $(\sin(\phi)+\sin(\lambda))^2$ is that $$(\cos(\phi)-\cos(\lambda))^2+(\sin(\phi)+\sin(\lambda))^2 = 2-2x+2y.$$ So, we just wish to show that $$\sqrt{(3x-y+3)^2+2-2x+2y}+x-3y-7 \le 0$$ for $\{(x,y) \in [-1,1]^2 : |x-y| \le 1\}$ with equality if and only if $x-y = 1$. Note that $|3x-y+3| = |2x+(x-y)+3| \le 6$ and $|2-2x+2y| = 2|1-(x-y)| \le 4$, so $\sqrt{(3x-y+3)^2+2-2x+2y} \le \sqrt{40}$. Therefore, we cannot have $y \ge 0$ and $x \le 0$.
Let's first start with $y \ge 0$. Then, as just explained, $x \ge 0$. We will show that $$\sqrt{(3x-y+3)^2+2-2x+2y}+x-7 \le 0$$ with equality if and only if $x=1,y=0$. Note the derivative of the term inside the square root with respect to $y$ is $2y+2-6(x+1)$, which is negative, so we choose $y=0$ to maximize. We then wish to show $$\sqrt{(3x+3)^2+2-2x} -7 \le 0$$ with equality if and only if $x = 1$. But this is equivalent to $(3x+3)^2+2-2x \le 49$, which is true for $x \le 1$ with equality at and only at $x=1$.
Now let's deal with the case $x \le 0$. As explained before, this means $y \le 0$. We show $$\sqrt{(3x-y+3)^2+2-2x+2y}+x-3y-7 \le 0$$ for any $x,y \in [-1,0]^2$ with equality if and only if $y=-1,x=0$. The derivative of the term inside the square root with respect to $y$ is, as before, $2y+2-6(x+1)$, which is negative, so we take $y=-1$ to maximize. We then obtain $$\sqrt{9x^2+22x+16}+x-4,$$ which is maximized at $x=0$, yielding the value $0$. We're done.
|
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|
Centers of circumcircle define an equilateral triangle Let $ABC$ be a triangle with side $a,b,c$ and angles $\alpha, \beta,\gamma$.
It holds that $\alpha, \beta, \gamma<\frac{2\pi}{3}$.
At the side $\overline{BC}$ there is an equilateral triangle $\triangle BCA''$ at the outter side, i.e. $A''$ is the point for which the points $A$ and $A''$ are on different sides of the line $BC$ and for which the triangle $\triangle BCA''$ is equilateral.
Let $A'$ be the center of the circumcircle of $\triangle BCA''$.
Similarily we define the points $B'$ and $C'$ at the lines $\overline{AC}$ and $\overline{AB}$ respectively.
The radius of the circumcircle of $BCA''$ is $\frac{a}{\sqrt{3}}$.
It holds that $$|A'B'|^2=\frac{1}{3}a^2+\frac{1}{3}b^2-\frac{2}{3}ab\cos \left (\frac{\pi}{3}+\gamma\right )$$
I want to show the following :
a) $\cos \left (\frac{\pi}{3}+\gamma\right )=\frac{1}{2}\cos (\gamma )-\frac{\sqrt{3}}{2}\sin (\gamma)$
b) $A'B'C'$ is an equilateral triangle.
$$$$
First I tried to draw the above and I get :
Is that correct?
I haven't really understood how we get the angle of $\frac{\pi}{3}+\gamma$. Which is this angle in the graph?
$$$$
**EDIT: **
We have that $$|A'B'|^2=\frac{1}{3}a^2+\frac{1}{3}b^2-\frac{2}{3}ab\cos \left (\frac{\pi}{3}+\gamma\right )=\frac{1}{3}a^2+\frac{1}{3}b^2-\frac{2}{3}ab \left (\frac{1}{2}\cos (\gamma )-\frac{\sqrt{3}}{2}\sin (\gamma)$
\right )$$
So do we use the cosine rule also for the other angles of the triangle to get the desired result?
Or do we show in an other way that the triangle is equilateral?
|
Join $B'C$ and $A'C$
$\angle A'CB'=\angle A'CB+\angle BCA+\angle ACB'$
$\angle A'CB'=30^{\circ}+\gamma+30^{\circ}$
$\angle A'CB'=60^{\circ}+\gamma$
The fact used above is that the line joining circumcenter of an equilateral triangle to any of the vertices bisects the angle corresponding to that vertex.
Now apply cosine rule on $A'CB'$
$|A'B'|^2=\frac{1}{3}a^2+\frac{1}{3}b^2-\frac{2}{3}ab\cos \left (\frac{\pi}{3}+\gamma\right )=\frac{1}{3}a^2+\frac{1}{3}b^2-\frac{2}{3}ab \left (\frac{1}{2}\cos (\gamma )-\frac{\sqrt{3}}{2}\sin (\gamma)
\right )=\frac{1}{3}(a^2+b^2-ab\cos\gamma)+\frac{2}{\sqrt 3}\Delta=\frac{1}{3}(a^2+b^2-\frac{b^2+a^2-c^2}{2})+\frac{2}{\sqrt 3}\Delta=\frac{1}{6}(a^2+b^2+c^2)+\frac{2}{\sqrt 3}\Delta$
Similarly
$|B'C'|^2=\frac{1}{3}b^2+\frac{1}{3}c^2-\frac{2}{3}bc\cos \left (\frac{\pi}{3}+\alpha\right )=\frac{1}{3}b^2+\frac{1}{3}c^2-\frac{2}{3}bc \left (\frac{1}{2}\cos (\alpha )-\frac{\sqrt{3}}{2}\sin (\alpha)
\right )=\frac{1}{3}(b^2+c^2-bc\cos\alpha)+\frac{2}{\sqrt 3}\Delta=\frac{1}{3}(b^2+c^2-\frac{b^2+c^2-a^2}{2})+\frac{2}{\sqrt 3}\Delta=\frac{1}{6}(a^2+b^2+c^2)+\frac{2}{\sqrt 3}\Delta$
Clearly above two side lengths are equal therefore.....
|
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|
Alternative approaches to maximize $y=x\sqrt{100-x^2}$ I could find three good approaches to find maximum of the function $y=x\sqrt{100-x^2}$. I will explain them briefly :
First: Finding $x$ satisfies $y'=0$ then plugging it in the function.
Second: Using the substitution $x=10\sin\theta$ (or $x=10\cos\theta)$ for
$\theta\in(0,\frac{\pi}2)$ to get $y=100\sin\theta\cos\theta=50\sin(2\theta)$ hence the maximum is $50$.
Third: Using AM-GM inequality: It is obvious that maximum occurs for $x>0$ So we can rewrite $y$ as $y=\sqrt{x^2(100-x^2)}$ . Now the sum of $x^2$ and $100-x^2$ is $100$ so the maximum of product happens when $x^2=100-x^2$ or $x^2=50$ Hence $y_{\text{max}}=50$.
Just for fun, can you maximize $y=x\sqrt{100-x^2}$ with other approaches?
|
$\,\\\;\;\;\;\;\;\;\; 2\,S \;=\; x \cdot \sqrt{100-x^2} \;=\; h \cdot 10 \;\le\; 5 \cdot 10$
|
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|
ARML $1994$ Polynomial Manipulation Question
(ARML $1994$)
If $x^5 + 5x^4 + 10x^3 + 10x^2 - 5x + 1 = 10$, and $x \neq -1$,
compute the numerical value of $(x + 1)^4$.
I came across this problem in a packet about polynomial manipulation and I couldn't find a way past expanding $(x+1)^4$ and subtracting it from the polynomial. I was wondering if any of you could explain it.
|
Remember your binomial coefficients, and notice that the left-hand side of the equation is equal to $(x+1)^5-10x$. So we can rewrite that equation as follows:
\begin{align*}
(x+1)^5-10x&=10\\
(x+1)^5-10(x+1)&=0\\
(x+1)\left[(x+1)^4-10\right]&=0 \, .
\end{align*}
Since $x \neq -1$, the first factor above is nonzero, so the second factor must be zero. That is, $(x+1)^4=10$.
|
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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.