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For positive $a$, $b$, $c$ with $a+b+c=1$, show that $(ab+bc+ca) \sum_{cyc}\frac{a}{b^2+b} \geq \frac34$
If $a,b,c > 0$ and $a+b+c = 1$, then prove that
$$\left(\frac{a}{b^2+b}+\frac{b}{c^2+c}+\frac{c}{a^2+a}\right)(ab+bc+ca)\geq\frac{3}{4}$$
It's been more than 35 years since I last touched algebra!!
|
Hint: Using generalised Holder’s inequality
$$\sum_{cyc}ab \cdot \sum_{cyc} \frac{a}{b^2+b}\cdot \sum_{cyc} a(b+1) \geqslant \left(\sum_{cyc}a\right)^3=1$$
Now it is enough to show $\sum_{cyc} a(b+1) \leqslant \frac43$ which is easy.
|
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|
Rational Decomposition
Find the partial fractions for $\frac{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}$
First, I recognize it's not a proper fraction. I also know the denominator is $(x+3)(3x+1)$. Now every time I try long division to convert to a proper fraction by hand I get $x^3-x^2+1+(\frac{-10x-3}{(x+3)(3x+1)})$, but this is wrong because I'm supposed to get $x^3-x^2+\frac{x}{(x+3)(3x+1)}$.
I can't find the partial fraction just yet if I can't convert this into a proper fraction. When I tried I got very close, but not exactly correct at all. I wish I could show my work, but it seems displaying long division or synthetic division isn't doable on this website.
Can anyone help? Or provide an alternative method that may not require so much long division?
|
The first step you need to follow is the long division method, which will for sure give you $$\frac{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}=\frac{(x^3-x^2)(3x^2+10x+3)+x}{(x+3)(3x+1)}=x^3-x^2+\frac{x}{(x+3)(3x+1)}$$Half work done!! Now we just need to break $\frac{x}{(x+3)(3x+1)}$ into partial fractions. That's easy,$$\frac{x}{(x+3)(3x+1)}=\frac{\frac{3}{8}(3x+1)-\frac{1}{8}(x+3)}{(x+3)(3x+1)}=\frac{3}{8(x+3)}-\frac{1}{8(3x+1)}$$Thus, we have$$\frac{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}=x^3-x^2+\frac{3}{8(x+3)}-\frac{1}{8(3x+1)}$$
|
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|
Find the Maximum and Minimum Values of $f(x) = \cos\sqrt{x+1} - \cos\sqrt{x}$ I saw today in a fb forum the following excersice:
Find the range of values of
$f(x)=\cos\sqrt{x+1}-\cos\sqrt{x}$
I have tried to find the mix and max of the function but failed. Any ideas?
|
To do an analytic check of the range, and possibly other properties, note you can combine the $2 \cos$ terms into a multiple of one $\sin$ term using several identities from List of trigonometric identities. First, note that $\cos(\theta) = \sin(\frac{\pi}{2} - \theta)$, so you get
$$f(x) = \cos\sqrt{x+1}-\cos\sqrt{x} = \sin\left(\frac{\pi}{2} - \sqrt{x+1}\right) - \sin\left(\frac{\pi}{2} - \sqrt{x}\right) \tag{1}\label{eq1}$$
Next, you can use the Arbitrary phase shift formula for a linear sum of $2$ $\sin$ values which gives
$$a\sin y+b\sin(y+\theta )=c\sin(y+\varphi ) \tag{2}\label{eq2}$$
where
$$c={\sqrt {a^{2}+b^{2}+2ab\cos \theta }} \tag{3}\label{eq3}$$
and
$$\varphi =\operatorname {atan2} \left(b\,\sin \theta ,a+b\cos \theta \right) \tag{4}\label{eq4}$$
where $\operatorname {atan2}$ is the principal arc-tangent angle in $(-\pi,\pi]$ (more details are at atan2)). For the case of \eqref{eq1}, you have $a = 1$, $b = -1$, $y = \frac{\pi}{2} - \sqrt{x+1}$ and $\theta = \sqrt{x+1} - \sqrt{x}$. As such, \eqref{eq3} gives
$$c = \sqrt{2 - 2\cos\left(\sqrt{x+1} - \sqrt{x}\right)} \tag{5}\label{eq5}$$
Since $\sqrt{x+1} - \sqrt{x}$ is a strictly decreasing function, from $1$ towards $0$, for $x \ge 0$ and $\cos$ is also a decreasing function in $(0,1]$, this means that at $x = 0$ you have $c = \sqrt{2 - 2\cos(1)}$ is it's maximum and, since $\cos(0) = 1$, its infimum is $0$ since $c \to 0$ as $x \to \infty$.
You can do similar appropriate checks for the range of $\varphi$ values in \eqref{eq4} and then determine, overall, the range of possible values for \eqref{eq1}.
|
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|
$|x^2-3x+2 | = mx$ has $x_1, x_2, x_3, x_4 $ $s(m) = \frac{1}{{x_1}^2} +\frac{1}{{x_2}^2} + \frac{1}{{x_3}^2 }+ \frac{1}{{x_4}^2}$ express $s(m)$
$|x^2-3x+2 | = mx$ has $x_1, x_2, x_3, x_4 $ 4 distinct solutions
$s(m) = \frac{1}{{x_1}^2} +\frac{1}{{x_2}^2} + \frac{1}{{x_3}^2 }+ \frac{1}{{x_4}^2}$
Express $s(m)$ in terms of $m$
$0 <m < 3-2\sqrt{2}$
$(x^2-3x+2 ) = \pm mx$
I get $x^2-(3+m)x+2 = 0$ and $x^2-(3-m)x+2 = 0$
i thought that i had to find the first
$\frac{1}{{x_1}^2} +\frac{1}{{x_2}^2} =\frac{({{x_1}+{x_2}})^2 - 2x_1x_2}{({x_1x_2})^2} $ from $x^2-(3+m)x+2 = 0$
and then find $ \frac{1}{{x_3}^2} +\frac{1}{{x_4}^2}$ from $x^2-(3-m)x+2 = 0$
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$$(x^2-3x+2)^2=m^2x^2$$
$$\implies x^4-6x^3+x^2(9+4-m^2)-12x+4=0$$
$$\iff x^4+(13-m^2)x^2+4=12x+6x^3$$
Replace $x^2=\dfrac1y\implies y=\pm\dfrac1{\sqrt x}$
$$\dfrac{1+(13-m^2)y+4y^2}{y^2}=\pm\dfrac{12y+6}{y^{3/2}}$$
$$(1+(13-m^2)y+4y^2)^2= y(12y+6)^2$$
$$\implies 16y^4+y^3(8(13-m^2)-144)+\cdots+1=0$$
Using Vieta's formula
$$\sum_{r=1}^4\dfrac1{x_r^2}=\sum_{r=1}^4y_r=\dfrac{8m^2+40}{16}$$
|
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Integral $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$
Prove that
$$\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx =\frac{\pi}{4} \ln2$$
I tried to use King's rule and to scale by $2$ and then to add the integrals, to get product of terms and use the result $$\int_{0}^{\frac{\pi}2} \ln(\sin{x})dx=\int_{0}^{\frac{\pi}2} \ln(\cos{x})dx=-\frac{\pi}2\ln2$$
but it didnt work. Any help?
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$$I=\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin(2x)})dx =\frac12 \int_0^\frac{\pi}{2} x'\ln(\sin x+\cos x+\sqrt{\sin (2x)})dx$$
$$\overset{IBP}=\frac12 \int_0^\frac{\pi}{2}x\,\frac{\sin x-\cos x}{\sqrt{\sin(2x)}}dx\overset{x=\arctan t}=\frac{1}{2\sqrt 2}\int_0^\infty \frac{\arctan t}{1+t^2}\frac{t-1}{\sqrt t}dt$$
$$I(a)=\int_0^\infty \frac{\arctan(at)}{1+t^2}\frac{t-1}{\sqrt t}dt\Rightarrow I'(a)=\int_0^\infty \frac{(t-1)\sqrt t}{(1+a^2 t^2)(1+t^2)}dt$$
$$\overset{t=x^2}=\frac{2}{1-a^2}\int_0^\infty \frac{1+a^2 x^2}{1+a^2 x^4}dx-\frac{2}{1-a^2}\int_0^\infty \frac{1+x^2}{1+x^4}dx=\frac{\pi}{\sqrt 2}\frac{1-\sqrt a}{\sqrt a (1+a)(1+\sqrt a)}$$
$$\Rightarrow I=\frac{\pi}{4}\int_0^1 \frac{1-\sqrt a}{\sqrt a (1+a)(1+\sqrt a)}da\overset{\sqrt a=x}=\frac{\pi}{2}\int_0^1 \frac{1-x}{(1+x^2)(1+x)}dx=\frac{\pi}{4}\ln 2 $$
|
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|
Given that $9^{2x} = 27^{x^2 - 5}$. Find the possible values of $x$. Given that $9^{2x} = 27^{x^2 - 5}$. Find the possible values of $x$.
I don't know how to approach this question.
|
As $9=3^2, 9^{2x}=(3^2)^{2x}=3^{4x}$
Similarly, $27^{x^2-5}=3^{3x^2-15}$
As $3^{4x}\ne0,$
$$\implies1=\dfrac{3^{3x^2-15}}{3^{4x}}=3^{3x^2-4x-15}$$
Now like Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$,
if $\displaystyle u^m=1,$
either $\displaystyle m=0 $
or $\displaystyle u=1$
or $\displaystyle u=-1,m$ is even
But here $u=3\ne\pm1$
|
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|
Find, analytically, the value of the following limit. How would one prove that$$\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k-1}}$$ converges (rather slowly) to $\frac {1}{\sqrt{2}}$, which appears obvious from numerical computation.
|
Another interesting way to prove the lower bound.
Consider:
$$A=\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k-1}}$$
$$I_k=\frac{1}{\sqrt{2k}+\sqrt{2k-1}}=\frac{2}{\sqrt{\pi}}\int_0^\infty e^{-(4k-1)x^2} e^{-4k x^2 \sqrt{1-1/(2k)}} dx$$
Since $(\sqrt{2k}+\sqrt{2k-1})^2=4 k-1+4 \sqrt{k(k-1/2)}$.
$$\sqrt{1-\frac{1}{2k}}= 1-\frac{1}{4k}-\frac{1}{32k^2}-O \left( \frac{1}{8k^3} \right)<1-\frac{1}{4k}$$
We have:
$$I_k > \frac{2}{\sqrt{\pi}} \int_0^\infty e^{-8kx^2} dx$$
$$A \geq \frac{2}{\sqrt{\pi}} \lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n \int_0^\infty e^{-8kx^2} dx= \\ =\frac{2}{\sqrt{\pi}} \lim_{n\to\infty}\frac{1}{\sqrt{n}} \int_0^\infty \frac{e^{-8 x^2}-e^{-8(n+1) x^2}}{1-e^{-8 x^2}} dx$$
$$A \geq \frac{2}{\sqrt{\pi}} \lim_{n\to\infty}\frac{1}{n} \int_0^\infty \frac{e^{-8 y^2/n}-e^{-8(1+1/n) y^2}}{1-e^{-8 y^2/n}} dy$$
$$A \geq \frac{2}{\sqrt{\pi}} \lim_{n\to\infty}\frac{1}{n} \int_0^\infty \frac{1-e^{-8 y^2}}{e^{8 y^2/n}-1} dy=\frac{1}{4 \sqrt{\pi}}\int_0^\infty \frac{1-e^{-8 y^2}}{y^2} dy$$
$$A \geq \frac{1}{\sqrt{2 \pi}}\int_0^\infty \frac{1-e^{-z^2}}{z^2} dz= \frac{1}{\sqrt{2}}$$
The last integral is easy to prove using integration by parts.
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|
shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{25}- \frac{8}{5})}^2} = \frac{\sqrt{52}}{25}$
but my answer is wrong. Where am i wrong?
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Another way:
Using coordinate geometry,
Any point on the circle $P(2\cos t,2\sin t)$
Distance of $P$ from the line $$\dfrac{|3(2\cos t)+4(2\sin t)-12|}5$$
Now $-\sqrt{6^2+8^2}\le 6\cos t+8\sin t\le?$
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|
Finding the best solution to an inconsistent system $A\mathbf{u} = \mathbf{b}$. Let $A = \begin{bmatrix}
-1 & 1 \\
2 & -1 \\
1 & 1
\end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}$.
1. Find a "best solution" to the inconsistent system $A\mathbf{u} = \mathbf{b}$.
2. Find the orthogonal projection of $\mathbf{b}$ onto the column space of $A$.
For the second question the column space of $A$ has vectors that are all linearly independent. We first find the projection matrix given by $P = A(A^TA)^{-1}A^T$. Lets first calculate $A^TA$
\begin{equation*}
A^TA = \begin{bmatrix}
-1 & 2 & 1 \\
1 & -1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 1 \\
2 & -1 \\
1 & 1
\end{bmatrix} = \begin{bmatrix}
6 & -2 \\
-2 & 3
\end{bmatrix}.
\end{equation*}
This means that
\begin{equation*}
(A^TA)^{-1} = \frac{1}{14}\begin{bmatrix}
3 & 2 \\
2 & 6
\end{bmatrix}.
\end{equation*}
Hence,
\begin{equation*}
\begin{split}
P = A(A^TA)^{-1}A^T &= \frac{1}{14}\begin{bmatrix}
-1 & 1 \\
2 & -1 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
3 & 2 \\
2 & 6
\end{bmatrix}
\begin{bmatrix}
-1 & 2 & 1 \\
1 & -1 & 1
\end{bmatrix} \\
&= \frac{1}{14}\begin{bmatrix}
-1 & 1 \\
2 & -1 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 4 & 5 \\
4 & -2 & 8
\end{bmatrix} \\
&= \frac{1}{14}\begin{bmatrix}
5 & -6 & 3 \\
-6 & 10 & 2 \\
3 & 2 & 13
\end{bmatrix}.
\end{split}
\end{equation*}
So the orthogonal projection of $\mathbf{b} = (1,2,0)$ onto the column space of $A$ is
\begin{equation*}
\frac{1}{14}\begin{bmatrix}
5 & -6 & 3 \\
-6 & 10 & 2 \\
3 & 2 & 13
\end{bmatrix}
\begin{bmatrix}
1 \\
2 \\
0 \\
\end{bmatrix} = \frac{1}{14}\begin{bmatrix}
-7 \\
14 \\
7
\end{bmatrix} = \begin{bmatrix}
-1/2 \\
1 \\
1/2
\end{bmatrix}.
\end{equation*}
Not sure what is meant by the "best solution" to the system. Any would help would be great!!!
|
Actually there are three lines $$-x+y=1, 2x-y=2, x+y=0~~~(1)$$ which make a triangle.
pre-multiply the given inconsistent equation $$AX=B,~~ A=\begin{bmatrix} -1 & 1\\ 2 & 1 \\ 1 & 1 \end{bmatrix}, B=\begin{bmatrix} 1 \\ 2\\ 0 \end{bmatrix} ~~~~(2)$$ by $A^{T}$ to get $$ A^T A X= A^T B$$ which is nothing but $${\cal A} X= {\cal B},~~ {\cal A}_{2 \times 2},~~ {\cal B}_{2,1} ~~~(3)$$ You can then solve (3) for two unknowns $(x_0, y_0)$. This point $P(x_0, y_0)$ is inside the triangle mentioned above. This solution is a representative approximate solution of the inconsistent equations (1),(2).
Here in your case $${\cal A}=\begin{bmatrix} 6 & -2 \\ -2 & 3 \end{bmatrix}, ~~{\cal B}=\begin{bmatrix} 3 \\ -1 \end{bmatrix} \Rightarrow X={\cal A}^{-1} {\cal B}\Rightarrow X=\begin{bmatrix} 1/2 \\ 0 \end{bmatrix} \Rightarrow ~~x_0=1/2, y_0=0.$$ So $(1/2,0)$ is representative solution of these inconsistent/ overdetermined equations (1) and (2). Further it will be interesting to see if it is also least square or the best solution to these equations.
|
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|
Finding the length of a Curve The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney. It is relatively
early in the book, so I would expect the integration to be easy.
Find the length of the curve:
$$ 9x^2 = 4y^3$$
from $(0,0)$ to $\left(2\sqrt{3},3\right)$.
Answer:
The formal for the length of a curve is:
$$ L = \int_a^b \sqrt{ 1 + {f'(x)}^2 } \, dx $$
In this case, we have:
\begin{align*}
a &= 0 \\
b &= 2\sqrt{3} \\
y^3 &= \frac{9x^2}{4} \\
f(x) &= \left( \frac{9x^2}{4} \right) ^ {\frac{1}{3}} \\
f'(x) &= \frac{1}{3} \left( \frac{18x}{4} \right) \left( \frac{9x^2}{4} \right) ^ {-\frac{2}{3}} \\
f'(x) &= \left( \frac{3x}{2} \right) \left( \frac{9x^2}{4} \right) ^ {-\frac{2}{3}} \\
\end{align*}
\begin{align*}
L &= \int_0^{2\sqrt{3}} \sqrt{ 1 + \left( \frac{9x^2}{4} \right) \left( \frac{9x^2}{4} \right) ^ {-\frac{4}{3}} } \, dx \\
L &= \int_0^{2\sqrt{3}} \sqrt{ 1 + \left( \frac{9x^2}{4} \right) ^ {-\frac{1}{3}} } \, dx \\
L &= \int_0^{2\sqrt{3}} \sqrt{ 1 + \left( \frac{4}{9x^2} \right) ^ {\frac{1}{3}} } \, dx \\
\end{align*}
The book's answer is $$ \frac{14}{3} $$
Using an online integral calculator, my integral did not match. What did I do wrong?
I used the following website to do the integration:
https://www.integral-calculator.com/
Their answer is:
$$ \left( \frac{4^\frac{1}{3}}{9^\frac{1}{3}x^{\frac{2}{3}}} + 1 \right) ^ \frac{3}{2} x + C $$
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From $9x^2=4y^3$, you can also get $x=\frac23y^{3/2}$. And, if $g(x)=\frac23x^{3/2}$, then\begin{align}\int\sqrt{1+\bigl(g'(x)\bigr)^2}\,\mathrm dx&=\int\sqrt{1+x}\,\mathrm dx\\&=\frac23(1+x)^{3/2}.\end{align}So, the length of the curve is$$\frac23(1+3)^{3/2}-\frac23(1+0)^{3/2}=\frac{16}3-\frac23=\frac{14}3.$$
|
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|
Tips on solving $a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$ $$a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$$
This is true with $a>b>0$, according to Wolfram Alpha, but I am not able to prove this.
I to simplify using the fact that $a^2(a+2b)=a^3+2ba^2>3b^3$, and then proving a stronger result:
$$3b^3(a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$$
$$\iff 3(a + b^2 + b)>(a^2 + a + b)$$
$$\iff 3a + 3b^2 + 3b>a^2 + a + b$$
But this is not true, e.g. $a=100, b=1$.
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COMMENT.-The points $(a,b)=(1,t)$ for $t\ge4$ does not satisfy the inequality.
|
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|
How is this expression derived? On Concrete Mathematics (page 7) we have the following:
$L_n = L_{n-1} + n = (\frac{1}{2}(n-1)n + 1 ) + n = \frac{1}{2}n(n+1)+1$
How is the last expression derived?
How do you convert $(\frac{1}{2}(n-1)n + 1 ) + n$ into $\frac{1}{2}n(n+1)+1$ ?
I've tried different things but was not able to solve it.
|
Multiply out, collect the first-degree terms, and then put $\frac12n$ back outside a parenthesis again:
$$ \begin{align} (\tfrac12 (n-1)n + 1) + n &=
\tfrac12 n^2 - \tfrac12 n + 1 + n \\&=
\tfrac12 n^2 + n - \tfrac12n + 1 \\&=
\tfrac12 n^2 + \tfrac12 n + 1 \\&=
\tfrac12n (n + 1) + 1
\end{align} $$
Or you could also do it as:
$$ \begin{align} (\tfrac12 (n-1)n + 1) + n &=
\tfrac12 n(n-1) + 1 + \tfrac 12n \cdot 2 \\&=
\tfrac12n(n-1+2) + 1 \\&=
\tfrac12n (n + 1) + 1
\end{align} $$
|
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|
Find the approximation of $\sqrt{80}$ with an error $\lt 0.001$ Find the approximation of $\sqrt{80}$ with an error $\lt 0.001$
I thought it could be good to use the function $f: ]-\infty, 81] \rightarrow \mathbb{R}$ given by $f(x) = \sqrt{81-x}$
Because this function is of class $C^{\infty}$, we can compute its Taylor expansion given by :
$$T^n_{0} = 9 - \frac{1}{2}(81)^{-1/2}(x) + \frac{1}{4}(81)^{-3/2}\frac{x^2}{2}-\frac{3}{8}(81)^{-5/2}\frac{x^3}{6}\ + \dotsm $$
By the Lagrange remainder, $\exists$ for each $n \in \mathbb{N}$, $c_n \in [0,1]$ such that :
$$R^n(1) = f^{n+1}(c) \frac{1-0^{n+1}}{(n+1)!} = f^{n+1}(c) \frac{1}{(n+1)!} \leq 9.\frac{1}{(n+1)!} \lt 10^{-3}$$
$=> n(+1)! \gt \frac{9}{10^{-3}} = 9000$
So we can take $n = 8$
The approximation seems a little bit tricky to calculate especially without a calculator. I'm wondering if everything above is correct ?
|
Since $8^2 = 64$ and $9^2 = 81$ and $81 - 64 = 17$, using $8.9$ as an estimate for $\sqrt {80}$ is more than just a lucky guess. Since
$\quad (8.9)^2 < 80 \text{ and } (9)^2 > 80$
we must have
$\quad 0 \lt \sqrt{80} - 8.9 \lt 0.1$
Define the function
$\tag 1 F(x) = \frac{8.9x + 80}{x+8.9}$
For $x \ne -8.9$ we have the following identity,
$\tag 2 F(x) - \sqrt{80} = (x - \sqrt{80}) \, (x + 8.9)^{-1} \, (8.9 - \sqrt{80}) $
(c.f. this link)
Since $(8.9 + 8.9)^{-1} \lt 0.1$ the following must be true
$$\tag 3 |F(8.9) - \sqrt{80}| \lt (0.1)^3 = 0.001$$
All that remains is to evaluate $F(8.9)$,
$\tag 4 F(8.9) = \frac{15921}{1780}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3316086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 6
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|
Normal equation question Let $C = \begin{bmatrix}
1 & -1 & 1 \\
1 & 1 & 3 \\
2 & 0 & 4
\end{bmatrix}$ and let $\mathbf{b} = \begin{bmatrix}
3 \\
5 \\
5
\end{bmatrix}$. Find the set of all solutions $\mathbf{x}$ to the normal equation}
\begin{equation*}
C^TC\mathbf{x} = C^T\mathbf{b}.
\end{equation*}
Hence, or otherwise, find the best approximation to $\mathbf{b}$ in $\text{col}(C)$.
Lets first calculate $C^TC$ which is
\begin{equation*}
\begin{bmatrix}
1 & 1 & 2 \\
-1 & 1 & 0\\
1 & 3 & 4
\end{bmatrix}
\begin{bmatrix}
1 & -1 & 1 \\
1 & 1 & 3 \\
2 & 0 & 4
\end{bmatrix} = \begin{bmatrix}
6 & 0 & 12 \\
0 & 2 & 2 \\
12 & 2 & 26
\end{bmatrix}.
\end{equation*}
Also
\begin{equation*}
C^T\mathbf{b} =\begin{bmatrix}
1 & 1 & 2 \\
-1 & 1 & 0\\
1 & 3 & 4
\end{bmatrix}
\begin{bmatrix}
3 \\
5 \\
5
\end{bmatrix} = \begin{bmatrix}
18 \\
2 \\
38
\end{bmatrix}.
\end{equation*}
Thus, to find $\mathbf{x}$ we solve
\begin{equation*}
\begin{bmatrix}
6 & 0 & 12 \\
0 & 2 & 2 \\
12 & 2 & 26
\end{bmatrix} \mathbf{x}= \begin{bmatrix}
18 \\
2 \\
38
\end{bmatrix}.
\end{equation*}
To do this we just use row reduction to get
\begin{equation*}
\begin{bmatrix}
6 & 0 & 12 & 18 \\
0 & 2 & 2 & 2 \\
12 & 2 & 26 & 38
\end{bmatrix} \sim \begin{bmatrix}
1 & 0 & 2 & 3 \\
0 & 1 & 1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{equation*}
Using $R$ we must have then $x_1 = 3-2x_3$, $v_2 = 1-x_3$. So
\begin{equation*}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} = \begin{bmatrix}
3 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}.
\end{equation*}
Hence, we see that
\begin{equation*}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}\in \left\{\begin{bmatrix}
3 \\
1 \\
0
\end{bmatrix}+c\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}:c\in \mathbb{R}\right\} = \begin{bmatrix}
3 \\
1 \\
0
\end{bmatrix}+\text{span}\left\{\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}\right\}.
\end{equation*}
|
*
*Solve the system with $C^T C x = C^T b$ with the Gaussian elimination (or the row echelon form) --- i.e. do not compute the inverse (which does not exist here)
*If $C^T C$ is singular, it means that you will get either no solution or an infinite number of solutions. The row echelon form should help you determine whether or not there is a solution.
|
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"url": "https://math.stackexchange.com/questions/3316834",
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"source": "stackexchange",
"question_score": "1",
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|
Find $y$ in $\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$ I would like to solve this equation for $y$. Any tips? It seems like you cant really do it analytically?
$$\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$$
|
Hint: After squaring one times we get
$$2\sqrt{4+(y-6)^2}\sqrt{16+(y-3)^2}=25-(y-6)^2-(y-3)^2$$
squaring again and simplfying we get
$$4 \left(46 y^2-450 y+975\right)=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3320253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
$\frac{1+m_v}{1+m_u}\leq \frac{1+u^T(M+I)^{-1} u}{1+v^T(M+I)^{-1}v} \leq \frac{1+m_u}{1+m_v}$ if $M$ is positive sym. PD & $u,v$ are $0-1$ vectors? Let $n$ be a positive integer. Let $m_u,m_v \in \{1,...,n-1 \}$.
Let $M$ be a $n \times n$ symmetric positive definite matrix with positive entries.
Let $u$ and $v$ be vectors of length $n$ with entries consisting $n-m_u$ (or $n-m_v$) $0$'s and $m_u $ (or $m_v$) $1$'s. Sort $u$ so that the first $n-m_u$ entires of $u$ are $0$'s and the last $m_u$ entries are $1$'s. Sort $v$ in the same way.
Is the following statement true for all $n$?
*
*If $m_v<m_u$, then
\begin{align*}
\frac{1+m_v}{1+m_u}
\leq
\frac{1+u^\top(M+I_n)^{-1} u}{1+v^\top(M+I_n)^{-1} v}
\leq
\frac{1+m_u}{1+m_v}
\end{align*}
*If $m_v>m_u$, then
\begin{align*}
\frac{1+m_u}{1+m_v}
\leq
\frac{1+u^\top(M+I_n)^{-1} u}{1+v^\top(M+I_n)^{-1} v}
\leq
\frac{1+m_v}{1+m_u}
\end{align*}
Note
This question was motivated by another question of mine. It contains somewhat lengthly motivation of why I would like to show the inequalities above.
Finding so far
I initially thought a sharper bound by $1$ might be possible, but it was not. Suppose $m_v<m_u$. It is not guaranteed that $u^\top(M+I_n)^{-1} u -v^\top(M+I_n)^{-1} v \geq 0$. For instance, consider the example provided here with the matrix
$$M =
\begin{bmatrix}
1 & 1 & 1\\
1 & 100 & 99\\
1 & 99 & 100\\
\end{bmatrix}, \\
$$
and the vectors $u = (0, 1, 1)$ and $v =(0, 0, 1)$.
This means that the sharper lower bound by $1$:
\begin{align*}
\frac{1+m_v}{1+m_u}
<
1
\leq
\frac{1+u^\top(M+I_n)^{-1} u}{1+v^\top(M+I_n)^{-1} v}
\end{align*}
is not possible. However, the proposed bounds by $\frac{1+m_v}{1+m_u}$ and $\frac{1+m_u}{1+m_v}$ still work with the $M$, $u$, and $v$ in the example above.
|
It doesn't hold.
Here is a counterexample for the right inequality in the first line.
Let $m_u = m+1$ and $m_v=m$. Let the matrix $M$ be diagonal with $M_{ii} = a$ for all diagonal entries, except for $M_{ii} = b$ for the $(m+1)$-last entry. If positive entries of the matrix are required, all other elements can be filled with extremely small positive numbers. We can show the effect with the diagonal matrix. Based on the obtained effect, a concrete counterexample without the diagonality conditions will be given below.
For the proposed diagonal matrix, the right inequality in $$
\begin{align*}
\frac{1+m_v}{1+m_u}
\leq
\frac{1+u^\top(M+I_n)^{-1} u}{1+v^\top(M+I_n)^{-1} v}
\leq
\frac{1+m_u}{1+m_v}
\end{align*}$$
demands that we need to have
$$
\begin{align*}
\frac{1+\frac{m}{1+a} + \frac{1}{1+b}}{1+\frac{m}{1+a}}
\leq
\frac{1+m+1}{1+m}
\end{align*}$$
Clearing denominators gives
$$
\begin{align*}
({1+\frac{m}{1+a} + \frac{1}{1+b}})(1+m) &\leq ({1+\frac{m}{1+a}})
({1+m+1})\\
m + \frac{m}{1+a} + \frac{1}{1+b} + m (\frac{m}{1+a} + \frac{1}{1+b}) &\leq m + \frac{m}{1+a} + 1 + \frac{m}{1+a} ( m+1) \\
\frac{1}{1+b} + \frac{m}{1+b} &\leq 1 + \frac{m}{1+a} \\
1+a + m(1+a) &\leq (1+a)(1+b) + m(1+b) \\
m(a-b) &\leq (1+a)b
\end{align*}$$
Now let $a>b$, then the inequality only holds for
$$
m \leq \frac{ (1+a)b}{a-b}
$$
However, a large enough $m$ can be easily constructed which violates this condition. The RHS doesn't have to be large: as an example, let $a=1000$, $b < 1$, the the RHS approximately equals $b$. So the violation can in those cases even be observed for small $m$.
With these considerations, we can now give a concrete counterexample without the diagonality conditions. Indeed, as we have just seen we have $m<<1$ for $a=1000$, $b << 1$, which will be violated even for $m=1$. This allows to construct the following
counterexample: Let
$$
M =
\begin{bmatrix}
1000 & 0.0001 & 0.0001 \\ 0.0001 & 0.1 & 0.0001 \\ 0.0001 & 0.0001 & 1000
\end{bmatrix}, \\
$$
and as in the other counterexample, $u = (0, 1, 1)$ and $v = (0, 0, 1)$.
Then, evaluating the numbers we need to have that
$$
\begin{align*}
\frac{1+u^\top(M+I_n)^{-1} u}{1+v^\top(M+I_n)^{-1} v}
= \frac{1+0.91}{1+0.001}
\leq
\frac{1+2}{1+1} = \frac{1+m_u}{1+m_v}
\end{align*}$$
which is clearly wrong.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evaluate $ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx $ $$ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx = ? $$
Attempt:
$$ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx = \int \frac{3x^{2}}{\sqrt{x^{3}-6} \sin^{2}\sqrt{x^{3}-6}} dx $$
$$ U = \sqrt{x^{3}-6} $$
then the integral is
$$ \int \frac{2U}{\sin^{2} U} dU $$
then using partial integration we will get the above equals:
$$ -2 U \cot U + \ln(\sin U) + C = -2 \sqrt{x^{3}-6} \cot(\sqrt{x^{3}-6}) + \ln (\sin \sqrt{x^{3}-6}) + C$$
But...
$$ \frac{d (-2 \cot(\sqrt{x^{3}-6}))}{dx} = \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} $$
|
$$u=\sqrt{x^3-6},du=\dfrac{3x^2}{2\sqrt{x^3-6}}dx$$
$$2\int\csc^2u=-2\cot u+K$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3323749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$. If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$.
How to prove it without using calculus?
I know if $a , b,c,d \in \mathbb R$ then $a^2+b^2+c^2+d ^2$ will be minimum when $a =b= c =d = \frac{4m+1}{4}$..So the minimum value would have been $4m^2 +2m +1/4$..But what to do in that case?
|
Taking $a = b = c = m$ and $d = m + 1$ gives that $a^2 + b^2 + c^2 + d^2 = 4m^2 + 2m + 1$, implying that the minimum value of the expression is at most $4m^2 + 2m + 1$. The calculus approach you mentioned, or maybe an application of the Cauchy-Schwarz inequality would lead to the fact that taking all the terms equal gives the minimum value of the expression, and the choice above is simply to ensure that all terms are integers in a way that is closest to all of them being equal.
Finally, to check that this works for any $a, b, c, d$ we assume first that $a \le b \le c \le d$. Then, $4d \geq 4m + 1$ which implies $d \geq m+1$ since $d$ is an integer. Thus, we can write
\begin{align*}
a^2 &+ b^2 + c^2 + d^2 \\
&= (a-m+m)^2 + (b-m+m)^2 + (c-m+m)^2 + (d-m-1+m+1)^2 \\
&= (a-m)^2 + (b-m)^2 + (c-m)^2 + (d - m-1)^2 + \\
& \qquad + 2m(a-m + b-m + c-m + d-m) -2m + 2(d - m - 1) \\
& \qquad + m^2 + m^2 + m^2 + (m+1)^2 \\
&= (a-m)^2 + (b-m)^2 + (c-m)^2 + (d - m-1)^2 + \\
& \qquad + 2m - 2m + 2(d - m - 1) \\
& \qquad + 4m^2 + 2m + 1 \\
&\geq 4m^2 + 2m + 1.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Verify the following limit using epsilon-delta definition: $ \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0$ Show that $$ \lim\limits_{(x,y)\to(0,0)}\dfrac{x^2y^2}{x^2+y^2}=0$$
My try:
We know that, $$ x^2\leq x^2+y^2 \implies x^2y^2\leq (x^2+y^2)y^2 \implies x^2y^2\leq (x^2+y^2)^2$$
Then, $$\dfrac{x^2y^2}{x^2+y^2}\leq x^2+y^2 $$
So we chose $\delta=\sqrt{\epsilon}$
|
HINT
\begin{align*}
0\leq x^{2} \leq x^{2} + y^{2} \Longleftrightarrow 0\leq \frac{x^{2}}{x^{2}+y^{2}} \leq 1 \Longleftrightarrow 0\leq \frac{x^{2}y^{2}}{x^{2}+y^{2}}\leq y^{2}
\end{align*}
Then apply the squeeze theorem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3325340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why any square can be written in this form? I have noticed that all squares, at least up to $19 ^ 2$ can be written as: $a^2 = 5k + p$, where $a \in \mathbb{Z}+\neq 1$ and $k \in \mathbb{Z}+$ and $p = \{0,\pm1\}$
Some examples:
$4^2 = 5 \cdot 3 + 1$
$13^2 = 5 \cdot 34 - 1 $
What is the intuitive and formal proof to see this?
|
Say we want to consider $b^2$. We look at $b$ modulo $5$, $b = 5c+d$, where $c$ is an integer and $d$ is one of $0$, $1$, $2$, $3$, or $4$. Then
$$ b^2 = (5c+d)^2 = 25c^2 + 10c d + d^2 \text{.} $$
The part "$25 c^2 + 10 c d$" is a multiple of $5$, so can be collected into $k$. This leaves $d^2$. We show in a table that $d^2$ is congruent to $0$ or $\pm 1$ modulo $5$ for each choice of $d$. \begin{align*}
&d & &d^2 \\
&0 & &0 \\
&1 & &1 \\
&2 & &4 \cong -1 \pmod{5} \\
&3 & &9 \cong -1 \pmod{5} \\
&4 & &16 \cong 1 \pmod{5} \text{.}
\end{align*}
In addition to showing that every $b$ squares to a multiple of $5$ or $\pm 1$ from a multiple of $5$, we can read from the table which of these cases occurs since $d \cong b \pmod{5}$.
|
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"language": "en",
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|
Prove this Ramanujan series
Prove how this series: $\frac{1}{1^3}\cdot\frac{1}{2} + \frac{1}{2^3}\cdot\frac{1}{2^2} + \frac{1}{3^3}\cdot\frac{1}{2^3} + \frac{1}{4^3}\cdot\frac{1}{2^4} +...= \frac{1}{6}(\log 2)^3+\frac{\pi ^2}{12}(\log 2) + (\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+..)$
I don't know how to begin but I know that the left-hand side is of the form $$\sum\frac{1}{n^3}\cdot\frac{1}{2^n}$$
|
For $0 < x < 1$ we have
$$
\sum_{k=1}^{\infty}x^k = \frac{x}{1-x}
$$
now calling $S_0 = \frac{x}{1-x}$ we have the relationships
$$
x S_3' = S_2\\
x S_2' = S_1\\
x S_1' = S_0
$$
after integration with null constants we have
$$
\left\{
\begin{array}{rcl}
S_1 & = & -\log (1-x) \\
S_2 & = & -\log (1-x) \log (x)-\text{Li}_2(1-x) \\
S_3 & = & -\log (1-x) \log ^2(x)-\text{Li}_2(1-x) \log (x)-\text{Li}_2(x) \log (x)+\text{Li}_3(x) \\
\end{array}
\right.
$$
making now $x = \frac 12$
$$
\cases{
S_1(\frac 12)=\log (2) \\
S_2(\frac 12)= \frac{1}{12} \left(\pi ^2-6 \log ^2(2)\right) \\
S_3(\frac 12)=\frac{1}{24} \left(4 \log ^3(2)-2\pi ^2\log (2)+21 \zeta (3)\right) \\
}
$$
|
{
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"url": "https://math.stackexchange.com/questions/3328676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to Evaluate proper Integral. Recently I stumbled upon an integral and its solution in a physics article but I couldn't understand how it was evaluated.I have plotted the function and it indicates that the value of the integral should be finite.
I want to ensure the answer is correct.Does anybody know how to do it?
$$\int_0^T d t' \frac{1}{(t'(T-t')) ^{3/2}}\exp \left[ -\frac{A}{t'}- \frac{B}{T-t'}\right]= \sqrt{ \frac{\pi}{T^{3}} } \frac{ \sqrt{A}+ \sqrt{B} }{ \sqrt{AB} }\exp\left[-\frac{(\sqrt{A}+\sqrt{B})^{2}}{T}\right] $$
|
Assuming $A,B,T>0$,
$$\begin{align}
I=&\int_{0}^{T}\frac{1}{(t(T-t))^{3/2}}\exp\left[-\frac{A}{t}-\frac{B}{T-t}\right]\,dt\\
\stackrel{x=1/t}{=}&\int_{1/T}^{\infty}\frac{x}{(Tx-1)^{3/2}}\exp\left[-Ax-\frac{Bx}{Tx-1}\right]\,dx\\
\stackrel{t=Tx}{=}&\;\frac{1}{T^{2}}\int_{1}^{\infty}\frac{t}{(t-1)^{3/2}}\exp\left[-\frac{1}{T}\left(At+\frac{Bt}{t-1}\right)\right]\,dt\\
\stackrel{x=t-1}{=}&\;\frac{e^{-(A+B)/T}}{T^{2}}\int_{0}^{\infty}\frac{x+1}{x^{3/2}}\exp\left[-\frac{1}{T}\left(Ax+\frac{B}{x}\right)\right]\,dx\\
\stackrel{t=\sqrt{x}}{=}&\;\frac{2e^{-(A+B)/T}}{T^{2}}\int_{0}^{\infty}\left(1+\frac{1}{t^{2}}\right)\exp\left[-\frac{1}{T}\left(At^{2}+\frac{B}{t^{2}}\right)\right]\,dt.
\end{align}$$
Then using
$$\int_{0}^{\infty}\exp\left[-ax^{2}-\frac{b}{x^{2}}\right]\,dx=\int_{0}^{\infty}\exp\left[-bx^{2}-\frac{a}{x^{2}}\right]\frac{1}{x^{2}}\,dx=\frac{\sqrt{\pi}e^{-2\sqrt{ab}}}{2\sqrt{a}} \tag{1}$$
we have
$$\int_{0}^{\infty}\left(1+\frac{1}{t^{2}}\right)\exp\left[-\frac{1}{T}\left(At^{2}+\frac{B}{t^{2}}\right)\right]\,dt=\frac{\sqrt{\pi T}e^{-2\sqrt{AB}/T}}{2}\left(\frac{1}{\sqrt{A}}+\frac{1}{\sqrt{B}}\right).$$
Thus
$$\begin{align}
I&=\frac{2e^{-(A+B)/T}}{T^{2}}\frac{\sqrt{\pi T}e^{-2\sqrt{AB}/T}}{2}\left(\frac{1}{\sqrt{A}}+\frac{1}{\sqrt{B}}\right)\\
&=\boxed{\sqrt{\frac{\pi}{T^{3}}}\left(\frac{1}{\sqrt{A}}+\frac{1}{\sqrt{B}}\right)\exp\left[-\frac{(\sqrt{A}+\sqrt{B})^{2}}{T}\right]}
\end{align}$$
Proof of $(1)$:
$$\begin{align}
\int_{0}^{\infty}\exp\left[-ax^{2}-\frac{b}{x^{2}}\right]\,dx
\stackrel{t=\sqrt{a}x}{=}&\;\frac{1}{\sqrt{a}}\int_{0}^{\infty}\exp\left[-t^{2}-\frac{ab}{t^{2}}\right]\,dt\\
\stackrel{x=\sqrt{ab}/t}{=}&\;\frac{\sqrt{ab}}{\sqrt{a}}\int_{0}^{\infty}\exp\left[-\frac{ab}{x^{2}}-x^{2}\right]\frac{1}{x^{2}}\,dx\\
={}&\;\frac{1}{2\sqrt{a}}\int_{0}^{\infty}\left(\frac{\sqrt{ab}}{x^{2}}+1\right)\exp\left[-\frac{ab}{x^{2}}-x^{2}\right]\,dx \tag{2}\\
\stackrel{t=x-\sqrt{ab}/x}{=}&\;\frac{1}{2\sqrt{a}}\int_{-\infty}^{\infty}\exp\left[-t^{2}-2\sqrt{ab}\right]\,dt\\
={}&\;\frac{e^{-2\sqrt{ab}}}{2\sqrt{a}}\int_{-\infty}^{\infty}e^{-t^{2}}\,dt\\
={}&\;\frac{\sqrt{\pi}e^{-2\sqrt{ab}}}{2\sqrt{a}},
\end{align}$$
where $(2)$ was obtained by averaging the previous two representations of the integral.
|
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|
Value of $(3\beta^2-4\beta)^{\frac{1}{3}}+(3\beta^2+4\beta+2)^{\frac{1}{3}}$ if $\beta$ is the root of $x^3-x-1=0$
If $\beta$ is the root of the equation $x^3-x-1=0$, find the value of
$$(3\beta^2-4\beta)^{\frac{1}{3}}+(3\beta^2+4\beta+2)^{\frac{1}{3}}.$$
This is what I tried:
$x=\beta$ is a root of $x^3-x-1=0,$
so getting $\displaystyle \beta^3-\beta-1=0\Rightarrow \beta^2=\frac{\beta+1}{\beta}.$
Now $$3\beta^2-4\beta = 3\bigg(\frac{\beta+1}{\beta}\bigg)-4\beta.$$
Don't know how to continue.
|
I assume that $\beta$ is the real root of $x^3-x-1=0$ so that the cube roots are well defined.
Let $\mu = 3\beta^2-4\beta$ and $\nu = 3\beta^2+4\beta+2$. We seek $\tau = \mu^{\frac{1}{3}}+\nu^{\frac{1}{3}}$. As in this question, we have
$$
\tau^3 = \mu+\nu+3(\mu\nu)^{\frac{1}{3}}\tau
$$
Hoping that $\mu\nu = \beta-\beta^2$ is a cube in $\mathbb Q(\beta)$, we find after some work that $\beta-\beta^2=(1-\beta^2)^3$. Therefore,
$$
\tau^3 = \mu+\nu+3(\mu\nu)^{\frac{1}{3}} = 6\beta^2+2+3(1-\beta^2)\tau
$$
Now $\tau=2$ is a root. The other roots are complex because the discriminant of the quotient quadratic is $-12\beta^2$.
Therefore, $\tau=2$ is the only real root and the answer is $2$.
|
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|
$\cos(x)=-\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$. Calculate $\sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x) \cos(y) - \sin(x) \sin(y)$. If $\cos(x)= -\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$ for $\frac{\pi}{2} < x < \pi$ and $\pi < y <\frac{3\pi}{2}$. What is the value of $\sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x) \cos(y) - \sin(x) \sin(y)$?
Solution:
If $\frac{\pi}{2} < x < \pi$ then right triangle with acute angle $x$ facing west at second quadrant, the hypotenuse must be positive and the only negative is the adjacent.
$\cos(x)= -\frac{24}{25}$ then $adjacent=-24$ and $front=7$ so $\sin(x) = \frac{7}{25}$.
If $\pi < y <\frac{3\pi}{2}$ then right triangle with acute angle $x$ facing west at third quadrant, the front and adjacent are negative. If $\tan(y) = \frac{9}{40}$ then the front is $-9$, the adjacent is $-40$, and the hypotenuse is $41$. So $\sin(y) = -\frac{9}{41}$ and $\cos(y)=-\frac{40}{41}$
The right triangle that we consider is the one with the adjacent side on the $x$-axis right..?
|
You can observe that
$$
\lvert\sin x\rvert=\sqrt{1-\cos^2x}=\sqrt{1-\frac{24^2}{25^2}}=\frac{7}{25}
$$
Since $\pi/2<x<\pi$, we conclude that $\sin x=7/25$.
Also
$$
\lvert\cos y\rvert=\sqrt{\frac{1}{1+\tan^2y}}=\frac{40}{41}
$$
and from $\pi<y<3\pi/2$ we conclude that $\cos y=-40/41$. Thus
$$
\sin y=\tan y\cos y=-9/41
$$
|
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|
Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$ Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$.
Let $P(x)=x^{2015}-x^{2014}=Q(x)(x-1)^3+ax^2+bx+c.$ If we put $x=1$ in $P(x)$ and $P'(x)$, we get $a+b+c=0$ and $2a+b=1$. Then: $c=a-1$. The second derivative won't help in finding $b$, so, what should I do? Thank you
|
Rewrite expression as
$$
\frac{x^{2015} - x^{2014}}{(x-1)^3} = \frac{x^{2014}}{(x-1)^2}
$$
so we have
$$
x^{2014} = Q(x) (x-1)^2 + ax + b.
$$
Now you need to find coefficients $a$ and $b$. Your argument with derivatives should work:
$$
1^{2014} = 1 = a + b
$$
$$
2014\cdot 1^{2013} = 2014 = a.
$$
Hence, $b = -2013$ and the final answer is
$$
x^{2015} - x^{2014} = x^{2014} (x-1) = Q(x) (x-1)^3 + (x-1)\cdot(2014x - 2013)
$$
|
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|
Find maxima and minima of $f(x) = x \cot x$ So I got a function
$$f(x)= x \cot x $$
I would like to find values of $x$ where $f'(x) = 0$
Applying product rule, we get:
$$f'(x) = \cot x - x \cdot csc^2 x $$
Setting equation to zero
$$\cot x - x \cdot csc^2 x = 0 $$
Now, I will try my best at simplifying equation above:
$$\frac{\cos x}{\sin x} - x \cdot \frac{1}{\sin^2x} = 0$$
Provided that $x ≠ 0$, multiply $\frac{cos x}{\sin x}$ by $\frac{\sin x}{\sin x}$
$$\frac{\cos x \sin x}{\sin^2 x} - \frac{x}{\sin^2x} = 0 \implies$$
$$\frac{\cos x \sin x - x }{\sin^2 x}= 0 $$
Multiplying both sides by $\sin^2 x$, we arrive at
$$\cos x \sin x - x = 0 $$
Now, assuming that all calculations above are correct, we can make some observations:
If $x ≥ 1 $ or $x ≤ -1$, then $\cos x \sin x + x ≠ 0 $. Thus if equation above has roots, then $x$ must be somewhere between $(-1,1)$
Equation above equals zero if $x = 0$, but we stipulated before that $x$ cannot equal $0$.
From here, I am stuck.
By looking at the graphing calculator, it's getting evident that there is
no value $x$ such that $f'(x) = 0$, but how do I show it mathematically?
|
Consider $g(x) = \cos x \sin x -x$, then $g(0)=0$ and $g(x)$ is monotone because $g'(x) = -2\sin^2(x)$.
|
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|
In $\triangle ABC$, if angle bisectors $AE$ and $CD$ meet at incenter $F$, and $|FE|=|FD|$, then the triangle is isosceles or $\angle B=60^\circ$ I was screwing around lately in GeoGebra and I realized something.
Draw a $\triangle ABC$, and let the bisectors for $\angle A$ and $\angle C$ meet sides $BC$ and $AB$ at points $E$ and $D$, respectively. If the angle bisectors meet at the incenter, $F$, and if $FD \cong FE$, then either $\triangle ABC$ must be isosceles or $\angle B$ must be $60^\circ$.
However I was unable to prove why that is. Any help would be appreciated.
|
Let $AB = c$, $BC = a$, $AC = b$. We will prove that if $EF = FD$, then $ABC$ is either isosceles or $\angle ABC = 60$.
We will first show that $\frac{DF}{FC} = \frac{c}{a+b}$. First, notice that by the angle bisector theorem, we have that $\frac{DF}{FC} = \frac{BD}{BC} = \frac{BD}{a}$. Now, note that $\frac{BD}{c - BD} = \frac{a}{b}$ by the angle bisector theorem, and expanding gives us $BD = \dfrac{\frac{a}{b} \cdot c}{1 + \frac{a}{b}}$. Thus, $\frac{BD}{a} = \dfrac{\frac{c}{b}}{\frac{b+a}{b}} = \frac{c}{b+a}$. Therefore, $\frac{DF}{DC} = \frac{c}{a+b+c}$.
Similarly, we have that $\frac{EF}{FA} = \frac{a}{b+c} \implies \frac{EF}{EA} = \frac{a}{a+b+c}$. Therefore, we have that, since $DF = EF$, $\frac{EA}{DC} = \frac{c}{a} \implies \frac{EA}{AB} = \frac{DC}{BC}$.
Now, note that by the Law of Sines, $\frac{EA}{AB} = \frac{\sin{\angle B}}{\sin \angle{AEB}}$, and $\frac{DC}{BC} = \frac{\sin{\angle{B}}}{\sin{\angle{BDC}}}$, which implies that $\angle AEB = \angle BDC$ or $\angle AEB + \angle BDC = 180$.
We now consider cases. In the first case, $\angle AEB = 180 - (\angle B + \angle BAE)$ and $\angle BDC = 180 - (\angle B + \angle DCB)$ \implies $\frac{\angle A}{2} = \frac{\angle C}{2}$ which implies $\angle BAC = \angle BCA$, and in the second case we have $360 - 2\angle B - (\angle DCB + \angle BAE) = 180$, and since $\angle DCB + \angle BAE = 90 - \frac{\angle{B}}{2}$ \implies $\frac{3\angle{B}}{2} = 90 \implies \angle{B} = 60$, so we are done.
|
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|
$\pi \cot\pi z\ne -\frac{1}{z}+\lim_{N\to\infty}\sum_{n=-N}^N\frac{1}{z+n}$ I know that
$$\pi\cot\pi z=\frac{1}{z}+\displaystyle\underbrace{\sum_{n=1}^\infty \left(\frac{1}{z-n}+\frac{1}{z+n}\right)}_{S}.$$
Then
$$\begin{align*}S&=\cdots +\frac{1}{z-1}+\frac{1}{z+1}+\frac{1}{z-2}+\frac{1}{z+2}+\cdots\\&=\cdots +\frac{1}{z-2}+\frac{1}{z-1}+\frac{1}{z+1}+\frac{1}{z+2}+\cdots\\&=\cdots +\frac{1}{z-2}+\frac{1}{z-1}+\frac{1}{z-0}+\frac{1}{z+0}+\frac{1}{z+1}+\frac{1}{z+2}+\cdots -\frac{2}{z}\\&=\displaystyle\lim_{N\to\infty}\displaystyle\sum_{n=-N}^N \frac{1}{z+n}-\frac{2}{z}.\end{align*}$$
Therefore,
$$\begin{align*}\pi\cot\pi z&=\frac{1}{z}+\displaystyle\lim_{N\to\infty}\displaystyle\sum_{n=-N}^N \frac{1}{z+n}-\frac{2}{z}\\&=-\frac{1}{z}+\displaystyle\lim_{N\to\infty}\displaystyle\sum_{n=-N}^N \frac{1}{z+n}.\end{align*}$$
But the correct result is
$$\pi\cot\pi z=\displaystyle\lim_{N\to\infty}\displaystyle\sum_{n=-N}^N \frac{1}{z+n}.$$
Where is the mistake?
|
Note that$$\sum_{n=-N}^N\frac1{z-n}=\frac1{z-N}+\cdots+\frac1{z-1}+\frac1z+\frac1{z+1}+\cdots+\frac1{z+N},$$instead of$$\sum_{n=-N}^N\frac1{z-n}=\frac1{z-N}+\cdots+\frac1{z-1}+\frac1z+\frac1z+\frac1{z+1}+\cdots+\frac1{z+N},$$which is what you wrote.
|
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|
Formula for calculating the odds per user of winning in a raffle each player can win once Am trying to write a program which gives each user in a raffle contest their odds so far at winning.
The rules of the game are simple:
A predefined number of tickets are sold for example 1000 each user can at most buy 50 tickets and a user can only win once in the game after that all his tickets become invalid or pulled out and a new draw begins giving the other users a better chance of winning. The game will have 10 winners only.
With the data collected from the software I would then know how many tickets were sold, the users that bought them and the amount of tickets that each user has. Now I just need a formula or pseudocode that would give each user their statistic probability of winning based on the data acquired, so that it be can used before and after each draw in the game to show each user their odds so far.
I have looked at similar questions asked here, but no one seems to want to address the part that if a person wins the rest of their tickets become invalid. Am not good with probability or understand those fancy notations, so I don't understand is such a thing possible to calculate per user.
Thanks for the help
Update
Testing my understanding of joriki second method:
lets say 10 tickets were sold to 4 users each bought A: 1, B: 2, C: 4, D: 3
and there will be 3 prizes given to users.
I calculated the total probability of being drawn for each user to be
A = $\frac{1}{10} + \frac{2}{10}*\frac{1}{8}*\frac{1}{6} + \frac{4}{10}*\frac{1}{6}*\frac{1}{2} + \frac{3}{10}*\frac{1}{7}*\frac{1}{4}$ = 0.1482
B = $\frac{1}{10}*\frac{2}{9}*\frac{2}{8} + \frac{2}{10} + \frac{4}{10}*\frac{2}{6}*\frac{2}{2} + \frac{3}{10}*\frac{1}{7}*\frac{1}{4}$ = 0.3817
C= $\frac{1}{10}*\frac{4}{9}*\frac{4}{8} + \frac{2}{10}*\frac{4}{8}*\frac{4}{6} + \frac{4}{10} + \frac{3}{10}*\frac{4}{7}*\frac{4}{4}$ = 0.6603
D= $\frac{1}{10}*\frac{3}{9}*\frac{3}{8} + \frac{2}{10}*\frac{3}{8}*\frac{3}{6} + \frac{4}{10}*\frac{3}{6}*\frac{3}{2} + \frac{3}{10}$ = 0.6500
Their total sum is 1.8403 and not 3 ? also is this considered the total probability of being drawn for the 3 draws or just for the first round of the game with the tickets becoming invalid
|
Let $N_i$ be the number of tickets that the $i$-th player bought. Then the total number of possible draws of $m$ tickets is the coefficient of $x^m$ in the polynomial
$$
P(x)=\prod_i\left(1-N_i x\right);
$$
the number of possible draws where the $j$-th player doesn't win is the coefficient of $x^m$ in
$$
Q_j(x)=\frac{P(x)}{1-N_j x}=\prod_{i\neq j}\left(1-N_i x\right);
$$
and the winning probability for the $j$-th player is $1$ minus the ratio of the two coefficients. Because you only need to keep coefficients out to the $m$-th power of $x$ during the calculation, you can compute the exact winning probabilities for all players in time proportional to the number of players and the square of the number of tickets to be drawn.
To demonstrate this (generating function) method for OP's example, where players A,B,C,D buy $1,2,4,3$ tickets respectively:
$$
\begin{eqnarray}
P(x)&=&(1-x)(1-2x)(1-4x)(1-3x) \\ &=&1 - 10x + 35x^2 - 50x^3 + 24 x^4; \\
Q_A(x)&=&\frac{P(x)}{1-x}=1-9x+26x^2-24x^3; \\
Q_B(x)&=&\frac{P(x)}{1-2x}=1-8x+19x^2-12x^3; \\
Q_C(x)&=&\frac{P(x)}{1-4x}=1-6x+11x^2-6x^3; \\
Q_D(x)&=&\frac{P(x)}{1-3x}=1-7x+14x^2-8x^3.
\end{eqnarray}
$$
What's nice is that these polynomials encode the probabilities of winning for every possible number of tickets. If one ticket is drawn, the probabilities come from the coefficients of $x^1$:
$$
p_A = 1-(-9)/(-10) = 1/10, p_B = 1-(-8)/(-10)=1/5, \\p_C=1-(-6)/(-10)=2/5, p_D=1-(-7)/(-10)=3/10,
$$
which are obviously correct and sum to $1$. If two tickets are drawn, the probabilities come from the coefficients of $x^2$:
$$
p_A = 1-26/35=9/35, p_B=1-19/35=16/35, \\p_C=1-11/35=24/35, p_D=1-14/35=3/5;
$$
these sum to $2$, as they should. Finally from the coefficients of $x^3$, we find the probabilities of winning when three tickets are drawn:
$$
p_A=1-(-24)/(-50)=13/25, p_B=1-(-12)/(-50)=19/25, \\p_C=1-(-6)/(-50)=22/25, p_D=1-(-8)/(-50)=21/25,
$$
which sum to $3$. (And from the coefficients of $x^4$, when four tickets are drawn, each player wins with probability $1-0/24=1$.)
In an actual application, where the number of players was much larger than the number of tickets to be drawn, each polynomial multiplication would be truncated after the $x^m$ term.
|
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|
Prove that solution of the equation $8x^3-4x^2-4x+1= 0$ has roots $\cos\frac{\pi}{7},\cos\frac{3 \pi}{7},\cos\frac{5 \pi}{7}$
Prove that solution of the equation $8x^3-4x^2-4x+1= 0$ has roots $\cos\frac{\pi}{7},\cos\frac{3 \pi}{7}\space \text{and}\space \cos\frac{5 \pi}{7}$.
How to even solve it ? I have no idea.
Because cubic have no formula like quadratic equations to obtain the roots and the expression like $\cos\frac{\pi}{7}$ is something that I don't think I can evaluate with my limited knowledge (I'm in highschool) of only $\cos 2x$ or $\cos 3x$ type. Since it has denominator of 7, I thought maybe the seventh root of unity might help me, but since only some of its multiples are used, I don't think so it will be of much use.
How do I solve this problem. Please help.
|
$$\cos\frac{\pi}{7}\cos\frac{3\pi}{7}\cos\frac{5\pi}{7}=\frac{8\sin\frac{\pi}{7}\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}}{8\sin\frac{\pi}{7}}=\frac{\sin\frac{8\pi}{7}}{8\sin\frac{\pi}{7}}=-\frac{1}{8}.$$
The rest is a similar and use the Viete's theorem.
|
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|
Proving that $5^n - 1$ is divisible by $4$ by mathematical induction. I have done it, but I am not sure that the inductive step is right. Can anybody please clear me about it?
Basic steps as:
Taking $n=1$: $p(1)=5-1=4$.
Inductive hypothesis: Assume the statement is true for $p(k)$. $5^k - 1$ is divisible by $4$.
Inductive steps: We must show $p(k+1)$ is true when $p(k)$ is true.
\begin{align*}
& 5^k -1 + 5^{k+1} -1\\
& 5^k -1 + 5.5^{k} -1\\
& (5^k -1) + 4
\end{align*}
|
The idea is $$5^{k+1} - 1 = 5\cdot 5^{k} - 1 = 4\cdot(5^k) + (5^k - 1)$$
We have $4\cdot(5^k)$ is obviously divisible by $4$ and $5^k - 1$ is divisible by $4$ by inductive hypothesis
|
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|
Prove $\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx$ How to prove
$$\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx\tag{1}$$
Here is how I came up with this relation:
In this solution @Kemono Chen elegantly proved
$$\int_0^a\frac{\ln(1+ax)}{1+x^2}dx=\int_0^1\frac{a\ln(1+a^2x)}{1+a^2x^2}dx=\frac12\arctan a\ln(1+a^2)\tag{2}$$
and while trying to prove the identity in (2) starting from RHS, I ended up with the relation in (1). So any straightforward method to prove (1)? Plus any good applications of (1)?
The transformation of the integral in (2) was done by @Jack D'Aurizio here.
I will post my proof in the answer section soon and I am tagging "harmonic number" as the proof involves it in case you are curious. Thanks
UPDATE: If we let $\frac{1-x}{x}\mapsto x$ in (1) then combine with (2) we have
$$\int_0^1\frac{1+a^2x}{1+a^2x^2}dx=\int_0^\infty\frac{\ln x}{a^2+(1+x)^2}=\frac1{2a}\arctan a\ln(1+a^2)\tag{3}$$
|
In the post body we have
$$\int_0^1\frac{a\ln(1+a^2x)}{1+a^2x^2}dx=\frac12\arctan a\ln(1+a^2)\tag{*}$$
and from this solution we have
\begin{align}
f(a)&=\frac12\arctan a\ln(1+a^2)=-\sum_{n=0}^\infty \frac{(-1)^n H_{2n}}{2n+1}a^{2n+1}\\
&=-\sum_{n=0}^\infty \frac{(-1)^n H_{2n+1}}{2n+1}a^{2n+1}+\sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)^2}a^{(2n+1)}\\
&=\int_0^1a\ln(1-x)\sum_{n=0}^\infty(-a^2x^2)^n-\int_0^1a\ln x\sum_{n=0}^\infty(-a^2x^2)^n\\
&=\int_0^1\frac{a\ln(1-x)}{1+a^2x^2}\ dx-\int_0^1\frac{a\ln x}{1+a^2x^2}\ dx\\
&=\int_0^1\frac{a\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}\ dx\tag{**}
\end{align}
From (*) and (**) we have
$$\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx$$
Note:
$-\frac{H_{2n+1}}{2n+1}=\int_0^1x^{2n}\ln(1-x)\ dx$
$\frac1{(2n+1)^2}=-\int_0^1 x^{2n}\ln x\ dx$
|
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|
Find the reflection of the point $(4,-13)$ in the line $5x+y+6=0$
Find The image(or reflection) of the point $(4,-13)$ in the line $5x+y+6=0$
Method 1
$$
y+13=\frac{1}{5}(x-4)\implies x-5y-69=0\quad\&\quad 5x+y+6=0\implies (3/2,-27/2)\\
(3/2,-27/2)=(\frac{x+4}{2},\frac{y-13}{2})\implies(x,y)=(-1,-14)
$$
Method 2
$m=\tan\theta=-5$
Ref$(\theta)$=$\begin{bmatrix}
\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta)
\end{bmatrix}$
$$
\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-25}{1+25}=\frac{-24}{26}=\frac{-12}{13}\\
\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{-10}{26}=\frac{-5}{13}\\
Ref(\theta)\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix}
\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta)
\end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix}
\dfrac{-12}{13} & \dfrac{-5}{13} \\ \dfrac{-5}{13} & \dfrac{12}{13}
\end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}\\
=\frac{1}{13}\begin{bmatrix}-48+65\\-20-156\end{bmatrix}=\frac{1}{13}\begin{bmatrix}17\\-176\end{bmatrix}
$$
Why am I not getting the required solution in Method two using matrix method ?
Thanx @ganeshie8 for the remarks, so in that case how do I find the operator for reflection of a point over the line not passing through the origin ?
|
Method 3. Translate the origin to the point $(\color{red}{\frac32,-\frac{27}{2}})$ and find the coordinates of the point ${\color{blue}{4\choose -13}}$ in the new system:
$${x'\choose y'}={x\color{red}{-\frac32}\choose y\color{red}{+\frac{27}{2}}}={\color{blue}4-\frac32\choose \color{blue}{-13}+\frac{27}{2}}={\frac52 \choose \frac{1}{2}}$$
Rotate it by $180^\circ$:
$${\begin{pmatrix}-1&0\\ 0&-1\end{pmatrix}}{\frac52\choose \frac12}={-\frac52\choose -\frac12}$$
Now translate it back:
$${x\choose y}={-\frac52+\frac32\choose -\frac12-\frac{27}{2}}={-1\choose -14}.$$
|
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|
Odd convergents of $\sqrt 2$ and Pythagorean triangles with consecutive legs I would like to prove the following:
Take any odd convergent of $\sqrt 2$. The denominator gives the hypotenuse of a triangle; the numerator split into two consecutive integers gives the other two sides.
For example, the first $10$ convergents of $\sqrt 2$ are:
$$1,\quad \frac 3 2,\quad \frac 7 5,\quad \frac {17} {12},\quad \frac {41} {29},\quad \frac {99} {70},\quad \frac {239} {169},\quad \frac {577} {408},\quad \frac {1393} {985},\quad \frac {3363} {2378}$$
By looking at odd convergents, we see that
*
*$7 = 3 + 4$ and $5^2 = 3^2 + 4^2$;
*$41 = 20 + 21$ and $29^2 = 20^2 + 21^2$;
*$239 = 119 + 120$ and $169^2 = 119^2 + 120^2$;
*and so on.
Now, the convergents are given by $h_n / k_n$ with
$$h_1 = 1 \qquad h_2 = 3 \qquad h_{n+2} = 2 h_{n+1} + h_n$$
$$k_1 = 1 \qquad k_2 = 2 \qquad k_{n+2} = 2 k_{n+1} + k_n$$
and I have noticed that the statement is equivalent to the fact that
$$h_{2n+1}^2 - 2 k_{2n+1}^2 = -1$$
One way to prove it is by solving both recurrence relations, substituting the closed formulas and checking that the equality above holds for any $n$. This is quite long and involves some tedious computations.
Another way is by induction, but I believe one needs to strengthen the inductive hypothesis, since the equality involves only odd indices whereas the recurrence relation of $h_n, k_n$ involves both odd and even indices. I'm not sure how to do that.
Are there any alternative ways to prove the statement? A geometric proof would be particularly interesting, if there is one.
|
Consecutive Legs
In this answer it is shown that
Theorem: Let $m$ and $n$ be positive integers so that
$$
\begin{align}
&m\gt n\\
&m+n\text{ is odd}\\
&m\text{ and }n\text{ are relatively prime}
\end{align}
$$
Then,
$$
\begin{align}
a &= m^2 - n^2\\
b &= 2mn\\
c &= m^2 + n^2
\end{align}
$$
gives all positive, relatively prime $a$, $b$, and $c$ so that
$$
a^2 + b^2 = c^2
$$
To get consecutive legs, we need
$$
m^2-2mn-n^2=\pm1\tag1
$$
which means
$$
\left(\frac mn\right)^2-2\frac mn-1=\pm\frac1{n^2}\tag2
$$
and consequently, for $\frac mn\gt3-\sqrt2$,
$$
\begin{align}
\left|\,\frac mn-1-\sqrt2\,\right|
&=\frac1{n^2}\frac1{\frac mn-1+\sqrt2}\\
&\lt\frac1{2n^2}\tag3
\end{align}
$$
which requires that $\frac mn$ be a convergent of the continued fraction for $1+\sqrt2=(2;2,2,2,2,\dots)$:
$$
\left\{\vphantom{\frac21}\right.\underset{\begin{array}{c}\downarrow\\(3,4,5)\end{array}}{\frac21},\underset{\begin{array}{c}\downarrow\\(21,20,29)\end{array}}{\frac52},\underset{\begin{array}{c}\downarrow\\(119,120,169)\end{array}}{\frac{12}5},\underset{\begin{array}{c}\downarrow\\(697,696,985)\end{array}}{\frac{29}{12}},\underset{\begin{array}{c}\downarrow\\(4059,4060,5741)\end{array}}{\frac{70}{29}},\underset{\begin{array}{c}\downarrow\\(23661,23660,33461)\end{array}}{\frac{169}{70}},\quad\dots\left.\vphantom{\frac21}\right\}\tag4
$$
Computing the Pythagorean Triples
Solving the linear recurrence satisfied by numerators and denominators, $x_n=2x_{n-1}+x_{n-2}$, we get
$$
\begin{align}
m_k&=\frac{\left(1+\sqrt2\right)^{k+1}-\left(1-\sqrt2\right)^{k+1}}{2\sqrt2}\\
n_k&=\frac{\left(1+\sqrt2\right)^k-\left(1-\sqrt2\right)^k}{2\sqrt2}
\end{align}\tag5
$$
Therefore, we get the Pythagorean Triples
$$
\begin{align}
a_k&=\frac{\left(1+\sqrt2\right)^{2k+1}+2(-1)^k+\left(1-\sqrt2\right)^{2k+1}}4\\
b_k&=\frac{\left(1+\sqrt2\right)^{2k+1}-2(-1)^k+\left(1-\sqrt2\right)^{2k+1}}4\\
c_k&=\frac{\left(1+\sqrt2\right)^{2k+1}-\left(1-\sqrt2\right)^{2k+1}}{2\sqrt2}
\end{align}\tag6
$$
Note that
$$
\begin{align}
a_k-b_k&=(-1)^k\\
a_k+b_k&=m_{2k+1}-n_{2k+1}\\
c_k&=n_{2k+1}
\end{align}\tag7
$$
$(7)$ makes sense since $\frac{m-n}n$ is a convergent for $\sqrt2$ when $\frac{m}n$ is a convergent for $1+\sqrt2$. This proves the relations mentioned in the question.
|
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|
Proving the inequality $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$ Prove that $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$. When does the inequality hold?
I really don't know how to prove the inequality and would like to know how.
I mainly tried to factorise the LHS-RHS fully but I could never properly do it:
https://imgur.com/user/Khansis/favorites/folder/7408635/math
|
Since the left side does not depend on changing signs of our variables, it's enough to prove our inequality for non-negatives $a$ and $b$.
Now, by C-S $$4(a^6+b^6)=2\cdot(1^2+1^2)(a^6+b^6)\geq2(a^3+b^3)^2.$$
Thus, it's enough to prove that:
$$2(a^3+b^3)\geq(a+b)(a^2+b^2)$$ or
$$2(a^2-ab+b^2)\geq a^2+b^2$$ or
$$(a-b)^2\geq0$$ and we are done!
|
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|
Solutions to a quadratic given 1 solution in form a+bi I was just really confused as to how I only ended up with 1 of 2 answers for the following question.
Given that $-2+bi$ is a solution of $x^2+ax+(3+a) $ find constants $a$ and $b$ given that they are real.
As soon as I saw that $-2+bi$ was a solution, I immediately jumped to $ -2-bi$ must also be a solution, by the fundamental theorem of algebra. By doing the sum and product of the solution a quadratic could be obtained
Sum $ (-2+bi)+(-2-bi)=-4$
Product $(-2-bi)(-2+bi)=(-2)^2-(-bi)^2=4+b^2$
Thus the quadratic $x^2+4x+(4+b^2)$ is obtained
Equation both sides of the equation
$\\x^2+4x+(4+b^2)=x^2+ax+(3+a)$
$a=4$
Therefore$ b={\sqrt 3}$ or $ b={-\sqrt 3}$
However, the solutions seem to suggest that an extra solution can be
$ b={0}, a=7$
Did I eliminate a solution by doing the sum and product of the solutions to find the quadratic? Or does it have to do with my working process
|
A brute force approach is to substitute $x=-2+bi$ into $x^2+ax+(3+a)$. Then
\begin{align}
x^2+ax+(3+a)
&= (-2+bi)^2+a(-2+bi)+(3+a) \\
&=(7-a-b^2)+b(a-4)i \\
\end{align}
where I have organized the last expression into real and imaginary parts (since $a,b$ are assumed real). If this last expression is to vanish as desired, then both real and imaginary parts are zero. From the latter, we deduce that $b=0$ or $a=4$. If $b=0$, then the condition $7-a-b^2=0$ yields $a=7$. If $a=4$, then this same condition implies $b^2=7-a=3$ and therefore $b=\pm \sqrt{3}$. This gives all three solutions.
|
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|
how do you differentiate x^(3/4) using first principle $$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac{3}{4}}-(x)^{\frac{3}{4}}\Bigr)}{h}$$
I understand the process till
$$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac34}-(x)^{\frac{3}{4}}\Bigr)}{h} * \frac{\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)}{\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)}$$
and post expansion
$$\lim_{h\to 0}\frac{\Bigl(h^3+3h^2x+3x^2h\Bigr)}{{h}\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)\Bigl((x+h)^{\frac{3}{2}}+(x)^{\frac{3}{2}}\Bigr)}$$
but beyond this i am unable to reduce to:
$\frac 34\cdot x^{\frac{-1}{4}}$
|
\begin{align*}
\lim_{h\to 0}\frac{\Bigl(h^3+3h^2x+3x^2h\Bigr)}{{h}\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)\Bigl((x+h)^{\frac{3}{2}}+(x)^{\frac{3}{2}}\Bigr)}&=\lim_{h\to0}\frac{h^3+3h^2x+3hx^2}{h}\lim_{h\to0}\frac1{(x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}}\lim_{h\to0}\frac{1}{(x+h)^{\frac{3}{2}}+(x)^{\frac{3}{2}}}\\&=3x^2\cdot\frac1{2x^{\frac34}}\cdot\frac1{2x^{\frac32}}\\&=\frac34x^{-\frac14}.
\end{align*}
|
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|
Given $a^4+1=\frac{a^2}{b^2}\left(4b^2-b^4-1\right)$, find $a^4+b^4$
Given $$a^4+1=\frac{a^2}{b^2}\left(4b^2-b^4-1\right)$$
Find $$a^4+b^4$$
My approach:
*
*Verify for different values of $a$ and $b$ to see that question is correct and yes, it turns out to be $2$ always for different values of $a$ and $b$.
*Treat the entire thing as quadratic in $a^2$, find its solution using wolfram or manually, and eventually we get the value of $a^4+b^4$ to be $2$.
Any better solution will be appreciated.
Thank you.
|
It is not true that we always have $a^4+b^4=2$. We can rewrite the equation as
$$
\left( a-\frac{1}{a}\right)^2+\left( b-\frac{1}{b}\right)^2=0.
$$
Now chose $a$ such that $a-1/a=i$ and $b-1/b=1$, i.e., with $a := \sqrt{ - \sqrt{3}i + 1}/\sqrt{2}$ and $b=(\sqrt{5} + 1)/2$. This comes from solving the quadratic equations $a^2-ia-1=0$ and $b^2-b-1=0$.
Then we have $i^2+1^2=0$ above satisfied and
$$
a^4+b^4=\frac{3\sqrt{5} - \sqrt{3}i + 6}{2}\neq 2.
$$
For real numbers we obtain $a^2=b^2=1$, hence $a^4+b^4=2$.
|
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|
We have the equation $2x^2-\sqrt{3}x-1=0$ and have to find $|x_1-x_2|$
We have the following quadratic equation:
$2x^2-\sqrt{3}x-1=0$ with roots $x_1$ and $x_2$.
I have to find $x_1^2+x_2^2$ and $|x_1-x_2|$.
First we have: $x_1+x_2=\dfrac{\sqrt{3}}{2}$ and $x_1x_2=-\dfrac{1}{2}$
So $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\dfrac{7}{4}$
Can someone help me with the second one?
I forgot to tell that solving the equation is not an option in my case.
|
Note that: if $a,b,c \in \mathbb{R}$ and $a\ne0$, if $ax^2+bx+c=0$, then $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$2x^2-\sqrt{3}x-1=0$,
solving, we get: $x_{1,2}=\frac{\sqrt{3}\pm\sqrt{3-4\cdot2\cdot(-1)}}{2\cdot2}=\frac{\sqrt{3}\pm\sqrt{11}}{4}$
The required expression, $x_1^2+x_2^2=(\frac{\sqrt{3}+\sqrt{11}}{4})^2+(\frac{\sqrt{3}-\sqrt{11}}{4})^2=\frac{3+2\sqrt{33}+11+3-2\sqrt{33}+11}{16}=\frac{7}{4}$.
The second required expression, $|x_1-x_2|=|\frac{\sqrt{3}+\sqrt{11}}{4}-\frac{\sqrt{3}-\sqrt{11}}{4}|=|\frac{\sqrt{3}+\sqrt{11}-\sqrt{3}+\sqrt{11}}{4}|=|\frac{\sqrt{11}}{2}|=\frac{\sqrt{11}}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $x^3 + y^3$ when $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $ $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $
$x^3 + y^3 =?$
my answer =
$(3 + \sqrt5)^3 = 47 + 32\sqrt5$
$(3 - \sqrt5)^3 = 47 - 32\sqrt5$
$x^3 + y^3 = \frac{47 + 32\sqrt5}{47 - 32\sqrt5} + \frac{47 - 32\sqrt5}{47 + 32\sqrt5} = 2*7329/-2911$
why my answer is wrong? please help me
|
Let us try another way:
$xy=1$
$x+y=\dfrac{(3+\sqrt{5})^2+(3-\sqrt {5})^2}{9-5}=\dfrac{2(9+5)}{4}=7$
$x^3+y^3=(x+y)^3-3xy(x+y)=?$
|
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|
What is the method to solve this system of equations? $\Huge{\text{Updated:}}$
In this question I got an answer for special non-zero values of $A$ and $B.$ I need to solve these polynomial equations for any arbitrary non-zero coefficients.
$\underline{\text{ I am looking for a solution that does not lead to the solution of a cubic equation.}}$
Here is the problem:
For all arbitrary non-zero coefficients $A$ and $B$, I am looking for a method to solve this system of equations, where $x\neq 0,y\neq 0,z\neq 0, u\neq 0, v\neq 0.$
$$\begin{cases}3z^2uB+3x+2uzA+xz^2A=0 &(1)\\
3y+2vzA+u^2A+2xzuA+yz^2A+3vz^2B+3x^2+3zu^2B=0 &(2)\\
3y^2+v^2A+2xuvA+2yzvA+3v^2zB+yu^2A+3vu^2B+3x^2y=0 &(3)\\
3xy^2+xv^2A+2yuvA+3v^2uB=0 &(4)\end{cases} $$
$\Huge{\text{My old attempts:}}$
$3y^2+2vzyA+u^2yA+2xzuyA+y^2z^2A+3vz^2yB+3x^2y+3zu^2yB-3y^2-v^2A-2xuvA-2yzvA-3v^2zB-yu^2A-3vu^2B-3x^2y=0\Longrightarrow (yz-v)(2xuA+3vzB+yzA+vA+3u^2B)=0$
Let, $v=yz$ we get from $(1)$ and $(4)$
$yz^2A+zvA=0 \Longrightarrow zy+v=0 \Longrightarrow 2v=0 \Longrightarrow v=0$ which is contradiction. So, $yz\neq v$
and $2xuA+3vzB+yzA+vA+3u^2B=0$ must be.
$\Huge{\text{My new attempts:}}$
$z(2xuA+3vzB+yzA+vA+3u^2B)-(3y+2vzA+u^2A+2xzuA+yz^2A+3vz^2B+3x^2+3zu^2B)=0 \Longrightarrow 3y+zvA+u^2A+3x^2=0$
$v(2xuA+3vzB+yzA+vA+3u^2B)-(3y^2+v^2A+2xuvA+2yzvA+3v^2zB+yu^2A+3vu^2B+3x^2y)=0 \Longrightarrow 3y+zvA+u^2A+3x^2=0$
Finally, I can construct a new system of equations: (If I didn't make any mistake)
$$\begin{cases}3z^2uB+3x+2uzA+xz^2A=0 &\\
3y+2vzA+u^2A+2xzuA+yz^2A+3vz^2B+3x^2+3zu^2B=0 &\\
3y^2+v^2A+2xuvA+2yzvA+3v^2zB+yu^2A+3vu^2B+3x^2y=0 &\\
3xy^2+xv^2A+2yuvA+3v^2uB=0 &\end{cases} \Longrightarrow \begin{cases}3z^2uB+3x+2uzA+xz^2A=0 &(1)\\
2xuA+3vzB+yzA+vA+3u^2B=0 &(2)\\
3y+zvA+u^2A+3x^2=0 &(3)\\
3xy^2+xv^2A+2yuvA+3v^2uB=0 &(4)\end{cases} $$
Here I am stuck. I've worked hard to make variables dependent on a single variable. Having a single variable requires a method. I can't find a method right now.
|
You can still eliminate. From $(1)$ we have
$$
y = \frac{ - 3x^2 - A(u^2 + vz)}{3},
$$
and then from $(3)$ we have
$$
x = \frac{uz( - 2A - 3Bz)}{Az^2 + 3}.
$$
Actually, the case of $Az^2+3=0$ leads to $Bz^3-2=0$.
Otherwise we have only two equations left in three variables $z,u,v$, namely $(2),(4)$ and can take the resultant. Then one can even specify something, e.g., $u=z$ and $v=-z$ to obtain a general solution. Here is one example:
$$
u=z,\; v=-z\; \text{ with $z$ such that } Bz^3 + Az^2 +1=0.
$$
Then $x=1$ and $y=-1$.
Edit: The general solution. After eliminating $x$ and $y$ as above we can eliminate $v$ by
$$
v:= \frac{u^2(A^4z^5 + 30A^3z^3 + 45A^2Bz^4 + 45A^2z + 27ABz^5 - 81B}{- A^4z^6 - 3A^3z^4 + 9A^2Bz^5 + 9A^2z^2 + 54ABz^3 + 27A + 81Bz},
$$
provided the denominator is nonzero. In this case we obtain a single equation, which gives all solutions, namely
$$
(AZ^2 + Bz^3 + 1)(2A^6z^6 + 90A^5z^4 + 162A^4Bz^5 + 270A^4z^2 + 108A^3B^2z^6 + 540A^3Bz^3 + 54A^3 + 1215A^2B^2z^4 - 486A^2Bz + 1458AB^3z^5 + 729B^4z^6 + 729B^2)=0.
$$
Note that this is independent of the variable $u$. In case that the denominator is zero, we have a special case "avoiding the cubic", but introducing a dependency of $A$ and $B$, which was not allowed.
|
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|
Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$ There is a trigonometric identity:
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\equiv 1\text{ when }A+B+C=\pi$$
It is easy to prove it in an algebraic way, just like that:
$\quad\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\\=\cos^2A+\cos^2B+\cos^2\left(\pi-A-B\right)+2\cos A\cos B\cos \left(\pi-A-B\right)\\=\cos^2A+\cos^2B+\cos^2\left(A+B\right)-2\cos A\cos B\cos \left(A+B\right)\\=\cos^2A+\cos^2B+\left(\cos A\cos B-\sin A\sin B\right)^2-2\cos A\cos B\left(\cos A\cos B-\sin A\sin B\right)\\=\cos^2A+\cos^2B+\cos^2A\cos^2B+\sin^2A\sin^2B-2\sin A\cos A\sin B\cos B-2\cos^2A\cos^2B+2\sin A\cos A\sin B\cos B\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+\left(1-\cos^2A\right)\left(1-\cos^2B\right)\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+1-\cos^2A-\cos^2B+\cos^2A\cos^2B\\=1$
Then, I want to find a geometric way to prove this identity, as $A+B+C=\pi$ and it makes me think of the angle sum of triangle. However, it is quite hard to prove it in a geometric way. Therefore, I hope there is someone who can help. Thank you!
|
I don't know if this counts as a proof, but following your suggestion, I used the cosine laws to obtain your result.
Suppose you have a triangle ABC as in the figure:
Since the angles $A+B+C=\pi$, these are the internal angles of a general triangle.
Using the law of cosines, you can write:
$a^2=b^2+c^2-2bc\cos A\\b^2=a^2+c^2-2ac\cos B\\c^2=a^2+b^2-2ab\cos C$
It follows from here that:
$
\cos A=\frac{a^2-b^2-c^2}{-2bc}\\
\cos B=\frac{b^2-a^2-c^2}{-2ac}\\
\cos C=\frac{c^2-a^2-b^2}{-2ac}
$
Now, to verify your formula, we have:
$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=\\
\left(\frac{a^2-b^2-c^2}{-2bc}\right)^2+\left(\frac{b^2-a^2-c^2}{-2ac}\right)^2+\left(\frac{c^2-a^2-b^2}{-2ac}\right)^2+2\left( \frac{a^2-b^2-c^2}{-2bc} \right)\left( \frac{b^2-a^2-c^2}{-2ac} \right)\left( \frac{c^2-a^2-b^2}{-2ac} \right)
$
It is now a matter of manipulation of the equation to show that this equals 1. Observe that the least common multiple of the first three terms is $4a^2b^2c^2$, which is equal to the product of the last term,
$
\frac{a^2(a^2-b^2-c^2)^2+b^2(b^2-a^2-c^2)^2+c^2(c^2-a^2-b^2)^2}{4a^2b^2c^2}-\frac{(a^2-b^2-c^2)(b^2-a^2-c^2)(c^2-a^2-b^2)}{4a^2b^2c^2}
$
expanding the products in the numerator, you can verify that:
$a^2(a^2-b^2-c^2)^2+b^2(b^2-a^2-c^2)^2+c^2(c^2-a^2-b^2)^2-(a^2-b^2-c^2)(b^2-a^2-c^2)(c^2-a^2-b^2)=4a^2b^2c^2$
so the fraction simplifies to
$
\frac{4a^2b^2c^2}{4a^2b^2c^2}=1
$
|
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|
Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f(-1)$.
Now since $x^{3}-x = x(x^{2}-1)=x(x-1)(x+1)$ is there any relation between the remainder of $f(x)$ divided by $x^{3}-x$ and remainder when $f(x)$ divided by $(x-1)$ and $(x+1)$?
|
$$x^{81} + x^{49} + x^{25} + x^{9} + x=\left(x^{81}-x\right)+\left(x^{49}-x\right)+\left(x^{25}-x\right)+\left(x^9-x\right)+5x.$$
Can you end it now?
|
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|
What is the smallest possible value of $q$ such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$?
If $p$ and $q$ are positive integers such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$ then the smallest possible value of $q$ is:
$(A)\quad 60;\quad (B)\quad 30;\quad (C)\quad 25;\quad (D)\quad 7$.
What is the correct way to solve this kind of problems? I have tried to simplify the given inequality:
$$0.70<\frac{p}{q}< \approx0.73 \quad\text{ or }\quad \frac{21}{30}<\frac{p}{q}<\frac{22}{30}$$
Thank you in advance!
|
If $7$ worked we would have
\begin{align}
&\frac{21}{30}<\frac{p}{7}< \frac{22}{30}\\
\iff&\frac{21*7}{30*7}<\frac{30p}{30*7}< \frac{22*7}{30*7}
\end{align}
So $7$ is a solution if and only if there is a multiple of $30$ between $21*7$ and $22*7$. Since $21*7=147$ and $22*7=154$ we have a multiple of $30,$ namely $150$. So $p=\frac{5}{7}$ is between those numbers.
To prove that this is the smallest just notice that every number smaller than $7$ other than $4$ is a divisor of $30$ so their fractions can be written as $\frac{p}{30}.$ And obviously it is impossible to have
\begin{align*}
\frac{21}{30}<\frac{p}{30}<\frac{22}{30}
\end{align*}
As for $4$ if $4$ was possible we would have
\begin{align}
&\frac{21}{30}<\frac{p}{4}<\frac{22}{30}\\
\iff & \frac{21*2}{30*2}<\frac{p*15}{4*15}<\frac{22*2}{30*2}
\end{align}
and there are no multiples of $15$ between $42$ and $44$.
|
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|
For positive real numbers $a,b,c$ prove that $ a^4 + b^4 + c^4 \ge abc(a+b+c)$ For positive reals $a,b,c$ prove that $$ a^4+b^4+c^4 \ge abc(a+b+c). $$
I tried to pls around trying to reorganize to get AM-GM but i couldn't
Thanks for the help in advance.
|
By AM_GM we get
\begin{align*}
\frac{2a^4+b^4+c^4}{4} & \geq a^2bc\\
\frac{2b^4+a^4+c^4}{4} & \geq b^2ac\\
\frac{2c^4+b^4+a^4}{4} & \geq c^2ab
\end{align*}
Now add these to get your inequality.
|
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|
Prove this inequality via weighted Jensen inequality The following inequality is derived from this difficult olympiad problem.
Suppose $a,b,c$ are three positive real numbers such that $abc = 8$, show that
$$\frac{1}{\sqrt{a + 1}} + \frac{1}{\sqrt{b + 1}} + \frac{1}{\sqrt{c + 1}} < 2$$
My idea: Note that $\sqrt{x}$ is a concave function on $(0,\infty)$. Let $x,y,z,p,q,r$ be any positive real numbers, then according to weighted Jensen inequality we have
$$\begin{aligned}\sqrt{x} + \sqrt{y} + \sqrt{z} &= \frac{\sqrt{p^2x}}{p} + \frac{\sqrt{q^2y}}{q} + \frac{\sqrt{r^2z}}{r}\\&\leq\sqrt{\left(\frac{1}{p} + \frac{1}{q} + \frac{1}{r}\right)(px + qy + rz)}\end{aligned}$$
Let $x = a,y = b,z = c = \frac{8}{ab}$. If we can find appropriate weights $p,q,r$ such that
$$\left(\frac{1}{p} + \frac{1}{q} + \frac{1}{r}\right)(px + qy + rz) < 4$$
then we are done.
Now if I ask Mathematica
Reduce[ForAll[a, a > 0, ForAll[b, b > 0, Exists[{p, q, r}, p > 0 && q > 0 && r > 0 && (1/p + 1/q + 1/r) (p/(1 + a) + q/(1 + b) + r/(1 + 8/(a b))) < 4]]]]
Mathematica returns True in less than 2 seconds(!!). However, it provides no more details other than a truth value.
If I ask instead
Reduce[ForAll[a, a > 0, ForAll[b, b > 0, Exists[{p, q, r}, p > 0 && q > 0 && r > 0 && (1/p + 1/q + 1/r) (p/(1 + a) + q/(1 + b) + r/(1 + 8/(a b))) < m]]], m]
Mathematical returns m >= 4, although using much longer time.
Which algorithm does Mathematica use to verify the proposition? It is certainly not cylindrical algebra, for CylindricalDecomposition takes much much longer.
Can we find simple expressions for $p,q,r$ that would complete the proof?
Update: I have found a way to carry out quantifier elimination by hand. Therefore the first part of this problem is solved.
|
The Contradiction method works!
Let $\frac{1}{\sqrt{a+1}}=p,$ $\frac{1}{\sqrt{b+1}}=q$ and $\frac{1}{\sqrt{c+1}}=r.$
Thus, $\{p,q,r\}\subset(0,1),$ $\frac{(1-p^2)(1-q^2)(1-r^2)}{p^2q^2r^2}=8$ and we need to prove that:
$$p+q+r<2.$$
Indeed, let $p+q+r\geq2,$ $r=kr'$ such that $k>0$ and $p+q+r'=2$.
Thus, $$p+q+kr'\geq2=p+q+r',$$ which gives $k\geq1.$
Thus, $$8=\frac{(1-p^2)(1-q^2)}{p^2q^2}\cdot\left(\frac{1}{k^2r'^2}-1\right)\leq\frac{(1-p^2)(1-q^2)}{p^2q^2}\cdot\left(\frac{1}{r'^2}-1\right),$$ which is a contradiction because we'll prove now that
$$8>\frac{(1-p^2)(1-q^2)}{p^2q^2}\cdot\left(\frac{1}{r'^2}-1\right).$$
Indeed, we need to prove that
$$8p^2q^2r'^2>(1-p^2)(1-q^2)(1-r'^2)$$ or
$$512p^2q^2r'^2>((p+q+r')^2-4p^2)((p+q+r')^2-4q^2)((p+q+r')^2-4r'^2)$$ or
$$512p^2q^2r'^2>(3p+q+r')(3q+p+r')(3r'+p+q)(p+q-r')(p+r'-q)(q+r'-p).$$
Now, if $(p+q-r')(p+r'-q)(q+r'-p)\leq0$, so our inequality is true, which says that it's enough to prove it for $(p+q-r')(p+r'-q)(q+r'-p)>0$.
Also, if $p+q-r'<0$ and $p+r'-q<0,$ so $p<0$, which is a contradiction.
Thus, we can assume that $p+q-r'=z>0,$ $p+r'-q=y>0$ and $q+r'-p=x>0$, which gives
$p=\frac{y+z}{2},$ $q=\frac{x+z}{2},$ $r'=\frac{x+y}{2}$ and we need to prove that
$$8(x+y)^2(x+z)^2(y+z)^2>xyz\prod_{cyc}(x+2y+2z),$$
which is obviously true after full expanding.
Done!
It's interesting that even the following is true.
Let $x$, $y$ and $z$ be non-negative numbers. Prove that:
$$125(x+y)^2(x+z)^2(y+z)^2\geq64xyz(x+2y+2z)(2x+y+2z)(2x+2y+z).$$
|
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|
Determining the radius of convergence of a series that I don't know how to express in sigma Question:
Problem 1: Consider the ODE: $(1+x^3)y'' -6xy = 0$
(a) Compute the first $3$ nonzero terms of power series expansion about $x = 0$ for two linearly independent solutions.
(b) Use the ratio test to determine the radius of convergence of the series. Could your result
have been predicted by inspection?
The solution for $y = a_0(1+x^3) + a_1(x + \frac{x^4}{2}- \frac{x^7}{14}+ \frac{x^{10}}{35}- \frac{x^{13}}{65}+\cdots)$
In order to use the ratio test $$ \lim_{n \rightarrow \infty}|\frac{a_{n+1}(x-x_0)^{n+1}}{a_n (x-x_0)^n}|$$
I need to know how to express $(x + \frac{x^4}{2}- \frac{x^7}{14}+ \frac{x^{10}}{35}- \frac{x^{13}}{65}+\cdots)$ in the sigma form. Am I right?
However, I cannot come up with a proper way to express the equation in a sigma form. Is there an alternate way to perform the ratio test?
|
You should get a coefficient recursion ($a_{-1}=a_{-2}=0$)
$$
x^n: (n+2)(n+1)a_{n+2}+(n-1)(n-2)a_{n-1}-6a_{n-1}=0
\\~\\\iff (n+1)[(n+2)a_{n+2}+(n-4)a_{n-1}]=0
\\~\\\implies
a_{n+3}=-\frac{n-3}{n+3}a_n.
$$
so that $a_2=0$ and thus $a_{3k+2}=0$ for all $k$ and also $a_6=0$ and thus $a_{3k}=0$ for $k\ge 2$.
The remaining part $\sum_ka_{3k+1}x^{3k+1}$ has radius of convergence equal $1$ by the quotient test.
From $(n+3)na_{n+3}=-n(n-3)a_n$ it follows that these products are constant with alternating sign, thus $a_{3k+1}=(-1)^{k-1}\frac{2}{(3k+1)(3k-2)}a_1$ giving the solution as
$$
y(x)=a_0(1+x^3)+2a_1\sum_{k=0}^\infty\frac{(-1)^{k-1}x^{3k+1}}{(3k+1)(3k-2)}.
$$
In the normalized form, $y''-\frac{6x}{x^3+1}y=0$ has analytical coefficients with convergence radius $1$, thus any solutions are also analytical with at least the same convergence radius, and at least one solution with exactly this radius.
|
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|
On the equation $\sqrt{x^2-9}=\frac{2(x+3)}{(x-3)^2}-x$ I'm trying to solve the equation
$$\sqrt{x^2-9}=\frac{2(x+3)}{(x-3)^2}-x \tag{1}$$
Attempt: I have rewritten the equation as
\begin{align*}
\sqrt{x^2-9} = \frac{2(x+3)}{\left (x-3 \right )^2}-x &\Leftrightarrow \sqrt{x^2-9} = \frac{2(x+3)-x\left ( x-3 \right )^2}{\left ( x-3 \right )^2} \\
&\Leftrightarrow \sqrt{x^2-9} = - \frac{x^3-6x^2+7x-6}{\left ( x-3 \right )^2} \\ &\Leftrightarrow \sqrt{x^2-9} = \frac{2}{x-3} -x + \frac{12}{\left ( x-3 \right )^2}
\end{align*}
This seems manageable (?) but I do not know how to proceed. The solution is $x=8-\sqrt{13}$ as suggested by Mr. Wolfy. I have no idea how to get it.
In the mean time if we square $(1)$ all 6th powers are simplified and we are left with
$$5x^4-96x^3+526x^2-105x+765=0$$
Trying to factorising it we get with some luck that
$$(x^2-16x+51)(5x^2-16x+15) =0$$
and we have to solve this. This is easy but we have to take into account the restriction
$$ -x^3+6x^2-7x+6\geq 0 $$
|
Hint:
Let $\sqrt{x^2-9}=3\tan2t\ge0,0\le2t<\dfrac\pi2$
$x=3\sec2t$
$$3(\tan2t+\sec2t)=\dfrac{6(\sec2t+1)}{9(\sec2t-1)^2}$$
$$3(\sin2t+1)=\dfrac{2(1+\cos2t)}{3(1-\cos2t)^2}$$
Set $\tan t=u$
$$\dfrac{9(1+u)^2}{1+u^2}=\dfrac{4(1+u^2)^2}{4(1+u^2)u^4}$$
$$\iff9(1+u)^2u^4=(1+u^2)^2$$
As $u=\tan t\ge0$
$$\implies3(1+u)u^2=1+u^2$$
$$3u^3+2u^2-1=0$$
Use Cardano's method
|
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Why is $\frac{d}{dt}(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} =\frac{m \vec u}{(1-\frac{u^2}{c^2})^{3/2}} \cdot \frac{d \vec u}{dt}$? Given that $\vec u(t)$, $u=|\vec u(t) |$ and $c$, $m$ are constants, how does one get from the LHS of the following equation to the RHS?
$$\frac{d}{dt}\left(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} =\frac{m \vec u}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \cdot \frac{d \vec u}{dt}$$
What is the differentiation rule used?
|
Leave $m$ out and start with,
$$\frac{\vec{u}\cdot\vec{u}} {\sqrt{1-\frac{u^2}{c^2}}}=\frac{u^2} {\sqrt{1-\frac{u^2}{c^2}}}$$
Take derivatives on both sides,
$$\frac{d}{dt}\left(\frac{\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} +\frac{\vec u}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \frac{d \vec u}{dt}
=\left[ \frac{1}{\sqrt{1-\frac{u^2}{c^2}}} + \frac 12 \frac{\frac{u^2}{c^2}}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \right]\frac{du^2}{dt}$$
Since $ du^2/dt = 2\vec{u}\cdot d\vec{u}/dt$, rearrange the RHS as,
$$ RHS = \frac{\vec u}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \frac{d \vec u}{dt}+\left[ \frac{1}{\sqrt{1-\frac{u^2}{c^2}}} + \frac{\frac{u^2}{c^2}}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \right]\vec{u}\cdot \frac{d\vec{u}}{st}$$
$$= \frac{\vec u}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \frac{d \vec u}{dt}+ \frac{\vec{u}}{\left(1-\frac{u^2}{c^2}\right)^{3/2}}\cdot \frac{d\vec{u}}{st}$$
Compare with the LHS to cancel the first term. Thus,
$$\frac{d}{dt}\left(\frac{\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} =\frac{\vec u}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \cdot \frac{d \vec u}{dt}$$
|
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|
Convergence at the boundary of $\sum_{n= 1}^\infty\frac{(n!)^3}{(3n)!}z^n $ We are given the complex series of functions $\sum_{n= 1}^\infty\frac{(n!)^3}{(3n)!}z^n $. It is a power series, and we can easily calculate that its disk of convergence is $D(0,27)$. How can we know if the series converges at some point of the boundary? Also, can we say to which function it converges? (I mean, can we express the sum function in an elementary way?)
|
Take some point on the boundary and then try to figure out whether the series you get by plugging in that point for $z$ converges. In the complex case you will mostly work with estimations then as you have infinitely many boundary points. Lets do it for your example:
Let $z \in \mathbb{C}$ with $\vert z \vert = 27$, i.e. $z$ lies on the boundary. Then we get
\begin{align*}\left\lvert \frac{(n!)^3}{(3n)!}z^n \right\rvert &= 27^n \cdot \frac{n}{3n} \cdot \frac{n}{3n-1} \cdot \frac{n}{3n-2} \cdot \frac{n-1}{3(n-1)} \cdot \frac{n-1}{3(n-1)-1} \cdots \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{1}\\ &= 27^n \cdot \frac{1}{3} \cdot \frac{n}{3n-1} \cdot \frac{n}{3n-2} \cdot \frac{1}{3} \cdot \frac{n-1}{3(n-1)-1} \cdots \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{1}\\ &= 9^n \cdot \frac{n}{3n-1} \cdot \frac{n}{3n-2} \cdot \frac{n-1}{3(n-1)-1} \cdots \frac{1}{2} \cdot \frac{1}{1}\\ &= \frac{3n}{3n-1} \cdot \frac{3n}{3n-2} \cdot \frac{3(n-1)}{3(n-1)-1} \cdots \frac{3}{2} \cdot \frac{3}{1},\end{align*} which yields $$\left\lvert \frac{(n!)^3}{(3n)!}z^n \right\rvert \geq 1,$$ as all the factors of our computation above are bigger than $1$. That means that the series cannot converge for any point on the boundary (Recall that if a series $\sum_{n = 1}^{\infty} a_n$ converges, then $a_n \rightarrow 0$ for $n \rightarrow \infty$).
Note that my argument can be generalized to the case of $\sum_{n = 1}^{\infty} \frac{(n!)^k}{(kn)!}$ for all natural numbers $k \geq 2$.
Most of the times you should not expect to be able to actually compute the value of some series. There are certainly series where this is possible (geometric ones, some generalized harmonic ones etc.), but that is generally non-trivial. As far as I know, one does not even know the value of the series $\sum_{n = 1}^{\infty}\frac{1}{n^3}$ for example.
|
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|
Finding $f\in\mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$ and $\deg(f)\leq 3$. What's wrong with my approach?
I'd like to find a polynomial $f(x) \in \mathbb{Q}[x]$ satisfying
$$f(\sqrt{2}+\sqrt{3})=\sqrt{2}$$
and $\deg(f) \leq 3$.
What I've been trying is the following:
Since $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$,
then $f(\sqrt{2}+\sqrt{3})-\sqrt{2}=0$.
so think of $g(x)=f(x+\sqrt{3})-x$ as a polynomial over $\mathbb{Q}(\sqrt{3})$.
And I've already known that $x^2 -2$ is irreducible over $\mathbb{Q}(\sqrt{3})$
$g(x)$ has a $\sqrt{2}$ as a root of itself, $x^2 -2$ divides $g(x)$ in $\mathbb{Q}(\sqrt{3})[x]$.
$g(x)$ must be of the form $(x^2 -2)(ex+f)$ where $e, f \in \mathbb{Q}(\sqrt{3})$
and also of the form $a(x+\sqrt{3})^3 +b(x+\sqrt{3})^2 +c(x+\sqrt{3}) +d -x$, where $f(x)=ax^3 +bx^2 +cx+d \in \mathbb{Q}[x]$.
after comparing the coefficients of two polynomials, I found that $f(x)=\frac{1}{4}x^3-\frac{9}{4}x$.
But the actual polynomial is $\frac{1}{2}x^3-\frac{9}{2}x$.
There must be a flaw in the above reasoning.
Where did I do a mistake? Could you point it out?
Thank you.
|
Let $e = g + \sqrt{3}h$, $f = j + \sqrt{3}k$.
$$\begin{eqnarray}(x^2 -2)(ex+f) &=& a(x+\sqrt{3})^3 +b(x+\sqrt{3})^2 +c(x+\sqrt{3}) +d -x \\
(g + \sqrt{3}h)x^3 + (j + \sqrt{3}k)x^2 - 2(g + \sqrt{3}h)x - 2(j + \sqrt{3}k) &=& a(x^3 + 3\sqrt{3}x^2 + 9x + 3\sqrt{3}) +b(x^2 + 2\sqrt{3}x + 3) + cx + \sqrt{3}c +d - x \\
(g + \sqrt{3}h)x^3 + (j + \sqrt{3}k)x^2 - 2(g + \sqrt{3}h)x - 2(j + \sqrt{3}k) &=& ax^3 + (b + 3\sqrt{3}a)x^2 + (9a + c - 1 + 2\sqrt{3}b)x + (3b + d + 3\sqrt{3}a + \sqrt{3}c) \\
\end{eqnarray}$$
So $$\begin{eqnarray}g &=& a \\
h &=& 0 \\
j &=& b \\
k &=& 3a \\
-2g &=& 9a+c-1 \\
-2h &=& 2b \\
-2j &=& 3b + d \\
-2k &=& 3a + c
\end{eqnarray}$$
Quickly reduces to $$\begin{eqnarray}h = b = j = d &=& 0 \\
g &=& a \\
k &=& 3a \\
c &=& -9a \\
2a &=& 1 \\
\end{eqnarray}$$
So there's no flaw in your reasoning: the flaw is in the part you didn't include in the question.
|
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|
How many incongruent solutions are there for $x^{2}\equiv49\:\left(10^{6}\right)$? How many solutions are there for $x^{2}\equiv49\:\left(10^{6}\right)$
?
I can see that $x=\pm7$ are two solutions, and I guess that $x$
is a solution to the given congruence iff $\left(\frac{x}{7}\right)^{2}\equiv1\:\left(10^{6}\right)$.
So as $\frac{x}{7}$ has to be an integer, we can write $x=7y$ and
then look for number of solutions to $y^{2}\equiv1\:\left(10^{6}\right)$
(right?). How can we continue from here?
EDIT:
The answer is 8.
I think it is somehow related to the fact that
$$\left(\mathbb{Z}/_{10^{6}}\mathbb{Z}\right)^{\times}\cong\left(\mathbb{Z}/_{2^{6}}\mathbb{Z}\right)^{\times}\times\left(\mathbb{Z}/_{5^{6}}\mathbb{Z}\right)^{\times}
$$
and that
*
*$y^{2}\equiv1\:\left(2^{6}\right)$ has 4 solutions
*$y^{2}\equiv1\:\left(5^{6}\right)$ has 2 solutions
But I don't see exactly why. Is that right?
|
Let $d=\gcd(x-7,x+7)$ then $d\mid (x+7)-(x-7)=14$ so $d=2$ or $d=1$ since $$2^6\cdot 5^6\mid (x-7)(x+7)$$ So we have this possibilities:
*
*$x-7=5^6a$ and $x+7 = 2^6b$ where $a,b$ are relatively prime
*$x-7=2\cdot 5^6a$ and $x+7 = 2^5b$ where $a,b$ are relatively prime
*$x+7=2\cdot 5^6a$ and $x-7 = 2^5b$ where $a,b$ are relatively prime
*$x+7=5^6a$ and $x-7 = 2^6b$ where $a,b$ are relatively prime.
Now solve this systems if $x<10^6$.
For the first case we have a following linear diophant equation $$ 2^6b -5^6a = 14$$
where $a\leq {999993\over 5^6}$ so $a\in \{0,1,...639\}$ and $b\leq {1000007\over 2^6}$ so $b\in \{0,1,...15625\}$. Clearly $a=2c$ so we have $$32\mid 125^2c+7\implies 32\mid c-17$$
so $c= 32k+17$ and $$b= 5^6k+8301\implies 0\leq k\leq 0 $$
So $k=0$ and $a=34$ and $b=8301$ and $x=531257$.
|
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Find the exact value of trigonometric expression: $ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$? How can I simplify this trigonometric expression?
$$ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$$
I used
$$\sin \frac{\pi}{15}=2 \sin \frac{\pi}{30} \cos \frac{\pi}{30}$$ and $$\cos\frac{\pi}{15}=2\cos^2\frac{\pi}{30}-1$$
But these give me more complicated expressions.
|
The identity can also be proved from: $\quad\cos36° = \cos 72° + {1\over2}$
$\cos 72° + {1\over2}= (2\cos^2 36° - 1)+ {1\over2}\;= \large{\phi^2\over2}-{1\over2} ={\phi \over 2}=\cos36°$
$T= \Large \frac{2\cos 6° \cos 12°}{1\;+\;2\cos 6° \sin 12°}$
$\Large{1\over T} = {1\over 2\cos 6° \cos 12°} + \normalsize \tan 12°$
$\begin{align} \tan 36° - \tan12°
&= {\sin36° \over \cos36°} - {\sin12° \over \cos12°} \cr
&= {\sin36° \cos12° - \cos36° \sin12° \over \cos12° \cos36°} \cr
&= {\sin24° \over \cos12° (\cos72° + 0.5)} \cr
&= {\sin24° \over \cos12°(\sin18° + \sin30°)} \cr
&= {\sin24° \over \cos12°(2 \sin24° \cos6°)} \cr
\tan36° &= {1\over 2\cos 6° \cos 12°} + \tan12°
\end{align}$
$$T = \cot 36° = \tan 54°$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve $(x^2 - 11x + 29)^{(6x^2 + x - 2)}=1$? This question comes from PURE MATHEMATICS 1 for As and A levels. This question is part of exercise 1 and it has 5 Answers : $1/2, - 2/3, 4, 6 $ and $7$ . The first 2 values $1/2$ and $-
2/3 $ I am able to find but the rest I can't. I get this first 2 values by make 1 to $(x^2 - 11x + 29)^0$ and this solving by transposition as base of power are same so they get cancelled and make it $(6x^2 + x - 2)=0.$
|
By inspection, $$a^b=1 \Rightarrow a=1\textrm{ or }b=0 \textrm{ or } a=-1 \textrm{ if } b \textrm{ is even}$$
So we can just solve three different cases.
Case 1: $a=1$.
\begin{align}x^2-11x+29&=1\\
x^2-11x+28&=0\\
(x-4)(x-7)&=0\\
x_1&=4\\
x_2&=7\end{align}
Case 2: $b=0$.
\begin{align}
6x^2+x-2&=0\\
(3x+2)(2x-1)&=0\\
x_3&=-\frac23\\
x_4&=\frac12
\end{align}
Case 3: $a=-1$ and $b$ is even.
\begin{align}
x^2-11x+29&=-1\\
x^2-11x+30&=0\\
(x-5)(x-6)&=0\\
x_5&=5 &\textrm{(inadmissible)}\\
x_6&=6
\end{align}
The possible answers are therefore $\boxed{x=-\frac23,\frac12,4,6,7}$.
|
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|
Factoring $3x^2+4x-4=0$ using the quadratic formula The correct answer is $(3x-2)(x+2)$ but I am getting $(x-\frac{2}{3})(x+2)$, why?
My calculations:
$x = \frac{-4\pm\sqrt{16-(-48)}}{6} = \frac{-4\pm8}{6}$, which gives the factors $(x-\frac{4}{6})(x+\frac{12}{6}) = (x-\frac{2}{3})(x+2)$.
|
Because if the roots of $ax^2+bx+c$ are $r_1$ and $r_2$, then$$ax^2+bx+c=a(x-r_1)(x-r_2).$$You forgot the $a$.
In your case, you get:\begin{align}3x^2+4x-4&=3\left(x-\frac23\right)(x+2)\\&=(3x-2)(x+2).\end{align}
|
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|
Let $t$ be a positive integer such that $2^t = a^b ±1$ for some integers $a$ and $b$, each greater than $1$. What are all the possible values of $t$? Let $t$ be a positive integer such that $2^t = a^b ±1$ for some integers $a$ and $b$, each greater than $1$. What are all the possible values of $t$?
I found that $t=3$ is a solution but I don’t know how to prove that is the only solution or if there is more
Taken from the 2008 IWYMIC
|
As mentioned in comments, this is a special case of Catalan's conjecture, now Mihăilescu's theorem, but this special case can be solved on its own without too much suffering.
If $b$ is even, then by letting $c = a^{b/2}$, we get either $2^t = c^2 + 1$ or $2^t = c^2 - 1$.
*
*The first case is eliminated modulo $4$: except when $t=1$ which doesn't work, $2^t \equiv 0 \pmod 4$, but $c^2 + 1$ can only be $1,2 \pmod 4$.
*In the second case, we factor $2^t = (c+1)(c-1)$, and the factors $c+1$ and $c-1$ must be powers of $2$. This is only possible when $c-1=2$ and $c+1=4$, giving us the known solution $t=3$.
If $b$ is odd, then $a^b \pm 1$ factors as two integers $a\pm 1$ and $\frac{a^b \pm 1}{a\pm 1}$, both of which must be powers of $2$. For the second factor, we have the congruence
$$
\frac{a^b \pm 1}{a \pm 1} = 1 + (\mp a) + (\mp a)^2 + \dots + (\mp a)^{b-1} \equiv 1 + 1 + \dots + 1 = b \pmod{a \pm 1}.
$$
This congruence is a key lemma used for example in 1960 by Cassels to solve some cases of Catalan's conjecture.
There are two cases:
*
*$a\pm 1$ is odd, and a power of $2$, therefore $a\pm1 = 1$, or
*$a\pm 1$ is even, and therefore $\frac{a^b\pm1}{a\pm1} \equiv b \pmod {a\pm1}$ means that $\frac{a^b \pm 1}{a\pm 1} \equiv b \pmod 2$. So $\frac{a^b \pm 1}{a\pm 1}$ is odd, and a power of $2$, therefore $\frac{a^b \pm 1}{a\pm 1} =1$.
As a result, from $2^t = a^b \pm 1$ we arrive at either $a\pm 1 = 1$ (which would mean $a=2$, but $2^t = 2^b\pm 1$ is impossible for positive $t,b$) or $a \pm 1 = 2^t$ (which would mean $a^b = a$, contradicting that $a,b>1$). In both cases, there are no further solutions.
|
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|
limit of a n-square root and series of exponents Again, I'm having trouble with the infinite limits:
$$ (1) .... \lim_{n \to \infty} \sqrt[n]{ a^n+b^n } $$ with $a,b$ positive reals.
and to show if the following series is divergent or convergent
$$ (2) ......\sum_{n=1}^{\infty} \frac{5^{n}-2^{n}}{7^n-6^n} $$
To be honest, don't have any idea how to approach them, at least for the (2) I may use their exponential representations, as follows:
$$\sum_{n=1}^{\infty} \frac{e^{n \ln 5}-e^{n\ln 2}}{e^{n\ln 7}-6^{n \ln 6}} $$
and in that case the series will diverge. But for (1) don't know.
Thanks in advance!
|
For the first part: Without loss of generality let $a \ge b$. Then by binomial expansion
$$
\sqrt[n]{ a^n+b^n } = a(1 + (b/a)^n)^{1/n} = a\bigg(1 + \frac{(b/a)^n}{n} + ...\bigg)
$$
Since $b/a <1$ each term after the first vanishes when $n \to \infty$. Hence
$$
\lim_{n \to \infty}\sqrt[n]{ a^n+b^n } = \max(a,b)
$$
For the second part:
$$
\sum_{n=1}^{\infty} \frac{5^{n}-2^{n}}{7^n-6^n}
= \sum_{n=1}^{5} \frac{5^{n}-2^{n}}{7^n-6^n} + \sum_{n=6}^{\infty} \frac{5^{n}-2^{n}}{7^n-6^n}
< C + \sum_{n=6}^{\infty} \frac{5^{n}}{(6+1)^n-6^n}
$$
where $C$ is the finite sum of the first five terms.
By binomial theorem,
$$(6+1)^n - 6^n> (6^n + n6^{n-1} + \cdots + 1) - 6^n = n6^{n-1} + \cdots + 1 > 6^n \text{ for $n \ge 6$}
$$
Hence
$$
\sum_{n=1}^{\infty} \frac{5^{n}-2^{n}}{7^n-6^n}
< C + \sum_{n=6}^{\infty} \frac{5^{n}} {6^n} < C + \sum_{n=1}^{\infty} \bigg(\frac{5} {6}\bigg)^n = C + 5
$$
which is convergent since $C$ is finite.
|
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Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter $a$. Question : Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter a.
I have solved the problem as follows
$\sqrt{a\left(2^{x}-2\right)+1}=1-2^{x}$
$a\left(2^{x}-2\right)+1=\left(1-2^{x}\right)^{2}=2^{2 x}+1-2 \cdot 2^{x}=2^{2 x}-2^{x+1}+1$
$a 2^{x}-2 a=2^{2 x}-2^{x+1}$
$2^{2 x}-(a-1) 2^{x}+2 a=0$
$y^{2}-(a-1) y+2 a=0$
$y=\frac{(a-1) \pm \sqrt{(a-1)^{2}-8 a}}{2}=\frac{(a-1) \pm \sqrt{a^{2}-10 a+1}}{2}$
$2^{x}=\frac{(a-1) \pm \sqrt{(a-5)^{2}-24}}{2}$
After this there are so many conditions on a. Do i need to check for each and every value ?
|
$$\sqrt{a(2^{x}-2)+1}=1-2^{x}\tag1$$
First of all, we have to have
$$a(2^{x}-2)+1\ge 0\qquad\text{and}\qquad 1-2^x\ge 0\tag2$$
Under $(2)$, we have
$$\begin{align}(1)&\implies a(2^x-2)+1=(1-2^x)^2
\\\\&\implies a(2^x-2)+1=1-2^{x+1}+2^{2x}
\\\\&\implies a\cdot 2^x-2a=2^{2x}-2\cdot 2^{x}
\\\\&\implies 2^{2x}+2^x(-2-a)+2a=0
\\\\&\implies (2^x-2)(2^x-a)=0
\\\\&\implies 2^x=2\quad \text{or}\quad 2^x=a\end{align}$$
Now, $2^x=2$ does not satisfy $(2)$.
When $2^x=a$, we have to have $0\lt a\le 1$ from $(2)$.
So, the answer is as follows :
*
*If $a\le 0$ or $a\gt 1$, then there is no $x$ satisfying $(1)$.
*If $0\lt a\le 1$, then $x=\log_2 a$.
|
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|
What is the principal root of $\sqrt{-i}$? Where is the mistake in this solution?
$$\sqrt{-i}=\sqrt {\cos \frac{3 \pi}2+i\sin \frac{3 \pi}2}=\cos \frac{3 \pi}4+i\sin \frac{3 \pi}4=-\frac{\sqrt 2}2+i \frac{\sqrt 2}2$$
WA gives me different result:
$$\frac{\sqrt 2}2-i \frac{\sqrt 2}2$$
Why the principal root must be equal to $\frac{\sqrt 2}2-i \frac{\sqrt 2}2$?
I used Moivre's formula. But I dont know. How can I must choose value of $k$ in Moivre's formula? I chose $k=0$.
$$z=r\left(\cos x+i\sin x\right)$$
$$z^{\frac 1n}=r^\frac1n \left( \cos \frac{x+2\pi k}{n} + i\sin \frac{x+2\pi k}{n}\right)$$
|
There are two square root of $-i$ namely, $$ \pm\frac {\sqrt 2}{2} (1-i)$$
You may verify it by squaring each one $$ (\pm\frac {\sqrt 2}{2} (1-i))^2= \frac {1}{2}(1-2i-1)=-i$$
One way to find them is of course the polar form of complex number $-i=e^{3i \pi /2}$ as mentioned in Kevin's answer.
The other way is looking at the unit circle and locate the square roots by dividing the angle $3\pi /2 $ by $2$ to find the first one and add $\pi $ to this angle to find the second one.
|
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|
On the proof of $2^b-1$ does not divide $2^a+1$.
Proposition. For any integer $a$ and $b$ where $b>2$, we never have $2^b-1|2^a+1$.
WLOG let $b<a$. The proof uses a critical fact that can be stated very simply as follows:
$2^a+1$ leaves a reminder of the form $2^r+1$ when divided by $2^b-1$, where $r\leq b$.
This again, directly follows by the following equation:
$$2^{r+kb}+1=2^r(2^{kb}-1)+(2^r+1).$$
I am digging some insight from these steps so that the above step is not just a “trick” to use. For example in the proposition the base $2$ can be replaced by any positive integer $x\geq 2$. And observe that the exponential function minus $1$(it is not true in general for $x^n+k$ where $k\neq -1$) satisfies the following functional equation:
One of the solution set for $f(x+y)=f(x)f(y)+f(x)+f(y)$ where $f:\mathbb Z_+\to \mathbb Z$ is the set of all exponential function minus $1$: $f(x)=x^n-1$, where the integer $x>1$ and $n>1$.
I am looking for a way to make the solution of the first line proposition not a “trick” to memorize, but a link to a theory or a more general result so the proof will follow naturally.
|
Claim: If $a,b,r, n\in \mathbb N$ if $a \equiv r \pmod b$ then $n^a \equiv n^r\pmod{n^b -1}$.
Pf: wolog $r < a$ so there is a $k > 0; k\in \mathbb N$ so that $a= kb + r$.
So $n^a = n^{a-b}(n^b-1)+ n^{a-b}\equiv n^{a-b}\pmod{n^b-1}$.
and by induction, for any $a-kb \ge b$
$n^a \equiv n^{a-(k-1)b} = n^{a-kb}(n^b-1) + n^{a-kb}\equiv n^{a-kb}\pmod{n^b - 1}$.
And so $n^a\equiv n^{n-kb}=n^r\pmod {2^b -1}$
.....
And that's all. Just because something is "tricky" doesn't mean it isn't valid.
==== old answer =====
$2^a + 1 \equiv m \pmod {2^b -1}$ means there is a $k$ so that $2^a+1 =k*(2^b-1) + m$.
To find so possible values for $k, m$ (and if $0 \le m < 2^a + 1$ then then $k$ and $m$ are unique) we notice.
$k*(2^b-1) + m = 2^k-k + m$.
If $k = 2^{a-b}$ we get $2^{a} + 1 = 2^{a-b}(2^b -1) + m = 2^a - 2^{a-b} + m$ and so
$2^{a-b} + 1 = m$ and
That's that. I'm not sure why that seems like "trick".
Or we couls simply note:
$2^a + 1 = 2^{b-a}2^b + 1 = 2^{b-1}(2^b - 1) + 2^{b-a} + 1\equiv 2^{b-a} +1\pmod{2^b -1}$.
or example in the proposition the base 2 can be replaced by any positive integer x≥2.
Sure.
If $b < a$ then
$x^a+1 = x^{a-b}(x^{b}-1) + x^{a-b} + 1 \equiv x^{a-b} + 1\pmod {x^b-1}$.
We can make that a theorem (never "theory".... math doesn't have "theories").
Theorem: for $a, b \in \mathbb N$ then for any $k\in \mathbb Z$ so that $a + kb \ge 0$ then $n^{a}\equiv n^{a+kb}\pmod {n^b -1}$.
Pf: If $a > b$ then $n^a = n^{a-b}(n^b -1) + n^{a-b}\equiv n^{a-b}\pmod {n^b-1}$ and by induction $n^a \equiv n^{a-kb}\pmod {n^b-1}$ for all $k; a-kb \ge 0$. And $n^{a+kb} \equiv n^{a}\pmod{n^b -1}$ for all $k \ge 0$ by induction.
|
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Is $\sum_{n=1}^{\infty} 1 = -\frac{3}{12}$ true? Here is how I derived this...
$$1+2+3+4+...=-\frac{1}{12}$$
$$2+4+6+8+...=2(1+2+3+4+...)=2(-\frac{1}{12})=-\frac{1}{6}$$
Thus $1+3+5+7+...=-\frac{1}{6}-(1+1+1+...) $ because $S_{odd}=S_{even}-(1+1+1+...)$
Also $S_{odd}=(1+2+3+4+...)-(2+4+6+8+...)=-\frac{1}{12}+\frac{1}{6}=\frac{1}{12}$
Thus $\frac{1}{12}=-\frac{1}{6}-(1+1+1+...)$
$$1+1+1+...=-\frac{3}{12}$$
What I think my mistake was (if I have one) is where I assume the sum of all numbers is half the sum of all even numbers; although, it should work since there is nothing two times infinity.
Please leave simple solution (im only 15 years old) to why this is false: Wikipedia says that $\sum_{n=1}^{\infty} 1 = \infty$.
Edit: I know now that the above notion is false. I recently watched a Numberphile video proving $\sum_{n=1}^{\infty} n = -\frac{1}{12}$. I followed the same line of reasoning to derive the untrue $\sum_{n=1}^{\infty} 1 = -\frac{3}{12}$ I'll admit; I was ignorant in believing so and I apologize for wasting your time.
Thanks for the answer kindly explaining why I was wrong. I guess I should watch this video to un-brainwash me. Sorry again.
|
First of all, you have to understand the following point:
There is no single canonical way of defining 'summing infinitely many things'.
As a result, there are several different notions of infinite sums, some of which are not even compatible to each other. For instance, the followings are some selected summability methods:
\begin{align*}
\begin{array}{|c|c|c|}
\hline
& \begin{array}{c}\textbf{Definition}\\ \scriptsize\text{(some notations are not standard)} \end{array} & \textbf{Examples} \\
%
\hline
\begin{array}{c}
\text{ordinary}\\
\text{summation}
\end{array}
%
& \displaystyle\sum_{n=1}^{\infty} a_n := \lim_{N\to\infty} \sum_{n=1}^{N} a_n
%
& \begin{array}{c}
\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \\
\displaystyle\sum_{n=1}^{\infty} (-1)^n = \mathtt{undefined} \\
\displaystyle\sum_{n=1}^{\infty} 1 = \infty
\end{array} \\
%
\hline
\begin{array}{c}
\text{Abel}\\
\text{summation}
\end{array}
%
& \displaystyle\sum_{n=1}^{\infty} a_n = \lim_{x\to1^-} \sum_{n=1}^{\infty} a_n x^n \quad \text{(A)}
%
& \begin{array}{cr}
\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} & \text{(A)} \\
\displaystyle\sum_{n=1}^{\infty} (-1)^n = -\frac{1}{2} & \text{(A)} \\
\displaystyle\sum_{n=1}^{\infty} 1 = \infty & \text{(A)}
\end{array} \\
%
\hline
\begin{array}{c}
\text{Dirichlet} \\
\text{regularization}
\end{array}
%
&
\begin{array}{c}
\displaystyle\sum_{n=1}^{\infty} a_n = \lim_{s\to0} D(a, s) \quad \text{(D)} \\
\scriptsize\text{where $D(a, s)$ is the Dirichlet series for $a = (a_n)_{n\geq 1}$}
\end{array}
%
& \begin{array}{cr}
\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} & \text{(D)} \\
\displaystyle\sum_{n=1}^{\infty} (-1)^n = -\frac{1}{2} & \text{(D)} \\
\displaystyle\sum_{n=1}^{\infty} 1 = -\frac{1}{2} & \text{(D)}
\end{array} \\
%
\hline
\end{array}
\end{align*}
Considering the plethora of different definitions, you must be very careful about which one you use. And usually in mathematics, any infinite summation is considered to be an ordinary one unless stated otherwise. So, make sure to explicitly state the method used in case it is different from the ordinary one.
Now let us take closer look on OP's question. Given the context, I suspect that OP is working under the Dirichlet regularization or Ramanujan summation. They are simply some systematic ways of assigning values to the symbol $\sum_{n=1}^{\infty} a_n$, whose exact detail requires advanced level of math. So let me sweep this under the rug, although we should remark one of the consequence of the definition:
Some summability methods do not necessarily follow all the familiar rules that hold for ordinary summation.
To emphasize this distinction, let's estrange us from the confusing old notation $a_1 + a_2 + a_3 + \cdots$ and instead adopt the following idiosyncratic one
$$ \mathtt{Sum?}[ a_1, a_2, a_3, \cdots ] \quad \text{or} \quad \mathtt{Sum?}[(a_n)_{n=1}^{\infty}] $$
for the summability method involved in OP. This will make easier to track the manipulation involved in and spot the flow in OP's computation. Also, we will assume that OP's summability method $\mathtt{Sum?}$ satisfies the following two properties
*
*$\mathtt{Sum?}$ is linear. In other words,
$$\mathtt{Sum?}[\alpha a + \beta b] = \alpha \, \mathtt{Sum?}[a] + \beta \, \mathtt{Sum?}[b] $$
holds for any constants $\alpha, \beta$ and sequences $a = (a_n)_{n=1}^{\infty}$, $b = (b_n)_{n=1}^{\infty}$ which are summable under $\mathtt{Sum?}$.
*$\mathtt{Sum?}[1,2,3,\cdots] = -\frac{1}{12}$, such as in Dirichlet regularization or Ramanujan summation.
The first one is satisfied by essentially any interesting summability methods, so we include this to our one as well. The second one was OP's starting point.
*
*OP's first step follows from linearity:
$$ \mathtt{Sum?}[2,4,6,\cdots] = 2 \, \mathtt{Sum?}[1,2,3,\cdots] = -\frac{1}{6}. $$
*The second step alwo follows from linearity:
\begin{align*}
\mathtt{Sum?}[1,3,5,\cdots]
&= \mathtt{Sum?}[2,4,6,\cdots] - \mathtt{Sum?}[1,1,1,\cdots] \\
&= -\frac{1}{6} - \mathtt{Sum?}[1,1,1,\cdots].
\end{align*}
*The next step is the problematic. What is attempted in this step can be rephrased as
\begin{align*}
\mathtt{Sum?}[1,3,5,\cdots]
&\stackrel{?}= \mathtt{Sum?}[1, 0, 3, 0, 5, 0, \cdots] \\
&= \mathtt{Sum?}[1, 2, 3, 4, 5, 6, \cdots] - \mathtt{Sum?}[0, 2, 0, 4, 0, 6, \cdots] \\
&\stackrel{?}= \mathtt{Sum?}[1, 2, 3, 4, 5, 6, \cdots] - \mathtt{Sum?}[2, 4, 6, \cdots]
\end{align*}
The issue is, we do not know whether inserting zeros to the sequence preserves the value of $\mathtt{Sum?}$ or not. Indeed, under the Dirichlet regularization, we have
\begin{align*}
&\left[ \text{Dirichlet reg. of $1+3+5+\cdots$} \right] \\
&= \lim_{s \to 0} \left[ \text{analytic continuation of } s \mapsto \sum_{n=1}^{\infty} \frac{2n-1}{n^s} \right] \\
&= \lim_{s \to 0} (2\zeta(s-1) - \zeta(s)) = \frac{1}{3},
\end{align*}
while we get
\begin{align*}
&\left[ \text{Dirichlet reg. of $1+0+3+0+5+\cdots$} \right] \\
&= \lim_{s \to 0} \left[ \text{analytic continuation of } s \mapsto \sum_{n=1}^{\infty} \frac{2n-1}{(2n-1)^s} \right] \\
&= \lim_{s \to 0} (1 - 2^{1-s}) \zeta(s) = \frac{1}{12}.
\end{align*}
So, we cannot expect that $\mathtt{Sum?}[1,3,5,\cdots]$ and $\mathtt{Sum?}[1, 0, 3, 0, 5, 0, \cdots]$ have the same value.
|
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A property of Nesbitt's inequality I think it's not new however I think it's interesting :
Let $a,b,c>0$ and $x>0$ then the function :
$$f(x)=\frac{a^x}{b^x+c^x}+\frac{b^x}{a^x+c^x}+\frac{c^x}{b^x+a^x}$$
$f(x)$ is increasing
My first try :
The derivative of $f(x)$ is :
$$f'(x)=\sum_{cyc}\frac{a^x\ln(a)(b^x+c^x)-a^x(b^x\ln(b)+c^x\ln(c))}{(b^x+c^x)^2}$$
We multiply by $x$ it gives :
$$f'(x)=\frac{1}{x}\Big(\sum_{cyc}\frac{a^x\ln(a^x)(b^x+c^x)-a^x(b^x\ln(b^x)+c^x\ln(c^x))}{(b^x+c^x)^2}\Big)$$
$x$ is positive so we have to prove :
$$\sum_{cyc}\frac{a^x\ln(a^x)(b^x+c^x)-a^x(b^x\ln(b^x)+c^x\ln(c^x))}{(b^x+c^x)^2}\geq 0$$
We put $a^x=u$ , $b^x=v$ and $c^x=w$ so the inequality is equivalent to :
$$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq 0$$
Now with the concavity of $\ln(x)$ and the inequality of Jensen's we have :
$$\ln\Big(\frac{v^2+w^2}{v+w}\Big)\geq \frac{v\ln(v)+w\ln(w)}{v+w}$$
So we have :
$$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq\sum_{cyc}\frac{u\ln(u)-u\Big(\ln\Big(\frac{v^2+w^2}{v+w}\Big)\Big)}{(v+w)}\geq 0$$
Or :
$$\sum_{cyc}\frac{u\ln(u)(v+w)-u(v\ln(v)+w\ln(w))}{(v+w)^2}\geq\sum_{cyc}\frac{u\Big(\ln\Big(\frac{u(v+w)}{v^2+w^2}\Big)\Big)}{(v+w)}\geq 0$$
Now we have to prove :
$$\sum_{cyc}\frac{u\Big(\ln\Big(\frac{u(v+w)}{v^2+w^2}\Big)\Big)}{(v+w)}\geq 0$$
Unfortunatly I'm stuck here ...
Second try :
For $a,b,c>0$ the function $g(x)=\frac{a^x}{b^x+c^x}$ is convex so the function $f(x)$ is convex as sum of convex functions .
Now we have with $a,b,c>0$ :
$$\sum_{cyc}\frac{a^2}{b^2+c^2}\geq \sum_{cyc}\frac{a}{b+c}$$
Fact wich have been proven by Michael Rozenberg (see my questions)
Remains to apply the three chord lemma to get the increase of the function $f(x)$
My question :
Can someone could find an alternative proof or prove this last inequality ?
There are generalizations of this facts ?
|
Alternative solution:
WLOG, assume $a \ge b \ge c = 1$. Denote $a^x = u, b^x = v$. Then $u \ge v \ge 1$.
We have
\begin{align*}
f'(x) &= A \ln a + B\ln b \\
&= A\ln \frac{a}{b} + (A + B) \ln b
\end{align*}
where
\begin{align*}
A &= {\frac {u}{v+1}} - {\frac {vu}{ \left( 1+u \right) ^{2}}} -{\frac {u}{
\left( u+v \right) ^{2}}},\\
B &={\frac {v}{1+u} -{\frac {vu}{ \left( v+1
\right) ^{2}}}} -{\frac {v}{ \left( u+v \right) ^{2}}}.
\end{align*}
We have
$$A \ge {\frac {u}{v+1}} - {\frac {vu}{ \left( 1+v \right) ^{2}}} -{\frac {u}{
\left( 1+v \right) ^{2}}} = 0.$$
Also, we have
$$A + B = \frac{u}{(1 + v)^2} - \frac{u}{(u + v)^2} + \frac{v}{(1 + u)^2} - \frac{v}{(u + v)^2} \ge 0.
$$
Thus, $f'(x) \ge 0$ for all $x > 0$.
We are done.
|
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|
Master Theorem: $T(n) = 3T(\frac{n}{3} − 2) + \frac{n}{2}$ I need to solve this recurrence using the Master Theorem; however, I don't know if its is possible since it doesn't follow the format
$T(n) = 3T(\frac{n}{3} − 2) + \frac{n}{2}$
$a = 3$
$b = 3$?
What happens to the -2 inside the recurrence?
|
Here's a start.
In
$T(n) = 3T(\frac{n}{3} − 2) + \frac{n}{2}
$,
replace $n$ by $n+c$.
I will choose $c$
so there is no offset.
$T(n+c)
= 3T(\frac{n+c}{3} − 2) + \frac{n+c}{2}
= 3T(\frac{n}{3}+\frac{c}{3} − 2) + \frac{n+c}{2}
$
so if
$c = c/3-2
$,
or
$c=-3$,
this becomes
$T(n-3)
= 3T(\frac{n}{3}-3) + \frac{n-3}{2}
$.
Let
$U(n) = T(n-3)$.
Then
$U(n)
=3U(n/3)+(n-3)/2
$.
Now solve this,
|
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|
Find maximum and minimum of : $\Omega=\sin x\sin y\sin z$ Question :
Let : $x,y,z>0$ and $x+y+z=\frac{π}{2}$
Compute the maximum and minimum of the
following expansion :
$\Omega=\sin x\sin y\sin z$
My attempt :
About maximum :
$\Omega=\sin x\sin y\cos (x+y)=\frac{1}{2}\sin x(\sin (2x+y)-\sin y)$
$≤\sin y(1-\sin y)$ as $\sin (2x+y)≤1$
Now :
Let define the function : $f(x)=x(1-x)$ then
$f(x)≤f(\frac{1}{2})=\frac{1}{4}$
So we get maximum $=\frac{1}{8}$
Now is my work correct ? Please if any one have
another method tell me
Thanks!
|
For $\Omega= (\sin x \sin y \sin z)$ by GM-AM
$$(\sin x \sin y \sin z)^{1/3} \le \frac{\sin x+\sin y+ \sin z}{3}\le \sin \frac{(x+y+z)}{3}=\frac{1}{2}.$$ The second inequality is due to Jensens Inequality for $x \in (0,\pi/2).$ as $f''(x)<0.$ So finally
$$0<\sin x \sin y \sin z \le \frac{1}{8},~ x,y,z >0, ~ x+y+z=\pi/2$$
So max of $\Omega$ is $1/8$ which is attained when $x=y=z=\pi/6$ but its min does not exist, however $\Omega >0$.
|
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|
Solution of a limit of a sequence $\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}$ I'm trying to solve the limit of this sequence without the use an upper bound o asymptotic methods:
$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\left(\frac{\infty-\infty}{\infty-\infty}\right)$$
Here there are my differents methods:
*
*assuming $f(n)=\sqrt{4n^2+1}, \,$$\ g(n)=2n$, $h(n)=\sqrt{n^2-1}$, $\ \psi(n)= n$ $$f(n)-g(n)=\frac{\dfrac{1}{g(n)}-\dfrac{1}{f(n)}}{\dfrac{1}{f(n)\cdot g(n)}}, \quad h(n)-\psi(n)=\frac{\dfrac{1}{\psi(n)}-\dfrac{1}{h(n)}}{\dfrac{1}{h(n)\cdot \psi(n)}}$$
I always have an undetermined form.
*I've done some rationalizations:
$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{4n^2+1}+2n}{\sqrt{4n^2+1}+2n}$$ where to the numerator I find $1$ and to the denominator an undetermined form. Similar situation considering
$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{n^2-1}+n}{\sqrt{n^2-1}+n}$$
*$$\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\frac{n\left(\sqrt{4+\dfrac{1}{n^2}}-2\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-1\right)}\rightsquigarrow \left(\frac{0}{0}\right)$$
At the moment I am not able to think about other possible simple solutions.
|
Hint:
$$\begin{align}
{\sqrt{4n^2+1}-2n\over\sqrt{n^2-1}-n}
&={\sqrt{4n^2+1}-2n\over\sqrt{n^2-1}-n}\cdot{\sqrt{4n^2+1}+2n\over\sqrt{4n^2+1}+2n}\cdot{\sqrt{n^2-1}+n\over\sqrt{n^2-1}+n}\\
&={(4n^2+1)-4n^2\over(n^2-1)-n^2}{\sqrt{n^2-1}+n\over\sqrt{4n^2+1}+2n}
\end{align}$$
|
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|
Let $\xi$ be the fifth primitive root of $1$ an let $\zeta= \xi + \frac{1}{\xi}$. Prove that $\zeta^2 + \zeta =1$ Let $\xi$ be the fifth primitive root of $1$ an let $\zeta= \xi + \frac{1}{\xi}$. Prove that $\zeta^2 + \zeta =1$
In my attempt I used $\xi = \cos(72k) + i\sin(72k)$ with $k=1,2,3,4$
Where I got $\zeta = \frac{1+\cos(144k)+i\sin(144k)}{\cos(72k)+i\sin(72k)}$
Which yielded $\zeta^2 + \zeta = 2\cos(72k)(2\cos(72k)+1)$
I don’t know what else I could do.
|
By definition, $\xi^5 = 1$ and we have the factorization $$\prod_{k=0}^4 (z - \xi^k) = z^5 - 1.$$ But since $(z - \xi^0) = (z - 1)$, it follows that $$\prod_{k=1}^4 (z - \xi^k) = \frac{z^5 - 1}{z-1} = 1 + z + z^2 + z^3 + z^4,$$ hence $$1 + \xi + \xi^2 + \xi^3 + \xi^4 = 0.$$ Dividing by $\xi^2$ yields $$\begin{align*}0
&= \xi^{-2} + \xi^{-1} + 1 + \xi + \xi^2 \\
&= (\xi^2 + 2 + \xi^{-2}) + (\xi + \xi^{-1}) - 1 \\
&= (\xi + \xi^{-1})^2 + (\xi + \xi^{-1}) - 1 \\
&= \zeta^2 + \zeta - 1. \end{align*}$$
|
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|
Geometry - Incircle I of $\triangle$ABC with chord MN intersecting AB at P. If BP = MC, find $\angle$AIC I recently found the following problem:
Circle with center $I$ is inscribed in triangle $ABC$ and touches the sides $A$C and $BC$ in points $M$ and $N$. The line $MN$ intersect the line $AB$ at $P$, as $B$ is between $A$ and $P$. If $BP = CM$ , find $\angle AIC$, in degrees.
This is from IWYMIC 2011
|
Let the side lengths of the triangle ABC be $a$, $b$ and $c$. Apply the sine rule to the triangles PBN and AIC,
$$\frac{\sin\alpha}{\sin\beta} = \frac{BN}{BP} = \frac{BN}{MC}
=\frac {\frac12 (a+c-b)}{{\frac12 (a+b-c)}} =\frac {a+c-b}{{a+b-c}}\tag{1}$$
$$\frac{\sin\alpha}{\sin(180-\beta)} = \frac{AM}{AP}= \frac{AM}{AB+MC}=\frac {\frac12 (b+c-a)}{{c+\frac12 (a+b-c)}} = \frac {b+c-a}{{a+b+c}}\tag{2}$$
where the incircle results,
$$BN = \frac 12 (a+c-b);\>\>\>MC = \frac 12 (a+b-c);\>\>\>
AM = \frac 12 (c+b-a)$$
are used. Combine (1) and (2),
$$\frac {a+c-b}{{a+b-c}} = \frac {b+c-a}{{a+b+c}}$$
which leads to $a^2+c^2=b^2$. Thus, ABC is a right triangle, with $\angle B = 90$. So,
$$\angle AIC = 180 - \frac{\angle A+\angle C}{2} = 180 - \frac 12 \angle B = 135^\circ$$
|
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|
Generating function for non-isomorphic regular graphs. Determine a generating function for the number of non-isomorphic (n−2)-regular graphs of order n, for n ≥ 2.
I've been staring at this for hours and can't find a place to start, any help would be appreciated.
|
We enumerate the complements, namely non-isomorphic 2-regular graphs.
These are sets of cycles and we find
$$\prod_{k\ge 3} \frac{1}{1-z^k}
= (1-z)(1-z^2) \prod_{k\ge 1} \frac{1}{1-z^k}.$$
The term $1/(1-z^k)$ is the OGF of zero, one, two, three
etc. instances of a cycle of order $k.$
This is the OGF
$$(1-z-z^2+z^3) \prod_{k\ge 1} \frac{1}{1-z^k}.$$
Using the partition function we get for $n\ge 3$
$$p(n) - p(n-1) - p(n-2) + p(n-3).$$
We obtain the sequence
$$1, 1, 1, 2, 2, 3, 4, 5, 6, 9, 10, 13,
\\ 17, 21, 25, 33, 39, 49, \ldots$$
which points us to OEIS A00843, where
these data are confirmed (indeed we have non-isomorphic 2-regular
graphs).
For the case where the graphs are labeled we again have sets of
cycles (with dihedral symmetry):
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}(\textsc{DHD}_{=3}(\mathcal{Z})
+ \textsc{DHD}_{=4}(\mathcal{Z})
+ \textsc{DHD}_{=5}(\mathcal{Z})
+ \cdots).$$
This gives the EGF
$$\exp
\left(\frac{1}{2} \frac{z^3}{3}
+\frac{1}{2} \frac{z^4}{4}
+\frac{1}{2} \frac{z^5}{5}+\cdots\right)
\\ = \exp\left(-\frac{1}{2} z - \frac{1}{2} \frac{z^2}{2}
+ \frac{1}{2} \log\frac{1}{1-z}\right)
\\ = \frac{1}{\sqrt{1-z}}
\exp\left(-\frac{z}{2}-\frac{z^2}{4}\right).$$
We get the sequence starting at $n\ge 3$:
$$1, 3, 12, 70, 465, 3507, 30016, 286884, 3026655,
\\ 34944085, 438263364, 5933502822, 86248951243, \ldots$$
which points us to OEIS A001205, where we
get confirmation of these data once more.
|
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|
Find the roots of $z^4-3z^2+1=0$ in polar form. Question :
Prove that the solutions of $z^4-3z^2+1=0$ are given by :
$$z=2\cos{36^\circ},2\cos{72^\circ},2\cos{216^\circ},2\cos{252^\circ}$$
My work :
First of all, i want ro find the roots with quadratic formula
$\begin{align}
&(z^2)^2-3z^2+1=0\\
&z^2=\dfrac{3\pm \sqrt{5}}{2}\\
&z_{1,2}=\pm\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\\
&z_{3,4}=\pm\dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}\\
\end{align}$
And i'm stuck. I don't know how to transform this complicated roots into polar form. Bcz, i'm not sure that the modulus for each number is exactly $2$?
Btw, i've found and read some possible duplicates
Here's one of the links :
Roots of $z^4 - 3z^2 + 1 = 0$.
But it seems doesn't answer my question...
Please give me a clear hint or another way to solve this without quadratic formula or something else.
|
Consider $z^5-1=0$
So the roots of $$0=\dfrac{z^5-1}{z-1}=z^4+z^3+z^2+z+1$$ are $e^{2i\pi r/5},r=1,2,3,4$
As $z\ne0,$ like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ , divide both sides by $z^2$
Replace $z+\dfrac1z=w\implies w^2=?$ to find $$w^2-2+w+1=0$$ whose roots are $$2\cos\dfrac{2r\pi}5;r=1,2$$
Similarly consider $\dfrac{z^5+1}{z+1}=0$
|
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|
When is $\small \gcd(2n^7\!+1,{3n^3\!+2})>1,$ i.e. when is $\frac{2n^{7}+1}{3n^{3}+2}$ reducible? How can I ind the values of $n\in \mathbb{N}$ that make the fraction $\frac{2n^{7}+1}{3n^{3}+2}$ reducible ?
I don't know any ideas or hints how I solve this question.
I think we must be writte $2n^{7}+1=k(3n^{3}+2)$ with $k≠1$
|
Suppose $p$ is prime.
If $p$ divides $2n^7+1$ & $3n^3+2$
then $p$ divides $2(2n^7+1)-(3n^3+2)=n^3(4n^4-3)$
then $p$ divides $4n^4-3$ ( See Footnote )
then $p$ divides $2(4n^4-3)+3(3n^3+2)= n^3(8n+9)$
then $p$ divides $8n+9$ (See Footnote)
then $p$ divides $9(3n^3+2)-2(8n+9)=n(27n^2-16)$
then $p$ divides $27n^2-16$ (See Footnote)
then $p$ divides $9(27n^2-16)+16(8n+9)=n(243n+128)$
then $p$ divides $243n+128$ (See Footnote)
then $p$ divides $9(243n+128)-128(8n+9)=1163$
then $p=1163$ because $p$ and $1163$ are both prime
then $8n+9\equiv 0 \mod 1163$ so $n\equiv 435 \mod 1163$
So $\gcd (2n^7+1, 3n^3+2)>1\implies n\equiv 435 \mod 1163.$ And we may verify that $n\equiv 435 \mod 1163\implies 2n^7+1\equiv 3n^3+2\equiv 0 \mod 1163\implies \gcd (2n^7+1,3n^3+2)>1.$
Footnote. Suppose $A,B,C, D$ are integers with $A>0$ and $C>0,$ and $p$ divides $n^A(Bn^C+D).$ Since $p$ is prime, $p$ divides $n^A$ or $p$ divides $Bn^C+D.$ Now if prime $p$ divides $n^A$ then $p$ divides $n$ and hence $p$ divides $2n^7$, BUT if $p$ also divides $2n^7+1$ then the prime $p>1$ divides $(2n^7+1)-(2n^7)=1,$ which is absurd. So instead, $p$ must divide $Bn^C+D.$
|
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|
Find $x,y,z$ for the given conditions $$4x^2+25y^2+9z^2-10xy-15yz-6zx=0$$
$$x+y+z=5$$
I tried two approaches
1) Substituting $x$ as $5-y-z$ in the first equation but didn't work out, I was getting $39y^2+19z^2-31yz-90y-70z+100=0$ which can't be factorized
2) First equation corresponds to $a^2+b^2+c^2-ab-bc-ca=0$, which means $a^3+b^3+c^3=3abc$, but didn't get a breakthrough.
I am stuck here, please help me.
|
Continuing your second method.
Denote: $a=2x,b=5y,c=3z$, then:
$$\begin{cases}4x^2+25y^2+9z^2-10xy-15yz-6zx=0\\
x+y+z=5\end{cases} \Rightarrow \\
\begin{cases}a^2+b^2+c^2-ab-bc-ca=0\\
\frac a2+\frac b5+\frac c3=5\end{cases} \Rightarrow \\
2a^2+2b^2+2c^2-2ab-2bc-2ca=0 \Rightarrow \\
(a-b)^2+(b-c)^2+(c-a)^2=0 \Rightarrow \\
a=b=c=\frac{150}{31} \Rightarrow \\
(x,y,z)=\left(\frac{75}{31},\frac{30}{31},\frac{50}{31}\right)$$
WA answer.
|
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|
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following.
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, show that $$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)=(1-q+s)^2+(p-r)^2.$$
I have tried putting
\begin{align}P(1)&=(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=1+p+q+r+s\\P(-1)&=(\alpha+1)(\beta+1)(\gamma+1)(\delta+1)=1-p+q-r+s.\end{align}
Then \begin{align}P(1)P(-1)&=(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)\\&=(1+p+q+r+s)(1-p+q-r+s)=(1+q+s)^2-(p+r)^2\end{align}
Somehow it does not match the statement given.
|
You correctly computed
$$
(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)=(1+q+s)^2-(p+r)^2
$$
only that this is not what was asked for.
But you were close: $\alpha^2+1 = (\alpha-i)(\alpha+i)$ etc, and therefore
$$
(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2) = P(i) P(-i)
$$
which expands to
$$
(1-ip-q+ir+s)(1+ip-q-ir+s) = (1-q+s)^2+(p-r)^2 \, .
$$
|
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|
$x$ and $y$-intercepts of an absolute value function $f(x)=-3|x-2|-1$ I am to find the $x$ and $y$-intercepts of the function $$f(x)=-3|x-2|-1.$$ The solution is provided in my book as
$(0, -7)$; no $x$ intercepts.
I cannot see how this was arrived at. I attempted to find the x intercepts and arrived at $2-\frac{1}{3}$ and $2+\frac{1}{3}$
My working:
$-3|x-2|-1=0$
$-3|x-2|=1$
$|x-2|=-\frac{1}{3}$
Then solve for both the negative and positive value of $\frac{1}{3}$:
Positive version:
$x-2=\frac{1}{3}$
$x=2+\frac{1}{3}$
Negative version:
$x-2=-\frac{1}{3}$
$x = 2-\frac{1}{3}$
How can I arrive at "$(0, -7)$; no x intercepts"? Where did I go wrong in my understanding?
|
The graph of the function $f(x) = -3|x - 2| - 1$ has vertex $(2, -1)$ and opens downward since the coefficient of $|x - 2|$ is negative. That tells you that the graph of the function cannot intersect the $x$-axis.
Remember that $|x|$ means the distance of the number $x$ from $0$. Therefore, $|x - 2|$ means the distance of the number $x - 2$ from $0$. However, a distance cannot be negative.
To solve for the $x$-intercepts, you set $f(x) = 0$, which yields
\begin{align*}
f(x) & = 0\\
-3|x - 2| - 1 & = 0\\
-3|x - 2| & = 1\\
|x - 2| & = -\frac{1}{3}
\end{align*}
which is impossible since $|x - 2| \geq 0$ for every real number $x$. Hence, there are no $x$-intercepts.
We know that $f(0) = -3|0 - 2| - 1 = -3|-2| - 1 = -3(2) - 1 = -6 - 1 = -7$. By symmetry, $f(4) = -7$, as you can check.
|
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|
Is this a valid way to prove that a sequence $\{a_n\}$ given by $a_n=\frac{1}{\sqrt{n+3}} + \frac{n-3n^2}{2n^2-1}$ converges? I'll be using the Archimedean Property to prove this.
Proof
Let $\epsilon>0$. Then $\epsilon^2/4>0$.
By the Archimedean Property, $\forall \epsilon^2/4>0, \exists N\in \mathbb{N}$ such that $1/N<\epsilon^2/4$.
($\star$) Since $1/N<\epsilon^2/4$ it follows that $2/\sqrt{N}<\epsilon$.
$|a_n+\frac{3}{2}|=|\frac{1}{\sqrt{n+3}} + \frac{n-3n^2}{2n^2-1}+\frac{3}{2}|=|\frac{1}{\sqrt{n+3}}+\frac{2n-3}{4n^2-2}|<|\frac{1}{\sqrt{n}}+\frac{n}{2n^2-1}|<\frac{2}{\sqrt{n}}$
$\frac{2}{\sqrt{n}}<\frac{2}{\sqrt{N}}\forall n\geq N$
By ($\star$), $|a_n+\frac{3}{2}|<\epsilon$ $\forall n\geq N$.
Therefore the sequence $\{a_n\}$ converges to $-\frac{3}{2}$.
|
I don't see why you need that
$\epsilon^2/4$ statement.
Once you have shown that
$|a_n+\frac{3}{2}|<\frac{2}{\sqrt{n}}
$,
to make
$|a_n+\frac{3}{2}|
\le \epsilon$,
just choose
$n$ so that
$\frac{2}{\sqrt{n}}
\lt \epsilon$
or
$n > 4/\epsilon^2$.
|
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|
Find $a\in \mathbb{R}$ for which $a\cdot \left(\frac{1}{1+x^2}\right)^2-3\cdot\frac{a}{1+x^2}+1=0$ will have all roots imaginary
Find $a\in \mathbb{R}$ for which $a\cdot \left(\frac{1}{1+x^2}\right)^2-3\cdot\frac{a}{1+x^2}+1=0$ will have all roots imaginary.
My attempt is as follows:-
Let $t=\frac{1}{1+x^2}$, and let's find out its range for which x is imaginary
$$t=\frac{1}{1+x^2}$$
$$(1+x^2)\cdot t=1$$
$$tx^2+t-1=0$$
$$D<0$$
$$0-4t(t-1)<0$$
$$t(t-1)>0$$
$$t\in (-\infty,0)\quad \cup \quad (1,\infty)$$
So for the equation $at^2-3at+1=0$, we have to find such values of a for which $t\in (-\infty,0) \cup (1,\infty)$. As $t$ should be real,so
$$D\geq 0\Leftrightarrow 9a^2-4a\geq 0\Leftrightarrow a(9a-4)\geq0$$
$$a\in \left(-\infty,0\right] \cup \left[\frac{4}{9},\infty\right)$$
But if we place $a=0$ in the quadratic equation in $t$, then $0+0+1=0$, which is not possible hence $a\in \left(-\infty,0\right) \cup \left[\frac{4}{9},\infty\right)$.
Now as we know that roots of quadratic equation $at^2-3at+1=0$ should lie in $(-\infty,0)\cup (1,\infty)$. So
Case 1 : When both roots are negative
$$af(0)>0$$
$$a>0$$
$0$ is greater than both the roots, so
$0>(a+b)/2$ where a and b are roots.
$$0>\frac{3a}{2a}$$
$$0>\frac{3}{2}$$
So $a\in \phi$ for first case
Case 2: When both roots are greater than $1$
$$af(1)>0$$
$$a(a-3a+1)>0$$
$$a(2a-1)<0$$
$$a\in \left(0,\frac{1}{2}\right)$$
$1$ should lie before the roots on the x-axis, so $1<\frac{a+b}{2}$
$$1<\frac{3a}{2a}$$
$$1<\frac{3}{2}$$
So $a\in \left(0,\frac{1}{2}\right)$ for the second case
Case 3: When one root is greater than $1$ and another is negative:
$$af(0)<0\quad \cap \quad af(1)<0$$
$$a<0\quad \cap\quad a(a-3a+1)<0$$
$$a<0\quad \cap \quad a(2a-1)>0$$
$$a\in \left(-\infty,0\right)$$
Hence $a\in \left(-\infty,0\right) \cup \left[\frac{4}{9},\frac{1}{2}\right)$
but answer is $a\in \left(-\infty,\frac{1}{2}\right)$
What mistake I am doing, I thought about it a lot but didn't get any breakthroughs. Please help me in this.
|
Hint: Factorizing your equation we get
$$x^4+x^2(2-3a)+1-2a=0$$
You can also write $$at^2-3at+1=0$$ where $t=\frac{1}{1+x^2}>0$
$a=0$ is impossible, so we get by the quadratic formula
$$t_{1,2}=-\frac{3}{2}\pm\sqrt{\frac{9}{4}-\frac{1}{a}}$$
If $$\frac{9}{4}-\frac{1}{a}<0$$ then the roots are imaginary.
|
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|
Taylor series of $\sinh{(x)}$ at $\ln{(2)}$. Determine the Taylor series of $\sinh{(x)}$ about $x = \ln{(2)}$.
Equating each derivative at $x = \ln{(2)}$ gives:
\begin{equation*}
\begin{split}
f'(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\
f''(\ln{(2)}) &= \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\
f^{(3)}(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\
f^{(4)}(\ln{(2)}) &= \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\
f^{(5)}(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4}
\end{split}
\end{equation*}
and so on. This means the Taylor series is
\begin{equation*}
\begin{split}
\sum_{n=0}^{\infty} \frac{f^{(n)}(\ln{(2)})}{n!}(x-\ln{(2)})^n &= f(\ln{(2)}) + \frac{f'(\ln{(2)})}{1!}(x-\ln{(2)}) + \frac{f''(\ln{(2)})}{2!}(x-\ln{(2)})^2 \\
&+ \frac{f^{(3)}(\ln{(2)})}{3!}(x-\ln{(2)})^3 + \frac{f^{(4)}(\ln{(2)})}{4!}(x-\ln{(2)})^4 + \frac{f^{(5)}(\ln{(2)})}{5!}(x-\ln{(2)})^5 + \ldots \\
&= \frac{3}{4}+\frac{5}{4\cdot1!}(x-\ln{(2)})^1+\frac{3}{4\cdot 2!}(x-\ln{(2)})^2 \\
&+\frac{5}{4\cdot 3!}(x-\ln{(2)})^3+\frac{3}{4\cdot 4!}(x-\ln{(2)})^4+\frac{5}{4\cdot 5!}(x-\ln{(2)})^5+\ldots
\end{split}
\end{equation*}
How do I find the general sum of this? Has it got something to do with the Maclaurin series which is $\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$.
|
Your computation leads naturally to:
$$\sinh x=\frac54\sum_{n=0}^{\infty} \dfrac{(x-\log2)^{2n+1}}{(2n+1)!}
+\frac34\sum_{n=0}^{\infty} \dfrac{(x-\log2)^{2n}}{(2n)!}\\
=\sum_{n=0}^{\infty} \left(1-\dfrac{(-1)^n}4\right)\dfrac{(x-\log2)^{n}}{n!}$$
The radius of convergence is infinite.
|
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|
How to write this as a multiplication: $4 x^2 y^2 - (x^2 + y^2 - z^2)^2$? I have the following polynomial:
$$ 4 x^2 y^2 - (x^2 + y^2 - z^2)^2$$
(which comes up, for example, in computing the area of a triangle using the cosine law).
I would like to convert this to a product.
Wolfram tells me it's
$$ -(x - y - z) (x + y - z) (x - y + z) (x + y + z)$$
How can I find this form if I don't already know it? What operations should I perform?
|
Note that you have a binomial formula:
$a^2-b^2=(a-b)(a+b)$
Where $a=2xy$ since $a^2=4x^2y^2$ and $b=x^2+y^2-z^2$ obviously.
This leads to $4x^2y^2-(x^2+y^2-z^2)^2=(2x^2y^2-x^2-y^2+z^2)(2x^2y^2+x^2+y^2-z^2)$
With additional use of binomial formulas you are able to factor this term.
|
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|
Algebraic proof of a combinatoric question (Combinatoric proof is given) I had a IMO training about double counting. Then, there is a problem which I hope there is a combinatoric proof. Here comes the problem:
For every positive integer $n$, let $f\left(n\right)$ be the number of all positive integers with exactly $2n$ digits, each having exactly $n$ of digits equal to $1$ and the other equal to $2$. Let $g\left(n\right)$ be the number of all positive integers with exactly $n$ digits, each of its digits can only be $1,2,3$ or $4$ and the number of $1$'s equals the number of $2$'s. Prove that $f\left(n\right)=g\left(n\right)$.
It is obvious to see that $f\left(n\right)=\binom{2n}{n}$, and $g\left(n\right)=\sum_{k\le\lfloor\frac{n}{2}\rfloor}\binom{n}{2k}\binom{2k}{k}2^{n-2k}$. However, it is hard to prove this in an algebraic way. I hope there are someone to prove it by algebraic way. Thank you!
Combinatoric proof
We can establish a one-to-one correspondence between $f\left(n\right)$ and $g\left(n\right)$. Let $F\left(n\right)$ be the set of all positive integers with exactly $2n$ digits, each having exactly $n$ of digits equal to $1$ and the other equal to $2$. Also, let $G\left(n\right)$ be the set of all positive integers with exactly $n$ digits, each of its digits can only be $1,2,3$ or $4$ and the number of $1$'s equals the number of $2$'s. Then, we can do this operation for all numbers in $F\left(n\right)$: For every two digits of the numbers in $F\left(n\right)$, $$\begin{cases}11\Rightarrow 1\\22\Rightarrow 2\\12\Rightarrow 3\\21\Rightarrow 4\end{cases}$$ Then all the numbers will change into a set which is totally same as $G\left(n\right)$, as we find that the difference between the number of $1$'s and $2$'s doesn't change at all. Therefore, we make a one-to-one correspondence between $F\left(n\right)$ and $G\left(n\right)$.
|
Here is a slightly different proof that
$$\sum_{k=0}^{\lfloor n/2\rfloor} {n\choose 2k} {2k\choose k}
2^{n-2k} = {2n\choose n}.$$
We observe that
$${n\choose 2k} {2k\choose k}
= \frac{n!}{(n-2k)! \times k! \times k!}
= {n\choose k} {n-k\choose n-2k}.$$
We get for our sum
$$\sum_{k=0}^{\lfloor n/2\rfloor} {n\choose k}
{n-k\choose n-2k} 2^{n-2k}
\\ = \sum_{k=0}^{\lfloor n/2\rfloor} {n\choose k}
[z^{n-2k}] (1+z)^{n-k} 2^{n-2k}
\\ = [z^n] (1+z)^n \sum_{k=0}^{\lfloor n/2\rfloor} {n\choose k}
z^{2k} (1+z)^{-k} 2^{n-2k}.$$
Now when $2k\gt n$ we get zero from the coefficient extractor,
which enforces the range, so we continue with
$$2^n [z^n] (1+z)^n \sum_{k\ge 0} {n\choose k}
z^{2k} (1+z)^{-k} 2^{-2k}
\\ = 2^n [z^n] (1+z)^n
\left(1+\frac{z^2}{2^2(1+z)}\right)^n
\\ = 2^n [z^n] \frac{(2^2+2^2z+z^2)^n}{2^{2n}}
= 2^n [z^n] \frac{(z+2)^{2n}}{2^{2n}}
= 2^n {2n\choose n} 2^{2n-n} \frac{1}{2^{2n}}
\\ = {2n\choose n}.$$
This is the claim.
|
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|
Alternative way to calculate $\int_0^1(x^4(1-x)^4)/(1+x^2)dx$ $$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx$$
$$=\int_0^1(x^8-4x^7+6x^6-4x^5+x^4)/(x^2+1)dx$$
$$=\int_0^1(x^6-4x^5+5x^4-4x^2-4/(x^2+1)+4)dx$$
$$=[1/7x^7-2/3x^6+x^5-4/3x^3-4\tan^{-1}x+4x]_0^1$$
$$I=22/7-\pi$$
Any other method to solve this problem?
|
A similar approach to @Quanto's is to define $I_n:=\int_0^1\frac{x^ndx}{1+x^2}$ so $I_0=\frac{\pi}{4}$ and $I_n+I_{n+2}=\frac{1}{n+1}$, and in particular $I_n-I_{n+4}=\frac{2}{(n+1)(n+3)}$. Hence $I_4=\frac{\pi}{4}-\frac{2}{3}$ and$$I_8+I_4-4(I_5+I_7)+6I_6=\underbrace{-\frac{2}{35}}_{I_8-I_4}+6\cdot\underbrace{\frac15}_{I_4+I_6}-4I_4-\frac{4}{6}\\=-\frac{2}{35}+\frac65-\pi+\frac83-\frac23=\frac{22}{7}-\pi.$$
|
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|
Convergence of $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$. Does $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ converge?
Dividing the top and bottom by $4^n$ gives
\begin{equation*}
\frac{2^n+5^n}{3^n+4^n} = \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1}.
\end{equation*}
Hence,
\begin{equation*}
\lim_{n\to\infty} \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1} = \frac{0+\infty}{0+1} = \infty.
\end{equation*}
Thus, $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ diverges.
Is this correct? Thanks.
|
$n>1$;
$a_n=\dfrac{2^n+5^n}{3^n+4^n} \gt$
$\dfrac{5^n}{4^n+4^n}=(1/2)(5/4)^n >$
$(1/2)(1+1/4)^n> (1/2)(1/4)n.$
Used : Binomial expansion
|
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|
If $A$ is a rotation matrix, then $||Ax||=||x||$. Attempt:
Let $$
A =
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix},
x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
$$
Then,
$$Ax =
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \cos\theta \cdot x_1 -\sin\theta\cdot x_2 \\ \sin\theta \cdot x_1 +\cos\theta\cdot x_2 \end{bmatrix} $$
$$
\begin{align}
||Ax|| &= \sqrt{(\cos\theta \cdot x_1 -\sin\theta\cdot x_2)^2 + (\sin\theta \cdot x_1 +\cos\theta\cdot x_2)^2} \\
&=\sqrt{(\cos^2\theta \cdot x^2_1 -2\cos\theta \cdot x_1\sin\theta\cdot x_2+ \sin^2\theta\cdot x^2_2) + (\sin^2\theta \cdot x^2_1 +2\sin\theta \cdot x_1\cos\theta\cdot x_2+ \cos^2\theta\cdot x^2_2)} \\
&=\sqrt{x^2_1+x^2_2}
\end{align}
$$
However, my text seems to suggest that this is only true for $0\leq\theta\leq\pi$ here:
So where in my attempt did I go wrong?
|
You didn't go wrong anywhere. You showed what the problem wanted you to show, and then some.
Note that some times the restrictions given in problems and exercises are necessary, and thus not using them means you're wrong somewhere. So you are right to be sceptical. But in this specific case the restriction $0\leq \theta\leq\pi$ is completely superfluous, so there is no issue.
|
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|
A homeomorphism of the unit disk that cannot be extended to the boundary of its domain I have a problem with the following exercise:
Let $D^2$ be the unit disc and S^1 be the unit circle. Show that the function
$
h: {D^2\setminus{S^1}} \to {D^2\setminus{S^1}} \\
h(re^{it})=
\begin{cases}
0 &\text{if}\, r=0 \\
r \cdot e^{i(t+\frac{2\pi r}{1-r})} &\text{else}
\end{cases}
$
cannot be extended to a homeomorphism $D^2 \to D^2$.
I started as follows:
Suppose that h can be extended to a homeomorphism $g: D^2 \to D^2$.
Define a sequence via
$
r_n=
\begin{cases}
\frac{n+1}{n} &\text{if n even} \\
\frac{n}{n+2} &\text{if n odd}
\end{cases}
$
This implies
$\frac{r_n}{1-r_n}= -(n+1)$ for n even and $\frac{r_n}{1-r_n}=\frac{n}{2}$ for n odd.
Hence
$e^{i\frac{2\pi r_n}{1-r_n}}$=1 if n even and $e^{i\frac{2\pi r_n}{1-r_n}}$=-1 if n odd. If I had $\lim_{n \to \infty} r_n = 1$ then for the sequence of points $(r_n,t)$ where $t=0$ I would get
$lim_{n \to \infty}g(r_n)\neq g(1)$ since $g(r_n)$ does not converge but alternates between 1 and -1. The problem is that currently the sequence $(r_n)$ does not converge, since the subsequences for even and odd n have different limits. Is it possible to define a sequence $(r_n)$ such that the conditions
$\lim_{n \to \infty} r_n = 1$
$
e^{i\frac{2\pi r_n}{1-r_n}}=
\begin{cases}
\ 1 \ \text{if n even} \\
\ -1 \ \text{if n odd} \\
\end{cases}
$
are satisfied?
|
Observe
$\quad \frac{2 \pi r}{1-r} = 1 \pi \text{ implies } r = \frac{1}{3}$
$\quad \frac{2 \pi r}{1-r} = 3 \pi \text{ implies } r = \frac{3}{5}$
$\quad \frac{2 \pi r}{1-r} = 5 \pi \text{ implies } r = \frac{5}{7}$
$\quad \frac{2 \pi r}{1-r} = 7 \pi \text{ implies } r = \frac{7}{9}$
etc.
Observe
$\quad \frac{2 \pi r}{1-r} = 2 \pi \text{ implies } r = \frac{1}{2}$
$\quad \frac{2 \pi r}{1-r} = 4 \pi \text{ implies } r = \frac{2}{3}$
$\quad \frac{2 \pi r}{1-r} = 6 \pi \text{ implies } r = \frac{3}{4}$
$\quad \frac{2 \pi r}{1-r} = 8 \pi \text{ implies } r = \frac{4}{5}$
etc.
|
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|
$\int \frac{\sin x}{1+2\sin x}dx$ calculate:
$$\int \frac{\sin x}{1+2\sin x}dx$$
I tried using $\sin x=\dfrac{2u}{u^2+1}$, $u=\tan \dfrac{x}2$ and after Simplification:
$$\int \frac{2u}{u^2+4u+1}×\frac{2}{u^2+1}du$$
and I am not able to calulate that.
|
Since $u=\tan\frac{x}{2}$ gives$$\int\frac{dx}{1+2\sin x}=\int\frac{2 du}{(u+2)^2-3}=\frac{-2}{\sqrt{3}}\operatorname{artanh}\frac{u+2}{\sqrt{3}}+C=\frac{-1}{\sqrt{3}}\ln\frac{\sqrt{3}+2+\tan\frac{x}{2}}{\sqrt{3}-2-\tan\frac{x}{2}}+C,$$we have$$\int\frac{\sin xdx}{1+2\sin x}=\frac12x+\frac{1}{2\sqrt{3}}\ln\frac{\sqrt{3}+2+\tan\frac{x}{2}}{\sqrt{3}-2-\tan\frac{x}{2}}+C.$$
|
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|
Find $x$ so that rational function is an integer Find all rational values of $x$ such that $$\frac{x^2-4x+4}{x^2+x-6}$$ is an integer.
How I attempt to solve this: rewrite as $x^2-4x+4=q(x)(x^2+x-6)+r(x)$. If we require that $r(x)$ be an integer then we can get some values of $r(x)$ by solving $x^2+x-6=0$, so that $x=-3$ or $x=2$. In the former case, $r=25$, and in the latter case $r=0$.
[I should note, however, that I'm not really convinced that this step is actually correct, because when $x$ is a root of $x^2+x-6$, the rational function above can have no remainder due to division by $0$. But I read about it as a possible step here and I can't explain it. Q1 How can this be justified?]
Now we want $(x^2+x-6)$ to divide $0$ (trivial) or $x^2+x-6$ to divide $25$. So we set $x^2+x-6=25k$ for some integer $k$ and solve. So $$x=\frac12 (\pm5\sqrt{4k+1}-1)$$
One of the trial-and-error substitutions for $k$ and then for $x$ gives $x=-8$, but that is just one number. Q2: So I'm wondering, how can we find all such $x$? The condition here is that $4k+1$ must be a perfect square, $k\ge 0$. Aren't there infinitely many perfect squares of this form?
I'd appreciate some clarifications about these two questions, Q1 and Q2.
|
$\frac{x^2-4x+4}{x^2+x-6}=\frac{x-2}{x+3}=1-\frac{5}{x+3}$, for this to be an integer, $\frac{5}{x+3}$ has to be an integer.
Say, $\frac{5}{x+3}=K$, where $K \in \Bbb Z$. This implies for each $K \in \Bbb Z$, $x=\frac{5}{K}-3$ would make the given expression an integer.
|
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|
What is the limit of $\lim_{n\to +\infty} n\left(\frac{e}{\left(1+\frac{1}{n}\right)^n}-1\right)$ I have read this post Solve the following limit: $\lim_{n->\infty} n(\frac{\frac{1}{n!}(\frac{n}{e})^n}{\frac{1}{(n+1)!}(\frac{n+1}{e})^{n+1}}-1)$ but I don't understand how do you get from $\log\left(1+\tfrac{1}{n}\right)^n = 1-\frac{1+o(1)}{2n}$ to $\left(1+\tfrac{1}{n}\right)^n = e -\frac{e+o(1)}{2n}$ and then to $\frac{e}{\left(1+\frac{1}{n}\right)^n} = 1+\frac{1+o(1)}{2n}$ from there is obviously simple but maybe there is a way to solve the limit in an easier fashion.
|
Let
$$a_n=\left(1+\frac{1}{n}\right)^n\implies \log(a_n)=n \log\left(1+\frac{1}{n}\right)$$ Now, using Taylor series
$$\log(a_n)=n \left(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)\right)=1-\frac{1}{2 n}+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ Continue with Taylor series
$$a_n=e^{\log(a_n)}=e\left(1-\frac{1}{2 n}+\frac{11}{24 n^2}\right)+O\left(\frac{1}{n^3}\right)$$
$$n\left(\frac{e}{\left(1+\frac{1}{n}\right)^n}-1\right)=n\left(\frac{e}{e\left(1-\frac{1}{2 n}+\frac{11}{24 n^2}\right)+O\left(\frac{1}{n^3}\right)}-1\right)$$ Now, long division to get
$$n \left(\frac{1}{2 n}-\frac{5}{24 n^2}+O\left(\frac{1}{n^3}\right) \right)=\frac{1}{2 }-\frac{5}{24 n}+O\left(\frac{1}{n^2}\right)$$ which gives the limit and also how it is approached.
Using your pocket calculator for $n=10$, the "exact" value would be $0.4802$ while the truncated expression given above leads to $\frac{23}{48}\approx 0.4792$.
|
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|
Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$ Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$
My attempt is as follows:
As $x>1,y>1$ , so $\log_e x>0, \log_e y>0$, hence we can apply $AM>=GM$
$$\dfrac{log_e x+log_e y}{2}>=\sqrt{\log_e x\log_e y}$$
As both the sides are positive, so we can square both the sides without breaking the inequality.
$$(\log_e x)^2+(\log_e y)^2+2\log_e x\log_e y>=4\log_e x\log_e y$$
Using the given condition $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$
$$(\log_e x^2)+(\log_e y^2)>=2\log_e x\log_e y$$
$$(\log_e x)+(\log_e y)>=\log_e x\log_e y$$
$$\log_e xy>=\log_e x^{\log_e y}$$
As $e>1$, so we can safely write $xy>=x^{\log_e y}$
But actual answer is $e^4$, I am not able to think of any other way. Please help me in this.
|
Let $\ln{x}=a$ and $\ln{y}=b$.
Thus, $a$ and $b$ are positives, $$a^2+b^2=2(a+b)$$ and by C-S
$$2(a+b)=\frac{1}{2}(1+1)(a^2+b^2)\geq\frac{1}{2}(a+b)^2.$$
Thus, $$a+b\leq4$$ and by AM-GM:
$$2\sqrt{ab}\leq a+b\leq4,$$ which gives $$ab\leq4.$$
Id est, $$x^{\ln{y}}=e^{\ln{x}\ln{y}}=e^{ab}\leq e^4.$$
The equality occurs for $x=y=e^2,$ which says that we got a maximal value.
C-S it's the following.
For any $a_i$ and $b_i$ we have:
$$(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\geq(a_1b_1+a_2b_2+...+a_nb_n)^2.$$
In our case $n=2$, $a_1=a_2=1$, $b_1=a$ and $b_2=b$.
Thus, $$2(a^2+b^2)=(1^2+1^2)(a^2+b^2)\geq(1\cdot a+1\cdot b)^2=(a+b)^2.$$
By the way, also, you can get the last inequality by the following way:
$$2(a^2+b^2)-(a+b)^2=(a-b)^2\geq0.$$
|
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|
Coefficient of $x^4y^3z^3$ in the expansion of $(5x+y-4z)^{10}$ The coefficient of $x^6y^4$ in the expansion of $(2x-3y)^{10}$ is
$$_{10}C_6 \cdot 2^6 \cdot (-3)^4$$
and as for the coefficient of $x^3y^4z^8$ in the expansion of $(x+y+z)^{15}$ is
$$_{15}C_3 \cdot _{15}C_4 \cdot _{15}C_8 = \frac{15!}{3!4!8!} $$
What would be the the coefficient if the case would be $x^4y^3z^3$ in the expansion of $(5x+y-4z)^{10}$
Thank You
With Respect Umer Selmani
|
We have that by trinomial expansion
$$(5x+y-4z)^{10} =\ldots+\frac{10!}{4!3!3!} (5x)^4y^3(-4z)^3+\ldots$$
therefore the coefficient for $x^4y^3z^3$ is equal to $-\frac{10!5^34^3}{4!3!3!}$.
|
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|
Least value of unit vector $|a+b|^2+|b+c|^2+|c+a|^2$ If $a, b, c$ are unit vectors, then least value of $|a+b|^2+|b+c|^2+|c+a|^2$ will be equal to
(1) 1
(2) 3
(3) 9
(4) 12
If am using the concept $a=c=-b$, I am getting the answer 4, but not matching with options provided
|
Writing in terms of inner product, you have that
\begin{align*}
|a+b|^2+|b+c|^2+|c+a|^2
&= 2(|a|^2 + |b|^2 + |c|^2) + 2 (a \cdot b + b \cdot c + c \cdot a) \\
&= 6 + 2 (a \cdot b + b \cdot c + c \cdot a)
\end{align*}
Thus, your problem amounts to minimizing the sum of inner products above. Considering that $a, b$ are fixed, notice that the choice of $c$ that minimizes the expression is taking $c$ at the opposite direction of $a + b$, since
$$
a \cdot b + b \cdot c + c \cdot a = a \cdot b + (a+b) \cdot c
$$
and we have
$$
|(a+b)\cdot c| = |a+b| \cos \theta
$$
with $\theta$ being the angle between $a+b$ and $c$. Thus, a true minimizer should satisfy that $a+b$ has the opposite direction of $c$ and the same will happen when you permute $a, b, c$. This implies that we want $a, b, c$ as vertices of an equilateral triangle. This means
$$
|a+b|^2+|b+c|^2+|c+a|^2 = 6 + 2 \cdot 3 \cdot \cos \frac{2\pi}{3} = 3.
$$
|
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|
Cosine of a $2 \times 2$ non-diagonalisable matrix Given
$$A = \begin{bmatrix}
\pi-1 & 1\\
-1 & \pi+1
\end{bmatrix}$$
I need to calculate its cosine, $\cos(A)$. Typically, I use diagonalisation to approach this type of problems:
$$\cos(A) = P \cos(D) P^{-1}$$
However, in this problem, the eigenvalues of the matrix are equal: $\lambda_1=\pi, \lambda_2=\pi$. Thus, there are no two linearly independent vectors and the method will not work.
Are there any alternative approaches?
Besides MacLaurin series expansion of $\cos(A)$, which does not work either since $A$ does not turn into a zero matrix at some point when multiplied by itself $n$ times.
|
When working this sort of thing out I usually take powers of the matrix and look for a pattern. In this case I found,
$$ M^0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$ M^1 = \begin{bmatrix} \pi -1 & 1 \\ -1 & \pi+1 \end{bmatrix}$$
$$ M^2 = \begin{bmatrix} \pi^2 - 2\pi & 2\pi \\ -2\pi & \pi^2 + 2\pi\end{bmatrix}$$
$$ M^3 = \begin{bmatrix} \pi^3 - 3\pi & 3\pi^2 \\ -3\pi & \pi^3 + 3\pi^2\end{bmatrix}$$
$$ M^4 = \begin{bmatrix} \pi^4 - 4\pi^3 & 4\pi^3 \\ -4\pi^3 & \pi^4 + 4\pi^3 \end{bmatrix},$$
This leads me to suspect that for $p\geq 0$ we have,
$$ \boxed{M^p = \begin{bmatrix} \pi^p - p \pi^{p-1} & p \pi^{p-1} \\ -p \pi^{p-1} & \pi^p + p \pi^{p-1} \end{bmatrix}}$$
This can be confirmed by a simple induction proof. Now we can evaluate $\cos(M)$ using the series definition of $\cos$
$$ \cos(M) = \sum_{p=0}^\infty (-1)^p\frac{M^{2p}}{(2p)!}$$
$$ = \sum_{p=0}^\infty \frac{(-1)^p}{(2p)!}\begin{bmatrix} \pi^{2p} - 2p \pi^{2p-1} & 2p \pi^{2p-1} \\ -2p \pi^{2p-1} & \pi^{2p} + 2p \pi^{2p-1} \end{bmatrix}$$
Bringing the summation inside the matrix entries we get the following,
$$ \cos(M) = \begin{bmatrix} \sum_{p=0}^\infty \frac{(-1)^p \pi^{2p}}{(2p)!} - \sum_{p=0}^\infty \frac{(-1)^p 2p \pi^{2p-1}}{(2p)!} & \sum_{p=0}^\infty \frac{(-1)^p 2p \pi^{2p-1}}{(2p)!} \\ - \sum_{p=0}^\infty \frac{(-1)^p 2p \pi^{2p-1}}{(2p)!} &
\sum_{p=0}^\infty \frac{(-1)^p \pi^{2p}}{(2p)!} + \sum_{p=0}^\infty \frac{(-1)^p 2p \pi^{2p-1}}{(2p)!} \end{bmatrix} $$
So we need to evaluate the following sums,
$$ S_1 = \sum_{p=0}^\infty \frac{(-1)^p \pi^{2p}}{(2p)!} = \cos(\pi),$$
and
$$ S_2 = \sum_{p=0}^\infty \frac{(-1)^p 2p \pi^{2p-1}}{(2p)!} = \frac{d}{d\pi} \sum_{p=0}^\infty \frac{(-1)^p \pi^{2p}}{(2p)!} = -\sin(\pi)$$
We now have
$$ \cos(M) = \begin{bmatrix} \cos(\pi) - \sin(\pi) & \sin(\pi) \\ -\sin(\pi) & \cos(\pi) + \sin(\pi) \end{bmatrix} $$
|
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|
compute $A^2$ and $A^6$. $A= \begin{bmatrix} -1 & 1 & 1 & -1\\ 1 & -1 & -1 & 1\\ 1 & -1 & -1 & 1\\ -1 & 1 & 1 & -1 \end{bmatrix} $ . Fuzhen Zhang's linear algebra, problem 3.11
$A= \begin{bmatrix}
-1 & 1 & 1 & -1\\
1 & -1 & -1 & 1\\
1 & -1 & -1 & 1\\
-1 & 1 & 1 & -1
\end{bmatrix} $
compute $A^2$ and $A^6$.
The answers are $-4A$ and $-2^{10} A$, respectively.
I have no clue how to calculate higher powers.
Thanks!
|
We can observe $$A_4 = \begin{bmatrix}A_2&-A_2\\-A_2&A_2\end{bmatrix}$$
where
$$A_2 = -\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$$
Which we see follow the same pattern (except minus sign). So this is nothing but Kronecker product of $A_2$ on $A_2$. Since it is smaller, we can investigate this $A_2$ more readily and see that it has eigenvalues $\lambda(A_2) = \{0,2\}$ and eigenvector (of course) $[1,-1]^T$.
Then the Kronecker laws of eigenvalue propagation tells us the eigenvalues for the $A_4$ shall be all possible products of $\{0,2\}$ on itself, these are the four combinations : $$\lambda(A_4) = \lambda(A_2 \otimes A_2) = \lambda(A_2) \otimes \lambda(A_2) = \{0,2\} \otimes \{0,2\} = \\\phantom{a}\\= \{0\cdot 0, 0\cdot 2, 2\cdot 0, 2\cdot 2\} = \{0,0,0,4\}$$So the eigenvalue $4$ is the only we need to worry about.
Now we can directly calculate what it will be. Multiply by $4$ times exponent (minus 1).
$$4\cdot (2 - 1) = 4$$
$$4\cdot (6 - 1) = 4\cdot 5 = 2 \cdot 10$$
And by law of exponents we know $4^{5} = 2^{10}$
|
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|
Prove that when x approaches to 1-, the function appraochs to negative infinity Prove that
$$\lim_{x\rightarrow1-}f(x):=\lim_{x\rightarrow1-}\frac{x+2}{2x^2-3x+1}=-\infty.$$
Using the basic definition.
This is the proof from the textbook:
Let $M\in R$ and assume M<0 without losing generality. As x converges to 1 from the left-hand side, $2x^2-3x+1$ is negative and approaches to 0 as x approaches to 1.
By observing that x has roots 1/2 and 1 and is a parabola opening upward, choosing $\delta\in(0,1)$ such that $1-\delta<x<1$ implies $2/M<2x^2-3x+1<0$;
That is, $-1/(2x^2-3x+1)>-M/2>0$. Since $0<x<1$ also implies $2<x+2<3$, it follows that $-(x+2)/(2x^2-3x+1)>-M$ ; that is,
$$f(x)=\frac{x+2}{2x^2-3x+1}<M$$
for all $1-\delta<x<1$.
I can follow the proof step by step but I have no idea why this question is proved in this way, the "choosing $\delta\in(0,1) $ such that... implies $2/M<2x^2-3x+1<0$" seems jump from no where. Could someone please tell me what's the reasoning behind each step? That is how poeople come up with the idea to solve the problem this way?
|
I like to let variables
go to zero.
So, in
$\lim_{x\rightarrow1-}\frac{x+2}{2x^2-3x+1}
$
I would let
$x = 1-y$.
Then
$x+2 = 1-y+2
=3-y$
and
$\begin{array}\\
2x^2-3x+1
&=2(1-y)^2-3(1-y)+1\\
&=2(1-2y+y^2)-3+3y+1\\
&=2-4y+2y^2-3+3y+1\\
&=-y+2y^2\\
&=y(-1+2y)\\
\end{array}
$
so
$\lim_{x\rightarrow1-}\frac{x+2}{2x^2-3x+1}
=\lim_{y\rightarrow 0}\frac{3-y}{y(-1+2y)}
$
and it is easy to see why this diverges.
Note that
$\lim_{y\rightarrow 0}\frac{3-y}{-1+2y}
=-3
$.
|
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|
find sum of $\sum_{n=1}^{\infty}\frac{(-1)^n}{n-(-1)^n}$ I want to find the exact sum of this expression: $\sum_{n=1}^{\infty}\frac{(-1)^n}{n-(-1)^n}$
I've already proved that it converges by condition.
Also, I think that it's sort of telescoping series,
because if I open it I get:
$\frac{-1}{2}+\frac{1}{1}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}+\frac{1}{5}...$
But I cant think of way to calculate it.
|
Hint. Note that if $N$ is even then
$$\begin{align}S_N&=\sum_{n=1}^{N}\frac{(-1)^n}{n-(-1)^n}=
-\frac{1}{2}+\frac{1}{1}-\frac{1}{4}+\frac{1}{3}+\dots-\frac{1}{N}+\frac{1}{N-1}\\&=
\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{N-1}-\frac{1}{N}=\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n},
\end{align}$$
and
$$S_{N+1}=S_N+\frac{(-1)^{N+1}}{N+1-(-1)^{N+1}}=\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n}
-\frac{1}{N+2}.$$
|
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|
Probability of getting a pair of symbols when throwing four dice marked with swords, helmets, and skulls I have 4 dices with those faces display:
1-2-3 = sword,
4-5 = helmet,
6 = skull
When rolling 4 dice, what are the probabilities of getting a PAIR of swords ? helmets ? skulls ?
I tried to solve this by using a regular dice at the beginning, by calculating the probability of doing at least one pair.
1 - (6/6)(5/6)(4/6)(3/6) = 26/36 = 13/18
But in our case I want a pair of:
*
*skulls with 1/6, so 13/18 * 1/6 = 13/108
*helmet with 1/3, so 13/18 * 1/3 = 13/54
*sword with 1/2, so 13/18 * 1/2 = 13/36
But I am pretty sure I am mistaken
|
I'm assuming you mean getting the probability of two swords, of two helmets, and of two skulls separately. Here, the individual probabilities are:
$P(Sword) = \frac{3}{6} = \frac{1}{2}$
$P(Helmet) = \frac{2}{6} = \frac{1}{3}$
$P(Skull) = \frac{1}{6}$
Here, we use combination instead of permutation for we do not care about the order of desired outcome appearing. Since we want the outcome to occur twice out of four rolls, our combination value is $4C2 = 6$. Hence,
$P(2 Swords) = \frac{1}{2}\times\frac{1}{2}\times(1-\frac{1}{2})\times(1-\frac{1}{2})\times6=\frac{3}{8}$
$P(2 Helmets) = \frac{1}{3}\times\frac{1}{3}\times(1-\frac{1}{3})\times(1-\frac{1}{3})\times6=\frac{8}{27}$
$P(2 Skulls) = \frac{1}{6}\times\frac{1}{6}\times(1-\frac{1}{6})\times(1-\frac{1}{6})\times6=\frac{25}{216}$
|
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|
From a point on a given circle, tangents are drawn to the ellipse. Need to find locus of chord of contact. From a point $O$ on the circle $x^2+y^2=d^2$, tangents $OP$ and $OQ$ are drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b$. Show that the locus of the midpoint of chord PQ is given by $$x^2+y^2=d^2\bigg[\frac{x^2}{a^2}+\frac{y^2}{b^2}\bigg]^2$$
I recognize that the locus of a chord whose midpoint is at $(h,k)$ is given by $\frac{xh}{a^2}+\frac{yk}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$
I also recognize that PQ is the chord of contact, but to find its equation using the chord of contact formula I would require the coordinates of point O which I do not have.
Here I am getting the equation in terms of $x,y,h,k$, but to find the locus I need the equation entirely in the form of $h,k$, right? So how do I eliminate $x,y$ from the equation of the locus of the midpoint?
|
Still not what you want, but uses Joachimsthals notations.
The locus is the midpoint of the two intersection points of $$s=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$$ and (from $s_1^2=s \cdot s_{11}$) $$(\frac{x(O)x}{a^2}+\frac{y(O)y}{b^2}-1)^2=(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1)(\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}-1)$$ which (since the square roots cancel in $\frac{x_1+x_2}{2}$ and $\frac{y_1+y_2}{2}$) is $$(x,y)=(\frac{x(O)}{\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}},\frac{y(O)}{\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}}),$$ where $x(O)^2+y(O)^2=d^2$ since $O$ is on that circle.
Writing $h=x(O), k=y(O)$ into M2
R=QQ[a,b,d]
S=R[h,k,x,y,MonomialOrder=>Eliminate 2]
I=ideal(h^2+k^2-d^2,(b^2*h^2+a^2*k^2)*x-a^2*b^2*h,(b^2*h^2+a^2*k^2)*y-a^2*b^2*k)
gens gb I
yields $$a^6b^6d^2(d^2(\frac{x^2}{a^2}+\frac{y^2}{b^2})^2-(x^2+y^2)).$$
|
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|
Evaluate $\lim_{x\to 0} \cot ^2 (x)-\frac{1}{x^2}$
Evaluate $\lim\limits_{x\to 0} \cot ^2 (x)-\dfrac{1}{x^2}.$
I was thinking of using L'Hôpital's Rule, but things got very, very ugly so I wasn't able to solve it. I know from experimentation that the limit is $-\dfrac{2}{3}$ though.
edit: i do not want to use a taylor expansion to solve this. That's too easy.
|
Unacceptably Easy Answer (Before Question Edit)
$$
\begin{align}
\cot^2(x)
&=\frac{\cos^2(x)}{\sin^2(x)}\\
&=\frac{\left(1-\frac12x^2+O\!\left(x^4\right)\right)^2}{\left(x-\frac16x^3+O\!\left(x^5\right)\right)^2}\\
&=\frac1{x^2}\frac{1-x^2+O\!\left(x^4\right)}{1-\frac13x^2+O\!\left(x^4\right)}\\
&=\frac1{x^2}\left(1-\frac23x^2+O\!\left(x^4\right)\right)\\[6pt]
&=\frac1{x^2}\color{#C00}{-\frac23}+O\!\left(x^2\right)
\end{align}
$$
A More Difficult Approach Employing L'Hôpital
$$
\begin{align}
\lim_{x\to0}\left(\cot^2(x)-\frac1{x^2}\right)
&=\lim_{x\to0}\left(\frac{\cos^2(x)}{\sin^2(x)}-\frac1{x^2}\right)\\
&=\lim_{x\to0}\frac{x^2\cos^2(x)-\sin^2(x)}{x^2\sin^2(x)}\\
&=\lim_{x\to0}\frac{x^2-x^2\sin^2(x)-\sin^2(x)}{x^2\sin^2(x)}\\
&=\lim_{x\to0}\frac{(x-\sin(x))(x+\sin(x))}{x^2\sin^2(x)}-1\\
&=\lim_{x\to0}\frac{x-\sin(x)}{x^3}\frac{x+\sin(x)}{x}\frac{x^2}{\sin^2(x)}-1\\
&=\underbrace{\lim_{x\to0}\frac{x-\sin(x)}{x^3}}_{\text{L'Hôpital}\,\times\,3}\underbrace{\left(1+\lim_{x\to0}\frac{\sin(x)}x\right)}_{\text{L'Hôpital}\,\times\,1}{\underbrace{\left(\lim\limits_{x\to0}\frac{\sin(x)}{x}\right)}_{\text{L'Hôpital}\,\times\,1}}^{-2}-1\\[3pt]
&=\frac16\cdot2\cdot1-1\\[3pt]
&=-\frac23
\end{align}
$$
|
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|
Lindquist Identity:$\big(|b|^{p-2}b-|a|^{p-2}a\big)(b-a)=\frac{|b|^{p-2}+|a|^{p-2}}{2}|b-a|^2+\frac{1}{2}\big(|b|^{p-2}-|a|^{p-2}\big)(|b|^2-|a|^2)$ When investigating the properties of the one-dimensional "French tower" function
$$
g: \mathbb{R} \to \mathbb{R}, \
z \mapsto \begin{cases}
| z |^{p - 2} z, & \text{if } z \ne 0, \\
0, & \text{else.}
\end{cases}
$$
for $p \in (1, \infty)$ these lecture notes (in German, Beispiel 13.5) mention a certain Lindquist identity, stating
$$
\big( | b |^{p - 2} b - | a |^{p - 2} a\big) (b - a)
= \frac{| b |^{p - 2} + | a |^{p - 2}}{2} | b - a |^2 + \frac{1}{2}\big(| b |^{p - 2} - | a |^{p - 2}\big)(| b |^2 - | a |^2).
$$
Where does this come from and is this related to the Swedish mathematician Anders Lindquist?
|
The RHS becomes
$$
| b |^{p} - | b |^{p - 2} a \cdot b - | a |^{p - 2} a \cdot b + | a |^{p}.
$$
The LHS is
\begin{align}
& \frac{1}{2} | b |^{p} - | b |^{p - 2} a \cdot b + \frac{1}{2} | b |^{p - 2} | a |^2 + \frac{1}{2} | a |^{p - 2} | b |^{2} - | a |^{p - 2} a \cdot b + \frac{1}{2} | a |^p \\
+ & \frac{1}{2} | b |^{p} - \frac{1}{2} | b |^{p - 2} | a |^2 - \frac{1}{2} | a |^{p - 2} | b |^2 + \frac{1}{2} | a |^p.
\end{align}
Now group similar terms to see the equality.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Coincidence? $\left(\frac 1e\right)^{\frac 1e}\approx \ln 2$ Is it a coincidence that
$$\color{lightgrey}{0.6922\cdots =}\left(\frac 1e\right)^{\frac 1e}\approx \ln 2\color{lightgrey}{=0.6931\cdots}$$
?
|
I'm not sure if this helps, but I have found that the Taylor series for these values end up being quite similar.
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\ldots,$$
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\ldots\,.$$
$\ln(2)=\ln(1+1)$ and $(\frac{1}{e})^\frac{1}{e}=e^\frac{-1}{e}$.
Therefore,
$$\ln(2)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots,$$
$$\left(\frac{1}{e}\right)^\frac{1}{e}=1-\frac{1}{e}+\frac{1}{2!e^2}-\frac{1}{3!e^3}+\frac{1}{4!e^4}-\frac{1}{5!e^5}+\ldots\,.$$
|
{
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"url": "https://math.stackexchange.com/questions/3434183",
"timestamp": "2023-03-29T00:00:00",
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|
Probability Question: Find $a$ and $b$ given expected value and density function The density function of $X$ is given by
\begin{align*}
f(x) =\begin{cases}
a+bx^2, & \text{if}\,\,0\leq x\leq 1\\
0, & \text{otherwise}
\end{cases}
\end{align*}
Once $\textbf{E}(X) = 4.25$, find $a$ and $b$
|
Using legitimacy,
$$\int\limits_{0}^{1} f(x) dx = 1$$
$$\int\limits_{0}^{1} (a + b x^{2}) dx = 1$$
$$ \bigg[ax + \frac{bx^{3}}{3}\bigg]_{0}^{1} = 1$$
$$ a + \frac{b}{3} = 1$$
$$ 3a + b = 3$$
Using expectation,
$$\int\limits_{0}^{1} x f(x) dx = 4.25$$
$$\int\limits_{0}^{1} (ax + b x^{3}) dx = 4.25$$
$$ \bigg[\frac{ax^{2}}{2} + \frac{bx^{4}}{4}\bigg]_{0}^{1} = 4.25$$
$$ \frac{a}{2} + \frac{b}{4} = 4.25$$
$$ 2a + b = 4 \times 4.25$$
$$ 2a + b = 17 $$
Can you take it from here?
EDIT: Turns out that after solving the equations for $a$ and $b$, $f$ takes negative values on $[0, 1]$, which makes it a wrong density function. Pointed out by @Michael Hoppe
|
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|
If $f(x)$ is a polynomial of degree $4$ and $f(n)=n+1$ for $n=1,2,3,4$. Find $f(5)$ Question: If $f(x)$ is a polynomial of degree $4$ and $f(n)=n+1$ for $n=1,2,3,4$. Find $f(5)$
If we construct $g(x)=f(x)-(x-2)(x-3)(x-4)(x-5)$, then is it possible to find f(5)?
|
$f(x)=x+1+c(x-1)(x-2)(x-3)(x-4)$ satisfies the hypothesis for any $c \neq 0$. So it is not possible to find $f(5)$ from the given information.
|
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"timestamp": "2023-03-29T00:00:00",
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|
taylor series approximations of $\ln x$
Compare the errors for the following methods for computing $\ln 2$. Which one is best?
$1.\; P_{n,1}(2)$
$2.\; -P_{n,1}(0.5)$
$3.\;P_{n,1}(\frac{4}{3})-P_{n,1}(\frac{2}{3})$
The Taylor series for $\ln x$ centred at $a=1$ is given by $P_{n,1}(x) = (x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\frac{(x-1)^4}{4}+O(x^5).$ The error for the $n$th term is given by $f(x)-P_{n,1}(x) =\dfrac{f^{(n+1)}(x_0)}{(n+1)!}(x-1)^{n+1},$ where $1<x_0<x$. Since $f^{(n+1)}(x_0)=\dfrac{n!}{x^{n+1}},$ it is decreasing for $x\geq 1$ and hence it has a maximum value of $n!$ at $x=1$. Thus, the absolute value of the maximum error is given by $\dfrac{n!}{(n+1)!}|(x-1)|^{n+1}=\dfrac{|(x-1)|^{n+1}}{n+1}$. So the error for the $n$th term of $P_{n,1}(2)$ is $\dfrac{1}{n+1}$ while the error for the $n$th term of $-P_{n,1}(0.5)$ is $\dfrac{0.5^{n+1}}{n+1}<\dfrac{1}{n+1}\;\forall n\geq 1.$
Note that $\ln \frac{4}{3} - \ln \frac{2}{3} = \ln 2$ so $|\ln 2-(P_{n,1}(\frac{4}{3})-P_{n,1}(\frac{2}{3}))|\leq|\ln \frac{4}{3}-P_{n,1}(\frac{4}{3})|+|\ln\frac{2}{3} -P_{n,1}(\frac{2}{3})|$. Hence the maximum error for the $n$th term is $\dfrac{(\frac{1}{3})^{n+1}}{n+1}+\dfrac{(\frac{1}{3})^{n+1}}{n+1}<\dfrac{0.5^{n+1}}{n+1}$ for $n=1.$ Assume the inequality for some $k\in\mathbb{Z}, k\geq 1.$ Then $\dfrac{2(\frac{1}{3})^{k+2}}{k+2}=\dfrac{2(\frac{1}{3})^{k+1}}{k+1}\cdot \dfrac{1}{3}\cdot \dfrac{k+1}{k+2}<\dfrac{0.5^{k+1}}{k+1}\cdot 0.5 \cdot \dfrac{k+1}{k+2}=\dfrac{0.5^{k+2}}{k+2}$ (by hypothesis). Thus, by induction, the error for $P_{n,1}(\frac{4}{3})-P_{n,1}(\frac{2}{3})$ is smaller than that for $-P_{n,1}(0.5)$ for all $n\geq 1.$
So the best method for computing $\ln 2$ is $P_{n,1}(\dfrac{4}{3})-P_{n,1}(\dfrac{2}{3}).$
Is there anything wrong with this approach? Can I simplify it more?
|
Indeed, there is something wrong in your approach.
The explicit formula for the remainder of Taylor's theorem is given by
$$R_{n,1}(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-1)^{k+1}$$ where $\xi \in (1,x)$. Not by $\dfrac{|(x-1)^{n+1}|}{(n+1)}$ as you mentioned.
Based on that, you can reconsider your inequalities.
|
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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.