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Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$ I'm having trouble computing the integral:
$$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$
I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.
Any suggestions are welcome. Thanks.
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$$= \frac{1}{2} \cdot \int \frac{\sin(x) + \sin(x)}{\sin(x) + \cos(x)} dx = \frac{1}{2} \int \frac{\sin(x) + \cos(x) + \sin(x) - \cos(x) }{\sin(x) + \cos(x)} dx $$ $$= \frac{x}{2} - \frac{1}{2} \int \frac{\cos(x) - \sin(x)}{\sin(x) + \cos(x)} dx$$ Let $u = \sin(x) + \cos(x)$ $$ \implies \frac{x}{2} - \frac{1}{2} \cdot \log \left| \sin(x) + \cos(x) \right| + C$$
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
Possible Duplicate:
Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Prove trigonometry identity?
If $A$, $B$, and $C$ are to be taken as the angles of a triangle, then I beg someone to help me the proof of
$$\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}.$$
Thanks!
|
Since $C = \pi - A - B$, what needs to be shown is
$$\sin(A) + \sin(B) + \sin(\pi - A - B) = 4\cos({A \over 2})\cos({B \over 2})\cos({\pi \over 2} - {A + B \over 2})$$
Equivalently, you need that
$$\sin(A) + \sin(B) + \sin( A + B) = 4\cos({A \over 2})\cos({B \over 2})\sin( {A + B \over 2})$$
Using the trig identity $\cos(x)\sin(y) = {1 \over 2}(\sin(x + y) + \sin(y - x))$, this is the same as
$$\sin(A) + \sin(B) + \sin( A + B) = 2\cos({A \over 2})\sin({A \over 2} + B) + 2\cos({A \over 2})\sin({A \over 2}) $$
Using it again on the first term on the right, this becomes
$$\sin(A) + \sin(B) + \sin( A + B) = \sin(A + B) + \sin(B) + 2\cos({A \over 2})\sin({A \over 2}) $$
This collapses into the sine double angle identity
$$\sin(A) = 2\cos({A \over 2})\sin({A \over 2}) $$
Reversing the above steps gives the desired result.
|
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How do I transform the equation based on this condition? If a and b are the roots of the equation $$2x^2-px+7=0$$ Then a-b is a root of ?
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So, $a+b=\frac{p}{2}$ and $ab=\frac{7}{2}$
If $y=a-b=>y^2=(a+b)^2-4ab$
$=>y^2=(\frac{p}{2})^2-4 \frac{7}{2}$
So, a-b is a root of $4y^2-p^2+56=0$
|
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Find the value of $x$ in the displayed figure Find $x$ in the following figure.
$AB,AC,AD,BC,BE,CD$ are straight lines.
$AE=x$, $BE=CD=x-3$, $BC=10$, $AD=x+4$
$\angle BEC=90^{\circ}$, $\angle ADC=90^{\circ}$
NOTE: figure not to scale.
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$∆BCE$ is right angle triangle.
Hence $BC^2 = BE^2 + EC^2$
$EC = \sqrt{(BC^2 - BE^2})= \sqrt{(100 - (x-3)^2)}$
$∆ACD$ is right angle triangle.
Hence $AC^2 = CD^2 + AD^2$
$(AE + EC)^2 = CD^2 + AD^2$
Substitute the values,
$(\sqrt{(100 - (x-3)^2)} + x)^2 = (x-3)^2 + (x+4)^2$
Then you can solve this equation easily for getting x value.
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Properties of Mediants If $\frac{a}{c} > \frac{b}{d}$, then the mediant of these two fractions is defined as $\frac{a+b}{c+d}$ and can be shown to lie striclty between the two fractions.
My question is can you prove the following property of mediants: if $|\frac{a}{c} - x| > |x - \frac{b}{d}|$ then $|b/d - mediant| < |mediant - x|$ for any $x$ that lies strictly between $\frac{a}{c}$ and $\frac{b}{d}$.
|
$|\frac{a}{c}-x|>|x-\frac{b}{d}|$
$=>(\frac{a}{c}-x)^2>(x-\frac{b}{d})^2$
$=>(\frac{a}{c}-\frac{b}{d})(\frac{a}{c}+\frac{b}{d}-2x)>0$
$=>2x<\frac{a}{c}+\frac{b}{d}=>-x>\frac{-1}{2}(\frac{a}{c}+\frac{b}{d})$
$\frac{b}{d}-mediant=\frac{b}{d}-\frac{a+b}{c+d}=\frac{bc-ad}{d(c+d)}=\frac{-c}{c+d}(\frac{a}{c}-\frac{b}{d})$
$mediant-x>\frac{a+b}{c+d} - \frac{1}{2}(\frac{a}{c}+\frac{b}{d})$
$2(mediant-x)>\frac{a+b}{c+d} - \frac{a}{c} + \frac{a+b}{c+d} - \frac{b}{d}$
$(mediant-x)>\frac{(bc-ad)(d-c)}{2cd(c+d)}=\frac{c-d}{2(c+d)}(\frac{a}{c}-\frac{b}{d})$
$|\frac{-c}{c+d}(\frac{a}{c}-\frac{b}{d})|>$ will be greater than |$\frac{c-d}{2(c+d)}(\frac{a}{c}-\frac{b}{d})|$
iff $|\frac{-c}{c+d}|>|\frac{c-d}{2(c+d)}|$
if $|-2c|>|c-d|$ when c+d>0 or if $|-2c|<|c-d|$ when c+d<0
if $2>|1-\frac{d}{c}|$ when c+d>0 or if $2<|1-\frac{d}{c}|$ when c+d<0
In the 1st case, $-2<1-\frac{d}{c}<2=>-3<-\frac{d}{c}<1=>-1<\frac{d}{c}<3$
c+d>0 which is true and d<3c
In the 2nd case, $1-\frac{d}{c}>2$ or $1-\frac{d}{c}<-2$
or $\frac{d}{c}<-1$ or $\frac{d}{c}>3$
or c+d<0 which is true or d>3c
So the conditions are $(c+d>0\ and\ d<3c)$ or $(c+d<0\ and\ d>3c)$ to make the inequality valid.
|
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Proving: $x\left ( 1-y \right )+y\left ( 1-z \right )+z\left ( 1-x \right )< 1$ What is the proof that:
$x\left ( 1-y \right )+y\left ( 1-z \right )+z\left ( 1-x \right )< 1$
if:
$0< x;y;z< 1$
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André’s solution is much nicer, but the problem can also be solved in a more routine fashion. Let $a=x+y$; then
$$\begin{align*}
x(1-y)+y(1-z)+z(1-x)&=x(1-a+x)+(a-x)(1-z)+z(1-x)\\
&=x^2-ax+a+(1-a)z\\
&=\left(x-\frac{a}2\right)^2+a+(1-a)z\;.\tag{1}
\end{align*}$$
If $a\le 1$, the third term is bounded above by $(1-a)\cdot1=1-a$, and since $0<x<a$, the first term is strictly bounded above by its value when $x=a$, so $(1)$ is strictly bounded above by $$\left(a-\frac{a}2\right)^2+a+1-a-\frac{a^2}4=1\;,$$ as desired.
If $a>1$, the third term is strictly bounded above by $0$, and $a-1<x<1$, so $(1)$ is strictly bounded above by
$$\left(a-1-\frac{a}2\right)^2+a-\frac{a^2}4=\left(\frac{a}2-1\right)^2+a-\frac{a^2}4=1\;,$$ again as desired.
The idea behind this approach is to see how the function varies with $z$ when $x$ and $y$ are held constant. Once quickly discovers that when $x+y=1$, its value is independent of $z$, when $x+y<1$ it increases as $z$ increases, and when $x+y>1$ it increases as $z$ decreases. This suggests fixing not $x$ and $y$, but $x+y$, evaluating the function at $z=1$ or $z=0$ to check that it’s bounded above by $1$, and then making sure that the bound is strict.
|
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|
Manipulating Exponents I'm doing my homework and there are a couple of things that I am having trouble grasping. All my homework asks is that I simplify the exponents. For example: 6^5 * 6^3 = 6^8
There are 2 problems I am unsure on what to do. They have to do with multiplying fractions:
(7^3)^5/6 = 7^5/2 is the answer but I don't understand how they got to that. I know there is a certain example I was shown where you had to get common demonators and such but if I apply that it doesn't turn out well. Same for this problem:
(7^3/5)^5/6 = 7^2
3^5/2 * 3^-2 = 3^1/2
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For any $x$, $a$, and $b$.
$(x^a)^b = x^{ab}$.
Hence for the first one, letting $x = 7$, $a = 3$, $b = \frac{5}{6}$, you have
$(7^3)^{\frac{5}{6}} = 7^{3 \cdot \frac{5}{6}} = 7^{\frac{5}{2}}$
For the second letting $x = 7$, $a = \frac{3}{5}$ and $b = \frac{5}{6}$, you have
$(7^{\frac{3}{5}})^{\frac{5}{6}} = 7^{\frac{3}{5} \cdot \frac{5}{6}} = 7^{\frac{3}{6}} = 7^{\frac{1}{2}}$.
For the last one, for any $x$, $a$ and $b$
$(x^a)(x^b) = x^{a + b}$.
Letting $x = 3$, $a = \frac{5}{2}$ and $b = -2$, you have
$(3^{\frac{5}{2}})(3^{-2}) = 3^{\frac{5}{2} + -2} = 3^{\frac{5}{2} + \frac{-4}{2}} = 3^{\frac{1}{2}}$.
|
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how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$
Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$
how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$
we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$
now, if $x=\sqrt[]{2+\sqrt{2}}$, then, $(x-\sqrt{2})(x+\sqrt{2})=\sqrt{2}$
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$x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\ldots }}}}}}$
$x^2 =2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\ldots }}}}}$
$x^2 =2+x$
$x^2-x-2=0$
$x^2-2x+x-2=0$
$(x-2)(x+1)=0$
we have, $x=2$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$ I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \frac{x(2x - 13) - 15(2x+3)}{x(2x + 3)} \lt 0$$
$$\iff \frac{2x^2 - 13x - 30x - 45}{2x(x + \frac{3}{2})} \lt 0 \iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 $$
$$\iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 \iff \frac{(x - \frac{43}{4})^2 - \frac{1939}{4}}{x(x+\frac{3}{2})} \lt 0$$
I don't know how to get any further, and I'm starting to get too high values to handle.
The next step as I can see would be to find an $x$ that makes $(x - \frac{43}{4})^2 = \frac{1939}{4}$. This, alongside the obvious ones for $x$ and $x+\frac{3}{2}$ (creating division by $0$), would help me find the possible values for $x$.
But how do I get the last step? Or am I already dead wrong?
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you have to resolve this inequality depending $x$ : $$\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$$
that is equivalent with: $\displaystyle \frac{x(2x-13)-15(2x+3)}{x(2x+3)} \lt 0$ so:
$\displaystyle \frac{2x^2-13x-30x-45}{2x^2+3x} \lt 0$ and now the simple form is :
$$\frac{2x^2-43x-45}{2x^2+3x} \lt 0$$
Now we have two equation . first: $2x^2-43x-45=0$ with : $\displaystyle x_{1,2}=-1 \mbox{and} \frac{45}{2}$
and second $2x^2+3x=0$ with $\displaystyle x_{1,2}=-\frac{3}{2} \mbox{and} 0$.
For that inequality to be satisfied it must: $\displaystyle 2x^2-43x-45 \lt0$ and $\displaystyle 2x^2+3x \gt 0$ (1) OR $2x^2-43x-45 \gt 0$ and $2x^2+3x \lt 0$ (2).
I attach a draw. it is easy now to view the solutions
for first case: $2x^2-43x-45 \lt 0$ when $\displaystyle x \in (-1, \frac{45}{2})$ and $2x^2+3x \gt 0$ equivalent with $x\in (-\infty, -\frac{3}{2}) \cup (0, \infty)$. Now the solution for this case is the intersection :
$$ (-1, \frac{45}{2}) \cap( (-\infty, -\frac{3}{2}) \cup (0, \infty))=(0, \frac{45}{2}) $$ and now the answer for second case is: $x^2-43x-45 \gt 0$ and $2x^2+3x \lt 0$. We have for first inequality $x\in (-\infty, -1) \cup (\frac{45}{2}, \infty)$ and for second inequality we have $x \in (-\frac{3}{2},0)$. So the answer for second case is :
$$(-\frac{3}{2},0) \cap((-\infty, -1) \cup (\frac{45}{2}, \infty))=(-\frac{3}{2},-1)$$
And the final answer is $$x \in (-\frac{3}{2}, -1) \cup (0, \frac{45}{2}).$$
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Integrating $ \int \frac{1}{1-\tan(x)}dx $ $$ \int \frac{dx}{1-\tan(x)} $$
Please help me to solve this problem as I'm trying this since last 1 day...
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An alternative is to use the substitution $t=\tan x$ and then expand into partial fractions.
$$\begin{equation*}
dt=\left( 1+\tan ^{2}x\right) dx=\left( 1+t^{2}\right) dx
\end{equation*}$$
$$\begin{equation*}
I=\int \frac{1}{1-\tan x}dx=\int \frac{1}{\left( 1-t\right) \left(
1+t^{2}\right) }\,dt.
\end{equation*}$$
Since
$$\begin{equation*}
\frac{1}{\left( 1-t\right) \left( 1+t^{2}\right) }=\frac{1}{2\left(
1-t\right) }+\frac{1}{2}\frac{t}{1+t^{2}}+\frac{1}{2}\frac{1}{1+t^{2}},
\end{equation*}$$
we have
$$\begin{eqnarray*}
I &=&\frac{1}{2}\int \frac{1}{1-t}\,dt+\frac{1}{2}\int \frac{t}{1+t^{2}}\,dt+
\frac{1}{2}\int \frac{1}{1+t^{2}}\,dt \\
&=&-\frac{1}{2}\ln \left\vert 1-t\right\vert +\frac{1}{4}\ln \left\vert
1+t^{2}\right\vert +\frac{1}{2}\arctan t+C \\
&=&-\frac{1}{2}\ln \left\vert 1-\tan x\right\vert +\frac{1}{4}\ln \left\vert
1+\tan ^{2}x\right\vert +\frac{x}{2}+C \\
&=&-\frac{1}{2}\ln \left\vert 1-\tan x\right\vert +\frac{1}{2}\ln \left\vert
\sec x\right\vert +\frac{x}{2}+C.
\end{eqnarray*}$$
This can be written as
$$\begin{eqnarray*}
I &=&-\frac{1}{2}\ln \left\vert \frac{1-\tan x}{\sec x}\right\vert +\frac{x}{
2}+C \\
&=&-\frac{1}{2}\ln \left\vert \cos x-\sin x\right\vert +\frac{x}{2}+C.
\end{eqnarray*}$$
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|
Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities Please help to solve the following inequality using rearrangement inequalities.
Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that
\begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\end{equation}
Thanks.
|
Homogenize the given problem into,
$$\frac{\sqrt{ab}}{a+3b}+\frac{\sqrt{ab}}{b+3a}\leq \frac 12.$$
Now note that, using the AM-GM inequality we have $a+3b\geq 2\sqrt{2b(a+b)},$ so that $$\dfrac{\sqrt{ab}}{a+3b}\leq \frac{\sqrt{a}}{2\sqrt{2(a+b)}}.$$
Hence it suffices to check that $\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{2(a+b)}}\leq 1,$ which is perfectly true from the Cauchy-Schwarz inequality.
$\Box$
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|
Simplify $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$.
Denest $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$.
I have tried completing square by several method but all failed. Can anyone help me please? Thank you.
p.s. I'm a poor question-tagger.
|
There is a formula. I'm not too sure how to prove it, but I know that there is a formula where you can denest $$\sqrt{\sqrt[3]{\alpha}+\sqrt[3]{\beta}}$$Into$$\pm\frac {1}{\sqrt{f}}\left(-\frac {s^2\sqrt[3]{\alpha^2}}{2}+s\sqrt[3]{\alpha\beta}+\sqrt[3]{\beta^2}\right)$$ where $$f=\beta-s^3\alpha$$ and $s$ is a real number solution to $f(x)=x^4+4x^3+8\frac {\beta}{\alpha}x-4\frac {\beta}{\alpha}$
So in this case, $\alpha=5$ and $\beta=-4$. So $s=-2$ and $f=-4-(-2)^3\times 5=36$ Therefore, we have$$\pm\frac {1}{6}\left(-\frac {4\sqrt[3]{25}}{2}-2\sqrt[3]{-20}+\sqrt[3]{16}\right)=\pm\frac {1}{3}\left(-\sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{2}\right)$$
Discard the negative value to get $\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}=\frac {1}{3}\left(-\sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{2}\right)$
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|
Inequality with unusual constraint $a,b,c\in (0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$ Suppose as in the title that $a,b,c$ are three real positive numbers in $(0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$. Then I was asked to prove that $$a+b+c\leq \frac 32.$$ I am not very good at inequalities, so can anybody help me with this task? Especially the first condition seems very unusual to me. Thanks.
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It is easy to see that the equality is achieved at a=b=c=1/2. Therefore, let's expand around that point, i.e. define $x,y,z$ by $a=x+1/2$, $b=y+1/2$, $c=z+1/2$. Then, the first equality becomes
$$1=2abc+a^2+b^2+c^2\\
\Longleftrightarrow 0=2xyz+xy+yz+xz+x^2+y^2+z^2+\frac{3}{2}(x+y+z)$$
Next, Jensen's inequality gives you that the geometric mean is smaller than the arithmetic mean, i.e.
$$(x^2y^2z^2)^{1/3}\le \frac{x^2+y^2+z^2}{3}\\
\Longleftrightarrow |xyz|\le \frac{x^2+y^2+z^2}{3}\sqrt{\frac{x^2+y^2+z^2}{3}}\le\frac{x^2+y^2+z^2}{6}$$
where the last inequality follows from the fact that $-1/2<x,y,z<1/2$. Therefore, going back to the equality above
$$0\ge -\frac{x^2+y^2+z^2}{3}+xy+yz+xz+x^2+y^2+z^2+\frac{3}{2}(x+y+z)\ge\\
\ge -\frac{x^2+y^2+z^2}{2}+xy+yz+xz+x^2+y^2+z^2+\frac{3}{2}(x+y+z)=\\
=\frac{2xy+2xz+2zy+x^2+y^2+z^2}{2}+\frac{3}{2}(x+y+z)\\
\Longleftrightarrow x+y+z\le -\frac{(x+y+z)^2}{3}\le 0$$
Thus,
$$a+b+c=3/2+x+y+z\le 3/2$$
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Calculate $\lim_{x\to\infty}\frac{\ln(x^2-x+1)}{\ln(x^{10}+x+1)}$ Please help me for solving $$\lim_{x\to\infty}\frac{\ln(x^2-x+1)}{\ln(x^{10}+x+1)}$$
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$\lim_{x\to\infty}\frac{\ln(x^2-x+1)}{\ln(x^{10}+x+1)}$=$\lim_{x\to\infty}\frac{\ln x^2(1-\frac{1}{x}+\frac{1}{x^2})}{\ln x^{10} (1+\frac{1}{x^9}+\frac{1}{x^{10}})}$=$\lim_{x\to\infty}\frac{\ln x^2+\ln(1-\frac{1}{x}+\frac{1}{x^2})}{\ln x^{10}+\ln(1+\frac{1}{x^9}+\frac{1}{x^{10}})}$=$\lim_{x\to\infty}\frac{\ln x^2+\ln 1}{\ln x^{10}+\ln 1}$=$\lim_{x\to\infty}\frac{\ln x^2}{\ln x^{10}}$=$\lim_{x\to\infty}\frac{2\ln x}{10\ln x}$=$\frac{1}{5}$.
|
{
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|
triangles and trigonometry A triangle has sides $a,b,c$ and angles $\alpha,\beta,\gamma$ such that:
$$ a \,\cos\beta + b \, \cos\gamma+ c \, \cos\alpha = \frac{a+b+c}{2}$$
Prove that the triangle is isosceles.
I tried writing $\cos$ in terms of the sides (using the cosine theorem),
for example $ \cos\alpha= \frac{a^2-b^2-c^2}{2ab}$.
I get the following equality:
$a^3(b-c)+b^3(c-a)+c^3(a-b)=0$
Maybe if I use the triangle inequality in a smart way, I can prove it, but I don't know.
|
To continue your proof, write $$a^3\left(b-c\right)+b^3\left(c-b+b-a\right)+c^3\left(a-b\right)=0$$ which implies $$\left(a^3-b^3\right)\left(b-c\right)=\left(b^3-c^3\right)\left(a-b\right).$$
If $b\neq a$ and $b\neq c$ then we can simplify the equation to $$a^2+ab+b^2=b^2+bc+c^2$$ which is equivalent to $$\left(a-c\right)\left(a+c-b\right)=0.$$ Since $a$, $b$, $c$ are sides of a triangle $a+c-b>0$ and thus we must have $a=c$.
|
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|
Inequality. $a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2} \leq 3\sqrt{3}$ Could you help me please with the following inequality
Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3.$ Prove that:
$$a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2} \leq 3\sqrt{3}.$$
|
$$a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2}=\sqrt{a}\sqrt{2ba+c^2a}+\sqrt{b}\sqrt{2bc+a^2b}+\sqrt{c}\sqrt{2ac+b^2c}$$
Using Cauchy-Schwarz we obtain :
$$\sqrt{a}\sqrt{2ba+c^2a}+\sqrt{b}\sqrt{2bc+a^2b}+\sqrt{c}\sqrt{2ac+b^2c} \leq \sqrt{(a+b+c)\left(a^2b+b^2c+c^2a+2ab+2bc+2ca\right)}=\sqrt{3\left(\sum{2ab}+\sum{c^2a}\right)}.$$
we have to show that:
$$\sum{2ab}+\sum{c^2a} \leq 9 \Leftrightarrow$$
$$2\left(\sum a\right)\left(\sum ab\right)+3\sum c^{2}a\le\left(\sum a\right)^{3}\Leftrightarrow$$
$$ 2\left(\sum a\right)\left(\sum ab\right)+3\sum c^{2}a\le 2\left(\sum a\right)\left(\sum ab\right)+\left(\sum a\right)\left(\sum a^{2}\right)\Leftrightarrow$$
$$3\sum c^{2}a\le\left(\sum a\right)\left(\sum a^{2}\right)\Leftrightarrow$$
$$2\sum c^{2}a\le\sum c^{3}+\sum ca^{2}$$
and using $AM-GM$ inequality we proved the desired inequality.
Original source can be check here.
|
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|
$(1+i)$ to the power $n$
Possible Duplicate:
Complex number: calculate $(1 + i)^n$.
I came across a difficult problem which I would like to ask you about:
Compute
$ (1+i)^n $ for $ n \in \mathbb{Z}$
My ideas so far were to write out what this expression gives for $n=1,2,\ldots,8$, but I see no pattern such that I can come up with a closed formula.
Then I remember that one can write any complex number $a+bi$ like:
$$(a+bi)=\sqrt{a^2+b^2} \cdot \left( \frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{a^2+b^2}}\cdot i\right)$$
and $\frac{a}{\sqrt{a^2+b^2}} = \cos(\phi)$ and $\frac{b}{\sqrt{a^2+b^2}} = \sin(\phi)$ where $\phi$ is $\arctan{\frac{b}{a}} $
So it becomes,
$(a+bi)=\sqrt{a^2+b^2} \cdot ( \cos(\phi)+\sin(\phi)\cdot i)$ Taking this entire thing to the power $n$ using De Moivre
$$(a+bi)^n=(\sqrt{a^2+b^2})^n \cdot ( \cos(n\phi)+\sin(n\phi)\cdot i)$$
Substituting my $a=1$ and $b=1$
$(1+i)^n=(\sqrt{2})^n \cdot ( \cos(n\cdot\frac{\pi}{4})+\sin(n\cdot\frac{\pi}{4})\cdot i)$
$\phi$ is 45 degrees hence $\frac{\pi}{4}$
But now I don't know how to continue further and I would really appreciate any help! Again, Im looking for a closed formula depending on n.
Best regards
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You got to this expression:
$$(1+i)^n=\left(\sqrt{2}\right)^n \left(\cos\left(n\frac{\pi}{4}\right)+i\sin\left(n\frac{\pi}{4}\right)\right)$$
Now, you should note that this expression (without $\left(\sqrt{2}\right)^n$) has periodicity of $8$, i.e. for $n=k$ and $n=k+8$ the value will be the same. So you can just look at $n=0,1,...,7$ and write the answer in the form:
$$(1+i)^n=\left(\sqrt{2}\right)^n\left\{ \begin{array}{cc} 1 & n\equiv0(\mod8)\\ \vdots & \end{array}\right.$$
|
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|
Multiplicative inverse of a quadratic algebraic number $\,a+b\sqrt 2$ Find the multiplicative inverse of $1+ 3\sqrt{2}$ in the ring $\mathbb{Q}(\sqrt{2})$ and use it to solve the equation $(1+3\sqrt{2})x=1-5\sqrt{2}$.
I think that the inverse is the conjugate, so it would be $1-3\sqrt{2}$, but then I don't know where to use in the equation that needs to be solved.
|
Let $a+b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ be the inverse of $1+3\sqrt{2}$, i.e. $(a+b\sqrt{2})(1+3\sqrt{2})=1$. Then
$$
1=a+6b+(3a+b)\sqrt{2},
$$
i.e.
$$
3a+b=0,\ a+6b=1.
$$
It follows that
$$
a=-\frac{1}{17},\ b=\frac{3}{17}.
$$
Now
$$
(1+3\sqrt{2})x=1-5\sqrt{2} \iff x=(1+3\sqrt{2})^{-1}(1-5\sqrt{2}),
$$
i.e.
$$
x=\frac{1}{17}(-1+3\sqrt{2})(1-5\sqrt{2})=-\frac{31}{17}+\frac{8}{17}\sqrt{2}.
$$
|
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|
A inequality proposed at Zhautykov Olympiad 2008 An inequality proposed at Zhautykov Olympiad 2008.
Let be $a,b,c >0$ with $abc=1$. Prove that:
$$\sum_{\mathrm{cyc}}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$
Set $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.
Our inequality becomes:
$$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \frac{3}{2}.$$
Now we use that: $z^2+x^2 \geq 2zx.$
$$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \sum_{\mathrm{cyc}}{\frac{2z^2}{z^2+x^2+2y^2}} \geq \frac{3}{2}.$$
Now applying Cauchy-Schwarz we obtain the desired result.
What I wrote can be found on this link: mateforum.
But now, I don't know how to apply Cauchy-Schwarz.
Thanks:)
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Since $\mathrm{LHS}$ of last inequality is homogeneous we can assume $x^2 + y^2 + z^2 = 1$. Then it becomes
$$
\mathrm{LHS} = 2\sum_{cyc} \frac {x^2} {1 + z^2} =:2I
$$
Now using Cauchy-Schwarz inequality we get
$$
1 = (x^2 + y^2 + z^2)^2 = \left(\sum_{cyc} x\sqrt{1 + z^2} \cdot \frac x {\sqrt{1 + z^2}}\right)^2 \leq\\
\left(\sum_{cyc} x^2(1 + z^2)\right) \cdot \left( \sum_{cyc} \frac {x^2}{1 + z^2} \right) = I \cdot \sum_{cyc} x^2(1 + z^2)
$$
To finish, let's note that CS inequality implies
$$
x^2\cdot z^2 + y^2 \cdot x^2 + z^2 \cdot y^2 \leq x^4 + y^4 + z^4
$$
and therefore
$$
\sum_{cyc} x^2(1 + z^2) = 1 + x^2 z^2 + y^2 x^2 + z^2 y^2 \leq 1 + \frac {(x^2 + y^2 + z^2)^2} 3 = \frac 4 3
$$
|
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|
A problem about multiples.
For any positive integers $a$, $ b$, if $ab+1$ is a multiple of $16$, then $a+b$ must be a multiple of $p$. Find the largest possible value of $p$.
I have no idea how to solve this. Please help. Thank you.
|
We can consider the problem in $\mathbb{Z}_{16}$.
Then $ab+1$ is a multiple of 16 means that $\overline{ab+1}=\overline{0}$, then $\overline{a}\overline{b}=\overline{ab}=\overline{15}$, in $\mathbb{Z}_{16}$, $\overline{15}$ only has four decompositions, means $\overline{15}=\overline{1}\cdot\overline{15}=\overline{3}\cdot\overline{5}=\overline{7}\cdot\overline{9}=\overline{11}\cdot\overline{13}$.
So $\overline{a+b}=\overline{a}+\overline{b}$ only has four cases, means $\overline{1}+\overline{15}=\overline{0}$, $\overline{3}+\overline{5}=\overline{8}$, $\overline{7}+\overline{9}=\overline{0}$ and $\overline{11}+\overline{13}=\overline{8}$.
So $\overline{a+b}=\overline{0}$ or $\overline{a+b}=\overline{8}$, $a+b$ must be a multiple of 8 (the case that $a+b$ is a multiple of 16 is include in this case), the least possible value of $p$ is 8.
|
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|
Correctness of Fermat Factorization Proof I have asked similar questions regarding this proof. But now I would like to know if my reformulation (after perseverance and different thinking) is correct.
Prove: An odd integer $n \in \mathbb{N}$ is composite $\iff$ $n$ can be written as $n = x^2 - y^2 s.t. y+1 < x$
Proof: $\leftarrow$
Let $n$ be odd and consider $n= x^2-y^2 s.t. y+1 < x$
We can factor the difference of two squares as: $(x+y)(x-y)$ and we note for $n$ to be composite (not prime) $(x-y) > 1$. Thus we have shown this direction.
$\rightarrow$
Let $n$ be odd an composite. By definition of composite we have $n=ab$ for some odd integers $a$ and $b$. Now, working backwards:
$\dfrac{4ab}{4} = \dfrac{(a^2 + 2b + b^2)}{4} - \dfrac{(a^2-2b +b^2)}{4} = (\dfrac{a+b}{2})^2 - (\dfrac{a-b}{2})^2$. Thus, we shall take $x = \dfrac{a+b}{2}$ and $y =\dfrac{a-b}{2}$. And we have $n = x^2 - y^2$.
How do I get $y+1 < x$ in this direction?
|
You need to specify that $0\le b\le a$ and that $b$ is not $1$. If $n$ is composite, we can certainly find such $a$ and $b$. Then we have
$$x-y=\frac{a+b}{2}-\frac{a-b}{2}=b\gt 1.$$
(If you want the inequality $y+1\gt x$, interchange the roles of $x$ and $y$.)
|
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|
Evaluate $\sum_{k=1}^n\lfloor \sqrt{k} \rfloor$ I know that $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$.
How can I use this fact to evaluate $\sum_{k=1}^n\lfloor \sqrt{k} \rfloor$ ?
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For $1\le r<\lfloor\sqrt n\rfloor$, there are exactly $2r+1$ summands of size $r$, namely for $r^2\le k<(r+1)^2$. The remaining $n+1-\lfloor\sqrt n\rfloor^2$ summands are $\lfloor\sqrt n\rfloor$ each. Therefore, with $m:=\lfloor\sqrt n\rfloor$,
$$\begin{align}\sum_{k=1}^n\lfloor\sqrt k\rfloor &= 2\sum_{r=1}^{m-1}r^2+\sum_{r=1}^{m-1}r+(n+1-m^2)m\\&=\frac{(m-1)m(2m-1)}{3}+\frac{(m-1)m}2+(n+1)m-m^3\\&=mn-\frac{m(m-1)(2m+5)}6.\end{align}$$
|
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prove $\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$ by mathematical induction How to prove
$$\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$$
by Mathematical induction，n$\ge $1
|
Here's an elementary proof, that requires no lengthy computations. Let
$$S_n=\sum_{k=n+1}^{2n}\frac1k$$
We show by induction that $S_n \le \frac{25}{36} - \frac{1}{4n+1}$ for all $n \ge 2$. To start with, $S_2 = \frac13+\frac14=\frac{25}{36} - \frac19$, so the hypothesis is true for $n=2$. Now suppose it is true for $n-1$. Then
$$\begin{align}
S_n &= S_{n-1} -\frac1n + \frac{1}{2n-1} + \frac{1}{2n}\\
&= S_{n-1} + \frac{1}{2n(2n-1)}\\
&\le \frac{25}{36} - \frac{1}{4n-3} + \frac{1}{2n(2n-1)}\\
&= \frac{25}{36} - \frac{1}{4n+1} - \frac{4}{(4n-3)(4n+1)} + \frac{1}{2n(2n-1)}\\
&< \frac{25}{36} - \frac{1}{4n+1}
\end{align}$$
because $2n(2n-1) > (4n-3)(4n+1)/4$.
|
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|
Finding the $n$-th derivative of $f(x) =e^x \sin x$, solving the recurrence relation I am trying to find a closed solution for the nth derivative of the function:
$f(x) = e^x \sin x$
So far I have been able to obtain the derivative as:
$f^{(n)}(x) = e^x S_n \sin x + e^x C_n \cos x$
The sequences S and C are defined as below:
$S_n = S_{n-1} - C_{n-1}$
$C_n = S_{n-1} + C_{n-1}$
$S_0 = 1$,
$C_0 = 0$
I have been able to further simplify this by combining the two equations and obtaining:
$C_n = 2S_{n-2}$
$S_n = S_{n-1} - 2 S_{n-3}$
However, I have no idea what to do now. Can anyone help me find the closed form solution?
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The recurrence $S_n=S_{n-1}-2S_{n-3}$ can be solved mechanically. Its auxiliary equation is $x^3-x^2+2=0$. By inspection $-1$ is a solution, so $x+1$ is a factor of the cubic: $$x^3-x^2+2=(x+1)(x^2-2x+2)\;.$$ The other roots are $$\frac{2\pm\sqrt{-4}}2=1\pm i\;,$$ so the solution is of the form $S_n=A(-1)^n+B(1+i)^n+C(1-i)^n$. From the initial values $S_0=S_1=1$ and $S_2=0$ we get
$$\left\{\begin{align*}
&A+B+C=1\\
&-A+B+C+Bi-Ci=1\\
&A+2Bi-2Ci=0
\end{align*}\right.$$
This system has the solution $A=0,B=C=\frac12$, so $$S_n=\frac12\left((1+i)^n+(1-i)^n\right)$$ and $$C_n=(1+i)^{n-2}+(1-i)^{n-2}\;.$$
Now $1+i=\sqrt2 e^{i\pi/4}$ and $1-i=\sqrt2 e^{-i\pi/4}$, so
$$\left\{\begin{align*}
&S_n=2^{(n-2)/2}\left(e^{in\pi/4}+e^{-in\pi/4}\right)=2^{n/2}\cos\frac{n\pi}4\\
&C_n=2^{(n-2)/2}\left(e^{i(n-2)\pi/4}+e^{-i(n-2)\pi/4}\right)=2^{n/2}\cos\frac{(n-2)\pi}4=2^{n/2}\sin\frac{n\pi}4\;.
\end{align*}\right.$$
|
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|
Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$ How to show the following equality?
$$\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$$
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It is well known that
$$\sum_{n=-\infty}^\infty f(n)= -\sum_{j=1}^k \operatorname*{Res}_{z=j}\pi \cot (\pi z)f(z) $$
Assume $a \neq 0$.
To find the residues of $g(z) := \pi \cot (\pi z)\frac{1}{a^2+n^2}$, we see
$$\frac{1}{a^2+n^2} = \frac{1}{(n+ia)(n-ia)}$$
so $g$ has poles at $z_1 = ia$ and $z_2 = -ia$. Their respective residues, $b_1$ and $b_2$ can be found:
$$b_1 = \operatorname*{Res}_{z=ia}\,g(z) = \lim_{z \to ia} \pi \cot (\pi z)\frac{(z-ia)}{(z+ia)(z-ia)} = \pi \cot (\pi i a)\frac{1}{2ia} = -\frac{\pi \coth (\pi a)}{2a}$$
$$b_2 = \operatorname*{Res}_{z=-ia}\,g(z) = \lim_{z \to -ia} \pi \cot (\pi z)\frac{(z+ia)}{(z+ia)(z-ia)} = -\pi \cot (-\pi i a)\frac{1}{2ia} = -\frac{\pi \coth (\pi a)}{2a}$$
And finally:
$$\sum_{k=-\infty}^\infty \frac{1}{a^2+k^2} = -(b_1+b_2)=\frac{\pi \coth (\pi a)}{a}$$
To change the starting number from $-\infty$ to $0$, we divide the series, as it is symmetrical (i.e. $g(n)=g(-n)$):
$$
\sum_{k=-\infty}^\infty \frac{1}{a^2+k^2}=
\frac{\pi \coth (\pi a)}{a}=\\
\sum_{k=-\infty}^{-1} \frac{1}{a^2+k^2}+\frac{1}{a^2}+\sum_{k=1}^\infty \frac{1}{a^2+k^2}=\\
\frac{1}{a^2}+2\sum_{k=1}^\infty \frac{1}{a^2+k^2}=\\
\frac{1}{a^2}+2\left(\sum_{k=0}^\infty \frac{1}{a^2+k^2}-\frac{1}{a^2}\right)=\\
2\sum_{k=0}^\infty \frac{1}{a^2+k^2}-\frac{1}{a^2}
$$
Thus
$$\sum_{k=0}^\infty \frac{1}{a^2+k^2} = \frac{\pi \coth (\pi a)}{2a}+\frac{1}{2a^2} = \frac{\pi a\coth (\pi a)+1}{2a^2}$$
|
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|
How to solve the recurrence relation of $f_n(x)=A_n(a,b)x+B_n(a,b)f(x)$ $$f(f(x))=ax+bf(x)$$
$$f(f(f(x)))=f_3(x)=af(x)+b(ax+bf(x))=abx+(a+b^2)f(x)$$
$$f(f(f(f(x))))=f_4(x)=abf(x)+(a+b^2)(ax+bf(x))=(a^2+ab^2)x+(2ab+b^3)f(x)$$
$$f_n(x)=A_n(a,b)x+B_n(a,b)f(x)$$
$$f_n(f(x))=A_n(a,b)f(x)+B_n(a,b)(ax+bf(x))$$
$$f_n(f(x))=aB_n(a,b)x+(bB_n(a,b)+A_n(a,b))f(x))$$
$$f_{n+1}(x)=A_{n+1}(a,b)x+B_{n+1}(a,b)f(x)$$
$A_2(a,b)=a$
$B_2(a,b)=b$
$A_3(a,b)=ab$
$B_3(a,b)=a+b^2$
$$A_{n+1}(a,b)=aB_n(a,b)$$
$$B_{n+1}(a,b)=A_n(a,b)+bB_n(a,b)$$
$$A_{n+2}(a,b)=aB_{n+1}(a,b)=aA_n(a,b)+abB_n(a,b))=aA_n(a,b)+bA_{n+1}(a,b)$$
$$A_{n+2}(a,b)=aA_n(a,b)+bA_{n+1}(a,b)$$
How can be found the closed form expression of $A_{n}(a,b)$?
Thanks a lot for answers
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I would like to share my solution for the problem
$$f(f(x))=ax+bf(x)$$
$f(x)=rx$ is a solution of the equation.
$$r^2x=(a+br)x$$
$$r^2-br-a=0$$
$$r_1=\frac{b+\sqrt{b^2+4a}}{2}$$
$$r_2=\frac{b-\sqrt{b^2+4a}}{2}$$
$$f_n(x)=r^nx$$
$$f_n(x)=A_n(a,b)x+B_n(a,b)f(x)=(A_n(a,b)+rB_n(a,b))x$$
$$A_n(a,b)+(\frac{b+\sqrt{b^2+4a}}{2})B_n(a,b)=(\frac{b+\sqrt{b^2+4a}}{2})^n$$
$$A_n(a,b)+(\frac{b-\sqrt{b^2+4a}}{2})B_n(a,b)=(\frac{b-\sqrt{b^2+4a}}{2})^n$$
$$B_n(a,b)=\cfrac{(\frac{b+\sqrt{b^2+4a}}{2})^n-(\frac{b-\sqrt{b^2+4a}}{2})^n}{\sqrt{b^2+4a}}$$
$$A_{n+1}(a,b)=aB_n(a,b)=a\cfrac{(\frac{b+\sqrt{b^2+4a}}{2})^n-(\frac{b-\sqrt{b^2+4a}}{2})^n}{\sqrt{b^2+4a}}$$
$$A_{n}(a,b)=a\cfrac{(\frac{b+\sqrt{b^2+4a}}{2})^{n-1}-(\frac{b-\sqrt{b^2+4a}}{2})^{n-1}}{\sqrt{b^2+4a}}$$
|
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|
Sequence of solutions to $x\sin x=1$
Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed.
Consider a sequence $x_n, n\ge1$ formed by positive solutions to $x \sin{x}=1$.
How can we find
$$\lim _{n\rightarrow \infty}(n(x_{2n+1}-2\pi n))= L$$
and
$$\lim _{n\rightarrow \infty}(n^3(x_{2n+1}-2\pi n- \frac{L}{n}))= L_2$$
?
|
Let $y$ be a variable tending to zero. Put
$$
\begin{array}{l}
a=y-\frac{5}{6}y^3, \\
b=y-\frac{5}{6}y^3+\frac{169}{120}y^5
\end{array}
$$
Using Taylor expansions, one finds that
$$
\begin{array}{l}
\sin(a)\bigg(\frac{1}{y}+a\bigg)=1-\frac{169}{120}y^4+O(y^5), \\
\sin(b)\bigg(\frac{1}{y}+b\bigg)=1+\frac{5021}{1680}y^6+O(y^7)
\end{array}
$$
so for small enough $y$ there will be a $c\in ]a,b[$ such that $\sin(c)\bigg(\frac{1}{y}+c\bigg)=1$. This $c$ will satisfy $c=y-\frac{5}{6}y^3+O(y^5)$.
Applying this to $y=\frac{1}{2\pi n}$ yields
$$
L_1=\frac{1}{2\pi}, \ L_2=\frac{-5}{6(2\pi)^3}=\frac{-5}{48\pi^3}
$$
|
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"url": "https://math.stackexchange.com/questions/211333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Limit of a Recurrence Sequence $a_0=c$ where $c$ is positive, with $a_n=\log{(1+a_{n-1})}$,Find
\begin{align}\lim_{n\to\infty}\frac{n(na_n-2)}{\log{n}}\end{align}
I'have tried Taylor expansion, but I can't find the way to crack this limit. Thanks alot for your attention!
|
You may find an asymptotic formula for $a_n$ by improving the accuracy in an adaptive manner.
Step 1. Since
$$0 < a_{n+1} = \log (1+a_n) < a_n,$$
it is a monotone decreasing sequence which is bounded. Thus it must converge to some limit, say $\alpha$. Then $\alpha = \log(1 + \alpha)$, which is true precisely when $\alpha = 0$. Therefore it follows that
$$a_n = o(1). \tag{1}$$
Before going to the next step, we make a simple observation: it is easy to observe that the function
$$\frac{x}{\log(1+x)}$$
is of class $C^3$. In particular, whenever $|x| \leq \frac{1}{2}$, we have
$$ \frac{x}{\log (1+x)} = 1+\frac{x}{2}-\frac{x^2}{12}+O(x^3). $$
This can be rephrased as
$$ \frac{1}{\log(1+x)} = \frac{1}{x}+\frac{1}{2}-\frac{x}{12}+O(x^2). \tag{2}$$
Here, we note that the bound, say $C > 0$, for the Big-Oh notation does not depend on $x$ whenever $|x| \leq \frac{1}{2}$.
Step 2.
By noting $(1)$, we fix a positive integer $N$ such that whenever $n \geq N$, we have $|a_n| \leq \frac{1}{2}$. Then by $(2)$,
$$ \frac{1}{a_{n+1}} - \frac{1}{a_n} = \frac{1}{2} + O(a_n), $$
where the bound for Big-Oh notation depends only on $N$. Indeed, we may explicitly choose a bounding constant as
$$C'=\frac{1}{12} + \frac{1}{2}C,$$
where $C$ is the same as in $(2)$. Thus if $n > m > N$, we then have
$$ \begin{align*}
\frac{1}{a_n}
&= \frac{1}{a_{m}} + \sum_{k=m}^{n-1} \left( \frac{1}{a_{k+1}} - \frac{1}{a_k} \right) \\
&= \frac{1}{a_{m}} + \sum_{k=m}^{n-1} \left( \frac{1}{2} + O(a_k) \right) \\
&= \frac{1}{a_{m}} + \frac{n-m}{2} + O((n-m)a_m).
\end{align*} $$
Thus we have
$$ \left|\frac{1}{n a_n} - \frac{1}{2}\right| \leq \frac{1}{n}\left(\frac{1}{a_m} + \frac{m}{2} + C'm a_m \right) + C' a_m.$$
Taking limsup as $n\to\infty$, we have
$$ \limsup_{n\to\infty}\left|\frac{1}{n a_n} - \frac{1}{2}\right| \leq C' a_m. $$
Since now $m$ is arbitrary, the right-hand side can be made as small as we wish. Thus the left-hand side must vanish, yielding
$$ \frac{1}{n a_n} = \frac{1}{2} + o(1),$$
or equivalently
$$ n a_n = 2 + o(1). \tag{3} $$
Step 3.
Let $N$ be as in the previous step. Then $(3)$ suggests that it is natural to consider
$$ \left( \frac{1}{a_{n+1}} - \frac{n+1}{2} \right) - \left( \frac{1}{a_{n}} - \frac{n}{2} \right) = -\frac{a_n}{12} + O(a_n^2). $$
Now from $(3)$, we have
$$ a_n = \frac{2}{n} + o\left(\frac{1}{n}\right) = 2(\log(n+1) - \log n) + o\left(\frac{1}{n}\right) = O\left(\frac{1}{n}\right).$$
Plugging this to the equation above, we have
$$ \left( \frac{1}{a_{n+1}} - \frac{n+1}{2} \right) - \left( \frac{1}{a_{n}} - \frac{n}{2} \right) = -\frac{1}{6}\left( \log(n+1) - \log n \right) + o\left(\frac{1}{n}\right). $$
Now for each $\epsilon > 0$, choose $m > N$ such that whenever $n > m$, the Small-Oh term is bounded by $\epsilon / n$. Then for such $n$ we have
$$ \left| \left( \frac{1}{a_{n+1}} - \frac{n+1}{2} + \frac{1}{6}\log(n+1) \right) - \left( \frac{1}{a_{n}} - \frac{n}{2} + \frac{1}{6}\log n \right) \right| \leq \frac{\epsilon}{n}. $$
Thus summing up from $m$ to $n-1$, we have
$$ \left| \frac{1}{a_{n}} - \frac{n}{2} + \frac{1}{6}\log n \right|
\leq \left| \frac{1}{a_{m}} - \frac{m}{2} + \frac{1}{6}\log m \right| + \epsilon (\log n - \log m). $$
Dividing both sides by $\log n$ and taking $n \to \infty$, we have
$$ \limsup_{n\to\infty} \frac{1}{\log n} \left| \frac{1}{a_{n}} - \frac{n}{2} + \frac{1}{6}\log n \right| \leq \epsilon. $$
Since this is true for every $\epsilon > 0$, it must vanish. Therefore we have
$$ \frac{1}{a_{n}} = \frac{n}{2} - \frac{1}{6}\log n + o(\log n). $$
In particular,
$$ \begin{align*}a_n
&= \left( \frac{n}{2} - \frac{1}{6}\log n + o(\log n) \right)^{-1} \\
&= \frac{2}{n} \left( 1 - \frac{1}{3n}\log n + o\left( \frac{\log n}{n} \right) \right)^{-1} \\
&= \frac{2}{n} + \frac{2}{3n^2} \log n + o\left( \frac{\log n}{n^2} \right).
\end{align*} $$
Therefore
$$ \frac{n(na_n - 2)}{\log n} = \frac{2}{3} + o(1)$$
and it follows that the limit is
$$\lim_{n\to\infty} \frac{n(na_n - 2)}{\log n} = \frac{2}{3}.$$
Further discussions. In fact, we can show that
$$ a_n = \frac{2}{n} + \frac{2}{3n^2} \log n + O\left( \frac{1}{n^2} \right). $$
More generally, we have the following proposition.
Proposition. Suppose $(a_n)$ is a sequence of positive real numbers converging to 0 and satisfying the recurrence relation $a_{n+1} = f(a_n)$.
*
*If $f(x) = x \left( 1 - (a + o(1)) x^m \right)$ for some real $a \neq 0$ and integer $m \geq 1$, then
$$ a_n = \frac{1}{\sqrt[m]{man}}(1 + o(1)). $$
*If $f(x) = x \left( 1 - a x^m + (b+o(1)) x^{2m} \right)$ for some some reals $a \neq 0$ and $b$, and integer $m \geq 1$, then
$$ a_n = \frac{1}{\sqrt[m]{man}} \left( 1 - \frac{(m+1) a^2 - 2 b}{2m^2a^2} \frac{\log n}{n} + o \left( \frac{\log n}{n} \right) \right). $$
*If $f(x) = x \left( 1 - a x^m + b x^{2m} + O(x^{3m}) \right)$ for some reals $a \neq 0$ and $b$, and integer $m \geq 1$, then
$$ a_n = \frac{1}{\sqrt[m]{man}} \left( 1 - \frac{(m+1) a^2 - 2 b}{2m^2a^2} \frac{\log n}{n} + O \left( \frac{1}{n} \right) \right). $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The limit of a nested radical Let $f(a,x)=\sqrt{a(a-1)+x}$, $f^{(n)}(a,x)$ denotes the $n^{th}$ iteration of $x$, where
\begin{align}f^{(1)}(a,x)=f(a,x),f^{(n)}(a,x)=f^{(n-1)}(a,f(a,x))\end{align}
Find
\begin{align}\lim_{n\to\infty}(2a)^{\frac{n}{2}}\sqrt{a-f^{(n)}(a,0)}\end{align}
This is a problem from a discussion in a forum, someone say that when $a=2$, the limit is equal to $\pi$, but I can't find the pattern behind this structure, Thanks for your attention.
PS: Expanding it shows
\begin{align}\lim_{n\to\infty}(2a)^{\frac{n}{2}}\sqrt{a-\sqrt{a(a-1)+\sqrt{a(a-1)+\cdots}}}\end{align}
where there are $n$ square root sign behind the minus sign, it is a $\infty \cdot 0$ problem.
|
Steps for solution ... With $a=2$ write $f(2,x) = f(x)$.
For the appropriate values of $\theta$ ...
$$\begin{align}
&2\cos\theta = \sqrt{2+2\cos(2\theta)}
\\
&f^{(n)}(2\cos\theta) = 2\cos\frac{\theta}{2^n}
\\
&f^{(n)}(0) = 2\cos\frac{\pi}{2^{n+1}}
\\
&\sqrt{2-2\cos\theta} = 2 \sin \frac{\theta}{2}
\\
&\sqrt{2-f^{(n)}(0)} = 2\sin\frac{\pi}{2^{n+2}}
\\
&\lim_{n\to\infty} 2^n\cdot 2\sin\frac{\pi}{2^{n+1}} = \frac{\pi}{2}
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Order of elements modulo p Let $p$ be prime. Suppose that $x\in Z$ has order 6 mod p. Prove that $(1-x)$ has order 6 mod p as well.
I know that I need to show that the order can't be 2 or 3 (4 and 5 are trivial cases), but I'm even having difficulty showing that $(1-x)^6$ is 1. I've expanded $x^{6}-1$ to get a nice congruence that might be useful...but I can't seem to apply it.
|
If $ord_px=6\implies p\mid(x^6-1)$ ,i.e, $p\mid(x^3-1)(x^3+1)$
If $p\mid(x^3-1), x^3\equiv 1\pmod p\implies ord_px=3\ne 6$
So, $p\mid(x^3+1)\implies p\mid (x+1)(x^2-x+1)$ and $x^3\equiv -1\pmod p$
If $p\mid(x+1), x\equiv -1\pmod p\implies x^2\equiv 1\pmod p\implies ord_px=2\ne 6$
So, $p\mid(x^2-x+1)$
(i) $ x^2\equiv x-1\pmod p$
So, $1-x=(-1)(x-1)\equiv x^3\cdot x^2\pmod p$ as $x^3\equiv -1\pmod p$
So, $1-x\equiv x^5\implies ord_p(1-x)=ord_p(x^5)$
We know, $ord_ma=d, ord_m(a^k)=\frac{d}{(d,k)}$ (Proof @Page#95)
So, $ord_m(x^5)=\frac{6}{(6,5)}=6$
(ii) $x-x^2\equiv1\pmod p$
$\implies 1-x\equiv x^{-1}$ dividing either side by $x$ as $(x,p)=1$
So, $ord_p(1-x)=ord_p(x^{-1})$
So, $ord_m(x^{-1})=\frac{6}{(6,-1)}=6$
Observation :
If $x^5\equiv x\pmod p, x^4\equiv 1\implies ord_px\mid 4$, but $ord_px=6$
If $x\equiv x^{-1}\pmod p,x^2\equiv 1\implies ord_px\mid 2$
In fact, $x^{-1}\equiv x^5\pmod p$ as $x^6\equiv1$
We also know, that are exactly $\phi(6)=2$ in-congruent elements that belong to order $6\pmod p$ if $6\mid \phi(p)\implies p\equiv 1\pmod 6$.
In that case, we can conclude $x,x^{-1}\equiv x^5\equiv(1-x)$ are those two.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Differentiating $x^2 \sqrt{2x+5}-6$ How do I differentiate this function: f(x)= $x^2 \sqrt{2x+5}-6$
I had: I had $2x\sqrt{2x+5} + x^2 \dfrac{1}{2\sqrt{2x+5}}$ but the correction model said it was I had $2x\sqrt{2x+5} + x^2 \dfrac{2}{2\sqrt{2x+5}}$
|
Using the chain rule where $\cfrac {df}{dx} = \cfrac {dg}{du} \cfrac {du}{dx} $ if $f(x) = g(u(x))$ and the product rule where $\cfrac {d}{dx} (uv) = v\cfrac {du}{dx} + u\cfrac {dv}{dx} $
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the equations of two lines that are tangent to the curve and parallel to a line. I have no idea how to do this as i have been absent from class for medical reasons.
the curve $y=3+x^2$
and the line $3x-y=-6$
The answer is a possibility of the following:
a) $ y=3x, y=3x+4$
b) $ y=12x-15, y=12x+17$
c) $ y=-3x+9, y=-3x-16$
d) $y=12x-13, y=12+19$
e) $y=-12x-16, y=-12x+8$
f) $y=3x+5, y=3x+1$
|
The equation of any line parallel to $3x-y=-6$ can be written as $3x-y=k$
Now, let this line touches the curve $y=3+x^2$ at $(a,b).$
So, $3a-b=k$ or $b=3a-k$ and $b=3+a^2$
Comparing the values of $b$ we get, $3a-k=3+a^2,$ or $a^2-3a+k+3=0$
Now, this is a quadratic equation in $a,$ for the tangency both the root should be same, i.e, discriminant$(-3)^2-4\cdot 1\cdot(k+3)$ must be $0\implies 9-4k-12=0,k=-\frac 3 4$
So, the equation of the tangent becomes $3x-y=-\frac 3 4,$ or $12x-4y=-3$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Partial Fractions - Calculus Evaluate the integral: $$\int \dfrac{9x^2+13x-6}{(x-1)(x+1)^2} dx$$
For some reason I cannot get the right answer. I split up the equation into three partial fractions but I cannot seem to find A, B, or C from the three subsequent equations. Thanks!
|
The motivation is to write $\dfrac{9x^2 + 13x - 6}{(x-1)(x+1)^2}$ as $$\dfrac{A}{(x+1)^2} + \dfrac{B}{x+1} + \dfrac{C}{x-1}$$
This gives us
\begin{align}
9x^2 + 13x - 6 & = A(x-1) + B(x^2-1) + C(x+1)^2\\
& = (B+C)x^2 + (2C+A)x + (C-A-B)
\end{align}
This gives us $$B+C = 9\\ 2C+A = 13\\ C - A - B = -6$$
Adding all the three equations, give us $4C = 16 \implies C = 4$. Hence, $A=B=5$.
Hence, we get that
$$\dfrac{9x^2 + 13x - 6}{(x-1)(x+1)^2} = \dfrac5{(x+1)^2} + \dfrac5{x+1} + \dfrac4{x-1}$$
Now you should be able to integrate and finish it off.
|
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|
Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.
Thanks :)
|
Let $$M=\dfrac{a^3}{a+b}+\dfrac{b^3}{b+c}+\dfrac{c^3}{c+a}$$ be the original expression, we introduce its "conjugate"
$$N=\dfrac{ab^2}{a+b}+\dfrac{bc^2}{b+c}+\dfrac{ca^2}{c+a}$$
Direct computing, and using $a^2+b^2+c^2\geq ab+bc+ca$, yields
$$M-N =\frac{a(a^2-b^2)}{a+b}+\frac{b(b^2-c^2)}{b+c}+\frac{c(c^2-a^2)}{c+a}\\=a(a-b)+b(b-c)+c(c-a)\\= a^2+b^2+c^2-ab-bc-ca\geq 0$$
Therefore $M\geq N$. Also using $2(x^2+y^2)\geq (x+y)^2$
$$M+N=\frac{a(a^2+b^2)}{a+b}+\frac{b(b^2+c^2)}{b+c}+\frac{c(c^2+a^2)}{c+a}\\ \geq \frac{a(a+b)+b(b+c)+c(c+a)}{2} \\ =\frac{a^2+b^2+c^2+ab+bc+ca}{2} \\ \geq ab+bc+ca$$
Hence we have both $M\geq N$ and $M+N\geq ab+bc+ca$. It follows that $M\geq \dfrac{ab+bc+ca}{2}$.
|
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|
Analyzing a function I am having a problem with the following function:
$f(x)=\sin^2(x)-\cos(3x)$
I need to examine the sign of $f'(x)$
I noticed that f is $2\pi$-periodic, therefore we need to analyze f(x) on $[0;2\pi]$
In addition to that $f(x)=-4\cos^3(x)-\cos^2(x)+3\cos(x)+1$
Hence $\forall x \in ]0;2\pi[, f'(x)=(-12\cos^2(x)-2\cos(x)+3)(-\sin(x))$
Let us solve $f'(x)=0$
We have $\forall x \in ]0;2\pi[ -\sin(x)=0 \Leftrightarrow x=\pi$
For the second equation, I know that there are two solution on $]0;2\pi[$ (since I visualized it on the graph), but I am unable to determine them by calculuation.
Please help.
Thank you in advance
|
$$
\begin{align}
f'(x)&=2 \cos x\sin x +3 \sin (3x) = 2 \cos x \sin x +3 (3 \sin x - 4 \sin^3 x) \\
&= \sin x \left( 2 \cos x + 9 -12 \sin^2 x \right) = \sin x \left( 2 \cos x + 9 -12 (1-\cos^2 x) \right)
\end{align}
$$
so that you have $\sin x=0$ and $12 \cos^2 +2 \cos x -3=0$.
|
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|
$(2^m -1)(2^n-1)$ divides $(2^{mn} -1)$ if and only if $\gcd(m,n) = 1$. If $\gcd(m,n) = 1$ then $(2^m-1)(2^n-1)$ divides $2^{mn} - 1$ because each of $2^m-1$, $2^n-1$ divide $2^{mn}-1$ and $\gcd(2^m-1, 2^n-1) = 2^{\gcd(m,n)}-1 = 1$. How about the converse? If $\gcd(m,n) > 1$, can one show that $(2^m-1)(2^n-1)$ does not divide $(2^{mn}-1)$?
|
Here's another proof I found. First we need the following:
Lemma If $m,n$ are positive integers with $m > n$, then $(2^m - 1)(2^n-1) \mid (2^{mn}-1)$ implies $(2^{m-n}-1)(2^n-1) \mid (2^{(m-n)n} - 1)$.
Proof of lemma. $(2^m-1)(2^n-1) \mid (2^{mn}-1)$ is equivalent to $(2^{mn} - 1)/(2^m-1) \equiv 0 \pmod{2^n-1}$. Using $2^m \equiv 2^{m-n} \pmod{2^n-1}$ we have
$(2^{mn} - 1)/(2^m-1) \equiv 2^{m(n-1)} + 2^{m(n-2)} + \cdots 2^m + 1 \equiv 2^{(m-n)(n-1)} + 2^{(m-n)(n-2)} + \cdots 2^{m-n} + 1 \equiv (2^{(m-n)n} - 1) / (2^{m-n}-1) \pmod{2^n-1}$.
Thus $(2^{(m-n)n} - 1) / (2^{m-n}-1) \equiv 0 \pmod{2^n-1}$ which is equivalent to the stated conclusion.
Now to our main task. Suppose $m$, $n$ are positive integers with $\gcd(m,n) = d$. By a well-known algorithm to compute the greatest common divisor, there is a sequence $(m_1,n_1), (m_2, n_2), \ldots, (m_k, n_k)$ such that: $(m_1, n_1) = (m, n)$, $(m_k, n_k) = (d, d)$, $m_i \not= n_i$ for $1 < i < k$ and
$(m_{j+1}, n_{j+1}) = (m_j-n_j,n_j)$ if $m_j > n_j$, $(m_{j+1}, n_{j+1}) = (m_j, n_j-m_j)$ if $n_j > m_j$ for $1 \leq j < k$.
If $(2^m-1)(2^n-1) \mid (2^{mn}-1)$ then using the lemma one shows by induction $(2^{m_j}-1)(2^{n_j}-1) \mid (2^{m_jn_j}-1)$ for all $j$ in the range $1 \leq j \leq k$. For $j =k$, this means $(2^d-1)(2^d-1) \mid (2^{d^2}-1)$ which is equivalent to $(2^{d^2}-1)/(2^d-1) \equiv 0 \pmod{2^d-1}$. But $(2^{d^2}-1)/(2^d-1) \equiv 2^{d(d-1)} + 2^{d(d-2)} + \cdots + 2^{d} + 1$. Since $2^d \equiv 1 \pmod{2^d-1}$, the latter sum is $\equiv d \pmod{2^d-1}$, and we conclude $d \equiv 0 \pmod{2^d-1}$, which happens only when $d = 1$. End of proof.
|
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|
How to factor a polynomial in real numbers? Which of the following polynomials can't be factored in real numbers?
Multiple choice question from an old test.
Got the following polynomials:
$x^8$
$(x-3)(x^2+x+1)$
$(x-2)^3(x^2-1)$
$x^8(x^2-1)$
$(x^2-1)^3$
$x(x-1)(x+1)$
Solution:
Solving $x^2+x+1$ isnt possible with real numbers hence D is negative?
There for answer must be $(x-3)(x^2+x+1)$
Edited right answer.
|
The only polynomial in that list that doesn't split into linear factors is the one with $x^2 + x + 1$ because its discriminant is $-3$. For the other ones,
\begin{gather}
\begin{aligned}
x^8 & = x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \\
(x-2)^3 (x^2-1) & = (x-2)(x-2)(x-2)(x-1)(x+1) \\
(x^2-1)^3 &= (x-1)(x+1)(x-1)(x+1)(x-1)(x+1) \\
\end{aligned}
\end{gather}
and so on. By the way, the answer $(x-3)(x^2+x+1)$ cannot be factored over the reals, but it can be factored over $\mathbb C$, because the complex numbers have this property that every polynomial splits. As an example,
$$
x^2 + x + 1 = \left( x - \frac{-1 + \sqrt 3 i}2 \right) \left( x - \frac{-1 - \sqrt 3 i}2 \right),
$$
where $i$ is defined so that $i^2 = -1$.
Hope that helps,
|
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"url": "https://math.stackexchange.com/questions/225924",
"timestamp": "2023-03-29T00:00:00",
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|
Area fractal pentagrams I When I saw this image I was a little curious.
How can I find the area of this fractal?
|
Each segment of the pentagram is the initiator of the fractal. Take its length to be 1. Now the generator consists of 2 line segments each of length $\frac{1}{3}$.
Hence on each iteration $n$ the area can be expressed as follows:
$$A_n=10\sum_{k=0}^{n}2^kS_k +S_{p}$$
Where $S_p$ is the area of the regular pentagon and:
$$S_k=\frac{1}{2}\frac{1}{3^{2k}}\sin\frac{\pi}{5}$$
is the area of the "k-th generation" petal.
$$10\sum_{k=0}^{n}2^kS_k=5\pi\sin\frac{\pi}{5}\sum_{k=0}^{\infty}\left(\frac{2}{9}\right)^k=\frac{45\pi}{7}\sin\frac{\pi}{5}$$
The area of the pentagon is:
$$S_p=\frac{t^2\sqrt{25+10\sqrt{5}}}{4}$$
where $t=2\sin\frac{\pi}{10}$
Finally,
$$A=\frac{45\pi}{7}\sin\frac{\pi}{5}+\sqrt{25+10\sqrt{5}}\left(\sin\frac{\pi}{10}\right)^2$$
Or equivalently
$$A=\frac{45\pi\sqrt{2(5-\sqrt{5})}}{28}+\frac{\sqrt{25+10\sqrt{5}}}{4\phi^2}$$
or any other way you wish to think of it.
|
{
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|
Area of triangle ABC inside circle Consider the following diagram:
$AB+AD=DE$, $\angle BAD= 60$, and $AE$ is $6$. How do we find the area of the triangle $ABC$?
|
From the above picture,
$x+y=z$ , $ y+z=6$ and $\frac x {y+z} = \cos 60^\circ = \frac 1 2$
After calculation, $x=3$, $y=\frac 3 2$ and $z = \frac 9 2$
$\angle ABE = 90^\circ$ and $\angle BAD = 60^\circ$
So, $\angle AEB = 30^\circ = \angle ACB$ (properties of a circle)
Now, $\cos 60^\circ = \frac {x^2 + y^2 - w^2}{2xy} = \frac 1 2$
After calculation, $w=DB=\frac {3 \sqrt 3}{2}$
$$\cos \angle ADB=\frac{y^2+w^2-x^2}{2yw}=\frac{(\frac 3 2)^2+(\frac {3 \sqrt 3}{2})^2-3^2}{2 \frac 3 2\frac {3 \sqrt 3}{2}}=0$$
So, $\angle ADB = 90^\circ$ and $\angle ABD = 30^\circ = \angle ACB$
So, $\triangle ABC $ is an isosceles triangle.
And, $CD = BD = \frac {3 \sqrt 3}{2}$
So, $\triangle ABC = \frac 1 2 \cdot \frac 3 2 \cdot (\frac {3 \sqrt 3}{2} + \frac {3 \sqrt 3}{2}) = \frac {9 \sqrt 3}{4}$
|
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|
Find the area of a triangle using analytic geometry Given are the points $P (1,0)$ and $Q (3,2)$. The points $P$ and $Q$ have the same distance to a certain line $l$, which intersects the positive x-axis in the point $A$ and the positive y-axis in the point $B$. The area of the triangle $ABO$ is minimal. Get the equation of $l$.
What I did:
First of all, I got the equation of line $PQ$, which is: $y=x-1$.
Then I got the bisector of $PQ$, because every point on that line is equidistant to $P$ and $Q$. The bisector: $y=-x+3$.
However, this turned out to be wrong, and I had a gut feeling that it would be wrong, since I did nothing with the statement 'The area of triangle of $ABO$ is minimal', because I thought (and still think) that there is just 1 line equidistant from both points.. Can anyone help me with this?
|
Let the equation of the line $l(AB)$ be $\frac x a+\frac y b=1$
So, $A(a,0)$ and $B(0,b)$ and $a>0$ and $b>0$.
The area of $\triangle ABO=\frac{ab}2$
The distance of $l:b x+ay-ab=0$ from $P(1,0)$ is $\frac{\mid b-ab\mid}{\sqrt{a^2+b^2}}$ and that of from $Q(3,2)$ is $\frac{\mid 3b+2a-ab\mid}{\sqrt{a^2+b^2}}$
So, $(b-ab)^2=(3b+2a-ab)^2$
$\implies 4(a+b)(ab-a-2b)=0\implies ab-a-2b=0$ as $ab>0$
So, $a=\frac{2b}{(b-1)}\implies b>1$
So, the area of $\triangle ABO=\frac{ab}2=\frac{b^2}{b-1}$
Its minimum value can be derived in the following ways:
(1) $\frac{b^2}{b-1}=\frac{b^2-1}{b-1}+\frac1{b-1}=b+1+\frac1{b-1}$ as $b\ne 1$
So, $\frac{b^2}{b-1}=b-1+\frac1{b-1}+2$
$=\left(\sqrt{b-1}-\frac1{\sqrt{b-1}}\right)^2+2+2\ge 4$ as $b-1>0$
So, the minimum are will be $4$ Square Unit for $\sqrt{b-1}=\frac1{\sqrt{b-1}}\implies b=2$
(2) $\frac{b^2}{b-1}=c$(say)
So,$c>0$ as $b>1$
So, $b^2-bc+c=0$ as $b$ is real, the discriminant $(c^2-4c)$ must be $\ge 0$
$\implies c\ge 4$ or $c\le 0$ which is impossible
So, the minimum area is $4$ Square Unit.
$\implies b^2-4b+4=0\implies b=2$
$\implies a=\frac{2b}{(b-1)}=4$
So, the equation of the line $l$ is $\frac x 4+\frac y 2=1$
|
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|
absolute value integrated For example I have the function $f(x) = |x^2-1| = \sqrt{(x^2-1)^2}$
$\int \sqrt{(x^2-1)^2}dx = \frac{x(x^2-3)\sqrt{(x^2-1)^2)}}{3 (x^2-1)}+constant$
But plotted it looks like this: Plot 1. There are values < 0.
I have to take the absolute value again to get the correct function.
Plot 2 Why do I have to do it again? I integrated $|x^2-1|$ and not $x^2-1$
Edit: Part 2
|
To evaluate $\int_a^b |x^2-1|dx$ there are 6 possibilities depending on where $a,b$ lie. I am assuming $a\leq b$ in the following, if $a>b$ then use $\int_a^b = -\int_b^a$.
If $a,b \in (-\infty,-1]$ or $a,b \in [1,\infty)$, then $\int_a^b |x^2-1|dx= \frac{1}{3} (b^3-3b-a^3+3a)$.
If $a,b \in (-1,1)$, then $\int_a^b |x^2-1|dx= \frac{1}{3} (a^3-3a-b^3+3b)$.
If $a\in (-\infty,-1], b \in (-1,1)$, then $\int_a^b |x^2-1|dx= \frac{1}{3} (4-a^3+3a-b^3+3b)$.
If $a\in (-1,1), b \in [1,\infty)$, then $\int_a^b |x^2-1|dx= \frac{1}{3} (4+a^3-3a+b^3-3b)$.
If $a \in (-\infty,-1]$, $b \in [1,\infty)$, then $\int_a^b |x^2-1|dx= \frac{1}{3} (8-a^3+3a+b^3-3b)$.
|
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|
Integral table contradiction? In a table of integrals, I see the following two formulas:
$\int \frac{dx}{(a+x)(b+x)} = \frac{1}{b-a}\ln\frac{a+x}{b+x}$, and
$\int \frac{dx}{ax^2+bx+c} = \frac{2}{\sqrt{4ac-b^2}}\tan^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}}$.
How can these both be true? It seems like if we expand $(a+x)(b+x)$ out to $x^2+(a+b)x+ab$, we can apply the 2nd equation to get
$\int \frac{dx}{(a+x)(b+x)} = \frac{2}{\sqrt{4ab-(a+b)^2}}\tan^{-1}\frac{2x+a+b}{\sqrt{4ab-(a+b)^2}}$,
which is surely not equivalent to $\frac{1}{b-a}\ln\frac{a+x}{b+x}$ (one involves a logarithm and the other involves an arctan, so no amount of algebraic fussing can reconcile them, can it?!)
|
You take the square root of a negative number $\left(-(a-b)^2\right)$, so the argument of arctan is complex. Using the following formula you should be able to turn one solution into the other:
$$\arctan(z) = {\ln(1 - iz) - \ln(1 + iz) \over 2}$$
|
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|
Express each power of the root $\alpha$ of $\frac{\mathbb Z_2[x]}{\langle x^3+x^2+1\rangle}$ as linear combinations of $1, \alpha$ and $\alpha^2$ There are $8$ elements in $\frac{\mathbb Z_2[x]}{\langle x^3+x^2+1\rangle} = GF(8)$
and this generates the set $\{0,1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1\}$
We're required to express $\alpha^1$ all the way up to $\alpha^7$ as linear combinations of $1, \alpha$ and $\alpha^2$
$\alpha^1 = \alpha$
$\alpha^2 = \alpha^2$
$\alpha^3 = \alpha^2 + 1$
$\alpha^4 = \alpha^2+\alpha+1$
$\alpha^5 = \alpha+1$
$\alpha^6 = \alpha^2+\alpha$
$\alpha^7 = 1$
I'm really not seeing where these combinations are coming from. Why is $\alpha^3 = \alpha^2+1$?
|
If $\alpha$ is a root of $x^3+x^2+1$, then $\alpha^3+\alpha^2+1=0$, so $\alpha^3=\alpha^2+1$ (since $1=-1$ over $\mathbb{Z}_2$).
From here:
$$\begin{array}{l} \alpha^4=\alpha\alpha^3=\alpha(\alpha^2+1)=\alpha^3+\alpha=\alpha^2+\alpha+1\\
\alpha^5=\alpha\alpha^4=\alpha^3+\alpha^2+\alpha=\alpha^2+1+\alpha^2+\alpha=\alpha+1\\
\alpha^6=\alpha\alpha^5=\alpha^2+\alpha\\
\alpha^7=\alpha\alpha^6=\alpha^3+\alpha^2=\alpha^2+1+\alpha^2=1 \end{array}$$
(using $1+1=0$ over $\mathbb{Z}_2$)
|
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|
Sum of the sum of the sum of the first $n$ natural numbers I have here another problem of mine, which I couldn't manage to solve.
Given that: $$x_n = 1 + 2 + \dots + n \\ y_n = x_1 + x_2 + \dots + x_n
\\ z_n = y_1 + y_2 + \dots + y_n $$
Find $z_{20}$.
I know the answer but I'm having a hard time reaching it. I recognized that $x_n$ is obviously $\dfrac{n(n + 1)}{2}$, but how to express the other two in a closed form to allow the calculation? I even tried writing the relations in a recursive way, without success. Is there an easy way to solve this one? I was not allowed to use calculators.
Thanks,
rubik
|
$$2y_n=2\sum_{1\le r\le n}x_1=\sum_{1\le r\le n}(r^2+r)=\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2=\frac{n(n+1)(n+2)}3$$
$\implies 6y_n=n^3+3n^2+2n$
$$6z_n=6\sum_{1\le r\le n}y_r=\sum_{1\le r\le n}(r^3+3r^2+2r)$$
$$=\left(\frac{n(n+1)}2\right)^2+3\frac{n(n+1)(2n+1)}6+2\frac{n(n+1)}2$$
$$=\frac{n(n+1)\{3n(n+1)+6(2n+1)+12\}}{12}=\frac{n(n+1)\{3n^2+15n+18\}}{12}=\frac{n(n+1)(n+2)(n+3)}4$$
So, $$z_n=\frac{n(n+1)(n+2)(n+3)}{24}$$
So, $$z_{20}=\frac{20(20+1)(20+2)(20+3)}{24}=5\cdot7\cdot11\cdot 23$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Rolling a fair die You and your friend play a game in which you and your friend take turns rolling a fair six-sided die and keep a running tally of the sum of the results of all rolls made. Play continues until either player wins if, after the player rolls, the number
on the running tally is a multiple of 7. Should you start first or it is better if you let your friend rolls the die first.
|
I thought I'd answer this question as another opportunity to learn some math that's way out of my league in hopes that some of my usual favorites will evaluate it and see if I'm on the right track.
The first thing I did was consider a few trials:
*
*Player 1 has a 0/6 chance of winning
*Player 2 has a 1/6 chance of winning
*Player 1 has a 1/6 chance of winning
*Player 2 has a 1/6 chance of winning
The probability of player 1 winning a 1 round game is 0. The probability of player 1 winning a 2 round game is $\mathbb{P} = \mathbb{P}(\text{Winning Round 1}) + \mathbb{P}(\text{Rounds 1 and 2 Losses}) \cdot \mathbb{P}(\text{Winning Round 3}))$.
I figured that continues infinitely with the outcomes becoming less and less likely as the number of trials approaches infinity.
So, I came up with this:
$$\mathbb{P}(\text{Player 1}) = \dfrac{0}{6} + \dfrac{6}{6} \cdot \dfrac{5}{6} \cdot \dfrac{1}{6} + \dfrac{6}{6} \cdot \left(\dfrac{5}{6}\right)^3 \cdot \dfrac{1}{6} + \ldots $$
$$\mathbb{P}(\text{Player 1}) = 0 + \dfrac{5}{6} \cdot \dfrac{1}{6} + \left(\dfrac{5}{6}\right)^3 \cdot \dfrac{1}{6} + \ldots = \dfrac{1}{6} \cdot \left( \dfrac{5}{6} + \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right) $$
So, then I had to figure out how to sum an infinite series: $\left( \dfrac{5}{6} + \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right) $
I did some research on how exactly that's done and the best I could come up with is to multiply the entire series by $\left(\dfrac{5}{6}\right)^2$ and subtract that from the original series. So, I think this is valid:
$$\left(\dfrac{5}{6}\right)^2 \cdot \left( \dfrac{5}{6} + \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right) = \left( \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right)$$
If I subtract the new series from the original, all of the elements but the first should cancel out and I should have this:
$$\mathbb{S} - \left(\dfrac{5}{6}\right)^2 \cdot \mathbb{S} = \dfrac{5}{6}$$
$$\mathbb{S} \cdot \left[1 - \left(\dfrac{5}{6}\right)^2\right] = \dfrac{5}{6}$$
$$\mathbb{S} = \dfrac{\dfrac{5}{6}}{\left[1 - \left(\dfrac{5}{6}\right)^2\right]}$$
If I plug that into my original formula, I get this:
$$\mathbb{P}(\text{Player 1}) = \dfrac{1}{6} \cdot \dfrac{\dfrac{5}{6}}{\left[1 - \left(\dfrac{5}{6}\right)^2\right]} = \overline{.45}$$
Because there are only two players this should be true: $\mathbb{P}(\text{Player 2}) = 1 - \mathbb{P}(\text{Player 1}) = \overline{.54} $
Unsure of my original conclusion, I thought I'd calculate the probability of Player 2 winning to verify.
$$\mathbb{P}(\text{Player 2}) = \dfrac{6}{6} \cdot \dfrac{1}{6} + \dfrac{6}{6} \cdot \left(\dfrac{5}{6}\right)^2 \cdot \dfrac{1}{6} + \dfrac{6}{6} \cdot \left(\dfrac{5}{6}\right)^4 \cdot \dfrac{1}{6} + \ldots $$
$$\mathbb{P}(\text{Player 2}) = \dfrac{1}{6} + \dfrac{1}{6} \cdot \left(\left(\dfrac{5}{6}\right)^2 \cdot \left(\dfrac{5}{6}\right)^4 + \left(\dfrac{5}{6}\right)^6 + \ldots \right) $$
To sum that series I thought: $\mathbb{S} = \dfrac{25}{36} + \left(\dfrac{25}{36}\right)^2 + \left(\dfrac{25}{36}\right)^3 + \ldots $
So, $\mathbb{S} - \dfrac{25}{36}\mathbb{S} = \dfrac{25}{36}$ and $\mathbb{S} = \dfrac{\dfrac{25}{36}}{1 - \dfrac{25}{36}}$
I plug that into the original formula and find that $\mathbb{P}(\text{Player 2}) = \overline{.54}$ which is what I expected from before.
|
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|
Possible values of $N$ Find the number of values of $N$ such that the below expression is an integer:
$(n+1)^2\over n+7$ is an integer
|
$(n+1)^2=n^2+2n+1=(n+7)(n-5)+36$
So, $\frac{(n+1)^2}{n+7}=n-5+\frac{36}{n+7}$
Assuming $n$ to be an integer, $(n+7)\mid36 \iff (n+7)\mid(n+1)^2$
So, $n+7$ can be any divisor of $36,$ namely $\pm1,\pm2,\pm3,\pm4,\pm6,\pm9,\pm12,\pm18,\pm36$
If we constrain $n$ to be non-negative i.e., if $n+7\ge 7,$ then $n+7$ can be $9,12,18,36$
|
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Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?
|
Here's a generating-function approach to both derive and prove the desired formula.
Write $T(n)$ as $T_n$, and let $f(z) = \sum_{n=1}^\infty T_n z^n$ be the ordinary generating function. Now use the recurrence relation and initial condition to obtain
$$T_1 z^1 + \sum_{n=2}^\infty (T_n-2T_{n-1}) z^n = z + \sum_{n=2}^\infty n z^n,$$
equivalently,
$$f(z)-2z f(z)=z + \left(z\sum_{n=1}^\infty n z^{n-1} - z\right) = z D_z \left(\sum_{n=1}^\infty z^n\right)=z D_z\left(\frac{1}{1-z}\right)=\frac{z}{(1-z)^2}.$$
Solving for $f(z)$ yields
$$f(z) = \frac{z}{(1-2z)(1-z)^2}.$$
Now use partial fraction decomposition to obtain
$$
f(z) = \frac{2}{1-2z}-\frac{1}{1-z}-\frac{1}{(1-z)^2}
= \sum_{n=0}^\infty \left(2 \cdot 2^n - 1 - (n+1)\right)z^n,
$$
which implies that
$$T_n = 2 \cdot 2^n - 1 - (n+1) = 2^{n+1} - n - 2.$$
|
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|
double absolute values I am having a little bit of problem with an inequality with nested absolute values:
$$|z^2-1| \ge |z+|1-z^2||$$
I've tried solving it by making three cases, $z\ge1$, $z\le-1$ and $z$ between $1$ and $-1$ and thus getting rid of absolute values for $z^2-1$ and $1-z^1$, and I am only left with 1 absolute value. But solutions at the end are not what they should be based on the graph. Here, $z$ is real, and WolframAlpha gives this solution.
What I am doing wrong?
|
$|z^2-1| \ge |z+|1-z^2||$
Case 1: Suppose $z \ge 1$. Then $|z^2 - 1| = z^2 - 1$ and $|1 - z^2| = z^2 - 1$:
$z^2-1 \ge |z+(z^2-1)||$
Also $z + (z^2 - 1) > 0$ so:
$z^2-1 \ge z+(z^2-1)$
$0 \ge z$
This is a contradiction.
Case 2: Suppose $z \le -1$. Then $|z^2 - 1| = z^2 - 1$ and $|1 - z^2| = z^2 - 1$:
$z^2-1 \ge |z+(z^2-1)||$
There is a root of $z^2 + z - 1$, so we must case on that.
Case 2a: Suppose $z \le -\frac{\sqrt5 + 1}{2}$, then:
$z^2-1 \ge z^2+z-1$
Also a contradiction.
Case 2b: Suppose $-\frac{\sqrt5 + 1}{2} \le z \le -1$, then
$z^2-1 \ge -z^2-z+1 \Rightarrow 2z^2 + z \ge 0$. $z \le -\frac{\sqrt5 + 1}{2}$ always satisfies this.
Case 3: Suppose $-1 \le z \le 1$, then
$1-z^2 \ge |z+1-z^2)|$
This has a root at $\frac{1-\sqrt5}{2}$, so we case there,
Case 3a: $\frac{1-\sqrt5}{2} \le z \le 1$
$1-z^2 \ge z+1-z^2)$
$0 \ge z$. $\frac{1-\sqrt5}{2} \le z \le 0$ satisfies this.
Case 3b: $-1 \le z \le \frac{1-\sqrt5}{2}$.
$1-z^2 \ge z^2-z-1$
$2z^2 - z \le 0$. This does not hold for negative $z$, so it is a contradiction.
We conclude that $z \le -\frac{\sqrt5 + 1}{2}$ or $\frac{1-\sqrt5}{2} \le z \le 0$.
The graphing method is definitely easier here. It also may be easier to consider the potential roots first and then use more cases instead of cases-with-subcases, though ultimately those are similar arguments.
|
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|
What am I doing wrong in calculating this determinant? I have matrix:
$$
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
And I want to calculate $\det{A}$, so I have written:
$$
\begin{array}{|cccc|ccc}
1 & 2 & 3 & 4 & 1 & 2 & 3 \\
2 & 3 & 3 & 3 & 2 & 3 & 3 \\
0 & 1 & 2 & 3 & 0 & 1 & 2 \\
0 & 0 & 1 & 2 & 0 & 0 & 1
\end{array}
$$
From this I get that:
$$
\det{A} = (1 \cdot 3 \cdot 2 \cdot 2 + 2 \cdot 3 \cdot 3 \cdot 0 + 3 \cdot 3 \cdot 0 \cdot 0 + 4 \cdot 2 \cdot 1 \cdot 1) - (3 \cdot 3 \cdot 0 \cdot 2 + 2 \cdot 2 \cdot 3 \cdot 1 + 1 \cdot 3 \cdot 2 \cdot 0 + 4 \cdot 3 \cdot 1 \cdot 0) = (12 + 0 + 0 + 8) - (0 + 12 + 0 + 0) = 8
$$
But WolframAlpha is saying that it is equal 0. So my question is where am I wrong?
|
$$
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
$$
P_1A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
$$
P_2P_1A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 0 & -1 & -2 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
$$
P_3P_2P_1A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 0 & -1 & -2 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
$$\det(P_3P_2P_1A)=\det(P_3).\det(P_2).\det(P_1).\det(A)=0$$
$$\det(P_3)\neq0,\det(P_2)\neq0,\det(P_1)\neq0$$
$$\implies \det(A)=0$$
|
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|
If $x$, $y$, $x+y$, and $x-y$ are prime numbers, what is their sum?
Suppose that $x$, $y$, $x−y$, and $x+y$ are all positive prime numbers. What is the sum of the four numbers?
Well, I just guessed some values and I got the answer.
$x=5$, $y=2$, $x-y=3$, $x+y=7$. All the numbers are prime and the answer is $17$.
Suppose if the numbers were very big, I wouldn't have got the answer.
Do you know any ways to find the answer?
|
Note that one of $x$ and $y$ has to be even, as if $x$ and $y$ are both odd, $x+y$ and $x-y$ are even, and there is only one even positive prime. As $x - y > 0$, we have $x > y$ and hence, as $2$ is the only even prime and the smallest prime, we have $y = 2$. No $x-2$, $x$ and $x+2$ are prime. But one of them is divisible by $3$: If $x$ has remainder 1 modulo 3, $x+2$ is divisible by 3, and if the remainder is 2, $x-2$ is. So, as $x-2$ is the smallest of the three numbers, we must have $x = 5$ (there is only one positive prime divisible by 3).
|
{
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|
Why does $\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}$? As much as it embarasses me to say it, but I always had a hard time understanding the following equality:
$$
\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}
$$
I always thought that the left-hand side of the above equation was equivalent to
$$
\frac{a}{\frac{b}{x}} = \frac{a}{b} \div \frac{x}{1} = \frac{a}{b} \times \frac{1}{x}
$$
What am I doing wrong, here?
|
$$
\frac{a}{\large\frac{b}{x}} = \large\frac {x}{\not x} \frac{a}{\frac{b}{\not x}} = x\cdot \frac{a}{b}
$$
Multiplying by $\dfrac xx = 1$ does not change the expression; but by multiplying numerator and denominator by $x$, the numerator becomes $ax$ and the denominator becomes $x\cdot \dfrac{b}{x} = b$
|
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|
Problem related with the similarity of matrices I came across this problem which says:
Let $A$ be a $2 \times 2$ matrix such that only $A$ is similar to itself.Then show that A is a scalar matrix, that is $A$ is of the form \begin{pmatrix}
a &0 \\
0 & a
\end{pmatrix}
?
My attempts:
Since $A$ is similar to itself,there exists an invertible matrix P such that A=
$P^{-1}AP$. Then I tried to solve it by choosing A and P of the form \begin{pmatrix}
a &b \\
c & d
\end{pmatrix}
and \begin{pmatrix}
x &y \\
z & w
\end{pmatrix}
respectively. But I could not get the desired result. Please help. Thanks everyone in advance for your time.
|
Alternatively, that $A$ is only similar to itself means $PA=AP$ for all invertible matrix $P$. Therefore
$$
\begin{pmatrix}a+cx&b+dx\\ cy&dy\end{pmatrix}=
\underbrace{\begin{pmatrix}1&x\\0&y\end{pmatrix}}_{P}
\underbrace{\begin{pmatrix}a&b\\c&d\end{pmatrix}}_{A}=
\begin{pmatrix}a&b\\c&d\end{pmatrix}
\begin{pmatrix}1&x\\0&y\end{pmatrix}=
\begin{pmatrix}a&ax+by\\c&cx+dy\end{pmatrix}
$$
for all $y\not=0$ and all $x$.
|
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|
Show $2(x+y+z)-xyz\leq 10$ if $x^2+y^2+z^2=9$ If $x,y,z$ are real and $x^2+y^2+z^2=9$, how can we prove that $2(x+y+z)-xyz\leq 10$?
Please provide a solution without the use of calculus. I know the solution in that way.
|
Here is an attempt. I believe it has to be reconsidered for negative numbers, though.
$0\leq (x+y+z-1)^2$
$0\leq x^2+y^2+z^2+1-2(x+y+z)+2xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
$2(x+y+z)-2xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\leq x^2+y^2+z^2+1$
EDIT: As pointed out my bound is the wrong way :( But I hope the first steps gives others some ideas.
That's almost what we want. We only need to bound the harmonic mean by the quadratic mean now.
$\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\leq \sqrt{\frac{x^2+y^2+z^2}{3}}$
In our case:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \sqrt{3}$
which is enough.
Hope there is no mistake :)
|
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|
If $(a,b)=1$ then prove $(a+b, ab)=1$.
Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.
$(a,b)=1$ means $a$ and $b$ have no prime factors in common
$ab$ is simply the product of factors of $a$ and factors of $b$.
Let's say $k\mid a+b$ where $k$ is some factor of $a$.
Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.
So $a(k-l)=b, \ a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].
But $(a,b)=1$ so $a$ cannot divide $b$.
We have a similar argument for $b$.
So $a+b$ is not divisible by any factors of $ab$.
Therefore, $(a+b, ab)=1$.
Would this be correct? Am I missing anything?
|
Here is another proof using Bezout's Identity:
$$
\begin{align}
ma+nb&=1\tag{1}\\
(n-m)b&=1-m(a+b)\tag{2}\\
(m-n)a&=1-n(a+b)\tag{3}\\
-(m-n)^2ab&=1-(m+n)(a+b)+mn(a+b)^2\tag{4}\\
1&=((m+n)-mn(a+b))\color{#C00000}{(a+b)}-(m-n)^2\color{#C00000}{ab}\tag{5}
\end{align}
$$
Explanation:
$(1)$: $(a,b)=1$ and Bezout's Identity
$(2)$: subtract $m(a+b)$ from both sides of $(1)$
$(3)$: subtract $n(a+b)$ from both sides of $(1)$
$(4)$: multiply $(2)$ and $(3)$
$(5)$: rearrange and collect terms of $a+b$ and $ab$ from $(4)$
Bezout's Identity and $(5)$ say that $(a+b,ab)=1$.
Henning Makholm and Bill Dubuque also give Bezout Identity based proofs. The coefficient of $ab$ is the same in all of our answers, but the coefficient of $a+b$ in mine looks different from theirs. They are the same however:
$$
\begin{align}
(m+n)-mn(a+b)
&=(m+n)\overbrace{(ma+nb)}^{1}-mn(a+b)\\
&=m^2a+n^2b+mn(a+b)-mn(a+b)\\
&=m^2a+n^2b
\end{align}
$$
|
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|
An inequality from the handbook of mathematical functions (by Abramowitz and Stegun) Prove that
$$\frac{1}{x+\sqrt{x^2+2}}<e^{x^2}\int\limits_x^{\infty}e^{-t^2} \, \text dt \le\frac{1}{x+\sqrt{x^2+\displaystyle\tfrac{4}{\pi}}}, \space (x\ge 0)$$
|
(The following argument is adapted from Dümbgen, ''Bounding Standard Gaussian Tail Probabilities.'')
Approximating $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$
Suppose we want to approximate $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$ with a function of the form $\dfrac{e^{-x^2}}{h(x)}.$ Let
$$\Delta(x) = \frac{e^{-x^2}}{h(x)} - \int_x^{\infty} e^{-t^2} \, dt.$$
Then, if $h(x) \to \infty$ as $x \to \infty$, then $\Delta(x) \to 0$ as $x \to \infty$. Because of this, we have the following.
*
*If $\Delta'(x) > 0$ for all $x \geq 0$ then $\Delta(x)$ increases to $0$. Therefore, $\dfrac{e^{-x^2}}{h(x)}$ is a lower bound on $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$ for $x \geq 0$.
*Similarly, if $\Delta'(x) < 0$ for all $x \geq 0$ then $\Delta(x)$ decreases to $0$. Therefore, $\dfrac{e^{-x^2}}{h(x)}$ is an upper bound on $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$ for $x \geq 0$.
We have
$$\Delta'(x) = \frac{e^{-x^2}}{h(x)^2} \left(h(x)^2 - 2xh(x) - h'(x) \right).$$
Thus the sign of $\Delta'(x)$ is determined by the sign of $f(x) = h(x)^2 - 2xh(x) - h'(x)$.
Given the bounds we're trying to show, let's consider functions of the form $h(x) = x + \sqrt{x^2 + c}$. Then $$f(x) = c - 1 - \frac{x}{\sqrt{x^2+c}}.$$ Thus $f(x)$ is decreasing on $[0, \infty)$.
The lower bound
To have $f(x) > 0$ for all $x \geq 0$, we need $$c > 1 + \frac{x}{\sqrt{x^2+c}}, \:\:\:\: x \geq 0.$$ The smallest value of $c$ for which this holds is $c = 2$. Therefore,
$$\frac{1}{x + \sqrt{x^2+2}} < e^{x^2} \int_x^{\infty} e^{-t^2} \, dt, \:\:\:\: x \geq 0,$$ and $2$ is the smallest value of $c$ for which this bound holds for functions of the form $h(x) = x + \sqrt{x^2 + c}$.
The upper bound
To have $f(x) < 0$ for all $x \geq 0$ we can take $c = 1$. However, we can do better this because $f(x)$ is decreasing. If we find a larger value of $c$ such that $\Delta(0)= 0$, then we will have $f(x) > 0$ on $[0, x_0)$ for some $x_0$ and then $f(x) < 0$ on $(x_0, \infty)$. Thus $\Delta(x)$ will initially increase from $0$ and then decrease back to $0$, giving us a tighter upper bound. Since $$\Delta(0) = 0 \Longleftrightarrow \frac{1}{\sqrt{c}} = \int_0^{\infty} e^{-t^2}\, dt = \frac{\sqrt{\pi}}{2},$$
we have $c = \dfrac{4}{\pi}$ yielding a tighter upper bound than $c = 1$.
Therefore, $$e^{x^2} \int_x^{\infty} e^{-t^2} \, dt \leq \frac{1}{x + \sqrt{x^2+\frac{\pi}{4}}}, \:\:\:\: x \geq 0,$$ and $\dfrac{\pi}{4}$ is the smallest value of $c$ for which this bound holds for functions of the form $h(x) = x + \sqrt{x^2 + c}$.
|
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|
How to get from $\frac{x}{x+1}\;$ to $\;1 - \frac{1}{x+1}$? Please show me how to manipulate $\dfrac{x}{x+1}\;\;$ to get $\;\;1 - \dfrac{1}{x+1}$
|
Sometimes when working with polynomial division, it helps to recall how we handle division of plain old integers:
Consider the fraction $\dfrac{17}{20}$. Note that $\dfrac{17}{20} = \dfrac{20 - 3}{20} = \dfrac{20}{20} - \dfrac{3}{20}$.
We can do the same for $\dfrac{x}{x+1}$:
$$\dfrac{x}{(x+1)}\; = \;\dfrac{(x+1) - 1}{(x+1)} \;=\; \dfrac{(x+1)}{(x+1)} - \dfrac{1}{(x+1)} \;=\; 1 - \dfrac{1}{(x+1)}$$
Also, since you'll likely be moving on to division of more complex polynomials very soon:
You can use "long division", just as you would for dividing, say $17$ by $20$. In this case you have $x$ as the "dividend" (numerator) and $(x + 1)$ the "divisor" (denominator):
So here, $1$ is your "quotient", and $-1$ is your "remainder", giving us: $\dfrac{x}{x+1} = 1 - \dfrac{1}{x+1}$
|
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|
How do we deal with recurrence relation characteristic equations that are not quadratic or have imaginary roots? Suppose we have $$H(n) = H(n-1)-H(n-2) \rightarrow x^2-x+1 \rightarrow r_1 = \frac{1+\sqrt{-3}}{2}, r_2 = \frac{1-\sqrt{-3}}{2}$$
or
$$H(n) = H(n-1)+H(n-2)+H(n-3) \rightarrow x^3-x^2-x-1=0$$
In either case, how would the recurrence relation be solved? Are there other techniques for complex roots/non-quadratics?
|
The first one repeats
$$ a,b,b-a,-a,-b,a-b, a,b,b-a,-a,-b,a-b, a,b,b-a,-a,-b,a-b, a,b,c,-a,-b,a-b,\ldots $$
This does follow, eventually, from the description
$$ H(n) = A \, r_1^n + B \, r_2^n $$
for some complex constants $A,B.$
Indeed,
$$
\left( \begin{array}{r}
A \\
B
\end{array}
\right) =
\left( \begin{array}{rr}
\frac{1}{2} - \frac{i}{2 \sqrt 3} & -\frac{1}{2} - \frac{i}{2 \sqrt 3} \\
\frac{1}{2} + \frac{i}{2 \sqrt 3} & -\frac{1}{2} + \frac{i}{2 \sqrt 3}
\end{array}
\right) \cdot
\left( \begin{array}{r}
a \\
b
\end{array}
\right)
$$
So, for the sequence
$$ 1,-1,-2,-1,1,2, 1,-1,-2,-1,1,2,\ldots $$
we have $a=1,b=-1,$ then $A=1,B=1$
and $$ H(n) = r_1^n + r_2^n. $$
For the sequence
$$ 1,1,0,-1,-1,0, 1,1,0,-1,-1,0,\ldots $$
we have $a=1,b=1,$ then $A=- \frac{i}{ \sqrt 3} ,B= \frac{i}{ \sqrt 3}$
and $$ H(n) = - \frac{i}{ \sqrt 3} \left( r_1^n - r_2^n \right). $$
The repetition of length 6 and the half-repetition with negation of length 3 can be seen from both roots satisfying $r_j^6 = 1$ and $r_j^3 = -1.$ Or, you can just start a sequence with symbols $a,b$ and confirm the pattern I gave first.
|
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|
Find the solutions of Boolean equations It's given 4 Boolean equations. I need to find the number of solutions of each.
$a)\ x_{1}x_{2}\oplus x_{2}x_{3}\oplus\ ...\ \oplus\ x_{n-1}x_{n}=1$
$b)\ x_{1}x_{2}\vee x_{2}x_{3}\vee\ ...\ \vee\ x_{n-1}x_{n}=1$
$c)\ x_{1}x_{2}\oplus x_{3}x_{4}\oplus\ ...\ \oplus\ x_{2n-1}x_{2n}=1$
$d)\ x_{1}x_{2}\vee x_{3}x_{4}\vee\ ...\ \vee\ x_{2n-1}x_{2n}=1$
I have solved $c)$ in the following way. First $2n-2$ variables can have any values. The sum of $\ x_{1}x_{2}\oplus x_{3}x_{4}\oplus\ ...\ \oplus\ x_{2n-3}x_{2n-2}$ can be $0$ or $1$, then the value of $x_{2n-1}x_{2n}$ depends on it so it is fixed. So the number of solutions is $2^{2n-2}$. Is it right?
Please give me solution or hint for the others.
|
Part d)
There are $4^n$ possible values for $(x_1, x_2, \ldots, x_{2n-1}, x_{2n})$ which
we can divide into $n$ $2$-bit vectors $(x_1,x_2), (x_3,x_4),\ldots, (x_{2n-1},x_{2n})$
each of which can take on $4$ values. Then,
$$x_{1}x_{2}\vee x_{3}x_{4}\vee\ ...\ \vee\ x_{2n-1}x_{2n}= 0$$
if and only if $(x_1,x_2) \neq (1,1)$ and $(x_3,x_4) \neq (1,1)$ and
$\cdots$ and $(x_{2n-1},x_{2n}) \neq (1,1)$. Thus, $3^n$ of the $4^n$ values
of $(x_1, x_2, \ldots, x_{2n-1}, x_{2n})$ result in the expression above having
value $0$.
Thus, the number of solutions to
$x_{1}x_{2}\vee x_{3}x_{4}\vee\ ...\ \vee\ x_{2n-1}x_{2n}= 1$
is $4^n-3^n$.
Part c)
Let $x_{2i-1}x_{2i} = y_i$ where we think of $Y_i$ as a
Bernoulli random variable with parameter $p = \frac{1}{4}$, and
$Z = Y_1+Y_2+\cdots+Y_{n}$ as a binomial random variable with
parameters $(n,p)$. Now, the value of $Z$ is an odd number
if and only if
$$x_{1}x_{2}\oplus x_{3}x_{4}\oplus\ ...\ \oplus\ x_{2n-1}x_{2n}=1$$
and we have that
$$\begin{align*}P\{Z ~\text{is an odd number}\}
&= \binom{n}{1}p(1-p)^{n-1}+\binom{n}{3}p^3(1-p)^{n-3} + \cdots\\
&= \frac{1}{2}\left[\sum_{j=0}^n \binom{n}{j}p^j(1-p)^{n-j}
- \sum_{j=0}^n \binom{n}{j}(-1)^jp^j(1-p)^{n-j}\right]\\
&= \frac{1}{2}\left[(p + (1-p))^n - ((1-p)-p)^n\right]\\
&= \frac{1}{2}\left[1 - (1-2p)^n\right]\\
&= \frac{1}{2} - \frac{1}{2^n}.\end{align*}$$
Turning back to Boolean variables, we conclude that
The
number of solutions to
$x_{1}x_{2}\oplus x_{3}x_{4}\oplus\ ...\ \oplus\ x_{2n-1}x_{2n}=1$
is $2^{2n-1}-2^{n-1}$.
I will leave it to you to try and apply these ideas to parts a) and b)
|
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|
How can I systematically find the roots of $ x^4 + 1?$ Is there some algorithm?
Possible Duplicate:
How to find the root of $x^4 +1$
What algorithms can be used for finding all roots of the given polynomial:
\begin{equation}
x^4 + 1 = 0
\end{equation}
|
$$x^4 + 1 = (x^2 + 1)^2 - 2x^2 = \\
(x^2+1+\sqrt2 x) (x^2+1-\sqrt2 x) =\\
(x^2 - x\sqrt 2+1)(x^2 + x\sqrt2+1)=0$$
*
*$x^2 - x\sqrt 2+1=0$
$$x_{1}={\sqrt2+\sqrt{-2}\over2}={\sqrt2\over2}(1+i)$$
$$x_{2}={\sqrt2-\sqrt{-2}\over2}={\sqrt2\over2}(1-i)$$
*$x^2 + x\sqrt2+1=0$
$$x_{3}={-\sqrt2+\sqrt{-2}\over2}=-{\sqrt2\over2}(1-i)$$
$$x_{4}={-\sqrt2-\sqrt{-2}\over2}=-{\sqrt2\over2}(1+i)$$
|
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|
Solutions for $ 2^i = 3^ j - 1 $ or $ 2^i = 3^ j + 1 $ Are there any solutions for $ 2^i = 3^ j - 1 $ or $ 2^i = 3^ j + 1 $,
for $i>3$ and $j>2$ ?
Thanks! $:)$
|
Gottfried Helms has dealt with $2^i=3^j+1$. For completeness we deal with $2^i =3^j-1$.
This has the solutions $i=j=1$ and $i=3$, $j=2$. We show there are no others.
First we deal with odd $j$. Note that
$$3^j-1=(3-1)\left(3^{j-1}+3^{j-2}+\cdots +1\right).$$
The term $3^{j-1}+3^{j-2}+\cdots +1$ is the sum of an odd number of odd numbers, so it is odd. It follows that if $j$ is odd, then $3^j-1$ can only be a power of $2$ if $j=1$.
Next we deal with even $j$, say $j=2k$. Then $3^j=3^{2k}-1=(3^k-1)(3^k+1)$. This can be a power of $2$ only if both $3^k-1$ and $3^k+1$ are powers of $2$. But the only two powers of $2$ that differ by $2$ are $2$ and $4$. It follows that $j=2$ and $i=3$.
|
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|
What is the value of D here? Number $S$ is obtained by squaring the sum of digits of a two digit number $D$. If the difference between $S$ and $D$ is $27$, then the two digit number $D$ is?
My thoughts:
Let the two digit number $D$ be $AB$.
And so $S=(A+B)^2$
If $\,S-D=27,\,$ then $\,(A+B)^2 -AB=27$
$$A^2 + 2AB + B^2 - AB=27$$
Now how to obtain the value of $D$ further?
|
I believe I see 2 solutions for this problem. First let's redefine $D$ algebraicly as $D=10A+B$. So our equation is
$$(A+B)^2-10A-B=27$$
I don't know if you know any modular arithmetic. You may at least be aware of the divisibility test for $9$. The sum of the digits of a number is closely related to its remainder when divided by $9$. So it is a reasonable guess that we try to determine the remainder of both sides when divided by $9$. This simplifies the equation to
$$(A+B)^2-A-B\equiv0\pmod9$$
$$(A+B)^2-(A+B)\equiv0\pmod9$$
$$(A+B)(A+B-1)\equiv0\pmod9$$
So we have the product of 2 consecutive integers has a remainder of $0$ when divided by $9$, or in other words this product is divisible by $9$. Since 2 consecutive numbers share no common factors besides $1$, either $A+B$ is divisible by $9$, or $A+B+1$ is divisible by $9$. $D=99$ is too large and $D=10$ is too small, leaving 2 possibilities.
Our first possibility is $A+B=9$. This gives us
$$S=9^2=81$$
$$D=S-27=81-27=54$$
Our second possibility is $A+B-1=9$ or $A+B=10$. This yields
$$S=10^2=100$$
$$D=S-27=100-27=73$$
Both values check out. $D$ is either $54$ or $73$.
|
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|
Prove that if $d$ divides $n$, then $2^d -1$ divides $2^n -1$ Prove that if $d$ divides $n$, then $2^d -1$ divides $2^n -1$.
Use the identity $x^k -1 = (x-1)*(x^{k-1} + x^{k-2} + \cdots + x +1)$
|
EDIT: Here's a visual proof. In binary base, $2^{d}-1 = \underbrace{1\cdots 1}_{d \text{ times}}$, and
\begin{align*}
2^n-1 &=\underbrace{1\cdots 1}_{n \text{ times}} = \underbrace{\underbrace{1\cdots 1}_{d \text{ times}} \cdots \underbrace{1\cdots 1}_{d \text{ times}}}_{n/d \text{ times}} \\
&=\underbrace{1\cdots 1}_{d \text{ times}}(1+1\underbrace{0\cdots 0}_{d}+1\underbrace{0\cdots 0}_{2d}+\cdots +1\underbrace{0\cdots 0}_{(n/d-1)d}) \\
&=(2^d - 1)(1+2^d + 2^{2d} + \cdots + 2^{(n/d-1)d}) \blacksquare
\end{align*}
Visualization apart, I just used the hint with $x=2^d-1, k= \frac{n}{d}$.
Original proof (it doesn't use the hint):
In general, $\gcd(2^a-1,2^b-1) = 2^{\gcd(a,b)}-1$. When $a|b$ we get your result.
The proof is by applying the Euclidean algorithm to compute $\gcd(a,b)$: Assuming $a\ge b$,
\begin{align*}
\gcd(2^a-1,2^b-1) &=\gcd(2^a - 1 - (2^b-1),2^b-1)=\gcd((2^{a-b}-1)2^b,2^b-1) \\
&=\gcd(2^{a-b}-1,2^b-1).
\end{align*}
So the pair $a,b$ is replaced by $b-a,a$. Repeating this ends with the pair $\gcd(a,b),\gcd(a,b)$, so the $\gcd$ is $\gcd(2^{\gcd(a,b)}-1,2^{\gcd(a,b)}-1)=2^{\gcd(a,b)}-1$.
More generally, $\gcd(a^n - 1, a^m-1)=a^{\gcd(n,m)-1}-1$ ($a$ is an integer or a variable!)
|
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|
How to compute the determinant of a tridiagonal matrix with constant diagonals? How to show that the determinant of the following $(n\times n)$ matrix
$$\begin{pmatrix}
5 & 2 & 0 & 0 & 0 & \cdots & 0 \\
2 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 2 & 5 & 2 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots & \vdots & \vdots \\
0 & \cdots & \cdots & 0 & 2 & 5 & 2 \\
0 & \cdots & \cdots & \cdots & \cdots & 2 & 5
\end{pmatrix}$$
is equal to $\frac13(4^{n+1}-1)$?
More generally:
How does one compute the determinant of the following tridiagonal matrix (where the three diagonals are constant)?
$$M_n(a,b,c) = \begin{pmatrix}
a & b & 0 & 0 & 0 & \cdots & 0 \\
c & a & b & 0 & 0 & \cdots & 0 \\
0 & c & a & b & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots& \vdots & \vdots \\
0 & \cdots & \cdots & 0 & c & a & b \\
0 & \cdots & \cdots & \cdots & \cdots & c & a
\end{pmatrix}$$
Here $a,b,c$ can be taken to be real numbers, or complex numbers.
In other words, the matrix $M_n(a,b,c) = (m_{ij})_{1 \le i,j \le n}$ is such that
$$m_{ij} = \begin{cases}
a & i = j, \\
b & i = j - 1, \\
c & i = j + 1, \\
0 & \text{otherwise.}
\end{cases}$$
There does not seem to be an easy pattern to use induction: the matrix is not a diagonal block matrix of the type $M = \bigl(\begin{smallmatrix} A & C \\ 0 & B \end{smallmatrix}\bigr)$ (where we could use $\det(M) = \det(A) \det(B)$ for the induction step), and there are no lines or columns with only one nonzero entry, so Laplace expansion gets complicated quickly.
Is there a general pattern that one could use? Or is the answer only known on a case-by-case basis? It's possible to compute the determinant by hand for small $n$:
$$\begin{align}
\det(M_1(a,b,c)) & = \begin{vmatrix} a \end{vmatrix} = a \\
\det(M_2(a,b,c)) & = \begin{vmatrix} a & b \\ c & a \end{vmatrix} = a^2 - bc \\
\det(M_3(a,b,c)) & = \begin{vmatrix} a & b & 0 \\ c & a & b \\ 0 & c & a \end{vmatrix} = a^3 - 2abc
\end{align}$$
But there is no readily apparent pattern and the computation becomes very difficult when $n$ gets large.
|
Let $M_n$ be the $n \times n$ matrix. Calculate the determinant by expanding along the first row and then by the second column, we get $ Det(M_n) = 5 Det(M_{n-1} ) - 4 Det(M_{n-2})$.
Let $Det(M_n) = D_n$, so $D_n$ satisfies the recurrence relation $D_n - 5 D_{n-1} + 4 D_{n-2} = 0$, with initial values $D_0 = 1, D_1 = 5$. The characteristic equation $x^2 - 5x + 4$ has roots $x= 4, 1$, so the solution has form $A4^n + B1^n$. Plugging in the initial values, we get $A= \frac {4}{3}, B= -\frac {1}{3}$, which yields the value $D_n = \frac {1}{3} (4^{n+1} - 1)$.
$D_0 = 1$ because it is the empty product, which by definition has the value 1. If you do not like to use $D_0 = 1$, you can just calculate $D_1 = 5$ and $D_2 = 5 \times 5 - 2 \times 2 = 21$ and then find the values of $A, B$.
|
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|
Why do we choose $3$ to be positive after $\sqrt{9 - x^2}$ in the following substitution? The integral $$\int \frac{\sqrt{9 - x^2}}{x^2}dx$$ is solved in my book by letting $x = 3\sin\theta$ where $-\frac {\pi}{2} \le \theta \le \frac {\pi}{2}$.
Then, $dx = 3\cos\theta\,d\theta$
and, $$\sqrt{9-x^2} = 3|\cos\theta| = 3\cos\theta$$
So, $$\int \frac{\sqrt{9 - x^2}}{x^2}dx = \int \cot^2 \theta \ d\theta = -\cot\theta - \theta + C$$
Returning to the original variable, $$\int \frac{\sqrt{9 - x^2}}{x^2}dx = -\frac {\sqrt{9 - x^2}}{x} - \sin^{-1}\left(\frac{x}{3}\right) + C$$
I don't understand why $\sqrt{9-x^2} = 3|\cos\theta| = 3\cos\theta \,$ instead of $\sqrt{9-x^2} = |3||\cos\theta| = |3|\cos\theta$. I feel like I have problems understanding this because I am not sure what is the purpose of the absolute value signs in this case, are they to indicate that, for example, $|\cos\theta| = \pm\cos\theta$? If that's the case, why do we choose $3$ to be positive instead of negative?
|
The core of your problem seems to be a confusion about the meaning of $\left| \cdot \right|$. $\left| x \right|$ is not the same as $\pm x$. Instead:
*
*$\left| x \right| = -x$ if $x \le 0$
*$\left| x \right| = x$ if $x \ge 0$
Now, because $\sqrt{\cdot}$ is a real-valued function, it can only return one of the quadratic roots of its argument. We've chosen that it's the nonnegative one, so $\sqrt{x^2}=\left|x\right|$, while the roots of $x^2-a$ are $\pm \sqrt{a}$.
|
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|
Find the limit of the expression $2^{n+1}\sqrt{2-t_n}$ as $n\to\infty$, where $t_n=\sqrt{2+t_{n-1}}$
Possible Duplicate:
Why is this series of square root of twos equal $\pi$?
Find the limit of the expression $2^{n+1}\sqrt{2-t_n}$ as $n\rightarrow\infty$, where $t_1=\sqrt{2}$, $t_2=\sqrt{2+\sqrt{2}}$, $t_3=\sqrt{2+\sqrt{2+\sqrt{2}}}$ and so on.
|
$L_n = 2^{n+1} \sqrt{2-t_n}$. Then $L_n = 2^{n+1} \sqrt{2 - \sqrt{2+t_{n-1}}}$. Now $$2 - \sqrt{2+t_{n-1}} = \dfrac{4 - (2+t_{n-1})}{2 + \sqrt{2+t_{n-1}}} = \dfrac{2-t_{n-1}}{2 + t_n} = \dfrac{2-t_{n-1}}{t_{n+1}^2}$$
Hence, $$L_n = 2^{n+1} \sqrt{\dfrac{2-t_{n-1}}{t_{n+1}^2}} = \dfrac{2L_{n-1}}{t_{n+1}}$$
We have $L_0 = 2 \sqrt{2}$. Hence,
$$L_n = \dfrac{2}{t_{n+1}} \cdot \dfrac{2}{t_{n}} \cdot \dfrac{2}{t_{n-1}} \cdots \dfrac{2}{t_{2}} L_0 = \dfrac{2^n}{\displaystyle \prod_{k=2}^{n+1} t_k} 2 \sqrt{2} = 2 \dfrac{2^{n+1}}{\displaystyle \prod_{k=1}^{n+1} t_k}$$
Hence, we need to evaluate $$S_n = \dfrac{\displaystyle \prod_{k=1}^n t_k}{2^n}$$
Recall that $t_n \uparrow 2$. Let $t_ n = 2 \cos(\theta_n)$. We then have
$$\cos(\theta_{n+1}) = \dfrac{t_{n+1}}2 = \dfrac{\sqrt{2+t_n}}2 = \dfrac{\sqrt{2+2 \cos(\theta_n)}}2 = \cos(\theta_n/2)$$
Hence, we get that
$$\cos(\theta_{n+1}) = \cos(\theta_1/2^n)$$
Hence, $$S_n = \dfrac{\displaystyle \prod_{k=1}^n t_k}{2^n} = \prod_{k=0}^{n-1} \cos(\theta_1/2^k) = \dfrac{\displaystyle \sin(\theta_1/2^{n-1})\prod_{k=0}^{n-1} \cos(\theta_1/2^k)}{\sin(\theta_1/2^{n-1})} = \dfrac1{2^n} \dfrac{\sin(2\theta_1)}{\sin(\theta_1/2^{n-1})}$$
Note that $\cos(\theta_1) = \dfrac1{\sqrt{2}}$, which implies $\theta_1 = \dfrac{\pi}4$. Hence,
$$S_n = \dfrac1{2^n \sin(\pi/2^{n+1})}$$ Hence, $S_n \to \dfrac2{\pi}$. Hence, $$L_n \to \pi$$
|
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|
$\tan(x) = x$. Find the values of $x$ How can I find the possible values of $x$ for:
$\tan(x)=x$
mathematically?
|
$$\frac{\sin(x)}{\cos(x)}=x$$
$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$$
$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}...$$
Your question is equivalent to solving the equation
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=x-\frac{x^3}{2!}+\frac{x^5}{4!}-\cdots$$
$$x^3\left(\frac{1}{3!}-\frac{1}{2!}\right)-x^5\left(\frac{1}{5!}-\frac{1}{4!}\right)+x^7\left(\frac{1}{7!}-\frac{1}{6!}\right)+\cdots=0$$
Evidently giving
$$x=0$$
The other solutions are given by the equation
$$\left(\frac{1}{3!}-\frac{1}{2!}\right)-x^2\left(\frac{1}{5!}-\frac{1}{4!}\right)+x^4\left(\frac{1}{7!}-\frac{1}{6!}\right)-\cdots=0$$
But I don't know if there is a way to get from that to a closed form.
|
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|
Find all polynomials $P$ such that $P(x^2+1)=P(x)^2+1$ Find all polynomials $P$ such that
$$P(x^2+1)=P(x)^2+1$$
|
Let $P(y)=\sum_{0\le r\le n}a_ry^r$
So, $$P(1+x^2)=\sum_{0\le r\le n}a_r(1+x^2)^r=a_0+a_1(1+x^2)+a_2(1+\binom 21x^2+x^4)+\cdots
+a_{n-1}(1+\binom {n-1}1x^2+\binom {n-1}2x^4+\cdots+\binom {n-1}{n-2}x^{2(n-2)}+\binom {n-1}{n-1}x^{2(n-1)})
+a_n(1+\binom n1x^2+\binom n2x^4+\cdots+\binom n{n-1}x^{2(n-1)}+\binom n nx^{2n})$$
$$=x^{2n}a_n+x^{2n-2}(a_n\binom n{n-1}+a_{n-1})+x^{2n-4}(a_n\binom n{n-2}+a_{n-1}\binom {n-1}{n-2}+a_{n-2})+x^2(a_n\binom n1+a_{n-1}\binom{n-1}1+\cdots+a_2\binom21+a_1)+\sum_{0\le r\le n}a_r$$
and $$\{P(x)\}^2+1=\{\sum_{0\le r\le n}a_rx^r\}^2+1$$
$$=a_n^2x^{2n}+x^{2n-1}2a_na_{n-1}
+x^{2n-2}(a_{n-1}^2+2a_na_{n-2})+x^{2n-3}2(a_na_{n-3}+a_{n-1}a_{n-2})
+x^{2n-4}(a_{n-2}^2+2a_na_{n-4}+2a_{n-1}a_{n-3})+\cdots+x^2(a_1^2+2a_0a_2)+\sum_{0\le r\le n}a_r^2+1$$
Comparing the coefficients of the different powers of $x$
$r=n\implies a_n=a_n^2\implies a_n=1$ as $a_n\ne0$
$r=n-1\implies 2a_na_{n-1}=0\implies a_{n-1}=0$
$r=n-2\implies a_n\binom n{n-1}+a_{n-1}=a_{n-1}^2+2a_na_{n-2}\implies a_{n-2}=\frac n2$
$r=n-3\implies 2(a_na_{n-3}+a_{n-1}a_{n-2})=0\implies a_{n-3}=0$
$r=n-4\implies a_n\binom n{n-2}+a_{n-1}\binom {n-1}{n-2}+a_{n-2}=a_{n-2}^2+2a_na_{n-4}+2a_{n-1}a_{n-3}\implies a_{n-4}=\frac {n^2}8=\frac1{2!}\left(\frac n2\right)^2$
$r=n-5\implies 2(a_na_{n-5}+a_{n-1}a_{n-4}+a_{n-2}a_{n-3})=0\implies a_{n-5}=0$
$r=n-6\implies 2(a_na_{n-6}+a_{n-1}a_{n-5}+a_{n-2}a_{n-4})+a_{n-3}^2=a_n\binom n{n-3}+a_{n-1}\binom{n-1}{n-3}+a_{n-2}\binom{n-2}{n-3}+a_{n-3}\implies a_{n-6}=\frac1{3!}\left(\frac n2\right)^3-\frac n3$
|
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|
Show that $\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=0$ I am asked to prove this statement $^{*}$. I am trying now, but it is getting to small and tiny steps that I even loose my way. my steps are as follows:
$$\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=0^{*}$$
$\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=\dfrac{(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}) \cdot (\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n})}{(\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n})}=\dfrac{(\sqrt[3]{n+\sqrt{n}})^2-(\sqrt[3]{n})^2}{(\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n})}=\dfrac{\sqrt[3]{n^2+2n\sqrt{n}+n}}{\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n}}=\dfrac{(n+\sqrt{n})^{\frac{2}{3}}-n^{\frac{2}{3}}}{(n+\sqrt{n})^{\frac{1}{3}}+n^{\frac{1}{3}}}= .. help = 0$ $$if \quad n\rightarrow \infty$$
|
Hint:
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
Apply for $a=\sqrt[3]{n+\sqrt n}$ and $b=\sqrt[3] n$
|
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|
Proving:$(1+x)(1+x^2)(1+x^3)\cdots(1+x^n)\ge(1+x^{\frac{n+1}{2}})^n$ How to prove that :
$$(1+x)(1+x^2)(1+x^3)\cdots (1+x^n)\ge(1+x^{\frac{n+1}{2}})^n$$
|
It's wrong of course! Try $n=3$ and $x=-1$.
For non-negative $x$ it's just Holder:
$$(1+x)(1+x^2)...(1+x^n)\geq\left(1+\sqrt[n]{x\cdot x^2\cdot...\cdot x^n}\right)^n=\left(1+x^{\frac{n+1}{2}}\right)^n$$
|
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|
Solve an equation involving the sine and the inverse tangent The equation is
$$ \sin\left(\frac{x}{x-1}\right) + 2 \tan^{-1}\left(\frac{1}{x+1}\right)=\frac{\pi}{2} $$
The answer is $0$, but I do not know how they got that.
|
$$2\tan^{-1}\left(\frac1{x+1}\right)=\cos^{-1}\left(\frac{1-\left(\frac1{x+1}\right)^2}{1+\left(\frac1{x+1}\right)^2}\right)=\cos^{-1}\frac{x^2+2x}{x^2+2x+2}$$
as $$2\tan^{-1}y=\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)$$ (Proof below)
$$\implies \sin^{-1}\left(\frac x{x-1}\right)=\frac\pi2-\cos^{-1}\left(\frac{x^2+2x}{x^2+2x+2}\right)=\sin^{-1}\left(\frac{x^2+2x}{x^2+2x+2}\right)$$
So, $$\frac x{x-1}=\frac{x^2+2x}{x^2+2x+2}$$
Now, solve for $x$
[Proof: $$\tan^{-1}y=z\implies y=\tan z, \cos 2z=\frac{1-\tan^2z}{1+\tan^2z}=\frac{1-y^2}{1+y^2}\implies 2z=\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)\implies 2\tan^{-1}y=\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)$$]
|
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|
How can I solve $\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$? The limit is: $$\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$$
The limit is the kind of $1^\infty$, since:
$$\lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x=\left(\frac{2\times\frac{\pi}{2}}{\pi}\right)^\infty=\left(\frac{\pi}{\pi}\right)^\infty=1^\infty$$
This is the farthest I went: $$L = \lim_{x\to\infty}\left(\frac{2\arctan(x)}{\pi}\right)^x$$
$$\ln(L) = \lim_{x\to\infty}\ln\left(\frac{2\arctan(x)}{\pi}\right)^x=\lim_{x\to\infty}x·\ln\left(\frac{2\arctan(x)}{\pi}\right)$$
Which becomes $0·\infty$
and I dont know where to go from here
|
There is a way to complete your argument, although the work is a little tedious. Observe that
$$
x \cdot \ln \left( \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) =
\frac{\ln \left( \dfrac{2 \cdot {\tan^{-1}}(x)}{\pi} \right)}{\left( \dfrac{1}{x} \right)}.
$$
As
$$
\lim_{x \to \infty} \ln \left( \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) = \ln \left( \lim_{x \to \infty} \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) = \ln(1) = 0
$$
and
$$
\lim_{x \to \infty} \frac{1}{x} = 0,
$$
we can apply l’Hôpital’s Rule to obtain
\begin{align}
\ln(L)
&= \lim_{x \to \infty} x \cdot \ln \left( \frac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) \\
&= \lim_{x \to \infty} \frac{\ln \left( \dfrac{2 \cdot {\tan^{-1}}(x)}{\pi} \right)}{\left( \dfrac{1}{x} \right)} \\
&= \lim_{x \to \infty} \frac{\dfrac{d}{dx} \left[ \ln \left( \dfrac{2 \cdot {\tan^{-1}}(x)}{\pi} \right) \right]}{\dfrac{d}{dx} \left[ \dfrac{1}{x} \right]} \\
&= \lim_{x \to \infty} \frac{\left[ \dfrac{1}{(1 + x^{2}) \cdot {\tan^{-1}}(x)} \right]}{\left( - \dfrac{1}{x^{2}} \right)} \\
&= - \frac{2}{\pi}.
\end{align}
Therefore,
$$
L = e^{\ln(L)} = e^{- 2/\pi}.
$$
|
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|
Prove quotient of factorials is integral If $n$ is an integer $\gt 0$, prove
$$\frac{(30n)!n!}{(15n)!(10n)!(6n)!}$$
is also an integer. I understand that a general approach is proving that the power of any prime factor is greater in the numerator than it is in the denominator, but I haven't been able to formulate this into a rigorous proof.
|
This is a gross brute force answer.
We will show that for any positive integer $D$:
$$0\leq\left\lfloor\frac{30n}{D}\right\rfloor + \left\lfloor\frac{n}{D}\right\rfloor - \left\lfloor\frac{15n}{D}\right\rfloor - \left\lfloor\frac{10n}{D}\right\rfloor - \left\lfloor\frac{6n}{D}\right\rfloor$$
This is enough to show your theorem because when $D=p^k$ is a prime power, this is the total number of multiples of $p^k$ in the numerator minus the total number of multiples of $p^k$ in the denominator.
Write $30n = Dq+r$ for some $0\leq r < D$. Then the right hand side is:
$$q + \left\lfloor\frac{q}{30}\right\rfloor - \left\lfloor\frac{q}{2}\right\rfloor - \left\lfloor\frac{q}{3}\right\rfloor - \left\lfloor\frac{q}{5}\right\rfloor$$
Writing $q=30p+s$ with $0\leq s<30$, we see this is:
$$30p +s + p - 15p -\left\lfloor\frac{s}{2}\right\rfloor - 10p - \left\lfloor\frac{s}{3}\right\rfloor - 6p- \left\lfloor\frac{s}{5}\right\rfloor\\=s-\left\lfloor\frac{s}{2}\right\rfloor - \left\lfloor\frac{s}{3}\right\rfloor - \left\lfloor\frac{s}{5}\right\rfloor$$
If this is non-negative for $s=0,...,29$ you are done. You can brute force from there.
Note
It seems like there should be some direct proof for:
$$0\leq s-\left\lfloor\frac{s}{2}\right\rfloor - \left\lfloor\frac{s}{3}\right\rfloor - \left\lfloor\frac{s}{5}\right\rfloor$$
for $s=0,\dots,29$. However, it is not true for $s=30$.
|
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|
Use Taylor Series Expansion in Calculating Integral I want to approximate this integral:
$$
I = \int_0^\infty {{e^{ - bx}}\ln \left( {{a_1}{e^{ - {b_1}x}} + {a_2}{e^{ - {b_2}x}}} \right)dx}
$$
where $b,{a_1},{a_2},{b_1},{b_2} > 0$ and $b_2>b_1$.
Here is my answer:
We can observe that there exists a constant $c$ so that if $x>c$, then $\frac{{{a_2}}}{{{a_1}}}{e^{ - \left( {{b_2} - {b_1}} \right)x}} \in \left( {0,1} \right)$; and if $0 \le x \le c$, then $\frac{{{a_1}}}{{{a_2}}}{e^{ - \left( {{b_1} - {b_2}} \right)x}} \in \left( {0,1} \right]$. Actually the value of $c$ is: $$c = \frac{{\ln \left( {{a_1}/{a_2}} \right)}}{{{b_1} - {b_2}}}$$
Using the Taylor Series Expansion $\ln \left( {1 + y} \right) = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{{y^n}}}{n}}$ with $y \in \left( {0,1} \right)$, we obtain:
$$
\begin{array}{l}
I = - \left( {\ln {a_1} - {b_1}} \right)\frac{{{e^{ - bc}}\left( {b + 1} \right)}}{{{b^2}}} + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\int_c^\infty {{e^{ - bx}}{{\left( {\frac{{{a_2}}}{{{a_1}}}} \right)}^n}{e^{ - n\left( {{b_2} - {b_1}} \right)x}}dx} } \\
- \left( {\ln {a_2} - {b_2}} \right)\frac{{\left( {1 - {e^{ - bc}}} \right)\left( {b + 1} \right)}}{{{b^2}}} + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\int_0^c {{e^{ - bx}}{{\left( {\frac{{{a_1}}}{{{a_2}}}} \right)}^n}{e^{ - n\left( {{b_1} - {b_2}} \right)x}}dx} } \\
= - \left( {\ln {a_1} - {b_1}} \right)\frac{{{e^{ - bc}}\left( {b + 1} \right)}}{{{b^2}}} + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}{{\left( {\frac{{{a_2}}}{{{a_1}}}} \right)}^n}\frac{{{e^{ - \left[ {n\left( {{b_2} - {b_1}} \right) + b} \right]c}}}}{{\left[ {n\left( {{b_2} - {b_1}} \right) + b} \right]}}} \\
- \left( {\ln {a_2} - {b_2}} \right)\frac{{\left( {1 - {e^{ - bc}}} \right)\left( {b + 1} \right)}}{{{b^2}}} + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}{{\left( {\frac{{{a_1}}}{{{a_2}}}} \right)}^n}\frac{{1 - {e^{ - \left[ {n\left( {{b_1} - {b_2}} \right) + b} \right]c}}}}{{\left[ {n\left( {{b_1} - {b_2}} \right) + b} \right]}}} \\
= - \frac{{\left( {b + 1} \right)}}{{{b^2}}}\left[ {\left( {\ln {a_1} - {b_1}} \right){e^{ - bc}} + \left( {\ln {a_2} - {b_2}} \right)\left( {1 - {e^{ - bc}}} \right)} \right]\\
+ \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\left( {{{\left( {\frac{{{a_2}}}{{{a_1}}}} \right)}^n}\frac{{{e^{ - \left[ {n\left( {{b_2} - {b_1}} \right) + b} \right]c}}}}{{\left[ {n\left( {{b_2} - {b_1}} \right) + b} \right]}} + {{\left( {\frac{{{a_1}}}{{{a_2}}}} \right)}^n}\frac{{1 - {e^{ - \left[ {n\left( {{b_1} - {b_2}} \right) + b} \right]c}}}}{{\left[ {n\left( {{b_1} - {b_2}} \right) + b} \right]}}} \right)}
\end{array}
$$
Since $\frac{{{a_2}}}{{{a_1}}}{e^{ - \left( {{b_2} - {b_1}} \right)c}} = \frac{{{a_1}}}{{{a_2}}}{e^{ - \left( {{b_1} - {b_2}} \right)c}} = 1$, we get
$$
\begin{array}{l}
I = - \frac{{\left( {b + 1} \right)}}{{{b^2}}}\left[ {\left( {\ln {a_1} - {b_1}} \right){e^{ - bc}} + \left( {\ln {a_2} - {b_2}} \right)\left( {1 - {e^{ - bc}}} \right)} \right]\\
+ \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\left( {\frac{{{e^{ - bc}}}}{{\left[ {n\left( {{b_2} - {b_1}} \right) + b} \right]}} + \frac{{{{\left( {\frac{{{a_1}}}{{{a_2}}}} \right)}^n} - {e^{ - bc}}}}{{\left[ {n\left( {{b_1} - {b_2}} \right) + b} \right]}}} \right)}
\end{array}
$$
Is that result converged?
|
Substitution and repeated integration by parts yields
$$
\int_0^\infty x^ne^{-kx}\mathrm{d}x=\frac{n!}{k^{n+1}}
$$
Assume $b_1\le b_2$ and $|a_2|\le|a_1|$, but not both equal, then
$$
\begin{align}
&\int_0^\infty e^{-bx}\log\left(a_1e^{-b_1x}+a_2e^{-b_2x}\right)\mathrm{d}x\\
&=\int_0^\infty e^{-bx}\left(\log(a_1)-b_1x+\log\left(1+\frac{a_2}{a_1}e^{-(b_2-b_1)x}\right)\right)\mathrm{d}x\\
&=\frac{\log(a_1)}{b}-\frac{b_1}{b^2}+\int_0^\infty e^{-bx}\log\left(1+\frac{a_2}{a_1}e^{-(b_2-b_1)x}\right)\mathrm{d}x\\
&=\frac{\log(a_1)}{b}-\frac{b_1}{b^2}+\int_0^\infty\left(\color{#C00000}{\sum_{k=1}^\infty(-1)^{k-1}\frac1k\frac{a_2^k}{a_1^k}e^{-(b+k(b_2-b_1))x}}\right)\mathrm{d}x\\
&=\frac{\log(a_1)}{b}-\frac{b_1}{b^2}+\sum_{k=1}^\infty(-1)^{k-1}\frac1k\frac{a_2^k}{a_1^k}\frac1{b+k(b_2-b_1)}\\
\end{align}
$$
where the series in red is derived from the Taylor series for $\log(1+x)$.
|
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|
Finding the integer solutions Find all integer solutions of
$$(a + b^2)(a^2 + b) = (a − b)^3.$$
Obviously $b = 0$ is one. But how to get other solutions?
|
On simplification, $$2b^3+(a^2-3a)b^2+b(a+3a^2)=0$$
If $$b\ne 0, 2b^2+(a^2-3a)b+(a+3a^2)=0 $$ which is a quadratic equation in $b$
As $b$ is integer, the discriminant must be perfect square.
So, $$(a^2-3a)^2-4\cdot2\cdot(a+3a^2)=a^4-6a^3-15a^2-8a=f(a)$$(say)
Clearly, $f(0)=0=f(-1)$ so, $a^4-6a^3-15a^2-8a=a(a+1)(a^2-7a-8)=a(a+1)^2(a-8)$
So, we need $a(a-8)$ to be perfect square $\implies a(a-8)\ge 0\implies a\le0$ or $a\ge 8$.
Now, $$a^2-8a=(a-4)^2-16\iff 4^2=(a-4)^2-(a^2-8a)$$
Observe that
$n^2-(n-1)^2=2n-1$ is odd $\ne 16$
$n^2-(n-2)^2\le 16 \iff n \le5$
$n^2-(n-3)^2=6n-9$ is odd
$n^2-(n-4)^2=8n-16\le 16\iff n\le \frac{25}8$
$4^2=5^2-3^2,4^2-0^2$
So, we need
$(i)a^2-8a=3^2$ or $(a-4)^2=5^2 $ both imply $a=-1,9$
or $(ii)a^2-8a=0\implies a=0,8$
There is no more solutions in integers.
|
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|
Summing $r(r+3)$ using induction We want to prove the following summation by induction:
$$\sum_{r=1}^{n}r(r+3)=\frac{1}{3}n(n+1)(n+5)$$
The problem is posted for a friend, but others can look at the solution if they want/need.
|
Let us denote the sum by $S_{n}$. We first establish the base case, that $S_{1}=1(1+3)=4$ is equal to $\frac{1}{3}\cdot 1\cdot 2\cdot 6=4$, which holds.
Then, assume that for some $k$, that $S_{k}=\frac{1}{3}k(k+1)(k+5)$. Obviously $S_{k+1}=S_{k}+(k+1)(k+4).$ Tidying up, we have
$$S_{k+1}=\frac{1}{3}(k+1)\left[k^{2}+5k+3k+12\right]=\frac{1}{3}(k+1)(k^{2}+8k+12)=\frac{1}{3}(k+1)(k+2)(k+6)$$
Which is as we expected. If $S_{k}$ is given by the formula above, then $S_{k+1}$ is. Therefore, since the claim holds for $n=1$, it holds for all natural $n$. $\square$
|
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|
Rational function partial fraction help? So I have to integrate
$$
\frac{3x+2}{(x)(x+1)^3}
$$ so I have the terms
$$
\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}
$$
... and then I have
$$
3x+2=A(x+1)^3 + B(x)(x+1)^2 +C(x)(x+1)^2+ D(x)(x+1)^2
$$
... here I tried to equal the quotients and I have $A=2$..how do I find the others?
|
You’re starting off on the wrong foot. When you put
$$\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}\tag{1}$$
back together into a single fraction, that fraction must have the same denominator as the original one. $(1)$ becomes
$$\frac{A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx}{x(x+1)^3}\;.$$
and you want to choose $A,B,C$, and $D$ so that
$$\frac{A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx}{x(x+1)^3}=\frac{3x+2}{x(x+1)^3}\;.$$
Two fractions with the same denominator are equal if and only their numerators are equal, so you require that
$$A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx=3x+2\;.\tag{2}$$
Setting $x=0$, you find that $A=2$. Setting $x=-1$, you find that $-D=-1$, so $D=1$. Thus, $(2)$ reduces to
$$2(x+1)^3+Bx(x+1)^2+Cx(x+1)+x=3x+2\;,$$ or
$$2(x+1)^3+Bx(x+1)^2+Cx(x+1)=2x+2\;.$$ Now just multiply out the lefthand side:
$$2x^3+6x^2+6x+2+Bx^3+2Bx^2+Bx+Cx^2+Cx=2x+2\;,$$ or
$$(B+2)x^3+(2B+C+6)x^2+(B+C+4)x=0\;.$$
The coefficients on the righthand side are all $0$, so all that remains is to solve
$$\left\{\begin{align*}
&B+2=0\\
&2B+C+6=0\\
&B+C+4=0
\end{align*}\right.$$
for $B$ and $C$. The first equation makes this easy: $B=-2$, and then $C=-2$ as well.
|
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|
Working with exponent on series Hi have this sequence:
$$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}$$
I understand that this is a Geometric series so this is what I've made to get the sum.
$$\sum\limits_{n=1}^\infty (-1)^n\frac{3^{n}\cdot 3^{-2}}{4^n}$$
$$\sum\limits_{n=1}^\infty (-1)^n\cdot 3^{-2}{(\frac{3}{4})}^n$$
So $a= (-1)^n\cdot 3^{-2}$ and $r=\frac{3}{4}$ and the sum is given by
$$(-1)^n\cdot 3^{-2}\cdot \frac{1}{1-\frac{3}{4}}$$
Solving this I'm getting the result as $\frac{4}{9}$ witch I know Is incorrect because WolframAlpha is giving me another result.
So were am I making the mistake?
|
The objective here is to transform your sum into a sum of the form:
$$\sum_{n=1}^\infty ar^{n-1}$$
$$\text{Transformation: }\quad\quad\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n} = \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\frac{(-3)^{n-1}}{4^{n-1}} = \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\left(\frac{-3}{4}\right)^{n-1}$$
Hence $a = -\dfrac{1}{12}$ and $r = -\dfrac{3}{4}.\quad$ Now use the fact that
$$\sum_{n=1}^\infty ar^{n-1} = \dfrac{a}{1 - r} = -\left(\frac{1}{12}\right)\cdot \left(\frac{1}{1 - (-\frac{3}{4})}\right)$$
Simpilfy, and then you are done!
|
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|
Weird question pertaining to HCF I encountered this question which seems weird/incomplete to me :
Q: H.C.F. of 3240, 3600 and a third number is 36, and their L.C.M. is
$2^4 \cdot 3^5 \cdot 5^2 \cdot 7^2$ . The third number is?
Can anyone please teach me concept wise how to solve it?
|
Find the prime power factorizations of the two given numbers. We get
$3240=2^3\cdot 3^4\cdot 5^1$ and
$3600=2^4\cdot 3^2\cdot 5^2$.
Let our unknown number be $n$. Because the LCM of $3240$, $3600$, and $n$ only involves the primes $2$, $3$, $5$, and $7$, we know that the prime power factorization of $n$ can involve no primes other than these.
So the only question is: how many of each?
Since the HCF of our three numbers is $36=2^2\cdot 3^2$, the highest power of $2$ that divides $n$ must be $2^2$.
The LCM has a $3^5$. Since the highest power of $3$ needed by our first two numbers is $3^4$, the highest power of $3$ that divides $n$ must be $3^5$.
Note that $5$ cannot divide $n$ since $5$ divides $3240$ and $3600$ but does not divide $36$.
Note also that since $7$ does not divide the first two numbers, the $7^2$ in the LCM must come from $n$.
It follows that $n=2^2\cdot 3^5\cdot 7^2$.
Remark: Your intuition about "not enough information" is reasonable. For example, if the HCF of the three numbers was $72$ instead of $36$, then the highest power of $2$ that divides $n$ could be $2^3$ or $2^4$, so $n$ would not be completely determined.
|
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|
How can I find a number $a$ such that this limit is 1 I want to find the number $a$ and $b$ such that $\lim_{x\to 0} \frac{\sqrt{ax+b}-2}{x}=1$.
First of all, I know that $b$ has to be 4, because the limit of the numerator has to be zero because the denominator is zero when we take its limit.
My problem is with the number
$a$. I need help.
Thanks a lot!
|
When $a=b=4$. We can calculate the binomial series of $\sqrt{ax+b}$ and we find that
$$\sqrt{ax+b} \sim \sqrt{b} + \frac{a}{2\sqrt{b}}x + \cdots$$
Using this approximation, we see that
$$\frac{\sqrt{ax+b}-2}{x} \sim \frac{\sqrt{b}-2}{x} + \frac{a}{2\sqrt{b}} + \cdots $$
where the tail "$+\cdots$" consists of terms divisible by $x$. Assuming that $\sqrt{b} \neq 2$, the limit is undefined as $x \to 0$. If $\sqrt{b} = 2$, i.e. $b=4$ then we have:
$$\frac{\sqrt{ax+4}-2}{x} \sim \frac{a}{4}-\frac{a^2}{64}x + \cdots $$
where the tail "$+\cdots$" consists of terms divisible by $x^2$. In this case, the limit as $x \to 0$ is $\frac{1}{4}a$ and so we need $a=4$. It follows that
$$\lim_{x \to 0}\left(\frac{\sqrt{4x+4}-2}{x}\right) = 1 \, . $$
|
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|
How do I solve this exponential equation? $5^{x}-4^{x}=3^{x}-2^{x}$ How do I solve this exponential equation?
$$5^{x}-4^{x}=3^{x}-2^{x}$$
|
$$5^x - 4^x = \int_4^5 x y^{x-1} \,dy$$
$$3^x - 2^x = \int_2^3 x y^{x-1} \,dy$$
$$= \int_4^5 x (y-2)^{x-1} \,dy$$
So the difference between $5^x - 4^x$ and $3^x - 2^x$ is
$$ \int_4^5 x (y^{x-1} - (y-2)^{x-1})\,dy$$
In the integrand here, since $y \rightarrow y^{x-1}$ is monotone whenever $x \neq 1$, the
expression $(y^{x-1} - (y-2)^{x-1})$ will either be always negative or always positive if $x \neq 1$, in which case the integral itself will be nonzero unless $x = 0$.
We conclude that as long
as $x \neq 0$ or $1$, $(5^x - 4^x) - (3^x - 2^x)$ is nonzero. Hence $3^x - 2^x = 5^x - 4^x$ only when $x = 0$ or $1$.
After writing all this out, I probably prefer the mean value theorem approach, but hey it's good to have more than one way of looking at a problem.
|
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|
Evaluate:: $ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 +\cdots + \frac 1n\right) $ How to evaluate the series:
$$ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 + \cdots + \frac 1n\right) $$
According to Mathematica, this converges to $ (\log 2)^2 $.
|
This is a special case of a more general result derived here.
$$S = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n+1} \sum_{k=1}^n \dfrac1k$$
Recall that $\dfrac1k = \displaystyle \int_0^1 x^{k-1} dx$ and $\dfrac1{n+1} = \displaystyle \int_0^1 y^n dy$.
Now use the following fact.
$$\sum_{k=0}^{\infty} \int_0^1 (-z)^k dz = \lim_{n \to \infty} \int_0^1 \dfrac{1 - (-z)^n}{1+z} dz$$
The sequence of functions $f_n(z) = \dfrac{1 - (-z)^n}{1+z}$ is dominated by the function $g(z) = \dfrac2{1+z}$ in the interval $[0,1]$, which is integrable. Hence, we can swap the limit and the integral to get that
$$\lim_{n \to \infty} \int_0^1 \dfrac{1 - (-z)^n}{1+z} dz = \int_0^1 \dfrac{dz}{1+z}$$
Hence,
$$S = \sum_{n=1}^{\infty} (-1)^{n+1} \int_0^1 y^n dy \left(\sum_{k=1}^n \int_0^1 x^{k-1} dx \right) = \sum_{n=1}^{\infty} (-1)^{n+1} \int_0^1 y^n dy \left(\int_0^1 \dfrac{1-x^n}{1-x} dx \right)$$
Hence,
$$S = \int_0^1 \int_0^1 \dfrac{\dfrac{y}{1+y} - \dfrac{xy}{1+xy}}{1-x} dy dx = \int_0^1 \int_0^1 \dfrac{y+xy^2-xy-xy^2}{(1+y)(1+xy)(1-x)} dx dy\\
=\int_0^1 \int_0^1 \dfrac{y}{(1+y)(1+xy)} dx dy = \int_0^1 \dfrac{\log(1+y)}{1+y} dy = \left. \dfrac{\log^2(1+y)}2 \right \vert_0^1 = \dfrac{\log^2(2)}2$$
The sum you are interested in is $2S$ and hence the answer is $\log^2(2)$.
|
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|
Show $\mathbb{Q}[\sqrt[3]{2}]$ is a field by rationalizing I need to rationalize $\displaystyle\frac{1}{a+b\sqrt[3]2 + c(\sqrt[3]2)^2}$
I'm given what I need to rationalize it, namely $\displaystyle\frac{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}$
But I fail to see the motivation behind this. Can someone give a more general explanation of how to come to this expression?
|
You rationalize by using Euclid's algorithm as used to prove Bézout's identity.
Let $\alpha = \sqrt[3]{2}$. We have obviously to assume $a+b\sqrt[3]2 + c(\sqrt[3]2)^2 = a+b\alpha + c \alpha^2 \ne 0$.
Consider the polynomial $0 \ne a + b x + c x^2 \in \mathbb{Q}[x]$. Since $x^3 - 2$ is irreducible in $\mathbb{Q}[x]$, for the gcd we have $(a + b x + c x^2, x^3 - 2) = 1$. Use Euclid's algorithm to find polynomials $u, v \in \mathbb{Q}[x]$ such that $(a + b x + c x^2) \cdot u + (x^3 - 2) \cdot v = 1$. Now evaluate this for $x = \alpha$ to obtain $(a + b \alpha+ c\alpha^2) \cdot u(\alpha) = 1$, so that
\begin{equation}
\frac{1}{a+b\sqrt[3]2 + c(\sqrt[3]2)^2}
=
\frac{1}{a + b \alpha+ c \alpha^2}
=
u(\alpha).
\end{equation}
|
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|
Discussion on even and odd perfect numbers. First of all thank you so much for answering my previous post. These are few interesting problems drawn from Prof. Gandhi lecture notes. kindly discuss:
1) If $n$ is even perfect number then $(8n +1)$ is always a perfect square.
2) Every odd perfect number has at least three different prime factors. This is by observation we can understand that. But, how to prove mathematically?
3) Every even perfect number (other than 6) can be expressible as sum of consecutive odd cubes.
Thanks in advance.
|
$1.$ By a theorem of Euler, any even perfect number is of the shape $2^{p-1}(2^p-1)$ where $p$ is prime. We don't need the primality part. Multiply by $8$, add $1$. We get $2^{2p+2}-2(2^{p+1})+1$, which is the square of $2^{p+1}-1$.
$2.$ We show that an odd perfect number cannot have only $2$ distinct prime factors (we don't deal with only $1$ prime factor, it's easier).
Let $N=p^aq^b$ where $p$ and $q$ are odd primes and $p\lt q$. Then the sum of the divisors of $N$ is
$$(1+p+\cdots+p^a)(1+q+\cdots+q^b).$$
Using the ordinary formula for the sum of a finite geometric series, this can be rewritten as
$$\frac{p^{a+1}-1}{p-1}\cdot \frac{q^{b+1}-1}{q-1}.$$
Divide by $N$. The result is
$$\frac{p-\frac{1}{p^a}} {p-1}\cdot \frac{q-\frac{1}{q^b}} {q-1} .$$
This is less than
$$\frac{p}{p-1}\cdot \frac{q}{q-1}.\tag{$1$}$$
We show that the product $(1)$ must be less than $2$. In particular, that shows the product cannot be $2$, so $N$ is in fact deficient.
It is easier to show that the reciprocal of Expression $(1)$ is greater than $\frac{1}{2}$. This reciprocal is
$$\left(1-\frac{1}{p}\right)\cdot \left(1-\frac{1}{q}\right).$$
But $p\ge 3$ and $q\ge 5$. So $1-\frac{1}{3}\ge \frac{2}{3}$ and $1-\frac{1}{q}\ge \frac{4}{5}$. So their product is $\ge \frac{8}{15}$, which is greater than $\frac{1}{2}$.
I have not thought about your Question $3$.
|
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Number theory fun problem Say $a,b > 2 $ are integers. Then we have that $2^a + 1$ is not divisible by $2^b - 1$.
Any thoughts on how to tackle this problem???
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Let $a = bq + r$ where $0 \leq r < b$. Hence, $$2^a+1 = 2^{bq} \cdot 2^r +1 = (2^{bq}-1)\cdot 2^r + 2^r+1$$
But $2^b-1$ divides $2^{bq}-1$ since $(x-1)$ divides $x^m-1$ for all $m \in \mathbb{N}$.
Hence, if $2^b-1$ divides $2^a+1$, then $2^b-1$ has to divide $2^r+1$. But $0 \leq r < b$. Hence, we need $$2^b-1 \leq 2^r + 1 \leq 2^{b-1} + 1$$
i.e. $$2^b - 2^{b-1} \leq 2$$
This is possible only if $b=1$ or $2$. But we are given that $b > 2$.
|
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|
$n$-digits numbers made of 1, 2, 3, such that none of two consecutive digits differ by more than one
We call a number to be good if none of two consecutive digits differ by more than one. How many good $n$-digits numbers made from digits $1$, $2$ and $3$ are there?
For example, $12232$ is good, but $12\textbf{31}2$ isn't.
My idea was to subtract bad numbers made of $1$, $2$ and $3$ (bad numbers are ones that aren't good) from $3^n$ (which represents number of $n$-digit numbers made of $1$, $2$ and $3$).
I tried with representing number $abcd$ ($n=4$) as sequence of 3 ($4-1$) pairs of digits: $(a,b),\text{ }(b,c),\text{ }(c,d)$. Then (I think) we could count number of ways to have at least one of the pairs $(1,3)$ and $(3,1) appearing at least once. But alas...
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Let the $n$ digit number be $a_1 a_2 \ldots a_n$. Let $N_1(n)$ be the number of $n$ digit numbers that are good starting with $1$, $N_2(n)$ be the number of $n$ digit numbers that are good starting with $2$ and , $N_3(n)$ be the number of $n$ digit numbers that are good starting with $3$.
We have
\begin{align}
N_1(n) & = N_1(n-1) + N_2(n-1)\\
N_2(n) & = N_1(n-1) + N_2(n-1) + N_3(n-1)\\
N_3(n) & = N_2(n-1) + N_3(n-1)
\end{align}
Hence, we have
$$\begin{bmatrix}N_1(n)\\ N_2(n)\\ N_3(n) \end{bmatrix} = \underbrace{\begin{bmatrix} 1 & 1 & 0\\ 1 & 1 & 1\\ 0 & 1 &1\end{bmatrix}}_A \begin{bmatrix} N_1(n-1)\\ N_2(n-1)\\ N_3(n-1) \end{bmatrix}$$
Hence,
$$\vec{N}(n+1) = A^n \vec{N}(1)$$
We have $$V = \dfrac12 \begin{bmatrix} 1 & -\sqrt2 & 1\\ -\sqrt2 & 0 & \sqrt 2\\ 1 & \sqrt2 & 1\end{bmatrix}$$ and $$\Lambda = \begin{bmatrix} 1-\sqrt2 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 +\sqrt2\end{bmatrix}$$ and $$A = V \Lambda V^T$$
Hence,
$$\vec{N}(n+1) = V \Lambda^n V^T \vec{N}(1)$$
where $$\vec{N}(1) = \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}$$
Hence, the answer is
\begin{align}
N_1(n) + N_2(n) + N_3(n) & = \begin{bmatrix} 1 & 1 & 1\end{bmatrix} V \Lambda^n V^T \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}\\
& = \dfrac14\begin{bmatrix} 2-\sqrt{2} & 0 & 2+\sqrt{2}\end{bmatrix} \begin{bmatrix} (1-\sqrt2)^{n-1} & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & (1 +\sqrt2)^{n-1}\end{bmatrix}\begin{bmatrix} 2-\sqrt{2} \\ 0 \\ 2+\sqrt{2}\end{bmatrix}\\
& = \dfrac12\begin{bmatrix} \sqrt{2}-1 & 0 & 1+\sqrt{2}\end{bmatrix} \begin{bmatrix} (1-\sqrt2)^{n-1} & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & (1 +\sqrt2)^{n-1}\end{bmatrix}\begin{bmatrix} \sqrt{2}-1 \\ 0 \\ 1+\sqrt{2}\end{bmatrix}\\
& = \dfrac{(1+\sqrt2)^{n+1} + (1-\sqrt2)^{n+1}}2
\end{align}
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The number of paths on a graph of a fixed length w/o repeatings Sorry for bad English.
Consider a graph $G$ with the adjacency matrix $A$. I know that the number of paths of the length $n$ is the sum of elements $A^n$.
But what if we can't walk through a vertex more than one times?
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[This is a counter-example to wece's answer, posted only as an answer because it's too long for a comment.]
wece's method is only correct for $A_2$. Counterexample: Let $A=\left[\begin{array}{ccccc}0&1&0&0&0\\1&0&1&0&0\\0&1&0&1&0\\0&0&1&0&1\\0&0&0&1&0\end{array}\right]$ representing a chain of 5 vertices connected by 4 edges. $A_1=A-diag(A)=A$.
$A_2=A\otimes A_1 =\left[\begin{array}{ccccc}1&0&1&0&0\\0&2&0&1&0\\1&0&2&0&1\\0&1&0&2&0\\0&0&1&0&1\end{array}\right]$.
In $A_2$ we see all the paths of length 2, the diagonal representing forbidden paths of length 2 from each vertex back to itself, the off-diagonal entries representing the 8 paths of length 2 (path from $a \rightarrow b$ distinct from $b \rightarrow a$). Then
$A_3 = A \otimes A_2 = \left[\begin{array}{ccccc}0&1&0&0&0\\1&0&1&0&0\\0&1&0&1&0\\0&0&1&0&1\\0&0&0&1&0\end{array}\right]\left[\begin{array}{ccccc}0&0&1&0&0\\0&0&0&1&0\\1&0&0&0&1\\0&1&0&0&0\\0&0&1&0&0\end{array}\right] = \left[\begin{array}{ccccc}0&0&0&1&0\\1&0&1&0&1\\0&1&0&1&0\\1&0&1&0&1\\0&1&0&0&0\end{array}\right]$,
which according to wece's method represents 10 paths of length 3, when there should be 3 such paths.
|
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$\det(\exp X)=e^{\mathrm{Tr}\, X}$ for 2 dimensional matrices I want to prove that for $X\in M_2(\mathbb{R})$ the formula $\det(\exp X)=e^{\mathrm{Tr}\, X}$ holds, writing $X$ in normal form gives $X=PJP^{-1}$, where $J$ is the Jordan matrix, now $\exp (PJP^{-1})=P(\exp J)P^{-1}$ and $\det P(\exp J)P^{-1}=\det \exp J$. If $J=\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ where we allow $a=b$, we get that $\det \exp J=e^{a+b}=e^{\mathrm{Tr}\, J}=e^{\mathrm{Tr}\, X}$, since $X$ and $J$ are similar. However, if $J=\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}$, then $\exp J=\exp\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \exp \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} e^a & e^{a+1} \\ e^a & e^{a} \end{pmatrix}$, so $\det \exp J=e^{2a}(1-e)$, which is not the result I want.
|
Your calculation of $\exp J$ is incorrect. $$\exp J = \exp \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \exp \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} e^a & 0 \\ 0 & e^a \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} e^a & e^a\\ 0 & e^a \end{pmatrix}$$
|
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Rolling three dice...am I doing this correctly? Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?
Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first
For 1) The first die, you have $\frac{6}{6}$. The second die needs to be equal to the first, so you have probability of $\frac{1}{6}$. Then the third die can't be equal to the first and second dice, so it's $\frac{5}{6}$.
All together you get $1 \cdot \frac{1}{6} \cdot \frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3 \cdot \left(1 \cdot\frac{1}{6} \cdot \frac{5}{6}\right)=\frac{5}{12}$$
Did I do this correctly?
Thank you.
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As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3\cdot6\cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $\frac{90}{216}=\frac{5}{12}$.
|
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How to prove the Mantel's theorem of graph theory 's bound is best possible? The theorem state that every graph of order $n$ and size greater than floor function $\lfloor \frac{n^2}{4} \rfloor$ contain a triangle.
I already know a proof of the number of the edge of graph $\leq\frac{n^2}{4}$ if it does not contains a triangle, but how to prove this bound is best possible if n is bigger than 100, in another word, how to prove there are infinite example satisfy the best bound?
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We can't use a smaller bound because we can show that for each natural number $n$, there exists a graph of order $n$ with exactly $\lfloor \frac{n^2}{4} \rfloor$ edges and no triangles. In response to your edit, this gives us an infinite number of examples.
So let $n$ be a natural number, and we will find a graph of order $n$ with no triangles and $\lfloor \frac{n^2}{4} \rfloor$ edges.
If $n$ is even, then $\lfloor \frac{n^2}{4} \rfloor = \frac{n^2}{4}$, and the graph $K_{n/2,n/2}$ has $\left( \frac{n}{2} \right) \left( \frac{n}{2} \right) = \frac{n^2}{4}$ edges and no triangles.
If $n$ is odd, then $n = 2m + 1$ for some integer $m$, so
\begin{align*}
\lfloor \frac{n^2}{4} \rfloor &= \lfloor \frac{4m^2 + 4m + 1}{4} \rfloor \\
&= \lfloor m^2 + m + \frac{1}{4} \rfloor \\
&= m(m + 1).
\end{align*}
Then the graph $K_{m,m+1}$ has order $m + m + 1 = n$, no triangles, and has $m(m + 1) = \lfloor \frac{n^2}{4} \rfloor$ edges.
|
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A series involving the harmonic numbers : $\sum_{n=1}^{\infty}\frac{H_n}{n^3}$ Let $H_{n}$ be the nth harmonic number defined by $ H_{n} := \sum_{k=1}^{n} \frac{1}{k}$.
How would you prove that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}?$$
Simply replacing $H_{n}$ with $\sum_{k=1}^{n} \frac{1}{k}$ does not seem like a good starting point. Perhaps another representation of the nth harmonic number would be more useful.
|
I will try to reduce the sum to an integral:
$$
\sum_{n=1}^\infty \frac{H_n}{n^3} = \sum_{n=1}^\infty H_n \frac{1}{\Gamma(3)} \int_0^\infty x^2 \mathrm{e}^{-n x} \mathrm{d} x = \frac{1}{2} \int_0^\infty x^2 \sum_{n=1}^\infty H_n \mathrm{e}^{-n x} \mathrm{d} x \tag{1}
$$
We now make use of the following generating function:
$$
\sum_{n=1}^\infty H_n z^n = \sum_{n=1}^\infty H_n \Delta_n \left(\frac{z^n}{z-1} \right)
$$
where $\Delta_n f_n = f_{n+1}-f_n$. We can now use summation by parts:
$$
\sum_{n=1}^m a_n \Delta_n b_n = b_{m+1} a_m - b_1 a_1 - \sum_{n=1}^{m-1} b_{n+1} \Delta_n a_n
$$
with $b_n = \frac{z^n}{z-1}$ and $a_n = H_n$, and using $\Delta_n H_n = \frac{1}{n+1}$, we get
$$
\sum_{n=1}^\infty H_n z^n = \sum_{n=1}^\infty H_n \Delta_n \left(\frac{z^n}{z-1} \right) = -1 - \sum_{n=1}^\infty \frac{z^{n+1}}{z-1} \frac{1}{n+1} = \frac{\log(1-z)}{z-1}
\tag{2} $$
Now, using $(2)$ in $(1)$:
$$
\sum_{n=1}^\infty \frac{H_n}{n^3} = -\frac{1}{2} \int_0^\infty x^2 \frac{\log\left(1-\mathrm{e}^{-x}\right)}{1-\mathrm{e}^{-x}} \mathrm{d}x \stackrel{t=\exp(-x)}{=} -\frac{1}{2} \int_0^1 \frac{\log(1-t)}{1-t} \frac{\log^2(t)}{t} \mathrm{d}t \tag{3}
$$
The latter integral can be evaluated using derivatives of the Euler beta function:
$$
\int_0^1 \frac{\log(1-t)}{1-t} \frac{\log^2(t)}{t} \mathrm{d}t = \lim_{\alpha \downarrow 0} \lim_{\beta \downarrow 0} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}^2}{\mathrm{d} \beta^2} \int_0^1 \left(1-t\right)^{\alpha-1} t^{\beta-1} \mathrm{d} t = \lim_{\alpha \downarrow 0} \lim_{\beta \downarrow 0} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}^2}{\mathrm{d} \beta^2} \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha + \beta)}
$$
Using
$$
\frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha + \beta)} = \left(\frac{1}{\alpha} + \frac{1}{\beta} \right) \frac{\Gamma(\alpha+1) \Gamma(\beta+1)}{\Gamma(\alpha + \beta+1)} = \left(\frac{1}{\alpha} + \frac{1}{\beta} \right) \left( 1 - \frac{\pi^2}{6} \alpha \beta + \left(\alpha \beta^2 + \beta \alpha^2\right) \zeta(3) - \frac{\pi^4}{360} \left(4 \alpha \beta^3 + \alpha^2 \beta^2 + 4 \alpha^3 \beta\right) + \cdots \right)
$$
Differentiating we get the result:
$$
\lim_{\alpha \downarrow 0} \lim_{\beta \downarrow 0} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}^2}{\mathrm{d} \beta^2} \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha + \beta)} = -\frac{\pi^2}{36}
$$
yielding with eq. $(3)$:
$$
\sum_{n=1}^\infty \frac{H_n}{n^3} = \frac{\pi^4}{72}
$$
|
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Geometry Problem and Isosceles Triangle $ABC$ is an isosceles triangle with $AB=AC$, $\angle BAC=96^\circ$. $D$ is a point such that $\angle DCA=48^\circ$, $AD=BC$ and angle $DAC$ is obtuse. What is the measure (in degrees) of $\angle DAC$?
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First, draw. Then calculate the angles you might already know and name the unknown ones. So, using that $ABC$ is isosceles, we have $\angle BAC=42^\circ$. Since $42+48=90$, we have $\cos 48^\circ=\sin 42^\circ$. Let $\delta:=\angle ADC$, which is $<90^\circ$ since $\angle DAC$ is obtuse. After finding $\delta$, we will have
$$\angle DAC=180^\circ-48^\circ-\delta\ .$$
Use the theorem of sines, for the triangles $ABC$ and $ACD$, and use the identity for $\sin(2x)$ with $x=48^\circ$:
$$\frac{\sin\delta}{\sin 48^\circ}=\frac{AC}{AD}=\frac{AC}{BC}=
\frac{\sin 42^\circ}{\sin 96^\circ}$$
From this, as $\sin 96^\circ=2\sin 48^\circ\cos 48^\circ=2\sin 48^\circ\sin 42^\circ $, we get that $\sin\delta=1/2$, that is $\delta=30^\circ$.
|
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Show that $7(3(2)^k + 2(5)^k) - 10(3(2)^{k-1} + 2(5)^{k-1}) = 3(2)^{k+1} + 2(5)^{k+1}$ $7(3(2)^k + 2(5)^k) - 10(3(2)^{k-1} + 2(5)^{k-1}) = 3(2)^{k+1} + 2(5)^{k+1}$
The problem is part of a proof. If you could also talk me through your thought process for solving this problem, I would greatly appreciate it. I've played around with several things like trying to get everything in terms of $a^{k+1}$ but I'm having no luck. Thanks!
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Assuming the missing right parenthesis goes at the end:
$$7(3(2)^k + 2(5)^k) - 10(3(2)^{k-1} + 2(5)^{k-1}) =\\21\cdot 2^k+14\cdot 5^k-30\cdot 2^{k-1} -20\cdot 5^{k-1}=\\ 21\cdot 2^k +14\cdot 5^k-15\cdot 2^k-4\cdot 5^k=\\ 6\cdot 2^k+10\cdot 5^k=\\3(2)^{k+1} + 2(5)^{k+1}$$
|
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Is it possible to have a rule which generates: 2, 4, 6, 8, 10, 12, 14, 16, -23? This is on Lagrange Interpolations . . .
Is it possible to have a rule which generates the sequence: 2, 4, 6, 8, 10, 12, 14, 16, -23?
The hint that he gave us is to use Summation Products, the only thing I can come up with is $\sum_{i=1}^{8}{2}$. I cannot come up with any formula for a Summation Product that would produce something like a 2 + 2 + 2 + . . .
I know how to use the equation for Lagrange Interpolation, but I cannot see how to apply it here . . .
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Thomas Andrews's solution is the best one, but you may also consider this simplistic but longer and bulky solution:
$$
\begin{array}{lccr} \\
f(x) & = & & a_1\times(-1)^{1-1}\times\frac{1}{9!}(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) \\
& & + & a_2\times(-1)^{2-1}\times\frac{2}{9!}(x-1)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) \\
& & + & a_3\times(-1)^{3-1}\times\frac{3}{9!}(x-1)(x-2)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) \\
& & + & a_4\times(-1)^{4-1}\times\frac{4}{9!}(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)(x-8)(x-9) \\
& & + & a_5\times(-1)^{5-1}\times\frac{5}{9!}(x-1)(x-2)(x-3)(x-4)(x-6)(x-7)(x-8)(x-9) \\
& & + & a_6\times(-1)^{6-1}\times\frac{6}{9!}(x-1)(x-2)(x-3)(x-4)(x-5)(x-7)(x-8)(x-9) \\
& & + & a_7\times(-1)^{7-1}\times\frac{7}{9!}(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-8)(x-9) \\
& & + & a_8\times(-1)^{8-1}\times\frac{8}{9!}(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-9) \\
& & + & a_9\times(-1)^{9-1}\times\frac{9}{9!}(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8) \\
\end{array} $$
Where $ a_1 = 2 $, $ a_2 = 4 $, $ a_3 = 6 $, $ a_4 = 8 $, $ a_5 = 10 $, $ a_6 = 12 $, $ a_7 = 14 $, $ a_8 = 16 $ and $ a_9 = -23$.
This may look very long if you are working analytically. But if you are implementing numerical methods in computer, this can be automated and greatly simplified by a simple software algorithm as a general solution to generate arbitrary discrete sequences.
|
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Solving an equation with fractional powers I was trying to find the maximum value for a function. I took the first derivative and arrived at this horrible expression:
$$ (x^2 + y^2)^\frac{3}{2} - y {\frac{3}{2}}(x^2 + y^2)^{\frac{1}{2}}2y = 0$$
How can I find the extrema by hand?
|
After some simplifying steps,
$$(x^2 + y^2)^\frac{3}{2} - y {\frac{3}{2}}(x^2 + y^2)^{\frac{1}{2}}2y = 0, \\
(x^2 + y^2)^\frac{3}{2} - 3y^2(x^2 + y^2)^{\frac{1}{2}} = 0, \\
(x^2 + y^2)^{\frac{1}{2}}\left(x^2+y^2-3y^2 \right)=0, \\
x^2+y^2=0 \Rightarrow x=0,\;y=0, \\
\text{or}\\
x^2-2y^2=0 \Rightarrow |x|=\sqrt{2}|y|.
$$
|
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Prove that $\sum_{j = 0}^{n} (-\frac{1}{2})^j = \frac{2^{n+1} + (-1)^n}{3 \times 2^n}$ whenever $n$ is a nonnegative integer. I'm having a really hard time with the algebra in this proof. I'm supposed to use mathematical induction (which is simple enough), but I just don't see how to make the algebra work.
$\sum_{j = 0}^{k} (-\frac{1}{2})^k + (-\frac{1}{2})^{k + 1} = \frac{2^{k+1} + (-1)^k}{3 \times 2^k}+(-\frac{1}{2})^{k + 1}$, by adding $(-\frac{1}{2})^{k + 1}$ to both sides.
I want to show that the right side is equal to:
$\frac{2^{k+1+1} + (-1)^{k+1}}{3 \times 2^{k+1}}$
Thank you!
|
OK let's tackle the one giving you grief. Just add the fractions and massage it:
$$
\begin{align*}
\frac{2^{k + 1} + (-1)^k}{3 \cdot 2^k} + \left(- \frac{1}{2}\right)^{k + 1}
&= \frac{2^{k + 1} + (-1)^k}{3 \cdot 2^k} + \frac{(-1)^{k + 1}}{2^{k + 1}} \\
&= \frac{2^{k + 2} + 2 \cdot (-1)^k + 3 \cdot (-1)^{k + 1}}{3 \cdot 2^{k + 1}} \\
&= \frac{2^{k + 2} + (2 - 3) \cdot (-1)^k}{3 \cdot 2^{k + 1}} \\
&= \frac{2^{k + 2} + (-1)^{k + 1}}{3 \cdot 2^{k + 1}}
\end{align*}
$$
|
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What is $q(x)$ and $r(x)$ when $(x^2-6x+9)q(x) + r(x) = x^3 -27$? I just failed this question on a test, so I would please like to get some feedback on where my thinking was wrong.
I need help with determining $q(x)$ and $r(x)$ when: $$(x^2-6x+9)q(x) + r(x) = x^3 -27$$
I know that: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
In this case $a=x$ and $b=\pm3$. By using $+3$ you get the calculation: $$(x^2-6x+9)(x-3) = x^3-9x^2+27x -27$$
To get the final expression: $x^3 -27$ you have to set $r(x)$ to be: $9x^2-27x$.
But this was wrong. Where is my line of thinking incorrect?
Thank you kindly for your help!
|
Long division!
$$\require{enclose}
\begin{array}{rll}
x\;+6\qquad\qquad\quad\\[-3pt]
x^2-6x+9\enclose{longdiv}{x^3\qquad\qquad\;\;-27}\\
\underline{x^3-6x^2+9x\phantom{\qquad\,}}\\[-3pt]
6x^2-\phantom{0}9x-27\\[-3pt]
\underline{6x^2-36x+54\;}\\[-3pt]
27x-81
\end{array}$$
So $q(x) = x + 6$ and $r(x) = 27 x - 81$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
CDF of $X+Y$,$X−Y$,$XY$ for $(X,Y)$ Chosen Uniformly Inside Triangle Let $(X,Y)$ be chosen uniformly on the triangle $\{(x,y)\in\mathbb R^2:x+y\leq1,x\geq0,y\geq0\}$. What is the joint density function of $(X,Y)$? Find the CDFs of $X+Y$, $X-Y$, and $XY$.
What I've tried:
$\displaystyle \frac{1}{Area\hspace{1mm} of\hspace{1mm} Triangle} =\frac{1}{\frac{1}{2}\cdot 1 \cdot 1}=2$,
Therefore the joint density function of $(X,Y)$ is:
$$
f_{X,Y}(x,y)=
\begin{cases}
2 & \text{inside the triangle} \\
0 & \text{elsewhere}
\end{cases}
$$
How do I then find the distributions? Thank you!
|
The definition of the cumulative distribution function of a random variable $X$ is the function given by $\displaystyle F_X(x)=P(X \leq x)$.
If $Z=X+Y$,
$\displaystyle F_{Z}(z)=P\{X + Y \leq z\} = \int_{0}^{z}\int_{0}^{z-x}2\,\mathrm dy\,\mathrm dx=2\int_{0}^{z}(z-x)\mathrm dx=z^2$
(Rather than integrating, it is easy to determine the CDF $F_{Z}(z)$ by
inspection:)
The PDF is the derivative of the CDF;
$\displaystyle \Rightarrow f_{Z}(z)=\frac{d}{dz}F_{Z}(z)=2z$.
If $Z=X-Y$,
$\displaystyle F_{Z}(z)=P\{X - Y \leq z\} =\mathbb P(Y\geq X-z)$
$\displaystyle \color{red}A\color{red}R\color{red}E\color{red}A\color{black} =\frac{1}{2}\left(\frac{1-z}{\sqrt{2}}\right)^2=\frac{(1-z)^2}{4}$
$\displaystyle \color{blue} A\color{blue}R\color{blue}E\color{blue}A\color{black} = \frac{1}{2}-\color{red}A\color{red}R\color{red}E\color{red}A\color{black}=\frac{1}{2}-\frac{(1-z)^2}{4}$
Therefore $\displaystyle F_{Z}(z)=2\left[\frac{1}{2}-\frac{(1-z)^2}{4}\right]=\frac{1}{2}(-z^2+2z+1)$
And differentiating to find the PDF:
$\displaystyle f_{Z}(z)=\frac{d}{dz}\left(\frac{1}{2}(-z^2+2z+1)\right)=1-z$.
If $Z=XY$,
$\displaystyle F_{Z}(z)=P\{XY \leq z\} =\mathbb P\left(y\leq \frac{z}{x}\right)$
The points of intersection of $\displaystyle y=\frac{z}{x}$ with the line $\displaystyle y=-x+1$ are:
$\displaystyle x=\frac{1\pm \sqrt{1-4z}}{2}$
The curve intersects the triangle if $z<\frac{1}{4}$.
$\color{blue}A\color{blue}R\color{blue}E\color{blue}A \color{black}=$QUADRILATERAL$+$INTEGRAL$+$TRIANGLE$ $
$\displaystyle=\frac{1}{4}\left(2z-\sqrt{1-4z}+1\right) + \int_{\frac{1-\sqrt{1-4z}}{2}}^{\frac{1+\sqrt{1-4z}}{2}}\frac{z}{x}dx+\frac{1}{8}\left(\sqrt{1-4z}-1\right)^2$
$\displaystyle =\frac{1}{4}\left(2z-\sqrt{1-4z}+1\right) + 2z\tan^{-1}\sqrt{1-4z} +\frac{1}{8}\left(\sqrt{1-4z}-1\right)^2$
$\displaystyle = \frac{1}{2}\left(1-\sqrt{1-4z}+4z\tan^{-1}\left(\sqrt{1-4z}\right)\right)$
Therefore, multiplying by 2, we have: $F_{Z}(z)=\left(1-\sqrt{1-4z}+4z\tan^{-1}\left(\sqrt{1-4z}\right)\right)$
And taking the derivative to find the PDF, we have:
PDF $=\frac{d}{dz}\left(1-\sqrt{1-4z}+4z\tan^{-1}\left(\sqrt{1-4z}\right)\right)$
$\displaystyle=4\tan^{-1}\left(\sqrt{1-4z}\right)$.
Further exercises related to the question:
The marginal distribution for $X$ is:
$\displaystyle f_{X}(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y')dy'=\int_{0}^{1-x}2dy=2(1-x)$
The marginal distribution for $Y$:
$\displaystyle f_{Y}(y)=\int_{-\infty}^{\infty}f_{X,Y}(x',y)dx'=\int_{0}^{1-y}2dx=2(1-y)$
The conditional PDF of $X$ given $Y$:
$\displaystyle f_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}=\frac{1}{1-y}$, $0\leq x \leq y$
Since the conditional PDF is uniform on $[0,1-Y]$, the conditional
expectation is simply: $\displaystyle\mathbb E[X|Y=y]=\frac{1-y}{2}$
The total expectation theorem yields:
$\displaystyle\mathbb E[X]=\int_{0}^{1}\frac{1-y}{2}f_{Y}(y)dy=\frac{1}{2}\int_{0}^{1}f_{Y}(y)dy-\frac{1}{2}\int_{0}^{1}yf_{Y}(y)dy$
$\displaystyle \mathbb E[X]=\frac{1}{2}-\frac{1}{2}\mathbb E[Y]$ and by symmetry, $\mathbb E[X]=\mathbb E[Y]\Rightarrow$
$\displaystyle\mathbb E[X]=\frac{1}{2}-\frac{1}{2}\mathbb E[X]\Rightarrow \mathbb E[X]=\frac{1}{3}$. $\blacksquare$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/318275",
"timestamp": "2023-03-29T00:00:00",
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|
positive integer value of $n$ for which $n^2-19n+99$ is a perfect square. Calculate positive integer value of $n$ for which $n^2-19n+99$ is a perfect square.
My Try:: Let $n^2-19n+99 = k^2$ where $k\in \mathbb{Z}$
$4n^2-76n+396 = 4k^2 $ or $(2n-19)^2-35 = (2k)^2$
$(2n-19)^2-(2k)^2 = 35$
now we have to take two perfect square whose difference is $ = 35$
so one pair is $(6^2,1^2)$
now my question is how can i calculate for other ordered pairs
Thanks
|
$(2n-19)^2-(2k)^2 = -35$
$\Rightarrow (2n-19-2k)(2n-19+2k)=35$
$35=5\times 7=35\times 1=-5\times -7=-35\times -1$
Check the possible pairs.
Another way: $n^2-19n+99 = k^2$ is a quadratic in n so for it to have integral(perhaps rational) roots its determinant is a square.
$\Rightarrow 19^2-4(99-k^2)=y^2,y\in \mathbb{N}$
$\Rightarrow19^2-4.99=y^2-4k^2\Rightarrow 361-396 \Rightarrow35=(2k)^2-y^2=(2k+y)(2k-y)$
$35=5\times 7=35\times 1=-5\times-7=-35\times -1$
$\Rightarrow 2k+y=5$
$\Rightarrow 2k-y=7$
$\Rightarrow 4k=12\Rightarrow k=3,y=-1$
or,
$\Rightarrow 2k+y=35$
$\Rightarrow 2k-y=1$
$\Rightarrow k=9,y=17$
or
$\Rightarrow 2k+y=-35$
$\Rightarrow 2k-y=-1$
$\Rightarrow k=-9,y=-17$
or
$\Rightarrow 2k+y=-5$
$\Rightarrow 2k-y=-7$
$\Rightarrow k=-3,y=1$
So we have all the solutions.
Check for other cases also when the values gets reversed.
We have $n=(19+y)/2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/318500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Finding the remainder. Let $a,b$ be positive integers such that $7$ divide $a^2+b^2$ .How to find the remainder when we divide $ab-1$ by $7$
|
Note that $7$ divides $a^2+b^2$ if and only if $7$ divides both $a$ and $b$.
You can prove this in two ways. One way is to calculate $a^2+b^2$ modulo $7$ for all possibilities. This is not as tedious as it sounds, since a square is congruent to $0$, $1$, $4$, or $2$ modulo $7$. No sum of these is congruent to $0$ modulo $7$ except $0+0$.
Or else we can appeal to the general theorem that if $p$ is a prime of the form $4k+3$, then $-1$ is not a quadratic residue of $p$. If $a^2+b^2\equiv 0\pmod{p}$, then $b^2\equiv -a^2\pmod{p}$. If $b\not\equiv 0\pmod{p}$, then multiplying both sides by the inverse of $b$ modulo $p$, we would obtain $(ab^{-1})^2\equiv -1\pmod{p}$, contradicting the fact that $-1$ is not a quadratic residue of $p$. Thus the modular inverse of $b$ cannot exist, and therefore $b\equiv 0\pmod{p}$. Similarly, $a\equiv 0\pmod{p}$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/318893",
"timestamp": "2023-03-29T00:00:00",
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|
Diagonalise a matrix and show the formula I have diagonlised P to get
$$P=\left(\begin{matrix}
-1 &0 &0\\
0 &0 &0\\
0 &0 &1
\end{matrix}\right)$$
however am unsure on how to proceed, would appreciate any help!
By diagonalising P by a transformation of similarity, show that
$$ e^{Pt} = (I_3 - P^2) + P \sinh( t) + P^2 \cosh(t)$$
when
$P$ is the matrix
$$P=\left(\begin{matrix}
0 &1 &0\\
0 &0 &1\\
0 &1 &0
\end{matrix}\right)$$
and where $I_3$ is the identity matrix of a $3\times3$
|
Given:
$$P=\left(\begin{matrix}
-1 &0 &0\\
0 &0 &0\\
0 &0 &1
\end{matrix}\right)$$
By diagonalising P by a transformation of similarity, show that
$$ e^{Pt} = (I_3 - P^2) + P \sinh( t) + P^2 \cosh(t)$$
If we diagonalize the matrix $P$, we arrive at:
$$P = S J S^{-1} = \begin{bmatrix} 1 & 1 & 1 \\ -1 & 0 & 1\\ 1 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix} 0 & -\frac{1}{2} & \frac{1}{2} \\ 1 & 0 & -1\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}$$
Note: $\cosh t = \frac{1}{2}(e^{-t} + e^{t}) ~ \text{and} ~ \sinh t = \frac{1}{2}(-e^{-t} + e^{t}) ~\text{and}~ P^{2} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$.
Using this diagonalization, the matrix exponential is given by:
$$\displaystyle e^{Pt} = e^{S J S^{-1} t} = \begin{bmatrix} 1 & -1 & 1 \\ -1 & 0 & 1\\ 1 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix} e^{-1} & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & e^{t}\end{bmatrix} \cdot \begin{bmatrix} 0 & -\frac{1}{2} & \frac{1}{2} \\ 1 & 0 & -1\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix} + \begin{bmatrix} 0 & 0 & \frac{1}{2}(e^{-t}+e^{t}) \\ 0 & \frac{1}{2}(e^{-t}+e^{t}) & 0\\ 0 & 0 & \frac{1}{2}(e^{-t}+e^{t})\end{bmatrix} + \begin{bmatrix} 0 & \frac{1}{2}(-e^{-t}+e^{t}) & 0 \\ 0 & 0 & \frac{1}{2}(-e^{-t}+e^{t})\\ 0 & \frac{1}{2}(-e^{-t}+e^{t}) & 0\end{bmatrix} = (I_3 - P^2) + P \sinh( t) + P^2 \cosh(t) = \begin{bmatrix} 1 & \frac{1}{2}(-e^{-t}+e^{t}) & -1 + \frac{1}{2}(e^{-t}+e^{t}) \\ 0 & \frac{1}{2}(e^{-t}+e^{t}) & \frac{1}{2}(-e^{-t}+e^{t})\\ 0 & \frac{1}{2}(-e^{-t}+e^{t}) & \frac{1}{2}(e^{-t}+e^{t})\end{bmatrix}$$
|
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
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