Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to Decompose into Partial Fractions Why is it useful to write:
$$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x+1}+\frac{}{x-1}$$
and not:
$$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x-1}$$
when decomposing into partial fractions?
| For example, try decomposing $\frac{1}{(x+1)^2(x-1)}$ using the second form:
\begin{align}
\frac{1}{(x+1)^2(x-1)} &= \frac{A}{(x+1)^2} + \frac{B}{x-1} \\
\frac{1}{(x+1)^2(x-1)} &= \frac{A(x-1) + B(x+1)^2}{(x+1)^2(x-1)}.
\end{align}
Then $B$ must be zero by looking at the $x^2$ term, and then $A$ must be zero as well. So as the commenter pointed out, it just doesn't work. But using the first form we get
$$\frac{1}{(x+1)^2(x-1)} = -\frac{1}{2 (x+1)^2}-\frac{1}{4 (x+1)}+\frac{1}{4 (x-1)}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Another simple limit problem L'Hôpital tells us that the following limit is $\frac{\pi}{2}$, but I'm wondering if there is a smart way of evaluating this limit without using L'Hôpital
$$ \lim_{x \to \frac{1}{4}}\frac{\log(\tan(\pi x))}{4x - 1} =?$$
| Completing the solution of the lab, we have using infinites series
$$
\frac{1}{1 + h} = 1 - h + h^2 - \ldots \quad |h|<1 \quad \Rightarrow
$$
$$
\int\frac{1}{1 + h}dh = \ln(1 + h) + C = h - \frac{h^2}{2} + \frac{h^3}{3} - \ldots
$$
For $h = 0 \ \Rightarrow \ C = 0$. Futhermore, $\tan x = x + x^3/3 + \ldots$. Thus,
$$
\lim_{y \to 0}\frac{\ln\biggl(1 + \tan \frac{\pi y}{4}\biggr)}{y} = \lim_{y \to 0}\frac{\tan\frac{\pi y}{4} - \frac{1}{2}\tan^2 \frac{\pi y}{4} + \ldots}{y} = \lim_{y \to 0} \frac{\frac{\pi y}{4} + O(y^2)\ldots}{y} = \frac{\pi}{4}
$$
Similarly,
$$
\lim_{y \to 0}\frac{\ln\biggl(1 - \tan \frac{\pi y}{4}\biggr)}{y} = -\frac{\pi}{4}
$$
Thus,
$$
\lim_{x \to 1/4}\frac{\ln(\tan(\pi x))}{4x - 1} = \frac{\pi}{2}
$$
| {
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Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$ The following result discussed by Ramanujan is very famous: $$\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}\tag {1}$$ and can be easily proved by cubing both sides and using $x = \sqrt[3]{2}$ for simplified typing.
Ramanujan established many such denesting of radicals such as $$\sqrt{\sqrt[5]{\frac{1}{5}} + \sqrt[5]{\frac{4}{5}}} = \sqrt[5]{1 + \sqrt[5]{2} + \sqrt[5]{8}} = \sqrt[5]{\frac{16}{125}} + \sqrt[5]{\frac{8}{125}} + \sqrt[5]{\frac{2}{125}} - \sqrt[5]{\frac{1}{125}}\tag {2}$$$$\sqrt[3]{\sqrt[5]{\frac{32}{5}} - \sqrt[5]{\frac{27}{5}}} = \sqrt[5]{\frac{1}{25}} + \sqrt[5]{\frac{3}{25}} - \sqrt[5]{\frac{9}{25}}\tag {3}$$$$\sqrt[4]{\frac{3 + 2\sqrt[4]{5}}{3 - 2\sqrt[4]{5}}} = \frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}\tag{4}$$$$\sqrt[\color{red}6]{7\sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}\tag{5}$$$$\sqrt[6]{4\sqrt[3]{\frac{2}{3}} - 5\sqrt[3]{\frac{1}{3}}} = \sqrt[3]{\frac{4}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{1}{9}}\tag{6}$$
$$\sqrt[8]{1\pm\sqrt{1-\left(\frac{-1+\sqrt{5}}{2}\right)^{24}}} = \frac{-1+\sqrt{5}}{2}\,\frac{\sqrt[4]{5}\pm 1}{\sqrt{2}}\tag{7}$$
with the last one found in Ramanujan's Notebooks, Vol 5, p. 300. Most of these radical expressions are units (a unit is an algebraic integer $\alpha$ such that $\alpha\beta = 1$ where $\beta$ is another algebraic integer).
For me the only way to establish these identities is to raise each side of the equation to an appropriate power using brute force algebra and then check the equality. However for higher powers (for example equation $(2)$ above) this seems very difficult.
Is there any underlying structure in these powers of units which gives rise to such identities or these are mere strange cases which were noticed by Ramanujan who used to play with all sorts of numbers as a sort of hobby? I believe (though not certain) that perhaps Ramanujan did have some idea of such structure which leads to some really nice relationships between units and their powers. I wonder if there is any sound theory of such relationships which can be exploited to give many such identities between nested and denested radicals.
| I found a PDF, where the authors have working algorithms to denest nested radicals like that of Ramanujan's, without the use of Galois theory.
https://www-old.cs.uni-paderborn.de/uploads/tx_sibibtex/DenestRamanujansNestedRadicals.pdf
I think this might help in offering an alternative way, regarding relations in the form of algorithms, to simplify and denest nested radicals as opposed to the helpful links given in the comments to your question. It is certainly detailed and interesting, so I hope it's of some use to you.
| {
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Area of the quadrilateral within a triangle Given the area of tringles $BEF=X,BFC=Y$ and $FDC=Z$, Can we find the area of the quadrilateral $AEFD$ in terms of $X,Y,Z$?
| Let $U = AEF$ and $V = ADF$. Note that $AEFD = U + V$.
$BCE\frac{AB}{BE} = ABC \Rightarrow (X+Y)\frac{AB}{BE} = X + Y + Z + U + V$
$BEF\frac{AB}{BE} = ABF \Rightarrow X\frac{AB}{BE} = X + U$
Dividing this two equations we get:
$\frac{X+Y}{X} = \frac{X + Y + Z + U + V}{X + U} \Rightarrow (X + U)\frac{X+Y}{X} = X + Y + Z + U + V \Rightarrow X + Y + U + U\frac{Y}{X} = X + Y + U + V \Rightarrow \boxed{U\frac{Y}{X} = Z + V}$
Similarly:
$BCD\frac{AC}{CD} = ABC \Rightarrow (Y + Z)\frac{AC}{CD} = X + Y + Z + U + V$
$CDF\frac{AC}{CD} = ACF \Rightarrow Z\frac{AC}{CD} = Z + V$
Dividing this two equations we get:
$\frac{Y + Z}{Z} = \frac{X + Y + Z + U + V}{Z + V} \Rightarrow (Z + V)\frac{Y + Z}{Z} = X + Y+ Z + U + V \Rightarrow Y + Z + V + V\frac{Y}{Z} = X + Y + Z + U + V\Rightarrow \boxed{V\frac{Y}{Z} = X + U}$
We have a nice system of two equations with two unknowns $U$ and $V$.
This can be easily solved:
$$U = \frac{XYZ + X^2Z}{Y^2-XZ}$$
$$V = \frac{XYZ + XZ^2}{Y^2-XZ}$$
The area of the quadrilateral is $U + V$:
$$AEFD = U + V = \frac{2XYZ + X^2Z + XZ^2}{Y^2-XZ}$$
| {
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Inequality question $$a,b,c,d\ge 0$$
$$a\le 1$$
$$a+b\le 5$$
$$a+b+c\le 14$$
$$a+b+c+d\le 30$$
Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$.
We can subtract inequalities to get the answer, but that is wrong... I can't think of any another method... Any hints or suggestions will be helpful.
| Let $(a_0,b_0,c_0,d_0)$ be a point in the domain where $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is maximized. It suffices to show that $a_0 = 1$, $a_0 + b_0 = 5$, $a_0 + b_0 + c_0 = 14$, and $a_0 + b_0 + c_0 + d_0 = 30$ for then you can solve for all four variables to get $(a_0,b_0,c_0,d_0) = (1,4,9,16)$.
We start with the last equation and proceed backwards. Note that if $a_0 + b_0 + c_0 + d_0$ were strictly less than $30$ we could increase $d_0$ slightly keeping the other variables fixed, and we'd stay in the domain yet $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ would increase, contradicting maximality of $\sqrt{a_0} + \sqrt{b_0} + \sqrt{c_0} + \sqrt{d_0}$. So we must have $a_0 + b_0 + c_0 + d_0 = 30$. Note that since $a_0 + b_0 + c_0 \leq 14$ this means $d_0 \geq 16$. In particular $d_0 > c_0$ since $c_0 \leq 14$.
Moving to the equation $a_0 + b_0 + c_0 \leq 14$, if strict inequality held here then one could replace $c_0$ by $c_0 + \epsilon$ and $d_0$ by $d_0 - \epsilon$ for some small $\epsilon$ and we would stay in the domain. However we must have $\sqrt{c_0 + \epsilon} + \sqrt{d_0 - \epsilon} > \sqrt{c_0} + \sqrt{d_0}$. To see why, squaring this inequality we see it's equivalent to
$$c_0 + d_0 + 2\sqrt{(c_0 + \epsilon)(d_0 - \epsilon)} > c_0 + d_0 + 2\sqrt{c_0d_0}$$
Subtracting $c_0 + d_0$ from both sides and squaring the result, we see this is the same as
$$(c_0 + \epsilon)(d_0 - \epsilon) > c_0d_0$$
Equivalently,
$$(d_0 - c_0)\epsilon - \epsilon^2 > 0$$
Since $d_0 \geq 16 > 14 \geq c_0$, this will hold if $\epsilon$ is small enough.
In summary, if we had $a_0 + b_0 + c_0 < 14$, then adjusting $c_0$ and $d_0$ as above would result in a new point $(a,b,c,d)$ in the domain for which $\sqrt{c} + \sqrt{d}$ is strictly larger than $\sqrt{c_0} + \sqrt{d_0}$. Since $a = a_0$ and $b = b_0$ are unchanged, $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is larger than $\sqrt{a_0} + \sqrt{b_0} + \sqrt{c_0} + \sqrt{d_0}$, contradicting the maximality of $\sqrt{a_0} + \sqrt{b_0} + \sqrt{c_0} + \sqrt{d_0}$.
We conclude $a_0 + b_0 + c_0 = 14$. To get the other two inequalities, you just iterate the above procedure to get $a_0 + b_0 = 5$, and then $a_0 = 1$.
| {
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Integral expression $P=x^3+x^2+ax+1$, ... I am trying to solve the following problem: Consider the integral expression $P=x^3+x^2+ax+1$, where $a$ is a rational number. At $a = ?$ the value of $P$ is a rational number for any x which satisfies the equation $x^2+2x-2=0$, and in this case the value of $P$ is ?
I have the key with the answers -4 and -1 but am wondering which steps do I take to get to that answer? So far I have only concluded that for $x^2+2x-2=0, x_1=-1+\sqrt3,x_2=-1-\sqrt3)$
| If $x^2 + 2x - 2 = 0$ then $$P-(x^2+2x-2)=P-0=P$$ which tells us that $$P = x^3 + (a-2)x +3$$
Now we need to plug in $x_1$ and $x_2$ in order to solve for $a$.
$$P=(-1+\sqrt{3})^3 + (a-2)(-1+\sqrt{3}) + 3 = (-1+3\sqrt{3}-9+3\sqrt{3})+(a-2)(-1+\sqrt{3}) + 3$$
$$=(-5) + 4\sqrt{3} + a(-1 + \sqrt{3})$$
$$=(-5) + (4+a)\sqrt{3} - a$$
Notice that we need to get rid of the $\sqrt{3}$ term in order to make $P$ rational. Thus the choice of $a=-4$ will work, since that zeros out the term with $\sqrt{3}$.
Apply the same idea with $x_2$ and that will give you your answer.
Note that we didn't have to subtract by $x^2+2x-2$ in the beginning, but by doing so we were able to get rid of the $x^2$ term, and thus allowed us to avoid doing a foil and a cube.
| {
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Show that $\frac{x^4 +7x^3+5}{4x+1}$ is big-theta($x^3$) I'm having trouble grasping how to set these types of problems. There are a lot of related questions but it's difficult to abstract a general procedure on finding constants that give the given function bounding constraints to make it big-theta(general function).
so $\frac{x^4 +7x^3+5}{4x+1}$ is $ \Theta (x^3) $
to show this, we need to find constants such that.
$$ |c_1|(x^3) \leq \frac{x^4 +7x^3+5}{4x+1} \leq |c_2|(x^3)$$
In addition, there also has to be a $k$ such that for all values $x >k $ the argument holds.
start with one inequality
$$ |c_1|(x^3) \leq \frac{x^4 +7x^3+5}{4x+1}$$
$$ = |c_1| \leq \frac{x^4 +7x^3+5}{4x^4+x^3}$$
$$ = |c_1| \leq \frac{x^4}{x^3(4x+1)} + \frac{7x^3}{x^3(4x+1)} + \frac{5}{x^3(4x+1)}$$
so basically for $x > 0$, $$ |c_1| \leq \frac{1}{4} + 0 + 0$$
I'm assuming after I take the limit as x goes to infinity, i could choose any $c_1$ less than or equal to $\frac{1}{4}$? The other way would then have the same procedure? What would I set $k$ to?
| Here is a nice simple method.
If $x>1$ then
$$\frac{x^4 +7x^3+5}{4x+1}<\frac{x^4+7x^4+5x^4}{4x}=\frac{13}{4}x^3$$
and
$$\frac{x^4 +7x^3+5}{4x+1}>\frac{x^4}{4x+x}=\frac{1}{5}x^3\ .$$
That is, we have shown that if $x>1$ then
$$\frac{1}{5}x^3<f(x)<\frac{13}{4}x^3\ .$$
| {
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How to find a function $\phi(x)$ such that $\sqrt{1+y^2} - \sqrt{1+x^2} \geq \phi(x) (y-x)$ for each $x,y\in \mathbb{R}$ How to find a function $\phi(x)$ such that $\sqrt{1+y^2} - \sqrt{1+x^2} \geq \phi(x) (y-x)$ for each $x,y\in \mathbb{R}$.
Here are some of my ideas:
Also by applying Mean Value theorem, we know that for every $x,y\in \mathbb{R}$ there exists a constant $c$ between $x,y$ such that
$$\sqrt{1+y^2} - \sqrt{1+x^2} = \frac{c}{\sqrt{1+c^2}}(y-x).$$
I know that $\phi(0) = 0$, by taking $x=0$; $y= \phi(0)$ and $y=-\phi(0)$. Also that $|\phi|$ is bounded above by $1$.
Could anyone give me a hint on how to continue? Thanks!
| For each x, use mean value theorem, $$\sqrt{1+y^2} - \sqrt{1+x^2} = \frac{c}{\sqrt{1+c^2}}(y-x), \lim_{y\rightarrow x} c(x,y)=x$$
When $y>x$, we get $$\phi(x)\leq \frac{c}{\sqrt{1+c^2}},
\forall y>x,\text{take limit } \phi(x)\leq \lim_{y\rightarrow x^+}\frac{c}{\sqrt{1+c^2}}=\frac{x}{\sqrt{1+x^2}}$$
When $y<x$, we get $$\phi(x)\geq \frac{c}{\sqrt{1+c^2}},
\forall y<x,\text{take limit } \phi(x)\geq \lim_{y\rightarrow x^-}\frac{c}{\sqrt{1+c^2}}=\frac{x}{\sqrt{1+x^2}}$$
So $\phi(x)=\frac{x}{\sqrt{1+x^2}}, x\in \mathbb{R}$.
| {
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Proof of an inequality
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c} + \frac{\sqrt{a+b+c}+\sqrt{b}}{c+a} + \frac{\sqrt{a+b+c}+\sqrt{c}}{a+b} \geq \frac{9+3\sqrt{3}}{2\sqrt{a+b+c}}$$
| As the inequality is homogeneous, we can set $a+b+c = 1$, say. Then we have to show the cyclic sum:
$$\sum_{cyc} \frac{1+\sqrt{a}}{1-a} = \sum_{cyc} \frac1{1- \sqrt a} \ge \frac{9+3\sqrt3}2$$
To show this, it is sufficient to show that
$$f(x) = \frac1{1-\sqrt x} - \frac{3+\sqrt 3}2 - k(\tfrac13-x) \ge 0$$
for some $k \in \mathbb R$ and $x \in (0, 1)$, as the inequality is equivalent to $f(a)+f(b)+f(c) \ge 0$.
We find that $k = \frac34(3+2\sqrt 3)$ works, as then
$$f(x) = \frac{(3 + 2 \sqrt3) \left(\sqrt3 - 3 \sqrt x \right)^2 (-3 + 2 \sqrt3 + 3 \sqrt x)}{36(1 - \sqrt x)} \ge 0, \quad \forall x \in (0, 1)$$
| {
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Solving Differential equations with Laplace transform $\displaystyle y''+4y'+3y=e^{-t}$, given $\displaystyle y(0)=y'(0)=1$
My Attempt:
Taking Laplace transforms on both sides
$\displaystyle $
$\displaystyle [s^2\bar y-sy(0)-y'(0)]+4[s\bar y-y(0)]+3\bar y=\frac{1}{s+1} $
$\displaystyle [s^2+4s+3]\bar y=\frac{1}{s+1}+s+5 $
$\displaystyle \bar y=\frac{s^2+6s+6}{(s+1)(s^2+4s+3)}$
Resolving into partial fractions,
$\displaystyle \frac{A}{s+1}+\frac{B}{(s+1)^2}+\frac{C}{s+3}=\frac{s^2+6s+6}{(s+1)(s^2+4s+3)}$
I get A=7/2; B=1/2 and C=1
Taking inverse,
$\displaystyle y=\frac{7}{2}e^{-t}+\frac{1}{2}te^{-t}+e^{-3t}$
The given answer is $\displaystyle y=\frac{7}{4}e^{-t}-\frac{1}{2}te^{-t}-\frac{3}{4}e^{-3t}$
I can't find where I am going wrong. Please help.
| The value of $C$ is wrong.
It should be $A=\frac{7}{4}, B=\frac{1}{2}, C=-\frac{3}{4}$
| {
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Calculate $\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$ I'm an eight-grader and I need help to answer this math problem.
Problem:
Calculate $$\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$$
This one is very hard for me. It seems unsolvable. How to calculate the series without using Wolfram Alpha? Please help me. Grazie!
| Hint :
Let
$$
S=\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots\tag1
$$
Dividing $(1)$ by $5^2$, we obtain
$$
\frac{S}{5^2}=\frac{1}{5^3}+\frac{3}{5^5}+\frac{5}{5^7}+\frac{7}{5^9}+\frac{9}{5^{11}}+\cdots\tag2
$$
Subtracting $(2)$ from $(1)$, we obtain
$$
S-\frac{S}{5^2}=\frac{1}{5}+\color{blue}{\text{infinite geometric progression}}
$$
| {
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Antiderivative of $\frac{1}{1+\sin {x} +\cos {x}}$ How do we arrive at the following integral
$$\displaystyle\int\dfrac{dx}{1+\sin {x}+\cos {x}}=\log {\left(\sin {\frac{x}{2}}+\cos {\frac{x}{2}}\right)}-\log {\left(\cos {\frac{x}{2}}\right)}+C\ ?$$
| You can use the standard substitution
\begin{align}
u=\tan{\frac{x}{2}}
\end{align}
Then
\begin{align}
\sin{x}=\frac{2u}{1+u^2}\\
\cos{x}=\frac{1-u^2}{1+u^2}\\
dx=\frac{2}{1+u^2}du\\
\end{align}
You will end up with a rational function as an integrand. Use partial fractions to integrate then.
Edit:
The evalaution of this integral isn't too tedious after all. Turns out that the $u^2$ terms cancel so we are left with a linear denominator.
\begin{align}
\int\frac{dx}{1+\sin x+\cos x}
&=2\int\frac{du}{1+u^2+2u+1-u^2}\\
&=\log(1+u)+c\\
&=\log\left(1+\tan{\frac{x}{2}}\right)+c\\
\end{align}
| {
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Let $a_n$ be the $nth$ term of the sequence $1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \dots$ Question:Let $a_n$ be the $nth$ term of the sequence $1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6\dots$, constructed by including the integer $k$ exactly $k$ times. Show that $a_n = \lfloor \sqrt{2n} + \dfrac{1}{2}\rfloor$.
Attempt:
Try to prove $k = a_n=\lfloor \sqrt{2n} + \dfrac{1}{2}\rfloor$ then $k = a_{n+k-1}=\lfloor \sqrt{2(n+k-1)} + \dfrac{1}{2}\rfloor$.
The inequality of $k = a_n=\lfloor \sqrt{2n} + \dfrac{1}{2}\rfloor$, is the following,
$$k \leq \sqrt{2n} + \dfrac{1}{2} < k + 1$$
$$k-\dfrac{1}{2} \leq \sqrt{2n} < k + \dfrac{1}{2}$$
$$\Big( k-\dfrac{1}{2} \Big)^2 \leq 2n < \Big(k +\dfrac{1}{2}\Big)^2$$
$$\dfrac{1}{2}\Big( k-\dfrac{1}{2} \Big)^2 \leq n < \dfrac{1}{2}\Big(k +\dfrac{1}{2}\Big)^2$$
$$\dfrac{1}{2}\Big( k-\dfrac{1}{2} \Big)^2 + k - 1 \leq n + k -1 < \dfrac{1}{2}\Big(k +\dfrac{1}{2}\Big)^2 + k - 1$$
$$\Big( k-\dfrac{1}{2} \Big)^2 + 2k - 2 \leq 2(n + k -1) < \Big(k +\dfrac{1}{2}\Big)^2 + 2k - 2$$
$$k^2 - k + \dfrac{1}{4} + 2k - 2 \leq 2(n+k - 1) < k^2 + k + \dfrac{1}{4} + 2k - 2$$
$$k^2 + k + \dfrac{1}{4} - 2 \leq 2(n+k - 1) < k^2 + 3k + \dfrac{9}{4} - \dfrac{9}{4} + \dfrac{1}{4} - 2$$
$$\Big( k+\dfrac{1}{2} \Big)^2 - 2 \leq 2(n+k - 1) < \Big( k+\dfrac{3}{2} \Big)^2 - 4$$
From here $2(n+k - 1) < \Big( k+\dfrac{3}{2} \Big)^2 - 4 \implies 2(n+k - 1) < \Big( k+\dfrac{3}{2} \Big)^2$ so I can take the square root of both sides. The problem is the left hand side, $\Big( k+\dfrac{1}{2} \Big)^2 - 2 \leq 2(n+k - 1)$ does not imply $\Big( k+\dfrac{1}{2} \Big)^2 \nleq 2(n+k - 1)$, thus we can't really get rid of the square.
It is easy to see that if we have $k = a_{n+k}=\lfloor \sqrt{2(n+k)} + \dfrac{1}{2}\rfloor$, the problems we encountered are avoided, but $k \neq a_{n+k}$.
| Let $t = \lfloor \sqrt{2n} + \dfrac{1}{2}\rfloor.$ By a look at the sequence $a_n,$ the $n$ for which $a_n=k$ satisfy $$k(k-1)/2+1 \le n < (k+1)k/2+1.\tag{1}$$
Now from the definition of $t$ we have $t \le \sqrt{2n}+1/2 < t+1$, from which after some manipulation we have
$$8\frac{t(t-1)}{2}+1 \le 8n < 8 \frac{t(t+1)}{2}+1.$$
This can now be divided by $8$, and the $1/8$ on the left side coming from division of $1$ by $8$ may be rounded up to $1$ since $n$ is an integer. (The $1/8$ on the right side may be rounded up to $1$ since that only increases the right side to an integer.)
This then gives
$$t(t-1)/2+1 \le n <t(t+1)/2+1,$$
showing that the $t$ defined via the floor function does the same job as the $k$ in the inequality $(1).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\int x^2 \sqrt{x^2-1} dx$ How do I evaluate the following indefinite integral?
$$\int x^2 \sqrt{x^2-1} dx$$
Through integration of parts, I have obtained
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int (x^2-1)^{3/2} dx $$
I've attempted evaluating the second term through substitution, where
$$ x = \sec(u)$$
However, I am stuck with
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int \tan^4(u) \sec (u) du $$
What would be my next step?
| For the domain $|x|>1$,
\begin{align}
\int x^2 \sqrt{x^2-1}\ dx=& \int\frac{x}{4\sqrt{x^2-1}}d\left[(x^2-1)^2\right]\\
\overset{ibp}=&\ \frac14 x(x^2-1)^{3/2}
+\frac18\int \frac{\sqrt{x^2-1}}x d(x^2)\\
\overset{ibp}= &\
\frac14 x(x^2-1)^{3/2}+\frac18 x(x^2-1)^{1/2}-\frac18
\tanh^{-1}\frac x{\sqrt{x^2-1}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Integrate a division of polynomials Hi I have the following integral:
$$\int \frac{2x}{x^2+6x+3}\, dx$$
I made some changes like:
$$\int \dfrac{2x+6-6}{x^2+6x+3}\, dx$$
then I have:
$$\int \dfrac{2x+6}{x^2+6x+3}\, dx -\int\dfrac{6}{x^2+6x+3}\, dx$$
and thus: $$\ln(x^2+6x+3)-\int\dfrac{6}{x^2+6x+3}\, dx$$
Ok, I have decomposed $$\frac{2x}{x^2+6x+3} $$ in: $$ \frac{3+\sqrt6}{\sqrt6(x+\sqrt 6+3)} + \frac{3-\sqrt6}{\sqrt6 (-x+\sqrt6-3)}$$
How can I integrate this expressions?
| A start: Note that $x^2+6x+3=0$ has the roots $\alpha=-3+\sqrt{6}$ and $\beta=-3-\sqrt{6}$. Thus $x^2+6x+3=(x-\alpha)(x-\beta)$.
Express $\frac{6}{(x-\alpha)(x-\beta)}$ as $\frac{A}{x-\alpha}+\frac{B}{(x-\beta)}$ (partial fractions).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find a value r so that the vector v is in the span of a set of vectors Find the value r so that,
$$v = \begin{pmatrix} 3 \\r \\-10\\14 \end{pmatrix}$$
is in the set,
$$ S= \text{span}\left(\begin{pmatrix} 3\\3\\1\\5
\end{pmatrix}, \begin{pmatrix} 0\\3\\4\\-3 \end{pmatrix}, \begin{pmatrix} 0\\0\\-3\\3 \end{pmatrix} \right) $$
What I've tried is this..., brute force and intuition. Since we're looking for a linear combination of the 3 vectors in the set S, we know that we can only use a multiple of 1 from the first vector, call this $c_1=1$. We still need $c_2$ and $c_3$ (combinations of the second vector and third vector, respectively). One can see that if we let $c_2=-2$, and $c_3=1$, we can get the bottom two entries of $v$. Hence, the only value that $r$ can be is $-3$ , so $r=-3$.
This is great..., but how about when the system is really complicated and trial and error won't suffice, how would I be able to calculate an arbitrary $r$ that is in a vector $\in$$\mathbb{R}^N$. With a set of vectors that span $\mathbb{R}^N$
| Your ad hoc solution is excellent. The general method consists in solving a system of linear equations, reducing it to row echelon form via Gaussian elimination. In your case you start with the matrix
$$
\begin{bmatrix}
0&0&3&3\\
0&3&3&r\\
-3&4&1&-10\\
3&-3&5&14\\
\end{bmatrix}
$$
and apply row elementary operations to get
$$
\begin{bmatrix}
0&0&3&|&3\\
0&3&3&|&r\\
0&1&6&|&4\\
3&-3&5&|&14\\
\end{bmatrix},\quad
\begin{bmatrix}
0&0&3&|&3\\
0&0&-15&|&r-12\\
0&1&6&|&4\\
3&-3&5&|&14\\
\end{bmatrix},\quad
\begin{bmatrix}
0&0&-15&|&r-12\\
0&0&3&|&3\\
0&1&6&|&4\\
3&-3&5&|&14\\
\end{bmatrix},\quad
\begin{bmatrix}
0&0&0&|&r+3\\
0&0&3&|&3\\
0&1&6&|&4\\
3&-3&5&|&14\\
\end{bmatrix}.
$$
This shows that there is a solution only when $r+3 = 0$, that is, $r = -3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/884043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to prove $(1-\frac1{36})^{25}\lt\frac12$? How to prove the inequality?
$(1-\frac1{36})^{25}\lt\frac12$
I'm in trouble.
Thank you very much for your help
| Here I present a basic proof only using the binomial expansion theorem without using the property of $(1-\frac{1}{x})^x$. Note that the inequality is equivalent to the following inequality
$$ \left(\frac{36}{35}\right)^{25}> 2.$$
By the binomial expansion theorem, we have
\begin{eqnarray}
\left(\frac{36}{35}\right)^{25}&=&\left(1+\frac{1}{35}\right)^{25}\\
&>&1+\binom{25}{1}\frac{1}{35}+\binom{25}{2}\frac{1}{35^2}+\binom{25}{3}\frac{1}{35}\\
&=&1+\frac{25}{35}+\frac{25\cdot 24}{2}\frac{1}{35^2}+\frac{25\cdot24\cdot23}{3\cdot 2}\frac{1}{35^3}\\
&=&1+\frac{5}{7}+\frac{12}{49}+\frac{92}{1715}\\
&=&1+\frac{1737}{1715}\\
&>&2.
\end{eqnarray}
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 10,
"answer_id": 0
} |
Calculating the following Integral Here is the question which I need help to calculate:
Knowing that $\displaystyle \int_2^3 (2f(x) - 3g(x)) dx = 2 $ and $\displaystyle \int_2^3 g(t)dt = -1$,
find the value of $$\int_2^3 (3f(s) + 2g(s)) ds$$
I do not even know where to start for such a question, any tips of where to start off would be greatly appreciated :)
| This is really essentially an algebra problem, once you use the given conditions and the linearity of the integral. This is one way to proceed: As $\int \limits_{x=2} ^{x=3} 2f - 3g = 2$, and $\int \limits_{x=2} ^{x=3} g = -1$, then
$$
\int \limits_{x=2} ^{x=3} f =
\frac{1}{2} \int \limits_{x=2} ^{x=3} 2f =
\frac{1}{2} \int \limits_{x=2} ^{x=3} (2f - 3g + 3g) =
\frac{1}{2} \int \limits_{x=2} ^{x=3} (2f - 3g) + \frac{3}{2} \int \limits_{x=2} ^{x=3} g =
\frac{1}{2} \cdot 2 + \frac{3}{2} \cdot (-1) = - \frac{1}{2}
.
$$
Hence,
$$
\int \limits_{x=2} ^{x=3} 3f + 2g
= 3 \int \limits_{x=2} ^{x=3} f + 2 \int \limits_{x=2} g
= - \frac{3}{2} + (-2) = - \frac{7}{2}.
$$
I see others beat me here - however, as I have it written up, perhaps a multitutde of solutions will be helpful, so I will post it anyways.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int x\sqrt{(a^2 - x^2)}dx$ I need to find $$\int x\sqrt{(a^2 - x^2)}dx$$
I tried putting $x=a cos(t)$ but I ended up getting a very complicated expression, so any tips?
| Let $y=a^2-x^2$ then $dy=-2xdx$ so
$$-\frac{1}{2}\int \sqrt{a^2-x^2} (-2x)dx=-\frac{1}{2}\int \sqrt{y} dy
=-\frac{1}{3}y^\frac{3}{2}= =-\frac{1}{3}(a^2-x^2)^\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate the integral $\int_{-1}^{1}\left\vert\, x^{3} - x\,\right\vert\,{\rm d}x$ I'm trying to solve:
$$\int_{-1}^{1}\left\vert\, x^{3} - x\,\right\vert\,{\rm d}x$$
I tried to solve this integral as follows:
solving $x^{3} - x = 0$ which implies $x = 0$ , $x = -1$ or $x = 1$. The problem is at which point will I break the integral ?. Since all of these points lie inside the interval of integration.
The answer is $1/2$. Any help will be appreciated.
| To solve this, notice that absolute value yields both a positive and a negative result.
Notice, when $x = -1, 1$, $f(x) = 0$
When $0 <x < 1$, $f(x) < 0$
And when $-1 <x < 0$, $f(x) > 0$
Therefore, we can say that, by spliting the integrals.
$$\int_{-1}^1|x^3-x|dx=\int_{-1}^0(x^3 - x)dx+\int_0^1(x-x^3)dx$$
Solving, we get:
$$\left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^0 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$
And
$$\left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{0}^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$
So,
$$\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $ \int \sqrt{\frac{1}{\theta^2}+ \frac{1}{\theta^4}} d\theta$ Any Ideas ! is this integrable function
| If you have known the formula of
$$ \int\frac{1}{\sqrt{1+t^2}} dt=\ln(t+\sqrt{1+t^2})+C,$$
you can try as follows (For convenience, I have set $t=\theta$):
\begin{align*}
\sqrt{\frac{1}{t^2}+\frac{1}{t^4}}dt&=\frac{1}{t^2}\cdot\sqrt{1+t^2}dt\\
&=-\sqrt{1+t^2}d\left(\frac{1}{t}\right)\\
&=d\left(\big(-\sqrt{1+t^2}\big)\cdot\frac{1}{t}\right)+\frac{1}{t}\cdot \frac{t}{\sqrt{1+t^2}}dt\\
&=d\left(-\frac{\sqrt{1+t^2}}{t}+\ln\Big(t+\sqrt{1+t^2}\Big)\right).
\end{align*}
As a result, we have that
$$\int\sqrt{\frac{1}{t^2}+\frac{1}{t^4}}dt=-\frac{\sqrt{1+t^2}}{t}+\ln\Big(t+\sqrt{1+t^2}\Big)+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\int_0^1\frac{1-x}{1-x^6}\ln^4x\,dx=\frac{16\sqrt{3}}{729}\pi^5+\frac{605}{54}\zeta(5)$ This integral comes from a well-known site (I am sorry, the site is classified due to regarding the OP.)
$$\int_0^1\frac{1-x}{1-x^6}\ln^4x\,dx$$
I can calculate the integral using the help of geometric series and I get the answer
\begin{align}
\sum_{n=0}^\infty\left(\frac{24}{(6n+1)^5}-\frac{24}{(6n+2)^5}\right)
&=\frac{1}{6^5}\left(\Psi^{(4)}\left(\frac{1}{3}\right)-\Psi^{(4)}\left(\frac{1}{6}\right)\right)\\
&=\frac{16\sqrt{3}}{729}\pi^5+\frac{605}{54}\zeta(5)
\end{align}
To be honest, I use Wolfram Alpha to calculate the sum of series. The problem is I don't think this is the correct way to calculate the integral because I use a machine to help me. I tried another way, I used partial fraction to decompose the integrand as
$$\frac{\ln^4x}{3(x+1)}+\frac{\ln^4x}{2(x^2+x+1)}-\frac{2x-1}{6(x^2-x+1)}\ln^4x$$
but none of them seemed easy to calculate. Could anyone here please help me to calculate the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
| You can use the geometric series to evaluate. In fact,
\begin{eqnarray}
I&=&\int_0^1\frac{(1-x)\ln^4x}{1-x^6}dx\\
&=&\int_0^1\sum_{n=0}^\infty(1-x)x^{6n}\ln^4xdx\\
&=&\sum_{n=0}^\infty(1-x)x^{6n}\ln^4xdx\\
&=&\sum_{n=0}^\infty\left(\frac{24}{(6n+1)^5}-\frac{24}{(6n+2)^5}\right)\\
&=&\frac{1}{324}\left(\sum_{n=0}^\infty\frac{1}{(n+1/6)^5}-\sum_{n=0}^\infty\frac{1}{(n+1/3)^5}\right)\\
&=&\frac{1}{324}(\zeta(5,\frac{1}{6})-\zeta(5,\frac{1}{3}))\\
&=&\frac{16\pi^5}{243\sqrt3}+\frac{605}{54}\zeta(5).
\end{eqnarray}
The (tedious) calculation of $\zeta(5,\frac{1}{6})$ and $\zeta(5,\frac{1}{3})$ is derived from two results from this (in Pages 1628-1629). First we have
$$ \zeta(5,\frac{1}{6})-\zeta(5,\frac{5}{6})=\frac{44\pi^5}{\sqrt3}, \zeta(5,\frac{1}{6})+\zeta(5,\frac{5}{6})=(2^5-1)(2^5-1)\zeta(5), $$
from which we obtain
$$ \zeta(5,\frac{1}{6})=\frac{22\pi^5}{\sqrt3}+3751\zeta(5). $$
Since
$$ \zeta(5,\frac{1}{6})+\zeta(5,\frac{1}{3})+\zeta(5,\frac{1}{2})+\zeta(5,\frac{2}{3})+\zeta(5,\frac{5}{6})+\zeta(5)=6^5\zeta(5),\zeta(5,\frac{1}{2})=31\zeta(5) $$
we have
$$ \zeta(5,\frac{1}{3})+\zeta(5,\frac{2}{3})=242\zeta(5). $$
Also
$$ \zeta(5,\frac{1}{3})-\zeta(5,\frac{2}{3})=\sum_{n=-\infty}^\infty\frac1{(n+\frac{1}{3})^5}=\frac{4\pi^5}{3\sqrt3}.$$
From this, we obtain
$$\zeta(5,\frac{1}{3})=\frac{2\pi^5}{3\sqrt3}+121\zeta(5). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/889110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Given that $w = e^{(2\pi/5)i}$, show that $w + \overline{w}$ is a root > of $z^2 + z - 1 = 0$ Hoping to get some help with the following problem
Given that $w = e^{(2\pi/5)i}$, show that $w + \overline{w}$ is a root
of $z^2 + z - 1 = 0$.
I've noticed that $w + \overline{w} = 2Re(w)$, but I'm not sure on how to proceed from there. $2\pi/5$ is a weird angle, so substituting a cosine doesn't seem to make much sense.
|
We may treat $ \ w \ = \ e^{i·2\pi/5} \ $ and $ \ \overline{w} \ = \ e^{-i·2\pi/5} \ $ as unit vectors in the complex plane, as seen in the graph above. Placing them head-to-tail, they then form two sides of an isosceles triangle in which their vector sum $ \ w + \overline{w} \ $ is the third side lying in the real-axis. We shall call the length of that side $ \ z_1 \ \ ; $ as the apex angle of this triangle has measure $ \ \pi/5 \ \ , $ we have $ z_1^2 \ = \ 1^2 + 1^2 - 2·1·1·\cos \frac{\pi}{5} \ = \ 2·(1 - \cos \frac{\pi}{5} )\ \ . $
If we employ the cosine-squared identity, we may also write
$$ \ \left(\cos \frac{2\pi}{5} \right)^2 \ = \ \frac12 · \left(1 \ + \ \cos \frac{4\pi}{5} \right) \ \ \Rightarrow \ \ \left(2 · \cos \frac{2\pi}{5} \right)^2 \ = \ 2 · \left(1 \ + \ \cos \frac{4\pi}{5} \right) \ \ , $$
and, since $ \ \cos \frac{4\pi}{5} = -\cos \frac{\pi}{5} \ \ , $
$$ \left(2 · \cos \frac{2\pi}{5} \right)^2 \ = \ 2 · \left(1 \ - \ \cos \frac{\pi}{5} \right) \ = \ z_1^2 \ \ . $$
(We also see this in that $ \ \cos \frac{2\pi}{5} \ $ is one-half of the base of the isosceles triangle.)
We can find the trigonometric values of interest from dissection of a unit regular pentagon (as mentioned on this site going back to here). By drawing in two diagonals of the pentagon, which has interior angles $ \ \frac{3\pi}{5} \ \ , $ we form a unit parallelogram and two sets of similar triangles.
The isosceles triangle we have been discussing appears here, for instance, as $ \ \Delta ABF \ \ , $ with unit sides and base $ \ x \ \ . $ The other type of isosceles triangle is represented by $ \ \Delta ABC \sim \Delta CFB \ \ . $ Between those triangles, we can set up the proportionality $$ \frac{x}{1} \ = \ \frac{1}{1 \ + \ x} \ \ \Rightarrow \ \ x·(1 \ + \ x) \ = \ 1 \ \ \rightarrow \ \ z^2 \ + \ z \ - \ 1 \ \ = \ \ 0 \ \ , $$
which is the equation for which we are seeking roots. So one of these is $ \ z_1 \ = \ 2 · \cos \frac{2\pi}{5} \ \ . $
We have demonstrated the proposition of the problem, but we can say rather more. A basic relation for the "golden ratio" $ \ \phi \ $ (some people prefer $ \ \tau \ $) is $ \ \frac{1}{\phi} + 1 \ = \ \phi \ \ ; $ returning to $ \ x·(1 \ + \ x) \ = \ 1 \ \ , $ we find $ \ \frac{1}{\phi}·\left(1 \ + \ \frac{1}{\phi} \right) \ = \ \frac{1}{\phi}·\phi \ = 1 \ \ . $ Applying the quadratic formula to $ \ z^2 + z - 1 \ = \ 0 \ \ , $ we obtain $ \ x \ = \ \frac{-1 \ \pm \ \sqrt5}{2} \ \ . $ The positive root of the quadratic equation then tells us that
$$ z_1 \ \ = \ \ 2 · \cos \frac{2\pi}{5} \ \ = \ \ \frac{1}{\phi} \ \ = \ \ \frac{-1 \ + \ \sqrt5}{2} \ \ \approx \ \ 0.61803 \ \ . $$
The other root of this quadratic equation is also contained in the golden-ratio relation:
$$ \frac{1}{\phi} + 1 \ = \ \phi \ \ \Rightarrow \ \ 1 - \phi \ = \ -\frac{1}{\phi} \ \ ; $$ $$ x·(1 \ + \ x) \ = \ 1 \ \ , \ \ \text{so} \ -\phi·(1 \ - \ \phi ) \ = \ -\phi·\left(-\frac{1}{\phi} \right) \ = 1 \ \ . $$
This corresponds to the unit vector sum $ \ v + \overline{v} \ \ , $ with
$ \ v \ = \ e^{i·4\pi/5} \ \ , $ in the graph at the top and $ \ \Delta ABC \ $ in the pentagon diagram. The second root is then
$$ z_2 \ \ = \ \ 2 · \cos \frac{4\pi}{5} \ \ = \ \ -\phi \ \ = \ \ \frac{-1 \ - \ \sqrt5}{2} \ \ \approx \ \ -1.61803 \ \ . $$
By similar reasoning, we can also find the roots of $ \ z^2 \ - \ z \ - \ 1 \ = \ 0 \ \ . $ Writing $ \ z·(z \ - \ 1) \ = \ 1 \ $ and the golden-ratio relation as $ \ \phi - 1 \ = \ \frac{1}{\phi} \ \ , $ we obtain
$$ u \ = \ e^{i·\pi/5} \ \ \rightarrow \ \ z'_1 \ \ = \ \ 2 · \cos \frac{\pi}{5} \ \ = \ \ \phi \ \ = \ \ \frac{+1 \ + \ \sqrt5}{2} \ \
$$
and
$$ s \ = \ e^{i·3\pi/5} \ \ \rightarrow \ \ z'_2 \ \ = \ \ 2 · \cos \frac{3\pi}{5} \ \ = \ \ -\frac{1}{\phi} \ \ = \ \ \frac{+1 \ - \ \sqrt5}{2} \ \ , $$
as depicted in the graph below.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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For a triangle $ABC$, $a^2+b^2+c^2=8R^2$ then it is a right triangle? $ABC$ is a triangle, $a^2+b^2+c^2=8R^2$ then how do we prove it is a right triangle?
| Assuming that $R$ means circumradius we can use the formula
$$R=\sqrt{\frac{a^2+b^2+c^2}{8(1+\cos A\cos B\cos C)}},$$
where $a$, $b$, and $c$ are the sidelengths and $A$, $B$, and $C$ are the angles. Plugging in $a^2+b^2+c^2=8R^2$, we reduce the above to
$$1+\cos A\cos B\cos C=1,$$
whence one of $\cos A$, $\cos B$, or $\cos C$ must be $0$ - meaning that we have a right triangle.
| {
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"url": "https://math.stackexchange.com/questions/889946",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Show $\forall x>0:\ln(1+x) > x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$ I'm reading a proof which aim to show that:
$$\forall x>0:\ln(1+x) > x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$$
the Taylor expansion of $\ln(1+x)$ is (not by chance):
$$x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}...$$
Now, the proof claims that the remainder, starting from $\frac{x^5}{5}$ is positive.
But, calculating the remainder using the Lagrange's form gives:
$$\frac{\ln(1+x)^{(6)}(z)}{6!}x^6 < 0$$
Because $$\ln(1+x)^{(6)} = -\frac{120}{(x+1)^6} < 0$$
Where is the mistake?
| You can use MVT to do. In fact, let
$$ f(x)=\ln(1+x)-(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}). $$
Then, by MVT, there is $c\in(0,x)$ such that
$$ f(x)-f(0)=f'(c)x=\frac{c^5}{1+c}x>0 $$
So $f(x)>f(0)$ for $x>0$ or
$$\ln(1+x)>x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying a trigonometric identity Simplify $1 + \tan^2x$
My attempt:
$$\begin{align}1 + \tan^2x&\\
&= \frac{1}{1} + \frac{\sin^2x}{\cos^2x}\\
&= \frac{1(\cos^2x)}{1(\cos^2x)} +\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x + \sin^2x}{\cos^2x\cos^2x}\\
&= \frac{\sin^2x}{\cos^2x}\\
&= \tan^2x\end{align}$$
The correct answer, however..is $sec^2x$ Wherever I went wrong, please show.
| $$1 + \tan^2 x \implies \sec^2 x - \tan^2 x + \tan^2 x \implies \sec^2 x$$
| {
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Area of the surface generated by revolving curve around y-axis
So I did something wrong in my solution because I'm not seeming to get the right answer.
$$\int_c^d 2\pi (4 \sqrt{9-y}\sqrt{1-\frac{4}{9-y}})~\mathrm{d}y$$
combine square roots and move out constants
$$\int_c^d 8\pi (\sqrt{5-y})~\mathrm{d}y$$
indefinite integral of $\sqrt{5-y}$ is $-\frac{2}{3}(5-y)^\frac{3}{2}$
move out constant $$-\frac{16}{3}\pi[(5-y)^\frac{3}{2} ]$$ from zero to $\frac{135}{16}$
$$-\frac{16}{3}\pi\left(-\frac{55}{16}^\frac{3}{2}-5^\frac{3}{2}\right)$$
Anyway this is supposed to come out as $-\frac{73\pi \sqrt{73}}{12}+\frac{208\pi \sqrt{13}}{3}$ and I can't even fathom how to put it in that form.
| $\sqrt{1-\dfrac{4}{9-y}}$ should be $\sqrt{1+\dfrac{4}{9-y}}$
| {
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any other method for evaluating $\int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx$? I tried below and its getting tedious :
$\begin{align}\\ \int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx &= \int\limits \frac{(2x-1)+ x^2-x+2 }{ \sqrt{x^2-x+1} } dx \\~\\ &= \int\limits \frac{(2x-1)dx}{ \sqrt{x^2-x+1} } + \int\limits \frac{ (x^2-x+1 )dx}{ \sqrt{x^2-x+1} } + \int\limits \frac{ dx }{ \sqrt{x^2-x+1} } \\~\\&\cdots\\~\\ \end{align}$
wolfram shows very much simplified answer :
http://www.wolframalpha.com/input/?i=%5Cint+%28x%5E2%2Bx%2B1%29%2F%28sqrt%28x%5E2-x%2B1%29%29
I'm wondering if there is any nice way to work this
| I'm adding a separate answer as it is markedly different from my other answer.
In fact, your way is not too tough either
$$\int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx =\cdots= \int\limits \frac{(2x-1)dx}{ \sqrt{x^2-x+1} } + \int\limits \sqrt{x^2-x+1}\ dx + \int\limits \frac{ dx }{ \sqrt{x^2-x+1} } $$
For the first integral, observe that $\displaystyle\frac{d(x^2-x+1)}{dx}=2x-1$
for the rest two as $\displaystyle x^2-x+1=\frac{(2x-1)^2+(\sqrt3)^2}4,$ set $2x-1=y$
and utilize $\#1,\#8$ of this
| {
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How prove this ineuality$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}\le\sqrt{xy+yz+zx+9}$ let $$x,y,z\in(-1,1), x+y+z=-xyz$$
show that
$$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}\le\sqrt{xy+yz+zx+9}$$
This problem is my frends ask me,I remenber this is old inequality,But Now I can't it
$$x+y+z+3+2\sum_{cyc}\sqrt{xy+x+y+1}\le xy+yz+xz+9$$
| Do as you said and we only have to prove
$$2\sum_{cyc}\sqrt{xy+x+y+1}\le xy+yz+xz+6-x-y-z$$
This is right, since $\displaystyle x+y+z=-xyz$ and $x,y,z,xy,yz,xz\in(-1;1)$, we have:
$2\sqrt{xy+x+y+1}
=2\sqrt{xy+x+y+1-x-y-z-xyz}
=2\sqrt{xy-xyz+1-z}
=2\sqrt{(1-z)(1+xy)}
\le 1-z+1+xy$
Do this with the other two and sum up, we have the inequality.
| {
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Prove $\sum\limits_{cyc}\frac{(b+c-a)^4}{a(a+b-c)}\geq ab+bc+ca$
if $a,b,c$ are positive real numbers,Prove:$$\sum\limits_{cyc}\frac{(b+c-a)^4}{a(a+b-c)}\geq ab+bc+ca$$
Additional info: Problem should be solved with AM-GM inequality only.
Things i have tried so far: Using AM-GM I can say $$\frac{((b+c-a)^2)^2}{a(a+b-c)}+a(a+b-c)\geq 2 (b+c-a)^2$$
is true.So i can conclude that:$$\sum\limits_{cyc}\frac{(b+c-a)^4}{a(a+b-c)}+\sum\limits_{cyc}a(a+b-c)\geq \sum \limits_{cyc}2(b+c-a)^2$$
So if I prove that $$\sum \limits_{cyc}2(b+c-a)^2-\sum\limits_{cyc}a(a+b-c)\geq ab+ac+bc$$
then the problem is solved.I stuck in here.I tried to write left side in expanded form but i was unsuccessful at the end to prove it.
UPDATE
I expanded left side and i got this:$$\sum \limits_{cyc}2(b+c-a)^2-\sum\limits_{cyc}a(a+b-c) = 5a^2+5b^2+5c^2-4ab-4bc-4ca$$
so I can rewrite the last inequality as $a^2+b^2+c^2\geq ab+bc+ca$ which is obviously true
| Check out again the last step, it become $a^2+b^2+c^2\geq ab+bc+ca$, and now that is easy
| {
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Solving $x^3+y^3=x^2y^2+1$ in non-negative integers I wanted to solve $x^3+y^3=x^2y^2+1$ in non-negative integers.
First I set $a=x+y$ and $b=xy$ to get $b^2+3ab+1=a^3$. View as a quadratic in $b$, the discriminant = $4a^3+9a^2-4$, which needs to be a perfect square.
Secondly, rearranging the quadratic in $b$ we get $4a^3+9a^2-4=(2b+3a)^2$.
So the discriminant is always a perfect square. Therefore we have (quadratic formula):
$b=\frac{-3a\pm (2b+3b)}{2}$ so $b\in \{b,-\frac{3a}{2}\}$.
Since we want $a,b\ge 0$, the only possibility is $a=b=0$ to give $x=y=0$.
This is the unique solution.
Note: I wasn't sure if it works, I never tried this way before.
Thanks!
| Introducing the variables $a$ and $b$ leads to the condition
$$4a^3+9a^2-4=(2b+3a)^2\ ,$$
which implies that
$$4a^3+9a^2-4$$
has to be a perfect square. A quick computer search ($|a|\leq 10^6$) produced the solutions
$$a=-2,\quad-1,\quad1,\quad 2,\quad 25\ .$$
This should be some material to work with.
(The logic in the second half of your argument is circular.)
| {
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Prove that the length of segment on tangent is constant for $y=\frac a2\ln{\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}}-\sqrt{a^2-x^2}$ Prove that the length of segment of tangent from point of tangency to the point where it cuts the y-axis is constant.
$$y=\frac a2\ln{\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}}-\sqrt{a^2-x^2}$$
After diffrentiating:
$$y'=-\sqrt{a^2-x^2}$$
So tangent is $$(y-k)=-\sqrt{a^2-h^2}(x-h)$$
Point where it cuts y-axis is $(0,k+h\sqrt{a^2-h^2})$
So the length is:
$$s=\sqrt{h^2+h^2(a^2-h^2)}=h\sqrt{1+h^2-a^2}$$ which isn't constant. I did everything correctly, what should be the correct way.
| Let the point of tangency be $P(x_1,y_1)$. It is easy to check that the slope of the tangent line is
$$ m=y'\big|_{x=x_1}=-\frac{\sqrt{a^2-x_1^2}}{x_1}. $$
Hence the equation of the tangent line is
$$ y-y_1=--\frac{\sqrt{a^2-x_1^2}}{x_1}(x-x_1) $$
which cuts y-axis at $Q(0,y_1+\sqrt{a^2-x_1^2})$. Now
$$ |PQ|=\sqrt{(x_1-0)^2+(y_1-(y_1+\sqrt{a^2-x_1^2}))^2}=a.$$
| {
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$2p-2$ as the sum of consecutive prime numbers Progress:
Let $p$ be a prime such that $p≡1$ (mod 6) then $2p-2$ can be written uniquely (up to the order of addends) as the sum of some consecutive prime numbers.
These are first ten examples:
$$2⋅7-2=12=5+7$$
$$2⋅13-2=24=11+13$$
$$2⋅19-2=36=17+19$$
$$2⋅31-2=60=29+31$$
$$2⋅37-2=72=5+7+11+13+17+19$$
$$2⋅43-2=84=41+43$$
$$2⋅61-2=120=59+61$$
$$2⋅67-2=132=13+17+19+23+29+31$$
$$2⋅73-2=144=71+73$$
$$2⋅79-2=156=17+19+23+29+31+37$$
I don't even have the slightest idea that how and from where to start the proof. Hope you guys can help me. Thanks in advance.
| Let's try the next prime that is $1 \pmod{6}$, i.e. $p = 97$.
We want either $2$ or $6$ consecutive primes which sum to $2 \cdot 97 - 2 = 192$.
Since $89,97,101$ are consecutive primes and $89+97 = 186 < 192 < 198 = 97+101$, we see that we cannot write $192$ as the sum of $2$ consecutive primes.
Since $19,23,29,31,37,41,43$ are consecutive primes and $19+23+29+31+37+41 = 180 < 192 < 204 = 23+29+31+37+41+43$, we see that we cannot write $192$ as the sum of $6$ consecutive primes.
Therefore, $p = 97$ is a counterexample to your conjecture.
| {
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How solve this equation $\sin x\cdot \sin20=2\sin(110-x) (\sin10)^2$ let
$0<x<90$, and such
$$\sin x\cdot \sin20=2\sin{(110-x)}(\sin10)^2$$
find the $x$
my idea: since
$$\sin x\cdot 2\sin10\cos10=2\sin(70+x)(\sin10)^2$$
so
$$\cot10=\dfrac{\sin(70+x)}{\sin x}$$
then How find it?
| How about this (assuming degrees throughout):
$$\begin{align*}
\sin x \sin 20 &= 2 \sin (110 - x) \sin^2 10 \\
&= 2 \cos (20 - x) \sin^2 10 \\
&= 2 (\cos 20 \cos x + \sin 20 \sin x) \sin^2 10 \\
&= (\cos 20 \cos x + \sin 20 \sin x) (1 - \cos 20).
\end{align*}
$$
Then
$$\begin{align*}
\tan x &= \frac{\cos 20 (1 - \cos 20)}{\sin 20 \cos 20} \\
&= \frac{1 - \cos 20}{\sin 20} \\
&= \tan 10,
\end{align*}
$$
so $x = 10^{\circ}$.
| {
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Show that $(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$ for some integers $A,B,C$ How do you solve this question?:$$(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$$ is for all $x$. Here $A$, $B$ and $C$ is constants. I need to know $A$, $B$ and $C$ to pass this. They are integers.
I got this far:
(LS = left side)
$$LS = 49\cos^2(x)+\sin^2(x)+14\sin(x)\cos(x)$$
And then i follow some steps online and got right side to
$$RS = (A+C)\cos^2(x)+(C-A)\sin^2(x)+2B\sin(x)\cos(x)$$
Only problem is that i get C to 49/2.
| $$(7 \cos{(x)}-\sin{(x)})^2=A \cos{(2x)}+B \sin{(2x)}+C \\ \Rightarrow 49 \cos^2{(x)}-14 \cos{(x)} \sin{(x)}+\sin^2{(x)}=A \cos{(2x)}+B \sin{(2x)}+C \ \ \ (*)$$
Use the formulas:
$$\cos{(2x)}=\cos^2{(x)}-\sin^2{(x)} \text{ and } \sin{(2x)}=2 \sin{(x)} \cos{(x)}$$
| {
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Verification of binomial coefficient congruence $\binom{jp}{j}\equiv j\binom{p}{j}\pmod{p^2}$ Let $j\ge 1$ be an integer and $p$ prime.
Is it true that
$$\binom{jp}{j}\equiv j\binom{p}{j}\pmod{p^2}$$
My work No, take $j>p$, then the RHS is zero, while the LHS need not be $\equiv 0$. For example, for $p=3$ and $j=5$, I find $\binom{jp}{j}=3003\equiv 6\pmod{p^2}$.
However, is the claim true for $j\le p$?
| $\binom{jp}{j} = \frac{jp}{j} \binom{jp-1}{j-1} = p \binom{jp-1}{j-1}$ and $j \binom{p}{j} = j \frac{p}{j} \binom{p-1}{j-1} = p \binom{p-1}{j-1}$. Now we want to show that $\binom{jp-1}{j-1} \equiv \binom{p-1}{j-1} \pmod{p}$.
Since $jp-i \equiv p-i \pmod{p}$, we have $$\binom{jp-1}{j-1} = \frac{(jp-1)\cdots(jp-j+1)}{(j-1)!} \equiv \frac{(p-1) \cdots (p-j+1)}{(j-1)!} = \binom{p-1}{j-1} \pmod{p}$$
Therefore $$\binom{jp}{j} = p \binom{jp-1}{j-1} \equiv p \binom{p-1}{j-1} = j \binom{p}{j} \pmod{p^2}. $$
Comment. $j \le p$ is used in the line expanding $\binom{jp-1}{j-1}$ because it makes sense only when the denominator is nonzero.
| {
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Prove that $\sum\limits_{\mathrm{cyc}}{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$ for $x, y, z > 0$
Let $x,y,z>0$. Prove that
$$\frac{1}{(x+2y)^2}+\frac{1}{(y+2z)^2}+\frac{1}{(z+2x)^2} \geq\frac{1}{xy+yz+zx}.$$
I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!
| Assume $z=min\{x,y,z\}$, we prove two inequalities
$$$$1. $$\sum_{cyc} \frac{1}{(2x+y)^2} \geqslant \frac{1}{9xy}+\frac{8}{9(x+z)(y+z)}$$
which is true by expanding
$$$$2.$$\frac{1}{9xy}+\frac{8}{9(x+z)(y+z)} \geqslant \frac{1}{xy+yz+zx}$$
which is equivalent to $(x+y)(y+z)(z+x) \geqslant 8xyz$
| {
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Why does this simplifies like that?
I don't understand how do we simplify this fraction ? How does 3^n becomes 3^(n+1) and how does 1-(1/3) becomes 2 ???
Is there a general rule for this ?
| Multiplying it by $3/3=1$ gives you
$$\frac{3^n}{1-(1/3)}\cdot \frac{3}{3}=\frac{3^n\cdot 3}{3(1-(1/3))}=\frac{3^{n+1}}{3-1}.$$
P.S. Multiplying a fraction by $1$ is sometimes helpful to make it simpler. For example, since $6/6=1$,
$$\frac{2/3}{5/6}=\frac{2/3}{5/6}\cdot \frac{6}{6}=\frac{(2/3)\cdot 6}{(5/6)\cdot 6}=\frac{4}{5}.$$
| {
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Limit of a rational function Calculate the limit
$$ \lim_{x \to 0} \frac{3x^{2} - \frac{x^{4}}{6}}{(4x^{2} - 8x^{3} + \frac{64x^{4}}{3} )}$$
I divided by the highest degree of x, which is $x^{4}$, further it gave
$$ \frac{-\frac{1}{6}}{\frac{64}{3}} = \frac{-1}{128}$$
which is wrong... what is my error?
| $$ \lim_{x \to 0} \frac{3x^{2} - \dfrac{x^{4}}{6}}{(4x^{2} - 8x^{3} + \dfrac{64x^4}{3} )}$$
$$=\lim_{x\to0}\frac{x^2\left(3-\dfrac{x^2}6\right)}{x^2\left(4-8x+\dfrac{64}3x^2\right)}$$
Cancel out $x^2$ as $x\ne0$ as $x\to0$
Then set $x=0$ as it is no longer of the form $\dfrac00$
| {
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What is the smallest $d$ such that $4$ has more than one distinct factorization in $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$? Or if there is no such $d$, how do I prove it?
Obviously there is no point to looking for this in an UFD. I've looked in other rings, and each time I think I found it, I divide one of the factors by $2$ and find a unit, e.g.,:
$$\frac{38 + 12 \sqrt{10}}{2} = 19 + 6 \sqrt{10}$$
$$\frac{8 + 2 \sqrt{15}}{2} = 4 + \sqrt{15}$$
$$\frac{102 + 20 \sqrt{26}}{2} = 51 + 10 \sqrt{26}$$
$$\frac{22 + 4 \sqrt{30}}{2} = 11 + 2 \sqrt{30}$$
$$\frac{70 + 12 \sqrt{34}}{2} = 35 + 6 \sqrt{34}$$
etc.
In general, it seems that if the "integer" part and the "surd" part have the same parity, it will turn out to just be a unit doubled.
| $$\left( \frac{\sqrt{65}+7}{2} \right) \left( \frac{\sqrt{65}-7}{2} \right) =4.$$
Since $65 \equiv 1 \bmod 4$, this is an algebraic integer, and clearly $2$ does not divide $\left( \frac{\sqrt{65}+7}{2} \right)$. We need to check that $2$ and $\left( \frac{\sqrt{65}+7}{2} \right)$ are irreducible in $\mathbb{Z} \left[ \frac{1+\sqrt{65}}{2} \right]$. Since they both have norm $4$, we need to check that there is no element in the ring with norm $\pm 2$. We need to show we can't have $N \left(\frac{a+b\sqrt{65}}{2} \right) = \pm 2$ or equivalently, we can't have $a^2-65 b^2 = \pm 8$. This relation is impossible modulo $5$.
How I found this: Let's think about how the ideal $(2)$ can factor into prime ideals.
*
*$(2)$ stays prime. Then $4 = 2 \times 2$ is the only factorization.
*$(2) = \mathfrak{p} \mathfrak{q}$ is a product of principal ideals, $\mathfrak{p} = (a)$ and $\mathfrak{q}=(b)$. Then $4 = \pm a^2 b^2$ is the unique prime factorization (and we can show the sign is $+$, but that isn't important.)
*$(2) = \mathfrak{p}^2$ with $\mathfrak{p}$ non principal. Again, $4 = 2 \times 2$ is the only factorization.
So the only option is that $2$ splits into distinct, non principal, primes. $2$ splits in the ring of integers of $\mathbb{Q}(\sqrt{d})$ (for $d$ square free) if and only if $d \equiv 1 \bmod 8$.
So we are looking for $d \equiv 1 \bmod 8$, square free, and class number $>1$. The first such $d$ is $65$ (I looked in this table) and it works.
After thinking a bit more, there are infinitely many such examples. Find $k$ such that $8k+5$ and $8k+13$ are both squarefree. There are infinitely many such pairs, see this answer. Take $D = (8k+5)(8k+13)$. It is also squarefree, since no $p^2$ divides either factor and, if $p$ divided both $8k+5$ and $8k+13$, then it would divide $(8k+13)-(8k+5)=8$, but $8k+5$ and $8k+13$ are odd.
Now, $\left( \frac{8k+9 + \sqrt{D}}{2} \right) \left( \frac{8k+9 - \sqrt{D}}{2} \right) = 4$. As before, it is obvious that $2$ doesn't divide $(8k+9 \pm \sqrt{D})/2$. Finally, we must check that there is no solution to $x^2 - D y^2 = \pm 8$. We break into cases, based on the prime factorization of $8k+5$:
Case 1 $8k+5$ is divisible by a prime $p$ which is $5 \bmod 8$. Then $\left( \frac{2}{p} \right) = \left( \frac{-2}{p} \right) = -1$, so $x^2 - D y^2 \equiv x^2 \bmod p$ cannot be $\pm 8$.
Case 2 $8k+5$ is divisible by a prime $p$ which is $3 \bmod 8$, and another prime $q$ which is $7 \bmod 8$. Then $\left( \frac{2}{p} \right) = -1$ and $\left( \frac{-2}{q} \right) = -1$, so $x^2 - D y^2 = 8$ is impossible modulo $p$ and $x^2 - D y^2 = -8$ is impossible modulo $q$.
Case 3 All prime divisors of $8k+5$ are either $1$ or $3$ mod $8$. But then $8k+5$ is $1$ or $3$ modulo $8$, a contradiction.
Case 4 All prime divisors of $8k+5$ are either $1$ or $7$ mod $8$. But then $8k+5$ is $1$ or $7$ modulo $8$, a contradiction.
After staring for a little bit, you will see that one of these cases must hold; QED.
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Use Taylor Series method to solve $y''-2xy+y=0$ I am doing some practice problems for solving second order ODEs, and I am a bit stuck on this one.
Here is what I have:
$y''-2xy'+y=0$
Let $y = \sum_{n=0}^{\infty} C_nx^n \implies y' = \sum_{n=0}^{\infty} nC_nx^{n-1} \implies y'' = \sum_{n=0}^{\infty} n(n-1)C_nx^{n-2} $
Substituting this into the ODE, and I get:
$$ \sum_{n=0}^{\infty} n(n-1)C_nx^{n-2} -2\sum_{n=0}^{\infty} nC_nx^{n}+ \sum_{n=0}^{\infty} C_nx^n = 0$$
Then getting each term to $x^n$ and starting each sum at $n=0$, I have:
$$ \sum_{n=0}^{\infty} [(n+2)(n+1)C_{n+2}-2 nC_n+ C_n]x^n = 0 $$
$$ \implies C_{n+2} = \frac{(2n-1)C_n}{(n+2)(n+1)}$$
I notice that this decouples into two series' for odd and even terms, but I am having trouble with determining the general formula for $C_n$ for each series:
For $n$ even:
When $n=0: C_2 = \frac{-C_0}{2} $
When $n=2: C_4 = \frac{3C_2}{4 \cdot 3} = \frac{-3C_0}{4!} $
When $n=4: C_6 = \frac{7C_4}{6 \cdot 5} = \frac{-7 \cdot 3C_0}{6!} $
When $n=6: C_8 = \frac{11C_6}{8 \cdot 7} = \frac{-11 \cdot 7 \cdot 3C_0}{8!} $
For $n$ odd:
When $n=1: C_3 = \frac{C_1}{3 \cdot 2} $
When $n=3: C_5 = \frac{5C_3}{5 \cdot 4} = \frac{5C_3}{5!} $
When $n=5: C_7 = \frac{9C_5}{7 \cdot 6} = \frac{9 \cdot 5C_1}{7!} $
When $n=7: C_9 = \frac{13C_7}{9 \cdot 8} = \frac{13 \cdot 9 \cdot 5C_1}{9!} $
I am mainly finding it difficult to determine the closed formula for the numerator in each series, so that I can calculate the radius of convergence of each one.
Thanks so much, any help is greatly appreciated.
| Starting from your last relation $$ C_{n+2} = \frac{(2n-1)C_n}{(n+2)(n+1)}$$ it seems that $$C_n=\frac{2^{n-\frac{9}{2}} \Gamma \left(\frac{n}{2}-\frac{1}{4}\right) \left({C_1}
\left((-1)^n-1\right) \Gamma \left(-\frac{1}{4}\right)-2 {C_0}
\left((-1)^n+1\right) \Gamma \left(\frac{1}{4}\right)\right)}{\pi \Gamma (n+1)}$$ which effectively separates the $C$'s for odd and even values because of the $\Big((-1)^n\pm 1\Big)$ terms which multiply the $C_0$ and $C_1$ terms.
This so gives $$C_{2n}=-\frac{2^{2 n-\frac{5}{2}} \Gamma \left(\frac{1}{4}\right) \Gamma
\left(n-\frac{1}{4}\right)}{\pi \Gamma (2 n+1)}C_0$$ $$C_{2n+1}=-\frac{ 2^{2 n-\frac{5}{2}} \Gamma \left(-\frac{1}{4}\right) \Gamma
\left(n+\frac{1}{4}\right)}{\pi \Gamma (2 n+2)}C_1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/913513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Help solve this Problem Given that $a +b +c =5$ and $ab +ac +bc=$5, what is the value of $a^2+b^2+c^2$?
| $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\Rightarrow a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=25-10=15$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/914147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Trigonometric integral: $\int_{25\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx$ Is it possible to evaluate the following in a closed form?
$$\int_{25\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx$$
I found the above definite integral at I&S but the solution is not given. I have tried various methods but none of them lead me to the solution, I am honestly out of ideas for this one.
Any help is appreciated. Thanks!
| We have to compute,
$$\int_{25\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx$$
$$=$$
$$\int_{25\pi/4}^{32\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx + \int_{32\pi/4}^{39\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx + \int_{39 \pi/4}^{46\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx +\int_{46\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx = I_1+I_2+I_3+I_4 $$
Now, in $I_2$ substitute, $u=x-\frac{7 \pi}{4}$ , in $I_3$ substitute, $u = x=\frac{7\pi}{2}$ and in $I_4$ substitute, $u=x-\frac{21\pi}{4}$ .
So after substitution limits in each will integral will become same and we can directly add them. So,
We have
$$\int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx + \frac{1}{(1+2^{\sin x+7\pi/2})(1+2^{\cos x+7\pi/2})}\,dx+ \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{(1+2^{\sin x+7\pi/4})(1+2^{\cos x+7\pi/4})}\,dx + \frac{1}{(1+2^{\sin x+21\pi/4})(1+2^{\cos x+21\pi/4})}\,dx = J_1+J_2$$
Notice that, $ \sin(a+7\pi/2) = -\cos(a) $ and $\cos(a+7\pi/2)=\sin(a) $
Using this for $J_1$ and $J_2$ we get
$$J_1 = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{\sin(x)}}$$
And $$ J_2 = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{\sin(x+7\pi/4)}} $$
Now, using well known result, that is
$$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$$
$$ I = J_1+J_2 = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{\sin(x)}} + \frac{1}{1+2^{\sin(x+7\pi/4)}} $$
Using that result, $I$ also equals
$$ I = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{-\sin(x)}} + \frac{1}{1+2^{-\sin(x+7\pi/4)}} $$
$$ 2I = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} 2 dx $$
$$ I = \frac{7\pi}{4} $$
Note::
In this solution, if i have written
$$ \int f(x) + \int g(x) + \cdots = I_1+I_2+\cdots $$
Then this means $$ I_1 = \int f(x) , I_2 = \int g(x) \cdots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/914652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Taylor polynomial of $\frac{1}{2-x}$ Can someone show how to find the Taylor polynomial of $\frac{1}{2-x}$?
I tried this: $\frac{1}{2-x}=\frac{1}{1-(x-1)}$ and then use that $ \ T_n(\frac{1}{1-x})=1+x+\dots +x^n.$ But this gives $1+(x-1)+\dots +(x-1)^n$ which is the wrong answer. Why?
| Here is how you find it about the point $x=0$.
$$ \frac{1}{2-x} = \frac{1}{2(1-x/2)} = \frac{1}{2}( 1+x/2+x^2/2^2+\dots\,. ). $$
Added: If you want to derive the Taylor series at $x=1$ then you can advance like this
$$ \frac{1}{2-x} = \frac{1}{1-(x-1)}. $$
Now let $t=x-1$ which gives
$$ T_n(\frac{1}{1-t}) = 1+t+\dots+t^n. $$
Substitute back $t=x-1$ gives what you have
$$T_n(\frac{1}{1-t}) = = 1+(x-1)+\dots+(x-1)^n. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/917815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Factoring $a^4(b-c)+b^4(c-a)+c^4(a-b)$ I was solving the question that wanted to factor $a^4(b-c)+b^4(c-a)+c^4(a-b)$. My idea was to factor a $(c-a)$ in first step.So $$b(a^4-c^4)+ac(c^3-a^3)+b^4(c-a)=a^4(b-c)+b^4(c-a)+c^4(a-b)$$
$$(c-a)(-ba^3-bc^3-a^2bc-abc^2+ac^3+c^3a+a^2c^2+b^4)$$
And i was not able to continue this. I put the polynomial in Wolfram Alpha and the result was $$(a-b)(a-c)(b-c)(a^2+a b+a c+b^2+b c+c^2)$$
but it did not provided any steps.I want to know is there any better way to factor this polynomial than my first idea or a general method?Does this Factorization can be used in here? $$a^2(c-b)+b^2(a-c)+c^2(b-a)=(a-b)(b-c)(c-a)$$
| Well, here is one way. First note that setting $a=b$ makes the polynomial zero, so $(a-b)$ must be a factor. By symmetry, $(b-c),(c-a)$ are also factors.
Then by degree considerations you must have a quadratic left, which can be found by division or comparing coefficients.
Then attempt to factor the quadratic, (which in this case is futile) and conclude.
| {
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"url": "https://math.stackexchange.com/questions/918976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Showing an Absolute Value Inequality Problem Proof I tried solving this question but it does not works for me.
Q.) Show that $\left|x + \frac1{x}\right| \ge 2$ for all $x \ne 0$
There are two ways to do. One is squaring and other is to use absolute value definition.
But I don't understand.
Please help.
|
One is squaring and other is to use absolute value definition. But I
don't understand.
So here you go:
1) One is squaring:
Since both sides of the inequality is positive, by squaring, it is equivalent to $$\left|x + \frac1{x}\right|^2 \ge 4$$
Notice that
$$\left|x + \frac1{x}\right|^2 = \left(x + \frac1{x}\right)^2 = x^2 + 2x\frac{1}{x}+\frac{1}{x^2} = x^2 - 2x\frac{1}{x}+\frac{1}{x^2} + 4 = \left(x - \frac1{x}\right)^2 + 4,$$
which is always greater or equal to $4$.
2) Other is to use absolute value definition:
By definition, $|a|$ equals $a$ if $a\ge 0$ and equals $-a$ if $a<0$. Thus:
*
*If $x>0$ then $x + \frac1{x} >0$, thus
$$\left|x + \frac1{x}\right| = x + \frac1{x} = \left( x + \frac1{x} - 2 \right) + 2 = \frac{(x-1)^2}{x}+2 \ge 2.$$
*If $x<0$ then $x + \frac1{x} <0$, thus
$$\left|x + \frac1{x}\right| = -x - \frac1{x} = \left( -x - \frac1{x} - 2 \right) + 2 = \frac{(x+1)^2}{-x}+2 \ge 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/919674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find positive integers $a,b$ such $a \mid b^2$, $b \mid a^2$, $(a+1) \mid (b^2+1)$. Find all postive integer $a,b$ such that
$$a \mid b^2, \quad b \mid a^2$$
and
$$(a+1) \mid (b^2+1)$$
It is clear that $a=b=1$ is a solution. It is also true that
$$(a,b)=(n^2,n^3),n\in N^{+}$$
is a solution. What other solutions exist?
| Let $b^2=ca$. By the given data $ca\mid a^4$ and $(a+1)\mid (ca+1)$ which is equivalent to $c\mid a^3$, $(a+1)\mid (c-1)$. Let $c=d(a+1)+1$, $d\in \mathbb{N}_0$. Now since $a^3\equiv-1\pmod{a+1}$, we have that $\frac{a^3}{c}\equiv-1\pmod{a+1}$ or $\frac{a^3}{c}=e(a+1)-1$ for $e\in\mathbb{N}$.
It follows that $a^3=(d(a+1)+1)(e(a+1)-1)$, which after multiplying and dividing by $a+1$ than we get $a^2-a+1=de(a+1)+(e-d)$ than it's $e-d\equiv a^2-a+1\equiv 3\pmod{a+1}$, so we have $e-d=k(a+1)+3$, $$de=a-2-k\tag{*}$$ for $k\in\mathbb{Z}$
Now we have 3 cases.
Case 1. $k\not\in\{-1,0\}$. From $(*)$, we have $de<\lvert e-d \rvert -1$, which is possible for $d=0$,so $c=1$ and $b^2=a$. Hence $(a,b)=(t^2,t)$
Case 2. $k=-1$. From $(*)$, we have that $a=d+1$. Now $c=a^2$ and $b^2=a^3$, and hence $(a,b)=(t^2,t^3)$.
Case 3. $k=0$. From $(*)$, we get $a=d^2+3d+2$. Now we have $c=d(a+1)+1=(d+1)^3$ and $b^2=ca=(d+1)^4(d+2)$. It follows that $d+2=t^2$ for $t\in\mathbb{N}$, which gives $(a,b)=(t^2(t^2-1),t(t^2-1)^2),t\geq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/920491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $3(\sin x-\cos x)^4 + 6(\sin x+ \cos x)^2 + 4(\sin^6 x + \cos^6 x) -13 = 0$
Q) Prove that $3(\sin \theta-\cos \theta)^4 + 6(\sin \theta+ \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) -13 = 0$
Source: Trigonometric Functions, Page 5.9, Mathematics XI - R.D. Sharma
My Attempt::
For writing convenience, let $\color{red}{s} = \sin \theta$ and $\color{blue}{c} = \cos \theta$
$
\begin{align}
\text{To Prove }&:\space 3(\color{red}{s}-\color{blue}{c})^4 + 6(\color{red}{s}+\color{blue}{c})^2 + 4(\color{red}{s^6} + \color{blue}{c^6}) - 13 = 0\\
\equiv \text{TP }&:\space 3(\color{red}{s}-\color{blue}{c})^4 + 6(\color{red}{s}+\color{blue}{c})^2 + 4(\color{red}{s^6} + \color{blue}{c^6}) = 13
\end{align}
$
$$
\require{cancel}
\begin{align}
\text{LHS }
&= 3\left[(\color{red}{s}-\color{blue}{c})^2\right]^2
+ 6(\color{red}{s} +\color{blue}{c})^2
+ 4(\color{red}{s^6} + \color{blue}{c^6})
\\
&= 3(\color{limegreen}{s^2 + c^2} - 2\color{red}{s}\color{blue}{c})^2
+ 6(\color{limegreen}{s^2 + c^2} + 2\color{red}{s}\color{blue}{c})
+ 4\left[(\color{red}{s^2})^3 + (\color{blue}{c^2})^3\right]
\\
&= 3(1 - 2\color{red}{s}\color{blue}{c})^2 + 6(1+2\color{red}{s}\color{blue}{c})
+ 4(\color{limegreen}{s^2 + c^2})(\color{red}{s^4}
- \color{red}{s^2}\color{blue}{c^2}
+ \color{blue}{c^4})
\\
&= 3(1 - 4\color{red}{s}\color{blue}{c} + 4\color{red}{s^2}\color{blue}{c^2}) + 6 (1+2\color{red}{s}\color{blue}{c}) + 4(\color{red}{s^4} - \color{red}{s^2} \color{blue}{c^2}+\color{blue}{c^4})
\\
&= 3
\cancel{- 12\color{red}{s}\color{blue}{c}}
+12\color{red}{s^2}\color{blue}{c^2}
+ 6
\cancel{+ 12\color{red}{s}\color{blue}{c}} + 4\color{red}{s^4}
-4\color{red}{s^2}\color{blue}{c^2} + 4\color{blue}{c^4}\\
&= 4\color{red}{s^4} + 8\color{red}{s^2}\color{blue}{c^2} + 4\color{blue}{c^4} + 9\\
&= 4(\color{red}{s^4} + 2\color{red}{s^2}\color{blue}{c^2} + \color{blue}{c^4}) + 9\\
&= 4(\color{limegreen}{s^2 + c^2})^2 + 9\\
&= 4 + 9 = 13 = \text{RHS} \tag{Q.E.D.}
\end{align}
$$
Thanks to @mathlove, I found and corrected the mistake in my attempt.
$\Huge\color{lightgrey}{☺}$ Although, a quicker alternate way will always be nice.
| Note that
*
*$(\sin x - \cos x)^4= (1-\sin 2x)^2 = 1 -2\sin 2x + \sin^2 2x;$
*$(\sin x + \cos x)^2= 1 + \sin 2x;$
*$(\sin x + \cos x)^6 = (\sin^2 x + \cos^2 x)^ 3 -3\sin x \cdot \cos x (\sin^2 x + \cos^2 x) = 1-\dfrac{3}{4}\sin^2 2x.$
There fore
$$ LHS = 3(1 -2\sin 2x + \sin^2 2x) + 6(1+\sin 2x) +4\left( 1-\dfrac{3}{4}\sin^2 2x\right)-13=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
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The supremum of a sequence of definite integrals I am interested to find the supremum of the following sequence of definite integrals:
$$
I_n=\int_0^\pi\sqrt{4\cos^2((2n+1)x)+4\cos((2n+1)x)\cos(x)+1}\,\textrm{d}x,\ n\ge 1.
$$
One of my ideas was to use the inequality $\cos^2(x)\le 1$ in order to obtain the following majorization:
$$
I_n\ge\int_0^\pi|2\cos((2n+1)x)\cos(x)+1|\,\textrm{d}x.
$$
Unfortunately, this majorization is not as strong as necessary.
| I managed to show that $I_n$ indeed converges.
\begin{align}
J_n &= \int_{0}^{\pi}\sqrt{4\cos^2 nx+4\cos nx\cos x+1}\,\textrm{d}x \\
&=\frac{1}{n}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\sqrt{4\cos^2 x+4\cos x\cos\frac{x}{n}+1}\,\textrm{d}x \\
&=\frac{1}{n}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\sqrt{4\cos^2 x+4\cos x\cos\frac{k\pi}{n}+1}\,\textrm{d}x+O\left(\frac{1}{n}\right) \\
&=\int_{0}^{\pi}\frac{1}{n}\sum_{k=0}^{n-1}\sqrt{4\cos^2 x+4\cos x\cos\frac{k\pi}{n}+1}\,\textrm{d}x+O\left(\frac{1}{n}\right) \\
&=\int_{0}^{\pi}\int_0^1\sqrt{4\cos^2 x+4\cos x\cos\pi y+1}\,\textrm{d}y\,\textrm{d}x+O\left(\frac{1}{n}\right)
\end{align}
So as $n\rightarrow \infty$, $$I_n\rightarrow \frac{1}{\pi}\int_{0}^{\pi}\int_0^\pi\sqrt{4\cos^2 x+4\cos x\cos y+1}\,\textrm{d}y\,\textrm{d}x=4.946...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Limit with trigonometric function $\lim_{x \to \frac{\pi}{4}} \frac{\tan^2(x)-1}{\cos(x)-\sin(x)}$ I have this limit, I have resolved it until a part but I'm stucked now.
$$\lim_{x \to \frac{\pi}{4}} \frac{\tan^2(x)-1}{\cos(x)-\sin(x)}$$
$$ \lim_{x \to \frac{\pi}{4}} \frac{\frac{\sin^2(x)}{\cos^2(x)}-\frac{\cos^2(x)}{\cos^2(x)}}{\cos(x)-\sin(x)}$$
$$ \lim_{x \to \frac{\pi}{4}} \frac{\frac{\sin^2(x)-\cos^2(x)}{\cos^2(x)}}{\frac{\cos(x)-\sin(x)}{1}}$$
$$\lim_{x \to \frac{\pi}{4}} \frac{ \sin^2(x)-\cos^2(x)}{\cos^2(x)(\cos(x)-\sin(x)}$$
$$\lim_{x \to \frac{\pi}{4}} \frac{\sin(x)-\cos(x)}{\cos^2(x)}$$
$$\lim_{x \to \frac{\pi}{4}} \frac{\sin(x)}{\cos^2(x)}-\frac{\cos(x)}{\cos^2(x)}$$
$$\lim_{x \to \frac{\pi}{4}} \frac{\sin(x)}{\cos^2(x)}-\frac{1}{\cos(x)}$$
And now I replace by $\frac{\pi}{4}$ and what I get is $0$.
What am I doing wrong?
| $$\frac{ \sin^2 x-\cos^2 x}{\cos x -\sin x} \neq \sin x-\cos x .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve the following inequality I have the following inequality $\frac {2x^2}{x+2} < x-2$. I tried to solve it the with the following steps.
step 1 $\frac {2x^2}{x+2} < x-2$
step 2 $\frac {2x^2}{x+2} - (x-2) < 0$
step 3 $\frac {2x^2}{x+2} - \frac {(x-2)(x+2)}{1(x+2)} < 0$
step 4 $\frac {2x^2}{x+2} - \frac {x^2-2^2}{x+2} < 0$
step 5 $\frac {2x^2 - x^2 + 4}{x+2} < 0$
step 5 $\frac {x^2 + 4}{x+2} < 0$
step 6 I used character study to get the result x > 2. But this is incorrect.
Where did I go wrong with this?. I feel that I made a mistake somewhere in step 2 but not sure what I did wrong.
Thanks!
| When you arrive at $$\frac{f(x)}{g(x)}<0$$ multiply $f(x)$ and $g(x)$ and create a table of values. Make sure $g(x) \neq 0$ though.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
How to solve this limits question I have a problem with this limit question.
$$\lim_{x \to \infty} \frac{x^3-4x}{7-2x^3}$$
How can the answer become $-\frac12$?
| Here are the steps
$$ \lim_{x\to\infty} \frac{x^3-4x}{7-2x^3}= \lim_{x\to\infty} \frac{\frac{x^3}{x^3}-\frac{4x}{x^3}}{\frac{7}{x^3}-\frac{2x^3}{x^3}}= \lim_{x\to\infty} \frac{1-\frac{4}{x^2}}{\frac{7}{x^3}-2} =\frac{1-0}{0-2}=-\frac{1}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/924973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
$\sin 2a + \sin 2b + \sin 2c - \sin 2(a+b+c) = 4 \sin (a+b) \sin (b+c) \sin (c+a)$ I am clueless about this homework question. Looking at it, I see I could use the compound angle and and sum formulas, and tried using them. Unfortunately, couldn't proceed beyond that. Help?
$$\sin 2a + \sin 2b + \sin 2c - \sin 2(a+b+c) = 4 \sin (a+b) \sin (b+c) \sin (c+a)$$
| Thanks to @labbhattacharjee. Here's a step by step version of their solution:
$$\sin 2a + \sin 2b + \sin 2c - \sin 2(a+b+c) = 4 \sin (a+b) \sin (b+c) \sin (c+a)$$
LHS:
$ = 2 \sin (a+b) \cos (a-b) + \sin 2c - \sin((2a+2b)+2c) $
$ = 2 \sin (a+b) \cos (a-b) + 2 \cos (a+b+2c) \sin (-a-b) $
$ = 2 \sin (a+b) \cos (a-b) + 2 \cos (a+b+2c) \sin (-a-b) $
$ = 2 \sin (a+b) \cos (a-b) - 2 \cos (a+b+2c) \sin (a+b) $
$ = 2 \sin (a+b) (\cos (a-b) - \cos (a+b+2c)) $
$ = 2 \sin (a+b) (-2 \sin (c+a) \sin (-b-c)) $
$ = 2 \sin (a+b) (2 \sin (c+a) \sin (b+c)) $
$ = 4 \sin (a+b) \sin (b+c) \sin (c+a) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/926161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $x+\frac{1}{2x}-\frac{1}{8x^3}<\sqrt{x^2+1}This is a math problem from the German Math Olympiad, but in this case I do not know where to start, probably because I do not have enough intuition regarding inequalities.
However, I tried to apply the standard AM-GM equation which didn't help, and in general I could not find a good up-/ downward estimation for the root term.
I also looked at the function plots and it seems that those three terms are very close to each other which seems to make this problem even more difficult to prove.
Any tips on how to begin? Please do not post a complete solution here.
BTW usually those problems provide an elegant solution based on non-university level knowledge, so I do not like to use an approach like a series expansion of the root term!
EDIT: Alright, the right side should've been obvious:
$$0<\frac{1}{4x^2}\Leftrightarrow \sqrt{x^2+1}<\sqrt{x^2+1+\frac{1}{4x^2}}=\sqrt{x^2+2x\frac{1}{2x}+\frac{1}{4x^2}}=x+\frac{1}{2x}$$
EDIT 2: So, I noted that squaring the left inequality (considering $x>0$ as well as $LHS>0$) gives me an inequality equivalent to $x\gt\sqrt{\frac{1}{8}}$.
Therefore, I only have to show that the Assumption $LHS\gt 0$ is implying exactly this.
I thus noted that $0=x+\frac{1}{2x}-\frac{1}{8x^3}$ has only one positive solution $\frac{\sqrt{\sqrt{3}-1}}{2}\gt\sqrt{\frac{1}{8}}$ -- but how do I show this as an inequality? I always get very confused about inequations with quadratic polynomials, especially regarding the direction of the inequality sign...
EDIT 3: Got it! So, let me state the whole proof ;)
To prove: $L:=x+\frac{1}{2x}-\frac{1}{8x^3}<\sqrt{x^2+1}=:R,\quad x>0$
a) Consider $L\leq0$. $x>0$ implies $R>0 \Rightarrow L<R$
b) Consider $L\gt0$. We get
$$
0<x+\frac{1}{2x}-\frac{1}{8x^3} \Rightarrow 0\lt x^4+\frac{1}{2}x^2-\frac{1}{8} = (x^2)^2 + 2\frac{1}{4}x^2+\frac{1}{4^2}-\frac{1}{4^2}-\frac{1}{8} = (x^2+\frac{1}{4})^2-\frac{3}{16}\\
\Leftrightarrow \frac{\sqrt{3}}{4}\lt x^2+\frac{1}{4} \Leftrightarrow x^2 \gt \frac{\sqrt{3}-1}{4}
\Leftrightarrow x \gt \sqrt{\frac{\sqrt{3}-1}{4}} \Rightarrow x\gt \sqrt{\frac{1}{8}}
$$
From that, we get
$$
\frac{1}{8x^2}\lt 1 \Leftrightarrow \frac{1}{8x^2}\frac{1}{8x^4}\lt \frac{1}{8x^4}
\Leftrightarrow (\frac{1}{8x^3})^2-\frac{1}{8x^4}\lt0 \\
\Leftrightarrow x^2+1+(\frac{1}{8x^3})^2-\frac{1}{8x^4}\lt x^2+1
$$
Noticing that $L^2=x^2+1+(\frac{1}{8x^3})^2-\frac{1}{8x^4}$ and $R^2=x^2+1$ gives you
$$
\Leftrightarrow L^2 \lt R^2 \Leftrightarrow L\lt R
$$
...quod erat demonstrandum.
Sorry it took so long, I am not used to write that much TeX ;)
BTW, can anyone leave a comment regarding the "beauty" of my proof? I tried to compose this final version as clean as possible!
| Just squaring everything seems the best way:
$$
\left(x+\frac1{2x}\right)^2
=x^2+1+\frac1{4x^2}
$$
and
$$
\begin{align}
\left(\color{#C00000}{x+\frac1{2x}}-\frac1{8x^3}\right)^2
&=\color{#C00000}{x^2+1+\frac1{4x^2}}-\frac2{8x^3}\left(\color{#C00000}{x+\frac1{2x}}\right)+\frac1{64x^6}\\
&=x^2+1-\frac1{8x^4}+\frac1{64x^6}
\end{align}
$$
Now you need to consider what happens to $x+\dfrac1{2x}-\dfrac1{8x^3}$ when $\dfrac1{64x^6}\gt\dfrac1{8x^4}$.
Now that the question seems answered, I might as well add the last little bit.
If $8x^2\lt1$, then
$$
\begin{align}
x+\frac1{2x}-\frac1{8x^3}
&=\frac{8x^4+4x^2-1}{8x^3}\\
&\le\frac{x^2+4x^2-1}{8x^3}\\
&\le\frac{\frac58-1}{8x^3}\\
&=-\frac3{64x^3}\\[4pt]
&\lt-\frac3{\sqrt8}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/926386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Can we determine the determinant? Could someone prove that this determinant is not zero?
$$\left|
\begin{array}{cccc}
1^n & 2^n & \cdots & (n+1)^n \\
2^n & 3^n & \cdots & (n+2)^n \\
\vdots & \vdots & & \vdots \\
(n+1)^n & (n+2)^n & \cdots & (2n+1)^n \\
\end{array}
\right|\neq0$$
How can we compute it?
| First we apply binomial theorem, e.g. write $2^n$ as $(1+1)^n$, then $2^n=1^n+C_n ^1 1^{n-1}1^1+ \cdots+C_n ^n 1^0 1^n$. And note that $\det(AB)=\det(A)\det(B)$ for square matrix $A,B$ with same order.
$$\left|
\begin{array}{cccc}
1^n & 2^n & \cdots & (n+1)^n \\
2^n & 3^n & \cdots & (n+2)^n \\
\vdots & \vdots & & \vdots \\
(n+1)^n & (n+2)^n & \cdots & (2n+1)^n \\
\end{array}
\right|
=
\left|
\begin{array}{cccc}
(0+1)^n & (0+2)^n & \cdots & (0+(n+1))^n \\
(1+1)^n & (1+2)^n & \cdots & (1+(n+1))^n \\
\vdots & \vdots & & \vdots \\
(n+1)^n & (n+2)^n & \cdots & (n+(n+1))^n \\
\end{array}
\right|
=
\left|
\begin{array}{cccc}
0 & 0 & \cdots & 1 \\
1^n & 1^{n-1} & \cdots & 1^0 \\
\vdots & \vdots & & \vdots \\
n^n & n^{n-1} & \cdots & n^0 \\
\end{array}
\right| \cdot
\left|
\begin{array}{cccc}
1 & 1 & \cdots & 1 \\
C_n^1 1^1 & C_n^1 2^{1} & \cdots & C_n ^1 (n+1)^1 \\
\vdots & \vdots & & \vdots \\
C_n^n 1^n & C_n^n 2^{n} & \cdots & C_n^n (n+1)^n \\
\end{array}
\right|
$$
Notice that this 2 matrix are nearly of the Vandermonde form. See here. Except that the first one should have all its columns reversed; the second one should extract the factors $C_n ^1, \dots,C_n ^n$. Hence,
$$
\left|
\begin{array}{cccc}
0 & 0 & \cdots & 1 \\
1^n & 1^{n-1} & \cdots & 1^0 \\
\vdots & \vdots & & \vdots \\
n^n & n^{n-1} & \cdots & n^0 \\
\end{array}
\right| \cdot
\left|
\begin{array}{cccc}
1 & 1 & \cdots & 1 \\
C_n^1 1^1 & C_n^1 2^{1} & \cdots & C_n ^1 (n+1)^1 \\
\vdots & \vdots & & \vdots \\
C_n^n 1^n & C_n^n 2^{n} & \cdots & C_n^n (n+1)^n \\
\end{array}
\right|= \\
(-1)^{n(n+1)/2}C_n ^1 C_n ^2 \cdots C_n ^n
\left|
\begin{array}{cccc}
1 & 0 & \cdots & 0 \\
1 & 1^{1} & \cdots & 1^n \\
\vdots & \vdots & & \vdots \\
1 & n^{1} & \cdots & n^n \\
\end{array}
\right| \cdot
\left|
\begin{array}{cccc}
1 & 1 & \cdots & 1 \\
1^1 & 2^{1} & \cdots & (n+1)^1 \\
\vdots & \vdots & & \vdots \\
1^n & 2^{n} & \cdots & (n+1)^n \\
\end{array}
\right|
\\ =(-1)^{n(n+1)/2}C_n ^1 C_n ^2 \cdots C_n ^n \prod_{0 \leq i <j \leq n}(j-i) \prod_{1 \leq i <j \leq n+1}(j-i)
\neq 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/927834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Does $\sum_1^\infty \frac{n}{n^2 + 4}$ converge or diverge? Is my solution correct? Does $\sum_1^\infty \frac{n}{n^2 + 4}$ converge or diverge?
I am confused because my friend insists the series converges conditionally. I think the series diverges. Here is my process and solution:
$\sum_1^\infty \frac{n}{n^2 + 4}$
Limit Comparison Test:
Let $\sum_1^\infty \frac{n}{n^2 + 4} = \sum_1^\infty a_n$
Let $\sum_1^\infty b_n = \sum_1^\infty \frac{1}{n}$
Since $$\lim \limits_{n \to \infty} | \frac{a_n}{b_n}| = \lim \limits_{n \to \infty} | \frac{\frac{n}{n^2 + 4}}{\frac{1}{n}}| = \lim \limits_{n \to \infty} \frac{n^2}{n^2 + 4} = \lim \limits_{n \to \infty} \frac{n^2}{n^2(1 + \frac{4}{n^2})} = \lim \limits_{n \to \infty} \frac{1}{1 + \frac{4}{n^2}} = 1$$
and $1 > 0$ and $1$ is a finite number,
we can say that the behavior of $\sum_1^\infty a_n$ is the same as the behavior of $\sum_1^\infty b_n$.
Since $\sum_1^\infty b_n$ diverges ( $\sum_1^\infty \frac{1}{n}$ diverges by p-series, $p=1$), $\sum_1^\infty a_n$ diverges.
So by the Limit Comparison Test, $\sum_1^\infty \frac{n}{n^2 + 4}$ diverges.
Right?
Also, this problem was an exercise in the Ratio/Root Test section. Both tests, however seem to fail. Can the convergence be solved with the Root Test or the Ratio Test?
| Write $\frac{n}{n^2 + 4} = \frac1{n + \frac{4}{n}}$. Note that $ n + \frac{4}{n} \le n + 4 $. Use comparison test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $p > 5$ is a prime number, then the last digit of $p^4-1$ is $0$. If $p > 5$ is a prime number, then the last digit of $p^4-1$ is $0$ (ex.: $7^4-1=2400$).
How do I prove this?
| By noting that the last digit of the fourth powers of $1$, $3$, $7$ and $9$ is $1$ and that the last digit of every prime number $p> 5$ is $1$, $3$, $7$ or $9$.
More formally, $p=i\pmod{10}$ with $i$ in $\{1,3,7,9\}$ hence $p^4=i^4\pmod{10}$. Since every $i$ in $\{1,3,7,9\}$ is such that $i^4=1\pmod{10}$, this yields $p^4=1\pmod{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/935417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Calculate $\pi$ By Hand? All over the internet the only hand equation i found was
$$\frac\pi4 = 1 - \frac13 + \frac15 - \frac17+\cdots.$$
But this takes something like a thousand iterations to get to four digits, is there a better way to calculate pi by hand?
| Jean-Claude Arbaut has reminded us of the identity
$$
\frac\pi4=4\arctan\frac15-\arctan\frac1{239}.
$$
Let us examine that. You learned in high school that $\tan\dfrac\pi4=1$, and that
\begin{align}
\tan(\alpha+\beta) & = \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \tag 1 \\[10pt]
& =\frac{c+d}{1-cd}
\end{align}
Thus
$$
\arctan c+\arctan d=\alpha+\beta=\arctan\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\arctan\frac{c+d}{1-cd}
$$
From $(1)$ we get
$$
\tan(\alpha+\beta+\gamma+\delta)=\frac{c+d+e+f-cde-cdf-cef-def}{1-cd-ce-cf-de-df-ef+cdef}
$$
where $c,d,e,f$ are the respective tangents of $\alpha,\beta,\gamma,\delta$, and hence
$$
\tan(4\alpha) = \frac{4\tan\alpha-4\tan^3\alpha}{1 - 6\tan^2\alpha+\tan^4\alpha}.
$$
Hence
$$
4\arctan c = \arctan\frac{4c-4c^3}{1-6c^2+c^4}.
$$
So
$$
4\arctan\frac15 = \arctan\frac{(4/5)-(4/5^3)}{1-(6/5^2)+ (1/5^4)} = \arctan\frac{480}{476} = \arctan\frac{120}{119}.
$$
Next we look at
\begin{align}
& 4\arctan\frac15 - \arctan\frac{1}{239} = \arctan\frac{120}{119} - \arctan\frac{1}{239} \\[15pt]
= {} & \arctan\frac{(120/119)-(1/239)}{1+(120/119)(1/239)} \\[15pt]
= {} & \arctan\frac{28561}{28561} = \arctan 1 = \frac\pi4.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/938414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 1
} |
Under what conditions will one solution of $ax^2+bx+c = 0$ be the reciprocal of the other? Under what conditions will one solution of
$ax^2+bx+c = 0$ be the reciprocal of the other?
| By the Quadratic Formula, given $ax^2+bx+c=0$ with $a\ne0$, we have $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ Let us assume that $$\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{2a}{-b-\sqrt{b^2-4ac}}.$$ Cross-multiplying, we have $$(2a)^2=\left(-b+\sqrt{b^2-4ac}\right)\left(-b-\sqrt{b^2-4ac}\right)$$ $$\implies 4a^2=\left(-b\right)^2-\left(\sqrt{b^2-4ac}\right)^2\implies 4a^2=b^2-\left(b^2-4ac\right)\implies 4a^2=4ac.$$ Since $a\ne 0$, we can divide both sides of this equation by $4a$ to obtain $a=c$. Therefore, the two roots of a quadratic will be reciprocals of one another precisely when $a=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do you show $(ac)^2+(bd)^2+(bc)^2+(ad)^2=1$ Show $(ac)^2+(bd)^2+(bc)^2+(ad)^2=1$ if $a^2+b^2=1$ and $c^2+d^2=1$.
I was thinking about solving for variables and plugging in, but that seems like too much work. Is there a simple trick or realization that will help me prove this more quickly?
| \begin{align}
& (a^2×c^2)+(b^2×d^2)+(b^2×c^2)+(a^2×d^2) \\
= {} & (a^2×c^2)+(b^2×c^2)+(b^2×d^2)+(a^2×d^2) \\
= {} & c^2(a^2+b^2) + d^2(a^2+b^2) \\
= {} & (a^2+b^2) (c^2+d^2) \\
= {} & 1\times1 \\
= {} & 1
\end{align}
Hope that makes more sense!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/943755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding minimum value of a trigonometric expression If $A+B+C=\pi$ then what is the minimal value of
$$\tan^{2}A/2+\tan^{2}B/2+\tan^{2}C/2$$
| Here $A,B,C$ are the angles of a $\triangle.$ So Here $$\displaystyle 0<\frac{A}{2},\frac{B}{2},\frac{C}{2}<\frac{\pi}{2}$$.
Now Using $\bf{A.M\geq G.M}\;,$ We Get
So we get $$\displaystyle \frac{\tan^2 \left(\frac{A}{2}\right)+\tan^2 \left(\frac{B}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{A}{2}\right)\cdot \tan^2 \left(\frac{B}{2}\right)} = \tan\left(\frac{A}{2}\right)\cdot \tan \left(\frac{B}{2}\right)\color{red}\checkmark$$
Similarly $$\displaystyle \frac{\tan^2 \left(\frac{B}{2}\right)+\tan^2 \left(\frac{C}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{B}{2}\right)\cdot \tan^2 \left(\frac{C}{2}\right)} = \tan \left(\frac{B}{2}\right)\cdot \tan \left(\frac{C}{2}\right)\color{red}\checkmark$$
Similarly $$\displaystyle \frac{\tan^2 \left(\frac{C}{2}\right)+\tan^2 \left(\frac{A}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{C}{2}\right)\cdot \tan^2 \left(\frac{A}{2}\right)} = \tan \left(\frac{C}{2}\right)\cdot \tan \left(\frac{A}{2}\right)\color{red}\checkmark$$
Now Add all Three, We Get
$$\displaystyle \tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{B}{2}\right)+\tan^2\left(\frac{C}{2}\right)\geq \tan\left(\frac{A}{2}\right)\cdot \tan\left(\frac{B}{2}\right)+\tan\left(\frac{B}{2}\right)\cdot \tan\left(\frac{C}{2}\right)+\tan\left(\frac{C}{2}\right)\cdot \tan\left(\frac{A}{2}\right)\color{blue}\checkmark\color{blue}\checkmark$$
Now Here $$\displaystyle \frac{A}{2}+\frac{B}{2}=\frac{C}{2}\Rightarrow \tan\left(\frac{A}{2}+\frac{B}{2}\right)=\tan \frac{C}{2}$$
So We Get $$\displaystyle \tan\left(\frac{A}{2}\right)\cdot \tan\left(\frac{B}{2}\right)+\tan\left(\frac{B}{2}\right)\cdot \tan\left(\frac{C}{2}\right)+\tan\left(\frac{C}{2}\right)\cdot \tan\left(\frac{A}{2}\right)=1$$
Put into $\color{blue}\checkmark\color{blue}\checkmark\;,$ We Get
$$\displaystyle \tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{B}{2}\right)+\tan^2\left(\frac{C}{2}\right)\geq 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/945353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to calculate indefinite integral involving infinite sums? I want to calculate the following integral:
$$ \int_{0}^{\infty}\left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot 4}-\frac{x^7}{2\cdot 4\cdot 6}+\cdots\right)\;\left(1+\frac{x^2}{2^2}+\frac{x^4}{2^2\cdot 4^2}+\frac{x^6}{2^2\cdot 4^2\cdot 6^2}+\cdots\right)\,\mathrm{d}x
$$
I have no idea how to start; any help is highly appreciated.
| Thank you, it was quite helpful; I know found an easier way to prove it:
Starting from $$\sum_{k=0}^{\infty} \frac{x^{2k+1}(-1)^k}{2^kk!}=xe^{-\frac{x^2}{2}}$$ we can rewrite the integral as follows:
$$ I=\int_{0}^{\infty}\left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot 4}-\frac{x^7}{2\cdot 4\cdot 6}+\cdots\right)\;\left(1+\frac{x^2}{2^2}+\frac{x^4}{2^2\cdot 4^2}+\frac{x^6}{2^2\cdot 4^2\cdot 6^2}+\cdots\right)\,\mathrm{d}x = \int_{0}^{\infty}xe^{-\frac{x^2}{2}}\cdot\ \sum_{k=0}^{\infty} \frac{x^{2k}}{2^{2k}{k!}^2}\mathrm{d}x = \sum_{k=0}^{\infty} \left( \frac{1}{2^{2k}{k!}^2}\int_{0}^{\infty}e^{-\frac{x^2}{2}}\cdot\ x^{2k+1}\mathrm{d}x \right)
$$
Know with the substitution $x\to \sqrt{2x}$ we obtain:
$$I=\sum_{k=0}^{\infty} \left( \frac{1}{2^{2k}{k!}^2}\int_{0}^{\infty}\frac{\sqrt2}{2\sqrt{x}}\cdot e^{-x}\cdot\ 2^{k+\frac{1}{2}}\cdot x^{k+\frac{1}{2}}\mathrm{d}x \right)=\sum_{k=0}^{\infty} \left( \frac{1}{2^{k}{k!}^2}\int_{0}^{\infty} e^{-x}\cdot\ x^{k}\mathrm{d}x \right)=\sum_{k=0}^{\infty} \frac{1}{2^{k}{k!}}=\sqrt e$$
Where I used that by the integral definition of the gamma function $\int_{0}^{\infty} e^{-x}\cdot\ x^{k}\mathrm{d}x=\Gamma (k+1)=k!$ and the well known series expansion $e^x=\sum_{k=0}^{\infty} \frac{x^k}{{k!}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/950818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
difference between runge kutta methods of same order I recently read about Runge-Kutta methods for solving differential equations. So far I understood the idea but up to know nobody could answer me following question:
If we consider the explicit RK methods of order 4, we get for example the classical RK method, the Gills-formula and the 3/8 rule. They all do the 'same' but what is the difference between them? Do they have different properties or do we just have those different ones for historical reasons?
Classical:
$
\begin{array}{c|cccc}
0 & 0 & 0 & 0 & 0\\
1/2 & 1/2 & 0 & 0 & 0\\
1/2 & 0 & 1/2 & 0 & 0\\
1 & 0 & 0 & 1 & 0\\
\hline
& 1/6 & 1/3 & 1/3 & 1/6\\
\end{array}$
3/8 Rule:
$\begin{array}{c|cccc}
0 & 0 & 0 & 0 & 0\\
1/3 & 1/3 & 0 & 0 & 0\\
2/3 & -1/3 & 1 & 0 & 0\\
1 & 1 & -1 & 1 & 0\\
\hline
& 1/8 & 3/8 & 3/8 & 1/8\\
\end{array}$
Gills-formula:
$\begin{array}{c|cccc}
0 & 0 & 0 & 0 & 0\\
1/2 & 1/2 & 0 & 0 & 0\\
1/2 & \frac{\sqrt{2}-1}{2} & \frac{2-\sqrt{2}}{2} & 0 & 0\\
1 & 0 & -\frac{\sqrt{2}}{2} & \frac{2+\sqrt{2}}{2} & 0\\
\hline
& 1/6 & \frac{2-\sqrt{2}}{2} & \frac{2+\sqrt{2}}{2} & 1/6\\
\end{array}$
| The "classical" or Heun-Kutta method is very simple in its equations/coefficients, easy to remember and easy to casually implement.
The 3/8 method (which was introduced first in the same paper of Kutta as the "classical" method) has slightly smaller coefficients in the leading error terms of the local error.
The Gill's method, according to an ancient tome on my desk, requires a minimum of storage spaces, has the highest accuracy among the 4th order methods and has a short code sequence in its implementation. (T.E. Shoup: Applied numerical methods for the micro-computer, Prentice Hall 1984, p. 116ff)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this differential equation: $x^2dy-y^2dx+xy^2(x-y)dy=0$
$$x^2dy-y^2dx+xy^2(x-y)dy=0$$
What I tried:
$$\frac{x^2}{y^2} \frac{dy}{dx}+x(x-y)\frac{dy}{dx}=1\\$$
Let $h=-1/x, \; k=-1/y,\; dh=1/x^2 \, dx, \; dk=1/y^2 \,dy$
$$\frac{dk}{dh}+\frac{(k-h)}{k^2} \frac{dk}{dh}=1\\
\frac{dk}{dh}+\frac hk=1+\frac1k\\
he^{\int-1/k^2\; dk}=\int\left(1+\frac1k\right)e^{\int-1/k^2\; dk}dk\\
he^{-y}=\int\frac{1-y}{y^2}e^{-y}dy=\int\left(\frac 1{y^2}-\frac1y\right)e^{-y}dy$$
Which probably is unsolvable?I tried using IBP on RHS.Dont use Ricatti Eqn(Not in my course)
Answer is:
$$\large\ln\left|\frac{x-y}{xy}\right|=\frac{y^2}2+\mathcal C$$
| It seems that the solution $y(x)$ cannot be expressed on a closed form. But a closed form can be found for the inverse function $x(y)$ :
$$x^2 - y^2\frac{\mathrm dx}{\mathrm dy} + xy^2(x-y) = 0 \implies \frac{dx}{dy} = \frac{y^2 + 1}{y^2}x^2 - yx$$
Considering the function inverse function $x(y)$, this is a Riccati ODE which can be solved thanks to the classical method to solve this kind of ODEs.
But, in this case, it is easier to proceed with a convenient change of function:
$$\text{Let: } x(y) = e^{-y^2/2}F(y) \implies \frac{dx}{dy} = e^{-y^2/2}F' - ye^{-y^2/2}F = \frac{y^2 + 1}{y^2}e^{-y^2}F^2 - ye^{-y^2/2}F\\
\,\\
\text{Now, } \int{\frac{F'}{F^2}}\mathrm dy = \int{\frac{y^2 + 1}{y^2}e^{-y^2/2}}\mathrm dy \implies -\frac{1}{F} = -\frac{1}{y}e^{-y^2/2} + C\\
\implies F = \frac{y}{e^{-y^2/2} + Cy}\\
\implies x(y) = \frac{ye^{-y^2/2}}{e^{-y^2/2} + Cy}= \frac{y}{1 + Cye^{y^2/2}}
$$
:: Source ::
| {
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"timestamp": "2023-03-29T00:00:00",
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How would you prove inequality $2^n \gt n^{10}$ using induction For the base case I can put a number such as $100$ for $n$ so $2^{100}\gt 100^{10}$.
Ok so now the induction hyp:
$2^{n+1} > (n+1)^{10}$ for $n \gt 101.$
where do I go from here? Also do I have to prove $2^{100}\gt 100^{10}$ ?
| We have $2^{n+1} = 2^n \cdot 2 \geq n^{10} \cdot 2$. From that we get
$$(1+n)^{10} =\sum_{k=0}^{10}\binom{10}{k}n^k =n^{10} + \sum_{k=0}^{9}\binom{10}{k}n^k $$ and substracting $n^{10}$ from both sides left is to show that
$$\sum_{k=0}^{9}\binom{10}{k}n^k \leq n^{10},$$ dividing by $n^{10}$ obtains
$$\frac{10n^9}{n^{10}} + \frac{45n^8}{n^{10}} +\frac{120n^7}{n^{10}} +\frac{210n^6}{n^{10}} +\frac{252n^5}{n^{10}} +\frac{210n^4}{n^{10}} +\frac{120n^3}{n^{10}} +\frac{45n^2}{n^{10}} +\frac{10n}{n^{10}} +\frac{1}{n^{10}} \leq 1 $$
which is
$$(*) =\frac{10}{n}+\frac{45}{n^2}+\frac{120}{n^3}+\frac{210}{n^4}+\frac{252}{n^5}+\frac{210}{n^6}+\frac{120}{n^7}+\frac{45}{n^8}+\frac{10}{n^9}+\frac{1}{n^{10}} \leq 1 $$ and since $n \geq 100$ we get
$$(*) \leq \frac{10}{100}+\frac{45}{100^2}+\frac{120}{100^3}+\frac{210}{100^4}+\frac{252}{100^5}+\frac{210}{100^6}+\frac{120}{100^7}+\frac{45}{100^8}+\frac{10}{100^9}+\frac{1}{100^{10}} \leq 1$$
which can be verfied by calculating it. This completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}$ May I know how can I calculate the following expression?
$$
\lim\limits_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}
$$
From calculator, the answer is $-\sqrt{6}$ , my approach is by using dividing numerator and denominator by using $x$. Which is,
$$
\lim\limits_{x\to-\infty}\frac{\frac{\sqrt{6x^2 - 2}}{x}}{\frac{x+2}{x}}
=
\lim\limits_{x\to-\infty}\frac{{\sqrt{\frac{6x^2 - 2}{x^2}}}}{\frac{x+2}{x}}=\sqrt{6}
$$
My answer is $\sqrt{6}$, is my working wrong or there are actually another approach? Thank you.
| Try looking at it this way: $$\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2}\cdot \sqrt{6-\frac{2}{x^2}}}{x+2}$$ Then since $x \rightarrow -\infty$ we have $x<0$ thus $\sqrt{x^2}=|x| = -x$ and we have
$$\lim_{x\rightarrow -\infty} \frac{-x\cdot \sqrt{6-\frac{2}{x^2}}}{x+2}=\lim_{x\rightarrow -\infty} \frac{-\sqrt{6-\frac{2}{x^2}}}{1+\frac{2}{x}}=-\sqrt6$$
| {
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"url": "https://math.stackexchange.com/questions/956767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Check the invertibility of a matrix given a parameter Find all values of $c$, if any, for which the given matrix is invertible (using row operations -> gotta try simplifying to I).
Here is the matrix :
\begin{pmatrix}c&1&0\\1&c&1\\0&1&c\end{pmatrix}
| $R_1,R_2,R_3$ are the first, second and third row
Let's assume that $c\neq0$
, as the case $c=0$ is very simple.
$\begin{pmatrix}c&1&0\\1&c&1\\0&1&c\end{pmatrix}$ $R_1\leftarrow \frac{R_1}c$
$\begin{pmatrix}1&1/c&0\\1&c&1\\0&1&c\end{pmatrix}$ $R_2\leftarrow R_2-R_1$
$\begin{pmatrix}1&1/c&0\\0&c-1/c&1\\0&1&c\end{pmatrix}$ $R_2\leftarrow R_2\cdot\frac{c}{c^2-1}$
$\begin{pmatrix}1&1/c&0\\0&1&\frac{c}{c^2-1}\\0&1&c\end{pmatrix}$ $R_3\leftarrow R_3-R_2$
$\begin{pmatrix}1&1/c&0\\0&1&\frac{c}{c^2-1}\\0&0&c-\frac{c}{c^2-1}\end{pmatrix}$ $R_1\leftarrow R_1-c\cdot R_2$
$\begin{pmatrix}1&0&\frac{c^2}{c^2-1}\\0&1&\frac{c}{c^2-1}\\0&0&\frac{c^3-2c}{c^2-1}\end{pmatrix}$
Can you do the last two steps by yourself ? :) :) :)
| {
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Prove that $\sum_{k=1}^nk^2{n\choose k}^2=n^2 \binom {2n-2}{n-1}$ Please help me / give a hand with combinational prove for:
$$ 1^2 \binom n 1 ^2 + 2^2 \binom n 2 ^2 + \dots + n^2 \binom n n ^2 = n^2 \binom {2n-2}{n-1}$$
| By way of enrichment here is another solution using basic complex
variables.
Start by restating the problem using the alternate binomial coefficient:
we seek to show that
$$\sum_{k=0}^n {n\choose k} k^2 {n\choose n-k}
= n^2 {2n-2\choose n-1}.$$
Introduce the integral representation
$${n\choose n-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n-k+1}} \; dz.$$
This gives the following integral for the sum:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n+1}}
\sum_{k=0}^n {n\choose k} k^2 z^k \; dz.$$
Now recall that
$$(1+z)^n = \sum_{k=0}^n {n\choose k} z^k$$
and hence
$$nz (1+z)^{n-1} = \sum_{k=0}^n {n\choose k} k z^k$$
and finally
$$z(n (1+z)^{n-1} + nz(n-1)(1+z)^{n-2})
= \sum_{k=0}^n {n\choose k} k^2 z^k.$$
Substituting this into the sum gives
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\left(n \frac{(1+z)^{2n-1}}{z^n}
+ n(n-1) \frac{(1+z)^{2n-2}}{z^{n-1}} \right)\; dz.$$
Extracting coefficients we finally obtain
$$n {2n-1\choose n-1} + n(n-1) {2n-2\choose n-2}
\\= n \frac{2n-1}{n} {2n-2\choose n-1}
+ n(n-1) \frac{n-1}{n} {2n-2\choose n-1}
\\ = (2n-1 + (n-1)^2) {2n-2\choose n-1}
= n^2 {2n-2\choose n-1}.$$
Apparently this method is due to Egorychev.
A trace as to when it appeared on MSE and by whom starts at this
MSE link.
| {
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Proof with combinatorial argument
Show with combinatorial argument that this is equal :
$$\dbinom{n}{k+1} = \dbinom{n-1}{k}+ \dbinom{n-2}{k} +...+ \dbinom{k}{k}$$
I have no idea how to do that so it would be really helpful if anyone knows some way.
| Suppose we want to distribute $n - k - 1$ identical balls into $k + 2$ distinct boxes. With stars and bars, there are $\binom{n - k - 1 + k + 1}{k + 1} = \binom{n}{k + 1}$ ways to do this.
We can also first distribute $0 \le i \le n - k - 1$ balls into the first box and then distribute remaining the balls into the remaining $k + 1$ boxes. The number of ways to do this is, using stars and bars for the remaining balls and boxes, is
$$\binom{(n - k - 1) + k}{k} + \binom{(n - k - 2) + k}{k} + \dots + \binom{(n - k - n) + k}{k}$$
$$= \binom{n - 1}{k} + \binom{n - 2}{k} + \dots + \binom{k}{k}$$
Since both ways of counting are equivalent, we have
$$\binom{n}{k + 1} = \binom{n - 1}{k} + \binom{n - 2}{k} + \dots + \binom{k}{k}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there any polygon exists such that sum of length of square of two adjacent sides is equal to another side/diagonal? In Right angle triangle we have $ a^2 + b^2 = c^2$ where $a^2 = (x_1-x_2)^2 + (y_1-y_2)^2 ,$ $b^2 = (x_3-x_2)^2 + (y_3-y_2)^2 $and $c^2 = (x_1-x_3)^2 + (y_1-y_3)^2$
And in Square we have
$ a^2 + b^2 = c^2$ $ d^2 + e^2 = c^2$ $ a^2 + d^2 = f^2$ $ b^2 + e^2 = f^2$ and $a=b=d=e , c=f$
where $a^2 = (x_1-x_2)^2 + (y_1-y_2)^2 ,$ $b^2 = (x_3-x_2)^2 + (y_3-y_2)^2 $ , $c^2 = (x_1-x_3)^2 + (y_1-y_3)^2$ , $d^2= (x_1-x_4)^2+(y_1-y_4)^2$ ,$e^2= (x_3-x_4)^2+(y_3-y_4)^2$ ,and $f^2= (x_2-x_4)^2+(y_2-y_4)^2$
Is there any other polygon exists which posses this property (sum of length of square of two adjacent sides is equal to another side/diagonal ) ? It's a sufficient condition to say if this property is satisfied by a polygon then it's a right angle triangle or square ?
| Perhaps you're looking for something like the hexagon $(3,4,12,84,3612,3613)$ where $3^2+4^2+12^2+84^2+3612^2=3613^2$. Polygons like this are found by matching side $C$ of one triple with side $A$ of another. There are infinite combinations with the number of sides ranging from $3$-to-$\infty$. I can show you formulas for finding them if you like.
There are also tiles composed of other triples with alternately matching sides B-with-B and C-with-C such as the isosceles triangle below. Little can be said of it except that the sums squares of sides adjacent to right angles add up to the squares of the sides opposite the right angles.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int$ $\frac{x^3dx}{\sqrt{1+x^2}}$ Use $u$ substitution:
$u = 1 + x^2$, $du = 2xdx$ , $dx =\frac {du}{2x}$
Write the given integral in terms of : $u$, $du$ and $dx$
$$\frac {1}{2}\int \frac {x^4}{x\sqrt{u}} du \implies \frac{1}{2} \int \frac {u+x^2 - 1}{x \sqrt{u}}$$
This works right I double checked my work. But it looks awfully complicated for a calculus problem than the other ones I have done.
| Set $\sqrt{1+x^2}=u\implies 1+x^2=u^2, xdx=u du$
$$\int\frac{x^3}{\sqrt{1+x^2}}dx=\int\frac{u^2-1}uudu$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$z^2=x^2+y^2$ Prove that $4\mid xyz$ ($xyz$ is divided by $4$) $z^2=x^2+y^2$ where $x,\ y,\ z$ - integers
Prove that $4\mid xyz$ ($xyz$ is divided by $4$)
All possible rest in divided by $4$ in this case is $1$. That's all I noticed.
| If an integer $n$ is not divisible by $4$, then it is equal to $1,2,3,5,6,$ or $7$ mod $8$. So $n^2$ is equal to $1$ or $4$ mod $8$.
Hence if neither $x$ nor $y$ is divisible by $4$, then $z^2 = x^2+y^2$ is equal to $0, 2,$ or $5$ mod $8$; in particular, $z^2$ is not equal to $1$ or $4$ mod $8$. And so $z$ must be divisible by $4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that $k^a=\sum_{m=1}^b\left ( c_m^a\prod_{n\neq m} \frac{k-c_n}{c_m-c_n} \right ).$ I used the following result in another post without providing proof (because I couldn't prove it):
$$k^a=\sum_{m=1}^b\left ( c_m^a\prod_{n\neq m} \frac{k-c_n}{c_m-c_n} \right ),$$
where $a$ and $b$ are non-negative integers, $b>a$, each $c_j\in\mathbb{C}$, each $|c_j|>0$, and each $c_j$ is unique. The product runs from $n=1$ to $b$ but skips $m$.
How can we prove this?
| Start by encoding the sum call it $S_b$ using residues. We have by
inspection that
$$S_b =
\sum_{m=1}^b
\mathrm{Res}
\left(f(z); z=c_m\right)$$
where
$$f(z) = \frac{z^a}{k-z}
\prod_{n=1}^b \frac{k-c_n}{z-c_n}$$
and $c_m\ne k$ and $b>a.$
We can therefore collect $S_b$ by integrating f(z) around a contour
that encloses the $b+1$ poles. We will then use the residue at infinity
to evaluate the sum of the residues inside the contour.
Now the residue at infinity of $f(z)$ is given by
$$\mathrm{Res}
\left(-\frac{1}{z^2} f\left(\frac{1}{z}\right); z=0\right).$$
The functional term becomes
$$-\frac{1}{z^2}
\frac{1}{z^a}\frac{1}{k-1/z}
\prod_{n=1}^b \frac{k-c_n}{1/z-c_n}
= -\frac{1}{z^2}
\frac{1}{z^a}\frac{z}{zk-1}
\prod_{n=1}^b \frac{z(k-c_n)}{1-zc_n}
\\ = -\frac{1}{z^2}
\frac{z^{b+1}}{z^a}\frac{1}{zk-1}
\prod_{n=1}^b \frac{k-c_n}{1-zc_n}
= \frac{1}{z^{a-b+1}}\frac{1}{1-zk}
\prod_{n=1}^b \frac{k-c_n}{1-zc_n}.$$
But we have $b>a$ and hence $b-a-1 > -1$ or $b-a-1\ge 0$ so the term
is in fact
$$ z^{b-a-1} \frac{1}{1-zk}
\prod_{n=1}^b \frac{k-c_n}{1-zc_n}.$$
and the residue at zero of the substituted function is zero.
This means that
$$S_b = -\mathrm{Res} \left(f(z); z=k\right)$$
which gives
$$k^a \prod_{n=1}^b \frac{k-c_n}{k-c_n} = k^a,$$
done.
Addendum I. As an alternative to using the residue at infinity we
could have used a circular contour enclosing all $b+1$ poles and
observed that the integral along this contour goes to zero since on
this circle we have
$$f(z) \in \Theta\left(R^{a-1} \times R^{-b}\right)
= \Theta\left(R^{a-b-1}\right)$$
and
$$\lim_{R\to\infty} 2\pi R \times R^{a-b-1}
= 2\pi \lim_{R\to\infty} R^{a-b} = 0$$
because $b>a.$
This has the advantage of working for $a$ a positive real parameter as opposed to a positive integer only.
Addendum II. The case $k=c_m$ for some $m$ gives
$$\sum_{m=1}^b\left(c_m^a\prod_{n\neq m} \frac{k-c_n}{c_m-c_n} \right)
= k^a$$ because all the products vanish except the one for $c_m = k.$
A similar calculation can be found at this MSE link.
| {
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How do I take the limit of this function? $$\lim_{x \to 2^+} \frac{3}{x-2} - \frac{6}{x^2-4}$$
I'm unable to simplify it, but when I put it into my calculator (i.ie i'll put 2.1, 2.0001, etc), I keep getting $15$. When I try to simplify it, I get nowhere at all. How can I make this answer be 15?
| \begin{align}
& \frac{3}{x-2} - \frac{6}{x^2-4} = \frac3{x-2} - \frac{6}{(x-2)(x+2)} = \frac{3(x+2)}{(x-2)(x+2)} - \frac{6}{(x-2)(x+2)} \\[10pt]
= {} & \frac{3(x+2)-6}{(x-2)(x+2)} = \frac{3x}{(x-2)(x+2)} = \underbrace{\frac{1}{x-2}}_{\text{This }\to\,+\infty}\cdot\underbrace{\frac{3x}{x+2}}_{\text{This }\to\,3/2}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $10^{x^2+x}+\log{x} = 10^{x+1}$? In one of my recent exam, I was ask to solve this:
$$
10^{x^2+x}+\log{x} = 10^{x+1}
$$
My attempt to solve it was:
$$
10^{x^2+x}+\log{x} = 10^{x+1} \\
\log{x}=10^{x+1}-10^{x^2+x} \\
\log{x} = 10^{x+1}(1-10^x) \\
\log(\log{x})=(x+1)+\log(1-10^x) \\
$$
At this point I got stuck because I don't know how to solve an equation with double logs.
| If $x > 1$, then $x^2 + x > x + 1> 0$, and $\log x > 0$, thus $LHS > RHS$, and if $0 < x < 1$, then $0 <x^2 + x < x + 1$, and $\log x < 0$, thus $LHS < RHS$. When $x = 1$, both sides equal to $100$. Thus the only answer is $x = 1$.
| {
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First four terms of the power series of $f(z) = \frac{z}{e^z-1}$? Attempt:
$$ e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$
$$ e^z - 1 = \sum_{n=0}^\infty \frac{z^n}{n!} -1$$
$$ e^z - 1 = z\sum_{n=0}^\infty \frac{z^n}{(n+1)!} $$
Thus
$$ \frac{z}{e^z-1} = \frac{1}{\sum_{n=0}^\infty \frac{z^n}{(n+1)!}}.$$
How to proceed from this form? $z$ is complex.
| $$\frac{z}{e^z - 1}=\frac{1}{1+S} = 1 - S + S^2 - S^3 + \ldots$$
where $S = \sum_{n=1}^{\infty} \frac{z^n}{(n+1)!}$
To get the first three terms in the series expansion at $z=0$ consider the first three terms in the series expansion of
$1 - \bar S + \bar S^2 $ where $\bar S = \sum_{n=1}^{2} \frac{z^n}{(n+1)!} = \frac{z}{2} + \frac{z^2}{6} $, that is
$$1 - (\frac{z}{2} + \frac{z^2}{6}) + (\frac{z}{2} + \frac{z^2}{6})^2 = 1 -\frac{z}{2} + \frac{z^2}{12} + \text{h.o.t.}$$
To get the first $5$ terms consider
$$1 - (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} ) + (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} )^2 -\\- (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} )^3 +(\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} )^4= \\=
1 - (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} ) + (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24})^2 - (\frac{z}{2} + \frac{z^2}{6} )^3 + (\frac{z}{2} )^4 + \text{h.o.t.} = \\ = 1 - \frac{z}{2} +\frac{z^2}{12} -\frac{z^4}{720} + \text{h.o.t.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimum volume cone. What would be the radius and the altitude of a right circular cone that circumscribes a sphere with a radius 8 cm if the volume of the cone is to be minimized?
Here is my rough sketch;
My idea is to write some characteristics of the cone as a function of the radius of the circle, minimize it with differential calculus and connect those characteristics to the base radius and height of the cone, however I'm stuck at step 1.
| Let $r,a$ be the radius and the altitude of a right circular cone respectively.
Then, since considering the vertical plane which cuts the cone in half gives us
$$BD=DE=r,AD=\sqrt{a^2+r^2},AE=\sqrt{a^2+r^2}-r$$
$$CB=CE=8,AC=\sqrt{8^2+\left(\sqrt{a^2+r^2}-r\right)^2},$$
we have
$$AB=CB+AC\Rightarrow a=8+\sqrt{8^2+\left(\sqrt{a^2+r^2}-r\right)^2}.$$
Since this leads
$$(a-8)^2=8^2+\left(\sqrt{a^2+r^2}-r\right)^2$$
$$\Rightarrow a^2-16a=(a^2+r^2)-2r\sqrt{a^2+r^2}+r^2$$
$$\Rightarrow r\sqrt{a^2+r^2}=r^2+8a\Rightarrow r^2(a^2+r^2)=(r^2+8a)^2$$
$$\Rightarrow r^2(a^2-16a)=64a^2\Rightarrow r^2=\frac{64a^2}{a^2-16a},$$
all you need to consider is the following function with only one variable for $a\gt 16$:
$$\frac{\pi r^2a}{3}=\frac{64\pi a^2}{3(a-16)}=f(a).$$
Since
$$f'(a)=\frac{64\pi}{3}\cdot\frac{a(a-32)}{(a-16)^2},$$
the minimum is attained when $a=32,r=8\sqrt 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving by induction $5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$ is divisible by $4$ I want to prove the following twice. Once by induction then again by any other method.
$$5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$$ is a multiple of 4 for all nonnegative integers n.
Let n=0 , since it is the first nonnegative integer
$$5^{3(0)}+2*5^{2(0)}-5^{0}-2 = 0 $$
Factoring gives us $(5-1)(5+1)(5+2)$
| Hint: for induction step
$$ 5^{3(n+1)}+2\cdot 5^{2(n+1)} -5^{n+1}-2 $$
$$= 5^{3n}\cdot(124+1)+2\cdot 5^{2n}\cdot(24+1) -5^{n}\cdot(4+1)-2$$
$$= (5^{3n}+2\cdot 5^{2n} -5^{n}-2) + (5^{3n}\cdot 124+2\cdot 5^{2n}\cdot 24 -5^{n}\cdot 4) $$
$$= (5^{3n}+2\cdot 5^{2n} -5^{n}-2) + 4\cdot (5^{3n}\cdot 31+2\cdot 5^{2n}\cdot 6 -5^{n}). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Exact value for a series I would like prove this equality
$$1-\frac{1}{n-1}+\frac{1}{n+1}-\frac{1}{2n-1}+\frac{1}{2n+1}-\frac{1}{3n-1}+\frac{1}{3n+1}+...
=\frac{\pi}{n\tan{\frac{\pi}{n}}}$$
| The series is equal to
$$1-2 \sum_{k=1}^{\infty} \frac1{k^2 n^2-1} - 1-\frac{2}{n^2} \sum_{k=1}^{\infty} \frac1{k^2-\frac1{n^2}}$$
Now, we can evaluate the sum
$$\sum_{k=-\infty}^{\infty} \frac1{k^2-\frac1{n^2}} $$
by the residue theorem. I will omit the proof; the sum is equal to
$$-\pi \sum_{\pm}\operatorname*{Res}_{z=\pm \frac1n} \frac{\cot{\pi z}}{z^2-\frac1{n^2}} = -\pi n\cot{\frac{\pi}{n}} $$
Therefore the sum is
$$1-\frac{2}{n^2} \frac12 \left (-\pi n\cot{\frac{\pi}{n}} + n^2 \right ) = \frac{\pi}{n} \cot{\frac{\pi}{n}}$$
as was to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum\limits_{n=1}^\infty \frac{n^2(n-1)}{2^n} = 20$ This sum $\displaystyle \sum_{n=1}^\infty \frac{n^2(n-1)}{2^n} $showed up as I was computing the expected value of a random variable.
My calculator tells me that $\,\,\displaystyle \sum_{n=1}^\infty \frac{n^2(n-1)}{2^n} = 20$.
How can I show that?
I know how to find the value of the sum $\,\displaystyle \sum_{n=1}^\infty \frac{n^2}{2^n},\,$ but I can't deal with $\displaystyle \sum_{n=1}^\infty \frac{n^3}{2^n}$
| As this post is intended to justify the answer of @Dr.Sonnhard Graubner, I make this CW.
The starting points are that, defining $S_k(m)=\sum_{n=1}^m\frac{n^k}{2^n},$ we can easily compute $S_0(m)$ and $S_k(1),$ and that we can use recursive relations to obtain $S_3(m)$ from $S_0(m)$ and $S_3(1).$
Thus we divide the post into four steps:
I. $k=0$
$S_0(m)=\sum_{n=1}^m\frac{1}{2^n}=1-\frac{1}{2^m}$ (Geometric series)
II. $k=1$
Since $S_1(m)-\frac{1}{2}=\frac{1}{2}(\sum_{n=1}^{m-1}\frac{n+1}{2^n})=\frac{1}{2}S_1(m-1)+\frac{1}{2}S_0(m-1),$ we find that $S_1(m)=\frac{1}{2}S_1(m-1)+1-2^{-m}=\frac{1}{2}(\frac{1}{2}S_1(m-2)+1-2^{-m+1})+1-2^{-m}=\cdots=\frac{1}{2^{m-1}}S_1(1)+\sum_{n=0}^{m-2}\frac{1}{2^{n}}(1-\frac{1}{2^{m-n}})=\frac{1}{2^m}+\frac{1-(1/2)^{m-1}}{1/2}-\frac{(m-1)}{2^m}=2-\frac{1}{2^m}(m+2).$
III. $k=2$
By similar token, one finds the recursive relation $S_2(m)=\frac{1}{2}S_2(m-1)+3-\frac{2m+3}{2^m}.$ Then we deduce the formular for $S_2(m):$
$$S_2(m)=\frac{1}{2^m}+\sum_{n=0}^{m-2}\frac{1}{2^n}(3-\frac{2(m-n)+3}{2^{m-n}})=\frac{1}{2^m}+6-\frac{12}{2^m}-\frac{1}{2^m}(m^2+4m-5)$$$$=6-\frac{1}{2^m}(m^2+4m+6).$$
IV. $k=3$
Again we have the resursive relation $S_3(m)=\frac{1}{2}S_3(m)+13-\frac{1}{2^m}(3m^2+9m+13),$ from which we deduce the formula:
$$S_3(m)=\frac{1}{2^{m-1}}S_3(1)+\sum_{n=0}^{m-2}\frac{1}{2^n}(13-\frac{3(m-n)^2+9(m-n)+13}{2^{m-n}})$$
$$=\cdots=\frac{1}{2^m}+26-\frac{52}{2^m}-\frac{m^3+6m^2+18m-25}{2^m}=26-\frac{m^3+6m^2+18m+26}{2^m}.$$
Here the dots mean a process of simplification.
Hence we see that $\lim_{m\rightarrow\infty}S_3(m)-S_2(m)=26-6=20.$
So I am wondering where do those $(1/2)^{m+1}$ terms in the answer by @Dr.Sonnhard Graubner appear.
Finally I would like to say something, in reply to @Did: Indeed no simple functional equation holds for $S_k(m),$ but one doesn't expect there to be one at all: instead one looks for a recursive relation, to which one is used in dealing with discrete relations, such as finding the partial sums.
Hope this helps still.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Does the series $\sum_{n=1}^\infty \frac{n+1}{n^3+10n}$ converge? Using the ratio test, we evaluate:
$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty}\left| \frac{(n+1) + 1}{(n+1)^3 + 10(n+1)} \cdot \frac{n^3+10n}{n+1} \right| = \lim_{n \to \infty} \left| \frac{n^4 + 2n^3 + 10n^2 + 20n}{(n+1)^4 + 10(n+1)^2} \right|$$
$$ < \lim_{n \to \infty} \left| \frac{n^4 + 2n^3 + 10n^2 + 20n}{n^4 + 10n^2} \right| = 1$$
Hence $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$ and the series converges.
Is this an appropriate solution?
| Hint. You may use the comparison test, writing
$$
\frac{n+1}{n^3+10n}=\frac{1+\frac1n}{n^2+10}\sim \frac{1}{n^2}
$$ as $n$ tends to $+\infty$ and conclude easily.
| {
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"url": "https://math.stackexchange.com/questions/972336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is a closed form for ${\large\int}_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx$? Some time ago I asked How to find $\displaystyle{\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$.
Thanks to great effort of several MSE users, we now know that
\begin{align}
\int_0^1\frac{\ln^3(1+x)\,\ln x}xdx=&\,\frac{\pi^2}3\ln^32-\frac25\ln^52+\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)\\&\,-\frac{21}4\zeta(3)\ln^22-12\operatorname{Li}_4\!\left(\tfrac12\right)\ln2-12\operatorname{Li}_5\!\left(\tfrac12\right)\tag1
\end{align}
Now, a natural follow-up to that question is to bump the power of the logarithm and to ask:
Question: What is a closed form for the next integral?
$$I=\int_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx\tag2$$
I think it is likely that $I$ has a closed form, because there are several very similar integrals having known closed forms:
$$\int_0^1\frac{\ln^2(1+x)\,\ln^2x}xdx=\frac{\pi^2\,\zeta(3)}3-\frac{29\,\zeta(5)}8\tag3$$
$$\int_0^1\frac{\ln^3(1-x)\,\ln^2x}xdx=12\zeta^2(3)-\frac{23\pi^6}{1260}\tag4$$
\begin{align}
\int_0^1\frac{\ln^3(1+x)\,\ln^2x}{x^2}dx=&\,\frac{3\zeta(3)}2+2\pi^2\zeta(3)+\frac{3\zeta(5)}2-\frac{21\zeta(3)}2\ln^22\\&\,-\frac{63\zeta(3)}2\ln2+\frac{23\pi^4}{60}-\frac{4\ln^52}5-\frac{3\ln^42}2\\&\,-4\ln^32+\frac{2\pi^2}3\ln^32+\frac{3\pi^2}2\ln^22-24\operatorname{Li}_5\!\left(\tfrac12\right)\\&\,-36\operatorname{Li}_4\!\left(\tfrac12\right)-24\operatorname{Li}_4\!\left(\tfrac12\right)\ln2\tag5
\end{align}
| partial solution
using the following identity: ( I can provide the proof if needed)
$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)$$
replace $x$ with $-x$ , then multiply both sides by $\ln^3x$ and integrate from $0$ to $1$, we have
\begin{align}
I&=\int_0^1\frac{\ln^2(1+x)\ln^3x}{1+x}\ dx=\sum_{n=1}^\infty (-1)^n\left(H_n^2-H_n^{(2)}\right)\int_0^1x^n\ln^3x\ dx\\
&=-6\sum_{n=1}^\infty \frac{(-1)^n}{(n+1)^4}\left(H_n^2-H_n^{(2)}\right)=6\sum_{n=1}^\infty \frac{(-1)^n}{n^4}\left(H_{n-1}^2-H_{n-1}^{(2)}\right)\\
&=6\sum_{n=1}^\infty \frac{(-1)^n}{n^4}\left(H_{n}^2-H_{n}^{(2)}-2\frac{H_n}{n}+\frac2{n^2}\right)\\
&=6\left(\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^4}-2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}-\frac{31}{16}\zeta(6)\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
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Prove that $\left(\sum_{k=1}^{n}k\right)^2=\sum_{k=1}^{n}k^3$ holds true for $n ≥ 1$ I've been trying to figure out this proof for way too long now, I'm just not sure where to begin for the inductive step. Any guidance would be greatly appreciated.
| To make this proof easier to follow, let's define the following: $$\begin{align*} A_n &= \sum_{k=1}^n k, \\ B_n &= A_n^2 = \biggl(\sum_{k=1}^n k\biggr)^2, \\ C_n &= \sum_{k=1}^n k^3. \end{align*}$$ Then the claim to be proven is that $B_n = C_n$ for all positive integers $n$. Here, we focus on the inductive step: that is, we wish to establish that if there exists a positive integer $n$ such that $B_n = C_n$, this implies that $B_{n+1} = C_{n+1}$. To this end, we first observe $A_{n+1} = (n+1) + A_n$: hence $$\begin{align*} B_{n+1} &= A_{n+1}^2 = ((n+1) + A_n)^2 \\ &= (n+1)((n+1) + 2A_n) + A_n^2 \\ &= (n+1)(A_n + (n+1) + A_n) + B_n \\ &= (n+1)(A_{n+1} + A_n) + C_n. \end{align*}$$ But note $$A_{n+1} + A_n = (n+1) + \sum_{k=1}^n k + \sum_{k=n}^1 ((n+1)-k) = (n+1) + \sum_{k=1}^n (n+1) = (n+1)^2,$$ thus $$B_{n+1} = (n+1)^3 + C_n = C_{n+1},$$ which completes the inductive step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/974091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Laurent series for $f(z) = (z^2-4z+5)^{-1}$ centered at $z_{0} = 1$ I know there's plenty of questions on Laurent series, but I haven't found any suitable for my case.
I have to find the series for $f(z) = \dfrac{1}{z^2-4z+5}$ for a disk centered at $z_{0} = 1$ with a radius $R = \sqrt{2}$ and later for an annulus centered also at $z_ {0} = 1$ and from radius $\sqrt{2}$ to $\infty$.
Figuring out how to find the series for the disk, I suppouse I will be able to find the one for the annulus.
I have rewritten the expression as:
\begin{equation}
f(z) = \frac{A}{z-2+i} + \frac{B}{z-2-i} = \frac{i/2}{z-2+i} + \frac{-i/2}{z-2-i}
\end{equation}
So I am able to find the series for each expression,
\begin{equation}
\frac{i/2}{z-1-1+i} = \frac{i/2}{(-1+i)(1+\frac{z-1}{-1+i})} = \frac{i}{2(-1+i)}\sum_{n=0}^{\infty} (-1)^{n} \left( \frac{z-1}{-1+i} \right) ^{n}
\end{equation}
The same for the other one,
\begin{equation}
\frac{-i/2}{z-1-1-i} = \frac{i}{2(1+i)}\sum_{n=0}^{\infty} (-1)^{n} \left( \frac{z-1}{-1-i} \right) ^{n}
\end{equation}
In both cases, the radius of convergence is $\sqrt{2}$
\begin{equation}
\left| \left( \frac{z-1}{-1-i} \right) \right| < 1 \rightarrow |z-1| < \left| -1-i \right| = \sqrt{(-1)^{2} + (-1)^{2}} = \sqrt{2}
\end{equation}
I think I haven't made any mistake until now. My problem is that I haven't found out a way to add the sums, if that is the operation that I should do, to obtain the result. To be more precisely, I don't even know where to start to get the result:
\begin{equation}
f(z) = \sum_{n=0}^{\infty} \frac{(z-1)^{n}}{(\sqrt{2})^{n+1}} \sin (n+1)\frac{\pi}{4} = \frac{1}{2}+\frac{1}{2}(z-1)+\frac{1}{4}(z-1)^{2}-\frac{1}{8}(z-1)^{4} + \mathcal{O}(z-1)^{5}
\end{equation}
Thanks for your time
| Hint: These and de Moivre should help: $$(-1+i)^{-1} = -\sqrt{2}^{-1}(\cos \pi/4 +i \sin \pi/4) $$
$$(-1-i)^{-1} = -\sqrt{2}^{-1}(\cos \pi/4 -i \sin \pi/4) $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\displaystyle 2<(1+x)(1+y)(1+z)<\frac{64}{ 27}$ If $x,y,z>0$ and $x+y+z=1$ then show that $\displaystyle2<(1+x)(1+y)(1+z)<\frac{64}{27}$.
I have solved the right hand first using AM-GM inequality,
$\displaystyle\frac{1+x+1+y+1+z}{3} > \sqrt[3]{(1+x)(1+y)(1+z)}$
$\displaystyle \frac{64}{27}>(1+x)(1+y)(1+z)$
But now I am stuck in proving the left hand side.How do I prove the left-hand side inequality?
| Expand it out. Four of the terms sum to 2, and the remaining terms are positive.
$$(1+x)(1+y)(1+z) = 1 + x + y + z + xy + xz + yx + xyz > 1 + x + y + z = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/978577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Closed-form of $\int_0^1 B_n(x)\psi(x+1)\,dx$ Is there a closed-form of the following integral?
$$I_n = \int_0^1 B_n(x)\psi(x+1)\,dx,$$
where $B_n(x)$ are the Bernoulli polynomials and $\psi(x)$ is the digamma function.
The motivation of the problem was this question. In this answer achille hui showed that for all $n>0$
$$\int_0^1 B_n\left(\left\{ \frac1x \right\}\right) \frac{dx}{x}
= -\int_0^1 B_n(u) \psi(1+u) du,$$
where $\{x\}$ is the fractional part of $x$.
By looking at specific $n$ values we could observe interesting patterns.
$$\begin{align}
I_0 & = 0\\
I_1 & = 1-\frac 12 \ln(2\pi) \\
I_2 & = -\frac 12 + 2 \ln A \\
I_3 & = \frac{1}{12} - \frac{3}{4\pi^2}\zeta(3) = \frac{1}{12} - \frac{1}{8}\frac{\zeta(3)}{\zeta(2)}\\
I_4 & = \frac{1}{45} - 4\zeta'(-3)\\
I_5 & = -\frac{13}{360} + \frac{15}{4\pi^4}\zeta(5) = -\frac{13}{360} + \frac{1}{24}\frac{\zeta(5)}{\zeta(4)}\\
I_6 & = -\frac{1}{252}-6\zeta'(-5)\\
I_7 & = \frac{47}{1260} - \frac{315}{8\pi^6}\zeta(7) = \frac{47}{1260} - \frac{1}{24}\frac{\zeta(7)}{\zeta(6)}\\
I_8 & = -\frac{8}{1575}-8\zeta'(-7)\\
I_9 & = -\frac{1703}{25200} -\frac{2835}{4\pi^8}\zeta(9) = -\frac{1703}{25200} -\frac{3}{40}\frac{\zeta(9)}{\zeta(8)}\\
I_{10} & = \frac{2461}{83160} - 10\zeta'(-9)\\
\dots
\end{align}$$
where $A$ is the Glaisher–Kinkelin constant, $\zeta(3)$ is the Apéry's constant and in general $\zeta$ is the Riemann zeta function and $\zeta'$ is its derivative.
| This is an answer for the odd case.
Proposition. Let $k=1,2,3,\ldots$. Then
$$
\int_0^1 B_{2k+1}(x)\: \psi (x+1) \:dx=(-1)^{k+1}\frac{(2k+1)!}{(2\pi)^{2k+1}}\pi \: \zeta(2k+1)-\sum_{j=0}^{2k}\!\frac{ {{2k+1}\choose j} B_j}{2k+1-j} \quad (*)
$$
The proof is here, observing that
$$\begin{align}
\int_0^1 B_{2k+1}(x)\: \psi (x+1) \:dx & = \int_0^{1} \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx. \tag1
\end{align}$$ A proof of $(1)$ may be found here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a compression of the frames over the subspaces
*
*Find a compression of the frames $B=\{(1,1,0),(1,-1,0),(1,1,1),(0,0,1),(0,1,-1)\}$ over the subspaces (a) $M=\{(x,y,z):x=0\}$, (b) $M=\{(x,y,z):x=y\}$
*Find an orthonormal basis $B$ of $\mathbb{R}^4$ such that the frame $B_1=\{(1,1),(1,-1),(\sqrt{\frac{4}{3}},\sqrt{\frac{2}{3}}),(-\sqrt{\frac{2}{3}},\sqrt{\frac{4}{3}}) \}$ is a compression of B.
Building blocks will be greatly appreciated
| $\textbf{Part a:}$
When $M=\{(x,y,z):x=0\}$, $P_MB=\begin{Bmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & -1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & -1 \end{Bmatrix}$
When $M=\{(x,y,z):x=y\}$ $B_M=\begin{Bmatrix} 1 & 0 \\ 1 & 0 \\ 0 & 1 \end{Bmatrix}$ and $B_{M^\perp}=\begin{Bmatrix} -1 \\ 1 \\ 0 \end{Bmatrix}$
$P_{M^\perp}B=\begin{Bmatrix} 0 & 1 & 0 & 0 & \frac{-1}{2} \\ 0 & -1 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \end{Bmatrix}$ and $P_MB=\begin{Bmatrix} 1 & 0 & 1 & 0 & \frac{1}{2} \\ 1 & 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 1 & -1 \end{Bmatrix}$
$\textbf{Part b}$
Let $B_1=\begin{Bmatrix} 1 & 1 & \sqrt{\frac{4}{3}} & -\sqrt{\frac{2}{3}} \\ 1 & -1 & \sqrt{\frac{2}{3}} & \sqrt{\frac{4}{3}}
\end{Bmatrix}$
These vectors in $\mathbb{R}^2$ have a norm of $\sqrt{2}$. Since $v=p_Mv+p_{M^\perp}v \iff \|v\|^2=\|p_Mv \|^2+\|p_{M^\perp}v\|^2 \Rightarrow \|v\| \leq \|p_Mv\|$. This implies that if $v_i \in \mathbb{R}^4$, and $p_Mv=b_i$, then $\|v\| \leq \sqrt{2}$. So we cannot construct an orthonormal basis in $\mathbb{R}^4$. However, we can construct an orthogonal basis.
$\begin{Bmatrix} \begin{Bmatrix} 1 & 1 & \sqrt{\frac{4}{3}} & -\sqrt{\frac{2}{3}} \\ 1 & -1 & \sqrt{\frac{2}{3}} & \sqrt{\frac{4}{3}}
\end{Bmatrix} \\ 0 & 0 & \frac{2+\sqrt{2}}{\sqrt{3}} & \frac{2-\sqrt{2}}{\sqrt{3}} \\
\alpha_1 & \alpha_2 & \alpha_3 & \alpha_4 \end{Bmatrix}$
By Gram-Schmidt, we can find the final orthonormal vector.
\begin{equation*}
\begin{split}
(\alpha_1,\alpha_2,\alpha_3, \alpha_4)
= & (1,0,0,0)-\frac{1}{2}(1, 1, \sqrt{\frac{4}{3}}, -\sqrt{\frac{2}{3}})-\frac{1}{2}( 1, -1, \sqrt{\frac{2}{3}}, \sqrt{\frac{4}{3}}) \\
= & (1,0,0,0)-(\frac{1}{2}, \frac{1}{2}, \sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{6}})-( \frac{1}{2}, -\frac{1}{2}, \sqrt{\frac{1}{6}}, \sqrt{\frac{1}{3}}) \\
= & (0,0,-\frac{\sqrt{2}+1}{\sqrt{6}},\frac{1-\sqrt{2}}{\sqrt{6}})
\end{split}
\end{equation*}
Hence, the orthonormal basis is
$\begin{Bmatrix} 1 & 1 & \sqrt{\frac{4}{3}} & -\sqrt{\frac{2}{3}} \\ 1 & -1 & \sqrt{\frac{2}{3}} & \sqrt{\frac{4}{3}} \\ 0 & 0 & \frac{2+\sqrt{2}}{\sqrt{3}} & \frac{2-\sqrt{2}}{\sqrt{3}} \\
0 & 0 & -\frac{\sqrt{2}+1}{\sqrt{6}} & \frac{1-\sqrt{2}}{\sqrt{6}}
\end{Bmatrix}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Assume that 495 divides the integer $\overline{273x49y5}$ where $x,y \in \{0,1,2...9\}$. Find $x$ and $y$. So, I know that $495 = 5\times 9\times 11$. So then, if that's the case, then the number $\overline{273x49y5}$ must be divisible by $495$ if and only if it is divisible by $5$ and $9$ and $11$.
Then, I know that $5$ divides it for any number $x$ and $y$ because it only has to divide the digits number which is $5$.
To see if it divides by $9$, the sum of the digits must be divisible by $9$, and as for $11$, the alternating sum of digits must be divisible by $11$.
So, to test it by $9$, then, $2+7+3+x+4+9+y+5 = 30+x+y$ has be divisble by $9$. So, the only numbers $x$ and $y$ that make $30+x+y$ divisible by $9$ are $x= \{0,1,2,3,4,5,6\}$ and $y=\{0,1,2,3,4,5,6\}$.
But $11$ is what gets me. To make it divisible by $11$, then the alternating sums must be divisible by $11$. So then, $-2+7-3+x-4+9-y+5= 12+x-y$ must be divided by $11$. Then, the only numbers $x$ and $y$ that make it divisible by $11$ must be $x=\{0,1,2,\ldots,8\}$ and $y=\{1,2,\ldots,9\}$, if I did my math right.
Next, I assume we must then find numbers in $x$ and $y$ that satisfy the division by $9$ and $11$. So, if that is the case, then I know that since the number must be divisible by $9$ and $11$, then the numbers $x$ and $y$ must fit between $0$ to $6$ because that's $6$ is the highest number $x$ or $y$ can be in the sequence to divide the number. So, how do I check to see which numbers both satisfy division of $9$ and $11$?
| $30+x+y$ must be a multiple of 9 which leads to:$$30+x+y=9a$$$$x+y=9a-30\tag{1}$$
$12+x-y$ must be a multiple of 11 which leads to:$$12+x-y=11b$$$$1+x-y=11c\text{ where c=b-1}\tag{2}$$
the biggest difference between two digits is 9 therefore the maximum value of $x-y$ is 9 which leads to (from (2)):$$11c\le1+9\le10$$therefore $c=0$ which leads to:$$y=x+1\tag{3}$$substitute (3) into (1) to get:$$2x+1=9a-30$$$$2x=9a-31$$Hopefully you can solve from here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding conditional probability from the joint PMF I have no idea how to get tabular to work on here so the table isn't rendering.
A joint PMF $p_{X, Y}[i, j]$ has the values shown in table. Determine the conditional PMF $p_{Y|X}$. Are the random variables independent?
$$
\begin{array}{|c|c|c|c|}
\hline
& j = 1 & j = 2 & j = 3\\
\hline
i = 1 & \tfrac{1}{10} & \tfrac{1}{10} & \tfrac{2}{10}\\
\hline
i = 2 & \tfrac{1}{20} & \tfrac{1}{20} & \tfrac{1}{10}\\
\hline
i = 3 & \tfrac{3}{10} & \tfrac{1}{20} & \tfrac{1}{20}\\
\hline
\end{array}
$$
When I follow the formula, the conditional probabilities aren't summing to one.
The $p_{Y|X = x_i}[j|x_i] = \frac{p_{X, Y}[x_i, j]}{p_X[x_i]}$ for all $j$ and similarly for $p_{X|Y = y_j}$.
\begin{alignat*}{4}
p_{Y|X = 1}[j|1] &=
\begin{cases}
\frac{3}{10}, & j = 1\\
\frac{3}{10}, & j = 2\\
\frac{6}{10}, & j = 3
\end{cases} & \qquad &
p_{Y|X = 2}[j|2] &&={}
\begin{cases}
\frac{3}{20}, & j = 1\\
\frac{3}{20}, & j = 2\\
\frac{3}{10}, & j = 3
\end{cases}\\
p_{Y|X = 3}[j|3] &=
\begin{cases}
\frac{9}{10}, & j = 1\\
\frac{3}{20}, & j = 2\\
\frac{3}{20}, & j = 3
\end{cases} & \qquad &
p_{X|Y = 1}[i|1] &&={}
\begin{cases}
\frac{3}{10}, & i = 1\\
\frac{3}{20}, & i = 2\\
\frac{9}{10}, & i = 3
\end{cases}\\
p_{X|Y = 2}[i|2] &=
\begin{cases}
\frac{3}{10}, & i = 1\\
\frac{3}{20}, & i = 2\\
\frac{3}{20}, & i = 3
\end{cases} & \qquad &
p_{X|Y = 3}[i|3] &&={}
\begin{cases}
\frac{6}{10}, & i = 1\\
\frac{3}{10}, & i = 2\\
\frac{3}{20}, & i = 3
\end{cases}
\end{alignat*}
| I'm not sure where you are getting those figures. The marginal probability of $X$ is the sum of the joint probabilities for all values of $Y$ at particular values of $X$.
$\begin{align}
\text{Thus:}
\\ \because p_X(i) & = \sum_{j=1}^3 p_{X,Y}(i,j)
\\ &= p_{X,Y}(i,1)+p_{X,Y}(i,2)+p_{X,Y}(i,3)
\\[1ex]
\therefore p_X(1) &= p_{X,Y}(1,1)+p_{X,Y}(1,2)+p_{X,Y}(1,3)
\\ &= \tfrac 1{10}+\tfrac 1{10}+\tfrac 2{10}
\\ &= \tfrac 2 5
\\ \text{Similarly:}
\\[2ex]
\because p_{Y\mid X}(j\mid i)&= \frac{p_{X,Y}(i, j)}{p_X(i)}
\\[2ex]
\therefore p_{Y\mid X}(j\mid 1)&= \frac{p_{X,Y}(1, j)}{p_X(1)}
\\ & = \frac{5}{2}\begin{cases}\frac 1{10} & :j=1\\\frac 1{10} &:j=2\\\frac 2{10} &:j=3\end{cases}
\\ & = \begin{cases}\frac 1{4} & :j=1\\\frac 1{4} &:j=2\\\frac 1{2} &:j=3\end{cases}
\end{align}$
You can do the rest now, right?
| {
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"source": "stackexchange",
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n! v.s. $a^{n}$ How can we know which one is faster without graphing? n! v.s. $a^{n}$ If we are given an arbitrary number a (a>1). How can we know which one is faster as n->INFINITY without graphing?
| For $n \rightarrow \infty$, $n!$ is always going to win. We have
\begin{align*}
\frac{a^n}{n!} & = \frac{a}{1} \frac{a}{2} \ldots \frac{a}{n-1} \frac{a}{n}
\end{align*}
Now, let $N$ be the smallest integer that satisfies $N \geq a$. Furthermore, let $n > N$. We have
\begin{align*}
\frac{a^n}{n!} & = \underbrace{\frac{a}{1} \frac{a}{2} \ldots \frac{a}{N-1}}_{>1} \times\underbrace{\frac{a}{N} \frac{a}{N+1} \ldots \frac{a}{n-1} \frac{a}{n}}_{<1}
\end{align*}
where the tail becomes smaller and smaller for increasing $n$. You can now verify that the tail of this product 'overwhelms' the head for $n$ sufficiently large.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is $\lim_{(x,y)\to(0,0)} \dfrac{x^3 y^4}{x^6+y^8}$? Why is this limit $0$. It seems to me that $y^4/(x^6+y^8)$ is a non limited function and so is $x^3/(x^6+y^8)$.
$$\lim_{(x,y)\to(0,0)} \dfrac{x^3 y^4}{x^6+y^8}$$
| Note that $\displaystyle \frac{x^3}{x^6+y^8}\cdot \frac{y^4}{x^6+y^8}\neq \frac{x^3y^4}{x^6+y^8}.$
What I would like to hint is: Compare the limits along the paths $x=y$ and $x^3=y^4$. That should do the trick.
| {
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"source": "stackexchange",
"question_score": "1",
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Using l'Hopital's rule to evaulate $\lim\limits_{x\to0}\left( (\csc x)^2-\frac{1}{x^2}\right)$ Use l'Hopital's rule to find the following limit
$$\lim\limits_{x\to0}\left( (\csc x)^2-\frac{1}{x^2}\right)$$
I tried differentiation but did not get the right answer. I know I have to put it into a single fraction.
| $ (\csc x)^2-\frac{1}{x^2}=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{x^2-\sin^2x}{x^2\sin^2x }$ ~ $\frac{x^2-\sin^2x}{x^4 }$
So using l'Hopital's rule
$$\lim\limits_{x\to0}\left( (\csc x)^2-\frac{1}{x^2}\right)=\lim\limits_{x\to0}\frac{2x-2\sin x\cos x}{4x^3}=\lim\limits_{x\to0}\frac{2x-\sin 2x}{4x^3}=\lim\limits_{x\to0}\frac{2-2\cos 2x}{12x^2}=\lim\limits_{x\to0}\frac{0+4\sin 2x}{24x}=\lim\limits_{x\to0}\frac{0+8\cos 2x}{24}=\frac13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/987326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Find one set of solutions for the following system: Find one set of solutions for the following system:
\begin{cases}
1+a^2+d^2=3+b^2+e^2=3+c^2+f^2 \\
1+ab+de=0 \\
ac+df=0 \\
bc+ef=0 \\
\end{cases}
| Setting $a = -d$ and $c = f$ and $ b = -e $, we are left with the top 2 equations.
Subbing in some values, these equations become,
$1 + 2d^2 = 3 + 2e^2 = 3 + 2c^2$ and $1 + 2de = 0$.
From this, $2de = -1 \iff d = \frac{-1}{2e}$.
Thus, we have $1 + \frac{1}{2e^2} = 3 + 2e^2 = 3 + 2c^2$.
Then, $1 = 4e^2 + 4e^4 \iff e^2 = \frac{1}{\sqrt{2}} - \frac{1}{2} \iff e = \pm\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}}$.
Now we can solve for $d$ and $c$. But we already have $d$ and $c = e$ from inspection.
We have $$\begin{equation} \begin{aligned} e = \pm\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}} \\ d = \frac{-1}{2e} \\ b = -e \\ a = -d \\ c = f = e\end{aligned} \end{equation} .$$
| {
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"source": "stackexchange",
"question_score": "3",
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Dividing a polynomial with $(x^2+1)^2$ I have been given that a polynomial $f(x)$ with real coefficients is divisible by $(x^2+1)$, and that when $f'(x)$ is divided by $(x^2+1)$, we get a remainder of $(x+1)$. I need to prove that $2f(x)+(x-1)(x^2+1)$ is divisible by $(x^2+1)^2$.
What I did was to express $f(x)$ as $f(x)=(x^2+1)g(x)$, for some polynomial $g$ (1),and $f'(x)=((x^2+1))q(x) + (x+1)$, for some polynomial $q$ (2).
From (1) I deduced another expression of $f'(x)$, (3): $$f'(x)=(x^2+1)g'(x) + 2x(g(x))$$
I then expressed the given expression as:
$$2f(x)+(x-1)(x^2+1) = 2(x^2+1)g(x) + (x-1)(x^2+1)$$ (4)
To find an expression for $g(x)$, I then equated (2) and (3) above, to get
$$ 2g(x) = (x^2+1)(q(x)-g'(x) + x+1)(1/x)$$
Letting $(q(x)-g'(x) + x+1)(1/x) = h(x)$, I simplified to:
$$2g(x)= (x^2+1)h(x)$$
then here I stopped, realizing that $h(x)$ is maybe not even a polynomial.
Any alternative method or advancement of this one will be very appreciated!
| Mark's answer is nicer than this, but it is possible to continue along the lines you started:
Since $f'(x)=(x^2+1)g'(x)+2x g(x)$, $2x g(x)$ has the same remainder $(x+1)$ when divided by $x^2+1$ as does $f'(x)$. That is,
$$
2xg(x)=h(x)(x^2+1)+x+1
$$
for some polynomial $h$. By setting $x=0$, we can see that $h(0)=-1$; thus $h(x)=xk(x)-1$ for some other polynomial $k$. So
\begin{align}
2xg(x)&=(xk(x)-1)(x^2+1)+x+1\\
&=xk(x)(x^2+1)-x^2+x
\end{align}
Dividing by $x$ gives
$$
2g(x)=k(x)-(x-1)
$$
which upon multiplying by $(x^2+1)$ yields the desired relation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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maximum, complex quadratic function, Is my solutions correct? I'm trying to compute $\max_{|z| \le 1} |(z+2)(z-1)|$.
Here's how I do it:
$\{z \in \mathbb{C} \ | \ |z| \le 1 \}$ is compact and $f(z) = (z+2)(z-1)$ is continuous, so it suffices to look for $\max_{|z| = 1} |(z+2)(z-1)|$.
Let $z = \cos \phi + i \sin \phi, \ \ 0 \le \phi \le 2 \pi$
Now, $|(z+2)(z-1)| = \sqrt{sin^2 \phi + (\cos \phi -1)^2} \sqrt{\sin^2 \phi + (\cos \phi +2)^2} = \sqrt{1+ 2 \cos \phi} \cdot 2 \sqrt{cos \phi +1} = 2 \sqrt{-2 cos^2 \phi - \cos \phi +1}$.
So we are left with a quadratic function with real coefficients and the leading coefficient is negative, so we can compute the vertex of the parabola $f(\cos \phi) = -2 cos^2 \phi - \cos \phi +1$ and $q =-\frac{ \Delta }{4a} = - \frac{9}{8}$ and $p = - \frac{b}{2a} = - \frac{1}{4}$.
So we have maximum value equal to $- \frac{9}{8}$ for $\cos \phi = - \frac{1}{4}$ (then $\sin \phi = \sqrt{1 - \frac{1}{16}} = - \frac{1}{4} \sqrt{15}$)
So it seems that $\max_{|z| \le 1} |(z+2)(z-1)| = \frac{9}{8}$ for $z = - \frac{1}{4} + i - \frac{1}{4} \sqrt{15}$.
Is it correct?
| The maximum will appear on the circle since your function is analytic on the domain.
$|(z+2)(z-1)| = \sqrt{sin^2 \phi + (\cos \phi -1)^2} \sqrt{\sin^2 \phi + (\cos \phi +2)^2} = \sqrt{2- 2 \cos \phi} \sqrt{4cos \phi +5} = \sqrt{-8 cos^2 \phi - 2\cos \phi +10}$.
So we are left with a quadratic function with real coefficients and the leading coefficient is negative, so we can compute the vertex of the parabola $f(\cos \phi) = -8 cos^2 \phi - 2\cos \phi +10$ and $p = - \frac{b}{2a} = - \frac{1}{8}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence to zero I need to prove that if $n \rightarrow \infty$ then this sum converges to zero.
$$
\sum_{k = [\frac{n}{2}]}^n
{n \choose k}
Q^k
\cdot
\left(
1-Q
\right)^{n-k}
$$
In this case $Q$ is constant and equals $e^{-1}$. But it seems that if this is true then it can be generalized for $Q < \frac{1}{2}$.
| What you write is the probability of $P(\sum_{k=1}^nX_n \geq \lfloor \frac{n}{2}\rfloor)$, where $X_i$'s are i.i.d Bernoulli variable with parameter $Q$.
$P(\sum_{k=1}^nX_n \geq \lfloor \frac{n}{2}\rfloor) = P(\dfrac{\sum_{k=1}^nX_n -nQ}{\sqrt{n}} \geq \dfrac{\lfloor \frac{n}{2}\rfloor -nQ}{\sqrt{n}})$ and
\begin{align}
\lim_{n\to \infty} \dfrac{\lfloor \frac{n}{2}\rfloor -nQ}{\sqrt{n}} = \begin{cases} +\infty & Q < \frac{1}{2}\\ 0 & Q = \frac{1}{2} \\ -\infty & Q > \frac{1}{2}\end{cases}
\end{align}
So by central limit theorem, your sum converges to $P( G \geq \lim_{n\to \infty} \dfrac{\lfloor \frac{n}{2}\rfloor -nQ}{\sqrt{n}})$, where $G$ is a centred normal distribution with variance $Q(1-Q)$. In particular we have
\begin{align}
P( G \geq \lim_{n\to \infty} \dfrac{\lfloor \frac{n}{2}\rfloor -nQ}{\sqrt{n}}) = \begin{cases} 0 & Q < \frac{1}{2}\\ \frac{1}{2} & Q = \frac{1}{2} \\ 1 & Q > \frac{1}{2}\end{cases}
\end{align}
| {
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Multiplying and adding fractions Why multiplying fractions is equal to multiply the tops, multiply the bottoms?
$$\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b \times d},$$
And why
$$\frac{a}{b}\times \frac{c}{c}=\frac{a}{b},$$
Also why
$$\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}.$$
I understand it, but I want a mathematical approach as a math student proves it. Also I want to know the mathematics topic of this question (number theory, logic, etc). A full answer is not necessary. Just a reference.
|
Complete Alternative Fraction Product Rule Proofs
In a previous post I showed how to create a legitimate elementary arithmetic proof of the fraction product rule using a field theorem about division. That proof was spread out amongst discussions of a number of issues, and some of the steps needed were not explicitly shown. I therefore have written out the complete proof below.
I will not duplicate the fundamentals required for this proof because they can be found in the my first post, which also contains another proof of the fraction product rule.
Theorems to Make Shorter Proofs
Theorem 2 (constant times a fraction)
Given b${\neq}$0,
\begin{equation*}
\boldsymbol{c}\cdot \frac{\boldsymbol{a}}{\boldsymbol{b}}=\frac{\boldsymbol{c}\cdot \boldsymbol{a}}{\boldsymbol{b}}.
\end{equation*}
Proof: (See my first post)
Notice that here we are assuming that the variables in the above theorem are non-negative integers. We will, of course, want these variables to eventually be other numbers, such as fractions. In fact, the theorem we are trying to prove, that
\begin{equation*}
\frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d},
\end{equation*}
can be viewed as part of that process. In particular, we are replacing the variable c in Theorem 2 (constant times a fraction) with a fraction.
Because we assume the multiplicative commutative law, in our proof we also consider the following as justified by Theorem 2 (constant times a fraction):
Given b${\neq}$0,
\begin{equation*}
\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \mathbf{c}=\frac{\boldsymbol{a}\cdot \boldsymbol{c}}{\boldsymbol{b}}.
\end{equation*}
Theorem 9 (fraction as product and inverse)
For b${\neq}$0,
\begin{equation*}
\frac{\boldsymbol{a}}{\boldsymbol{b}}=\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}.
\end{equation*}
Proof:
$\frac{\boldsymbol{a}\cdot \boldsymbol{c}}{\boldsymbol{b}}=a\cdot \frac{\boldsymbol{c}}{\boldsymbol{b}}$ (Theorem 2, Constant Times a Fraction)
$\frac{\boldsymbol{a}\mathbf{\cdot}\mathbf{1}}{\boldsymbol{b}}=a\cdot \frac{\mathbf{1}}{\boldsymbol{b}}$ (substituting c=1)
$\frac{\boldsymbol{a}}{\boldsymbol{b}}=a\cdot \frac{\mathbf{1}}{\boldsymbol{b}}$ (multiplicative identity)
Theorem 10: (multiplicative inverse as a fraction)
If a${\neq}$0, then the multiplicative inverse of a is the fraction $\frac{\mathbf{1}}{\boldsymbol{a}}$ .
Proof:
$a\cdot \frac{1}{a}=\frac{a\cdot 1}{a}$ (Theorem 2, Constant Times a Fraction)
$=\frac{a}{a}$ (multiplicative identity)
$=1$ (Law 1)
Theorem 11: (product of inverses)
Given a${\neq}$0 and b${\neq}$0,
\begin{equation*}
\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{a}}=\frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}}
\end{equation*}
Comment: note that this theorem and its proof is very similar to the fraction product law itself and the proof presented in my previous post. In this proof, factor $a\cdot b$ is first introduced followed by factor $\frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}}$ . In the proof of the fraction product rule in my first post, $\frac{a\cdot b}{a\cdot b}$ was introduced as the multiplicative identity and then separated during the rest of the proof.
Proof:
$a\cdot b\cdot \left(\frac{1}{b}\cdot \frac{1}{a}\right)=a\cdot b\cdot \frac{1}{b}\cdot \frac{1}{a}$ (removing unnecessary parenthesis)
$=a\cdot \left(b\cdot \frac{1}{b}\right)\cdot \frac{1}{a}$ (associative law)
$=a\cdot 1\cdot \frac{1}{a}$ (Theorem 10, multiplicative inverse as a fraction)
$=a\cdot \frac{1}{a}$ (multiplicative identity)
$=1$ (Theorem 10, multiplicative inverse as a fraction)
$=a\cdot b\cdot \frac{1}{a\cdot b}$ (Theorem 10, multiplicative inverse as a fraction)
Since $a\cdot b\cdot \left(\frac{1}{b}\cdot \frac{1}{a}\right)=a\cdot b\cdot \frac{1}{a\cdot b}$ , we next multiply both sides of this equality by $\frac{1}{a\cdot b}$ . This is possible because we are given that a${\neq}$0 and b${\neq}$0, and by the lemma we know that $a\cdot b\neq 0$. The result is
$\frac{1}{a\cdot b}\cdot a\cdot b\cdot \left(\frac{1}{b}\cdot \frac{1}{a}\right)=\frac{1}{a\cdot b}\cdot a\cdot b\cdot \frac{1}{a\cdot b}$
$1\cdot \left(\frac{1}{b}\cdot \frac{1}{a}\right)=1\cdot \frac{1}{a\cdot b}$ (Theorem 10, multiplicative inverse as a fraction)
$\left(\frac{1}{b}\cdot \frac{1}{a}\right)=\frac{1}{a\cdot b}$ (multiplicative identity)
$\frac{1}{b}\cdot \frac{1}{a}=\frac{1}{a\cdot b}$ (removing unnecessary parentheses)
Proof 3 of the Fraction Product Rule
With Theorems 9, 10, and 11, we can prove the fraction product rule as shown next.
$\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\right)\cdot \left(\boldsymbol{c}\cdot \frac{\mathbf{1}}{\boldsymbol{d}}\right)$ (Theorem 9, fraction as product and inverse)
$=\boldsymbol{a}\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \boldsymbol{c}\right)\cdot \frac{\mathbf{1}}{\boldsymbol{d}}$ (multiplicative associative law)
$=\boldsymbol{a}\cdot \left(\boldsymbol{c}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\right)\cdot \frac{\mathbf{1}}{\boldsymbol{d}}$ (multiplicative commutative law)
$=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{d}}\right)$ (multiplicative associative law)
$=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}\right)$ (Theorem 11, product of inverses)
$=\frac{\left(a\cdot c\right)\cdot 1}{b\cdot d}$ (Theorem 2, constant times a fraction)
$=\frac{\left(a\cdot c\right)}{b\cdot d}$ (multiplicative identity)
$=\frac{a\cdot c}{b\cdot d}$ (removing unnecessary parentheses)
Q.E.D.
Proof 4 of the Fraction Product rule
Methodology
Using the step-by-step form used in Proof 3 of the Fraction Product Rule makes it especially easy to make shorter proofs by collecting the transformations effected by contiguous proof lines and turning them into a theorem. Using Proof 3 as an example, we collect the first 5 lines to create Theorem 12.
Theorem 12 (product of fractions as a product of inverses)
Given b${\neq}$0 and d${\neq}$0
\begin{equation*}
\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}\right)
\end{equation*}
Proof:
$\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\right)\cdot \left(\boldsymbol{c}\cdot \frac{\mathbf{1}}{\boldsymbol{d}}\right)$ (Theorem 9, fraction as product and inverse)
$=\boldsymbol{a}\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \boldsymbol{c}\right)\cdot \frac{\mathbf{1}}{\boldsymbol{d}}$ (multiplicative associative law)
$=\boldsymbol{a}\cdot \left(\boldsymbol{c}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\right)\cdot \frac{\mathbf{1}}{\boldsymbol{d}}$ (multiplicative commutative law)
$=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{d}}\right)$ (multiplicative associative law)
$=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}\right)$ (Theorem 11, product of inverses)
Q.E.D.
Next, we consolidate the remaining lines into a theorem. We need to show
\begin{equation*}
\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}\right)=\frac{\boldsymbol{a}\cdot \boldsymbol{c}}{\boldsymbol{b}\cdot \boldsymbol{d}}
\end{equation*}
but this is just an application of
Theorem 9 (fraction as product and inverse):
For b${\neq}$0,
\begin{equation*}
\frac{\boldsymbol{a}}{\boldsymbol{b}}=\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}.
\end{equation*}
Thus, our Proof 4 of the Fraction Product Rule requires only two steps.
Proof 4
Given b${\neq}$0 and d${\neq}$0,
$\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}\right)$ (Theorem 12, product of fractions as a product of inverses)
$=\frac{\boldsymbol{a}\cdot \boldsymbol{c}}{\boldsymbol{b}\cdot \boldsymbol{d}}$ (Theorem 9, fraction as product and inverse
Q.E.D.
Results like this suggest to me that, if we are allowed to use theorems in a proof, then the complexity should not be measured by the number of steps. This is because we can easily hide complexity within a theorem. If we must prove a theorem strictly from axioms, then the proofs can be fairly long. However, perhaps this is the best way to provide a simple measure of the complexity of a proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/990953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 4
} |
Prove $\lim_{n \rightarrow \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0 $ Prove $\lim_{n \rightarrow \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0 $
Let $\epsilon > 0$ be arbitrary. I want to find $N$ such that $n \in \mathbb{N}$ guarantees $ \left | \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} - 0 \right | < \epsilon$.
$\Leftrightarrow \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} < \epsilon$
$\Leftarrow \frac{1}{n} \cdot \frac{4}{3} < \epsilon \Leftrightarrow n > \frac{4}{3} \cdot \frac{1}{\epsilon} \Rightarrow n \geq N$
Take $N = \left \lfloor \frac{4}{3} \cdot \frac{1}{\epsilon} \right \rfloor + 1$
In the above proof, I'm confused as to how $\frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} < \epsilon$
$\Leftarrow \frac{1}{n} \cdot \frac{4}{3} < \epsilon$. How did $\frac{3 + \frac{1}{n}}{4 - \frac{1}{n}}$ become $\frac{4}{3} $?
| Observe that for all $n \geq 1$, $3 + \dfrac{1}{n} \leq 4$, and $4 - \dfrac{1}{n} \geq 3$. Thus: $\dfrac{3+\dfrac{1}{n}}{4 - \dfrac{1}{n}} \leq \dfrac{4}{3}$. Thus given $\epsilon > 0$, choose $N_{0} =\lfloor\dfrac{4}{3\epsilon}\rfloor + 1$, then if $n > N_0$, we have: $\dfrac{1}{n}\cdot \dfrac{3+\dfrac{1}{n}}{4 - \dfrac{1}{n}} < \dfrac{4}{3n} < \epsilon$, and the conclusion follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is there any way to simplify following summation? Is there any way to simplify following summation?
$$\sum_{k=1}^n \frac{1}{k^2(k+1)^2}$$
| Consider
\begin{align}
\frac{1}{k^2 (k+1)^2} = \left(\frac{1}{k} - \frac{1}{k+1}\right)^{2} = \frac{1}{k^2} + \frac{1}{(k+1)^2} - \frac{2}{k} + \frac{2}{k+1}
\end{align}
for which
\begin{align}
\sum_{k=1}^{n} \frac{1}{k^2 (k+1)^2} &= \sum_{k=1}^{n} \frac{1}{k^2} + \sum_{k=2}^{n+1} \frac{1}{k^2} + 2 \sum_{k=2}^{n+1} \frac{1}{k} - 2 \sum_{k=1}^{n} \frac{1}{k} \\
&= -1 - \frac{1}{(n+1)^2} + 2 \sum_{k=1}^{n+1} \frac{1}{k^2} + \frac{2}{n+1} - 2 \\
&= -2 - \left( \frac{1}{n+1} - 1\right)^2 + \sum_{k=1}^{n+1} \frac{2}{k^2}\\
&= - \frac{3n^2 + 4n +2}{(n+1)^2} + \sum_{k=1}^{n+1} \frac{2}{k^2} \\
&= \frac{\pi^2}{6} - \frac{3n^2 + 4n +2}{(n+1)^2} - \psi_{1}(n+2)
\end{align}
where $\psi_{1}(x)$ is the derivative of the digamma function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Posible combinations of available rooms I have $x$ people to divide over different type of rooms.
Possible rooms:
*
*$1$ person bedroom normal view;
*$1$ person bedroom sea view;
*$2$ person bedroom
For $x = 4$
Possible combinations are
$$(0,0,2),
(0 ,2, 1),
(2 ,0, 1),
(1 ,1, 1),
(4 ,0, 0),
(3 ,1, 0),
(2 ,2, 0),
(1 ,3 ,0),
(0 ,4, 0)$$
Note : Plenty of rooms are available.
How can I get all posible combinations?
| Are you interested in the number of compositions, or the list of compositions itself?
The number of compositions can be obtained using generating functions. I assume, as in your example, that a two-person bedroom cannot be occupied by a single person. The number of compositions for $n$ people is the coefficient of $y^n$ in the expansion of
$$
\frac{1}{(1-y)^2(1-y^2)}=\frac{(1+y)^2}{(1-y^2)^3}=(1+2y+y^2)\sum_{j=0}^\infty\frac{(-3)(-2)\ldots(-j-2)}{j!}(-y^2)^j.
$$
The expression on the left comes from interpreting $\frac{1}{1-y}$ as $1+y+y^2+\ldots$ and $\frac{1}{(1-y^2)}$ as $1+y^2+y^4+\ldots$ and realizing that the coefficient of $y^n$ when the three factors are multiplied out is exactly the number of whole-number solutions to $a+b+2c=n.$ The middle expression comes from writing $\frac{1}{1-y}$ as $\frac{1+y}{(1-y)(1+y)}.$ The expression on the right is obtained using the binomial theorem.
With some further manipulations, the generating function becomes
$$
(1+2y+y^2)\sum_{j=0}^\infty\binom{j+2}{j}y^{2j}=(1+2y+y^2)\sum_{j=0}^\infty\binom{j+2}{2}y^{2j}.
$$
You can see that when $n$ is even, the coefficient of $y^n$ is $\binom{n/2+2}{2}+\binom{n/2+1}{2}$. When $n$ is odd, the coefficient is $2\binom{(n-1)/2+2}{2}.$
These answers can be understood using stars-and-bars: if $n$ is even, pair up the people. So there are $n/2$ pairs. Either we have an even number of people in all three types of room, in which case there are $\binom{n/2+2}{2}$ compositions, or there is an odd number of people in the single-type rooms, in which case we split one of the pairs, putting one member of the pair in a normal room and the other in a sea-view room, and then find that there are $\binom{n/2+1}{2}$ compositions for the remaining pairs.
If $n$ is odd, we have $(n-1)/2$ pairs, and one single person. That single person goes either in a normal room or a sea-view room, and there are $\binom{(n-1)/2+2}{2}$ compositions for the pairs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.