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Summation of general series One of the problems in Donald Knuth's Art of Programming is phrased as follows:
Find and prove a simple formula for the sum $$\sum\limits_{n=0}^k\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}.$$
I have very little experience with summations. My method was to try to find some function such that
$$\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}=f(x)-f(x-1).$$
However, I have no idea how to do that systematically. Fortunately, partial fraction decomposition happens to give a function in that form, with
$$f(x)=\frac{(-1)^x(x+1)}{4(x+1)^2+1}$$
which is the solution in the back of the book. I was wondering how you might go about finding this solution. This specific problem is volume 1, chapter 1.2.1, problem 11.
|
Since
$$
x^4+4=(x^2+2-2x)(x^2+2+2x),
$$
we get
$$
f(x)=\frac{x^3}{x^4+4}=\frac{1}{4}\frac{(x^2+2-2x+x^2+2+2x)^2}{(x^2+2-2x)(x^2+2+2x)}.
$$
Let $a(x)=4x^2+1$. Then
$$
4f(2n+1)=\frac{(a(n)+a(n+1))^3}{a(n)a(n+1)}=\frac{a(n)^2}{a(n+1)}+\frac{a(n+1)^2}{a(n)}+3(a(n)+a(n+1)).
$$
Hence
$$
\sum (-1)^nf(2n+1)=\sum (-1)^n\left\{ \frac{a(n)^2}{a(n+1)}+\frac{a(n+1)^2}{a(n)}+3(a(n)+a(n+1))\right\}
$$
As there is another answer appeared, I should stop here. Rearranging the sum, one can finish.
|
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|
Solve $y' = x^4y+x^4y^4$ Solve the differential equation $$y' = x^4y+x^4y^4.$$
I'm not sure how to deal with the $x^4y^4$ term. So far I have only encountered differential equations where the exponent of $y$ was at most one.
Could someone please share a general strategy/hint?
(This is test preparation, not homework.)
|
$$\frac{y'}{y+y^4}=x^4\tag{1}$$
and since
$$\int\frac{dz}{z+z^4}=\frac{1}{3}\log\frac{z^3}{1+z^3}\tag{2}$$
by partial fraction decomposition, by integrating both terms in $(1)$ we get:
$$\log\left(1+\frac{1}{y^3}\right)=-\frac{3}{5}x^5+C,\tag{3}$$
hence:
$$ 1+\frac{1}{y^3}= K e^{-\frac{3}{5}x^5},$$
$$ y= \frac{1}{\sqrt[3]{K e^{-\frac{3}{5}x^5}-1}}.\tag{4}$$
|
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|
solution of $y' + y^2 = \varphi^2(x)$ I need to solve differential equation in the interval $[-\pi/2,\pi/2]$
\begin{eqnarray}
y''(x) = y(x)\sin^2x
\end{eqnarray}
Trying $y(x) = \exp(\psi(x))$ yields,
\begin{eqnarray}
\zeta'(x) + \zeta^2(x) = \sin^2x \hspace{2cm} \zeta(x) = \psi'(x)
\end{eqnarray}
This equation seems to simpler than the original second order equation but
still I can't find way to solve this. Even if the equation is solved for function other than $\sin(x)$ with some important properties of $\sin(x)$ preserved, I will consider myself fortunate.
Is it possible to solve for any $\varphi(x)$ such that,
\begin{eqnarray}
\zeta'(x) + \zeta^2(x) = \varphi^2(x)
\end{eqnarray}
Where, $\varphi(x)$ is a monotonic function some interval $[a,b]$, with exactly one inflation point at $(a+b)/2$ and derivative vanishing at endpoints?
\begin{eqnarray}
\varphi'(x) >= 0 \\
\varphi''(x)|_{x=\frac{a+b}{2}} = 0\\
\varphi'(x)|_{x=a,b} = 0
\end{eqnarray}
|
Let
$$
v = y’
$$
then
$$
y’’ = \frac{dv}{dx} = \frac{dy}{dx} \frac{dv}{dy} = v \frac{dv}{dy}
\\ \frac{y''}{y} = \frac{v}{y} \frac{dv}{dy} = \frac{dv^2}{dy^2} = \sin^2{x}
$$
This might be used as the basis of a series solution. For example let
$$
v = (a_0 + a_1y + a_2y^2 + ...)^{-1}
$$
Then the LHS becomes
$$
\frac{dv^2}{dy^2} = \frac{v}{y} \frac{dv}{dy} =-\frac{(a_1 + 2a_2y + 3a_3y^2...)}{y(a_0 + a_1y + a_2y^2 + ...)^3}
$$
and
$$
v = \frac{dy}{dx}= (a_0 + a_1y + a_2y^2 + ...)^{-1}
$$
can be integrated to obtain
$$
x = a_0y + \tfrac{1}{2}a_1y^2 + \tfrac{1}{3}a_2y^3 + ...
$$
where the constant of integration has been set to $0$ to fix the $x$ origin where $y=0$, and $a_0$ is determined by the gradient of $y(x)$ at $x=0$:
$$
a_0 = 1/y'(0)
$$
We now have
$$
-\frac{(a_1 + 2a_2y + 3a_3y^2...)}{y(a_0 + a_1y + a_2y^2 + ...)^3} = \sin^2{(
a_0y + \tfrac{1}{2}a_1y^2 + \tfrac{1}{3}a_2y^3 + ...)}
$$
Expressing the RHS as a Taylor series and equating coefficients of $y^n$:
$$
a_1 =0, a_2 = 0, a_3= 0, a_4=-\tfrac{1}{4} a_0^5
$$
and so on.
So as a first approximation
$$
x \approx a_0y + \tfrac{1}{5}a_4y^5 = a_0y - \tfrac{1}{20}a_0^5y^5
$$
|
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|
How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$
show that
$$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$
it's well know that
$$(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e$$
But I found this $e=2.718>\dfrac{8}{3}=2.6666\cdots$
so how to prove this inequality by hand?
Thank you everyone solve it,I want see don't use $e=2.718$,because a most middle stundent don't know this value.
before I have use this well know
$$(1+\dfrac{1}{2n+1})(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e\cdot\dfrac{33}{34}\approx 2.638<\dfrac{8}{3}$$ to solve this, But Now we don't use $e=2.718$.
to prove this inequality by hand
|
\begin{align}
(1+\dfrac{1}{16})^{16} &= \sum_{k=0}^{16} {16 \choose k}(\frac{1}{16})^k \\
& = 2 + \frac{15}{32} + \frac{35}{256} + \sum_{k=4}^{16} {16 \choose k}(\frac{1}{16})^k \\
& \leq 2 + \frac{15}{32} + \frac{35}{256} +\sum_{k=4}^{16} \frac{1}{k!}\\
& \leq 2+ \frac{15}{32} + \frac{35}{256} + e - 1 - 1- \frac{1}{2} - \frac{1}{6}\\
& = e - \frac{2}{3} + \frac{155}{256} \\
& \leq 2.719 - 0.666 + 0.606 = 2.659
\end{align}
I used the fact ${n \choose k} \leq \dfrac{n^k}{k!}$ and $e \geq \sum_{k=0}^{16}\dfrac{1}{k!}$. In addition, $e< 2.719, \frac{2}{3} > 0.666, \frac{155}{256} < 0.606$
Added: for a proof which doesn't use the value of $e$, we could use
\begin{align}
\sum_{k=4}^{16} \frac{1}{k!} \leq \frac{1}{4!}(1 + \frac{1}{5} + \frac{1}{5\times6} +\frac{10}{5\times 6\times 7}) = \frac{269}{7!} < \frac{39}{6!}< \frac{7}{5!} = \frac{7}{120} < 0.06
\end{align}
Then we have $2 + \frac{155}{256} + \frac{7}{120} < 2 + 0.606 + 0.06 = 2.666$
|
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|
Quadratic formula and factoring are leading to different answers $$x^{ 2 }-2x-15=0$$
By factoring, I get:
$$(x-5)(x+3)$$
Which has the solutions:
$$x=5, x=-3$$
However when I use the quadratic formula (which is what the book saids to use), I get
$$\frac { 2 \pm \sqrt { 4-(4\cdot1\cdot(-15)) } }{ -2 } =$$
$$\frac { 2\pm 8 }{ -2 } $$
Which I evaluate to be $$x=-5, x=3$$
Where am I going wrong?
|
$$ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$\implies x^2-2x-15=0\implies x=\frac{2\pm\sqrt{(-2)^2-4\cdot1(-15)}}{2\cdot1}=\frac{2\pm8}2=5,-3$$
|
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|
Dividing by $\sqrt n$ Why is the following equality true?
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$
|
Suppose that $n>0$.
$$\frac{\sqrt{n}}{ \sqrt{n+\sqrt{n+\sqrt{n}}} } = \frac{1}{ \frac{1}{\sqrt{n}} \left( \sqrt{n+\sqrt{n+\sqrt{n}}} \right) } = \frac{1}{ \sqrt{1+ \frac{1}{n} \left( \sqrt{n + \sqrt{n}} \right) } } = \frac{1}{\sqrt{1+ \sqrt{ \frac{1}{n} + \frac{1}{n^2} \sqrt{n} } } } = \frac{1}{\sqrt{1+ \sqrt{ \frac{1}{n} + \sqrt{ \frac{1}{n^3}} } } } $$.
|
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|
How does $2^n + 2^n = 2^{n+1}$? What property of exponents can be used to show that
$$2^n + 2^n = 2^{n+1}$$
Does this work for all constants raised to a variable exponent?
|
I still remember trying to make my friends understand how we can take $2^n$ outside, when I was in middle school.
$$2^n + 2^n \\
= 2(2^{n-1} + 2^{n-1}) \\
= 2^2(2^{n-2} + 2^{n-2}) \\
= 2^3(2^{n-3} + 2^{n-3})\\
\dots\\ \dots
$$
$$=2^n(2^{n-n} + 2^{n-n})\\
= 2^n(2^0 + 2^0)\\
= 2^n(1 + 1)\\
= 2^n\cdot 2^1\\
= 2^{n+1}$$
But really, all they had to understand was that multiplication is repeated addition (something which they knew but didn't know how to apply):
$$a \times b = \sum^a b = \underbrace{b + b + b + \dots}_{a \text{ times}} $$
So, naturally,
$$2^n + 2^n = 2\times 2^n = 2^{n+1}$$
Now, there are an infinite number of similar equations to this one you find interesting:
$$ 3^n + 3^n + 3^n = 3^{n +1}\\
4^n + 4^n + 4^n + 4^n = 4^{n+1}\\
5^n + 5^n + 5^n + 5^n + 5^n = 5^{n+1}\\
\dots \\ \dots$$
Generally,
$$ \sum^a a^n = a^{n+1} \quad \forall\space n>0$$
Please note, $a^n + a^n = a^{n+1} $ is only true if $a = 0 ,2$ and $n>0$
$$[\because 0^n + 0^n = 0 + 0 = 0 = 0^{ \text{potatoes} } = 0^{ \text{tomatoes} } = 0^{n+1}]$$
|
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|
Finding the derivative of $y=12x^4\sqrt[3]{x^2}-2e^x+9$ Let
$$
y=12x^4\sqrt[3]{x^2}-2e^x+9
$$
How can we find $y^\prime$?
|
$$\require{cancel} y=12x^4\underbrace{\sqrt[3]{x^2}}_{x^\frac23}-2e^x+9=12x^\frac{14}3-2e^x+9\\\frac{dy}{dx}=\cancelto{4}{12}\left(\frac{14}{\cancel{3}}\right)(x^\frac{11}3)-2e^x+0\\=56x^\frac{11}3-2e^x$$
|
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|
Tricky geometry proof If a,b,c belong to the interval $(0,1)$ and $ab + ca + bc = 1$, prove that
$$\frac{a}{1-a^2} + \frac{b}{1-b^2} + \frac{c}{1-c^2}\ge\frac{3^{3/2}}{2}$$
How would you go about solving such a problem?
|
Another way: $(a+b+c)^2 \ge 3(ab+bc+ca) =3 \implies s=\dfrac{a+b+c}3 \ge \dfrac1{\sqrt3}$.
Now use $t \mapsto \dfrac{t}{1-t^2}$ is convex and Jensen's inequality to get
$$LHS \ge 3\frac{s}{1-s^2} \ge \frac{3\sqrt3}2=RHS$$
|
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|
Factorization of $z^4 +1 = (z^2 - \sqrt 2z+1)(z^2 + \sqrt 2 z+1)$ for complex $z$ How can I get this equation from LHS to RHS by using the four roots of $z^4 +1 = 0$ are $z=\pm\sqrt{\pm i}$
$$z^4 +1 = (z^2 - \sqrt2 z+1)(z^2 + \sqrt2 z+1)$$
|
If the roots are known then $x^4+1=(x+\sqrt i)(x-\sqrt i)(x+\sqrt{-i})(x-\sqrt{-i})$. Now difference of two squares will mean that $(x+\sqrt i)(x-\sqrt i)=x^2-i$ therefore multiplying with the other possible choices is a better idea. $(x+\sqrt i)(x+\sqrt{-i})$ gives $x^2+\sqrt{2}x+1$ therefore pairing the roots that way gives your form.
|
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|
How find the value $\beta$ such $\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^2}$ Find all positive real number $\beta$,there are infinitely many relatively prime integers $(p,q)$ such that
$$\left|\dfrac{p}{q}-\sqrt{2}\right|<\dfrac{\beta}{q^2}$$
maybe this problem background is Hurwitz's theorem:
$$\left|\sqrt{2}-\dfrac{p}{q}\right|<\dfrac{1}{\sqrt{5}q^2}$$
so I guess my problem ? $\beta\ge\dfrac{1}{\sqrt{5}}$
and this problem is Germany National Olympiad 2013 last problem (1), see: http://www.mathematik-olympiaden.de/aufgaben/52/4/A52124b.pdf
|
Proof sketch. First write
$$\left|\frac{p}{q}-\sqrt{2}\right| = \frac{|p^2-2q^2|}{q^2\left|\frac{p}{q} + \sqrt{2}\right|}$$
The Pell equation $$p^2 - 2q^2 = 1$$
is known to have infinitely many solutions (should be proven) so
$$\left|\frac{p}{q}-\sqrt{2}\right| = \frac{|p^2-2q^2|}{q^2\left|\frac{p}{q} + \sqrt{2}\right|} = \frac{1}{q^2}\frac{1}{\left|\sqrt{2 + \frac{1}{q^2}} + \sqrt{2}\right|}$$
for infinitely many $p,q$ so all $\beta \geq \frac{1}{\sqrt{8}}$ are possible. Now try to show that it fails for $\beta < \frac{1}{\sqrt{8}}$. Let $p^2 - 2q^2 = k$ then $p/q = \sqrt{2 + k/q^2}$ and by inserting this into the inequality above show that it cannot hold for large enough $q$.
|
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|
If $\sec \theta+\tan \theta= \sqrt{3}$ then the positive value of $\sin \theta$ If $\sec \theta+\tan \theta=\sqrt{3}$ then the positive value of $\sin \theta$
Note:
$1/\cos\theta+\sin\theta/\cos \theta=\sqrt{3}$
$\sin\theta=\sqrt{3}\cos \theta-1$
squaring on both sides we get
$\sin^2\theta=$
|
Since $1 = \sec^2 \theta - \tan^2 \theta = (\sec \theta + \tan \theta)(\sec \theta - \tan \theta)$, we have $\sec \theta - \tan \theta = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$.
Now, we have two equations:
(1) $\sec \theta + \tan \theta = \sqrt{3}$
(2) $\sec \theta - \tan \theta = \dfrac{\sqrt{3}}{3}$
Adding the two gives $\sec \theta = \dfrac{2\sqrt{3}}{3}$. Subtracting the 2nd from the 1st gives $\tan \theta = \dfrac{\sqrt{3}}{3}$.
Can you find $\sin \theta$ from this? Note that this method doesn't yield any extraneous solutions.
|
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|
Prove $a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$ if $a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$
if $a,b,c,d$ are positive real numbers and $$a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$$
Prove $a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$
Things i have done: from assumption $a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$ I can conclude that $$a^2+b^2-ab=c^2+d^2-cd$$
Powering both sides to two gives $$a^4+b^4+a^2b^2-2a^3b-2ab^3=c^4+d^4+c^2d^2-2c^3d-2cd^3$$
And $$a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4 \rightarrow 2a^4+2b^4+6a^2b^2-2ab^3-2a^3b=2c^4+2d^4+6c^2d^2-2cd^3-2c^3d$$
I can't continue any more.
|
Hint:
$$\left(a^2+b^2+(a-b)^2\right)^2=2\left(a^4+b^4+(a-b)^4\right).$$
I wish I knew a smart, non brute-force, way to show that.
|
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|
Require help with Inequality problems I am unable to find the solution for below Inequality problems.
1) $2/x<3$
The answer seems to be x belong to $(-\infty,0)\cup (2/3,\infty)$
2) $\dfrac{x+4}{x-3}<2$
The answer seems to be x belongs to $(-\infty,3)\cup (10,\infty)$
Progress
This is how I solved the second problem, but it does not seem to work.
$$x+4<2x-6$$
$$x-6>4$$
$$x>10$$
But of course it's incorrect
|
1) $$\frac{2}{x}<3$$
$\frac{2}{x}$ is not defined at $x=0$.
*
*For $x > 0$ :
$\frac{2}{x}<3 \Rightarrow 2<3x \Rightarrow x>\frac{2}{3}$
$\{x>0\} \cap \{x>\frac{2}{3} \}=\{x>\frac{2}{3}\}$
*
*For $x < 0$ :
$\frac{2}{x}<3 \Rightarrow 2>3x \Rightarrow x<\frac{2}{3}$
But since $\{x<\frac{2}{3}\} \cap \{x<0\}=\{x<0\}$, at this case the inequality stands for $x<0$.
Therefore the inequality stands for $x \in \left ( -\infty, 0 \right ) \cup \left (\frac{2}{3} , +\infty \right )$.
2) $$\frac{x+4}{x-3}<2$$
$\frac{x+4}{x-3}$ is not defined at $x=3$.
*
*For $x>3$ :
$(x-3)\frac{x+4}{x-3}<2(x-3) \Rightarrow x+4<2x-6 \Rightarrow x>10$
$\{x>3\} \cap \{x>10\}=\{x>10\}$
*
*For $x<3$ :
$(x-3)\frac{x+4}{x-3}>2(x-3) \Rightarrow x+4>2x-6 \Rightarrow x<10$
$\{x<3\} \cap \{x<10\}=\{x<3\}$
Therefore, the inequality stands for $x \in \left ( -\infty, 3 \right ) \cup \left (10, +\infty \right )$.
|
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|
Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we have $x^3-y^3\ge 0$ or $(x-y)(x^2+xy+y^2)\ge 0$. Since $x\ge y$ we can divide by $x-y$ to get $x^2+xy+y^2\ge 0$.
Is it right?
Thanks for your help!
|
Another approach: note that $-2|xy| \leq xy \leq 2|xy|$, so we have
$$x^2 - 2|xy| + y^2 \leq x^2 + xy + y^2 \leq x^2 + 2|xy| + y^2$$
or equivalently,
$$(|x|-|y|)^2 \leq x^2 + xy + y^2 \leq (|x| + |y|)^2$$
Since the left hand side is nonnegative, the result follows.
|
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|
What is the Hilbert curve's equation? The Hilbert curve has always bugged me because it had no closed equation or function that I could find. What is its equation or function? For example, if I wanted to find the Hilbert's curve point at 4/7, how would I find it?
|
As pointed out by almagest, there is a formula for Hilbert's space filling curve in Space-Filling Curves by Hans Sagan. The following formula appears as formula 2.4.3 on page 18 of the text.
If we write $t\in[0,1)$ in its base four expansion,
$$t=0_{\dot 4}q_1q_2q_3\ldots,$$
then
$$h(t) = \sum_{j=1}^{\infty} \frac{(-1)^{e_{0j}}}{2^j}\text{sign}(q_j)
\left(
\begin{array}{c}
(1-d_j)q_j - 1 \\
1-d_jq_j
\end{array}
\right),
$$
where $e_{kj}$ denotes the number of $k$s preceding $q_j$ and $d_j=e_{0j}+e_{3j} \mod 2$ and $\text{sign}(x)$ is $1$ if $x$ is positive, $0$ if $x$ is zero and $-1$ if $x$ is negative (though $q_j$ is never negative).
While it takes a bit to digest, it's not too outrageously complicated and quite easy to program. If the denominator of $t$ is a power of 2, then the base four expansion will terminate and the computation is exact. For any other rational number, the base four expansion is periodic and the value can still be computed exactly using a geometric series. For irrational numbers, the sum can be truncated to obtain a good decimal approximation.
Consider, for example, your question about $t=4/7$. We have the base four expansion
$$\frac{4}{7} = 0_{\dot 4}\overline{210}.$$
Now,
$$
h(0_{\dot 4}210) =
\left(
\begin{array}{c}
1/2 \\
1/2 \\
\end{array}
\right) + \left(
\begin{array}{c}
0 \\
1/4 \\
\end{array}
\right) + \left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) =
\left(
\begin{array}{c}
1/2 \\
3/4 \\
\end{array}
\right),
$$
$$
h(0_{\dot 4}000210) =
\left(
\begin{array}{c}
1/16 \\
1/16 \\
\end{array}
\right) + \left(
\begin{array}{c}
1/32 \\
0 \\
\end{array}
\right) + \left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) =
\left(
\begin{array}{c}
3/32 \\
1/16 \\
\end{array}
\right),
$$
$$
h(0_{\dot 4}000000210) =
\left(
\begin{array}{c}
1/128 \\
1/128 \\
\end{array}
\right) + \left(
\begin{array}{c}
0 \\
1/256 \\
\end{array}
\right) + \left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) =
\left(
\begin{array}{c}
1/128 \\
3/256 \\
\end{array}
\right),
$$
and
$$
h(0_{\dot 4}000000000210) =
\left(
\begin{array}{c}
1/1024 \\
1/1024 \\
\end{array}
\right) + \left(
\begin{array}{c}
1/2048 \\
0 \\
\end{array}
\right) + \left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) =
\left(
\begin{array}{c}
3/2048 \\
1/1024 \\
\end{array}
\right).
$$
Looking at the columns, it's not too hard to spot the pattern and we have
$$
h(4/7) =
\left(
\begin{array}{c}
1/2 \\
1/2 \\
\end{array}
\right) \sum _{k=0}^{\infty } \frac{1}{8^k}+\left(
\begin{array}{c}
1/32 \\
1/4 \\
\end{array}
\right) \sum _{k=0}^{\infty } \frac{1}{64^k} =
\left(
\begin{array}{c}
38/63 \\
52/63 \\
\end{array}
\right).
$$
Finally, here are some approximations to Hilbert's curve obtained by passing a polygonal curve the points $h(p/2^k)$ for increasing values of $k$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 0
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|
How to simplify $\sqrt{\tan^2\theta}\sin\theta\cos^2\theta$? I am solving $\int_{7}^{14}\frac{\sqrt{x^2-49}}{x^4}$ and got the integral down to $\frac{1}{343}\int_{0}^{\frac{\pi}{3}}\sqrt{\tan^2\theta}\sin\theta\cos^2\theta$ and wolfram simplified $\sqrt{\tan^2\theta}\sin\theta\cos^2\theta$ to $\sin^2(\theta)\cos(\theta)$ and I cannot figure out how to do this, thanks for all the help in advance.
|
Note that $\tan^2 \theta = (\tan\theta)^2$, and likewise for $\cos^2 \theta$, and on the interval of integration, we have that $\tan\theta \geq 0$.
$$\sqrt{\tan^2 \theta}\sin \theta \cos^2 \theta = \tan\theta\cdot \sin \theta \cos^2 \theta$$
$$= \frac{\sin\theta}{\cos\theta}\cdot \sin\theta \cos^2 \theta $$
$$ = \sin^2 \theta \cos\theta$$
Now put $u = \sin\theta \implies du = \cos \theta d\theta$ and use the power-rule to integrate.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
What is $\sqrt{-x^3}$? What is $\sqrt{-x^3}$, assuming $x \in \mathbb R$ and $x < 0$? It seems as if there are two possibilities:
$\sqrt{-x^3} = \sqrt{-x\times x \times x} = \sqrt{-x \times x^2} = x\sqrt{-x}$
$\sqrt{-x^3} = \sqrt{(-x)^3} = \sqrt{(-x)\times (-x)^2} = -x\sqrt{-x}$
But I get the feeling I'm not doing the math properly in one of them.
|
$\sqrt {-x^3} = \sqrt {-x}^3 = \sqrt {x(-1)}^3 = \sqrt {xi^2}^3 = \sqrt{x}^3 \times \sqrt{i^2}^3 = \sqrt {x}^3 \times i^3 = -i\sqrt{x}^3 = -i\sqrt{x^3}$
$\because i^3 = -\sqrt{-1} =$ {$-i : i = \sqrt {-1}$}, assuming we are proposing that $\sqrt {-x^3} = \sqrt {(-x)^3}$. Otherwise we would write the term as not $\sqrt {x(-1)}^3$ but instead $\sqrt {x^3(-1)}$ which is equal to $i\sqrt{x^3}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Question about G.C.D. Let,
$$a_{n}=n^2+20$$
$$d_{n}=\gcd(a_{n},a_{n+1})$$
where $n$ is a positive integer. Find the set of all values attained by $d_{n}$
I tried,
$d_{n}=\gcd(n^2+2n+21,n^2+20)$
$=\gcd(n^2+2n+21,2n+1)$
$=\gcd(n^2+20+2n+1,2n+1)$
$=\gcd(n^2+20,2n+1)$
However, after this I'm stuck.
Please Help!
Thanks!
|
If $d$ divides both $n^2+20,(n+1)^2+20$
$d$ must divide $(n+1)^2+20-(n^2+20)=2n+1$
$d$ must divide $2(n^2+20)-n(2n+1)=40-n$
$d$ must divide $2(40-n)+(2n+1)=81$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Proof - Area of a cyclic quadrilateral So we have a cyclic quadrilateral, as depicted below:
I have a conjecture that the area of this cyclic quadrilateral equals $$ \dfrac{\sqrt{(a+b+c-d)(a+b+d-c)(a+c+d-b)(b+c+d-a)}}{4} $$
I want to prove this. I know that the area of triangle ABC equals $\dfrac{1}{2}ab\sin(B)$ and the area of triangle ACD equals $\dfrac{1}{2}cd\sin(D)$. Seeing as $\sin(B) = \sin(D)$, I figured that the area of the quadrilateral equals $\dfrac{1}{2}(ab+cd)\sin(B)$.
But I'm stuck here. I have no clue how to get to the result I want. Can anyone provide an insight (and tell me whether or not my work up until now is valid)?
|
The Inscribed Angle Theorem shows that opposite angles of a cyclic quadrilateral are supplementary. Thus,
$$
D=\pi-B
$$
Therefore, $\cos(D)=-\cos(B)$ and $\sin(D)=\sin(B)$.
The Law of Cosines says
$$
e^2=\overbrace{a^2+b^2-2ab\cos(B)}^{\text{as a side of $\triangle ABC$}}=\overbrace{c^2+d^2+2cd\cos(B)}^{\text{as a side of $\triangle ADC$}}
$$
Solving for $\cos(B)$, we get
$$
\cos(B)=\frac12\frac{a^2+b^2-c^2-d^2}{ab+cd}
$$
Furthermore,
$$
\begin{align}
\frac14\sin^2(B)
&=\frac{(2ab+2cd)^2-(a^2+b^2-c^2-d^2)^2}{16(ab+cd)^2}\\
&=\frac{\left((c+d)^2-(a-b)^2\right)\left((a+b)^2-(c-d)^2\right)}{16(ab+cd)^2}\\
&=\frac{((c+d+a-b)(c+d+b-a))((a+b+c-d)(a+b+d-c))}{16(ab+cd)^2}\\
&=\frac{(s-a)(s-b)(s-c)(s-d)}{(ab+cd)^2}
\end{align}
$$
where $s=\frac{a+b+c+d}2$ is the semi-perimeter.
The area of ${\large\unicode{x23E5}}ABCD\ \ $ is the sum of the area of $\triangle ABC=\frac12ab\sin(B)$ and the area of $\triangle ADC=\frac12cd\sin(D)$
$$
\frac12(ab+cd)\sin(B)=\sqrt{(s-a)(s-b)(s-c)(s-d)}
$$
which, as Christian Blatter mentions, is called Brahmagupta's Formula.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Integral $\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}dx$ I have a problem with the following integral:
$$
\int_{0}^{1}\ln\left(\,3 + x \over 3 - x\,\right)\,
{{\rm d}x \over \,\sqrt{\,x\left(\,1 - x\,\right)\,}\,}
$$
The first idea was to use the integration by parts because
$$
\int{{\rm d}x \over \,\sqrt{x\left(\,1 - x\,\right)\,}\,}
=\arcsin\left(\,2x - 1\,\right) + C
$$
but what would be the next step is not clear. Another idea would be expand
$\ln\left(\,\cdot\right)$ into Taylor series but it seems to be even worse option.
So, what are the other options?
|
Let $x = \sin(t)^2$ and $s = 2t$, we have
$$\int_0^1 \log\left(\frac{3+x}{3-x}\right)\frac{dx}{\sqrt{x(1-x)}}
= \int_0^{\pi/2} \log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right)\frac{2\sin t\cos tdt}{
\sqrt{\sin(t)^2(1-\sin(t)^2)}}\\
= 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right) dt
= \int_0^{\pi}\log\left(\frac{3 + \frac{1-\cos s}{2}}{3-\frac{1-\cos s}{2}}\right) ds\\
= \int_0^{\pi}\left(\log(7-\cos s) - \log(5+\cos s)\right) ds
$$
Notice for any $a > 1$, we have
$$\frac{1}{\pi}\int_0^\pi \log(a \pm \cos s)ds = \cosh^{-1}(a) = \log\left(\frac{a + \sqrt{a^2-1}}{2}\right)\tag{*1}$$
The integral we desired is simply
$$\pi\left( \cosh^{-1}(7) - \cosh^{-1}(5)\right)
= \pi \log\left(\frac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\approx 1.072804016182156$$
I'm sure the identity in $(*1)$ has a name but I can't remember what it is. Let us
prove it!
Notice for any $b > 1$, the function $\log(b+z)$ is analytic over and inside the unit circle $S^1$ in $\mathbb{C}$.
By Residue theorem, we have
$$\frac{1}{2\pi i}\int_{S^1} \log(b + z) \frac{dz}{z} = \log(b)$$
If one parametrize the unit circle by $z = e^{i\theta}$, we get
$$\frac{1}{2\pi}\int_0^{2\pi} \log(b + e^{i\theta}) d\theta = \log(b)$$
Take the real part on both sides, this leads to
$$\begin{align}
&\frac{1}{2\pi}\int_0^{2\pi} \log(b^2 + 1 + 2b\cos\theta) d\theta = \log(b^2)\\
\iff &
\frac{1}{2\pi}\int_0^{2\pi} \log\left(\frac{b+b^{-1}}{2} + \cos\theta\right)d\theta = \log\left(\frac{b}{2}\right)\end{align}$$
Substitute $\displaystyle\;\frac{b+b^{-1}}{2}\;$ by $a$, we have
$\displaystyle\;\frac{b-b^{-1}}{2} = \sqrt{a^2-1}$ and it is clear $(*1)$ follows.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 1
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|
Compound interest problem with increasing deposits An Investor starts with an initial investment : $A$
He earns a steady profit of 10 percent per year.
But every year he adds additional amount which increases by 15 percent every year.
At the end of first year he adds an amount $x$
That means he adds 1.15$x$ at the end of 2nd year. And adds $(1.15)^{n-1}x$ at the end of $n$th year
What is amount of funds after $n$ years, $n>2$
|
If the initial amount is $A$, at the end of the first year the money amount is $\frac{11}{10}A+x$.
At the end of the second year, the money amount is:
$$\frac{11}{10}\left(\frac{11}{10}A+x\right)+\frac{23}{20}x,$$
while at the end of the $n$-th year it is:
$$\begin{eqnarray*}A\cdot\left(\frac{11}{10}\right)^n + x\cdot\sum_{k=1}^{n}\left(\frac{11}{10}\right)^k\left(\frac{23}{20}\right)^{n-k}&=&A\cdot\left(\frac{11}{10}\right)^n+x\cdot\left(\frac{23}{20}\right)^n\sum_{k=1}^{n}\left(\frac{22}{23}\right)^k\\&=&A\cdot\left(\frac{11}{10}\right)^n+22\,x\cdot\left(\frac{11}{10}\right)^n\left(\left(\frac{23}{22}\right)^n-1\right).\end{eqnarray*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/929468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Rewriting $1/(v^2-(mg/k))$ as two fractions I'm looking at the solution someone gave to me and I'm having a bit of trouble following one of the steps. This step in particular,
$\frac{1}{k}\frac{1}{v^2-\frac{mg}{k}}=\frac{1}{2k\sqrt{mg/k}}(\frac{1}{v-\sqrt{mg/k}}-
\frac{1}{v+\sqrt{mg/k}})$. All letters are constants except for $v$. Any help is appreciated.
|
$$\frac{1}{k}\frac{1}{v^2-\frac{mg}{k}}=\frac{1}{k}\frac{1}{v+\sqrt\frac{mg}{k}}\frac{1}{v-\sqrt\frac{mg}{k}}=$$
$$\ =\frac{1}{k}(\frac{A}{v+\sqrt\frac{mg}{k}}+\frac{B}{v-\sqrt\frac{mg}{k}})$$
$$\text{now you bring every term in the brackets under a common denominator}$$
$$\text{so you get:}\begin{cases}
A+B=0\\
-A+B= \sqrt\frac{k}{mg}
\end{cases} \iff \begin{cases}
A=-\frac{1}{2\sqrt\frac{mg}{k}}\\
B= \frac{1}{2\sqrt\frac{mg}{k}}
\end{cases} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/931188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$? Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align}$$
I was able to evaluate the limit as follows:
$$\begin{align}
\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ] &\sim \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{\dfrac x2 + \dfrac{x^2} 8}{\dfrac x2 + \dfrac{x^2} 8}} - 1\right ] =\\
&= \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{1 + \frac{2x^2}{4x - x^2}} - 1\right ] \sim\\
&\sim \lim_{x \to 0} \frac{2x^2}{12x^2 - 3x^3} = \frac 16
\end{align}$$
What are other ways to evaluate it? Maybe pure algebraically? I tried to rationalize the denominator, but got stuck at some point...
|
Multiply $\frac{1-\sqrt{1-x}}{\sqrt{x+1}-1}$ by $\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}$ and simplify to get $\frac{1+\sqrt{x+1}-\sqrt{1-x}-\sqrt{x+1} \sqrt{1-x}}{x}$
Expanding this, you only need keep two terms and you get
$$ \frac{1+\sqrt{x+1}-\sqrt{1-x}-\sqrt{x+1} \sqrt{1-x}}{x} = \frac{x+x^2/2 + \mathcal{O}(x^3)}{x}$$
then your limit becomes:
$$\lim_{x\rightarrow 0} \frac{1}{x}\left((1+x/2+\mathcal{O}(x^2))^{1/3}-1\right)=\lim_{x\rightarrow 0}\frac{1}{x}\left(x/6+\mathcal{O}(x^2)\right)=\frac{1}{6}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of the infinite series $\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \dots$ We can find the sum of infinite geometric series but I am stuck on this problem.
Find the sum of the following infinite series:
$$\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots$$
|
Using binomial expansion, we have:
$(1-x)^{-2/3} = 1 + \dfrac{\frac{2}{3}}{1!}x + \dfrac{\frac{2}{3} \cdot \frac{5}{3}}{2!} x^2 + \dfrac{\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3}}{3!}x^3 + \dfrac{\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3} \cdot \frac{11}{3}}{3!}x^4 + \cdots$
$(1-x)^{-2/3} = 1 + \dfrac{2}{3}x + \dfrac{2 \cdot 5}{3 \cdot 6} + \dfrac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}x^3 + \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}x^4 + \cdots$
Plug in $x = \dfrac{1}{2}$ to get:
$(1-\frac{1}{2})^{-2/3} = 1 + \dfrac{2}{3}\cdot\dfrac{1}{2} + \dfrac{2 \cdot 5}{3 \cdot 6}\cdot\dfrac{1}{2^2} + \dfrac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\cdot\dfrac{1}{2^3} + \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\cdot\dfrac{1}{2^4} + \cdots$
$2^{2/3} = 1 + \dfrac{2}{6} + \dfrac{2 \cdot 5}{6 \cdot 12}+ \dfrac{2 \cdot 5 \cdot 8}{6 \cdot 12 \cdot 18}+ \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{6 \cdot 12 \cdot 18 \cdot 24} + \cdots$
Finally, subtract $1$ and divide both sides by $2$ to get:
$\dfrac{2^{2/3} - 1}{2} = \dfrac{1}{6} + \dfrac{5}{6 \cdot 12}+ \dfrac{ 5 \cdot 8}{6 \cdot 12 \cdot 18}+ \dfrac{ 5 \cdot 8 \cdot 11}{6 \cdot 12 \cdot 18 \cdot 24} + \cdots$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is $5^2x^3-x^5 = x^3(x-5)(x+5)$ or $-x^3(5-x)(5+x)$ Geogebra's Factor function says that
$5^2x^3-x^5$
is
$-x^3(x-5)(x+5)$
but from what I do, it is positive, $x^3(5+x)(5-x)$
Note the x isnt in the same position
Am I wrong?
|
No, you're not wrong (but the title doesn't reflect the question):
$$
5^2x^3-x^5=x^3(5^2-x^2)=x^3(5-x)(5+x)
$$
You can use $5-x=-(x-5)$ to rewrite it as
$$
-x^3(x-5)(x+5)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/940607",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
"answer_count": 4,
"answer_id": 3
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|
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left(x^2+12\right)^3},\frac{3 \left(x^4+162
x^2+9477\right)}{\left(x^2+27\right)^4},\frac{4 \left(x^6+324 x^4+44928
x^2+2847744\right)}{\left(x^2+48\right)^5},\frac{5 \left(x^8+564 x^6+141750
x^4+19912500 x^2+1388390625\right)}{\left(x^2+75\right)^6},\frac{6 \left(x^{10}+900
x^8+366120 x^6+87829920 x^4+13038019200
x^2+998326798848\right)}{\left(x^2+108\right)^7}, \dots \right)$$
|
Given that $\frac{6\sqrt{3}}{\pi}\cdot\frac{1}{(x^2+3)^2}$ is the pdf of a random variable $X$, we have:
$$\varphi_X(t)=\mathbb{E}[e^{itX}]=(1+\sqrt{3}\,|t|)\,e^{-\sqrt{3}\,|t|}$$
hence the characteristic function of $Y=X_1+\ldots+X_n$, where $X_1,\ldots,X_n$ are independent random variables with the same distribution of $X$, is:
$$\varphi_Y(t) = (1+\sqrt{3}|t|)^n\,e^{-n\sqrt{3}\,|t|}$$
and the pdf of $Y$ is given by:
$$f_Y(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}(1+\sqrt{3}\,|t|)^n\,e^{-n\sqrt{3}\,|t|}e^{-itx}\,dt=\frac{1}{\pi}\int_{0}^{+\infty}(1+\sqrt{3}\,t)^n\,e^{-n\sqrt{3}\,t}\cos(tx)\,dt$$
that can be computed through the residue theorem or by integration by parts.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Is the union of all elliptic curves $\mathbb{R}^2$? An elliptic curve
could be written as
$$y^2 = x^3 + a x + b \;.$$
Q1. Is it the case that every point $p \in \mathbb{R}^2$ lies
on some elliptic curve?
Q2. And what is the natural generalization of this question to higher
dimensions $d$, $p \in \mathbb{R}^d$?
I ask these questions (obviously!) naively. Thanks for enlightening me!
|
Not sure about the second question, but here is my take on the first.
Let $p = (p_1, p_2)$.
Suppose $p_2^2 \neq p_1^3$, then $p$ lies on $y^2 = x^3 + (p_2^2 - p_1^3)$; note, here $\Delta = -432(p_2^2 - p_1^3)^2 \neq 0$, so $y^2 = x^3 + (p_2^2 - p_1^3)$ is non-singular.
If $p_2^2 = p_1^3$, then $p$ lies on $y^2 = x^3 + p_1x - p_1^2$, but $\Delta = -16(4p_1^3 + 27p_1^4) = -16p_1^3(4+27p_1)$ which is non-zero unless $p_1 = 0$ or $p_1 = -\frac{4}{27}$. However, as $p_1^3 = p_2^2$, $p_1^3 \geq 0$ so $p_1 \geq 0$ so $p_1 \neq -\frac{4}{27}$.
If $p_2^2 = p_1^3$ and $p_1 = 0$, then $p = (0, 0)$. As $p$ lies on $y^2 = x^3 + ax + b$, $b = 0$; note that $a$ can be chosen arbitrarily. For the curve $y^2 = x^3 + ax$, $\Delta = -64a^3$, so as long as $a \neq 0$, the curve is non-singular, so $y^2 = x^3 + x$ will do.
In summary, for every $p = (p_1, p_2) \in \mathbb{R}^2$ there is an elliptic curve $L_p$ such that $p \in L_p$. In particular, one choice of $L_p$ is
$$L_p = \begin{cases}
y^2 = x^3 + (p_2^2 - p_1^3) & \text{if}\ p_2^2 \neq p_1^3\\
y^2 = x^3 + p_1x - p_1^2 & \text{if}\ p_2^2 = p_1^3, p \neq (0, 0)\\
y^2 = x^3 + x & \text{if}\ p = (0, 0).
\end{cases}$$
|
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|
Multiple answer for integration of a function? Q. $\int \left(\frac{sin2x}{sin^4x+cos^4x}\right)\:dx$
My method:
$$\int \:\left(\frac{sin2x}{sin^4x+cos^4x}\right)\:dx=\int \:\:\left(\frac{sin2x}{\left(cos^2x-sin^2x\right)^2+2sin^2\left(x\right)cos^2\left(x\right)}\right)\:dx\:$$
=> $$\int \:\left(\frac{2sin2x}{2\left(cos^2\left(2x\right)\right)+sin^2\left(2x\right)}\right)=\int \:\left(\frac{2sin2x}{cos^2\left(2x\right)+1}\right)$$
Now let $t=cos\left(2x\right)$
therefore $-dt=2sin\left(2x\right)$
So we have the integration as:
$$\int \:\left(\frac{-dt}{t^2+1}\right)=-\tan ^{-1}\left(t\right)=-\tan \:^{-1}\left(cos\left(2x\right)\right)$$
But my text provides the answer as: $\tan \:^{-1}\left(\tan ^2\left(x\right)\right)$. Am I wrong somewhere?
|
$$\frac{\sin2x}{\sin^4x+\cos^4x}=2\frac{\tan x\sec^2x}{1+\tan^4x}$$
Set $u=\tan^2x$ to find $I=\arctan(\tan^2x)=\arctan\dfrac{1-\cos2x}{1+\cos2x}=\arctan1-\arctan(\cos2x)$
$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^22x}2=1-\frac{1-\cos^22x}2$$
Set $\cos2x=u$
|
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|
Equation simplification, can't get it right $$\frac{1}{\frac{x-1}{x+2}}-\frac{2}{x^2-1}$$ should be simplified into $$\frac{x^2+3x}{x^2-1} \quad .$$
However, when I try to do it (tried several times), I fail to get it done right:
$$\frac{1}{\frac{x-1}{x+2}}-\frac{2}{x^2-1} = 1 * \frac{x+2}{x-1} - \frac{2}{x^2-1} $$ $$=\frac{x(x+2)-2}{x^2-1} = \frac{x^2+2x-2}{x^2-1}$$
What am I doing wrong?
Thanks a lot for the help
|
$$\begin{align}\frac{x+2}{x-1}-\frac{2}{x^2-1}&=\frac{x+2}{x-1}-\frac{2}{(x-1)(x+1)}\\&=\frac{(x+2)(x+1)}{(x-1)(x+1)}-\frac{2}{(x-1)(x+1)}\\&=\frac{(x+2)(x+1)-2}{(x-1)(x+1)}\\&=\frac{x^2+3x}{x^2-1}.\end{align}$$
|
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Partitions without twice odd numbers and where every odd number appears at most once Let $A=\{2,6,10,14,\ldots\}$ be the set of integers that are twice an odd number.
Prove that, for every positive integer $n$, the number of partitions of $n$ in which no odd number appears more than once is equal to the number of partitions of $n$ containing no element of $A$.
I can't seem to find the generating function for either of these.
|
So I'm currently solving this problem with the AOPS intermediate counting and probability course, and if you are too, you should probably use the message board but here is an explanation for the two values:
We start by making a generating function for partitions when no odd number appears more than once.
For the even numbers that will be in the partition, we can first form the power series for each individual number. For $2,$ we have $((x^2)^0+(x^2)^1+(x^2)^2+(x^2)^3+ \dots)$ where the highest power represents the number of $2s$ there are, so in $(x^2)^3,$ there will be a $2+2+2$ in our partition. Similarly for $4$ we have $((x^4)^0+(x^4)^1+(x^4)^2+(x^4)^3+ \dots).$ And for $6$ we have $((x^6)^0+(x^6)^1+(x^6)^2+(x^6)^3+ \dots).$ And so on and so forth for all even numbers.
Multiplying the power series together we have $((x^2)^0+(x^2)^1+(x^2)^2+(x^2)^3+ \dots) \cdot ((x^4)^0+(x^4)^1+(x^4)^2+(x^4)^3+ \dots) \cdot ((x^6)^0+(x^6)^1+(x^6)^2+(x^6)^3+ \dots) \cdot \dots$
Then for the odd numbers in our partition, we are limited by the problem condition: no odd number appears more than once. Thus, $1$ can only appear $0$ times or $1$ time so $((x^1)^0+(x^1)^1$, $3$ can also only appear $0$ or $1$ times, giving us $((x^3)^0+(x^3)^1).$ This applies to all odd numbers so we have:
$$((x^1)^0+(x^1)^1) \cdot ((x^3)^0+(x^3)^1) \cdot ((x^5)^0+(x^5)^1) \dots.$$
Now all we need to do is to multiply the two functions. So our generating function for partitions of any number when no odd number appears more than once is done:
\begin{align*}
\left(((x^2)^0+(x^2)^1+(x^2)^2+(x^2)^3+ \dots) \cdot ((x^4)^0+(x^4)^1+(x^4)^2+(x^4)^3+ \dots) \cdot ((x^6)^0+(x^6)^1+(x^6)^2+(x^6)^3+ \dots) \cdot \dots \right) \cdot \left( ((x^1)^0+(x^1)^1) \cdot ((x^3)^0+(x^3)^1) \cdot ((x^5)^0+(x^5)^1) \dots \right)
&= \left( (1+x^2+x^4+x^6+ \dots)(1+x^4+x^8+x^{12}+ \dots)(1+x^6+x^{12}+x^{18}+\dots) \dots \right) \cdot \left( (1+x)(1+x^3)(1+x^5) \dots \right) \\
&= \prod_{n=1}^{ \infty } \frac{1}{1-x^{2k}} \cdot \prod_{n=0}^{ \infty } (1+x^{2k+1}). \\
\end{align*}
Now we need to make the generating function for when the partition contains no element of $A.$
To do this, we can take the generating function for partitions with no restrictions, and then remove the power series for numbers that are part of set $A.$
Thus we have $$\frac{ (1+x+x^2+ \dots) \cdot (1+x^2+x^4+ \dots) \cdot (1+x^3+x^6+ \dots) \cdot (1+x^4+x^8+ \dots) \cdot (1+x^5+x^10+ \dots) \cdot (1+x^6+x^{12}+ \dots) \dots}{(1+x^2+x^4+ \dots) \cdot (1+x^6+x^{12}+ \dots) \dots} = \prod_{n=1}^{ \infty } \frac{1}{1-x^n} \prod_{n=0}^{ \infty } (1-x^{4n+2}). $$
|
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Is it true that $\left\lfloor\sum_{s=1}^n\operatorname{Li}_s\left(\frac 1k \right)\right\rfloor\stackrel{?}{=}\left\lfloor\frac nk \right\rfloor$ While studying polylogarithms I observed the following.
Let $n>0$ and $k>1$ be integers. Is the following statement true?
$$\left\lfloor \sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right) \right\rfloor \stackrel{?}{=} \left\lfloor \frac{n}{k} \right\rfloor $$
If it is, then how could we prove it? If not, give a counterexample.
|
We use the series expansion for the polylogarithm.
Note that
$$\sum_{s=1}^{n}\operatorname{Li_{s}}\left(\frac{1}{k}\right) =
\sum_{s=1}^{n}\sum_{j=1}^{\infty}\frac{1}{j^sk^j} = \sum_{j=1}^{\infty}\sum_{s=1}^{n}\frac{1}{j^{s}k^j}$$
we are able to interchange the summations because each term is positive, which implies that the sum is absolutely convergent.
Now separating out the first term gives
$$\frac{n}{k} + \sum_{j=2}^{\infty}\sum_{s=1}^{n}\frac{1}{j^sk^j}$$
We may then use the formula for a geometric sum to bring this to
$$\frac{n}{k} + \sum_{j=2}^{\infty}\frac{1}{j}\frac{1 - \frac{1}{j^{n+1}}}{\left(1 - \frac{1}{j}\right)k^j} = \frac{n}{k} + \sum_{j=2}^{\infty}\frac{j^{n+1} - 1}{j^{n+1}\left(j-1\right)k^j} = $$
$$\frac{n}{k} + \sum_{j=2}^{\infty}\frac{1}{\left(j-1\right)k^j} - \sum_{j=2}^{\infty}\frac{1}{j^{n+1}\left(j-1\right)k^j}$$
Now since clearly
$$\sum_{j=2}^{\infty}\frac{1}{\left(j-1\right)k^{j}} > \sum_{j=2}^{\infty}\frac{1}{j^{n+1}\left(j-1\right)k^j}$$
we have
$$\frac{n}{k} < \sum_{s=1}^{n}\operatorname{Li}_{s}\left(\frac{1}{k}\right) <
\frac{n}{k} + \sum_{j=2}^{\infty}\frac{1}{\left(j-1\right)k^j} =
\frac{n}{k} + \frac{1}{k}\sum_{j=1}^{\infty}\frac{1}{jk^j} =
\frac{n}{k} + \frac{\log\left(\frac{k}{k-1}\right)}{k}$$
where we have used one of the series for $\log$ given here. The series is valid since $k \geq 2$.
Thus
$$\frac{n}{k} < \sum_{s=1}^{n}\operatorname{Li}_{s}\left(\frac{1}{k}\right) < \frac{n}{k} + \frac{\log(k) - \log(k-1)}{k}$$
Now for $n$ and $k$ natural numbers, the smallest possible difference between
$\left \lfloor \frac{n}{k}\right \rfloor + 1$ and $\frac{n}{k}$ is $\frac{1}{k}$.
Thus the result will hold if
$$\frac{\log(k) - \log(k-1)}{k} < \frac{1}{k}$$
or in other words if
$$\log(k) - \log(k-1) < 1$$
It's easy to check that $\log(x) - \log(x-1)$ is monotonically decreasing for $x > 1$. Also $\log(2) - \log(1) = \log(2) < 1$. It follows that the result holds for all natural numbers $k \geq 2$.
|
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Evaluate this square root $\sqrt{6 + 2\sqrt{5}} + \sqrt{6 - 2\sqrt{5}}$
I have no clue where to begin. I would appreciate a hint, the answer should be
$2\sqrt{5}$
In general, how do you evaluate
$\sqrt{a + b} + \sqrt{a - b}$?
Thanks!
|
Set $r=\sqrt{6 + 2\sqrt{5}} + \sqrt{6 - 2\sqrt{5}}$ and observe that $r>0$; then
\begin{align}
r^2
&=6+2\sqrt{5}+2\sqrt{(6 + 2\sqrt{5})(6 - 2\sqrt{5})} + 6 - 2\sqrt{5}\\
&=12+2\sqrt{36-20}\\
&=12+2\sqrt{16}\\
&=12+8=20
\end{align}
Thus $r=\sqrt{r^2}=\sqrt{20}=2\sqrt{5}$.
More generally, if $a>b$ and $r=\sqrt{a+b}+\sqrt{a-b}$, then
$$
r^2=2a+2\sqrt{a^2-b^2}
$$
which can be simplified if $a^2-b^2$ is a perfect square.
There is the general formula
$$
\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}
$$
with its companion
$$
\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}
$$
(supposing $a>0$ and $a^2-b\ge0$), but it's in general better to look for a perfect square; in $6+2\sqrt{5}$ you have a double product $2\sqrt{5}$, so writing $6=1+5$ seems the best:
$$
6+2\sqrt{5}=1+2\sqrt{5}+(\sqrt{5})^2=(1+\sqrt{5})^2
$$
|
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Trigonometric Integral $\int _0^{\pi/4}\:\frac{dx}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$ Q. $$\int _0^{\pi/4}\:\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$$
My method:
$$\int_0^{ \pi /4} \frac{dx}{(\cos^2x+\sin^2x)^2-3\cos^2x \sin^2x}=\int_0^{ \pi /4} \frac{dx}{1-3\cos^2x\sin^2x} $$
Dividing numerator and denominator by $\cos^2x$ we have:
$$\int_0^{\pi /4}\frac{\sec^2x}{\sec^2x-3\tan^2x}dx=\int_0^{\pi /4}\frac{\sec^2x}{1-2\tan^2x}dx=\int _0^1 \frac{dt}{1-2t^2}=\int _0^1 \frac{1}{2}\frac{dt}{\frac{1}{2}-t^2}=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}t}{1+\sqrt{2}t}\right|_0^1=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}}{1+\sqrt{2}}\right|$$
But when we do the same integration by dividing the initial term by $\sec^4x$ and solving it yields an answer $$\frac{\pi }{2}$$
Am I wrong somewhere?
|
Basically, you make a change of variable $t=\tan(x)$; doing so, you have $$\int _0^{\frac{\pi }{4}}\:\left(\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}\right)\:dx=\int _0^1\frac{t^2+1}{t^4-t^2+1}dt$$ and $$\int\frac{t^2+1}{t^4-t^2+1}dt=\tan ^{-1}\left(\frac{t}{1-t^2}\right)$$
I must say that I do not see where the $3$ disappeared.
|
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Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$.
$$z^{10}+\frac{1}{z^{10}}=?$$
For the solution:
*
*the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
*converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ and $z_2=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}$
*How do I proceed?
|
$z^2+z+1 = 0 \implies z^2+z = -1 $
$z^2= -z-1 \implies z^3 = -z^2-z = -(z^2+z)= 1$
$z^{10} = z^3\cdot z^3\cdot z^3\cdot z= z \implies$
$z^{10} + z^{-10} = z + 1/z = \frac{z^2 +1 }z$ but $z^2+1 = -z$ so this results in $-z/z = -1$
|
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|
Divisibility property of $(a+b)^n-a^n-b^n$ Let $n$ be a natural number of the form $n=6k+1$ (while $k$ is a positive integer).
Show that $(a^2+ab+b^2)^2$ divides $(a+b)^n-a^n-b^n$ for all integer numbers $a,b$ (such that $a^2+ab+b^2\ne0$).
|
if $\omega$ is a complex cube root of unity, then both $\omega$ and $1+\omega$ are sixth roots of unity. hence for $k \ge 1$$\omega$ is a root of:
$$
(x+1)^{6k+1} - x^{6k+1} - 1 = 0 \tag{1}
$$
likewise for $\omega^2$
but these are exactly the roots of
$$
x^2+x+1=0
$$
this shows that $a^2+ab+b^2$ is a factor of $(a+b)^{6k+1}-a^{6k+1}-b^{6k+1}$
however the derivative of the LHS of (1) is $(6k+1)\left((x+1)^{6k}-x^{6k}\right)$
which is also satisfied by $\omega, \omega^2$, so each is, in fact, a double root of the equation (1). hence the result
|
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Theory number problem I need to prove that there are infinitely many natural numbers $n$ for which $2n^2+3$ and $n^2+n+1$ are relatively prime.
This is not true for every $n$ (for example, $n=4$), I tried to check for odd $n$ but can't find way to prove it (I tried induction). Any ideas?
|
Any common divisor of $2n^2+3$ and $n^2+n+1$ divides $2(n^2+n+1)-(2n^2+3)$, which is $2n-1$.
Any common divisor of $2n^2+3$ and $2n-1$ divides $2n^2+3-n(2n-1)$, which is $n+3$.
And any common divisor of $2n-1$ and $n+3$ divides $7$.
Now produce infinitely many numbers $2n^2+3$ that are not divisible by $7$.
Remark: We have in essence used the Euclidean Algorithm for polynomials. If we chase down details, we find that the gcd is $1$ except when $n\equiv 4\pmod{7}$.
|
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Why does $ \frac{b^n-a^n}{b-a}=\sum_{k=1}^nb^{n-k}a^{k-1}$? Trying to work through the answer in this question:
The inequality $b^n - a^n < (b - a)nb^{n-1}$
|
First consider a sample of the equation to be shown. Consider
\begin{align}
a^{2} + ab + b^{2} &= \frac{(b-a)(b^{2} + ab + a^{2})}{b-a} \\
&= \frac{1}{b-a} \, [ b^{3} -a b^{2} + a b^{2} - a^{2} b + a^{2} b - a^{3} = \frac{b^{3}- a^{3}}{b-a}.
\end{align}
Now apply the same logic to the form
\begin{align}
b^{n-1} + a b^{n-2} + \cdots + a^{n-2} b + a^{n-1}
\end{align}
for which the result is
\begin{align}
\frac{b^{n} - a^{n}}{b-a} = \sum_{k=0}^{n-1} b^{n-k-1} \, a^{k}.
\end{align}
|
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Is this the correct procedure for Integral Partial Fraction. $∫ (x^3+x^2+x+3)/((x^2+1)(x^2+3))$
The first step I did is distributed the denominator so that I can find out if I should use synthetic division. Which after doing this I discovered that the denominator's exponent is greater therefore I do not use synthetic division
= $∫ (x^3+x^2+x+3)/(x^4+4x^2+3)$ Therfore I solve.
$∫ (x^3+x^2+x+3)/((x^2+1)(x^2+3))$ =∫ $(Ax +B)/(x^2+1) + (Cx+D)/(x^2+3)$ then I distributed $(x^2+1)(x^2+1)$ on both sides and got.
$∫ (x^3+x^2+x+3) $= $(Ax +B)(x^2+3)+( Cx+D)(x^2+1)$
Then I solve for B.
x=0
$0+0+0+3 = (A(0)+B)(0+3)+(0+0)(0+1) =$
3=B If this is right how does one find A,CD? Do I make x =0 and B =3 and e verything else 0?
|
We do indeed have $$(Ax +B)(x^2+3)+( Cx+D)(x^2+1) = x^3 + x^2 + x + 3$$
If you put $x=0$, you get $$(A(0) +B)(0 + 3) + (C(0) + D)(0 + 1) = 3 \iff 3B + D = 3$$
Try using $x = (-1)$ and $x = 1$, and apply the same process.
Then use any other constant that hasn't been used yet, say $x = 2$, using the same process. That will guarantee that you have four equations to solve for four unknowns: A, B, C, D.
|
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Mathematical Induction Question, Proof Help Prove using Mathematical Induction that for all natural numbers ($n>0$):
$$
\frac 1 {\sqrt{1}} + \frac 1 {\sqrt{2}} + \cdots + \frac 1 {\sqrt{n}} \ge \sqrt{n}.
$$
Proof by Induction:
Let P(n) denote 1/ √1 + 1/ √2 + … + 1/ √n ≥ √n
Base Case: n = 1, P(1) = 1/√1 ≥ √1
The base cases holds true for this case since the inequality for P(1) holds true.
Inductive Hypothesis: For every n = k > 0 for some integer k
P(k) = 1/ √1 + 1/ √2 + … + 1/ √k ≥ √k, p(k) holds true for any integer k
Inductive Step:
P(k + 1)) = 1/ √1 + 1/ √2 + … + 1/ √k + 1/ √(k + 1) ≥ √k + √(k+1)
√k + √(k+1) > √(k+1) (this is where I got stuck)
|
You know that
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}} \geq \sqrt{k}$$$$
and want to prove that:
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k+1}} \geq \sqrt{k+1}$$$$
Add $\sqrt{k+1}-\sqrt{k}$ to both sides of first inequality, you get:
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}} +\sqrt{k+1}-\sqrt{k}\geq \sqrt{k+1}$$
But:
$$\sqrt{k+1}-\sqrt{k}=\frac{(\sqrt{k+1}+\sqrt{k})(\sqrt{k+1}-\sqrt{k})}{(\sqrt{k+1}+\sqrt{k})}=\frac{k+1-k}{(\sqrt{k+1}+\sqrt{k})}=\frac{1}{(\sqrt{k+1}+\sqrt{k})}\leq \\ \leq \frac{1}{\sqrt{k+1}}$$
So:
$$\sqrt{k+1} \leq \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}} +\sqrt{k+1}-\sqrt{k} \leq \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k+1}}$$
|
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|
A long nasty limit problem: $\lim_{x \to \infty}\left[8e\,\sqrt[\Large x]{x^{x+1}(x-1)!}- 8x^2-4x \ln x - \ln^2 x - (4x + 2 \ln x) \ln 2\pi\right]$ Does the following limit admit a closed-form?
$$\lim_{x \to \infty}\left[8e\,\sqrt[\Large x]{x^{x+1}(x-1)!}- 8x^2-4x \ln x - \ln^2 x - (4x + 2 \ln x) \ln 2\pi\right]$$
My professor gives this question as a challenge problem in the end of calculus class but I can't solve it. I tried to use a Stirling's approximation for the first term and I got
$$8e\,\sqrt[\Large x]{x^{x+1}(x-1)!} = 8e\cdot x\sqrt[\Large x]{x!} \sim 8 x^2\cdot \sqrt[\Large 2x]{2\pi x}$$but I don't know how to use it to solve the original problem. Any idea? Any help would be appreciated. Thanks in advance.
|
Taking the logarithm of the first term divided by $ 8e $:
$$ \frac 1 x\left[x\ln x + \ln\left(x!\right)\right], $$
Stirling-approximate the second term to get
$$ \ln\left(x!\right) = x\ln x - x + \frac 1 2\ln x + \frac 1 2 \ln 2\pi + \frac 1 {12x} + \mathcal{O}(x^{-1}). $$
The first term is therefore
$$\begin{align} &8\exp\left[1 + 2 \ln x - 1 + \frac 1 2\frac{\ln x}{x} + \frac 1 2 \frac{\ln 2\pi}{x} + \frac 1 {12x^2} + \mathcal{O}(x^{-2})\right]\\
&=8x^2 \cdot \exp\left[\frac 1 2\frac{\ln x}{x} + \frac 1 2 \frac{\ln 2\pi}{x} + \frac 1 {12x^2} + \mathcal{O}(x^{-2})\right];
\end{align}$$
now expand the exponential to order $ \mathcal{O}(x^{-2}) $ and you get
$$\begin{align} &8x^2 \bigg[1 + \bigg(\frac 1 2\frac{\ln x}{x} + \frac 1 2 \frac{\ln 2\pi}{x} + \frac 1 {12x^2}\bigg) +\frac 1 2\bigg(\frac 1 2\frac{\ln x}{x} + \frac 1 2 \frac{\ln 2\pi}{x} \bigg)^2 + \mathcal{O}(x^{-2}) \bigg]\\
& =8x^2 \bigg[1 + \frac 1 2\frac{\ln x}{x} + \frac 1 2 \frac{\ln 2\pi}{x} + \frac 1 {12x^2} +\frac 1 8\frac{\ln^2 x}{x^2} + \frac 1 8 \frac{\ln^2 2\pi}{x^2} + \frac 1 4 \frac{\ln x\ln 2\pi}{x^2} + \mathcal{O}(x^{-2}) \bigg]\\
& = 8x^2 +4x\ln x + 4x\ln 2\pi + \frac 2 {3} + \ln^2 x
+ \ln^2 2\pi + 2\ln x\ln 2\pi + \mathcal{O}(1),
\end{align}$$
which cancel term by term with the rest of the expression to give $ \ln^2 2\pi +\dfrac{2}{3} $.
|
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|
Evaluate $\int_{0}^{\infty} \mathrm{e}^{-x^2-x^{-2}}\, dx$ I have to find
$$I=\int_{0}^{\infty} \mathrm{e}^{-x^2-x^{-2}}\, dx $$
I think we could use
$$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2} $$ But I don't know how.
Thanks.
|
Consider
\begin{align}
x^{2} + \frac{1}{x^{2}} = \left( x - \frac{1}{x} \right)^{2} +2
\end{align}
for which
\begin{align}
I = \int_{0}^{\infty} e^{-\left(x^{2} + \frac{1}{x^{2}}\right)} \, dx = e^{-2} \, \int_{0}^{\infty} e^{-\left(x - \frac{1}{x}\right)^{2}} \, dx.
\end{align}
Now make the substitution $t = x^{-1}$ to obtain
\begin{align}
e^{2} I = \int_{0}^{\infty} e^{- \left( t - \frac{1}{t} \right)^{2}} \, \frac{dt}{t^{2}}.
\end{align}
Adding the two integral form leads to
\begin{align}
2 e^{2} I = \int_{0}^{\infty} e^{- \left( t - \frac{1}{t} \right)^{2}} \left(1 + \frac{1}{t^{2}} \right) \, dt = \int_{-\infty}^{\infty} e^{- u^{2}} \, du = 2 \int_{0}^{\infty} e^{- u^{2}} \, du = \sqrt{\pi},
\end{align}
where the substitution $u = t - \frac{1}{t}$ was made. It is now seen that
\begin{align}
\int_{0}^{\infty} e^{-\left(x^{2} + \frac{1}{x^{2}}\right)} \, dx = \frac{\sqrt{\pi}}{2 e^{2}}.
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}
\end{align}
I know the final answer is $\frac{9}{2}$. After multiplying by the conjugate, I see where the $9$ in the numerator comes from. I just can't remember how I solved the rest of the problem.
|
$\bf{My\; Solution::}$ Given $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-7x}\right) = \lim_{x\rightarrow \infty}x\left\{\left(1+\frac{2}{x}\right)^{\frac{1}{2}}-\left(1-\frac{7}{x}\right)^{\frac{1}{2}}\right\}$
Now Using Binomial Expansion::
$\displaystyle \lim_{x\rightarrow \infty}\left\{\left(1+\frac{1}{2}\cdot \frac{2}{x}+\frac{1}{2}\cdot -\frac{1}{2}\cdot \frac{4}{2x^2}......+\infty\right)-\left(1-\frac{1}{2}\cdot \frac{7}{x}-\frac{1}{2}\cdot -\frac{1}{2}\cdot \frac{49}{2x^2}.+\infty\right)\right\}$
So $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-7x}\right) = 1+\frac{7}{2} = \frac{9}{2}$
|
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|
Evaluate $ \int_0^1 \sum_{k=0}^\infty (-x^4)^k dx = \int_0^1 \frac{dx}{1+x^4} $ I have read this thread and I found in some comments the above named equality. I couldn't follow the transformation, which are done to get from the left to the right side at that point in particular.
Can someone help me and show how it's done?
|
On the left, reverse the order of summation and integration to get
$$\sum_{k=0}^{\infty} (-1)^k \, \int_0^1 dx \, x^{4 k} = \sum_{k=0}^{\infty} \frac{(-1)^k}{4 k+1} = 1 - \frac15 + \frac19 - \frac1{13} + \cdots$$
This sum is equal to
$$\int_0^1 \frac{dx}{1+x^4}$$
To evaluate this, observe that
$$1+x^4 = (1+\sqrt{2} x + x^2)(1-\sqrt{2} x+x^2)$$
so that we may use partial fractions in the integrand as follows:
$$\frac1{1+x^4} = \frac{A x+B}{x^2-\sqrt{2} x+1} + \frac{C x+D}{x^2+\sqrt{2} x+1}$$
where one may deduce that $A=-1/(2 \sqrt{2})$, $B=1/2$, $C=1/(2 \sqrt{2})$, and $D=1/2$. We rewrite the resulting expression for ease of integration:
$$\begin{align}\int_0^1 \frac{dx}{1+x^4} &= \frac1{2} \int_0^1 dx \, \left (\frac{-\frac1{\sqrt{2}} \left (x-\frac1{\sqrt{2}} \right )+\frac12}{\left (x-\frac1{\sqrt{2}} \right )^2+\frac12} + \frac{\frac1{\sqrt{2}} \left (x+\frac1{\sqrt{2}} \right )+\frac12}{\left (x+\frac1{\sqrt{2}} \right )^2+\frac12} \right ) \\ &= \frac1{4 \sqrt{2}} \left [\log{\left (\frac{1+\frac1{\sqrt{2}}}{1-\frac1{\sqrt{2}}} \right )} + 2 \arctan{(\sqrt{2}-1)} + 2 \arctan{(\sqrt{2}+1)} \right ] \\ &= \frac1{4 \sqrt{2}} \left [\log{\left (\frac{1+\frac1{\sqrt{2}}}{1-\frac1{\sqrt{2}}} \right )} + \pi\right ]\end{align}$$
Thus,
$$\begin{align}\sum_{k=0}^{\infty} \frac{(-1)^k}{4 k+1} = 1 - \frac15 + \frac19 - \frac1{13} + \cdots &= \frac1{4 \sqrt{2}} \left [\pi+\log{\left (\frac{1+\frac1{\sqrt{2}}}{1-\frac1{\sqrt{2}}} \right )} \right ]\\ &=\frac1{4 \sqrt{2}} \left [\pi + 2 \log{\left (\sqrt{2}+1\right )} \right ]\end{align}$$
ADDENDUM
I may as well illustrate how to get the sum of another, similar series almost for free from the above series. Note that
$$\int_1^{\infty} \frac{dx}{1+x^4} = \int_0^1 dx \frac{x^2}{1+x^4} = \sum_{k=0}^{\infty} \frac{(-1)^k}{4 k+3} $$
However,
$$\begin{align}\int_1^{\infty} \frac{dx}{1+x^4} &= \int_0^{\infty} \frac{dx}{1+x^4} -\int_0^{1} \frac{dx}{1+x^4} \end{align} $$
The first integral on the right may be evaluated using a contour integral and the residue theorem. I prefer to use a quarter-circle in the upper-right quadrant; the only enclosed pole is at $z=e^{i \pi/4}$. By the residue theorem, we have
$$(1-i) \int_0^{\infty} \frac{dx}{1+x^4} = \frac{i 2 \pi}{4 e^{i 3 \pi/4}} \implies \int_0^{\infty} \frac{dx}{1+x^4} = \frac{\pi}{2 \sqrt{2}}$$
Thus,
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{4 k+3} = \frac13 - \frac17 + \frac1{11} - \frac1{15} + \cdots = \frac1{4 \sqrt{2}} \left [\pi - 2 \log{\left (\sqrt{2}+1\right )} \right ] $$
|
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|
Calculating Moments The following problem is from a Schaum book on statistics. While I thought I did it right, I did not come up with the right answer. Therefore, I am thinking I did something wrong.
Problem: Find (a) the moment generating function of the random variable
$ x = \begin{cases}
\, \frac{1}{2} &\text{ prob. 1/2} \\
-\frac{1}{2} &\text{ prob. 1/2} \\
\end{cases}
$
and (b) the first four moments about the origin.
Answer:
The moment generating function for $x$ is:
$M_x(t) = E(e^{tx})$
Since $x$ can have only two values, this gives us the following function:
$
M_x(t) = \frac{1}{2}e^{\frac{t}{2}} + \frac{1}{2}e^{-\frac{t}{2}}
$
Now to find the first four moments about the origin, I compute the first four
derivatives of $M_x(t)$.
$ M_x'(t) = \frac{1}{4}e^{\frac{t}{2}} - \frac{1}{4}e^{-\frac{t}{2}} $
$ M_x''(t) = \frac{1}{8}e^{\frac{t}{2}} + \frac{1}{8}e^{-\frac{t}{2}} $
$ M_x'''(t) = \frac{1}{16}e^{\frac{t}{2}} - \frac{1}{16}e^{-\frac{t}{2}} $
$ M_x''''(t) = \frac{1}{32}e^{\frac{t}{2}} + \frac{1}{32}e^{-\frac{t}{2}} $
Now, let $u_1, u_2, u_3, u_4$ be the first four moments of $x$.
$ u_1 = M_x'(0) = \frac{1}{4}e^{\frac{0}{2}} - \frac{1}{4}e^{-\frac{0}{2}} $
$ u_1 = \frac{1}{4} - \frac{1}{4} = 0 $
$ u_2 = M_x''(0) = \frac{1}{8}e^{\frac{0}{2}} + \frac{1}{8}e^{-\frac{0}{2}} $
$ u_2 = \frac{1}{8} + \frac{1}{8} = \frac{1}{4} $
$ u_3 = M_x'''(0) =
\frac{1}{16}e^{\frac{0}{2}} - \frac{1}{16}e^{-\frac{0}{2}} $
$u_3 = \frac{1}{16} - \frac{1}{16} = 0 $
$ u_4 = M_x''''(0) =
\frac{1}{32}e^{\frac{0}{2}} + \frac{1}{32}e^{-\frac{0}{2}} $
$ u_4 = \frac{1}{32} + \frac{1}{32} = \frac{1}{16} $
However, the books answer is:
$ u_1 = 0 $
$ u_2 = 1 $
$ u_3 = 0 $
$u_4 = 1 $
I do not understand what I am doing wrong and I am hoping that somebody here
can tell me.
Thanks
Bob
|
You work seems correct. For example, without using the moment generating function
$$
E[X^2] = (1/2)^2 (1/2) + (-1/2)^2(1/2) = 1/8+1/8 = 1/4.
$$
It seems the book chose to change the random variable to $\tilde{x}$ where $$\tilde{x} = \begin{cases} 1 & \text{ with probability $1/2$} \\ -1 & \text{ with probability $1/2$} \end{cases}$$
|
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|
$\Sigma_{k=m+1}^{\infty}\frac{1}{10^k} \leq \frac{1}{10^m}$ Is it true that $\Sigma_{k=m+1}^{\infty}\frac{1}{10^k} \leq \frac{1}{10^m}$ for $m \in \mathbb{N}$?
|
$$\sum\limits_{k=m+1}^\infty \dfrac{1}{10^k} = \dfrac{1}{10^{m+1}}\sum\limits_{\ell=0}^\infty \dfrac{1}{10^\ell} = \dfrac{1}{10^{m+1}} \cdot \frac{1}{1-1/10} = \dfrac{1}{10^{m+1}}\cdot \frac{10}{9} = \frac{1}{9 \cdot 10^m} < \dfrac{1}{10^m}$$
|
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|
Solving $2\cos2x-4\sin x\cos x=\sqrt{6}$ How I solve the following equation for $0 \le x \le 360$:
$$
2\cos2x-4\sin x\cos x=\sqrt{6}
$$
I tried different methods. The first was to get things in the form of $R\cos(x \mp \alpha)$:
$$
2\cos2x-2(2\sin x\cos x)=\sqrt{6}\\
2\cos2x-2\sin2x=\sqrt{6}\\
R = \sqrt{4} = 2 \\
\alpha = \arctan \frac{2}{2} = 45\\
\therefore \cos(2x + 45) = \frac{\sqrt6}{2}
$$
which is impossible. I then tried to use t-substitution, where:
$$
t = \tan\frac{x}{2}, \sin x=\frac{2t}{1+t^2}, \cos x =\frac{1-t^2}{1+t^2}
$$
but the algebra got unreasonably complicated. What am I missing?
|
You have made a mistake in the third step.
$$
2\cos2x-2(2\sin x\cos x)=\sqrt{6}\\
2\cos2x-2\sin2x=\sqrt{6}\\
R = \sqrt{8} = 2\sqrt{2} \\
\alpha = \arctan -\frac{2}{2} = -\frac{\pi}{4}\\
\therefore \cos(2x - \frac{\pi}{4}) = \frac{\sqrt3}{2}
$$
Hope this help.
|
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|
How to simplify $\sqrt[3]{29\sqrt{2}-45}-\sqrt[3]{29\sqrt{2}+45}$ I in trouble simplifying this:
$$\sqrt[3]{29\sqrt{2}-45}-\sqrt[3]{29\sqrt{2}+45}$$
couldn't find a solution. Can you help?
|
HINT:
Let $$\sqrt[3]{29\sqrt{2}-45}-\sqrt[3]{29\sqrt{2}+45}=a$$
$$a^3=29\sqrt{2}-45-(29\sqrt{2}+45)-3a(-7)$$
as $(29\sqrt{2}-45)(29\sqrt{2}+45)=-7^3$
$$\iff a^3-21a+90=0$$ whose only real root is $-6$
|
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|
Calculate this sum $\sum_{x=1}^{{n}/{4}} \frac{1}{x}$ How can I calculate this sum
$$\sum_{x=1}^{{n}/{4}} \frac{1}{x}$$
for $n$ an odd integer?
Added later: For example, WolframAlpha outputs 4 - Pi/2 - Log[8] for HarmonicNumber[1/4], what are the steps that produce that? Can they be generalized for any odd number?
|
The series could be defined by
\begin{align}
S_{m} = \sum_{n=1}^{\lfloor m/4 \rfloor} \frac{1}{n}
\end{align}
where $\lfloor x \rfloor$ represents the largest integer of $x$. Now, consider the series
\begin{align}
H_{n} = \sum_{k=1}^{n} \frac{1}{k}
\end{align}
which are known as the Harmonic numbers. From this it is seen that the series in question is given by
\begin{align}
S_{m} = H_{\lfloor m/4 \rfloor}.
\end{align}
Alternatively, since $\psi(x+1) + \gamma = H_{x}$ then $S_{m} = \gamma + \psi(\frac{m}{4} + 1)$, where $\gamma$ is the Euler-Mascheroni constant and $\psi(x+1)$ is the digamma function.
Examples of values:
\begin{align}
S_{4} &= \sum_{n=1}^{\lfloor 4/4 \rfloor} \frac{1}{n} = 1 \\
S_{5} &= \sum_{n=1}^{\lfloor 5/4 \rfloor} \frac{1}{n} = \sum_{n=1}^{1} \frac{1}{n} = 1.
\end{align}
Now, making use of
$$J_{m} = \sum_{k=1}^{m/4} \frac{1}{k} = \gamma + \psi\left( \frac{m+4}{4} \right)$$
then
\begin{align}
J_{1} = \gamma + \psi\left( \frac{5}{4} \right) = 4 + \gamma + \psi\left( \frac{1}{4} \right)
\end{align}
It is known that $\psi\left(\frac{1}{4} \right) = - \gamma - \frac{\pi}{2} - 3\ln 2$ which leads to the result
\begin{align}
J_{1} = 4 - \frac{\pi}{2} - 3 \ln 2.
\end{align}
For the case of $S_{3}$ it is seen that
\begin{align}
J_{3} &= \gamma + \psi\left( \frac{7}{4} \right) = 4 + \gamma + \psi\left( \frac{3}{4} \right) = 4 + \frac{\pi}{2} - 3 \ln 2.
\end{align}
|
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|
Code is not cyclic for any q I have code $C$ over $F_p$ with generator matrix which looks like
$G =
\begin{pmatrix}
0 &0& 0& 1& 0& 1& 1 &1\\
1& 0 &0& 0 &1 &0 &1& 1\\
1& 1& 0& 0& 0& 1& 0& 1\\
1 &1& 1& 0 &0 &0& 1 &0\end{pmatrix}$
I need to show that this code is not cyclic for any $p$.
I constructed vector which looks like
$$(c_2+c_3+c_4, c_3+c_4, c_4, c_1, c_2, c_1+c_3, c_1+c_2+c_4, c_1+c_2+c_3)$$ and its shifts but it didn't help me a lot.
Does somebody have any ideas how it can be proven?
|
A base for your code is a base for $\ker G$. So the code has dimension 4, and a base is
$$
v_1=\begin{pmatrix}0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ -1\end{pmatrix}
\quad v_2=\begin{pmatrix}-1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0\end{pmatrix}
\quad v_3=\begin{pmatrix}0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0\end{pmatrix}
\quad v_4=\begin{pmatrix}0 \\ 0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0\end{pmatrix}
$$
The code was cyclic if and only if the base, shifted by one, is still a base of the code, but
$$
v_2=\begin{pmatrix}-1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0\end{pmatrix}
\quad v_3=\begin{pmatrix}0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0\end{pmatrix}
\quad v_4=\begin{pmatrix}0 \\ 0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0\end{pmatrix}
\quad v_1=\begin{pmatrix}0 \\ 0 \\ 0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1\end{pmatrix}
$$
Is a base, so it is cyclic for every p
|
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|
how to solve $\int\frac{1}{1+x^4}dx$ i want find the answer and method of solve of $\int\frac{1}{1+x^4}dx$.
I know $$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$$,
How I can use this to solve of that integration.
|
Note the following:
*
*$2 = (1 + x^{2}) + (1-x^{2})$.
*$\displaystyle \int \frac{1}{1+x^{4}} = \frac{1}{2} \int\frac{2}{1+x^{4}} = \frac{1}{2} \int\frac{(1+x^{2})+(1-x^{2})}{1+x^{4}} = \frac{1}{2} \int\frac{1+x^{2}}{1+x^{4}} + \frac{1}{2}\int\frac{1-x^{2}}{1+x^{4}} = \frac{1}{2} I_{1} + \frac{1}{2}I_{2}$.
*$\displaystyle I_{1} =\int\frac{1+x^{2}}{1+x^{4}} = \int\frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}} =\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2} +2} = \int\frac{1}{t^{2}+2}.$
*Note. We have made the substituion $t =x - \frac{1}{x}$. The integral $I_{2}$ can be evaluated in the same manner.
|
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|
How do I solve this fraction addition problem? $4\frac{2}{9} + -9\frac{1}{2}$ yeilds result of $-5\frac{13}{18}$ but WolframAlpha says the answer is $-5\frac{5}{18}$
fixed.
|
In mixed fractions (just as in decimal ones), the sign applies to both whole and fractional parts, so
$$4\frac{2}{9} + -9\frac{1}{2}=4\frac{2}{9}-\left(9+\frac{1}{2}\right)=(4-9)+\left(\frac{2}{9}-\frac{1}{2}\right)$$
Since
$$4-9=-5$$
and
$$\frac{2}{9}-\frac{1}{2}=\frac{2\times2}{9\times2}-\frac{1\times9}{2\times9}=\frac{4-9}{18}=\frac{-5}{18}$$
the final answer is
$$-5+\frac{-5}{18}=-\left(5+\frac{5}{18}\right)=-5\frac{5}{18}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find inverse laplace transform $$
F(s) = \dfrac{6s+9}{s^2-10s+29}
$$
How do you solve the inverse Laplace transform of this above equation?
|
$$
\underbrace{s^2-10s+29 = (s-5)^2+2^2}_{\text{completing the square}} = t^2+2^2
$$
So
$$
\frac{6s+9}{s^2-10s+29} = \frac{6(s-5)+39}{(s-5)^2+2^2} = \frac{6t+39}{t^2+2^2} = 6\frac{t}{t^2+2^2} + \frac{39}{2}\cdot\frac{2}{t^2+2^2}
$$
Now look at each term separately.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solve for $\int \sqrt{x}(\sqrt{x}-2x)^2 dx $ $\int \sqrt{x}(\sqrt{x}-2x)^2 dx $ so I solved this using U-substitution where $u= \sqrt{x}$ so my $du2\sqrt{x}=dx$ then it will be $2 \int u^2(u-2u^2)^2$ and just expand then distribute the $u^2$ so i got $\frac{8u^7}{7}-\frac{4u^6}{3}+\frac{2u^5}{5}+c$ then just substitue $\sqrt{x}$ to all of the $u$ is that right? My friend was debating with me with the $du=dx$ thing he said that it's suppose to be equal. Like in this equation where $\int x^2\sqrt{x^3-5}$ the $du=dx$ here is $du = 3x^2dx$ so it will be $\frac{du}{3}=x^2dx$. He keeps on debating that U substitution SHOULD be equal like in my second example. Where he has the $x^2dx$ really in the equation. Is this true? can't I really just equal it to $dx$ like $u= \sqrt{x}$ $du2\sqrt{x}=dx$
|
Not answering your question, but another way to solve is with partial integration, is to observe that $$\sqrt{x}(\sqrt{x}-2x)^2=\sqrt{x}\left(\sqrt{x}(1-2\sqrt{x})\right)^2=\sqrt{x}\cdot x\cdot(\sqrt{x}-2x)^2=x\dfrac{(1-2\sqrt{x})^2}{\sqrt{x}}$$
Now $$\left((1-2\sqrt{x})^3\right)'=3(1-2\sqrt{x})^2\dfrac{-2}{2\sqrt{x}}=-3\dfrac{(1-2\sqrt{x})^2}{\sqrt{x}}$$ Hence the given integral can be written as $$-\dfrac{1}{3}\int x \cdot\left((1-2\sqrt{x})^3\right)'dx$$ and the result follows with partial integration.
Concerning your question, no your friend is not correct. There is no rule stipulating the equality that he insists on.
|
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|
Integration $I_n=\int_{0}^{1}\frac{dx}{(x^n+1)(\sqrt[n]{x^n+1})}$ $$I_n=\int_{0}^{1}\frac{dx}{(x^n+1)\large\sqrt[n]{\normalsize x^n+1}}$$
Could someone help me through this problem?
|
Let $\displaystyle v=v(x)=\left(\frac{x^n+1}{x^n}\right)^{\Large \frac{1}{n}}$.
Then we have: $\displaystyle v(0)=+\infty \ , \ v(1)=2^{\large \frac{1}{n}} \ , \ \frac{1}{x^n+1} = 1 - v^{-n} \ , \ x=\frac{1}{\left(v^n-1\right)^{\large \frac{1}{n}}}$ and $$\displaystyle dx = - \frac{v^{n-1}}{\left(v^n-1\right)^{1+\large \frac{1}{n}}} dv$$
Thus,
$$\displaystyle \int_0^1{\frac{dx}{(x^n+1)\large \sqrt[n]{x^n+1}}} = \int_{2^{\large \frac{1}{n}}}^{+\infty} \left(1 - v^{-n}\right)^{\large\frac{n+1}{n}} \frac{v^{n-1}}{\left(v^n-1\right)^{1+\large \frac{1}{n}}} dv =\int_{2^{\large \frac{1}{n}}}^{+\infty} \frac{dv}{v^2} = 2^{\large -\frac{1}{n}}$$
Please doublecheck my working!
|
{
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|
Trigonometry graphs sinusoidal waves i need help on this questions. I couldn't figure how to determine for both question A and B.
But i have the answers for them, i just don't understand how the amplitude is 3 and so on.
|
To find the amplitude, take half the difference of the maximum and minimum values. In this case, the amplitude is
$$a = \frac{1 - (-5)}{2} = \frac{6}{2} = 3$$
If you subtract the amplitude from the maximum value, you will find the average value. In this case, the average value of the function is $1 - 3 = -2$. If there were no vertical shift, the average value of the function would be zero. Therefore, the vertical shift is
$$d = 0 - (-2) = 2$$
Since the sine function assumes its average value at $0$, there is no phase shift. Hence, $c = 0$.
Observe that the graph shows that $3/4$ of a period is $1/2$. Hence, one period is
$$\frac{\dfrac{1}{2}}{\dfrac{3}{4}} = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}$$
Since $\sin t$ has period $2\pi$, you can find the frequency by setting $bt = 2\pi$, where $t$ is the period of the sinusoidal function. Here $t = 2/3$. Thus,
\begin{align*}
\frac{2}{3}b & = 2\pi\\
b & = 2\pi \cdot \frac{3}{2}\\
b & = 3\pi
\end{align*}
Substituting for $a$, $b$, and $d$ yields $x = 3\sin(3\pi t) - 2$.
We can check this by substituting $0$ and $0.5$ for $t$. If $t = 0$, then
$$x = 3\sin(0) - 2 = -2$$
If $t = 0.5$, then
$$x = 3\sin\left(\frac{3\pi}{2}\right) - 2 = 3(-1) - 2 = -3 - 2 = -5$$
Both values agree with the graph.
|
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|
Solve $\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$
Find the set of integer solutions $(x,y)$ to
$$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$$
I can't seem to multiply both sides by the expression in the denominator.
Nor can I simplify and cancel any terms. How should I it?
|
$$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y \iff \frac{x+y}{x-y}> x+y$$
Case 1: $x+y > 0$
In this case, we have $\dfrac1{x-y}> 1 \iff 0 < x-y< 1$ which is not possible.
Case 2: $x+y < 0$
In this case, we have $\dfrac1{x-y} < 1 \iff x-y < 0$ or $x-y > 1$.
Now $x+y < 0, \; x-y < 0$ gives $x < 0, \;x < y < -x$
and $x+y < 0, \; x-y > 1$ gives $y < 0, \;1+y< x < -y$
Hence we can write the possible solutions also as integer pairs satisfying $\{|y| < -x\} \cup \{|x| < -y-\frac12\}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}$ for $0 < x < 1$ I stumbled upon this question while doing practice inequalities questions, and I do not know how to start...
Problem: Prove that
\begin{align*}
\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}
\end{align*}
for $0 < x < 1$.
I thought possibly of having an intermediate equality, for example
\begin{align*}
\sqrt{\frac{2x^2-2x+1}{2}}\ge\text{something}\ge\frac{1}{x+\frac{1}{x}}
\end{align*}
where the "something" is simple, but I could not deduce anything...any help would be appreciated, thanks!
|
This inequality is true for all $x>0$. Indeed$,$
$$ \frac{(x+\frac{1}{x})^2(2x^2-2x+1)}{2}-1$$
$$={\frac {2 \ \left( 2\ x-1 \right) ^{2}{x}^{2}+ \left( x-1
\right) ^{4} \left( x+1 \right) ^{2}+{x}^{2} \left( x-1 \right) ^{2}
\left( x+1 \right) ^{2}}{2{x}^{2}}} \geqq 0$$
|
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|
For the series $S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2$...... Problem :
For the series $$S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2+\frac{1}{(1+3+5+7)}(1+2+3+4)^2+\cdots $$ Find the nth term of the series.
We know that nth can term of the series can be find by using $T_n = S_n -S_{n-1}$
$$S_n =1+ \sum \frac{(\frac{n(n+1)}{2})^2}{(2n-1)^2}$$
$$\Rightarrow S_n =\frac{n^4+5n^2+2n^3-4n+1}{(2n-1)^2}$$
But I think this is wrong, please suggest how to proceed thanks..
|
Consider the series
\begin{align}
S_{n} = 1 + \frac{(1+2)^{2}}{1+3} + \frac{(1+2+3)^{2}}{1+3+5} + \cdots + \frac{(1+2+\cdots+n)^{2}}{1+3+\cdots+(2n-1)}.
\end{align}
This series is seen as
\begin{align}
S_{n} &= 1 + \frac{1}{2^2}\binom{3}{2}^{2}+ \frac{1}{3^{2}} \binom{4}{2}^{2}+ \cdots + \frac{1}{n^{2}} \binom{n+1}{2}^{2} \\
&= \sum_{r=1}^{n} \frac{1}{r^{2}} \binom{r+1}{2}^{2} \\
&= \frac{1}{4} \sum_{r=1}^{n} (r+1)^{2} = \frac{1}{4} \sum_{r=2}^{n+1} r^{2} \\
&= \frac{1}{4} \left[ -1 + \frac{(n+1)(n+2)(2n+3)}{6} \right] \\
&= \frac{n}{24} \left(2 n^{2} + 9 n + 13 \right)
\end{align}
Making use of this formula it is quickly seen that
\begin{align}
S_{1} &= 1 \\
S_{2} &= 1 + \frac{3^{2}}{2^{2}} = 1 + \frac{(1+2)^{2}}{(1+3)}
\end{align}
|
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|
Prove that $\int_0^\infty \frac{\ln x}{x^n-1}\,dx = \Bigl(\frac{\pi}{n\sin(\frac{\pi}{n})}\Bigr)^2$ This question inspired me to ask the following.
Prove that
$$I_n = \int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2,$$
for $\Re(n)>1$.
For some cases there is a nice specific form of $I_n$. For example
$$\begin{align}
I_2 & = \frac{\pi^2}{4} \\
I_3 & = \frac{4\pi^2}{27} \\
I_4 & = \frac{\pi^2}{8} \\
I_5 & = \frac{2\left(5+\sqrt 5\right)\pi^2}{125} \\
I_6 & = \frac{\pi^2}{9} \\
I_7 & = \frac{2\pi^2}{49\left(1-\sin\left(\frac{3\pi}{14}\right)\right)} \\
I_8 & = \frac{\left(2+\sqrt 2\right)\pi^2}{32}
\end{align}$$
|
$$
\begin{align}
\int_0^\infty\frac{\log(x)}{x^n-1}\mathrm{d}x
&=\int_{-\infty}^\infty\frac{x}{e^{nx}-1}e^x\,\mathrm{d}x\\
&=\int_0^\infty\frac{x}{e^{nx}-1}e^x\,\mathrm{d}x+\int_0^\infty\frac{x}{1-e^{-nx}}e^{-x}\,\mathrm{d}x\\
&=\int_0^\infty x(e^{(1-n)x}+e^{(1-2n)x}+e^{(1-3n)x}+\dots)\,\mathrm{d}x\\
&+\int_0^\infty x(e^{-x}+e^{(-1-n)x}+e^{(-1-2n)x}+\dots)\,\mathrm{d}x\\
&=\frac1{(n-1)^2}+\frac1{(2n-1)^2}+\frac1{(3n-1)^2}+\dots\\
&+1+\frac1{(n+1)^2}+\frac1{(2n+1)^2}+\frac1{(3n+1)^2}+\dots\\
&=\frac1{n^2}\sum_{k\in\mathbb{Z}}\frac1{\left(k+\frac1n\right)^2}\\
&=\frac{\pi^2}{n^2}\csc^2\left(\frac\pi{n}\right)
\end{align}
$$
where the last step uses the derivative of $(7)$ from this answer
|
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|
Proving $a^ab^bc^c\ge(abc)^{(a+b+c)/3}$ for positive real numbers. Prove that
$$a^ab^bc^c\ge(abc)^{(a+b+c)/3}$$
where $a,b,c\in\mathbb{R^+}$
I tried using powered AM-GM but didn't get anything. please give me a hint to solve it.
|
Take the $\log$ of both sides. This is equivalent to:
$$
a\log a + b\log b + c\log c \ge \frac{a+b+c}3 (\log a + \log b + \log c)
$$
now use the rearrangement inequality twice to get:
$$
a\log a + b\log b + c\log c \ge b\log a + c\log b + a\log c ;\\
a\log a + b\log b + c\log c \ge c\log a + a\log b + b\log c;\\
a\log a + b\log b + c\log c = a\log a + b\log b + c\log c.
$$
The sum of these inequalities is the right result.
Alternative:
$$
\frac{a+b+c}3 \frac{\log a + \log b + \log c}3
\le \frac{a+b+c}3 \log \frac{a+b+c}3
$$using the concavity of $\log$;
$$
\le \frac{a\log a + b\log b + c\log c }3
$$using convexity of $x\to x\log x$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Number of $0$ in the end of $11^n-1$ $n$ is integer, calculate number of $0$ in the end of $11^n-1$(i.e. largest integer $m$ such that $10^m|11^n-1$).
The original question was $n=100$ and I could only choose $m$ from 1 to 5. I calculated the $11^{100}-1$ mod $100000$ and got the answer, I'm looking for much better and general approach.
|
The binomial expansion used for $n=100$ in the answer to the linked question can also be applied to general $n$. Let us set
$$11^n-1= (10+1)^n-1= \sum_{i=0}^n {n \choose i} 10^i -1$$
The sum can be expanded as
$$ 11^n-1= 10 \cdot n+ 100 \cdot \frac{n(n-1)}{2!}+ 1000 \cdot \frac{n(n-1)(n-2)}{3!}....$$
Now, as correctly hypothesized by Martingalo in his comment, if $n$ is not a multiple of $10$ it is clear that the sum is divisible only by $10$ and not by higher powers of $10$. So, the quantity $11^n-1$ ends with only one zero.
On the other hand, if $n$ is a multiple of $10$, we can write it in the form $n=j \cdot 10^k$ (where $j$ is not divisible by $10$ and then $k$ is the number of times that the factor $10$ is contained in $n$). Then, in the sum above:
*
*the first term $10 \cdot n$ contains the factor $10$ exactly $k+1$ times;
*the second term $100 \cdot \frac{n(n-1)}{2!}$ also contains the factor $10$ exactly $k+1$ times (in fact, multiplying $n$ by the odd number $n-1$ and dividing by $2$ we get that the binomial coefficient ${n \choose 2}$ contains the factor $10$ one time less as compared to $n$, i.e. $k-1$ times);
*the third term $1000 \cdot \frac{n(n-1)(n-2)}{3!}$ contains the factor $10$ exactly $k+3$ times (in fact, looking again at the binomial coefficient, we can note that multiplying $n$ by the even number $(n-1)(n-2)$ and dividing by $2 \cdot 3$, we clearly obtain a number that contains the factor $10$ the same number of times of $n$, i.e. $k$ times;
*all successive terms contain the factor $10 $ at least $4$ times.
Therefore, taking only the first and second term of the sum, we have
$$10 \cdot n+ 100 \cdot \frac{n(n-1)}{2!}=10n+50n^2-50n=50n^2-40n=10n(5n-4)$$
and reminding that $n$ is of the form $n=j \cdot 10^k$ with $j$ not divisible by $10$, we get
$$10n(5n-4)=10^{k+1} j(5j\cdot 10^k-1)$$
which clearly contains the factor $10$ exactly $k+1$ times. So, in this case, the whole expansion of $11^n-1$ also contains the factor $10$ exactly $k+1$ times, and then ends with $k+1$ zeros.
|
{
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|
Show that complex numbers are vertices of equilateral triangle 1)Show if $|z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$ then $z_1,z_2,z_3$ are vertices of equilateral triangle inscribed in a circle of radius.
I thought I can take use from roots of unity here, since $|z_1|=|z_2|=|z_3|=1$ they lie at circle at radius $1$ but I don't know how to take advantage from $z_1+z_2+z_3=0$
2)Let $z=\cos\alpha+i\sin\alpha$ where $\alpha \in 0,2\pi$ then find $\arg(z^2-z)$
I come to this siutation $\displaystyle z^2-z=-2\sin{\frac{1}{2}x}(\sin{\frac{3}{2}x}+i\cos{\frac{3}{2}x})=-2\sin{\frac{1}{2}x}(\cos(\frac{\pi}{2}-{\frac{3}{2}x})+i\sin({\frac{\pi}{2}-\frac{3}{2}x}))$ so $\displaystyle 0\le\frac{\pi}{2}-\frac{3}{2}x\le2\pi$ so $\displaystyle\frac{\pi}{3}\ge x \ge - \pi$ so $\displaystyle\arg(z^2-z) =[-\pi,\frac{\pi}{3}]$ ???
|
Let: $z_1 =e^{ia} ; z_2 = e^{ib}; z_3 = e^{ic}$
$ z_1 +z_2 = e^{i\frac{a+b}{2}}*(e^{i\frac{a-b}{2}} + e^{-i\frac{(a-b)}{2}}) = e^{i\frac{a+b}{2}}*2*cos(\frac{a-b}{2}) = -z_3 $
=> $|2*cos(\frac{a-b}{2})| = |-z_3| = |z_3| = 1$ ,
If $ cos(\frac{a-b}{2}) =\frac{1}{2} $ -> $a = b \pm \frac{2\pi}{3}$ $mod(2\pi)$
here without loss of generality you can assume a= b+ $\frac{2\pi}{3}$ $ mod(2\pi)$ (the other case is the same)
you get : $\frac{a+b}{2} = c+\pi$ $ mod(2\pi)$ -> b+ $\frac{\pi}{3} = c + \pi$ $ mod(2\pi)$ -> $ b = c + \frac{2\pi}{3} $ $ mod(2\pi)$
You get your equilateral triangle, since you proved that you can rotate of $\frac{2\pi}{3}$ to pass from one point to another. The other cases are exactly the same.
As for 2) , I would use : $z= e^{ia}$
$z^2 - z = e^{2ia} - e^{ia}$ = $e^{\frac{3}{2}ia}*2i*sin(\frac{a}{2}) $ = $ e^{(\frac{3}{2}a + \frac{\pi}{2})i}*2*sin(\frac{a}{2}) $. The sign of the sin is the only thing you have take into account to evaluate correctly the argument. If it is negative, you add $\pi$, else you already have your argument
|
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|
Prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1)$ Using induction
prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1) \forall n \in \mathbb{N}$
Attempt:
Let $n =1$ so $3(1)-2 = 1$ and $\frac{1}{2}(3(1)-1)=1$
Assume true at $n=k$ so $3k-2 = \frac{k}{2}(3k-1)$
What do I do next? Here's where I'm stuck:
Let $n=k+1$ So $3(k+1) -2 = \frac{k+1}{2}(3(k+1)-1)$
|
Notice that you have an aritmetic progression with initial term $1$ and difference $3$. So, $a_1 = 1$, and $a_k = 1 + 3(k-1)$. Then, $3n - 2 = 1 + 3(k-1) \implies k = n$, so $3n - 2$ is the $n$th term of the progression. This way: $$S_n = n\frac{a_1 + a_n}{2} = \frac{n}{2}(3n-1).$$
|
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|
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots+\frac{1}{1331}=\frac{p}{q}$; is $p$ divisible by $1997$? if $p,q\in \mathbb{N}$ and
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots+\frac{1}{1331}=\frac{p}{q}$$
why is $p$ divisible by $1997$?
|
Write $$1-\frac 12+\frac 13- \dots+\frac 1{1331}=\left(1+\frac 12+\frac 13+ \dots +\frac 1{1331}\right)-\left(1+\frac 12+ \dots +\frac 1{665}\right)$$
Using $1=2\cdot \frac 12$, and $ \frac 12=2\cdot \frac 14$, and $ \frac 13=2\cdot \frac 16$ etc to eliminate the negative fractions.
Then the sum becomes $$\frac 1{666}+\frac 1{667}+\dots +\frac 1{1331}$$
Now note that $\cfrac 1{n}+\cfrac 1{N-n}=\cfrac N{n(N-n)}$ and apply to $1997=666+1331=667+1330= \dots = 998+999$
Finally $1997$ is a prime, and all the factors in the denominators are $\lt 1997$, so there is no factor which will cancel the $1997$ in the numerator.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Using Dirichlet's hyperbola method and Dirichlet's formula Dirichlet Hyperbola Method. For $x \geq 2$:
$$ \sum_{n \leq x} \frac{d(n)}{n} = \frac{1}{2} \log^2 x + 2\gamma \log x + \gamma^2 + O(\frac{\log x}{\sqrt{x}})$$
I know already that the summation of 1/n can be written $\log x + \gamma + O(1/x)$ and summation of $d(n)$ can be written $$ \sum_{n \leq x} d(n) = x \log x+(2 \gamma-1)x + O(\sqrt{x})$$ if this helps?
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Following the notation from the
Planetmath Article
we note that
$$\frac{\tau(n)}{n} = 1/n * 1/n =
\sum_{d|n} \frac{1}{d} \left(\frac{n}{d}\right)^{-1}.$$
Substituting this into the formula for the method we obtain
$$\sum_{n\le x} \frac{\tau(n)}{n} =
\sum_{a\le\sqrt{x}} \sum_{b\le x/a}
\frac{1}{a} \frac{1}{b}
+ \sum_{b\le\sqrt{x}} \sum_{a\le x/b}
\frac{1}{a} \frac{1}{b}
- \sum_{a\le\sqrt{x}} \sum_{b\le\sqrt{x}}
\frac{1}{a} \frac{1}{b}
\\ =
\sum_{a\le\sqrt{x}} \frac{1}{a} \sum_{b\le x/a}
\frac{1}{b}
+ \sum_{b\le\sqrt{x}} \frac{1}{b} \sum_{a\le x/b}
\frac{1}{a}
- \sum_{a\le\sqrt{x}} \frac{1}{a} \sum_{b\le\sqrt{x}}
\frac{1}{b}
\\ =
\sum_{a\le\sqrt{x}} \frac{1}{a} (\log x - \log a + \gamma)
+ \sum_{b\le\sqrt{x}} \frac{1}{b} (\log x - \log b + \gamma)
- (\log\sqrt{x} + \gamma)^2
\\ = 2\log x (\log\sqrt{x} + \gamma)
+ 2\gamma (\log\sqrt{x} + \gamma)
- 2 \sum_{q\le\sqrt{x}} \frac{\log q}{q}
- (\log\sqrt{x} + \gamma)^2
\\ = 2\log x (\log\sqrt{x} + \gamma)
+ \gamma^2
- 2 \sum_{q\le\sqrt{x}} \frac{\log q}{q}
- \frac{1}{4} \log^2 x
\\ = \log^2 x + 2\gamma\log x
+ \gamma^2
- 2 \sum_{q\le\sqrt{x}} \frac{\log q}{q}
- \frac{1}{4} \log^2 x.$$
To evaluate the remaining sum term we could use the fact that
$$\left(\frac{1}{2} \log^2 x\right)' = \frac{\log x}{x}$$
but we need the constant term which is given in terms of the
Stiltjes constants
as $$\sum_{q=1}^n \frac{\log q}{q} \sim
\frac{1}{2} \log^2 n + \gamma_1.$$
This finally yields the formula
$$\log^2 x + 2\gamma\log x
+ \gamma^2
- 2 \left(\frac{1}{2} \log^2\sqrt{x} + \gamma_1\right)
- \frac{1}{4} \log^2 x
\\ = \log^2 x + 2\gamma\log x
+ \gamma^2
- 2 \left(\frac{1}{8} \log^2 x + \gamma_1\right)
- \frac{1}{4} \log^2 x
\\ = \log^2 x + 2\gamma\log x
+ \gamma^2
- \frac{1}{4} \log^2 x - 2\gamma_1
- \frac{1}{4} \log^2 x
\\ = \frac{1}{2} \log^2 x + 2\gamma\log x
+ \gamma^2 - 2\gamma_1.$$
The equalities are plus implicit lower order terms.
Addendum. The bound on the error terms follows from the harmonic number asymptotics $H_n = \log n + \gamma + \frac{1}{2n} + \cdots$ We get from the first term $\sum_{a\le \sqrt{x}} \frac{1}{a} \frac{1}{2x/a} = \frac{1}{2x} \sum_{a\le \sqrt{x}} 1$ which is $O(1/\sqrt{x}).$ The second term is the same. The third term contributes $\log\sqrt{x} \times 1/\sqrt{x}$ which is $O(\log x/\sqrt{x}).$ We see that the third lower order term dominates the first two, which gives exactly the formula proposed by the OP.
|
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|
derivative of $y=x^3\sqrt{x}-\frac{1}{x^2\sqrt{x}}$ $y=x^3\sqrt{x}-\dfrac{1}{x^2\sqrt{x}}$
Both terms require the product rule, right? My try:
$x^3\dfrac{1}{2}x^{-1/2}+3x^2x^{1/2}-\dfrac{-1}{2}x^{-3/2}x^{-2}--2x^{-3}x^{-1/2}$
What am I doing wrong? The correct answer is: $y\;'=3.5x^2\sqrt{x}+\dfrac{2.5}{x^3\sqrt{x}}$ and I don't see how what I got can reduce to this.
|
$$\ y=x^3\sqrt x-\frac{1}{x^2\sqrt x}=x^{\frac{7}{2}}-x^{-\frac{5}{2}}$$
$$\ y'=\frac{7}{2}x^{\frac{5}{2}}+\frac{5}{2}x^{-\frac{3}{2}}=$$
$$=\frac{7}{2}x^2\sqrt x+\frac{5}{2}\cdot\frac{1}{x\sqrt x}$$
|
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|
How to prove $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2$? Prove the inequality for a triangle with sides $a,b,c$ we have $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} <2$$
Trial: Since $a,b,c$ are sides of a triangle I know $a+b>c,b+c>a,a+c>b$
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That $a,b,c$ are sides of a triangle is equivalent to $a=x+y$, $b=y+z$ and $c=z+x$. Therefore we can rewrite our inequality as follows:
$$
\dfrac{x+y}{x+y+2z}+\dfrac{y+z}{2x+y+z}+\dfrac{z+x}{x+2y+z} <2
$$
Now, we have $\dfrac{x+y}{x+y+2z}<\dfrac{x+y}{x+y+z}$, $\dfrac{y+z}{2x+y+z}<\dfrac{y+z}{x+y+z}$ and $\dfrac{z+x}{x+2y+z}<\dfrac{z+x}{x+y+z}$ and therefore:
$$
\dfrac{x+y}{x+y+2z}+\dfrac{y+z}{2x+y+z}+\dfrac{z+x}{x+2y+z} <\dfrac{x+y}{x+y+z}+\dfrac{y+z}{x+y+z}+\dfrac{z+x}{x+y+z} =2
$$
|
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|
Proving Identity of Combination The lecturer had given two questions of proving that are
$$\binom{r}{r}+\binom{r+1}{r}+...+\binom{n}{r}=\binom{n+1}{r+1}\text{for }n\geq{r}\geq{1} $$
$$\binom{r}{0}+\binom{r+1}{1}+...+\binom{r+k}{k}=\binom{r+k+1}{k}\text{for }r,k\geq{1}$$
I tried to use the induction to prove these two identites but the lecturer said these two proving questions should be related to the identity which is
$$\binom{m+n}{r}=\binom{m}{0}\binom{n}{r}+...+\binom{m}{r}\binom{n}{0}$$
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$\bf{My\; Solution::}$ Given $\displaystyle \binom{r}{r}+\binom{r+1}{r}+\binom{r+2}{r}+\cdot \cdot \cdot \cdot \cdot \cdot\cdot +\binom{n}{r} = \binom{n+1}{r+1}\;,$
Where $n\geq r \geq 1.$
Coefficeint of $x^r$ in
$\displaystyle \left\{(1+x)^{r}+(1+x)^{r+1}+(1+x)^{r+1}\cdot \cdot \cdot \cdot \cdot \cdot \cdot+(1+x)^{n}\right\} = \frac{(1+x)^{n+1}-(1+x)^r}{(1+x)-1}=\frac{(1+x)^{n+1}-(1+x)^r}{x}$
(Using the formula $\displaystyle \left (r+r^2+..............+r^{n} = \frac{r^{n+1}-1}{r-1} \right)$
So Coeff. of $x^{r}$ in $\displaystyle \frac{(1+x)^{n+1}-(1+x)^r}{x}$
So Coeff. of $x^{r+1}$ in $\displaystyle (1+x)^{n+1}-(1+x)^r = \binom{n+1}{r+1}$
|
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|
Poisson Integral is equal to 1
Show
$$
\int_{-\pi}^{\pi}P(r, \theta)d\theta = 1
$$
Let $\alpha(r) = \frac{r^2 - 1}{2r}$ and $\gamma(r) = -\big(\frac{r^2 + 1}{2r}\big)$.
Then
$$
\frac{1}{2\pi}
\int_{-\pi}^{\pi}\frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}d\theta
= \frac{\alpha}{2\pi}
\int_{-\pi}^{\pi}\frac{1}{\cos(\theta) + \gamma}d\theta
$$
where
$$
\frac{r^2 - 1}{2r}\frac{1}{\cos(\theta) - \frac{1}{2r} - \frac{r^2}{2r}} = \frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}
$$
Next, let $z = e^{i\theta}$. Then $d\theta = \frac{-i}{z}dz$. Since $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$, $\cos(\theta) = \frac{z + z^{-1}}{2}$.
$$
= \frac{-i\alpha}{\pi}\int_C\frac{1}{z^2 + 2z\gamma + 1}dz
$$
and $C$ is the contour oriented counter clockwise with simple poles at $z = -\gamma\pm\sqrt{\gamma^2 - 1}$. Let $f(z) = \frac{1}{z^2 + 2z\gamma + 1}$. Then
$$
= \Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z)
\tag{1}
$$
The only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$.
Then
$$
2\pi i\lim_{z\to z_j}\Bigg[\big(z + \gamma - \sqrt{\gamma^2 - 1}\big)
\frac{1}{\big(z_j + \gamma + \sqrt{\gamma^2 - 1}\big)
\big(z + \gamma - \sqrt{\gamma^2 - 1}\big)}\Bigg] =
\frac{\pi i}{\sqrt{\gamma^2 - 1}}
$$
Now, we can substitute $\frac{\pi i}{\sqrt{\gamma^2 - 1}}$ for $2\pi i\sum\text{Res}$ in equation (1).
\begin{align*}
\Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z)
&= \frac{\alpha}{\sqrt{\gamma^2 - 1}}\\
&= \frac{r^2 - 1}{2r\sqrt{\frac{(r^2 + 1)^2}{4r^2} - 1}}\\
&= \frac{r^2 - 1}{\sqrt{(r^2 + 1)^2 - 4r^2}}\\
&= \frac{r^2 - 1}{\sqrt{r^4 - 2r^2 + 1}}\\
&= \frac{(r - 1)(1 + r)}{(r - 1)(r + 1)}\\
&= 1
\end{align*}
I have been unable to convince myself that the only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$. I know it is the case because if I use the other pole, the integral becomes $-1$
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From your definition, $\gamma = -\frac{r^2+1}{2r}$ so $-\gamma>1$ for all positive $r$. So $-\gamma+\sqrt{\gamma^2-1}>-\gamma>1$ so that root cannot possibly lie in $|z|<1$. Are you sure you do not have a sign error somewhere between lines 3 and 4 of your derivation?
|
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|
Can two pythagoras triplet have a common number If I have a pythagoras triplet $(a,b,c)$ such that
$$a^2+b^2=c^2$$
then is there another triplet $(a,d,e)$ possible such that
$$a^2+d^2=e^2, \; b\neq d$$
|
You can easily find nonprimitive triples that share an element: start with your two favorite triples (say $3^2 + 4^2 = 5^2$ and $5^2 + 12^2 = 13^2$). We'll get triples with $15$ in each: multiply the first triple by $5^2$ and the second by $3^2$. Then we have $5^2(3^2 + 4^2) = 5^2\cdot5^2$ and $3^2(5^2 + 12^2) = 3^2\cdot13^2$. Hence, $15^2 + 20^2 = 25^2$ and $15^2 + 36^2 = 39^2$. However, you can also find primitive triples sharing an element, for example $20^2 + 21^2 = 29^2$ and $20^2 + 99^2 = 101^2$.
Given an arbitrary triple $a^2 + b^2 = c^2$, there may or may not be another triple $a^2 + d^2 = e^2$ with $e\neq c$. To see this, note that all pythagorean triples are of the form $a = n^2 - m^2$, $b = 2mn$, $c = n^2 + m^2$, where $n,m\in\mathbb{N}$. There is only one pair $(n,m)$ such that any of these three numbers is $3$: $n = 2$, $m = 1$ (this is simple to see if you look at the way differences of squares behave and eliminate the possibility that $n^2 + m^2 = 3$).
|
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|
Proving $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\ldots\left(1+\frac{1}{n^3}\right)<3$ for all positive integers $n$
Prove that $\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\ldots\left(1+\dfrac{1}{n^3}\right)<3$ for all positive integers $n$
This problem is copied from Math Olympiad Treasures by Titu Andreescu and Bogdan Enescu.They start by stating that induction wouldn't directly work here since the right hand side stays constant while the left increases.They get rid of this problem by strengthening the hypothesis.$$\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\ldots\left(1+\dfrac{1}{n^3}\right)\le3-\dfrac{1}{n}$$
and then proceed by induction.
The problem is that I can't find a motivation for the above change.I mean,we could have subtracted a lot of things from the RHS but what should nudge us to try $\dfrac{1}{n}$?The rest of the proof is quite standard,but I can't see how I am supposed to have thought of it.Is it just experience?Or is it a standard technique?A little guidance and motivation will be appreciated.
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Since:
$$ 1+\frac{1}{k^3}=\left(1+\frac{1}{k}\right)\left(1-\frac{1}{k}+\frac{1}{k^2}\right) = \left(1-\frac{1}{k^2}\right)\left(1+\frac{1}{k(k-1)}\right) $$
and:
$$ \prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right)=\frac{1}{2} $$
we have:
$$ \prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right)=\prod_{k=2}^{+\infty}\left(1+\frac{1}{k(k-1)}\right)=\prod_{k=1}^{+\infty}\left(1+\frac{1}{k(k+1)}\right),$$
but since $1+x < e^x$ and $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$ it follows that:
$$ \prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right)<\exp\sum_{k=1}^{+\infty}\frac{1}{k(k+1)}=e<3.$$
With the same trick we can also prove the stronger:
$$ \prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right) < \frac{3}{2}\sqrt{e} <\frac{5}{2}.$$
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|
Derive an explicit formula for a power series $\sum_{n=1}^\infty n^2x^n$ Could anyone help me find an explicit formula for:
$$
\sum_{n=1}^\infty n^2x^n
$$
We're supposed to use: $$\sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2} \qquad |x| <1 $$
|
What if we didn't know
$$ \sum_{n=1}^\infty n x^n = \frac{x}{(1-x)^2}, |x| < 1 \text{?} $$
Note that $n^2 = (n+2)(n+1)-3(n+1) + 1$. (Where did this come from? We'll get to that.) So
$$ S(z) = \sum_{n=1}^\infty \left( (n+2)(n+1)-3(n+1) + 1 \right) z^n \text{.} $$
Assuming each of these converges on the same interval as the given series (which we will check later) we can rewrite this
$$ S(z) = \sum_{n=1}^\infty (n+2)(n+1)z^n - 3 \sum_{n=1}^\infty (n+1)z^n + \sum_{n=1}^\infty z^n \text{.} $$
From back to front: \begin{align*}
\sum_{n=1}^\infty z^n &= -1 + \sum_{n=0}^\infty z^n \\
&= -1 + \frac{1}{1-z} \text{,} \\
- 3 \sum_{n=1}^\infty (n+1)z^n &= - 3 \sum_{n=1}^\infty \frac{\mathrm{d}}{\mathrm{d}z} z^{n+1} \\
&= - 3 \frac{\mathrm{d}}{\mathrm{d}z} \sum_{n=1}^\infty z^{n+1} \\
&= - 3 \frac{\mathrm{d}}{\mathrm{d}z} \sum_{n=2}^\infty z^n \\
&= - 3 \frac{\mathrm{d}}{\mathrm{d}z} \left( -1-z + \sum_{n=0}^\infty z^n \right) \\
&= - 3 \frac{\mathrm{d}}{\mathrm{d}z} \left( -1-z + \frac{1}{1-z} \right) \\
&= - 3 \left( 0 -1 + \frac{1}{(1-z)^2} \right) \\
&= 3 + \frac{-3}{(1-z)^2} \text{, and} \\
\sum_{n=1}^\infty (n+2)(n+1)z^n &= \sum_{n=1}^\infty \frac{\mathrm{d}^2}{\mathrm{d}z^2} z^{n+2} \\
&= \frac{\mathrm{d}^2}{\mathrm{d}z^2} \sum_{n=1}^\infty z^{n+2} \\
&= \frac{\mathrm{d}^2}{\mathrm{d}z^2} \sum_{n=0}^\infty z^{n+3} \\
&= \frac{\mathrm{d}^2}{\mathrm{d}z^2} \left(z^3 \sum_{n=0}^\infty z^{n} \right) \\
&= \frac{\mathrm{d}^2}{\mathrm{d}z^2} \frac{z^3}{1-z} \\
&= \frac{2(z^3 - 3z^2 + 3z)}{(1-z)^3} \\
&= \frac{6z}{1-z} + \frac{6z^2}{(1-z)^2} + \frac{2z^3}{(1-z)^3} \text{.}
\end{align*}
Adding up the three pieces,
$$ S(z) = \frac{z(1+z)}{(1-z)^3} \text{.} $$
In each case, we used the geometric series $\frac{1}{1-z} = \sum_{n=0}^\infty z^n$, which converges on $(-1,1)$. By the ratio test, the given series also converges on $(-1,1)$. At $z = -1$ and at $z = 1$, the terms of the series don't converge to zero so neither endpoint is in the interval of convergence. Therefore, the resulting function agrees with the series on its interval of convergence.
So where did $n^2 = (n+2)(n+1)-3(n+1) + 1$ come from? I hope from the above it is obvious that we wanted the coefficients to be "constants time increasing products of $n+1$, $n+2$, ... of whatever length was needed" so that we could replace the "$(n+k)\cdots(n+1)z^{n}$" with $\frac{\mathrm{d}^k}{\mathrm{d}z^k} z^{n+k}$, exchange differentiation and summation, adjust the index, cancel the missing terms from the geometric series, then replace with the geometric series.
(An alert reader will notice I haven't said where the identity came from, only why it was useful.) Notice that we want coefficients that are "rising factorials" in the index $n$ (strictly in $n+1$). This is a job for Stirling numbers. And there is a lot of fun structure to discover in those numbers. However, we might be thinking "isn't this a little too hard?". Maybe. We can go another way without seeing the cool set of numbers doing the work for us.
\begin{align*}
S(z) &= \sum_{n=1}^\infty n^2 z^n \\
&= \sum_{n=1}^\infty n(n+1-1) z^n \\
&= \sum_{n=1}^\infty \left( n(n+1) z^n - n z^n \right) \\
&= \sum_{n=1}^\infty \left( \frac{\mathrm{d}}{\mathrm{d}z} n z^{n+1} - n z^n \right) \\
&= \sum_{n=1}^\infty \left( \frac{\mathrm{d}}{\mathrm{d}z} (n+2-2) z^{n+1} - n z^n \right) \\
&= \sum_{n=1}^\infty \left( \frac{\mathrm{d}}{\mathrm{d}z} (n+2) z^{n+1} - \frac{\mathrm{d}}{\mathrm{d}z} 2 z^{n+1} - n z^n \right) \\
&= \sum_{n=1}^\infty \left( \frac{\mathrm{d}^2}{\mathrm{d}z^2} z^{n+2} - 2 \frac{\mathrm{d}}{\mathrm{d}z} z^{n+1} - (n+1-1) z^n \right) \\
&= \sum_{n=1}^\infty \left( \frac{\mathrm{d}^2}{\mathrm{d}z^2} z^{n+2} - 2 \frac{\mathrm{d}}{\mathrm{d}z} z^{n+1} - (n+1) z^n + z^n \right) \\
&= \sum_{n=1}^\infty \left( \frac{\mathrm{d}^2}{\mathrm{d}z^2} z^{n+2} - 2 \frac{\mathrm{d}}{\mathrm{d}z} z^{n+1} - \frac{\mathrm{d}}{\mathrm{d}z} z^{n+1} + z^n \right) \\
&= \sum_{n=1}^\infty \left( \frac{\mathrm{d}^2}{\mathrm{d}z^2} z^{n+2} - 3 \frac{\mathrm{d}}{\mathrm{d}z} z^{n+1} + z^n \right) \text{,}
\end{align*}
where our new friends, the Stirling numbers have reappeared as the coefficients (the same coefficients in the identity starting this Answer). You should be able to split thisinto three sums, exchange differentiation and summation in two of them, adjust the index in all three, cancel the missing terms from the geometric series, then replace with the geometric series, as was done in the work shown above.
That's great and all, but surely there's a way to avoid so much writing? Having done all the above, we now know what we are trying to do -- replace the two copies of $n$ in "$n^2$" with a sum of constant multiples of derivatives, so just start there, the same as we start with the correct form for partial fractions and solve for the coefficients. We require (starting with the power of $z$ we have in the general term and incrementing the power in parallel with incrementing the repetition of the derivative)
\begin{align*}
n^2 z^n &= A z^n + B \frac{\mathrm{d}}{\mathrm{d}z} z^{n+1} + C \frac{\mathrm{d}^2}{\mathrm{d}z^2} z^{n+2} \\
&= A z^n + B (n+1) z^{n} + C(n+2)(n+1) z^{n} \\
&= \left( A + B (n+1) + C(n+2)(n+1) \right) z^{n} \\
&= \left( A + Bn + B + Cn^2 + 3Cn + 2C \right) z^{n} \\
&= \left( (A + B + 2C) + (B + 3C)n + Cn^2 \right) z^{n} \text{.}
\end{align*}
For two polynomials to agree, they agree in each coefficient, so $C = 1$, $B+3C = 0$, so $B = -3$, and $A+B+2C = 0$, so $A = 1$. And, surprise(?) we immediately get the Stirling numbers again. And, using the methods shown above, we know what to do with the sums obtained from each term in the first line.
|
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|
Number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
Find number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
$\bf{My\; Try::}$ Let $f(x) = (6-x)^4+(8-x)^4\;,$ and we have to find real values of $x$ for
which $f(x) = 16$. Now we will form Different cases.
$\bf{\bullet \; }$ If $x<6$ or $x>8\;,$ Then $f(x)>16$
$\bf{\bullet\; }$ If $6 <x<8\;,$ Then $f(x)<16$.
$\bf{\bullet \; }$ If $x=6\;,x=8\;,$ Then $f(x) = 16$
So Solutions of the above equation are $x=6$ and $x=8$
Can we solve it using Derivative or Algebraic way
Thanks
|
I guess by Algebraic you mean this way
$$(6-x)^4+(8-x)^4 = 16$$
put $7-x=t$
(reason behind this substitution is not just because the later is easy to expand using binomial formula but the magic it does).
$$(t-1)^4+(t+1)^4 = 16$$
$$(t^4-4t^3+6t^2-4t+1)+(t^4+4t^3+6t^2+4t+1)=16$$
$$2(t^4+6t^2+1)=16$$
$$t^4+6t^2+1=8$$
Well, its quartic but it's easy to find it's roots
put $p=t^2$
$$ p^2+6p-7=0$$
$$\implies p=1,p-7$$
Now $p=-7$ will not yield any real solution so, we ignore that
$$\implies t=\pm1$$
$$x=7-t\implies x=6,8$$
Hence, only two real roots, other two being complex.
|
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|
Simplifying the derivative of $f(x)= \frac{e^x - e^{-x}}{e^x+e^{-x}}$ I was having some trouble on simplifying the derivative because I didn't know if it's correct. The original function is $$f(x)= \frac{e^x - e^{-x}}{e^x+e^{-x}}$$ What would the simplified derivative be with no negative exponents?
So far I got
$$f'(x)= \frac{(e^x+e^{-x})^2 - (e^x-e^{-x})^2}{(e^x+e^{-x})^2}$$
is this correct?
|
What you have thus far is correct.
Observe that the numerator of
$$f'(x) = \frac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$
is a difference of squares. Thus,
\begin{align*}
f'(x) & = \frac{[(e^x + e^{-x}) + (e^x - e^{-x})][(e^x + e^{-x}) - (e^x - e^{-x})]}{(e^x + e^{-x})^2}\\
& = \frac{(2e^x)(2e^{-x})}{(e^x + e^{-x})^2}\\
& = \frac{4}{(e^x + e^{-x})^2}
\end{align*}
If you do not want any negative exponents, multiply the numerator and denominator by $e^{2x}$ to obtain
\begin{align*}
f'(x) & = \frac{4e^{2x}}{(e^x + e^{-x})^2e^{2x}}\\
& = \frac{4e^{2x}}{(e^{2x} + 2 + e^{-2x})e^{2x}}\\
& = \frac{4e^{2x}}{e^{4x} + 2e^{2x} + 1}\\
& = \frac{4e^{2x}}{(e^{2x} + 1)^2}
\end{align*}
Alternatively, multiply the numerator and denominator of
$$f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$
by $e^x$ to obtain
$$f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$$
then differentiate to obtain
\begin{align*}
f'(x) & = \frac{2e^{2x}(e^{2x} + 1) - 2e^{2x}(e^{2x} - 1)}{(e^{2x} + 1)^2}\\
& = \frac{2^{4x} + 2e^{2x} - 2e^{4x} + 2e^{2x}}{(e^{2x} + 1)^2}\\
& = \frac{4e^{2x}}{(e^{2x} + 1)^2}
\end{align*}
which agrees with the previous result.
|
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|
Solving trigonometric system of equations What are the solutions for this system of equations when $\alpha \in \mathbb{R}$ is considered a constant and $0 \leq x < 2\pi$.
$$ I) \ (y - \cos x)\sin x + (\alpha - \sin x) (-\cos x) = 0$$
$$ II) \ (y - \cos x) = 0$$
My attempt:
From II) we know that $y = \cos x$.
If we apply that in I), we get:
$$(\alpha - \sin x) (-\cos x) = 0 \Leftrightarrow$$
$$- \cos x = 0 \vee \alpha = \sin x$$
Now the first pair of solutions is $(\frac{\pi}{2},0),(\frac{3\pi}{2},0)$. However I fail to express the other solutions in terms of $\alpha$. Any help would be very much appreciated.
|
$$ \begin{cases}
(y−\cos{x})\sin{x}+(α−\sin{x})(−\cos{x}) = 0 \\
y − \cos{x} = 0
\end{cases} \tag{1} $$
First, substitute all instances of $(y−\cos{x})$ with $0$.
$$ \begin{cases}
\color{red}{(0)}\sin{x}+(α−\sin{x})(−\cos{x}) = 0 \\
\color{red}{(0)} = 0
\end{cases} \tag{2} $$
Now, we are left with one equation.
$$ (α−\sin{x})(−\cos{x}) = 0 \tag{3} $$
We need to find the zeros of each term.
$$ α−\sin{x} = 0 \tag{4} $$
$$ -\sin{x} = -a \tag{4.1} $$
$$ \sin{x} = a \tag{4.2} $$
$$ x = \arcsin{a} \\ \operatorname{or} \\ x = \pi - \arcsin{a} \tag{4.3} $$
We have found one set of $x$ values, in terms of $a$. Next, we zero the second term.
$$ −\cos{x} = 0 \tag{5} $$
$$ \cos{x} = 0 \tag{5.1} $$
$$ x = \pm\arccos{0} \tag{5.2} $$
$$ x = \pm\frac{\pi}{2} \tag{5.3} $$
$-\pi/2$ is outside our interval of $x$, $[0,2\pi)$, but $-\frac{\pi}{2}$ is coterminal with $\frac{3\pi}{2}$; thus
$$ x = \frac{\pi}{2} \\ \operatorname{or} \\ x = \frac{3\pi}{2} \tag{5.4} $$
Finally, we have found all values of $x$.
$$ x = \{ \arcsin{a} , \pi - \arcsin{a} , \frac{\pi}{2} , \frac{3\pi}{2} \} \tag{6} $$
Next, solve the second equation in the original system for $y$.
$$ y = \cos{x} \tag{7} $$
Then, substitute each value of $x$ to find the corresponding value of $y$.
$$ y = \cos{( \arcsin{a} )} \ | \ x = \arcsin{a} \tag{8} $$
$$ y = \sqrt{ 1 - a^2 } \ | \ x = \arcsin{a} \tag{8.1} $$
$$ y = \cos{( \pi - \arcsin{a} )} \ | \ x = \pi - \arcsin{a} \tag{9} $$
$$ y = -\sqrt{ 1 - a^2 } \ | \ x = \pi - \arcsin{a} \tag{9.1} $$
$$ y = \cos{ \frac{\pi}{2} } \ | \ x = \frac{\pi}{2} \tag{10} $$
$$ y = 0 \ | \ x = \frac{\pi}{2} \tag{10.1} $$
$$ y = \cos{ \frac{3\pi}{2} } \ | \ x = \frac{3\pi}{2} \tag{11} $$
$$ y = 0 \ | \ x = \frac{3\pi}{2} \tag{11.1} $$
In conclusion, the values of $x$ and $y$ that satisfy the system of equations are
$$ ( \arcsin{a} , \sqrt{ 1 - a^2 } ) , \\ ( \pi - \arcsin{a} , -\sqrt{ 1 - a^2 } ) , \\ ( \frac{\pi}{2} , 0 ) , \\ ( \frac{3\pi}{2} , 0 ) $$
Also, note that the system is only solvable when $-1 \le a \le 1$. Depending on the value of $a$, the $x$ value of the first point may fall outside the interval, so a coterminal value of $x$ may be required.
|
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|
Give a combinatorial argument Give a combinatorial argument to show that $$\binom{6}{1} + 2 \binom{6}{2} + 3\binom{6}{3} + 4 \binom{6}{4} + 5 \binom{6}{5} + 6 \binom{6}{6} = 6\cdot2^5$$
Not quite where to starting proving this one. Thanks!
|
For whatever it's worth, here's a standard probabilistic argument: The number of ways to get $k$ heads in six trials when tossing a coin is $\dbinom 6 k$. If all $2^6=64$ sequences are equally likely (as they are when the coin is "fair", i.e. gives heads and tails equally often) then the probability of exactly $k$ successes in $6$ trials is $\dbinom 6 k \cdot \dfrac 1 {64}$. So the average number of successes in $6$ trials is
$$
\frac{\binom{6}{1} + 2 \binom{6}{2} + 3\binom{6}{3} + 4 \binom{6}{4} + 5 \binom{6}{5} + 6 \binom{6}{6}}{64}.
$$
But the average number of successes in $6$ trials, with probability $1/2$ of success on each trial, is $6\cdot\frac 1 2=3$, so the sum above equals $3$.
On way of showing that the average number of successes with six trials is three is to observe that the average number of successes in one trial is $1/2$, so you're just adding up $1/2$ six times, getting $3$.
|
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|
$a^2+ab+b^2 \equiv 0 \pmod n$ if and only if $ a\equiv b\equiv 0 \pmod n$ Let $n$ be a prime number with $n \equiv -1 \mod 6$ and $a,b$ be positive integers. I want to prove:
$$a^2+ab+b^2 \equiv 0 \mod n \iff a\equiv b\equiv 0 \mod n$$
|
$a^2+ab+b^2\equiv0 \pmod n\to a^3-b^3\equiv0 \pmod n\to (ab')^3\equiv1\pmod n\to 3\mid n-1\; or\; a\equiv b \pmod 3$ where $b'$ is the inverse of $b$ $\pmod n$. Because $3\nmid n-1$ we have $a\equiv b \pmod n$ so we have $n\mid 3a^2$ we have $(n,3)=1$ so we have $n\mid a^2$ and $a\equiv 0 \pmod n$.
|
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|
Calculate the area between the function $\sqrt{x}$ and $x^2$ in the interval $[0 , 2]$. Calculate the area between the function
$\sqrt{x}
$and
$x^2$
in the interval [0 , 2].
Ok, $0<x<1$, $\sqrt{x}$$>x^2$
and $1<x<2$, $\sqrt{x}$$<x^2$
$$ Area = \int_{0}^1(\sqrt{x}-x^2)dx + \int_{1}^2(x^2-\sqrt{x})dx$$
I know that $$ \int(\sqrt{x})dx = \frac 23x^\frac 32$$
and $$ \int(x^2)dx = \frac 13x^3 $$
So the Area is $$ \frac {10}{3} - \frac {{4}\sqrt{2}}{3} $$
Is this correct? I'm not sure if I'm doing the right thing
|
To begin to answer your question one can first check have a graph of the equations handy, or understand that $y=\sqrt{x}$ raises faster than $y=x^2$ in the interval from $[0,1]$, after that $y=x^2$ is above the $y=\sqrt{x}$ from $[1,2]$. So know to setup the integrals:
\begin{align*}A_{\mathrm{tot}}&=A_1+A_2\\A_1&=\int_{0}^1(\sqrt{x}-x^2)dx\\A_2&=\int_1^2(x^2-\sqrt{x})dx\\ A_1&=[\frac{2}{3}x^{\frac{3}2}-\frac{x^3}3]\Bigg\vert_0^1=\frac{1}3 \\ A_2&=[\frac{x^3}{3}-\frac{2}{3}x^\frac{3}2]\Bigg\vert_1^2=\frac{9-4\sqrt2}{3}\\A_\mathrm{tot}&=\frac{10-4\sqrt{2}}{3}\end{align*}
|
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|
What is the probability of two out of three events happening? All events are independent.
$$\Pr(A) = \frac{9}{10}$$
$$\Pr(B) = \frac{9}{10}$$
$$\Pr(C) = \frac{6}{10}$$
What is the probability of at least two events happening?
I'd like to use negation, to negate the possibility that event no event happen plus the probability that only one happens.
$$D = \text{at least two events happen}$$
$$\Pr(D) = 1-\Pr(\text{none happens})-\Pr(\text{exactly one happens})$$
$$\Pr(D) = 1 - \left(\frac{1}{10}\cdot \frac{1}{10}\cdot\frac{4}{10}\right) - \left(\frac{1}{10}\cdot\frac{6}{10}\cdot\frac{1}{10} + \frac{9}{10}\cdot\frac{1}{10}\cdot\frac{4}{10} + \frac{9}{10}\cdot\frac{1}{10}\cdot\frac{4}{10}\right) = 0.918$$
The answer seems a little larger, I can't convince myself that I'm right.
|
Can confirm that is the right answer, calculated it directly and obtained the same number.
|
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|
Find the maximum value of the expression Let a,b,c,d be positive real numbers. Show that
$$\frac{(ab+bc+ca)}{(a^3+b^3+c^3)}+\frac{(ab+bd+da)}{(a^3+b^3+d^3)}+\frac{(ac+cd+da)}{(a^3+c^3+d^3)}+\frac{ (bc+cd+db)}{(b^3+c^3+d^3)} \le \frac{(a^2+b^2)}{(ab)^{3/2}} + \frac{(c^2+d^2)}{(cd)^{3/2}} +\frac{(a^2+c^2)}{(ac)^{3/2}} + \frac{(b^2+d^2)}{(bd)^{3/2}} +\frac{ (a^2+d^2)}{(ad)^{3/2}} + \frac{(b^2+c^2)}{(bc)^{3/2}}$$
|
Note that $\displaystyle \frac{a^2+b^2}{(ab)^{3/2}} \ge 2\frac{a^2+b^2}{a^3+b^3}$ by AM-GM. Using that, we get
$$\frac{a^2+b^2}{(ab)^{3/2}}+\frac{b^2+c^2}{(b c)^{3/2}}+\frac{c^2+a^2}{(c a)^{3/2}} \ge 2\left(\frac{a^2+b^2}{a^3+b^3} + \frac{b^2+c^2}{b^3+c^3} + \frac{c^2+a^2}{c^3+a^3} \right) \tag{1}$$
Using Cauchy-Schwarz inequality on the RHS of $(1)$ above, we have
$$\frac{a^2+b^2}{(ab)^{3/2}}+\frac{b^2+c^2}{(b c)^{3/2}}+\frac{c^2+a^2}{(c a)^{3/2}} \ge 2\frac{(a+b+c)^2}{a^3+b^3+c^3} \ge 6\frac{ab+bc+ca}{a^3+b^3+c^3} \tag{2}$$
Now cyclically summing four similar expressions using $(2)$, we get
$$\sum_{sym} \frac{a^2+b^2}{(ab)^{3/2}} \ge 2\sum_{sym} \frac{ab+bc+ca}{a^3+b^3+c^3}$$
which is tighter than the inequality in your problem...
|
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|
How to evaluate $\lim\limits_{x\to 0} \frac{\sin x - x + x^3/6}{x^3}$ I'm unsure as to how to evaluate:
$$\lim\limits_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}$$
The $\lim\limits_{x\to 0}$ of both the numerator and denominator equal $0$. Taking the derivative of both ends of the fraction we get:
$$\lim\limits_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2}$$
But I don't know how to evaluate this?
Many thanks for any help.
|
Using L'Hospital twice,
$$
\lim_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}=\lim_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2}=\lim_{x\to 0} \frac{x - \sin x }{6x}=\frac{1}{6}\lim_{x\to 0} \left(1-\frac{\sin x}{x}\right)=0
$$
|
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|
Manipulating roots of a cubic Given that $A,B,C$ are the roots of the equation $x^3-5x^2+x+1$, how do I find the value of $$\dfrac{A}{B+C}+\dfrac{B}{A+C}+\dfrac{C}{A+B}$$ I know the Vieta's formulas but I am not able to manipulate the above expression into something known. And taking the LCM doesn't help. Please help me out. Thank you.
|
By Vieta's, you know that $A+B+C = 5$. Therefore, we have
$\dfrac{A}{B+C}+\dfrac{B}{A+C}+\dfrac{C}{A+B} = \dfrac{A}{5-A}+\dfrac{B}{5-B}+\dfrac{C}{5-C}$
Also, you perform the following manipulation:
$x^3-5x^2+x+1 = 0$
$5x^2-x^3 = x+1$
$\dfrac{5-x}{x} = \dfrac{x+1}{x^3}$
$\dfrac{x}{5-x} = \dfrac{x^3}{x+1}$
$\dfrac{x}{5-x} = x^2-x+1+\dfrac{1}{x+1}$
Now take the sum of both sides over $x = A,B,C$ and apply Vieta's.
Alternatively, if $x = A,B,C$ are the roots of the equation $f(x) = x^3-5x^2+x+1 = 0$, then $x = \dfrac{A}{5-A}, \dfrac{B}{5-B}, \dfrac{C}{5-C}$ are the roots of $f\left(\dfrac{5x}{x+1}\right) = 0$. Apply Vieta's to that equation.
|
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can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods???
i only know quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ i tried many algebraic manipulations and i get $(x^2+1)^2=3(x^3+x)$, so can we have solution using only high school methods? i guess i have to do an algebraic substitution to reduce $x^3+x$ to polynomial of degree 2?? i also know $$(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1r_2$$ where $r_1,r_2$ are roots of polynomial of degree 2
|
Hint: $$\color{#C00}{x^4}\color{darkmagenta}{-3x^3}\color{royalblue}{+2x^2}\color{darkmagenta}{-3x}\color{#C00}{+1},\tag1$$ is a symmetric polynomial of degree $4$. I will describe here the basic procedure to solve such polynomial equations. Plugging in $x=0$ yields $1$, thus $0$ isn't a root and therefore $x\neq0$ which means that $x^2\neq0$. Thus it is legal to divide both sides by $x^2$, we should get $$x^2-3x+2-\dfrac3x+\dfrac1{x^2}=0\implies \left(x^2+\dfrac1{x^2}\right)-3\left(x+\dfrac1x\right)+2=0.\tag2$$ Now note that $$\left(x+\dfrac1x\right)^2=\left(x^2+\dfrac1{x^2}\right)+2\implies\left(x^2+\dfrac1{x^2}\right)=\left(x+\dfrac1x\right)^2-2.\tag3$$ From $(2)$ and $(3)$ we get $$\left(x+\dfrac1x\right)^2-3\left(x+\dfrac1x\right)=0.\tag4$$ This certainly looks like a quadratic, to convince yourself of that you can use the substitution $u\leadsto\left(x+\tfrac1x\right)$, and thus $(4)$ takes the usual form $$u^2-3u=0.$$ I'm sure you can finish it off from here.
|
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|
Evaluating $\int e^{x\sin x+\cos x}\left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right)dx$ Evaluate
$$\displaystyle \int e^{x\sin x+\cos x}\left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right)dx$$
$\bf{My\; Try::}$
Let $$\begin{align}I &= \int e^{x\sin x+\cos x}\left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right)dx\\
&=\int e^{x\sin x+\cos x}\left(x^2\cos x+\frac{\cos x-x\sin x}{x^2\cos^2 x}\right)dx\\
&= \int x\cdot e^{x\sin x+\cos x}\left(x\cos x\right)dx+\int e^{x\sin x+\cos x}\left(\frac{\cos x-x\sin x}{x^2\cos^2 x}\right)dx\\
\end{align}$$
Now Let $x\sin x+\cos x = t\;,$ Then $x\cos x\,dx = dt$ and Integration by parts for $\bf{1^{st}}$ Integral
So $$\displaystyle I = x\cdot e^{x\sin x+\cos x}-\int e^{x\sin x+\cos x}dx+\int e^{x\sin x+\cos x}\left(\frac{\cos x-x\sin x}{x^2\cos^2 x}\right)dx$$
Now I do not understand how to solve after that.
|
Let $$\begin{align}I&= \int e^{x\sin x+\cos x}\left[\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right]\mathrm d x\\
&= \int e^{x\sin x+\cos x}\left[\frac{\cos x - x \sin x + x^4 \cos^3 x}{x^2\cos^2x}\right]\mathrm d x\tag{1}\\
&=\int e^{x\sin x+\cos x} \left[ \frac{1}{x^2 \cos x} -\frac{\sin x}{x\cos ^2x}+x^2\cos x\right]\mathrm d x\tag{2}\\
&=\int e^{x\sin x+\cos x} \left[ \frac{1}{x^2 \cos x} -\frac{\sin x}{x\cos ^2x}+1+x^2\cos x-1\right]\mathrm dx\tag{3}\\
&=\int e^{x\sin x+\cos x} \left[ \left(\frac{1}{x^2 \cos x} -\frac{\sin x}{x\cos^2x}+1\right) +(x\cos x)\left(x-\frac{1}{x\cos x}\right)\right] \mathrm dx \tag{4} \\
\end{align}$$
Notice that
$$\frac{ \mathrm d }{ \mathrm d x }\left[x-
\frac{1}{x\cos x}\right]=\frac{1} { x^2\cos x } -\frac{\sin x}{x\cos^2x}+1$$
$$\frac{\mathrm d}{\mathrm dx}(x\sin x+\cos x)=x\cos x$$
Now use
$$\int e^t[ f ( t ) + f ' ( t ) ] \mathrm d t = e^t\cdot f(t ) +C$$
to get
$$\int e^{x\sin x+\cos x}\left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right)\mathrm d x=e^{x\sin x+\cos x}\left(x-\frac{1}{x\cos x}\right)+C$$
$\text{ Explanations }\\
1 .\text{ Rearranging terms}\\
2 .\text{ Seperating out terms}\\
3 .\text{ Adding zero}\\
4 .\text{ Grouping terms and factor }$
|
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|
Reduce $\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}$ to a homogenous equation by substituting $x=X-1$ and $y=Y+3$
By substituting $x=X-1$ and $y=Y+3$ reduce the differential equation
$$
\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}
$$
to a homogeneous equation and hence find the general solution in terms of $x$ and $y$.
All I have so far is:
$$
\frac{dy}{dx}=\frac{4X-Y}{2X+Y}
$$
Then I thought about doing something like:
$$
X=x+1\\
dX=dx
$$
But I wasn't sure whether that was correct or useful here. Can somebody help point me in the right direction? The answer listed is $(y-x-4)^3(y+4x+1)^2=A$ where $|A|=e^c$.
|
Notice that $$\frac{dY}{dX}=\frac{4X-Y}{2X+Y}= \frac{4 -\frac{Y}{X}}{2 + \frac{Y}{X}}.$$
So, let $V= Y/X$, so $Y=VX\implies dY/dX = V + X(dV/dX)$. With these, we substitute:
$$V + X\frac{dV}{dX} = \frac{4 -V}{2 +V}\implies \frac{2+V}{4-3V-V^2}dV = \frac{dX}{X}.$$
This can be broken up using partial fractions. After we will exploit the properties of logarithms to get an answer in the form of $$(1-V)^3(V+4)^2X^5=e^C.$$
Since $V=Y/X$, we substitute that back in, simplify, and then substitute $X=x+1$, and $Y=y-3$, and simplify again. It's more of an algebra exercise!
|
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|
Optimization problem - Trapezoid under a parabola recently I've been working on a problem from a textbook about Optimization. The result that I get is $k = 8$, even thought the answer from the textbook is $k = \frac{32}{3}$
The problem follows:
--
The x axis interepts the parabola $12-3x^2$ at the points $A$ and $B$, and also the line $y = k$ (for $0 < k < 12$) at the points C and D. Determine $k$ in a way that the trapezoid $ABCD$ has a maximum area.
--
My solution was this
--
The trapezoid area is
$$A_{T} = \frac{(B+b) \cdot h}{2} = \frac{4+2 \cdot \sqrt{\frac{12-k}{3}} \cdot k}{2}$$
$$A_{T}' = 0 \therefore \frac{\sqrt{12-k}}{\sqrt{3}} - \frac{k}{\sqrt{3} \cdot 2 \cdot \sqrt{12-k}} = 0$$
$$2(12-k)-k=0 \therefore k = 8$$
--
Where did I go wrong on?
Thank you guys!
|
The formula for the area should read
$$A_{T} = \frac{(B+b) \cdot h}{2} = \frac{\left(4+2 \cdot \sqrt{\frac{12-k}{3}}\right) \cdot k}{2}=\left(2+\sqrt{\frac{12-k}{3}}\right) \cdot k.$$
Now do the same you have done with this function.
|
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|
Proving a Binomial Sum How do I prove that $$\sum_{r=0}^{n-1}\left[ r \binom{n}{r} \binom{n}{r+1}\right]=n \binom{2n-1}{n-2}$$ without induction?
I've tried manipulating $(1+x)^n$ and the binomial coefficients, but to no avail. Also, how can one explain this identity combinatorially?
|
By way of enrichment here is another algebraic proof using basic
complex variables.
We seek to compute
$$\sum_{r=0}^n r {n\choose r} {n\choose r+1}.$$
Introduce the integral representation
$${n \choose r+1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+2}} \; dz.$$
We use this to obtain an integral for the sum. Note that when
$r+2>n+1$ or $r>n-1$ the pole at zero disappears which means that the
integral is zero. Therefore we obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{r=0}^n r {n\choose r} \frac{(1+z)^n}{z^{r+2}}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^2}
\sum_{r=0}^n r {n\choose r} \frac{1}{z^r}\; dz.$$
Note that
$$x \frac{d}{dx} (1+x)^n = \sum_{r=0}^n r {n\choose r} x^r
= nx (1+x)^{n-1}.$$
Returning to the integral this gives
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^2}
\times n \frac{1}{z} \left(1+\frac{1}{z}\right)^{n-1} \; dz
\\ = \frac{n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^3}
\frac{(1+z)^{n-1}}{z^{n-1}} \; dz
\\ = \frac{n}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n-1}}{z^{n+2}} \; dz.$$
This last integral can be evaluated by inspection and the result is
$$n\times {2n-1\choose n+1} = n {2n-1\choose n-2}.$$
|
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|
Integration of $1/(x^4 \sin x +2x)$ $$\int\frac{1}{x^4 \sin x +2x} dx\ $$ How to evaluate this integral. How to go about evaluating these integrals?
|
Hint:
$\int\dfrac{1}{x^4\sin x+2x}dx$
$=\int\dfrac{1}{2x\left(1+\dfrac{x^3\sin x}{2}\right)}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{3n-1}\sin^nx}{2^{n+1}}dx$
$=\int\left(\dfrac{1}{2x}+\sum\limits_{n=1}^\infty\dfrac{(-1)^nx^{3n-1}\sin^nx}{2^{n+1}}\right)dx$
But this approach is only suitable for the very limited ranges of $x$ : $\left|\dfrac{x^3\sin x}{2}\right|\leq1$
|
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|
Area of $\triangle ABC$ whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is The maximum area of a triangle whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is
$\bf{My\; Try::}$ Area of $\displaystyle \triangle ABC = \frac{1}{2}ab\sin C = \frac{1}{2}ab\cdot \sqrt{1-\cos^2 C}\;,$ bcz Largest side has largest angle.
Now Using $\displaystyle \cos C = \frac{a^2+b^2-c^2}{2ab}.$ So we get area of $\displaystyle \triangle ABC = \frac{1}{2}ab\cdot \sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}$
Now How can i find Maximum area.
|
As the above user said a better approach would be to look at the area of a triangle as $\frac12 ab \sin{x}$, where $a$ and $b$ are the lengths of two adjacent sides and $x$ is the measure of the angle between these two consecutive sides. The maximum value of $\sin{x}$ occurs when $x=\pi/2$ which gives a value of 1. We now try to maximize the length of the legs and get that the maximum area is $2\cdot 1/2=1$. We make sure that we can have 2 and 1 as leg lengths by checking if $c$ fits in the inequality that was given in the problem, and indeed $\sqrt{5}$ is between 2 and 3.
|
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|
Improper Integral of $\int\frac{dx}{(2x-1)^3}$ Improper Integral of
$$\int_{-\infty}^0\frac{dx}{(2x-1)^3}$$
from Anton Calculus 8th Edition, page 576, question 9. Answer is $-\frac{1}{4}$ but I'm finding $-1$
The integral, substituting $u= (2x-1)$:
$$\frac{1}{2}\cdot\frac{-2}{(2x-1)^2}+C$$
Definite solution to use with limit:
$$ \frac{1}{2}\cdot\frac{-2}{(2x-1)^2)}\Big|^0_a = \frac{1}{2}\cdot\left(-2+\frac{2}{(2a-1)^2}\right)$$
Then solving limit:
$$\lim_{x \to {-\infty}} \frac{1}{2}\cdot\left(-2+\frac{2}{(2a-1)^2}\right) = \frac{1}{2}\cdot(-2+0) = -1$$
Any leads?
Thank you.
|
You made correct substitution
$$u= 2x-1\iff du=2dx$$
$$\int\frac{dx}{(2x-1)^3}dx=\frac12\int\frac{du}{u^3}dx=-\frac{1}{4u^2}+C$$
$$\int\frac{dx}{(2x-1)^3}=-\frac{1}{4(2x-1)^2}+C$$
I hope you can take it from here!
|
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|
Does $x,y,z>0$ and $x+y+z=1$ imply $\left(1+\frac 1x\right)\left(1+\frac 1y \right)\left(1+\frac 1z \right)\ge 64$? If $x,y,z$ are positive real numbers such that $x+y+z=1$ then is it true that
$\left(1+\dfrac 1x\right)\left(1+\dfrac 1y \right)\left(1+\dfrac 1z \right)\ge 64$ ?
|
By AM-GM, we have
$$1+x=\frac{1}3+\frac13+\frac13+x\ge 4\sqrt[4]{\frac{x}{3^3}}.$$
So
$$(1+x)(1+y)(1+z)\ge 4^3\sqrt[4]{\frac{xyz}{3^9}}.$$
We then need only show that
$$\sqrt[4]{\frac{xyz}{3^9}}\ge xyz,$$
or that
$$1\ge 3^3xyz.$$
The last inequality is clear thanks to AM-GM.
|
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|
Maximisation problem I am trying the following question:
If$$a+b+c+d=0,\;a^2+b^2+c^2+d^2=1$$ Then what is the maximum value of
$ab+bc+cd+da?$
By the rearrangement inequality I can get $ab+bc+cd+da\leq 1$ but I am having trouble bringing the first condition into this. I have tried squaring it but that does not look like it leads anywhere. By playing with some numbers it looks like the answer might be $-½$. Help?
|
$ab+bc+cd+da = (a+c)(b+d) = -(a+c)^2 \leq 0$, and $\text{max} = 0$ when $a = -c$,
and so $b = -d$. Thus $a^2+b^2+c^2+d^2 = 2(a^2+b^2) = 1 \to a^2+b^2 = \dfrac{1}{2}$
|
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|
Evaluation of $\lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try}::$
$\bullet\; $If $x\in \mathbb{Z}\;,$ and $x\rightarrow \infty\;,$ Then $x^2<x^2+x+1<x^2+2x+1$
So $x<\sqrt{x^2+x+1}<(x+1)\;,$ So $\lfloor x^2+x+1 \rfloor = x\;,$ bcz it lies between two integers.
So $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right) = \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right)\cdot \frac{\left(\sqrt{x^2+x+1}+x\right)}{\left(\sqrt{x^2+x+1}+x\right)}$
So $\displaystyle \lim_{x\rightarrow \infty} \frac{x+1}{\sqrt{x^2+x+1}+x} = \frac{1}{2}$
$\bullet\; $ If $x\notin \mathbb{Z}$ and $x\rightarrow \infty\;,$ Then How can i solve the above limit in that case,
Help me, Thanks
|
As you mention, when $x$ is a large integer, the difference is roughly $1/2$. On the other hand, when $\sqrt{x^2+x+1}$ is an integer, which happens for unboundedly large $x$, the difference is $0$. This gives two different sequences tending to infinity with two different limits of your function $\sqrt{x^2+x+1}-\lfloor\sqrt{x^2+x+1}\rfloor$, showing that it doesn't tend to a limit. In fact, this shows that its $\liminf$ is $0$, and you can similarly show that the $\limsup$ is $1$.
|
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|
Can -3 and 2 be eigenvalues of the following matrix? Can $-3$ and $2$ be eigenvalues of and nxn matrix B such that $A = B^{2}+B-6I$ and A's determinant is $0$?
So this is what I concluded:
At first glance, it can be seen that the matrix $A$ can be factored into two different terms.
\begin{align*} A = B^{2}+B-6I = (B+3I)(B-2I) \end{align*}
\begin{align*}(B+3I)(B-2I) = B^{2}-2IB+3IB-6I^{2} \end{align*}
\begin{align*} 2IB=2B, 3IB=3B, I^{2}=I.\end{align*}
\begin{align*} (B+3I)(B-2I) = B^{2}-2IB+3ID-6I^{2} = B^{2}+B-6I = A. \end{align*}
Given that the $det(A) = 0$, let the matrix $C = B+3I$, the matrix $D = B-2I$, and
\begin{align*}
det(A) = det(CD)=
\begin{vmatrix}CD\end{vmatrix}
=
\begin{vmatrix} C\end{vmatrix}
\begin{vmatrix} D\end{vmatrix}
= 0.
\end{align*}
In turn, this suggests that either \begin{align*} \begin{vmatrix} C \end{vmatrix}, \begin{vmatrix} D \end{vmatrix}, or (
\begin{vmatrix}C\end{vmatrix} and \begin{vmatrix}D\end{vmatrix}) = 0.\end{align*}
So my question is... is this the correct way of proving this so far?
|
Your reasoning starts out fine, up to "In turn, this suggests that either
$$
|C|,|D|, \textrm{ or } |C| \textrm{ and } |D| = 0.''
$$
After that, I'm not sure what you're trying to do, but it is something circular/unnecessary.
To start over from your last correct assertion, you now know that
$$
\det[(B + 3I)(B-2I)] = 0
$$
which tells you that either
$$
\det(B+3I) = 0
$$
or
$$
\det(B-2I) = 0
$$
(or both). If the first statement is true, then $-3$ is an eigenvalue of $B$. This is (equivalent to) the definition of eigenvalue. Similarly, if the second statement is true, then $2$ is an eigenvalue. So, at least one of those two things is true, and thus both could be eigenvalues. (But it isn't required that $B$ have both as eigenvalues, consider $B = \begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}$; so the correct statement is "at least one of $2$ or $-3$ is an eigenvalue of $B$")
|
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|
Inequality: $(a^3+3b^2+5)(b^3+3c^2+5)(c^3+3a^2+5) \ge 27(a+b+c)^3$ Proving inequality for positive real $a,b,c > 0$: $$ (a^3+3b^2+5)(b^3+3c^2+5)(c^3+3a^2+5) \ge 27(a+b+c)^3$$
|
Use AM-GM inequality we have $$a^3+2=a^3+1+1\ge 3a$$
so
$$a^3+3b^2+5\ge 3a+3b^2+3=3(a+b^2+1)$$
so
$$LHS\ge 27(a+b^2+1)(1+b+c^2)(a^2+1+c)$$
Use Holder inequality we have
$$(a+b^2+1)(1+b+c^2)(a^2+1+c)\ge (a+b+c)^3$$
By done!
|
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|
Proving the sum of the reciprocals squared converges I'm investigating the Basel Problem, and the sum to consider is:
$\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}$
How can I show this converges? Using graphs/computer software is also fine, but how would I do it? Is there a way using calculus?
I've looked on Google for a while and I found a few proofs, but I didn't understand any of them.
|
Note that
$$
\begin{alignat}{8}
&\frac{1}{1^2}+&&\frac{1}{2^2}+&&\frac{1}{3^2}+&&\frac{1}{4^2}+&&\frac{1}{5^2}+&&\frac{1}{6^2}+&&\frac{1}{7^2}+&&\dots\\
\leq &\frac{1}{1^2}+&&\frac{1}{2^2}+&&\frac{1}{2^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\dots
\end{alignat}
$$
where there are $2^n$ copies of $\frac{1}{(2^n)^2}$.
But $\frac{1}{(2^n)^2}=\frac{1}{4^n}$, so $2^n$ copies of it is $\frac{2^n}{4^n}=\frac{1}{2^n}$. So the sum is bounded above by $\sum_{k=0}^\infty \frac{1}{2^n}=2$.
|
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Compute $\lim\limits_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left (\frac{i^2}{n^2} \right)$ Compute the given limit
$$
\lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left (\frac{i^2}{n^2} \right)
$$
The sum is:
Can someone please show me the steps to complete this problem? The answer I arrived at was 0 but the homework program is telling me that it's wrong.
Thank you.
|
Hint1:
$$\begin{split}
\lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left(\frac{i^2}{n^2} \right)
&= \lim_{n \to \infty} \frac{981}{n+5} \cdot \frac{\sum_{i=1}^{n}i^2}{n^2}\\
\end{split}$$
Hint2:
$$ \sum_{i=1}^{n}i^2 = \frac{n\cdot (n+1) \cdot 2n+1)}{6}$$
$$\begin{split}
\lim_{n \to \infty }\frac{981}{n+5}\sum_{i=1}^{n} \left(\frac{i^2}{n^2} \right)
&= \lim_{n \to \infty} \frac{981}{n+5} \frac{\sum_{i=1}^{n}i^2}{n^2}\\
&= \lim_{n \to \infty}\frac{981}{n+5} \frac{n(n+1)(2n+1)}{6 \cdot n^2}\\
&= \lim_{n \to \infty} \frac{n}{n} \cdot \frac{981 (2n^2+3n+1)}{(6n^2 +30n)}\\
&= \lim_{n \to \infty}\frac{n^2}{n^2} \cdot \frac{981\cdot(2+\frac{3}{n} + \frac{1}{n^2})}{6(1 + \frac{30}{n})} \\
&= \lim_{n \to \infty}\frac{327\cdot(2+\frac{3}{n} + \frac{1}{n^2})}{2(1 + \frac{30}{n})} &= 327
\end{split}$$
|
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|
Why is $(1-\cot 37^\circ)(1-\cot 8^\circ)=2.00000000\cdots$? Apparently,
$$(1-\cot 37^\circ)(1-\cot 8^\circ)=2.00000000000000000\cdots$$
Since it is a $2.0000000000\cdots$ instead of $2$, it isn't exactly $2$.
Why is that?
|
Alternatively,
$$\cot(x+y) = \frac{\cos(x+y)}{\sin(x+y)} = \frac{\cos x \cos y - \sin x \sin y}{\sin x \cos y + \cos x \sin y} = \frac{\cot x \cot y - 1}{\cot x + \cot y}$$
Now if $x = 37^\circ$, $y = 8^\circ$, then $\cot(x + y) = \cot 45^\circ = 1$ and
$$\begin{align} \ \ \ \ 1 & = \frac{\cot x \cot y - 1}{\cot x + \cot y} \\ \cot x + \cot y & = \cot x \cot y - 1 \\
2 + \cot x + \cot y & = \cot x \cot y + 1 \\
1 - \cot x - \cot y + \cot x \cot y & = 2 \\
(1 - \cot x)(1 - \cot y) & = 2 \\
(1 - \cot 37^\circ)(1 - \cot 8^\circ) & = 2
\end{align}$$
|
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|
Evaluating $\int [(5x^3-3x^2+7x-3)/(x^2+1)^2] dx$ Can I get hints on how to factor?
Should I break down by term, i.e. $$\frac{5x^3}{(x^2+1)^2}\text{...?}$$
|
Since the degree of the numerator is less than the degree of the denominator, you can immediately apply the method of partial fractions to decompose the integrand: You're looking for numbers $A, B, C, D$ such that,
$$\frac{5x^3 - 3x^2 + 7x - 3}{(x^2 + 1)^2} = \frac{Ax + B}{(x^2 + 1)^2} + \frac{Cx + D}{x^2 + 1}$$
which you can find, for example, by first clearing denominators.
Then, you can deal with each of the pieces separately, each of which is a standard exercise: To handle the terms
$$\frac{Ax}{(x^2 + 1)^2} \qquad \text{and} \qquad \frac{Cx}{x^2 + 1},$$
use a $u$-substitution, and to handle the other two, use a trigonometric substitution.
The partial fractions decomposition of the rational expression here turns out to be $$\frac{2x}{(x^2 + 1)^2} + \frac{5x-3}{x^2 + 1},$$ so $B = 0$. In particular this means you don't have to evaluate $$\int \frac{dx}{(x^2 + 1)^2},$$ which is easily the most involved of the four integrals.
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.