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Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides.the length of each perpendicular is equal to a units.The distance between their feet being equal to b units.Find the area of the rhombus
| Set the $A,B,C,D \,\,$ vertices of the rhombus in $(m,0) $, $(0,n)\,\, $, $(-m,0)\,\, $, and $(0, -n) \,\, $, respectively. The slope of the side $AB\,\, $ is $-n/m \,\, $. So the slope of the perpendicular to $AB\,\, $, drawn from its opposite vertex $C $, is $m/n\,\, $. Since this perpendicular line passes through $(-n,0) \,\, $, its equation is $y=m/n x +m^2n \,\, \,\, $. The $(x_0,y_0 ) \,\, $ coordinates of the crossing point between this perpendicular line and $AB $ (let us call this foot point $E $) can be found by solving
$$ m/n x +m^2/n=-n/m x +n$$
which gives
$$x_0=\frac {m (n^2-m^2)}{n^2+m^2}$$
and then
$$y_0=\frac {2n m^2}{n^2+m^2}$$
Since the distance between $E $ and $C $ is $a $, we have
$$\left ( \frac {m (n^2-m^2)}{(n^2+m^2)}+m \right)^2+\frac {4n^2 m^4}{(n^2+m^2)^2}=a^2$$
which reduces to
$$\frac {4n^2 m^2}{n^2+m^2}=a^2$$
Also note that, as a result of the symmetry of the problem, the distance between $E $ and its symmetric point on side $CD\,\, $ (i.e. the foot of the perpendicular drawn from vertex $A $ to $CD\,\, $) is twice the distance between $E $ and the origin. Since this distance is given as equal to $b $, we have
$$\frac {m^2 (n^2-m^2)^2}{(n^2+m^2)^2}+\frac {4n^2 m^4}{(n^2+m^2)^2}=\frac {b^2}{4}$$
which reduces to
$$m^2=\frac {b^2}{4}$$
and then $m=b/2 \,\, $. Substituting in the equation above, this also leads to
$$\frac{4n^2 b^2/4}{n^2+b^2/4}=a^2$$
$$n=\frac {ab}{2\sqrt {b^2-a^2}} $$
Since the area $A $ of the rhombus is $2mn \,\, $, it follows
$$A= 2 \frac {b}{2} \cdot \frac{ab}{2\sqrt {b^2-a^2}}$$
$$=\frac {ab^2}{2 \sqrt {b^2-a^2}} $$
Note that, when $n=m \,\, $, that is to say the rhombus is a square, $a $ corresponds to the side of the square and $b $ to its diagonal, so in this case $a=\sqrt {2}n \,\, $ and $b=2n \,\, $. Accordingly, the formula reduces to $A=2n^2\,\, $, which is the area of a square with vertices in $(n,0)\,\, $, $(0,n) \,\, $, $(-n,0) \,\, $, $(0, -n) \,\, $.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying fraction with factorials: $\frac{(3(n+1))!}{(3n)!}$ I was trying to solve the limit:
$$\lim_{n \to \infty} \sqrt[n]{\frac{(3n)!}{(5n)^{3n}}}$$
By using the root's criterion for limits (which is valid in this case, since $b_n$ increases monotonically):
$$L= \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
Now I realise using Sterling's formula would make everything easier, but my first approach was simplifying the factorial after applying the criterion I mentioned before. So, after a few failed attempts I looked it up on Mathematica and it said that $\frac{(3(n+1))!}{(3n)!}$ (which is one of the fractions you have to simplify) equals $3(n+1)(3n+1)(3n+2)$. Since I can't get there myself I want to know how you would do it.
Just so you can correct me, my reasoning was:
$$\frac{(3(n+1))!}{(3n)!} = \frac{3\cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 4 \cdot (...) \cdot 3 \cdot (n+1)}{3 \cdot 1 \cdot 3 \cdot 2 \cdot 3 \cdot 3 \cdot (...) \cdot 3 \cdot n } = $$
$$= \frac{3^n(n+1)!}{3^{n}n!} = \frac{(n+1)!}{n!} = n+1$$
Which apparently isn't correct. I must have failed at something very silly. Thanks in advance!
| $$ (3(n+1))! \neq 3 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 4 \cdots 3 \cdot (n+1) $$ To make it clear what the problem is, let's write the right-hand side with brackets: $$ (3 \cdot 2) \cdot (3 \cdot 3) \cdot (3 \cdot 4) \cdots (3 \cdot (n+1)) $$ That's just multiplying all the positive multiples of 3 less than $ 3(n+1) $ together; the factorial is defined as multiplying together all positive integers less than $ 3(n+1) $. So the correct expansions are \begin{align*}
(3(n+1))! &= 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots (3n-2) \cdot (3n-1) \cdot 3n \cdot (3n+1) \cdot (3n+2) \cdot 3(n+1) \\
(3n)! &= 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots (3n-2) \cdot (3n-1) \cdot 3n
\end{align*}
which clearly have the quotient Mathematica gave you.
| {
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Ask about beautiful properties of $e$ One of students asked me about "some beautiful properties (or relation) of $e$". Then I list like below
\begin{align}
& e \equiv \lim_{x \to \infty} \left(1+\frac{1}{x} \right)^x\\[10pt]
& e = \sum_{k=0}^\infty \frac{1}{k!}\\[10pt]
& \frac{d}{dx} (e^x) = e^x\\[10pt]
& e^{ix} = \cos x + i \sin x \quad \text{(Euler)}\\[10pt]
& e^{i \pi} + 1 = 0
\end{align} After this, he asked me for more relation or properties. I said
I'll think and answer ...
Now I want help to add some relation, properties, or visual things (like proof without words)
Please help me to add something more. Thanks in advance.
***The class was math. 1. engineering
| Here are some more relations which might be pleasing.
From section 1.3 of Mathematical Constants by S.R. Finch:
*
*A Wallis-like infinite product is
\begin{align*}
e=\frac{2}{1}\cdot\left(\frac{4}{3}\right)^{\frac{1}{2}}
\cdot\left(\frac{6\cdot 8}{5\cdot 7}\right)^{\frac{1}{4}}
\cdot\left(\frac{10\cdot 12\cdot 14\cdot 16}{9\cdot 11\cdot 13\cdot 15}\right)^{\frac{1}{8}}\cdots
\end{align*}
*From Stirling's formula we derive
\begin{align*}
e=\lim_{n\rightarrow \infty}\frac{n}{(n!)^{\frac{1}{n}}}
\end{align*}
*Another continued fraction representation is
\begin{align*}
e&=2+\frac{\left.2\right|}{\left|2\right.}
+\frac{\left.3\right|}{\left|3\right.}
+\frac{\left.4\right|}{\left|4\right.}
+\frac{\left.5\right|}{\left|5\right.}+\cdots\\
&=2+\frac{2}{2+\frac{3}{3+\frac{4}{4+\frac{5}{5+\cdots}}}}
\end{align*}
$$ $$
In section Intriguing Results in Real Infinite Series by D.D. Bonar and M.J. Khoury we find
*
*Gem 89 (American Math Monthly 42:2 pp. 111-112)
\begin{align*}
e=\frac{1}{5}\left(\frac{1^2}{0!}+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\cdots\right)
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it possible to evaluate $\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$? Let's say we have the following limit:
$$\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$$
Would the following solution be correct?
The solution is incorrect, please see the correction of @YvesDaoust
\begin{align}
\lim_{x \rightarrow 0}(-1+\cos x)^{\tan x}
&= \lim_{x \rightarrow 0}\left((1-\cos x)^{\tan x}\cdot (-1)^{\tan x}\right) \\
&= \lim_{x \rightarrow 0}\left(1-\left(1-2\sin^2\left(\frac{x}{2}\right)\right)^{\tan x}\right) \cdot \lim_{x \rightarrow 0}(-1)^{\tan x} \\
&= \lim_{x \rightarrow 0}\left(2\sin^2 \left(\frac{x}{2}\right)\right)^{\tan x} \cdot 1\\
&= \lim_{x \rightarrow 0}2^{\tan x} \cdot \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\tan x} \\
&= 1 \cdot \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\tan x} \\
&= \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\frac{\sin x}{\cos x}} \\
&= \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{\frac{4\sin (\frac{x}{2})\cdot \cos (\frac{x}{2})}{\cos x}} \\
&= \lim_{x \rightarrow 0}\left(\sin \left(\frac{x}{2}\right)^{\sin \left(\frac{x}{2}\right)}\right)^{\frac{4\cos\frac{x}{2}}{\cos x}} \\
&= \left(\lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{\sin \left(\frac{x}{2}\right)}\right)^{\lim_{x \rightarrow 0}\frac{4\cos\frac{x}{2}}{\cos x}} \\
&= \left(\lim_{u \rightarrow 0}u^u\right)^{4} \\
&= 1^4 \\
&= 1 \\
\end{align}
The result seems to be correct, but the way leading to it seems to be quite lengthy. Am I doing something redundant?
| You can reduce 1 step that's looking odd.
You can write simply.
$$\lim_{x\to 0} \left[-1\left(1-\cos x \right)\right]^{\tan x}$$
$$\lim_{x\to 0} \left[-1\left(2\sin^2 \frac x2\right)\right]^{\tan x}$$
$$\lim_{x\to 0} \left(-2\right)^{\tan x} . \left(\sin^2 \frac x2 \right)^{\tan x}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $\alpha $ and $\beta$ are roots of equation $a\tan\theta +b \sec\theta=c$. Prove that $\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$ If $\alpha $ and $\beta$ are roots of equation $a\tan\theta +b \sec\theta=c$. Prove that $\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$
i have converted tan to sin and cos and reached to $\sin^2\theta(a^2-c^2+2ac) + c^2-b^2-ac=0$. how do i proceed
thanks
| We can write the quadratic equation as: $$a\tan \theta + b\sec \theta = c $$ $$\Rightarrow a\tan \theta + b\sqrt {\tan^2 \theta + 1} = c $$ $$\Rightarrow (a^2-b^2)\tan^2 \theta -2ac \tan \theta +(c^2-b^2)=0$$
Now $$\tan (\alpha +\beta ) =\frac {\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}. $$ Can you take it from here? Hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$ Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$
My Attempt,
$$L.H.S=8\cos^3(\frac {\pi}{9}) - 6\cos(\frac {\pi}{9})$$
$$=2\cos(\frac {\pi}{9}) [4\cos^2(\frac {\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2+2\cos(\frac {2\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2\cos(\frac {2\pi}{9})-1]$$.
What should I do further?
| Hint:
$$\cos(3x) = 4\cos^3(x) - 3\cos(x).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to show that $F(x)=F(xq)+xq F(xq^2)$ Define$$\begin{align*}F(x) & =1+\dfrac {xq}{1-q}+\dfrac {x^2q^4}{(1-q)(1-q^2)}+\dfrac {x^3q^9}{(1-q)(1-q^2)(1-q^3)}+ \cdots\\ & =1+\sum\limits_{n=1}^{\infty}\dfrac{x^nq^{n^2}}{(q;q)_{n}}\end{align*}$$
Then $F(x)$ also satisfies $F(x)=F(xq)+xq\cdot F(xq^2)$.
Question: How would you prove $F(x)=F(xq)+xq\cdot F(xq^2)$?
I attempted substituting $x=xq$ into $F(x)$ to get$$\begin{align*}F(xq) & =1+\dfrac {xq^2}{1-q}+\dfrac {x^2q^6}{(1-q)(1-q^2)}+\dfrac {x^3q^{12}}{(1-q)(1-q^2)(1-q^3)}+\cdots\\ & =1+\sum_{n=1}^{\infty}\dfrac {x^nq^{n^2+n}}{(q;q)_n}\end{align*}$$
And$$\begin{align*}F(xq^2) & =1+\dfrac {xq^3}{1-q}+\dfrac {x^2q^8}{(1-q)(1-q^2)}+\dfrac {x^3q^{15}}{(1-q)(1-q^2)(1-q^3)}+\cdots\\ & =1+\sum\limits_{n=1}^{\infty}\dfrac {x^nq^{n^2+2n}}{(q;q)_n}\end{align*}$$
So $xq\cdot F(xq^2)=xq+xq\sum\limits_{n=1}^{\infty}\dfrac {x^nq^{n^2+2n}}{(q;q)_n}$. However, adding them together,$$F(xq)+xq\cdot F(xq^2)=1+\sum\limits_{n=1}^{\infty}\dfrac {x^nq^{n^2+n}}{(q;q)_n}+xq+xq\sum\limits_{n=1}^{\infty}\dfrac {x^nq^{n^2+2n}}{(q;q)_n}$$Which I don't think equals $F(x)$.
| Using the convention $(q;q)_0 = 1$, we can write
$$F(x) = \sum_{n = 0}^\infty \frac{x^n q^{n^2}}{(q;q)_n}$$
So
$$F(xq) = \sum_{n = 0}^\infty \frac{x^n q^{n^2 + n}}{(q;q)_n}$$
and
$$xq F(xq^2) = xq\sum_{n = 0}^\infty \frac{x^n q^{n^2+2n}}{(q;q)_n} = \sum_{n = 0}^\infty \frac{x^{n+1}q^{n^2+2n+1}}{(q;q)_n} = \sum_{n = 1}^\infty \frac{x^nq^{n^2}}{(q;q)_{n-1}} = \sum_{n = 1}^\infty \frac{x^nq^{n^2}}{(q;q)_n}(1-q^n)$$
Thus
$$F(xq) + xqF(xq^2) = 1 + \sum_{n = 1}^\infty \frac{x^nq^{n^2}}{(q;q)_n}(q^n + (1-q^n)) = 1 + \sum_{n = 1}^\infty \frac{x^nq^{n^2}}{(q;q)_n} = F(x)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The Formula to this Sequence Series What is the formula for following sequence?
$$\frac12 + \frac12 \cdot \frac34 + \frac12 \cdot \frac34 \cdot \frac56 + ... + \frac12 \cdot \frac34 \cdot \frac56 ... \frac{2n - 1}{2n}$$
This is a question from my Calculus class about sequences.
| $$
\begin{align}
&\, 1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1) = (2n-1)!! \,,\quad\, 2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n) = (2n)!! \\[2mm]
&\, S_{\small n} = \frac{1}{2}+\frac{1\,\cdot\,3}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[6mm]
&\, \color{red}{1+S_{\small n}} = 1+\frac{1}{2}+\frac{1\,\cdot\,3}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[3mm]
&\, \qquad = \frac{1\,\cdot\,3}{2}+\frac{1\,\cdot\,3}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[3mm]
&\, \qquad = \frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[3mm]
&\, \qquad = \,\cdots\, = \frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n+1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} = \color{red}{\frac{(2n+1)!!}{(2n)!!}} \,\Rightarrow\, S_{\small n}=\frac{(2n+1)!!}{(2n)!!}-1 \\[6mm]
&\, 1+\lim_{n\rightarrow\infty}S_{\small n} = \lim_{n\rightarrow\infty}\frac{(2n+1)!!}{(2n)!!}\color{red}{\,\longrightarrow\,\infty} \qquad \{{\small\text{divergent}}\}
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the determinant of order $100$ Find the determinant of order $100$:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}$$
I think I should be using recurrence relations here but I'm not entirely sure how that method works. I tried this:
Multiplying the first row by $(-1)$ and adding it to all rows:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
0 &0 &0 &\ldots &0 &-6 &6\\
0 &0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &0 &-6 &\ldots &0 &0 &6\\\
0 &-6 &0 &\ldots &0 &0 &6\\
-6 &0 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
Applying Laplace's method to the first column
$$D=5\begin{vmatrix}
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}+6\begin{vmatrix}
5 &5 &\ldots &5 &5 &-1\\
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
I can see that this one is $D$ but of order $99$...Is this leading anywhere? How would you solve this?
| Using this matrix determinant lemma I am getting $(1+5\cdot100\cdot(-\frac{1}{6}))(-6)^{100}$.
Note about general size, one needs to think about the determinant of anti-diagonal matrix to compute it. My guess is that it equals $(1+5n(-\frac{1}{6}))(-1)^{\frac{n(n-1)}{2}}(-6)^{n}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the maximal value of the expression $\overline{abc}-(a^3+b^3+c^3)?$ Let $\overline{abc}$ be $3$-digits number. What is the maximal value of the expression $\overline{abc}-(a^3+b^3+c^3)?$
Computer gives the answer $396$ for the number $620$ but I have no idea how to do it by hand.
| You can write $\overline{abc} = 100a + 10 b + c$. Then $$\overline{abc} - (a^3 + b^3 + c^3) = a(100-a^2) + b(10 - b^2) + c(1-c^2).$$ You just need to find the digits $a,b,c$ that maximize each term individually.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculate: $\lim_{x\to 0}{\frac{\tan x-\sin x}{x^3}}$
Calculate: $\lim{\frac{\tan x-\sin x}{x^3}}$ as $x\to 0$
I have solved it using a way that gives a wrong answer but I can't figure out why:
$$\lim{\frac{\tan x-\sin x}{x^3}}\\
=\lim{\frac{\tan x}{x^3}-\lim\frac{\sin x}{x^3}}\\
=\lim{\frac{1}{x^2}}\cdot\lim{\frac{\tan x}{x}-\frac{1}{x^2}}\cdot\lim{\frac{\sin x}{x}}\\
=\lim{\frac{1}{x^2}}-\lim{\frac{1}{x^2}}
=0$$
The answer using the standard method gives $\frac{1}{2}$.
It would be great if someone could explain to me why this method is wrong.
| 1) With l'Hospital:
$$\lim_{x\to0}\frac{\tan x-\sin x}{x^3}\stackrel{\text{l'H}}=\lim_{x\to0}\frac{\frac1{\cos^2x}-\cos x}{3x^2}\stackrel{\text{l'H}}=\lim_{x\to0}\frac{\frac{2\sin x}{\cos^3x}+\sin x}{6x}=$$
$$=\lim_{x\to0}\frac16\frac{\sin x}x\left(\frac2{\cos^3x}+1\right)=\frac16\cdot1\cdot(2+1)=\frac12$$
2) Without l'Hospital:
$$\lim_{x\to0}\frac{\tan x-\sin x}{x^3}=\lim_{x\to0}\frac{\sin x}x\frac{1-\cos x}{x^2}\frac1{\cos x}=1\cdot\frac12\cdot1=\frac12$$
Where we used
$$\frac{1-\cos x}{x^2}=\frac{\sin^2x}{(1+\cos x)x^2}=\frac1{1+\cos x}\left(\frac {\sin x}x\right)^2\xrightarrow[x\to0]{}\frac12\cdot1^2=\frac12$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the matrix of a linear transformation3. Let $f$$ \begin{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{pmatrix}
\end{pmatrix}
$=
$ \begin{pmatrix}
x_1+2x_2+x_3 \\
x_1-2x_2 \\
x_1+x_3 \\
3x_1-4x_2 \\
\end{pmatrix}
$ be a linear map from $\Bbb{R}^3 \to \Bbb{R}^4$
Let $v_1= \begin{pmatrix}
1 \\
0 \\
1 \\
\end{pmatrix}
$ $v_2= \begin{pmatrix}
0 \\
1 \\
0 \\
\end{pmatrix}
$ $v_3= \begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}
$ and $w_1= \begin{pmatrix}
1 \\
0 \\
1 \\
0\\
\end{pmatrix}
$ $w_2= \begin{pmatrix}
0 \\
1 \\
0 \\
1\\
\end{pmatrix}
$
$w_3= \begin{pmatrix}
2 \\
0 \\
0 \\
0\\
\end{pmatrix}
$
$w_4= \begin{pmatrix}
0 \\
0 \\
0 \\
2\\
\end{pmatrix}
$ Let $B=(v_1,v_2,v_3)$ and $ C=(w_1,w_2,w_3,w_4)$
Find the matrix $M_{B,C}(f)$
Hi so this is the question . What i did was i took a vector $v_1$ i applied f to it and got a vector like $(2,1,2,3)^T$ then i tried to find the coefficeients of the linear combination from $f(v_1)=aw_1+bw_2+cw_3+dw_4$ and solved the system of equation. I put the solutions as coloumns in my matrix. I repeated this 3 times and got a matrix like
$ \begin{pmatrix}
2 & 0 & 1 \\
1 &-2 & 1 \\
1 &-1 &0 \\
2 &-1 &1\\
\end{pmatrix}
$
My question is how can i check if what i did was correct ?
| Here is another way to do it:
$B$ is the representation of the principle vectors of B in the standard basis.
$B^{-1}$ is then the representation of the principle vectors in the standard basis as represented in the basis of $B.$
We can do a similar exercise for $C.$
$fB$ would take the principle vectors in $B$ to a matrix in the standard basis of $\mathbb R^4.$
Then we have to convert a matrix in the standard basis into the basis for $C.$
$C^{-1}fB$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$ I need to find the following limit: $$\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$$
I started this way: $$\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=e^{\frac{1}{x}\cdot \ln\left[\frac {\sin x}{x}\right]}$$
So it's enough to find: $$\lim_{x\to 0^+}\frac{\ln(\frac {\sin x}{x})}{x}$$
I tried to use L'Hôpital's rule but it got me nowhere. Any ideas?
| Hints:
*
*Write
$$\left(\frac{\sin x }{x} \right)^\frac{1}{x} = \exp\left( \frac{1}{x} \ln \left(\frac{\sin x }{x} \right)\right).$$
*$\exp$ is continuous, $\lim_{x\to 0} \frac{\sin x }{x}=1$ and $\ln 1=0$.
*Evaluate $\lim_{x\to 0}\frac{1}{x} \ln \left(\frac{\sin x }{x} \right)$ by using L'Hospital.
For step 3 use $$\ln \left(\frac{\sin x }{x} \right)' =\frac{\cos x}{\sin x} - \frac{1}{x} = \frac{x\cos x - \sin x}{x \sin x}.$$
For $x\to 0+$ we have an expression like $0/0$ again. Now apply L'Hospital again:
$$\lim_{x\to 0+}\frac{x\cos x-\sin x}{x\sin x}=\lim_{x\to 0+}\frac{(\cos x-x\sin x)-\cos x}{\sin x+x\cos x}=-\lim_{x\to 0+}\frac{x\sin x}{\sin x+x\cos x}$$
$$=-\lim_{x\to 0+}\frac{\sin x+x\cos x}{\cos x+(\cos x-x\sin x)}=-\lim_{x\to 0+}\frac{\sin x+x\cos x}{2\cos x-x\sin x}=0.$$
So we get
$$ \lim_{x\to 0+} \left(\frac{\sin x }{x} \right)^\frac{1}{x}= \lim_{x\to 0+} \exp\left( \frac{1}{x} \ln \left(\frac{\sin x }{x} \right)\right) = \exp \left( \lim_{x\to 0+} \ln \left(\frac{\sin x }{x}\right) \right) = \exp 0 =1.$$
| {
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"url": "https://math.stackexchange.com/questions/2127280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomials $P(x,y)$ with nonnegative integer coefficients such that $P(x,y) \equiv 1 \text{ (mod } x+y-1)$ and $P(1,1) = n$. In 1971 Richard Guy sent a letter to Neil Sloane outlining some integer sequences. One of these sequences, A279196, was added to the OEIS by Neil only in December of 2016:
A279196: Number of polynomials $P(x,y)$ with nonnegative integer coefficients such that $P(x,y) \equiv 1 \text{ (mod } x+y-1)$ and $P(1,1) = n$.
The first few terms of the sequence are:
1, 1, 2, 5, 13, 36, 102, 295, 864
In a recent lecture (see 9:50), Neil said:
I don't even know what the [...] polynomials are for the first few values, so it might be interesting to look into this.
I tried to address this in an ad hoc manner, and I managed to find all of the examples for $n \leq 3$ and three examples for $n = 4$:
$a(1) = 1$ with:
*
*$1$
$a(2) = 1$ with:
*
*$x + y = (x + y - 1) + 1$
$a(3) = 2$ with:
*
*$xy + x + y^2 = (y + 1) (x + y - 1) + 1$ and
*$xy + y + x^2 = (x + 1) (x + y - 1) + 1$
And for $n = 4$:
*
*$x^2 + y^2 = (x + y + 1)(x + y - 1) + 1$
*$x^2y + x^2 + xy^2 + y = (xy + x + 1)(x + y - 1) + 1$
*$x^2y + xy^2 + x + y^2 = (xy + y + 1)(x + y - 1) + 1$
What is an algorithm or technique to enumerate polynomials for some arbitrary $n$?
Could anyone provide an example of, say, the remaining two polynomials when $n=4$? All thirteen polynomials for $n=5$?
| I think I figured it out (but I would still love to see a proof that this technique is exhaustive!)
It seems that (at least the small) values can be built up recursively:
To create a list of polynomials for $a(n)$, for each term (on the right) and polynomial (on the left) in $a(n-1)$, take a term from the left-hand side and move it to the factor of the right hand side.
For example, to create a polynomial for $n = 4$, take the first term/factor pair for $n = 3$ (highlighted below in bold) and move the term to be inside of the factor:
n = 4 (example 1)
$\boldsymbol{xy} + x + y^2 = \boldsymbol{(y + 1)}(x + y - 1) + 1$
Leads to:
$x^2y + xy^2 + x + y^2 = \boldsymbol{(xy + y + 1)}(x + y - 1) + 1$
Remaining examples:
n = 4 (example 2)
$xy + \boldsymbol{x} + y^2 = \boldsymbol{(y + 1)}(x + y - 1) + 1$ and
$xy + \boldsymbol{y} + x^2 = \boldsymbol{(x + 1)}(x + y - 1) + 1$
Leads to:
$x^2 + y^2 = \boldsymbol{(x + y + 1)}(x + y - 1) + 1$
n = 4 (example 3)
$xy + x + \boldsymbol{y^2} = \boldsymbol{(y + 1)}(x + y - 1) + 1$
Leads to:
$xy^2 + xy + x + y^3 = \boldsymbol{(y^2 + y + 1)}(x + y - 1) + 1$
n = 4 (example 4)
$\boldsymbol{xy} + y + x^2 = \boldsymbol{(x + 1)} (x + y - 1) + 1$
Leads to:
$x^2y + xy^2 + x + y^2 = \boldsymbol{(xy + y + 1)}(x + y - 1) + 1$
n = 4 (example 5)
$xy + y + \boldsymbol{x^2} = \boldsymbol{(x + 1)} (x + y - 1) + 1$
Leads to:
$ x^3 + x^2y + xy + y = \boldsymbol{(x^2 + x + 1)}(x + y - 1) + 1$
| {
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} |
Find the maximum of the value a such $\sum_{cyc}\sqrt{\frac{x}{ax+y+z}}\le 3\sqrt{\frac{1}{2+a}}$ Find the maximum of the value $a$ such foy any real postive numbers $x,y,z$ have
$$\sqrt{\dfrac{x}{ax+y+z}}+\sqrt{\dfrac{y}{ay+z+x}}+\sqrt{\dfrac{z}{az+x+y}}\le3\sqrt{\dfrac{1}{2+a}}$$
I conjecture $a>0?$
| It's wrong for $a\rightarrow0^+$. Try $x\rightarrow+\infty.$
By the way, just by Jensen your inequality is true for all $a\geq\frac{3}{4}.$
Also, by Vasc's LCF Theorem it's enough to prove your inequality for $z=y$
and since our inequality is homogeneous, we can assume $y=z=1$,
which gives a minimal value of $a$, for which our inequality is true: $a=\frac{1}{3}$
For $a=\frac{1}{3}$ we get the following interesting inequality.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\sqrt{\frac{a}{a+3b+3c}}+\sqrt{\frac{b}{b+3c+3a}}+\sqrt{\frac{c}{c+3a+3b}}\leq\frac{3}{\sqrt7}$$
The equality occurs also for $a=b=1$ and $c=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Second-order linear differential equation of the form $x^2 y'' + (ax-b)y' - ay =0$ I need to solve the following differential equation
\begin{equation}
x^2 y'' + (ax-b)y' - ay =0
\end{equation}
with $a,b>0$, $x\geq 0$ and $y(0)=0$. The power series method will fail since there is a singularity at $x=0$, while the form of the equation does not conform with the Frobenius method.
What other methods can I try in order to solve this?
| $$x^2 y'' +(ax-b)y'-ay=0
$$
Start by removing the leading behavior for small $x$. Near $x=0$ the leading terms in the equation are
$$
-by'-ay \approx 0
$$
(this is valid as long as in the end, as $x\to 0$, $y''$ does not grow as fast as $y'/x^2$ or $y/x^2$, and $y'$ does not grow as fast as $\frac{y}x$).
The solution to the approximate equation is $y=e^{-\frac{a}{b}x}$ which motivates the substitution
$$
y = u(x) e^{-\frac{a}{b}x} \\
y' = u' e^{-\frac{a}{b}x} - \frac{a}{b}ue^{-\frac{a}{b}x} \\
y'' = u''e^{-\frac{a}{b}x}-2\frac{a}{b}u'e^{-\frac{a}{b}x}+\frac{a^2}{b^2}ue^{-\frac{a}{b}x}
$$
and the differential equation becomes
$$
x^2u'' - \left( 2\frac{a}{b}x^2 -ax+b\right)u'
+\left( \frac{a^2}{b^2}x^2-\frac{a^2}{b}x+a-a \right)u=0
$$
with $u(0)=0$.
Now if we let $u(x) = \sum_0^\infty u_kx^k$
$$ \sum_0^\infty k(k-1)u_kx^k-2\frac{a}{b}\sum_0^\infty ku_kx^{k+1}
+ a\sum_0^\infty ku_kx^k-b\sum_0^\infty ku_kx^{k-1} \\
+\frac{a^2}{b^2}\sum_0^\infty u_kx^{k+2}-\frac{a^2}{b}\sum_0^\infty u_kx^{k+1}=0
$$
And in each term we can substitute an offset index to always get $x^k$; this gives
$$
\left[ k(k-1) +ak\right]u_k -\left[2\frac{a}{b}(k-1)+\frac{a^2}{b}\right]u_{k-1} -b(k+1)u_{k+1}+
\frac{a^2}{b^2}u_{k-2}=0
$$
and this is to be solved with $u_{-1}=u_0=0$, there is one free parameter $u_1$.
$$
u_2 = \frac{au_1}{b} \\
$$
and so forth. So the solutions are this series, which is well behaved near zero, times $e^{-\frac{a}{b}x}$.
All of the coefficients are proportional to $u_1$. The overall solution has one free parameter (a scale factor that we will continue to call $u_1$) and is then
$$
y =u_1 e^{-ax/b} \left[
x+\frac{a}{b}x^2+\frac23 \frac{2^2}{b^2} x^3+ \frac{5a^3+3a^2}{12b^2}x^4
+\frac{17a^4+39a^3+36}{60b^4}x^5 + \cdots\right]\\ =
u_1 \left[ x + \frac{a^2}{6b^2}x^3+\frac{a^3+3a^2}{12b^3}x^4
+ \frac{3a^4+16a^3+24a^2}{60b^4}x^5+ \right]
$$
As to whether this series is convergent for all positive $x$, or has a finite radius of convergence, that depends on the values of $a$ and $b$.
| {
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Power series expansion of a holomorphic function One must show that equality$$\sum_{n=0}^\infty n^2z^n = \frac{z(z+1)}{(1-z)^3} $$ holds for $z\in\mathbb C$ and $|z|<1$.
I tried to expand $f(z):=\frac{z(z+1)}{(1-z)^3}$ into power series where $ z_0=0$ is the center and coefficients given by $a_n=\frac{1}{2\pi i}\int_{|z|<1}\frac{f(z)}{(z-0)^{n+1}}dz$, because f is holomorphic on the open domain $D=${$z:|z|<1$}.
Is there a more easy way to show this equality?
| $\sum_{n \ge 0} n^2z^n$ is absolutely convergent for $|z| < 1$ and so the following manipulations are valid for $|z| < 1$ :
$(1-z)^3(z+4z^2+9z^3+16z^4+\ldots)
\\ = (1-z)^2(z-z^2+4z^2-4z^3+9z^3-9z^4+16z^4+\ldots)
\\ = (1-z)^2(z+3z^2+5z^3+7z^4+\ldots)
\\ = (1-z)(z-z^2+3z^2-3z^3+5z^3-5z^4+7z^4+\ldots)
\\ = (1-z)(z+2z^2+2z^3+2z^4+\ldots)
\\ = z-z^2+2z^2-2z^3+2z^3-2z^4+2z^4+\ldots
\\ = z+z^2
$
This proves that $z+4z^2+9z^3+16z^4+\ldots = \frac {z+z^2}{(1-z)^3}$ when $|z| <1 $ (and $1-z \neq 0$ but this check is redundant)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the locus of the point of intersection of the tangents of the ellipse $b^2x^2+a^2y^2=a^2b^2$ at the points $\phi$ and $\frac{\pi}{2}+\phi$. - "a" and "b" are constants.
- "x" and "y" are variables.
- pi=22/7
- ø= an angle
I try this question at tangent method. But perhaps I do some mistake.
I take the tangents when intersect make angle $\phi$ and $\dfrac{\pi}{2}+\phi$ and then calculate for finding locus. But I failed at that way.
| Let $(x',y')$ be the point of intersection, then the equation of the polar line (i.e. the chord in this case) is
$$\frac{x'x}{a^2}+\frac{y'y}{b^2}=1 \tag{1}$$
End points of the chord:
$$(a\cos \phi,b\sin \phi) \: \text{ and } \: (-a\sin \phi,b\cos \phi)$$
Hence the equation of the chord is
\begin{align*}
\frac{y-b\sin \phi}{x-a\cos \phi} &=
\frac{b\sin \phi-b\cos \phi}{a\cos \phi+a\sin \phi} \\
(y-b\sin \theta)(a\cos \phi+a\sin \phi) &=
(x-a\cos \phi)(b\sin \phi-b\cos \phi) \\
b(\cos \phi-\sin \phi)x+a(\cos \phi+\sin \phi)y &=
ab(\cos^2 \phi+\sin^2 \phi) \\
\frac{(\cos \phi-\sin \phi)x}{a}+\frac{(\cos \phi+\sin \phi)y}{b} &= 1 \tag{2}
\end{align*}
Comparing $(1)$ and $(2)$,
$$\begin{pmatrix} x' \\ y' \end{pmatrix} =
\begin{pmatrix}
a(\cos \phi-\sin \phi) \\
b(\cos \phi+\sin \phi)
\end{pmatrix}$$
which lies on the ellipse
$$\frac{x^2}{2a^2}+\frac{y^2}{2b^2}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the inequality $\sin(x)\sin(3x) > \frac{1}{4}$
Find the range of possible values of $x$ which satisfy the inequation $$\sin(x)\sin(3x) > \frac{1}{4}$$
SOURCE : Inequalities (PDF)( Page Number 6; Question Number 306)
One simple observation is that both $x$ and $3x$ have to positive or negative simultaneously. I tried expanding $\sin(3x)$ by the regular indentity as :
$$\sin(x) \times \big(3\sin(x)-4\sin^3(x)\big) > \frac {1}{4}$$
$$\implies \sin^2(x)\times\big(3-4\sin^2(x)\big) >\frac{1}{4}$$
I do not find any way of proceeding. Wolfram Alpha gives 4 sets of answers. Do I have to observe this problem "case-by-case"? Can this question be solved without calculus ? Can anyone provide a hint to what should be done ?
Thanks in Advance ! :)
| $\require{cancel}$
\begin{eqnarray}
\sin x\sin3x &>&\dfrac14\\
2\sin x\sin3x&>&\dfrac12\\
\cos2x-\cos4x&>&\dfrac12\\
\cos2x-2\cos^22x+1&>&\dfrac12~~~~~~~~~~~~~~~~~~,~~~~~\cos2x=t\\
4t^2-2t-1&<&0
\end{eqnarray}
with $\Delta=20$ so $t=\dfrac{2\pm2\sqrt{5}}{8}=\dfrac{1\pm\sqrt{5}}{4}$ are roots and then
$$\dfrac{1-\sqrt{5}}{4}<t<\dfrac{1+\sqrt{5}}{4}$$
$$-0.309=\dfrac{1-\sqrt{5}}{4}<\cos2x<\dfrac{1+\sqrt{5}}{4}=0.809$$
thus $\color{blue}{18^\circ<x<54^\circ}$ or $\color{blue}{126^\circ<x<162^\circ}$ or $\color{blue}{198^\circ<x<234^\circ}$ or $\color{blue}{306^\circ<x<342^\circ}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort?
I will keep it short and take only an extract (most important part) of
the old task.
$$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$
What I have done is a lot work and time consuming, I have "simply" solved it. But I think with a lot less work, there would be an easier and faster way. It's just I cannot see it : /
If anyone wants see, here is my long solution which I'm not happy with:
$$\frac{n(n+1)(2n+1)+6(n+1)^2}{6}=\frac{(n^2+2n+n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{(2n^3+n^2+2n^2+n)+6n^2+12n+6}{6} = \frac{(n^2+3n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+3n^2+n+6n^2+12n+6}{6}=\frac{2n^3+3n^2+6n^2+9n+4n+6}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+9n^2+13n+6}{6}=\frac{2n^3+9n^2+13n+6}{6}$$
| Since you want $n+1$ to appear as a factor at the end, I would just leave that as a factor:
$$
\frac{n(n+1)(2n+1)+6(n+1)^2}6 = \frac {n(2n+1) + 6(n+1)} 6 (n+1).
$$
That becomes
$$
\frac {2n^2 + n + 6n + 6} 6 (n+1) = \frac{2n^2 + 7n+6} 6 (n+1) = \frac{(n+2)(2n+3)} 6 (n+1)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$ Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$
My Attempt:
$$L.H.S=\sec^2 20^\circ + \sec^2 40^\circ +\sec^2 80^\circ$$
$$=\dfrac {1}{\cos^2 20°} +\dfrac {1}{\cos^2 40°} +\dfrac {1}{\cos^2 80°}$$
$$=\dfrac {\cos^2 40°.\cos^2 80°+\cos^2 20°.\cos^2 80°+\cos^2 20°.\cos^2 40°}{\cos^2 20°.\cos^2 40°.\cos^2 80°}$$.
I got paused here. Please help to prove this..
| Let $z = \cos 20^\circ\,$ then by the triple angle formula $\frac{1}{2}=\cos 60^\circ = 4 z^3 - 3 z$ $\iff 8z^3-6z-1=0$.
By the double angle formula $\cos 40^\circ = 2 z^2 - 1\,$ and $\cos 80^\circ = 2(2z^2-1)^2-1=8z^4-8z^2+1\,$. But $8z^4=z\cdot8z^3=z(6z+1)$ per the previous equation, so $\cos 80^\circ = -2z^2+z+1\,$.
Then the equality to prove becomes:
$$
\frac{1}{z^2}+\frac{1}{(2z^2-1)^2}+\frac{1}{(2z^2-z-1)^2} = 36
$$
$$
\iff \quad (2z^2-1)^2 (2z^2-z-1)^2 + z^2 (2z^2-z-1)^2 + z^2 (2z^2-1)^2 - 36 z^2(2z^2-1)^2 (2z^2-z-1)^2 = 0
$$
After expanding and routine simplifications, the above reduces to:
$$
576 z^{10} - 576 z^9 - 1024 z^8 + 880 z^7 + 740 z^6 - 452 z^5 - 265 z^4 + 82 z^3 + 41 z^2 - 2 z - 1 = 0
$$
It can be verified by Euclidian division that the latter polynomial has $8z^3-6z-1$ as a factor:
$$
(8 z^3 - 6 z - 1) \cdot (72 z^7 - 72 z^6 - 74 z^5 + 65 z^4 + 28 z^3 - 17 z^2 - 4 z + 1)
$$
Therefore the equality holds, which completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Computing: $\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$ Find the following limit:
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$$
I'm not sure whether my solution is correct.
$t:=\frac{1}{x}$
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}=\lim_{t\to 0}\frac{\sqrt{1-\cos^2 t}\left(3^t-5^{-t}\right)}{\log_2(1+t^2+t^3)}$$
$$=\lim_{t\to 0}\frac{\frac{\sqrt{1-\cos^2t}}{\sqrt t^2}\cdot t\cdot\left(\frac{3^t-1}{t}\cdot t+(-t)\frac{(-5)^{-t}+1}{-t}\right)}{\log_2(1+t^2+t^3)}$$
$$=\frac{1}{2}(\ln 3+\ln 5)\left[\lim_{t\to 0}\log_2(1+t^2+t^3)^\frac{1}{t^2}\right]^{-1}=\frac{1}{2}(\ln3+\ln 5)\left(e^{{\lim_{t\to 0}\frac{t^2+t^3}{t^2}}^{-1}}\right)^{-1}=\frac{\ln3+\ln5}{2e}$$
| $\begin{array}\\
\dfrac{\sqrt{1-\cos^2\frac{1}{x}}
(3^\frac{1}{x}-5^\frac{-1}{x})}{\log_2(1+x^{-2}+x^{-3})}
&=\dfrac{\sin(1/x)
(e^\frac{\ln 3}{x}-e^\frac{-\ln 5}{x})}{(1/\ln 2)\ln(1+x^{-2}+x^{-3})}\\
&=\dfrac{(1/x+O(1/x^3))
((1+\ln 3/x+O(1/x^2)-(1-\ln 5/x+O(1/x^2))}{(1/\ln 2)(x^{-2}+x^{-3}+O(x^{-4})}\\
&=\dfrac{(1/x+O(1/x^3))
((\ln 3+\ln 5)/x+O(1/x^2))\ln 2}{x^{-2}(1+O(x^{-3}))}\\
&=\dfrac{1+O(1/x))
((\ln 3+\ln 5)+O(1/x))\ln 2}{1+O(1/x)}\\
&=(\ln 2(\ln 15)+O(1/x))(1+O(1/x))\\
&=(\ln 2)(\ln 15)+O(1/x)\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\frac{\sum_{k=0}^6 \csc^2(a+\frac{k\pi}{7})}{7\csc^2(7a)}$ The question is to evaluate $$\frac{\sum_{k=0}^{6}\csc^2(a+\frac{k\pi}{7})}{7\csc^2(7a)}$$ where $a=\pi/8$ without looking at the trigonometric table.
I tried to transform the $\csc^2$ term to $\cot^2$ term and use addition formula but it made the problem too cumbersome.I also tried to manipulate the numerator in the form so that it telescopes but couldnot succeed.I am not in need of a full solution.Any idea or hint to proceed shall be highly appreciated.Thanks.
| The result of the summation is:
$\frac{\sum_{k=0}^{6}\csc^2(a+\frac{k\pi}{7})}{7\csc^2(7a)}=7$.
The summation, after some calculation, is reduced to:
$\frac{(A E F+B D F+C D E)+D E F}{7 D E F}=7$
where
$A=4-2\sqrt{2}+(2\sqrt{2}-2)cos(\frac{\pi}{7})$,
$B=-4+2\sqrt{2}+(2-2\sqrt{2})sin(\frac{\pi}{14})$,
$C=-4+2\sqrt{2}+(-2+2\sqrt{2})sin(\frac{3\pi}{14})$,
$D=2\sqrt{2} cos(\frac{\pi}{7})+ sin(\frac{3\pi}{14})+2$,
$E= cos(\frac{\pi}{7})- 2\sqrt{2} sin(\frac{\pi}{14})-2$,
$F=\sqrt{2} sin(\frac{3\pi}{14})+ sin(\frac{\pi}{14})-2$.
By substituting and grouping we get:
$\frac{6+\frac{1}{8}}{7 \frac{1}{8}}=7$.
| {
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} |
If $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3 − m − 6 \not = 0$. Prove, by contradiction, that, if $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3 − m − 6 \not = 0$.
Proposition: If $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3 − m − 6 \not = 0$.
Hypothesis: $n$ is a positive integer such that $n^3 − n − 6 = 0$.
Conclusion: For every positive integer $m$ with $m \not = n$, $m^3 − m − 6 \not = 0$.
My workings
A (Hypothesis): $n$ is a positive integer such that $n^3 − n − 6 = 0$.
A1 ($\neg B$): There exists a positive integer $m$ with $m \not = n$ such that $m^3 − m − 6 = 0$.
A2: $n^3 - n = 6$
$\implies n(n^2 - 1) = 6$ where $n \in \mathbb{Z}^+$
A3: $m^3 - m = 6$
$\implies m(m^2 - 1) = 6$ where $m \in \mathbb{Z}^+$ and $m \not = n$.
A4: $n(n^2 - 1) = m(m^2 - 1)$ where $m \not = n$.
$\implies \dfrac{n(n^2 - 1)}{m(m^2 - 1)} = 1$ where $m \not = n$.
Since $m,n \in \mathbb{Z}^+$, $(n^2 - 1) \ge 0$ and $(m^2 - 1) \ge 0$. Therefore, $n(n^2 - 1) = m(m^2 - 1)$ iff $n = m$.
Contradiction.
$Q.E.D.$
I would greatly appreciate it if people could please take the time to provide feedback on the correctness of my proof. Given that $m \not = n$, did I construct my proof correctly? Or is it incorrect to structure it this way if $m \not = n$?
| Let $f(x)=x^3-x-6.$ Let $n$ be any real number such that $f(n)=0$. Then $f(x)=(x-n)(x^2+Ax+B)$ for all $x,$ for some constants $A, B.$
$$\text {So }\quad x^3-x-6=x^3+x^2(A-n)+x(B-nA)+(-nB)$$ for all $x.$ This requires $0=A-n$ and $-1=B-nA ,$ so $A=n$ and $B=nA-1=n^2-1.$
Now if $f(m)=0$ with $m\ne n$ then $0=m^2+Am+B=m^2+nm+(n^2-1),$ implying $m=(-n\pm \sqrt {4-3n^2}\;)/2.\;$ In order for $m$ to be a real number we must have $4-3n^2\geq 0,$ which implies $|n|\leq 2/\sqrt 3.$
But if $|n|\leq 2/\sqrt 3$ then $|n^3-n|\leq |n|^3+|n|\leq 8/(3\sqrt 3\;)+2/\sqrt 3<6.$
So $f(n)=0$ has at most one real-number solution.
| {
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"url": "https://math.stackexchange.com/questions/2141843",
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Probabilty of a certain run in a 1000 coin flips So a run of $x$ heads has the probability $0.5^x$.
Thefore a run of $9$ would be $0.5^9 = 0.001953125$.
But what is the chance of that run of $9$ occuring in $1000$ coinflips?
Can anyone please explain how to work this out?
|
Here is an answer based upon generating functions. We start with a generating function for words of a two character alphabet $V=\{T,H\}$ which counts words with no consecutive equal characters at all.
These words are called Smirnov or Carlitz words. See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information. (You might also find this answer helpful.)
The generating function $A(z)$ counting Smirnov words over a two character alphabet is according to the reference
\begin{align*}
A(z)=\left(1-\frac{2z}{1+z}\right)^{-1}
\end{align*}
The coefficient of $z^n$ of $A(z)$ gives the number of Smirnov words of length $n$, i.e. the number of words with no consecutive equal heads and tails.
Since there is no restriction to the distribution of tails, we can replace each character "T" in a Smirnov word by one or more "T"s, which means to replace
\begin{align*}
z\longrightarrow z+z^2+z^3+\cdots=\color{blue}{\frac{z}{1-z}}
\end{align*}
in the corresponding generating function $A(z)$.
Based upon $A(z)$ we obtain this way a generating function $B(z)$ with
\begin{align*}
B(z)&=\left(1-\frac{z}{1+z}-\frac{\color{blue}{\frac{z}{1-z}}}{1+\color{blue}{\frac{z}{1-z}}}\right)^{-1}\\
&=\left(1-\frac{z}{1+z}-z\right)^{-1}\\
\end{align*}
The coefficient of $z^n$ of $B(z)$ gives the number of words of length $n$ with no consecutive equal heads.
We next allow a distribution of heads with runs up to length $8$. This implies a replacement of each character "H" by one up to eight "H"s, which means
\begin{align*}
z\longrightarrow z+z^2+z^3+\cdots+z^8=\frac{z(1-z^8)}{1-z}
\end{align*}
Based upon $B(z)$ we obtain this way a generating function $C(z)$ with
\begin{align*}
C(z)&=\left(1-\frac{\color{blue}{\frac{z(1-z^8)}{1-z}}}{1+\color{blue}{\frac{z(1-z^8)}{1-z}}}-z\right)^{-1}\\
&=\frac{(1+z+z^2)(1+z^3+z^6)}{1-z+z^2+z^3+z^4+z^5+z^6+z^7+z^8+z^9}
\end{align*}
The coefficient of $z^n$ of $C(z)$ gives the number of words of length $n$ having no runs of heads of length $9$.
Finally in order to get the number of all words of length $n$ which have a run of heads of length $9$, we take all words and subtract those counted by $C(z)$.
We obtain
\begin{align*}
D(z)&=\frac{1}{1-2z}-C(z)\\
&=\frac{z^9}{(1-2z)(1-z+z^2+z^3+z^4+z^5+z^6+z^7+z^8+z^9)}\\
&=z^9+3z^{10}+\color{blue}{8}z^{11}+20z^{12}+48z^{13}+112z^{14}+\cdots\tag{1}
\end{align*}
whereby the last line was obtained with some help of Wolfram Alpha.
We observe (blue marked coefficient) in (1) there are $8$ words of length $11$ having a run of (at least) $9$ consecutive heads. These are
\begin{array}{ccc}
HHHHHHHHH\color{blue}{TT}&HHHHHHHHH\color{blue}{HT}&HHHHHHHHH\color{blue}{HH}\\
\color{blue}{T}HHHHHHHHH\color{blue}{T}&HHHHHHHHH\color{blue}{TH}\\
\color{blue}{TT}HHHHHHHHH&\color{blue}{T}HHHHHHHHH\color{blue}{H}\\
&\color{blue}{HT}HHHHHHHHH\\
\end{array}
The probability of runs of heads of length $9$ in sequences of $10,100$ and $1000$ coin flips is according to WA
\begin{align*}
2^{-10}\cdot[z^{10}]D(z)&=\frac{3}{1024}\doteq 0.00292969\\
2^{-100}\cdot[z^{100}]D(z)&\doteq 0.0875589\\
2^{-1000}\cdot[z^{1000}]D(z)&\doteq 0.624241\\
\end{align*}
| {
"language": "en",
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"answer_count": 1,
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} |
Solve system using $1+j+j^{2}=0$ Exercise :
Solve the following system:
$$
(S):\begin{cases}
A+B+C=2^{n} & \\
A+j\;B+j^{2}\;C=(−1)^{n}\;j^{2n} & \\
A+j^{2}\;B+j\;C=(−1)^{n}\;j^{n} & \\
\end{cases}
$$ Using the coefficients $1,j,j^{2}$ and $1+j+j^{2}=0$
Solution :
\begin{aligned}
A&=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}j^{2n}+(-1)^{n}j^{n}\right) \\
&=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}2\cos\left( \dfrac{2n\pi}{3}\right)\right)\\
B&=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}j^{2n+2}+(-1)^{n}j^{n+1}\right) \\
&=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}2\cos\left( \dfrac{(2n+1)\pi}{3}\right)\right)\\
C&=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}j^{2n+1}+(-1)^{n}j^{n+2}\right) \\
&=\dfrac{1}{3}\left( 2^{n}+(-1)^{n}2\cos\left( \dfrac{(2n-1)\pi}{3}\right)\right)\\
\end{aligned}
My question i can't get the same solution as the book :
so here is my attempt using Gaussian elimination (row reduction )
$$\fbox{$\forall j \in [|2,n|]\quad L_j \longleftarrow L_j-\dfrac{a_{1,j}}{a_{1,1}}L_{1} $}$$
$$
S \iff
\begin{cases}
A+B+C=2^n \\
(j-1)B+(j^{2}-1)C=(-1)^{n}j^{2n}-2^n\\
(j^{2}-1)B+(j-1)C=(-1)^{n}j^{n}-2^n
\end{cases}
\iff
\begin{cases}
A+B+C=2^n \\
(j-1)B+(j^{2}-1)C=(-1)^{n}j^{2n}-2^n\\
-2C=(-1)^{n}j^{n}-2^n-\dfrac{1}{(j-1)}\left((-1)^{n}j^{2n}-2^n \right)
\end{cases}
$$
*
*Could someone elaborate the calculation
| Hint: In order to get $A$ add all equations note that the coefficients of B and C will be equal to zero. So you directly get $A$. Then Multiply the second equation with $j$ and subtract the second and third equation, don't forget to plug in $A$.
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
How to prove this conjecture $x_{n}(n\ge 4)$ can't be an integer.
Define sequence $x_{1}=1$,and $x_{n+1}=1+\dfrac{n}{x_{n}},n\ge 1$.
I found
$$x_{2}=2,x_{3}=2,x_{4}=1+\dfrac{3}{x_{3}}=\dfrac{5}{2}, x_{5}=1+\dfrac{4}{2.5}=\dfrac{13}{5},x_{6}=1+\dfrac{5}{x_{5}}=\dfrac{38}{13}$$
$$x_{7}=1+\dfrac{6}{x_{6}}=\dfrac{58}{19},x_{8}=1+\dfrac{7}{x_{7}}=\dfrac{191}{58},\cdots$$
I conjecture: $x_{n}\notin Z$ for $n\ge 4$; in other words, this sequence has only three integer terms.
| This answer uses that for $n\ge 4$,
$$\frac{1+\sqrt{4n-3}}{2}\lt x_n\lt \frac{1+\sqrt{4n+1}}{2}\tag1$$
The proof for $(1)$ is written at the end of this answer.
Multiplying $(1)$ by $2$, subtracting $1$ and squaring, we get
$$4n-3\lt 4(x_n^2-x_n)+1\lt 4n+1$$
Then, subtracting $1$ and dividing by $4$, we get
$$n-1\lt x_n^2-x_n\lt n$$
Since $x_n^2-x_n$ is not an integer, we have that $x_n$ is not an integer.
So, your conjecture is true.
Let us prove $(1)$ by induction.
For $n=4$, since $13\lt 16\lt 17$, we get
$$\frac{1+\sqrt{13}}{2}\lt \frac{1+\sqrt{16}}2=x_4\lt \frac{1+\sqrt{17}}{2}$$
Supposing that $(1)$ holds for some $n\ge 4$ gives
$$\begin{align}x_{n+1}-\frac{1+\sqrt{4(n+1)-3}}{2}&=1+\frac{n}{x_n}-\frac{1+\sqrt{4n+1}}{2}\\\\&\gt 1+\frac{2n}{1+\sqrt{4n+1}}-\frac{1+\sqrt{4n+1}}{2}=0\end{align}$$
and
$$\begin{align}\frac{1+\sqrt{4(n+1)+1}}{2}-x_{n+1}&=\frac{1+\sqrt{4n+5}}{2}-1-\frac{n}{x_n}\\\\&\gt \frac{1+\sqrt{4n+5}}{2}-1-\frac{2n}{1+\sqrt{4n-3}}\\\\&=\frac{1+\sqrt{4n+5}}{2}-1-\frac{2n(1-\sqrt{4n-3})}{1-(4n-3)}\\\\&=\frac{(n-1)\sqrt{4n+5}+1-n\sqrt{4n-3}}{2(n-1)}\\\\&=\frac{\sqrt{4n+5}-3}{(n-1)\sqrt{4n+5}+1+n\sqrt{4n-3}}\\\\&\gt 0\qquad\quad\blacksquare\end{align}$$
| {
"language": "en",
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"answer_count": 2,
"answer_id": 1
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Solve $\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}}$ I tried first without L'Hôpital's rule:
$$\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}} =
\frac{\sqrt{x}}{\sqrt[3]{x}} \cdot \frac{1+\frac{x}{\sqrt x}}{1+ \sqrt[3] x} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\frac{\sqrt x+x}{\sqrt x}}{\frac{\sqrt[3]{x}+x}{\sqrt[3]{x}}} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\sqrt[3]{x}(\sqrt x +x)}{\sqrt x (\sqrt[3]{x}+x)} = \frac{\sqrt x \sqrt[3]{x}(\sqrt x +x)}{\sqrt[3] x \sqrt x (\sqrt[3]x +x )} = \frac{\sqrt x +x }{\sqrt[3] x +x }$$
That didn't work, so then I tried L'Hôpital's:
$$\frac{\frac{1}{2}(x)^{-\frac{1}{2}}+1}{1+\frac{1}{3}(x)^{-\frac{1}{3}}} = \frac{\frac{1}{2\sqrt x }+1}{1+\frac{1}{{3\sqrt[3]{x}}}} = \frac{\frac{1+2\sqrt x}{2\sqrt x}}{\frac{1+3\sqrt[3]{x}}{3\sqrt[3]{x}}} = \frac{3\sqrt[3]{x}(1+2\sqrt x)}{2\sqrt x (1+3\sqrt[3]{x})} = \frac{3\sqrt[3]{x}+3\sqrt[3]{x}2\sqrt{x}}{2\sqrt{x} + 2\sqrt{x}3\sqrt[3]{x}} = ??? $$
How do I solve this? If you can solve this with and without L'Hôpital's, please do so.
| Hint:
$$\frac{\sqrt{x}+x}{x+\sqrt[3]{x}}=\frac{\frac1{\sqrt x}+1}{1+\frac1{\sqrt[3]{x^2}}}$$
It's that simple!
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $f(n+1)=(-1)^{n+1}n-2f(n)$ and $f(1)=f(1986)$ find $f(1)+f(2)+f(3)+...+f(1985)$ I got a solution but it was really messy with f(1)=$\frac{1987}{2^{1985}+1}$ and an explicit formula for $f(x)$. Then the whole mess was a crocodile (With many $2^{1985}$ and $4^{1985}$ terms) when I solved it so I assume I made a mistake somewhere. (The answer must be simpler because it's an ARML problem)
| Here's a fully worked version. The sum we are looking for is ($N = 1986$)
$$
\begin{align}
S = \sum_{n = 1}^{N - 1}f(n) &= \sum_{n = 1}^{N - 1}\left((- 1)^n(n-1) - 2f(n-1)\right)\\
&= \sum_{n = 1}^{N - 1}(-1)^n(n-1) - 2\sum_{n = 1}^{N - 1}f(n-1)\ .
\end{align}
$$
If we look at the first sum on the right side and write it out, knowing $N$ is even, we get
$$
\begin{align}
\sum_{n = 1}^{N - 1}(-1)^n(n-1) &= 0 + 1 - 2 + 3 - 4 + \dots + (N - 3) - (N - 2)\\
&=0 + (1 - 2) + (3 - 4) + \dots + ((N - 3) - (N - 2))\\
&=1 - \frac{N}{2}\ .
\end{align}
$$
Looking at the sum on the right side, if we substitute $n \rightarrow n + 1$ then
$$
\begin{align}
\sum_{n = 1}^{N - 1}f(n-1) = \sum_{n = 0}^{N - 2}f(n) = \sum_{n = 1}^{N - 1}f(n) + f(0) - f(N-1) = S + f(0) - f(N-1)\ .
\end{align}
$$
Again knowing $N$ is even in our case we can calculate
$$
\begin{gather}
f(1) = -2f(0)\quad \Rightarrow \quad f(0) = -\frac{1}{2}f(1)\ , \\
f(N) = -(N-1) - 2f(N-1)\quad \Rightarrow\quad f(N - 1) = -\frac{f(N) + (N - 1)}{2}\ .
\end{gather}
$$
Substituting into the equation above we have
$$
\sum_{n = 1}^{N - 1}f(n-1) = \sum_{n = 1}^{N - 1}f(n) + \frac{1}{2}\left(f(N) - f(1) - (N - 1)\right)\ .
$$
Substituting back into $S$, and noting that $f(1) = f(N)$, we finally get
$$
\begin{align}
S &= 1 - \frac{N}{2} - 2S + \left(f(1) - f(N)\right) + N - 1\\
\Rightarrow 3S &= \frac{N}{2}\\
\Rightarrow S &= \frac{N}{6}\ .
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Can anyone explain how the complex matrix representation of a quaternions is constructed? I am reading some properties of quaternionic matrices and I am unable to understand how can we got such matrix representation. please help in this regards.
| The ring of quaternions $\mathbb{H}$ is isomorphic to the
ring of matrices with complex entries of the form $A =\begin{pmatrix}
x & y
\\
- \bar{y} & \bar{x}
\end{pmatrix} $
For a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ the isomorphism is given by:
$$
z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} \quad \mapsto \quad \mathbf{Z}= a\mathbf{U}+b\mathbf{I}+c\mathbf{J}+d\mathbf{K} \quad a,b,c,d \in \mathbb{R}
$$
with:
$$
\mathbf{U}=
\left(
\begin{array}{ccccc}
1&0 \\
0 &1
\end{array}
\right)
\qquad
\mathbf{I}=
\left(
\begin{array}{ccccc}
i&0 \\
0 &-i
\end{array}
\right)
\qquad
\mathbf{J}=
\left(
\begin{array}{ccccc}
0&1 \\
-1 &0
\end{array}
\right)
\qquad
\mathbf{K}=
\left(
\begin{array}{ccccc}
0&i \\
i &0
\end{array}
\right)
$$
We can easy see that:
$$
\mathbf{I}^2=\mathbf{J}^2=\mathbf{K}^2=\mathbf{I}\mathbf{J}\mathbf{K}=-\mathbf{U}
$$
and
$$
z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} \quad \mapsto \quad \mathbf{Z}=
\left(
\begin{array}{ccccc}
a+ib&c+id \\
-c+id &a-ib
\end{array}
\right)
$$
with
$$
\mbox{det}(\mathbf{Z})=
\left |
\left(
\begin{array}{ccccc}
a+ib&c+id \\
-c+id &a-ib
\end{array}
\right)
\right |=
a^2+b^2+c^2+d^2=|z|^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148802",
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"source": "stackexchange",
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Other Idea to show an inequality $\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$ $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$$
I want to prove this by Induction
$$n=1 \checkmark\\
n=k \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}\geq \sqrt k\\
n=k+1 \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt {k+1}}\geq \sqrt {k+1}$$ so $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}+\dfrac{1}{\sqrt {k+1}}\geq \sqrt k+\dfrac{1}{\sqrt {k+1}}$$now we prove that $$\sqrt k+\dfrac{1}{\sqrt {k+1}} >\sqrt{k+1} \\\sqrt{k(k+1)}+1 \geq k+1 \\ k(k+1) \geq k^2 \\k+1 \geq k \checkmark$$ and the second method like below ,
and I want to know is there more Idia to show this proof ? forexample combinatorics proofs , or using integrals ,or fourier series ,....
Is there a close form for this summation ?
any help will be appreciated .
| By the generalized mean inequality the harmonic mean is no larger than the quadratic mean:
$$
\require{cancel}
\cfrac{n}{\cfrac{1}{\sqrt{1}}+\cfrac{1}{\sqrt{2}}+\cdots+\cfrac{1}{\sqrt{n}}} \;\le\; \sqrt{\frac{(\sqrt{1})^2+(\sqrt{2})^2+\cdots+(\sqrt{n})^2}{n}} = \sqrt{\frac{\cancel{n}(n+1)}{2\,\cancel{n}}}
$$
$$
\implies \quad \cfrac{1}{\sqrt{1}}+\cfrac{1}{\sqrt{2}}+\cdots+\cfrac{1}{\sqrt{n}} \;\ge\; \sqrt{\frac{2\,n^2}{n+1}} \;\ge\; \sqrt{n}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
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Is this limit $\lim\limits_{x \to\, -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = 0$? Is this limit $\lim\limits_{x \to\, -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = 0$?
It's in indeterminant form, 0/0 when $x$ approaches $-8$. So I used LHopital's rule and got $$-\frac{3x^{\frac{2}{3}}}{2\sqrt{1-x}}$$ plug in $-8$ it is $-2(-1)^{2/3}$ which is imaginary. I used wolframalpha, the answer is $0$. So, which is correct?
| With substiuation $x+8=t$ we have
$$\lim_{x\to-8}\dfrac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}=\lim_{t\to0}\frac{\sqrt{9-t}-3}{\sqrt[3]{t-8}+2}=\frac{\lim_{t\to0}\frac{\sqrt{9-t}-3}{t}}{\lim_{t\to0}\frac{\sqrt[3]{t-8}+2}{t}}=\frac{(\sqrt{9-t})'\Big|_{t=0}}{(\sqrt[3]{t-8})'\Big|_{t=0}}=\frac{-\frac16}{\frac{1}{12}}=\color{blue}{-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$ \lim_{n \to \infty} \frac{\sqrt{n}(\sqrt{1} + \sqrt{2} + ... + \sqrt{n})}{n^2} $ How do I find the following limit?
$$ \lim_{n \to \infty} \frac{\sqrt{n}(\sqrt{1} + \sqrt{2} + ... + \sqrt{n})}{n^2} $$
Can limit be find by Riemann sums?
$$\lim_{n\to \infty}\sum_{k=1}^{n}f(C_k)\Delta{x} = \int_{a}^{b}f(x)\,dx$$
I'm not sure what $f(C_k)$ is.
| By Stolz we have $$ \lim\limits_{n \to \infty} \frac{\sqrt{n}(\sqrt{1} + \sqrt{2} + ... + \sqrt{n})}{n^2}= \lim\limits_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + ... + \sqrt{n}}{n^\frac{3}{2}}= \lim\limits_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n^3}-\sqrt{(n-1)^3}}=$$
$$= \lim\limits_{n \to \infty} \frac{\sqrt{n}\left(\sqrt{n^3}+\sqrt{(n-1)^3}\right)}{n^3-(n-1)^3}=\lim\limits_{n \to \infty} \frac{\sqrt{n}\left(\sqrt{n^3}+\sqrt{(n-1)^3}\right)}{3n^2-3n+1}=\frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
$\lim \limits_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$ find the limit :
$$\lim_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$$
my try :
$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\tan 2x}}{3^{\sin 2x}-3^{2\tan x}}=$$
$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\frac{\sin 2x}{\cos2x}}}{3^{\sin 2x}-3^{\frac{2\sin x}{cosx}}}=$$
$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\sin 2x}+2^{\cos 2x}}{3^{\sin 2x}-3^{2\sin x}+3^{\cos x}}=?$$
now ?
| Hint. One may write, by using a Taylor series expansion, as $x \to 0$,
$$
\begin{align}
\frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}&= \frac {2^{\tan 2x}\cdot \left(2^{2\sin x-\tan 2x}-1\right)}{3^{2\tan x}\cdot \left(3^{\sin 2x-2\tan x}-1\right)}=\frac {2^{\tan 2x}\cdot \left(-x^3 \ln 8+o(x^3)\right)}{3^{2\tan x}\cdot \left(-x^3 \ln 9+o(x^3)\right)} \to \frac{\ln 8}{\ln 9}.
\end{align}
$$
Remark. Observe that, in general,
$$
\frac{a^m}{a^n}=a^{m-n}\neq a^{\frac{m}{n}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$S=1+2i+3i^2+4i^3+\dots+(n+1)i^n$; $4\mid n$; closed form for $S$ $$S=1+2i+3i^2+4i^3+\dots+(n+1)i^n$$
where $4\mid n$. How can I simplify this exprerssion so as to obtain a general expression?
| Let $n=4m$.
$$\begin{align}
S&=\boxed{\begin{array}
&\;\;\;\;1&+2i&+3i^2&+4i^3\\
+5i^4&+6i^5&+7i^6&+8i^7\\
+\vdots\\
+(4m-3)i^{4m-4}&+(4m-2)i^{4m-3}&-(4m-1)i^{4m-2}&+4mi^{4m-1}\\
+(4m+1)i^{4m}\\
\end{array}}\\\\
&=\boxed{\begin{array}
&\;\;\;1&+2i&-3&-4i\\
+5&+6i&-7&-8i\\
+\vdots\\
+(4m-3)&+(4m-2)i&-(4m-1)&-4mi\\
+(4m+1)\\
\end{array}}\\\\
&=\;\;\;\;\;\;[1-3+5-7+\cdots+(4m-3)-(4m-1)]\\
&\;\;\;\;+i\;[2-4+6-8+\cdots +(4m-2)-4m]\\
&\;\;\;\;\;\;\;\;\;+(4m+1)\\
&=[\overbrace{-2-2-\cdots-2}^m]+i[\overbrace{-2-2\cdots-2}^m]+(4m+1)\\
&=-2m+i(-2m)+(4m+1)\\
&=2m+1-i2m\\
&=\left(\frac n2+1\right)-i\frac n2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to calculate this sum. I am trying to solve an excercise and i come across the following sum
$ \sum_{k=1}^n \frac{(k+1)(k^3-2k+2)}{k(k+2)} $
I put it in Wolfram alpha and it says that it is equal to:
$\frac{n(2n^4+6n^3+2n^2+3n+11)}{6(n+1)(n+2)}$
but how can i prove this?
| You may notice that
$$ \frac{(k+1)(k^3-2k+2)}{k(k+2)} = \left(\frac{1}{k}-\frac{1}{k+2}\right)+2\binom{k}{2} \tag{1}$$
hence by creative telescoping and the hockey stick identity
$$ \sum_{k=1}^{n}\frac{(k+1)(k^3-2k+2)}{k(k+2)} = 2\binom{n+1}{3}+\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}.\tag{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x}$ without using L'Hopital's rule I tried:
$$\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x} = \\
\frac{e^{x}(1+e^{-2x}-\frac{2}{e^x})}{x(x-2)} = \frac{e^x(1+e^{-2x})-2}{x(x-2)} = \frac{e^x(1+e^{-2x})}{x(x-2)} - \frac{2}{x(x-2)} = \\
\frac{1+e^{-2x}}{x} \cdot \frac{e^x}{x-2} - \frac{2}{x(x-2)} = ???$$
What do I do next?
| Aside from not carelessly turning $x(x+2)$ into $x(x-2)$, partial fractions is the wrong tool for this job. If you are allowed to used Taylor series, you can see that
$$
e^x + e^{-x} - 2 = x^2 + O(x^x)$$
so you get the expression $$\frac{x^2}{x(x+2)} = \frac{x}{x+2} \to 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divisible by $4$ so $8|(n^3 - 1)(n^3 + 1)$. I'm stuck at proving divisibility by $9$ and $7$
| Hint -
We have 504 =$(2^3-1)2^3(2^3+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 5
} |
Max and Min$ f(x,y)=x^3-12xy+8y^3$ First I solved $f_y=0$ then plugged in my variable into $f_x$ to get an output and then plugged that output back into $f_y$ to get a point and did this again for $f_x$ I think I screwed up on the $(-7,\sqrt{3})$
$f_y=-12x+24y^2$
$-12x+24y^2=0$
$-12x=24y^2$
$x=2y^2$
$f_x(2y^2,y)=3(2y^2)^2-12y$
$3(4y^4)-12y=0$
$12y^4 - 36y=0$
$12y(y^3-3=0) \to y=0$ or $y=\sqrt{3}$
$f_y(x,0)=-12x=0 \to (0,0)$
$f_y(x,\sqrt{3}) = -12x+24(3)$
$f_y(x.sqrt{3})= -12x-84=0$
$f_y(x,\sqrt{3})=-12x=84$
$x=-7 \to (-7,\sqrt{3})$
$f_y=-12x+24y^2=0$
$-12x=24y^2$
$x=2y^2$
$f_x(2y^2,y)=3(2y^2)^2-12y=0$
$12y^4-12y=0$
$12y(y^3-1)=0$
$y=0$ or $y=1$
$f_y(x,o)= -12x=0 \to x=0$
$f_y(x,1)=-12x+24=0$
$f_y(x,1)=x=2$
Critical points: $(2,-1),(0,0),(-7,\sqrt{3})$
$D=(6x)(48y)-(-12)^2$
$D(0,0)=-144 so (0,0) is a saddle point
$D(,2,-1)= Min
D(-7,\sqrt{3})= Saddle point
| $\hskip 2.1in$
Given function $$f(x,y) = x^3 - 12xy + 8y^3$$ $$ \Rightarrow f_x = 3x^2 - 12y, f_y = 24y^2 -12x = 0 $$
Which on solving gives two stationary points $(2,1)$ and $(0,0)$.
Now, consider hessian matrix $$H = \begin{bmatrix} 6x & -12 \\ -12 & 48y \end{bmatrix} $$
For $(2,1)$, $|H| > 0$ and $f_{xx} ( 2,1) > 0$, implying that there is a minima at that point.
For $(0,0)$, $|H| <0$, implying there is a saddle point at $(0,0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
If $f_k(x)=\frac{1}{k}\left (\sin^kx +\cos^kx\right)$, then $f_4(x)-f_6(x)=\;?$ I arrived to this question while solving a question paper. The question is as follows:
If $f_k(x)=\frac{1}{k}\left(\sin^kx + \cos^kx\right)$, where $x$ belongs to $\mathbb{R}$ and $k>1$, then $f_4(x)-f_6(x)=?$
I started as
$$\begin{align}
f_4(x)-f_6(x)&=\frac{1}{4}(\sin^4x + \cos^4x) - \frac{1}{6}(\sin^6x + \cos^6x) \tag{1}\\[4pt]
&=\frac{3}{12}\sin^4x + \frac{3}{12}\cos^4x - \frac{2}{12}\sin^6x - \frac{2}{12}\cos^6x \tag{2}\\[4pt]
&=\frac{1}{12}\left(3\sin^4x + 3\cos^4x - 2\sin^6x - 2\cos^6x\right) \tag{3}\\[4pt]
&=\frac{1}{12}\left[\sin^4x\left(3-2\sin^2x\right) + \cos^4x\left(3-2\cos^2x\right)\right] \tag{4}\\[4pt]
&=\frac{1}{12}\left[\sin^4x\left(1-2\cos^2x\right) + \cos^4x\left(1-2\sin^2x\right)\right] \tag{5} \\[4pt]
&\qquad\quad \text{(substituting $\sin^2x=1-\cos^2x$ and $\cos^2x=1-\sin^2x$)} \\[4pt]
&=\frac{1}{12}\left(\sin^4x-2\cos^2x\sin^4x+\cos^4x-2\sin^2x\cos^4x\right) \tag{6} \\[4pt]
&=\frac{1}{12}\left[\sin^4x+\cos^4x-2\cos^2x\sin^2x\left(\sin^2x+\cos^2x\right)\right] \tag{7} \\[4pt]
&=\frac{1}{12}\left(\sin^4x+\cos^4x-2\cos^2x\sin^2x\right) \tag{8} \\[4pt]
&\qquad\quad\text{(because $\sin^2x+\cos^2x=1$)} \\[4pt]
&=\frac{1}{12}\left(\cos^2x-\sin^2x\right)^2 \tag{9} \\[4pt]
&=\frac{1}{12}\cos^2(2x) \tag{10}\\[4pt]
&\qquad\quad\text{(because $\cos^2x-\sin^2x=\cos2x$)}
\end{align}$$
Hence the answer should be ...
$$f_4(x)-f_6(x)=\frac{1}{12}\cos^2(2x)$$
... but the answer given was $\frac{1}{12}$.
I know this might be a very simple question but trying many a times also didn't gave me the right answer. Please tell me where I am doing wrong.
| Hint: let $f(x):=f_4(x)-f_6(x)$. Then show that $f'(x)=0$ for all $x$. Hence $f$ is constant. Furthermore: $f(0)=\frac{1}{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Triangular inequality extended Let $a,b,c,d,e,f\in\mathbb{R}$ all positive or zero, such that $a\leq c+d$ and $b\leq e+f$ show that:
$$\sqrt{a^2+b^2}\leq\sqrt{c^2+e^2}+\sqrt{d^2+f^2}$$
Some hint or auxiliar inequality that would help? I've done many attempts but write them will take a lot of time, I tried to use the artihmetic and geometric mean, minkowski but no one works (or fits) can you help me?
| Just $$\sqrt{c^2+e^2}+\sqrt{d^2+f^2}\geq\sqrt{(c+d)^2+(e+f)^2}\geq\sqrt{a^2+b^2}$$
If I don't see the triangle inequality I can make the following.
By C-S
$$\sqrt{c^2+e^2}+\sqrt{d^2+f^2}=\sqrt{c^2+e^2+d^2+f^2+2\sqrt{(c^2+e^2)(d^2+f^2)}}\geq$$
$$\geq\sqrt{c^2+e^2+d^2+f^2+2(cd+ef)}=\sqrt{(c+d)^2+(e+f)^2}\geq\sqrt{a^2+b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$ I'm trying to find $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$.
*
*I tried to use the squeeze theorem, failed.
*I tried to use a sequence defined recursively: $a_{n+1} = {a_n} + \frac{1}{\sqrt{(n+1)^2 +n+1}}$. It is a monotone growing sequence, for every $n$, $a_n > 0$. I also defined $f(x) = \frac{1}{\sqrt{(x+1)^2 +x+1}}$. So $a_{n+1} = a_n + f(a_n)$. But I'm stuck.
How can I calculate it?
| Hint:
$$n\frac{1}{\sqrt{n^2+n}}\leq\sum\frac{1}{\sqrt{n^2+n}}\leq n\frac{1}{\sqrt{n^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
how to sum this series of combinations I have to calculate this
\begin{align}
& (1-p)^{N-1}+\frac{1}{2}C_{N-1}^{N-2}(1-p)^{N-2}p+\frac{1}{3}C_{N-1}^{N-3}(1-p)^{N-3}p^{2}+...+\frac{1}{N-2}C_{N-1}^{1}(1-p)p^{N-2}\\
& +\frac{1}{N}p^{N-1}
\end{align}
where $C_{n}^{m}=\frac{n!}{m!(n-m)!}$.
Can anyone help to sum this up? Thank you.
| The given sum can be written as:
\begin{align*}
\sum_{k=1}^{N} \frac 1k \binom{N-1}{N-k} p^{k-1}(1-p)^{N-k} = (1-p)^{N-1} \sum_{k=1}^{N} \frac 1k \binom{N-1}{N-k} \left(\frac{p}{1-p}\right)^{k-1}
\end{align*}
Set $r=\frac{p}{1-p}$ and observe that
\begin{align*}
\sum_{k=1}^{N} \frac 1k \binom{N-1}{N-k} r^k &= \sum_{k=1}^{N} \binom{N-1}{N-k} \int_0^r x^{k-1} \mathrm{d}x \\
&= \int_0^r \sum_{k=1}^{N} \binom{N-1}{N-k} x^{k-1} \mathrm{d}x \\
&= \int_0^r \sum_{k=1}^{N} \binom{N-1}{k-1} x^{k-1} \mathrm{d}x \\
&= \int_0^r \sum_{k=0}^{N-1} \binom{N-1}{k} x^{k} \mathrm{d}x \\
&= \int_0^r (1+x)^{N-1} \mathrm{d}x \\
&= \frac{(1+r)^N - 1}{N}
\end{align*}
Thus, the required sum is $$(1-p)^{N-1} \left(\frac{1-p}{p}\right) \frac{\left(\frac{1}{1-p}\right)^N - 1}{N} = \frac{1-(1-p)^N}{pN}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\int \sec^3x dx$ in disguise I found this integral,
$$\int \sqrt{x^2+1}dx$$
on a problem list and I think it is a sneaky way of hiding a $\int \sec^3xdx$ problem but I am not sure if what I did was correct though, because of what happens at the end.
So what I did was use trig-substitution and let $u=\tan\theta$ and $du=\sec^2\theta d\theta$
$$\int \sqrt{x^2+1}dx=\int\sqrt{\tan^2\theta+1}\sec^2\theta d\theta=\int \sec^3\theta d\theta$$
$$= \int (\tan^2\theta+1)\sec\theta d\theta = \int \tan^2\theta \sec\theta +\int \sec\theta d\theta$$
I will label $I_s=\int \sec\theta d\theta = \ln|\sec\theta+\tan\theta|$ so I have
$$\int \sec^3xdx=\int\tan^2x\sec x dx+I_s$$
and use integration by parts to solve $\int tan^2\theta \sec\theta d\theta$ that remains. I let $u=\sec\theta$ so $du=\sec\theta\tan\theta d\theta$ and $dv=\tan^2\theta\, d\theta$ such that $v=\tan\theta -\theta$
$$\int\tan^2x\sec x dx =\sec\theta (\tan\theta -\theta) - \int (\tan\theta-\theta)\sec\theta\tan\theta d\theta$$
$$= \frac 12 \sec\theta (\tan\theta -\theta) + \frac 12\int \theta \sec\theta\tan\theta \, d\theta$$
Again I use integration by parts, this time on $\frac 12\int \theta \sec\theta\tan\theta \, d\theta$ and let $u=\theta$ so $du=d\theta$ and $dv=\sec\theta\tan\theta \,d\theta$ so $v=\sec\theta$
$$\frac 12\int \theta \sec\theta\tan\theta \, d\theta = \frac 12 \theta \sec\theta - \frac 12 I_s$$
$$= \frac 12(\theta\sec\theta - I_s)$$
and use this result to get
$$\int\tan^2x\sec x dx = \frac 12 \sec\theta (\tan\theta -\theta) + \frac 12(\theta\sec\theta - I_s)$$
$$= \frac 12 \sec\theta\tan\theta - \frac 12 I_s$$
Putting everything back together and including an arbitrary constant $C$ i get
$$\int \sec^3\theta\,d\theta = \int \tan^2\theta\sec\theta d\theta + I_s$$
$$= \frac 12 \sec\theta\tan\theta - \frac 12 I_s + I_s + C$$
$$=\frac 12\sec\theta\tan\theta +\frac 12 \ln|\sec\theta+\tan\theta|+C$$
and substituting back into $x$-variable terms using $\tan\theta = x$ I have $\sec\theta = \sqrt{x^2+1}$ so
$$\int \sqrt{x^2+1}dx =\frac 12x\sqrt{x^2+1} +\frac 12 \ln|\sqrt{x^2+1}+x|+C$$
The reason why I am not sure about this is because when substituting back I have $\sec\theta = \sqrt{x^2+1}$ which is the same as the integrand $\sqrt{x^2+1}$ that became $\sec^3\theta$ in the initial substitutions. What is going on here?
Thank you all for the precious help!
| It is just how integration works, you plugged in $x=\tan\theta$ to get $\sqrt{1+x^2}$ to equal $\sec\theta$, your answer is completely correct. Just because $\sec\theta=\sqrt{x^2+1}$ does not mean that $\int\sqrt{1+x^2}dx=\int\sec\theta d\theta$ as you recognized because of the chain rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$
My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\\=a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+4\\=(a+b)^2-2ab+\bigg(\dfrac{1}{a}+\dfrac{1}{b}\bigg)^2-\dfrac{2}{ab}+4\\=4-2ab-\dfrac{2}{ab}+1+\dfrac{1}{a^2b^2}\\=4-2\bigg(\dfrac{a^2b^2+1}{ab}\bigg)+\dfrac{a^2b^2+1}{a^2b^2}\\=4-\bigg(\dfrac{a^2b^2+1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)\\=4-\bigg(ab+\dfrac{1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)$
Seems to using Cauchy-Schwartz. Please help.
| I just differentiate.
$$(ab+\dfrac1{ab})(2-\dfrac1{ab})=(x+\dfrac1x)(2-\dfrac1x)=f(x)$$
and $x\leq\dfrac14$
$f'(x)=(1-\dfrac1{x^2})(2-\dfrac1x)+\dfrac1{x^2}(x+\dfrac1x)
=\dfrac{2x^3-2x+2}{x^3}
=2*\dfrac{x^3-x+1}{x^3}>0$
$LHD=4-minf(x)=4-f(\dfrac14)=4-(\dfrac14+4)(-2)=\dfrac{25}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
} |
Prove by induction that $n^4+2n^3+n^2$ is divisible by 4
I'm trying to prove by induction that $n^4+2n^3+n^2$ is divisible by 4.
I know that P(1) it's true. Then $ n=k, P(k):k^4+2k^3+k^2=4w$ it's true by the hypothesis of induction.
When I tried to prove $n=k+1, P(k+1):(k+1)^4+(k+1)^3+(k+1)^2 = 4t$,
$$k^4+4k^3+6k^2+4k+1+2k^3+6k^2+6k+2+k^2+2k+1 = 4t.$$
$$(k^4+2k^3+k^2)+4k^3+6k^2+4k+1+6k^2+6k+2+2k+1 = 4t.$$
Can I replace $(k^4+2k^3+k^2)$ of the previous expression by $4w$ and then continue the proof? Is this correct?
Thanks.
| You are not doing it wrongly but it's easier, with $f(n)=n^4+2n^3+n^2$ , to observe that $f(n)=(n+1)^2n^2,$ so $$f(n+1)-f(n)=(n+2)^2(n+1)^2-(n+1)^2n^2=$$ $$=(n+1)^2 ((n+2)^2-n^2)=$$ $$=(n+1)^2(4n+4)=4(n+1)^3$$ which is a multiple of $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find an indefinite integral of $\int \left | \sin x + \cos x\right | dx$ So, I've already showed that $\sin x + \cos x = \sqrt{2}\sin (x+\frac{\pi}{4})$ and $\sin x + \cos x>0$ on the interval $\left ( -\frac{\pi}{4}+\pi k; \frac{3\pi}{4}+ \pi k\right )$, $\sin x + \cos x<0$ on the interval $\left ( \frac{3\pi}{4}+\pi k; \frac{7\pi}{4}+ \pi k\right )$.
So an integral on the "positive" interval is equal to $\sin x - \cos x +C_{1}$, on the "negative" one is equal to $\cos x - \sin x +C_{2}$.
After that I found the limits of $\sin x - \cos x +C_{1}$ and $\cos x - \sin x +C_{2}$ at the point $\frac{3\pi}{4}$ and found that the constants should be equal $C_{1}=0$ and $C_{2}=2\sqrt{2}$ for example.
But I don't know how to merge these two expression and get a final answer. I guess it can't be just $sgn(\sin x + \cos x)(\sin x - \cos x) + C $?
| By a shift and a scaling, this is the same as integrating $|\sin x|$.
Over the first period ($0$ to $\pi$) we have $\displaystyle\int_0^x\sin t\,dt=1-\cos x$, which is $2$ for the full period. For other values of $x$, you add $2$ as many times as necessary to return $x$ in the first period. In other words,
$$\int_0^x\sin t\ dt=1-\cos(x\bmod\pi)+2\left\lfloor\frac x\pi\right\rfloor.$$
For the initial integral, after (optionally) adjusting the integration constant
$$\sqrt2\left(\frac1{\sqrt2}-\cos\left(\left( x+\frac\pi4\right)\bmod\pi\right)+2\left\lfloor\frac x\pi+\frac14\right\rfloor\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the limit of function using an epilson delta proof Find $\lim\limits_{x \to 4} \frac{\sqrt{x}-2}{x-4}$ using an epilson delta proof.
I multiplied and divided by the rational conjugate to get
$\frac{\sqrt{x}+2}{x-4}=\frac{1}{\sqrt{x}+2}$ So the limit is probably $\frac{1}{4}$.
I can't seem to figure out how to turn $|\frac{1}{\sqrt{x}+2}-\frac{1}{4}|$ into something of the form $|x-4||f(x)|$ where f(x) is some function that I can then find the maximum value over [3,5] to find the value of delta.
|
(answer to the edited question)
We claim that $$\lim_{x\to 4}\frac{1}{\sqrt{x}+2}=\frac{1}{4}.$$
Note that $\sqrt{x}+2>1$ and so
$$0<\frac{1}{\sqrt{x}+2}<1. \tag 1$$
Also, $$x-4=(\sqrt{x}+2)(\sqrt{x}-2).\tag 2$$
Now,
$$\begin{align}
\bigg|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\bigg|&=\bigg|\frac{2-\sqrt{x}}{4(\sqrt{x}+2)}\bigg|\\
&=\frac{|2-\sqrt{x}|}{|4(\sqrt{x}+2)|}\\
&=\frac{|\sqrt{x}-2|}{4(\sqrt{x}+2)}\\
&=\frac{1}{4}\cdot\frac{1}{\sqrt{x}+2}\cdot |\sqrt{x}-2|\quad\text{then use $(1)$ to get}\\
&<\frac{|\sqrt{x}-2|}{4}.
\end{align}$$
But using $(2)$ and $(1)$, we get
$$|\sqrt{x}-2|=\bigg|\frac{x-4}{\sqrt{x}+2}\bigg|<|x-4|.$$ Hence,
$$\bigg|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\bigg|<\frac{|x-4|}{4}.$$
Finally, we pick $\epsilon>0$ and take $\delta=4\epsilon$ (you can also take $\delta$ such that $0<\delta<4\epsilon$). If $0<|x-4|<\delta$ then $$\bigg|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\bigg|<\frac{|x-4|}{4}<\frac{\delta}{4}=\epsilon.$$ This proves that
$$\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}=\lim_{x\to 4}\frac{1}{\sqrt{x}+2}=\frac{1}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine the eigenvector and eigenspace and the basis of the eigenspace The yellow marked area is correct, so don't check for accuracy :)
$A=\begin{pmatrix} 0 & -1 & 0\\ 4 & 4 & 0\\ 2 & 1 & 2
\end{pmatrix}$ is the matrix.
Characteristic polynomial is $-\lambda^{3}+6\lambda^{2}-12\lambda+8=0$
The (tripple) eigenvalue is $\lambda=2$.
Calculate the eigenvectors now:
$\begin{pmatrix}
-2 & -1 & 0\\ 4 & 2 & 0\\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0
\end{pmatrix}$
We get the equations:
$I: -2x-y=0 \Leftrightarrow y = -2x$
$II: 4x+2y=0$
$III: 2x+y=0 \Leftrightarrow 2x-2x=0 \Leftrightarrow 0=0$
We see that in every eequation $z$ is unknown, so we can choose an arbitrary $z$.
$x\begin{pmatrix}
1\\
-2\\
z
\end{pmatrix}$ and this is the eigenspace...?
And what is the basis of this eigenspace? Can I just set $x=1$ and some value for $z$? So this would be a correct basis of the eigenspace: $\begin{pmatrix}
1\\
-2\\
3
\end{pmatrix}$?
Now we need three linearly independent eigenvectors but I couldn't find them as I always got linearly dependent vectors...
I need a detailled, not too complicated answer that explains it well and I will give that answer a nice bounty (up to 200 rep) because I couldn't find another site explaining this correctly to me and I'm really in need of it.
| You must solve $(A - 2I)\vec{x} = \vec{0}$, i.e.
$$
\begin{pmatrix}
-2 & -1 & 0\\ 4 & 2 & 0\\ 2 & 1 & 0
\end{pmatrix}
\begin{pmatrix}x\\ y\\ z\end{pmatrix}
=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}
$$
Note that all equations are multiple of each other, so let's leave the last one to use and eliminate all others. You get the constraint $2x+y = 0$. So Your solutions will have to have $y = -2x$, in other words,
$$
\begin{pmatrix}x\\ y\\ z\end{pmatrix}
= \begin{pmatrix}x\\ -2x\\ z\end{pmatrix}
= x \begin{pmatrix}1\\ -2\\ 0\end{pmatrix}
+ z \begin{pmatrix}0\\ 0\\ 1\end{pmatrix}
$$
and so you see that there are 2 eigenvectors that can be used as a basis.
As for the number of independent eigenvectors in the basis not equal the multiplicity of the root of the characteristic equation, this has to do with the fact that for your matrix, the geometric and algebraic multiplicities don't match for the eigenvalue $\lambda = 2$. You can read more about it here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Is this matrix diagonalisable or not? We will only be looking at one specific eigenvalue, eigenspace (there are three eigenvalues in total, I know two of them are fine and last one seems not but as I'm not sure I need to ask you).
We have matrix $A=\begin{pmatrix}
3 & -1 & 0\\
2 & 0 & 0\\
-2 & 2 & -1
\end{pmatrix}$
The eigenvalues are $\lambda_{1}=1, \lambda_{2}=2, \lambda_{3}=-1$
We will only concentrate on $\lambda_{3}=-1$ because I have checked the others already and I'm very sure they are fine (they show that the matrix is diagonalisable).
We insert $\lambda_{3}=-1$ here $\begin{pmatrix}
3-\lambda & -1 & 0\\
2 & -\lambda & 0\\
-2 & 2 & -1-\lambda
\end{pmatrix} \begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}$
Then we get $\begin{pmatrix}
4 & -1 & 0\\
2 & 1 & 0\\
-2 & 2 & 0
\end{pmatrix} \begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}$
$I: 4x-y=0 \Leftrightarrow y=4x$
$II: 2x+y=0$
$III: -2x+2y=0$
We see that there is no value for $z$, so we can choose an arbitrary $z$. We get the eigenspace $E_{A}(-1)=\left\{ \left. \begin{pmatrix}
x\\
4x\\
z
\end{pmatrix} \right | x,z \in \mathbb{R} \right\}$
Two variables mean the basis is made up by $2$ vectors and this means the dimension of the eigenspace is $2$ and this means the matrix is not diagonalisable because $\lambda_{3}=-1$ is just a single eigenvalue?
Can you please tell me if this is correct?
| Another way of saying that is- no matter how many distinct eigenvalues an by n matrix has, it is diagonalizable if and only if there are n independent Eigenvectors. Of course, eigenvectors corresponding to distinct eigenvalues are necessarily independent so if an n by n matrix has n distinct eigenvalues then it must be diagonalizable- but it is the eigenvectors that are important!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Summation of $n/3^n$ $$\sum_{n=1}^\infty \frac{n}{3^n}$$
How do you find the sum?
I don't know how to start this problem and no other website I found talks about a problem like this.
| Approach 1:
$$\sum_{n=1}^\infty \frac{n}{3^n} = \frac{1}{3}\left.\sum_{n=1}^\infty \frac{d}{dx}x^{n}\right|_{x=1/3} = \frac{1}{3} \left.\frac{d}{dx} \frac{x}{1-x} \right|_{x=1/3} = \frac{1}{3} \cdot \frac{1}{(1-1/3)^2}= \frac{3}{4}.$$
Approach 2 (no derivatives):
\begin{align}
S &:= \sum_{n=1}^\infty \frac{n}{3^n}\\
S/3 &= \sum_{n=1}^\infty \frac{n}{3^{n+1}} = \sum_{n=2}^\infty \frac{n-1}{3^n}
\end{align}
Subtract the above two equations.
\begin{align}
2S/3 = S - S/3 &= \frac{1}{3} + \sum_{n=2}^\infty \frac{1}{3^n} = \frac{1}{2}
\\
S &= \frac{3}{4}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Noob question about $\int \frac{1}{\sin(x)}dx$ I manually integrate $\int \frac{1}{\sin(x)}dx$ as $$\int \frac{\sin(x)}{\sin^{2}(x)}dx = -\int \frac{1}{\sin^{2}(x)}d\cos(x) = \int \frac{1}{\cos^{2}(x) - 1}d\cos(x).$$
After replacing $u = \cos(x)$,
$$\int \frac{1}{u^{2} - 1}du = \int \frac{1}{u^{2} - 1}du = \frac {1} {2} \int \left(\frac {1} {u - 1} - \frac {1} {u+1}\right) du = \frac {1} {2} \ln\left(\frac {u-1} {u+1}\right) + C.$$
Substitute back to obtain
$$\frac {1} {2} \ln\left(\frac {\cos(x)-1}{\cos(x)+1}\right) + C.$$
The problem is that this solution is incorrect (I guess) because for example http://www.integral-calculator.com/ gives another solution
$$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$
And all other online solvers gives equivalent solution to
$$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$
The question is there I made a mistake?
Update: some of you may say that in complex space my answer is right but not so fast:
Take wolfram solver: integrate 1/sinx
The we get: $-ln(cot(x) + csc(x)) + C$ It is easy to see that it is equvalent to $$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$
$-\ln(\cot(x) + \csc(x)) + C = -\ln(\frac {\cos(x)} {\sin(x)} + \frac {1} {\sin(x)})$
then
$-\ln(\frac {\cos(x)} {\sin(x)} + \frac {1} {\sin(x)}) = -\frac {1}{2} \ln(\frac {(1+\cos(x))^{2}} {\sin^{2}x}) = -\frac {1}{2} \ln(\frac{1+\cos(x)+\cos(x)+\cos^{2}(x)} {1-\cos^{2}x}) = -\frac {1}{2} \ln(\frac {(1+\cos(x))(\cos(x)+\cos^{2})(x)} {1-\cos^{2}x}) = -\frac {1}{2} \ln(\frac {1+\cos(x)} {1-\cos(x)}) = \frac {1}{2} \ln(\frac {1-\cos(x)} {1+\cos(x)})$
| The integral of $1/x$ is not $\ln(x)$, but $\ln|x|$. See What is the integral of 1/x?
Then, in your case $\frac{1}{2}\int du \left( \frac{1}{u-1} - \frac{1}{u+1} \right) = \frac{1}{2}\ln\left( \frac{|u-1|}{|u+1|} \right) + C$. When you put back $u=\cos(x)$, the expression $|u-1|$ becomes $|\cos(x)-1|=1-\cos(x)$ and $|u+1|$ becomes $|\cos(x)+1|=1+\cos(x)$. Notice, you can forget about taking the absolute value, because $1-\cos(x) \ge 0$ and $1+\cos(x)\ge 0$. Finally
$\int dx \frac{1}{\sin(x)} = \frac{1}{2}\ln\left( \frac{1-\cos(x)}{1+\cos(x)} \right) + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Incorrectly solving the determinant of a matrix Compute $det(B^4)$, where $B =
\begin{bmatrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
1 & 2 & 1
\end{bmatrix}
$
I created
$C=\begin{bmatrix}
1 & 2 \\
1 & 1 \\
\end{bmatrix}
$ and
$
D= \begin{bmatrix}
1 & 1 \\
1 & 2 \\
\end{bmatrix}
$
$det(B) = -det(C) -2det(D) = -(1-2) - 2(2-1) = -(-1)-2(1)=1-2=1, 1^4=1$
However, the correct answer is 16. I'm confused on where I made my wrong turn
| Use row reduction to calculate the determinant:
\begin{align}
\begin{vmatrix}1&0&1\\1&1&2\\1&2&1\end{vmatrix}=\begin{vmatrix}1&0&1\\0&1&1\\0&2&0\end{vmatrix}=\begin{vmatrix}1&0&1\\0&1&1\\0&0&-2\end{vmatrix}
\end{align}
hence $\;\det B=1\cdot1\cdot(-2)=-2$, so $\;\det B^4=(\det B)^4=16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Menelaus theorem & collinear points From vertex C of the right triangle ABC height CK is dropped and in triangle ACK bisector CE is drawn. Line that passes through point B parallel to CE meets CK at point F. Prove that line EF divides segment AC in halves.
So far I have:
Construct point M on AC such that AM=MC. WE want to show that M, E, and F are collinear. So we want to prove using Menelaus' that
$\displaystyle\frac{AM}{CM} \cdot \displaystyle\frac{CF}{KF} \cdot \displaystyle\frac{KE}{AE} = 1$.
We know AM=MC so that cancels out to 1. From the angle bisector theorem in triangle ACK we know that $\displaystyle\frac{KE}{AE} = \displaystyle\frac{CK}{CA}$.
So from this we have
$\displaystyle\frac{CF}{KF} \cdot \displaystyle\frac{CK}{CA}=1$. Using the fact that $\triangle CKE$ is similar to $\triangle BFK$ we know $\displaystyle\frac{CK}{KF}=\displaystyle\frac{EK}{KB}$
Now we obtain that $\displaystyle\frac{CF}{CA} \cdot \displaystyle\frac{EK}{KB} = 1$
FROM HERE I AM STUCK.
| Let the line through point $E$ and orthogonal to $CA$ intersect $CA$ at point $L$. Let the line through point $F$ and orthogonal to $BC$ intersect $BC$ at point $N$. Since $AB$ is orthogonal to $CK$ and $CE$ is the angle bisector of angle $\angle \, ACK$,
$$LE = KE$$
Let $\angle \, BAC = \alpha = \angle \, CAK$. Then $\angle \, ACK = 90^{\circ} - \alpha$ and $\angle \, BCK = \alpha$ which means that $$\angle \, ACE = \frac{1}{2} \, \angle \, ACK = 45^{\circ} - \frac{\alpha}{2}$$ since $CE$ is the angle bisector of angle $\angle \, ACK$. Consequently $$\angle \, BEC = \angle \, ACE + \angle \, CAE = 45^{\circ} - \frac{\alpha}{2} + \alpha = 45^{\circ} + \frac{\alpha}{2}$$ Since $BF$ is parallel to $CE$
$$\angle \, EBF = \angle \, BEC = 45^{\circ} + \frac{\alpha}{2}$$ However, $$\angle \, EBN = 180^{\circ} - \angle \, CBA = 180^{\circ} - (90^{\circ} - \angle \, BAC) = 90^{\circ} + \alpha$$ and so
$$\angle \, NBF = \angle \, EBN - \angle \, EBF = 90^{\circ} + \alpha - 45^{\circ} - \frac{\alpha}{2} = 45^{\circ} + \frac{\alpha}{2} = \angle \, EBF$$ which means that $BF$ is the angle bisector of $\angle \, EBN$ and thus
$$NF = KF$$ Since triangles $AEL$ and $CFN$ are right and $$\angle \, EAL = \angle \, BAC = \alpha = \angle \, BCK = \angle \, FCN$$ the two triangles $AEL$ and $CFN$ are similar and thus
$$\frac{LE}{EA} = \frac{NF}{FC}$$ Take point $P$ on line $CF$ so that $F$ is the midpoint of $CP$. Then $FC = FP$. Therefore,
$$\frac{KE}{EA} = \frac{LE}{EA} = \frac{NF}{FC} = \frac{KF}{FP}$$
which by the Thales' intercept theorem implies that $EF$ is parallel to $AP$. Since $M$ is the intersection point of $EF$ with $CA$, the segment $FM$ is parallel to $AP$ and $F$ is the midpoint of $CP$ by construction, so $M$ is the midpoint of $AC$.
Edit. To use Menelaus' theorem, simply get to the point where we have proven that
$$\frac{LE}{EA} = \frac{NF}{CF}$$ as explained above. After that use the fact that $LE = KE$ and $NF = FK$ and apply Menelaus' theorem as follows
$$1 = \frac{CF}{FK} \cdot \frac{KE}{EA} \cdot \frac{AM}{MC} =
\frac{CF}{NF} \cdot \frac{LE}{EA} \cdot \frac{AM}{MC}$$ However $$\frac{LE}{EA} = \frac{NF}{CF}$$ is equivalent to $$ \frac{CF}{NF} \cdot \frac{LE}{EA} = 1$$ which reduces the identity
$$1 =
\frac{CF}{NF} \cdot \frac{LE}{EA} \cdot \frac{AM}{MC}$$ to
$$1 =
\frac{AM}{MC}$$ which is possible if and only if $AM = MC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
prove that $\lim_{x \to 1}\frac{x^{2}-x+1}{x+1}=\frac{1}{2}$ Please check my proof
Let $\epsilon >0$ and $\delta >0$
$$0<x<\delta \rightarrow \frac{x^{2}-x+1}{x+1}-\frac{1}{2}<\epsilon $$
$$\frac{2x^{2}-2x+2-x+1}{2x+2}<\epsilon $$
$$\frac{2x^{2}-3x+1}{2x+2}<\epsilon $$
since $\frac{2x^2-3x+1}{2x+2} <\frac{2x^{2}-3x+2}{2}$
then $\frac{2x^{2}-3x+1}{2}< \epsilon $
$2x^{2}-3x+1<2\epsilon $
choose $2\epsilon =\delta $
then $\frac{2x^{2}-3x+1}{2}<\frac{2\epsilon }{2}=\epsilon $
by transitivity of inequality
$\frac{2x^{2}-3x+1}{2x+2}<\epsilon $
| There was an error in the second line of the development.
Note that
$$\begin{align}
\left|\frac{x^2-x+1}{x+1}-\frac12\right|&=\left|\frac{2x^2-3x+1}{2(x+1)}\right|\\\\
&=\left|\frac{(2x-1)(x-1)}{2(x+1)}\right|
\end{align}$$
Now, bound $|x-1|$ by something somewhat arbitrary such as $0<x<2$. Then $1<x+1<3$ and $-1<2x-1<3$ so that $\frac{|2x-1|}{2(x+1)}<\frac32$ and for $|x-1|<1$
$$\begin{align}
\left|\frac{x^2-x+1}{x+1}-\frac12\right|&=\frac32 |x-1|\\\\
&<\epsilon
\end{align}$$
whenever $|x-1|<\delta=\min\left(1,\frac{2}{3}\epsilon\right)$. And we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove that the spectrum of $K_n$ is $((n-1)^1, (-1)^{(n-1)})$ I am trying to prove that the spectrum of the complete graph $K_n$ is $((n-1)^1, (-1)^{(n-1)})$ (where superscripts denote multiplicities of eigenvalues, not exponents). I have part of the proof but having trouble completing it.
The adjacency matrix $A(K_n)$ is the $n \times n$ matrix:
$$ A(K_n) =\begin{pmatrix}
0 & 1 & 1 & \ldots & 1 & 1 \\
1 & 0 & 1 & \ldots& 1 & 1 \\
1 & 1 & \ddots & \ldots & \ldots & \vdots \\
\vdots & \ldots & \ldots & \ddots & 1 & 1 \\
1 & 1& \ldots & 1 & 0 & 1 \\
1 & 1 & \ldots & 1 & 1 & 0 \\
\end{pmatrix}
$$
The eigenvalues of A ($\lambda_1, \lambda_2, \ldots \lambda_n$) can be found by solving:
$$det(\lambda I - A) = 0$$
$$\begin{vmatrix}
\lambda & -1 & -1 & \ldots & -1 & -1 \\
-1 & \lambda & -1 & \ldots & -1 &- 1 \\
-1 & -1 & \ddots & \ldots & \ldots & \vdots \\
\vdots & \ldots & \ldots & \ddots & -1 & -1 \\
-1 & -1& \ldots & -1 & \lambda & -1 \\
-1 & -1 & \ldots & -1 & -1 & \lambda \\
\end{vmatrix} = 0
$$
I understand that one way to show $det(\lambda I - A) = 0$ for a given $\lambda$ is to show that the matrix $(\lambda I - A)$ is linearly dependent ie. one row is a linear combination of the others. From this it follows that $\lambda$ is an eigenvalue.
For $\lambda = (n-1)$ we have the matrix
$$\begin{pmatrix}
(n-1) & -1 & -1 & \ldots & -1 & -1 \\
-1 & (n-1) & -1 & \ldots & -1 &- 1 \\
-1 & -1 & \ddots & \ldots & \ldots & \vdots \\
\vdots & \ldots & \ldots & \ddots & -1 & -1 \\
-1 & -1& \ldots & -1 & (n-1) & -1 \\
-1 & -1 & \ldots & -1 & -1 & (n-1) \\
\end{pmatrix}
$$
It can be readily seen that any particular row is a linear combination of all of the other rows, specifically that $Row_i = \sum_{j \neq i} (-1)Row_j$, hence $\lambda = (n-1)$ is an eigenvalue.
Also for $\lambda = (-1)$ we have the matrix:
$$\begin{pmatrix}
-1 & -1 & -1 & \ldots & -1 & -1 \\
-1 & -1 & -1 & \ldots & -1 &- 1 \\
-1 & -1 & \ddots & \ldots & \ldots & \vdots \\
\vdots & \ldots & \ldots & \ddots & -1 & -1 \\
-1 & -1& \ldots & -1 & -1 & -1 \\
-1 & -1 & \ldots & -1 & -1 & -1 \\
\end{pmatrix}
$$
Clearly every row is identical hence each of the rows is a linear combination of any of the other rows, so the matrix is linearly dependent and $\lambda = (-1)$ is an eigenvalue also.
However I do not know how to show their multiplicities. (Can we appeal to the fact that the eigenvalue $(n-1)$ reduces the rank by 1 hence has multiplicity 1, while the eigenvalue $(-1)$ reduces the rank by $(n-1)$, hence has multiplicity $(n-1)$?)
Also I feel there may be a much simpler proof?
| Hint: Try to use induction with the Laplace expansion of the determinant to get the characteristic polynomial of $A(K_n)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit without l'Hopital or Taylor series: $\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}$ find the limit without l'Hôpital and Taylor rule :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}=?$$
My Try :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}\\=\lim\limits_{x \to 0}\frac{x\cos x \sin x- \sin x\sin x}{x^3\sin x}=\\\lim\limits_{x \to 0}\frac{x\sin 2x- \sin^2 x}{2x^3\sin x}=$$?
what now ?
| By a scaling of the variable, $$L:=\lim_{x\to 0}\frac{x\cos x-\sin x}{x^3}=\lim_{x\to 0}\frac{3x\cos3x-\sin3x}{27x^3}.$$
Then by the triple angle formulas,
$$3x\cos3x-\sin3x=3x\cos x(1-4\sin^2x)-3\sin x+4\sin^3x\\
=(3-4\sin^2x)(x\cos x-\sin x)-8x\cos x\sin^2x,$$
so that
$$L=\lim_{x\to0}(3-4\sin^2x)\cdot\frac L{27}-\lim_{x\to0}\frac{8x\cos x}{27x}\cdot\lim_{x\to0}\frac{\sin^2x}{x^2}.$$
Using the $\text{sinc}$ limit, we can conclude
$$\frac89L=-\frac8{27}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Power series representation of $ f(x) = \frac{(x+2)^2}{x^2+1} $ I'm trying to find the power series representation of $ f(x) = \frac{(x+2)^2}{x^2+1} $
Here is what I tried:
$$
f(x) = \frac{((x+2)^2}{x^2+1}=(x+2)^2\frac{1}{1-(-x^2)}
$$
$$
\sum_{n=1}^\infty(-x^2)^{n-1}=1-x^2+x^4-x^6+x^8-+...
$$
$$
f(x)=(x+2)^2[1-x^2+x^4-x^6+x^8-+...] =(x+2) -(x+2)x^2+(x+2)x^4-(x+2)x^6+(x+2)x^8-+...
$$
I'm not sure if this is correct.
| I think it's better to write
$$f(x) = \frac{(x+2)^2}{x^2+1}=\frac{x^2+4x+4}{x^2+1}=1+\frac{4x+3}{x^2+1}=1+(4x+3) (1-x^2+x^4-x^6+x^8+\cdots)=4+4x-3x^2-4x^3+\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.
I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a >0$. Then...
$x_2 = 2 + \frac{1}{2} = 2.5$
$x_3 = 2.5 + \frac{1}{2.5} = 2.9$
$x_4 = 2.9 + \frac{1}{2.9} = 3.2448$
$x_5 = 3.2448 + \frac{1}{3.2448} = 3.5530$
$x_6 = 3.5530 + \frac{1}{3.5530} = 3.8344$
$x_7 = 3.8344 + \frac{1}{3.8344} = 4.0952$
$x_8 = 4.0952 + \frac{1}{4.0952} = 4.3394$
$x_9 = 4.3394 + \frac{1}{4.3394} = 4.5698$
$x_{10} = 4.5698 + \frac{1}{4.5698} = 4.7887$
but others have said it converges so I'm confused on whether it converges or diverges? Can someone please explain.
| As others have shown, $x_{n+1} \geq x_n > 0$ for all $n \in \mathbb{N}$. The important point that we will be using is the fact that each term is strictly positive (i.e., non-zero).
Suppose that $(x_n)$ is convergent. Let $L$ be its limit. $L$ is either zero or non-zero. We shall first show that $L \neq 0$. Consider the sequence $(x_{n+1}^2)$. If $L = 0$, notice that $\lim_{n\to\infty}x_{n+1}^2 = L^2 = 0$. For all $n$,
$$x_{n+1}^2 = \left(x_n + \frac{1}{x_n}\right)^2 = x_n^2 + \frac{1}{x_n^2} + 2 \geq 2$$
implying that $\lim_{n\to\infty}x_{n+1}^2 \geq 2$, which is a contradiction. Hence, $L \neq 0$.
Since $L \neq 0$,
$$
\lim_{n\to\infty}x_n = \lim_{n\to\infty}x_{n+1} = \lim_{n\to\infty}\left(x_n + \frac{1}{x_n}\right) = \lim_{n\to\infty}x_n + \frac{1}{\lim_{n\to\infty} x_n}
$$
The last equality is true because $L \neq 0$ and $x_n \neq 0$. Hence,
$$
L = L + \frac{1}{L} \Leftrightarrow \frac{1}{L} = 0
$$
which is also a contradiction. Hence $(x_n)$ is divergent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of an infinite sequence I want to know if
$$
B=\sqrt{2+\sqrt{4+\sqrt{8+\sqrt{16+...}}}}
$$
converges to a finite value or not. $B$ can be written as
$$
B=\sqrt{2^1+\sqrt{2^2+\sqrt{2^3+\sqrt{2^4+...}}}}.
$$
However, I have no idea the next step. Any suggestion, idea, or comment is welcome, thanks!
| We have
$$B=\sqrt{2^1+\sqrt{2^2+\sqrt{2^3+\sqrt{2^4+\cdots}}}}<\sqrt{2^{2^0}+\sqrt{2^{2^1}+\sqrt{2^{2^2}+\sqrt{2^{2^3}+\cdots}}}}$$
We then note that
$$A=\sqrt{2^{2^0}+\sqrt{2^{2^1}+\sqrt{2^{2^2}+\sqrt{2^{2^3}+\cdots}}}}\\A=\sqrt{2+A\sqrt2}\implies A=\frac{\sqrt2+\sqrt{10}}2$$
Thus, $B$ is bounded above, so it converges.
One may see $A$ is bounded above since
$$A_1=\sqrt{2^{2^0}}=\sqrt2=\frac{\sqrt2+\sqrt2}2<\frac{\sqrt2+\sqrt{10}}2$$
$$A_2=\sqrt{2^{2^0}+\sqrt{2^{2^1}}}=\sqrt{2+\sqrt2A_1}<\sqrt{2+\sqrt2\frac{\sqrt2+\sqrt{10}}2}=\frac{\sqrt2+\sqrt{10}}2$$
$$A_3=\sqrt{2^{2^0}+\sqrt{2^{2^1}+\sqrt{2^{2^2}}}}=\sqrt{2+\sqrt2A_2}<\sqrt{2+\sqrt2\frac{\sqrt2+\sqrt{10}}2}=\frac{\sqrt2+\sqrt{10}}2$$
And on with the induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Surds question grade 9 I am a student and I need help answering this question.
Simplify:
$\frac{6}{\sqrt{28}}$-$\frac{9}{\sqrt{63}}$
What I did:
$\frac{6}{\sqrt{28}}$- $\frac{9}{\sqrt{63}}$
=$\frac{6}{\sqrt{7×4}}$ - $\frac{9}{\sqrt{7×4}}$
= $\frac{6}{2\sqrt{7}}$ - $\frac{9}{3\sqrt{7}} $
$\frac{6}{2\sqrt{7}}$ × $\frac{2\sqrt{7}}{2\sqrt{7}}$ = $\frac{12\sqrt{7}}{14}$
$\frac{9}{3\sqrt{7}}$ × $\frac{3\sqrt{7}}{3\sqrt{7}}$ = $\frac{27\sqrt{7}}{63}$
($\frac{12\sqrt{7}}{14}$)/2 = $\frac{6\sqrt{7}}{7}$
($\frac{27\sqrt{7}}{63}$)
/2 = $\frac{3\sqrt{7}}{7}$
$\frac{6\sqrt{7}}{7}$ - $\frac{3\sqrt{7}}{7}$
= $\frac{3\sqrt{7}}{7}$
Thankyou and help is appreciated.
| Hint: $\frac{6}{\sqrt{28}} = \frac{3}{\sqrt{7}} = \frac{9}{\sqrt{63}}$
You made some typos in your work.
| {
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Simplification of $\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50}$ There was a post on this web site an hour ago asking for the sum of
\begin{equation*}
\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50}
\end{equation*}
expressed as a single binomial coefficient. (Four choices were provided in the post.) The post seems to have been deleted. I think it is worth keeping on this web site.
| Solution
For any positive integer $n$ and any nonnegative integer $r \leq n$,
\begin{equation*}
\binom{n}{r} = \binom{n}{n - r} .
\end{equation*}
According to Vandermonde's Identity,
\begin{align*}
&\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50} \\
&\qquad = \binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\
&\qquad = \sum_{r=0}^{49} \binom{50}{r} \binom{50}{49-r} \\
&\qquad = \binom{100}{49} .
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ I am supposed to proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n$ being a natural number (integer, $n > 0$).
What I have so far:
Basis: $n = 1$
\begin{align}
3^{2 \cdot 1-1} + 2^{2 \cdot 1-1} & = 3^1 + 2^1\\
& = 5
\end{align}
Assumption: $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n = k \in \mathbb{N}$.
$5 \mid (3^{2n-1} + 2^{2n-1}) \implies 3^{2n-1} + 2^{2n-1} = 5m$, $m \in \mathbb{Z}$
Proof: Let $n = k + 1$
\begin{align}
3^{2 \cdot (k+1)-1} + 2^{2 \cdot (k+1)-1} & = 3^{2k+2-1} + 2^{2k+2-1}\\
& = 3^{2k+1} + 2^{2k+1}\\
& = 3^{2k} \cdot 3^1 + 2^{2k} \cdot 2^1\\
& = 3^{2k} \cdot 3 + 2^{2k} \cdot 2
\end{align}
And here I got stuck. I don't know how to get from the last line to the Assumption. Either I am overlooking a remodeling rule or I have used a wrong approach.
Anyway, I am stuck and would be thankful for any help.
| $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
3^{2n + 1} + 2^{2n + 1} & =
9 \times 3^{2n - 1} + 4 \times 2^{2n - 1} =
\pars{10 - 1}3^{2n - 1} + \pars{5 - 1}\times 2^{2n - 1}
\\[5mm] & =
\color{#f00}{5}\pars{2 \times 3^{2n - 1} - 2^{2n - 1}} -
\pars{\color{#f00}{3^{2n - 1} + 2^{2n - 1}}}
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2195460",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Compute : $\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$ Question: Compute this integral
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
My Approach:
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
$$=\int\frac{x+2}{\sqrt{x^2+5x}+6}\times \frac{{\sqrt{x^2+5x}-6}}{{\sqrt{x^2+5x}-6}}~dx$$
$$\int\frac{(x+2)(\sqrt{x^2+5x})}{x^2+5x-36}~dx~~- \underbrace {~\int\frac{(6x+12)}{x^2+5x-36}~dx~}_{\text{This one I know how to deal with} }$$
$$\text{Now:} ~\int\frac{(x+2)(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$=\frac{1}{2}\int\frac{(2x+5-1)(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$=\frac{1}{2}\int\frac{(2x+5)(\sqrt{x^2+5x})}{x^2+5x-36}~dx~~- \frac{1}{2}\int\frac{(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$\Big( \text{Let} ~ x^2+5x=t \implies (2x+5)~dx = dt \Big)$$
$$ \underbrace{\frac{1}{2}\int \frac{\sqrt{t}}{t-36}~dt}_{\text{I can deal with this}} ~~- \frac{1}{2}\int \frac{\sqrt{x^2+5x}}{x^2+5x-36}~dx$$
Now I'm stuck. I am unable to calculate: $$ \int \frac{\sqrt{x^2+5x}}{x^2+5x-36}~dx$$
P.S.: I am high school student so please try to use elementary integrals only; i.e. integration by parts and substitution. I don't know how to use complex numbers in integration, multiple integrals, error function, etc. (I don't know if it can be used here or not, just clarifying.)
As answered by @Kanwaljit Singh: Finally I have to compute:
$$\int \frac{1}{\sqrt{x^2+5x}-6}$$
But if I was able to compute it, I would have done it in the very first step, id est ;
$$\int \frac{x+2}{\sqrt{x^2+5x}+6}~dx =
\frac{1}{2}\int \frac{2x+5-1}{\sqrt{x^2+5x}+6}~dx
\\ \frac{1}{2}\int \frac{2x+5}{\sqrt{x^2+5x}+6}~dx ~- \frac{1}{2}\int \frac{1}{\sqrt{x^2+5x}+6}~dx
\\ \Big( \text{Let} ~ x^2+5x=t \implies (2x+5)~dx = dt \Big)
\\ \underbrace{\frac{1}{2}\int \frac{1}{t+6}~dt}_{\text{Doable}} ~-~\frac{1}{2}\int \frac{1}{\sqrt{x^2+5x}+6}~dx
\\ \int \frac{1}{\sqrt{x^2+5x}+6}~dx $$
Reached to a similar step by a short path.
But how do I compute this one?
A screenshot of this question:
| all of these answers are so stupendously complicated yo! here's a much better method:
as stated, we need to deal with integral of 1/(sqrt(x^2+5x)+6)
write x^2+5x= (x+5/2)^2 - (5/2)^2
now substitute x+5/2 = 5y/2
now, our main focus after taking the constants out becomes the integral of
1/(sqrt(y^2 - 1) + 12/5)
put arcsecy= z or y= secz so integral after getting rid of constants is
secz.tanz/(tanz + a) where a = 12/5.
now, here comes the genius step:
secz.tanz/(tanz + a)= secz - asecz/(tanz + a). else this prob was extremely hard. Obviously we already know integral of secz.
For secz/(tanz + a), write it as 1/(sinz + acosz). now this is the standard form that can be solved by either of the 2 methods:
*
*Let 1= rcosq and a= rsinq and hence forming a sine term in the denominator.
*Or even better, go for u= tan(z/2) substitution (also referred to as the universal substitution) and you will get an easy standard rational function that even a 4 yo can integrate!
| {
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Proof that $y^2=x^3+21$ has no integral solution with elementary methods? I tried to prove that $$y^2=x^3+21$$ has no integral solution in the way as shown here : http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf
My work so far : $x$ cannot be even because we would have $y^2\equiv 5\mod 8$ , which is a contradiction. Hence, $x$ is odd and $y$ is even. We have $$x\equiv x^3=y^2-21\mod 8$$, which gives the possibilities $x=3\ ,\ y=0\ or \ 4$ and $x=7\ ,\ y=2\ or\ 6$ modulo $8$.
Now I tried various factorizations to get a contradiction with quadratic residues :
$1)$ : $y^2+6=x^3+27=(x+3)(x^2-3x+9)$
$2)$ : $y^2-13=x^3+8=(x+2)(x^2-2x+4)$
$3)$ : $y^2-20=x^3+1=(x+1)(x^2-x+1)$
but in no case I found a contradiction with quadratic residues.
Can this particular equation be shown to be non-solveable in integers in the described way ?
How can I find systematically a suitable factorization in general leading to the desired contradiction, assuming the equation actually has no integral solution ?
| This will not be a complete solution. Note that from your equations 1) and 3), two of the three possible cases of $x \pmod{3}$ can be eliminated:
If $x \equiv 0 \pmod{3}$, then from 1), $y^2 + 6 \equiv 0 \pmod{27} \longrightarrow y^2 \equiv 21 \pmod{27}$, which is impossible.
If $x \equiv 2 \pmod{3}$, then from 3), $y^2 - 20 \equiv 0 \pmod{9} \longrightarrow y^2 \equiv 2 \pmod{9}$, which is impossible.
Now for the apparently more difficult case of $x \equiv 1 \pmod{3}$. You have already shown that $x \equiv 3 \pmod{4}$, which means that $x \equiv 7 \pmod{12}$ is the only remaining possibility. Unfortunately, I can’t see how to carry this argument any further.
| {
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Show that for $p$ to be odd prime and $p \equiv 3$ mod $4$, then $x^2+y^2 = p$ has no integer solution Show that for $p$ to be odd prime and $p \equiv 3$ mod $4$, then $x^2+y^2 = p$ has no integer solution. I have no idea how can i apply quadratic reciprocity to the equation $x^2+y^2 = p$ or should use other method.
| Simplest way is to note that if $x = 2k + m; m = 0,1$ and $y= 2j + n; n=0,1$ then $x^2 + y^2 = 4(k^2 + j^2 + km + jn) + m^2 + n^2 \equiv m^2 + n^2 \equiv 0,1,2 \mod 4 \not \equiv 3\mod 4$
$p$ being prime has nothing to do with it.
... or $x \equiv 0, \pm 1, 2 \mod 4 \implies x^2 \equiv 0,1 \mod 4$ so $x^2 + y^2 \equiv 0,1,2 \mod 4$.
Not sure which is easier.
| {
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$\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $\left|x\right|$ Im trying to prove that $f_{n}\left(x\right)=\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $f(x) = \left|x\right|$ in $[-1,1]$.
So for evary $\varepsilon$ exists $N \in \mathbb{N}$ s.t for all $n>N$ and for all $x \in [-1,1]$ $\left|f_{n}\left(x\right)-f\left(x\right)\right| \le \varepsilon$
And this how i tried to continue :
$\left|f_{n}\left(x\right)-f\left(x\right)\right|=\left|\sqrt{x^{2}+\frac{1}{n}}-\left|x\right|\right|=\left|\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right|=\left|\frac{\left(\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right)\left(\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}\right)}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|=\left|\frac{x^{2}+\frac{1}{n}-x^{2}}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|=\left|\frac{\frac{1}{n}}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|$
Which lead me to the same problem of $\left|\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right|$
any advice ?
| If $a,b\ge 0,$ then $\sqrt {a+b} \le \sqrt a + \sqrt b.$ Proof: Square both sides. We conclude
$$\sqrt {x^2+1/n} \le \sqrt {x^2}+\sqrt {1/n},$$
which will lead to your result.
| {
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Using 3 equations, find $2a-b+c$. From the problem, I extract the following 3 equations.
*
*$a+b+c=10$
*$ab+bc+ca=31$
*$abc=30$
The question is to find $(2a-b+c)$.
Using the equation $(a+b+c)^2 = a^2 + b^2 + c^2 +2 (ab+bc+ca)$, I found $a^2+b^2+c^2=38$.
Using the equation $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2 + b^2+c^2-ab-bc-ca)$, I found $a^3+b^3+c^3=160$.
However, I was stuck here. Please let me know some directions. Thank you.
| By Viete $a$, $b$ and $c$ are roots of the following equation:
$$x^3-10x^2+31x-30=0$$ or
$$(x-2)(x-5)(x-3)=0$$
and the rest for you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202303",
"timestamp": "2023-03-29T00:00:00",
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Trouble integrating $e^{ax}\cos bx$ and $e^{ax}\sin bx$ I have to integrate the following indefinite integrals $$\int e^{ax}\sin bxdx~~;~\int e^{ax}\cos bx$$ The procedure I used is the same for both integrals:
*
*Make the change of variables $bx=t$
*Use partial integration twice with $dv=e^{at/b}dt$ until the initial integral reappears and then solve.
This is the full procedure for the first integral. For the 2nd integral the result I obtain is the same except for the signs. $$\int e^{ax}\sin bxdx$$ Change of variables $bx=t$ $$\frac{1}{b}\int e^{at/b}\sin tdt$$ Partial integration $$dv=e^{at/b}dt~;~v=\frac{b}{a}e^{at/b}~;~u=\sin t~;~du=\cos tdt$$ Which leads to $$\frac{1}{a}e^{at/b}\sin t-\frac{1}{a}\int e^{at/b}\cos tdt$$ Partial integration again $$dv=e^{at/b}~;~v=\frac{b}{a}e^{at/b}~;~u=\cos t~;~du=-\sin tdt$$ Which leads to $$\frac{1}{b}I=\frac{1}{a}e^{at/b}\sin t-\frac{b}{a^2}e^{at/b}\cos t-\frac{b}{a^2}\int e^{at/b}\sin tdt$$ $$\frac{a^2+b^2}{ba^2}I=\frac{1}{a}e^{at/b}\sin t-\frac{b}{a^2}e^{at/b}\cos t$$ $$(a^2+b^2)I=bae^{at/b}\sin t-b^2e^{at/b}\cos t$$ $$I=\frac{ba}{a^2+b^2}e^{at/b}\sin t-\frac{b^2}{a^2+b^2}e^{at/b}\cos t$$ $$I=\frac{ba}{a^2+b^2}e^{ax}\sin bx-\frac{b^2}{a^2+b^2}e^{ax}\cos bx+C$$
The correct result is the one I obtained but divided by $b$ and for the love of God I can't figure out where I made the mistake.
| Let, $I=\displaystyle\int{e^{ax}\sin{bx}}\ dx$
now by applying by parts rule
\begin{align*}
I&= e^{ax}\int \sin{bx}\ dx-\int\left[\frac{d}{dx}(e^{ax})\cdot\int\sin{bx}\ dx\right]\ dx\\
&= e^{ax}\left(-\frac{\cos{bx}}{b}\right)-\int{ae^{ax}}\cdot\left(-\frac{\cos{bx}}{b}\right)\ dx\\
&=-\frac{e^{ax}\cos{bx}}{b}+\frac{a}{b}\int{e^{ax}\cos{bx}}\ dx\\
&=-\frac{e^{ax}\cos{bx}}{b}+\frac{a}{b}\left[e^{ax}\int\cos{bx}\ dx-\int\left\{\frac{d}{dx}(e^{ax})\cdot\int\cos{bx}\ dx\right\}dx\right]+k_1\\
&=-\frac{e^{ax}\cos{bx}}{b}+\frac{a}{b}e^{ax}\cdot\frac{\sin{bx}}{b}-\frac{a}{b}\int ae^{ax}\cdot\frac{\sin{bx}}{b}\ dx+k_1\\
or,\ I&=-\frac{e^{ax}\cos{bx}}{b}+\frac{a}{b^2}e^{ax}\sin{bx}-\frac{a^2}{b^2}I+k_1\\
or,\ I+\frac{a^2}{b^2}I&=\frac{e^{ax}}{b^2}\left(a\sin{bx}-b\cos{bx}\right)+k_1\\
or,\ I&=\frac{e^{ax}}{a^2+b^2}\left(a\sin{bx}-b\cos{bx}\right)+k
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to prove that $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$? I have a series $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \quad n \ge 1$$
For example, $a_3 = \frac{1}{4}+\frac{1}{5}+\frac{1}{6}$.
I need to prove that for $n \ge 1$:
$$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$$
I guess one could say that:
$$
\sum_{i=1}^n\frac{1}{n+i} \le \sum_{i=1}^n\frac{n}{n+1}
$$
However, I'm not sure this is rigorous enough (for example, in $\sum_{i=1}^n\frac{1}{n+i}$ how do we really know that the index goes from $1$ to $n$) and I think this needs to be proven via induction.
So the base case is:
$$a_1 = \frac{1}{2} \le \frac{1}{2} = \frac{n}{n+1}$$
The step: suppose $a_n \le \frac{n}{n+1}$ then let's prove that $$a_{n+1} \le \frac{n+1}{n+2}$$
The above can be developed as:
$$
\frac{1}{n+1}+\frac{1}{n+3}+...+\frac{1}{2(n+1)} \le \frac{n}{n+1}+\frac{1}{n+2}
$$
This is where I get stuck. If I could somehow prove that the number of terms to the left $\le$ the terms to the left I would be golden. Or maybe there's another way.
| Your inequality can be greatly improved. The sequence given by $a_n = H_{2n}-H_{n}$ is increasing (it is enough to compute $a_{n+1}-a_n$) hence $H_{2n}-H_n\leq \lim_{n\to +\infty}\left(H_{2n}-H_n\right) = \color{red}{\log 2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
find a power series for $f(x)=\ln(x^2+1)$. (a) Find a power series representation for
$f(x) = \ln(1 + x)$.
$$f(x) = \sum _{n=1}^{\infty } \frac{\left(-1\right)^{n-1}x^n}{n}$$
What is the radius of convergence, $R$?
$R = 1$
(b) Use part (a) to find a power series for
$f(x) = x \ln(1 + x)$.
$$f(x)=\sum _{n=2}^{\infty } \frac{\left(-1\right)^n x^n}{n-1}$$
What is the radius of convergence, $R$?
$R = 1$
(c) Use part (a) to find a power series for
$f(x) = \ln(x^2 + 1).$
$f(x)=\sum _{n=1}^{\infty}$ ?? <-- This is the answer I'm having trouble finding out.
What is the radius of convergence, $R$?
$R = 1$
| From (a),
\begin{equation}
f(x) = \ln (1 + x) = \sum_{n=1}^\infty \frac{\left(-1\right)^{n-1}\left(x^n\right)}{n}.
\end{equation}
Then, (c) becomes
\begin{equation}
\ln (x^2 + 1) = \ln(1 + x^2) = f(x^2) = \sum_{n=1}^\infty \frac{\left(-1\right)^{n-1}\left(x^{2n}\right)}{n}.
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How have I computed the integral $\int \sin^{3}(3x)\cos(3x)\,dx$ incorrectly? $\displaystyle \int \sin^{3}(3x)\cos(3x)\,dx $
$u = 3x $
$du = 3\,dx $
$3\displaystyle\int \sin^{3}(u)\cos(u)\,du$
$3\displaystyle\int \sin^{2}(u)\sin(u)\cos(u)\,du$
$3\displaystyle\int (1 - \cos^{2}(u))\sin(u)\cos(u)\,du$
$v = \cos(u)$
$dv = -\sin(u)\,du $
$3\displaystyle\int (1 - v^{2})(-1)v\,dv$
$ 3\displaystyle\int (-v + v^{3})\,dv$
$ 3\left[-\frac{v^{2}}{2} + \frac{v^{4}}{4}\right]$
$3\left[ -\frac{\cos^{3}{3x}}{2} + \frac{\cos^{4}(3x)}{4}\right] $
$= -3 \frac{\cos^{2}(3x)}{2} + 3\frac{\cos^{4}(3x)}{4} + C$
wolfram states this as incorrect, what have I done wrong?
Thank you
| You determined $du=3\,dx$, but substituted $dx = 3\,du$. So your answer is off by a factor of nine.
$$
\int \sin^3(3x) \cos(3x)\,dx = \int \sin^3(u) \cos(u)\cdot \frac{1}{3}\,du
= \frac{1}{3}\int \sin^3(u) \cos(u)\,du
$$
Then proceed as you did. You'll get
$$
\int \sin^3(3x) \cos(3x)\,dx =- \frac{1}{6} \cos^2(3x) -\frac{1}{12}\cos^4(3x)+C
$$
The other answers are suggesting substituting $u=\sin(3x)$ in one fell swoop. Then $du = 3 \cos(3x)\,dx$, so
$$
\int \sin^3(3x) \cos(3x)\,dx = \frac{1}{3}\int u^3\,du = \frac{1}{3}\cdot\frac{1}{4}u^4 + C = \frac{1}{12}\sin^4(3x)+C
$$
Both are correct, although the second solution is more compact and concise. In the comments you asked if the two are “equivalent.” Yes, but only in the antiderivative sense. Using the trig identities:
\begin{align*}
- \frac{1}{6}\cos^2(3x) + \frac{1}{12}\cos^4(3x)
&= \frac{1}{12}\cos^2(3x)\left(\cos^2(3x) - 2\right) \\
&= \frac{1}{12}\left(1-\sin^2(3x)\right)\left(1-\sin^2(3x)-2\right) \\
&= \frac{1}{12}\left(1-\sin^2(3x)\right)\left(-1-\sin^2(3x)\right) \\
&= \frac{1}{12}\left(\sin^2(3x)-1\right)\left(\sin^2(3x)+1\right) \\
&= \frac{1}{12}\left(\sin^4(3x)-1\right) = \frac{1}{12}\sin^4(3x) - \frac{1}{12}
\end{align*}
You can see the two antiderivatives differ by a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving $\sqrt[3]{{x\sqrt {x\sqrt[3]{{x \ldots \sqrt x }}} }} = 2$ with 100 nested radicals I have seen a book that offers to solve the following equation:
$$\underbrace {\sqrt[3]{{x\sqrt {x\sqrt[3]{{x \ldots \sqrt x }}} }}}_{{\text{100 radicals}}} = 2$$
The book also contains the answer:
$$x = {2^{\left( {\frac{{5 \times {6^{50}}}}{{3 \times ({6^{50}} - 1)}}} \right)}}$$
How did they get the answer for such equation? I tried to obtain the recurrence relation, but could not find the way to get the above answer.
EDIT
$${u_{100}} = 2,$$
$$\sqrt[3]{{x{u_{99}}}} = 2,$$
$$x{u_{99}} = {2^3},$$
$${u_{99}} = \sqrt {x{u_{98}}} ,$$
$$x\sqrt {x{u_{98}}} = {2^3},$$
$${x^2}x{u_{98}} = {({2^3})^2},$$
$${x^3}{u_{98}} = {2^6},$$
$${u_{98}} = \sqrt[3]{{x{u_{97}}}},$$
$${x^3} \times \sqrt[3]{{x{u_{97}}}} = {2^6},$$
$${x^9}x{u_{97}} = {({2^6})^3},$$
$${x^{10}}{u_{97}} = {2^{18}},$$
$${u_{97}} = \sqrt {x{u_{96}}} ,$$
$${x^{10}}\sqrt {x{u_{96}}} = {2^{18}},$$
$${x^{20}}x{u_{96}} = {({2^{18}})^2},$$
$${x^{21}}{u_{96}} = {2^{36}},$$
$${u_{96}} = \sqrt[3]{{x{u_{95}}}},$$
$${x^{21}} \times \sqrt[3]{{x{u_{95}}}} = {2^{36}},$$
$${x^{63}}x{u_{95}} = {2^{108}}$$
$${x^{64}}{u_{95}} = {2^{108}},$$
$${u_{95}} = \sqrt {x{u_{94}}} ,$$
$${x^{64}}\sqrt {x{u_{94}}} = {2^{108}},$$
$${x^{128}}x{u_{94}} = {2^{216}},$$
$${x^{129}}{u_{94}} = {2^{216}},$$
$$ \ldots $$
but I still have no idea how to find a generalized formula which allows to obtain the answer.
| Essentially what is written there is just a big product of $x^{a_n}$ where $a_n$ changes from term to term. The terms $a_n$ follow the following sequence:
$$\frac{1}{3}, \frac{1}{3\cdot 2}, \frac{1}{3^2\cdot 2}, \frac{1}{3^2\cdot 2^2},\ldots, \frac{1}{3^{50}\cdot 2^{50}}$$
The equation you have now is
$$\prod_{n=1}^{100} x^{a_n} = 2$$
Which can be simplified to
$$x^{\left(\sum_{n=1}^{100} {a_n}\right)}= 2$$
The way the sequence $a_n$ is created allows you to say that $a_{2m-1}=2\cdot a_{2m}$, which gives
$$x^{3\left(\sum_{n=1}^{50} \frac{1}{6^n}\right)}= 2$$
This will now lead you directly to the solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Identify $\lim\limits_{x \to +\infty } x^2 \left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)$
Identify $$\lim\limits_{x \to +\infty } x^2 \left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)$$
My Try :
$$\sqrt{x^4+x+1}=\sqrt{x^4(1+\frac{1}{x^3}+\frac{1}{x^4})}=x^2\sqrt{(1+\frac{1}{x^3}+\frac{1}{x^4})}$$
Now : $$\frac{1}{x^3}+\frac{1}{x^4}=z$$
$$(1+z)^{\frac{1}{n}}= 1 + \frac1n x + \frac{1 - n}{2n^2}x^2 + \frac{2n^2 - 3n + 1}{6n^3}x^3 + O(x^4)$$
$$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac12 z - \frac{1}{8}z^2 + + O(z^3)$$
$$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac12 (\frac{1}{x^3}+\frac{1}{x^4}) - \frac{1}{8}(\frac{1}{x^3}+\frac{1}{x^4})^2 + O(z^3)$$
$$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac{1}{2x^3}+\frac{1}{2x^4} - \frac{x^2+2x+1}{8x^8} + O(z^3)$$
And :
$$\sqrt{x^4+x+1}=\sqrt{x^4(1+\frac{1}{x^3}+\frac{5}{x^4})}=x^2\sqrt{(1+\frac{1}{x^3}+\frac{5}{x^4})}$$
So :
$$(1+(\frac{1}{x^3}+\frac{5}{x^4}))^{\frac{1}{2}}= 1 + \frac{1}{2x^3}+\frac{5}{2x^4} - \frac{x^2+10x+25}{8x^8}+ O(z^3)$$
$$\lim\limits_{x \to +\infty }=x^4( 1 + \frac{1}{2x^3}+\frac{1}{2x^4} - \frac{x^2+2x+1}{8x^8} + O(z^3)-( 1 + \frac{1}{2x^3}+\frac{5}{2x^4} - \frac{x^2+10x+25}{8x^8}+ O(z^3)))$$
$$\lim\limits_{x \to +\infty }=x^4(\frac{1}{2x^4}-\frac{5}{2x^4})=x^4(\frac{-4}{2x^4})=-2$$
is it right ?
| $$
\begin{aligned}
\lim _{x\to \infty }\left(x^2\:\left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)\right)
& = \lim _{x\to \infty }\left(-\frac{4x^2}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}}\right)
\\& = \lim _{x\to \infty }\left(-\frac{4x^2}{x^2\sqrt{1+\frac{x}{x^4}+\frac{1}{x^4}}+x^2\sqrt{1+\frac{x}{x^4}+\frac{5}{x^4}}}\right)
\\& \approx_\infty \lim _{x\to \infty }\left(-\frac{4x^2}{2x^2}\right)
\\& = \color{red}{-2}
\end{aligned}
$$
In this case you can solve it with just an approximation without using the Taylor expansion, anyway your procedure is right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve $\frac{2}{\sin x \cos x}=1+3\tan x$ Solve this trigonometric equation given that $0\leq x\leq180$
$\frac{2}{\sin x \cos x}=1+3\tan x$
My attempt,
I've tried by changing to $\frac{4}{\sin 2x}=1+3\tan x$, but it gets complicated and I'm stuck. Hope someone can help me out.
| Multiply by LHS and RHS by $\sin x\cos x$ :
$$3\sin ^2x +\sin x\cos x -2 =0 \implies 3\cos^2 x-\sin x\cos x -1 =0$$
$$\implies \frac32 (2\cos ^2 x -1) -\frac{2\sin x \cos x }{2}=\frac{-1}{2} \implies 3\cos {2x} - \sin{2x}=-1$$
Now, squaring both the sides, you get :
$$5 \sin^2{2x}-\sin{2x}-4=0 \implies \sin {2x} =1~ \text{or}~ -4/5$$
Now you may proceed !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Find UMVUE of $6\theta^2$ given $f(x\mid\theta) = \frac{1}{2\theta^2} e^{\frac{-\sqrt{x}}{\theta}} I_{(0,\infty)}(x)$ Given $X_1, X_2,\ldots, X_n$ are i.i.d rvs with pdf $f(x\mid\theta) = \frac{1}{2\theta^2} e^{\frac{-\sqrt{x}}{\theta}} I_{(0,\infty)}(x)$ for $\ \theta > 0$.
(a) Find the UMVUE of $\ 6\theta^2$, and the variance of UMVUE.
(b) Find the CRLB of the UMVUE
(c) Find the variance of MME of $6\theta^2$ based on first moment.
My thought: For part (a), I got the same answer using Lehmann-Scheffer theorem as that of NCh's answer below. I would now present my computation for variance of UMVUE $= \frac{3}{2n^2+n} (\sum_{i=1}^{n} \sqrt{X_i})^2$.
We have: $Var(\frac{3}{2n^2+n} (\sum_{i=1}^{n} \sqrt{X_i})^2) = \frac{9}{(2n^2+n)^2} (E(Z^4) - E^2(Z^2))$ where $Z = \sum_{i=1}^{n} \sqrt{X_i}$. Now, $E(Z^2) = \theta^2(4n^2+2n)$. In addition, $Z$ is the sum of $n$ i.i.d r.vs which follow $N(\ 2n\theta, \ 2n\theta^2)$, so $E(Z^4) = (2n\theta)^4 + 6(2n\theta)^2(2n\theta^2)^2 + 3(2n\theta^2)^4$.
Thus, $\frac{9}{(2n^2+n)^2} (E(Z^4) - E^2(Z^2)) = \frac{9}{(2n^2+n)^2} (16n^4\theta^4 + 96n^4\theta^6 + 48\theta^8n^4 - 16n^4\theta^4 - 16n^3\theta^4 - 4n^2\theta^4)$
(b) To find the CRLB, I need first to verify $\ E\ (\overline{X}\frac{\partial}{\partial\theta}\log f(\overline{X}\mid\theta)) = 12\theta,$ as $\overline{X}$ is an unbiased estimator of $6\theta$. Now, $\log f(\overline{X}\mid\theta) = \frac{-\sqrt{\overline{X}}}{\theta} - \log{2\theta^2}$, so $E\ (\overline{X}\frac{\partial}{\partial\theta}\log f(\overline{X}\mid\theta)) = E(\frac{\overline{X}^{3/2} - 2\theta\overline{X}}{\theta^2}) = -12\theta + \frac{1}{\theta^2}E(\overline{X}^{3/2}).$ But I got stuck here since I cannot compute $\ E(\overline{X}^{3/2})$.
(c) The MME of $6\theta^2$ based on first moment is $\overline{X}$. Thus, we only need to compute $Var(\overline{X}) = \frac{Var(X_1)}{n} = \fbox{$\frac{84\theta^4}{n}$}$ (since $E(X^2) = 120\theta^4$)
My question: Could someone please help with parts (a) and (b) above? Also, is my answer for part (c) correct?
| It is no need to compute conditional expectations. Let $$T=\overline{\sqrt{X}}=\frac{\sum_{i=1}^n \sqrt{X_i}}{n}$$
You have already found its expectation:
$$\mathbb ET=\mathbb E\sqrt{X_1}=2\theta.$$
Let us find UMVUE of $6\theta^2$ as $\theta^*=c(n)T^2$. Firstly find $\mathbb E T^2$:
$$
\mathbb ET^2 = \text{Var}\,T+\left(\mathbb ET\right)^2 = \frac{\text{Var}\sqrt{X_1}}{n}+4\theta^2=\frac{\mathbb EX_1-4\theta^2}{n}+4\theta^2 = \frac{2\theta^2}{n}+4\theta^2.
$$
And if we take $$
c(n) = \frac{6}{\frac2n+4}=\frac{3n}{1+2n}$$
we'll obtain the desired result:
$$
\mathbb E\theta^* = \mathbb E\left(\frac{6}{\frac2n+4}T^2\right) = \frac{6}{\frac2n+4} \times \left(\frac{2\theta^2}{n}+4\theta^2\right)=6\theta^2.
$$
And $\theta^*$ is UMVUE since it is a function of complete sufficient statistics.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the general formula of the sequence $ a_n = 7a_{n-2} + 6a_{n-3} $ Given the equation $ a_n = 7a_{n-2} + 6a_{n-3} $, and $ a_0 = a_1 = a_2 = 1 $, how do I find the general equation?
I have tried to express this sequence in matrix form, $ Q = Xb $ as follows
$$ Q = \begin{bmatrix} a_{n+1} \\ a_{n-2} \\ a_{n-3} \\ \end{bmatrix}
\hspace{1cm}
X =\begin{bmatrix}0 & 7 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}
\hspace{1cm}
b = \begin{bmatrix} a_{n} \\ a_{n-1} \\ a_{n-2} \\ \end{bmatrix} $$
I attempt diagonalize $X$ in order to find $X^n$, but the eigenvalues of $X$ seem to be not real
Can anyone see if my method is correct?
| Try instead $$Q = A_n=\begin{bmatrix} a_{n} \\ a_{n-1} \\ a_{n-2} \\ \end{bmatrix}
\hspace{1cm}
X =\begin{bmatrix} 0&7 & 6 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix}
\hspace{1cm}
b = A_{n-1}=\begin{bmatrix} a_{n-1} \\ a_{n-2} \\ a_{n-3} \end{bmatrix}$$
Hence $A_n=X^n A_0=X^{n-3}A_3$
The matrix $X$ can be diagonalized as
$$ X = \begin{bmatrix} 9&4 & 1 \\ 3 & -2 & -1 \\1 & 1 & 1 \\ \end{bmatrix}
\begin{bmatrix} 3&0 & 0 \\ 0 & -2 &0 \\0 & 0 & -1 \\ \end{bmatrix}
\frac{1}{20}\begin{bmatrix} 1&3 & 2 \\ 4 & -8 &-12 \\-5 & 5 & -30 \\ \end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the minimum value of $f$ on $\{(a,b,c)\in \mathbb{R^3}\mid a+b+c=1\}$ where $f(a,b,c)=\int_{0}^1 (a+bx+cx^2)^2dx$. Find the minimum value of $f$ on $\{(a,b,c)\in \mathbb{R^3}\mid a+b+c=1\}$ where $f(a,b,c)=\int_0^1 (a+bx+cx^2)^2 \, dx$.
What I did is
$$
\int_0^1 (a+bx+cx^2)^2 \, dx=a^2+a \left( b+\frac{2c}{3} \right) + \frac{b^2}{3} + \frac{bc}{2}+\frac{c^2}{5}.
$$
Then we consider the function $F(a,b,c)=a^2+a(b+\frac{2c}{3})+\frac{b^2}{3} + \frac{bc}{2} + \frac{c^2}{5}-\lambda(a+b+c-1)$. Thus, we can consider $\nabla F=0$ with $a+b+c=1$ and try to find $a,b$ and $c$.
However, since the function and the constraint seem to be related, I am curious whether there is some other way to see the minimum without computing Lagrange multiplier. Thank you.
| Considering a quadric
$$x^2+\frac{y^2}{3}+\frac{z^2}{5}+\frac{yz}{2}+\frac{2zx}{3}+xy=\lambda$$
which touches the plane $x+y+z=1$ at $(a,b,c)$.
The tangent plane will be
$$ax+\frac{by}{3}+\frac{cz}{5}+
\frac{cy+bz}{4}+\frac{az+cx}{3}+\frac{bx+ay}{2}=\lambda$$
Comparing coefficients:
$$a+\frac{b}{2}+\frac{c}{3}=
\frac{a}{2}+\frac{b}{3}+\frac{c}{4}=
\frac{a}{3}+\frac{b}{4}+\frac{c}{5}=\lambda$$
gives $$(a,b,c)=3\lambda(1,-8,10)$$
and $$\lambda=\frac{1}{9}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum value of $a^2c^3+b^2d^3$. Let $a,b,c,d$ be non negative real numbers such that $a^5 + b^5 \le 1$ and $c^5 + d^5 \le 1$. Find the maximum possible value of $a^2c^3 + b^2d^3$.
I tried using AM, GM and some other basic inequalities but they were of no use. Need some hints.
| It can be done easily with Holder's inequality, with $p=5/2, q=5/3$ and $u=(a^2,b^2), v=(c^3,d^3)$ then:
$$\begin{align}a^2c^3+b^2d^3 &= |(a^2c^3,b^2d^3)|_1\\&\leq |(a^2,b^2)|_p |(c^3,d^3)|_q \\&= (a^5+b^5)^{2/5}(c^5+d^5)^{3/5}\leq 1
\end{align}$$
It now suffices to find a case where it is equal to $1$, which is pretty easy to do.
A more elementary approach is to use AM/GM and get that:
$$\frac{2}{5}a^5+\frac{3}{5}c^5= \frac{1}{5}(a^5+a^5+c^5+c^5+c^5)\geq \sqrt[5]{a^{10}c^{15}} = a^2c^3$$
Similarly:
$$\frac{2}{5}b^5+\frac{3}{5}d^5\geq b^2d^3$$
Adding you get:
$$1 = \frac{2}{5}+\frac{3}{5}\geq \frac{2}{5}(a^5+b^5)+\frac{3}{5}(c^5+d^5)\geq a^2c^3+b^2d^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can anything interesting be said about this fake proof? The Facebook account called BestTheorems has posted the following. Can anything of interest be said about it that a casual reader might miss?
Note that \begin{align}
\small 2 & = \frac 2{3-2} = \cfrac 2 {3-\cfrac2 {3-2}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\[10pt]
\text{and } \\
\small 1 & = \frac 2 {3-1} = \cfrac 2 {3 - \cfrac 2 {3-1}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\
\text{So } & 2=1.
\end{align}
| Simple, the problem is the assumption the three dots aka the ellipsis, in the first equation equal the ellipsis in the second. They are not.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by Taylor's series method that $|\sin{x}-(x-\frac{x^3}{3!}+\frac{x^5}{5!})|<\frac{1}{7!}$ for $x\in [-1,1]$ Prove by Taylor's series method that $$|\sin{x}-(x-\frac{x^3}{3!}+\frac{x^5}{5!})|<\frac{1}{7!}$$ for $x\in [-1,1]$
Attempt:
Let $f(x)=\sin{x}$, then $f^n(x)=\sin{(\frac{n\pi}{2}+x)}, ~x\in \mathbb{R}$
Then $$f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\frac{x^4}{4!}f^{iv}(0)+\frac{x^5}{5!}f^{v}(0)+\frac{x^6}{6!}f^{vi}(0)+\frac{x^7}{7!}f^{vii}(\theta x)\\=x-x^3/3!+x^5/5!+\frac{x^7}{7!}f^{vii}(\theta x)$$
then $$|\sin{x}-(x-\frac{x^3}{3!}+\frac{x^5}{5!})|=\frac{1}{7!}|f^{vii}(\theta x)|$$
| By the Taylor expansion with Lagrange remainder,
$$
\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\cos\xi
$$
for some $\xi\in(-1,1)$ (actually $\xi$ can be chosen between $0$ and $x$, but it's not fully relevant; the only fact we need is that $\xi\ne0$ and $\xi\in[-1,1]$). Therefore
$$
\left|\sin x-x+\frac{x^3}{3!}-\frac{x^5}{5!}\right|
=\left|\frac{x^7}{7!}\cos\xi\right|
$$
Now $|x^7\cos\xi|<1$, for $x\in[-1,1]$, because $\lvert\cos\xi\rvert<1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Expand binomially to prove trigonometric identity Prompt: By expanding $\left(z+\frac{1}{z}\right)^4$ show that $\cos^4\theta = \frac{1}{8}(\cos4\theta + 4\cos2\theta + 3).$
I did the expansion using binomial equation as follows
$$\begin{align*}
\left(z+\frac{1}{z}\right)^4 &= z^4 + \binom{4}{1}z^3.\frac{1}{z} + \binom{4}{2}z^2.\frac{1}{z^2} + \binom{4}{3}z^3.\frac{1}{z}+\frac{1}{z^4}\\
&=z^4+4z^2+6+\frac{4}{z^2}+\frac{1}{z^4}\\
&=z^4+\frac{1}{z^4}+4\left(z^2+\frac{1}{z^2}\right) + 6. -(eqn 1)
\end{align*}
$$
I'm not sure how to go on about rest of the problem.
[update]
Reading comments, I tried assuming $z = e^{i\theta}$
$2\cos\theta = e^{i\theta} + e^{-i\theta}$
$(2\cos\theta)^4 = (e^{i\theta} + e^{-i\theta})^4$
$=e^{4i\theta} + e^{-4i\theta} + 8(e^{2i\theta}+e^{-2i\theta})+6$ (from eqn 1)
| hint
Take now $z=e^{i\theta} $
and use Euler identity
$$\cos (x)=\frac {e^{ix}+\frac {1}{e^{ix}}}{2} $$
| {
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Mass of solid inside paraboloid $z=x^2+y^2$ and beneath plane $z=1$ I need to find the mass of the solid inside paraboloid $z=x^2+y^2$ and beneath plane $z=1$ if the density is 1.
Here's what I tried:
$$
M=\int\int_R\int^1_{r^2}r^2dzdA=\int\int_Rr^2z|^1_{r^2}dA=\int\int_R(r^2-r^4)dA=\int^{2\pi}_0\int^1_0(r^2-r^4)rdrd\theta=\int^{2\pi}_0\frac{r^4}{4}-\frac{r^6}{6}|^1_0=\int^{2\pi}_0\frac{1}{12}d\theta=\frac{1}{6}\pi
$$
I have a feeling I'm not correct and if so, where did I go wrong?
| To find the mass, the quantity we should be integrating is $1$. You appear to be integrating $r^2$!
\begin{align*} \int_{\theta = 0}^{\theta = 2\pi} \int_{r = 0}^{r = 1} \int_{z = r^2}^{z = 1} 1\times r dz dr d\theta = \int_{\theta = 0}^{\theta = 2\pi} \int_{r = 0}^{r = 1} (r - r^3)dr d\theta = \int_{\theta = 0}^{\theta = 2\pi} \frac 1 4 d\theta = \frac \pi 2. \end{align*}
(The volume element, as always, is $dV = rdrd\theta dz$.)
To double check, we could try changing the order of integration:
$$ \int_{\theta = 0}^{\theta = 2\pi} \int_{z = 0}^{z=1} \int_{r = 0}^{r = \sqrt {z}} 1\times rdr dz d\theta = \int_{\theta = 0}^{\theta = 2\pi} \int_{z = 0}^{z=1} \frac z 2 dz d\theta = \int_{\theta = 0}^{\theta = 2\pi} \frac 1 4 d\theta = \frac \pi 2 .$$
| {
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Can someone please check if my reasoning for this proof is valid or not? I have already seen the other questions about this proof. I'm just trying a different sort of method, though I'm not sure if it's valid or not.
Context for the main question: Prove by induction $2^n\gt n^3$ for $n\ge10$
Obviously, the base case works for n=10
$1024=2^{10}\gt1000=10^3$
The induction hypothesis: Assume $P_n$ is true $\rightarrow$ $2^n\gt n^3$
I want to then prove that $2^{n+1}\gt (n+1)^3$
Now, using the induction hypothesis:
$2^n\gt n^3$
multiply both sides by 2
$2^{n+1}\gt 2n^3$
Using the fact that $n\ge10$ this implies that $n^3\ge10n^2$
$2n^3=n^3 +n^3\gt n^3 +10n^2=n^3 +3n^2 +7n^2$
Using the fact that $7\gt 1$ this implies that $7n\gt n$ since n is positive.
$n^3 +3n^2 +7n^2\gt n^3 +3n^2 +n^2$
Once again, using $n\ge 10$ this implies $n^2\ge 10n$
$n^3 +3n^2 +n^2\gt n^3 +3n^2 +10n=n^3 +3n^2 +3n+7n$
Again $7\gt 1$
$n^3 +3n^2 +3n+7n\gt n^3 +3n^2 +3n +n$
Using $n\ge 10$ one last time
$n^3 +3n^2 +3n +n\gt n^3 +3n^2 +3n+10$
Since $10\gt 1$
$n^3 +3n^2 +3n+10\gt n^3 +3n^2 +3n+1=(n+1)^3$
Thus, through the chain of inequalities, I have proved that $2^{n+1}\gt (n+1)^3$.
QED
Sorry if there are any errors in my reasoning. Thank you for reading and feedback.
| It is ok, but I think you can go faster.
If you want to prove that:
$$2n^3>(n+1)^3\Leftrightarrow 2>\left(\frac{n+1}{n}\right)^3=\left(1+\frac{1}{n}\right)^3$$
You can do:
$$n\ge10\to\frac{1}{n}\le0.1\to \left(1+\frac{1}{n}\right)^3\le (1.1)^3=1.331<2$$
| {
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Suppose $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$. What are $x,y,z,t$? Let $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$.
What are $x,y,z,t$?
| Consider
$$
(x^2+y^2)(z^2+t^2)-(xt+yz)^2=0
$$
Expanding the left-hand side you get
$$
x^2z^2+\color{red}{y^2z^2}+\color{red}{x^2t^2}+y^2t^2
-\color{red}{x^2t^2}-2xyzt-\color{red}{y^2z^2}=0
$$
that simplifies to
$$
(xz-yt)^2=0
$$
so $xz=yt$.
Now consider the linear system in the unknowns $z$ and $t$:
\begin{cases}
yz+xt=6 \\
xz-yt=0
\end{cases}
With standard methods you get
$$
z=\frac{6y}{x^2+y^2}=\frac{3}{2}y,
\qquad
t=\frac{6x}{x^2+y^2}=\frac{3}{2}x
$$
Now any pair of numbers $x,y$ with $x^2+y^2=4$ will do.
Alternatively, consider the complex numbers
$$
A=x+yi,\qquad B=z+ti
$$
Then the hypotheses yield $|A|=2$, $|B|=3$ and that the imaginary part of $AB$ is $6$, that is
$$
AB=C+6i
$$
with $C$ real. Then $\bar{A}\bar{B}=C-6i$ and multiplying gives
$$
A\bar{A}B\bar{B}=C^2+36
$$
Thus $36=C^2+36$ and so $C=0$. Therefore
$$
AB=6i
$$
and
$$
z+ti=\frac{6i}{x+yi}=\frac{6i}{x-yi}{4}=\frac{3}{2}i(x-yi)=\frac{3}{2}(y+xi)
$$
whence again $z=\frac{3}{2}y$ and $t=\frac{3}{2}x$.
| {
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Solve $a_n=7a_{n-1}-12a_{n-2}+3^n$ $a_0=1\:a_1=2$
Using generating functions I get:
$f\left(x\right)-1-2x=7xf\left(x\right)-7x-12x^2f\left(x\right)+\frac{1}{1+3x}-1-3x$
$f\left(x\right)=\frac{3}{14\left(1+3x\right)}-\frac{13}{2\left(3x-1\right)}\:+\:\frac{40}{7\left(4x-1\right)}$
Now how can I express $a_n$ from this?
| Hint (without generating functions): eliminate $3^n$ between the recurrence relations for $n$ and $n+1$, then you get a $3^{rd}$ order linear homogeneous recurrence:
$$\require{cancel}
a_{n+1}-3 a_n=7 a_n-12a_{n-1}+\cancel{3^{n+1}} - 3(7 a_{n-1}-12 a_{n-2}+ \cancel{3^n}) \\[3px]
\iff a_{n+1} = 10 a_n - 33 a_{n-1}+36 a_{n-2}
$$
The characteristic polynomial of the latter is $t^3-10t^2+33t-36=(t-3)^2(t-4)$ so the solution of the recurrence will be of the form $a_n=(a+b\cdot n)\cdot 3^n+c \cdot 4^n$ where constants $a,b,c$ remain to be determined from the initial conditions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Binomial series sum of the form $\sum^{k}_{r=0}(-1)^r(2)^{k-r}\binom{20}{r}\binom{20-r}{20-k}$ the value of
$$2^k\binom{20}{0}\binom{20}{20-k}-2^{k-1}\binom{20}{1}\binom{19}{20-k}+2^{k-2}\binom{20}{2}\binom{18}{20-k} \cdots+ +(-1)^k\binom{20}{k}\binom{20-k}{20-k}$$
options:
$(a)\;\; 7$
$(b)\;\;8$
$(c)\;\; 10$
$(d)\;\; 20$
Attempt: $$\sum^{k}_{r=0}(-1)^r(2)^{k-r}\binom{20}{r}\binom{20-r}{20-k} = (2)^{k}\sum^{k}_{r=0}\left(-\frac{1}{2}\right)^r\frac{20!}{r!\times (20-r)!}\cdot \frac{(20-r)!}{(20-k)!\times (k-r)!}$$
$$ = \frac{(20)!(2)^{k}}{(20-k)!(k)!}\sum^{k}_{r=0}\left(-\frac{1}{2}\right)^r\frac{k!}{r!\cdot (k-r)!} = \frac{(20)!(2)^{k}}{(20-k)!(k)!}\sum^{k}_{r=0}\left(-\frac{1}{2}\right)^r\binom{k}{r}$$
$$ = \frac{(20)!(2)^{k}}{(20-k)!(k)!}\times \frac{1}{2^k} = \binom{20}{k}$$
none of the options is match.
please help me solve it, thanks
| The only possibility is D with k=1 or 19.
| {
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Find a parameterisation for $(x-8)^2 + y^2 = 16$ starting at the point $(4,0)$ and moving clockwise once around the circle. Find a parameterisation for $(x-8)^2 + y^2 = 16$ starting at the point $(4,0)$ and moving clockwise once around the circle.
My work
We can describe the circle using polar coordinates:
$x(\theta) = \rho\cos(\theta)$, $y(\theta) = \rho\sin(\theta).$
$(x-8)^2 + y^2 = 16$ is a circle with radius of $4$.
$\therefore \rho = 4$
Let $t = \theta \ \forall \ t \in [0, 2\pi]$ be our parameter.
$\therefore (x(t), y(t)) = (4\cos(t), 4\sin(t))$
$(x(t), y(t)) = (4\cos(\pi-t), 4\sin(\pi-t))$ represents the circle of radius $4$ and starting at $\pi$ and moving clockwise.
However, this circle is still centred at $(0, 0)$. To centre the circle at $(8,0)$, we must have the parameterisation $(x(t), y(t)) = (8 + 4\cos(\pi - t), 4\sin(\pi-t)) \ \forall \ t \in [0, 2\pi]$.
We can test values some values to get an indication of any errors in our parameterisation.
$(x(0), y(0)) = (4, 0)$
$(x(\pi/2), y(\pi/2)) = (8, 4)$
$(x(\pi), y(\pi)) = (12, 0)$
$(x(3\pi/2), y(3\pi/2)) = (8, -4)$
$(x(2\pi), y(2\pi)) = (4, 0)$
I would greatly appreciate it if people could please take the time to review my reasoning and solution for correctness.
| Yes, the work is correct.
Below is a parametric plot of $\left\{8 + 4 \cos \left( \pi - t \right), 4 \sin \left(t\right)\right\}$ for $0\le t < 2\pi$.
To show that your parameterization starts at $(4,0)$ and proceeds in a clockwise manner, a sequence of values was created:
$$
\begin{array}{cccc}
k & t & x(t) & y(t) \\\hline
1 & 0 & 4 & 0 \\
2 & \frac{\pi }{6} & 8-2 \sqrt{3} & 2 \\
3 & \frac{\pi }{3} & 6 & 2 \sqrt{3} \\
4 & \frac{\pi }{2} & 8 & 4 \\
5 & \frac{2 \pi }{3} & 10 & 2 \sqrt{3} \\
6 & \frac{5 \pi }{6} & 8+2 \sqrt{3} & 2 \\
7 & \pi & 12 & 0 \\
8 & \frac{7 \pi }{6} & 8+2 \sqrt{3} & -2 \\
9 & \frac{4 \pi }{3} & 10 & -2 \sqrt{3} \\
10 & \frac{3 \pi }{2} & 8 & -4 \\
11 & \frac{5 \pi }{3} & 6 & -2 \sqrt{3} \\
12 & \frac{11 \pi }{6} & 8-2 \sqrt{3} & -2 \\
1 & 0 & 4 & 0 \\
2 & \frac{\pi }{6} & 8-2 \sqrt{3} & 2 \\
3 & \frac{\pi }{3} & 6 & 2 \sqrt{3} \\
4 & \frac{\pi }{2} & 8 & 4 \\
5 & \frac{2 \pi }{3} & 10 & 2 \sqrt{3} \\
6 & \frac{5 \pi }{6} & 8+2 \sqrt{3} & 2 \\
7 & \pi & 12 & 0 \\
8 & \frac{7 \pi }{6} & 8+2 \sqrt{3} & -2 \\
9 & \frac{4 \pi }{3} & 10 & -2 \sqrt{3} \\
10 & \frac{3 \pi }{2} & 8 & -4 \\
11 & \frac{5 \pi }{3} & 6 & -2 \sqrt{3} \\
12 & \frac{11 \pi }{6} & 8-2 \sqrt{3} & -2 \\
\end{array}
$$
The $k$ values are plotted against the circle.
| {
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Find parametar λ if plane and line don't intersect How can I find λ, if I have line $3\lambda x-2=3y+1=\lambda z$ and plane $\lambda x-3y+2z-3=0$ and they don't intersect. The given solution says that λ=3, but I don't have any idea how can I come to that solution. Small hint would be helpful...
| Hint:
The line is the intersection of the two planes
$$
3 \lambda x - 2 = 3 y + 1 \iff 3 \lambda x - 3 y = 3 \\
3y + 1 = \lambda z \iff 3 y - \lambda z = -1
$$
Then we take the plane
$$
\lambda x - 3 y + 2z - 3 = 0
$$
into the consideration of common intersections.
This is equivalent to analyzing the solutions of the inhomogeneous linear system $A u = b$ with the augmented matrix
$$
[A | b] =
\left[
\begin{array}{rrr|r}
3 \lambda & -3 & 0 & 3 \\
0 & 3 & -\lambda & - 1 \\
\lambda & -3 & 2 & 3
\end{array}
\right]
$$
Solution:
This can be transformed into
$$
\left[
\begin{array}{rrr|r}
0 & 6 & -6 & -6 \\
0 & 3 & -\lambda & - 1 \\
\lambda & -3 & 2 & 3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
0 & 1 & -1 & -1 \\
0 & 3 & -\lambda & -1 \\
\lambda & -3 & 2 & 3
\end{array}
\right]
\to \\
\left[
\begin{array}{rrr|r}
0 & 1 & -1 & -1 \\
0 & 0 & 3 -\lambda & 2 \\
\lambda & -3 & 2 & 3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
\lambda & -3 & 2 & 3 \\
0 & 1 & -1 & -1 \\
0 & 0 & 3 -\lambda & 2
\end{array}
\right]
$$
We see that the choice $3 - \lambda = 0$ leads to an inconsistency in the last equation ($0 \cdot x + 0 \cdot y + 0 \cdot z = 2$), and thus no solution of the system.
| {
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Compute $\prod_{k=5}^{\infty}\frac{k^4-17k^2+16}{k^4-8k^2+16}$ Compute the product: $$\prod_{k=5}^{\infty}\frac{k^4-17k^2+16}{k^4-8k^2+16}$$
I was able to factor in the following manner:
$$ \frac{k^4-17k^2+16}{k^4-8k^2+16}=\frac{(k-1)(k+1)(k-4)(k+4)}{(k-2)^2(k+2)^2}$$
but what should I do now?
| As you noted we have $$\prod_{k\geq5}\frac{k^{4}-17k^{2}+16}{k^{4}-8k+16}=\prod_{k\geq0}\frac{\left(k+4\right)\left(k+6\right)\left(k+1\right)\left(k+9\right)}{\left(k+3\right)^{2}\left(k+7\right)^{2}}$$ and now using the identity $$\prod_{k\geq0}\frac{\left(k+a\right)\left(k+b\right)\left(k+c\right)\left(k+d\right)}{\left(k+e\right)\left(k+f\right)\left(k+g\right)\left(k+h\right)}=\frac{\Gamma\left(e\right)\Gamma\left(f\right)\Gamma\left(g\right)\Gamma\left(h\right)}{\Gamma\left(a\right)\Gamma\left(b\right)\Gamma\left(c\right)\Gamma\left(d\right)},\,a+b+c+d=e+f+g+h$$ which follows from the definition of gamma as an infinite product, we have $$\prod_{k\geq5}\frac{k^{4}-17k^{2}+16}{k^{4}-8k+16}=\frac{\Gamma^{2}\left(3\right)\Gamma^{2}\left(7\right)}{\Gamma\left(4\right)\Gamma\left(6\right)\Gamma\left(1\right)\Gamma\left(9\right)}=\color{red}{\frac{1}{14}}$$ as wanted.
| {
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} |
Using side length of triangle, find radius of touching circle Question:
The side lengths of a triangle are equal to lengths $8, 9$ and $10$. Find the exact value of the radius of the circle passing through the endpoints of the longest side and the midpoint of the shortest side.
I am very confused as to where to begin to solve the problem. If I solve the angles of the triangle, how will that help me to find the radius of the circles?
| Let the triangle be $\triangle ABC$ with $AB=8$, $AC=9$, $BC = 10$, and let the midpoint of $AB$ be $M$. Then, our goal is to find the radius of the excircle of $\triangle BCM$. By the cosine law, we know that $$\cos\angle B=\frac{AB^2+BC^2-AC^2}{2\cdot AB\cdot BC}=\frac{83}{160}$$
and similarly by the cosine law, $$\cos \angle B = \frac{BM^2+BC^2-MC^2}{2\cdot BM\cdot BC}=\frac{116-MC^2}{80}\implies MC = \sqrt{\frac{149}{2}}$$
Hence, by the sine law, the radius of the excircle of $\triangle BCM$ is $$R=\frac{MC}{2\sin \angle B} = \sqrt{\frac{149}{2}} \cdot \frac{80}{\sqrt{18711}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The maximum possible area of a yard in terms of $x$ A total of $x$ feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of $x$?
(A) $\frac{x^2}{9}$
(B) $\frac{x^2}{8}$
(C) $\frac{x^2}{4}$
(D) $x^2$
(E) $2x^2$
I do not know the meaning of level rectangular in mathematics, Also I do not know how to divide the problem into smaller ones so that I can solve it in $2.5$ minutes. Could anyone help me please?
| Suppose that the side of your rectangle whose opposite side is not bordered by the fence has a length of $s$. Then the two adjacent sides have length $\frac{x-s}{2}$. The area of the rectangle can then calculated to be $s\cdot\frac{x-s}{2}=\frac{sx-s^2}{2}$, so we must maximize $sx-s^2$. With some manipulation we obtain $$sx-s^2=-s^2+2s\cdot \frac{x}{2}-\frac{x^2}{4}+\frac{x^2}{4}=\frac{x^2}{4}-(s-\frac{x}{2})^2$$Now since $x$ is fixed, $\frac{x^2}{4}$ is fixed, too. Hence, we must only minimize $(s-\frac{x}{2})^2$ in order to maximize the area of the rectangle. Since a square is always nonnegative, $(s-\frac{x}{2})^2$ is minimal when it's equal to $0$ which implies $s=\frac{x}{2}$.
Therefore the maximum area of your rectangle is equal to
$$\frac{x}{2}\cdot\frac{x-\frac{x}{2}}{2}=\frac{x^2}{8}$$
which we get by chucking $s=\frac{x}{2}$ into the formula for the area of the rectangle we derived above. Note that this solution does not make use of calculus at all.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the minimum of the value $k$ such
Let $x\geq0$, $y\geq0$, $z\ge 0$ such that
$$\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}\le\sqrt{k(x^2+y^2+z^2)+(6-k)(xy+yz+xz)}$$
Find the minimum of the real value $k$
I square the side
$$2(x^2+y^2+z^2)+2\sum_{cyc}\sqrt{(x^2+y^2)(y^2+z^2)}\le k(x^2+y^2+z^2)+(6-k)(xy+yz+xz)$$
$$(k-2)(x^2+y^2+z^2)+(6-k)(xy+yz+xz)\ge 2\sum_{cyc}\sqrt{(x^2+y^2)(y^2+z^2)}$$
Following it seem hard to work
and other hand I think $k_{min}=4\sqrt{2}$,because I let $x=y=1,z=0$,then we have
$$2+\sqrt{2}\le\sqrt{2k+(6-k)}\Longrightarrow k\ge 4\sqrt{2}$$
| Yes! For $k=4\sqrt2$ your inequality is true.
Indeed, by C-S
$$\left(\sum_{cyc}\sqrt{x^2+y^2}\right)^2\leq\sum_{cyc}(x+y+(\sqrt2-1)z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}.$$
Thus, it remains to prove that:
$$\sum_{cyc}(x+y+(\sqrt2-1)z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}\leq\sum_{cyc}\left(4\sqrt2x^2+(6-4\sqrt2)xy\right)$$ or
$$(1+\sqrt2)(x+y+z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}\leq\sum_{cyc}\left(4\sqrt2x^2+(6-4\sqrt2)xy\right),$$
which is fifth degree homogeneous inequality.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality it's $f(w^3)\geq0$, where $f$ is a linear function,
which says that it's enough to prove the last inequality for an extremal value of $w^3$,
which happens in the following cases.
*
*$y=z$.
Since our inequality is homogeneous, we can assume $y=z=1$, which gives
$$(x-1)^2x(x+4\sqrt2)\geq0;$$
2. $w^3=0$.
Let $z=0$ and $y=1$.
We need to prove that
$$(x-1)^2(2(\sqrt2-1)^2x^2+5(3\sqrt2-4)x+2(\sqrt2-1)^2)\geq0,$$
which is obvious.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Sine-related infinite products How do you find the infinite product for functions which share the same roots?
$$\frac{\sin(\pi x)}{\pi x}=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\dotsb .$$
However, this function has the exact same roots, but clearly doesn't have the same infinite product:
$$\frac{(3+\cos x)\sin(\pi x)}{\pi x}.$$
Any ideas? Thanks in advance. :)
| They don't have the same roots: $\cos{z}+3=0$ has complex solutions: suppose $x,y$ are real. Then
$$ \cos{(x+iy)} = \cos{x}\cosh{y}+i\sin{x}\sinh{y}, $$
so if this is $3$, we need $\sin{x}=0$ and $\cosh{y} = 3/\cos{x}$. $\sin{x}=0 \implies x = n\pi$ for some integer $n$, so $\cos{x}=(-1)^n$. $\cosh{y}$ is positive, so $n$ is even, and $x=2k\pi$ for some integer $k$. $\arg\cosh{3} = \log{(3+\sqrt{3^2-1})} = \log{(3+2\sqrt{2})} $, so $\cos{z}+3$ is zero when
$$ z = 2k\pi + i\log{(3+2\sqrt{2})}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2234058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the value of $(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a+1)$ if $\log_{b}a+\log_{c}b+\log_{a}c=13$ and $\log_{a}b+\log_{b}c+\log_{c}a=8$ Let $a,b$, and $c$ be positive real numbers such that
$$\log_{a}b + \log_{b}c + \log_{c}a = 8$$
and
$$\log_{b}a + \log_{c}b + \log_{a}c = 13.$$
What is the value of
$$(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a + 1) ?$$
I tried to convert the entire thing to fractional logs and multiply the expression and add the two equations but it did not help.
| Since I prefer working with exponents rather than logarithms, here is how you could tackle it that way, using your quite possibly better intuition at handling powers:
$$x + y + z = 8$$
$$u + v + w = 13$$
With:
$$ a^x = b\,,\, b^y = c\,,\, c^z = a $$
and
$$ b^u = a\,,\, c^v = b\,,\, a^w = c $$
From this you can see that:
$$ a^{x\,u} = b^u = a \; \Rightarrow \; x u = 1$$
So the second equation is actually:
$$ 1/x + 1/y + 1/z = 13 $$
If you add the fractions:
$$ yz + xz + xy = 13 \, x\,y\,z $$
Since:
$$ (1+x)(1+y)(1+z) = xyz + xz + yz + xy + x + y + z + 1 $$
We need to figure out what is $xyz$. But:
$$ a^{x\,y\,z} = b ^{y\,z} = c^z = a \; \Rightarrow \; x y z = 1 $$
Now we have all the information we need and the result is $23$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
How to differentiate $\left(\frac{d}{dt}\left(x\right)\right)^2$? How does
$\frac{d}{dt}\left(\frac{1}{2}m\left(\frac{d}{dt}\left(x\right)\right)^2+\frac{1}{2}kx^2\right)$ = 0
equals to
m$\frac{d^2}{dt^2}\left(x\right)$ + kx = 0
| \begin{align}\frac{d}{dt}\left(\frac{1}{2}m\left(\frac{d}{dt}\left(x\right)\right)^2+\frac{1}{2}kx^2\right) &= 0 \\ \Longrightarrow \frac{1}{2} m \frac{d}{dt}\big(\frac{d}{dt}x\big)^2+\frac{1}{2}k \frac{d}{dt}x^2 &= 0 \\ \Longrightarrow m \frac{d}{dt} x\frac{d}{dt}\frac{d}{dt}x+kx \frac{d}{dt}x &= 0 ~\text{ (chain rule)}\\ \Longrightarrow m \frac{d}{dt}\frac{d}{dt}x+k x &= 0 ~\text{ (assuming that $\dot x \neq 0$)} \\ \Longrightarrow m\frac{d^2}{dt^2}x + kx &= 0\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Do the following series converge or diverge? Justify. $\sum_{k=0}^{\infty} \frac{3k^2 + 1}{k^3 + k^2 + 5}$ $$\sum_{k=0}^{\infty} \frac{3k^2 + 1}{k^3 + k^2 + 5}$$
Can I do this using direct comparison test?
for $k \in [1, \infty), a_k = \frac{3k^2 + 1}{k^3 + k^2 + 5} \geq 0$
for $k \in [1,\infty), a_k = \frac{3k^2 + 1}{k^3 + k^2 + 5} \geq \frac{3k^2}{k^3 + k^3 + 5k^3} = \frac{3}{7k} = b_k$
Consider $\frac{3}{7}\sum_{n=1}^{\infty} \frac{1}{k}$. This is a p-series with $p = 1$. By the p-series test $\sum b_k$ diverges, therefore by the comparison test $\sum a_k$ diverges too.
My textbook does this using limit comparison test wondering if I can do it using direct comparison test too. Is this right?
| It's much quicker with equivalents:
$$\frac{3k^2+1}{k^3+k^2+5}\sim_\infty\frac{3k^2}{k^3}=\frac3k,\quad\text{which diverges}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the last three digits of $383^{101}$ We have to find out $383^{101} \equiv ? \pmod {1000}$.
I know that $383^2 ≡ 689 \pmod {1000}$
$383^5≡143 \pmod {1000}$
I know that $ϕ(1000)=400 >101 $ from Euler.
It definitely can't help me.
I don't know how to continue.
I can't use the Chinese Remainder Theorem.
| The units digit of $383^1$ is $3$.
The units digit of $383^2$ is $9$ ($3 \times 3 = 9$).
The units digit of $383^3$ is $7$ ($3 \times 9 = 27$).
The units digit of $383^4$ is $1$ ($3 \times 7 = 21$).
Calculating,
$\; 383^4 \equiv 721 \pmod{1000}$
Calculating,
$\; 383^{101} \equiv 383 \cdot 721^{25} \equiv 383 \cdot (1 + 720)^{25} \equiv 383 \cdot (1 + 2^4 \cdot 3^2 \cdot 5^1)^{25} \equiv$
$\quad\quad\quad\quad \; 383 \; \bigr(1 + \binom{25}{1}\,2^4 \cdot 3^2 \cdot 5^1 + \binom{25}{2}\,2^8 \cdot 3^4 \cdot 5^2 + 0\bigr) \equiv$
$\quad\quad\quad\quad \; 383 \; \bigr(1 + 0 + 0 + 0 \bigr) \equiv 383 \pmod{1000}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Applying Stokes' theorem - what surface?
$\def\d{\mathrm{d}}$Determine the integral $$\oint_L \mathbf{A} \cdot \,\d\mathbf{r},$$
where $$\mathbf{A} = \mathbf{e}_x(x^2-a(y+z))+\mathbf{e}_y(y^2-az)+\mathbf{e}_z(z^2-a(x+y)),$$
and $L$ is the curve given by the intersection between the cylinder $$\begin{cases}(x-a)^2+y^2=a^2 \\z\geq0\end{cases}$$
and the sphere $$x^2+y^2+z^2=R^2, \quad (R^2>4a^2)$$
The orientation is such that at $x=0$ the tangent to the curve is parallel with $-\mathbf{e}_y$.
Attempted solution:
Let's apply Stokes' theorem. First, let me introduce a graphical representation of the problem.
The path $L$ will then, as seen from above, be the following:
A simple calculation shows $\nabla \times \textbf{A} = (0,0,a)$. Here comes my problem...
Question
What surface am I looking to take a surface integral over? Is it the whole cylinder or just the "top"? How can I determine?
| The surface can be ANY surface whose border is the curve $L$.
In this case, a possibility would be the part of the sphere inside the cylinder, which you can parametrize as follows:
\begin{cases}
x= x\\
y=y \quad \quad \quad \quad \quad \quad \text{with} \quad (x,y)\mid (x-a)^2+y^2\le a^2\\
z= \sqrt{R^2-x^2-y^2}\\
\end{cases}
Applying Stokes' theorem to finish the question yields
\begin{align}
\oint_L \textbf{A}\cdot d\textbf{r} = \iint_S \nabla\times \textbf{A}\cdot d\textbf{S} &= \iint_\limits{(x-a)^2+y^2\le a^2}\pmatrix{0\\0\\a}\cdot \pmatrix{1\\0\\\frac{-x}{\sqrt{R^2-x^2-y^2}}} \times \pmatrix{0\\1\\\frac{-y}{\sqrt{R^2-x^2-y^2}}}\; dA
\\ &= \iint_\limits{(x-a)^2+y^2\le a^2}\pmatrix{0\\0\\a}\cdot \pmatrix{*\\*\\1}\; dA
\\ &= \iint_\limits{(x-a)^2+y^2\le a^2}a\; dA = \pi a^3
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Am I missing something quite obvious from understanding this proof?
Theorem. There exists a nonempty set of rational numbers which is bounded above in $\mathbb{Q}$ but has no least upper bound in $\mathbb{Q}$.
Here is the relevant part of the proof that I am going to ask about.
Let $k=\frac{a}{b}\in\mathbb{Q}$ be an upper bound for the set $S=\{q\in\mathbb{Q}:q^2<2\}$. Suppose that $k^2<2$. Define $\delta=2-k^2>0$. Fix $N\in\mathbb{N}$ such that
$$
N\geq\max\{2a+1,3a/(b^2\delta)\}
$$
Since $N\geq 2a+1$, we have $N^2\geq N(2a+1)\geq 2Na+1$ and since $N>3a/(b^2\delta)$, then $N^2b^2\delta>3Na\geq 2Na+1$ which forces
$$
\frac{2Na+1}{N^2b^2}<\delta
$$
Define $l=\frac{Na+1}{Nb}$. Then
$$
l^2=k^2+\frac{2Na+1}{N^2b^2}<k^2+\delta=2
$$
What is my question?
No matter how hard I think about this part of the proof I see nowhere that $N\geq 2a+1$ is used to show that $l^2<2$. I am sure that I am missing something quite obvious but I can not figure it out. Perhaps the thinking of the author of the proof was different to mine. Why was it necessary to fix $N\in\mathbb{N}$ such that
$$
N\geq\max\{2a+1,3a/(b^2\delta)\}
$$
and not just $N> 3a/(b^2\delta)$? In other words, I see $N\geq 2a+1$ unnecessary.
$k=\frac{a}{b}\in\mathbb{Q}$ is an upper bound in $\mathbb{Q}$ for $S$. We know that $\sqrt{2}\approx1.41421356237$ is the supremum of $S$ in real numbers so $k\geq 1.5$ or $k\geq 1.42$ or $k\geq 1.415$ and so on which forces $a>1$.
$l^2=k^2+\frac{2Na+1}{N^2b^2}$ and I can approximate $\frac{2Na+1}{N^2b^2}$ as follows
$$
\begin{align*} \frac{2Na+1}{N^2b^2}&\leq\frac{2Na+N}{N^2b^2}&&\text{since }N\geq 1\\ &=\frac{2a+1}{Nb^2}&&\\ &<\frac{2a+a}{Nb^2}&&\text{since }a>1\\ &=\frac{3a}{Nb^2}&& \end{align*}
$$
So I need $\frac{3a}{Nb^2}<\delta$ which implies $N>\frac{3a}{b^2\delta}$.
|
The author presents a proof purely based upon calculations within $\mathbb{Q}$. It does not rely on an embedding of $\mathbb{Q}$ in $\mathbb{R}$ and the knowledge that $\mathbb{R}$ is complete.
Insofar is the approach of the author somewhat different than yours.
Let's revisit the author's proof. We want to show there does not exist a rational number $k$ which is a least upper bound of $S$, the set of rational numbers having square less than $2$.
In order to do so we show that whenever we want to take a least upper bound $k=\frac{a}{b}\in\mathbb{Q}$ of $S$ with $k^2<2$, we find a rational number $l$ in $S$ with $l>k$ contradicting the assumption $k$ is a least upper bound.
We consider the distance $\delta=2-k^2=2-\frac{a^2}{b^2}$ from the square of $k$ to $2$ and construct a rational number $l\in S$ having a smaller distance.
We take a natural number $N$ with
\begin{align*}
N\geq\max\{2a+1,3a/(b^2\delta)\}
\end{align*}
Since $N\geq 2a+1$ and since $N>\frac{3a}{b^2\delta}=\frac{3a}{b^2\left(2-\frac{a^2}{b^2}\right)}=\frac{3a}{2b^2-a^2}$ we obtain
\begin{align*}
N^2b^2\delta&=N^2b^2\left(2-\frac{a^2}{b^2}\right)=N^2(2b^2-a^2)\\
&>3Na\\
&\geq 2Na+1
\end{align*}
which implies
\begin{align*}
\frac{2Na+1}{N^2b^2}<\delta
\end{align*}
If we now define $l:=\frac{Na+1}{Nb}>\frac{Na}{Nb}=k$ we obtain
\begin{align*}
l^2&=\frac{(Na+1)^2}{N^2b^2}=\frac{N^2a^2+2Na+1}{N^2b^2}\\
&=\frac{a^2}{b^2}+\frac{2Na+1}{N^2b^2}\\
&<\frac{a^2}{b^2}+\delta\\
&=2
\end{align*}
and the claim follows.
Note: In accordance with OP's comment below we don't need the following part of the author's proof:
\begin{align*}
N^2\geq N(2a+1)=2Na+N>2Na+1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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