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How to prove that the sum of reciprocals of one plus perfect powers is $\frac{\pi^{2}}{3}-\frac{5}{2}$ Let $S$ be Set of perfect powers without duplicates $1,4,8,9,\dots$ (http://oeis.org/A001597 ) How to prove the following?
$$\sum_{s\in S}\frac{1}{s+1}=\frac{\pi^{2}}{3}-\frac{5}{2}$$ (starting with $s=4$ ) I found this formula in the book "Mathematical Constants" by Steven R. Finch on page 113.
| Let $S$ be the set of perfect powers without $1$ . That is, $S = \{4,8,9,16,25,27,\ldots\}$. Consider the sum $$
S_{N} = \sum_{s \in S} \frac 1{s^{N}-1}
$$
which converges for $N\geq 2$ because $\sum_{n=1}^\infty \frac 1{n^N}$ converges, and the above sequence is thus a subsequence of a convergent sequence. For $N=1$, a separate argument can be created by "working backwards" from what we do below so I won't really emphasize that point.
We can evaluate this sum using some clever ideas. The first is to consider the set of non-powers $T$ (insist on $1 \notin T$) and its relation to $S$. Of course it is the complement of $S$, but there is a deeper relation.
Indeed, let $s \in S$. We can find $k \geq 2$ such that $s$ is a perfect $k$-th power. Let $K$ be the largest number such that $s$ is a $K$th perfect power. Then, $s^{\frac 1K}$ is a positive integer that has to be a non-power by maximality of $k$. Thus, every $s \in S$ is uniquely of the form $t^K$ where $K \geq 2$ and $t \in T$. On the other hand, if $t \in T$ and $K \geq 2$, obviously $t^K \in S$. Therefore, we may write $$
\sum_{s \in S} \frac 1{s^N-1} = \sum_{K \geq 2} \sum_{t \in T} \frac{1}{t^{NK}-1}
$$
Now we use a very nice trick : the identity $\frac{1}{n-1} = \frac 1{n} + \frac 1{n(n-1)}$ gives that $$
\sum_{K \geq 2} \sum_{t \in T} \frac{1}{t^{NK}-1} = \sum_{K \geq 2} \sum_{t \in T} \frac 1{t^{NK}} + \sum_{K \geq 2} \sum_{t \in T} \frac 1{t^{NK}(t^{NK} - 1)} \tag{*}
$$
However, observe that $$
\sum_{K \geq 2} \sum_{t \in T} \frac 1{t^{NK}} = \sum_{t \in T} \sum_{ K \geq 2} \frac 1{t^{NK}} = \sum_{t \in T} \frac 1{t^N(t^N-1)}
$$
Therefore, combining this with $(*)$ gives $$
\sum_{K \geq 2} \sum_{t \in T} \frac{1}{t^{NK}-1} = \sum_{t \in T} \frac 1{t^N(t^N-1)} + \sum_{K \geq 2} \sum_{t \in T} \frac 1{t^{NK}(t^{NK} - 1)}
$$
However, take a very careful look at the RHS here. We are actually summing the quantity $\frac 1{v^N(v^N-1)}$, first for $v \in T$, and then for numbers of the form $v^K$ for $v \in T, k \geq 2$ : which we know to be equal to $S$!
That is, we in fact, have $$
\sum_{t \in T} \frac 1{t^N(t^N-1)} + \sum_{K \geq 2} \sum_{t \in T} \frac 1{t^{NK}(t^{NK} - 1)} = \sum_{t \in T} \frac 1{t^N(t^N-1)} + \sum_{s \in S}\frac 1{s^N(s^N-1)} = \sum_{k=2}^{\infty} \frac 1{k^N(k^N-1)}
$$
We have obtained the identity $$
S_N = \sum_{k=2}^{\infty} \frac{1}{k^N(k^N-1)}
$$
Let's put $N=1$ first. Then, we get by telescoping, $$
S_1 = \sum_{k=2}^{\infty} \frac{1}{k(k-1)} = \sum_{k=2}^{\infty} \left(\frac 1{k-1} - \frac 1{k} \right)\\ = 1 - \frac 12 + \frac 12 - \frac 13+ \ldots = 1
$$
This is a proof of the first identity in Finch's book. The proof of the second identity follows by the evaluation of $S_2$. We write by the telescoping identity $$
S_2 = \sum_{k=2}^{\infty} \frac{1}{k^2(k^2-1)} = \sum_{k=2}^{\infty} \left(\frac 1{k^2-1} - \frac 1{k^2}\right) = \sum_{k=2}^{\infty} \frac 1{k^2-1} - \sum_{k=2}^{\infty} \frac 1{k^2}
$$
We know that $\sum_{k=2}^{\infty} \frac 1{k^2} = \frac{\pi^2}{6}-1$. What about $\sum_{k=2}^{\infty} \frac 1{k^2-1}$? For that, perform partial fractions and notice yet another telescoping occuring.
$$
\sum_{k=2}^{\infty} \frac 1{k^2-1} = \frac 12\sum_{k=2}^{\infty} \frac 2{k^2-1} = \frac 12\sum_{k=2}^{\infty} \left(\frac{1}{k-1} - \frac 1{k+1}\right) \\ = \frac 12 \left(1 - \frac 13 + \frac 12 - \frac 14 + \frac 13 - \frac 15 + \frac 14 - \frac 16 + \ldots\right) \\ = \frac 12\left(1+\frac 12\right) = \frac 34
$$
That is, we obtain $$
S_2 = \frac{3}{4} + 1 - \frac{\pi^2}{6} = \frac{7}{4} - \frac{\pi^2}{6}
$$
We are finally in a position to finish: (and I need to , because merely typing the word telescoping has made my voice hoarse) $$
\sum_{s \in S} \frac 1{s+1} = \sum_{s \in S} \left(\frac{1}{s-1} - \frac{2}{s^2-1}\right) = S_1 - 2S_2 = \frac{\pi^2}{3} - \frac 72 + 1 = \frac{\pi^2}{3} - \frac 52
$$
as desired.
Note that the evaluation of higher $S_N$ is possible, because $$
S_N = \sum_{k=2}^{\infty} \frac 1{k^N-1} - \zeta(N) + 1
$$
One uses partial fraction decomposition, and the definition of the Digamma function like has been done here, to obtain $$
S_N = 1 - \zeta(N) - \frac 1N \sum_{\omega^N = 1}\omega \psi(2-\omega)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Proving using Mathematical Induction I'm Stuck on the last step, this is proving using mathematical induction, a lecture from my Elementary number theory class.
The question goes to,
Prove that $\sum_{k=1}^n \frac{1}{k^2}=\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n}$ whenever $n$ is a positive integer.
This is my Attempt,
Step 1: Base Case ($n=1$)
$$\sum_{k=1}^n \frac{1}{k^2}=\frac{1}{1^2}\le2-\frac{1}{1}$$
$$=1\le1$$, Therefore Base case is true.
Step 2: Induction Hypothesis
Suppose $\sum_{k=1}^n \frac{1}{k^2}=\frac{1}{n^2}\le2-\frac{1}{n}$ is true for $n=m$
$\implies$ $\sum_{k=1}^m \frac{1}{k^2}=\frac{1}{m^2}\le2-\frac{1}{m}$ $\forall m \in \mathbb{N}$
Step 3: $n = m+1$
$\sum_{k=1}^{m+1} \frac{1}{k^2}=\sum_{k=1}^m \frac{1}{k^2}+\frac{1}{(m+1)^2}\le2-\frac{1}{m}+\frac{1}{(m+1)^2}$
I'm Stuck on this step
| Starting from the induction hypothesis which is $\sum_{k=1}^m \frac{1}{k^2}\le2-\frac{1}{m}$:-
Thus:-
$$\sum_{k=1}^m \frac{1}{k^2}\le2-\frac{1}{m}$$
$$\sum_{k=1}^{m+1} \frac{1}{k^2}\le2-\frac{1}{m}+\frac{1}{(m+1)^{2}}$$
As $m\ge1$ so:-
$$\Rightarrow m^2+2m\le m^2+2m+1$$
$$\Rightarrow \frac{m(m+2)}{(m+1)^2}\le 1$$
$$\Rightarrow \frac{(m+1)+1}{(m+1)^2}\le \frac{1}{m}$$
$$\Rightarrow \frac{1}{(m+1)^2} + \frac{1}{m+1}\le \frac{1}{m}$$
$$\Rightarrow \frac{1}{(m+1)^2} -\frac{1}{m} \le -\frac{1}{m+1}$$
$$\Rightarrow 2-\frac{1}{m}+\frac{1}{(m+1)^2} \le 2 -\frac{1}{m+1}$$
Now making use of the 2nd equation:-
$$\sum_{k=1}^{m+1} \frac{1}{k^2}\le2-\frac{1}{m}+\frac{1}{(m+1)^{2}}\le2 -\frac{1}{m+1}$$
Thus completing the induction step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $\frac{\sqrt[7]{x-\sqrt2}}{2}-\frac{\sqrt[7]{x-\sqrt2}}{x^2}=\frac{x}{2}\sqrt[7]{\frac{x^3}{x+\sqrt2}}$ Solve the equation $$\dfrac{\sqrt[7]{x-\sqrt2}}{2}-\dfrac{\sqrt[7]{x-\sqrt2}}{x^2}=\dfrac{x}{2}\sqrt[7]{\dfrac{x^3}{x+\sqrt2}}$$ We have $x\ne0;-\sqrt2$.
Let's multiply both sides of the equation by $2x^2\ne0$ to get $$x^2\sqrt[7]{x-\sqrt2}-2\sqrt[7]{x-\sqrt2}=x^3\sqrt[7]{\dfrac{x^3}{x+\sqrt2}}$$$$(x^2-2)\sqrt[7]{x-\sqrt2}=x^3\sqrt[7]{\dfrac{x^3}{x+\sqrt2}}$$ Let's multiply both sides of the equation by$\sqrt[7]{x+\sqrt2}\ne0$ to get $$(x^2-2)\sqrt[7]{x^2-2}=x^3\sqrt[7]{x^3}$$ $$(x^2-2)^8=x^{24}$$
| Hint
Write the equation as
$$\frac{\sqrt[7]{x-\sqrt2}}{2}-\frac{\sqrt[7]{x-\sqrt2}}{x^2}=\frac{x}{2}\sqrt[7]{\frac{x^3(x-\sqrt2)}{x^2-2}}$$
In the transformation above, take care of the situation $x=\sqrt 2$.
$$\frac{\sqrt[7]{x-\sqrt2}}{2}-\frac{\sqrt[7]{x-\sqrt2}}{x^2}=\frac{x}{2}\sqrt[7]{\frac{x^3(x-\sqrt2)}{x^2-2}}$$
$$\left(\sqrt[7]{x-\sqrt2}\right)\left(\frac{1}{2}-\frac{1}{x^2}-\frac{x}{2}\sqrt[7]{\frac{x^3}{x^2-2}}\right)=0$$
$$\frac{1}{2x^2}\left(\sqrt[7]{x-\sqrt2}\right)\left(x^2-2-x^3\sqrt[7]{\frac{x^3}{x^2-2}}\right)=0$$
$$\frac{1}{2x^2}\left(\sqrt[7]{x-\sqrt2}\right)\left(x^2-2-x^3\sqrt[7]{\frac{x^3}{x^2-2}}\right)=0$$
$$\frac{x^2-2}{2x^2}\left(\sqrt[7]{x-\sqrt2}\right)\left(1-\frac{x^3}{x^2-2}\sqrt[7]{\frac{x^3}{x^2-2}}\right)=0$$
$$\left(\frac{x^2-2}{2x^2}\right)\left(\sqrt[7]{x-\sqrt2}\right)\left(1-\left(\frac{x^3}{x^2-2}\right)^{8/7}\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
From a question I asked before this, I have trouble actually with the numbers manipulating part.
Using trigo identity, $\sin^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{12} = 1$ so , $\cos^2 \frac{\pi}{12} = 1- \sin^2 \frac{\pi}{12}$
To find $\cos \frac{\pi}{12} = \sqrt{1- \sin^2 \frac{\pi}{12}}$
$\sin^2 \frac{\pi}{12} = (\frac{\sqrt{3} -1}{2 \sqrt{2}})^2 = \frac{(\sqrt{3}-1)^2}{(2\sqrt{2})^2} = \frac{2- \sqrt{3}}{4}$
$\cos \frac{\pi}{12} = \sqrt{1-(\frac{\sqrt{3} -1}{2 \sqrt{2}})^2} $
$\cos \frac{\pi}{12} = \sqrt{1- \frac{2-\sqrt{3}}{4}}$
$\cos \frac{\pi}{12} = \frac{\sqrt{2+\sqrt{3}}}{2}$
What is wrong with my steps?
| Let $$\sqrt{2+\sqrt{3}}=\sqrt{x}+\sqrt{y}\implies x+y=2. xy=3/4 \implies x=3/2, y=1/2.$$
So $$\cos(\pi/12)=\frac{\sqrt{3}+1}{2\sqrt{2}}$$
OP is right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
A formal power series such that $f(f(x))=x$ Let
$$f(x)=\sum_{n=1}^\infty a_n x^n$$
be a formal power series with no constant term such that $f(f(x))=x$. We find that
$$f(f(x))=a_1^2x+(a_1a_2+a_1^2a_2)x^2+(a_1a_3+2a_1a_2^2+a_3a_3)x^3+\dots$$
so $a_1^2=1$. If $a_1=1$, you need all the other terms to be zero, however if $a_1=-1$ we get a family of nontrivial solutions. Let $a_2=a$, and requiring the higher coefficients of $f(f(x))$ to be zero we can find $a_3=2a^2$, $a_4=-\frac{11}2a^3$, $a_5=\frac{11}2a^4$, $a_6=\frac{105}4a^5$, $a_7=-\frac{279}2 a^6$...
Is there a closed form for these numbers?
| The claim that the coefficients are unique given the quadratic term is incorrect. Before going into that, here's some background on Anne's answer. We consider the simpler question of how to determine solutions to $f(f(x)) = x$ where $f$ is a Mobius transformation $f(x) = \frac{ax + b}{cx + d}$. Over a field $K$ the group of Mobius transformations is isomorphic to the projective general linear group $PGL_2(K)$, with the isomorphism given by sending a $2 \times 2$ matrix $\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$ to $\frac{ax + b}{cx + d}$. So the problem reduces to finding matrices $M$ squaring to a scalar multiple of the identity.
So write $M^2 = c$ for some scalar $c$, so that the eigenvalues of $M$ are $\pm \sqrt{c}$ (over an algebraically closed field, say $\mathbb{C}$). Since we only need to work up to scale we might as well divide $M$ by $c$, or equivalently assume WLOG that $c = 1$, so $M^2 = 1$ and $M$ has eigenvalues $\pm 1$. If $M$ is conjugate to a nontrivial Jordan block then it squares to another nontrivial Jordan block so can't square to $1$; hence $M$ is diagonalizable. If $M$ has eigenvalues $\{ 1, 1 \}$ or $\{ -1, -1 \}$ then it's a scalar multiple of the identity; otherwise its eigenvalues are $\{ 1, -1 \}$. This implies that $\text{tr}(M) = 0$ and $\det(M) = -1$, so $M$ has the form $\left[ \begin{array}{cc} a & b \\ c & -a \end{array} \right]$ where $\det(M) = -a^2 - bc = -1$. This gives $bc = 1 - a^2$.
Now we add the constraint that as a formal power series $f(x)$ should have no constant term. This means $f(0) = 0$ which gives $b = 0$. Then $a = \pm 1$ and we can take $a = 1$ WLOG, which gives $M = \left[ \begin{array}{cc} 1 & 0 \\ c & -1 \end{array} \right]$, hence
$$f(x) = \frac{x}{cx - 1} = -x - cx^2 - c^2 x^3 - \dots $$
as in Anne's answer (with $a = -c$).
On the other hand Klaus's answer implies that the resulting power series cannot be unique, since we can conjugate by an invertible (with respect to composition) formal power series. With a little more effort (showing that we can even arrange for the first $n$ coefficients of this series past $x$ to vanish) we can show that the first $n$ coefficients never uniquely determine the others. So you did something funny with your calculations but I'm not sure what.
Generally, it's known that every formal power series of finite order (with respect to composition) is conjugate to $f(x) = \zeta x$ for $\zeta$ some root of unity (above we have $\zeta = -1$, and in general we probably need to work over an algebraically closed field). This implies:
Classification: If $f(x)$ is a formal power series satisfying $f(0) = 0$ and $f(f(x)) = x$, then either $f(x) = x$ or $\boxed{ f(x) = g(-g^{-1}(x)) }$ where $g(x) = x + \dots$.
Given $g(x)$, the coefficients of $g^{-1}(x)$ can be computed using Lagrange inversion. This gives solutions depending on an infinite number of parameters, namely the higher coefficients $g_i$ of $g(x)$, and the first $n$ coefficients of $f$ depend only on the first $n$ coefficients of $g$. For example, if $g(x) = x + g_2 x^2 + \dots$ then $g^{-1}(x) = x - g_2 x^2 + \dots$ which gives $f(x) = -x + 2 g_2 x^2 + \dots$.
To give a relatively explicit example, take $g(x) = x - ax^2$. Then $g^{-1}(x) = \frac{1 - \sqrt{1 - 4ax}}{2a}$ by the quadratic formula (we have to take the minus sign so that $g^{-1}(0) = 0$); this is a version of the generating function of the Catalan numbers. If $y - ay^2 = x$ then $g(-y) = -y - ay^2 = x - 2y$, which gives
$$\boxed{ \begin{align*} f(x) &= x - \frac{1 - \sqrt{1 - 4ax}}{a} \\
&= -x - 2ax^2 - 4a^2 x^3 - 10a^3 x^4 - \dots \\
&= -x - \sum_{n=2}^{\infty} \frac{2}{n} {2n-2 \choose n-1} a^{n-1} x^n. \end{align*} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Possible "clever" ways to solve $x^4+x^3-2x+1=0$, with methodological justification
Solve the quartic polynomial :
$$x^4+x^3-2x+1=0$$
where $x\in\Bbb C$.
Algebraic, trigonometric and all possible methods are allowed.
I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, this equation doesn't require general formula. We need some substitutions here.
I realized there is no any rational root, by the rational root theorem.
The harder part is, WolframAlpha says the factorisation over $\Bbb Q$ is impossible.
Another solution method can be considered as the quasi-symmetric equations approach. (divide by $x^2$).
$$x^2+\frac 1{x^2}+x-\frac 2x=0$$
But the substitution $z=x+\frac 1x$ doesn't make any sense.
I want to ask the question here to find possible smarter ways to solve the quartic.
| We can look for a difference of squares factorization. Completing the square gives
$$\left( x^2 + \frac{1}{2} x + c \right)^2 - \left( 2c + \frac{1}{4} \right) x^2 - (c + 2) x - (c^2 - 1)$$
and we want to find a value of $c$ such that the discriminant of the quadratic on the right is equal to zero. This gives
$$\Delta = (c + 2)^2 - 4 \left( 2c + \frac{1}{4} \right) \left( c^2 - 1) \right) = - 8c^3 + 12c + 5$$
which happily has a rational root $c = - \frac{1}{2}$ (I guess we must be essentially using the resolvent cubic here). This gives us a factorization
$$\left( x^2 + \frac{1}{2} x - \frac{1}{2} \right)^2 + \frac{3}{4} (x - 1)^2$$
which gives a difference of squares factorization
$$\left( x^2 + \frac{1}{2} x - \frac{1}{2} + \frac{i \sqrt{3}}{2} (x - 1) \right) \left( x^2 + \frac{1}{2} x - \frac{1}{2} - \frac{i \sqrt{3}}{2} (x - 1) \right)$$
and we can use the quadratic formula from here; if you want to know what the roots end up looking like you can ask WolframAlpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 3
} |
Partial Fraction of $\frac{1-x^{11}}{(1-x)^4} $ for Generating Function The original question involves using generating functions to solve for the number of integer solutions to the equation $c_1+c_2+c_3+c_4 = 20$ when $-3 \leq c_1, -3 \leq c_2, -5 \leq c_3 \leq 5, 0 \leq c_4$.
Using generating functions I was able to get it into the rational polynomial form:
$$f(x) = {\left(\frac{1}{1-x}\right)}^3\left(\frac{1-x^{11}}{1-x}\right) = \frac{1-x^{11}}{{(1-x)}^4}$$
I was also able to determine that the sequence could be represented in two factors:
$${\left(1+x^1+x^2+x^3+\cdots\right)}^3\left(1+x^1+x^2+\cdots+x^{10}\right)$$
However, to find the coefficient on $x^{31}$ to solve the problem, I figured I would have to get the term $\frac{1-x^{11}}{{(1-x)}^4}$ into a more typical generating function summation form. Thus, I endeavored to find the partial fraction decomposition of the term, however, I can't seem to do it at all.
How would I find the partial fraction decomposition of $\frac{1-x^{11}}{{(1-x)}^4}$?
Or is there a better method in using ${(1+x^1+x^2+x^3+\ldots)}^3(1+x^1+x^2+\ldots+x^{10})$ in order to find the coefficient on $x^{31}$?
Thank you very much for your help, I've been trying this partial fraction for a while now and Wolfram alpha doesn't seem to be giving me an answer that is of much value.
| I would use the
generalized binomial theorem
in the form
$
\dfrac1{(1-x)^s}
=\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^k
$.
Then
$\begin{array}\\
\dfrac{1-x^a}{(1-x)^s}
&=\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^k
-x^a\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^k\\
&=\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^k
-\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^{k+a}\\
&=\sum_{k=0}^{\infty}
\binom{s+k-1}{s-1}x^k
-\sum_{k=a}^{\infty}
\binom{s+k-a-1}{s-1}x^{k}\\
&=\sum_{k=0}^{a-1}
\binom{s+k-1}{s-1}x^k
+\sum_{k=a}^{\infty}
\binom{s+k-1}{s-1}x^k
-\sum_{k=a}^{\infty}
\binom{s+k-a-1}{s-1}x^{k}\\
&=\sum_{k=0}^{a-1}
\binom{s+k-1}{s-1}x^k
+\sum_{k=a}^{\infty}
\left(\binom{s+k-1}{s-1}-\binom{s+k-a-1}{s-1}\right)x^k
\\
&=\sum_{k=0}^{a-1}
\binom{s+k-1}{s-1}x^k
+\sum_{k=a}^{\infty}
\left(\dfrac{(s+k-1)!}{(s-1)!k!}-\dfrac{(s+k-a-1)!}{(s-1)!(k-a)!}\right)x^k
\\
&=\sum_{k=0}^{a-1}
\binom{s+k-1}{s-1}x^k
+\sum_{k=a}^{\infty}
\left(\dfrac{(s+k-1)!}{(s-1)!k!}-\dfrac{(s+k-a-1)!}{(s-1)!(k-a)!}\right)x^k
\\
\end{array}
$
You can do more manipulation if you want.
| {
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"url": "https://math.stackexchange.com/questions/4546911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\sum \limits_{j=0}^{n+1- (a+b)} \binom{a+j-1}{a -1} \binom{n- a-j}{b -1} =\binom{n}{a+b-1} $ I have run into quiete a tricky sum, I am reasonably sure I know what is sums to but I have been unable to prove it
$$
\forall a,b,n \in \mathbb{N}, \ n\geq a+b \geq 2:\quad
\sum \limits_{j=0}^{n+1- (a+b)}
\binom{a+j-1}{a -1}
\binom{n- a-j}{b -1}
=\binom{n}{a+b-1}
$$
I have tried to evaluate it for quiete a number of particular values and it seems to hold. I've tried to prove it by induction however that didn't seem to lead anywhere but I could be wrong.
| We seek to verify that with $n\ge a+b\ge 2$
$$\sum_{j=0}^{n+1-a-b} {a+j-1\choose a-1} {n-a-j\choose b-1}
= {n\choose a+b-1}.$$
The LHS is
$$\sum_{j=0}^{n+1-a-b} {a+j-1\choose a-1}
{n-a-j\choose n+1-a-b-j}
\\ = [z^{n+1-a-b}] (1+z)^{n-a}
\sum_{j\ge 0} {a+j-1\choose a-1} \frac{z^j}{(1+z)^j} $$
Here we have extended to infinity because the coefficient extractor
enforces the upper limit. Continuing,
$$[z^{n+1-a-b}] (1+z)^{n-a}
\frac{1}{(1-z/(1+z))^a}
\\ = [z^{n+1-a-b}] (1+z)^{n}
= {n\choose n+1-a-b} = {n\choose a+b-1}.$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The convergence speed of $ \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x $? I have already known how to prove
\begin{equation*}
\lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \sqrt{\frac{\pi}{2n}}
\end{equation*}
with Wallis's formula
\begin{equation*}
\quad \frac{\pi}{2}=\frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot 8 \cdots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdot 9 \cdots}
\end{equation*}
But the method I used was considered not to be universal.
How to prove that
\begin{equation*}
\lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \frac{\sqrt{2 \pi}}{2} \cdot \frac{1}{n^{\frac{1}{2}}}-\frac{\sqrt{2 \pi}}{8} \cdot \frac{1}{n^{\frac{3}{2}}}+\frac{\sqrt{2 \pi}}{64} \cdot \frac{1}{n^{\frac{5}{2}}}
\end{equation*}
And is
\begin{equation*}
\lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \frac{\sqrt{2 \pi}}{2} \cdot \frac{1}{n^{\frac{1}{2}}}-\frac{\sqrt{2 \pi}}{8} \cdot \frac{1}{n^{\frac{3}{2}}}+ \dots +
(-1)^{k}\cdot\frac{\sqrt{2 \pi}}{2^{\frac {k(k+1)}{2}}} \cdot \frac{1}{n^{\frac{2k+1}{2}}}
\end{equation*}
true? Are there any more powerful tools, like numerical methods to calculate the integration?
| You need to have an equivalent of the factorial, or equivalently, of the gamma function. A well known is the Stirling's approximation but you can go further as $$ {\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}\left(1+{\frac {1}{12n}}+{\frac {1}{288n^{2}}}-{\frac {139}{51840n^{3}}}-{\frac {571}{2488320n^{4}}}+\cdots \right).}$$ where the coefficients corresponds to the "Stirling series".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Computing the eigenvalues and eigenvectors of a $ 3 \times 3$ with a trick The matrix is: $
\begin{pmatrix}
1 & 2 & 3\\
1 & 2 & 3\\
1 & 2 & 3
\end{pmatrix}
$
The solution says that
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}
=
\begin{pmatrix}
6 \\
6 \\
6\end{pmatrix}$
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
-1
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$
$ B\cdot
\begin{pmatrix}
3 \\
0 \\
-1
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$
Thus the eigenvalues are $ \lambda_{1}=6,\lambda_{2}=0 $
My question is, how can I easily find
$\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$ and $\begin{pmatrix} 6 \\
6 \\
6\end{pmatrix}$?
Is there any way to see it "quickly"?
| The range of the given matrix is spanned by one vector:
$$
\left[\begin{array}{c}1 \\ 1 \\ 1\end{array}\right]
$$
Therefore, any eigenvector (which must be non-zero by definition) must be a scalar multiple of this vector, or it must be in the null space of the given matrix. The above is an eigenvector of the given matrix. The null space is spanned by
$$
\left[\begin{array}{r}2 \\ -1 \\ 0\end{array}\right],\left[\begin{array}{r}3 \\ 0 \\ -1 \end{array}\right]
$$
You can easily rewrite this null space in terms of the vectors given in your statement of the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Spivak, Ch. 22, "Infinite Sequences", Problem 1(iii): How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$? The following is a problem from Chapter 22 "Infinite Sequences" from Spivak's Calculus
*
*Verify the following limits
(iii) $\lim\limits_{n\to \infty} \left
[\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$
The solution manual says
$$\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right
]$$
$$=\lim\limits_{n\to \infty} \left [\left
(\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right )+\left
(\sqrt[4]{n}-\sqrt[4]{n+1}\right )\right ]$$
$$=0+0=0$$
(Each of these two limits can be proved in the same way that
$\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was proved in the
text)
How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right ]=0$?
Note that in the main text, $\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was solved by multiplying and dividing by $(\sqrt{n+1}+\sqrt{n})$ to reach
$$0<\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2\sqrt{n}}<\epsilon$$
$$\implies n>\frac{1}{4\epsilon^2}$$
| \begin{align}
\sqrt[8]{n^2+1}-\sqrt[4]{n+1}&=\big(n^2+1)^{1/8}-(n+1)^{2/8} \\
&=\big(n^2+1)^{1/8}-(n^2+2n+1)^{1/8}\\
&=\frac{\big(1+\tfrac{1}{n^2}\big)^{1/8}-1-\Big(\big(1+\tfrac{2}{n}+\tfrac{1}{n^2}\big)^{1/8}-1\Big)}{\tfrac{1}{n^{1/4}}}
\end{align}
The function $f(x)=x^{1/8}$, $x>0$ has a finite derivative at $x=1$. Letting $h=\frac{1}{n^2}$, we obtain that
$$\frac{f(1+h)-f(1)}{h^{1/8}}=h^{7/8}\frac{f(1+h)-f(1)}{h}\xrightarrow{h\rightarrow0}0\cdot f'(1)=0.$$
Similarly, letting $h=\frac{1}{n}$ we obtain
$$\frac{f(x+2h+h^2)-f(1)}{h^{1/4}}=(2h^{3/4}+h^{7/4})\frac{f(1+2h+h^2)-f(1)}{2h+h^2}\xrightarrow{h\rightarrow0}0\cdot f'(1)=0$$
Hence
$$\lim_{n\rightarrow}\sqrt[8]{n^2+1}-\sqrt[4]{n+1}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Product of Inverse Hankel Matrix Consider $H_n$, the $n\times n$ Hankel matrix of the Catalan numbers starting from $2$:
$$H_n = \begin{bmatrix}
2 & 5 & 14 & 42 & 132\\
5 & 14 & 42 & 132 & 429\\
14 & 42 & 132 & 429 & 1430 & \cdots\\
42 & 132 & 429 & 1430 & 4862\\
132 & 429 & 1430 & 4862 & 16796\\
&&\vdots\end{bmatrix}$$
It is known that $\text{det}(H_n) = n + 1$. (see Hankel Matrix)
Consider the column vector,
$$c_n = \begin{bmatrix}1 \\ 2 \\ 5 \\ 14 \\ \vdots \end{bmatrix}$$
that contains the first $n$ Catalan numbers.
I have found a pattern that I have checked up to $n=240$, that $$(c_n)^T(H_n)^{-1}(c_n) = \frac{n}{n+1}$$
Is there any method I can take to prove this, or is there a counterexample? Note also that this product is the only non-zero eigenvalue of $(c_n)(c_n)^T(H_n)^{-1}$.
| In the provided link, it says that
if $S_{n+1}=\begin{pmatrix} 1 & v^t\\
v & H_n \\
\end{pmatrix}$, where $v^t=(1,2,5,\ldots)$, then $\det(S_{n+1})=1$.
Notice that $$\begin{pmatrix} 1 & -v^tH_n^{-1}\\
0 & Id \\
\end{pmatrix}\begin{pmatrix} 1 & v^t\\
v & H_n \\
\end{pmatrix}\begin{pmatrix} 1 & 0^t\\
-H_n^{-1}v & Id \\
\end{pmatrix}=\begin{pmatrix} 1-v^tH_n^{-1}v & 0^t\\
0 & H_n \\
\end{pmatrix}.$$
Now, $\det\begin{pmatrix} 1 & -v^tH_n^{-1}\\
0 & Id \\
\end{pmatrix}=\det\begin{pmatrix} 1 & 0^t\\
-H_n^{-1}v & Id \\
\end{pmatrix}=1$, which implies
$$1=\det\begin{pmatrix} 1 & v^t\\
v & H_n \\
\end{pmatrix}=(1-v^tH_n^{-1}v).\det(H_n)=(1-v^tH_n^{-1}v).(n+1)$$
Therefore, $v^tH_n^{-1}v=1-\frac{1}{n+1}=\frac{n}{n+1}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proof that $\frac{xy^2}{x^2+y^6}$ is unbounded on $\mathbb{R}^2$ I am trying to prove that $\frac{xy^2}{x^2+y^6}$ is unbounded. Is the following correct?
Consider arbitrary $M \in \mathbb{R} \setminus\{0\}$ (the case in which $M=0$ is trivial). Set $y:=\sqrt{\frac{|x|}{2}}$. Thus, for any $x$ $\in \mathbb{R} \setminus \{0\}$, we have that $|x|>y^2$, and so that $\frac{xy^2}{x^2+y^6}>\frac{y^4}{x^2+y^6}$. We now argue that there exists an $x \in \mathbb{R}$ such that $\frac{xy^2}{x^2+y^6}>\frac{y^4}{x^2+y^6}>M$.
Note that $\frac{y^4}{x^2+y^6}$ is monotonically increasing as $x \rightarrow 0$. Also, note that $\frac{y^4}{x^2+y^6}$ is bounded above by $\frac{1}{y^2}=\frac{2}{x}$. Thus, as $x \rightarrow 0$, $\frac{y^4}{x^2+y^6}$ becomes arbitrarily large (as the upper bound towards which it is monotonically increasing becomes arbitrarily large). Thus, there must exist an $x \in \mathbb{R}$ such that $\frac{xy^2}{x^2+y^6}>M$, for $(x,y):=(x, \sqrt{\frac{x}{2}})$.
| Let $x=\frac{1}{n^3}$, $y=\frac{1}{n}$, where $n$ is a positive integer. Then
$$
\frac{xy^2}{x^2+y^6}=\frac{n}{2},
$$
which $\to \infty$ when $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to derive $\lim\limits_{x,n→∞}\prod\limits_{k=0}^{n-1}\left(1-\frac kx\right)=1-\frac eα\frac1x$ where $α=2e-4$? Background
$\def\d{\mathrm{d}}\def\e{\mathrm{e}}\def\i{\mathrm{i}}$
The following two equations are well known,
$$\begin{align*}
\left(1+\frac1x\right)^x &= \sum_{n=0}^\infty {x \choose n}\frac1{x^n}, \\[1ex]
\e &= \sum_{n=0}^\infty\dfrac{1}{n!}.
\end{align*}$$
Subtract them, which will produce the product in the question, and then perform a backward derivation (see end for details) from
$$\lim_{x\to\infty}x\left[\left( 1+\dfrac{1}{x} \right)^{x}-\e \right] = -\dfrac{\e}{2},$$
yields the result in the question.
P.S. The limit above comes from a small exercise, its solution steps are:
① Reciprocal substitution, i.e. $x=\dfrac1t$;
② Use L'Hôpital's rule once;
③ Utilize the derivation formula for compound logarithmic functions, i.e. $(u^v)' = u^v \left( \ln u^{v'} + \dfrac{u'}{u}v \right)$;
④ Use L'Hôpital's rule twice.
Q
How to solve
$$\lim_{\substack{n\to\infty\\x\to\infty}}\prod_{k=0}^{n-1}\left(1-\frac kx\right)$$
in a forward approach? (while requiring the retention of the "exact form" in the title)
Actually, what I'd like to ask more is: Is there a general solution for products of this form? In other words, is it possible to assume that the result is in the form containing {$\e, \alpha$} and find it?
Side note
The specific process of backward derivation
$$\begin{align*}
\Sigma_1-\Sigma_2&=\sum_{n=0}^1\,(1-1) + \sum_{n=2}^{\infty}\left( \frac{x(x-1)\cdots(x-(n-1))}{n!}\frac{1}{x^n}-\frac{1}{n!} \right) \\
&=\sum_{n=2}^{\infty}\dfrac{1}{n!}\left[ 1\left( 1-\dfrac{1}{x} \right)\cdots\left( 1-\dfrac{n-1}{x} \right) - 1 \right] \\
&=\sum_{n=2}^{\infty}\dfrac{1}{n!}\left[ {\color{teal}{\prod_{k=0}^{n-1}\left( 1-\dfrac{k}{x} \right) - 1}} \right],
\end{align*}$$
Substitute into the equation in the question, immediately get
$$\lim_{x\to\infty} x\,\left(\Sigma_1-\Sigma_2\right) = {\color{teal}{-\dfrac{\e}{2}\dfrac{1}{\e-2}}}\sum_{n=2}^{\infty}\dfrac{1}{n!}=-\dfrac{\e}{2}.$$
The result is consistent with the known conclusions from the normal calculation, so the equation in question is verified.
| The product $$ \prod_{k=0}^{n-1} \left( 1 - \frac{k}{x} \right) $$
can be seen as
$$ \prod_{k=0}^{n-1} \left( 1 - \frac{k}{x} \right) = \frac{x (x-1) (x-2)\cdots (x-n+1)}{x^n} = \frac{x!}{x^n \, (x-n)!}. $$
Using $n! \approx \sqrt{2 \pi n} \, n^n \, e^{-n}$ then
$$ \prod_{k=0}^{n-1} \left( 1 - \frac{k}{x} \right) = \frac{x!}{x^n \, (x-n)!} \approx \left( 1- \frac{n}{x}\right)^{n-x-1/2} \, e^{-n}. $$
If $ n \to \infty$ then
$$ \lim_{n \to \infty} \, \prod_{k=0}^{n-1} \left( 1 - \frac{k}{x} \right) = 0. $$
If $x \to \infty$ then, by use of $\lim_{x \to \infty} \left(1 - \frac{n}{x}\right)^{-x} = e^{n}$,
$$ \lim_{x \to \infty} \, \prod_{k=0}^{n-1} \left( 1 - \frac{k}{x} \right) = e^{-n} \, \lim_{x \to \infty} \left( 1 - \frac{n}{x}\right)^{n-1/2} \, \left(1 - \frac{n}{x}\right)^{-x} = 1. $$
Additional note:
Consider the limit
$$ \lim_{x \to \infty} \, x \, \left[ \left(1 - \frac{a}{x}\right)^x - e^{-a} \right]$$
as follows.
\begin{align}
\left(1 - \frac{a}{x}\right)^x - e^{-a} &= e^{x \, \ln\left(1 - \frac{a}{x}\right)} - e^{-a} \\
&= e^{x \, \left( - \frac{a}{x} \right) - \frac{a^2}{2 \, x^2} + \cdots} - e^{-a} \\
&= e^{-a - \frac{a^2}{2 \, x} - \cdots} - e^{-a} = e^{-a} \, \left( -1 + e^{\frac{a^2}{2 \, x} + \cdots} \right) \\
&= e^{-a} \, \left( -1 + 1 + \left( -\frac{a^2}{2 \, x} - \frac{a^3}{3! \, x^2} + \cdots \right) + \frac{1}{2!} \, \left( -\frac{a^2}{2 \, x} - \frac{a^3}{3! \, x^2} + \cdots \right)^2 + \cdots \right) \\
&= e^{-a} \, \left( -\frac{a^2}{2 \, x} + \frac{a^3 \, (3 a - 4)}{4! \, x^2} + \mathcal{O}\left(\frac{1}{x^3}\right) \right) \\
&= -\frac{a^2 \, e^{-a}}{2 \, x} \, \left( 1 + \frac{2 a (4 - 3 a)}{4! \, x} + \mathcal{O}\left(\frac{1}{x^2}\right) \right).
\end{align}
This leads to
\begin{align}
\lim_{x \to \infty} \, x \, \left[\left(1 - \frac{a}{x}\right)^x - e^{-a} \right] &= \lim_{x \to \infty} \, -\frac{a^2 \, e^{-a}}{2} \, \left( 1 + \frac{2 a (4 - 3 a)}{4! \, x} + \mathcal{O}\left(\frac{1}{x^2}\right) \right) \\
&= -\frac{a^2 \, e^{-a}}{2}.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Points that satisfy the equation $|z + 3| + |z + 1| = 4$ I am trying to find the points that satisfy the equation $$|z+3|+|z+1|=4$$
Substituting the value of $z$ and evaluating its modulus gives me $$\sqrt{(x+3)^2+y^2}+\sqrt{(x+1)^2+y^2}=4$$
What I tried to do is to square both sides giving me $$a+b+2\sqrt{ab}=16$$ $$a=(x+3)^2+y^2,\ b=(x+1)^2+y^2$$
and I know what follows is going to be a lengthy and time-consuming process. Is there any faster or easier method to solve this problem?
| How to obtain the equation without getting bogged down in square root radicals:
Start with
$|z+3|+|z+1|=4$
and note that
$|z+3|^2-|z+1|^2=(x+3)^2+y^2-(x+1)^2-y^2=4x+8.$
Thus from the difference of squares factorization we must accept
$|z+3|-|z+1|=(4x+8)/4=x+2.$
So
$|z+3|=(1/2)[(|z+3|+|z+1|)+(|z+3|-|z+1|)]=(4+x+2)/2=(x+6)/2.$
Squaring then gives
$|z+3|^2=(x+3)^2+y^2=(x+6)^2/4$
$4x^2+24x+36+4y^2=x^2+12x+36$
$3x^2+12x+4y^2=0.$
Completing the square in the $x$ variable gives
$3(x+2)^2+4y^2=12,$
which we recognize as a standard form for an ellipse centered at $(-2,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Logarithm with negative base and negative argument in an alternating geometric sequence formula I was trying to find the number of terms "$n$" of the sequence
{$a_k$} = {$5, -1, 0.2, -0.04, ..., -0.00000256 $}, where $a_k$ = $a_1⋅r^{k-1}$ = $5⋅(-\frac{1}{5})^{k-1}$.
My problem comes up when I set $k=n$ in the previous formula to find, of course, the number of terms of the sequence $a_n$ = $5⋅(-\frac{1}{5})^{n-1}$. (Notice that $a_n$ = $-0.00000256$).
Then:
$\rightarrow$ $-0.00000256$ = $5⋅(-\frac{1}{5})^{n-1}$
$\rightarrow$ $-0.000000512$ = $(-\frac{1}{5})^{n-1}$
$\rightarrow$ $\log_{(-\frac{1}{5})} (-0.000000512)$ = $n-1$
$\rightarrow$ $n$ = $1 + \log_{(-\frac{1}{5})} (-0.000000512)$
$\rightarrow$ $n=1+9=10$
(technically)
I don't know how to get rid of negative base and negative argument. However, if they were positive the answer
will be $n$ = $10$. I don't know how that undefined logarithm came up.
| So you have
$$
256 \times 10^{-8} = -5 \cdot \left( \frac{1}{-5} \right)^{n-1}.
$$
Note $256=2^8$ so this is equivalent to
$$
\begin{split}
0.2^8 (-5)^{n-1} &= -5 \\
\frac{1}{5^8} \times (-5)^{n-2} &= 1 \\
5^{n-10} \times (-1)^{n-2} &= 1
\end{split}
$$
so sounds like $n=10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Multiple solutions for trig function with period Find all angles with limits $−\pi \leq \theta \leq \pi$ which satisfy $\sin 4 \theta = 1$
The working out is given as:
If $\sin 4 \theta = 1$ then $4 \theta = \pi/2 + 2 k \pi$, so $\theta = \pi/8 + k \pi/2$
For $−\pi \leq \theta \leq \pi$ we have $\theta = \pi/8, 5 \pi/8, −3 \pi/8, −7\pi/8$
How did they find the other values of $5 \pi/8, −3 \pi/8$ and $−7 \pi/8$?
Was it just using the unit circle?
| Alternative approach:
You have that $4\theta$ must be congruent to $\pi/2$, within a modulus of $(2\pi).$
So, form the following sequence
*
*$4\theta = \pi/2 \implies $
$\theta = \pi/8.$
*$4\theta = 2\pi + \pi/2 = 5\pi/2 \implies $
$\theta = 5\pi/8.$
*$4\theta = 9\pi/2\implies $
$\theta = 9\pi/8.$
*$4\theta = (13)\pi/2\implies $
$\theta = (13)\pi/8.$
*$4\theta = (17)\pi/2\implies $
$\theta = (17)\pi/8.$
*$4\theta = (21)\pi/2\implies $
$\theta = (21)\pi/8.$
*$\cdots$
At this point, you stop and take stock:
the candidate values for distinct solutions are the elements in the following set:
$$\left\{~\frac{\pi}{8}, ~\frac{5\pi}{8}, ~\frac{9\pi}{8}, ~\frac{13\pi}{8}, ~\frac{17\pi}{8}, ~\frac{21\pi}{8}, \cdots ~\right\}. \tag1 $$
Now, you (again) stop and consider.
If you examine the values in (1) above, considering that you are only interested in values that are distinct, with respect to a modulus of $(2\pi)$, you realize that
$$\frac{\pi}{8} \equiv \frac{17\pi}{8} \pmod{2\pi}, ~~~\frac{5\pi}{8} \equiv \frac{21\pi}{8} \pmod{2\pi}. \tag2 $$
Further, you should also realize at this time, that the ongoing pattern will recur. That is, you should realize that the following infinite sequence of angles
$$\frac{25\pi}{8}, ~\frac{29\pi}{8}, ~\frac{33\pi}{8}, ~\frac{37\pi}{8}, \cdots $$
will not be yielding any distinct angles, with respect to the $(2\pi)$ modulus.
Therefore, you realize that there are only $(4)$ distinct solutions, within a modulus of $(2\pi)$. These solutions are given by the set
$$\left\{~\frac{\pi}{8}, ~\frac{5\pi}{8}, ~\frac{9\pi}{8}, ~\frac{13\pi}{8} ~\right\}. \tag3 $$
At this point, you are close to your goal.
(3) above represents all distinct satisfying values $\theta$ such that $0 \leq \theta \leq 2\pi.$
However, this isn't good enough. The problem requires you to identify all distinct values $\theta$ such that $-\pi \leq \theta \leq \pi.$
If you examine (3) above, you realize that the first two values (from the left) are okay, but that the next two values need adjustment. What this means is that the two values of $~\dfrac{9\pi}{8}~$ and $~\dfrac{13\pi}{8}~$ need to be adjusted.
The adjustment needed is that these two values must be re-expressed in their modulus $(2\pi)$ equivalents that are within the range $-\pi \leq \theta \leq \pi.$
The easiest way to do this is to subtract $(2\pi)$ from each of the two out of range angles.
So:
*
*$\dfrac{9\pi}{8} - 2\pi = \dfrac{-7\pi}{8}.$
*$\dfrac{13\pi}{8} - 2\pi = \dfrac{-3\pi}{8}.$
Therefore, the refined expression of the 4 distinct solutions, from (3) above, are expressed as
$$\left\{~\frac{\pi}{8}, ~\frac{5\pi}{8}, ~\frac{-7\pi}{8}, ~\frac{-3\pi}{8} ~\right\}. \tag4 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding range of variables with defined relations Q: Given that $\{x,y,z\}\in\mathbb{R}$ , and
$$\begin{cases}
x+y+z=6\\
xy+yz+zx=7
\end{cases}$$ and ; determine range of $x$ ,$y$ , and $z$.
I thought that defining $x$ in terms of $y$ and $z$ : $x=6-y-z$ , using it in the second provided equation , then assume it to be a quadratic in $y$ or $z$ and then using $D\ge 0$ could solve it ; am i right?
Can someone provide a simpler solution; i will be grateful for it.
| As you are looking for something different, here is a mainly geometrical approach.
Due to relationship:
$$\underbrace{(x+y+z)^2}_{36}=x^2+y^2+z^2+\underbrace{2(xy+yz+zw)}_{14}$$
the initial issue is equivalent to the system:
$$\begin{cases}x^2+y^2+z^2&=&22\\x+y+z&=&6\end{cases}$$
Therefore, the locus is the intersection of the sphere centered in $(0,0,0)$ with radius $\sqrt{22}$ and a plane .
As a consequence , the set of solutions is a circle whose center is $C(x_c=2,y_c=2,z_c=2)$ (due to symmetry in coordinates $x,y,z$) and radius $\sqrt{10}=\sqrt{22-12}$ by Pythagoras ; indeed, length OC = $\sqrt{12}$).
A parametrization of the circle is as follows:
$$\begin{pmatrix}x\\y\\z\\\end{pmatrix}=\underbrace{\begin{pmatrix}2\\2\\2\\\end{pmatrix}}_C+\sqrt{10}\cos\theta \underbrace{\begin{pmatrix} \ \ \ \tfrac{1}{\sqrt{2}}\\ -\tfrac{1}{\sqrt{2}}\\0\\\end{pmatrix}}_U+\sqrt{10}\sin\theta \underbrace{\begin{pmatrix}\ \ \ \tfrac{1}{\sqrt{6}}\\ \ \ \ \tfrac{1}{\sqrt{6}} \\ -\tfrac{2}{\sqrt{6}}\end{pmatrix}}_V$$
(please note that vectors $U$ and $V$ constitute an orthonormal basis of the vector plane with equation $x+y+z=0$ parallel to affine plane $x+y+z=6$).
Let us now take a look at coordinate
$$z=2+2\tfrac{\sqrt{10}}{\sqrt{6}}\sin \theta$$
whose values are taken in the interval:
$$z \in [2-2\sqrt{\tfrac53},2+2\sqrt{\tfrac53} ]\tag{1}$$
Coordinate ranges for $x$ and $y$ are identical, on account of symmetry.
| {
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"answer_count": 1,
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} |
Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)? Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)?
My answer is not possible because for $a^2$ to be equal to $b$ means that the argument of $b$ is twice of arg(a) and for $b^2$ to be equal to $a$ means that arg(a) = 2.arg(b) but the answer is it is possible.
How is it possible when arg(b) = 2.arg(a) and arg(a) = 2.arg(b) contradict each other?
| There are the number of real solutions of the system
$$a^2-b^2=(a^2-b^2)^2-4a^2b^2\\2ab=4ab(a^2-b^2)\tag1$$ which come from
the equalities $$(a+bi)^2=(a^2-b^2)+2abi\\((a^2-b^2)+2abi)^2=(a^2-b^2)^2-4a^2b^2+4ab(a^2-b^2)i$$
The solution of system $(1)$ is easy and leaves to the only solutions, the two non real roots of unity.
| {
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"source": "stackexchange",
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"answer_count": 5,
"answer_id": 4
} |
Uniform convergence of the derivative function $h_n'(x)$ I am self-learning Real Analysis from the text Understanding Analysis, by Stephen Abbott. I would like for someone to
(1) Verify my proof for part (a) of this exercise problem.
(2) Do you have any clues for part (b) without giving away the entire solution/proof?
[Abbott 6.3.2] Consider the sequence of functions :
\begin{equation*}
h_{n}( x) =\sqrt{x^{2} +\frac{1}{n}}
\end{equation*}
(a) Compute the pointwise limit of $\displaystyle ( h_{n})$ and then prove that the convergence is uniform on $\displaystyle \mathbf{R}$.
Proof.
Fix $\displaystyle x\in \mathbf{R}$. We know that, if $\displaystyle \lim a_{n} =a$, then $\displaystyle \lim \sqrt{a_{n}} =\sqrt{\lim a_{n}} =\sqrt{a}$. Thus:
\begin{equation*}
\begin{array}{ c l }
\lim _{n\rightarrow \infty } h_{n}( x) & =\lim _{n\rightarrow \infty }\sqrt{x^{2} +\frac{1}{n}}\\
& =\sqrt{\lim _{n\rightarrow \infty }\left( x^{2} +\frac{1}{n}\right)}\\
& =\left[\lim _{n\rightarrow \infty } x^{2} +\lim _{n\rightarrow \infty }\frac{1}{n}\right]^{( 1/2)}\\
& =\sqrt{x^{2}}\\
& =|x|
\end{array}
\end{equation*}
Consider the expression:
\begin{align*}
|h_{n}( x) -h_{m}( x) | & =\left| \sqrt{x^{2} +\frac{1}{n}} -\sqrt{x^{2} +\frac{1}{m}}\right| & \\
& =\frac{\left| \left( x^{2} +\frac{1}{n}\right) -\left( x^{2} +\frac{1}{m}\right)\right| }{\left| \sqrt{x^{2} +\frac{1}{n}} +\sqrt{x^{2} +\frac{1}{m}}\right| } & \\
& =\frac{\left| \frac{1}{n} -\frac{1}{m}\right| }{\sqrt{x^{2} +\frac{1}{n}} +\sqrt{x^{2} +\frac{1}{m}}} \\
& \leq \frac{\left| \frac{1}{n} -\frac{1}{m}\right| }{\frac{1}{\sqrt{n}} +\frac{1}{\sqrt{m}}} & \left\{\because x^{2} \geq 0\right\}\\
& =\frac{\left| \frac{1}{\sqrt{n}} -\frac{1}{\sqrt{m}}\right| \left(\frac{1}{\sqrt{n}} +\frac{1}{\sqrt{m}}\right)}{\left(\frac{1}{\sqrt{n}} +\frac{1}{\sqrt{m}}\right)} & \\
& =\left| \frac{1}{\sqrt{n}} -\frac{1}{\sqrt{m}}\right| &
\end{align*}
Pick an arbitrary $\displaystyle \epsilon >0$. Since $\displaystyle \frac{1}{\sqrt{n}}\rightarrow 0$, and convergent sequences are Cauchy, there exists $\displaystyle N( \epsilon ) >0$, such that for all $\displaystyle n >m\geq N$,
\begin{equation*}
\left| \frac{1}{\sqrt{n}} -\frac{1}{\sqrt{m}}\right| < \epsilon
\end{equation*}
Consequently, by Cauchy criterion for uniform convergence of a sequence of functions, $\displaystyle ( h_{n})$ converges uniformly on $\displaystyle \mathbf{R}$ to $\displaystyle h$.
(b) Note that each $\displaystyle h_{n}$ is differentiable. Show that $\displaystyle g( x) =\lim h_{n} '( x)$ exists for all $\displaystyle x$ and explain how we can be certain that the convergence is not uniform on any neighbourhood of zero.
Proof.
By Chain rule of differentiation, we have:
\begin{equation*}
h_{n} '( x) =\frac{x}{\sqrt{x^{2} +\frac{1}{n}}}
\end{equation*}
Moreover,
\begin{equation*}
\lim h_{n} '( x) =\lim _{n\rightarrow \infty }\frac{x}{\sqrt{x^{2} +\frac{1}{n}}} =\frac{x}{|x|}
\end{equation*}
| Hint: Uniform Convergence preserves certain properties from the sequence of functions to the limit function.
| {
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} |
Solve the equation $\sqrt{x^2+x+1}+\sqrt{x^2+\frac{3x}{4}}=\sqrt{4x^2+3x}$ Solve the equation $$\sqrt{x^2+x+1}+\sqrt{x^2+\dfrac{3x}{4}}=\sqrt{4x^2+3x}$$
The domain is $$x^2+\dfrac{3x}{4}\ge0,4x^2+3x\ge0$$ as $x^2+x+1>0$ for every $x$. Let's raise both sides to the power of 2: $$x^2+x+1+x^2+\dfrac{3x}{4}+2\sqrt{(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)}=4x^2+3x\\2\sqrt{(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)}=2x^2+\dfrac{5x}{4}$$ Let's raise both sides to the power of 2 again but this time the roots should also satisfy $A:2x^2+\dfrac54x\ge0$:$$4(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)=(2x^2+\dfrac54x)^2$$ I came at $$x(2x^2+\dfrac{87}{16}x+3)=0$$ I obviously made a mistake as the answer is $x=-4$, but is there an easier approach?
| HINT
I would start with multiplying both sides by the number $2$:
\begin{align*}
\sqrt{x^{2} + x + 1} + \sqrt{x^{2} + \frac{3x}{4}} = \sqrt{4x^{2} + 3x} & \Longleftrightarrow 2\sqrt{x^{2} + x + 1} + \sqrt{4x^{2} + 3x} = 2\sqrt{4x^{2} + 3x}\\\\
& \Longleftrightarrow 2\sqrt{x^{2} + x + 1} = \sqrt{4x^{2} + 3x}\\\\
\end{align*}
Can you take it from here?
| {
"language": "en",
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Find the volume of the solid that lies above the cone : $z^2 = x^2 + y^2$ and inside the sphere $x^2 + y^2 + (z-2)^2 = 4$ Using spherical coordinates find the volume of the solid that lies above the cone: $z^2 = x^2 + y^2$ and inside the sphere $x^2 + y^2 + (z-2)^2 = 4$
I'm aware that there's a similar question here using a different method.
Assuming it's the same problem I should get the same sol. but my attempts with spherical coor. yield a different answer so I need to know where the mistake is.
My attempt :
The solid is bounded above by the sphere centered at $(0,0,2)$ and below by the circle at which the sphere and the cone intersect: $x^2+y^2=2z$ (at $z=2$) whose projection on the $xy$-plane is $x^2+y^2=4$
So the limits are given by : $$ Q= \{ 0 \leq \rho \leq 2 \;,\; 0 \leq \phi \leq \frac{\pi}{2} \;,\; 0 \leq \theta \leq 2\pi \}$$
$$
\begin{align}
V &= \underset{Q}{\iiint} \rho^2 \sin(\phi) \;d\rho \;d\phi \;d\theta \\
&= \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^{2} \rho^2 \sin(\phi) \;d\rho \;d\phi \;d\theta \\
&= \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \sin(\phi) \left[ \frac{1}{3} \rho^3 \right]_0^2 \;d\phi \;d\theta
= \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \frac{8}{3} \sin(\phi) \;d\phi \;d\theta \\
&= \int_0^{2\pi} \frac{8}{3} \bigg[ \sin(\phi) \bigg]_0^{\frac{\pi}{2}} \;d\theta \\
&= \frac{8}{3} \int_0^{2\pi} d\theta \\
&= \frac{16}{3}\pi
\end{align}
$$
Edit: Thank you for correcting me on the limits for $\rho$. But I imagined the problem to be computing the volume of the hemisphere $z = \sqrt{4-x^2-y^2}+2$.
Mainly because it's supported by the answered questions in my textbook. The book claims the answer to be $\dfrac{16}{3}\pi$ but without showing the steps. Therefore I interpreted "the solid that lies above the cone" in the question as " (strictly) above but not inside".
Hence,
$$
\begin{align}
V &= \underset{Q}{\iiint} \rho^2 \sin(\phi) \;d\rho \;d\phi \;d\theta \\
&= \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \int_2^{4\cos(\phi)} \rho^2 \sin(\phi) \;d\rho \;d\phi \;d\theta \\
&(\cdots) \Rightarrow \; V = \frac{16}{3}\pi
\end{align}
$$
I attached an image visualizing my understanding of the problem.
| As a check, the volume is the sum of a half-circle and a cone with a circular base, i.e.
$$ V= \frac{2\pi r^3}3 + \frac13 (\pi r^2)r=\pi r^3 \overset{r=2} = 8\pi$$
| {
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How to solve $20(x-\lfloor x\rfloor)=x+\lfloor x\rfloor+\left\lfloor x+\frac{1}{2}\right\rfloor$ analytically? How to solve this analytically?
$$20(x-\lfloor x\rfloor)=x+\lfloor x\rfloor+\left\lfloor x+\frac{1}{2}\right\rfloor$$
where $\lfloor .\rfloor$ is the floor function.
I attempted to solve this equation numerically for the first 7 numbers, after which no solution exists. However, solving numerically is a pain.
How can I solve this analytically?
My working (an example for the first 3-4 reals )
$$19x-21\lfloor x\rfloor =\left\lfloor x+\frac{1}{2}\right\rfloor$$
*
*When $x$ is between $0$ and $0.5$, there is a single solution at $x=0$.
*When $x$ is between $0.5$ and $1$, we have $19x=1$, which means no solution in the given interval
*Next, solving on $[1,1.5]$, we have $19x-21=1$, which gives us another solution in the given interval.
And so on, until no solutions occur for two or 3 tries, at which point all solutions have been obtained.
I haven't yet found all the elements as it would obviously take forever. (I know there are 7 solutions as I graphed these on Desmos to confirm my idea.)
Any suggestions?
| Denote $\{x\}$ the fractional part of $x$, then
$$\Longleftrightarrow 19\{x\} = 3[x] +\left[\{x\} +\frac{1}{2} \right] $$
We deduce that
$$-1<3[x]<19 \Longleftrightarrow 0\le [x] \le6$$
Case 1: If $\{x\} < \frac{1}{2}$, then
$$19\{x\} = 3[x] \Longleftrightarrow [x] < \frac{1}{3}\cdot\frac{19}{2} \Longleftrightarrow [x] \le 3$$
then for $[x] =n \in \{0,1,2,3 \}$, we have
$$ \{x\} = \frac{3n}{19}\Longleftrightarrow x = n +\frac{3n}{19} \qquad \text{for } n= 0,1,2,3 \tag{1}$$
Case 2: If $\{x\} \ge \frac{1}{2}$, then
$$19\{x\} = 3[x] +1 \Longleftrightarrow [x] \ge \frac{1}{3}\left(\frac{19}{2}-1\right) \Longleftrightarrow [x] \ge 3$$
then for $[x] =n \in \{3,4,5,6 \}$, we have
$$ \{x\} = \frac{3n+1}{19}\Longleftrightarrow x = n +\frac{3n+1}{19} \qquad \text{for } n= 3,4,5 \tag{2}$$
Attention: We need to remove the case $n = 6$ as in this case, $\{x\} = \frac{3*6+1}{19} = 1$ that cannot occur ($\{x\}$ must be in $[0,1)$).
From $(1),(2)$, we have 7 solutions
$$x \in \left\{0, \frac{22}{19}, \frac{44}{19}, \frac{66}{19}, \frac{67}{19}, \frac{89}{19}, \frac{111}{19} \right\}$$
| {
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} |
Triple integral in spherical coordinate, where am I wrong $$ \iiint_{D} z\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z $$
D is given by $x^{2}+y^{2}+z^{2}\leq 2z$
I try to use $ \left\{\begin{matrix}
x=r\sin \phi \cos \theta \\
y=r\sin \phi \sin \theta \\
z=r\cos\phi
\end{matrix}\right. $ while the Jacobian is $r^2 \sin \phi$ and $ \left\{\begin{matrix} 0\leq \theta \leq 2\pi\\
0\leq \phi \leq \frac{\pi}{2} \\0\leq r \leq 1\end{matrix}\right. $
$\Rightarrow$
$$\int^{2\pi}_0 \mathrm{d}\theta\int^{\pi/2}_0\mathrm{d}\phi\int_0^{2\cos\phi}r^4\sin^2\phi \cos \phi\mathrm{d}r $$
$\Rightarrow$
$$2\pi\int^{\pi/2}_0\cos^6\phi-\cos^8\phi \ \mathrm{d}\phi$$
$$I_{n}=\int_{0}^{\frac{\pi}{2}} \cos ^{n} x d x\Rightarrow I_{n}=\frac{n-1}{n} I_{n-2}\text{ integral by part }$$
$$I_{2 m}=\frac{2 m-1}{2 m} \cdot \frac{2 m-3}{2 m-2} \cdot \cdots \cdot \frac{3}{4} \cdot \frac{1}{2} I_{0}\\
I_{2 m+1}=\frac{2 m}{2 m+1} \cdot \frac{2 m-2}{2 m-1} \cdot \cdots \cdot \frac{4}{5} \cdot \frac{2}{3} I_{1}\\
I_{0}=\int_{0}^{\frac{\pi}{2}} d x=\frac{\pi}{2}, I_{1}=\int_{0}^{\frac{\pi}{2}} \cos x d x=1\\$$
$\Rightarrow$
$$2\pi\int^{\pi/2}_0\cos^6\phi-\cos^8\phi \ \mathrm{d}\phi=\frac{\pi^2}{4}$$
but the answer is $\frac{8\pi}{3}$, where am I wrong.
I used wolframalpha to calculate $\int^{\pi/2}_0\mathrm{d}\phi\int_0^{2\cos\phi}r^4\sin^2\phi \cos \phi\mathrm{d}r $ the answer is $\frac{\pi}{8}$, am I wrong at first?
| Firstly, I have a typo it is $r\leq2\cos \phi$.
Secondly, it is
$$\int^{2\pi}_0 \mathrm{d}\theta\int^{\pi/2}_0\mathrm{d}\phi\int_0^{2\cos\phi}r^5\sin\phi \cos \phi\mathrm{d}r $$
and it is easy to integrate
$\Rightarrow$
$$-2\pi \int_0^{\pi/2}\frac{2^6}{6}\sin \phi \cos^7\phi d\phi\Rightarrow-\frac{2^7\pi}{6}\int_0^{\pi/2} \cos^7\phi d\cos\phi\\ \Rightarrow-\frac{2^7\pi}{6\times2^3}\cos^8 \phi \bigg|^{\pi/2}_0=\frac{8\pi}{3}$$
| {
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Seeking for other methods to evaluate $\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx$ for $n\geq 2$. Inspired by my post, I go further to investigate the general integral and find a formula for
$$
I_n=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx =-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right] \tag*{}
$$
Let’s start with its partner integral
$$
I(a)=\int_0^{\infty}\left(x^n+1\right)^a d x
$$
and transform $I(a)$, by putting $y=\frac{1}{x^n+1}$, into a beta function
$$
\begin{aligned}
I(a) &=\frac{1}{n} \int_0^1 y^{-a-\frac{1}{n}-1}(1-y)^{-\frac{1}{n}-1} d y \\
&=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right)
\end{aligned}
$$
Differentiating $I(a)$ w.r.t. $a$ yields
$$
I^{\prime}(a)=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right)\left(\psi(-a)-\psi\left(-a-\frac{1}{n}\right)\right)
$$
Then putting $a=-1$ gives our integral$$
\begin{aligned}
I_n&=I^{\prime}(-1) \\&=\frac{1}{n} B\left(1-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(1)-\psi\left(1-\frac{1}{n}\right)\right] \\
&=-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right]\end{aligned}
$$
For examples,
$$
\begin{aligned}& I_2=-\frac{\pi}{2} \csc \frac{\pi}{2}\left[\gamma+\psi\left(1-\frac{1}{2}\right)\right]=\pi \ln 2,\\ & I_3=-\frac{\pi}{3} \csc \left(\frac{\pi}{3}\right)\left[\gamma+\psi\left(\frac{2}{3}\right)\right]=\frac{\pi \ln 3}{\sqrt{3}}-\frac{\pi^2}{9} ,\\ &I_4=-\frac{\pi}{4} \csc \left(\frac{\pi}{4}\right)\left[\gamma+\psi\left(\frac{3}{4}\right)\right]=\frac{3 \pi}{2 \sqrt{2}}\ln 2-\frac{\pi^2}{4 \sqrt{2}},\\
& I_5=-\frac{\pi}{5} \csc \left(\frac{\pi}{5}\right)\left[\gamma+\psi\left(\frac{4}{5}\right)\right]=-\frac{2 \sqrt{2} \pi}{5 \sqrt{5-\sqrt{5}}}\left[\gamma+\psi\left(\frac{4}{5}\right)\right], \\
& I_6=-\frac{\pi}{6} \csc \left(\frac{\pi}{6}\right)\left[\gamma+\psi\left(\frac{5}{6}\right)\right]=\frac{2 \pi}{3} \ln 2+\frac{\pi}{2} \ln 3-\frac{\pi^2}{2 \sqrt{3}},
\end{aligned}
$$
Furthermore, putting $a=-m$, gives
$$\boxed{I(m,n)=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{(x^n+1)^m} dx = \frac{1}{n} B\left(m-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(m)-\psi\left(m-\frac{1}{n}\right)\right] }$$
For example,
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(x^6+1\right)}{(x^6+1)^5} dx & =\frac{1}{6} B\left(\frac{29}{6}, \frac{1}{6}\right)\left[\psi(5)-\psi\left(\frac{29}{6}\right)\right] \\
& =\frac{1}{6} \cdot \frac{21505 \pi}{15552} \cdot\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right) \\
& =\frac{21505 \pi}{93312}\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right)
\end{aligned}
$$
Are there any other methods?
Your comments and alternative methods are highly appreciated.
| For large $n$, a simple asymptotic behaviour of the integral can be deduced.
Let us first examine the structure of the integrand
$$y(x)=\frac{\ln \left(x^n+1\right)}{x^n+1} $$
Take the derivative of the function and set the result to zero to get the point $x=x_{m}$ where $y(x)$ reaches his maximum
$$1+(x_{m})^{n}=e$$
Putting this into $y(x)$ gives
$$y_{max}=y(x_{m})=\frac{\ln e}{e}=\frac{1}{e}$$
$y_{max}$ is independent of $n$
For integrals whose integrand, $y(x)$, has a sharp maximum a simple asymptotic formula exists.
$$I\approx \sqrt{2\pi\frac{y^{3}(x_{m})}{\left|y''(x_{m}) \right|}}$$
Here $y''(x_{m})$ is the second derivative of $y(x)$ at $x=x_{m}$.
I will skip elementary computations and write down final result
$$I_{n}\approx \frac{\sqrt{2\pi}}{n(e-1)^{1-\frac{1}{n}}}$$
Below is a few numerical examples to show approximation errors produced by the last formula
$n=10$, the approximation error about $0.03$
$n=20$, the approximation error about $0.01$
$n=30$, the approximation error about $0.007$
The higher n is, the higher the accuracy of the last formula.
| {
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} |
general solution beta $(\sin(\alpha+\beta)+\cos(\alpha +2\beta)\sin(\beta))^2 = 4\cos(\alpha)\sin(\beta)\sin(\alpha+ \beta);\tan(\alpha)=3\tan(\beta)$ We need to find general solution of $\beta$ for
$$(\sin(\alpha+\beta)+\cos(\alpha +2\beta)\sin(\beta))^2 = 4\cos(\alpha)\sin(\beta)\sin(\alpha+ \beta)$$
$$\tan(\alpha)=3\tan(\beta)$$
I took the first equation and managed to simplify it up till here:
$$(\cos(\beta)\sin(\alpha+2\beta))^2=4\cos(\alpha)\sin(\beta)\sin(\alpha+ \beta)$$
What should I do next? I tried to use $2\sin(\beta)\sin(\alpha+\beta)=\cos(\alpha)-\cos(\alpha+2\beta)$ in the RHS in the hopes of cancelling or doing something about the $\alpha+2\beta$ terms. But I cannot seem to boil it down further.
Thanks!
| By request, here's a sketch of the solution I mentioned in a comment to the question. (There's probably a quicker path.)
I'll ignore the case $\beta=\pm\pi/2$.
Now, start by expanding the sides of OP's first equation, express them in terms of $p:=\tan\alpha$ and $q:=\tan\beta$, and then apply the substitution $p=3q$ from OP's second equation:
$$\begin{align}
\left(\sin(\alpha+\beta) + \cos(\alpha + 2\beta) \sin\beta\right)^2 &=
\left( 2\sin\alpha\cos^2\beta + 2 \cos\alpha \cos\beta \sin\beta-\sin\alpha \right)^2 \cos^2\beta \\[4pt]
&= \frac{(2\tan\alpha + 2 \tan\beta-\tan\alpha\sec^2\beta)^2}{\sec^4\beta} \cos^2\alpha\cos^2\beta\\[4pt]
&= \frac{(2p + 2q-p(1+q^2))^2}{(1+q^2)^2}\cos^2\alpha\cos^2\beta\\[4pt]
&= \frac{(p + 2q-pq^2)^2}{(1+q^2)^2}\cos^2\alpha\cos^2\beta \\[4pt]
&\overset{p=3q}{=}\;\frac{q^2(5-3q^2)^2}{(1+q^2)^2}\cos^2\alpha\cos^2\beta \tag1\\[4pt]
4 \cos\alpha \sin\beta \sin(\alpha+\beta) &=
4 \sin\beta (\sin\alpha\cos\beta + \cos\alpha \sin\beta)\cos\alpha \\[4pt]
&= 4\tan\beta\,(\tan\alpha + \tan\beta) \cos^2\alpha \cos^2\beta \\[4pt]
&= 4q(p+q)\cos^2\alpha \cos^2\beta \\[4pt]
&\overset{p=3q}{=} 16q^2\cos^2\alpha \cos^2\beta \tag2
\end{align}$$
So, OP's first equation, $(1)=(2)$, becomes
$$\begin{align}
\frac{q^2(5-3q^2)^2}{(1+q^2)^2}\cos^2\alpha\cos^2\beta &\;=\; 16q^2\cos^2\alpha\cos^2\beta \\[4pt]
q^2(5-3q^2)^2 &\;=\; 16q^2(1+q^2)^2\\[6pt]
q^2 (7 q^2-1) (q^2+9) &\;=\; 0 \tag3
\end{align}$$
Therefore, ignoring non-real values, we have $\tan\beta=0$ or $\tan\beta=\pm1/\sqrt{7}$; and, respectively, $\tan\alpha=0$ or $\tan\alpha=\pm3/\sqrt{7}$. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find All Complex solutions for $z^3+3i\bar{z}=0$ Find All Complex solutions for $z^3+3i \bar z =0$.
I tried substituting $z=a+bi$ and simplifying as much as I can
and this is what I ended up with: $a^3-2b^2a+3b+i(3ba^2-b^3+3a)=0$.
I just did not understand how do I get the values of $z$ from this equation
| This is a commentary on all of the existing answers, rather than a suggested method of solution. It seems fashionable to state that there are $9$ solutions, when in reality there are $5$ solutions $z$ to the equation $$z^3-3i\overline z = 0$$
Writing $z=x+iy$, we obtain the following polynomial equations by taking real and imaginary parts of the given equation:
\begin{align*}
&x^3-3xy^2+3y = 0 \\
&3x^2y+3x-y^3 = 0
\end{align*}
By Bézout's Theorem, the corresponding projective curves have $9$ intersections, counting multiplicity. However, we are seeking real solutions $(x,y)$ in the affine plane, so there could be fewer than $9$ solutions to the equation $z^3-3i\overline z = 0$.
By standard polynomial algebra, this system is equivalent to
\begin{align*}
&9x-8y^7+51y^3 = 0 \\
&64y^9-432y^5+81y=0
\end{align*}
Factoring $y$ out of the second equation and performing the substitution $Y=y^4$, we obtain
\begin{equation*}
64Y^2-432Y+81
\end{equation*}
The roots are $$Y = \frac{27}8 \pm \frac94\sqrt2$$
These are positive real numbers, and so the solutions for $y$ are
\begin{equation*}
y = 0, \pm\left(\frac{27}8 \pm \frac94\sqrt2\right)^{1/4}
\end{equation*}
For each $y$, the coordinate $x$ is uniquely determined and real. This means that there are $5$ solutions to the equation,
$$z^3-3i\overline z = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$ Inspired by my post, I decided to investigate the integral in general
$$
I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$$
by the powerful substitution $x=\frac{1-t}{1+t} .$
where $n$ is a natural number greater $1$.
Let’s start with easy one
\begin{aligned}
I_1 &=\int_0^1 \frac{\ln \left(\frac{2 t}{1+t}\right)}{1+t^2} d t \\ &=\int_0^1 \frac{\ln 2+\ln t-\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2+\int_0^1 \frac{\ln t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2-G-\frac{\pi}{8} \ln 2 \\
&=\frac{\pi}{8} \ln 2-G\end{aligned}
By my post
$$I_2= \frac{\pi}{4} \ln 2-G $$
and
$$\begin{aligned}I_4 &=\frac{3 \pi}{4} \ln 2-2 G
\end{aligned}
$$
$$
\begin{aligned}
I_3=& \int_0^1 \frac{\ln (1-x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x \\=& \frac{\pi}{8} \ln 2-G+\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x-G
\\ =& \frac{\pi}{8} \ln 2-\frac{4G}{3} +\frac{\pi}{6} \ln (2+\sqrt{3})
\end{aligned}
$$
where the last integral refers to my post.
Let’s skip $I_5$ now.
$$
I_6=\int_0^1 \frac{\ln \left(1-x^6\right)}{1+x^2} d x=\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x+I_3\\
$$
$$\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x = \int_0^1 \frac{\ln (1+x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x\\=\frac{\pi}{8}\ln 2+ \frac{1}{2} \int_0^{\infty} \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x-G \\= \frac{\pi}{8}\ln 2+ \frac{1}{2}\left( \frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G \right)- G \\= \frac{\pi}{8} \ln 2+\frac{\pi}{3} \ln (2+\sqrt{3})-\frac{5}{3} G $$
Hence $$I_6= \frac{\pi}{4} \ln 2+\frac{\pi}{2} \ln (2+\sqrt{3})-3 G$$
How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x?$
| We can proceed to $I_{8}$ now.
$$
\int_0^1 \frac{\ln \left(1-x^8\right)}{1+x^2} d x=\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1+x^4\right)}{1+x^2} dx\\ \qquad\qquad =\frac{3 \pi}{4} \ln 2-2 G+ \int_0^1 \frac{\ln \left(1+x^4\right)}{1+x^2} dx.
$$
In my post, two beautiful formula were found.
$$\boxed{\begin{align}
&\int_0^\infty \frac{\ln(1+x^{4m})}{1+x^2}dx
=2\pi \ln \bigg( 2^m \prod_{k=1}^m \cos\frac{(2k-1)\pi}{8m}\bigg)\
\\
&\int_0^\infty \frac{\ln(1+x^{4m+2})}{1+x^2}dx
= \pi\ln2 + 2\pi \ln \bigg( 2^m \prod_{k=1}^m \cos\frac{k\pi}{2(2m+1)}\bigg)
\end{align}}$$
As $$
\int_0^1 \frac{\ln \left(1+x^n\right)}{1+x^2} d x=\frac{1}{2}\left[\int_0^{\infty} \frac{\ln \left(1+x^n\right)}{1+x^2} d x-n G\right]
$$
Putting $m=1$ into the first formula in the box yields $$
\int_0^1 \frac{\ln \left(1+x^4\right)}{1+x^2} d x=\frac{1}{2}\left[2 \pi \ln \left(2 \cos \frac{\pi}{8}\right)-4 G\right]
$$
Hence
$$\boxed{\int_0^1 \frac{\ln \left(1-x^8\right)}{1+x^2} d x = \frac{3 \pi}{4} \ln 2-4 G+\pi \ln (\sqrt{2+\sqrt{2}}) }$$
Similarly, We can go further to $I_{12}$ now.
$$
\int_0^1 \frac{\ln \left(1-x^{12}\right)}{1+x^2} d x=\int_0^1 \frac{\ln \left(1-x^6\right)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1+x^6\right)}{1+x^2} dx\\ \qquad\qquad = \frac{\pi}{4} \ln 2+\frac{\pi}{2} \ln (2+\sqrt{3})-3 G + \int_0^1 \frac{\ln \left(1+x^6\right)}{1+x^2} dx.
$$
Putting $m=1$ into the second formula in the box yields
$$\boxed{\int_0^1 \frac{\ln \left(1-x^{12}\right)}{1+x^2} d x= \frac{\pi}{4} \ln 2+\frac{\pi}{2} \ln (2+\sqrt{3})-3 G + \frac{\pi}{2} \ln 6-3G= \frac{\pi}{4} \ln (72(7+4 \sqrt{3}))-6 G }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
Prove that $(x^2−1) \bmod 8$ $\in \{ 0 , 3 , 7 \}, \forall x \in \mathbb{Z}$. It must be verified that for all $x \in \mathbb{Z}$ it holds that $x^2 - 1 \bmod{8} \in \{0, 3, 7\}$. First some definitions. Using the following theorem a definition for $\bmod$ is provided:
Theorem. For all $a \in \mathbb{Z}$ and $d \in \mathbb{N}$ unique integers q, r exist satisfying $a = q \cdot d + r \wedge 0 \leq r <d$
The integers $q$, $r$ correspond to the quotient and remainder, respectively, of $a$, these being defined as:
$a = (a / d) \cdot d + a \bmod d \wedge 0 \leq a \bmod d < d$
With $q = (a/d)$ and $r = a \bmod d$. To verify the claim I use induction. I use a brute force approach where I search for all numbers $x$ such that $x^2 - 1 \bmod 8 = 0$, $x^2 - 1 \bmod 8 = 3$ or $x^2 - 1 \bmod 8 = 7$. First I search for all $x$ such that $x^2 - 1 \bmod 8 = 0$. If $x^2 - 1 \bmod 8 = 0$ then 8 is a divisor of $x^2 - 1$, i.e.: $8 \mid x^2 - 1$. Now I search for some other $x$ such that 8 is a divisor of $x^2 - 1$. Let $f(x) = x^2 - 1$ and $8 \mid f(x)$ then $8 \mid f(x + a)$ for some $a \in \mathbb{N}$. I compute this a:
$8 \mid f(x) = 8 \mid x^2 - 1 \implies 8 \mid (x + a)^2 - 1 = 8 \mid x^2 + a^2 + 2ax - 1 = 8 \mid x^2 - 1 + a^2 + 2ax$
The implication holds for e.g.: $a = 4$ since:
$8 \mid x^2 - 1 + a^2 + 2ax \implies 8 \mid x^2 - 1 + 16 + 8x = 8 \mid f(x) \wedge 8 \mid 16 + 8x = true$
So it can be concluded that if $8 \mid f(x)$ then $8 \mid f(x + 4)$. In fact it may be concluded that if $8 \mid f(x)$ then $8 \mid f(x + 4k), \forall x \in \mathbb{Z} \wedge k \in \mathbb{N}$, since:
$8 \mid (x + 4k)^2 - 1 = 8 \mid x^2 - 1 + 16k^2 + 8kx = 8 \mid f(x) \wedge 8 \mid 16k^2 + 8kx$
So it holds that $8 \mid f(x) \implies 8 \mid f(x + 4k)$. By inspection one finds that $8 \mid f(1) = 8 \mid 1^2 - 1 = 8 \mid 0 = true$ and so $f(x) \bmod 8 = 0$ for all $x \in \{1, 5, 9, 13, ...\}$. Since $f(x)$ is symmetric it holds that $f(-x) = f(x)$ and so it follows that $f(x) \bmod 8 = 0$ for all $x \in \{-1, -5, -9, -13, ...\}$. Upon further inspection one finds that $8 \mid f(3) = 8 \mid 3^2 - 1 = 8 \mid 8 = true$ and so it follows that $x^2 - 1 \bmod 8 = 0$ for all $x \in \{3, 7, 11, 15, ...\}$ and all $x \in \{-3, -7, -11, -15, ...\}$
Next I search for all x such that $x^2 - 1 \bmod 8 = 3$, or equivalently $x^2 - 4 \bmod 8 = 0$. Using the same approach I find that $x^2 - 1 \bmod 8 = 3$ for all $x \in \{2, 6, 10, 14, ...\}$ and all $x \in \{-2, -6, -10, -14, ...\}$. Finally $x^2 - 1 \bmod 8 = 7$ for all $x \in \{0, 4, 8, 12, ...\}$ and $x \in \{0, -4, -8, -12, ...\}$. And so one concludes that for all $x \in \mathbb{Z}$ it holds that $x^2 - 1 \bmod{8} \in \{0, 3, 7\}$.
This question comes from a course in discrete mathematics in computer science. I feel that my approach is overkill and that I'm doing something wrong and that there must be some cleverer way of solving the problem. If anyone can help with a better, cleaner, approach, or point out errors, it will be greatly appreciated :).
| First of all, if $x^2 - 1 \equiv y \pmod 8$, then $$(x+4)^2 - 1 = (x^2 - 1) + 8(x + 2) \equiv y \pmod 8;$$ that is to say, $x^2-1$ and $(x+4)^2 - 1$ have the same remainder upon division by $8$, for all integers $x$.
Hence it suffices to test $x \in \{0, 1, 2, 3\}$. We have $$\begin{align}
0^2 - 1 &\equiv 7 \pmod 8, \\
1^2 - 1 &\equiv 0 \pmod 8, \\
2^2 - 1 &\equiv 3 \pmod 8, \\
3^2 - 1 &\equiv 0 \pmod 8.
\end{align}$$
This concludes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $2^\frac{2x-1}{x-1}+2^\frac{3x-2}{x-1}=24$, find all values of $x$ that satisfy this As title suggests, the problem is as follows:
Given that $$2^\frac{2x-1}{x-1}+2^\frac{3x-2}{x-1}=24$$ find all values of $x$ that satisfy this.
This question was shared in an Instagram post a few months ago that I came across today. Examining it at first, it seems there are many ways to solve this. I'll show my own approach here, please let me know if there are any issues in my solution and please share your own solution too!
Here's my approach for the problem:
Let $a=2^\frac{2x-1}{x-1}$ and $b=2^\frac{3x-2}{x-1}$
We then get $a+b=24$.
Now notice that:
$$(3x-2)-(2x-1)=x-1$$
That gives us a motivation to perform division with $a$ and $b$ (as the denominator and numerator of the exponent will be equal, hence reducing the exponent) thus:
$$\frac{b}{a}=\frac{2^\frac{3x-2}{x-1}}{2^\frac{2x-1}{x-1}}$$
$$\frac{b}{a}=2^\frac{x-1}{x-1}=2$$
$$b=2a$$
Therefore:
$$2a+a=24$$
$$3a=24$$
$$a=8$$
$$2^\frac{2x-1}{x-1}=8$$
$$2^\frac{2x-1}{x-1}=2^3$$
$$\frac{2x-1}{x-1}=3$$
$$2x-1=3x-3$$
Thus, $x=2$
| Here's my resolution :)
$$
2^\frac{2x-1}{x-1} + 2^\frac{3x-2}{x-1} = 24 \\
\left(2^\frac{1}{x-1}\right)^{2x} \ \left(2^\frac{1}{x-1}\right)^{-1} + \left(2^\frac{1}{x-1}\right)^{3x} \ \left(2^\frac{1}{x-1}\right)^{-2} = 24 \\
2^\frac{2x-1}{x-1} \ \left[ 1 + 2^\frac{x-1}{x-1} \right] = 24\\
2^\frac{2x-1}{x-1} \cdot 3 = 24\\
\log_2 \ 2^\frac{2x-1}{x-1} = 8 \\
\frac{2x-1}{x-1} \cdot \log_22= \log_2 8\\
2x-1=3(x-1)\\
x=2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4587013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Write an equation for a sphere passing through a circle and tangent to a plane I'm trying to solve this task:
'''Write an equation for a sphere passing through a circle $x^2 + y^2 = 11$ and tangent to a plane $x + y + z - 5 = 0$.'''
Center of the sphere should moves only in axis Z, so it has coordinates $(0, 0, \alpha)$. I also found that
$\alpha^2 = R^2 - 11$, where R - radius of a sphere.
I came to this system:
*
*$x_0^2 + y_0^2 + (z_0 - \sqrt(R^2 - 11))^2 = R^2$
*$x_0 + y_0 + z_0 = 5$
where $x_0, y_0, z_0$ are coordinates of a touch point of a sphere with the plane.
The correct answer is two spheres:
$x^2 + y^2 + (z + 1)^2 = 12$
$x^2 + y^2 + (z + 4)^2 = 27$
I can't figure out how to find these two values of the parameter $\alpha$. Could somebody please explain how to do it?
Thanks in advance.
| If the center of the sphere is $(x_0, y_0, z_0)$, and its radius is $R$ then
its equation is
$ (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = R^2 $
Set $z = 0 $ , then you get
$ (x - x_0)^2 + (y - y_0)^2 + z_0^2 = R^2 $
Comparing this with $ x^2 + y^2 = 11 $ we deduce that $x_0 = y_0 = 0 $ and
$ 11 = R^2 - z_0^2 $
Since the sphere is tangent to $ x+ y+z = 5 $, then the distance of the center which is (0, 0, z_0) to the plane is equal to $R$ , thus
$ R = \dfrac{| 0 + 0 + z_0 - 5 |} {\sqrt{3}} $
Squaring
$ R^2 = \dfrac{(z_0 - 5)^2}{3} $
Thus
$ 11 + z_0^2 = \dfrac{(z_0 - 5)^2 }{3} $
So that,
$ 33 + 3 z_0^2 = z_0^2 - 10 z_0 + 25 $
From which,
$ 2 z_0^2 + 10 z_0 + 8 = 0 $
Dividing through by $2$,
$ z_0^2 + 5 z_0 + 4 = 0 $
Factoring,
$ (z_0 + 4)(z_0 + 1) = 0 $
Therefore, $z_0 = -4$ or $z_0 = -1$, and corresponding to that,
$R^2 = 27 $ or $ R^2 = 12 $
And the equations of the two spheres are
$ x^2 + y^2 + (z + 4)^2 = 27 $
and
$ x^2 + y^2 + (z + 1)^2 = 12 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding $x_1x_2+x_1x_3+x_2x_4+x_3x_4$ without explicitly finding the roots of $x^4-2x^3-3x^2+4x-1=0$
The equation $x^4-2x^3-3x^2+4x-1=0$ has $4$ distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1\lt x_2\lt x_3\lt x_4$ and product of $2$ roots is unity, then find the value of $x_1x_2+x_1x_3+x_2x_4+x_3x_4$
This question has an answer on this link but I am trying to solve it without explicitly finding the roots because the question tells us that the product of $2$ roots is unity. I want to use it.
My Approach:
Using Descartes rule, I can see that there is one negative root and three positive roots.
Also, at $x=0, 1, -1$, the value of the polynomial is negative.
Thus, $x_1\lt-1, x_4\gt1$ and $x_2,x_3$ lies between $0$ to $1$.
Thus, I am concluding that $x_2x_4=1$ and $x_1x_3=-1$ (because product of roots is $-1$)
How to conclusively reject the case $x_3x_4=1$?
For $\alpha\gt1, \beta\gt1$, $x_1=-\beta, x_2=\frac1\alpha, x_3=\frac1\beta, x_4=\alpha$
Sum of roots$=-\beta+\frac1\alpha+\frac1\beta+\alpha=2\implies\frac1\beta-\beta=2-(\alpha+\frac1\alpha)$
Sum of product of roots taken $3$ at a time$=-\frac1\alpha+\frac1\beta-\alpha-\beta=-4\implies\frac1\beta-\beta=-4+(\alpha+\frac1\alpha)$
Therefore, $\alpha+\frac1\alpha=3, \frac1\beta-\beta=-1$
Multiplying these two, $\frac\alpha\beta-\alpha\beta+\frac1{\alpha\beta}-\frac\beta\alpha=-3$
The question asks us to find $\frac\alpha\beta-\frac\beta\alpha$, that means $-3+\alpha\beta-\frac1{\alpha\beta}$
Can we conclude this approach?
| For $ \ f(x) \ = \ x^4 - 2x^3 - 3x^2 + 4x-1 \ \ , \ $ the "depressed" polynomial is $ \ f \left(x + \frac12 \right) $ $ \ = \ f(y) \ = \ y^4 - \frac92·y^2 + \frac{1}{16} \ \ , \ $ which, as an even function, has symmetrically-arranged zeroes given by $ \ y^2 \ = \ \frac14·( \ 9 \pm \sqrt{80} \ ) \ \ . \ $ We can approximate the locations as
$$ y^2 \ \ = \ \ \frac14·\left( \ 9 \ \pm \ 9·\sqrt{1 \ - \ \frac{1}{81}} \ \right) \ \ \approx \ \ \frac94·\left( \ 1 \ \pm \ \left[ \ 1 \ - \ \frac{1}{2·81} \ \right] \ \right) \ \ \approx \ \ \frac{1}{8·9} \ \ , \ \ \frac92 \ \ . $$
The four zeroes of $ \ f(x) \ $ are thus estimated by
$$ x_1 \ \approx \ \frac12 - \frac{3}{\sqrt2} \ < \ 0 \ \ \ , \ \ \ x_2 \ \approx \ \frac12 - \frac{1}{6\sqrt2} \ \ \ , \ \ \ x_3 \ \approx \ \frac12 + \frac{1}{6\sqrt2} \ \ \ , \ \ \ x_4 \ \approx \ \frac12 + \frac{3}{\sqrt2} \ > \ 2 \ \ . $$
What is clear from this is that $ \ x_3·x_4 \ $ cannot be equal to $ \ 1 \ $ and that the only product of two zeroes than can be is $ \ \mathbf{x_2}·x_4 \ \ . \ $ [The exact values are in fact $ \ -\phi \ \ , \ \ 2 - \phi \ \ , \ \ \frac{1}{\phi} \ = \ \phi - 1 \ \ , \ \ \phi + 1 \ \ , \ $ but we don't need to know that in order to resolve this particular issue.]
To return to the main question, if we label the four zeroes of $ \ f \left(x + \frac12 \right) \ $ as $ \ -\gamma \ , \ -\delta \ , \ +\delta \ , \ +\gamma \ \ , \ $ then their product is $ \ \gamma^2·\delta^2 \ = \ \frac{1}{16} \ \Rightarrow \ \gamma·\delta \ = \ \frac14 \ \ . \ $ (The estimates above actually happen to fit this nicely.) We may then express the zeroes of $ \ f(x) \ $ as (using your notation)
$$ x_1 \ \ = \ \ \frac12 \ - \ \gamma \ \ = \ \ -\beta \ \ \ , \ \ \ x_2 \ \ = \ \ \frac12 \ - \ \frac{1}{4 \ \gamma} \ \ = \ \ \frac{1}{\alpha} \ \ \ , $$ $$ x_3 \ \ = \ \ \frac12 \ + \ \frac{1}{4 \ \gamma} \ \ = \ \ \frac{1}{\beta} \ \ \ , \ \ \ x_4 \ \ = \ \ \frac12 \ + \ \gamma \ \ = \ \ \alpha \ \ . $$
We can establish from these results that
$$ \alpha·\frac{1}{\alpha} \ \ = \ \ \left( \ \gamma \ + \ \frac12 \ \right)·\left( \ \frac12 \ - \ \frac{1}{4 \ \gamma} \ \right) \ \ = \ \ \frac{4·\gamma^2 \ - \ 1}{8 \ \gamma} \ \ = \ \ 1 $$
$$ \Rightarrow \ \ 4·\gamma^2 \ - \ 8·\gamma \ - \ 1 \ \ = \ \ 0 \ \ \Rightarrow \ \ \gamma \ \ = \ \ \frac{8 \ + \ \sqrt{64 \ + \ 16}}{8} \ \ = \ \ 1 \ + \ \frac{ \sqrt5}{2} \ \ , $$
where we have retained only the positive solution (we obtain the same result from $ \ \beta·\frac{1}{\beta} \ \ ) $
$$ \Rightarrow \ \ \alpha \ \ = \ \ \frac{3 + \sqrt5}{2} \ \ ( \ = \ x_4 \ ) \ \ \ , \ \ \ \beta \ \ = \ \ \frac{1 + \sqrt5}{2} \ \ ( \ = \ -x_1 \ ) \ \ $$
$$ \Rightarrow \ \ \alpha·\beta \ \ = \ \ \frac{8 \ + \ 4\sqrt5}{4} \ \ = \ \ \ 2 \ + \ \sqrt5 \ \ \Rightarrow \ \ \frac{1}{\alpha·\beta} \ \ = \ \ \frac{ 2 \ - \ \sqrt5 }{4 \ - \ 5} \ \ = \ \ \sqrt5 \ - \ 2 $$
$$ \Rightarrow \ \ \alpha·\beta \ - \ \frac{1}{\alpha·\beta} \ \ = \ \ 4 \ \ \Rightarrow \ \ x_1 x_2 \ + \ x_1 x_3 \ + \ x_2 x_4 \ + \ x_3 x_4 \ \ = \ \ -3 \ + \ 4 \ \ = \ \ 1 \ \ . $$
[Using the exact value of the zeroes written above in terms of the Golden Ratio $ \ \phi \ \ , \ $ we verify that
$$ (-2 \phi \ + \ \phi^2) \ + \ (-\phi^2 \ + \ \phi) \ + \ (2 \phi \ - \ \phi^2 \ + \ 2 \ - \ \phi) \ + \ (\phi^2 \ - \ 1) \ \ = \ \ 1 \ \ . \ ] $$
So while we have succeeded in finding the value of the sum of the specified pair-products of zeroes without explicitly determining all of the zeroes, it appears to be necessary to characterize those zeroes to a certain extent to evaluate the sum by your approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Sum binomial coefficients $$\sum_{k=0}^{n} {\frac{k^2+k}{3^{k+2}} {n \choose k}}=?$$
What I've tried:
$$(k^2+k){n \choose k}=k(k+1){\frac{n!}{k!(n-k)!}}$$
$$k^2+k = k^2-k+2k=k(k-1)+2k$$ =>
$$
\begin{align}
(k^2+k){n \choose k} &= k(k-1){\frac{n!}{k!(n-k)!}}+2k{\frac{n!}{k!(n-k)!}}\\&={\frac{n!}{(k-2)!(n-k)!}}+2{\frac{n!}{(k-1)!(n-k)!}}\\&=n(n-1){n-2 \choose k-2}+2n{n-1 \choose k-1}
\end{align}
$$
and
$${\frac{1}{3^{k+2}}}={\frac{1}{9}}({\frac{1}{3}})^k$$
So I have
$$\sum_{k=0}^{n} {\frac{n(n-1){n-2 \choose k-2}+2n{n-1 \choose k-1}}{9*3^k}}$$
And I'm stuck... Can anyone help me?
| As in the first hint
$$(\ x(1+x)^n\ )''= \left(\sum_{k=0}^n {n\choose k}x^{k+1}\right)''= \sum_{k=1}^n k(k+1){n\choose k}x^{k-1}= f(x)$$
So if you take $x= \frac{1}{3}$ :
$$\frac{1}{27}f(\frac{1}{3})= \frac{1}{3^3}\sum_{k=1}^n k(k+1){n\choose k}\frac{1}{3^{k-1}}= \sum_{k=1}^n \frac{k(k+1)}{3^{k+2}}{n\choose k}= \sum_{k=0}^n \frac{k(k+1)}{3^{k+2}}{n\choose k}- 0$$
And differentiate :
$$f(x)= (\ (1+x)^n+ nx(1+x)^{n-1}\ )'= 2n(1+x)^{n-1}+ n (n-1)x(1+x)^{n-2}$$
You can compute $f(\frac{1}{3})= 2n(1+\frac{1}{3})^{n-1}+ n (n-1)\frac{1}{3}(1+\frac{1}{3})^{n-2}= 2n(\frac{4}{3})^{n-1}+ n(n-1)\frac{1}{3}(\frac{4}{3})^{n-2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find all three complex solutions of the equation $z^3=-10+5i$ Let $z\in \mathbb{C}$. I want to calculate the three solutions of the equation $z^3=-10+5i$. Give the result in cartesian and in exponential representation.
Let $z=x+yi $.
Then we have $$z^2=(x+yi)^2 =x^2+2xyi-y^2=(x^2-y^2)+2xyi$$
And then $$z^3=z^2\cdot z=[(x^2-y^2)+2xyi]\cdot [x+yi ] =(x^3-xy^2)+2x^2yi+(x^2y-y^3)i-2xy^2=(x^3-3xy^2)+(3x^2y-y^3)i$$
So we get
$$z^3=-10+5i \Rightarrow (x^3-3xy^2)+(3x^2y-y^3)i=-10+5i \\ \begin{cases}x^3-3xy^2=-10 \\ 3x^2y-y^3=5\end{cases} \Rightarrow \begin{cases}x(x^2-3y^2)=-10 \\ y(3x^2-y^2)=5\end{cases}$$
Is everything correct so far? How can we calculate $x$ and $y$ ? Or should we do that in an other way?
| There is another way
$$z^3+10-5i=0$$
Let's replace the complex number with a constant $a$
$$z^3+a=0$$
$$z^3+(a^{1/3})^3=0$$
Using the Sum of Cubes
$$(z+a^{1/3})(z^2-a^{1/3}z+a^{2/3})=0$$
We get $z=-a^{1/3}$ as a solution
$$z^2-a^{1/3}z+a^{2/3}=0$$
$$z=\frac{-(-a^{1/3})±\sqrt{(-a^{1/3})^2-4(1)(a^{2/3})}}{2(1)}=\frac{a^{1/3}±\sqrt{-3a^{2/3}}}{2}=\frac{a^{1/3}±\sqrt3ia^{1/3}}{2}$$
$$=\left(\frac{1±i\sqrt3}{2}\right)a^{1/3}=a^{1/3}e^{±\pi i/3}$$
Now we find the cube root of $a$ using polar form
$$(10-5i)^{1/3}=\left(\sqrt{10^2+5^2}\exp\left(i\arctan\left(-\frac{5}{10}\right)\right)\right)^{1/3}$$$$=\left(5\sqrt5\exp\left(-i\arctan\left(\frac{1}{2}\right)\right)\right)^{1/3}=\sqrt{5}\exp\left(\frac{-i}{3}\arctan{\frac{1}{2}}\right)$$
WE GET THE SOLUTION
$${z=\sqrt{5}\exp\left(\frac{-i}{3}\arctan{\frac{1}{2}}\right)e^{k\pi i/3}}, k=-1,0,1$$
NOTE:
Instead of using the sum of cubes, we can notice the following
$$z^3=-a$$
$$z^3=ae^{k\pi i}, k=-1,1,3$$
Take the cube root on both sides
$$z=(ae^{k\pi i})^{1/3}=a^{1/3}e^{k\pi i/3}, k=-1,1,3$$
Then find the cube root and finish
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Does $\sum\limits_{n = 1}^{\infty} \frac{3^n + 4^n}{2^n + 5^n}$ converge? My Attempt
First, check the limit
$$\lim_{n \to \infty} \frac{3^n + 4^n}{2^n + 5^n} =
\lim_{n \to \infty} \frac{\left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n}{\left(\frac{2}{5}\right)^n + 1} = 0.$$
So, we cannot conclude anything.
I used Comparison Test and Ratio Test.
Consider that
$$\frac{3^n + 4^n}{2^n + 5^n} < 2\frac{5^n}{2^n + 5^n}.$$
If I can proof the convergence of the series on the right-side, then it's done by comparison test.
I used ratio test, in order to proof the convergence of
$$\sum_{n= 1}^{\infty} \frac{5^n}{2^n + 5^n}.$$
$$\lim_{n \to \infty} \frac{55^n}{2(2^n) + 5(5^n)} \frac{2^n + 5^n}{5^n} = 5 \lim_{n \to \infty} \frac{2^n + 5^n}{2(2^n) + 5(5^n)} = 1.$$
I didn't find the way to proof.
Any suggestion? Thanks in advanced.
Solution (@abiessu & @Thomas Andrew)
Consider that
$$\frac{3^n + 4^n}{2^n + 5^n} < 2\frac{4^n}{2^n + 5^n}.$$
Proof this series converge.
$$\sum_{n= 1}^{\infty} \frac{4^n}{2^n + 5^n}.$$
Proof (Ratio Test)
$$\lim_{n \to \infty} \frac{44^n}{2(2^n) + 5(5^n)} \frac{2^n + 5^n}{4^n} = 4 \lim_{n \to \infty} \frac{2^n + 5^n}{2(2^n) + 5(5^n)} = \frac{4}{5}.$$
The series converge.
Hence
$$\sum_{n = 1}^{\infty} \frac{3^n + 4^n}{2^n + 5^n}$$
converge.
| My preferred method would be to note that
$$
\frac{3^n+4^n}{2^n+5^n}
\leq \frac{4^n+4^n}{0+5^n}
= 2 \cdot \left(\frac{4}{5}\right)^n.
$$
So, the series converges by the comparison test and the geometric series criterion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as:
$\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \int_0^{2 \pi} \frac{1-\cos \theta \cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & \int_0^{2 \pi} \frac{d x}{1-\cos ^2 \theta \cos ^2 x}-\cos \theta \int_0^{2 \pi} \frac{\cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & 4 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-\cos ^2 \theta} d x+\int_0^{2 \pi} \frac{d(\cos \theta \sin x)}{\sin ^2 \theta+\cos ^2 \theta \sin ^2x} \\= & 4 \int_0^{\frac{\pi}{2}} \frac{d\left(\tan x\right)}{\sin ^2 \theta+\tan ^2 x}+\frac{1}{\sin \theta}\left[\tan ^{-1}\left(\frac{\cos ^2 \theta \sin x}{\sin \theta}\right)\right]_0^{2 \pi} \\= & \frac{4}{\sin \theta}\left[\tan ^{-1}\left(\frac{\tan x}{\sin \theta}\right)\right]_0^{\frac{\pi}{2}} \\= & \frac{4}{\sin \theta} \cdot \frac{\pi}{2} \\= & \frac{2 \pi}{\sin \theta}\end{aligned}\tag*{} $
Is there another simpler method to evaluate the integral?
Your comments and alternative methods are highly appreciated.
| Splitting the integral interval into two gives another solution.
$$
\begin{aligned}
\int_0^{2 \pi} \frac{1}{1+a \cos x} d x & =\int_0^\pi \frac{1}{1+a \cos x} d x+\int_\pi^{2 \pi} \frac{1}{1+a \cos x} d x \\
& =\int_0^\pi \frac{1}{1+a \cos x} d x+\int_0^\pi \frac{1}{1-a \cos x} d x \\
& =2 \int_0^\pi \frac{1}{1-a^2 \cos ^2 x} d x \\
& =4 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-a^2} d x \\
& =4 \int_0^{\frac{\pi}{2}} \frac{d(\tan x)}{\tan ^2 x+\left(1-a^2\right)} \\
& =\frac{4}{\sqrt{1-a^2}}\left[\tan ^{-1}\left(\frac{\tan x}{\sqrt{1-a^2}}\right)\right]_0^{\frac{\pi}{2}} \\
& =\frac{2 \pi}{\sqrt{1-a^2}}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Sum of two subspaces: representing it with equations I found the following excercise:
Let $W_1 = \{(x_1, ..., x_6) : x_1 + x_2 + x_3 = 0, x_4 + x_5 + x_6 = 0 \}$. Let $W_2$ be the span of $S := \{(1, -1, 1, -1, 1, -1), (1, 0, 2, 1, 0, 0), (1, 0, -1, -1, 0, 1), (2, 1, 0, 0, 0, 0)\}$.
Give a base, a dimension and an equation representation of $W_1 + W_2$
I'm new to the concept of sum of subspaces. But as I understand it, the first step would be to note that any $\textbf{x} = (x_1, ..., x_6) \in W_1$ satisfies
\begin{equation*}
\begin{cases}
x_1 = -x_2 - x_3 \\
x_4 = -x_5 - x_6
\end{cases}
\end{equation*}
so that its general form is
\begin{equation*} \textbf{x} =
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \\
x_6
\end{pmatrix} = \begin{pmatrix}
-x_2 - x_3 \\
x_2 \\
x_3 \\
-x_5 - x_6 \\
x_5 \\
x_6
\end{pmatrix}
\end{equation*}
We also know any $\textbf{y} \in W_1$ is of the general form
\begin{align*} \textbf{y} =
\begin{pmatrix}
y_1 \\
y_2 \\
y_3 \\
y_4 \\
y_5 \\
y_6
\end{pmatrix} = \begin{pmatrix}x + y +z + 2w \\
-x + w \\
x + 2y - z \\
-x + y - z \\
x\\
-x + z\end{pmatrix}
\end{align*}
Then for generals $\mathbf{x}, \mathbf{y}$ we have
\begin{align*}
\textbf{x} + \textbf{y} &= \begin{pmatrix}
x + y +z + 2w + (-x_2 - x_3)\\
-x + w + x_2\\
x + 2y - z + x_3\\
-x + y - z + (-x_5 - x_6)\\
x + x_5\\
-x + z + x_6
\end{pmatrix}
\end{align*}
One can then conclude
$$W_1 + W_2 = \Big\{\big(x + y + z + 2w - x_2 - x_3\big), \big(-x + w + x_2 \big), \big(x +2y - z + x_3 \big), \big(-x + y - z - x_5 - x_6 \big), \big(x + x_5 \big), \big(-x + z + x_6 \big) \mid x, y, z, x_2, x_3 \in \mathbb{R} \Big\}$$
But what would be an representation via equations of this system? I'm very new to linear algebra so go easy on me!
| In contrast to Jose's and Anne's answers using dimension, let me show you the rote method you could use. From your step here:
\begin{equation*} \mathbf{x} =
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \\
x_6
\end{pmatrix} = \begin{pmatrix}
-x_2 - x_3 \\
x_2 \\
x_3 \\
-x_5 - x_6 \\
x_5 \\
x_6
\end{pmatrix},
\end{equation*}
you could then find a spanning set for $W_1$ like so:
\begin{align*}
\mathbf{x} &= \pmatrix{-x_2\\x_2\\0\\0\\0\\0} + \pmatrix{-x_3\\0\\x_3\\0\\0\\0} + \pmatrix{0\\0\\0\\-x_5\\x_5\\0} + \pmatrix{0\\0\\0\\-x_6\\0\\x_6} \\
&= x_2\pmatrix{-1\\1\\0\\0\\0\\0} + x_3\pmatrix{-1\\0\\1\\0\\0\\0} + x_5\pmatrix{0\\0\\0\\-1\\1\\0} + x_6\pmatrix{0\\0\\0\\-1\\0\\1}.
\end{align*}
The vector $\mathbf{x}$ is arbitrary in $W_1$, so we've shown that
$$W_1 \subseteq \operatorname{span}\left\{\pmatrix{-1\\1\\0\\0\\0\\0},\pmatrix{-1\\0\\1\\0\\0\\0},\pmatrix{0\\0\\0\\-1\\1\\0},\pmatrix{0\\0\\0\\-1\\0\\1}\right\}.$$
Equality holds, because every vector in the spanning set also lies in $W_1$. It's also easy to see the above is linearly independent, so we have a basis, but this is not necessary to observe!
Next, once you have spanning sets for $W_1$ and $W_2$, you can form a spanning set for $W_1 + W_2$ by unioning the sets; every vector in $W_1 + W_2$ is the sum of a vector in $W_1$ (a linear combination of the first spanning set) and a vector in $W_2$ (a linear combination of the second spanning set), so the union will span the sum. That is,
$$W_1 + W_2 = \operatorname{span}\left\{\pmatrix{-1\\1\\0\\0\\0\\0},\pmatrix{-1\\0\\1\\0\\0\\0},\pmatrix{0\\0\\0\\-1\\1\\0},\pmatrix{0\\0\\0\\-1\\0\\1}, \pmatrix{1\\-1\\1\\-1\\1\\-1},\pmatrix{1\\0\\2\\1\\0\\0},\pmatrix{1\\0\\-1\\-1\\0\\1},\pmatrix{2\\1\\0\\0\\0\\0}\right\}.$$
This is already a technically correct description of $W_1 + W_2$, but it's usually best to reduce the spanning set down to a basis. There are two standard ways to do this, both involving row-reduction:
*
*Place them as rows in a matrix, row-reduce down to row-echelon form (reduced, if you prefer) and keep only the non-zero rows. These rows will be linearly independent, but retain the same span as the original set, thus producing a basis, or
*Place them as columns in a matrix, and row-reduce down to row-echelon form. Note the columns where the pivots (leading $1$s) appear, and retain only the vectors from the original set that you placed in those columns.
Either way, we get an element in the basis of $W_1 + W_2$ for every pivot in the row-reduced matrix. And, no matter which method you use, you will get $6$ pivots, which tells you that the dimension of $W_1 + W_2$ is $6$, and must be all of $\Bbb{R}^6$. Indeed, if you apply the first method, reducing until reduced row-echelon form, you should find that you get back the standard basis!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is there other method to evaluate $\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x, \textrm{ where }n\in N?$ Letting $x\mapsto \frac{1}{x}$ transforms the integral into
$\displaystyle I=\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x=-\int_0^1 \frac{x^{2 n-1} \ln x}{x^n+1} d x \tag*{} $
Splitting the integrand into two pieces like
$\displaystyle I=- \underbrace{\int_0^1 x^{n-1} \ln x d x}_{J} + \underbrace{\int_0^1 \frac{x^{n-1} \ln x}{x^n+1} d x}_{K} \tag*{} $
For the integral $J,$ letting $z=-n\ln x$ transforms $J$ into
$\displaystyle J= -\frac{1}{n^2} \int_0^{\infty} z e^{-z} d z =-\frac{1}{n^2}\tag*{} $
For integral $K$, using the series for $|x|<1,$
$\displaystyle \ln (1+x)=\sum_{k=0}^{\infty} \frac{(-1)^k}{k+1} x^{k+1},\tag*{} $
we have
$\displaystyle \begin{aligned}K& =-\frac{1}{n} \sum_{k=0}^{\infty} \frac{(-1)^k}{k+1} \int_0^1 x^{n(k+1)-1} d x \\& =-\frac{1}{n^2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2} \\& =-\frac{1}{n^2}\left[\sum_{k=1}^{\infty} \frac{1}{k^2}-2 \sum_{k=1}^{\infty} \frac{1}{(2 k)^2}\right] \\& =-\frac{1}{2 n^2} \cdot \frac{\pi^2}{6} \\& =-\frac{\pi^2}{12 n^2}\end{aligned}\tag*{} $
Putting them back yields
$\displaystyle \boxed{ I=\frac{1}{12 n^2}\left(12-\pi^2\right)}\tag*{} $
Is there alternative method? Comments and alternative methods are highly appreciated.
| Thanks to @Quanto’s short solution and @Claude’s generalisation using hypergeometric function. I am going to use Feynman’s Integration technique to prove the generalisation
$$
\boxed{I=\int_1^{\infty} \frac{\ln x}{x^m\left(1+x^n\right)} d x=\frac{1}{4 n^2}\left[\zeta\left(2, \frac{m+n-1}{2 n}\right)-\zeta\left(2, \frac{m+2 n-1}{2 n}\right)\right]}
$$
Letting $x=t^{-\frac{1}{n}}$ transforms the integral into
$$
I=-\frac{1}{n^2} \int_0^1 \frac{t^{\frac{m-1}{n}} \ln t}{1+t} d t= -\frac{1}{n^2}\left.\frac{\partial}{\partial b} J(b)\right|_{b=\frac{m-1}{n}},
$$
where $J(b)=\int_0^1 \frac{t^b}{1+t} d t.$
Using power series, we have
$$
J(b) =\int_0^1 \frac{t^b}{1+t} d t=\sum_{k=0}^{\infty}(-1)^k \int_0^1 t^{k+k} d t=\sum_{k=0}^{\infty} \frac{(-1)^k}{b+k+1}
$$
Differentiating both sides w.r.t. $b$ and putting $b=\frac{m-1}{n}$ gives
$$
\begin{aligned}
\int_1^{\infty} \frac{\ln x}{x^m\left(1+x^n\right)} d x & = \frac{1}{n^2}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{\left(k+\frac{m+n-1}{n}\right)^2} \\
& =\frac{1}{4 n^2}\left[\zeta\left(2, \frac{m+n-1}{2 n}\right)-\zeta\left(2, \frac{m+2 n-1}{2 n}\right)\right],
\end{aligned}
$$
which matches exactly @Cluade’s answer
$$\color{blue}{J=\frac 1{4n^2}\left(\psi ^{(1)}\left(\frac{m+n-1}{2 n}\right)-\psi ^{(1)}\left(\frac{m+2
n-1}{2 n}\right)\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4609287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the minimum of $\sqrt{\cos x+3}+\sqrt{2\sin x+7}$ without derivative How do we find the minimum of
$$f(x)=\sqrt{\cos x+3}+\sqrt{2\sin x+7}$$
without using derivatives?
This problem is probably related to circles of Apollonius.
I have tried AM-GM and Cauchy-Schwarz inequality but I can't work it out.
Anyway, I have solved it in a more geometric way. Here's my answer.
Firstly we can do some identical transformation.
$$f(x)=\dfrac{\sqrt{2}}{2}(\sqrt{(\cos x+1)^2+(\sin x)^2+4}+\sqrt{(\cos x)^2+(\sin x+2)^2+9})$$
So that it makes sense in geometry.
$P(\cos x,\sin x)$ is on the circle $x^2+y^2=1$, and the value of $f(x)$ equals to sum of the distance from $A(0,-2)$ to $P$ and from $B(-1,0)$ to $P$.
In other words:
$$f(x)=\dfrac{\sqrt{2}}{2}(\sqrt{|PB|^2+4}+\sqrt{|PA|^2+9}).$$
And here we can use Minkowski inequality.
$$f(x)\geq \dfrac{\sqrt{2}}{2} \sqrt{(|PA|+|PB|)^2+25}$$
When $P$,$A$,$B$ is collinear, $RHS$ gets the minimum. Meanwhile, $LHS = RHS$.
Therefore, $f(x)_{min}=\sqrt{15}$.
| Because $$\sqrt{\cos x+3}+\sqrt{2\sin x+7}=$$
$$=\sqrt{15}+\left(\sqrt{\cos x+3}-2\sqrt{\frac{3}{5}}+\frac{5\sqrt5}{24\sqrt3}\left(\cos^2x-\frac{9}{25}\right)\right)+$$
$$+\left(\sqrt{2\sin x+7}-3\sqrt{\frac{3}{5}}+\frac{5\sqrt5}{24\sqrt3}\left(\sin^2x-\frac{16}{25}\right)\right)\geq\sqrt{15}.$$
The equality occurs for $(\cos{x},\sin{x})=\left(-\frac{3}{5},-\frac{4}{5}\right).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 1
} |
Solving $f\left( n \right) = \sum\limits_{i > j \ge 0} {{}^{n + 1}{C_i}{}^n{C_j}} $
My approah is as follow
Let $f\left( n \right) = \sum\limits_{i > j \ge 0} {{}^{n + 1}{C_i}{}^n{C_j}} $
$f\left( n \right) = \sum\limits_{i = i}^n {{}^{n + 1}{C_i}} \sum\limits_{j = 0}^i {{}^i{C_j}} \Rightarrow f\left( n \right) = \sum\limits_{i = i}^{n + 1} {{2^i}.{}^{n + 1}{C_i}} $
${\left( {1 + 2x} \right)^{n + 1}} = {}^{n + 1}{C_0} + {}^{n + 1}{C_1}\left( {2x} \right) + ... + {}^{n + 1}{C_{n + 1}}{\left( {2x} \right)^{n + 1}} = {3^{n + 1}}\left\{ {x = 1} \right\}$
Not able to proceed
| The summand is the coefficient of $x^{i-j}$ from the following expression:
$$
(1+x)^{n+1}\left(1+\frac{1}{x}\right)^{n}
=
\frac{(1+x)^{2n+1}}{x^{n}}
$$
Since $i-j>1$, we just need to sum the coefficients of $x^{n+1},...,x^{2n+1}$ from the numerator on RHS.
$$
\binom{2n+1}{n+1}+\binom{2n+1}{n+2}+...+\binom{2n+1}{2n+1}=2^{2n}
$$
Seems like all the options are correct except (B)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integrating $\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}$
I'm struggling with the integral, $$\int\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}dx.$$
I tried it as follows:
Substituting $x-2 = \frac1t \implies dx = \frac{-dt}{t^2}.$
$$\therefore \int\frac{dx}{(x-2)^4 \sqrt{x^2 + 6x + 2}} = \int \frac{- dt}{\frac{t^2}{t^4} \sqrt{(\frac1t + 2)^2 + 6 (\frac1t + 2) + 2}} = \int \frac{-t^3}{\sqrt{18t^2 + 10t + 1}}\ dt$$
How to continue from here?
| Substitute $y=x-2$ to rewrite the integral as
$$I_4= \int\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}dx
=\int \frac{1}{y^4 \sqrt{y^2 + 10y + 18}}dy $$
and then integrate by parts to get the reduction formula
$$18(n-1)I_n=-\frac{\sqrt{y^2 + 10y + 18}}{y^{n-1}}-5(2n-3)I_{n-1}-(n-2)I_{n-2}
$$
Apply the formula three times to reduce it to
$$I_4=\frac1{54} \left(-\frac1{y^3} + \frac{25}{36y^2}-\frac{101}{216y}\right)\sqrt{y^2 + 10y + 18}-\frac{355}{11664}I_1
$$
where
$$I_1= \int\frac{1}{y \sqrt{y^2 + 10y + 18}}dy
=-\frac1{\sqrt{18}}\tanh^{-1}\frac{\sqrt{18}\sqrt{y^2 + 10y + 18} }{5y+18}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove $|f(y) − f(x)| \leq f(|y − x|)$ if $|y − x| ≤ 1/2$ given $f(x)=-x\log_2 x$
How do I prove that whenever $|y − x| ≤ 1/2$ it follows that $|f(y) − f(x)| \leq f(|y − x|)$ given $f(x)=-x\log_2 x$?
where $x,y\in[0,1]$
The graph of $f(x)=-x\log_2 x$ function is
$$
f'(x)=-\log x-\frac{x}{x\ln 2}=-\log x-\frac{1}{\ln 2}=0\implies \log x=-\frac{1}{\ln 2}\\
\frac{\ln x}{\ln 2}=-\frac{1}{\ln 2}\implies \ln x=-1\implies x=1/e\approx 0.3679
$$
I was only able to write the following proof:
When $0\le x\le y\le 1$,
\begin{align}
|f(x)-f(y)|&=|-x\log x+y\log y|\\
&=|-x\log x+\frac{x}{\ln 2}+y\log {y}-\frac{y}{\ln 2}+\frac{y}{\ln 2}-\frac{x}{\ln 2}|\\
&\leq |-(x\log x-\frac{x}{\ln 2})+(y\log y-\frac{y}{\ln 2})|+\frac{|y-x|}{\ln 2}\\
&=|\int_x^y \log tdt|+\frac{|y-x|}{\ln 2}=-\int_x^y \log tdt+\frac{|y-x|}{\ln 2}\\
&=-\int_x^{x+(y-x)} \log tdt+\frac{|y-x|}{\ln 2}\\
&\leq -\int_0^{y-x} \log tdt+\frac{|y-x|}{\ln 2}\\
&=-\Big(t\log t-t\Big)_0^{y-x}+\frac{|y-x|}{\ln 2}\\
&=-(y-x)\log(y-x)+(y-x)+\frac{|y-x|}{\ln 2}\\
&=f(|y-x|)+|y-x|+\frac{|y-x|}{\ln 2}
\end{align}
My Attempt
Thanks @Balajisb for the hint.
If $f$ is a concave function then $f(a+b)\leq f(a)+f(b)$ for all $a,b>0$
$$
-y\log y=-(y-x+x)\log(y-x+x)\le-(y-x)\log(y-x)-x\log x\\
-y\log y-(-x\log x)\le -(y-x)\log(y-x)\\
f(y)-f(x)\le f(y-x)
$$
$$
D_{\log x}\in(0,\infty]\implies x,y,y-x\geq 0\implies 1\ge y>x>0\\
$$
$$
1>y-x>0\implies -(y-x)\log(y-x)>0
$$
$$
f(y)-f(x)\le f(|y-x|) \;\forall\;(x,y)\;|\;y>x>0\\f(x)-f(y)\le f(|y-x|) \;\forall\;(x,y)\;|\;x>y>0
$$
Therefore,
$$
|f(y)-f(x)|\le f(|y-x|) \;\forall\;(x,y)\;|\;y>x>0\;\&\;f(y)>f(x)\\|f(y)-f(x)|\le f(|y-x|) \;\forall\;(x,y)\;|\;y<x<0\;\&\;f(y)<f(x)
$$
In order to prove $|f(y)-f(x)|\le f(|y-x|) \;\forall\;x,y>0$ we need to also consider the cases $y>x>0\;\&\;f(y)<f(x)$ and $x>y>0\;\&\;f(x)<f(y)$. So I think that's where the condition $y-x\leq 1/2$ lies in.
Case 1 : $y>x>0\;\&\;f(y)<f(x)$
$$
-y\log y<-x\log x\implies y\log y>x\log x\\
x\log x-y\log y<0\\x\log x-\frac{x}{\ln 2}-y\log y+\frac{y}{\ln 2}-\frac{y-x}{\ln 2}<0\\
\int_y^x \log t dt-\frac{y-x}{\ln 2}>0\\
y-x<\ln 2\int_y^x \log t dt=-\ln 2\int_x^y \log t dt=-\ln 2\int_x^y \frac{\ln t}{\ln 2}dt=-\int_x^y \ln t dt\\
<-\int_0^1 \log t dt=-1\times -1=1
$$
How do I obtain the condition $y-x\leq 1/2$ in this case ?
| Here is a proof.
WLOG, assume that $x \le y$.
We need to prove that
$$|-y\ln y + x\ln x| \le -(y - x)\ln(y - x). \tag{1}$$
Using the identity for $u \ge 0$ (easy to prove)
$$u\ln u =\int_0^1 \frac{u(u - 1)}{1+(u-1)t}\,\mathrm{d} t,$$
(1) is written as
$$\left|\int_0^1 \left(\frac{y(1-y)}{1+(y-1)t} - \frac{x(1-x)}{1+(x-1)t}\right)\,\mathrm{d} t\right|
\le \int_0^1 \frac{(y-x)(1 - (y-x))}{1 + (y-x - 1)t}\,\mathrm{d} t.$$
It suffices to prove that
$$\int_0^1 \left|\frac{y(1-y)}{1+(y-1)t} - \frac{x(1-x)}{1+(x-1)t}\right|\,\mathrm{d} t
\le \int_0^1 \frac{(y-x)(1 - (y-x))}{1 + (y-x - 1)t}\,\mathrm{d} t.$$
It suffices to prove that, for all $t \in [0, 1]$ and $0 \le x \le y \le 1$ with $y - x \le 1/2$,
$$\left|\frac{y(1-y)}{1+(y-1)t} - \frac{x(1-x)}{1+(x-1)t}\right|
\le \frac{(y-x)(1 - (y-x))}{1 + (y-x - 1)t}$$
or
$$\left(\frac{y(1-y)}{1+(y-1)t} - \frac{x(1-x)}{1+(x-1)t}\right)^2
\le \left(\frac{(y-x)(1 - (y-x))}{1 + (y-x - 1)t}\right)^2$$
or (clearing the denominators and simplifying)
$$xy[1 - 2(y - x)]t^2 + (2x^2 - xy - 2y^2 + 2y)t(1 - t) + 2(1-y)(1-t)^2 \ge 0 \tag{2}$$
which is true (the proof is given at the end).
We are done.
Proof of (2):
It suffices to prove that $2x^2 - xy - 2y^2 + 2y \ge 0$.
If $0 \le x \le 1/3$, then
$$2x^2 - xy - 2y^2 + 2y
\ge 2x^2 - xy - 2y\cdot (x + 1/2) + 2y = 2x^2 + y(1 - 3x) \ge 0.$$
If $1/3 < x < 1/2$, then
$$2x^2 - xy - 2y^2 + 2y
\ge 2x^2 - xy - 2y\cdot (x + 1/2) + 2y = 2x^2 - y(3x - 1) $$
$$\ge 2x^2 - (x + 1/2)(3x - 1) = \frac12(1 + x)(1 - 2x) \ge 0.$$
If $1/2 \le x \le 1$, then $2x^2 - xy - 2y^2 + 2y
= x(2x - y) + 2y(1 - y) \ge 0$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4614700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Solve the equation $\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$ Solve the equation $$\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$$
We have $$D_x:\begin{cases}1-2x>0\\6x^2-5x+1>0\\1-3x>0\\1-3x\ne1\\4x^2-4x+1>0\iff(2x-1)^2>0\iff x\ne\dfrac12\end{cases}\iff x\in(-\infty;0)\cup(0;\dfrac{1}{3})$$
Also the quadratic $6x^2-5x+1$ factors as $(2x-1)(3x-1)$. The equation then becomes $$\log_{1-2x}(2x-1)(3x-1)-\log_{1-3x}(2x-1)^2=2\\\log_{1-2x}(2x-1)(3x-1)-2\log_{1-3x}(1-2x)=2,$$ as $\log_{1-3x}(2x-1)^2=2\log_{1-3x}|2x-1|,$ but we know from $D_x$ that $2x-1<0$,
$$\log_{1-2x}(2x-1)+\log_{1-2x}(3x-1)-\dfrac{2}{\log_{1-2x}(1-3x)}=2$$ I don't know what to do next.
| We have that for $1-2x>0$, $1-3x>0$, $1-2x\neq 1$, $1-3x\neq 1$
$$\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2 $$
$$\iff \frac{\log((1-2x)(1-3x))}{\log (1-2x)}-2\frac{\log (1-2x)}{\log (1-3x)}=2$$
$$\iff \frac{\log (1-3x)}{\log (1-2x)}-2\frac{\log (1-2x)}{\log (1-3x)}=1$$
then by $u= \frac{\log (1-3x)}{\log (1-2x)}$ we obtain
$$u-\frac 2 u =1 \implies u = \frac{\log (1-3x)}{\log (1-2x)}=2 \implies x=\frac14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find minimum value of $(ab-2a+4)^2 + (bc-2b+4)^2+ (ca-2c+4)^2$ where $0 \leq a,b,c \leq2$ Find minimum value of $(ab-2a+4)^2 + (bc-2b+4)^2+ (ca-2c+4)^2$ where $0 \leq a,b,c \leq2$
For this kind of problems it's usually easy to guess that the minimum value is obtained at $a=b=c$ or when the answer is at the boundary.
When $a=b=c$ we can easily find minimum = 27. When we are at the boundary, we let $c=2$ we can see the expression is greater than or equal to 24. But this solution seems hacky and i'd like to find a better way.
| For $(a,b.c)=(2,1,0)$ we'll get a value $24$.
We'll prove that it's a minimal value.
For $a=2$ by C-S we obtain:
$$\sum_{cyc}(ab-2a+4)^2=(2b)^2+(bc-2b+4)^2+16=$$
$$=\frac{1}{2}(1+1)((2b)^2+(bc-2b+4)^2)+16\geq\frac{1}{2}(2b+bc-2b+4)^2+16=$$
$$=\frac{1}{2}(bc+4)^2+16\geq8+16=24,$$ which says that it's enough to solve our problem for $\{a,b,c\}\subset[0,2).$
Now, let $a=\frac{2x}{x+1},$ $b=\frac{2y}{y+1}$ and $c=\frac{2z}{z+1},$ where $x$, $y$ and $z$ are non-negatives.
Thus, we need to prove that:
$$\sum_{cyc}\left(\frac{4xy}{(x+1)(y+1)}-\frac{4x}{x+1}+4\right)^2\geq24$$ or
$$2\sum_{cyc}\frac{(xy+y+1)^2}{(x+1)^2(y+1)^2}\geq3$$ or
$$\sum_{cyc}(x^2y^2z^2+2x^2y^2z+x^2y^2-2x^2y+2x^2z+x^2+2x+1)\geq0,$$ which is true because
$$\sum_{cyc}(x^2y^2z^2+2x^2y^2z+x^2y^2-2x^2y+2x^2z+x^2+2x+1)>$$
$$>\sum_{cyc}(x^2y^2-2x^2y+x^2)=\sum_{cyc}x^2(y-1)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to derive a closed form of a recursion (maybe using generating functions) Let $a_0=9$ and consider the following recurrence relation: $$a_n=36(n+1)2^{n-2}+2a_{n-1},$$ I'm looking for the closed form of $\{a_n\}.$
I have tried using generator functions:
\begin{align*}
f(x)&=\sum_{n=0}^\infty a_nx^n\\
&=9 +\sum_{n=1}^\infty \left(4.5(n+1)2^{n+1}+2a_{n-1}\right)x^n\\
&=9 +4.5\sum_{n=1}^\infty (n+1)2^{n+1}x^n+2\sum_{n=1}^\infty a_{n-1}x^n\\
&=9 +\frac{4.5}{x}\sum_{n=1}^\infty (n+1)2^{n+1}x^{n+1} + 2x\sum_{n=1}^\infty a_{n-1}x^{n-1}\\
&=9 +\frac{4.5}{x}\sum_{n=2}^\infty n2^{n}x^{n} + 2xf(x)
\end{align*}
What should I do next? Is there a quicker way to solve this?
Thanks in advance
| We are given that
$$
a_n = 36 (n+1) 2^{n-2} + 2a_{n-1}, \tag{0}
$$
we find that, for any positive integer $n$, we have
\begin{align}
a_n &= 36 (n+1) 2^{n-2} + 2a_{n-1} \\
&= 36 \left( 2^{n-2} \right) (n+1) + 2^1 a_{n-1} \tag{1} \\
&= 36 (n+1)2^{n-2} + 2 \left( 36 (n-1+1) 2^{n-1-2} + 2 a_{n-2} \right) \\
& \qquad \mbox{[using $n-1$ in place of $n$ in (0) above ]} \\
&= 36 \left( 2^{n-2} \right) \big( (n+1) + n \big) + 2^2 a_{n-2} \tag{2} \\
&= 36 \left( 2^{n-2} \right) \big( (n+1) + n \big) + 2^2 \left( 36 (n-2+1) 2^{n-2-2} + 2a_{n-2-1} \right) \\
& \qquad \mbox{[ using $n-2$ in place of $n$ in (0) above ]} \\
&= 36 \left( 2^{n-2} \right) \big( (n+1) + n + (n-1) \big) + 2^3 a_{n-3} \tag{3} \\
&= 36 \left( 2^{n-2} \right) \big( (n+1) + n + (n-1) \big) + 2^3 \left( 36 (n-3+1) 2^{n-3-2} + 2a_{n-3-1} \right) \\
&\qquad \mbox{[ using $n-3$ in place of $n$ in (0) above ]} \\
&= 36 \left( 2^{n-2} \right) \big( (n+1) + n + (n-1) + (n-2) \big) + 2^4 a_{n-4} \tag{4} \\
&= \ldots \\
&= 36 \left( 2^{n-2} \right) \big( (n+1) + n + (n-1) + (n-2) + \ldots 2 \big) + 2^n a_0 \\
&\qquad \mbox{[ using the pattern suggested by (1), (2), (3), and (4) above ]} \\
&= 36 \left( 2^{n-2} \right) \big( 1 + 2 + \ldots + (n+1) -1 \big) + 2^n (9) \\
&= 9 \left( 2^n \right) \left( \frac{(n+1)(n+2)}{2} - 1 +1 \right) \\
&= 9 \left( 2^n \right) \frac{(n+1)(n+2)}{2} \\
&= 9 (n+1)(n+2) \left( 2^{n-1} \right).
&=
\end{align}
Finally, we can use induction to verify that this formula is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $8^x+3\cdot2^{2-x}=1+2^{3-3x}+3\cdot2^{x+1}$ Solve the equation $$8^x+3\cdot2^{2-x}=1+2^{3-3x}+3\cdot2^{x+1}$$ The given equation is equivalent to $$2^{3x}+\dfrac{12}{2^x}=1+\dfrac{8}{2^{3x}}+6\cdot2^x$$ If we put $a:=2^x>0$, the equation becomes $$a^3+\dfrac{12}{a}=1+\dfrac{8}{a^3}+6a$$ which is $$a^6-6a^4-a^3+12a^2-8=0$$ The LHS factors as $(a+1)(a-2)(a^4+a^3-3a^2-2a+4)$, which is in no case obvious. Let's say that we find the roots $1$ and $-2$, then how do we show that $(a^4+a^3-3a^2-2a+4)$ does not factor any more? Taking these into consideration, I believe there is an another approach. Any ideas would be appreciated.
| Hint
Let's start from
$$a^3+\dfrac{12}{a}=1+\dfrac{8}{a^3}+6a$$
Now rewrite it as
$$\left[a^3-\left(\frac2a\right)^3\right]-6\left[a-\frac 2a\right]-1=0$$
$$\left(a-\frac2a\right)\left[a^2+\frac{4}{a^2}-4\right]-1=0$$
$$\left(a-\frac2a\right)^3-1=0$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\lim\limits_{x\to \infty}[f(x)-f(x-1)]\overset{?}{=}e$ Let :
$$f\left(x\right)=\int_{0}^{\lfloor x\rfloor}\prod_{n=1}^{\lfloor x\rfloor}\frac{\left(y+2n\right)\ln\left(y+2n-1\right)}{\left(y+2n-1\right)\ln\left(y+2n\right)}dy$$
Conjecture:
$$\lim_{x\to \infty}f(x)-f(x-1)=e$$
We have for $x=150000$:
$$f(x)-f(x-1)\approx 2.714446$$
I cannot proceed further because Desmos is a bit unfriendly.
How i come up with this conjecture :
Have a look to Andersson's inequality https://www.sciencedirect.com/science/article/pii/S0893965905003666 :
$$\int_{0}^{1}f_{1}\left(x\right)f_{2}\left(x\right)...f_{n}\left(x\right)dx\ge\frac{2n}{n+1}\left(\int_{0}^{1}f_{1}\left(x\right)dx\right)\left(\int_{0}^{1}f_{2}\left(x\right)dx\right)...\left(\int_{0}^{1}f_{n}\left(x\right)dx\right)$$
here :
$$f_{n}\left(x\right)=\frac{\left(x+2n\right)\ln\left(x+2n-1\right)}{\left(x+2n-1\right)\ln\left(x+2n\right)}$$
It doesn't fullfilled the constraint $f_n(0)=0$ and the convexity for $n$ small on $x\in[0,1]$ .
On the other hand the use of the floor function is just a pratical graph point of view on Desmos (free software) enlightening the probable existence of an asymptote .
Does it converge? If yes, is it $e$?
A counter-example is also welcome!
Ps : Something weird should be :$\lim_{x\to\infty}f(x)-f(x-1)=\frac{egg}{gg}$
| Here is a weaker result which is still enough to show that the limit is not $e$:
Claim. We have
$$ \lim_{n\to\infty} \frac{f(n)}{n} = \sqrt{3} + 2\log(1+\sqrt{3}) - \log 2 \approx 3.04901 $$
Proof. Let $n$ be a positive integer. Then
$$ f(n) = \int_{0}^{n} \prod_{j=1}^{n} \frac{\log(y+2j-1)}{\log(y+2j)} \frac{y+2j}{y+2j-1} \, \mathrm{d}y. $$
Substituting $y = nt$, the integral is recast as
\begin{align*}
f(n)
&= n \int_{0}^{1} g_n(t) \, \mathrm{d}t,
\qquad g_n(t) := \prod_{j=1}^{n} \frac{\log(nt+2j-1)}{\log(nt+2j)} \frac{nt+2j}{nt+2j-1}. \tag{*}
\end{align*}
Using the inequality $\frac{1}{1-x} \leq \exp(x + x^2) $ for $x \in [0, \frac{1}{2}]$, we find that
\begin{align*}
g_n(t)
&= \prod_{j=1}^{n} \frac{\log(nt+2j-1)}{\log(nt+2j)} \frac{1}{1 - \frac{1}{nt+2j}} \\
&\leq \exp\left[ \sum_{j=1}^{n} \left( \frac{1}{nt+2j} + \frac{1}{(nt+2j)^2} \right) \right] \\
&\leq \exp\left[ \int_{0}^{n} \frac{1}{nt+2s} \, \mathrm{d}s + \sum_{j=1}^{\infty} \frac{1}{(2j)^2} \right] \\
&= \exp\left[ \frac{1}{2} \log \left(\frac{t+2}{t}\right) + \frac{\zeta(2)}{4} \right].
\end{align*}
This proves that $g_n(t) \leq Ct^{-1/2}$ uniformly in $n$, and so, we can apply the dominated convergence theorem provided $g_n(t)$ converges pointwise as $n \to \infty$. However, for each fixed $t \in (0, 1]$,
\begin{align*}
g_n(t)
&= \prod_{j=1}^{n} \biggl( 1 + \frac{\log(1 - \frac{1}{nt + 2j})}{\log(nt+2j)} \biggr) \frac{1}{1 - \frac{1}{nt+2j}} \\
&= \exp \left[ \sum_{j=1}^{n} \biggl( \frac{1}{nt + 2j} + \mathcal{O}\biggl( \frac{1}{n \log n} \biggr) \biggr) \right] \\
&= \exp \left[ \sum_{j=1}^{n} \frac{1}{t + 2(j/n)} \frac{1}{n} + \mathcal{O}\left( \frac{1}{\log n} \right) \right] \\
&\to \exp\left( \int_{0}^{1} \frac{\mathrm{d}s}{t + 2s} \right)
= \sqrt{\frac{t + 2}{t}}.
\end{align*}
Therefore, by the dominated convergence theorem,
$$ \lim_{n\to\infty} \frac{f(n)}{n}
= \int_{0}^{1} \sqrt{\frac{t + 2}{t}} \, \mathrm{d}t
= \boxed{\sqrt{3} + 2\log(1+\sqrt{3}) - \log 2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4625047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\frac{1}{x^2}$ is continuous. I wanted to prove that the function $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $f(x)=\frac{1}{x^2}$ is continuous for all $x\in\mathbb{R}$ excluding x=0 of course. The proof goes as follows.
Let $\epsilon>0$ be given arbitrary and choose $\delta=min(1,\frac{c}{2},\frac{c^4\epsilon}{4(1+2|c|)})$. Assume that $\forall x.c\in\mathbb{R}: 0<|x-c|<\delta$. It follows:
$$|f(x)-f(c)|=|\frac{1}{x^2}-\frac{1}{c^2}|=|\frac{x^2-c^2}{x^2c^2}=|\frac{|x-c||x+c|}{x^2c^2}|=|\frac{|x-c||(x-c)+2c|}{x^2c^2}|\leq\frac{|x-c|(|x-c|+2|c|)}{x^2c^2}$$ $$ \leq\frac{|x-c|(1+2|c|)}{x^2c^2}<\frac{|x-c|4(1+2|c|)}{c^2c^2}<4\delta\frac{1+2|c|}{c^4}\leq\epsilon$$
Is this proof valid and in particular is the delta I chose fine and if not provide some tips please.
| Let $\epsilon>0$ is arbitrary. We will consider $c>0$ (for $c<0$ we can follow similar logic).
$$|f(x)-f(c)|=|\frac{1}{x^2}-\frac{1}{c^2}|=|\frac{x^2-c^2}{x^2c^2}|=|x-c|\frac{|x+c|}{x^2c^2}$$ From here, we need to find bounds for $x \text{ and }|x+c|$ than do not depend on $x$. Let $\delta_1=0.5c$. Then $|x-c|<\delta_1 \implies 0.5c < x < 1.5c, 1.5c <x+c < 2.5c$ We have for $|x-c|<\delta_1$:
$$|f(x)-f(c)|=|x-c|\frac{|x+c|}{x^2c^2} < \delta_1 \frac{2.5c}{0.25c^4}=\frac{10\delta_1}{c^3}$$
Next, pick $\large{\delta=min\{\delta_1, \frac{\large{c^3}\epsilon}{10}\}}$ then for $|x-c| < \delta$ we have $|f(x)-f(c)| < \epsilon$
| {
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Complex binomial series $$\sum_{r=0}^n \left[\frac{r+1}{r+2} (^n C_r) (x^r)\right] $$
Can someone help me evaluate this summation? I could solve till a certain extent but was then stuck. Would really help! so i started by differentiating $x(x+1)^n$ and got the coefficient of $(r+1)$. then I integrated the function I got above to get the coefficient of $(r+2)$ in the denominator but it was going a bit lengthy and i even got an incorrect expression at the end. so just wanted to know if there's any fallacy in my method or there is another elegant method too.
The summation can also be written as:
$(1+x)^n$ - $$\sum_{r=0}^n \frac {^nC_r}{r+2} $$
is there a way to evaluate the second summation?
| Using $$\frac{r+1}{r+2} \binom{n}{r} = \frac{1}{n+1} \cdot\frac{(r+1)^2}{r+2} \binom{n+1}{r+1} = \frac{1}{(n+1)(n+2)}\cdot (r+1)^2\binom{n+2}{r+2}$$ and re-indexing, your sum is $$\frac{1}{(n+1)(n+2)}\sum_2^{n+2} (r-1)^2 \binom{n+2}{r} x^{r-2} \\=\frac{1}{(n+1)(n+2)}\sum_2^{n+2} r(r-1)\binom{n+2}{r} x^{r-2} - \frac{1}{(n+1)(n+2)}\sum_2^{n+2}(r-1)\binom{n+2}{r}x^{r-2} \\ = \frac{1}{(n+1)(n+2)}\frac{d^2}{dx^2}(1+x)^{n+2} - \frac{1}{(n+1)(n+2)} \frac{d}{dx}\frac{(1+x)^{n+2}-1-(n+2)x}{x} \\ \vdots$$
| {
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Calculate $\frac{2}{\alpha^2}-\frac{1}{(\alpha +1)^2}$ if $\alpha$ be a root of $x^2+(1-\sqrt3)x+1-\sqrt3=0$ If $\alpha$ is a root of $x^2+(1-\sqrt3)x+1-\sqrt3=0$ calculate
$$\frac{2}{\alpha^2}-\frac{1}{(\alpha +1)^2}$$
What I have done:
$$\begin{aligned}
\alpha^2+\alpha+1&=\sqrt3(\alpha+1)\\
&\implies\alpha^4+2\alpha^3+3\alpha^2+2\alpha+1=3\alpha^2+6\alpha+3\\
&\implies\alpha^2(\alpha+1)^2=\alpha^2+4\alpha+2\\
&\implies\frac{2}{\alpha^2}-\frac{1}{(\alpha +1)^2}=1
\end{aligned} $$
Is there any easy way to calculate?
| Another way (not necessarily easier than OP’s one).
$\sqrt3(\alpha+1)=\alpha(\alpha+1)+1$
$\dfrac{\sqrt3}{\alpha}=1+\dfrac1{\alpha(\alpha+1)}$
$\left(\dfrac{\sqrt3}{\alpha}\right)^2=\left(1+\dfrac1\alpha-\dfrac1{\alpha+1}\right)^2$
$\dfrac3{\alpha^2}=1+\dfrac1{\alpha^2}+\dfrac1{(\alpha+1)^2}\;\;$ (by simplifying twice products)
$\dfrac2{\alpha^2}-\dfrac1{(\alpha+1)^2}=1$
| {
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Proving that ${(1-\frac{2}{x^2})}^x < \frac{x-1}{x+1}$ for any $x > 2$. Proof that ${\left(1-\dfrac{2}{x^2}\right)}^x\!< \dfrac{x-1}{x+1}$ for any $x > 2$.
any ideas?
| Let $f(x)=\ln\frac{x-1}{x+1}-x\ln(1-2/x^2)$. It suffices to show that $f(x)>0$ for $x>2$, as $\ln$ is an increasing function. Using the series for $\ln(1+a)=a-a^2/3+a^3/3-a^4/4+\dots$, we get the asymptotic expansion
$$
\ln \frac{x-1}{x+1}=\ln \left(1-\frac1x\right)-\ln \left(1+\frac1x\right)=-\frac{2}{x}-\frac{2}{3x^3}-\frac{2}{5x^5}-\frac{2}{7x^7}-\dots
$$
and
$$
-x\ln\left(1-\frac2{x^2}\right)=\frac2x+\frac{2^2}2 \frac{1}{x^3}+\frac{2^3}3 \frac{1}{x^5}+\frac{2^4}4 \frac{1}{x^7}+\dots
$$
hence
$$
f(x)=\sum_{k=1}^\infty \frac 1{x^{2k+1}}\left(\frac{2^{k + 1}}{k + 1} - \frac2{2k+1}\right)>0
$$
since
$$
\frac{2^{k + 1}}{k + 1} - \frac2{2k+1}=2\frac{2^k(2k+1)-(k+1)}{(k+1)(2k+1)}>0.
$$
(Note that the inequality $f(x)>0$ holds for all $x>\sqrt 2$.)
| {
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Find intersection points of $x^2 - 3xy+ 2y^2 - x + 1 = 0$ and $y = \alpha x + \beta$ This question comes from exercise $1.3$ in Rational Points on Elliptic Curves (Silverman & Tate). I am self studying and trying to work some of the exercises. This one is giving me some trouble.
Let $C$ be the conic given by the equation
$$x^2 - 3xy+ 2y^2 - x + 1 = 0.$$
Let $L$ be the line $y = \alpha x + \beta$. Suppose that the intersection $L \cap C$ contains the point $\left(x_0, y_0\right)$. Assuming that the intersection consists of two distinct points, find the second point of $L \cap C$ in terms of $\alpha, \beta, x_0, y_0$.
We know the line intersections the conic at point $P = (x_0, y_0)$, so by the group law it also intersects at point $P^2$. I think my understanding of this is algebraically correct, but I don't know how to translate it to my analytic understanding and write $P^2$ in terms of$\alpha, \beta, x_0,$ and $y_0$.
Substituting $y$ yields
\begin{align*}
x^2 -3x(\alpha x +\beta) + 2(\alpha x +\beta)^2 -x + 1 &= 0\\
x^2 -3\alpha x^2 -3x\beta + 2\alpha^2x^2 + 4\alpha x\beta +2\beta^2 -x + 1 &= 0\\
\end{align*}
which doesn't seem to lead anywhere.
| Let's write your quadratic as
$$
(1 -3\alpha+2\alpha^2) x^2 -(1+3\beta - 4\alpha \beta)x +2\beta^2 + 1 = 0
$$
Now, by Vieta's formulas, the sum of the roots of the quadratic $ax^2 + b x + c$ is $-b/a$. So for this quadratic we have
$$
x_0 + x_1 = \frac{1+3\beta - 4\alpha \beta }{1 -3\alpha+2\alpha^2}
$$
Next we use the line equation to get $y_0 + y_1$:
$$
y_0 + y_1=\alpha (x_1 + x_0) +2\beta = \alpha \frac{1+3\beta - 4\alpha \beta}{1 -3\alpha+2\alpha^2}+2\beta = \frac{\alpha + 2\beta -3\alpha\beta}{1 -3\alpha+2\alpha^2}
$$
Solving then gives
$$
x_1 = \frac{1+3\beta - 4\alpha \beta }{1 -3\alpha+2\alpha^2} - x_0\\
y_1 = \frac{\alpha + 2\beta -3\alpha\beta}{1 -3\alpha+2\alpha^2} - y_0.
$$
Note that the denominator here is the coefficients of the quadratic form in the conic. I'm not sure of an intuitive way to get the coefficients in the numerator, though.
| {
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Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{3^n}\ge 1+\frac{2n}3$ Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{3^n}\ge 1+\frac{2n}3$
We start with $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac1{3^n}\ge 1+\frac{2n}3$$ for all positive integers.
I have finished the basic step, the hypothesis is easy too, but I do not know what to do for n+1:
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac1{3^n}+\frac1{3^{n+1}} \ge 1+\frac{2n}3 +\frac{1}{3^{n+1}}$$
Can you help me from here?
| The sum on the LHS is
$$S_n=\sum_{k=1}^{3^n} \frac1k$$
The hypothesis is that $S_n\ge 1+\frac{2n}{3}$
Base case $n=0$. As $S_0=1\ge1+0=1$, the case $n=0$ is true.
Assume true for $n$, prove
$$S_n+\sum_{k=3^n+1}^{3^{n+1}}\frac1k\ge 1+\frac{2n}{3}+\frac{2}{3}$$
As
$$\sum_{k=3^n+1}^{3^{n+1}}\frac1k\ge\sum_{k=3^n+1}^{3^{n+1}}\frac1{3^{n+1}}=\frac{2\cdot3^n}{3^{n+1}}=\frac23$$
is true at $n=0$, the induction holds.
| {
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Solving $|x-1|^{\log_2(4-x)} \le|x-1|^{\log_2 (1+x)}$ Let us solve
$|x-1|^{\log_2(4-x)}\le|x-1|^{\log_2 (1+x)}......(*)$
Let $4-x>0 ~\& ~1+x>0$.......(1)
Case 1: $|x-1|\le 1\implies 0\le x\le 2$........(2)
We get $\log_2(4-x) \color{red}{\ge} \log_2(1+x)\implies 4-x \ge 1+x \implies x\le 3/2$....(3)
The overlap of (1,2,3) gives $x\in [0,3/2]$........(4)
Case 2: $|x-1|\ge 1 \implies x\le 0 ~or~ x\ge 2$....(5)
We get $\log_2(4-x) \color{red}{\le} \log_2(1+x)\implies 4-x <1+x \implies x\ge 3/2$....(6)
Taking overlap of (1,5,6), we get $x \in [2,4)$.
So the final solution is : $[0,3/2] \cup [2,4).$
Now the question is whether this solution is complete and how else this (*) could be solved?
| $|x-1|^{\log_2(4-x)}\le|x-1|^{\log_2 (1+x)}......(1)$
Note that $x=0,1,2$ are already roots.
Taking $\log_a$ both side ( $a\in (0,1)$ or $a\in(1,\infty))$, let us take $a=2$. We get
$$\log_2(4-x) \log_2|x-1|\color{red}{\le} \log_2(1+x)\log_2|x-1|\quad(2)$$
Since, $\log_2(4-x)$ and $\log_2(1+x)$ have to be real we declare
$$(4-x)>0~\&~ (1+x)>0\quad(3)$$
Case 1: $|x-1|<1\implies 0<x< 2, \log_2|x-1|\color{red}{<}0\quad(4)$,
From (2) we get
$$\log_2(4-x) \color{red}{\ge} \log_2(1+x) \implies 4-x \ge 1+x \implies x \le 3/2\quad(5)$$
Taking intersection of (3,4,5) and inclusion of $x=0,1$; we get the solution as
$x\in [0,3/2]\quad (6)$
Case 2: $|x-1|> 1 \implies x<0 ~or~x > 2, \log_2|x-1|\color{red}{>}0 \quad (7)$
This time from (2), we get
$$\log_2(4-x) \color{red}{\le} \log_2(1+x) \implies 4-x \le 1+x \implies x\ge 3/2 \quad (8)$$
Finally, intersection of (3,7,8) and inclusion of $x=2$, gives the solution as $2\le x< 4 \quad (9)$
So the final solution lies in $[0,3/2] \cup [2,4)$.
Note that by choosing the base e.g., $1/2$ wont change the final answer.
| {
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Finding the maximum area of a triangle when the distance from a point to three vertices is fixed (1).$P$ is inside $\Delta ABC$ and $$PA=2,PB=7,PC=11.$$Let the area of the triangle be $S$, then find The maximum value of $S$.
(2).$P$ is inside $\Delta ABC$ and $$PA=x,PB=y,PC=z.$$Let the area of the triangle be $S$, then find The maximum value of $S$.
My approach: for (1),considering the fixed $B,C$, we can easily know that the triangle area is largest when the $AP$ is perpendicular to $BC$. By symmetry, we can know that $P$ is pendicular.
Of course, this is not rigorous. How can we show that when the area reaches the maximum, $P$ must be pendicular without partial derivative.
| For the first case
$PA = 2, PB = 7 , PC = 11 $
So we let $P$ be at the origin $(0,0)$ , and we draw three circles $A,B,C$ of radii $2, 7, 11$ respectively. Now we can select point $A$ to be at $(2, 0)$.
If point $B$ is fixed at $(x_2, y_2)$ , and we vary point $C$ along the perimeter of circle $C$, then to obtain the maximum area of $\triangle ABC$, point $C$ must have the maximum possible distance from the line segment $AB$, and this can only happen if the extension of segment $PC$ is perpendicular to segment $AB$, which is what you stated in the question. Extending this result to all three vertices, we deduce that in the maximum area triangle, $PA$ is perpendicular to $BC$ , and $PB$ is perpendicular to $AC$ , and $PC$ is perpendicular to $AB$. Hence point $P$ must be the orthocenter of $\triangle ABC$.
Now for the given values of radii, if $A$ is at $(2, 0)$, then $BC$ lies parallel to $y$ axis, i.e.
$B = (x, \sqrt{49 - x^2} )$
$ C = (x, -\sqrt{ 121 - x^2} )$
And we have to determine $x$ such that $B$ is perpendicular to $AC$
$ AC = (x - 2, - \sqrt{121 - x^2} ) $
Hence, by using the dot product
$ B \cdot AC = x (x - 2) - \sqrt{ (49 - x^2) (121 - x^2) } = 0 $
Hence,
$ x^2 - 2 x = \sqrt{ (49 - x^2)(121 - x^2) } $
Squaring
$ x^4 + 4 x^2 - 4 x^3 = (49)(121) - (170 x^2) + x^4 $
So that,
$ 4 x^3 - 174 x^2 + 5929 = 0 $
Solving gives the following solutions:
$-5.5 , 6.31346652052679 , 42.6865334794732 $
The valid root is the negative one: $-5.5$
Hence, $ BC = \sqrt{ 49 - (-5.5)^2 } + \sqrt{ 121 - (-5.5)^2 } = \sqrt{18.75} + \sqrt{90.75} $
Therefore, the maximum area is $ \frac{1}{2} (7.5) ( \sqrt{18.75} + \sqrt{90.75} ) = 30 \sqrt{3} $
| {
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Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge? If so, what is its sum?
| A proof without words by Matt Hudleson
| {
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Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!
Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
$2 \times 0 = 0.$ So it is divisible by $3.$
Induction: Assume that for an arbitrary natural number $n$,
$n^3+ 2n$ is divisible by $3.$
Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use
the induction hypothesis. Got it
$$( n + 1 )^3+ 2( n + 1 ) = ( n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$
$$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying
and regrouping}\}$$
$$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out
the 3}\}$$
which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$
by the induction hypothesis. What?
Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$
| The driving force behind induction is that you show that a base case (when $n = 1$, for example). Then you show that the hypothesis being true at some $k$ implies that it holds at $k+1$. Then, since you have verified the hypothesis at $n = 1$, you have it at $n = 2$. Then, since it holds at $n = 2$, it holds at $n = 3$, and so on. Note that the domain over which the hypothesis holds should be defined in the hypothesis itself.
Now, for your specific case, let's see how this works.
First, for the base case ($n = 1$), we see the following:
$$1^3 + 2 \cdot 1 = 3.$$
That is clearly divisible by $3$, so we have our base case.
Now assume that for all $k \geq 1$, $k^3 + 2k$ is divisible by $3$.
Then for $n = k + 1$, we have:
$$(k+1)^3 + 2(k+1) = k^3 + 3k^2 + 3k + 1 + 2k + 2 = k^3 + 3k^2 + 5k + 3$$
The $3k^2 + 3$ portion is clearly divisible by $3$, so we need only show that $k^3 + 5k$ is divisible by $3$. From the assumption above, we know that $k^3 + 2k = 3m$ for some positive integer $m$. Then,
$$k^3 + 5k = k^3 + 2k + 3k = 3m + 3k = 3(m + k),$$
so the hypothesis holds at $n = k+1$.
Thus, we have for all $n \geq 1$, $n^3 + 2n$ is divisible by $3$.
As the others have suggested, there are certainly other ways of showing the $(k+1)$th case, but hopefully this overall form helps you see how mathematical induction works.
| {
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Simplification of expressions containing radicals As an example, consider the polynomial $f(x) = x^3 + x - 2 = (x - 1)(x^2 + x + 2)$ which clearly has a root $x = 1$.
But we can also find the roots using Cardano's method, which leads to
$$x = \sqrt[3]{\sqrt{28/27} + 1} - \sqrt[3]{\sqrt{28/27} - 1}$$
and two other roots.
It's easy to check numerically that this expression is really equal to $1$, but is there a way to derive it algebraically which isn't equivalent to showing that this expression satisfies $f(x) = 0$?
| Pardon my skepticism, but has anyone so much as breadboarded Blömer '92 or Landau '93 in all these 18 years? For lack of same, people still publish ugly surdballs, e.g.,
$$\vartheta _3\left(0,e^{-6 \pi }\right)=\frac{\sqrt[3]{-4+3 \sqrt{2}+3 \sqrt[4]{3}+2 \sqrt{3}-3^{3/4}+2 \sqrt{2}\, 3^{3/4}} \sqrt[4]{\pi }}{2\
3^{3/8} \sqrt[6]{\left(\sqrt{2}-1\right) \left(\sqrt{3}-1\right)} \Gamma \left(\frac{3}{4}\right)}$$
(J. Yi / J. Math. Anal. Appl. 292 (2004) 381–400, Thm 5.5 vi) instead of
$$\vartheta _3\left(0,e^{-6 \pi }\right)=\frac{\sqrt{2+\sqrt{2}+\sqrt{2} \sqrt[4]{3}+\sqrt{6}} \,\sqrt[4]{\pi }}{2\ 3^{3/8} \Gamma
\left(\frac{3}{4}\right)}\quad .$$
And why do both papers trot out the same old Ramanujan denestings instead of new and interesting ones? E.g.,
$$\sqrt{2^{6/7}-1}=\frac{2^{8/7}-2^{6/7}+2^{5/7}+2^{3/7}-1}{\sqrt{7}}$$
or
$$\sqrt[3]{3^{3/5}-\sqrt[5]{2}}=\frac{2^{2/5}+\sqrt[5]{3}+2^{3/5} 3^{2/5}-\sqrt[5]{2}\, 3^{3/5}}{5^{2/3}}$$
or
$$\frac{\sqrt[3]{1+\sqrt{3}+\sqrt{2}\, 3^{3/4}}}{\sqrt[6]{\sqrt{3}-1}}=\frac{\sqrt{1+\sqrt{3}+\sqrt{2} \sqrt[4]{3}}}{\sqrt[6]{2}}\quad ?$$
These results were found by two young students of mine who would very much like to know values of q and b in Bill Dubuque's structure theorem which effect the denesting
$$\sqrt[3]{-\frac{106}{25}-\frac{369 \sqrt{3}}{125}+\frac{3 \sqrt{3} \left(388+268 \sqrt{3}\right)}{100 \sqrt[3]{2}\,
5^{2/3}}}=\frac{3}{5^{2/3}}-\frac{1+\sqrt{3}}{\sqrt[3]{10}}+\frac{1}{5} \sqrt[3]{2} \left(3+2 \sqrt{3}\right)\quad.$$
Thanks in advance.
| {
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How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)?
| Note that $$2\cdot \dfrac{2\pi}{5} + 3\cdot \dfrac{2\pi}{5} = 2\pi,$$
therefore $$\cos\left(2\cdot \dfrac{2\pi}{5}\right) = \cos\left(3\cdot \dfrac{2\pi}{5}\right).$$
Put $\dfrac{2\pi}{5} = x$. Using the formulas
\begin{equation*}
\cos 2x = 2\cos^2 x - 1, \quad \cos 3x = 4\cos^3 x - 3\cos x,
\end{equation*}
we have
\begin{equation*}
4x^3 - 2x^2 -3x + 1 = 0 \Leftrightarrow (x - 1)(4x^2 + 2x - 1) = 0.
\end{equation*}
Because $\cos \dfrac{2\pi}{5} \neq 1$, we get
\begin{equation*}
4x^2 + 2x - 1 = 0.
\end{equation*}
Solving the above quadratic equation for $x$ gives us $\cos \dfrac{2\pi}{5} = \dfrac{-1 \pm \sqrt{5}}{4}$. Because $\cos \dfrac{2\pi}{5} > 0$, we take the positive sign, giving us $\cos \dfrac{2\pi}{5} = \dfrac{-1 + \sqrt{5}}{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Continued Fraction expansion of $\tan(1)$ Prove that the continued fraction of $\tan(1)=[1;1,1,3,1,5,1,7,1,9,1,11,...]$. I tried using the same sort of trick used for finding continued fractions of quadratic irrationals and trying to find a recurrence relation, but that didn't seem to work.
| We use the formula given here: Gauss' continued fraction for $\tan z$ and see that
$$\tan(1) = \cfrac{1}{1 - \cfrac{1}{3 - \cfrac{1}{5 -\dots}}}$$
Now use the identity
$$\cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{x-1}}}$$
To transform $$\cfrac{1}{a - \cfrac{1}{b - \cfrac{1}{c - \dots}}}$$ to
$$\cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-2 + \cfrac{1}{1 + \cfrac{1}{c-2 + \dots}}}}}$$
to get the expansion for $\displaystyle \tan(1)$
The above expansion for $\tan(1)$ becomes
$$ \cfrac{1}{1-1 + \cfrac{1}{1 + \cfrac{1}{3-2 + \cfrac{1}{1 + \cfrac{1}{5-2 + \dots}}}}}$$
$$ = 1 + \cfrac{1}{3-2 + \cfrac{1}{1 + \cfrac{1}{5-2 + \dots}}}$$
$$= 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{3 + \cfrac{1}{1 + \cfrac{1}{5 + \dots}}}}}$$
To prove the transformation,
let $\displaystyle x = b - \cfrac{1}{c - \dots}$
Then
$$ \cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{x-1}}}$$
$$ = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-1 + \cfrac{1}{c - \dots}}}}$$
Applying the identity again to
$$\cfrac{1}{b-1 + \cfrac{1}{c - \dots}}$$
we see that
$$\cfrac{1}{a-\cfrac{1}{x}} = \cfrac{1}{a-1 + \cfrac{1}{1 + \cfrac{1}{b-2 + \cfrac{1}{1 + \cfrac{1}{c-1 + \cfrac{1}{d - \dots}}}}}}$$
Applying again to $\cfrac{1}{c-1 + \cfrac{1}{d - \dots}}$ etc gives the required CF.
| {
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Suggest a tricky method for this problem Find the value of:
$$ 1+ \biggl(\frac{1}{10}\biggr)^2 + \frac{1 \cdot 3}{1 \cdot 2} \biggl(\frac{1}{10}\biggr)^4 + \frac{1\cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \biggl(\frac{1}{10}\biggr)^6 + \cdots $$
|
Hint: What is the generalized binomial expansion of $\left( 1-2 \times
\left(\frac{1}{10}\right)^2 \right) ^{-\frac{1}{2}}$?
| {
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If $4^x + 4^{-x} = 34$, then $2^x + 2^{-x}$ is equal to...? I am having trouble with this:
If $4^x + 4^{-x} = 34$, then what is $2^x + 2^{-x}$ equal to?
I managed to find $4^x$ and it is:
$$4^x = 17 \pm 12\sqrt{2}$$
so that means that $2^x$ is:
$$2^x = \pm \sqrt{17 \pm 12\sqrt{2}}.$$
Correct answer is 6 and I am not getting it :(. What am I doing wrong?
| Let $a=2^x$. Then $2^{-x}=1/a$ and $$\left(a+\frac1a\right)^2=a^2+2+1/a^2=2+(2^x)^2+(2^{-x})^2=2+4^x+4^{-x}=2+34=36.$$ This means that $a+1/a=6$ (it should be 6 or -6, since these are the square roots of $36$ but since $a$ and $1/a$ are positive, it is 6).
| {
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$n$th derivative of $e^{1/x}$ I am trying to find the $n$'th derivative of $f(x)=e^{1/x}$. When looking at the first few derivatives I noticed a pattern and eventually found the following formula
$$\frac{\mathrm d^n}{\mathrm dx^n}f(x)=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}$$
I tested it for the first $20$ derivatives and it got them all. Mathematica says that it is some hypergeometric distribution but I don't want to use that. Now I am trying to verify it by induction but my algebra is not good enough to do the induction step.
Here is what I tried for the induction (incomplete, maybe incorrect)
$\begin{align*}
\frac{\mathrm d^{n+1}}{\mathrm dx^{n+1}}f(x)&=\frac{\mathrm d}{\mathrm dx}(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}\\
&=(-1)^n e^{1/x} \cdot \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} (-2n+k) x^{-2 n+k-1}\right)-e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 (n+1)+k}\\
&=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}((-2n+k) x^{-2 n+k-1}-x^{-2 (n+1)+k)})\\
&=(-1)^{n+1} e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}(2n x-k x+1) x^{-2 (n+1)+k}
\end{align*}$
I don't know how to get on from here.
| How's this?
$$\left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} \left(-2n+k\right) x^{-2 n+k-1}\right) - \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$
$$= \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} \left(-2n+k\right) x^{-2\left(n+1\right)+k+1}\right) - \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$
$$= \left(\sum _{k'=1}^{n} \left(k'-1\right)! \binom{n}{k'-1} \binom{n-1}{k'-1} \left(-2n+k'-1\right) x^{-2\left(n+1\right)+k'}\right) - \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$
$$= \left(\sum _{k'=0}^{n} \left(k'-1\right)! \binom{n}{k'-1} \binom{n-1}{k'-1} \left(-2n+k'-1\right) x^{-2\left(n+1\right)+k'}\right) - \left(\sum _{k=0}^{n} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =$$
$$= \sum _{k=0}^{n} \left(\left(k-1\right)! \binom{n}{k-1} \binom{n-1}{k-1} \left(-2n+k-1\right) - k! \binom{n}{k} \binom{n-1}{k}\right) x^{-2 \left(n+1\right)+k}$$
Then
$$\left(k-1\right)! \binom{n}{k-1} \binom{n-1}{k-1} \left(-2n+k-1\right) - k! \binom{n}{k} \binom{n-1}{k} =$$
$$= \frac{\left(k-1\right)!n!\left(n-1\right)!\left(-2n+k-1\right)}{\left(n-k+1\right)!\left(k-1\right)!\left(n-k\right)!\left(k-1\right)!} - \frac{k!n!\left(n-1\right)!}{\left(n-k\right)!k!k!\left(n-k-1\right)!} =$$
$$= \frac{n!\left(n-1\right)!\left(-2n+k-1\right)k}{\left(n-k+1\right)!\left(n-k\right)!k!} - \frac{n!\left(n-1\right)!\left(n-k\right)\left(n-k+1\right)}{\left(n-k\right)!k!\left(n-k+1\right)!} =$$
$$= \frac{n!\left(n-1\right)!}{\left(n-k+1\right)!\left(n-k\right)!k!} \left(\left(-2n+k-1\right)k - \left(n-k\right)\left(n-k+1\right)\right) =$$
$$= \frac{-n\left(n+1\right)n!\left(n-1\right)!}{\left(n-k+1\right)!\left(n-k\right)!k!} =$$
$$= -\frac{\left(n+1\right)!}{\left(n-k+1\right)!} \binom{n}{k} =$$
$$= -k! \binom{n+1}{k} \binom{n}{k}$$
| {
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How to know that $a^3+b^3 = (a+b)(a^2-ab+b^2)$ Is there a way of go from $a^3+b^3$ to $(a+b)(a^2-ab+b^2)$ other than know the property by heart?
| It's a homogenization of the cyclotomic factorization $\rm\ x^3 + 1 = (x+1)\ (x^2 - x + 1)\:.\ $ Recall that the homogenization of a degree $\rm\:n\:$ polynomial $\rm\ f(x)\ $ is the polynomial $\rm\ y^n\ f(x/y)\:.\ $ This maps $\rm\ x^{k}\ \to\ x^{k}\ y^{n-k}\ $ so the result is a homogeneous polynomial of degree $\rm\:n\:.\ $ While this cyclotomic factorization is rather trivial, other cyclotomic homogenizations can be far less trivial.
For example, Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations:
$$\begin{array}{rl}
x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\
\frac{x^6 + 3^2}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\
\frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\
\frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\
\end{array}$$
| {
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Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$ Using congruences, show that the following is always an integer for every integer
value of $n$:
$$\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n.$$
| HINT $\displaystyle\rm\quad \frac{n^5}5\: +\: \frac{n^3}3\: +\: \frac{7\:n}{15}\ =\ \frac{n^5-n}5\: +\: \frac{n^3-n}3\: +\: n\ \in \mathbb Z\ $ by Fermat's Little Theorem.
| {
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Simple limit, wolframalpha doesn't agree, what's wrong? (Just the sign of the answer that's off) $\begin{align*}
\lim_{x\to 0}\frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}
&=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{4+x}}-\frac{\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
&=\lim_{x\to 0}\frac{\frac{2-\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
&=\lim_{x\to 0}\frac{2-\sqrt{4+x}}{2x\sqrt{4+x}}\\
&=\lim_{x\to 0}\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{(2x\sqrt{4+x})(2+\sqrt{4-x})}\\
&=\lim_{x\to 0}\frac{2 \times 2 + 2\sqrt{4-x}-2\sqrt{4-x}-((\sqrt{4-x})(\sqrt{4-x})) }{2 \times 2x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\
&=\lim_{x\to 0}\frac{4-4+x}{4x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\
&=\lim_{x\to 0}\frac{x}{x(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\
&=\lim_{x\to 0}\frac{1}{(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\
&=\frac{1}{(4\sqrt{4+0} + 2\sqrt{4+0}\sqrt{4-0})}\\
&=\frac{1}{16}
\end{align*}$
wolframalpha says it's negative. What am I doing wrong?
| Certainly for $x \gt 0,\frac{1}{\sqrt{4+x}}-\frac{1}{2} \lt 0$ so the limit should be negative. Between the fifth and sixth limit you flipped a sign under the sqrt in the numerator and that changes the sign of the total thing
| {
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How to find the highest power of a prime $p$ that divides $\prod \limits_{i=0}^{n} 2i+1$?
Possible Duplicate:
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
Given an odd prime $p$, how does one find the highest power of $p$ that divides
$$\displaystyle\prod_{i=0}^n(2i+1)?$$
I wrote it down all paper and realized that the highest power of $p$ that divides this product will be the same as the highest power of $p$ that divides $(\lceil\frac{n}{2}\rceil - 1)!$
Since
$$10! = 1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10$$ while
$$\prod_{i=0}^{4} (2i+1) = 1\times 3\times 5\times 7\times 9$$
Am I in the right track?
Thanks,
Chan
| Note that $\displaystyle \prod_{i=1}^{n} (2i-1) = \frac{(2n)!}{2^n n!}$.
Clearly, the highest power of $2$ dividing the above product is $0$.
For odd primes $p$, we proceed as follows.
Note that the highest power of $p$ dividing $\frac{a}{b}$ is nothing but the highest power of $p$ dividing $a$ - highest power of $p$ dividing $b$.
i.e. if $s_p$ is the highest power of $p$ dividing $\frac{a}{b}$ and $s_{p_a}$ is the highest power of $p$ dividing $a$ and $s_{p_b}$ is the highest power of $p$ dividing $b$, then $s_p = s_{p_a}-s_{p_b}$.
So the highest power of $p$ dividing $\displaystyle \frac{(2n)!}{2^n n!}$ is nothing but $s_{(2n)!}-s_{2^n}-s_{n!}$.
Note that $s_{2^n} = 0$.
Now if you want to find the maximum power of a prime $q$ dividing $N!$, it is given by
$$s_{N!} = \left \lfloor \frac{N}{q} \right \rfloor + \left \lfloor \frac{N}{q^2} \right \rfloor + \left \lfloor \frac{N}{q^3} \right \rfloor + \cdots$$
(Look up this stackexchange thread for the justification of the above claim)
Hence, the highest power of a odd prime $p$ dividing the product is $$\left ( \left \lfloor \frac{2N}{p} \right \rfloor + \left \lfloor \frac{2N}{p^2} \right \rfloor + \left \lfloor \frac{2N}{p^3} \right \rfloor + \cdots \right ) - \left (\left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots \right)$$
| {
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$T(1) = 1 , T(n) = 2T(n/2) + n^3$? Divide and conquer $T(1) = 1 , T(n) = 2T(n/2) + n^3$? Divide and conquer, need help, I dont know how to solve it?
| Hmm, possibly another way of heuristics is instructive.
First write the undisputable elements of the sequence:
$$\begin{array} &
a(1) &=&a(2^0) & = & 1 \\\
a(2)&=&a(2^1) &=&10 &= & 2^3 + 2*1 &=& 2*(4^1+1) \\\
a(4)&=&a(2^2) &=&84 &=& 4^3 + 2*10 &=& 4*(4^2+4^1+1) \\\
a(8)&=&a(2^3) &=&680 &=& 8^3 + 2*84 &=& 8^3+2*4^3+4*2^3+8*1^3\\\
& & & & &=& 8*(4^3+4^2+4^1+1) \\\
\ldots &=&a(2^k)&=& \ldots
\end{array} $$
It is obvious how this can be continued, because at the exponent k we get always $8^k$ plus two times the previous, thus the weighted sum of all powers of 8 which can be expressed as consecutive powers of 4:
$$ a(2^k) = 2^k*(4^k+4^{k-1} \ldots +4^0)= 2^k*\frac{4^{k+1}-1}{4-1} $$
Now the step "divide" can be taken: the above gives also a meaningful possibility for interpolation of the non-explicitely defined elements. If we allow base-2 logarithms for k we get for
$$\begin{array} &
& a(2^k) &= & 2^k*\frac{4^{k+1}-1}{4-1} \\\
& &= & 2^k*\frac{4*(2^{k})^2-1}{3} \\\
\text{assuming }& k&=& \frac{\log(n)}{\log(2)} \\\
& a(n) &=& n*\frac{4*n^2-1}{3} \\\
& &=& n^3 + \frac{(n-1)n(n+1)}{3!} \\\
& &=& n^3 + 2*\binom{n+1}{3} \\\
\end{array} $$
where the expression in the fourth line is the same as Fabian's result.
| {
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Show that 13 divides $2^{70}+3^{70}$
Show that $13$ divides $2^{70} + 3^{70}$.
My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat's little theorem. So how can I separate the two?
Thanks!
| $2^{12} \equiv 1 \pmod{13}$ and $3^{12} \equiv 1 \pmod{13}$ by Fermat's Little Theorem.
Hence, $2^{72} \equiv 1 \pmod{13}$ and $3^{72} \equiv 1 \pmod{13}$
$2^{72} \equiv 1 \pmod{13} \Rightarrow 2^{72} \equiv 40 \pmod{13} \Rightarrow 2^{70} \equiv 10 \pmod{13}$
$3^{72} \equiv 1 \pmod{13} \Rightarrow 3^{72} \equiv 27 \pmod{13} \Rightarrow 3^{70} \equiv 3 \pmod{13}$
Hence, you get the result.
| {
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Representability as a Sum of Three Positive Squares or Non-negative Triangular Numbers Let $r_{2,3}(n)$ and $r_{t,3}(n)$ denote the number of ways to write $n$ as a sum of three positive squares (A063691) and as a sum of three non-negative triangular numbers (A008443), respectively. I have noticed that $r_{2,3}(8k+3) = r_{t,3}(k)$ for $k \geq 1$. For example, $r_{2,3}(11) = 3$ because $11 = 3^2 + 1^2 + 1^2 = 1^2 + 3^2 + 1^2 = 1^2 + 1^2 + 3^2$ and $r_{t,3}(1) = 3$ because $1 = 1 + 0 + 0 = 0 + 1 + 0 = 0 + 0 + 1$, where $0$ and $1$ are triangular numbers.
Is this identity well-known? If so, where can I find its proof?
A proof should follow from showing that the coefficient of the $q^{8k + 3}$ of the $q$-series of $(\sum_{n \geq 1} q^{n^{2}})^{3}$ is equal to the corresponding coefficient of the $q$-series of $\frac{1}{8} \theta^{3}_{2}(q^{4})$, where $\theta_2(q) = \theta_2(0,q)$ is a Jacobi theta function.
| It is easy enough to prove something like this for squares and triangular numbers since modulo 4, squares are 0 or 1, so any three squares adding to $8k+3$ must each be odd, and the equation
$$k = \frac{a(a+1)}{2} + \frac{b(b+1)}{2} + \frac{c(c+1)}{2}$$
implies and is implied by
$$8k+3 = (2a+1)^2 + (2b+1)^2 + (2c+1)^2 .$$
| {
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Multiply: $(4 + x)(x^2 + 2x +3)$ How would I solve this:
Multiply: $(4 + x)(x^2 + 2x +3)$
| This question, as you probably know, requires the use of the distributive property. To use JavaMan's suggestion $$(a + b) \cdot c = a \cdot c + b \cdot c$$
Let "a + b" be your $4 + x$ and let "c" be your $x^2 + 2x + 3$
Then we need to multiply $a \cdot c$, or $4 \cdot (x^2 + 2x + 3)$, and add it to
$b \cdot c$, which is $x \cdot (x^2 + 2x + 3)$
So $$ a\cdot c + b \cdot c = [4\cdot (x^2 + 2x +3)] + [x \cdot (x^2 + 2x + 3)]$$
After taking these steps, combine like terms and write the result in order of decreasing exponents (for convention's sake)
| {
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exponential equation $$\sqrt{(5+2\sqrt6)^x}+\sqrt{(5-2\sqrt6)^x}=10$$
So I have squared both sides and got:
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2\sqrt{1^x}=100$$
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2=100$$
I don't know what to do now
| You've already seen that $(5-2\sqrt{6})(5+2\sqrt{6})=1$ when you squared both sides. This means that $5-2\sqrt{6}=\frac{1}{5+2\sqrt{6}}$, so your last equation can be rewritten as $$\left(\frac{1}{5+2\sqrt6}\right)^x+(5+2\sqrt6)^x+2=100$$
or, letting $y=(5+2\sqrt{6})^x$,
$$\frac{1}{y}+y+2=100$$
so
$$1+y^2+2y=100y$$
which is quadratic in $y$. Solve this for $y$, then use that solution to solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/28157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Integral $\int{\sqrt{25 - x^2}dx}$ I'm trying to find $\int{\sqrt{25 - x^2} dx}$
Now I know that $\int{\frac{dx}{\sqrt{25 - x^2}}}$ would have been $\arcsin{\frac{x}{5}} + C$, but this integral I'm asking about has the rooted term in the numerator.
What are some techniques to evaluate this indefinite integral?
| Since you already know that
$$\int{\frac{dx}{\sqrt{25 - x^2}}}=\arcsin{\frac{x}{5}} + C$$
you can actually skip the trigonometric substitution part and solve by partial integration:
$$\begin{array}{lcl}\int{\sqrt{25 - x^2} dx} & = & x\sqrt{25 - x^2} - \int{\frac{x (-2x)dx}{2\sqrt{25 - x^2}}} \\
& = & x\sqrt{25 - x^2} - \int{\frac{-x^2 dx}{\sqrt{25 - x^2}}} \\
& = & x\sqrt{25 - x^2} - \int{\frac{25-x^2 dx}{\sqrt{25 - x^2}}} + \int{\frac{25 dx}{\sqrt{25 - x^2}}} \\
& = & x\sqrt{25 - x^2} - \int{\sqrt{25 - x^2} dx} + 25\arcsin{\frac{x}{5}} + C \; .
\end{array}$$
Or after rearranging
$$\int{\sqrt{25 - x^2} dx} = \frac{1}{2} x\sqrt{25 - x^2} + \frac{25}{2}\arcsin{\frac{x}{5}} + C \; .$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding maxima and minima of a function A couple problems are giving me trouble in finding the relative maxima/minima of the function. I think the problem stems from me possibly not finding all of the critical numbers of the function, but I don't see what I missed.
Given $f(x)= 5x + 10 \sin x$, I calculated the derivative as $5 + 10 \cos x$, and found the first critical number by this work:
$$5+ 10 \cos x=0$$
$$\frac{5}{5}+10 \cos x= 0-5 \Rightarrow 10 \cos x= -5$$
$$\frac{10 \cos x}{10}= \frac{-5}{10}\Rightarrow \cos x= -\frac{1}{2}$$
$$x= \arccos(-\frac{1}{2}) = \text{First critical number is }\frac{2\pi}{3}$$
That gave me the maxima of the formula, since $$f(\frac{2\pi}{3})= 5(\frac{2\pi}{3})+10 \sin(\frac{2\pi}{3})= \frac{10\pi}{3}+5\sqrt3$$
However, I need the other critical number to calculate the minima. Should I look for the value of $\arccos(\frac{1}{2})$?
| $\cos(x)=-1/2$ if and only if there exists an integer $n$ such that $x=2n\pi+2\pi/3$ or $x=2n\pi+4\pi/3$. Hence any of these real numbers $x$ may be (and in fact, is) a relative maximum or a relative minimum of $f$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Express $z$ in terms of $x$ and $y$, i.e., find $z= f(x,y)$ I've been banging my head against the wall for a while now:
$x = s^2 - t^2$
$y = s + t$
$z = s^2 + 3t$
Express $z$ in terms of $x$ and $y$.
| $$x=s^{2}-t^{2}=(s+t)(s-t)$$
so
$$s+t=\frac{x}{s-t}$$
$$s-t=\frac{x}{s+t}=\frac{x}{y}$$
$$(s+t) + (s-t) = 2s=\frac{x}{s-t}+\frac{x}{y}$$
$y=s+t$, so $t=y-s$ and therefore:
$$2s=\frac{x}{2s-y}+\frac{x}{y}=x(\frac{1}{2s-y}+\frac{1}{y})$$
$$2s=x(\frac{y}{2sy-y^2}+\frac{2s-y}{2sy-y^2})=x(\frac{2s}{2sy-y^2})$$
$$1=\frac{x}{2sy-y^2}$$
$$2sy-y^2=x$$
$$2sy=x+y^2$$
$$s=\frac{x+y^2}{2y}$$
From $z=s^2+3t$ we have:
$$z=(\frac{x+y^2}{2y})^{2}+3t$$
$y=s+t$ so $t=y-\frac{x+y^2}{2y}$ and finally:
$$z=(\frac{x+y^2}{2y})^{2}+3(y-\frac{x+y^2}{2y})$$
Pretty sure this is correct...
| {
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How to calculate $\int_0^{2\pi} \sqrt{1 - \sin^2 \theta}\;\mathrm d\theta$ How to calculate:
$$ \int_0^{2\pi} \sqrt{1 - \sin^2 \theta}\;\mathrm d\theta $$
| \begin{align}
\int_0^{2\pi} \sqrt{1 - \sin^2 \theta} d\theta &= \int_0^{2\pi} \sqrt{\cos^2 \theta} d\theta \\ &= \int_0^{2\pi} | \cos \theta | d\theta
\\ &= 4 \int_0^{\frac{pi}{4}} \cos \theta d\theta
\\ &= 4
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Finding $A^n$ for a matrix I have a matrix $$
A =
\left[ {\begin{array}{cc}
1 & c \\
0 & d \\
\end{array} } \right]
$$
with $c$ and $d$ constant. I need to find $A^n$ ($n$ positive) and then need to prove that formula using induction.
I would like to check that the formula I derived is correct:
$$
A^n =
\left[ {\begin{array}{cc}
1 & c^{n-2}(dc + c) \\
0 & d^n \\
\end{array} } \right]
$$
If this is correct, how can I prove this? I suppose I can write $A^{n+1} = A^n A$, which would be
$$
\left[ {\begin{array}{cc}
1 & c^{n-2}(dc + c) \\
0 & d^n \\
\end{array} } \right]
\left[ {\begin{array}{cc}
1 & c \\
0 & d \\
\end{array} } \right]
$$
But then what would I do?
Thanks.
| Letting $a_n$ be the upper right hand corner of $A^n$, and assuming it is obvious that the lower right corner of $A^n$ is $d^n$, and the left column is $[1,0]^T$, we get:
$$A^{n+1} = A^n A =
\left[ {\begin{array}{cc}
1 & a_n \\
0 & d^n \\
\end{array} } \right]
\left[ {\begin{array}{cc}
1 & c \\
0 & d \\
\end{array} } \right]$$
This gives us $a_{n+1} = c + d a_n$, with $a_1=c$.
In particular, then $a_n = c + cd + cd^2 + ... + cd^{n-1} = c\frac{d^n-1}{d-1}$.
So:
$$A^n =
\left[ {\begin{array}{cc}
1 & c\frac{d^n-1}{d-1} \\
0 & d^n \\
\end{array} } \right]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Permutation/Combinations in bit Strings I have a bit string with 10 letters, which can be {a, b, c}. How many bit strings can be made that have exactly 3 a's, or exactly 4 b's?
I thought that it would be C(7,2) + C(6,2), but that's wrong (the answer is 24,600).
| First note that
Number of ways of event $A$ or event $B = $ (Number of ways of event $A$) + (Number of ways of event $B$) - (Number of ways of event $A$ and $B$ occurring simultaneously)
In our problem, event $A$ is the number of strings containing exactly $3$ $a$'s and event $B$ is the number of strings containing exactly $4$ $b$'s
We are interested in the number of strings containing exactly $3$ $a$'s and in the number of strings containing exactly $4$ $b$'s.
Number of strings containing exactly $3$ $a$'s is obtained as follows. Of the $10$ letters in our string, if we want exactly $3$ $a$'s, each of the remaining $7$ has $2$ choices, namely, it should be either $b$ or $c$ and hence we have $2^7$ choices for the remaining $7$. Once we have this, we can now arrange them in $\frac{10!}{3! \times 7!}$ ways. Hence, number of ways of this event is $\frac{10!}{3! \times 7!} \times 2^7$.
Similarly, number of strings containing exactly $4$ $b$'s is obtained as follows. Of the $10$ letters in our string, if we want exactly $4$ $b$'s, each of the remaining $6$ has $2$ choices, namely, it should be either $a$ or $c$ and hence we have $2^6$ choices for the remaining $6$. Once we have this, we can now arrange them in $\frac{10!}{4! \times 6!}$ ways. Hence, number of ways of this event is $\frac{10!}{4! \times 6!} \times 2^6$.
Number of strings containing exactly $3$ $a$'s and $4$ $b$'s is $\frac{10!}{3! 4! 3 !}$.
Hence, the total number of strings containing exactly $3$ $a$'s or exactly $4$ $b$'s is $$\frac{10!}{3! \times 7!} \times 2^7 + \frac{10!}{4! \times 6!} \times 2^6 - \frac{10!}{3! 4! 3 !} = 24,600$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Exponents in the denominator? I'm having trouble understanding exponents in the denominator.
For example: I have the expression: $\displaystyle 1 - \frac{1}{3^n} + \frac{2}{3^{n+1}}$.
I know that this simplifies to $\displaystyle 1 - \frac{1}{3^{n+1}}$, but how/why? Can someone please list the steps?
My understanding is that the exponent $(n+1)$ in the expression $x^{n+1}$ means that $x^{n+1} = x x^n$, but how does this fit with the above problem?
| You should get a common denominator for the last two terms by multiplying the second term by $3/3$. You have
$$
\begin{align*}
1-\frac{1}{3^n}+\frac{2}{3^{n+1}} &= 1-\frac{3}{3^{n+1}}+\frac{2}{3^{n+1}}\\
&= 1-\frac{1}{3^{n+1}}
\end{align*}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
| $\begin{aligned} & \hspace{0.5in} \begin{aligned}\displaystyle \sum_{1 \le k \le n}k^2 & = \sum_{1 \le k \le n}~\sum_{1 \le r \le k}r =\sum_{1 \le r \le n}~\sum_{r \le k \le n}r \\& = \sum_{1 \le r \le n}~\sum_{1 \le k \le n}r-\sum_{1 \le r \le n}~\sum_{1 \le k \le r-1}r \\& = n\sum_{1 \le r \le n}r-\frac{1}{2}\sum_{1 \le r \le n}r(r-1) \\& =\frac{1}{2}n^2(n+1)-\frac{1}{2}\sum_{1 \le r \le n}r^2+\frac{1}{2}\sum_{1 \le r \le n}r \\& =\frac{1}{2}n^2(n+1)-\frac{1}{2}\sum_{1 \le k \le n}k^2+\frac{1}{4}n(n+1) \end{aligned} \\& \begin{aligned}\implies\frac{3}{2}\sum_{1 \le k \le n}k^2 & = \frac{1}{2}n^2(n+1)+\frac{1}{4}n(n+1) \\& = \frac{1}{4}n(n+1)(2n+1) \end{aligned}\\& \implies \hspace{0.15in} \displaystyle \sum_{1 \le k \le n}k^2 = \frac{1}{6}n(n+1)(2n+1).\end{aligned}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Problem finding zeros of complex polynomial I'm trying to solve this problem
$$ z^2 + (\sqrt{3} + i)|z| \bar{z}^2 = 0 $$
So, I know $ |z^2| = |z|^2 = a^2 + b ^2 $ and $ \operatorname{Arg}(z^2) = 2 \operatorname{Arg} (z) - 2k \pi = 2 \arctan (\frac{b}{a} ) - 2 k\pi $ for a $ k \in \mathbb{Z} $. Regarding the other term, I know $ |(\sqrt{3} + i)|z| \bar{z}^2 | = |z|^3 |\sqrt{3} + i| = 2 |z|^3 = 2(a^2 + b^2)^{3/2} $ and because of de Moivre's theorem, I have $ \operatorname{Arg} [(\sqrt{3} + i ) |z|\bar{z}^2] = \frac{\pi}{6} + 2 \operatorname{Arg} (z) - 2Q\pi $.
Using all of this I can rewrite the equation as follows
$$\begin{align*}
&|z|^2 \Bigl[ \cos (2 \operatorname{Arg} (z) - 2k \pi) + i \sin (2 \operatorname{Arg}(z) - 2k \pi)\Bigr]\\
&\qquad \mathop{+} 2|z|^3 \Biggl[\cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right) + i \sin \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right)\Biggr] = 0
\end{align*} $$
Which, assuming $ z \neq 0 $, can be simplified as
$$\begin{align*}
&\cos (2 \operatorname{Arg} (z) - 2k \pi) + i \sin (2 \operatorname{Arg} (z) - 2k \pi) \\
&\qquad\mathop{+} 2 |z|\Biggl[\cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q \pi \right) + i \sin \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right)\Biggr] = 0
\end{align*} $$
Now, from this I'm not sure how to go on. I tried a few things that got me nowhere like trying to solve
$$ \cos (2 \operatorname{Arg}(z) - 2k \pi) = 2 |z| \cos \left(\frac{\pi}{6} + 2 \operatorname{Arg} (z) -2Q\pi\right) $$
I'm really lost here, I don't know how to keep going and I've looked for error but can't find them. Any help would be greatly appreciated.
| The relation is equivalent to $z^2=-(\sqrt{3}+i)|z|\overline{z}^2$. $z=0$ is a solution, so in the following $z \neq 0$. Take modulus of both sides and denote $r=|z|=|\overline{z}|$. Then $r^2=2r^3$, which means $r=\frac{1}{2}$.
The relations turns to $z^2+\frac{1}{2}(\sqrt{3}+i)\overline{z}^2=0$. Multiply by $z^2$ and get $z^4+\frac{1}{2}(\sqrt{3}+i)\frac{1}{16}=0$. Write it in trigonometric form
$$ z^4=\frac{1}{16}\left(-\frac{\sqrt{3}}{2}-i\frac{1}{2}\right)=\frac{1}{16}(\cos \frac{7\pi}{6}+\sin \frac{7\pi}{6})$$.
From here on it is just the extraction of complex roots.
[edit] I did not answer your question, as to how to continue your calculations, but I can say from experience that in most complex numbers problems the substitution $z=a+bi$ gets you in more troubles in the end, than working with the properties of complex conjugate, modulus and trigonometric form. You can see that in my solution no great computational problems were encountered.
| {
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Deriving the rest of trigonometric identities from the formulas for $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, and $\cos (A-B)$ I am trying to study for a test and the teacher suggest we memorize $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, $\cos (A-B)$, and then be able to derive the rest out of those. I have no idea how to get any of the other ones out of these, it seems almost impossible. I know the $\sin^2\theta + \cos^2\theta = 1$ stuff pretty well though. For example just knowing the above how do I express $\cot(2a)$ in terms of $\cot a$? That is one of my problems and I seem to get stuck half way through.
| Since $\displaystyle\cot(2a) = \frac{\cos(2a)}{\sin(2a)}$, you would have (assuming you know the addition formulas for sines and cosines):
$$\begin{align*}
\cos(2a) &= \cos(a+a) = \cos(a)\cos(a) - \sin(a)\sin(a)\\
&= \cos^2(a) - \sin^2(a);\\
\sin(2a) &= \sin(a+a) = \sin(a)\cos(a) + \cos(a)\sin(a)\\
&= 2\sin(a)\cos(a),
\end{align*}$$
and therefore
$$\begin{align*}
\cot(2a) &= \frac{\cos(2a)}{\sin(2a)} = \frac{\cos^2(a) - \sin^2(a)}{2\sin(a)\cos(a)}\\
&= \frac{1}{2}\left(\frac{\cos^2(a)}{\sin(a)\cos(a)}\right) - \frac{1}{2}\left(\frac{\sin^2(a)}{\sin(a)\cos(a)}\right)\\
&= \frac{1}{2}\left(\frac{\cos(a)}{\sin(a)} - \frac{\sin(a)}{\cos(a)}\right)\\
&= \frac{1}{2}\left(\cot(a) - \tan(a)\right)\\
&= \frac{1}{2}\left(\cot(a) - \frac{1}{\cot(a)}\right)\\
&= \frac{1}{2}\left(\frac{\cot^2(a)}{\cot(a)} - \frac{1}{\cot(a)}\right)\\
&= \frac{1}{2}\left(\frac{\cot^2(a) - 1}{\cot (a)}\right).
\end{align*}$$
P.S. Now, as it happens, I don't know the formulas for double angles, nor most identities involving tangents, cotangents, etc. I never bothered to memorize them. What I know are:
*
*The definitions of tangent, cotangent, secant, and cosecant in terms of sine and cosine;
*That sine is odd ($\sin(-x) = -\sin(x)$) and cosine is even ($\cos(-x)=\cos(x)$);
*The addition formulas for sine and cosine;
*The values of sine and cosine at $0^{\circ}$, $30^{\circ}$, $45^{\circ}$, $60^{\circ}$, and $90^{\circ}$.
(I can derive $\sin^2\theta + \cos^2\theta = 1$ from the above, but in all honesty that one comes up so often that I do know it as well). I do not know the addition or double angle formulas for tangents nor cotangents, so the above derivation was done precisely "on the fly", as I was typing. I briefly thought that I might need to $\cos(2a)$ with one of the following equivalent formulas:
$$\cos^2(a)-\sin^2(a) = \cos^2(a) + \sin^2(a) - 2\sin^2(a) = 1 - 2\sin^2(a)$$
or
$$\cos^2(a) - \sin^2(a) = 2\cos^2(a) - \cos^2(a) - \sin^2(a) = 2\cos^2(a) - 1,$$
if the first attempt had not immediately led to a formula for $\cot(2a)$ that involved only $\cot(a)$ and $\tan(a) = \frac{1}{\cot(a)}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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For which $n$ is $ \int \limits_0^{2\pi} \prod \limits_{k=1}^n \cos(k x)\,dx $ non-zero? I can verify easily that for $n=1$ and $2$ it's $0$, $3$ and $4$ nonzero, $4$ and $5$ $0$, etc. but it seems like there must be something deeper here (or at least a trick).
| Hint: Start as anon did with
$$
\int_0^{2\pi}\prod_{k=1}^n\cos(kx)\,\mathrm{d}x
=\int_0^{2\pi}e^{-i\frac{n(n+1)}{2}x}\prod_{k=1}^n(1+e^{i2kx})\,\mathrm{d}x\tag{1}
$$
which would be $2\pi$ times the coefficient of $x^{n(n+1)/2}$ in
$$
\prod_{k=1}^n(1+x^{2k})\tag{2}
$$
$(2)$ is the number of ways to write $n(n+1)/2$ as the sum of distinct even integers $\le2n$.
So $(1)$ is non-zero precisely when you can write $n(n+1)/2$ as the sum of distinct even integers $\le2n$ (a much simpler problem).
Claim: $\dfrac{n(n+1)}{2}$ can be written as the sum of distinct even integers no greater than $2n$ in at least one way precisely when $n\in\{0,3\}\pmod{4}$.
Proof: By induction.
If $n\in\{1,2\}\pmod{4}$, then $\dfrac{n(n+1)}{2}$ is odd, and so cannot be written as the sum of even integers.
Suppose that $n\in\{0,3\}\pmod{4}$ and $\dfrac{n(n+1)}{2}$ can be written as the sum of distinct even integers no greater than $2n$. Then
$$
\frac{(n+4)(n+5)}{2}=\frac{n(n+1)}{2}+(2n+4)+(2n+6)
$$
Thus, if the statement is true for $n$, it is true for $n+4$.
Once we note that $\dfrac{3(3+1)}{2}=2+4$ and $\dfrac{4(4+1)}{2}=4+6$, we are done. QED
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $ 30 \mid ab(a^2+b^2)(a^2-b^2)$ How can I prove that $30 \mid ab(a^2+b^2)(a^2-b^2)$ without using $a,b$ congruent modulo $5$ and then
$a,b$ congruent modulo $6$ (for example) to show respectively that $5 \mid ab(a^2+b^2)(a^2-b^2)$ and
$6 \mid ab(a^2+b^2)(a^2-b^2)$?
Indeed this method implies studying numerous congruences and is quite long.
| The intention of this answer is to show you that trying all possibilities is in fact not that long. (Of course, it si more elegant, when you find a solution which avoids trying all possibilities.) Let me start by plotting 5x5 table with all possibilites for the remainders of a and b. (I do not know of a good way of making tables here - I tried something anyway.)
$$
\begin{array}{c|ccccc}
b \backslash a & 0 & 1 & 2 & 3 & 4 \\
\hline
0 & & & & & \\
1 & & & & & \\
2 & & & & & \\
3 & & & & & \\
4 & & & & & \\
\end{array}
$$
If we rewrite our expression as $ab(a-b)(a+b)(a^2+b^2)$, we see that all possibilities where $a=0$ or $b=0$ are ok (marked by $\circ$).
$$
\begin{array}{c|ccccc}
b \backslash a & 0 & 1 & 2 & 3 & 4 \\
\hline
0 & \circ & \circ & \circ & \circ & \circ \\
1 & \circ & & & & \\
2 & \circ & & & & \\
3 & \circ & & & & \\
4 & \circ & & & & \\
\end{array}
$$
Also possibilities where a=b are ok (since $a-b\equiv 0\pmod 5$ and so are those where $a=5-b$ (since $a+b\equiv 0\pmod 5$), hence we can omit both diagonals (marked by $\bullet$).
$$
\begin{array}{c|ccccc}
b \backslash a & 0 & 1 & 2 & 3 & 4 \\
\hline
0 & \circ & \circ & \circ & \circ & \circ \\
1 & \circ & \bullet & & & \bullet \\
2 & \circ & & \bullet & \bullet & \\
3 & \circ & & \bullet & \bullet & \\
4 & \circ & \bullet & & & \bullet \\
\end{array}
$$
There are only 8 possibilities left, and since the roles of a and b are symmetric, we only have to try: (1,2), (1,3), (4,2), (4,3). In all these cases $a^2+b^2\equiv 0 \pmod 5$.
| {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\
&\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\
&\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\
&\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\
&\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0
\end{align}
| You may be able to see more easily the correspondences between the equations and the graph through the following picture which is from the link I got after a curious search on Google(link broken now):
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/54506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "466",
"answer_count": 10,
"answer_id": 7
} |
Determining which values to use in place of x in functions When solving partial fractions for integrations, solving x for two terms usually isn't all that difficult, but I've been running into problems with three term integration.
For example, given
$$\int\frac{x^2+3x-4}{x^3-4x^2+4x}$$
The denominator factored out to $x(x-2)^2$, which resulted in the following formulas
$$
\begin{align*}
x^2+3x-4=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2}\\
x^2+3x-4= A(x-2)(x-2)^2+Bx(x-2)^2+Cx(x-2)\\
x^2+3x-4= A(x-2)^2+Bx(x-2)+Cx\\\\
\text{when x=0, }A=-1
\text{ and x=2, }C=3
\end{align*}
$$
This is where I get stuck, since nothing immediately pops out at me for values that would solve A and C for zero and leave some value for B. How do I find the x-value for a constant that is not immediately apparent?
| The point of these equations involving $A$, $B$, $C$, and $x$ is that there are unique values of $A$, $B$, and $C$ that make the decomposition work, but they are supposed to be valid for every value of $x$. In particular, they should be true when $x = 0$ and $x = 2$. These values happen to be convenient because they cause all but one of the unknowns to vanish, thus allowing you to easily solve for the remaining unknown. Once you know $A$ and $C$, however, your equation looks like
$$
\begin{align*}
x^2 + 3x - 4 = -(x-2)^2 + Bx(x-2) + 3x,
\end{align*}
$$
since $A = -1$ and $C = 3$ are the unique solutions. I re-emphasize that they do not depend on the choice of $x$; we just chose convenient values for $x$ to help us discover $A$ and $C$.
To finish, we can just use any value for $x$, leaving $B$ as the only unknown. I'll choose $x = 1$, simply because I'm not very imaginative:
$$
\begin{align*}
1^2 + 3 \cdot 1 - 4 &= -(1 - 2)^2 + B \cdot 1 (1 - 2) + 3 \cdot 1\\
0 &= -B + 2\\
B &= 2.
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/57114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that all even integers $n \neq 2^k$ are expressible as a sum of consecutive positive integers How do I prove that any even integer $n \neq 2^k$ is expressible as a sum of positive consecutive integers (more than 2 positive consecutive integer)?
For example:
14 = 2 + 3 + 4 + 5
84 = 9 + 10 + ... + 15
n = sum (k + k+1 + k+2 + ...)
n ≠ 2^k
| The sum of the integers from $1$ to $n$ is $n(n+1)/2$. The sum of the integers from $k$ to $k+n$ is then
$$\begin{align*}
k+(k+1)+\cdots+(k+n) &= (n+1)k + 1+\cdots+n\\
& = (n+1)k + \frac{n(n+1)}{2} \\
&= \frac{(n+1)(2k+n)}{2}.\end{align*}$$
Therefore, $a$ can be expressed as the sum of consecutive integers if and only if $2a$ can be factored as $(n+1)(2k+n)$.
Suppose that $a$ is a power of $2$. Then $2a$ is a power of $2$, so $(n+1)(2k+n)$ must be a power of $2$. If we want to avoid negatives, and also avoid the trivial expression as a sum with one summand, we must have $n\geq 1$ and $k\gt 0$. But the parities of $n+1$ and of $2k+n$ are opposite, so this product cannot be a power of $2$ unless either $n+1=1$ (which requies $n=0$) or $2k+n=1$ (which requires $k=0$). Thus, no power of $2$ can be expressed as a sum of at least two consecutive positive integers. In particular, $8$, $16$, $32$, etc cannot be so expressed.
On the other hand, suppose that $a$ is even but not a power of $2$. If we can write $2a = pq$ with $p\gt 1$ and odd, $q$ even, and $q\geq p+1$, then setting $n=p-1$ and $k=(q-p+1)/2$ gives the desired decomposition. If this cannot be done, then every time we factor $2a$ as $pq$ with $p\gt 1$ odd, we have $q\lt p+1$. Then we can set $n=q-1$ and $k = (p+1-q)/2$.
Thus, the powers of $2$ are the only even numbers that are not expressible as the sum of at least two consecutive positive integers.
Added. The OP has now excluded powers of $2$, but has also required that the sum contains strictly more than two summands; i.e., $k\gt 0$ and $n\gt 1$. With the above decompositions, the only case in which we could have $n=1$ is if $2a=pq$ with $p$ odd, $p\gt 1$, and $q=2$. But this is impossible, since $a$ is assumed to be even, and this leads to $2a = 2p$ with $p$ odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/59131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 3
} |
Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
| It may be helpful to recognize that both the RHS and LHS represent the sum of the entries in a the multiplication tables. The LHS represents the summing of Ls (I'll outline those shortly), and the RHS, the summing of the sum of the rows [or columns])$$\begin{array}{lll}
\color{blue}\times&\color{blue}1&\color{blue}2\\
\color{blue}1&\color{green}1&\color{red}2\\
\color{blue}2&\color{red}2&\color{red}4\\
\end{array}$$
Lets begin by building our multiplication tables with a single entry, $1\times1=1=1^2=1^3$. Next, we add the $2$s, which is represented by the red L [$2+4+2 = 2(1+2+1)=2\cdot2^2=2^3$].
So the LHS (green 1 + red L) currently is $1^3+2^3$, and the RHS is $(1+2)+(2+4)=(1+2)+2(1+2)=(1+2)(1+2)=(1+2)^2$.
$$\begin{array}{llll}
\color{blue}\times&\color{blue}1&\color{blue}2&\color{blue}3\\
\color{blue}1&\color{green}1&\color{red}2&\color{maroon}3\\
\color{blue}2&\color{red}2&\color{red}4&\color{maroon}6\\
\color{blue}3&\color{maroon}3&\color{maroon}6&\color{maroon}9\\
\end{array}$$
Next, lets add the $3$s L. $3+6+9+6+3=3(1+2+3+2+1)=3\cdot3^2=3^3$. So now the LHS (green 1 + red L + maroon L) currently is $1^3+2^3+3^3$, and the RHS is $(1+2+3)+(2+4+6)+(3+6+9)=(1+2+3)+2(1+2+3)+3(1+2+3)=(1+2+3)(1+2+3)=(1+2+3)^2$.
By now, we should see a pattern emerging that will give us direction in proving the title statement.
Next we need to prove inductively that $\displaystyle\sum_{i=1}^n i = \frac{n(n+1)}{2}$, and use that relationship to show that $1+2+3+\dots+n+\dots+3+2+1 = \dfrac{n(n+1)}{2}+ \dfrac{(n-1)n}{2} = \dfrac{n((n+1)+(n-1))}{2}=\dfrac{2n^2}{2}=n^2$
Finally, it should be straight forward to show that:
$$\begin{array}{lll}
(\sum^n_{i=1}i+(n+1))^2 &=& (\sum^n_{i=1}i)^2 + 2\cdot(\sum^n_{i=1}i)(n+1)+(n+1)^2\\
&=& \sum^n_{i=1}i^3 + (n+1)(\sum^n_{i=1}i + (n+1) + \sum^n_{i=1}i)\\
&=& \sum^n_{i=1}i^3 + (n+1)(n+1)^2\\
&=& \sum^n_{i=1}i^3 + (n+1)^3\\
&=& \sum^{n+1}_{i=1}i^3\\
\end{array}$$
and, as was already pointed out previously, $$(\sum_{i=1}^1 i)^2 = \sum_{i=1}^1 i^3=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/62171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "67",
"answer_count": 16,
"answer_id": 3
} |
Limit of difference of two square roots I need to find the limit, not sure what to do.
$\lim_{x \to \infty} \sqrt{x^2 +ax} - \sqrt{x^2 +bx}$
I am pretty sure I have to divide by the largest degree which is x^2 but that gets me some weird numbers that don't seem to help.
| Applying the formula $x^2-y^2=(x-y)(x+y)$ we get
$\sqrt{x^2+ax}-\sqrt{x^2+bx}= \frac{x^2+ax-x^2-bx}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\frac{x(a-b)}{x\left(\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}\right) }=\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}$
now you can take the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/62418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Sum of a series of minimums I should get sum of the following minimums.Is there any way to solve it?
$$\min\left\{2,\frac{n}2\right\} + \min\left\{3,\frac{n}2\right\} + \min\left\{4,\frac{n}2\right\} + \cdots + \min\left\{n+1, \frac{n}2\right\}=\sum_{i=1}^n \min(i+1,n/2)$$
| Suppose first that $n$ is even, say $n=2m$. Then $$\min\left\{i,\frac{n}2\right\}=\min\{i,m\}=\begin{cases}i,&\text{if }i\le m\\
m,&\text{if }i\ge m.
\end{cases}$$
Thus, $$\begin{align*}
\sum_{i=2}^{n+1}\min\left\{i,\frac{n}2\right\} &= \sum_{i=2}^m i + \sum_{i=m+1}^{n+1} m\\
&= \frac{m(m+1)}2-1 + (n+1-m)m\\
&= \frac12\left(\frac{n}2\right)\left(\frac{n}2+1\right)-1+\left(\frac{n}2\right)\left(\frac{n}2+1\right)\\
&= \frac{3n(n+2)}{8}-1\\
&=\frac{3n^2+6n-8}8.
\end{align*}$$
If $n$ is odd, say $n=2m+1$, $$\min\left\{i,\frac{n}2\right\}=\min\left\{i,m+\frac12\right\}=\begin{cases}i,&\text{if }i\le m\\
m+\frac12,&\text{if }i> m.
\end{cases}$$
Thus, $$\begin{align*}
\sum_{i=2}^{n+1}\min\left\{i,\frac{n}2\right\} &= \sum_{i=2}^m i + \sum_{i=m+1}^{n+1} \left(m+\frac12\right)\\
&= \frac{m(m+1)}2-1 + (n+1-m)\left(m+\frac12\right)\\
&= \frac12\left(\frac{n-1}2\right)\left(\frac{n+1}2\right)-1+\left(\frac{n}2\right)\left(\frac{n+3}2\right)\\
&= \frac{3n^2+6n-1}{8}-1\\
&= \frac38 (n^2+2n-3).
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/68873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A question on partitions of n Let $P$ be the set of partitions of $n$. Let $\lambda$ denote the shape of a particular partition. Let $f_\lambda(i)$ be the frequency of $i$ in $\lambda$ and let $a_\lambda(i) := \# \lbrace j : f_\lambda(j) \geq i \rbrace$.
For example: $n=5,~ \lambda=(1,1,3),~ f_\lambda(1)=2,~ a_\lambda(1)=2$ (added: since $f_\lambda(1)$ and $f_\lambda(3)$ are both at least 1). It is easy to see that for a fixed $\lambda$, $\sum_k f_\lambda(k)=\sum_k a_\lambda(k)$. But, I am having trouble showing:
For a fixed $k$, $$\sum_\lambda f_\lambda(k)=\sum_\lambda a_\lambda(k)$$
Thanks for the help!
| Here's a proof using generating functions; I haven't given much thought yet to how it could be translated into a combinatorial argument.
The generating function for the number of partitions is
$$p(x)=(1+x+x^2+\dotso)(1+x^2+x^4+\dotso)(1+x^3+x^6+\dotso)\dots=\prod_m\frac1{1-x^m}\;.$$
To count the number of times a part $k$ occurs in the partitions of $n$ (the left-hand side of the equation), we can replace the $k$-th factor by one including this count:
$$
\begin{align}
f_k(x)
&=\left(0+1x^k+2(x^k)^2+3(x^k)^3+\dotso\right)\prod_{m\neq k}\frac1{1-x^m}\\
&=\left(x^k\frac{\mathrm d}{\mathrm d(x^k)}\left(1+x^k+(x^k)^2+(x^k)^3+\dotso\right)\right)\prod_{m\neq k}\frac1{1-x^m}\\
&=\left(x^k\frac{\mathrm d}{\mathrm d(x^k)}\frac1{1-x^k}\right)\prod_{m\neq k}\frac1{1-x^m}\\
&=\frac{x^k}{(1-x^k)^2}\prod_{m\neq k}\frac1{1-x^m}\\
&=\frac{x^k}{1-x^k}\prod_m\frac1{1-x^m}\\
&=p(x)\frac{x^k}{1-x^k}\;.\\
\end{align}
$$
To count the number of at-least-$k$-fold occurrences of parts (the right-hand side of the equation), consider this generation function (which I'll first write out for $k=2$ to illustrate the idea and then generalize):
$$\begin{align}
g_2(x,y)
&=\left(1+x+y(x^2+x^3+\dotso)\right)\left(1+x^2+y((x^2)^2+(x^2)^3+\dotso)\right)\dots\\
&=\left(1+x+\frac{yx^2}{1-x}\right)\left(1+x^2+\frac{y(x^2)^2}{1-x^2}\right)\dots\\
&=\frac{1-x^2+yx^2}{1-x}\frac{1-(x^2)^2+y(x^2)^2}{1-x^2}\dots\\
&=p(x)\left(1-x^2+yx^2\right)\left(1-(x^2)^2+y(x^2)^2\right)\dots\;,\\
g_k(x,y)
&=\prod_m\left(\sum_{l=0}^{k-1}(x^m)^l+y\sum_{l=k}^\infty(x^m)^l\right)\\
&=p(x)\prod_m\left(1-(x^k)^m+y(x^k)^m\right)\;.
\end{align}
$$
Every factor of $y$ tracks the at-least-$k$-fold use of a part. What we want is the total count of factors of $y$ included in the coefficient of $x^n$; that is, we want to count $j$ times the coefficient of $y^jx^n$. This we can get by differentiating with respect to $y$ and then setting $y$ to $1$; the coefficient of $x^n$ in the result will be the desired count. But
$$\begin{align}
\left.\frac{\mathrm d}{\mathrm dy}g_k(x,y)\right|_{y=1}
&=\left.\frac{\mathrm d}{\mathrm dy}p(x)\prod_m\left(1-(x^k)^m+y(x^k)^m\right)\right|_{y=1}\\
&=\left.p(x)\sum_{l=1}^\infty(x^k)^l\prod_{m\neq l}\left(1-(x^k)^m+y(x^k)^m\right)\right|_{y=1}\\
&=p(x)\sum_{l=1}^\infty(x^k)^l\\
&=p(x)\frac{x^k}{1-x^k}\;,\\
\end{align}$$
which coincides with the result for the left-hand side.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/69244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
} |
Find x to keep the equality $\sqrt[2x+1]{\sqrt[11-4x]{(-2x)^{3x}}}=\sqrt[3x-1]{7x+2}$ $$\mbox{ Find }x \in \mathbb{Q} \mbox{ to keep the equality: } \sqrt[2x+1]{\sqrt[11-4x]{(-2x)^{3x}}}=\sqrt[3x-1]{7x+2}$$
I tried to write the roots using powers:
\begin{align*}\sqrt[2x+1]{\sqrt[11-4x]{(-2x)^{3x}}}=\sqrt[3x-1]{7x+2}&\Rightarrow
[(-2x)^{\frac{3x}{11-4x}}]^{\frac{1}{2x+1}}=(7x+2)^{\frac{1}{3x-1}}\\
&\Rightarrow (-2x)^{\frac{\frac{3x}{11-4x}}{2x+1}}=(7x+2)^{\frac{1}{3x-1}}\\
&\Rightarrow (-2x)^{\frac{3x}{(11-4x)(2x+1)}}=(7x+2)^{\frac{1}{3x-1}}\\
&\Rightarrow (-2x)^{\frac{3x}{-8x^2+18x+11}}=(7x+2)^{\frac{1}{3x-1}}
\end{align*}
I hope I did it right until this point. But I've stuck here. Can someone help me?
Thanks.
| Take logs and then use your favourite numerical root-finding algorithm. Hint: the logarithms are only both real for $-\frac{2}{7} < x < 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/69780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Help solving this integral $\int_{-\infty}^{\infty} \frac{x^2 e^{-x^2/2}}{a+bx^2}dx$ So, I've got an integral in the following form:
$$\int_{-\infty}^{\infty} \frac{x^2 e^{-x^2/2}}{a+bx^2}dx$$
where $b<0$ and $a\in\mathbb{R}$.
I've tried substituting $y=x^2$ (after changing changing lower limit to 0 and multiplying by 2 of course) and $z=y+a$ but there is that pesky square root in the denominator...
Anyone with better ideas? Is this thing even soluble?
| Consider the function
$$\mathcal{I}(a)=\int_{-\infty}^{+\infty} \frac{a}{a^2+x^2}e^{-(a^2+x^2)}dx.$$
Integration by parts gives
$$\mathcal{I}(a)=\left[\tan^{-1}\left(\frac{x}{a}\right) e^{-(a^2+x^2)}\right]_{-\infty}^{+\infty}-\int_{-\infty}^{+\infty}\tan^{-1}\left(\frac{x}{a}\right)(-2x)e^{-(a^2+x^2)}dx$$
$$=\int_{-\infty}^{+\infty}\tan^{-1}\left(\frac{x}{a}\right)2xe^{-(a^2+x^2)}dx.$$
Now differentiate $\mathcal{I}$ with respect to $a$ and obtain
$$\frac{d\,\mathcal{I}}{da}=\int_{-\infty}^{+\infty}\left[-\frac{x}{a^2+x^2}\right]2xe^{-(a^2+x^2)}+\tan^{-1}\left(\frac{x}{a}\right)\left[(-2a)2xe^{-(a^2+x^2)}\right]dx$$
$$=-\int_{-\infty}^{+\infty}\frac{2x^2}{a^2+x^2}e^{-(a^2+x^2)}dx-2a\mathcal{I}(a) $$
$$=-\int_{-\infty}^{+\infty}\left(\frac{2x^2}{a^2+x^2}+2a\frac{a}{a^2+x^2}\right)e^{-(a^2+x^2)}dx $$
$$=-2\int_{-\infty}^{+\infty}e^{-(x^2+a^2)}dx=-2\sqrt{\pi}e^{-a^2}.$$
Equipped with the fact $\lim\limits_{a\to\infty}\mathcal{I}(a)=0$, we arrive at
$$\mathcal{I}(a)=\int_{+\infty}^a -2\sqrt{\pi}e^{-u^2}du= \pi \,\mathrm{erfc}(a),$$
where $\mathrm{erfc}$ is the complementary error function. Note this agrees as $a\to0$ because of the distributional fact that $a/(a^2+x^2)\to\delta(x)$. This implies
$$\int_{-\infty}^{+\infty}\frac{1}{x^2+a}e^{-x^2}dx=\pi e^a\frac{\mathrm{erfc}\left(\sqrt{a}\right)}{\sqrt{a}}.$$
Finally, observe that
$$\int_{-\infty}^{+\infty}\frac{x^2}{a+bx^2}e^{-x^2/2}dx=\frac{1}{b}\int_{-\infty}^{+\infty}\left(1-\frac{a}{a+bx^2}\right)e^{-x^2/2}dx$$
$$=\frac{1}{b}\left(\sqrt{2\pi}-\frac{a}{\sqrt{2}b}\int_{-\infty}^{+\infty}\frac{1}{\frac{a}{2b}+x^2}e^{-x^2}dx\right)$$
$$=\frac{1}{b}\left(\sqrt{2\pi}-\sqrt{\frac{a}{b}}\pi\exp\left(\frac{a}{2b}\right)\mathrm{erfc}\left(\sqrt{\frac{a}{2b}}\right)\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Variance of sample variance? What is the variance of the sample variance? In other words I am looking for $\mathrm{Var}(S^2)$.
I have started by expanding out $\mathrm{Var}(S^2)$ into $E(S^4) - [E(S^2)]^2$
I know that $[E(S^2)]^2$ is $\sigma$ to the power of 4. And that is as far as I got.
| Showing the derivation of $E(\left[{1\over2}(X-Y)^2-\sigma^2\right]^2) = (\mu_4+\sigma^4)/2$ of user940:
LHS:
$E(\left[{1\over2}(X-Y)^2-\sigma^2\right]^2) = E(\frac{1}{4}(X-Y)^4 - (X-Y)^2 \sigma^2 + \sigma^4) = E(\frac{1}{4}(X-Y)^4) - 2\sigma^2\sigma^2 + \sigma^4 = E(\frac{1}{4}(X-Y)^4) - \sigma^4 = \frac{1}{4}E(X^4 -4X^3Y +6X^2Y^2 -4XY^3 + Y^4) -\sigma^4 = \frac{1}{4}(2E(X^4) -8E(X)E(X^3) +6 E(X^2)(X^2)) - \sigma^4 = \frac{1}{2}(E(X^4)-4E(X)E(X^3) +3 E(X^2)(X^2) - 2\sigma^4)$
I use the fact that $E((x-y)^2) = 2\sigma^2$ here.
RHS:
$\require{cancel} (\mu_4+\sigma^4)/2 = \frac{1}{2}(E((X-\mu)^4) + \sigma^4) = \frac{1}{2}(E((X-E(X))^4) + \sigma^4) = \frac{1}{2}(E(X^4 -4X^3E(X) + 6X^2E(X)^2 -4XE(X)^3 + E(X)^4) + \sigma^4) = \frac{1}{2}(E(X^4 -4X^3E(X) + 6X^2E(X^2) - 6X^2\sigma^2 -4XE(X)(E(X^2)-\sigma^2) + (E(X^2)-\sigma^2)^2) + \sigma^4) =
\frac{1}{2}(E(X^4) -4E(X)^3E(X) + 6E(X)^2E(X^2) - 6E(X)^2\sigma^2 -4E(X)^2(E(X^2)-\sigma^2) + (E(X^2)-\sigma^2)^2 + \sigma^4) =
\frac{1}{2}(E(X^4) -4E(X)^3E(X) + 6E(X)^2E(X^2) - \cancel{6E(X)^2\sigma^2} -4E(X^2)E(X^2) +\cancel{4E(X^2)\sigma^2 +4E(X^2)\sigma^2} - 4\sigma^4 + E(X^2)^2-\cancel{2E(X^2)\sigma^2} + \sigma^4 + \sigma^4) =
\frac{1}{2}(E(X^4) -4E(X)^3E(X) + 3E(X)^2E(X^2) - 2\sigma^4)$
I use the fact that $E(x) = \mu$ and that $E(x)^2 = E(x^2) - \sigma^2$
Now LHS = RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/72975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "101",
"answer_count": 6,
"answer_id": 2
} |
Isolate a variable in a polynomial function How would I go about isolating $y$ in this function? I'm going crazy right now because I can't figure this out.
The purpose of this is to allow me to derive $f(x)$ afterwards.
$$ x = \frac{y^2}{4} + 2y .$$
| Point the Zeroth: ignore points the first and points the second (in the sense that they aren't really the 'right' way of proceeding; they are presented so you can see that they are not the right path to take).
Point the First: $y$ is not a function of $x$; if you plot this equation on the plane, you'll have a parabola,
$$x = \frac{y^2}{4} + 2y = \left(\frac{y}{2}\right)^2 + 2\left(\frac{y}{2}\right)(2) + 2^2 - 2^2 = \left(\frac{y}{2} + 2\right)^2 - 4.$$
This parabola opens right, so it is not the graph of a function of $x$.
Point the Second: You can break up the graph into two functions by using the quadratic formula:
$$ \frac{y^2}{4} + 2y - x = 0$$
gives
$$y^2 + 8y - 4x = 0,$$
so
$$y = \frac{-8+\sqrt{64+16x}}{2},\quad\text{or}\quad y = \frac{-8-\sqrt{64+16x}}{2}.$$
We would then need to find the derivatives of each of these two separately, and for any given value of $x$ and $y$, determine which of the two formulas to use. They are not hard, but they are somewhat annoying.
If $y = -4 + \frac{1}{2}\sqrt{64+16x}$, then
$$\frac{dy}{dx} = \frac{1}{4}(64+16x)^{-1/2}(16) = \frac{4}{\sqrt{64+16x}}.$$
Similarly, if $y=-4-\frac{1}{2}\sqrt{64+16x}$, then
$$\frac{dy}{dx} = -\frac{4}{\sqrt{64+16x}}.$$
Point the Third: What you really want to do here is implicit differentiation, which is a way of handling all of these difficulties without having to solve for $y$ first, and without having to worry about "which formula" to use later. Explicitly, from
$$x = \frac{y^2}{4} + 2y,$$
take derivatives on both sides, using the Chain Rule and remembering that $y$ is an (implicit) function of $x$, so that $y'$ needs to be left indicated (we don't know what it is right now):
$$\begin{align*}
x & = \frac{y^2}{4} + 2y\\
\frac{d}{dx}x &= \frac{d}{dx}\left( \frac{y^2}{4} + 2y\right)\\
1 &= \frac{2y}{4}y' + 2y'\\
1&= \frac{y}{2}y' + 2y'\\
1 &= y'\left(\frac{y}{2} + 2\right).\end{align*}$$
Solving for $y'$ gives an implicit definition for $\frac{dy}{dx}$ in terms of $y$ and $x$ (though in this case, $x$ plays no role):
$$y' = \frac{1}{\frac{y}{2}+2} = \frac{2}{y+4}.$$
Point the Fourth: Alternatively, since you have $x$ explicitly as a function of $y$, use the Inverse Function Theorem: taking derivatives with respect to $y$, we have:
$$\frac{dx}{dy} = \frac{1}{2}y + 2,$$
so
$$\frac{dy}{dx} = \frac{1}{\quad\frac{dx}{dy}\quad} = \frac{1}{\frac{1}{2}y + 2} = \frac{2}{y+4}.$$
Point the Fifth: So, do these "implicit formulas" give the same answer as the "explicit ones" we got in Point the Second? Yes!
If $y = \frac{-8+\sqrt{64+16x}}{2}$, then $y+4 = \frac{\sqrt{64+16x}}{2}$, so
$$\frac{2}{y+4} = \frac{2}{\frac{1}{2}\sqrt{64+16x}} = \frac{4}{\sqrt{64+16x}},$$
same answer as in Point the Second; and if $y=\frac{-8-\sqrt{64+16x}}{2}$ then $y+4 = -\frac{1}{2}\sqrt{64+16x}$, so
$$\frac{2}{y+4} = \frac{2}{-\frac{1}{2}\sqrt{64+16x}} = -\frac{4}{\sqrt{64+16x}},$$
again, same answer as in Point the Second.
But using implicit differentiation (or in cases like this, when $x$ is an explicit function of $y$, the inverse function theorem) is much easier than first solving for $y$, possibly requiring breaking up the original implicit function into several different explicit functions, and then differentiating. If you were trying to work with the Folium of Descartes ($x^3+y^3=3xy$), you would have to consider three different formulas, each involving a sum of cubic roots that has square roots inside the radicals; if you were trying to work with a function like $y = \sin(x+y)$, you would have a hard time solving for $y$, but using implicit differentiation is pretty easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/73544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $\sum\limits_{n=1}^\infty\frac{(n-1)!}{n\prod\limits_{i=1}^n(a+i)}=\sum\limits_{k=1}^\infty \frac{1}{(a+k)^2}$ for $a>-1$? A problem on my (last week's) real analysis homework boiled down to proving that, for $a>-1$,
$$\sum_{n=1}^\infty\frac{(n-1)!}{n\prod\limits_{i=1}^n(a+i)}=\sum_{k=1}^\infty \frac{1}{(a+k)^2}.$$ Mathematica confirms this is true, but I couldn't even prove the convergence of the original series (the one on the left), much less demonstrate that it equaled this other sum; the ratio test is inconclusive, and the root test and others seem hopeless. It was (and is) quite a frustrating problem. Can someone explain how to go about tackling this?
| Since at least J. M. asked for it, here's another solution for the case
when $a$ is a natural number.
I'll use the forward difference operator $\Delta$, defined by
$\Delta f(n) = f(n+1) - f(n)$, and the falling factorial defined
by
$$
n^{\underline{a}} =
\begin{cases}
n(n-1)(n-2) \dots (n-a+1), & a > 0, \\
1, & a=0 \\
\frac{1}{(n+1)(n+2) \dots (n+|a|)}, & a < 0,
\end{cases}
$$
and satisfying $\Delta n^{\underline{a}} = a n^{\underline{a-1}}$.
The summand, which I'll denote by $F_a(n)$, can be rewritten as
$$
F_a(n)
= \frac{(n-1)!}{n\prod_{i=1}^n(a+i)}
= \frac{(n-1)! a!}{n (a+n)!}
= \frac{a!}{n \cdot n(n+1)(n+2) \dots (n+a)}
$$
$$= \frac{(a-1)!}{n} \left( -(-a) (n-1)^{\underline{-(a+1)}}\right)
= -\frac{(a-1)!}{n} \Delta\left( (n-1)^{\underline{-a}}\right).
$$
Using the rule $\Delta(f(n)g(n)) = \Delta f(n) \, g(n+1) + f(n) \Delta g(n)$,
we get
$$
F_a(n)
= - \Delta\left( \frac{(a-1)!}{n} (n-1)^{\underline{-a}}\right)
+ \Delta\left( \frac{(a-1)!}{n} \right) \, n^{\underline{-a}}
$$
$$
= - \Delta\left( \frac{(a-1)!}{n \cdot n (n+1) \dots (n+a-1)} \right)
+ (a-1)! \left( \frac{1}{n+1} - \frac{1}{n} \right) \frac{1}{(n+1)\dots (n+a)}
$$
$$= - \Delta\left( \frac{(a-1)!}{n \cdot n(n+1) \dots (n+a-1)} \right)
+ F_{a-1}(n+1) - (a-1)! \Delta\left( \frac{(n-1)^{\underline{-a}}}{-a} \right).
$$
Summing over $n \ge 1$ gives (because of telescoping in the sums-of-deltas)
$$
\sum_{n=1}^{\infty} F_a(n)
= \frac{(a-1)!}{1 \cdot a!} + \sum_{n=1}^{\infty} F_{a-1}(n+1)
- \frac{(a-1)!}{a} 0^{-\underline{a}}
$$
$$
= \frac{1}{a} + \sum_{m=2}^{\infty} F_{a-1}(m)
- \frac{1}{a^2}
$$
$$
= \sum_{m=1}^{\infty} F_{a-1}(m) - \frac{1}{a^2}
$$
(since $F_{a-1}(1) = 1/a$).
Finally, since $F_0(n) = 1/n^2$, we obtain after using this result to work our way down $n$ steps that
$$
\sum_{n=1}^{\infty} F_a(n)
= \sum_{n=1}^{\infty} F_{a-1}(n) - \frac{1}{a^2}
= \dots = \sum_{n=1}^{\infty} F_0(n) - \left( \frac{1}{a^2} + \dots + \frac{1}{1^2} \right)
= \sum_{n=a+1}^{\infty} \frac{1}{n^2}
= \sum_{k=1}^{\infty} \frac{1}{(k+a)^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/75681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 1
} |
Prove the identity $ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$ $$ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$$
Class themes are: Generating functions and formal power series.
| Here is a slightly different proof that is simpler than the other one
I posted earlier.
Suppose we seek to verify that
$$\sum_{q\ge 0} {p+q\choose q} {m+2p\choose m-2q}
= 2^{m-1} \frac{2p+m}{m} {m+p-1\choose p}.$$
This is
$$\sum_{q\ge 0} {p+q\choose q} {m+2p\choose 2p+2q}.$$
We introduce
$${m+2p\choose 2p+2q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-2q+1}} \frac{1}{(1-z)^{2p+2q+1}} \; dz.$$
This integral controls the range, being zero when $2q\gt m$ and we
may extend the range of $q$ to infinity. We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} \frac{1}{(1-z)^{2p+1}}
\sum_{q\ge 0} {p+q\choose q} \frac{z^{2q}}{(1-z)^{2q}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} \frac{1}{(1-z)^{2p+1}}
\frac{1}{(1-z^2/(1-z)^2)^{p+1}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1-z}{z^{m+1}}
\frac{1}{((1-z)^2-z^2)^{p+1}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1-z}{z^{m+1}}
\frac{1}{(1-2z)^{p+1}}
\; dz.$$
Extracting coefficients we get
$$2^m {m+p\choose p} - 2^{m-1} {m-1+p\choose p}
\\ = 2^{m-1} {m-1+p\choose p}
\left(2\frac{m+p}{m} - 1 \right)
\\ = 2^{m-1} {m-1+p\choose p}
\frac{m+2p}{m}.$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
How to add cosines with different phases using phasors So, here's a question:
$ \cos( \omega t ) + 2 \cos( \omega t + \frac{\pi}{4} ) + 3 \cos( \omega t + \frac{\pi}{2} ) $
To add these together, I figure there should be at least 2 ways:
1) Cosine addition laws:
$$
\cos( \omega t ) +
2 \left(
\cos( \omega t ) \cos( \frac{\pi}{4} ) -
\sin( \omega t ) \sin( \frac{\pi}{4} )
\right)
+ 3 \left(
\cos( \omega t ) \cos( \frac{\pi}{2} ) -
\sin( \omega t ) \sin( \frac{\pi}{2} )
\right) \\
=\cos( \omega t )
\left(
1 + \sqrt{2}
\right)
-
\sin( \omega t )
\left(
3 + \sqrt{2}
\right)
$$
2) Phasors / complex addition
$$
1 \angle 0 + 2 \angle 45 ^\circ + 3 \angle 90^\circ
$$
$$
= 1 + \sqrt{2} + j \sqrt{2} + j 3
$$
$$
= 1 + \sqrt{2} + j ( 3 + \sqrt{2} )
$$
Which has
$ A = \sqrt{ 14 + 8 \sqrt{2} } \approx 5.03 $
$ \phi = \arctan{ \left( \frac{ 3 + \sqrt{2} }{ 1 + \sqrt{2} } \right) } \approx 1.07 rad \approx 61 ^\circ $
Thus answer is $ 5 \angle 61^\circ $, or $5 \cos( \omega t + 1.07 )$
If you graph them, $5 \cos( \omega t + 1.07 )$ produces the same graph as $ \cos( \omega t )
\left( 1 + \sqrt{2} \right) - \sin( \omega t ) \left( 3 + \sqrt{2} \right) $
So how can you convert between them?
| You have
$$ \tan\phi = \frac{\sin\phi}{\cos\phi}= \frac{3+\sqrt{2}}{1+\sqrt{2}} $$
Since $0<\phi<\pi/2$ we know that $\sin\phi,\cos\phi>0$. Therefore, $\sin\phi$ and $\cos\phi$ are equal to
$$ \sin\phi = \frac{3+\sqrt{2}}{\sqrt{(3+\sqrt{2})^2+(1+\sqrt{2})^2}}
=\frac{3+\sqrt{2}}{\sqrt{14+8\sqrt{2}}}$$
and
$$ \cos\phi = \frac{1+\sqrt{2}}{\sqrt{(3+\sqrt{2})^2+(1+\sqrt{2})^2}}
=\frac{1+\sqrt{2}}{\sqrt{14+8\sqrt{2}}}$$
Therefore
$$
\begin{split}
\cos(\omega t)(1+\sqrt{2}) - \sin(\omega t)(3+\sqrt{2}) &=
\sqrt{14+8\sqrt{2}}\left( \cos(\omega t)\cos\phi - \sin(\omega t) \sin\phi \right)\\&
= \sqrt{14+8\sqrt{2}} \cos(\omega t+\phi)
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/79063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Taking the derivative of $\frac1{x} - \frac1{e^x-1}$ using the definition Given $f$:
$$
f(x) = \begin{cases}
\frac1{x} - \frac1{e^x-1} & \text{if } x \neq 0 \\
\frac1{2} & \text{if } x = 0
\end{cases}
$$
I have to find $f'(0)$ using the definition of derivative (i.e., limits). I already know how to differentiate and stuff, but I still can't figure out how to solve this. I know that I need to begin like this:
$$
f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{\frac1{h} - \frac1{e^h-1}-\frac1{2}}{h}
$$
But I don't know how to do this. I feel like I should, but I can't figure it out. I tried distributing the denominator, I tried l'Hôpital's but I get $0$ as the answer, while according to what my prof gave me (this is homework) it should be $-\frac1{12}$. I really don't know how to deal with these limits; could someone give me a few tips?
| A good strategy in such problems is to massage the problem into recognizable limits. (EDIT: I like this approach mainly because it avoids Taylor expansion and l'Hôpital's rule. This is, however, not the simplest approach.)
We can "simplify" given function as follows:
$$
\begin{eqnarray*}
\frac{\frac{1}{x} - \frac{1}{e^x - 1} - \frac{1}{2}}{x}
&=&
\frac{(2-x)(e^x - 1) - 2x}{2x^2 (e^x - 1)}
\\ &=&
\frac{\color{Blue}{(2-x)}(e^x - 1)- \color{Blue}{(2-x)} \frac{2x}{2-x}}{\color{Red}{2} \ \color{Magenta}{x^2} \color{Green}{(e^x - 1)}}
\\ &=&
\frac{\color{Blue}{2-x}}{\color{Red}{2}} \cdot \frac{\color{Magenta}{x}}{\color{Green}{e^x - 1}} \cdot \frac{e^x - 1 - \frac{2x}{2-x}}{\color{Magenta}{x^3}}. \tag{1}
\end{eqnarray*}
$$
The first two factors both approach $1$ as $x \to 0$. Let us concentrate on the third factor. By Taylor expansion (or the formula for summing a geometric series), we have (for $|x| < 1$),
$$
\begin{eqnarray*}
\frac{2x}{2-x}
=
\frac{x}{1 - x/2}
&=&
x + x \left(\frac{x}{2} \right) +x \left(\frac{x}{2} \right)^2 + x \left( \frac x 2 \right)^3 + \cdots
\\ &=&
\color{Red}{x + \frac{x^2}{2}} + \color{Blue}{x \left(\frac{x}{2} \right)^2 + x \left( \frac x 2 \right)^3 + \cdots}
\\ &=&
\color{Red}{x + \frac{x^2}{2}} + \color{Blue}{x \left(\frac{x}{2} \right)^2 \frac{1}{1 - \frac x 2}} \quad\quad \text{(summing the GP)}
\\ &=&
\color{Red}{x + \frac{x^2}{2}} \color{Blue}{+\frac{x^3}{2(2-x)}}. \tag{2}
\end{eqnarray*}
$$
(Though I used infinite GPs to obtain the final expression, one could verify it directly as well. In particular, the two expressions are equal for all $x \neq 2$, not just $|x| < 1$.) Plugging $(2)$ in $(1)$, we have
$$
\begin{eqnarray*}
\frac{e^x - 1 - \frac{2x}{2-x}}{x^3}
&=&
\frac{e^x - 1 \color{Red}{- x - \frac{x^2}{2}} \color{Blue}{-\frac{x^3}{2(2-x)}}}{x^3}
\\ &=&
\frac{e^x - 1 \color{Red}{- x - \frac{x^2}{2}}}{x^3} \color{Blue}{-\frac{1}{2(2-x)}}
\end{eqnarray*}
$$
Once again, the second term has an easy limit of $\frac14$. The first term
$$
\frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}
$$
is also a standard limit. This limit can be evaluated using, say, the l'Hôpital's rule or using the Taylor expansion of $e^x$; it's value is $\frac{1}{3!}$. Plugging in both these limits, we can get the final answer to be
$$
\frac{1}{3!} - \frac{1}{4} = -\frac{1}{12}.
$$
Bonus! If you wish to avoid Taylor expansion and l'Hôpital's rule even further, I will mention an "elementary" ways to evaluate limits such as
$$
\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} \quad\text{and}\quad \lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}
$$
assuming that these limits exists! (I stress that this is not a complete proof; yet I present it because I find the technique interesting.)
I will show the idea for the first limit, and leave the second one as an exercise. Suppose $A \stackrel{\text(def)}{=} \lim \limits_{x \to 0} \frac{e^x - 1 - x}{x^2}$ exists. Then $e^x = 1 + x + A x^2 + o(x^2)$. Therefore, by squaring: $$e^{2x} = (1 + x + A x^2 + o(x^2))^2 = \color{Red}{1} + \color{Blue}{2x} + \color{DarkGreen}{x^2 (1+2A)} + o(x^2) .$$ On the other hand, making the substitution $x \to 2x$ in the definition of the limit, we have $e^{2x} = \color{Red}{1} + \color{Blue}{2x} + \color{DarkGreen}{4A x^2} + o(x^2)$.
Equating the dominant terms in these two expressions, we must have $\color{DarkGreen}{4A} = \color{DarkGreen}{1 + 2A}$, which gives $A = \frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/80078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How can one prove that $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$? How can one prove that $\sqrt[3]{\left ( \frac{a^4+b^4}{a+b} \right )^{a+b}} \geq a^ab^b$, $a,b\in\mathbb{N^{*}}$?
| This is expanding on a comment by Bill, the following might work:
You need
$$ (a+b)\ln\sqrt[3]{(\frac{a^4+b^4}{a+b})} \geq a \ln(a) + b \ln(b) \,.$$
Or
$$ (\ln\sqrt[3]{(\frac{a^4+b^4}{a+b})} \geq \frac{a}{a+b} \ln(a) + \frac{b}{a+b} \ln(b) \,.$$
Now, if I remember right, the Jensen inequality for Log reads:
$$\frac{a}{a+b} \ln(a) + \frac{b}{a+b} \ln(b) \leq \ln (\frac{a^2+b^2}{a+b}) \,.$$
Thus, you only need to show
$$\left( \frac{a^2+b^2}{a+b} \right)^3 \leq \frac{a^4+b^4}{a+b} \,.$$
Or
$$(a^2+b^2)^3 \leq (a+b)^2(a^4+b^4) \,.$$
EDIT
After a long calculation, this reduces to
$$a^6+3a^4b^2+3a^2b^4+b^6 \leq a^6+a^2b^4+2a^5b+2ab^5+a^2b^4+b^6$$
or
$$a^4b^2+a^2b^4 \leq a^5b+ab^5$$
After canceling $ab$ this follows imediatelly form the AM-GM.: $a^3b \leq \frac{a^4+a^4+a^4+b^4}{4}$ and $ab^3 \leq \frac{a^4+b^4+b^4+b^4}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/80550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
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