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Evaluate $\sum\limits_{1 \le j\le k\le n} jk $ I am trying to solve an exercise from Concrete Mathematics but I seem to be stuck on the sum $$\sum_{1 \le j\le k\le n} jk $$ How to proceed? I have tried using Iverson's bracket condition like
$$ [1\le j\le k\le n] = [1\le j \le n][j\le k \le n] $$
but I am not sure how to write the sum as multiple sums.
|
An alternative is to set $j-1=a, k-j=b, n-k=c$. Then we are looking for
$$ S(n)=\sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}(a+1)(a+b+1)=\sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}(1+2a+b+a^2+ab) $$
where
$$ \sum_{d\geq 0}x^d = \frac{1}{1-x},\qquad \sum_{d\geq 0}dx^d = \frac{x}{(1-x)^2},\qquad \sum_{d\geq 0}d^2 x^d = \frac{x(1+x)}{(1-x)^3} $$
and
$$ \sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}\!\!\!1 = [x^{n-1}]\frac{1}{(1-x)^3},\qquad \sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}a=[x^{n-1}]\frac{x}{(1-x)^4} $$
$$ \sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}\!\!\!a^2 = [x^{n-1}]\frac{x(1+x)}{(1-x)^5},\qquad \sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}ab=[x^{n-1}]\frac{x^2}{(1-x)^5} $$
so
$$ S(n) = [x^{n-1}]\left(\frac{1}{(1-x)^3}+\frac{3x}{(1-x)^4}+\frac{x(1+x)}{(1-x)^5}+\frac{x^2}{(1-x)^5}\right)$$
or
$$ S(n) = [x^{n-1}]\frac{2x+1}{(1-x)^5}=\binom{n+3}{4}+2\binom{n+2}{4}=\color{red}{\frac{n(n+1)(n+2)(3n+1)}{24}}. $$
An interesting and more practical way is to prove first that $S(n)$ is a fourth-degree polynomial in the $n$ variable, then find its coefficients through Lagrange interpolation, once computed $S(1),S(2),S(3),S(4),S(5)$.
|
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|
Multiplication of matrices --$P^{-1}Q^{n}P^{-1}$ If $P$ be a square matrix - $\begin{pmatrix} √3/2 &1/2 \\ 1/2&√3/2 \end{pmatrix}$
and $A$ be another square matrix -$\begin{pmatrix} 1&1 \\ 0&1 \end{pmatrix}$. Let $Q$ be the matrix given by $Q=PAP^{-1}$ then the question is to find the value of $P^{-1}(Q^{n})P^{-1}$ for some natural number $n$. Here $P^{-1}$ represents inverse of matrix $P$.
Since $A$ have determinant unity, its inverse matrices are given by the adjoints. I calculated $Q$ and then took its powers but could not see a pattern from which $ Q^n$ can be predicted. I tried many times but failed. Please help me out. Thanks.
|
Hint: $Q^n = P A^n P^{-1}$ so the problem reduces to calculating $A^n$. Write:
$$
A = \begin{pmatrix} 1&1 \\ 0&1 \end{pmatrix} = \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} + \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}=I+A_1
$$
Note that $I,A_1$ commute, $I^k=I$ and $A_1^2=0$ so by binomial expansion:
$$
\require{cancel}
A^n = (I+A_1)^n = I^n + \binom{n}{1} I^{n-1} A_1 + \cancel{\binom{n}{2} I^{n-2} A_1^2 + \cdots} = I + n A_1 = \begin{pmatrix} 1&n \\ 0&1 \end{pmatrix}
$$
|
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By knowing, that $a_{n+1}=(n+3)a_n$, how can I find $a_n$? $$a_1=1$$
$$a_{n+1}=(n+3)a_n$$
How can I get to the answer of this, which is:
$$a_n=\frac{(n+2)!}{6}$$
|
write it as follows
$$\frac { { a }_{ n+1 } }{ { a }_{ n } } =n+3\\ \frac { { a }_{ 2 } }{ { a }_{ 1 } } =4,\frac { { a }_{ 3 } }{ { { a } }_{ 2 } } =5,\frac { { a }_{ 4 } }{ { a }_{ 3 } } =6,...\frac { { a }_{ n } }{ { a }_{ n-1 } } =n+2\\ \\ \frac { { a }_{ 2 } }{ { a }_{ 1 } } \cdot \frac { { a }_{ 3 } }{ { { a } }_{ 2 } } \cdot \frac { { a }_{ 4 } }{ { a }_{ 3 } } \cdot ...\cdot \frac { { a }_{ n } }{ { a }_{ n-1 } } =\frac { \left( n+2 \right) ! }{ 6 } $$
|
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Find the limit of $x(\sqrt{x^2+1}- \sqrt[3]{x^3+1})$ as $x\to +\infty$.
Calculate $\lim_{x\to +\infty} x(\sqrt{x^2+1}- \sqrt[3]{x^3+1})$.
First thing came to my mind is to simplify this to something easier. So multiply the numerator and the denominator by something like we used to do when two square roots involves. But I am trying to find that suitable term but it seems out of my reach. Can anybody help me to solve this ? Any hint or help would be nice . Thanks.
|
Hint :
Use that
$$a-b=\frac{a^6-b^6}{(a^2+ab+b^2)(a^3+b^3)}$$
where $a=\sqrt{x^2+1}$ and $b=\sqrt[3]{x^3+1}$ to have
$$\begin{align}\lim_{x\to\infty}x(a-b)&=\lim_{x\to\infty}\frac{(a^6-b^6)x}{(a^2+ab+b^2)(a^3+b^3)}\\\\&=\lim_{x\to\infty}\frac{((x^2+1)^3-(x^3+1)^2)x}{(x^2+1+\sqrt{x^2+1}\sqrt[3]{x^3+1}+(x^3+1)^{2/3})((x^2+1)^{3/2}+x^3+1)}\end{align}$$
Now divide top and bottom by $x^5$.
|
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|
Find all $f:\mathbb{N} \to \mathbb{N}$ such that $f(m)^2 + f(n) \mid (m^2+n)^2;\; \forall n,m \in \mathbb{N}$
Find all $f:\mathbb{N} \to \mathbb{N}$ such that $f(m)^2 + f(n) \mid (m^2+n)^2;\; \forall n,m \in \mathbb{N}$
I tried to write $k(f(m)^2 + f(n)) = (m^2 + n) ^2$ and tried to find if there is any fixed $k$ but failed. I can't find a way to start.
|
(Note the obvious fact that the identity does satisfy the requirements.)
This is a partial answer, whose result suffices to show that $f(n) = n$ for all $n \leq 100000$.
Letting $n = m^2$, we obtain $f(m)^2 + f(m^2) \mid 4m^4$, and in particular $f(1)^2 + f(1) \mid 4$. It's easy to check that $f(1) = 1$ is the only possibility, since $f(1) > 1$ means $f(1)^2 + f(1) > 4$.
Letting $n=1$, we obtain $f(m)^2 + 1 \mid (m^2 + 1)^2$.
So if $m^2+1$ is prime, then $f(m)^2+1 = m^2+1$ or $(m^2+1)^2$; so $f(m) = m$ or $f(m)^2+1$ is square (a contradiction).
Therefore if $m^2+1$ is prime, then $f(m) = m$.
Similarly, if $m^2+2$ is prime, then $f(m) = m$ (since the other option is $f(m)^2+2 = (m^2+2)^2$, a contradiction since no squares differ by $2$).
Similarly, if $m^2+3$ is prime, then $f(m) = m$ or $f(m)^2 + 3 = (m^2 + 3)^2$. The second case is no longer an instant contradiction, but it requires that $f(m) = 1$ and $m^2 + 3 = 2$, which is a contradiction.
We now generalise this, by showing that if there is a prime between $m^2$ and $m^2 + m$, and if $f(n) = n$ for all $n < m$, then $f(m) = m$.
Suppose there is a prime $p$ between $m^2$ and $m^2+m$.
Then let $n = p-m^2$, so by inductive hypothesis (since $n < m$) we have that $f(n) = n$.
We have $f(m)^2 + p-m^2 \mid p^2$, so one of the following three is true:
*
*$f(m)^2 + p - m^2 = 1$ (so $f(m)^2 = 1+m^2-p = 1-n < 0$, a contradiction),
*$f(m)^2 +p-m^2 = p$ (so $f(m) = m$), or
*$f(m)^2 + p - m^2 = p^2$.
This latter case ("Case 3") is the one we want to disprove, of course.
Case 3 is false
Case 3 implies that $p(p-1) = (f(m) + m)(f(m)-m)$; since $p$ is prime, that means $p$ divides one of those brackets. It can't divide $f(m)-m$, since that would make $f(m) + m$ greater than $p$ and therefore the right-hand side would be too big to equal the left-hand side; hence $p \mid f(m) + m$.
Lemma: $p = f(m)+m$, not just $p \mid f(m)+m$.
Proof: If $p \not = f(m) + m$, then we have $f(m) + m \geq 2p > 2m^2$; so $f(m) > 2m^2 - m$. Then $$p(p-1) = f(m)^2 - m^2 > (2m^2-m)^2 - m^2 = 4m^3(m-1)$$
but $p \in (m^2, m^2+m)$, and $p-1 \in [m^2, m^2+m-1)$, so $p(p-1)$ cannot be greater than or equal to $4m^3(m-1)$, let alone $f(m)^2 - m^2$.
Hence Case 3 implies that $f(m) = p-m$; since $p(p-1) = (f(m)+m)(f(m)-m)$, we must also have $p-1 = f(m)-m$, and hence $m = 1-m$, a contradiction.
The above shows that if there is a prime between $m^2$ and $m^2 + m$, and if $f(n) = n$ for all $n < m$, then $f(m) = m$.
We have already shown that $f(1) = 1$, but the question of whether there is a prime between $m^2$ and $m^2 + m$ is a stronger version of Legendre's conjecture which is unproven. There is probably a better way.
|
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Find the limit of $\frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$
Let $\{x_n\}_{n\geq 1}$ be defined as $x_n = \frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$. Then $\lim\limits_{n\to\infty} x_n$ is?
I want to find limit of this problem by a very specific method . I am uploading the pic of that method and my attempt.
Please guide me as to how to take this method ahead.
|
Alternate solution:
$$x_n < \frac{n}{n^3+1} + \frac{2n}{n^3+1} + \cdots + \frac{n\cdot n}{n^3+1} = \frac{n(1+2+\cdots + n)}{n^3+1} = \frac{n\cdot n(n+1)/2}{n^3+1}.$$
The limit of the last expression is $1/2.$ From below we have
$$\frac{n}{n^3+n} + \frac{2n}{n^3+n} + \cdots + \frac{n\cdot n}{n^3+n} = \frac{n\cdot n(n+1)/2}{n^3+n} < x_n.$$
The last fraction also has limit $1/2.$ By the squeeze theorem the desired limit is $1/2.$
|
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How many positive integers are divisors of at least one of $24^5, 20^6$, and $45^7$? I have a vague idea of prime factorizing 24, 20, and 45, then distributing the exponent, find how many numbers are divisible by them (factors), and then subtract the overcounts. How do I do this?
|
$24=2^3\times 3\\
\implies 24^5=2^{15}\times 3^5$
Positive divisors of $24^5=16 \times 6=96$
$20=2^2\times 5\\
\implies 20^6=2^{12}\times 5^6$
Positive divisors of $20^6=13 \times 7=91$
$45=3^2\times 5\\
\implies 45^7=3^{14}\times 5^7$
Positive divisors of $45^7=15 \times 8=120$
Positive divisors of both $24^5$ and $20^6=13$ (as $2^{12}$ is the only common factor)
Positive divisors of both $24^5$ and $45^7=6$
Positive divisors of both $20^6$ and $45^7=7$
Positive divisors of $24^5,20^6,45^7=1$
Required answer
$96+91+120-(13 + 6 + 7) + (1)$
|
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I have done this induction proof for the Fibonacci-sequence Would you please let me know how do I improve this proof?
The numbers $$F_{0},F_{1},F_{2},...$$ are defined as follows (this is a definition
by mathematical induction, by the way): $$F_{0} = 0, F_{1} = 1, F_{n+2} =
F_{n+1} + F_{n} \ for \ n=0,1,2,...$$
Prove that for any $$n≥0$$ we have
$$F_{n} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{n−1}$$
proof
Let P(n) be the proposition that
$$F_{n} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{n−1}$$
And we know
$$F_{0} = 0, F_{1} = 1, F_{n+2} =
F_{n+1} + F_{n} \ for \ n=0,1,2,...$$
Then set up the base cases P(0),P(1),P(2) as
$$F_{0} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{0−1},F_{1} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{1−1},F_{2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{2−1}$$
we know from the fibonacci sequence and the given information, $$F_{0} = 0,F_{1} = 1,F_{2} = 1$$
Therefore, P(0),p(1),p(2) is true.
Assume that there is an $$k, (k \in \mathbf{Z})$$, then p(k) is
$$F_{k} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k−1}$$
And because $$F_{n+2}>F_{n+1}>F_{n} \ and \ F_{n+2} =
F_{n+1} + F_{n} \ for \ n=0,1,2,...$$
And then by PMI, $$p(k) \implies p(k+1) \implies p(k+2)$$ so that
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k+2−1}=\left(\frac{1 + \sqrt {5}}{2}\right)^{k+1}$$ and because
$$F_{k+2} ＝ F_{k+1} ＋ F_{k}$$
Therefore $$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}+\left(\frac{1 + \sqrt {5}}{2}\right)^{k-1}$$
Therefore $$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}+\left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(\frac{1 + \sqrt {5}}{2}\right)^{-1}$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(1+\left(\frac{1 + \sqrt {5}}{2}\right)^{-1}\right)$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(1+\frac{2}{1 + \sqrt {5}}\right)$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(\frac{1 + \sqrt {5}}{1 + \sqrt {5}}+\frac{2}{1 + \sqrt {5}}\right)$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(\frac{3+ \sqrt {5}}{1 + \sqrt {5}}\right)$$
because $$\left(\frac{3+ \sqrt {5}}{1 + \sqrt {5}}\right)=\left(\frac{1 + \sqrt {5}}{2}\right)$$
Therefore $$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k+1}$$
In conclusion, $$p(k) \implies p(k+1) \implies p(k+2)$$ is true.
By PMI, $$p(n) \implies p(n+1) \implies p(n+2) \ and p(n)$$ is true.
|
Base cases are fine.
At the inductive hypothesis you must assume that $P(k)$ and $P(k-1)$ are true. You have only said to assume $P(k)$
You could use "Strong induction" and assume that for all $i\le k, P(i)$ is true.
And then you seem to spin a while, to get to the point.
Show that $P(k+1)$ is true based on the assumption $P(k)$ and $P(k-1)$ are true
let $\phi = \frac {1 + \sqrt 5}{2}$
Show that
$F_{k-1} < \phi^{k-2} , F_{k} < \phi^{k-1} \implies F_{k+1}<\phi^{k}$
$F_{k+1} = F_k + F_{k-1}$
$F_k + F_{k-1}<\phi^{k-1} + \phi^{k-2}$
$F_{k+1}<\phi^{k-2} (\phi+1)$
I say $\phi^2 = \phi+1$
$\left(\frac {1+\sqrt 5}{2}\right)^2 = \frac {6+2\sqrt 5}{4} = 1+\frac {1+\sqrt 5}{2}$
$F_{k+1}<\phi^{k}$
QED
|
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If n is a positive multiple of 6, show that ${n\choose1}-3{n\choose3}+3^2{n\choose5}-\cdots=0$ If n is a positive multiple of 6, show that
$${n\choose1}-3{n\choose3}+3^2{n\choose5}-\cdots=0$$
$${n\choose1}-\frac{1}{3}{n\choose3}+\frac{1}{3^2}{n\choose5}-\cdots=0$$
I know that $\frac{1}{2}[(1+x)^n-(1-x)^n]$ will isolate the odd terms in the expansion, but I'm not sure how to make the series alternate and have the corresponding powers.
|
As you already stated,
$$(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$$
and
\begin{align*}
\frac{1}{2}\left((1+x)^n - (1-x)^n\right) &= \sum_{k \geq 0} \binom{n}{2k+1} x^{2k+1}\\
&= \binom{n}{1}x + \binom{n}{3}x^3 + \binom{n}{5}x^5 + \ldots
\end{align*}
Now, to get alternate powers of odd terms, what rings a bell, is that $i^{4k+1} = i$ and $i^{4k+3} = -i$. So we use, this to get the required result as
\begin{align*}
\frac{1}{2}\left((1+ix)^n - (1-ix)^n\right) &= \binom{n}{1}ix + \binom{n}{3}(ix)^3 + \binom{n}{5}(ix)^5 + \ldots \\
&= i\left(\binom{n}{1}x - \binom{n}{3}x^3 + \binom{n}{5}x^5 + \ldots\right) \\
\end{align*}
Thus,
$$\sum_{k\geq 0} \binom{n}{2k+1} (-1)^k x^{2k} = \frac{i}{2x}\left((1-ix)^n - (1+ix)^n\right)$$
Substituting $x=\sqrt(3)$ yields
$$\left(\binom{n}{1} - \binom{n}{3}3 + \binom{n}{5}3^2 + \ldots\right) = \frac{i}{2\sqrt{3}} ((1-\sqrt(3)i)^n - (1+\sqrt(3)i)^n) = \frac{i}{2} (e^{-in\pi/3}-e^{in\pi/3})$$
Since, $n = 6m$, on RHS we get $\frac{i}{2\sqrt{3}} (e^{-2im\pi} - e^{2im\pi}) = 0$
Similarly, substitute $x=\frac{1}{\sqrt{3}}$ to get the next result.
|
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Any hints on how to show that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $ I am trying to prove that $$\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $$
My attempt was to suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10}$, and show that it was absurd.
Here's what I did:
Suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10} $, then $2 - \sqrt{3} \geq \frac{2\sqrt{3}}{10}$ thus:
$$\frac{20 -10 \sqrt{3}}{10} \geq \frac{2 \sqrt{3}}{10} \implies 20 -\sqrt{300} \geq 2 \sqrt{3} \implies 10 - \sqrt{\frac{300}{4}}\geq \sqrt{3} \implies 10 - \sqrt{75} \geq \sqrt{3} \implies \frac{10}{\sqrt{3}}- \sqrt{25} \geq 1 \implies \frac{10}{\sqrt{3}}- 5 \geq 1$$
Now I elevate everything to the square:
$$\frac{100}{3} - \frac{100}{\sqrt{3}} + 25 \geq 1 \implies \frac{100}{3} - \frac{100 \sqrt{3}}{3} + \frac{75}{3} \geq 1 \implies \frac{100(1-\sqrt{3}) +75}{3} \geq \frac{3}{3}$$
All is left to show is that $$100(1-\sqrt{3}) +75 \geq 3$$
And here, I don't know where to go, and it doesn't seem as I simplified the problem.
Any hints of help will be appreciated
|
The inequality is iff
\begin{aligned}
20-10\sqrt{3}<2\sqrt{3}\iff 20<12\sqrt{3}\iff 400<144\times3=432
\end{aligned}
which is clearly true.
|
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Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}$ if $\measuredangle C = 90^{\circ}$
In$\triangle ABC$, $\angle C$ is a right angle. Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}.$ Legs are named in traditional way.
My Work
As $r = \frac{A}{s}$. So, here $r = \frac{ab}{a+b+c}$. Is there any way to prove this is equal to $\frac{a+b-c}{2}$?. Or else how to solve this is also the second part?
|
$$r=\frac{2S}{a+b+c}=\frac{ab}{a+b+c}=\frac{ab(a+b-c)}{a^2+2ab+b^2-c^2}=\frac{a+b-c}{2}$$
$$r_c=\frac{2S}{a+b-c}=\frac{ab}{a+b-c}=\frac{ab(a+b+c)}{a^2+2ab+b^2-c^2}=\frac{a+b+c}{2}$$
|
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How to show that $\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx=3\pi$
Consider
$$I=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx
\qquad J=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx$$
I want to show that $I=3\pi$ and that $I=J$.
First, we noticed that $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$
So it gives us an idea to try and factorise $x^4-x^2+1$ but cannot find any factors.
Integrate $I$ (We try some substitutions to see where it will get us to)
$x=\sqrt{u}$ then $dx={2\over \sqrt{u}}du$
$$I=16\cdot{1\over 2}\int_{0}^{\infty}{u^{3/2}\over (1-u+u^2)^4}\mathrm du$$
$u=\tan(y)$ then $du=\sec^2(y)dy$
$$I=8\int_{0}^{\pi/2}{\tan^{3/2}(y)\over (1-\tan(y)+\tan^2(y))^4}{\mathrm dy\over \cos^2(y)}$$
then simplified down to
$$I=128\int_{0}^{\pi/2}{\cos^6(y)\tan^{3/2}(y)\over (2-\sin(2y))^4}\mathrm dy$$
we further simplified down to
$$I={128\over 2^{3/2}}\int_{0}^{\pi/2}{\cos^3(y)\sin^{3/2}(2y)\over (2-\sin(2y))^4}\mathrm dy$$
Not so sure what is the next step.
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Note that the integrands are even, so $$I=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}dx
$$ and $$J=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}dx.
$$ Now let us consider the semicircular contour of radius $R$ centred in the origin on the upper plane of the complex plane. It is not difficult to note that the integral over the semicircumference vanish as $R\rightarrow\infty
$ so $$\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}dx=2\pi i\left(\underset{x=\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}\right)+\underset{x=-\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}\right)\right)
$$ $$=2\pi i\left(\frac{247+77i\sqrt{3}}{54\left(-\sqrt{3}+3i\right)}+\frac{247-77i\sqrt{3}}{54\left(\sqrt{3}+3i\right)}\right)=6\pi
$$ hence $$I=\color{red}{3\pi}.
$$ In a similar manner $$\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}dx=2\pi i\left(\underset{x=\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}\right)+\underset{x=-\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}\right)\right)
$$ $$=2\pi i\left(\frac{13+i5\sqrt{3}}{3\left(-\sqrt{3}+3i\right)}+\frac{13-i5\sqrt{3}}{3\left(\sqrt{3}+3i\right)}\right)=6\pi
$$ hence $$\color{blue}{I=J}.$$
|
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|
Difficult trigonometry problem to find the minimum value Find the minimum value of $5\cos A + 12\sin A + 12$.
I don't know how to approach this problem. I need help.
I'll show you how much I got...
$$5\cos A +12\sin A + 12 = 13(5/13\cos A +12/13\sin A) + 12$$
|
As we know that,
$−\sqrt{a^2+b^2} \le a \cosθ + b \sinθ \le \sqrt{a^2+b^2}$
So for $5 \cos θ+12 \sin θ$ we have,
$−\sqrt{5^2+12^2} \le 5cosθ+12sinθ \le \sqrt{5^2 +12^2}$
= $−13 ≤ 5cosθ +12sinθ ≤ 13$
Now adding 12 we get,
= $−13 + 12 ≤ 5 \cosθ + 12 \sinθ + 12 ≤ 13 + 12$
= $− 1 ≤ 5 \cosθ + 12 \sinθ + 12 ≤ 25$
So minimum value = −1
And maximum value = 25
|
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|
How to i solve this Exponential equation How to solve this exponential equation?
$$7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}$$
|
$$7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}$$
$$21 \cdot 3^{x} -25\cdot 5^{x}= 81\cdot3^{x}- 125\cdot5^{x}$$
$$100\cdot 5^{x}= 60\cdot3^{x}$$
$$\left(\frac{5}{3}\right)^x=\frac{3}{5}=\left(\frac{5}{3}\right)^{-1}$$
$$x=-1$$
|
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|
The digits 0-7 are used to form 4-digit numbers... Restrictions for all questions: No repetition and the thousands place cannot be 0 (ex. 0123, 0555, etc.)
A. How many 4-digit numbers are possible?
B. How many of these 4-digit numbers are odd?
C. How many of these 4-digit numbers are even?
D. How many of these 4-digit numbers are greater than 3100?
E. How many of these 4-digit numbers are less than 3100?
My Answers:
A. 1680, 1470 or 1344.
B 840, 735, or 672
C. Same as B because $B + C = A$
D. 900 (listed down)
E. 450 (listed down)
The main problem of this equation is that we're not allowed to have 0 as our thousands, which makes a seemingly huge contradiction to... everything! My friend and I listed it all down and we're not even sure if it's right or not.
If we work the digits left to right, we get an answer, but if we work right to left, we get a totally different answer!
Is there even a possible formula/equation for this or is this question a lost cause and we really have to list down everything?
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How many of these four-digit numbers can be formed from the set $\{0, 1, 2, 3, 4, 5, 6, 7\}$ if no two digits are the same?
Since $0$ cannot be placed in the thousands place, we have seven choices for the thousands digit, which leaves seven choices for the hundreds digit (since we can now use $0$ but cannot select the number we used for the thousands digit), six choices for the tens digit (since we cannot use our choices for the thousands digit or hundreds digit), and five choices for the units digit (since we cannot use our choices for the thousands digit, hundreds digit, or tens digit). Hence, there are $7 \cdot 7 \cdot 6 \cdot 5 = 1470$ such numbers.
Of these four-digit numbers, how many are odd?
We must first impose the restriction that the number is odd, which leaves us with four choices for the units digit ($1$, $3$, $5$, or $7$). Doing so eliminates one of the seven permissible choices for the thousands digit, leaving six choices for the thousands digit. We are now left with six choices for the tens digit and five choices for the units digit. Hence, there are $4 \cdot 6 \cdot 6 \cdot 5 = 720$ odd numbers.
How many of the four-digit numbers are even?
Subtract the number of odd numbers from the total.
Edit: We could also make a direct calculation. We consider two cases, depending on whether or not the last digit is zero.
*
*The last digit is zero. We have one choice for the units digit. Since any of the other digits may be used for the thousands digit, we have seven choices for the thousands digit. This leaves us with six choices for the hundreds digit and five choices for the tens digit, so there are $7 \cdot 6 \cdot 5 \cdot 1 = 210$ such numbers.
*The last digit is not zero. Since the number is even, this leaves us with three choices for the units digit ($2$, $4$, or $6$). Since we cannot use either $0$ or the units digit in the thousands place, we are left with six choices for the thousands digit. This leaves us with six choices for the hundreds digit and five choices for the tens digit, so there are $6 \cdot 6 \cdot 5 \cdot 3 = 540$ such numbers.
Since the two cases are mutually exclusive, there are a total of $210 + 540 = 750$ four-digit even numbers.
How many of the four-digit numbers are greater than $3100$?
Notice that this condition imposes a restriction on the thousands digit. If the thousands digit is $3$, it imposes a restriction on the hundreds digit. If the thousands digit is $3$ and the hundreds digit is $1$, there are no further restrictions since it is not possible to obtain $3100$ because no digit may be used more than once.
We consider cases:
*
*The number is at least $4000$. We have four choices for the thousands digit ($4$, $5$, $6$, or $7$). This leaves us with seven choices for the hundreds digit, six choices for the tens digit, and five choices for the units digit. Hence, there are $4 \cdot 7 \cdot 6 \cdot 5 = 840$ possible numbers in this case.
*The number $n$ satisfies $3200 \leq n < 4000$. We have one choice for the thousands digit. Since $3$ has been used for the thousands digit, this leaves us with five choices for the hundreds digit ($2$, $4$, $5$, $6$, or $7$). Once the hundreds digit has been selected, we are left with six choices for the tens digit and five choices for the units digit. There are $1 \cdot 5 \cdot 6 \cdot 5 = 150$ choices in this case.
*The number $n$ satisfies $3100 < n < 3200$. We have choice for the thousands digit and one choice for the hundreds digit. Since $3$ and $1$ cannot be used, we have six choices for the tens digit and five choices for the units digit. Thus, there are $1 \cdot 1 \cdot 6 \cdot 5 = 30$ choices in this case.
Since the three cases are disjoint, there are a total of $840 + 150 + 30 = 1020$ four-digit numbers greater than $3100$.
How many of these four-digit numbers are less than $3100$?
Since $3100$ is not a possible outcome, subtract the number of four-digit number greater than $3100$ from the total.
|
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|
Finding maxima and minima of $f(x,y)= x^4 + y^4 - 2x^2 - 2y^2 + 4xy$ For $f(x,y)=x^4+y^4-2x^2-2y^2+4xy$, I need to find maxima or minima. There are three critical points: $(0,0),(\sqrt2, -\sqrt2),(-\sqrt2,\sqrt2)$
So at $(\sqrt2, -\sqrt2)$, $f$ has minimum value, $-8$ and at $(-\sqrt2,\sqrt2),$ it has same minimum value, $-8$. At $(0,0)$, after inspecting, I get that it has neither maxima or minima. But when we substitute $(0,0),$ we get $0\le f \le8,$ so should not we get $(0,0)$ as a point of maxima?
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It's obvious that a maximal value does not exist.
We'll prove that $-8$ is a minimal value.
Let $x=\sqrt2a$ and $y=-\sqrt2b$.
Hence, we need to prove that
$$4a^4+4b^4-4a^2-4b^2-8ab+8\geq0$$ or
$$a^4+b^4+2\geq(a+b)^2,$$
which is AM-GM and C-S:
$$a^4+b^4+2\geq2\sqrt{(a^4+1)(1+b^4)}\geq2(a^2+b^2)\geq(a+b)^2$$
|
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|
System of equations - conics
What would be the best way to solve the following the following system of equations? $$x^2+y^2+6y+5=0$$ $$x^2+y^2-2x-8=0$$
This is how I did it, but I am hoping there is a simpler way.
Subtracting the two equations, I get $$6y+2x+13=0$$which simplifies to $$x=-3y-\dfrac{13}2$$I substitute it into the first equation where I get $$9y^2+39y+\dfrac{169}4+y^2+6y+5=0$$ and solving for $y$ I get $$y=-\dfrac{45\pm\sqrt{1552.5}}{20}$$which simplifies to $$y=-\dfrac{9\pm{\sqrt{62.1}}}{4}$$ Plugging this back into the equation at the beginning of this highlighted section I get $$\dfrac{-53\pm{3\sqrt{62.1}}}{2}=-2x$$or $$x=\dfrac{53\pm{3\sqrt{62.1}}}{4}$$ So my final answer is $$\big(\dfrac{53\pm{3\sqrt{62.1}}}{4},-\dfrac{9\pm{\sqrt{62.1}}}{4})$$
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After the substitution, we have
$$10y^2 + 45 y + \dfrac{189}{4} \implies y_{1, 2} = -\dfrac{9}{4} \pm \dfrac{3~ \sqrt{\dfrac{3}{5}}}{4}$$
This leads to
$$ x_{1, 2} = \dfrac{1}{4} \mp \dfrac{9~ \sqrt{\dfrac{3}{5}}}{4}$$
|
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|
Evaluation of given indefinite integral Evaluate the given integral
$$\int e^x \bigg[\frac{2-x^2}{(1-x)\sqrt{1-x^2}} \bigg]dx$$
I was trying to convert it to $\int e^x (f(x)+f'(x))dx=e^x \cdot f(x)+C$ but did not succeed in algebraic manipulations. Could someone hint me to something so that I could proceed?
|
$$\begin{aligned}
\frac{2-x^2}{(1-x)\sqrt{1-x^2}}
&=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{1-x^2}{(1-x)\sqrt{1-x^2}} \\
&=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{\sqrt{1-x^2}}{(1-x)} \\
&=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{\sqrt{1+x}}{\sqrt{1-x}} \\
\end{aligned}$$
The only thing left is to notice that,I'll leave that to you.
$$\left(\frac{\sqrt{1+x}}{\sqrt{1-x}}\right)'=\frac{1}{(1-x)\sqrt{1-x^2}}$$
|
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|
Given the positive real numbers $0\le a,b,c\le 2$ and $a+b+c=3$. Prove that $a^3+b^3+c^3\le 9$ Given the positive real numbers $$0\le a,b,c\le 2$$ and $$a+b+c=3$$. Prove that $$a^3+b^3+c^3\le 9$$
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Assume without loss of generality that $a$ is the maximum of $a,b,c$. This gives us that $$ 3a \ge a+b+c=3 \iff 2 \ge a \ge 1 \tag{1}$$
From the condition $0 \le b,c \le a \le 2$. Note that we have that $$a^3+b^3+c^3 \le a^3+(b+c)^3=a^3+(3-a)^3=9\left(a-\frac{3}{2} \right)^2+\frac{27}{4} \tag{2}$$
Which follows from the fact that $b,c \ge 0$ and $ a+b+c=3$.
Note that as we have $1 \le a \le 2$ from $(1)$ we have $$\left(a-\frac{3}{2}\right)^2 \le \frac{1}{4} \tag{3}$$
Thus $$a^3+b^3+c^3 \le 9\left(a-\frac{3}{2} \right)^2+\frac{27}{4} \le \frac{9}{4}+\frac{27}{4}= 9 $$
From $(2)$ and $(3)$. Thus, our proof is done. Equality is true when $a=2, b=1, c=0$.
|
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|
Find all matrices $X$ such that $ABXB^tA^t=I$ Find all matrices $X$ such that:
$$ABXB^tA^t=I$$ if
$A=\begin{pmatrix}
1 &-2 &2\\
3 &-5 &6\\
-1 &2 &-1
\end{pmatrix}$ and $B=\begin{pmatrix}
-3 &-2 &-2\\
2 & 1 &1\\
6 &3 &4
\end{pmatrix}$.
So I managed to get that $AB=\begin{pmatrix}
5 &2 &4\\
17 &7 &13\\
1&1&0
\end{pmatrix}$ and $B^tA^t=(AB)^t=\begin{pmatrix}
5 &17 &1\\
2 &7 &1\\
4 &13 &0
\end{pmatrix}$
So we have $\begin{pmatrix}
5 &2 &4\\
17 &7 &13\\
1&1 &0
\end{pmatrix}\cdot X\cdot\begin{pmatrix}
5 &17 &1\\
2&7&1\\
4&13&0
\end{pmatrix}=I$
Now how do I get $X$ from here?
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If either $AB$ or $B^TA^T$ is singular, there is no chance to get a solution.
Therefore the $3 \times 3$ matrices, have to be nonsingular (and they are, I just checked).
So, to get $X$ you need to invert both of them and get to $X=B^{-1}A^{-1}A^{-T}B^{-T} = B^{-1}(A^TA)^{-1}B^{-T}$
Usually you would decompose $AB$ into $LR$ or $QR$ decompositions and pull them to the right hand side.
|
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|
Prove that $\sin\frac{\pi}n·\sin\frac{2\pi}n···\sin\frac{(n-1)\pi}n=\frac{n}{2^{n-1}}$ How to prove that
$$
\sin\dfrac{\pi}n·\sin\dfrac{2\pi}n···\sin\dfrac{(n-1)\pi}n=\dfrac{n}{2^{n-1}}
$$
using the roots of $(z+1)^n-1=0$?
My rough idea is to solve $(z+1)^n-1=0$ and use De Moivre's Theorem to find the product of roots to prove the equality.
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Clearly $z=0$ is a root of $(z+1)^n-1$, so we need to exclude it to get something nontrivial for the product. Dividing and taking the limit,
$$ \lim_{z \to 0} \frac{(z+1)^n-1}{z} = n, $$
so the product of the remaining roots is $(-1)^{n-1} n$. Now we have to find expressions for the roots. We have roots
$$ z_k+1 = e^{2\pi ik/n}, \quad k \in \{1,2,\dotsc,n-1\}. $$
Therefore, rearranging and applying the formula for sine,
$$ z_k = e^{2\pi i k/n} -1 = e^{\pi i k/n} (e^{\pi i k/n}-e^{-\pi ik/n}) = e^{\pi i k/n} 2i\sin{\left( \frac{\pi k}{n} \right)}. $$
Hence, we have
$$ \begin{align}
(-1)^{n-1}n &= \prod_{k=1}^{n-1} z_k \\
&= \prod_{k=1}^{n-1} e^{\pi i( k/n+1/2)} 2\sin{\left( \frac{\pi k}{n} \right)} \\
&= 2^{n-1} \exp{\left( \pi i \left(\frac{n-1}{2}+\frac{1}{n} \sum_{k=1}^{n-1} k \right) \right)} \prod_{k=1}^{n-1} \sin{\left( \frac{\pi k}{n} \right)}.
\end{align} $$
The result now follows, since
$$ \exp{\left( \pi i \left(\frac{n-1}{2}+\frac{1}{n} \sum_{k=1}^{n-1} k\right) \right)} = \exp{\left( \pi i (n-1) \right)} = (-1)^{n-1}. $$
|
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|
How could I know that $X^4+1$ is $(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1)$? I thought that $X^4+1$ was irreducible, but in fact, $$X^4+1=(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1).$$
In general, how can I have the intuition of such a factorisation if I don't know it ?
|
There's a sort of completion of the square that goes like this:
\begin{align}
x^4+1 & = \underbrace{(x^4+2x^2 + 1)}_\text{This is a square.} - \underbrace{(2x^2)}_\text{So is this.} \\[10pt]
& = \left( x^2+ 1 \right)^2 - (\sqrt 2\ x)^2 \\[10pt]
& = (x^2 + 1 - \sqrt 2\ x)(x^2 + 1 + \sqrt2\ x).
\end{align}
|
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|
Proof in Arithmetic Progression My maths teacher at school asked a question which I am finding difficult to crack down.
We are given that $a^2 , b^2$ and $c^2$ are in AP. We need to prove that $\frac{a}{b+c} , \frac{b}{a+c}$ and $\frac{c}{a+b}$ are in AP.
This is what I tried.
Let the common difference of the AP be d . So,
$b^2 - a^2 = d \implies b-a = \frac{d}{a+b} ........(1)$
Similarly
$c^2 - b^2 = d \implies c-b = \frac{d}{b+c}. ........(2)$
Also,
$a^2 - c^2 = -2d \implies a-c = \frac{-2d}{a+c} .........(3) $
Now adding the three equations,
$$0 = \frac{d}{a+b} + \frac{d}{b+c} - \frac{2d}{a+c}
\implies \frac{2d}{a+c} = \frac{d}{a+b} + \frac{d}{b+c}$$
$$\implies \frac{2}{a+c} = \frac{1}{a+b} + \frac{1}{b+c}$$
So, $\frac{1}{a+b}, \frac{1}{a+c}$ and $\frac{1}{b+c}$ are in AP. How should I go further? Or if I am going wrong anywhere, please tell.
|
$AP1$: $$a^2\\ b^2\\ c^2$$
$$\text{Common Difference, }\quad d=b^2-a^2=c^2-b^2\qquad \qquad $$
$XP2$:
$$P=\frac a{b+c}\color{lightgrey}{\cdot\frac{(c-b)}{(c-b)}}=\frac {a(c-b)}{c^2-b^2}=\frac {a(c-b)}d\\
Q=\frac b{c+a}\color{lightgrey}{\cdot\frac{(c-a)}{(c-a)}}=\frac {b(c-a)}{c^2-a^2}=\frac {b(c-a)}{2d}\\
R=\frac c{a+b}\color{lightgrey}{\cdot\frac{(b-a)}{(b-a)}}=\frac {c(b-a)}{b^2-a^2}=\frac {c(b-a)}d\\$$
If $XP2$ is an AP, then $P+R=2Q$.
Testing both sides:
$$\text{LHS}=P+R=\frac {a(c-b)+c(b-a)}d=\frac {bc-ba}d=2\cdot\frac {b(c-a)}{2d}=2Q=\text{RHS}$$
Hence $XP2$ is also an $AP$, if $AP1$ is an $AP$.
NB - I had worked this out independently but credit to the hint given by @Lozenges in the comments.
|
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|
Can a factor of $x^2+4$ for odd $x$ be $\equiv 3\mod 4$? When $x$ is odd, then $x^2+4 \equiv 5 \mod 8$. Can any factor of $x^2+4$ be $\equiv 3\mod 4$?
Examples:
$$7^2+4=53$$
$$11^2+4= 5^3$$
$$31^2+4= 5 \cdot 193$$
$$47^2+4=2213$$
$$89^2+4=5^2 \cdot 317$$
All of the prime factors above are $\equiv 1 \mod 4$.
Is there no counter example and if so, how to prove this?
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If $p$ is an odd prime divisor of $x^2+4$, then we have $-4 \equiv x^2 \pmod p$
So $1=\left(\frac{-4}{p}\right)=\left(\frac{4}{p}\right) \left(\frac{-1}{p}\right) = \left(\frac{-1}{p}\right)$ which means that $p \equiv 1 \pmod 4$ by the first supplement to quadratic reciprocity.
|
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|
Whats is the limit of $\lim_{(x,y)\to(a,a)}\frac{x^3-y^3}{x^2-y^2}$ for $a\ne0$? What is $\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^3-y^3 }{x^2-y^2}$ for $a\ne 0$? I have found that the limit coould be $\frac{3a}{2}$ to proof it I have come to the point where $\left|\frac{x^3-y^3 }{x^2-y^2}-\frac{3a}{2}\right| = \left|\frac{x^2+xy+y^2 }{x+y}-\frac{3a}{2}\right|\le \left|x+y-\frac{3a}{2}\right|$. I have to go to something like $\sqrt{(x-a)^2+(y-a)^2}$ however I have no idea what to do from here.
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$$\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^3-y^3 }{x^2-y^2}=\lim^{}_{(x,y) \rightarrow (a,a)} \frac{(x-y)(x^2+xy+y^2) }{(x-y)(x+y)}=\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^2+xy+y^2}{x+y}=\frac{3a}{2}$$
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Let $x_n$ be a sequence in a complete metric space $(X,d)$ such that $d(x_n,x_{n+1}) \le \frac {1}{n^2}$. Is $x_n$ convergent? I have a proof that $x_n$ is convergent. Here it is:
Let $\epsilon \gt 0$ be given and let $m \gt n$ where $n,m \in \Bbb N$.
Then $d(x_n,x_m)\le d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})+...+d(x_{m-1},x_m) \le \frac 1{n^2} +\frac 1{(n+1)^2}+...\frac 1{(m-1)^2}=\sum_{k=1}^{m-1}\frac 1{k^2}-\sum_{k=1}^{n-1} \frac {1}{k^2}.$
But since we know that $\sum \frac 1{n^2}$ is convergent. hence it's sequence of partial sums is Cauchy. So there exists $N \in \Bbb N$ such that $m-1, n-1 \ge N \Rightarrow \sum_{k=1}^{m-1} \frac 1{k^2} - \sum_{k=1}^{n-1} \frac {1}{k^2} \lt \epsilon$.
Hence $d(x_n,x_m) \le \epsilon$ whenever $m,n \gt N$. So $x_n$ is a Cauchy sequence in a complete metric space hence it is convergent.
QED.
My question is that is it possible to show that $x_n$ is Cauchy without using convergence of $\sum \frac 1{n^2}$?
I mean from "$d(x_n,x_m) \le \frac 1{n^2} +\frac 1{(n+1)^2}+...\frac 1{(m-1)^2}$", what manipulations can be done without introducing $\sum \frac 1{n^2}$?
For example : In case of the expression $\frac 1{n!} + \frac 1{(n+1)!} + ... + \frac 1{(m-1)!}$, we can use the inequality $2^n \lt n! (\text {for} \; n \gt 3)$ such that $\frac 1{n!} + \frac 1{(n+1)!} + ... + \frac 1{(m-1)!} \lt \frac 1{2^n}+\frac 1{2^{n+1}}+...+ \frac 1{2^{m-1}} \lt \frac 1{2^{n-1}}$.
Then we can choose $n \gt 1-\log_2\epsilon$ so that $\frac 1{2^{n-1}} \lt \epsilon$.
I tried this : $\frac 1{n^2} +\frac 1{(n+1)^2}+...\frac 1{(m-1)^2} \lt \frac {m-n}{n^2}.$ But this $m-n$ in numerator is bugging me.
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How about
\begin{align*}
d(x_n,x_m) \le \frac 1{n^2} +\frac 1{(n+1)^2}+...\frac 1{(m-1)^2}&\le \frac 1{n(n-1)} +\frac 1{(n+1)n}+...\frac 1{(m-1)(m-2)}\\
&\le \frac 1{n-1} -\frac 1{m-1}\le\frac 1{n-1},
\end{align*}
but this is in fact one of the approaches to show $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent.
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Is the sequence $\left\{\frac{2^n}{n!}\right\}$ convergent? If so, what is the limit? Is the sequence $\left\{\frac{2^n}{n!}\right\}$ convergent? If so, what is the limit?
$$ \frac{2^n}{n!} - 0 = \frac{2^n}{n!} < \frac {2^n}{n} <\; ? < \epsilon$$
I dont know how to simplify $\frac{2^n }{ n}$.
I cannot just do $\frac{2^n}{n}<\epsilon$ right ?
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$\frac{2^n}{n!} = \frac{2}{1} \cdot \frac{2}{2} \cdot\frac{2}{3} \cdot\frac{2}{4} \cdot\frac{2}{5} \dots \frac{2}{n} < \frac{2}{1} \cdot \frac{2}{2} \cdot\frac{2}{3} \cdot\frac{2}{4} \cdot\frac{2}{4} \dots \frac{2}{4} = \frac{8}{6}\cdot(\frac{1}{2})^{n-3}$
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Least Positive Residue How would I find the least positive residue of
say
$6! \bmod 7$
or
$12! \bmod 13$
I just learned modular arithmetic and my book doesn't explain what least positive residues are so I'm a bit lost.
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Do you already know this?: $a\equiv b \bmod m \implies ac\equiv bc \bmod m$.
You can use this to keep the numbers small as you investigate these results:
$4! = 1\cdot 2\cdot 3\cdot 4 \equiv 24\equiv 3 \bmod 7$
$5! = 4!\cdot 5 \equiv 3\cdot 5 \equiv 15 \equiv 1 \bmod 7$
$6! = 5!\cdot 6 \equiv 1\cdot 6 \equiv 6 \bmod 7$
(and to reassure yourself that this is working, $6!=720 \equiv 20\equiv 6 \bmod 7\ \ \checkmark$)
We can shortcut a little with some precalculation for the $\bmod 13$ question:
$2\cdot 7=14\equiv 1\bmod 13$,
$3\cdot9=27\equiv 1\bmod 13$,
$4\cdot 10 = 5\cdot 8=40\equiv 1 \bmod 13$,
giving us:
$10! \equiv 1^5\cdot 6\equiv 6 \bmod 13$
and so on.
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Prove using induction the inequality.
$\forall$:n∈${N}$
$\binom{2n}{n}$ $\ge \frac{4^n}{2n+1}$
I tried with no any success...
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$$n=1 :\binom{2}{1}=2 \geq \dfrac{4}{3}
\\n=k \to \binom{2k}{k}\geq \dfrac{4^k}{2k+1} \\
n=k+1 \to \binom{2(k+1)}{k+1}\geq \dfrac{4^{k+1}}{2(k+1)+1}$$
$$\binom{2(k+1)}{k+1}=\dfrac{(2k+2)!}{(k+1)!(k+1)!}=\dfrac{(2k+2)(2k+1)(2k)!}{(k+1)^2k!}=\\
\dfrac{(2k+2)(2k+1)}{(k+1)^2} \binom{2k}{k}\geq \dfrac{4.4^k}{2k+3}=4.\dfrac{2k+1}{2k+3}\dfrac{4^k}{2k+1}\\$$if we prove the $\dfrac{(2k+2)(2k+1)}{(k+1)^2} \geq 4\dfrac{2k+1}{2k+3}$ ,we proved the original relation
$$\dfrac{(2k+2)(2k+1)}{(k+1)^2} \geq 4\dfrac{2k+1}{2k+3} \to \\
\dfrac{(2k+2)}{(k+1)^2} \geq 4\dfrac{1}{2k+3}\\
\dfrac{2}{(k+1)} \geq 4\dfrac{1}{2k+3}\\\\2(2k+3) \geq 4(k+1) \\4k+6 \geq 4k+4\\6\geq 4 \checkmark$$
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Proving that a limit goes to infinity Looking for some help in proving why the following infinity goes to $\infty$ as n approaches $\infty$
$$\frac{(1-\frac{1}{n})^n}{1-(1-\frac{1}{n})}$$
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$$\dfrac{(1-\dfrac{1}{n})^n}{1-(1-\dfrac{1}{n})}=\\
\dfrac{\binom{n}{0}1^{n}{(-\dfrac{1}{n})}^{0}+\binom{n}{1}1^{n-1}{(-\dfrac{1}{n})}^{1}+\binom{n}{2}1^{n}{(-\dfrac{1}{n-2})}^{2}+...+\binom{n}{n}1^{0}{(-\dfrac{1}{n})}^{n}
}{1-(1-\dfrac{1}{n})}=\\
\dfrac{1+n{(-\dfrac{1}{n})}+\dfrac{n(n-1)}{2}{(-\dfrac{1}{n})}^{2}+\dfrac{n(n-1)(n-2)}{6}{(-\dfrac{1}{n})}^{3}+...+1{(-\dfrac{1}{n})}^{n}
}{1-(1-\dfrac{1}{n})}=\\
\dfrac{1-1+\dfrac{(n-1)}{2}{(\dfrac{1}{n})}+\dfrac{(n-1)(n-2)}{6}{(-\dfrac{1}{n^2})}+...+1{(-\dfrac{1}{n})}^{n}
}{\dfrac{1}{n}}=\\$$
so apply limit
$$\lim_{n \to \infty }\dfrac{1-1+\dfrac{(n-1)}{2}{(\dfrac{1}{n})}+\dfrac{(n-1)(n-2)}{6}{(-\dfrac{1}{n^2})}+...+1{(-\dfrac{1}{n})}^{n}
}{\dfrac{1}{n}}=\\
\lim_{n \to \infty }\dfrac{\dfrac{(n-1)}{2n}-\dfrac{(n-1)(n-2)}{6n^2}+...+1{(-\dfrac{1}{n})}^{n}
}{\dfrac{1}{n}}=\\
\lim_{n \to \infty }n({\dfrac12-\dfrac16+\dfrac{1}{24}-\dfrac{1}{120})
}=\to \infty\\
$$
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Show that $|z - w| \geq \big||z|-|w|\big|$
Show that $|z - w| \geq \big||z|-|w|\big|$ given $z = x + iy$ and $w = u + iv$.
So far I have,
$$|z - w| \geq \big||z|-|w|\big|$$
$$\sqrt{(x - u)^2 + (y - v)^2} \geq |\sqrt{x^2 + y^2} -\sqrt{u^2 + v^2}|$$
At this point I get a little 'undone'.
I tried a few different routes, but none of them have come up with the desired result.
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Other answers address your issue in a better way but I think it will be nice if a proof by your line of thought is given.
$$\sqrt{(x - u)^2 + (y - v)^2} \geq |\sqrt{x^2 + y^2} -\sqrt{u^2 + v^2}|$$
Since both sides are $>0$,
$$(x - u)^2 + (y - v)^2 \geq x^2 + y^2 - u^2 + v^2 -2\sqrt{x^2 + y^2}\sqrt{u^2 + v^2}$$
$$(x^2 + u^2 -2xu) + (y^2 + v^2 - 2yv) \geq x^2 + y^2 + u^2 + v^2 -2\sqrt{x^2 + y^2}\sqrt{u^2 + v^2}$$
$$-xu - yv \geq -\sqrt{x^2 + y^2}\sqrt{u^2 + v^2}$$
$$\sqrt{x^2 + y^2}\sqrt{u^2 + v^2} -xu - yv \geq 0$$
$$\sqrt{x^2u^2 + x^2v^2 + y^2u^2 + y^2v^2} -xu - yv \geq 0$$
$$x^2u^2 + x^2v^2 + y^2u^2 + y^2v^2 \geq x^2u^2 + y^2v^2 + 2xyuv $$
$$x^2v^2 + y^2u^2 \geq 2xyuv \tag{*}$$
If you remember (*) is golden AM-GM inequality.
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Shifted Legendre Polynomial We know that the Legendre Polynomial is $P_n(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n}[(x^2-1)^n]$. The Shifted Legendre Polynomial $\tilde P_n(x)$ is defined as $P_n(2x-1)$. Can you please tell me how to get $\tilde P_n(x) = \frac{1}{n!}\frac{d^n}{dx^n}[(x^2-x)^n]$? Thanks.
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$$P_n(X) = \frac{1}{2^nn!}\frac{d^n}{dX^n}[(X^2-1)^n]$$
With $X=(2x-1)$
$$\tilde P_n(x)=P_n(2x-1)=P_n(X)=\frac{1}{2^nn!}\frac{d^n}{d(2x-1)^n}[\left((2x-1)^2-1\right)^n]$$
$d(2x-1)=2dx \quad\to\quad d(2x-1)^n=2^n dx^n$
$\left((2x-1)^2-1\right)^n=(4x^2-4x)^n=4^n(x^2-x)^n$
$$\tilde P_n(x)=\frac{1}{2^n n!}\frac{d^n}{2^n dx^n}[4^n(x^2-x)^n]=\frac{4^n}{2^n 2^n n!}\frac{d^n}{dx^n}[(x^2-x)^n]$$
$$\tilde P_n(x) = \frac{1}{n!}\frac{d^n}{dx^n}[(x^2-x)^n]$$
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Validating the inequality. $x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x)<x-\frac{x^2}{2}+\frac{x^3}{3}$, $x>0$
Here I can only see that the right side of second inequality i.e. $x-\frac{x^2}{2}+\frac{x^3}{3}$ comes in the expansion of $\log(1+x)$.
We have done the Lagrange's mean value theorem and intermediate value theorem, do these have anything to do with the inequality.
Kindly provide some hint.
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defining $$f(x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\ln(1+x)$$ for $x=0$ we get $$f(0)=0$$ and $$f'(x)=1-x+x^2-\frac{1}{x+1}=\frac{(1-x+x^2)(1+x)-1}{x+1}=\frac{x^3}{1+x}>0$$ for $x>0$ thus our function is monotonously increasing and we get $$f(x)>0$$ for all $$x>0$$
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Find the limit: $\lim_{x\to\infty}x\sin(\tan\frac1x)$ How do I find the limit:
$$\lim_{x\to\infty}x\sin(\tan\frac1x)$$
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Rewrite the limit as
\begin{align}
L:=\lim\limits_{x\rightarrow\infty}x\sin\left(\tan\frac{1}{x}\right)=\lim\limits_{x\rightarrow\infty}\frac{\sin\left(\tan\frac{1}{x}\right)}{\frac{1}{x}}
\end{align}
Since both the numerator and the denominator tend to $0$ in the limit, this is an indeterminate form and we can apply L'Hospital's Rule.
\begin{align}
L&=\lim\limits_{x\rightarrow\infty}\frac{\sin\left(\tan\frac{1}{x}\right)}{\frac{1}{x}}\\
&=\lim\limits_{x\rightarrow\infty}\frac{-\frac{1}{x^2}\sec^2\left(\frac{1}{x}\right)\cos\left(\tan\frac{1}{x}\right)}{-\frac{1}{x^2}}\\
&=\lim\limits_{x\rightarrow\infty}\sec^2\left(\frac{1}{x}\right)\cos\left(\tan\frac{1}{x}\right)\\
\end{align}
By continuity of all the functions involved, we can pull the limits into the functions. Since $\frac{1}{x}\rightarrow 0$ as $x\rightarrow\infty$, we have
$$\lim\limits_{x\rightarrow\infty}x\sin\left(\tan\frac{1}{x}\right) = 1$$
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integration by parts with $\frac{2}{\sqrt{2\pi}} \int^\infty_0 z^2e^{\frac{-z^2}{2}}dz$ Can someone please show how to calculate this with integration by parts $(\int udv = uv - \int vdu)$? I found an example in the book is not clear and confusing.
$$\frac{2}{\sqrt{2\pi}} \int^\infty_0 z^2e^{\frac{-z^2}{2}}dz$$
They start with u = z, dv = $ze^{\frac{-z^2}{2}} dz$, v = $ -e^{\frac{-z^2}{2}}$, du = dz
then:
$$\int^\infty_0 udv = -ze^{\frac{-z^2}{2}} - \int^\infty_0 -e^{\frac{-z^2}{2}} dz = -ze^{\frac{-z^2}{2}} + \int^\infty_0 e^{\frac{-z^2}{2}} = -ze^{\frac{-z^2}{2}} + \frac{\sqrt{2\pi}}{2}$$
I have no problem with the most right part of integration $\int^\infty_0 e^{\frac{-z^2}{2}}$ i know it is equal to$\frac{\sqrt{2\pi}}{2}$ but what about the $-ze^{\frac{-z^2}{2}}$ how to deal with it?
how can I continue from here?
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With $\dfrac{z^2}{2}=u$ and
$$\frac{2}{\sqrt{2\pi}} \int^\infty_0 z^2e^{\frac{-z^2}{2}}dz=\frac{2}{\sqrt{\pi}} \int^\infty_0 u^\frac12e^{-u}du=\frac{2}{\sqrt{\pi}} \Gamma(\frac32)=\frac{2}{\sqrt{\pi}} \frac{\sqrt{\pi}}{2}=1$$
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How to integrate this $\int \frac{\sqrt{x}}{1+x^4}dx$? How to integrate this ? $$\int \frac{\sqrt{x}}{1+x^4}dx$$
Any hint or idea on how to proceed?
Edit: Here is the final answer using Wolfram Alpha
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You can use the substitution $x=t^2$, $dx=2t\ dt$, so you get
$$\int \frac{2t^2}{(1+t^8)} \ dt$$
although that one is a rational integral which involves many calculations.
To solve that rational integral, we first have to decompose $p(t)=1+t^8$ in irreducible polynomials. As it's clear $p(t)$ doesn't have real roots, its decomposition would only have polynomials of degree $2$. The easiest way to solve this is finding the $8$-th roots of $-1$, which are:
$$\xi_k=e^\frac{(2k-1)i}{8}, k\in\{0,1,...,7\}$$
If we notice that $\bar\xi_7=\xi_0$, $\bar\xi_6=\xi_1$, $\bar\xi_5=\xi_2$ and $\bar\xi_4=\xi_3$, that will quickly lead us to a factorization of $p(t)$, because
$$(t-\xi)(t-\bar\xi)=(t^2-2Re \xi +|\xi|^2)$$
As $|\xi_k|=1$, we get:
\begin{equation}\begin{split} p(t)=&(t-\xi_0)\cdots(t-\xi_7)=\\=&(t^2-2\cos \frac\pi8t+1)(t^2-2\cos \frac{3\pi}8t+1)(t^2-2\cos \frac{5\pi}8t+1)(t^2-2\cos \frac{7\pi}8t+1)=\\=&(t^2-2\cos \frac\pi8t+1)(t^2-2\cos \frac{3\pi}8t+1)(t^2+2\cos \frac{3\pi}8t+1)(t^2+2\cos \frac{\pi}8t+1 )\end{split}\end{equation}
because $\cos(\pi-\alpha)=-\cos(\alpha)$.
(from now on, I simply explain the process. The calculations are way too messy.)
Now, we have to write $\frac{2t^2}{1+t^8}$ as a partial fractions sum, that is, finding $A,B,...H$ so that:
$$\frac{2t^2}{1+t^8}=\frac{At+B}{t^2-2\cos \frac\pi8t+1}+\frac{Ct+D}{t^2-2\cos \frac{3\pi}8t+1}+\frac{Et+F}{t^2+2\cos \frac{3\pi}8t+1}+\frac{Gt+H}{t^2+2\cos \frac{\pi}8t+1}$$
and integrate each term using the general formula
$$\int\frac{mt+n}{at^2+bt+c}dt = \frac{m}{2a}\ln\left|at^2+bt+c\right|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan\frac{2at+b}{\sqrt{4ac-b^2}}$$
To get the final result given by Wolfram Alpha, remember to substitute back $t=\sqrt{x}$.
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The following equation to solve :$ \tan x+\cot x=\sqrt{2}(\cos x+\sin x)$ The following equation to solve :
$$ \tan x+\cot x=\sqrt{2}(\cos x+\sin x)$$
My try:
$$\frac{2}{\sin 2x}=\sqrt{2}(\cos x+\sin x)$$
$$\left(\frac{2}{\sin 2x}\right)^2=(\sqrt{2}(\cos x+\sin x))^2$$
$$\left(\frac{2}{\sin 2x}\right)^2=2(1+\sin 2x)$$
$$2\sin^2 2x +2\sin ^3 2x=4$$
$$2\sin^2 2x +2\sin ^3 2x-4=0$$
$t=\sin 2x$
$$2t^3+2t^2-4=(t-1)(t^2+2t+4)$$
$$\sin 2x =1\\$$
is it right ?
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Excluding $\sin x\cos x=0$, you can rewrite
$$\sqrt2(\cos x+\sin x)\sin x\cos x=1$$
or
$$\sin\left(x+\frac\pi4\right)\sin(2x)=1.$$
For this product to be $1$, both factors must be $1$ or $-1$.
Then
$$x+\frac\pi4=k\pi+\frac\pi2,\\2x=l\pi+\frac\pi2,$$
where $k$ and $l$ have the same parity. Then as $l=2k$, $l$ and $k$ are even and
$$x=2n\pi+\frac\pi4.$$
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Polynomial system If there are 3 numbers $x,y,z$ satisfying
$f=x+y+z=3$ , $g=x^2+y^2+z^2=5$ , $h=x^3+y^3+z^3=7$ then prove that they also satisfy
$x^4+y^4+z^4=9$ but not $x^5+y^5+z^5=11$
I dont know how to tackle this to be honest, i have started trying to write $x^4+y^4+z^4-9$ as $r_1\times f + r_2\times g+ r_3\times h$ where $r_1,r_2,r_3$ in $C[x,y,z]$ but then i realised that i cant possibly do that by hand.This was an exercise on my first computational algebra course so we havent really learnt anything much yet. I would like a hint if possible.
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I give here an answer advocating the simplicity of using Newton (or Newton-Girard) formulas. Using the notations of the Wikipedia article in all generality:
Let $e_0=1, e_1=x+y+z, e_2=xy+yz+zx, e_3=xyz.$
Let $p_1=x+y+z$, $p_2=x^2+y^2+z^2$, $p_3=x^3+y^3+z^3$.
Then:
$$\begin{cases}
e_1&=&p_1\\
2e_2&=&p_1e_1-p_2\\
3e_3&=&p_1e_2-p_2e_1+p_3
\end{cases} \ \ \ \text{Using the given values:} \ \ \ \begin{cases}
e_1&=&3\\
2e_2&=&3e_1-5\\
3e_3&=&3e_2-5e_1+7
\end{cases}$$
from which $e_1=3, e_2=2, e_3=-\frac23$.
Using Vieta's formulas, $x,y,z$ are solutions of the third degree equation:
$$\tag{1}t^3-3t^2+2t+\frac23=0$$
(whose roots are one real, and two complex conjugate other roots).
Any root $t$ of (1), verifies $t^4=3t^3-2t^2-\frac23t$. When one writes this relationship for $t=x$, $t=y$ and $t=z$.
$$\tag{2}\begin{cases}x^4=3x^3-2x^2-\frac23x\\y^4=3y^3-2y^2-\frac23y\\z^4=3z^3-2z^2-\frac23z\\\end{cases}$$
Adding these equations, one obtains $x^4+y^4+z^4=9.$
Multiplying the equations in (2) resp. by $x$,$y$ and $z$, and adding again, one obtains $x^5+y^5+z^5=\dfrac{29}{3}$ ($\neq 11$!).
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Evaluate $1+\left(\frac{1+\frac12}{2}\right)^2+\left(\frac{1+\frac12+\frac13}{3}\right)^2+\left(\frac{1+\frac12+\frac13+\frac14}{4}\right)^2+...$ Evaluate:
$$S_n=1+\left(\frac{1+\frac12}{2}\right)^2+\left(\frac{1+\frac12+\frac13}{3}\right)^2+\left(\frac{1+\frac12+\frac13+\frac14}{4}\right)^2+...$$
a_n are the individual terms to be summed.
My Try :
\begin{align}
&a_1=1\\
&a_2=\left(\frac{3}{4}\right)^2=\frac{9}{16}\\
&a_3=\left(\frac{11}{18}\right)^2\\
&a_4=\left(\frac{25}{48}\right)^2
\end{align}
now :?
|
Recall that the multiple zeta values are defined by the series
$$
\zeta(s_1,\ldots,s_k):=\sum_{n_1>\ldots>n_k\geq 1}\frac{1}{n_1^{s_1}\ldots n_k^{s_k}}.
$$
The sum $S$ can be expressed as a linear combination of multiple zeta values. We have
$$
\begin{align*}
S&=\sum_{n=1}^\infty \sum_{k_1,k_2=1}^n\frac{1}{n^2k_1k_2}\\
&=\left(2\sum_{n>k_1>k_2}+\sum_{n>k_1=k_2}+2\sum_{n=k_1>k_2}+\sum_{n=k_1=k_2}\right)\frac{1}{n^2k_1k_2}\\
&=2\zeta(2,1,1)+\zeta(2,2)+2\zeta(3,1)+\zeta(4).
\end{align*}
$$
Each of these multiple zeta values is a rational multiple of $\pi^4$. The expressions have been tabulated for instance on the MZV data mine:
$$
\begin{align*}
\zeta(2,1,1)&=\frac{\pi^4}{90},\\
\zeta(2,2)&=\frac{\pi^4}{120},\\
\zeta(3,1)&=\frac{\pi^4}{360},\\
\zeta(4)&=\frac{\pi^4}{90}.
\end{align*}
$$
So we get
$$
S=\frac{17\pi^4}{360}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2162328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Limit of a infinite series $$\frac{2}{2} + \frac{2\cdot 5}{2\cdot 9} + \frac{2\cdot 5\cdot 10}{2\cdot 9\cdot 28} + \cdots + \frac{2\cdot 5\cdot 10 \cdots (n^2+1)}{2\cdot 9\cdot 28\cdots (n^3+1)}\tag1$$
For this series $(1)$, how would one go about applying the comparison test to check for convergence or divergence?
|
For $n\ge 3$ we get
$$\frac{n^2+1}{n^3+1}<\frac{n^2+1}{n^3}=\frac1n+\frac1{n^3}\le\frac13+\frac1{27}=\frac4{27}<\frac12$$
Then
$$\sum_{k=1}^n\prod_{j=1}^k\frac{j^2+1}{j^3+1}<1+\frac{5}{9}+\sum_{k=3}^n\frac59\left(\frac12\right)^{k-2}$$
So, by comparison test, we get that the given series converges.
|
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"timestamp": "2023-03-29T00:00:00",
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|
equation in D'(\R) we consider in $\mathcal{D}'(\mathbb{R})$ the equation
$$2 x T'' - T' =\delta$$.
I begin to solve the homogeneous equation $2x T'' - T'=0$
i search an solution $T=x^r$ with $r \in \mathbb{R}$, then i found that:
$T_h(x)= C_1 x^{1/2} + C_2 x$
my question is: can you give me please an indication to found an particular solution for the on homogeneous equation?
Thank's for the hel
Ma question est: pouvez vous me donner une indication pour trouver une solution particulière à l'équation non homogène? S'il vous plaît.
Merci d'avance.
|
Let's pose $S = T'$. Then:
$$2 x T'' - T' =\delta \Rightarrow 2x S' - S = \delta \Rightarrow \\
\Rightarrow 2xS' = \delta + S \Rightarrow 2x \frac{dS}{dx} = \delta + S \Rightarrow \\
\frac{dS}{\delta + S} = \frac{dx}{2x}.$$
Integrating both side...
$$\int\frac{dS}{\delta + S} = \int \frac{dx}{2x} \Rightarrow \log(\delta+S) = \frac{1}{2}\log(x) + c.$$
Imposing $c = \frac{1}{2}\log(a)$ then:
$$\int\frac{dS}{\delta + S} = \int \frac{dx}{2x} \Rightarrow \log(\delta+S) = \frac{1}{2}\log(ax) \Rightarrow \delta + S = \sqrt{ax} \Rightarrow \\
S(x) = \sqrt{ax} + \delta.$$
Let's find $T(x)$:
$$T(x) = \int \left(\sqrt{ax} + \delta\right)dx = \frac{2}{3}\sqrt{a}x^{\frac{3}{2}} + \delta x + b.$$
Now, we can check if this work:
$$T'(x) = \sqrt{ax} + \delta, T''(x) = \frac{1}{2}\sqrt{\frac{a}{x}},$$
and hence:
$$2 x T'' - T' =\delta \Rightarrow 2x\frac{1}{2}\sqrt{\frac{a}{x}} - \sqrt{ax} + \delta = \delta \Rightarrow \\
\Rightarrow \sqrt{ax} - \sqrt{ax} + \delta = \delta \Rightarrow 0 = 0.$$
Finally, we said that the solution is:
$$T(x) = \frac{2}{3}\sqrt{a}x^{\frac{3}{2}} + \delta x + b.$$
Since $a$ and $b$ are constants, then we can pose:
$$\begin{cases}
C_1 = \frac{2}{3}\sqrt{a}\\
C_2 = b
\end{cases},$$
thus obtaining
$$T(x) = C_1 x^{\frac{3}{2}} + \delta x + C_2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2167703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Number of ways in which 3 people can throw a normal die to have a total score of 11
Number of ways in which 3 people can throw a normal die to have a total score of 11
My approach:
The answer can be obtained by finding the coefficient of $x^{11}$ in the expansion of $(x+x^2+x^3+x^4+x^5+x^6)^3$.
General term $T=\frac{3}{a!b!c!d!e!f!} x^{a+2b+3c+4d+5e+6f}$
where $a+b+c+d+e+f=3$ and $a+2b+3c+4d+5e+6f=11$
But I can't figure out how to solve further.
|
The approach using generating functions is also fine. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{11}](x^1+x^2+x^3+x^4+x^5+x^6)^3}&=[x^{11}]x^3(1+x+x^2+x^3+x^4+x^5)^3\\
&=[x^8]\left(\frac{1-x^6}{1-x}\right)^3\tag{1}\\
&=[x^8](1-x^6)^3\sum_{n=0}^\infty\binom{-3}{n}(-x)^n\tag{2}\\
&=[x^8](1-3x^6)\sum_{n=0}^\infty\binom{n+2}{2}x^n\tag{3}\\
&=([x^8]-3[x^2])\sum_{n=0}^\infty\binom{n+2}{2}x^n\tag{4}\\
&=\binom{10}{2}-3\binom{4}{2}\tag{5}\\
&=45-3\cdot6\\
&=\color{blue}{27}
\end{align*}
Comment:
*
*In (1) we use the geometric series expansion and apply the rule
\begin{align*}
[x^p]x^qA(x)=[x^{p-q}]A(x)\tag{6}
\end{align*}
*In (2) we apply the binomial series expansion.
*In (3) we expand $(1-x^6)^3$ and observe we do only need two terms of it since we are looking for powers up to $x^8$. We also use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q
\end{align*}
*In (4) we apply the rule (6) twice again.
*In (5) we select the coefficient of $x^8$ and $x^2$ accordingly.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2168095",
"timestamp": "2023-03-29T00:00:00",
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|
How to find an upper bound on the number of solutions of $y^3=x^2+4^k$ I have solved the first two parts of this question but I am struggling with the remaining section. I can't see any meaningful way to reuse what I did before and/or find a way forward.
Just to be clear it is part c) that I am stuck on.
Would anyone be able to help me thanks!
|
The following is heavily based on Theorem 3.3 in Keith Conrad's excellent "Examples of Mordell’s Equation".
$$y^2+4^k=x^3$$
Observe that $4^k$ is always a square, so we factor in $\mathbb{Z}[i]$
$$(y-2^ki)(y+2^ki)=y^2+2^{2k}=x^3$$
If $y-2^ki$ and $y+2^ki$ are cubes then: $y-2^ki=(m+ni)^3$, so $2^k=n(3m^3-n^2)$ there are only $2(k+1)$ possible factorization. Each factorization represent at most one possible solutions, so there are at most $2k+2$ solutions.
Look at the equation modulo 2: $y^2+4^k \equiv x^3 \pmod 2$ so $y \equiv x \pmod 2$.
Assume both $x$ and $y$ are odd and let $δ$ be a common divisor. So $δ$ also divides $(y+2^ki) - (y-2^ki)=2^{k+1}i$. Therefore, the norm of $δ$, $N(δ)$, divides $N(2^{k+1}i)=2^{k+1}$. However it also divides $N(y+2^ki)=y^2+4^k=x^3$ which is odd. Thus the norm of $δ$ is 1 and $(y+2^ki),(y-2^ki)$ are relatively prime. Because $\mathbb{Z}[i]$ is UFD, if a product of two relatively prime factors is a cube then the factors must also be cube up to units. Every unit in $\mathbb{Z}[i]$ is a cube so it can be absorbed inside the other factors and we don't need to worry about it. So $(y+2^ki),(y-2^ki)$ are indeed cubes.
Assume both x and y are even and let $x=2x',y=2y'$. The equation becomes $4{y'}^2+2^{2k}=8{x'}^3$, if k is greater than zero then we can divide by four and get ${y'}^2+2^{2k-2}=2{x'}^3$. If $k$ is greater than one, look modulo $2$ and get ${y'}^2\equiv 0 \pmod 2$ so $4|{y'}^2$ and thus $4|2{x'}^3$ and $2|x'$. Let $x'=2x'',y=2y''$, so $4{y''}^2+2^{2k-2}=16{x''}^3$. If $2^{2k-2}$ is not $1$, divide by $4$, get ${y''}^2+2^{2k-4}=4{x''}^3$. $y''$ must be divisible by $4$ if $2^{2k-4}$ is not one, let $y''=2y'''$ and divide by $4$. ${y'''}^2+2^{2k-6}={x''}^3$ continue like we started until the power of four get to one.
So we have $y^2=x^3-1,y^2=2x^3-1,y^2=4x^3-1$ as possible ending state. $y^2=2x^3-1$ is already dealt with in Conrad's notes (and this answer is getting long). In $y^2=4x^3-1$, $-1$ is not a quadratic residue modulo 4 and therefore has no solution. $y^2=x^3-1$ have only one integer solution and is reached when $k \equiv 0 \pmod 3$.
It is possible to do better - we can divide by $2^6$ as many time we want because we will get a rational point. $\frac{y^2}{2^6}+4^{k-3}=\frac{x^3}{2^6}$ becomes $\left(\frac{y}{2^3}\right)^2+4^{k-3}=\left(\frac{x}{2^2}\right)^3$. So we ask about rational points instead on that curve and the only thing that matters is $k \pmod 3$.
$$
\begin{array}{clcr}
\text{Curve} & \text{Mordell-Weil group}& \text{Number of rational points} & \text{Easier bound} & \text{Overall bound} \\
\hline
y^2=x^3-2^{2\cdot0} & \mathbb{Z}/{2}\mathbb{Z} & 1 & 2k+3 & 1 \\
y^2=x^3-2^{2\cdot1} & \mathbb{Z} & \infty & 2k+2 & 2k+2 \\
y^2=x^3-2^{2\cdot2} & \text{Trivial} & 0 & 2k+2 & 0 \\
\end{array}
$$
See 0,1,2. I will admit I don't know how to prove these facts about Mordell-Weil groups at all yet.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why do I get this matrix the wrong way round? The question is: The set B = {${1+t^2, t+t^2, 1+2t+t^2}$} is a basis for P2. Find the coordinate vector of $p(t)=1+4t+7t^2$ relatvive to B.
I made a matrix:
\begin{bmatrix}
1 & 0 & 1 & 1\\ 0 & 1 & 2 & 4\\
0 & 1 & 0 & 6
\end{bmatrix}
When I row reduce it I get:
\begin{bmatrix}
1 & 0 & 0 & -1\\ 0 & 1 & 0 & 6\\
0 & 0 & 1 & 2
\end{bmatrix}
But the solutions (and MATLAB) say it's:
\begin{bmatrix}
1 & 0 & 0 & 2\\ 0 & 1 & 0 & 6\\
0 & 0 & 1 & -1
\end{bmatrix}
I've tried it many times and I still end up with the same matrix. Does anyone know what I've done wrong?
Thanks
|
Here is one sequence of steps to row reduce the matrix.
\begin{bmatrix}
1 & 0 & 1 & 1\\ 0 & 1 & 2 & 4\\
0 & 1 & 0 & 6
\end{bmatrix}
Subtract the second row from the third:
\begin{bmatrix}
1 & 0 & 1 & 1\\ 0 & 1 & 2 & 4\\
0 & 0 & -2 & 2
\end{bmatrix}
Add the third to the second:
\begin{bmatrix}
1 & 0 & 1 & 1\\ 0 & 1 & 0 & 6\\
0 & 0 & -2 & 2
\end{bmatrix}
Divide the third by $2$:
\begin{bmatrix}
1 & 0 & 1 & 1\\ 0 & 1 & 0 & 6\\
0 & 0 & -1 & 1
\end{bmatrix}
Add the third to the first:
\begin{bmatrix}
1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 6\\
0 & 0 & -1 & 1
\end{bmatrix}
Multiply the third by $-1$:
\begin{bmatrix}
1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 6\\
0 & 0 & 1 & -1
\end{bmatrix}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2171604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Solve an integral $\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$ Solve an integral $$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$$
Using the third substitution of Euler,
$$\sqrt{x^2-1}=(x-1)t\Rightarrow x=\frac{1+t^2}{1-t^2},dx=\frac{4t}{(1-t^2)^2}$$
we get an integral $$\int\frac{1-t^2}{1+t^4}dt=\frac{1}{2}\int\frac{1-\sqrt 2 t}{t^2-\sqrt 2 t+1}dt+\frac{1}{2}\int\frac{1+\sqrt 2t}{t^2+\sqrt 2t+1}dt$$
Using substitutions $$u=t^2-\sqrt 2t+1$$ and $$v=t^2+\sqrt 2t+1$$
we get $$\int\frac{1-t^2}{1+t^4}dt=-\frac{\sqrt 2}{4}\ln|u|+\frac{\sqrt 2}{4}\ln|v|=-\frac{\sqrt 2}{4}\ln|t^2-\sqrt 2t+1|+\frac{\sqrt 2}{4}\ln|t^2+\sqrt 2t+1|$$
From $$x=\frac{1+t^2}{1-t^2}\Rightarrow t=\sqrt{\frac{x-1}{x+1}}\Rightarrow$$
$$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx=$$$$-\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}-\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}+\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+c$$
Is there another, quicker method to solve this integral, rather than Euler substitution.
|
$I=\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$
Put $x=\frac{1}{t}\implies dx=-\frac{1}{t^2}dt \implies
I=-\int\frac{t}{(t^2+1)\sqrt{1-t^2}}dt
$ Put$ 1-t^2=y^2\implies -t dt=y dy. \implies
I=\int\frac{y}{(2-y^2)y}dy
=\frac{1}{2\sqrt2}ln\frac{\sqrt2+y}{\sqrt2-y}+C
=\frac{1}{2\sqrt2}ln\frac{\sqrt2+\sqrt{1-t^2}}{\sqrt2-\sqrt{1-t^2}}+C
=\frac{1}{2\sqrt2}ln\frac{\sqrt2x+\sqrt{x^2-1}}{\sqrt2x-\sqrt{x^2-1}}+C
$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of series $\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$ Find the radius of convergence and the sum of power series
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$$
Radius of convergence is $R=1$, and the interval of convergence is $-1<x<1$.
I am having trouble in finding the sum.
Here is what I have tried.
$$\sum_{n=0}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=\sum_{n=1}^{+\infty}(n+1)x^{2n+1}\int_0^1t^{2n}dt=\int_0^1\left(\sum_{n=1}^{+\infty}(n+1)x^{2n+1}t^{2n}\right)dt$$
$$=x\int_0^1\left(\sum_{n=1}^{+\infty}(n+1)(xt)^{2n}\right)dt$$
$$\sum_{n=1}^{+\infty}(n+1)(xt)^{2n}=\sum_{n=1}^{+\infty}n(xt)^{2n}+\sum_{n=1}^{+\infty}(xt)^{2n}$$
$$\sum_{n=1}^{+\infty}(xt)^{2n}=\frac{(xt)^2}{1-(xt)^2}$$
How to find the sum of $$\sum_{n=1}^{+\infty}n(xt)^{2n}?$$
EDIT:
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=\left(\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}\right)'$$
$$\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}=\frac{1}{2}x^2\sum_{n=1}^{+\infty}\frac{x^{2n}}{2n+1}$$
$$=\frac{1}{2}x^2\sum_{n=1}^{+\infty}x^{2n}\int_0^1t^{2n}dt=\frac{1}{2}x^2\int_0^1\left(\sum_{n=1}^{+\infty}(xt)^{2n}\right)dt$$
$$=\frac{1}{2}x^2\int_0^1\frac{(xt)^2}{1-(xt)^2}=\frac{1}{2}x^4\int_0^1\frac{t^2}{1-(xt)^2}dt$$
$$=\frac{1}{2}x^4\cdot\frac{1}{x^3}(-x-\frac{1}{2}\ln|x-1|+\frac{1}{2}\ln|x+1|)\Rightarrow$$
$$\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}=\frac{1}{2}x\left(-x-\frac{1}{2}\ln|x-1|+\frac{1}{2}\ln|x+1|\right)\Rightarrow$$
$$\left(\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}\right)'=-x-\frac{1}{4}\left(\ln|x-1|+\frac{x}{x-1}\right)+\frac{1}{4}\left(\ln|x+1|+\frac{x}{x+1}\right)\Rightarrow$$
Finally,
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=-x-\frac{1}{4}\left(\ln|x-1|+\frac{x}{x-1}\right)+\frac{1}{4}\left(\ln|x+1|+\frac{x}{x+1}\right)$$
Question: Is this correct?
|
Your work is good. Here's another way to do it.
Let,
$$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n+1}}{2n+1}$$
Clearly we have,
$$f'(x)=\sum_{n=1}^{\infty} x^{2n}=\frac{1}{1-x^2}-1$$
It follows,
$$f(x)=\tanh^{-1}(x)-x$$
Now let,
$$g(x)=\sum_{n=1}^{\infty} n\frac{x^{2n+1}}{2n+1}$$
Clearly we have,
$$2g(x)+f(x)=\sum_{n=1}^{\infty} x^{2n+1}$$
$$2g(x)+\tanh^{-1}(x)-x=\frac{x}{1-x^2}-x$$
From which we get,
$$g(x)=\frac{x}{2-2x^2}-\frac{1}{2}\tanh^{-1}(x)$$
We are interested in $g(x)+f(x)$ this is,
$$\sum_{n=1}^{\infty}\frac{n+1}{2n+1}x^{2n+1}=\frac{1}{2}\tanh^{-1}(x)+\frac{x}{2-2x^2}-x$$
$$=\frac{1}{4}\ln (x+1)-\frac{1}{4}\ln (1-x)+\frac{x}{2-2x^2}-x$$
Edit
Note another way:
Assume $|x|<1$ and $|t|<1$.
$$\int_{0}^{x} t^{2n} dt=\frac{x^{2n+1}}{2n+1}$$
So that
$$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n+1}}{2n+1}$$
$$=\int_{0}^{x} \sum_{n=1}^{\infty} t^{2n} dt$$
$$=\int_{0}^{x} \left(\frac{1}{1-t^2}-1 \right) dt $$
$$=\text{tanh}^{-1}(x)-x$$
It also follows that,
$$\frac{1}{2}x^2\sum_{n=1}^{+\infty}\frac{x^{2n}}{2n+1}$$
$$=\frac{1}{2}xf(x)$$
$$=\frac{1}{2}x\text{tanh}^{-1}(x)-\frac{1}{2}x^2$$
So this way works too. Just continue with your work or the previous solution.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Maximum value of equation $ab+bc+ca$ if $a+2b+c=4$ If $a,b,c$ ${\in}$ $\mathbb{R} $ such that
$a+2b+c=4$, then find the $max$ value of
$ab+bc+ca$.
I always get stuck with max, min questions. We cannot apply AM:GM here, I have not studied much calculus yet.
Can you do this with graphs or by plane algebra?
I don't really know what tag to put for this!
|
Substitution of $b=\frac{4-a-c}{2}$ into $ab+bc+ca$ gives:
$$\begin{align}-\frac{a^2}{2} + 2 a -\frac{c^2}{2} + 2 c
& = \tfrac{1}{2}\left( 4a-a^2+4c-c^2\right) \\[3pt]
& = \tfrac{1}{2}\left( 4-\left(2-a \right)^2+4-\left(2-c \right)^2 \right) \\[3pt]
& = 4-\tfrac{1}{2}\left(2-a \right)^2-\tfrac{1}{2}\left(2-c \right)^2
\end{align}$$
And this is clearly maximal when $a=c=2$; yielding a maximal value of $4$.
|
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"url": "https://math.stackexchange.com/questions/2174458",
"timestamp": "2023-03-29T00:00:00",
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|
Stewart's Theorem In triangle $ABC$, angle $\angle A = 90^\circ$. Let $M$ be a point on the hypotenuse $BC$. Prove that $MB^2\left(AC^2\right)+MC^2\left(AB^2\right)=MA^2\left(BC^2\right)$.
Can someone please provide guidance as to how to go about this proof?
|
Note that $AQ = AC\cdot \frac{BM}{BC}$ and $AP = AB\cdot \frac{CM}{BC}$
Then $AM^2 = AP^2 + AQ^2 = \left(AC\cdot \frac{BM}{BC}\right)^2 + \left(AB\cdot \frac{CM}{BC}\right)^2$
So $AM^2\cdot BC^2 = BM^2\cdot AC^2 + CM^2\cdot AB^2$ as required.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of transition probabilities of an infinite Markov chain I have a Markov chain with state space $S = \left\{ 1,2,\dots \right\}$ with transition matrix as follows
$\begin{bmatrix}
0 & 1 & 0 & 0 & 0 & 0 & \dots \\
0 & 0 & 1 & 0 & 0 & 0 & \dots \\
0 & 0 & \frac{2}{3} & \frac{1}{3} & 0 & 0 & \dots \\
0 & 0 & \frac{2}{3} & 0 & \frac{1}{3} & 0 & \dots \\
0 & 0 & \frac{2}{3} & 0 & 0 & \frac{1}{3} & \dots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{bmatrix}
$
I am trying to find $\lim_{n\to\infty} P^n(4,7)$
and I believe I should be using this theorem:
$\textbf{Theorem 7}$ Let $X_n, n > 0$, be an irreducible positive recurrent
Markov chain having stationary distribution $\pi$. If the chain is aperiodic,
$(55) \lim_{n\to\infty} P^n(x,y) \space x,y \in S$
If the chain is periodic with period d, then for each pair $x, y$ of states in
$S$ there is an integer $r, 0 \leq r < d$, such that $P^n(x, y) = 0$ unless
$n = md + r$ for some nonnegative integer $m$, and
$(56) \lim_{m\to\infty} P^{md+r}(x,y) = d\pi(y)$
The problem is that it does not satisfy the conditions of being irreducible and positive recurrent so I cannot use the theorem. However I see that if I consider only the block not containing states $1,2$ ie
$\begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 & 0 & \dots \\
\frac{2}{3} & 0 & \frac{1}{3} & 0 & \dots \\
\frac{2}{3} & 0 & 0 & \frac{1}{3} & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{bmatrix}
$
Then this is irreducible, positive recurrent and would be aperiodic and can use the theorem. Would it be valid to apply the theorem to this block matrix so that I can find
$\lim_{n\to\infty} P^n(4,7)$?
|
Yes it is appropriate to consider the the state transition matrix
$$\begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 & 0 & \dots \\
\frac{2}{3} & 0 & \frac{1}{3} & 0 & \dots \\
\frac{2}{3} & 0 & 0 & \frac{1}{3} & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{bmatrix}
$$
since the original chain gets in this chain in two steps. So we can forget about the first two states, 1 and 2, $P_1$ and $P_2$ will be $0$ on the long run because the chain will never get (back) to these states.
In the case of the transition matrix above, it is easy to calculate the stationary probabilities:
So, we have the stationary probabilities in a row vector $\pi=[P_3\ P_4 \ P_5\cdots]$ and we have the equation for the stationary probabilities
$$[P_3\ P_4 \ P_5\cdots]\begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 & 0 & \dots \\
\frac{2}{3} & 0 & \frac{1}{3} & 0 & \dots \\
\frac{2}{3} & 0 & 0 & \frac{1}{3} & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{bmatrix}=[P_3\ P_4 \ P_5\cdots].$$
In the case of the row vector and the firs column we get
$$\frac23(P_3+P_4+\cdots)=\frac23=P_3.$$
Then in the case of the row vector and the second column we have
$$P_3\frac13=\frac29=P_4$$
and so on.
So we have for the row vector of the stationary probabilities
$$\left[0 \ 0\ \frac23\ \ \frac29\ \ \frac2{27}\ \cdots\right].$$
Here $P_1$ and $P_2$ were included.
If I understood the question well then
$$\lim_{n\to\infty} p^n(4,7)=\frac2{3^5}=\pi(7)$$
because
$$\lim_{n\to\infty} \begin{bmatrix}
0&1&0&0&\cdots\\
0&0&1&0&\cdots\\
\frac{2}{3} & \frac{1}{3} & 0 & 0 & \dots \\
\frac{2}{3} & 0 & \frac{1}{3} & 0 & \dots \\
\frac{2}{3} & 0 & 0 & \frac{1}{3} & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{bmatrix}^n
=\begin{bmatrix}
0&0&\frac23&\frac2{3^2}&\frac2{3^3}&\frac2{3^4}&\frac2{3^5}\cdots\\
0&0&\frac23&\frac2{3^2}&\frac2{3^3}&\frac2{3^4}&\frac2{3^5}\cdots\\
0&0&\frac23&\frac2{3^2}&\frac2{3^3}&\frac2{3^4}&\frac2{3^5}\cdots\\
0&0&\frac23&\frac2{3^2}&\frac2{3^3}&\frac2{3^4}&\boxed{\color{red}{\frac2{3^5}}}\cdots\\
\vdots & \vdots & \vdots & \vdots & \vdots& \vdots & \ddots
\end{bmatrix}.$$
|
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|
Factorise the expression Factorise of the expression $$ a^4+b^4-c^4-2a^2b^2+4abc^2~ \\(S:(a + b - c) (a + b + c) (a^2+ b^2 + c^2 - 2 a b))$$
I tried to solve this problem:
$$ a^4+b^4-c^4-2a^2b^2+4abc^2\rightarrow\\\
(a^2-b^2)^2-c^4+4abc^2\rightarrow\\\
(a^2-b^2-c^2)(a^2-b^2+c^2)+4abc^2\\\
((a-b)(a+b)-c^2)((a-b)(a+b)+c^2)+4abc^2$$ but I stopped here. Can anyone help me?
|
.You should actually do:
\begin{split}
a^4+b^4-c^4-2a^2b^2+4abc^2 & = (a^4+b^4 + 2a^2b^2) - (4a^2b^2 + c^4 - 4abc^2) \\
& = (a^2+b^2)^2 - (c^2-2ab)^2 \\
& = (a^2+b^2-c^2+2ab)(a^2+b^2+c^2-2ab) \\
& = ((a+b)^2 - c^2) (a^2+b^2+c^2-2ab) \\
& = (a+b+c)(a+b-c)(a^2+b^2+c^2-2ab)
\end{split}
In the first step, I added and subtracted $4a^2b^2$, and in the second step grouped them into squares. From then, it is just the formula $x^2-y^2 = (x+y)(x-y)$ applied twice.
In your attempt, the $4abc^2$ was just left hanging. That shoul have been dealt with by trying to bring it into the expansion of a square term (which I did).
|
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|
Solve $BAB^{-1}=J$ for matrix $B$ I have 3 Matrices : $A , B$ and $J.$ I know matrices $A$ and $J. J$ is the real jordan normal form of $A.$ The relationship between the three matrices is given by $BAB^{-1} =J.$
How can I find the matrix $B.$
For example,
$A= \begin{bmatrix} 0 & 1 & 0 \\
0 & 0 & 1 \\
1 &-1 & 1 \end{bmatrix}$
$J= \begin{bmatrix} 0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \end{bmatrix}$
$BAB^{-1}=J$ can be written as $BA-JB=0$. That's as far as I got.
Thanks for your help.
|
$\det(A-\lambda I) = 0$
$-(\lambda-1)(\lambda+i)(\lambda-i)$
Complex eigenvalues
Here is a little trick
$(B^{-1} A B)^2 = J^2\\
B^{-1} A^2 B = J^2$
And $A^2$ has real eigenvalues and the same matrices $B, B^{-1}$
$A^2 = \begin{bmatrix} 0&0&1\\1&-1&1\\1&0&1\end{bmatrix}$
and $J^2 = \begin{bmatrix}-1&0&0\\0&-1&0\\0&0&1\end{bmatrix}$
And we know that $-1,-1,1$ will be the eigenvalues of $A^2.$ Find the associated vectors.
$A^2 \begin{bmatrix}-1&0&1\\0&1&1\\1&0&1\end{bmatrix} = \begin{bmatrix}-1&0&1\\0&1&1\\1&0&1\end{bmatrix} J^2$
$B = \begin{bmatrix}-1&0&1\\0&1&1\\1&0&1\end{bmatrix}$
And I will leave it to you to find $B^{-1}$
|
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|
Solve the the given equation: $\sqrt{3x^2+x+5} = x-3$ We have to find the number of solution for the given equation:
$$\sqrt{3x^2+x+5} = x-3.$$
There are two solution one is
By using graph we get one solution
By squaring both sides we get no solution
I want to know which solution is correct
|
Hint:
The given equation is equivalent to the system:
$$
\begin{cases}
3x^2+x+5=(x-3)^2\\
x-3\ge 0
\end{cases}
$$
Note that the system has no solutions because the roots of the second degree equations :
$$
x=\frac{-7\pm\sqrt{79}}{4}
$$
are less than $3$.
And this is in accord with the fact that the graphs of the two functions
$$
y=\sqrt{3x^2+x+5} \qquad y=x-3
$$
have no common points (in your graph you have the wrong function $x=3$).
|
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|
then minimum number of number of roots of $p(x) = 0$ is let $p(x)=x^6+ax^5+bx^4+x^3+bx^2+ax+1.$ given that $x=1$ is a one rot of $p(x)=0$
and $-1$ is not a root. then minimum number of number of roots of $p(x) = 0$ is
Attempt: $x=0$ in not a root of $p(x)=0.$
So $\displaystyle \left(x^3+\frac{1}{x^3}\right)+a\left(x^2+\frac{1}{x^2}\right)+b\left(x+\frac{1}{x}\right)+1=0$
So $\displaystyle \left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)+a\left(x+\frac{1}{x}\right)^2-2+b\left(x+\frac{1}{x}\right)+1=0$
So $\displaystyle t^3+at^2+(b-3)t+1=0,$ where $\displaystyle \left(x+\frac{1}{x}\right) = t$ and $|t|\geq 2$
Could some help me to solve it, thanks
|
We assume you want real roots (complex roots are 6 exactly).
$x=1$ is one of the root of the main equation so $t=2$ is the root of the second, write
$$\displaystyle t^3+at^2+(b-3)t+1-2a=0~~;~~~t\neq0~~(x\neq-1)$$
$$(t-2)(t^2+mt+n)=0~~;~~~t\neq0$$
this concludes that $n=a-\dfrac12$ and $m=a+2$.
$t\neq0$ says $n=a-\dfrac12\neq0$ and for $t^2+mt+n=t^2+(a+2)t+(a-\dfrac12)=0$ we see
$$t_1=\dfrac{-(a+2)+\sqrt{a^2+6}}{2}~~~,~~~t_2=\dfrac{-(a+2)-\sqrt{a^2+6}}{2}$$
if $a>-\dfrac14$ then $t_2<-2$ thus $t+\dfrac1t=t_2$ has two roots.
if $a<-\dfrac52$ then $t_1>2$ thus $t+\dfrac1t=t_1$ has two roots.
for $-\dfrac52<a<-\dfrac14$ is not determined if the eqution has other root(s) or not.
|
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|
geometric progression Help me to solve this question :
The sum of first two terms of a positive geometric progression is $9$ times the sum of the following two terms. Given that the sum of the first four terms is $\frac{320}{9}$.
Find
*
*the first term and the common ratio
*the sum of fifth term to infinity
I know that:
$$a+ar=9(ar^2+ar^3)\tag 1$$
And:
$$a+ar+ar^2+ar^3=\frac{320}{9}\tag 2$$
But I am unsure where to continue from here.
|
You're almost there. First, the common ratio $r$ must be positive (since all terms are positive), that is, $\color\red{r>0}$. Start with equation $(1)$ by dividing by $a$ to get $$1+r=9(r^2+r^3).$$ Next, apply factoring to get $$1+r=9r^2(1+r).$$ Then cancel $1+r$ (since $r>0$ and so $1+r\neq 0$) to get $$1=9r^2$$ so that $9r^2-1=0$ and since $r>0$, so $r\neq -\frac{1}{3}$ and hence $$r=\frac{1}{3}.$$ Equation $(2)$ is equivalent to $$\frac{a(1-r^4)}{1-r}=\frac{320}{9}.$$ Hence,
$$a=\frac{320}{9}\cdot\frac{1-r}{1-r^4}=\frac{320}{9}\cdot\frac{2/3}{80/81}=\frac{320}{9}\cdot\frac{27}{40}=24.$$ This answer part 1.
For part 2, we have
$$\begin{align}
24(1/3)^4+24(1/3)^5+24(1/3)^6+\dots=\frac{24(1/3)^4}{1-\frac{1}{3}}=\frac{4}{9}.
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
a surface integral problem $F=(y-z,z-x,x-y)$ $S$ be the portion of surface defined by $x^2+y^2+z^2=1$ and $x+y+z\ge 1$. We want to evaluate $\int_S curl(F)\cdot dS$.
I have found $curl(F)=(-2,-2,-2)$ and the normal vector of $S$ is $(x,y,z)$. Thus the integral becomes $\int_S -2(x+y+z) dS$. However, if I use the polar coordinate of $S$, then the boundary of $\phi$ and $\theta$ is a big problem.
I still consider using Stoke's Theorem. It then should be $\int_A F\cdot dr$ where $A$ is the circle defined by the unit ball and the plane $x+y+z=1$. However I can not find a way to paramettrize this circle. That is a big problem.
|
You can also use the divergence theorem here, it will save you the heavy computations you are not comfortable with:
If $S_2$ denotes the part of the plane $x+y+z=1$ inside the sphere $x^2+y^2+z^2=1$, and $E$ the 3D-region bounded by $S$ and $S_2$, then
$$
\iint_S \nabla \times F\cdot dS+ \iint_{S_2} \nabla \times F\cdot dS= \iiint_E \nabla\cdot \nabla \times F\; dV = 0
$$
Since the plane $x+y+z=1$ has normal unitary vector $n=-\frac{\pmatrix{1\\1\\1} }{\sqrt{3}}$, it follows that
$$
\iint_S \nabla \times F\cdot dS= - \iint_{S_2} \nabla \times F\cdot dS = - \iint_{S_2}\pmatrix{-2\\-2\\-2}\cdot\pmatrix{-1\\-1\\-1}\frac{dS}{\sqrt{3}}=\frac{-6}{\sqrt{3}}\;A(S_2)=\frac{-6}{\sqrt{3}}\; \frac{2\pi}{3}=\frac{-4 \pi}{\sqrt{3}}
$$
Note: Perhaps the most difficult part is finding the area of $S_2$ ($A(S_2)$). The projection of $S_2$ in the $xy$ plane is the ellipse $x^2+y^2+(1-x-y)^2=1$, which has area $A'=\frac{2\pi}{3\sqrt{3}}$. It is not difficult to see that $A(S_2)=\sqrt{3}A'$. Alternatively, you can use the arguments described here to show that the disc $S_2$ has radius $\sqrt{\frac{2}{3}}$.
|
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|
Explanation why $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ = $\frac{-\pi^2}{12}$ with Euler's $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ Can somebody explain why the sum is $\frac{-\pi^2}{12}$. Can you somehow use the zeta function? If so, how?
|
Let's add them together:
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}+\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{(-1)^n+1}{n^2}$$
and for odd $n$, $(-1)^n+1=0$ and for even $n$, $(-1)^n+1=2$, so this rewrites to
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}+\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{2}{(2n)^2}$$
which simplifies to
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}+\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac12\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{12}$$
Thus,
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}=\frac{\pi^2}{12}-\frac{\pi^2}{6}=-\frac{\pi^2}{12}$$
|
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|
Find $1^2+3^2+...99^2$ given $1^1+2^2+...+100^2$ and $1^1+2^2+...+50^2$ We know that $1^1+2^2+...+100^2=338350$ and $1^1+2^2+...+50^2=42925$. Find $1^2+3^2+...99^2$.
I don't know really where to start. I tried to find a pattern in the sequences, but there was none. Can I substitute values for the equations?
|
As mentioned in the comments the sum of the series can be derived in the following way
$$\sum_{i=1}^{2n}i^2=1^2+2^2+3^2+...+(2n)^2=\frac{2n(2n+1)(4n+1)}{6}$$
Similarly,
$$\sum_{i=1}^{n}{(2i)}^2=2^2+4^2+...+(2n)^2=4\sum_{i=1}^{n}{i}^2=\frac{2n(n+1)(2n+1)}{3}$$
Subtacting gives required sum
$$\sum_{i=1}^{n}{(2i-1)}^2=1^2+3^2+...+(2n-1)^2=\frac{2n(2n+1)(4n+1)}{6}-\frac{4n(n+1)(2n+1)}{6}=\frac{2n(2n+1)(4n+1-2n-2)}{6}=\frac{n(2n+1)(2n-1)}{3}$$
Put $n=50$ to get required solution.
which gives
$${50(101)33}=166650$$
|
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|
Given 4 equations, is there a unique solution that can fit a model with 2, 3, or 5 variables? Given the a data set with 4 equations with x1, x2 as main drivers.
A) Can there be a unique multilinear model s(x1,x2) = B0+ B1x1+ B2x2 that perfectly fits the data.
Even before putting the equations into MAtLab, I assumed there could NOT be a unique solution because there are more equations than variables.
B) I followed the same logic when asked to fit the data to the equation: B0 + B1x1 + B2x2 + B3x1x2
Since there are three variables and still 4 equations, there could not be a unique solution...Is this logic correct?
and when asked for to fit the data to the model: B0 + B1x1 + B2x2 + B3x1x2 + B4(x1^2) + B4 (x2^2)..I was not sure because now there are more variables than equations. There could be a unique solution I believe.
x1= avg. temp
x2=median income
Additional info: the data set Year, Avg. Temp, Median Income, and Total Sales
2009 86.92 30.11 27.93
2010 88.51 31.48 28.29
2011 88.01 32.03 29.70
2012 87.05 33.34 31.09
|
A
Problem specification: Start with a sequence of $m=4$ measurements $\left\{ x_{k}, y_{k}, z_{k} \right\}_{k=1}^{m}$. Use the method of least squares to find the best trial function
$$
z(x,y) = b_{0} + b_{1}x + b_{2}y.
$$
That is, find the solution vector $b$ defined as
$$
b_{LS} = \left\{ b \in \mathbb{C}^{m} \colon \lVert \mathbf{A}b - z \rVert_{2}^{2} \text{ is minimized} \right\}.
$$
Your problem has full column rank, so the least squares solution will be unique (the null space $\mathcal{N}\left( \mathbf{A} \right)$ is trivial.
$$
\begin{align}
\mathbf{A} b & = z \\
\left[ \begin{array}{ccc}
1 & x_{1} & y_{1} \\
1 & x_{2} & y_{2} \\
1 & x_{3} & y_{3} \\
1 & x_{4} & y_{4}
\end{array} \right]
%
\left[ \begin{array}{c}
b_{0} \\ b_{1} \\ b_{2}
\end{array} \right]
%
&=
%
\left[ \begin{array}{c}
z_{1} \\ z_{2} \\ z_{3} \\z_{4}
\end{array} \right]
%
\end{align}
$$
Because the matrix has full column rank, we may solve directly with the normal equations:
$$
b_{LS} = \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} z.
$$
B
The new trial function
$$
z(x,y) = a_{00} + a_{10}x + a_{01}y + a_{20} x^{2} + a_{20} x y + a_{20} y^{2}
$$
is complete through second order and involves finding $n=6$ coefficients. We no longer have full column rank and the solution is not unique.
$$
\begin{align}
\mathbf{A} a & = z \\
\left[ \begin{array}{ccc}
1 & x_{1} & y_{1} & x_{1}^{2} & x_{1}y_{1} & y_{1}^{2} \\
1 & x_{2} & y_{2} & x_{2}^{2} & x_{2}y_{2} & y_{2}^{2} \\
1 & x_{3} & y_{3} & x_{3}^{2} & x_{3}y_{3} & y_{3}^{2} \\
1 & x_{4} & y_{4} & x_{4}^{2} & x_{4}y_{4} & y_{4}^{2} \\
\end{array} \right]
%
\left[ \begin{array}{c}
a_{00} \\ a_{10} \\ a_{01} \\ a_{20} \\ a_{20} \\ a_{20}
\end{array} \right]
%
&=
%
\left[ \begin{array}{c}
z_{1} \\ z_{2} \\ z_{3} \\z_{4}
\end{array} \right]
%
\end{align}
$$
The general least squares solution for this problem is
$$
a_{LS} =
\color{blue}{\mathbf{A}^{\dagger}z} +
\color{red}{\left(
\mathbf{I}_{6} - \mathbf{A}^{\dagger} \mathbf{A}
\right) \zeta}
, \qquad \zeta \in\mathbb{C}^{6}.
$$
where blue vectors are in a $\color{blue}{range}$ space, and red vectors are in a $\color{red}{null}$ space.
|
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|
Why are vectors of a regular hexagon cos60 of each other? In a regular hexagon OABCDE $a$ , $b$ denote respectively the position vectors of $A$, $B$ with respect to $O$
I would have assumed that $\overrightarrow {OA} = \overrightarrow {AB} = \overrightarrow {BC} $ since it's a regular hexagon. And, therefore, $\overrightarrow {OC}$ should equal $\overrightarrow {OA} + \overrightarrow {AB} + \overrightarrow {BC} $ and therefore $\overrightarrow {3AB}$.
However, in the literature I was reading, $\overrightarrow{OC}$ is defined as $\overrightarrow {OA}\cos60 + \overrightarrow {AB} + \overrightarrow {BC} \cos60 $, and thus $\frac12 \overrightarrow{AB} + \overrightarrow {AB} + \frac12 \overrightarrow {AB} $ — ultimately $\overrightarrow{2AB}$. Why is this?
An image is included below.
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The actual vectors are:
$$\overrightarrow{OA}=\begin{pmatrix}\frac12\\\frac{\sqrt{3}}2\end{pmatrix}$$
$$\overrightarrow{AB}=\begin{pmatrix}1\\0\end{pmatrix}$$
$$\overrightarrow{BC}=\begin{pmatrix}\frac12\\\frac{\sqrt{3}}2\end{pmatrix}$$
So we have $|\overrightarrow{OA}|=|\overrightarrow {AB}|=|\overrightarrow{BC}|$.
And $\overrightarrow{OA}+\overrightarrow {AB}+\overrightarrow{BC}=\begin{pmatrix}\frac12\\\frac{\sqrt{3}}2\end{pmatrix}+\begin{pmatrix}1\\0\end{pmatrix}+\begin{pmatrix}\frac12\\\frac{-\sqrt{3}}2\end{pmatrix}=\begin{pmatrix}2\\0\end{pmatrix}=2\overrightarrow {AB}$.
The literature uses $OA \cos 60^\circ$ to mean $|\overrightarrow{OA}| \cos 60^\circ$, and not $\overrightarrow{OA} \cos 60^\circ$, to mean the length of $\overrightarrow{OA}$, which is $1$.
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Complex number equation to cartesian equation I need to find the cartesian equation of the loci of z given the below equation:
$\frac{(z-j)\cdot(z-j)^*}{|3-4j|}=5$
So if $z=x+jy$
Then $(z-j)^*=(x+jy-j)^*$
$=[x+j(y-1)]^*$
$=x-j(y-1)$
$=x-jy+y$
Substituting back into original equation
$\frac{(x+jy-j)\cdot(x-jy+j)}{|3-4j|}=5$
$\frac{(x+j(y-1))\cdot (x-j(y-1))}{|3-4j|}=5$
$\sqrt{x^2+(y-1)^2}\cdot \sqrt{x^2-(y-1)^2}=5|3-4j|$
This is where I'm getting a little lost...
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For any complex number $w = a + bj$,
$$|w| = \sqrt{a^2 + b^2}$$
$$ww^* = |w|^2 = a^2 + b^2$$
hence if $z=x + yj$, then
\begin{align*}
z-j &= x +yj - j = x + (y-1)j\\[4pt]
\implies\; (z-j)(z-j)^*&=|z-j|^2\\[4pt]
&= |x + (y-1)j|^2\\[4pt]
&= x^2 + (y-1)^2\\[4pt]
\end{align*}
Also, $|3-4j| = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5,\,$ hence
\begin{align*}
&\frac{(z-j)(z-j)^*}{|3-4j|}=5\\[6pt]
\implies\; &\frac{x^2 + (y-1)^2}{5}=5\\[6pt]
\implies\; &x^2 + (y-1)^2=25\\[6pt]
\end{align*}
which is the equation of a circle in the $xy$-plane, centered at $(0,1)$, with radius $5$.
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Solving $\int{\sqrt{x^2+1}} \ \mathrm{d}x$ using integration by parts, given that $\int{\frac{1}{\sqrt{x^2+1}} \ \mathrm{d}}x=\ln(x+\sqrt{x^2+1})+k$. Here is the question once again:
Deduce the integral $\int{\sqrt{x^2+1}} \ \mathrm{d}x$ using integration by parts, given the following identity: $$\int{\frac{1}{\sqrt{x^2+1}} \ \mathrm{d}}x=\ln(x+\sqrt{x^2+1})+k$$
I know that the integral can be solved by using trigonometric substitutions, but I am looking for another way as that's not what is asked in the question.
Here is what I have done so far:
$$\int{\sqrt{x^2+1}} \ \mathrm{d}x=x \cdot \sqrt{x^2+1}-\underbrace{\int{\frac{x^2}{\sqrt{x^2+1}} \ \mathrm{d}x}}_{\text{$I_1$}}$$
$$I_1=x^2 \cdot \ln{(x+\sqrt{x^2+1})}-\underbrace{\int{\ln{(x+\sqrt{x^2+1})}\cdot 2x \ \mathrm{d}x}}_{\text{$I_2$}}$$
Solving $I_2$ using integration by parts gives an expression which includes $I_1$ so I got into a loop. I'm out of ideas so I would be glad if you could help me out, thank you for your time.
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Notice that
$$\int \sqrt {x^2 + 1} \ \Bbb d x = \int (x')\sqrt {x^2 + 1} \ \Bbb d x = x \sqrt {x^2 + 1} - \int x \frac {x} {\sqrt {x^2 + 1}} \ \Bbb d x = \\
x \sqrt {x^2 + 1} - \int \frac {x^2 + 1 -1} {\sqrt {x^2 + 1}} \ \Bbb d x = x \sqrt {x^2 + 1} - \int \sqrt {x^2 + 1} - \frac 1 {\sqrt {x^2 + 1}} \ \Bbb d x = \\
- \int \sqrt {x^2 + 1} \ \Bbb d x + x \sqrt {x^2 + 1} + \ln (x + \sqrt {x^2 + 1})$$
whence it follows that
$$2 \int \sqrt {x^2 + 1} \ \Bbb d x = x \sqrt {x^2 + 1} + \ln (x + \sqrt {x^2 + 1}) $$
so that
$$\int \sqrt {x^2 + 1} \ \Bbb d x = \frac 1 2 x \sqrt {x^2 + 1} + \frac 1 2 \ln (x + \sqrt {x^2 + 1}) + C$$
where $C \in \Bbb R$ is an arbitrary integration constant added at the end of the calculations.
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Given $x^9 = e$ and $x^{11} = e$ prove $x = e$. Full Problem: Prove that for any element $x$ in a group $G$ that satisfies
$$x^9 = e \\
x^{11} = e,$$
where $e$ is the identity element, that $x$ itself must be $e$.
Is this as simple as showing that
*
*$x^{11} = x^{9} \cdot x^{2} = e \cdot x^2 \Rightarrow x^2 = e$
*$x^{9} = x^{2} \cdot x^{7} = e \cdot x^7 \Rightarrow x^7 = e$
*$x^{7} = x^{2} \cdot x^{5} = e \cdot x^5 \Rightarrow x^5 = e$
*$x^{5} = x^{2} \cdot x^{3} = e \cdot x^3 \Rightarrow x^3 = e$
*$x^{3} = x^{2} \cdot x = e \cdot x \Rightarrow x = e$
Therefore, $x = e$.
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$$e=(x^{11})^5=x^{55}=x^{54}\cdot x=(x^9)^{6}\cdot x=e\cdot x=x$$
As you can see, this thus follows because there is integer solutions to $11x-9y=1$, which is true because $11$ and $9$ are relatively prime.
Your approach is doing much the same, using a slow form of the Euclidean algorithm to show that $11$ and $9$ are relatively prime:
$$11=9\cdot 1 + 2\\
9=2\cdot 1 + 7\\
7=2\cdot 1 + 5\\
5=2\cdot 1 + 3\\
3=2\cdot 1 + 1$$
You could have skipped a lot of steps by doing the equivalent of $9=2\cdot 4 + 1$, as other answers have suggested.
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Factoring a quartic polynomial over the integers with roots that are not integers The quartic polynomial
$$
1728(x - 3) - x^2(12^2 - x^2)
$$
factors "nicely" as
$$
(x^2 - 12x + 72) (x^2 + 12x - 72) = (x^2 - 12x + 72)(x - 6\sqrt{3} + 6)(x + 6\sqrt{3} + 6) \, .
$$
(Note that $1728 = 3(24^2)$.) How is this factorization obtained?
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We have
$$
(x^2-12x+72)(x^2+12x-72)=x^4 - 144x^2 + 1728x - 5184=x^2(x^2-12^2)+1728(x-3),
$$
The factorization is of the form $(a^2+b)(a^2-b)=a^4-b^2$, with $a=x$ and $b=12x-72$.
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|
$3\mid a^3+b^3+c^3$ if only if $3\mid a+b+c $
Prove the following equivalence: $3\mid (a^3 + b^3 + c^3) $ if and only if $3\mid (a + b + c) $.
My try:
I know $a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) + 3abc$, but I can't seem to proceed from here.
Thanks all!
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Very short with Lil' Fermat:
for any $x\in\mathbf Z$, $\;x^3\equiv x\mod 3$. Hence $\;a^3+b^3+c^3\equiv a+b+c\mod 3$.
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How can I prove this trigonometric equation with squares of sines? Here is the equation:
$$\sin^2(a+b)+\sin^2(a-b)=1-\cos(2a)\cos(2b)$$
Following from comment help,
$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$
$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \cos^2 b + \cos^2 a \sin^2 b$$
I am stuck here, how do I proceed from here?
Edit: from answers I understand how to prove,but how to prove from where I am stuck?
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$\sin^2(a+b)+\sin^2(a−b)
$
$=\sin^2a\cos^2b+\cos^2a\sin^2b+\sin^2a\cos^2b+\cos^2a\sin^2b
$
$=2(\sin^2a\cos^2b+\cos^2a\sin^2b)
$
$=2\left ({\left (\dfrac{(1-\cos(2a)}{2}\right )(\cos^2b)+\left (\dfrac{\cos(2a)+1}{2}\right )(\sin^2b)}\right )
$
$=1+\cos(2a)\sin^2b-\cos(2a)\cos^2b
$
$=1-\cos(2a)\cos(2b)
$
Hence proved.
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Prove that if $x \equiv 5 \pmod{10}$, then $y \equiv 0 \pmod{7}$
Let $x,y$ be positive integers satisfying $2x^2-y^2 = 1$. Prove that if $x \equiv 5 \pmod{10}$, then $y \equiv 0 \pmod{7}$.
I wasn't sure how to use the fact that $x,y$ are positive integers satisfying $2x^2-y^2 = 1$. We could use the theory of Pell's equations to find the solutions, but that would get complicated. Is there a simpler way?
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If $(x,y)$ is a solution of $2x^2-y^2=1$ such that $x>5$, $y>0$ and $x\equiv 5(10)$ then
$$(x',y')=(99 x-70 y,-140 x+99 y)$$
is also a solution with $0<x'<x$ , $x' \equiv 5(10)$, and $y' \equiv y(7)$
The process of repeatedly applying this transformation to a solution $(x,y)$ will stop at the solution $(5,7)$ thereby proving the result
It easy to check that $(x',y')$ is a solution. Next
$$x\sqrt{2}-y=\frac{1}{x\sqrt{2}+y}<\frac{1}{5\sqrt{2}}$$
gives, after some algebra,
$$10 y> 10\sqrt{2}x-\sqrt{2} > 14x$$
the last inequality is true since $x\geq 10$
Therefore $99x-70y<x$
Also, $x\sqrt{2}-y>0$ gives $99x-70y>0$
Finally,
$$99 x-70 y\equiv 99*5\equiv 5(10)$$
and
$$-140 x+99 y\equiv y(7)$$
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The minimum value of $x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$ is The minimum value of $$f(x)=x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$$ is
(a)-1
(b)9
(c)6
(d)1
Apart from trying to obtain $1$, which in this case is simple and $f(2)=1$ is there a standard method to approach such problems.
Please keep in mind that this is an objective question in one of the competitive exam and you get around 5 mins to solve it.
Also this is asked in elementary section, so only knowledge of basic calculus and polynomials is assumed.
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Let $a,b,c,d,e,k\in\Bbb R$,
$$(x^4-ax^2)^2+(bx^2-cx)^2+(dx-e)^2+k=\\ x^8- 2 a x^6+(a^2+ b^2) x^4 - 2 b c x^3 + (c^2 + d^2) x^2 - 2 d e x+e^2+k$$
By identification we easily obtain
$$a=4 \implies b = \sqrt{3} \implies c=\frac{6}{\sqrt{3}} \implies d= \sqrt{2} \implies e= \frac{4}{\sqrt{2}} \implies k=1$$
this implies that
$$x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9 = \\ \big(x^4-4x^2\big)^2+\Big(\sqrt{3}x^2-\frac{6}{\sqrt{3}}x\Big)^2+\Big(\sqrt{2}x-\frac{4}{\sqrt{2}}\Big)^2+1 \\
= \big(x^2(x^2-4)\big)^2+3\big(x(x-2)\big)^2+2\big(x-2\big)^2+1>0 $$
Hence, $-1$ can not be the minimal value.
As you have already noticed, it is easy to observe that $f(2)=1$ and since you know that exactly one answer is correct (see comments under OP), it must be the answer (d).
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Prove $\sum_{i=0}^{n-1} \binom{2i}{i}\frac{1}{i+1}\frac{1}{2^{2i}} = 2(1 - \frac{1}{2^{2n}}\binom{2n}{n})$ How can I prove $\sum_{i=0}^{n-1} \binom{2i}{i}\frac{1}{i+1}\frac{1}{2^{2i}} = 2(1 - \frac{1}{2^{2n}}\binom{2n}{n})$ without induction? I think I have to use Taylor series at some point.
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We may notice that by setting $C_i=\frac{1}{i+1}\binom{2i}{i}$ and $A_i=\frac{C_i}{4^i}$ we have
$$ f(x) = \sum_{i\geq 0}A_i x^i = 2\frac{1-\sqrt{1-x}}{x}\tag{1} $$
hence
$$ \frac{f(x)}{1-x}=\sum_{i\geq 0}(A_0+\ldots+A_i)x^i = 2 \frac{1-\sqrt{1-x}}{x(1-x)} \tag{2} $$
and the wanted sum is the coefficient of $x^{n-1}$ in the RHS of $(2)$, or the coefficient of $x^n$ in
$$ \frac{2}{1-x}-\frac{2}{\sqrt{1-x}}\tag{3} $$
whose Taylor series is well-known and leads to
$$ \sum_{i=0}^{n-1}A_i = 2-\frac{2}{4^n}\binom{2n}{n}\tag{4} $$
as wanted.
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Computing $ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$ $$ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$$
My idea for this was to break each numerator into its own fraction as follows
$$ \int_0^1 \left(\frac{1}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{5x^3}{\sqrt{x}}\right)dx$$
$$ \int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2})\ dx $$
$$ \int_0^1 2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2} $$
Not really sure where to go from there. Should I sub 1 in for the x values and let that be the answer?
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We put $t=\sqrt{x}$, so $dt=\frac{1}{2\sqrt{x}}dx$.
$$\int_0^12(1+3t^2+5t^6)$$
$$=2\left[t+t^3+\frac{5t^7}{7}\right]^1_0$$
$$=2\left[\sqrt{x}+\sqrt{x}^3+\frac{5\sqrt{x}^7}{7}\right]^1_0$$
$$=\frac{38}{7}$$
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Different method of proving $\int_{0}^{1}{ dx\over x}\ln\left({1-\sqrt{x}\over 1+\sqrt{x}}\cdot{1+\sqrt[3]{x}\over 1-\sqrt[3]{x}}\cdots\right)=-\pi^2$ Given the integral $(1)$
$$\text{Prove that}\ \int_{0}^{1}{\mathrm dx\over x}\ln\left({1-\sqrt{x}\over 1+\sqrt{x}}\cdot{1+\sqrt[3]{x}\over 1-\sqrt[3]{x}}\cdot{1-\sqrt[5]{x}\over 1+\sqrt[5]{x}}\right)=-\pi^2\tag1$$
An attempt:
$u=x^2,x^3 \text{and}\ x^5$, then $(1)$ becomes
$$\int_{0}^{1}{\mathrm du\over u}\ln\left[\left({1-u\over 1+u}\right)^2\left({1+u\over 1-u}\right)^3\left({1-u\over 1+u}\right)^5\right]\tag2$$
Simplify to
$$\int_{0}^{1}{\mathrm du\over u}\ln\left({1-u\over 1+u}\right)^4\tag3$$
Apply $\ln\left({1+u\over 1-u}\right)$ series, then we have
$$-\sum_{n=0}^{\infty}{1\over (2n+1)}\int_{0}^{1}u^{2n}\mathrm du\tag4$$
$$-8\sum_{n=0}^{\infty}{1\over (2n+1)^2}=-\pi^2\tag5$$
Looking for another method of proving $(1)$
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Let us generalize to
$$\int^1_0 \frac{\mathrm d x}{x} \log \left(\frac{1-x^a}{1+x^a}\right)$$
Let $x^a = y $ which implies $\mathrm d x = \frac{1}{a}y^{\frac{1}{a}-1} \mathrm d y$
\begin{align}\int^1_0 \frac{\frac{1}{a}y^{\frac{1}{a}-1}\mathrm dy}{y^{\frac{1}{a}}} \log \left(\frac{1-y}{1+y}\right) &= \frac{1}{a}\int^1_0 \frac{\mathrm dy}{y} \log \left(\frac{1-y}{1+y}\right) \\&=\frac{1}{a}\int^1_0 \frac{\log(1-y)}{y} \mathrm dy- \frac{1}{a}\int^1_0 \frac{\log(1+y)}{y} \mathrm dy \\&= \frac{\mathrm{Li}_2(-1)-\mathrm{Li}_2(1)}{a} \\ &= -\frac{\pi^2}{4a}\end{align}
Now consider the general case
\begin{align}\int^1_0 \frac{\mathrm d x}{x} \log \left(\prod_{k=1}^{n}\frac{1-x^{a_k}}{1+x^{a_k}}\right) &=\sum_{k=1}^n\int^1_0 \frac{\mathrm d x}{x} \log \left(\frac{1-x^{a_k}}{1+x^{a_k}}\right) \\&=-\frac{\pi^2}{4}\sum_{k=1}^n\frac{1}{a_k} \end{align}
So we get the result for $a_k \geq 1$
$$\int^1_0 \frac{\mathrm d x}{x} \log
\left[\prod_{k=1}^{n}\frac{1-x^{1/a_k}}{1+x^{1/a_k}}\right]
=-\frac{\pi^2}{4}\sum_{k=1}^n a_k$$
Similarly
$$\int^1_0 \frac{\mathrm d x}{x} \log
\left[\prod_{k=1}^{n}(-1)^{k-1}\frac{1-x^{1/a_k}}{1+x^{1/a_k}}\right]
=\frac{\pi^2}{4}\sum_{k=1}^n a_k (-1)^k$$
In your case we have the sequence
$$a_k = 2, 3 , 5 $$
Hence
\begin{align}\int_{0}^{1}{\mathrm dx\over x}\ln\left({1-\sqrt{x}\over 1+\sqrt{x}}\cdot{1+\sqrt[3]{x}\over 1-\sqrt[3]{x}}\cdot{1-\sqrt[5]{x}\over 1+\sqrt[5]{x}}\right)&= \int_{0}^{1}{\mathrm dx\over x}\log \prod^3_{k=1}(-1)^{k-1}\left(\frac{1-x^{1/a_k}}{1+x^{1/a_k}}\right)\\&=\frac{\pi^2}{4}\sum_{k=1}^3 (-1)^k a_k \\&= \frac{\pi^2(-2+3-5)}{4} = -\pi^2\end{align}
Perhaps if we are able to interchange the limit and the integral and under some conditions on $a_k$
$$\int^1_0 \frac{\mathrm d x}{x} \log
\left[\prod_{k=1}^{\infty}\frac{1-x^{1/a_k}}{1+x^{1/a_k}}\right]
=-\frac{\pi^2}{4}\sum_{k=1}^{\infty} a_k$$
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If $a$ and $b$ are odd perfect squares, then $a + b$ is not a perfect square.
Prove: If $a$ and $b$ are odd perfect squares, then $a + b$ is not a perfect square.
Proof by Contradiction:
If $a$ and $b$ are odd perfect squares then $a = (2k+1)^2$ and $b = (2r + 1)^2$.
Assume $a + b$ is a perfect square.
\begin{align}a + b &= (2k+1)^2 + (2r + 1)^2\\
&= 4(k^2 + r^2) + 4(k + r) + 2\\
&= 4(k + r)^2 + 4(k + r) -2kr + 2\\
&= z^2 + 2z -2kr + 2 \hspace{0.5cm} \text{where } z = 2(k+r)\\
&= (z+1)^2 -2kr + 1\\
&= (z+1)^2 - q \hspace{0.5cm} \text{where } q = 2kr + 1\end{align}
Hence, $a+b$ is in inexpressible as a perfect square when $a, b$ are odd.
Is this a valid proof? I'm not sure how to approach it any other way.
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You could stop at $4(k^2 + r^2) + 4(k + r) + 2$ and note that this is divisible by $2$, but not $4$. Therefore it cannot be a square.
|
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The functional equation $\frac{f(x)}{f(1-x)} = \frac{1-x}{x}$ One set of solutions to this is $f(x) = \frac{c}{x}$ for constant $c$. Are these the only solutions?
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Let $f$ be a solution to the equation, and let $g(x):=(x+ \frac{1}{2})f(x + \frac{1}{2})$. Then $g$ is an even function (i.e.
$ g(x)=g(-x)$). Conversly, let $h$ be any even function. Then by setting $f(x):=\frac{h\left(x-\frac{1}{2}\right)}{x}$, we get $xf(x)=(1-x)f(1-x)$. Thus the set of all solutions to this equation is exactly the set of functions of the form:
$$f(x):=\frac{h\left(x-\frac{1}{2}\right)}{x}$$
Where $h$ is an even function.
Some examples:
\begin{align*}
f(x) &= \frac{c}{x} \\
\\
f(x) &= \frac{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}{x} = 1-x \\
\\
f(x) &=\frac{\cos\left(x-\frac{1}{2}\right)}{x} \\
\end{align*}
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|
Show that if $gcd(n,42) = 1$ then $n^6 \equiv 1 \pmod {42} $ Show that if $gcd(n,42) = 1$ then $n^6 \equiv 1 \pmod {42} $
I notice that $a \equiv 1 \pmod{2}$, $a^2\equiv 1 \pmod{3}$, $a^6 \equiv 1 \pmod{7}$
In the first case $a=-1,1,-3,3$ In order for the second condition to hold $a=-1,1$ which would imply then that $a^6 \equiv 1 \pmod{2}$ and $a^6 \equiv 1 \pmod{3}$, and $a^6\equiv 1 \pmod{7}$ so by the chineese remainder theorem you can conclude that $a^6 \equiv 1\pmod{42}$
|
Fermat's little theorem gives immediately $n^6\equiv 1\mod 7$
Since $n$ is odd, we have $n^6\equiv 1\mod 2$
Since $n$ is not divisble by $3$, we have $n^6\equiv 1\mod 3$ because of $n^2\equiv 1\mod 3$
The chinese remainder theorem gives $n^6\equiv 1\mod 42$
|
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|
Evaluating the indefinite integral $\int \frac{x^3}{\sqrt{4x^2 -1}}~dx$.
Find $$\int {x^3\over \sqrt{4x^2 -1}}\,dx.$$
Let
$2x = \sec u$, $2 =\sec (u) \tan(u) u^{'}(x).$ Then
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \, du\\
&= \frac1{16}\int {\sec^4 u} \, du\\
&= \frac1{16}\left(\tan u + {\tan^3 u\over 3} \right) + C\\
&= \frac1{16}\left(\tan (\sec^{-1} 2x) + {\tan^3 (\sec^{-1} 2x)\over 3} \right) + C.
\end{align*}$$
Given answer : $$\dfrac{\left(2x^2+1\right)\sqrt{4x^2-1}}{24}+C$$
Why is my answer incorrect ?
|
By the above diagram, we can simplify the answer obtained by the author.
$$
\begin{array}{l}
\begin{aligned}
I &=\frac{1}{16}\left(\tan \theta+\frac{\tan ^{3} \theta}{3}\right)+C \\
&=\frac{1}{16}\left[\sqrt{4 x^{2}-1}+\frac{\left(4 x^{2}-1\right)^{\frac{3}{2}}}{3}\right]+C
\\&=\frac{\sqrt{4 x^{2}-1}}{48}\left[3+4 x^{2}-1\right]+C \\
\\&=\frac{\left(2 x^{2}+1\right) \sqrt{4 x^{2}-1}}{24}+C
\end{aligned} \\
\end{array}
$$
|
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|
Winning strategy of Peter such that quadratic equations has Rational roots Peter picked three non zero real numbers $a,b,c$ and Alan arranges these numbers as coefficients of Quadratic equation $ax^2+bx+c=0$.
Peter wins if this Quadratic has distinct Rational roots, else Alan wins.
How can we prove that Peter always has winning strategy.
Since $a,b,c$ are non zero reals we have roots of quadratic as
$$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
Now Peter can win game only if $$b^2-4ac=k^2$$ where $k$ is Integer. But even though $b^2-4ac=k^2$ is integer, how can we say roots are rational?
|
It's enough to find one set of real numbers that Peter can choose to win. For example, $\{1, 2, -3\}$ will work. We can check that:
\begin{align}
x^2 + 2x - 3 &= (x+3)(x-1) \\
x^2 - 3x + 2 &= (x-2)(x-1) \\
2x^2 + x - 3 &= (2x+3)(x-1) \\
2x^2 - 3x + 1 &= (2x-1)(x-1) \\
-3x^2 + x + 2 &= -(3x+2)(x-1) \\
-3x^2 + 2x + 1 &= -(3x+1)(x-1)
\end{align}
and so both roots of all six polynomials Alan can form are rational and distinct.
I found this by brute force, but in retrospect it seems obvious why this worked and how we can find lots more such triples. Since $1+2+(-3)=0$, $x=1$ will always be a root of the quadratic equation, and factoring out $x-1$ must produce a second rational root.
|
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|
Computing a sequence in the Cantor set Let $\eta:2^\omega\to 2^\omega$ be such that:
If $x \neq$ all 1's and $n$ is least such that $x_n=0$, then $\eta(x)=y$ where $$y_i=\begin{cases}x_i &\text{for } i>n\\ 1-x_i&\text{for }i\leq n.\end{cases}$$
I am interested in starting with the particular $x$ defined below. I believe this point corresponds to the number $1/4$ in the middle-thirds Cantor set $C\subseteq [0,1]$. Because $\sum_{k=1}^\infty \frac{2}{9^k}=1/4$.
\begin{align}x=&0101010101...\\
\eta(x)=&1101010101...\\
\eta^2(x)=&0011010101...\\
\eta^3(x)=&1011010101...\\
\eta^4(x)=&0111010101...
\end{align}
QUESTION: Can you give a formula (closed form or recursive) for $\eta^n(x)$?
It's obvious that all the numbers will correspond to fractions in $C$, which might make for a nice equation(?). It's also true that you will never get the same number twice.
Here are the first few numbers:
$\frac{1}{4}$
$\frac{11}{12}$
$\frac{11}{108}$
$\frac{83}{108}$
$\frac{35}{108}$
$\frac{107}{108}$
$\frac{11}{972}$
...
maybe a pattern?
|
Applying $\eta$ to $y = 0000\dots$ instead would correspond to counting in binary, but backwards:
\begin{align}
y &= 0000\dots \\
\eta(y) &= 1000\dots \\
\eta^2(y) &= 0100\dots \\
\eta^3(y) &= 1100\dots \\
\eta^4(y) &= 0010\dots
\end{align}
This is closely related to sequence A030109 in the OEIS, and won't have a nice closed formula; you could write down the recurrence
\begin{align}
\eta^{2n}(y) &= \frac13 \eta^n(y), \\
\eta^{2n+1}(y) &= \frac13 \eta^n(y) + \frac23.
\end{align}
When we start with $x$ instead of $y$, there's a trailing infinite sequence of $\dots1010101\dots$ that periodically throws us off. So the sequence $\eta^n(x)$ can be naturally broken up into block of length $4^k$ for $k=0, 1, 2, \dots$: in each of these blocks, we are copying $\eta(y)$ on the first $2k$ bits, followed by $110101\dots$.
More precisely, when $n = 1 + 4 + 4^2 + \cdots + 4^{k-1} + j$, for $1 \le j \le 4^k$, we have
\begin{align}
\eta^n(x) &= \eta^{j-1}(y) + \frac{2}{3^{2k+1}} + \sum_{i=k+1}^\infty \frac{2}{3^{2i}} \\
&= \eta^{j-1}(y) + \frac{11}{4 \cdot 3^{2k+1}}.
\end{align}
For example:
*
*If $n=5 = 1 + 4$, then $k=1$ and $j=4$, so we have $\eta^{n}(x) = \eta^{3}(y) + \frac{11}{108} = \frac{8}{9} + \frac{11}{108} = \frac{107}{108}$.
*If $n=6 = 1 + 4 + 1$, then $k=2$ and $j=1$, so we have $\eta^{n}(x) = \eta^{0}(y) + \frac{11}{972} = \frac{11}{972}$.
|
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|
Rationalise the Denominator Rationalise the denominator and simplify fully:
$$\dfrac{6}{\sqrt{7} + 2}$$
I got the answer $\dfrac{2 \sqrt{7}}{3}$, but didn't get the mark. Is that not fully simplified?
I did $\displaystyle \frac{6}{\sqrt{7} + 2} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{6 \sqrt{7}}{(7 + 2)} = \frac{6 \sqrt{7}}{9} = \frac{2 \sqrt{7}}{ 3}$ .
|
$$\frac{6}{\sqrt{7}+2}=\frac{6(2-\sqrt{7})}{(2+\sqrt{7})(2-\sqrt{7})} = \frac{12-6\sqrt{7}}{4-7} = \frac{3(4-2\sqrt{7})}{-3} = 2\sqrt{7}-4$$
Note that in general $$\frac{a}{b+c\sqrt{d}} = \frac{a(b-c\sqrt{d})}{(b+c\sqrt{d})(b-c\sqrt{d})} = \frac{ab-ac\sqrt{d}}{b^2-c^2d}$$
Notice that the denominator is now rational, i.e. in particular for this case it has no surds in it.
|
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|
If $f(x+1)+f(x-1)=4x^2-2x+10$ then what is $f(x)$
If $$f(x+1) +f(x-1)= 4x^2 -2x +10$$ then what is $f(x)$
What is strategy of solving this kind of problems ?
Thank you for help
|
First, you can see that
$$f(x)=-f(x-2) + 4(x-1)^2-2(x-1)+10$$
$$f(x-2) = -f(x-4) + 4(x-3)^2-2(x-3)+10$$
$$f(x-4) = -f(x-6) + 4(x-5)^2 - 2(x-5)+10$$
$$\dots$$
You can see that you need a basis or some additional constraint to express $f(x)$, because now a solution is not unique. Once you find a solution $f_0(x)$, the general form of the answer would be $f(x) = f_0(x) + g(x)$, where $g(x)$ is a function satisfying $$g(x) = -g(x-2)$$
For example for $g(x) = \sin\left(\dfrac{\pi x}{2}\right)$ and $f_0(x) = 2x^2-x+3$ we can obtain a new solution
$$f_1(x) = 2x^2-x+3 + \sin\left(\dfrac{\pi x}{2}\right)$$
|
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Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = y$
Let $x,y$ be integers. Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = \pm y$.
The given condition is equivalent to $x^2+y^2 \equiv 0 \pmod{xy}$. How do we continue from here to prove that $x = \pm y$?
|
$$x^2 + y^2 = kxy$$
Let $d =gcd(x,y)$. Then $x=da, y=db$ for some relatively prime $a,b$.
The equation becomes
$$a^2+b^2=kab $$
with gcd$(a,b)=1$. This implies $a=\pm 1$ and $b=\pm1$. Indeed, if $a$ or $b$ are not $\pm1$, then $a$ or $b$ is divisible by a prime $p$. Then the prime $p$ also divides $kab$ and hence divides both $a$ and $b$.
|
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Find a,b and c. Any method? Given that $$\frac{1}{a+\frac{1}{b+\frac{1}{c+1}}}=\frac{16}{38}$$, find $a,b,c$.
I've been figuring for this quite some time? Is it possible to solve?
|
Here is a solution ; sorry for hand writing
$$\begin{align*}
\frac{1}{a+\frac{1}{b+\frac{1}{c+1}}}&=\frac{16}{38}\\
a+\frac{1}{b+\frac{1}{c+1}}&=\frac{38}{16}= 2+\frac{3}{8}\\
\text{So}\quad a=2\\
\frac{1}{b+\frac{1}{c+1}}&=\frac{3}{8}\\
b+\frac{1}{c+1}&=\frac{8}{3}=2+\frac{2}{3}\\
\text{So}\quad b=2\\
\frac{1}{c+1}&=\frac{2}{3}\\
c+1=\frac{3}{2}=1+\frac{1}{2}\\
\text{So}\quad c=\frac{1}{2}\\
\end{align*}$$
|
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Limit of a sum 1: $\lim _{n\to \infty }\sum _{k=1}^n\sqrt{n^4+k} \cdot \sin \frac{2k\pi }{n}$ $$\lim _{n\to \infty }\sum _{k=1}^n\sqrt{n^4+k} \cdot \sin \frac{2k\pi }{n}$$
It looks like a Riemann sum, but I don't know how to approach it. Any hint would be appreciated. (Without Taylor expansion)
EDIT: The answer is $- \frac{1}{4 \pi}$
|
I don't have time to finish right now; I'll leave this as Community Answer, so that anyone can feel free to edit it; if no one does, I'll get back to it later.
The inhomogeneity between $n^4$ and $k$ will a priori prevent you from seeing this as a Riemann sum: since there is no way to massage $\sqrt{n^4+k}$ in order to make a factor $\sqrt{1+\frac{k}{n}}$ appear.
However, what you can do is "guess and then use inequalities and the squeeze theorem to prove the guess was right." I know you do not want to use Taylor expansions, so I will not (but it's hidden in the "guess" part).
First, rewrite
$$
\sum_{k=1}^n \sqrt{n^4+k}\sin\frac{2\pi k}{n}
= n^3 \frac{1}{n}\sum_{k=1}^n \sqrt{1+\frac{k}{n^4}}\sin\frac{2\pi k}{n}
$$
*
*Guess part: Now, we are going to use the fact that
$$\lim_{x\to0} \frac{\sqrt{1+x}-1}{x} = \lim_{x\to0} \frac{1}{\sqrt{1+x}+1} = \frac{1}{2}\tag{i}$$, so we expect $\sqrt{1+\frac{k}{n^4}}-1$ to "behave like $\frac{k}{2n^4}$" and (ii) $$\frac{1}{n}\sum_{k=1}^n\sin\frac{2\pi k}{n} \xrightarrow[n\to\infty]{} \int_0^1 \sin(2\pi x)\, dx = 0 \tag{ii}$$
(Riemann sum).
Combining the two, it is "reasonable" to expect that
$$\begin{align}
n^3 \frac{1}{n}\sum_{k=1}^n \sqrt{1+\frac{k}{n^4}}\sin\frac{2\pi k}{n}
&=
n^3 \frac{1}{n}\sum_{k=1}^n \left(\sqrt{1+\frac{k}{n^4}}-1\right)\sin\frac{2\pi k}{n} + n^3\cdot\frac{1}{n}\sum_{k=1}^n\sin\frac{2\pi k}{n} \\
&\approx
n^3 \frac{1}{n}\sum_{k=1}^n \frac{k}{2n^4}\sin\frac{2\pi k}{n} + n^3\cdot\frac{1}{n}\sum_{k=1}^n\sin\frac{2\pi k}{n} \\
&= \frac{1}{2}\cdot\frac{1}{n}\sum_{k=1}^n \frac{k}{n}\sin\frac{2\pi k}{n} + n^3\cdot\frac{1}{n}\sum_{k=1}^n\sin\frac{2\pi k}{n} \\
&\xrightarrow[n\to\infty]{} \frac{1}{2}\int_0^1 x\sin(2\pi x)\, dx+\frac{1}{2}\int_0^1 \sin(2\pi x)\, dx = \frac{1}{2}\int_0^1 x\sin(2\pi x)\, dx \\&= \boxed{-\frac{1}{4\pi}}
\end{align}$$
This is our (very handwavy) guess; let us prove it.
*Proving the guess:
|
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Rational points on a circle with a center having non rational coordinates Reflection of a rational point with respect to line having its equation with rational coefficient
gives a rational point
What is the number of rational points on a circle with centre $(\sqrt 2, \sqrt 3)$ and with any radius
In this the centre is not rational then also how they decided the answer as 'at most 1'
Original source : https://i.stack.imgur.com/3aMvD.gif
|
Let $(a,b)$ and $(c,d)$ be distinct rational points on the circle.
$$\begin{array}{rcl}
(a-\sqrt2)^2 + (b-\sqrt3)^2 &=& (c-\sqrt2)^2 + (d-\sqrt3)^2 \\
a^2 + b^2 - 2a\sqrt2 - 2b\sqrt3 &=& c^2 + d^2 - 2c\sqrt2 - 2d\sqrt3\\
a^2 + b^2 - c^2 - d^2 &=& 2(a-c)\sqrt2 + 2(b-d)\sqrt3\\
(a^2 + b^2 - c^2 - d^2)^2 &=& 8(a-c)^2 + 12(b-d)^2 + 8(a-c)(b-d)\sqrt6 \\
\sqrt6 &=& \dfrac{(a^2 + b^2 - c^2 - d^2)^2 - 8(a-c)^2 - 12(b-d)^2}{8(a-c)(b-d)}\\
\end{array}$$
Contradicting the fact that $\sqrt6$ is irrational, unless $a=c$ or $b=d$.
I believe you can deal with the cases:
*
*Case 1: $a=c$ and $b \ne d$
*Case 2: $a \ne c$ and $b=d$
To complete the proof.
|
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Summation calculus: $\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}}$ How can I solve this?
$$\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}}$$
Actually I tried many direction, but failed.
Please give me some right direction.
$$\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}} = \sum_{k=1}^n \frac{2^{2^{k-1}}}{(1-2^{2^{k-1}})(1+2^{2^{k-1}})}=\cdots$$
|
Let $$S = \sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}} = \sum^{n}_{k=1}\bigg[\frac{(1+2^{2^{k-1}})-1}{1-2^{2^k}}\bigg] = \sum^{n}_{k=1}\bigg[\frac{1}{1-2^{2^{k-1}}}-\frac{1}{1-2^{2^k}}\bigg]$$
which is nothing but Telescopic Sum
So $$S = -1-\frac{1}{1-2^{2^{n}}}$$
|
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|
Does the series $\sum_{n=1}^{\infty} \frac{(-1)^n - (-1)^{n+1}}{n+1}$ converges absolutely or conditionally?
Does the following series converges absolutely or conditionally? $$\sum_{n=1}^{\infty} \frac{(-1)^n - (-1)^{n+1}}{n+1}$$
$$\sum_{n=1}^{\infty} \frac{(-1)^n - (-1)^{n+1}}{n+1}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n+1} - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$$
Both $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n+1}$$ and $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$$ are convergent by alternating test, hence $\sum_{n=1}^{\infty} \frac{(-1)^n}{n+1} - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$ is convergent.
$\sum_{n=1}^{\infty} \frac{(-1)^n - (-1)^{n+1}}{n+1}$ is convergent.
$$\sum_{n=1}^{\infty}| \frac{(-1)^n - (-1)^{n+1}}{n+1} = \sum_{n=1}^{\infty} \frac{2}{n+1}$$ which is divergent by comparison test with $$\sum_{n=1}^{\infty} \frac{1}{n}$$
Hence conditionally convergent.
Is this correct ?
|
Hint:
$$(-1)^n-(-1)^{n+1}=(-1)^n+(-1)^{n+2}=(-1)^{n}\left[1+1^2\right]=2\cdot(-1)^n\ldots$$
|
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|
ABC is Isosceles right triangle with $AB=BC$ $P$ , $Q$ are points on $AC$ such that $AP ^2 + CQ^2 = PQ^2$ What is the value of angle $PBQ (x) $?
ABC is Isosceles right triangle with $AB=BC$
$P$ , $Q$ are points on $AC$ such that
$$AP ^2 + CQ^2 = PQ^2$$
What is the value of angle $PBQ (x) $?
Thank you for help
|
Let $AC=1, AP=x, CQ=y.$ Then, $x^2+y^2 = (1-x-y)^2\Rightarrow 2x+2y-1=2xy$ or $y = \dfrac{3}{2x+2}-1=\dfrac{1-2x}{2-2x}.$ By theorem of Cosine, $BP^2 = \dfrac{1}{2}+x^2 - 2x\dfrac{1}{\sqrt{2}}\cos\dfrac{\pi}{4} = x^2-x+\dfrac{1}{2}$ and similarly $BQ^2 = y^2-y+\dfrac{1}{2}.$ Therefore, $$\cos\angle PQB = \dfrac{BP^2+BQ^2-PQ^2}{2BP\cdot BQ} = \dfrac{1-x-y}{2BP\cdot BQ} = \dfrac{2x^2-2x+1}{2-2x}\cdot \dfrac{1}{2\sqrt{(x^2-x+\frac{1}{2})\frac{2x^2-2x+1}{4(x-1)^2}}}=\dfrac{1}{\sqrt{2}},$$
So $\angle PBQ = \dfrac{\pi}{4}.$
|
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How to solve the partial fraction decomposition $\frac{x^3+5x^2+3x+6}{2x^2+3x}$. I have the following integral:
$$\int\frac{x^3+5x^2+3x+6}{2x^2+3x}dx$$
I'm trying to use partial fraction decomposition but I'm getting stuck at the following formula:
$$\int\frac{(x+6)(1+5x+x^2)}{x(2x+3)}-\frac{x+27}{2x+3}dx$$
I can't necessarily guarantee that this problem has a nice solution. But is there a way to circumvent the fact that $(1+5x+x^2)$ has a somewhat messy root? (I'm getting $\frac{5+-\sqrt{21}}{2}$). Or do I have to use that value?
|
We first deal with the improper fraction by division and then resolve it into partial fractions, $$
\begin{array}{l} \displaystyle
\frac{x^{3}+5 x^{2}+3 x+6}{x(2 x+3)}=\frac{1}{2} x+\frac{7}{4}+\frac{-
\frac{9}{4} x+6}{2 x^{2}+3 x} \\
\displaystyle \frac{-\frac{9}{4} x+6}{x(2 x+3)} \equiv \frac{A}{x}+\frac{B}{2 x+3} \\
\displaystyle -\frac{9}{4} x+6 \equiv A(2 x+3)+B x
\end{array}
$$
Putting $x=0$ yields $$6=3 A \Rightarrow A=2$$
Putting $ \displaystyle x=-\frac{3}{2}$ yields $$ \displaystyle \quad \frac{27}{8}+6=-\frac{3}{2} B \Rightarrow B=-\frac{25}{4}$$
Now we can conclude that
$$
\begin{aligned}
I &=\int\left[\frac{1}{2} x+\frac{7}{4}+\frac{1}{2 x}-\frac{25}{4(2 x+3)}\right] d x \\
&=\frac{x^{2}}{4}+\frac{7}{4} x+\frac{1}{2} \ln |x|-\frac{25}{8} \ln |2 x+3|+C
\end{aligned}
$$
|
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|
partial fractions for a function I need help finding the partial fraction decomposition for this function, I am just lost on it, here it is:
$(x^2 + x + 1)/(2x^4+3x^2+1)$. the help is appreciated. thank you.
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The key to it is how to factor the denominator $2x^4+3x^2+1$. Denote $y=x^2$ and you're left with a quadratic $2y^2+3y+1$ whose roots are $-1$ and $-1/2$. So
$$2x^4+3x^2+1=(y+1)(2y+1)=(x^2+1)(2x^2+1)$$
Now we're looking for
$${x^2+x+1\over 2x^4+3x^2+1}={ax+b\over x^2+1}+{cx+d\over 2x^2+1}$$
Multiply both sides by $x^2+1$ and substitute $x=i$ to get
$${i\over -1}=ai+b$$
and so $a=-1$ and $b=0$
Multiply both sides by $2x^2+1$ and substitute $x=i/\sqrt{2}$ to get
$${-{1\over 2}+{i\over\sqrt{2}}+1\over -{1\over 2}+1}=c{i\over\sqrt{2}}+d\iff {2i\over\sqrt{2}}+1=c{i\over\sqrt{2}}+d$$
and this yields $c=2$ and $d=1$ and so
$${x^2+x+1\over 2x^4+3x^2+1}={2x+1\over 2x^2+1}-{x\over x^2+1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2237858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15. Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15.
My instinct was to find the primitive root and then use a theorem to directly show the number of incongruent solutions, which follows from knowing the primitive root. But, apparently there are no primitive solutions to $mod 15.$ So what's another option?
|
To make things more interesting, and slightly less susceptible to exhaustive search, we can show that $x^2\equiv 1\bmod 91$ has four incongruent solutions and find them.
The analogy is that both $15=3\cdot 5$ and $91=7\cdot 13$ are composite numbers, the product of two odd primes.
So in my example, $x^2\equiv 1\bmod 91$ $\implies$ $x^2\equiv 1\bmod 7$ and $x^2\equiv 1\bmod 13$. Thus
$x\equiv \pm1\equiv \{1,6\}\bmod 7 \\
x\equiv \pm1\equiv \{1,12\}\bmod 13 $
The Chinese Remainder theorem says that each combination of values from these two equations will have distinct results $\bmod 91$, so that proves the existence of those four roots. Then we can also find them:
$
(1 \bmod 7, 1 \bmod{13}) \to 1\bmod {91}\\
(1 \bmod 7, 12 \bmod{13}) \to 64\bmod {91}\\
(6 \bmod 7, 1 \bmod{13}) \to 27\bmod {91}\\
(6 \bmod 7, 12 \bmod{13}) \to 90\bmod {91}
$
giving $x^2\equiv 1 \implies x \equiv \{1,27,64,90\} \bmod 91$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2239351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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|
prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$ Here is my question: Prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$.
They gave me a hint that I should consider the triple $3x+2z+1, 3x+2z+2, 4x+3z+2$, but I honestly do not know how to start.
Do I need contradiction?
|
Given the odd number $x$,
$$x^2=2y^2-1$$
which is just a Pell equation, then we get the Pythagorean triple,
$$\Big(\frac{x-1}2\Big)^2+\Big(\frac{x+1}2\Big)^2=y^2$$
The difference $d$ between the addends, of course, is $d=1$. This yields,
$$3^2+4^2=5^2\\20^2+21^2=29^2\\119^2+120^2=169^2$$
and so on.
P.S. If interested, a nice infinite counterpart is $a^2+b^2=(b+1)^2$ with solution,
$$(2m+1)^2+(2m^2+2m)^2 = (2m^2+2m+1)^2$$
which yields,
$$3^2+4^2=5^2\\5^2+12^2=13^2\\7^2+24^2=25^2$$
ad infinitum.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2241186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
}
|
Factor $ab^3-ac^3+bc^3-ba^3+ca^3-cb^3$
a) Use the remainder theorem to prove that $(a+b+c)$ is a factor of $(a^3+b^3+c^3-3abc)$ . Then find the other factor.
b) Hence factor $(ab^3-ac^3+bc^3-ba^3+ca^3-cb^3)$
So far I have managed to find the other remainder being $(a^2+b^2+c^2-ab-ac-bc)$ but I don't understand how to use this to hence factor the expression.
Also I was reading a solution which factorised the expression without using part a) but how does the second last line turn into the last line?
\begin{align}&\color{white}=(bc^3−cb^3)−(ac^3−ca^3)+(ab^3−ba^3)\\
&=bc(c^2−b^2)−ac(c^2−a^2)+ab(b^2−a^2)\\
&=(c−b)(c+b)bc−ac(c−a)(c+a)+ab(b−a)(b+a)\\
&=−(a−c)(b−c)(a−b)(a+b+c)\end{align}
Thanks in advance
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The polynomial
$$f(a,b,c)=ab^3-ac^3+bc^3-ba^3+ca^3-cb^3$$
is an alternating function of $a$, $b$, $c$. That is it changes sign
whenever you swap two variables:
$$f(a,b,c)=-f(b,a,c)=-f(c,b,a)=-f(a,c,b).$$
This means that $a-b$, $a-c$ and $b-c$ are all factors (if a polynomial
changes sign when you swap $a$ and $b$ then $a-b$ is a factor). Therefore
$$f(a,b,c)=(a-b)(a-c)(b-c)g(a,b,c)$$
for some polynomial $g(a,b,c)$. As $f$ is alternating, $g$ is symmetric:
$$g(a,b,c)=g(b,a,c)=g(c,b,a)=g(a,c,b).$$
It's clear that $g$ is homogeneous of $a$, $b$ and $c$ of degree one.
So $g(a,b,c)=k(a+b+c)$ for some constant $k$. One only needs to
compute one coefficient to see that $k=-1$. Therefore
$$f(a,b,c)=-(a-b)(a-c)(b-c)(a+b+c).$$
As a follow-up you might try to factorise
$$ab^4-ac^4+bc^4-ba^4+ca^4-cb^4$$
or
$$a^2b^3-a^2c^3+b^2c^3-b^2a^3+c^2a^3-c^2b^3.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2241289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
What is the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt 5)$? What is the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt 5)$ ?
I proved that $\mathbb Q(\sqrt 3+\sqrt 5)=\mathbb Q(\sqrt 3,\sqrt 5)\supset \mathbb Q(\sqrt 5)$, and thus that $\mathbb Q(\sqrt 3+\sqrt5 )/\mathbb Q(\sqrt 5)$ has degree $2$. Since the minimal polynomial of $\sqrt 3+\sqrt 5$ is $X^4-16X^2+4$, the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt 5)$ must be a polynomial of order 2 that divide $X^4-16X^2+4$. It's roots are $\pm\sqrt{8\pm 2\sqrt{15}}$. But I tried to make all combinaisons $(X\pm\sqrt{8\pm2\sqrt{15}})(X\mp\sqrt{8\mp2\sqrt{15}})$, but I didn't get any polynomial of order $2$ in $\mathbb Q(\sqrt 5)$ that canceled $\sqrt 3+\sqrt 5$. Where is my mistakes ?
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Its conjugate is $\sqrt5-\sqrt3$.
Therefore, the minimal polynomial is:
$$\begin{array}{rcl}
(x-(\sqrt3+\sqrt5))(x-(\sqrt5-\sqrt3))
&=& (x-\sqrt5)^2 - (\sqrt3)^2 \\
&=& x^2-2\sqrt5x+5-3 \\
&=& x^2-2\sqrt5x+2
\end{array}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2242092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
Intermediate Value theorem inequality problem Let $h:[0,1]\rightarrow \mathbb{R}$ be continuous. Prove that there exists $w\in[0,1]$ such that $$h(w)=\frac{w+1}{2}h(0)+\frac{2w+2}{9}h\left(\frac{1}{2}\right)+\frac{w+1}{12}h(1).$$
I tried several things with this problem. I first tried a new function
\begin{align*}
g(x)&=h(x)-\frac{x+1}{2}h(0)+\frac{2x+2}{9}h\left(\frac{1}{2}\right)+\frac{x+1}{12}h(1)\\
&=h(x)-(x+1)\left(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)\right).
\end{align*}
Evaluating $g(0)$ and $g(1)$ didn't really work.
Since I want to find a point $x_0$ where $g(x_0)>0$ and another $x_1$ where $g(x_1)<0$, I started thinking how I can make $g(x_0)>0$,or $$h(x_0)-(x_0+1)\left(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)\right)>0,$$
or$$\frac{h(x_0)}{x_0+1}>\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1).$$
Since the LHS is the gradient from the point $(-1,0)$ to $(x_0,h(x_0))$, I need to somehow prove that that $$\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)$$ is always less than the steepest gradient from $(-1,0)$ to $(x_0,h(x_0))$. However, I am stuck here. Any help?
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Let $g:[0,1]\rightarrow \mathbb{R}$ be defined by
$$
g(w)=h(w) - \left(\frac{w+1}{2}h(0)+\frac{2w+2}{9}h\left({\small{\frac{1}{2}}}\right)+\frac{w+1}{12}h(1)\right)
$$
It suffices to show that $g(0),g(1/2),g(1)$ cannot all be positive, and cannot all be negative.$\\[9pt]$
\begin{align*}
\text{Let}\;\, a &= h(0),\;b=h(1/2),\;c=h(1)\\[9pt]
\text{Then}\;\,g(0) &= h(0) - \left(\frac{1}{2}h(0)+\frac{2}{9}h\left({\small{\frac{1}{2}}}\right)+\frac{1}{12}h(1)\right)\\[0pt]
&=a - \left(\frac{1}{2}a+\frac{2}{9}b+\frac{1}{12}c\right)\\[0pt]
&=\frac{1}{2}a-\frac{2}{9}b-\frac{1}{12}c\\[9pt]
g\left({\small{\frac{1}{2}}}\right) &= h\left({\small{\frac{1}{2}}}\right) - \left(\frac{3}{4}h(0)+\frac{1}{3}h\left({\small{\frac{1}{2}}}\right)+\frac{1}{8}h(1)\right)\\[0pt]
&=b - \left(\frac{3}{4}a+\frac{1}{3}b+\frac{1}{8}c\right)\\[0pt]
&=-\frac{3}{4}a+\frac{2}{3}b-\frac{1}{8}c\\[9pt]
g(1) &= h(1) - \left(h(0)+\frac{4}{9}h\left({\small{\frac{1}{2}}}\right)+\frac{1}{6}h(1)\right)\\[0pt]
&=c - \left(\frac{1}{2}a+\frac{2}{9}b+\frac{1}{12}c\right)\\[0pt]
&=-a-\frac{4}{9}b+\frac{5}{6}c\\[9pt]
\text{Let}\;\,{\mathbf u} &= \langle g(0),g(1/2),g(1) \rangle\\[4pt]
{\mathbf v} &= \langle 18,8,3 \rangle\\[9pt]
\text{Then}\;\,
{\mathbf u}
\cdot
{\mathbf v}
&=
\langle g(0),g(1/2),g(1) \rangle
\cdot
\langle 18,8,3 \rangle\\[6pt]
&=
(18)g(0)
+
(8)g(1/2)
+
(3)g(1)
\\[4pt]
&=
(18)\left(\frac{1}{2}a-\frac{2}{9}b-\frac{1}{12}c\right)
+
(8)\left(-\frac{3}{4}a+\frac{2}{3}b-\frac{1}{8}c\right)
+
(3)\left(-a-\frac{4}{9}b+\frac{5}{6}c\right)
\\[4pt]
&=0\qquad\text{[identically]}
\end{align*}
It follows that $g(0),g(1/2),g(1)$ cannot all be positive, and cannot all be negative, as was to be shown.
|
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"url": "https://math.stackexchange.com/questions/2245549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Hard limit to solve involving exponential and square root How to find this limit:
$$\lim_{x \to 0}\frac{a(1-e^{-x})+b(e^x-1)}{\sqrt{a(e^{-x}+x-1)+b(e^x-x-1)}}.$$
Where $a$ and $b$ are integer constants. I've tried substitution, L'Hôpital's rule, rationalizing the denominator and all the techniques I knew about limits, but I can't find the answer.
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(assuming limit from the right). Use Taylor Series: $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\implies e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\ldots$$
so that $$\lim_{x \to 0}\frac{a(1-e^{-x})+b(e^x-1)}{\sqrt{a(e^{-x}+x-1)+b(e^x-x-1)}}$$ $$=\lim_{x \to 0}\frac{a(x-\frac{x^2}{2}+\ldots)+b(x+\frac{x^2}{2}+\ldots)}{\sqrt{a(\frac{x^2}{2}-\frac{x^3}{3!}+\ldots)+b(\frac{x^2}{2}+\frac{x^3}{3!})}}\cdot\dfrac{\frac{1}{x}}{\frac{1}{x}}$$ $$= \lim_{x \to 0}\frac{a(1-\frac{x}{2}+\ldots)+b(1+\frac{x}{2}+\ldots)}{\sqrt{a(\frac{1}{2}-\frac{x}{3!}+\ldots)+b(\frac{1}{2}+\frac{x}{3!})}}=\dfrac{a+b}{\sqrt{\frac{1}{2}(a+b)}}=\boxed{\sqrt{2(a+b)}}$$
If it's the limit from the right, it's just the negative of this. Note that we used that $\frac{1}{x}=\frac{1}{\sqrt{x^2}}$ (principle).
Alternatively, we could have used L'Hopital just fine: $$I=\lim_{x \to 0}\frac{a(1-e^{-x})+b(e^x-1)}{\sqrt{a(e^{-x}+x-1)+b(e^x-x-1)}}$$
$$= \lim_{x \to 0}\frac{ae^{-x}+be^x}{\frac{1}{2}\left(a(e^{-x}+x-1)+b(e^x-x-1)\right)^{-\frac{1}{2}}\cdot \left(a(-e^{-x}+1)+b(e^x-1)\right)}=2(a+b)\lim_{n\to0}\frac{\sqrt{a(e^{-x}+x-1)+b(e^x-x-1)}}{a(1-e^{-x})+b(e^x-1)}=2(a+b)\frac{1}{I}\implies I^2=2(a+b)\implies \boxed{I=\sqrt{2(a+b)}}$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/2247107",
"timestamp": "2023-03-29T00:00:00",
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|
Help with Induction to show an Equality of Linear Operators I am trying to solve the following:
Let $X$ be a finite dimensional complex inner product space. Let $A,B \in \mathcal{L}(X)$, the space of linear mappings from $X$ to $X$. Use induction to show for any $n\geq1$, $$A^n-B^n=\sum_{m=0}^{n-1}A^m(A-B)B^{n-m-1}$$
For the base cases for $n=1,2$, this is straightforward to show, since for $n=1$ we have $A^0(A-B)B^0=A-B$, and for $n=2$ we have $A^0(A-B)B+A(A-B)B^0=AB-B^2+A^2-AB=A^2-B^2$.
Now we case assume the induction hypothesis: Suppose for $n=k$, it is true that $$A^k-B^k=\sum_{m=0}^{k-1}A^m(A-B)B^{k-m-1}$$. I attempted the following: Using the hypothesis, we can write $$A^k=\sum_{m=0}^{k-1}A^m(A-B)B^{k-m-1}+B^k$$ Thus $$A^{k+1}=\sum_{m=0}^{k-1}A^{m+1}(A-B)B^{k-m-1}+AB^k$$ Similarly for $B$, we have $$B^k=A^k-\sum_{m=0}^{k-1}A^m(A-B)B^{k-m-1}$$ Thus $$B^{k+1}=A^kB-\sum_{m=0}^{k-1}A^m(A-B)B^{k-m}$$ Combining, we have $$A^{k+1}-B^{k+1}=\sum_{m=0}^{k-1}A^{m+1}(A-B)B^{k-m-1}+AB^k-A^kB+\sum_{m=0}^{k-1}A^m(A-B)B^{k-m}$$ But I am unsure how to proceed. Any help is appreciated.
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We know that $A^n-B^n=\sum_{m=0}^{n-1}A^m(A-B)B^{n-m-1}$ holds for $n = 1$. Assume it holds for $n = k$, then for $n = k+1$, we have
\begin{align}
& A^{(k+1)}-B^{(k+1)} = A^{(k+1)}-B^{(k+1)}+A^kB-A^kB \\
=\ & A^kB-B^{(k+1)} + A^{(k+1)}-A^kB = (A^k-B^k)B + A^k(A-B) \\
=\ & \left(\sum_{m=0}^{k-1}A^m(A-B)B^{k-m-1}\right)B+A^k(A-B) \\
=\ & \sum_{m=0}^{k-1}A^m(A-B)B^{(k+1)-m-1}+A^k(A-B)B^{(k+1)-k-1}\\
=\ & \sum_{m=0}^{(k+1)-1}A^m(A-B)B^{(k+1)-m-1}
\end{align}
So $A^n-B^n=\sum_{m=0}^{n-1}A^m(A-B)B^{n-m-1}$ holds for $n = k+1$, thus it holds for all positive integer $n$.
|
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"url": "https://math.stackexchange.com/questions/2248523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to show that $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}}$ is like $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n}$ if $n\rightarrow\infty$? How to show that $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}}$ is like $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n}$ if $n\rightarrow\infty$?
I tried show that difference between terms of sums $\rightarrow 0$
$|\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}} - \frac{2^{\frac{k}{n}}}{n}| = 2^\frac{k}{n}(\frac{1}{n+\frac{1}{k}} - \frac{1}{n})= 2^\frac{k}{n}(\frac{n-n-\frac{1}{k}}{n^2+\frac{n}{k}}) = -2^\frac{k}{n} \frac{1}{n^2k+n}\rightarrow 0$ if $n\rightarrow \infty$
But i'm not sure that is right method, because difference of terms can accumulete and it would be $\neq 0$
Which correct method i must use to prove it?
|
Let $f(x)$
be a function such that,
for $x \ge 0$,
$f(x) \ge 0$
and
$f(x)$ is bounded for
$0 < x \le 1$.
Let
$M = \max_{0 < x \le 1} f(x)
$.
This problem is the case
$f(x) = 2^x$,
so $M = 2$.
Let
$g(n)
=\sum\limits_{k=1}^n \frac{f(k/n)}{n+\frac{1}{k}}
= \frac{1}{n} \sum\limits_{k=1}^n \frac{f(k/n)}{1 + \frac{1}{kn}}
$.
Since
$1 \ge \frac1{1+x}
\ge 1-x
$
for
$0 \le x \le 1$,
$g(n)
\le \frac{1}{n} \sum\limits_{k=1}^n f(k/n)
$
and
$\begin{array}\\
g(n)
&\ge \frac{1}{n} \sum\limits_{k=1}^n f(k/n)(1 - \frac{1}{kn})\\
&= \frac{1}{n} \sum\limits_{k=1}^n f(k/n) - \frac{1}{n} \sum\limits_{k=1}^n \frac{f(k/n)}{kn}\\
&= \frac{1}{n} \sum\limits_{k=1}^n f(k/n) - \frac{1}{n} \sum\limits_{k=1}^n \frac{f(k/n)}{k^2}\\
&\ge \frac{1}{n} \sum\limits_{k=1}^n f(k/n) - \frac{f(1)}{n} \sum\limits_{k=1}^n \frac{1}{k^2}\\
&\ge \frac{1}{n} \sum\limits_{k=1}^n f(k/n) - \frac{2M}{n}
\qquad\text{since } \sum\limits_{k=1}^n \frac{1}{k^2}
\lt 1+\sum\limits_{k=2}^n \frac{1}{k(k-1)} < 2\\
\end{array}
$
Therefore
$0
\le \frac{1}{n} \sum\limits_{k=1}^n f(k/n)
-\frac{1}{n} \sum\limits_{k=1}^n \frac{f(k/n)}{1 - \frac{1}{kn}}
\le \frac{2M}{n}
$.
Letting $n \to \infty$,
we see that the limits
of the two sums
are the same.
Note that the function
inside being $2^{k/n}$
does not matter,
only that it is bounded.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
$\cos^2(\frac{\pi}{101})+\cos^2(\frac{2\pi}{101})+\cos^2(\frac{3\pi}{101})+...+\cos^2(\frac{100\pi}{101})=?$
Find the value: $$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$
My attempt:I've tried it by considering the sum
$$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$
along with
$$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots +\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$ which gives $ 100$ as resultant but failed to separate the sum of
$$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots\\ \dots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$ at last.
I tried the next approach by using de Movire's theorem but failed to separate the real and imaginary part.
I've invested a great amount of time in the so it would be better if someone please come up with an answer.
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Let $x_{k} = \cos^{2}(k\pi/n)$ where $k = 1, 2, \dots, n$. Then we can see that $z_{k} = 2x_{k} - 1 = \cos(2k\pi/n)$ and hence these are the roots of the equation $P_{n}(z) = 1$ where $P_{n}(z)$ is a polynomial such that $P_{n}(\cos x) = \cos nx$. These polynomials are famous by the name Chebyshev polynomials and satisfy the recurrence $$P_{n + 1}(z) = 2zP_{n}(z) - P_{n - 1}(z)$$ and using the above recurrence we can easily show (via induction) that if $n\geq 2$ then the coefficient of $z^{n - 1}$ in $P_{n}(z)$ is $0$ and hence the sum of roots $z_{k} = \cos(2k\pi/n)$ is $0$. Also note that $z_{n} = 1$ so that $\sum_{k = 1}^{n - 1} z_{k} = -1$ and hence $$\sum_{k = 1}^{n - 1}x_{k} = \frac{n - 2}{2}$$ Putting $n = 101$ we get the desired sum as $99/2$.
In case the recurrence satisfied by $P_{n} (z) $ seems mysterious one has to note that $$\cos(n+1)x+\cos(n-1)x=2\cos x\cos nx$$ which means that $$P_{n+1}(\cos x) +P_{n-1}(\cos x)=2\cos x P_{n} (\cos x) $$ and replacing $\cos x$ by $z$ we get the desired recurrence.
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"url": "https://math.stackexchange.com/questions/2254887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Have I correctly taken this function $f(x) = \frac{x^{2}}{1+2x}$ and $f ' (x) $and turned them into power series? $f(x) = \dfrac{x^{2}}{1+2x}$
To turn this into a power series I recall the similar looking geometric series,
$f(x) = \sum\limits_{n=1}^{\infty} x^{n} = \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^5 +\cdots$
And not I can mutate my original function to fit the form the geometric series:
$f(x) = \dfrac{x^{2}}{1-(-2x)}$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-2x)} = 1 + (-2x) + (-2x)^{2} + (-2x)^{3} + (-2x)^{4} +\cdots$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-2x)} = 1 - 2x + 4x^2 - 8x^3 + 16x^4 +\cdots$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-x^{2})} = x^2( 1 - 2x + 4x^2 - 8x^3 + 16x^4 +\cdots)$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-x^{2})} = x^2 - 2x^3 + 4x^4 -8x^5 + 16x^{6} +\cdots)$
And for $'(x)$
$f'(x) = 2x - 6x^2 + 16x^3 -40x^4 + 16*6x^{5} +\cdots$
Is this right or wrong?
Thank you
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Your approach based upon the geometric series expansion
\begin{align*}
\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots
\end{align*}
is fine and appropriate. Just the calculations are not correct in all aspects.
We obtain
\begin{align*}
f(x)&=\frac{x^2}{1+2x}\\
&=x^2\cdot\frac{1}{1-(-2x)}\\
&=x^2\left(1-2x+4x^2-8x^3+16x^4-\cdots\right)\\
&=x^2-2x^3+4x^4-8x^4+16x^6-\cdots\tag{1}
\end{align*}
and from (1) we derive the derivative of $f$
\begin{align*}
f^\prime(x)=2x-6x^2+16x^3-32x^3+72x^5-\cdots
\end{align*}
in accordance with your calculation.
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"timestamp": "2023-03-29T00:00:00",
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Given the square, calculate $\tan{\alpha}.$ Problem: Given the square $ABCD$, let $M$ be the midpoint on the side $|CD|$ and designate $\alpha=\angle AMB$. Calculate $\tan{\alpha}.$
Attempt: We can, without compromising generality, assume that the side of the square is equal to 1. So drawing a figure we get
I know the following:
1) That $|AM|=|BM|=\sqrt{1^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2}.$
2) The area of $ABM$ can be expressed in two ways. One way with normal geometry for triangle and another way is by using the areakit involving $\sin{\alpha}$. So: $$\begin{array}{lcl}
A_1 & = & \frac{1\cdot 1}{2} = \frac{1}{2} \\
A_2 & = & \frac{|AM|\cdot|BM|\cdot\sin{\alpha}}{2} = \frac{5}{2}\cdot\sin{\alpha} \\
\end{array}$$
3) I know that $\tan{\alpha}=\frac{\sin{\alpha}}{\cos{\alpha}},$ so finding $\sin{\alpha}$ and $\cos{\alpha}$ and dividing these two will solve this problem.
Setting $A_1=A_2$ yields the equation $$\frac{5}{2}\sin{\alpha}=\frac{1}{2} \ \Longleftrightarrow \ \sin{\alpha} = \frac{1}{5}.$$
Using the law of cosines in the triangle $ABM$ I get
$$\begin{array}{lcl}
|AB|^2 & = & |AM|^2+|BM|^2 -2|AM||BM|\cos{\alpha} \\
1 & = & \sqrt{5}-\sqrt{5}\cos{\alpha} \ \Leftrightarrow \ \cos{\alpha} = \frac{\sqrt{5}-1}{\sqrt{5}} \\
\end{array}$$
And finally we have $$\tan{\alpha}=\frac{\sin{\alpha}}{\cos{\alpha}}=\frac{\frac{1}{5}}{\frac{\sqrt{5}-1}{\sqrt{5}}} = \frac{5+\sqrt{5}}{20}.$$
But it's not correct. Any idea where I'm making the mistake in my attempt, and is there an easier way of solving this problem?
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You have
$$
\tan\frac{\alpha}{2}=\frac{1/2}{1}=\frac{1}{2}
$$
Then use
$$
\tan2\beta=\frac{2\tan\beta}{1-\tan^2\beta}
$$
With $\beta=\alpha/2$, we get
$$
\tan\alpha=\frac{1}{1-1/4}=\frac{4}{3}
$$
Using your tools it can be done as well; set $r=AM$, for simplicity. Then
$$
\frac{1}{2}=\frac{1}{2}r^2\sin\alpha
$$
from the area and
$$
1=r^2+r^2-2r^2\cos\alpha
$$
from the cosine law.
Thus
$$
\sin\alpha=\frac{1}{r^2}\qquad \cos\alpha=\frac{2r^2-1}{2r^2}
$$
Finally
$$
\tan\alpha=\frac{1}{r^2}\frac{2r^2}{2r^2-1}=\frac{2}{2r^2-1}
$$
Since $r=\sqrt{1+\frac{1}{4}}=\sqrt{5}/2$, we have
$$
\tan\alpha=\frac{2}{5/2-1}=\frac{4}{3}
$$
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"timestamp": "2023-03-29T00:00:00",
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Show that fourier transformation is a circle I've calculated the fourier transformation of a function and got $\hat{f}(w) = \frac{1}{\rho+iw}$.
Now I must show, that this is equal to a circle with radius $r$ and middlepoint $m$, where $r = m = \frac{1}{2\rho}$
When I plot the function, I see the circles, but I'm not sure how to show that.
I thought that I must split it into the real and imaginary part and got:
$Re(\hat{f}(w))=\frac{1}{2\rho+2iw}+\frac{1}{2\rho-2iw}$
$Im(\hat{f}(w))=\frac{1}{2i\rho-2w}-\frac{1}{2i\rho+2w}$
Am I on the right way? Can anyone help me to get to $\frac{1}{2\rho}$ for the radius and middlepoint ?
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Note that
$$
\hat{f}(\omega) = \frac{1}{\rho + i \omega}\frac{\rho - i \omega}{\rho - i \omega} = \frac{\rho - i\omega}{\rho^2 + \omega^2}
$$
Therefore
$$
x = \Re(\hat{f}(\omega)) = \frac{\rho}{\rho^2 + \omega^2} ~~~\mbox{and}~~~
y = \Im(\hat{f}(\omega)) = -\frac{\omega}{\rho^2 + \omega^2}
$$
We then have
\begin{eqnarray}
\left(x - \frac{1}{2\rho} \right)^2 + y^2 &=& \left(\frac{\rho}{\rho^2 + \omega^2} - \frac{1}{2\rho} \right)^2 + \left(- \frac{\omega}{\rho^2 + \omega^2} \right)^2 \\
&=& \frac{1}{(\rho^2 + \omega^2)^2} \left[ \left(\rho - \frac{\rho^2 + \omega^2}{2\rho} \right)^2 + \omega^2\right] \\
&=& \frac{1}{(\rho^2 + \omega^2)^2} \left[ \left(\frac{2\rho^2 - \rho^2 - \omega^2}{2\rho} \right)^2 + \omega^2\right] \\
&=& \frac{1}{4\rho^2(\rho^2 + \omega^2)^2} \left[ (\rho^2 - \omega^2) + 4\rho^2 \omega^2\right] \\
&=& \frac{1}{4\rho^2(\rho^2 + \omega^2)^2} \left[ \rho^4 - 2\rho^2\omega^2 + \omega^4 + 4\rho^2 \omega^2\right] \\
&=& \frac{1}{4\rho^2(\rho^2 + \omega^2)^2} (\rho^2 + \omega^2)^2 \\
&=& \frac{1}{4\rho^2}
\end{eqnarray}
So the locus of $\hat{f}(\omega)$ on the complex plane is a circle of radius $\frac{1}{2\rho}$ centered at $x = \frac{1}{2\rho}$, $y = 0$
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"timestamp": "2023-03-29T00:00:00",
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|
Are there three integers such the nontrivial sums are squares and three are squares of consecutives?
Do there exist such three different positive integers $a, b, c$ that the following numbers: $a+b, b+c, c+a, a+b+c$ are all squares of integers and three of them ($a+b, b+c, c+a, a+b+c$) are squares of consecutive integers?
My first thought is to order the numbers $a,b,c$. Let $a<b<c$ and $a+b=k^2$, $b+c=l^2$, $c+a=m^2$, $a+b+c=n^2$. Then $n^2>l^2>m^2>k^2$ so $n>l>m>k$. Now we can consider two variants:
(i) $\quad n=l+1=k+2$
(ii) $\quad l=k+1=l+2$
I don't know what to do from this place...
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In the previous post, i miss understood the question.
We need to find $a,b,c$ such that $a+b,a+c,b+c,a+b+c$ all are square number and three of them are consecutive square numbers.
Which means with $d,e$ integers $a+b,a+c,b+c,a+b+c$ are equal to $(d-1)^2,d^2,(d+1)^2 ,e^2$. since $a,b,c$ are positive so $a+b+c >a+b,a+c,b+c$ which means that $a+b+c =(d+1)^2 $ or $a+b+c = e^2$, and since the other three equations are symmetric it does not matter how to order them, so with out loss of generality let :
$a+b=(d-1)^2$ and $a+c = d^2$ and $b+c=(d+1)^2$ and $a+b+c=e^2$.
Summing togther the first three equations we get $a+b+a+c+b+c = 2(a+b+c)= 2e^2=(d-1)^2+d^2 +(d+1)^2$.
Solving the equation $2e^2 = (d-1)^2 + d^2 +(d+1)^2$ we arrive at $2e^2=3d^2+2$
Solving this equation in Wolfram for integers gives that for any positive integer number $n$ the following formulas :
$d_n = \frac{\left(5+2 \sqrt{6}\right)^n-\left(5-2 \sqrt{6}\right)^n}{\sqrt{6}}$ and $e_n = \frac{1}{2} \left(\left(5-2 \sqrt{6}\right)^n+\left(5+2 \sqrt{6}\right)^n\right) $
Now substituting these values instead of $d,e$ in the equations $a+b=(d-1)^2,a+c=d^2,b+c=(d+1)^2,a+b+c=e^2$ gives that for any positive integer number $n$ the following formulas :
$a_n=\sqrt{\frac{2}{3}} \left(5-2 \sqrt{6}\right)^n+\frac{1}{12} \left(5-2
\sqrt{6}\right)^{2 n}-\sqrt{\frac{2}{3}} \left(5+2
\sqrt{6}\right)^n+\frac{1}{12} \left(5+2 \sqrt{6}\right)^{2 n}-\frac{1}{6}$
$b_n=\frac{1}{12} \left(5-2 \sqrt{6}\right)^{2 n}+\frac{1}{12} \left(5+2
\sqrt{6}\right)^{2 n}+\frac{5}{6}$
$c_n=-\sqrt{\frac{2}{3}} \left(5-2 \sqrt{6}\right)^n+\frac{1}{12} \left(5-2
\sqrt{6}\right)^{2 n}+\sqrt{\frac{2}{3}} \left(5+2
\sqrt{6}\right)^n+\frac{1}{12} \left(5+2 \sqrt{6}\right)^{2 n}-\frac{1}{6}$.
For the first few integers $n$ we get the following triples :
$n=0$ => $\{0,1,0\}$
$n=1$ => $\{0,9,16\}$
$n=2$ => $\{720,801,880\}$
$n=3$ => $\{77616, 78409, 79200\}$.
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"timestamp": "2023-03-29T00:00:00",
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