Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Two inequalities Let$n$ be an integer, $n\geq 2$ and $a_1,a_2,...,a_n$ positive real numbers.
For each$\{2,3,...,n\}$ we define the numbers as follows:
$s_k=a_1a_2...a_k)^{1/k}$.
a.) Pro thafor all$k$ in the set $\{2,3,...,n-1\}$ we have $s_{k+1}\geq s_k$
b.) Prove, that $$eq i<j\leq n} (\sqrt{a_i}-\sqrt{a_j})^2$$ (CALCULUS allowed)
| a) $s_{k+1} - s_k = a_{k+1} - (k+1)(a_1a_2...a_{k+1})^{\frac{1}{k+1}} + k(a_1a_2...a_k)^{\frac{1}{k}}$.
Apply AM-GM inequality we have: $a_{k+1} + k(a_1a_2...a_k)^{\frac{1}{k}} \geq (k+1)(((a_1a_2...a_k)^{\frac{1}{k}})^ka_{k+1})^{\frac{1}{k+1}} = (k+1)(a_1a_2...a_{k+1})^{\frac{1}{k+1}}$, this means: $s_{k+1} \geq s_k$.
b) Assume without loss of generality that $0 < a_1 \leq a_2 \leq ...\leq a_n$, then:
$\displaystyle \max_{1\leq i < j\leq n} \left(\sqrt{a_i} - \sqrt{a_j}\right)^2 = \left(\sqrt{a_n} - \sqrt{a_1}\right)^2 = a_1 - 2\sqrt{a_1a_n} + a_n$. Thus we need to prove:
$a_2 + a_3 + a_4 + ... + a_{n-1} - n(a_1a_2...a_n)^{\frac{1}{n}} + 2\sqrt{a_1a_n} \geq 0 $
Consider the function $f(a_2, a_3,...,a_{n-1}) = a_2 + a_3 +...+ a_{n-1} - n(a_1a_2...a_n)^{\frac{1}{n}} + 2\sqrt{a_1a_n}$ on the interval $[a_1, a_n]$. We now find the critical points of $f$. Setting all partial derivatives of $f$ w.r.t $a_2$, $a_3$,...,$a_{n-1}$ to $0$ we have:
$\dfrac{\partial f}{\partial a_2} = 0 \iff a_2^{n-1} = a_1a_3...a_n$
$\dfrac{\partial f}{\partial a_3} = 0 \iff a_3^{n-1} = a_1a_2a_4...a_n$
....
$\dfrac{\partial f}{\partial a_{n-1}} = 0 \iff a_{n-1}^{n-1} = a_1a_2...a_{n-2}a_n$.
So: $a_2^n = a_3^n = ...= a_{n-1}^n = a_1a_2...a_n$.
Thus: $a_2 = a_3 = ... = a_{n-1} = x$. Solve for $x$ and get:
$x^{n-1} = a_1\cdot x^{n-3}\cdot a_n \to x = \sqrt{a_1a_n}$.
Thus the critical point of $f$ is: $(a_2,a_3,...,a_{n-1}) = (x,x,...x)$.
Thus: $f_{min} = f(x,x,...x) = n\sqrt{a_1a_n} - n(a_1a_n(a_1a_n)^{\frac{n-2}{2}})^{\frac{1}{n}} = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/781031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
A square number and a positive cubic number which differ by six Is there a square number and a positive cubic number (both positive integers) which differ by six? If not, how do we prove this?
| Here is a proof that $y^2 = x^3 - 6$ has no solutions, taken from here.
First note modulo $8$ that $y^2$ is one of $0,1, 4$ and $x^3$ is one of
$0, 1, 3, 5, 7$.
Thus, we must have $y^2 \equiv 1$ and $x^3 \equiv 7$.
I.e., $y$ is odd and $x \equiv 7 \pmod 8$.
Now write
$$
y^2 - 2 = (x - 2)(x^2 + 2x + 4)
$$
Since $x - 2 \equiv 5 \pmod 8$, a prime $p \; \mid \; x - 2$ congruent to $3$ or $5$ mod $8$.
Then $p \; \mid \; y^2 - 2$,
so $2$ is a perfect square modulo $p$.
But by a property of the Legendre symbol,
letting $p = 8k \pm 3$, we have
$$
\left( \frac{2}{p} \right)
= (-1)^{\frac{p^2 - 1}{8}}
= (-1)^{8k^2 \pm 6k + 1} = -1,
$$
which is a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/781115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Ground plan of Backward direction (<=) - Let $p$ be an odd prime. Prove $x^{2} \equiv -1 \; (mod \, p)$ has a solution $\iff p\equiv 1 \; (mod 4)$ Apply the identity $p-i \equiv -i \mod p$ for $i=1, \ldots$ to the pink factors
$ \begin{align}
\color{seagreen}{ (p-1)! } = 1\times 2\times\cdots\times \dfrac{p-1}{2} & \times \quad \color{magenta}{\dfrac{p+1}{2} \times\cdots\times(p-2)\times(p-1)}\
\\
= \quad \dfrac{p-1}{2}! \quad & \times \quad \color{magenta}{ \dfrac{p-1}{2} \times\cdots\times(2)\times(1) \times \quad (-1)^{ \dfrac{p-1}{2}} }
\\
& \equiv \color{seagreen}{ ((\frac{p-1}{2})!)^{2} \cdot (-1)^{ \frac{p-1}{2} } } \mod p .
\end{align} $
(1) Why start with $(p - 1)!$ ? How can you prefigure this?
(2) Then how can you prefigure to write $(p - 1)!$ with $ \dfrac{p-1}{2} \times \quad \color{magenta}{ \dfrac{p+1}{2} }$ in the middle?
(3) How can you prefigure to use the trick $p-i \equiv -i \mod p $ ?
Now Wilson's Theorem is $\color{seagreen}{ (p-1)! } \equiv -1 \mod p$ therefore $
\color{seagreen}{ ((\frac{p-1}{2})!)^{2} \cdot (-1)^{ \frac{p-1}{2} } } \equiv -1 \mod p$. Implying $(\dfrac{p-1}{2}!)^{2}\equiv(-1)^{\frac{p+1}{2}} \quad (\#)$.
If $p\equiv 1 \mod 4$ then there exists an natural number n such that $4n = p - 1$. Hence $\dfrac{p-1}{2} = \dfrac{(4n + 1)-1}{2}$, an even number.
$\frac{p-1}{2}$ even implies
$(\frac{p-1}{2}!)^{2}\equiv-1$, by reason of $(\#)$. Thence $x=\frac{p-1}{2}!$ solves $x^{2}+1\equiv 0 \mod p$ .
Origin - Elementary Number Theory, Jones,
p71, Theorem 4.6
| Here is problem I liked on math.SE: look at Primes congruent to 1 mod 6
a prime $p$ can be written in the form $p = a^2 + ab + b^2 $ for some $a,b \in \mathbb{Z}$ iff $p \equiv 1 \mod 6$
One part of the solution says it is elementary to show there exists an integer $d$ such that $d^2 \equiv -3 \mod p$. This was not so "elementary to me", but we can prove it using your factorial identity.
Since $p = 6k+1$ we have $6$ divides $ p-1 $ which is the size of $\mathbb{Z}_p^\times$. In particular
$$ \frac{p-1}{3} \in \mathbb{Z}$$
We have a cube root of unity $ z $
Then - mysteriously - we set $z = \frac{-1+d}{2}$ so that $z^3 =
\left( \tfrac{-1+d}{2} \right)^3 = \tfrac{-1+3d-3d^2 + d^3}{8} \equiv -1 (\mod p)$.
Notice the algebraic identity - regardless of taking mod $ p$
$$ -9+3d-3d^2 + d^3 = (d^2 + 3)(d-3) \equiv 0 \mod p$$
Therefore either $d \equiv 3 \mod p$ - corresponding to $z=1$ or $d^2 \equiv - 3 \mod p$.
We can't really say what the cube root of unity would be in terms of the factorial...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solutions to a particular equation of perfect squares Can we prove (or is it otherwise known) that for positive integers $a>b>c>1$, the only solutions to the equation:
$$a^2+1=b^2(c^2+1)$$
is given by the following family for any fixed $c$:
$$ a_1 = 4 c^3 + 3 c, \; \; \; b_1 = 4 c^2 + 1,$$
$$ a_2 = 16 c^5 + 20 c^3 + 5 c , \; \; \; b_2 = 16 c^4 + 12 c^2 + 1,$$
$$ a_{n+2} = (4 c^2 + 2) a_{n+1} - a_n, \; \; b_{n+2} = (4 c^2 + 2) b_{n+1} - b_n $$
| Yes. Given $c$, this is a "Pell equation" $a^2 - (c^2+1) b^2 = -1$
whose general solution is
$$
a_n + b_n \sqrt{c^2+1} = (c + \sqrt{c^2+1})^{2n-1}
$$
because $c + \sqrt{c^2+1}$ is clearly the fundamental unit in
${\bf Z}[\sqrt{c^2+1}]$.
I just answered this on stackexchange.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/788857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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how can we get from $x \equiv 4 \pmod 9$ to $x \equiv 1 \pmod 3$ I want to solve the system:
$$ x \equiv 1 \pmod 3 , x \equiv 2 \pmod 5, x \equiv 3 \pmod 7, x \equiv 4 \pmod 9, x \equiv 5 \pmod {11}$$
The numbers $3,5,7,9,11$ are not pairwise coprime,especially the numbers $3,9$ are not coprime.
According to my notes, $x \equiv 4 \pmod 9 \Rightarrow x \equiv 4 \pmod 3 \Rightarrow x \equiv 1 \pmod 3$
But..how can we get from the relation $x \equiv 4 \pmod 9$ to this one: $x \equiv 1 \pmod 3$ ?
| $x\equiv4(\text{mod }9)$ is equivalent to:
$x-4=9k (\text{ for some integer k})\implies x-3-1=3(3k)\implies x-1=3(3k+1)$
or, $x\equiv1(\text{mod }3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/788956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Recurrence with multiplication Let $\{a_{n}\}$ be a sequence of nonnegative numbers such that $a_{n} = 2^{n}a_{n - 1}^{3/2}$. If $a_{1}$ is sufficiently small, why must $a_{n} \rightarrow 0$ as $n \rightarrow \infty$?
| If $a_{n_0}=0$ for some positive integer $n_0$ then $a_n=0$ for every $n\geq n_0$ and we are done.
So let us suppose that $a_n>0$ for every $n$. Let $b_n=\ln(a_n)$. We have
$$
b_n=n\ln2+\frac{3}{2}b_{n-1}
$$
So
$$
\left(\frac{2}{3}\right)^nb_n=\left(\frac{2}{3}\right)^{n-1}b_{n-1}+ (\ln2)n\left(\frac{2}{3}\right)^n
$$
So, adding from $n=2$ to $n=m$ and cancelling $(2/3)$ we get
$$
\left(\frac{2}{3}\right)^{m-1}b_m= b_{1}+ (\ln2)\sum_{k=2}^{m}k\left(\frac{2}{3}\right)^{k-1}
$$
But $$\sum_{k=2}^{m}k\left(\frac{2}{3}\right)^{k-1}=8+2(3+n)\left(\frac{2}{3}\right)^{m-1}$$
Thus
$$
b_m=\left(\frac{3}{2}\right)^{m-1}\ln(2^8a_1)+ (\ln4)(3+m)
$$
So, if $2^8a_1<1\iff a_1<\dfrac{1}{256}$, we have $\ln(2^8a_1)<0$ and $\lim\limits_{m\to\infty}b_m=-\infty$, therefore $\lim\limits_{m\to\infty}a_m=0$.$\qquad\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/790747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find all non-singular $3 \times 3$ matrices, such as $A$ and $A^{-1}$ elements are non-negative Task is to describe all non-singular $3 \times 3$ matrices $A$ for which holds: all elements of $A$ and $A^{-1}$ is non-negative.
As I discovered linear algebra is the most problematic part of math for me. Expect to get better with a bit of your help.
I have trouble even approaching the problem, would you provide a slight hint where to go?
| We know that $A A^{-1}= I$.
Then suppose $X=A^{-1}$ => Elements of X could be derived from following set of linear systems:
$\begin{pmatrix}
a & 0 & 0 & | & 1\\
a_{21} &a_{22} &a_{23} & | & 0\\
a_{31} &a_{32} &a_{33} & | & 0\\
\end{pmatrix} => x_{1,1}= \frac 1 a$
$ \begin{pmatrix}
a & 0 & 0 & | & 0\\
a_{21} &a_{22} &a_{23} & | & 1\\
a_{31} &a_{32} &a_{33} & | & 0\\
\end{pmatrix} => x_{2,1}= 0$
$\begin{pmatrix}
a & 0 & 0 & | & 0\\
a_{21} &a_{22} &a_{23} & | & 0\\
a_{31} &a_{32} &a_{33} & | & 1\\
\end{pmatrix} => x_{3,1}= 0$
This derivations prove second bullet from Jyrki's answer.
Also, in order to satisfy requirements, we need: $|A| > 0$,
$Minor_{i,j} \ge 0 $ if $i+j = 0 (mod 2)$
$Minor_{i,j} \le 0 $ if $i+j = 1 (mod 2)$
Then, A is:
$\begin{pmatrix}
a & 0 & 0\\
0 &b &0\\
0 &0 &c\\
\end{pmatrix}, a,b,c > 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/791394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving implicit equation for x or y I want to solve the following equation for $x$ or $y$ (does not matter wich one) analytically.
$$\sqrt[3]{x+y} + \sqrt[3]{x-y} = 1$$
Wolframalpha returns following solution, but I could not think of a way how to get there:
$$ x+1 \neq 0 \qquad y = \frac{(x+1) \sqrt{8 x-1}}{3 \sqrt{3}}$$
Is there a nice 'tool' I do not know for solving this?
| Suppose $a+b+c=0$ and that $a,b,c$ are the roots of the equation $$f(x)=x^3-px^2+qx-r=0$$
Then $p=a+b+c=0, r=abc$ and adding $f(a)+f(b)+f(c)=0$ we obtain:$$a^3+b^3+c^3+q(a+b+c)-3r=0$$ whence $$a^3+b^3+c^3=3abc$$
Now let $a=\sqrt[3]{x+y}, b=\sqrt[3]{x-y}, c=-1$ to obtain:$$2x-1=-3\sqrt[3]{x^2-y^2}$$ You can then cube this and isolate $y$ to solve, which should give the answer you are looking for.
Note this method does not make it obvious that $x=-1$ is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/799562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Binomial Sum Related to Fibonacci: $\sum\binom{n-i}j\binom{n-j}i=F_{2n+1}$ How would I prove
$$
\sum\limits_{\vphantom{\large A}i\,,\,j\ \geq\ 0}{n-i \choose j} {n-j \choose i}
=F_{2n+1}
$$
where $n$ is a nonnegative integer and $\{F_n\}_{n\ge 0}$ is a sequence of Fibonacci numbers?
Thank you very much! :)
| Here is a solution using two complex variables.
Suppose we seek to evaluate in terms of Fibonacci numbers
$$\sum_{p,q\ge 0} {n-p\choose q} {n-q\choose p}.$$
We use the integrals
$${n-p\choose q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(1-z)^{q+1} z^{n-p-q+1}} \; dz$$
and
$${n-q\choose p} =
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{(1-w)^{p+1} w^{n-p-q+1}} \; dw.$$
These correctly control the range
so we may let $p$ and $q$ go to infinity to get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(1-z) z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{(1-w) w^{n+1}}
\sum_{p,q\ge 0} \frac{z^{p+q} w^{p+q}}{(1-w)^p (1-z)^q}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(1-z) z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{(1-w) w^{n+1}}
\\ \times \frac{1}{1-zw/(1-w)} \frac{1}{1-zw/(1-z)}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n+1}}
\frac{1}{1-w-zw} \frac{1}{1-z-zw}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2} (1+z)}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n+1}}
\frac{1}{w-1/(1+z)} \frac{1}{w-(1-z)/z}
\; dw \; dz.$$
We evaluate the inner integral using the fact that the residues of the
function in $w$ sum to zero. We have two simple poles. We get for the
first pole at $w=(1-z)/z$
$$\frac{z^{n+1}}{(1-z)^{n+1}}
\frac{1}{(1-z)/z-1/(1+z)}
= \frac{z^{n+1}}{(1-z)^{n+1}}
\frac{z(1+z)}{(1-z)(1+z)-z}
\\ = \frac{z^{n+2}}{(1-z)^{n+1}}
\frac{1+z}{(1-z)(1+z)-z}.$$
Substituting this expression into the outer integral we see that the
pole at $z=0$ is canceled making for a contribution of zero.
For the second pole at $w=1/(1+z)$ we get
$$(1+z)^{n+1} \frac{1}{1/(1+z)-(1-z)/z}
= (1+z)^{n+1} \frac{z(1+z)}{z-(1-z)(1+z)}.$$
This yields the contribution (taking into account the sign flip from
the sum of residues)
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2} (1+z)}
(1+z)^{n+1} \frac{z(1+z)}{1-z-z^2} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
(1+z)^{n+1} \frac{1}{1-z-z^2} \; dz.$$
We evaluate this using again the fact that the residues sum to
zero. There are simple poles at $z=-\varphi$ and $z=1/\varphi.$
These yield
$$\left(\frac{1-\varphi}{-\varphi}\right)^{n+1}
\frac{1}{-1+2\varphi} +
\left(\frac{1+1/\varphi}{1/\varphi}\right)^{n+1}
\frac{1}{-1-2/\varphi}
\\= \frac{1}{\sqrt{5}} \frac{1}{\varphi^{2n+2}}
- \frac{1}{\sqrt{5}} \varphi^{2n+2}.$$
Taking into account the sign flip this is obviously Binet / de Moivre
for $${\large F_{2n+2}}.$$
Remark. If we want to do this properly we also need to verify that
the residue at infinity in both cases is zero. For example in the
first application we use the formula for the residue at infinity
$$\mathrm{Res}_{z=\infty} h(z)
= \mathrm{Res}_{z=0}
\left[-\frac{1}{z^2} h\left(\frac{1}{z}\right)\right]$$
which in the present case gives for the inner term in $w$
$$- \mathrm{Res}_{w=0} \frac{1}{w^2} w^{n+1}
\frac{1}{1/w-1/(1+z)} \frac{1}{1/w-(1-z)/z}
\\ = - \mathrm{Res}_{w=0} w^{n+1}
\frac{1}{1-w/(1+z)} \frac{1}{1-w(1-z)/z}$$
which is zero by inspection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/801730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 5
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How to show if $m$ and $n$ are coprime and $m-n$ is odd, then $(m^2-n^2)^2$, $(2mn)^2$, $(m^2+n^2)^2$ have a common factor? We know that the Pythagorean triples can be generated by the Euclid's formula $a=m^2-n^2$, $b=2mn$, $c=m^2+n^2$ for any positive integers $m,n$ and $m>n$.
I am trying to prove the statement:
The triple generated by Euclid's formula is primitive if and only if $m$ and $n$ are coprime and $m-n$ is odd.
I could prove the $\Rightarrow$ direction. Now I am trying to show the $\Leftarrow$ proof.
I am trying to prove by contradiction. If $m$ and $n$ are coprime and $m-n$ is odd, let's assume the triple generated is not primitive. Then there is a factor $k$ in each term. But since $m$ and $n$ are coprime and $m-n$ is odd, we can deduce that $(m^2-n^2)^2$ is odd, $(2mn)^2$ is even and $(m^2+n^2)^2$ is odd. Then $k$ can only be odd. Let $k=2p+1$ where $p$ is an integer. But then I am not sure how to proceed to obtain a contradiction.
Any helps are greatly appreciated. Many thanks!
| Informal outline Suppose for example that $(m^2-n^2)^2$ and $(2mn)^2$ are not relatively prime. Then there is a prime $p$ that divides $m^2-n^2$ and $2mn$. Since $m$ and $n$ have opposite parity, $p\ne 2$.
Since $p$ divides $2mn$, and $p\ne 2$, it follows that $p$ divides one of $m$ or $n$, say $m$. Since $p$ divides $m^2-n^2$, it follows that $p$ divides $n^2$, and hence $n$. This contradicts the fact that $m$ and $n$ are relatively prime.
Remark: We started, as you did, from the assumption that the numbers $(m^2-n^2)^2$, $(2mn)^2$, and $(m^2+n^2)^2$ are not pairwise relatively prime. However, a triple $(x,y,z)$ such that $x^2+y^2=z^2$ is usually called primitive if $x$, $y$, and $z$ are pairwise relatively prime. It makes no real difference, since $x^2$, $y^2$, and $z^2$ are pairwise relatively prime if and only if $x$, $y$, and $z$ are.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$
Let $a,b,c>0$ such that
$$\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}=\dfrac{1}{3}.$$
Show that
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}.$$
My try: since
$$\dfrac{1}{2+a^2}=\dfrac{1}{2}\left(1-\dfrac{a^2}{a^2+2}\right)$$
so
$$\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}=\dfrac{7}{3}$$
we only prove
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}$
and I want use Cauchy-Schwarz inequality to prove it,But I can't works,such
$$\left(\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}\right)(a^2+2+b^2+2+c^2+2)\ge (a+b+c)^2$$
$$(a^2+b^2+c^2+6)\ge \dfrac{3}{7}(a^2+b^2+c^2+2ab+2bc+2ac)$$
$$\Longrightarrow 4(a^2+b^2+c^2)+42\ge 6(ab+bc+ac)$$
and let $$p=a+b+c,q=ab+bc+ac,r=abc$$
so
$$2p^2+21\ge 7q$$
and we only prove
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+bc+ac}{abc}=\dfrac{q}{r}\le\dfrac{3}{\sqrt{7}}$$
maybe this is not true.
But this not usefull to solve this problem .
Thank you
| Let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>\frac{3}{\sqrt7}$, $\frac{1}{a}=\frac{kx}{\sqrt7}$, $\frac{1}{b}=\frac{ky}{\sqrt7}$ and $\frac{1}{c}=\frac{kz}{\sqrt7}$, where $k>0$ and $x+y+z=3$.
Hence, $k>1$ and $\frac{1}{3}=\sum\limits_{cyc}\frac{1}{a^2+2}=\sum\limits_{cyc}\frac{1}{\frac{7}{k^2x^2}+2}>\sum\limits_{cyc}\frac{1}{\frac{7}{x^2}+2}=\sum\limits_{cyc}\frac{x^2}{7+2x^2}$,
which is a contradiction because we'll prove now that $\sum\limits_{cyc}\frac{x^2}{7+2x^2}\geq\frac{1}{3}$.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that
$$4x^2y^2x^2+\sum\limits_{cyc}(8x^2y^2+7x^2)-49\geq0$$ or
$$4w^6+8(9v^4-6uw^3)u^2+7(9u^2-6v^2)u^4-49u^6\geq0$$
or $f(w^3)\geq0$, where $f(w^3)=2w^6-24u^3w^3+7u^6-21u^4v^2+36u^2v^4$.
We see that $f'(w^3)=4w^3-24u^3<0$, which says that it's enough to prove that
$f(w^3)\geq0$ for a maximal value of $w^3$, which happens for equality case of two variables.
Let $y=x$ and $z=3-2x$.
Hence, $\sum\limits_{cyc}\frac{x^2}{7+2x^2}\geq\frac{1}{3}\Leftrightarrow(x-1)^2(2x-1)^2\geq0$.
Done!
Also we can use the Vasc's LCF Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/806224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 0
} |
Strictly decreasing sequence Let $$a_n=\sum_{k=1}^n\sqrt{k^2+1}$$
Define the sequence $b_n=\dfrac{2a_n}{n} - n$.
I need to show that the sequence $\langle b_n\rangle_{n=1}^\infty$ is strictly decreasing.
| Try to show $b_{n+1} - b_n < 0$:
$b_{n+1} - b_n = \left[\dfrac{2a_{n+1}}{n+1} - (n+1)\right] - \left[\dfrac{2a_n}n - n\right]$
$ = \dfrac1{n(n+1)}\left[2n \sum_{k=1}^{n+1}\sqrt{k^2+1} - 2(n+1) \sum_{k=1}^n\sqrt{k^2+1} - n(n+1)\right]$
$ = \dfrac1{n(n+1)} \left[2n \sqrt{(n+1)^2 + 1} - 2 \sum_{k=1}^n\sqrt{k^2+1} - n(n+1)\right]\qquad\qquad\mbox{(1)}$
$ < \dfrac1{n(n+1)} \left[2n \sqrt{\left(n + 1 + \dfrac1{2n}\right)^2} - 2 \sum_{k=1}^n \left(k + \dfrac1{2k}\right) - n(n+1)\right]$
$ < \dfrac1{n(n+1)} \left[2n \left(n + 1 + \dfrac1{2n}\right) - n(n+1) - \ln(n+1) - n(n+1)\right]$ ... [Ref: http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Integral_test to show $\sum_{k=1}^n\dfrac1k > \ln(n+1)$]
$ = \dfrac1{n(n+1)} \left[1 - \ln(n+1)\right]$
$ < 0$ for all $n \geq 2$.
For $n = 1$ case: calculate: $b_2 - b_1 = \sqrt5 - \sqrt2 - 1 < 0$.
Revised part to fix mistake GL5 found, following from (1):
$ < \dfrac1{n(n+1)} \left[2n \sqrt{\left(n + 1 + \dfrac1{2(n+1)}\right)^2} - 2 \sum_{k=1}^n \left(k + \dfrac1{2(k+1)}\right) - n(n+1)\right]$
$ < \dfrac1{n(n+1)} \left[2n \sqrt{\left(n + 1 + \dfrac1{2(n+1)}\right)^2} - n(n+1) - \sum_{k=2}^{n+1} \dfrac1{k} - n(n+1)\right]$
$ < \dfrac1{n(n+1)} \left[2n \left(n + 1 + \dfrac1{2(n+1)}\right) - n(n+1) - \ln(n+1) + 1 - \dfrac1{n+1} - n(n+1)\right]$
$ = \dfrac1{n(n+1)} \left[2 - \dfrac2{n+1} - \ln(n+1)\right]$
$ < 0$ for all $n \geq 4$.
Initial cases for $1 \leq n \leq 3$ to be done by individual calculation. Having 3 initial cases is inelegant. There might be a way to tighten up the inequalities to reduce it.
Mick
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/807786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
odd/even binomial coefficient identity For all n\geq1
:
$$\left(\begin{matrix}2n\\
0
\end{matrix}\right)
+\left(\begin{matrix}2n\\
2
\end{matrix}\right)
+\left(\begin{matrix}2n\\
4
\end{matrix}\right)
+ \ldots
+\left(\begin{matrix}2n\\
2k
\end{matrix}\right)
+ \ldots
+\left(\begin{matrix}2n\\
2n
\end{matrix}\right)
$$
is equal to
$$\left(\begin{matrix}2n\\
1
\end{matrix}\right)
+\left(\begin{matrix}2n\\
3
\end{matrix}\right)
+\left(\begin{matrix}2n\\
5
\end{matrix}\right)
+ \ldots
+\left(\begin{matrix}2n\\
2k+1
\end{matrix}\right)
+ \ldots
+\left(\begin{matrix}2n\\
2n-1
\end{matrix}\right)
.$$
MY SOLUTION
Binomial Formula : (a+b)^{2n}
= \sum_{k=0}^{2n}
\left(\begin{matrix}2n\\
k
\end{matrix}\right)
a^{k}
b^{n-k}
let a=1
, let b=-1
, let n=2
(1+(-1))^{2(n)}
= \left(\begin{matrix}2n\\
0
\end{matrix}\right)
-\left(\begin{matrix}2n\\
1
\end{matrix}\right)
+...\mbox{-}\left(\begin{matrix}2n\\
2n-1
\end{matrix}\right)
+
\left(\begin{matrix}2n\\
2n
\end{matrix}\right)
(1+(-1))^{2(4)}
= \left(\begin{matrix}2(2)\\
0
\end{matrix}\right)-\left(\begin{matrix}2(2)\\
1
\end{matrix}\right)+\left(\begin{matrix}2(2)\\
02
\end{matrix}\right)-\left(\begin{matrix}2(2)\\
2(2)-1
\end{matrix}\right)+
\left(\begin{matrix}2(2)\\
2(2)
\end{matrix}\right)
0=1-4+6-4+1
0=0
Thus for all n\geq1
:
\left(\begin{matrix}2n\\
0
\end{matrix}\right)
+\left(\begin{matrix}2n\\
2
\end{matrix}\right)
+\left(\begin{matrix}2n\\
4
\end{matrix}\right)
+ \ldots
+\left(\begin{matrix}2n\\
2k
\end{matrix}\right)
+ \ldots
+\left(\begin{matrix}2n\\
2n
\end{matrix}\right)
is equal to
\left(\begin{matrix}2n\\
1
\end{matrix}\right)
+\left(\begin{matrix}2n\\
3
\end{matrix}\right)
+\left(\begin{matrix}2n\\
5
\end{matrix}\right)
+ \ldots
+\left(\begin{matrix}2n\\
2k+1
\end{matrix}\right)
+ \ldots
+\left(\begin{matrix}2n\\
2n-1
\end{matrix}\right)
.
Can anyone please give feedback and tell me if my solution is correct.
| Hint:
From binomial formula $$(1+x)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}x^k$$ for $x=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find a formula for $0 · 1 · 2 + 1 · 2 · 3 + 2 · 3 · 4 + \dots +n(n + 1)(n + 2)$, for $n \in \mathbb N$ $$\sum\limits_{i=1}^n i(i + 1)(i + 2)$$
$$\sum\limits_{i=1}^n i^3 + 3i^2 + 2i$$
$$\sum\limits_{i=1}^n i^3 + 3\sum\limits_{i=1}^ni^2 + 2\sum\limits_{i=1}^ni$$
$$= (\frac14)n^4 + (\frac12)n^3 + (\frac14)n^2 + n^3 + 3(\frac{n^2}2) + (\frac{n}2) + n^2 + n$$
$$ =n^4 + 3n^3 + 8n^2 + 3n$$
Why does what I did above not work?
| $$
0 \cdot 1 \cdot 2 + 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \cdots +n(n + 1)(n + 2)
=
3!\ \sum_{k=1}^{n+2} \binom{k}{3}
=
6 \binom{n+3}{4}
$$
The key sum is the sum of a column of Pascal's triangle, which is found by induction using Pascal's rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Proving a Pellian connection in the divisibility condition $(a^2+b^2+1) \mid 2(2ab+1)$ I'm trying to prove that all integer solutions $a > b \ge 0$ to the divisibility condition in the title, namely
$$(a^2+b^2+1) \mid 2(2ab+1),$$
are given by
$$(a,b)=(1,0),(4,1),(15,4),(56,15),(209,56),\dots$$
where $a_n$ is in http://oeis.org/A001353, and $b_{n+1}=a_n$. This sequence is related to the Pell equation $x^2-3y^2=1$, in that the sequence $0,1,4,15,56,\dots$ give the $y$ values in the solutions $(x,y)$.
Related MSE posts of possible interest include Proving there are an infinite number of pairs of positive integers $(m,n)$ such that $\frac{m+1}{n}+\frac{n+1}{m}$ is a positive integer and Conjecture on integer solutions to the equation $ (ab + 1) \mid (a^{2}+b^{2})$. Vieta jumping doesn't seem to apply (since it's best used to derive a contradiction, or prove a constant value). Any help/pointers would be appreciated.
| Your link does not like me. Solutions for hard to find such equations. So easy.
the equation: $\frac{m+1}{n}+\frac{n+1}{m}=a$
Can be solved using the equation Pell:
$p^2-(a^2-4)s^2=1$
Solutions have the form:
$n=2(p-(a+2)s)s$
$m=-2(p+(a+2)s)s$
Solutions have the form:
$n=\frac{2p(p+(a-2)s)}{a-2}$
$m=\frac{2p(p-(a-2)s)}{a-2}$
If there is a solution of the following equation Pell: $p^2-(a^2-4)s^2=4$
We use its solutions and the formula is:
$n=\frac{p-(a-2)s+2}{2(a-2)}$
$m=\frac{p+(a-2)s+2}{2(a-2)}$
For your case.
To solve the equation in general form: $\frac{X^2+aX+Y^2+bY+c}{XY}=j$
In the case where a square:
$t=\sqrt{(b+a)^2+4c(j-2)}$
We use the solutions of Pell's equation:
$p^2-(j^2-4)s^2=1$
Solutions can be written.
$X=\frac{(b+a\pm{t})}{2(j-2)}p^2+(t\mp{(b-a)})ps-\frac{(b(3j-2)+a(6-j)\mp{(j+2)t})}{2}s^2$
$Y=\frac{(b+a\pm{t})}{2(j-2)}p^2+(-t\mp{(b-a)})ps-\frac{(b(6-j)+a(3j-2)\mp{(j+2)t})}{2}s^2$
It must be remembered that the numbers $p,s$ may be any mark.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that $7\mid(3^{2n+1}+2^{n+2})$ for all $n\in\mathbb{N}$
Prove that the following is true for every $n∈ℕ$:
$$7\mid(3^{2n+1}+2^{n+2}).$$
I've noticed
$$3^{2n+1}+2^{n+2} =3^{2n} \cdot 3+2^{n} \cdot 4.$$
Any suggestions how to continue from there to get something like $7k$ for $k\in\mathbb{N}$.
Thank you in advance!
| \begin{align}3^{2n+1}+2^{n+2}=3\cdot 9^n + 4\cdot 2^n&=7\cdot 9^n -4(9^n-2^n)\\
&=7\cdot 9^n-4(9-2)(9^{n-1}+9^{n-2}\cdot2+\cdots+2^{n-1})\\
&=7[ 9^n-4(9^{n-1}+9^{n-2}\cdot 2\cdots+2^{n-1})]\end{align}
Which is divisible by $7$
Here is a solution based on congruences,
$$3\cdot 9^n +4\cdot 2^n \equiv 3\cdot 2^n +4\cdot 2^n \equiv 7\cdot 2^n \equiv 0\text{(mod 7)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How to evaluate $\lim\limits_{n\to\infty}(n^3)\cdot(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2})$?
How to evaluate
$\lim\limits_{n\to\infty}(n^3\cdot(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}))$?
My though is, that since $\lim\limits_{n\to\infty}n^3=\infty$, the limit must be $\infty$, if the right hand side is greater than or equal to $1$. Otherwise it's a fraction, and there might be many possible solutions.
However, I'm far for being confident about the solution. What am I missing here? What's the "trick"?
| Hint: \begin{align}\\&\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}
\\\\=&\dfrac{n^2+\sqrt{n^4+1}-2n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}
\\\\=&\dfrac{\sqrt{n^4+1}-n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}
\\\\=&\dfrac{n^4+1-n^4}{\left(\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}\right)\cdot\left(\sqrt{n^4+1}+n^2\right)}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/811777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $(4/5)^{\frac{4}{5}}$ is irrational Prove that $(4/5)^{\frac{4}{5}}$ is irrational.
My proof so far:
Suppose for contradiction that $(4/5)^{\frac{4}{5}}$ is rational.
Then $(4/5)^{\frac{4}{5}}$=$\dfrac{p}{q}$, where $p$,$q$ are integers.
Then $\dfrac{4^4}{5^4}=\dfrac{p^5}{q^5}$
$\therefore$ $4^4q^5=5^4p^5$
I've got to this point and now I don't know where to go from here.
| A slightly modified approach.
$(\frac{4}{5})^\frac{4}{5} = (\frac{4}{5})^{1-\frac{1}{5}} = (\frac{4}{5})^1 \times(\frac{5}{4})^\frac{1}{5}$
Since the first factor is rational we need only show that the latter term $(\frac{5}{4})^\frac{1}{5}$ is irrational.
Let us assume that the latter is rational and write (where $p$ and $q$ are coprime)
$\frac{p}{q}=(\frac{5}{4})^\frac{1}{5} \implies 4\times p^5 = 5 \times q^5$
The L.H.S is even and this implies that $q$ is even (sub $q=2^5\times n$ in above); $q^5$ then must have a factor of $2^5$. This would then require $p$ to be even and thus violates the coprime assumption and gives the required contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/812778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
if A diagonalizable then show that $a=0$ Let $A= \begin{pmatrix}
2 & 0 & 0\\
a & 2 & 0 \\
b & c & -1\\
\end{pmatrix}$
if A diagonalizable then show $a=0$.
$P_A(x)=|xI-A|=(x-2)^2(x+1)=x^3-3x^2+4$
since A diagonalizable there is a P matrix such as $A=P^{-1}DP$
$D= \begin{pmatrix}
2 & 0 & 0\\
0 & 2 & 0 \\
0 & 0 & -1\\
\end{pmatrix}$
from cayley hamil. $A^3-3A^2+4I=0$ how do we continue?
| A matrix is diagonalisable if and only if its minimal polynomial is the product of distinct factors. Since the characteristic polynomial is $(x-2)^2(x+1)$, we see that $A$ is diagonalisable if and only if the polynomial $(x-2)(x+1)$ is the minimal polynomial.
Calculating this we get
$$\begin{pmatrix}
0&0&0 \\
a&0&0\\
b&c&-3\\
\end{pmatrix}
\begin{pmatrix}
3&0&0 \\
a&3&0\\
b&c&0\\
\end{pmatrix}
=
\begin{pmatrix}
0&0&0 \\
3a&0&0\\
ac&0&0\\
\end{pmatrix}
$$
And this is the zero matrix of and only if $a=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/814792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Values of $n$ for which $\lfloor 2 x\rfloor +\lfloor 4 x\rfloor +\lfloor 8 x\rfloor +\lfloor 20 x\rfloor =n$ has a solution $$\lfloor 2 x\rfloor +\lfloor 4 x\rfloor +\lfloor 8 x\rfloor +\lfloor 20 x\rfloor =n$$
How would you find the values of $n$ for which the equation has a solution under the condition that $n \leq 500$? I haven't really worked with equations involving floors before.
| Let's make a change of variables, that $x \mapsto \tfrac{x}{40}$, so our new function reads:
$$ f(x) = \lfloor \tfrac{x}{20} \rfloor +
\lfloor \tfrac{x}{10} \rfloor +
\lfloor \tfrac{x}{5}\rfloor +
\lfloor \tfrac{x}{2} \rfloor $$
This is the sum of 4 step functions and we notice this function behaves kind of like a line:
$$
\begin{array}{ccc}
f(x + 40) &=&
\lfloor \tfrac{x + 40}{20} \rfloor +
\lfloor \tfrac{x + 40}{10} \rfloor +
\lfloor \tfrac{x + 40}{5}\rfloor +
\lfloor \tfrac{x + 40}{2} \rfloor \\\\
&=&
(\lfloor \tfrac{x}{20} \rfloor + 2)+
(\lfloor \tfrac{x + 40}{10} \rfloor + 4)+
(\lfloor \tfrac{x + 40}{5}\rfloor + 8)+
(\lfloor \tfrac{x + 40}{2} \rfloor + 20) \\\\
&=& \lfloor \tfrac{x}{20} \rfloor +
\lfloor \tfrac{x }{10} \rfloor +
\lfloor \tfrac{x }{5}\rfloor +
\lfloor \tfrac{x }{2} \rfloor + 34 \\\\
&=& f(x) + 34
\end{array}
$$
Notice how the jumps are distributed like the lines on a ruler:
Here is a list of the values they obtain with some repetitions
0, 0, 1, 1, 2, 3, 4, 4, 5, 5, 8, 8, 9, 9, 10, 11, 12,
12, 13, 13, 17, 17, 18, 18, 19, 20, 21, 21, 22, 22, 25, 25, 26, 26,
27, 28, 29, 29, 30, 30
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Ordered pair satisfying the equation $x^2 + 6x + y^2 = 4$ How many ordered pairs of integers $(x,y)$ satisfy the equation $x^2+6x + y^2 = 4$?
I can't approach this problem. Please help me.
| By completing the square we get $(x+3)^2+y^2=13$.
If $x$ and $y$ are integers, so are $x+3$ and $y$ so we require two square numbers that add to $13$. The only option is $9$ and $4$. We can choose which is $9$ and which is $4$ as well as whether we use the positive or negative square roots. Hence $8$ possibilities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Integral $\int_0^\infty \frac{\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}{\sqrt{\alpha^2+x^2}}\sin (\gamma x)\,dx$ I am having trouble showing this equality is true$$
\int_0^\infty \frac{\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}{\sqrt{\alpha^2+x^2}}\sin (\gamma x)\,dx=\sqrt\frac{\pi}{2}\frac{\gamma \exp\big(-\alpha\sqrt{\gamma^2+\beta^2}\big)}{\sqrt{\beta^2+\gamma^2}\sqrt{\beta+\sqrt{\beta^2+\gamma^2}}},
$$
$$
\mathcal{Re}(\alpha,\beta,\gamma> 0).
$$
I do not know how to approach it because of all the square root functions.
It seems if $x=\pm i\alpha \ $ we may have some convergence problems because of the denominator. Perhaps there are ways to solve this using complex methods involving the branch cut from the square root singularity. I just do not know what to choose $f(z)$ for a suitable complex function to represent the integrand.
I also tried differentiating under the integral signs w.r.t $\alpha,\beta,\gamma$ but it did not simplify anything. Thanks. How can we calculate this integral?
| Replace $\alpha$, $\beta$ and $\gamma$ with $a$, $b$ and $c$ respectively.
With the substitution $x=a\sinh t$, the integral can be written as:
$$\begin{aligned}
I & = \sqrt{2a}\int_0^{\infty} e^{-ab\cosh t}\sin(ac\sinh t)\sinh \left(\frac{t}{2}\right)\,dt \\
&=-\sqrt{2a}\Im\left(\int_0^{\infty} e^{-ab\cosh t}\cos\left(ac\sinh t+\frac{it}{2}\right)\,dt \right)
\end{aligned}$$
Thanks to sir O.L. for evaluating the final integral here: Integral: $\int_0^{\infty} e^{-ab\cosh x}\cos\left(ac\sinh(x)+\frac{ix}{2}\right)\,dx$
The result is hence,
$$\begin{aligned}
I & = -\sqrt{2a}\Im\left(e^{-\frac{i}{2}\arctan\frac{c}{b}}\sqrt{\frac{\pi}{2a\sqrt{b^2+c^2}}}e^{-a\sqrt{b^2+c^2}}\right) \\
&=\sqrt{2a}\sqrt{\frac{\pi}{2a\sqrt{b^2+c^2}}}e^{-a\sqrt{b^2+c^2}} \sin\left(\frac{1}{2}\arctan\frac{c}{b}\right) \\
&=\sqrt{\frac{\pi}{2}}\sqrt{\frac{1}{\sqrt{b^2+c^2}}}e^{-a\sqrt{b^2+c^2}}\frac{c}{\sqrt{\left(\sqrt{ b^2+c^2}+b \right)\sqrt{b^2+c^2}}} \\
&=\boxed{\sqrt{\dfrac{\pi}{2}}\dfrac{c\exp\left(-a\sqrt{b^2+c^2}\right)}{\sqrt{b^2+c^2}\sqrt{\sqrt{b^2+c^2}+b}}}
\end{aligned}$$
I used the following to simplify the above expression
$$\begin{aligned}
\sin\left(\frac{1}{2}\arctan\frac{c}{b}\right) &=\sqrt{\frac{1-\cos\left(\arctan\frac{c}{b}\right)}{2}}\\
&= \frac{1}{\sqrt{2}}\sqrt{1-\frac{b}{\sqrt{b^2+c^2}}}\\
&= \frac{1}{\sqrt{2}}\sqrt{\frac{\sqrt{b^2+c^2}-b}{\sqrt{b^2+c^2}}}=\frac{1}{\sqrt{2}}\frac{c}{\sqrt{\left(\sqrt{ b^2+c^2}+b \right)\sqrt{b^2+c^2}}}
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
Find the row echelon form of a 4x4 matrix I want to row reduce the following matrix into an echelon form:
\begin{pmatrix}
-(-\beta+\alpha)^{1/2} & 0 & 1 & 0\\
0& -(-\beta+\alpha)^{1/2}& 0 & 1\\
-\beta & \alpha & -(-\beta+\alpha)^{1/2}&0 \\
\alpha & -\beta & 0 & -(-\beta+\alpha)^{1/2}
\end{pmatrix}
where $\beta = k(1/M + 1/m)$, $\alpha = k/m$
I have no idea. I have been sitting here in the library for two hours doing calculations like $-\beta + \alpha = -k/M$, trying to add rows together, having the $-(-\beta+\alpha)^{1/2} = -i\sqrt{k/M}$, letting $-(-\beta+\alpha)^{1/2} = \gamma$ and I just don't know anymore
| Set $\gamma=-(-\beta+\alpha)^{1/2}$; then your matrix becomes
$$
\begin{pmatrix}
\gamma & 0 & 1 & 0\\
0& \gamma & 0 & 1\\
-\beta & \alpha & \gamma &0 \\
\alpha & -\beta & 0 & \gamma
\end{pmatrix}
$$
Now, multiply the first row by $\gamma^{-1}$:
$$
\begin{pmatrix}
1 & 0 & \gamma^{-1} & 0\\
0& \gamma & 0 & 1\\
-\beta & \alpha & \gamma &0 \\
\alpha & -\beta & 0 & \gamma
\end{pmatrix}
$$
Add to the third row the first multiplied by $\beta$; then add to the fourth row the first multiplied by $-\alpha$:
$$
\begin{pmatrix}
1 & 0 & \gamma^{-1} & 0\\
0& \gamma & 0 & 1\\
0 & \alpha & \gamma+\beta\gamma^{-1} &0 \\
0 & -\beta & -\alpha\gamma^{-1} & \gamma
\end{pmatrix}
$$
Multiply the second row by $\gamma^{-1}$:
$$
\begin{pmatrix}
1 & 0 & \gamma^{-1} & 0\\
0 & 1 & 0 & \gamma^{-1}\\
0 & \alpha & \gamma+\beta\gamma^{-1} &0 \\
0 & -\beta & -\alpha\gamma^{-1} & \gamma
\end{pmatrix}
$$
Add to the third row the second row multiplied by $-\alpha$; add to the fourth row the second row multiplied by $\beta$:
$$
\begin{pmatrix}
1 & 0 & \gamma^{-1} & 0\\
0 & 1 & 0 & \gamma^{-1}\\
0 & 0 & \gamma+\beta\gamma^{-1} &-\alpha\gamma^{-1} \\
0 & 0 & -\alpha\gamma^{-1} & \gamma+\beta\gamma^{-1}
\end{pmatrix}
$$
Now observe that
$$
\gamma+\beta\gamma^{-1}=\frac{\gamma^2+\beta}{\gamma}=\alpha\gamma^{-1}
$$
Thus the matrix you got is
$$
\begin{pmatrix}
1 & 0 & \gamma^{-1} & 0\\
0 & 1 & 0 & \gamma^{-1}\\
0 & 0 & \alpha\gamma^{-1} &-\alpha\gamma^{-1} \\
0 & 0 & -\alpha\gamma^{-1} & \alpha\gamma^{-1}
\end{pmatrix}
$$
Multiply the third row by $\alpha^{-1}\gamma$:
$$
\begin{pmatrix}
1 & 0 & \gamma^{-1} & 0\\
0 & 1 & 0 & \gamma^{-1}\\
0 & 0 & 1 & -1 \\
0 & 0 & -\alpha\gamma^{-1} & \alpha\gamma^{-1}
\end{pmatrix}
$$
Finally, add to the fourth row the third row multiplied by $\alpha\gamma^{-1}$:
$$
\begin{pmatrix}
1 & 0 & \gamma^{-1} & 0\\
0 & 1 & 0 & \gamma^{-1}\\
0 & 0 & 1 & -1 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$
The reduced row echelon form is obtained now by adding to the first row the third row multiplied by $-\gamma^{-1}$:
$$
\begin{pmatrix}
1 & 0 & 0 & \gamma^{-1}\\
0 & 1 & 0 & \gamma^{-1}\\
0 & 0 & 1 & -1 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$
Of course, this assumes that $\gamma\ne0$ (that is $\alpha\ne\beta$) and $\alpha\ne0$. Depending on your assumptions, these cases should be examined separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is it possible to find the limit without l'Hopital's Rule or series Is it possible to find $$\lim_{x\to0}\frac{\sin(1-\cos(x))}{x^2e^x}$$
without using L'Hopital's Rule or Series or anything complex but just basic knowledge (definition of a limit, limit laws, and algebraic expansion / cancelling?)
| We can proceed as follows: $$\begin{aligned}L &= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{x^{2}e^{x}}\\
&= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{1 - \cos x}\cdot\frac{1 - \cos x}{x^{2}e^{x}}\\
&= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{1 - \cos x}\cdot\frac{1 - \cos x}{x^{2}}\cdot \frac{1}{e^{x}}\\
&= \lim_{t \to 0}\frac{\sin t}{t}\cdot\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot \lim_{x \to 0}\frac{1}{e^{x}}\text{ by putting }t = 1 - \cos x \to 0 \text{ as } x \to 0\\
&= 1\cdot\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot 1\\
&= \lim_{x \to 0}\frac{2\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x^{2}}\\
&= 2\cdot 1\cdot\frac{1}{4} = \frac{1}{2}\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/821350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Solving a separable differential equation Solve the differential equation:
$$y'=\frac{1-y^2}{1-x^2}$$
My book says the solution is: $$y=\frac{x+c}{cx+1},$$ where $c$ is a constant.
It's been ten minutes I tried to verify if it was correct but I'm pretty sure the book is wrong. Can someone confirm it?
| $$y'=\frac{cx+1-c(x+c)}{(cx+1)^2}=\frac{1-c^2}{(cx+1)^2},$$
$$\frac{1-y^2}{1-x^2}=\frac{\frac{(cx+1)^2-(x+c)^2}{(cx+1)^2}}{(1-x^2)}=\frac{(c^2-1)x^2+(1-c^2)}{(cx+1)^2(1-x^2)}=\frac{1-c^2}{(cx+1)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/822680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Integrating $(3x + 1) / x^\frac{1}{2}$ I am trying to integrate the following equation, but my answer is different from the textbook and I cannot see where I am going wrong:
\begin{align} \int_1^2\frac{ 3x + 1}{x^{1/2}}dx
&= \int_1^2 \frac{3x}{x^{1/2}}dx + \int_1^2 \frac{1}{x^{1/2}}dx \\
&= \int_1^2 3x^{1/2}dx + \int_1^2 x^{-1/2}dx\\
&= 2x^{3/2}\vert_1^2 + 2x^{1/2}\vert_1^2
\end{align}
The book has an answer $6\sqrt{2} - 4$, but according to what I have just done the answer is roughly $7.07 - 4 = 3.07$.
| Continuing from where you left off, which is correct so far...
\begin{align} 2x^{3/2}\vert_1^2 + 2x^{1/2}\vert_1^2&=[2(2\sqrt{2})-2(1)]+[2(\sqrt{2})-2(1)] \\
&=4\sqrt{2}-2+2\sqrt{2}-2 \\
&=6\sqrt{2}-4\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to arrange $e^3,3^e,e^{\pi},\pi^e,3^{\pi},\pi^3$ in the increasing order? For these six numbers, $e^3,3^e,e^{\pi},\pi^e,3^{\pi},\pi^3$, how to arrange them in the increasing order?
This problem is taken from the today test: National Higher Education Entrance Examination.
I think we can consider
$$f(x)=\dfrac{\ln{x}}{x}$$
or perhaps other methods, so I am looking forward to seeing other methods to solve this problem. Thank you.
| We start by assuming that somehow we know that
$$ \color{green}{e < 3 < \pi}$$.
This immediately gives the relations
\begin{align*}
\color{blue}{e^3 }&\color{blue}{< e^\pi < 3^\pi}\\
\color{blue}{3^e }&\color{blue}{< \pi ^ e < \pi^3}
\end{align*}
Next, we can consider $$f(x) = \frac {\ln x}x$$ as you stated.
Differentiating with respect to $x$ we have
$$ f'(x) = -\frac{\ln x}{x^2} +\frac 1 {x^2} = \frac{1-\ln x}{x^2} < 0\quad \forall x > e.$$
Therefore, for values of $x$ larger than $e$, this function is decreasing, which gives
\begin{align*}
\frac{\ln 3}{3} &> \frac{\ln \pi}{\pi} &\implies\\
\pi \ln 3 &> 3 \ln \pi &\implies\\
\color{blue}{3^\pi} &\color{blue}{> \pi^3},
\end{align*}
and similarly $\color{blue}{e^3 > 3^e}$ and $\color{blue}{e^\pi > \pi^e}$
The two inequalities we have left are $e^\pi$ vs. $\pi^3$ and $\pi^e$ vs. $e^3$. They're taking longer than I thought...
Edit, some years later. (I was browsing through MSE and came across this question again.)
Suppose we can prove that $\color{green}{6 > e + \pi}$. Then we can show the following:
\begin{align*}
6 &> e + \pi & \Leftrightarrow \\
3 - e &> \pi - 3 & \Leftrightarrow \\
3\cdot\left[1 - \frac{e}{3}\right] &> \pi - 3 & \Leftrightarrow
\end{align*}
For all $x > 0$, we have $\log x > 1 - \frac{1}{x}$. From the preceding, we therefore see that $3 \log(3/e) > 3\cdot\left[1 - \frac{e}{3}\right] > \pi - 3$. We thus have
\begin{align*}
3 \log(3/e) &> \pi - 3 &\Leftrightarrow\\
3 \log(3) &> \pi &\Leftrightarrow\\
3 ^{3} &> e^{\pi}.
\end{align*}
Since $\pi^3 > 3^3$, we conclude $\pi^3 > e^{\pi}$. To verify the initial inequality, we can pick our favourite upper bounds for $e$ and $\pi$. Archimedes' $\pi < {22}/{7}$ will do for $\pi$, while for $e$ we can use the fact that
$$
e = 3 - \sum_{k=2}^{\infty} \frac{1}{k!(k-1)k} < 3 - \frac{1}{4}.
$$
For the last inequality, we start by showing that $\color{green}{\pi > \frac{e^2}{2e - 3}} \implies \color{blue}{\pi^e > e^3}$ as long as $\color{green}{2e > 3}$:
\begin{align*}
\pi &> \frac{e^2}{2e - 3} &\Leftrightarrow \\
\left(2e - 3\right) \pi &> e^2 &\Leftrightarrow \\
-\frac{e^2}{\pi} + e &> 3 - e &\Leftrightarrow\\
e\left(1 - \frac{e}{\pi}\right) &> 3 - e.
\end{align*}
We now use the same trick as before, that for all $x$, $\log x > 1 - \frac{1}{x}$
\begin{align*}
e\left(1 - \frac{e}{\pi}\right) &> 3 - e &\Rightarrow \\
e\log {\pi}/{e} &> 3-e &\Leftrightarrow\\
e\log \pi &> 3 &\Leftrightarrow\\
\pi^e &> e^3.
\end{align*}
To finish, we just need to show the initial inequality. The function $f : x \mapsto \frac{x^2}{2x -3}$ is decreasing for $\frac{3}{2}<x<3$, so we'll need a lower bound for $e$. From the series $e = \sum_{k=0}^{\infty} \frac{1}{k!}$ we see $e > {8}/{3}$. For $\pi$, we need a lower bound, and we can again channel Archimedes who proved $\pi > \frac{223}{71}$. We then have
\begin{align*}
\pi > \frac{223}{71} &\operatorname{?} \frac{\frac{64}{9}}{\frac{16}{3} - 3} > \frac{e^2}{2e-3}\\
\pi > \frac{223}{71} &\operatorname{?} \frac{64}{21} > \frac{e^2}{2e-3},
\end{align*}
and the conclusion follows. The final ordering of the six numbers is therefore $\color{blue}{3^e< e^3 < \pi^e < e^\pi < \pi^3 < 3^\pi}$.
As an aside, we've used a number of the properties of $e$ and $\log$ during the derivations, so an incorrect estimate for $e$ could easily throw the order off. However, we haven't used any of the specific properties of $3$ or $\pi$. Therefore, for any numbers $x$, $y$, such that
\begin{align*}
e &< x < y,\\
\frac{e^2}{2e - y} &< x < 2 y - e,\\
y &< 2e,
\end{align*}
the ordering $x^e< e^x < y^e < e^y < y^x < x^y$ will hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
$ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ $\implies$ the triangle is equilateral? If in a triangle $ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ , then is the triangle equilateral ?
| This is a 1990 Russian maths contest problem
$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, so
$$a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $$
$$\Longleftrightarrow$$
$$a^4+b^4+c^4-2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)=0\tag{1}$$
There are several ways to show (1)
In fact
$$a^4+b^4+c^4-2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)$$
$$=(a+b+c)\Big(abc-(-a+b+c)(a-b+c)(a+b-c)\Big)$$
It is easy to show that
$$abc-(-a+b+c)(a-b+c)(a+b-c)=0\Longleftrightarrow a=b=c$$
another method to prove (1)
one can use Schur's inequality
$$\sum a^2(a-b)(a-c)\geq0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Help needed in solving a system of DE The system of DE is:
$$\frac{dI}{db}=-\frac{b}{c}\frac{dJ}{db}-\frac{2ab+1}{2c}J$$
$$\frac{dJ}{db}=\frac{b}{c}\frac{dI}{db}-\frac{2ab-1}{2c}I$$
Assume that $a$ and $c$ are constants and both $I$ and $J$ tends to zero as $b$ tends to $\infty$.
To be honest, I have never solved a system of DE before. I usually use W|A if I am in a situation like this but in this case, even W|A fails. I am not sure if it is even possible to solve the above system.
Any help is appreciated. Many thanks!
| Removing mixed derivative terms is possible by inserting the second equation's $dJ/db$ into the first equation and vice versa:
$$
\frac{dI}{db} = -\frac{b^2}{c^2} \frac{dI}{db} +
\frac{b}{c}\frac{2ab-1}{2c} I -
\frac{2ab+1}{2c} J \\
\frac{dJ}{db} = -\frac{b^2}{c^2} \frac{dJ}{db} -
\frac{2ab-1}{2c} I -
\frac{b}{c}\frac{2ab+1}{2c} J
$$
rearranging gives two coupled equations
$$
\frac{dI}{db} +
\left(-\frac{c^2}{c^2+b^2}\frac{b}{c}\frac{2ab-1}{2c}\right) I =
\left(-\frac{c^2}{c^2+b^2}\frac{2ab+1}{2c}\right) J \quad (*) \\
\frac{dJ}{db} +
\left(\frac{c^2}{c^2+b^2}\frac{b}{c}\frac{2ab+1}{2c}\right) J =
\left(-\frac{c^2}{c^2+b^2}\frac{2ab-1}{2c}\right) I \quad (**)
$$
which one can turn into a vector equation:
$$
\frac{d}{db}
\left[
\begin{array}{c}
I \\
J
\end{array}
\right]
=
\left[
\begin{array}{cc}
\alpha & \beta \\
\gamma & \delta
\end{array}
\right]
\left[
\begin{array}{c}
I \\
J
\end{array}
\right]
$$
with the matrix
$$
A =
\left[
\begin{array}{cc}
\alpha & \beta \\
\gamma & \delta
\end{array}
\right]
=
\left[
\begin{array}{cc}
\frac{b}{c^2+b^2} \frac{2ab-1}{2} &
\frac{c}{c^2+b^2} \frac{2ab+1}{2} \\
-\frac{c}{c^2+b^2} \frac{2ab-1}{2} &
-\frac{b}{c^2+b^2} \frac{2ab+1}{2}
\end{array}
\right]
$$
This is now cast into a form like it is used for dynamical systems, there could be useful tools to get more analytic information.
A graph of the directional vector field would be nice now, with stationary points etc.
The case $a = c = 1$
Here the matrix reduces to
$$
A =
\left[
\begin{array}{cc}
\frac{b}{1+b^2} \frac{2b-1}{2} &
\frac{1}{1+b^2} \frac{2b+1}{2} \\
-\frac{1}{1+b^2} \frac{2b-1}{2} &
-\frac{b}{1+b^2} \frac{2b+1}{2}
\end{array}
\right]
$$
This is the given candidate for $I$:
$$
I^c(b) =
\frac{e^{-\sqrt{1 + b^2}}}{\sqrt{1 + b^2}\sqrt{b + \sqrt{b^2+1}}}
$$
with derivative
$$
\frac{dI^c}{db}(b) =
-\frac{e^{-\sqrt{1+b^2}}
\left(2 b^3 + \left(2 \sqrt{1+b^2} + 3 \right) b^2 +
\left(3 \sqrt{1+b^2}+2\right) b + 1 \right)}
{2 (1 + b^2)^{3/2} \left(\sqrt{1+b^2}+ b \right)^{3/2}}
$$
from equation $(*)$ we get
$$
J^c(b) = \frac{2(1+b^2)}{2b+1}\left(
-\frac{e^{-\sqrt{1+b^2}}
\left(2 b^3 + \left(2 \sqrt{1+b^2} + 3 \right) b^2 +
\left(3 \sqrt{1+b^2}+2\right) b + 1 \right)}
{2 (1 + b^2)^{3/2} \left(\sqrt{1+b^2}+ b \right)^{3/2}}
- \frac{2(1+b^2)}{b(2b-1)}
\frac{e^{-\sqrt{1 + b^2}}}{\sqrt{1 + b^2}\sqrt{b + \sqrt{b^2+1}}}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/824304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$ Please help me to prove this trigonometric equation.
$\cos \left( 36^\circ \right)-\cos \left( 72^\circ \right) = \frac{1}{2}$
Thank you.
| As $\displaystyle\cos(180^\circ-y)=-\cos y$
$$S=\cos36^\circ-\cos72^\circ=\cos36^\circ+\cos(180^\circ-108^\circ)=\cos36^\circ+\cos108^\circ$$
Now multiplying the numerator & the denominator by $\displaystyle2\sin\frac{(108-36)}2^\circ,$
$$S=\frac{2\sin36^\circ\cos36^\circ+2\sin36^\circ\cos108^\circ}{2\sin36^\circ}$$
Using Werner Formula & $\sin2x$ formula,
$$S=\frac{\sin72^\circ+\sin144^\circ-\sin72^\circ}{2\sin36^\circ}=\frac{\sin(180^\circ-36^\circ)}{2\sin36^\circ}$$
Hope you can take it home from here
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$\frac{1}{\sin^2(x)}+\frac{1}{\cos^2(x)} = \cos^2(x)+\sin^2(x) ?$ This is a simple question. Since $\cos(\theta)^2 + \sin(\theta)^2 = 1$
Can I take the inverse of this $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \frac{1}{1}$?
Finally getting $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \cos(\theta)^2 + \sin(\theta)^2$
If this is incorrect thinking can someone please put me on the right path?
Thanks
| The (multiplicative) inverse of $\sin^2\theta+\cos^2\theta$ is
$$\frac{1}{\sin^2\theta+\cos^2\theta},$$
and in general
$$\frac{1}{a+b}\neq \frac{1}{a}+\frac{1}{b},$$
for 'numbers' $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
formula for number triangles Hi, I have a triangle starting from $0$ and going up by one on the bottom row until there are $r$ items on the bottom row and there are $r$ rows a number is formed by adding the two numbers towards the fat end of the triangle together.
Example:
$$x \downarrow\ $$
$$
\begin{matrix}12\end{matrix}\\
\begin{matrix}4 & 8\end{matrix}\\
\begin{matrix}1 & 3 & 5\end{matrix}\\
\begin{matrix}0 & 1 & 2 & \boxed{3}\end{matrix}
$$
$$\begin{matrix}& & r - 1 \uparrow\end{matrix}$$
$\therefore$ when $r = 4$, $x = 12$.
When $r = 8$
$$
\begin{matrix}0\ \ & 1\ \ & 2\ \ & 3\ \ & 4\ \ & 5\ \ & 6\ \ & 7\ \ \end{matrix}\\
\begin{matrix}1\ \ & 3\ \ & 5\ \ & 7\ \ & 9\ \ & 11\ & 13\ \end{matrix}\\
\begin{matrix}4\ \ & 8\ \ & 12\ & 16\ & 20\ & 24\ \end{matrix}\\
\begin{matrix}12\ & 20\ & 28\ & 36\ & 44\ \end{matrix}\\
\begin{matrix}32\ & 48\ & 64\ & 80\ \end{matrix}\\
\begin{matrix}80\ & 112 & 144 \end{matrix}\\
\begin{matrix}192 & 256 \end{matrix}\\
\begin{matrix}448\end{matrix}\\
$$
$\therefore x = 448$
How do I work out $x$ from $r$? What is the formula? Any help by suggesting a new viewpoint or formula(e) would be greatly appreciated. Thank you.
| Consider the first two 'columns'. Notice any patterns?
$$0 \space 1\\1 \space 3 \\4 \space 8\\ 12\space 20$$
What is the rule?
$$a_{i+1} = a_i + b_i$$
$$b_{i+1} = a_{i+1} + 2 (b_i - a_i) = 3b_i - a_i$$
Try find a formula for $a_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Using Cylindrical Coordinates to find volume of a solid How do I use Cylindrical Coordinates to find volume of a solid in the first Octant that is bounded by the cylinder $x^2 + y^2 = 2y$, the half cone $z = \sqrt{x^2 + y^2}$, and the $xy$-plane.
I have drawn the region of integration and obtained this:
$\int_0^2 \int_0^\sqrt{2y-y^2}\int_0^\sqrt{x^2 + y^2} dzdxdy$
Is this correct and from here were do I apply the cylindrical coordinates?
| The question makes only sense if the solid is the volume outside the half cone.
Otherwise the volume is infinite, because (almost) half of the cylinder which extends
to $+\infty$ in $z$ is in it. So we have to integrate over $x^2+y^2\ge z^2$.
For constant $z$ the section through the solid
is the area inside the circle $x^2+y^2-2y=0$ defined by the cylinder, which is inside the circle $x^2+(y-1)^2=1$,
but outside the circle $z^2=x^2+y^2$ defined by the cone. And since we're supposed to look only at one octant,
$z\ge 0, x\ge 0, y\ge 0$. The intersection of the 2 circles vanishes if the radius of
the circle at (0,0) is too large, which means $z\le 2$.
$$
V=\int_0^2 dz \int\int_{x^2+y^2\ge z^2, x^2+(y-1)^2\le 1} dx dy
$$
[Note that the lower limit in the partial integral $\int_0^{\sqrt{2y-y^2}}dx$ of the OP is wrong.]
In cylindrical coordinates $r^2=x^2+y^2$, $x=r\cos\theta, y=r\sin\theta$
we find the limits of $\theta=[\theta_0,\pi/2]$ where the circle $x^2+y^2=r^2$ intersects
the circle $x^2+(y-1)^2=1$:
$$
r^2\cos^2\theta_0+(r\sin\theta_0-1)^2=1;
$$
$$
r^2-2r\sin\theta_0+1=1;
$$
$$
\to \theta_0=\arcsin (r/2);
$$
So with the usual Jacobi Determinant factor $r$ in cylindrical coordinates
$$
V=\int_0^2 dz \int_z^2 r dr \int_{\arcsin(r/2)}^{\pi/2}d\theta
$$
$$
=\int_0^2 dz \int_z^2 r dr (\pi/2 - \arcsin(r/2))
$$
Here $\int r\arcsin (r/2) dr= (r^2/2-1)\arcsin(r/2)+r\sqrt{1-r^2/4}/2$, so
$$
V=\int_0^2 dz \left[\frac{\pi}{2} \frac{r^2}{2}-\left((\frac{r^2}{2}-1)\arcsin\frac{r}{2}+\frac{r\sqrt{1-r^2/4}}{2}\right)\right]\mid_{r=z}^2
$$
$$
=\int_0^2 dz \left[\frac{\pi}{2} - \frac{\pi}{2} \frac{z^2}{2}+\left((\frac{z^2}{2}-1)\arcsin\frac{z}{2}+\frac{z\sqrt{1-z^2/4}}{2}\right)\right]
$$
and this is evaluated via
$$
\int (\frac{z^2}{2}-1)\arcsin(z/2) dz = \frac{z^3}{6}\arcsin\frac{z}{2}-z\arcsin\frac{z}{2}+\frac{z^2-10}{9}\sqrt{1-z^2/4}
$$
and
$$
\int \frac{z}{2}\sqrt{1-z^2/4} dz= -\frac{1}{12}(4-z^2)^{3/2}.
$$
$$
V=16/9.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/830530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the integrals $\int \sin{x} \cot^2{x} \,dx$ and $\int \cos{x} \cot^2{x} \,dx$. Can you please show how to evaluate the integrals $$\int \sin{x} \cot^2{x} \,dx$$ and $$\int \cos{x} \cot^2{x} \,dx.$$
I know that $\cot x=\dfrac{\cos x}{\sin x}$, which can simplify the integrands a bit.
$$\sin x\cot^2x = \frac{\cos^2x}{\sin x}\\
\cos x\cot^2x = \frac{\cos^3x}{\sin^2 x}$$
But I still do not know where to continue from there.
| Hint :
\begin{align}
\int\sin x\cot^2x\ dx&=\int\sin x\cdot\frac{\cos^2x}{\sin^2x}\ dx\\
&=\int\frac{\cos^2x}{\sin x}\ dx\\
&=\int\frac{1-\sin^2x}{\sin x}\ dx\\
&=\int \frac1{\sin x}\ dx-\int\sin x\ dx\\
&=\color{red}{\int \frac{\sin x}{1-\cos^2 x}\ dx}-\int\sin x\ dx\tag1
\end{align}
and
\begin{align}
\int\cos x\cot^2x\ dx&=\int\cos x\cdot\frac{\cos^2x}{\sin^2x}\ dx\\
&=\int\cos x\cdot\frac{1-\sin^2x}{\sin^2x}\ dx\tag2
\end{align}
For $(1)$ take substitution $\color{red}{u=\cos x}$ and for $(2)$ take substitution $v=\sin x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/830609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How would I factor $a^3+b^3+c^3-6abc$ How would I factor the polynomial $a^3+b^3+c^3-6abc$? The values are homogenous, so so must be the factors. I don't know where to go from there.
| Try solving the cubic for $c=x+y$ so that $c^3=(x^3+y^3)+3xy(x+y)=(x^3+y^3)+3xyc$
Whence $x^3+y^3=-a^3-b^3$ and $xy=2ab$ so that $x^3y^3=8a^3b^3$ and $x^3, y^3$ are roots of the quadratic $$z^2+(a^3+b^3)z+8a^3b^3$$ so that $$x^3,y^3=\frac {-a^3-b^3\pm \sqrt {a^6+b^6-30a^3b^3}}2$$
And use this to build your factors $(c-x-y)(c-\omega x-\omega^2 y)(c-\omega^2 x - \omega y)$ where $\omega$ is a non-trivial cube root of $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/831254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is there a simpler/better proof of this simple trigonometric property? The sine function has the following nice property : for any
$x,y$, we have $\sin(x)+\sin(y)=\sin(x+y)$ iff at least one of
$x,y,x+y$ is $0$ modulo $2\pi$.
I sketch below my current proof of it, which I find somewhat
unsatisfying. Does anyone know a better proof ?
Let $\xi=\sin(x)$ and $\eta=\sin(y)$. Then
$\sin(x+y)=\xi\cos(y)+\eta\cos(x)$, so
$$
\begin{array}{lcl}
\sin^2(x+y)&=&\xi^2(1-\eta^2)+\eta^2(1-\xi^2)+2\xi\eta\cos(x)\cos(y) \\
&=& \xi^2+\eta^2-2\eta^2\xi^2+2\xi\eta\cos(x)\cos(y)
\end{array} \tag{1}$$
Whence
$$
\big(\sin^2(x+y)-(\xi^2+\eta^2-2\eta^2\xi^2)\big)^2=
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \tag{2}
$$
If $\sin(x)+\sin(y)=\sin(x+y)$, it follows then that $A(\xi,\eta)=0$
where
$$
\begin{array}{lcl}
A(\xi,\eta)&=&\big((\xi+\eta)^2-(\xi^2+\eta^2-2\eta^2\xi^2)\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& \big(2\xi\eta-2\eta^2\xi^2\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& \big(2\xi\eta\big(1-\eta\xi\big)\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& 4(\xi\eta)^2 \Bigg(\big(1-\eta\xi\big)^2-(1-\xi^2)(1-\eta^2) \Bigg) \\
&=& 4(\xi\eta)^2 \Bigg(\big(1-2\eta\xi+\eta^2\xi^2\big)-\big(1-\xi^2-\eta^2+\eta^2\xi^2\big) \Bigg) \\
&=& 4(\xi\eta)^2 \big(\xi-\eta\big)^2 \\
\end{array} \tag{3}$$
So one of $\xi,\eta,\xi-\eta$ must be zero. A little more case analysis then shows
that in the end one of $x,y,x+y$ must be
zero modulo $2\pi$ as wished.
| Well here is (yet another) solution, this using only real variables.
By the sum formula we have
$$\sin x(1-\cos y)+\sin y(1-\cos x)=0$$
multiplying by $(1+\cos x)(1+\cos y)$ we get
$$\sin x \sin y [\sin y(1+\cos x)+\sin x(1+\cos y)]=0$$ so
either $\sin x=0$ or $\sin y=0$ or
$$\sin y(1+\cos x)+\sin x(1+\cos y))=0$$
which is equivalent to
$$\sin x +\sin y +\sin(x+y) =0$$
so $$\sin(x+y) =0$$
Then it is easy to derive the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/834321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Prove summation related to cycles Let $b_r(n,k)$ be the number of n-permutations with $k$ cycles, in which numbers $1,2,\dots,r$ are in one cycle.
Prove that for $n \geq r $ there is:
$$
\sum_{k=1}^{n} {b_r(n,k)x^k=(r-1)!\frac{x^\overline{n}}{(x+1)^\overline{r-1}}}
$$
| As Brian Scott points out I have the wrong interpretation of the
question. We want permutations where $1,2,\ldots r$ are on one cycle
plus possibly some other elements, say $q.$
We obtain
$$b_r(n, k) = \sum_{q=0}^{n-r} {n-r\choose q}
\frac{(r+q)!}{r+q} \left[n-r-q\atop k-1\right].$$
Recall the bivariate generating function of the Stirling numbers of
the first kind, which is
$$G(z, u) = \exp\left(u\log\frac{1}{1-z}\right).$$
Observe that when we multiply two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the product of the two generating functions is the generating
function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
In the present case we have
$$A(z) = \sum_{q\ge 0} (r+q-1)! \frac{z^q}{q!}
= (r-1)! \sum_{q\ge 0} {r+q-1\choose q} z^q
\\ = (r-1)! \frac{1}{(1-z)^r}$$
and
$$B(z) = \sum_{q\ge 0} \left[q\atop k-1\right] \frac{z^q}{q!}
= \frac{1}{(k-1)!}
\left(\log\frac{1}{1-z}\right)^{k-1}.$$
We get for the sum
$$\sum_{k=1}^n b_r(n, k) x^k
\\ = (n-r)! [z^{n-r}] (r-1)! \frac{1}{(1-z)^r}
\sum_{k=1}^n x^k
\frac{1}{(k-1)!}
\left(\log\frac{1}{1-z}\right)^{k-1}.$$
We can certainly extend the sum to infinity as we are extracting the
coefficient on $[z^{n-r}]:$
$$x (n-r)! [z^{n-r}] (r-1)! \frac{1}{(1-z)^r}
\sum_{k=1}^\infty x^{k-1}
\frac{1}{(k-1)!}
\left(\log\frac{1}{1-z}\right)^{k-1}
\\ = x (n-r)! [z^{n-r}] (r-1)! \frac{1}{(1-z)^r}
\exp\left(x\log\frac{1}{1-z}\right)
\\ = x (n-r)! [z^{n-r}] (r-1)! \frac{1}{(1-z)^r} \frac{1}{(1-z)^x}
\\ = x (n-r)! [z^{n-r}] (r-1)! \frac{1}{(1-z)^{x+r}}.$$
This yields
$$x (n-r)! (r-1)! {x+r-1+n-r\choose n-r}
\\ = x (n-r)! (r-1)! {x+n-1\choose n-r}
\\ = (r-1)! \times x\times (x+n-1)^{\underline{n-r}}
= (r-1)! \times x\times (x+r)^{\overline{n-r}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/834829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the orthogonal trajectories of the family of curves given by $x^2 + y^2 + 2Cy =1$. Find the orthogonal trajectories of the family of curves given by $$x^2 + y^2 + 2Cy =1.$$
The ordinary differential equation for the family of curves is given by $y'=\frac{2xy}{x^2-y^2-1}$.Therefore, the differential equation for the orthogonal curves is given by $y'=\frac{1-x^2-y^2}{2xy}$.
This is an exact differential equation. So, solving by the standard method for exact differential equation gives $x-x^3/3+xy^2=C$. But this is not the correct answer according to the answers given at back of the book.
The answer given at the back of the book is $x^2 - y^2 - Cx +1 = 0$.
Can someone please find out at which step I am maing a mistake or provide a solution that leads to the correct answer?
| For the given family of curves
$$
x^2+y^2+2cy=1
$$
which can be written as
$$
\frac{1-x^2-y^2}{2y} = c
$$
differentiating both sides
$$
\frac{dy}{dx}=\frac{2xy}{x^2-y^2-1}
$$
and the family of curves orthogonal to this will be
$$
-\frac{dx}{dy}=\frac{2xy}{x^2-y^2-1}
$$
or,
$$(x^2-y^2-1)dx+(2xy)dy=0
$$
this is not an exact DE, and hence will have to converted to one,
$$
\frac{\partial{M}}{\partial{y}}=-2y; \frac{\partial{N}}{\partial{x}}=2y
$$
$$
\frac{1}{N}\bigg(\frac{\partial{M}}{\partial{y}}-\frac{\partial{N}}{\partial{x}}\bigg)=-\frac{2}{x}
$$
$$
I.F=e^{\int{-\frac{2}{x}}dx}=\frac{1}{x^2}
$$
after multiply with the Integrating Factor the DE becomes exact
$$
\bigg(1-\frac{y^2}{x^2}-\frac{1}{x^2}\bigg)dx+\frac{2y}{x}dx=0
$$
solving for this,the orthogonal family of curves is
$$
x^2+y^2+1=cx
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/834911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculation of $ \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots(1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $ Calculation of $\displaystyle \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots (1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $
My Trial After simplification, we get $$\displaystyle \lim_{x\rightarrow 1}\frac{(1-x^{n+1})\cdot(1-x^{n+2})\cdot(1-x^{n+3})\cdots(1-x^{2n})}{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)}$$
Now I did not understand How can I solve after that, Help me
Thanks
| By L'Hospital's rule,
\begin{align*}
\lim_{x \to 1} \frac{1 - x^{n + k}}{1 - x^k} &= \lim_{x \to 1} \frac{-(n + k) x^{n + k}}{-k x^k} = \frac{n + k}{k}
\end{align*}
Hence the desired limit is
$$\frac{n + 1}{1} \cdot \frac{n + 2}{2} \cdot \frac{n + 3}{3} \cdots \frac{n + n}{n} = \frac{(2n)!}{(n!)^2}$$
Alternatively, factor out a common term of $1 - x$ to find
$$\frac{1 - x^{n + k}}{1 - x^k} = \frac{1 + x + x^2 + \dots + x^{n + k - 1}}{1 + x + x^2 + \dots + x^{k - 1}}$$
Now directly evaluate at $1$ to get $n + k$ in the numerator and $k$ in the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/836841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$\alpha$ and $\beta$ are solution of $a \cdot \tan\theta + b \cdot \sec\theta = c$ show $ \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2}$ If $\alpha$ and $\beta$ are the solution of $$a \cdot \tan\theta + b \cdot \sec\theta = c$$, then show that $$ \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2}$$
I did the following:
$$a \cdot \tan\theta + b \cdot \sec\theta = c$$
Squaring both sides, $$a^2 \cdot \tan^2\theta + b^2 \cdot \sec \theta + 2 \cdot a \cdot b \cdot \tan \theta \cdot \sec \theta = c^2$$
This implies $$(a^2 + b^2) \cdot \sin^2 \theta + 2 \cdot a \cdot b \cdot \sin \theta + (b^2 - c^2) = 0$$
Hence I get $$\sin \alpha + \sin \beta = \frac{-2ab}{a^2 + c^2}$$ and $$\sin \alpha \cdot \sin \beta = \frac{b^2 - c^2}{a^2 - c^2}$$
But I am not able to solve it further.
| Using Double-Angle Formula,
$$\sec\theta=\dfrac{1+t^2}{1-t^2},\tan\theta=\dfrac{2t}{1-t^2}$$ where $t=\tan\dfrac\theta2$ to form a Quadratic Equation in $t$ whose roots will be $$\tan\dfrac\alpha2,\tan\dfrac\beta2$$
Can you find $$\tan\left(\dfrac\alpha2+\dfrac\beta2\right)$$ and subsequently, $$\tan(\alpha+\beta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/838082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculus Area Cubic curve Why area bounded between the "line $AB$" and the "cubic curve" and area bounded between the "line $BC$" and the "cubic curve" is $16$ times?
| Given is
$$
f(x) = x^3
$$
For the point $A$, $x=x_o$, we have a tangent
$$
3 x_o^2
$$
So that line is given by
$$
y = \Big(3 x - 2 x_o \Big) x_o^2
$$
The intersection is given by the equation
$$
x^3 - 3 x x_o^2 + 2 x_o^3 = 0
$$
which can be written as
$$
\Big( x - x_o \Big)^2 \Big( x + 2 x_o \Big) = 0
$$
As for point $A$ we have $x = x_o$, we get $x = - 2 x_o$ for point $B$.
For the point $B$, $x =-2 x_o$, we have a tangent
$$
12 x_o^2
$$
So that line is given by
$$
y = 4 \Big(3 x - 4 x_o \Big) x_o^2
$$
The intersection is given by the equation
$$
x^3 - 12 x x_o^2 - 16 x_o^3 = 0
$$
which can be written as
$$
\Big( x + 2 x_o \Big)^2 \Big( x - 4 x_o \Big) = 0
$$
As for point $B$ we have $x = - 2x_o$, we get $x = 4 x_o$ for point $C$.
We now have two surfaces
$$
S_{BA} = \int_{-2x_o}^{+x_o} \Big( x^3 - 3 x x_o^2 + 2 x_o^3 \Big) d x
= x_o^2 \int_{-2}^{+1} \Big( \xi^3 - 3 \xi + 2 \Big) d \xi
$$
and
$$
S_{BC} = \int_{-2x_o}^{+4x_o} \Big( - x^3 + 12 x x_o^2 + 16 x_o^3 \Big) d x
= x_o^2 \int_{-2}^{+4} \Big( - \xi^3 + 12 \xi + 16 \Big) d \xi
$$
The ratio of these surfaces is constant and is given by
$$
\frac{ \int_{-2}^{+4} \Big( - \xi^3 + 12 \xi + 16 \Big) d \xi }
{ \int_{-2}^{+1} \Big( \xi^3 - 3 \xi + 2 \Big) d \xi } = \frac{108}{27/4} = 16
$$
So the ratio of these surfaces is
$$
16
$$
for any point $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/838480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \sqrt{2\pi}$ I wish to show
$$\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \sqrt{2\pi}$$
I've tried substitution and integration by parts to get a recursive formula for the integral, but so far it hasn't worked. Any help would be appreciated, thanks.
| I will present you the main steps to a final solution and the details on each step can be your job.
First notice that a simple substiution $n=2k$ gives
$$\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \lim_{k\rightarrow \infty} \int_0^{\pi/2} 2^{2k} \sqrt{2k} \sin^{2k}(x) \cos^{2k-2}(x) \; dx$$
Now elaborating with the new expression in the following way we have
$$2^{2k-1} \sqrt{2k}\cdot2\int_0^{\pi/2}\sin^{2(k+1/2)-1}(x) \cos^{2(k-1/2)-1}(x) \; dx$$
\begin{eqnarray}
&=& 2^{2k-1}\sqrt{2k}\cdot\beta(k+1/2,k-1/2) \\
&=& 2^{2k-1}\sqrt{2k}\frac{\Gamma(k+1/2)\Gamma(k-1/2)}{\Gamma(2k)} \\
&=& 2^{2k}\sqrt{2k}\frac{2k}{2k-1}\cdot\frac{\Gamma(k+1/2)^2}{\Gamma(2k+1)} \\
&=& 2^{2k}\sqrt{2k}\frac{2k}{2k-1}\cdot\frac{(\frac{(2k-1)!!\sqrt{\pi}}{2^{k}})^{2}}{(2k)!} \\
&=& \pi\cdot\sqrt{2k}\frac{2k}{2k-1}\cdot\frac{(2k-1)!!}{(2k)!!} \\
&=& \pi\sqrt{\frac{2k}{2k+1}}\frac{2k}{2k-1}\cdot\sqrt{\frac{(2k-1)!!^2(2k+1)}{(2k)!!^2}} \\
&=& \;\pi\sqrt{\frac{2k}{2k+1}}\frac{2k}{2k-1}\sqrt{\prod_{j=1}^{k}\frac{(2j-1)}{(2j)}\frac{(2j+1)}{(2j)}} \\
& \rightarrow & \pi\cdot\sqrt{1}\cdot1\sqrt{\prod_{j=1}^{\infty}\frac{(2i-1)}{(2i)}\frac{(2i+1)}{(2i)}} \\
&=&\pi\cdot\sqrt{\frac{2}{\pi}} \\
&=& \sqrt{2\pi},\;k\rightarrow \infty
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/839343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integration by parts or substitution for $\int x \cos (x^2) dx$ With $$\int x \cos (x^2) dx$$ I can use substitution to solve this integral.
$$u=x^2, dx =\frac{du}{2x}$$
$$\int x \cos (x^2) dx = \int x \cos (x^2) \frac{du}{2x} = \frac{1}{2} \int \cos(u) = \frac{1}{2}\sin(u)=\frac{1}{2}sin(x^2)+C$$
But can't I also solve it with integration by parts and should get the same result?
$$\begin{array} \int x\cos (x^2) dx &= \frac{1}{2}x\sin(x^2) - \frac{1}{2}\int 1\sin(x^2)dx
\\
&= \frac{1}{2}x\sin(x^2) + \frac{1}{4}\cos(x^2)+C
\end{array}$$
Or am I mixing up things?
| $$\int \sin(x^2)\,dx \neq -\frac{1}{2}\cos(x^2)+C$$
$$\int \cos(x^2)\,dx \neq \frac{1}{2}\sin(x^2)+C$$
You can confirm that by differentiating the RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/839823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Local minimum of $f(x) = 4x + \frac{9\pi^2}{x} + \sin x$ What's the minimum value of the function
$$f(x) = 4x + \frac{9\pi^2}{x} + \sin x$$
for $0 < x < +\infty$? The answer should be $12\pi - 1$, but I get stuck with the expression involving both $\cos x$ and $x^2$ in the derivative. Taking the derivative, we have:
$$f'(x) = 4 - \frac{9\pi^2}{x^2} + \cos x.$$
In order to find the local extrema of the function, we set $f'(x) = 0$. Therefore,
\begin{align}
4 - \frac{9\pi^2}{x^2} + \cos x &= 0 \\
4x^2 - 9\pi^2 + x^2 \cos x &= 0 \\
x^2 (4 + \cos x) &= 9\pi^2.
\end{align}
However, I'm not sure what to do from here or if, indeed, I'm doing it right at all. Any help would be appreciated.
| By AM-GM, $4x + \dfrac{9\pi^2}{x} \ge 2\sqrt{4x \cdot \dfrac{9\pi^2}{x}} = 12\pi$, with equality iff $4x = \dfrac{9\pi^2}{x}$, i.e. $x = \dfrac{3\pi}{2}$.
Also, $\sin x \ge -1$, with equality iff $x = \dfrac{3\pi}{2} + 2\pi k$ for some integer $k$.
Therefore, $f(x) = 4x + \dfrac{9\pi^2}{x} + \sin x \ge 12\pi - 1$ with equality iff $x = \dfrac{3\pi}{2}$.
Note: This solution does use the fact that the domain is $x > 0$, as stated in the question.
| {
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Show that $\lim_{x \rightarrow 1} \frac{x^4-2x+1}{x-1} + \sqrt{x} =3$
Show that $\lim_{x \rightarrow 1} \frac{x^4-2x+1}{x-1} + \sqrt{x} =3$ from the definition (using $\epsilon-\delta$)
Why can't I do something like this?
We want:
$|\frac{x^4-2x+1}{x-1} + \sqrt{x}-3| = |\frac{x^4-2x+1}{x-1} + \frac{x-9}{\sqrt{x}+3}| \le |\frac{x^4-2x+1}{x-1}| + |\frac{x-9}{\sqrt{x}+3}| \le |x^3+x^2+x-1| + |x-9| \le |x|^3 + |x|^2+|x-1| + |x-9| < \epsilon$
Now suppose $|x-1| < 1$, then $|x|\le 2$ and $|x-9| \le 9$, so
$$|x|^3 + |x|^2+|x-1| + |x-9| < 2^3+2^2+9+|x-1| < \epsilon$$
Now take $\delta = \min\{1, \epsilon -21\}$.
But this clearly is wrong because $\delta$ needs to be positive for every $\epsilon>0$, what part of my working is incorrect? (I know how to get the answer properly, just wondering why this way is incorrect, I'm quite new to $\epsilon-\delta$ proofs)
| You went off the track when you introduced the term
$$\left|\frac{x-9}{\sqrt x+3}\right|\ .$$
As $x\to1$, this term will tend to $2$ and therefore it can't be made arbitrarily small. Something like
$$\eqalign{\left|\frac{x^4-2x+1}{x-1}+\sqrt{x}-3\right|
&\le\left|\frac{x^4-2x+1}{x-1}-2\right|+\left|\sqrt{x}-1\right|\cr
&=\left|x-1\right|\left|x^2+2x+3\right|+\left|\frac{x-1}{\sqrt x+1}\right|\cr}$$
would have a better chance of working.
| {
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Algebraic manipulation of floors and ceilings I am trying to solve the summation
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \sum_{i=3}^{n} \left\lceil\frac{i-2}{2}\right\rceil
$$
I will list some of the simplifications that I've found so far, and then get around to asking my question. Please feel free to point out any logical errors that I have (in all likelihood) made.
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \sum_{i=1}^{n-2} \left\lceil\frac{i}{2}\right\rceil
$$
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \left(\sum_{i=1}^{\left\lceil\frac{n-2}{2}\right\rceil} i+ \sum_{i=1}^{\left\lfloor\frac{n-2}{2}\right\rfloor} i\right)
$$
Since the upper bound in both of the inner summations will be an integer, I have:
$$
\sum_{n=3}^{\infty} \frac{1}{2^n} \left(\frac{\left\lceil\frac{n-2}{2}\right\rceil\left(\left\lceil\frac{n-2}{2}\right\rceil + 1\right)}{2}+\frac{\left\lfloor\frac{n-2}{2}\right\rfloor\left(\left\lfloor\frac{n-2}{2}\right\rfloor + 1\right)}{2}\right)
$$
How do I simplify a floor times a floor and a ceiling times a ceiling algebraically?
I'm thinking that the identity
$$
n = \left\lceil\frac{n}{2}\right\rceil + \left\lfloor\frac{n}{2}\right\rfloor
$$
might come in handy, but I have very little experience manipulating floors and ceilings algebraically. Let me be clear: I'm not looking for an answer to the summation, just guidance in simplifying this current step.
| Start with the generating function for $\displaystyle a_i=\left\lceil \frac{i-2}{2}\right\rceil\; , i\ge 3$
which turns out to be
\begin{align*}
G(x) &= \frac{x^3}{1-x-x^2+x^3}
\end{align*}
and for the sum of the coefficients, it's
\begin{align*}
H(x) &= \frac{1}{1-x}\, G(x)
\end{align*}
and the required answer is $\displaystyle H\left(\frac{1}{2}\right)$
| {
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Existence of some type matrix Is there square matrix $A$ of size $3$ with real entries such that
$$
\operatorname{tr}(A)=0\text{ and }A^2+A^T=I.
$$
I have proved that there is not with size $2$ using definition of "trace", but for size $3$ it becomes complicated.
Here is the sketch of the proof for $2$.
$$
a_{11}+a_{22}=0,\\
a_{11}^2+a_{12}a_{21}+a_{11}=1,\\
a_{11}a_{12}+a_{12}a_{22}+a_{21}=0,\\
a_{21}a_{11}+a_{21}a_{22}+a_{12}=0,\\
a_{12}a_{21}+a_{22}^2+a_{22}=1.
$$
Putting $a_{11}=-a_{22}$ it is easy to see that above inequalities can't be true together.
Thanks!
| The answer is no.
If we substitute $A^T=I-A^2$ into the transpose of $A^2+A^T=I$, namely $(A^T)^2+A=I$, we get $$(I-A^2)^2+A=I$$
Hence $I-2A^2+A^4+A=I$, or $$A^4-2A^2+A=0 ~~~~~~(\star)$$
must be satisfied by our matrix $A$. This has just four roots: $0,1,\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}$.
Now, consider $B=QAQ^{-1}$, which is $A$ in Jordan canonical form. By multiplying $(\star)$ on the left and right by $Q, Q^{-1}$ respectively, we must also have $$B^4-2B^2+B=0 ~~~~(\dagger)$$
$B$ has the same eigenvalues as $A$, which must sum to $0$ (counted by multiplicity) by the first condition. Each block of $B$ must separately verify $(\dagger)$, hence the only possible nonzero $B$, up to reordering of the three blocks, is $$B=\left(\begin{smallmatrix}1&0&0\\0&\frac{-1+\sqrt{5}}{2}&0\\0&0&\frac{-1-\sqrt{5}}{2}\end{smallmatrix}\right)$$
Now, we plug $A=Q^{-1}BQ$ into $A^2+A^T=I$, to get $$Q^{-1}B^2Q+Q^TB^T(Q^{-1})^T=I$$
Multiplying on the left and right by $Q$ and $Q^{-1}$ respectively, and set $R=QQ^T$, to get
$$B^2+RBR^{-1}=I$$
It turns out that $$I-B^2=\left(\begin{smallmatrix}0&0&0\\0&\frac{-1+\sqrt{5}}{2}&0\\0&0&\frac{-1-\sqrt{5}}{2}\end{smallmatrix}\right)$$
But since $RBR^{-1}=(I-B^2)$, $B$ and $I-B^2$ must have the same eigenvalues. They have two in common, but not all three -- $B$ has $1$ while $I-B^2$ has $0$. Hence no such $A$ exists.
| {
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Show that $\sum_{k=0}^{\infty}k^2x^k= \frac{x(1+x)}{(1-x)^3}$ for $0 < |x| < 1$ Show that $$\sum_{k=0}^{\infty}k^2x^k= \frac{x(1+x)}{(1-x)^3}$$ for $0 < |x| < 1$.
(This is appendix question A.1-3 from Introduction to Algorithms by Cormen)
| To find the required sum you could just find the sum of the geometric progression $1,x,x^2,x^3,... \space where \space |x|\lt 1,$ and then double differentiate it .$$\sum_{k=0}^\infty x^k=\frac 1{1-x}\\=>\frac d{dx}\left(\sum_{k=0}^\infty x^k\right)=\frac d{dx}\left(\frac 1{1-x}\right)\\=>\sum_{k=0}^\infty kx^{k-1}=\frac 1{(1-x)^2}\\=>\sum_{k=0}^\infty kx^k=\frac x{(1-x)^2}\\=>\frac d{dx}\left(\sum_{k=0}^\infty kx^k\right)=\frac d{dx}\left(\frac x{(1-x)^2}\right)\\=>\sum_{k=0}^\infty k^2x^{k-1}=\frac{(1-x)^2+2(1-x)x}{(1-x)^4}\\=>\sum_{k=0}^\infty k^2x^{k-1}=\frac{1+x}{(1-x)^3}\\=>\sum_{k=0}^\infty k^2x^k=\frac {x(1+x)}{(1-x)^3}$$
| {
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Sum of two odd squares What is the characterization of numbers expressible as a sum of two odd squares? I showed that it must be congruent to $2$ mod $4$, but obviously this is insufficient since not all numbers $2$ mod $4$ can be written this way.
| Analyzing this question, I came to the startling discovery:
If we define $S(X)$ as "X is the sum of two squares", then, for any integer $X$,
$$
S(X) \iff S(2X)
$$
We'll get back to that later. For this question, we have $X = a^2 + b^2$, where $a$ and $b$ are odd integers.
Let
$$
c = {a + b\over 2}\\
d = {a - b\over 2}
$$
Since $a$ and $b$ are both odd, $c$ and $d$ must both be integers. (We don't care if $d$ is negative, because $d^2 = (-d)^2 = |d|^2$.) Furthermore, since $a = c + d$, one of $c, d$ must be odd, and the other even.
Now to calculate:
$$
\begin{align}
c^2 + d^2 &= \left({a+b\over 2}\right)^2 + \left({a-b\over 2}\right)^2\\
&= {1\over 4}(a^2 + 2ab + b^2 + a^2 - 2ab + b^2)\\
&= {1\over 2}(a^2 + b^2)
\end{align}
$$
So, if $X = a^2 + b^2$ with both $a, b$ odd, then $X = 2(c^2 + d^2)$ with one of $c,d$ odd and the other even.
If $c$ is odd, then $c = 2i+1$ and
$$c^2 = 4i^2 + 4i + 1 = 4i(i+1) + 1 \equiv 1 \mod 4$$
If $d$ is even, then $d = 2j$ and
$$d^2 = 4j^2 \equiv 0 \mod 4
$$
So
$$X = 2 (c^2 + d^2) \equiv 2 \mod 8$$
Thus, the answer to your question: if $X$ is the sum of two odd squares, then $X \equiv 2 \mod 8$.
To continue with the proof of my original conjecture:
Whenever $X = a^2 + b^2$, we can let $e = a + b$, $f = a-b$, and
$$
e^2 + f^2 = (a^2 + 2ab + b^2) + (a^2 - 2b + b^2) = 2(a^2 + b^2) = 2X
$$
Thus, we have shown $S(x)\implies S(2X)$.
If $S(2X)$, then $2X$ (being even) must the the sum of two odd squares or two even squares:
$2X = e^2 + f^2$. We can do as above:
$$
a = {e+f\over 2}\\
b = {e-f\over 2}
$$
And we have $X = a^2 + b^2$. Since $e, f$ are both odd or both even, $a, b$ must both be integers.
Thus, $S(2X) \implies S(X)$ for any integer $X$.
As shown above, if $X$ is an odd number that is the sum of two squares, then we must have $X \equiv 1 \mod 4$. Thus, if $X \equiv 3 \mod 4$, then $X$ cannot be the sum of two squares. Combine this with the above:
$$
\neg S(X) \implies \exists i,j \mid X = (4i+3)\cdot 2^j
$$
Any integer that is not the sum of two squares is of the form $(4i+3)\cdot 2^j$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding Integral of $y = e^{ax}\sin (bx)$ I was solving questions like $ \int e^{2x} \sin x dx. $ I decided to find the general term for $$\int e^{ax} \sin (bx) $$ which can be directly used in questions like (above).
I hope it helps others :) (Just wanted to share my work. I will appreciate feedback/suggestions or any alternate method available)
| $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \large \textbf{Method -1} $
$$
\begin{align}
I & = \sin (bx) \left( \cfrac{1}{a} e^{ax} \right) - {\int} \left( b\cos (bx) \left(\cfrac{e^{ax}}{a} \right) \right) \\
& = \sin (bx) \left( \cfrac{1}{a} e^{ax} \right) - \cfrac{b}{a} \left[ (\cos (bx)) \cfrac{e^{ax}}{a} + b\sin (bx) \cfrac{e^{ax}}{a} \right] \\
&= \sin (bx) \left( \cfrac{1}{a} (e^{ax}) \right) - \cfrac{b}{a} \left[ \cfrac{1}{a} (\cos (bx) ^{e^{ax}}) + \cfrac{b}{a} (I) \right] \\
& = \sin (bx) \left(\cfrac{1}{a} e^{ax} \right) - \cfrac{b}{a^2} \cos (bx) e^{ax} - \cfrac{b^2}{a^2} (I) \end{align}$$
Now,
$$\begin{align}
I + \cfrac{b^2}{a^2} (I) & = \sin (bx) \cfrac{1}{a} (e^{ax}) - \cfrac{b}{a^2} (\cos(bx)) e^{ax} \\
I \left(\cfrac{a^2+b^2}{a^2}\right) & = e^{ax} \left(\cfrac{\sin (bx)}{a} - \cfrac{b}{a^2} \cos(bx) \right) \\
I \left(\cfrac{a^2+b^2}{\require{cancel}\cancel{a^2}}\right)& = \cfrac{e^{ax}}{\require{cancel}\cancel{a^2}} \left[a \sin (bx) – b\cos (bx) \right] \\
I (a^2 + b^2) & = e^{ax} \left[ a\sin bx - b \cos bx \right] \end{align}$$
Thus, we have our answer:
$$\begin{align}
I &= \cfrac{e^{ax}}{a^2 + b^2} \left[a\sin bx - b \cos bx \right] + \color{grey}{\rm C} \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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limits using $ \epsilon - \delta $ to prove two variable function I'm trying to use the $ \epsilon - \delta $ argument to prove $\lim_{(x,y) \rightarrow (1.1)} \frac{2xy}{x^2+y^2} =1$. I know that I need to show that $\forall \epsilon>0, \exists \delta>0$ s.t. for all (x,y) in the domain of f, $| \frac{2xy}{x^2+y^2} -1| < \epsilon$ whenever $0< \sqrt{(x-1)^2 + (y-1)^2} <\delta$.
I've made $| \frac{2xy}{x^2+y^2} -1|$ to $\frac{(x-y)^2}{x^2 + y^2} $, but I can't get it to look like $\sqrt{(x-1)^2 + (y-1)^2}$..... what should I do from now? :(
| What about substitution $x=s+1$, $y=t+1$?
You get
$$\frac{2xy}{x^2+y^2}-1=
\frac{2(s+1)(t+1)}{(s+1)^2+(t+1)^2}-1=
\frac{2(st+s+t+1)}{(s+1)^2+(t+1)^2}-1=
\frac{2(st+s+t+1)-s^2-t^2-2s-2t-2}{(s+1)^2+(t+1)^2}=
\frac{2st-s^2-t^2}{(s+1)^2+(t+1)^2}=
\frac{-(s-t)^2}{(s+1)^2+(t+1)^2}
$$
Now you want to choose $\delta>0$ in such way that $|s|,|t|<\delta$ implies
$$\frac{(s-t)^2}{(s+1)^2+(t+1)^2} < \varepsilon.$$
If you are able to find a $\delta_1>0$ such that $|s|,|t|<\delta_1$ implies
$$(s+1)^2+(t+1)^2\ge \frac12$$
then you only need to find $\delta_2>0$ such that $|s|,|t|<\delta_2$ implies
$$(s-t)^2<\frac\varepsilon2.$$
Then you can choose $\delta=\min\{\delta_1,\delta_2\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A simple conditional probability problem Assume that two fair dice are rolled one at a time. Given that the sum of the two numbers that occured was at least $7$, compute the probability that it was equal to $7$.
I tried computing the probability as follows:
$$P \left( X+Y=7| X+Y \geq 7 \right)=\frac{P \left(X+Y=7 \right)}{P \left(X+Y \geq 7 \right)}$$
And then I examined all different pairs whose sum is $7$ and then those whose sum is greater or equal to $7$. I am getting $3/12$ but the right answer is $6/21$. Could you please help me understand what I am doing wrong?
Thank you.
| The total number of possible outcomes when rolling two fair dice is $6^2=36$.
Examining the denominator of your conditional probability, it is easier to deal with the case when the sum of the two dice ($X+Y$) is less than $7$, and subtract this result from $36$.
Now, there are $5$ outcomes (pairs ($5$,$1$),($4$,$2$),($3$,$3$),($2$,$4$),($1$,$5$)) for the sum to be $6$, $4$ outcomes for the sum to be $5$, and so on, until the minimum value (which is $2$, where each die shows a $1$).
So the total number of outcomes where $X+Y < 7$ is $5+4+3+2+1=15$.
As a result the total number of outcomes where $X+Y \geq 7$ is $36-15=21$, leading to
$$P(X+Y\geq 7)=\frac{21}{36}$$
Next, we count the number of total outcomes where $X+Y=7$, which is $6$ (($6$,$1$)($5$,$2$),($4$,$3$),($3$,$4$),($2$,$5$),($1$,$6$)), so that
$$P(X+Y=7)=\frac{6}{36}$$
We thus obtain
$$P(X+Y=7|X+Y\geq 7)=\frac{(P(X+Y=7) \cap (X+Y\geq7))}{P(X+Y\geq7)}=\frac{P(X+Y=7)}{P(X+Y\geq7)}\\=\frac{6}{36}\times\frac{36}{21}=\frac{6}{21}=\frac{2}{7}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Producing a CDF from a given PDF So I have this PDF:
$$
f(x)=
\begin{cases}
x + 3 & \text{ for } -3 \leq x < -2\\
3 - x & \text{ for } 2 \leq x < 3\\
0 & \text{ otherwise}
\end{cases}
$$
To make this a CDF, I have integrated the PDF from $-\infty$ to some value, $x$.
$$
F(x)=
\begin{cases}
\frac{x^2}{2} + 3x + \frac{9}{2} & \text{ for } -3 \leq x <-2\\
\frac{1}{2} & \text{ for } -2 \leq x<2\\
\frac{-x^2}{2} + 3x + \frac{7}{2} & \text{ for } 2 \leq x<3
\end{cases}
$$
My friend argues that the first term in this CDF which is $(x^2/2 + 3x + 9/2)$ should actually be $(x^2/2 + 3x)$. But isn't this impossible? At $x = -3$, the CDF must be $0$, am I correct?. This is only true in the case where the first term is $(x^2/2 + 3x + 9/2)$.
If someone could shed light on this topic, that would be much appreciated.
| Let's put it this way
$$\int_{-\infty}^Y f(x)=
\begin{cases}
0 & -\infty < Y < -3 \\
Y^2+3Y+{9\over 2} & -3\le Y <-2 \\
{1\over 2} & -2\le Y <2 \\
{1\over 2} +\int_2^Y(3-x)\, dx & 2\le Y < 3 \\
1 & Y\ge 3
\end{cases}$$
The first term is INDEED including the 9/2, this is because it is
$$\int_{-3}^Y f(x)\,dx\int_{-3}^Y (x+3)\,dx$$
By the fundamental theorem of calculus, this is the antiderivative at the top limit, Y, minus the antiderivative at the bottom limit, -3. The value of $x^2/2+3x$ at $x=-3$ is $-9/2$ so you get the result with the +9/2 because you subtract the bottom limit and subtracting a negative gives a positive.
| {
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Solving a quadratic trigonometric equation? The equation is $6 \cos^2x+\cos x=1$,
My work:
$6x^2+x-1=0$
$(3x-1)(2x+1)$
$3x-1=0 ∨ 2x+1=0$
$x=\frac{1}{3} ∨ x= \frac{-1}{2}$
But I do not know how to progress further.
| I would not use the same letter, $x$, to refer to two different things.
$$
6\cos^2 x + \cos x - 1 = 0
$$
$$
6u^2 + u - 1 = 0
$$
$$
(3u-1)(2u+1)=0
$$
$$
u = \frac 1 3\text{ or } u = \frac{-1}{2}
$$
$$
\cos x=\frac 1 3 \text{ or }\cos x = \frac{-1}2
$$
$$
\Big(x = \left(\arccos\frac 1 3\right) + 2\pi n \text{ or } \left(\pi-\arccos\frac 1 3\right) + 2\pi n \Big)\text{ or }\Big( x = \frac{2\pi}3+2\pi n \text{ or }x=\frac{4\pi}3+2\pi n \Big)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $0\leq a \leq 3; 0\leq b \leq 3$ and the equation $x^2 +4+3 cos(ax+b)=2x$ has at least one solution , then find the value of a+b Problem : If $0\leq a \leq 3; 0\leq b \leq 3$ and the equation $x^2 +4+3 cos(ax+b)=2x$ has at least one solution , then find the value of a+b.
Solution :
We can write the given equation : $x^2 +4+3 cos(ax+b)=2x$ as $x^2-2x+4 =-3cos(ax+b)$
Since the L.H.S. of this problem is always $+ve$ $\forall x \in \mathbb{R}$ , and we know that $ -1 \leq cosx \leq 1$
But I am not getting any clue how to proceed further please help.
Thanks
| If we complete the square we get
$$x^2 -2x +4 = (x-1)^2 +3 \geq 3$$
Further we note that $-3\cos(ax+b) \leq 3$. So it must be that we have $x^2-2x+4 = 3 = -3\cos(ax+b)$. Further, from the completing the square formula, we see that in fact we need $x=1$.
So it reduces down to the problem of solving $-3\cos(a+b) = 3$ or $\cos(a+b) = -1$
Hence $a+b = (2n+1)\pi$ for any integer $n$.
EDIT: Since $0 \leq a \leq 3$, $0 \leq b \leq 3$, we get $0 \leq a+b \leq 6$. Hence the only viable solution would be $\pi$.
| {
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Compute integral: $\int_{0}^{\pi/2}\log(a^2\sin^2 x+b^2\cos^2 x )dx$ This is a integral for a calculus exam, and I have no idea how to solve it.
$$\int_0^{\frac{\pi}{2}} \log \big( a^2 \sin^{2}(x)+b^2 \cos^{2}(x) \big) \, \mathrm{d}x$$
| Without loss of generality, let $b>a>0$. Noting
$$ \sin^2x=\frac{1-\cos(2x)}{2}, \cos^2x=\frac{1+\cos(2x)}{2} $$
we have
\begin{eqnarray*}
I&=&\int_0^{\frac{\pi}{2}} \log \big(\frac{a^2+b^2}{2}+\frac{b^2-a^2}{2}\cos(2x)\big) \mathrm{d}x\\
&=&\frac{1}{2}\int_0^{\pi} \log \big(\frac{a^2+b^2}{2}+\frac{b^2-a^2}{2}\cos x\big) \mathrm{d}x\\
&=&\frac{1}{2}\int_0^{\pi} \log \left(\frac{a^2+b^2}{b^2-a^2}+\cos x\right) \mathrm{d}x+\frac{\pi}{2}\log\frac{b^2-a^2}{2}.
\end{eqnarray*}
Note, for $a>1$
$$\int_0^{\pi} \log \left(\alpha+\cos x\right) \mathrm{d}x=\pi\log\frac{a(a+\sqrt{a^2-1})}{2a} $$
and hence we have
\begin{eqnarray*}
I&=&\pi\log\frac{a+b}{2}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/860066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Ratio of sides of triangle $ABC$ If in a triangle $\Delta ABC$ with $a$, $b$ and $c$ as sides
$$\begin{align}\left(Cot\frac{A}{2}\right)^2 +\left(2Cot\frac{B}{2}\right)^2+\left(3Cot\frac{C}{2}\right)^2=\left(\frac{6s}{7r}\right)^2\end{align} \tag{1}$$ where $r$ is inradius and $s$ is SemiPerimeter, then find the ratio $a:b:c$
My try: If $\Delta$ is the area of the triangle we have
$$Cot\frac{A}{2}=\frac{s(s-a)}{\Delta}$$ and $$sr=\Delta$$
Using these in $(1)$ we get
$$ \frac{s^2(s-a)^2}{\Delta^2}+\frac{4s^2(s-b)^2}{\Delta^2}+\frac{9s^2(s-c)^2}{\Delta^2}=\frac{36s^4}{49\Delta^2}$$
$\implies$
$$49\left((s-a)^2+4(s-b)^2+9(s-c)^2\right)=36s^2$$
$\implies$
$$650s^2-98s(a+4b+9c)+49(a^2+4b^2+9c^2)=0$$
since for a triangle $s$ need to be unique, the above equation must have Discriminant Zero $\implies$
$$49(a+4b+9c)^2-650(a^2+4b^2+9c^2)=0$$
$\implies$
$$601a^2+1816b^2+1881c^2=392ab+3528bc+882ac$$
I am stuck up here, please help me.
| You have proved that the proposed condition is equivalent to
$$
(s-a)^2+4(s-b)^2+9(s-c)^2=\frac{36}{49}s^2\tag 1
$$
Note that, by the Cauchy-Schwarz inequality, we have
$$\eqalign{
s&=(s-a)+\frac{1}{2}(2(s-b))+\frac{1}{3}(3(s-c))\cr
&\leq\sqrt{(s-a)^2+4(s-b)^2+9(s-c)^2}\sqrt{1+\frac{1}{4}+\frac{1}{9}}\cr
&\leq\frac{7}{6}\sqrt{(s-a)^2+4(s-b)^2+9(s-c)^2}\cr
}$$
So $(1)$ corresponds to the case of equality in the Cauchy-Schwarz inequality, hence
$$
\frac{s-a}{1}=\frac{2(s-b)}{\frac{1}{2}}=\frac{3(s-c)}{\frac{1}{3}}
$$
that is
$$
s-a=\lambda,\quad s-b=\frac{\lambda}{4},\quad s-c=\frac{\lambda}{9}
$$
for some $\lambda>0$. Adding these equalities we get: $s=\frac{49}{36}\lambda$,
that is
$$
a=\frac{13}{36}\lambda,\quad b=\frac{40}{36}\lambda,\quad c=\frac{45}{36}\lambda
$$
Finally $a:b:c=13:40:45$.$\qquad\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proof that if $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$ Question: Prove that if
$$a_n=\left\{
\begin{array}{ll}
a_1=1\\
a_{n+1}=1+\frac{1}{1+a_n}
\end{array}
\right.$$
then $a_n$ converges, and then find $\lim_{n \to \infty}a_n$.
I found that $\lim_{n \to \infty}a_n=\sqrt{2}$, and that $1\leq a_n<2$, but the convergence seems difficult.
My idea is to show that $a_{2n}$ is decreasing, and $a_{2n-1}$ is increasing, but I don’t see how.
I’d be thankful for any hint.
| Most inelegant method I guess:
$a_{n+2}=1+\cfrac{1}{1+a_{n+1}}=1+\cfrac{1}{1+1+\cfrac{1}{1+a_n}}=1+\cfrac{1+a_n}{3+2a_n}
$
So $a_{n+2}-a_n=2\cfrac{2-a_n^2}{3+2a_n}$
Clearly from definition $a_n>1$
Also if $a_n>\sqrt{2}$ then $a_{n+2}>1+\cfrac{1+\sqrt{2}}{3+2}=\cfrac{6}{5}+\cfrac{\sqrt{2}}{5}>\cfrac{4\sqrt{2}}{5}+\cfrac{\sqrt{2}}{5}=\sqrt{2}$
if $a_n<\sqrt{2}$ then $a_{n+2}<1+\cfrac{1+1}{3+2\sqrt{2}}=1+\cfrac{2(3-2\sqrt{2})}{9-8}=7-4\sqrt{2}=5\times\cfrac{7}{5}-4\sqrt{2}<5\times \sqrt{2}-4\sqrt{2}=\sqrt{2}$
Now $a_1=1<\sqrt{2}$, $a_2=1.5>\sqrt{2}$ So by induction, the sequences $a_1,a_3,a_5,\ldots$ and $a_2,a_4,\ldots$ are monotone increasing/decreasing respectively, and is bounded above/below by $\sqrt{2}$. These two sequences clearly converges to the same limit, and so $(a_n)$ converges to the common limit which is $\sqrt{2}$ (so in fact $\sqrt{2}$ is supremum and infimum of the sequences)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 7
} |
Solve the System of Equations in Real $x$,$y$ and $z$ Solve for $x$,$y$ and $z$ $\in $ $\mathbb{R}$ if
$$\begin{align} x^2+x-1=y \\
y^2+y-1=z\\
z^2+z-1=x \end{align}$$
My Try:
if $x=y=z$ then the two triplets $(1,1,1)$ and $(-1,-1,-1)$ are the Solutions.
if $x \ne y \ne z$ Then we have
$$\begin{align} x(x+1)=y+1 \\
y(y+1)=z+1\\
z(z+1)=x+1 \end{align}$$
Multiplying all we get $$xyz=1 \tag{1}$$ and adding all we get
$$x^2+y^2+z^2=3 \tag{2}$$
Now from Original Equations
$$\begin{align} x^2=y+1-x\\
y^2=z+1-y\\
z^2=x+1-z \end{align}$$ Multiplying all and Using $(1)$ we get
$$(y+1-x)(z+1-y)(x+1-z)=1 $$ $\implies$
$$xy+yz+zx-3=(x-y)(y-z)(z-x) \tag{3}$$ I am unable to proceed further..
| $\textbf{1)}$ Substituting Eq. 1 into Eq. 2 gives $xy(x+1)=z+1$, and then substituting Eq. 2 into Eq. 3 gives $xyz(x+1)=x+1$; so $x=-1$ or $xyz=1$. By symmetry, we have that $y=-1$ or $xyz=1$ and $z=-1$ or $xyz=1$. Since $x=-1, y=-1, z=-1$ is a solution, all other solutions must satisfy $xyz=1$.
Since $x=1,y=1, z=1$ is clearly a solution, it is left to show that there are no other solutions satisfying $xyz=1$.
$\textbf{2)}$ If $x\ge1$, then $y=x(x+1)-1\ge2-1=1$ and $z=y(y+1)-1\ge2-1=1$; so $xyz=1\implies x=1, y=1, z=1$. Therefore we can assume $x<1$ and, by symmetry, $y<1$ and $z<1$.
$\textbf{3)}$ If $x, y, z >0$, then $0<x, y, z <1\implies xyz<1$; so two of the variables must be negative, and we can assume that $x<0, y<0, z>0$. Since $z<1$, $xy=\frac{1}{z}>1$. However, this is impossible since $xy=x(x^2+x-1)=x^3+x^2-x$, and $g(x)=x^3+x^2-x$ has a maximum for $x<0$ given by $g(-1)=1$
since $g^{\prime}(x)=(3x-1)(x+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/864430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Determining the best possible constant $k$, for an Integral Inequality If $f : [0,\infty) \to [0,\infty)$ is an integrable function, then what is the best possible constant $k$, for which the following ineqality holds:
$$\int_0^{\infty}f(x)dx \leq k\left(\int_0^{\infty}\sqrt{x}f(x)dx\right)^{1/2}\cdot\left(\int_0^{\infty}f^{2}(x)dx\right)^{1/4}$$
For example, $k=2$ is a bound, since, $$\int_0^{\infty}f(x)dx = \int_0^{y}f(x)dx + \int_y^{\infty}f(x)dx < \sqrt{y}\left(\int_0^{\infty}f^2(x)dx\right)^{1/2} + \frac{1}{\sqrt{y}}\int_0^{\infty}\sqrt{x}f(x)dx$$
and setting, $ y = \dfrac{\displaystyle \int_0^{\infty}\sqrt{x}f(x)dx}{\left(\displaystyle\int_0^{\infty} f^2(x)dx\right)^{1/2}}$, establishes the inequality for the case $k=2$, but it is not strict.
How to improve the bound $k$ and also determine the function where equality holds for the best constant $k$ ?
Thank you.
| Suppose
$$
\int_0^\infty\sqrt{x}f(x)\,\mathrm{d}x=A\tag{1}
$$
and
$$
\int_0^\infty f^2(x)\,\mathrm{d}x=B\tag{2}
$$
Then, letting $g(x)^2=A^{1/2}B^{-3/4}f(AB^{-1/2}x)$, we get
$$
\int_0^\infty\sqrt{x}g(x)^2\,\mathrm{d}x=\int_0^\infty g(x)^4\,\mathrm{d}x=1\tag{3}
$$
and
$$
\int_0^\infty f(x)\,\mathrm{d}x=A^{1/2}B^{1/4}\int_0^\infty g(x)^2\,\mathrm{d}x\tag{4}
$$
Thus, maximizing $\int_0^\infty g(x)^2\,\mathrm{d}x$ given $(3)$ yields the best constant for the given inequlity.
The variation of $(4)$ is stationary when
$$
\int_0^\infty g(x)\,\delta g(x)\,\mathrm{d}x=0\tag{5}
$$
Equations $(3)$ imply
$$
\int_0^\infty\sqrt{x}g(x)\,\delta g(x)\,\mathrm{d}x=\int_0^\infty g(x)^3\,\delta g(x)\,\mathrm{d}x=0\tag{6}
$$
To insure $(5)$ holds whenever $(6)$ does, we must have that $g(x)^2=0$ or $g(x)^2=a-b\sqrt{x}$.
Thus, consider $g(x)^2=a-b\sqrt{x}$ on $\left[0,\frac{a^2}{b^2}\right]$ and $g(x)^2=0$ for $x\gt\frac{a^2}{b^2}$.
$$
\begin{align}
\int_0^{a^2/b^2}\sqrt{x}g(x)^2\,\mathrm{d}x
&=\int_0^{a^2/b^2}\left(a\sqrt{x}-bx\right)\,\mathrm{d}x\\
&=\frac{a^4}{6b^3}\tag{7}
\end{align}
$$
$$
\begin{align}
\int_0^{a^2/b^2}g(x)^4\,\mathrm{d}x
&=\int_0^{a^2/b^2}\left(a^2-2ab\sqrt{x}+b^2x\right)\,\mathrm{d}x\\
&=\frac{a^4}{6b^2}\tag{8}
\end{align}
$$
$$
\begin{align}
\int_0^{a^2/b^2}g(x)^2\,\mathrm{d}x
&=\int_0^{a^2/b^2}\left(a-b\sqrt{x}\right)\,\mathrm{d}x\\
&=\frac{a^3}{3b^2}\tag{9}
\end{align}
$$
Solving $(7)$ and $(8)$ given $(3)$ yields
$$
(a,b)=\left(6^{1/4},1\right)\tag{10}
$$
Plugging $(10)$ into $(9)$ yields an optimal constant of
$$
\left(\frac83\right)^{1/4}\doteq1.277886208492545\tag{11}
$$
attained by the function
$$
\left(6^{1/4}-\sqrt{x}\right)\left[0\le x\le6^{1/2}\right]\tag{12}
$$
where $\left[\dots\vphantom{6^{1/2}}\right]$ are Iverson brackets.
The Slick Answer
Suppose $f(x)\ge0$ and
$$
\int_0^\infty\sqrt{x}f(x)\,\mathrm{d}x=\int_0^\infty f(x)^2\,\mathrm{d}x=1\tag{13}
$$
Then letting $u(x)=\left(6^{1/4}-\sqrt{x}\right)\left[0\le x\le6^{1/2}\right]$
$$
\begin{align}
0
&\le\int_0^\infty(f(x)-u(x))^2\,\mathrm{d}x\\
&=\int_0^\infty\left(f(x)^2-2f(x)u(x)+u(x)^2\right)\,\mathrm{d}x\tag{14}
\end{align}
$$
Applying $(13)$ to $(14)$, dividing by $2$, and rearranging, we get
$$
\begin{align}
1
&\ge\int_0^\infty f(x)u(x)\,\mathrm{d}x\\
&\ge\int_0^\infty f(x)\left(6^{1/4}-\sqrt{x}\right)\,\mathrm{d}x\\
&=6^{1/4}\int_0^\infty f(x)\,\mathrm{d}x-1\tag{15}
\end{align}
$$
Therefore,
$$
\int_0^\infty f(x)\,\mathrm{d}x\le\left(\frac83\right)^{1/4}\tag{16}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/865907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Global Max and Min Hi can someone please help me figure out how to start this question?
Temperature within a circular region $R = \{(x, y)| x^2 + y^2 \le 49\}$ is given by $T(x,y) = 4x^2 -4xy + y^2$
Find the global minimum and maximum temperatures in the region and the points where these extreme temperatures occur.
I'm supposed to use Lagrange multipliers and parametrization which I don't really understand. If someone can please help me get started I would really appreciate it.
| We use Lagrange multipliers. Let, $F(x,y) = 4x^2 - 4xy + y^2 + \lambda(x^2 + y^2 - 49)$. Thus,
$$
\begin{cases}
\dfrac{\partial F}{\partial x} = 8x - 4y + 2\lambda x = 0 \qquad (1)\\
\dfrac{\partial F}{\partial y} = 8y - 4x + 2\lambda y = 0 \qquad(2)\\
\dfrac{\partial F}{\partial \lambda} = x^2 + y^2 -49= 0 \qquad (3)
\end{cases}
$$
From equations (1) and (2), we have,
$$
\dfrac{2x - 4y}{y} = \lambda = \dfrac{2y -4x}{x} \quad \Rightarrow \quad x^2 = y^2
$$
Thus, from equation (3), $2x^2 = 49 \ \Rightarrow \ x = \pm \frac{7}{\sqrt{2}}$ and $y = \pm \frac{7}{\sqrt{2}}$
The points of maximum and minimum are $(-\frac{7}{\sqrt{2}},-\frac{7}{\sqrt{2}})$ and $(\frac{7}{\sqrt{2}},\frac{7}{\sqrt{2}})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Interview riddle On the Mathematics chat we were recently talking about the following problem @Chris'ssis had to solve during an interview :
$$3\times 4=8$$
$$4\times 5=50$$
$$5\times 6=30$$
$$6\times 7=49$$
$$7\times 8=?$$
We have not managed to solve it so far, all we know is the solution (which was given after we had given up) :
$224$
How do we find this solution ?
| Easy, just define
$$\begin{array}{rcl}a \times b &=&
\hspace{10.5pt}(a-4)(b-5)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/72 + \\&& 25(a-3)(b-4)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/18 + \\&& 15(a-3)(b-4)(a-4)(b-5)(a-6)(b-7)(a-7)(b-8)/8 \hspace{5.25pt}+ \\&& 49(a-3)(b-4)(a-4)(b-5)(a-5)(b-6)(a-7)(b-8)/36 + \\&&\hspace{5.5pt}7(a-3)(b-4)(a-4)(b-5)(a-5)(b-6)(a-6)(b-7)/18\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "48",
"answer_count": 15,
"answer_id": 0
} |
How to calculate $\int_{-\infty}^\infty\frac{x^2+2x}{x^4+x^2+1}dx$? I want to calculate the following integral: $$I:=\displaystyle\int_{-\infty}^\infty\underbrace{\frac{x^2+2x}{x^4+x^2+1}}_{=:f(x)}dx$$ Of course, I could try to determine $\int f(x)\;dx$ in terms of integration by parts. However, I don't think that's the way one should do this. So, what's the trick to calculate $I$?
| $$\int_{-\infty}^\infty\frac{x^2+2x}{x^4+x^2+1}dx$$
$$f(x)=\frac{x^2+2x}{x^4+x^2+1}$$
$$\int_{-\infty}^\infty\frac{x^2+2x}{x^4+x^2+1}dx=2\pi i\sum_{\Im x_i>0} \text{Res}(f(x);x_i)$$
$$x_i=\left\{\sqrt[3]{-1},(-1)^{2/3}\right\}$$
so we get $$\left\{\frac{1}{12} \left(3-5 i \sqrt{3}\right),\frac{1}{4} i \left(\sqrt{3}+i\right)\right\}$$ as two residues.
$$
\therefore 2\pi i*-\frac{i}{2 \sqrt{3}}=\dfrac{\pi}{\sqrt{3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to compute power series by composition Is it possible to compute the power series of every function (e.g. around $0$) just by composing of the power series of its arguments?
For example:
The power series of $\sin(x^2)$ around $0$ is the composition of power series of $\sin(x)$ and $x^2$:
$compose(x-\frac{x^3}{6}+\frac{x^5}{120}+..., x^2) = x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}+...$
If the answer to my question is "yes", then how would you compute the power series of $\sqrt{\cos(x)}$? Because $\sqrt{x}$ doesn't have a power series expansion around $0$ and you can't simply compose the power series of $\sqrt{x}$ and that one of $\cos(x)$ to find the expansion of $\sqrt{\cos(x)} = 1-\frac{x^2}{4}-\frac{x^4}{96}+O\left(x^5\right)$.
| Use the series for $\sqrt{1+x}$ and $\cos x - 1$
$$\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3\pm\dots$$
$$\cos x - 1 = -\frac{1}{2}x^2+\frac{1}{24}x^4 \pm \dots$$
Now substitute the second into the first:
$$\sqrt{\cos x} =\sqrt{1 + (\cos x-1)} \\=
1+\frac{1}{2}\left(-\frac{1}{2}x^2+\frac{1}{24}x^4\pm \dots\right)
-\frac{1}{8}\left(-\frac{1}{2}x^2+\frac{1}{24}x^4 \pm\dots\right)^2+ \dots\\
= 1-\frac{1}{4}x^2-\frac{1}{96}x^4 \pm \dots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/867976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Suggestion for Computing an Integral Let $$A=\left\{(x,y,z)\in \mathbb R^3:\dfrac{x^2}{2}+\dfrac{y^4}{4}+\dfrac{z^6}{6}\leq1\right\}.$$
Then I want to compute the following integral:
$$\frac{1}{\operatorname{vol}(A)}\displaystyle\int_{\partial A}^{}\!\frac{1}{\sqrt{x^2+y^6+z^{10}}}\, d(x,y,z)$$
Should I use spherical coordinates or something like that?
I am a beginner to this topic, so any help would be nice.
Edit: I put the factor $\frac{1}{\operatorname{vol}(A)}$ before the integral (that's exactly my task now), but there should be no different concerning our problem...
$$I := \frac{1}{\lambda_3(A)} \int\limits_{\partial A} \frac{1}{\sqrt{x^2 + y^6 + z^{10}}} \, dS_{\partial A}$$
| Let us express the surface element in terms of $y,z$:
$$dS=\sqrt{1+\left(\frac{\partial x}{\partial y} \right)^2+\left(\frac{\partial x}{\partial z} \right)^2} \,dy dz,\tag{1}$$
where
$$\frac{x^2}{2}+\frac{y^4}{4}+\frac{z^6}{6}=1. \tag{2}$$
Now from (2) it follows that
$$\frac{\partial x}{\partial y}=-\frac{y^3}{x},\qquad \frac{\partial x}{\partial z}=-\frac{z^5}{x},$$
and therefore the formula (1) transforms into
$$dS=\frac{\sqrt{x^2+y^6+z^{10}}}{|x|}dydz.$$
The surface integral we want to compute (without the inverse volume prefactor) then becomes
$$I_S=4\sqrt{2}\iint_D\frac{dydz}{\sqrt{\displaystyle1-\frac{y^4}{4}-\frac{z^6}{6}}},$$
where $\displaystyle D=\left\{(y,z)\in\mathbb{R}^2:\frac{y^4}{4}+\frac{z^6}{6}\leq1,y\geq0,z\geq0\right\}$. Next define
$$a=\frac{y^2}{2},\qquad b=\frac{z^3}{\sqrt{6}}\qquad
\Longleftrightarrow \qquad y=\sqrt{2a},\qquad z=6^{\frac16}b^{\frac13}$$
and rewrite the integral as
$$I_S=8\cdot 6^{-5/6}\iint_{D'}\frac{a^{-1/2}b^{-2/3}dadb}{\sqrt{1-a^2-b^2}},$$
with $$D'=\{(a,b)\in\mathbb{R}^2:a^2+b^2\leq 1,a\geq0,b\geq0\}.$$
Now it becomes helpful to introduce polar coordinates $a=r\cos\theta,b=r\sin\theta$ and rewrite the integral as
$$I_S=8\cdot 6^{-5/6}\underbrace{\int_0^1\frac{r^{-1/6}dr}{\sqrt{1-r^2}}}_{I_1}\;\underbrace{\int_0^{\pi/2}(\cos\theta)^{-1/2}(\sin\theta)^{-2/3}d\theta}_{I_2}.$$
It is not difficult to express $I_{1,2}$ in terms of gamma functions:
$$I_1=-6\sqrt{\pi}\frac{\Gamma(\frac{5}{12})}{\Gamma(-\frac{1}{12})},\qquad
I_2=\frac12\frac{\Gamma(\frac14)\Gamma(\frac16)}{\Gamma(\frac5{12})},$$
and therefore
$$\boxed{\displaystyle I_S=-24\sqrt{\pi}\cdot 6^{-5/6}\cdot\frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{6})}{\Gamma(-\frac{1}{12})}}\tag{3}$$
The volume of $A$ can be computed very similarly. Indeed,
\begin{align}\operatorname{vol}A&=\iiint_A dx dy dz=\\&=8\sqrt2\iint_D \sqrt{\displaystyle1-\frac{y^4}{4}-\frac{z^6}{6}} \,dy dz=\\&=
16\cdot 6^{-5/6}\iint_{D'}a^{-1/2}b^{-2/3}\sqrt{1-a^2-b^2}\,dadb=\\
&=16\cdot 6^{-5/6}\cdot I_2\cdot \int_0^1 r^{-1/6}\sqrt{1-r^2}dr=\\
&=16\cdot 6^{-5/6}\cdot I_2\cdot\frac{\sqrt\pi}{4}\frac{\Gamma(\frac5{12})}{\Gamma(\frac{23}{12})},
\end{align}
so that
$$\boxed{\displaystyle I=\frac{I_S}{\operatorname{vol}A}=\frac{11}{12}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/871950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
How to prove $\int_0^\pi \frac{dx}{2+2\sin x+\cos x}=\log3$? How can we prove that: $$\int_0^\pi \frac{dx}{2+2\sin x+\cos x}=\log3$$
I don't have any ideas, the $f(\pi-x)$ thing doesn't work as well. Please help :)
| $$
\begin{aligned}
&\int_0^\pi\frac{1}{2+2\sin x+\cos x}dx
\\=& \int_0^\pi \frac{dx}{2[\cos^2(x/2)+\sin^2(x/2)]+4\cos(x/2)\sin(x/2) + \cos^2(x/2)-\sin^2(x/2)}\\
=&\int_0^\pi\frac{dx}{3\cos^2(x/2)+4\cos(x/2)\sin(x/2)+\sin^2(x/2)}\\
=&\int_0^\pi \frac{\sec^2(x/2)dx}{\tan^2(x/2)+4\tan(x/2)+3}\\
=&\int_0^\infty \frac{2du}{u^2+4u+3}
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How find this integral $I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\frac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$
Find the value:
$$I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\dfrac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$$
I use computer have this reslut
$$I=-\dfrac{4\pi}{\sqrt{2}}C+\dfrac{13\pi^3}{24\sqrt{2}}+\dfrac{9}{2}\dfrac{\pi\ln^2{2}}{\sqrt{2}}-\dfrac{3}{2}\dfrac{\pi^2\ln{2}}{\sqrt{2}}$$
where $C$ is Catalan constant.
My idea: let $$\tan{x}=t,\sin{2x}=\dfrac{2t}{1+t^2},dx=\dfrac{1}{1+x^2}$$
then
$$I=\int_{0}^{\infty}\ln^2{(1+t^4)}\dfrac{\dfrac{2}{1+t^2}}{2-\left(\dfrac{2t}{1+t^2}\right)^2}\cdot\dfrac{dt}{1+t^2}=\int_{0}^{\infty}\dfrac{\ln^2{(1+t^4)}}{(t^2-1)^2}dt$$
Then I can't.Thank you
| You have made a mistake with the denominator, which cannot be zero. See, $2-sin(2x)^2$ is always bigger or equal than $1$.
I think that the mistake was in expanding $2-(\frac{2t}{1+t^2})^2$: you have changed that initial $2$ for a $1$ and then completed the square. Repeat that expansion slowly.
| {
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Proving $\frac{p-1}{2}!\equiv (-1)^t$ where $t$ is the number of integers which are not quadratic squares
Prove that $\frac{p-1}{2}!\equiv (-1)^t$ where $t$ is the number of integers $0<a<\frac{p}{2}$ which are not quadratic squares $\pmod p$ ($p\equiv3\bmod4$)
I don't know really from where to start (we know from wilson that $(p-1)!\equiv -1\pmod p$ but that won't help here as far as I see). The lecturer suggested lookigng at $p-n$ where $1\le n\le p-1$. I'd be glad for any hint.
| I believe that theorem only holds for $p \equiv 3 \pmod{4}$.
In fact for $p = 5$ then $((p-1)/2)! \equiv 2$ and for $p = 13$ the factorial yields $5$, not a power of $-1$.
Not let $p \equiv 3 \pmod{4}$.
$$(p-1)! \equiv \left( 1·2·\ldots ·\frac{p-1}{2}\right)\left( \frac{p+1}{2}\ldots (p-1)\right)$$
But $p-a \equiv (-1)a$ So:
$$(p-1)! \equiv \left( 1·2·\ldots ·\frac{p-1}{2}\right)\left( \left( p-\frac{p-1}{2}\right)\ldots (p-1)\right) \equiv \left( 1·2·\ldots ·\frac{p-1}{2}\right)^2(-1)^{\frac{p-1}{2}} \equiv \left(\frac{p-1}{2}!\right)^2(-1)^{\frac{p-1}{2}}$$
Since $(p-1)! \equiv -1$ then:
$$(-1)^{\frac{p+1}{2}} \equiv \left(\frac{p-1}{2}!\right)^2 \equiv \prod_{k=1}^{\frac{p-1}{2}}k^2$$
Now, $p\equiv 3 \pmod{4}$ thus $(-1)^\frac{p+1}{2} = 1$:
$$1 \equiv \left(\frac{p-1}{2}!\right)^2 \equiv \prod_{k=1}^{\frac{p-1}{2}}k^2$$
Note that since $a^2 \equiv (p-a)^2$ then $\left(\frac{p-1}{2}!\right)^2$ is the product of all quadratic residues $\text{mod } p$.
Let $G = \lbrace k^2 \mod p \text{ such that } k^2 \text{ mod } p < p/2 \rbrace$. i.e. $G$ is the set of the quadratic residues lower than $p/2$.
Note that $(p-1)/2 - \vert G\vert$ is the amount of quadratic non-residues lower than $p/2$.
But there are $(p-1)/2$ quadratic residues $\text{mod } p$ so $(p-1)/2 - \vert G\vert$ is also the amount of quadratic residues bigger than $p/2$.
Let $s = (p-1)/2 - \vert G\vert$
Let $r_k = \begin{cases} k^2 \mod p,&\text{if } k^2 \in G
\\ -k^2 \mod p,&\text{if } k^2 \not\in G
\end{cases}
$
So our product becomes:
$$\prod_{k=1}^{\frac{p-1}{2}}k^2 = (-1)^s\prod_{k=1}^{\frac{p-1}{2}}r_k$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute limit of a function Compute: $$\lim_{x \rightarrow 0^+} \frac{\arctan(e^x+\arctan x)-\arctan(e^{\sin x}+\arctan(\sin x))}{x^3}$$
WolframAlpha tells me it's 1/6. Any nice idea how to rewrite that expression? Thanks!
| We can proceed as follows $$\begin{aligned}L\, &= \lim_{x \to 0^{+}}\frac{\tan^{-1}(e^{x} + \tan^{-1}x) - \tan^{-1}(e^{\sin x} + \tan^{-1}(\sin x))}{x^{3}}\\
&= \lim_{x \to 0^{+}}\dfrac{\tan^{-1}\left(\dfrac{e^{x} + \tan^{-1}x - e^{\sin x} - \tan^{-1}(\sin x)}{1 + (e^{x} + \tan^{-1}x)(e^{\sin x} + \tan^{-1}(\sin x))}\right)}{x^{3}}\\
&= \lim_{t \to 0^{+}}\dfrac{\tan^{-1}t}{t}\cdot\dfrac{t}{x^{3}}\\
&= \lim_{x \to 0^{+}}\frac{t}{x^{3}}\\
&= \lim_{x \to 0^{+}}\dfrac{\left(\dfrac{e^{x} + \tan^{-1}x - e^{\sin x} - \tan^{-1}(\sin x)}{1 + (e^{x} + \tan^{-1}x)(e^{\sin x} + \tan^{-1}(\sin x))}\right)}{x^{3}}\\
&= \frac{1}{2}\lim_{x \to 0^{+}}\frac{e^{x} + \tan^{-1}x - e^{\sin x} - \tan^{-1}(\sin x)}{x^{3}}\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{e^{x} - e^{\sin x}}{x^{3}} + \frac{\tan^{-1}x - \tan^{-1}(\sin x)}{x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{e^{\sin x}(e^{x -
\sin x} - 1)}{x^{3}} + \dfrac{\tan^{-1}\left(\dfrac{x - \sin x}{1 + x\sin x}\right)}{x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{e^{\sin x}(e^{x -
\sin x} - 1)}{x^{3}} + \lim_{x \to 0^{+}}\dfrac{\tan^{-1}\left(\dfrac{x - \sin x}{1 + x\sin x}\right)}{x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{(e^{x -
\sin x} - 1)}{x^{3}} + \lim_{v \to 0^{+}}\dfrac{\tan^{-1}v}{v}\cdot\frac{v}{x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{u \to 0^{+}}\frac{(e^{u} - 1)}{u}\cdot\frac{u}{x^{3}} + \lim_{v \to 0^{+}}\dfrac{\tan^{-1}v}{v}\cdot\frac{v}{x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{x - \sin x}{x^{3}} + \lim_{x \to 0^{+}}\frac{x - \sin x}{(1 + x\sin x)x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{x - \sin x}{x^{3}} + \lim_{x \to 0^{+}}\frac{x - \sin x}{x^{3}}\right)\\
&= \lim_{x \to 0^{+}}\frac{x - \sin x}{x^{3}}\\
&= \lim_{x \to 0^{+}}\frac{1 - \cos x}{3x^{2}}\text{ (by L'Hospital's Rule)}\\
&= \lim_{x \to 0^{+}}\frac{1}{3}\cdot\frac{2\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x^{2}}\\
&= \frac{1}{3}\cdot 2\cdot\frac{1}{4} = \frac{1}{6}\end{aligned}$$ We have used subsitutions $$t = \dfrac{e^{x} + \tan^{-1}x - e^{\sin x} - \tan^{-1}(\sin x)}{1 + (e^{x} + \tan^{-1}x)(e^{\sin x} + \tan^{-1}(\sin x))}, u = x - \sin x, v = \dfrac{x - \sin x}{1 + x\sin x}$$ and each of these variables $t, u, v$ tends to $0$ as $x \to 0^{+}$ and hence we have $$\lim_{t \to 0^{+}}\frac{\tan^{-1}t}{t} = 1,\,\lim_{u \to 0^{+}}\frac{e^{u} - 1}{u} = 1,\, \lim_{v \to 0^{+}}\frac{\tan^{-1}v}{v} = 1$$ The above derivation shows that the normal rules of algebra of limits coupled with few standard limit formulas can handle very complicated expressions and the use of LHR can be minimized to cases where it is really necessary.
| {
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Proving Fibonacci inequality I didn't see a question regarding this particular inequality, but I think that I have shown by induction that, for $n>1$. I am hoping someone can verify this proof.
$$\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}<F_{n+1}<\left(\frac{1+\sqrt{5}}{2}\right)^{n}$$
Here we state that $F_0=0, F_1=1$. The base case is established since
$$\left(\frac{1+\sqrt{5}}{2}\right)<F_{3}<\left(\frac{1+\sqrt{5}}{2}\right)^{2}$$
$$1.61803...<2<2.61803...$$
Now assume that $n=k$ and that the following holds:
$$\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}<F_{k+1}<\left(\frac{1+\sqrt{5}}{2}\right)^{k}$$
Starting with the left side of the inequality, and looking at $F_{k+2}=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+2}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{k+2}$
$$F_{k+2}>\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+2}=\frac{2}{\sqrt{5}+\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+2}>\frac{2}{3+\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+2}=\left(\frac{2}{1+\sqrt{5}}\right)^2\left(\frac{1+\sqrt{5}}{2}\right)^{k+2}=\left(\frac{1+\sqrt{5}}{2}\right)^{k}$$
So the left side didn't need the induction hypothesis, just some careful analysis. Now the right side of the inequality. By the induction hypothesis,
$$\left(\frac{1+\sqrt{5}}{2}\right)^{k}>F_{k+1} \Rightarrow \left(\frac{1+\sqrt{5}}{2}\right)^{k+1}>\left(\frac{1+\sqrt{5}}{2}\right)F_{k+1} $$
But since $F_{k+1}=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k+1}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{k+1}$
$$\left(\frac{1+\sqrt{5}}{2}\right)^{k+1}>\left(\frac{1+\sqrt{5}}{2}\right)F_{k+1}=F_{k+2} $$
Therefore,
$$\left(\frac{1+\sqrt{5}}{2}\right)^{k}<F_{k+2}<\left(\frac{1+\sqrt{5}}{2}\right)^{k+1}$$
I feel as though the analysis is good. Is there any holes or misstatements?
| I think your proof would be clearer if you wrote things in terms of the golden ratio $\phi=\dfrac{1+\sqrt{5}}{2}$. As a demonstration of why that's helpful, let me give a non-inductive proof of the upper bound. Note that this inequality is equivalent by the Binet formula to
$$ \frac{1}{\sqrt{5}}\left(\phi^{n+1}-(-\phi)^{-n-1}\right)< \phi^{n}.$$ Dividing out $\phi^{n-1}/\sqrt{5}$ gives
$$\phi^2+(-1)^n \phi^{-2n}<\phi\sqrt{5}.$$
But $\phi^2=\phi+1$, $\sqrt{5}=2\phi-1$, and $\phi^{-2} = (\phi^2-\phi)\phi^{-2} =1-\phi^{-1}\cdot(\phi^2-\phi) = 2-\phi.$ So the inequality becomes
$$\phi+1+(\phi-2)^n<2\phi^2-\phi =\phi+2\implies (\phi-2)^n < 1.$$ But $-1<\phi-2 <0$, so it follows that $|\phi-2|<1$ and so $(\phi-2)^n\leq|\phi-2|^n<1$ for $n\geq 2$. This completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Polynomial representation Why is the polynomial $P(x)$ represented as
$$ P(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots+ a_2 x^2 + a_1 x + a_0 \text{ ?}$$
A polynomial can be $5x^4 + 3x^3 + 7x^2 + 10x -2$ and it is not necessarily $a_n x^n + a_{n-1} x^{n-1}+\cdots$ I.e if $ a_n x^n$ is $ 5x^4$, doesn't $a_{n-1} x^{n-1} $ mean $4x^3$?
| $$a_{n-1}$$ is not the same as $$a_n -1 $$
The $-1$ is in the subscript in $a_{n-1}$.
Think of it as a function $a$, and we are looking at $a(1), a(2), \dots, a(n-1), a(n)$, so the polynomial is
$$a(n) x^n + a(n-1) x^{n-1} + \dots + a(1) x + a(0)$$
So for instance, if we take $a(m) = m^2$ (can also be written as $a_{m} = m^2$), then for $n=3$ the polynomial will be
$$ 9x^3 + 4x^2 + x $$
| {
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How prove this inequality $\frac{3(x^2+y^2+z^2)}{(x+y+z)^2+2(yz+xz+xy)}\ge\sum\limits_{cyc}\frac{x^2}{x^2+(y+z)^2}$ Question:
let $x,y,z\ge 0$.prove or disprove
$$\dfrac{3(x^2+y^2+z^2)}{(x+y+z)^2+2(yz+xz+xy)}\ge\sum\limits_{cyc}\dfrac{x^2}{x^2+(y+z)^2}$$
My idea: let $x+y+z=1$,
then we can only
$$\sum_{cyc}\dfrac{x^2}{x^2+(y+z)^2}=\sum_{cyc}\dfrac{x^2}{2x^2-2x+1}\le\dfrac{3(x^2+y^2+z^2)}{1+2(xy+yz+xz)}$$
then I can't.Thank you
| It's wrong! Try $x=4$ and $y=z=9$.
| {
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Find $\tan x $ if $\sin x+\cos x=\frac12$ It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $.
I tried all identities I know but I have no idea how to proceed. Any help would be appreciated.
| With no quadratic steps, only using trig identities:
Multiplying through by $\frac1{\sqrt{2}}$:
$$\begin{align}
\sin(x)\frac1{\sqrt{2}}+\cos(x)\frac1{\sqrt{2}}&=\frac{1}{\sqrt{8}}\\
\sin(x)\cos(\pi/4)+\cos(x)\sin(\pi/4)&=\frac{1}{\sqrt{8}}\\
\sin(x+\pi/4)&=\frac{1}{\sqrt{8}}
\end{align}$$
Given that $x$ is in $[0,\pi]$, $x$ must be in $(\pi/2,3\pi/4)$ because
*
*if $x$ is in $[0,\pi/2]$ then both $\sin(x)$ and $\cos(x)$ are nonnegative, and $\max(\sin(x),\cos(x))\geq\frac{1}{\sqrt{2}}>\frac12$
*if $x$ is in $[3\pi/4,\pi]$ then $\cos(x)$ is negative and outweighs $\sin(x)$
So $x+\pi/4$ is in $(3\pi/4,\pi)$, and: $$\begin{align}
\sin(x+\pi/4)&=\frac{1}{\sqrt{8}}\\
x+\pi/4&=\pi-\arcsin\left(\frac{1}{\sqrt{8}}\right)\\
x&=\pi-\arcsin\left(\frac{1}{\sqrt{8}}\right)-\pi/4\\
\tan(x)&=\tan\left(\pi-\arcsin\left(\frac{1}{\sqrt{8}}\right)-\pi/4\right)\\
\tan(x)&=-\tan\left(\arcsin\left(\frac{1}{\sqrt{8}}\right)+\pi/4\right)\\
\tan(x)&=-\frac{\tan\left(\arcsin\left(\frac{1}{\sqrt{8}}\right)\right)+1}{1-\tan\left(\arcsin\left(\frac{1}{\sqrt{8}}\right)\right)}\\
\tan(x)&=-\frac{\frac{\frac{1}{\sqrt{8}}}{\sqrt{1-\frac{1}{8}}}+1}{1-\frac{\frac{1}{\sqrt{8}}}{\sqrt{1-\frac{1}{8}}}}\\
\tan(x)&=-\frac{\frac{1}{\sqrt{8}}+\sqrt{\frac78}}{\sqrt{\frac78}-\frac{1}{\sqrt{8}}}\\
\tan(x)&=\frac{1+\sqrt{7}}{1-\sqrt{7}}\\\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding common ratio from two sums I'm struggling with this very basic question on the binomial theorem:
The sum of the first and second terms of a geometric progression is 12, and the sum of the third and fourth term is 48. Find the two possible values of the common ratio and the corresponding values of the first term.
So I tried approaching this in two ways. The first:
$$
u_1 + u_2 = a + ar = 12 \\
u_3 + u_4 = ar^2 + ar^3 = 48 \\
4a(1 + r) = ar^2 + ar^3 \\
4 + 4r = r^2 + r^3 \\
r^3 + r^2 - 4r - 4 =0
$$
Okay, no solution there. So then I tried:
$$
S_2 = \frac{a(r^2 - 1)}{r-1} = 12 \\
S_4 = \frac{a(r^4 - 1)}{r-1} = 48 \\
\frac{4a(r^2 - 1)}{r-1} = \frac{a(r^4 - 1)}{r-1} \\
4r^2 - 4 = r^4 - 1 \\
r^4 - 4r^2 + 3 =0
$$
Am I missing something obvious here?
| Continuing with the notation you've used, given $a + ar = 12$, $ar^2 +ar^3 = 48$, we see that
$ar^2 + ar^3 = r^2(a + ar)$
$\implies 48 = r^2*12$
$\implies r = \pm 2$
Back-substituting this gives us $r = 2$ and $a = 4$ or $r=-2$ and $a = -12$.
| {
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Show that $e^{iy} = 1 + iy + \frac{\mu_1 y^2}{2}$ for all $y \in \mathbb{R}$ with $|\mu_1| \leq 1$ How to show the expansions
\begin{gather*}
e^{iy} = 1 + iy + \frac{\mu_1 y^2}{2}\\
e^{iy} = 1 + iy - \frac{1}{2}y^2 + \frac{\mu_2 |y|^3}{3!}
\end{gather*}
where $y \in \mathbb{R}$ and $|\mu_1| \leq 1$ and $|\mu_2| \leq 1$ for all $y$?
I attempts to consider the remainder of Taylor expansion of one real-variable function: $e^{iy} = \cos y + i\sin y$, and their Taylor expansions with Lagrange remainders are
\begin{gather*}
\cos y = 1 - \frac{y^2}{2}\cos (\theta_1 y) y^2\\
\sin y = y - \frac{1}{2}\sin (\theta_2 y) y^2
\end{gather*}
So we may put $\mu_1 = -[\cos (\theta_1 y) + i \sin (\theta_2 y)]$. But here we only have $|\mu_1| \leq 2$, which does not satisfy the requirement...
Any help will be appreciated!
| You made a mistake in your second expansion. The correct one is
\begin{align*}
\cos y = 1 - \cos(\theta_1y)\frac{y^2}{2}, &&
\sin y = y - \sin(\theta_2 y)\frac{y^3}{6}
\end{align*}
then
\begin{align*}
e^{i y} &= \cos y + i \sin y \\
& = 1 + i y - \cos(\theta_1y)\frac{y^2}{2}- i\sin(\theta_2 y)\frac{y^3}{6}
\end{align*}
which is closer to your answer ;)
| {
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$a,b,c \geq 0$,prove that $a^2+b^2+c^2+abc+5 \geq3(a+b+c) $
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$a^2+b^2+c^2+abc+5 \geq3(a+b+c).$$
I'm certain that this problem could be solved by using dirchlet's theory.but I do not know how to apply it exactly.
| I am not sure if you can apply "dirchlet's theory". But another way is:
for $f(x)=px^2+qx+r, \Delta=q^2-4pr$, if $p>0 \cap\Delta \le 0 \implies f(x) \ge 0$
WLOG,let $ a \ge b \ge c$
$a^2+b^2+c^2+abc+5 \geq3(a+b+c) \iff a^2+(bc-3)a+b^2+c^2+5-3(b+c) \ge 0 \iff \Delta_a =(bc-3)^2-4(b^2+c^2+5-3(b+c)) \le 0 \iff (4-c^2)b^2+(6c-12)b+4c^2-12c+11 \ge 0 \iff 4-c^2>0 \cap \Delta_b=(6c-12)^2-4(4-c^2)(4c^2-12c+11) \le 0 \iff 4-c^2>0 \cap (c-1)^2(c-2)(c+1) \le 0 $
which is true. the "=" will hold $(c=1, \Delta_b=0) \cap (b=1,\Delta_a=0) \cap (a-1)^2=0 \implies a=b=c=1$
it remains $c^2 \ge 4$ or $c\ge 2$ case:
$abc\ge c^2a\ge4a ,abc\ge c^2b \ge 4b ,abc\ge 4c \implies abc\ge \dfrac{4(a+b+c)}{3} \implies a^2+b^2+c^2+abc+5 \geq3(a+b+c) \iff a^2+b^2+c^2+5 \ge \dfrac{5(a+b+c)}{3} \iff (a^2+ \dfrac{25}{36} \ge \dfrac{5a}{3} )\cap( b^2+ \dfrac{25}{36}\ge \dfrac{5b}{3}) \cap (c^2+ \dfrac{25}{36}\ge \dfrac{5c}{3}) \cap (5-3\times\dfrac{25}{36}=\dfrac{35}{12}>0)$
which means the inequality is always true and only ">" is hold.
| {
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Area for an ellipsoid "Calculate the area for the rotation ellipsoid you get by rotating the ellipsoid $\frac{x^2}{2}+y^2 = 1$ around the x-axis."
I solved for x:
$$ y = \pm \sqrt{1-\frac{x^2}{2}} $$
Then did $ y = 0$ to get the integration limits, $\pm \sqrt(2)$.
So I've ended up with an integral I don't know if it's correct, and even if it is, I can't solve it.
$$ 4\pi \int_{\sqrt{2}}^{\sqrt{2}} \sqrt{1-\frac{x^2}{2}} \sqrt{1-\frac{x}{2\sqrt{1-\frac{x^2}{2}}}}dx$$
| Yours is not correct. Since $\frac{dy}{dx}=-\frac{x}{2y}$ and $4y^2=2(2-x^2)$, we have
$$\begin{align}2\pi\int_{-\sqrt 2}^{\sqrt 2}\sqrt{1-\frac{x^2}{2}}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx&=2\pi\int_{-\sqrt 2}^{\sqrt 2}\sqrt{1-\frac{x^2}{2}}\sqrt{1+\left(-\frac{x}{2y}\right)^2}\ dx\\&=2\pi\int_{-\sqrt 2}^{\sqrt 2}\sqrt{\frac{\color{red}{2-x^2}}{2}\cdot \frac{2(2-x^2)+x^2}{2\color{red}{(2-x^2)}}}\ dx\\&=2\pi\int_{-\sqrt 2}^{\sqrt 2}\sqrt{\frac{4-x^2}{4}}\ dx\end{align}$$
I hope this helps!
| {
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Does $x^2+ x^4/(3\cdot4) + x^6/(3\cdot4\cdot5\cdot6) + \cdots$ have any compact form? Is there any compact form for the following series
$$F_1(x) = x^2+ \frac{x^4}{3\cdot4} + \frac{x^6}{3\cdot4\cdot5\cdot6} + \cdots$$
$$F_2(x) = x+ \frac{x^3}{2\cdot3} + \frac{x^5}{2\cdot3\cdot4\cdot5} + \cdots$$
$$F_3(x) = F_2(x) - F_1(x)$$
| Yes,
$F_1(x) = 2(\cosh x - 1)$;
$F_2(x) = \sinh x$.
(see Taylor series).
| {
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How to evaluate the integral $\int e^{2x} \sin^2x\ dx$ How can I evaluate the following integral?
$$\int e^{2x} \sin^2x\ dx$$
| Here are the steps
\[
\int e^{2x}\sin^{2} x\ dx=\frac{1}{2}\int e^{2x}(1-\cos(2x))\ dx
\]
Let $u=2x$
\[
\frac{d}{dx}u=2\Rightarrow du=2\ dx \Rightarrow \frac{1}{2}du=dx
\]
Then
\[
\frac{1}{2}\int e^{2x}(1-\cos(2x))\ dx= \frac{1}{4}\int e^{u}(1-\cos u)\ du = \frac{1}{4}\int e^{u}-e^{u}\cos u\ du
\]
\[
= \frac{1}{4}\int e^{u}\ du- \frac{1}{4}\int e^{u}\cos u\ du= \frac{1}{4}e^{u}- \frac{1}{8}e^{u}(\sin u+\cos u) +C
\]
\[
=\frac{2}{8}e^{u}- \frac{1}{8}e^{u}(\sin u+\cos u) +C=-\frac{1}{8}e^{2x}(\sin(2x)+\cos(2x)-2)+C
\]
Thus
\[
\int e^{2x}\sin^{2} x\ dx =-\frac{1}{8}e^{2x}(\sin(2x)+\cos(2x)-2)+C
\]
| {
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Finding the number of solutions to $x+2y+4z=400$ My question is how to find the easiest way to find the number of non-negative integer solutions to $$x+2y+4z=400$$
I know that I can use generating functions, and think of it as partitioning $400$ with $x$ $1$'s, $y$ $2$'s, and $z$ $4$'s. The overall generating function is: $$(1+x+x^2 + \cdots x^{400})(1+x^2+x^4+\cdots x^{200})(1+x^4+x^8+\cdots x^{100})$$
And then from this I have to calculate the coefficient of $x^{400}$, which I don't know how to do. If there's an easier way to do, I'd love to know.
| You were on the right track, but instead of truncating the factors, just consider the coefficient of $x^{400}$ in:
$$(1+x+x^2+x^3+\ldots)(1+x^2+x^4+x^6+\ldots)(1+x^4+x^8+x^{12}+\ldots)=\frac{1}{(1-x)(1-x^2)(1-x^4)},\tag{1}$$
then write the RHS of $1$ as a sum of terms like $\frac{A}{(1-\xi x)^k}$, with $\xi\in\{1,-1,i,-i\}$, and exploit the identities:
$$\frac{1}{1-\xi x}=\sum_{k=0}^{+\infty}(\xi x)^k,$$
$$\frac{1}{(1-\xi x)^2}=\sum_{k=0}^{+\infty}(k+1)(\xi x)^k,$$
$$\frac{1}{(1-\xi x)^3}=\sum_{k=0}^{+\infty}\frac{(k+2)(k+1)}{2}(\xi x)^k$$
to recover the final expression:
$$[x^n]\frac{1}{(1-x)(1-x^2)(1-x^4)}=\frac{2n^2+14n+21}{32}+\frac{(7+2n) (-1)^n}{32}+\frac{1}{8} \cos\left(\frac{n \pi }{2}\right)+\frac{1}{8} \sin\left(\frac{n \pi }{2}\right)$$
that if $8\mid n$ becomes:
$$[x^n]\frac{1}{(1-x)(1-x^2)(1-x^4)}=\frac{1}{16}(n+4)^2.$$
| {
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Conditional extreme value of a function Let $x,y,z$ be the positive real numbers, if $x^2+y^2+z^2=1$, then how can we find the minimal value of this function $f(x,y,z)=\dfrac{xz}{y}+\dfrac{yz}{x}+\dfrac{xy}{z}$.
| Hint First, note that $$\dfrac{xz}{y}+\dfrac{yz}{x}+\dfrac{xy}{z}=xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right).$$ Using $a+b+c\geq 3\sqrt[3]{abc}$, try to estimate lower bounds for $xyz$ and $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$. If you are stuck in the second, consider $$(x^2+y^2+z^2)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right).$$ Show that the conditions for minimizing $xyz$ and $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$ can happen at the same time.
| {
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Solve the equation $(x^2-9)+\sqrt{2-x}=0$ Solve the equation $(x^2-9)+\sqrt{2-x}=0$
*
*$(x+3)(x-3)+\sqrt{2-x}=0$
Conditions: $x\neq\pm3 \wedge x\leq2$
*$(x+3)(x-3) = -\sqrt{2-x}$
*$(x+3)^2(x-3)^2 = 2-x$
*$x^4-18x^2+81 = 2-x$
*$x^4-18x^2+x+79 = 0$
I'm positive this isn't going to the right direction.
Please help.
| The answer will be $x\approx -2.6175$. (see here. You can find its closed form if you want.)
Here, notice that we have $-3\le x\le 2$ because of $9-x^2=\sqrt{2-x}\ge 0$.
| {
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Matrix Multiplication Understanding I have to multiply two 3x3 matrices. I don't understand the answer.
$$A=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$$
$$B=\begin{pmatrix}1&0&-1\\ 0&1&-1\\ 0&0&1 \end{pmatrix}$$
The answer given online is $$AB=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$$
I don't understand how this is so. I have said that:
$$AB=\begin{pmatrix}(1)(1)+(0)(0)+(1)(0) & (1)(0)+(0)(1)+(1)(0) & (1)(-1)+(0)(-1)+(1)(1)\\(0)(1)+(1)(0)+(0)(0) & (0)(0)+(1)(1)+(1)(0) & (0)(-1)+(1)(-1)+(1)(1) \\ (0)(1)+(0)(0)+(1)(0) & (0)(0)+(0)(1)+(1)(0) & (0)(-1)+(0)(-1)+(1)(1) \end{pmatrix}$$
$$AB=\begin{pmatrix}1 & 0 & (1)(-1)+(0)(-1)+(1)(1)\\0 & 1 & (0)(-1)+(1)(-1)+(1)(1) \\ 0 & 0 & 1 \end{pmatrix}$$
$$AB=\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
Can anyone tell me where I am going wrong?
| I agree with your work, and if they posted $AB=\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}$ as the answer for the product, it seems they're incorrect.
Another fast way to see that their answer is not feasible is to notice that they are writing $AB=A$. Since $A$ is an upper triangular matrix with no zeros on the main diagonal, its determinant is clearly nonzero, so it is invertible. Then we can take $AB=A$ and multiply on the left with the inverse:
$$A^{-1}AB=A^{-1}A\implies B=I$$
But $B$ is obviously not the identity matrix.
| {
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Number of elements of order $2$ in $S_n$
How many elements of order $2$ are there in $S_n$?
Using combinatorics I arrived at this:
For $n$ even ($n=2k$) there are ${n\choose2}+{n\choose 2}{n-2\choose 2}\dfrac{1}{2!}+{n\choose 2} {n-2\choose 2}{n-4\choose 2}\dfrac{1}{3!}+\cdots+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\cdots{2\choose 2}\dfrac{1}{k!}$.
For $n$ odd ($n=2k+1$) there are ${n\choose 2}+{n\choose 2}{n-2\choose 2}\dfrac{1}{2!}+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\dfrac{1}{3!}+\cdots+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\cdots{3\choose 2}\dfrac{1}{k!}$
But how do I find the sums?
Seems like I have to use induction. But not quite upto there.
Thanks for the help!!
| it is probably not any help, but i think your sum may be written $f(\sqrt{2})$ where
$$
f(x) = \sqrt{2}^{-n} \sum_{k=1}^{\lfloor \frac{n}2 \rfloor} \frac1{k!}\frac{d^{2k}}{dx^{2k}}x^n
$$
| {
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Prove $\frac{a^2+b^2+c^2}{ab+bc+ca} + 8\frac{abc}{(a+b)(b+c)(c+a)} \ge 2$ Let $a,b,c>0$, prove that
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 2.$$
I tried using the equality $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ and the Schur inequality but it's not very helpful.
Thanks.
| I would like to use @math 110 idea with a little difference when we prove that
$$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}\ge 2$$
Since $\dfrac{a^2+b^2+c^2}{ab+bc+ac}=\dfrac{(a+b+c)^2}{ab+bc+ac}-2$, we can rewrite the inequality as
$\dfrac{(a+b+c)^2}{ab+bc+ac}\ge 4-\dfrac{4ac}{(a+c)^2}=\dfrac{4(a^2+ac+c^2)}{(a+c)^2}$
Or
$4(a^2+ac+c^2)(ab+bc+ac)\le (a+c)^2(a+b+c)^2$
Using AM-GM, we have
$4(a^2+ac+c^2)(ab+bc+ac)\le (a^2+ac+c^2+ab+bc+ac)^2
=[(a+c)^2+b(a+c)]^2
=(a+c)^2(a+b+c)^2$
| {
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How many ways can 5 dice produce a total of 20?
How many ways can $5$ dice produce a total of $20$?
I set up the equation $x_1+x_2+x_3+x_4+x_5 = 20$. The total possible number of combinations is $\binom{19}4$. From there I subtracted the number of possibilities where $1$ of the variables is greater than $6$, which I got as $5\times\binom{13}4$. I also subtracted the possibilities where $2$ variables is greater than $6$, which I used $10 \times \binom74$. I got the $10$ from the number of ways I can choose $2$ of the variables to be greater than $6$ out of the $5$ total variables.
So I have $$\binom{19}4 -5\times\binom{13}4 -10\times\binom74$$
However, I get a negative answer, which can't be right. Can anyone point a flaw in my logic?
| The maximum of the five dice is either $4, 5$, or $6$. Reasoning:
If the maximum of all five is $6$, then the maximum of the remaining four is $4, 5$ or $6$, because $6+3+3+3+3 < 20.$ If the maximum of all five is $5$, then the maximum of the remaining four is $4$ or $5$, because $5+3+3+3+3 < 20.$ If the maximum of all five is $4$, then they're all $4$. The maximum cannot be $3$ or less because $3+3+3+3+3 < 20.$
If the top two dice are $6,6$, the sum of the other three is $8$. Five combinations with all three $\leq 6$: $$(6,1,1), (5,2,1), (4,3,1), (4,2,2), (3,3,2).$$
If the top two dice are $6,5$, the sum of the other three is $9$. Five combinations with all three $\leq 5$: $$(5,3,1), (5,2,2), (4,4,1), (4,3,2), (3,3,3).$$
If the top two dice are $6,4$, the sum of the other three is $10$. Two combinations with all three $\leq 4$: $$(4,4,2), (4,3,3).$$
If the top two dice are $5,5$, the sum of the other three is $10$. Four combinations with all three $\leq 5$: $$(5,4,1), (5,3,2), (4,4,2), (4,3,3).$$
If the top two dice are $5,4$, the sum of the other three is $11$. Just one combination with all three $\leq 4$: $$(4,4,3).$$
Finally, there's $(4,4,4,4,4).$
So, there are $18$ combinations of $5$ dice that add up to $20$.
Edit: If order matters, then you'll need to consider the distinct permutations of each combination above. For each combination, there are $5!/n_i$ permutations, where
$$n_i =
\begin{cases}
2! = 2 & \text{if there is one pair}\\
2!2! = 4 & \text{if there are two pair}\\
3! = 6 & \text{if there are three of a kind}\\
3!2! = 12 & \text{if it's a full house}\\
5! = 120 & \text{if it's a yahtzee}\\
\end{cases}
$$
It's straightforward to verify that
$$\sum_{i=1}^{18} \frac{5!}{n_i} = 651.$$
| {
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Find the derivative: Do I use the Quotient Rule, Product Rule, or Chain Rule? I have to find the derivative of:
$$y=\frac{(5x^6-1)}{x^2}$$
I keep on getting this problem wrong. Should I use the quotient rule
$$\frac{f(x)g'(x) - g(x)f'(x)}{g(x)^2}$$
However, my answer is:
$$y'= 30x^2 +\frac{6}{x^4} - 120{x^3}$$
Am I utilizing the wrong method, or have I just evaluated the problem incorrectly?
| $$\frac{dy}{dx}=\frac{(5x^6-1)'x^2-(5x^6-1) (x^2)'}{x^4}=\frac{30x^5 \cdot x^2-2x(5x^6-1)}{x^4}=\frac{30x^7-10x^7+2x}{x^4}=\frac{20x^7+2x}{x^4}=20x^3+\frac{2}{x^3}$$
| {
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Show that each integer of the form $a^2+b^2$ has all the factors of this form, where $(a, b)$ are distinct integers and relatively prime
Show that each integer of the form $a^2+b^2$ has all the factors of this form, where $(a, b)$ are distinct integers and relatively prime
Progress
If $a^2+b^2$ is prime then it is already proved, since every prime is a factor of its self and $1=0^2+1^2$, but I don't know how to prove for the rest of the case. Thanks!
| Assuming that you are talking about numbers of the form $n=a^2+b^2$ with $\gcd(a,b)=1$, then every odd prime $p$ that divides $n$ has the property:
$$\left(\frac{-1}{p}\right)=1,\tag{1}$$
since $(1)$ is a consequence of $a^2+b^2\equiv 0\pmod{p}$. However, $(1)$ is equivalent to
$$ p\equiv 1\pmod{4},\tag{2}$$
since the Legendre symbol $\left(\frac{-1}{p}\right)$ equals $(-1)^{\frac{p-1}{2}}$.
Now it is possible to prove that any prime $\equiv 1\pmod{4}$ is the sum of two coprime squares - see, for instance, this similar question in which I proved that every prime $p\equiv 1\pmod{3}$ is represented by the quadratic form $a^2+3b^2$ through Fermat's descent - and since the numbers represented by $a^2+b^2$ are a semigroup, due to the Lagrange's identity:
$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2\tag{3},$$
we have that every divisor of $n$ is represented by the quadratic form $a^2+b^2$. Notice that $(3)$ is just equivalent to the multiplicativity of the norm on the ring $\mathbb{Z}[i]$ (the gaussian integers).
| {
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A cycle as a product of transpositions Can someone please explain how a cycle $(1234)$ can be written as a product of transpositions: $(14)(13)(12)$? And how they can be multiplied to (1234)? Thanks in advance.
| $$(14)=\begin{pmatrix}1&2&3&4 \\ 4&2&3&1\end{pmatrix}$$
$$(13)=\begin{pmatrix}1&2&3&4 \\ 3&2&1&4\end{pmatrix}$$
$$(12)=\begin{pmatrix}1&2&3&4 \\ 2&1&3&4\end{pmatrix}$$
$$(14)(13)=(14)\circ(13)=\begin{pmatrix}1&2&3&4 \\ 4&2&3&1\end{pmatrix}\circ\begin{pmatrix}1&2&3&4 \\ 3&2&1&4\end{pmatrix}=\begin{pmatrix}1&2&3&4 \\ 4&2&1&3\end{pmatrix}$$
$$(14)(13)(12)=((14)\circ(13))\circ(12)$$
$$(14)(13)(12)=\begin{pmatrix}1&2&3&4 \\ 4&2&1&3\end{pmatrix}\circ\begin{pmatrix}1&2&3&4 \\ 2&1&3&4\end{pmatrix}=\begin{pmatrix}1&2&3&4 \\ 4&1&2&3\end{pmatrix}=(1432)$$
| {
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Is there an easier way to find $\frac{\mathrm d^9}{\mathrm dx^9}(x^8\ln x)$ than using the product rule repeatedly?
Find $\dfrac{\mathrm d^9}{\mathrm dx^9}(x^8\ln x)$.
I know how to solve this problem by repeatedly using the product rule, but I was wondering if there is a short cut. Thanks.
| For any analytic function $f$,
$$\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} f(x) = f(x+z) $$
Thus $\dfrac{d^n f}{dx^n}$ is $n!$ times the coefficient of $z^n$ in the Maclaurin series of $f(x+z)$.
In this case $f(x+z) = (x+z)^8 \ln(x+z)$. Now $$
\eqalign{(x+z)^8 &= \sum_{j=0}^8 {8 \choose j} x^{8-j} z^j \cr
\ln(x+z) &= \ln(x) + \ln(1 + z/x) = \ln(x) + \sum_{k=1}^\infty (-1)^{k-1} \dfrac{z^k}{k x^k}\cr}$$
Thus what you want is
$$\eqalign{ 9! &\left( {8 \choose 0} x^8 \dfrac{1}{9 x^9} - {8 \choose 1} x^7 \dfrac{1}{8 x^8} + \ldots + {8 \choose 8} x^0 \dfrac{1}{1 x^1}\right)\cr
&=\dfrac{9!}{x} \left(\dfrac{1}{9}- \dfrac{8}{8} + \dfrac{28}{7} - \dfrac{56}{6} + \dfrac{70}{5} - \dfrac{56}{4} + \dfrac{28}{3} - \dfrac{8}{2} + \dfrac{1}{1}\right) = \dfrac{9!}{9x} = \dfrac{8!}{x} = \dfrac{40320}{x}}$$
How did this work out so nicely? You might notice that
$$ { 8 \choose j} \dfrac{1}{9-j} = \dfrac{8!}{j! (9-j)!} = \dfrac{1}{9} {9 \choose j}$$
and the $\dfrac{(-1)^j}{9} {9 \choose j}$ and $\dfrac{(-1)^{9-j}}{9} {9 \choose 9-j}$ terms cancel out for $j = 1 \ldots 4$, leaving just the $j=0$ term.
| {
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Solving these two equations simultaneously I'm having a hard time to solve these two equations simultaneously. I'm arriving to a very long equation..
$$x_0^2+y_0^2=(7\sqrt{2})^2=98$$
$$\sqrt{25+(x_0+2)^2}+\sqrt{4+(y_0-5)^2}=7\sqrt{2}$$
| Use polar coordinates $(x,y) = (-r \sin \theta, r \cos \theta)$ to get $r=\sqrt{98}$ and
$$ \sqrt{98 \cos^2 \theta - 2 \sqrt{98} (5) \cos \theta + (-2)^2 + (5)^2} + \\
\sqrt{98 \sin^2 \theta+2 \sqrt{98} (-2) \sin\theta + (-2)^2 +(5)^2 } = \sqrt{98} $$
Numerically there are two solutions at $\theta=28.27437°$ and $\theta=45°$ for the solutions
$$ (x_0,y_0) = ( -4.6893, 8.7184 ) \\ (x_0,y_0) = ( -7, 7 ) $$
| {
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How to find $\lim_{x\to 0}\frac{\tan 3x}{\tan 5x}$? I am asked to find the following limit:
$$ \lim_{x \to 0} \frac{\tan 3x}{\tan 5x}$$
My problem is in simplifying the function. I followed two different approaches to solve the problem. But both seems incorrect.
Apprach 1)
Since $\tan \theta = \frac{sin \theta}{cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$, we have:
$$\frac{\tan 3x}{\tan 5x} = \frac{\sin 3x \times \cos 5x}{\sin 5x \times \cos 3x}$$
This approach does not work well, because I cannot simplify more.
Approach 2)
Since $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \times \tan y)}$, we have:
$$\frac{\tan 3x}{\tan 5x} = \frac{\tan 3x}{\tan 3x + \tan 2x} = \frac{\tan 3x}{\frac{\tan 2 + \tan3}{1 - \tan 2 \times \tan 3}}$$
What am I doing wrong?
| Using L'Hopital's:
$$\lim \limits_{x \to 0} \frac{\tan 3x}{\tan 5x} = \lim \limits_{x \to 0} \frac{3 \sec^2 3x}{5 \sec^2 5x} = \frac{3}{5}$$
Without L'Hopital's:
$$\lim \limits_{x \to 0} \frac{\tan 3x}{\tan 5x} = \frac{3}{5} \lim \limits_{x \to 0} \frac{\cos 5x}{\cos 3x} \cdot \frac{\sin 3x}{3x} \cdot \frac{5x}{\sin 5x} = \frac{3}{5}$$
This uses the fact that $\lim \limits_{u \to 0} \frac{\sin u}{u} = 1$. This approach derives the identities $\lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac ab$ and $\lim_{x \to 0} \frac{\tan ax}{\tan bx} = \frac ab$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Differentiate with product rule Question: differentiate $x(x^2 +3x)^3$
I have gotten to the point where i've used the product rule and i've gotten
$$(x^2 + 3x)^3 + x\cdot(3x+9)(x^2 + 3x)^2$$
but now that it comes to the simplifying im completely at a loss, any help would greatly appreciated
| Using the product rule is not enough.
$$(x(x^2 + 3x)^3)^\prime = (x)^\prime \cdot (x^2 + 3x)^3 + x \cdot ((x^2 + 3x)^3)^\prime$$
$$= (x^2 + 3x)^3 + x \cdot ((x^2 + 3x)^3)^\prime$$
For the rigth part you need to use the chain rule.
$$= (x^2 + 3x)^3 + x \cdot 3(x^2 + 3x)^2\cdot(x^2+3x)^\prime$$
$$= (x^2 + 3x)^3 + x \cdot 3(x^2 + 3x)^2\cdot(2x+3)$$
Now you can simplifying the result:
$$= (x^2 + 3x)^2 \cdot \left((x^2 + 3x) + x \cdot 3\cdot(2x+3)\right)$$
$$= (x^2 + 3x)^2 \cdot \left(x^2 + 3x + 6x^2 + 9x\right)$$
$$= (x^2 + 3x)^2 \cdot \left(7x^2 + 12x\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Make $n$ cents with $1$-cent, $2$-cent, and $3$-cent coins I encountered the following problem in Herber Wilf's book Generatingfunctionology:
Prove that, in country that has $1$-cent, $2$-cent, and $3$-cent coins
only, the number of ways of changing $n$ cents is exactly the integer
nearest to $ \frac{(n+3)^2}{12} $.
Of course, the solution provided was one that makes use of a generating function. However, it is unbelievably tedious. Is there a way to do this without making use of a generation function?
For reference, here is the solution given in the book.
If $f(n)$ is that number, then
$$ \begin {gather*}
\sum_{n\ge 0} f(n) x^n =
\frac {1}{(1-x)(1-x^2)(1-x^3)}\qquad\qquad\qquad\qquad \\
= \frac{1}{6(1-x)^3} + \frac{1 }{4(1-x)^2} +
\frac {17}{72(1-x)} + \frac {1}{8(1+x)} + \frac {1}{9(1-\omega x)} +
\frac {1}{9(1-\overline{\omega}x)}.
\end {gather*} $$
If we expand each of
the fractions on the right, we find the formula $$ f(n) = \frac {1}{6}
\dbinom {n+2}{2} + \frac {1}{4} (n+1) + \frac {17}{72} + \frac
{(-1)^n}{8} + \frac {2}{9} \cos \left( \frac {2n \pi}{3} \right), $$
which can be rewritten as
$$ f(n) = \frac {(n+3)^2}{12} + \frac {-7 + 9 (-1)^n
+ 16 \cos \left( \frac {2n\pi}{3} \right)}{72}. $$
The second
fraction cannot exceed $ \frac {32}{72} < \frac {1}{2} $ in absolute
value, so $f(n)$ is the unique integer who distance from
$ \frac {(n+3)^2}{12} $ is less than $\frac{1}{2}$, as required. $\Box$
| I'll only show how to obtain a linear recurrent relation from the problem, it has been explained at various places how to solve these.
Let:
*
*$a(n)$ be solutions (ways how to change $n$) using only $1$c,
*$b(n)$ solutions using only $1,2$c and at least one $2$c,
*$c(n)$ solutions using all three coins and at least one $3$c.
Clearly each solution fits exactly into one of the 3 categories, therefore $f(n)=a(n)+b(n)+c(n)$.
We have $a(n)=1$. For $b(n)$, we remove one $2$c, and either we get something with another $2$c left or not. Therefore $b(n)=b(n-2)+a(n-2)$. Similarly $c(n)=c(n-3)+b(n-3)+a(n-3)$. Of course, this works only for $n>3$, so we have to add that $a(1)=a(2)=a(3)=b(2)=b(3)=c(3)=1$ and $b(1)=c(1)=c(2)=0$. This linear recurrence should have the desired solution.
However, as you see, it's much simpler to just use OGF for this kind of problems.
| {
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If $a_{1}\;a_{2},a_{3}$ are the Roots of cubic eq. , Then $1000\left(a^2_{1}+a^2_{2}+a^2_{3}\right)$
If $a_{1}\;a_{2},a_{3}$ are three real values of $a$ which satisfy the equation $$\displaystyle \int_{0}^{1}\left(\sin x+a\cdot \cos x\right)^3dx-\frac{4a}{\pi-2}\int_{0}^{1}x\cdot \cos xdx = 2.$$ Then value of $\displaystyle 1000\left(a^2_{1}+a^2_{2}+a^2_{3}\right) = $
$\bf{My\; Trial::}$ Let $\displaystyle I = \int_{0}^{1}\left(\sin x+a\cdot \cos x\right)^3dx=\int_{0}^{1}\left(\sin^3 x+a^3\cos^3 x+3a\sin^2 x\cdot \cos x+3a^2\sin x\cdot \cos^2 x \right)dx$
So $\displaystyle I = -\int_{0}^{1}(1-\cos^2 x)\cdot (\cos x)^{'}dx+a^3\int_{0}^{1}(1-\sin^2 x)\cdot (\sin x)^{'}dx+3a\int_{0}^{1}(\sin x)^2\cdot (\sin x)^{'}dx-3a^2\int_{0}^{1} (\cos x)^2\cdot (\cos x)^{'}dx$
and Let $\displaystyle J = \int_{0}^{1}x\cdot \cos xdx = \left[x\cdot \sin x+\cos x\right]_{0}^{1} = \left(\sin 1+\cos 1-1\right)$
Now How can I solve after that
Help me
Thanbks
| Continuing from JimmyK4542's answer, we find that $$C_3 = \displaystyle\int_0^1 \cos^3 x\,dx=\frac{1}{12} (9 \sin (1)+\sin (3))$$ $$C_2 = \displaystyle\int_0^1 3\sin x \cos^2 x\,dx=1-\cos ^3(1)$$ $$C_1 = \displaystyle\int_0^1 3\sin^2 x \cos x\,dx - \dfrac{4}{\pi - 2}\int_0^1 x\cos x\,dx=\sin ^3(1)-\frac{4 (-1+\sin (1)+\cos (1))}{\pi -2}$$ $$C_0 =\displaystyle-2+\int_0^1 \sin^3 x\,dx=\frac{1}{12} (-16-9 \cos (1)+\cos (3))$$ and $$a_1^2+a_2^2+a_3^2 = (a_1+a_2+a_3)^2-2(a_1a_2+a_2a_3+a_3a_1) = \dfrac{C_2^2-2C_1C_3}{C_3^2}$$ can be computed. However, the result is far to be simple; the numerical value is $4.02458872035$.
But, as I said in my comment, two of the roots of the cubic equation are complex $$a_1=-1.29971 + 0.712738 i$$ $$a_2=1.29971 - 0.712738 i$$ $$a_3=1.28923$$
Could it be possible that there are typo's in the expression ?
| {
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If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$ if $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$
My try :
$2pp' = (b^2-a^2)\sin 2\theta$
$p'^2 + pp'' = (b^2-a^2)\cos 2\theta$
Thats it ! it doesn't simplify no matter what I try. Any help ?
| Since
$$p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta$$
we have
$$p=\mu (a^2 \cos^2 \theta+b^2 \sin^2 \theta)^{1/2}, \mu=\pm 1$$
$$p + p'' =\frac{\mu a^2b^2(\cos^4\theta+\sin^4 \theta+2\cos^2\theta \sin^2\theta)}{(a^2 \cos^2 \theta+b^2 \sin^2 \theta)^{3/2}}=\frac{a^2b^2}{p^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/907124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometry and triangle proof Question:
Prove that in an acute angle triangle ABC:
$$\tan A\tan B +\tan A \tan C + \tan B \tan C \geq 9$$
I have no idea where to even begin this question. Please help me!
| Since for an acute triangle $ABC$ we have
$$\begin{align}\tan A+\tan B&=(\tan (A+B))\cdot (1-\tan A\tan B)\\&=(\tan(\pi -C))\cdot (1-\tan A\tan B)\\&=(-\tan C)\cdot (1-\tan A\tan B),\end{align}$$
we have$$\tan A+\tan B+\tan C=\tan A\tan B\tan C.$$
Hence, we have, by AM-GM inequality,$$\tan A+\tan B+\tan C\ge 3\sqrt[3]{\tan A\tan B\tan C}$$
$$\iff \tan A\tan B\tan C\ge 3\sqrt[3]{\tan A\tan B\tan C}$$
$$\iff \tan A\tan B\tan C\ge 3\sqrt 3.$$
Hence, we have, by AM-GM inequality,
$$\begin{align}\tan A\tan B+\tan B\tan C+\tan C\tan A&\ge 3\sqrt[3]{\tan^2A\tan^2B\tan^2C}\ge 9.\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What happens to a function when it is undefined? If I have the function
$$f(x) = {x^2 - 2 \over x + \sqrt 2}$$
this is undefined for $x = -\sqrt 2$, am I correct? Since the denominator would be zero.
But the numerator is a difference of squares $x^2-2=(x - \sqrt 2)(x+\sqrt 2)$.
Now the second factor is equal to my denominator, so those cancel and I am left with $(x - \sqrt 2)$, and this is defined at $x=-\sqrt 2$, is it not? If I put $x = -\sqrt 2$, the result is simply $-\sqrt 2 - \sqrt 2=-2\sqrt 2$. What am I doing wrong here?
| It is important to realise what you are doing when we say that a common factor "cancels" from the numerator and denominator. We are actually dividing both the numerator and denominator by that common factor. As you have already pointed out: you can't divide by zero.
It is true that
$$\frac{x^2-2}{x+\sqrt{2}} \equiv \frac{(x-\sqrt{2})(x+\sqrt{2})}{x+\sqrt{2}}$$
Neither the left hand side, nor the right hand side are defined when $x=-\sqrt{2}$. Both are indeterminate forms. However, if we assume that $x\neq -\sqrt{2}$ then we may divide both the numerator and the denominator by $x+\sqrt{2}$ to give
$$\frac{x^2-2}{x+\sqrt{2}} = \frac{(x-\sqrt{2})(x+\sqrt{2})}{x+\sqrt{2}} = x-\sqrt{2}$$
The next step to understand this is the idea of the limit. If we take the limit as $x$ tends to $-\sqrt{2}$, we assume that $x \neq-\sqrt{2}$, but that it gets closer and closer. Even though the original function is not defined when $x = -\sqrt{2}$, we can look what happens to its value as $x \to -\sqrt{2}$. In that case we have
$$\lim_{x \to -\sqrt{2}}\left(\frac{x^2-2}{x+\sqrt{2}}\right) = \lim_{x \to -\sqrt{2}}\left(\frac{(x-\sqrt{2})(x+\sqrt{2})}{x+\sqrt{2}}\right) = \lim_{x \to -\sqrt{2}}\left(x-\sqrt{2}\right) = -2\sqrt{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Summing various rearrangements of $1-\frac12+\frac13-\frac14+\cdots$ Show that the series $1-\frac12+\frac13-\frac14+\ldots$ converges to $\log2$ but the rearranged series:
*
*$1-\frac12-\frac14-\frac16-\frac18+\frac13-\frac{1}{10}-\frac{1}{12}-\frac{1}{14}-\frac{1}{16}+\frac15-\ldots$ converges to $0$.
*$1+\frac13-\frac12+\frac15+\frac17-\frac14+\ldots$ converges to $\frac32\log2$
*$1+\frac13-\frac12-\frac14+\frac15+\frac17-\frac16+\ldots$ converges to $\log2$
*$1+\frac13+\frac15-\frac12+\frac17+\frac19+\frac{1}{11}-\frac14+\ldots$ converges to $\frac12\log12$
I have absolutely no idea on how to approach these kind of problems. Please help me out from the very start. I consulted books but I didn't understand. I have knowledge on tests for convergence and absolute convergence and the statement of Riemann's theorem.
| Assuming that I have correctly understood the patterns occurring, I get:
(1)
$$\begin{eqnarray*}\sum_{j=0}^{+\infty}\left(\frac{1}{2j+1}-\frac{1}{8j+2}-\frac{1}{8j+4}-\frac{1}{8j+6}-\frac{1}{8j+8}\right)&=&\sum_{j=0}^{+\infty}\left(-\frac{1}{8j+2}+\frac{3}{8j+4}-\frac{1}{8j+6}-\frac{1}{8j+8}\right)\\&=&\int_{0}^{1}\frac{-x+3x^3-x^5-x^7}{1-x^8}\,dx\\&=&0.\end{eqnarray*}$$
(2)
$$\begin{eqnarray*}\sum_{j=0}^{+\infty}\left(\frac{1}{4j+1}+\frac{1}{4j+3}-\frac{2}{4j+4}\right)&=&\int_{0}^{1}\frac{1+x^2-2x^3}{1-x^4}\,dx\\&=&\frac{3}{2}\log 2.\end{eqnarray*}$$
(3)
$$\begin{eqnarray*}\sum_{j=0}^{+\infty}\left(\frac{1}{4j+1}+\frac{1}{4j+3}-\frac{1}{4j+2}-\frac{1}{4j+4}\right)&=&\int_{0}^{1}\frac{1+x^2-x-x^3}{1-x^4}\,dx\\&=&\log2.\end{eqnarray*}$$
(4)
$$\begin{eqnarray*}\sum_{j=0}^{+\infty}\left(\frac{1}{6j+1}+\frac{1}{6j+3}+\frac{1}{6j+5}-\frac{3}{6j+6}\right)&=&\int_{0}^{1}\frac{1+x^2+x^4-3x^5}{1-x^6}\,dx\\&=&\frac{1}{2}\log 12.\end{eqnarray*}$$
All the integrals can be computed through simple fractions decomposition.
| {
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Proof by induction (exponents)
Use proof by induction and show that the formula holds for all positive integers:$$1+3+3^2+\dots+3^{n-1}=\frac{3^n -1}2$$
The confusing step in my opinion is the first expression: $3^{n-1}$, when I have to show for $k+1$. Any solutions?
| For n+1 it is
$1+3+3^2+\ldots+3^{n-1}+3^n=\frac{3^{n+1}-1}{2}$
$\left( 1+3+3^2+\ldots +3^{n-1}\right) +3^n=\frac{3\cdot 3^{n}-1}{2}$
$\left( 1+3+3^2+\ldots +3^{n-1}\right) +3^n=\frac{(1+2)\cdot 3^{n}-1}{2}$
$\left( 1+3+3^2+\ldots +3^{n-1}\right) +3^n=\frac{1\cdot 3^{n}-1}{2}+\frac{2\cdot 3^{n}}{2}$
$\left( 1+3+3^2+\ldots +3^{n-1}\right) +3^n=\left( \frac{1\cdot 3^{n}-1}{2}\right) + 3^n$
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.