Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
How do I solve $\int\frac{\sqrt{x}}{\sqrt{x}-3}\,dx?$ $$
\int \frac{\sqrt{x}}{\sqrt{x}-3}dx
$$
What is the most dead simple way to do this?
My professor showed us a trick for problems like this which I was able to use for the following simple example:
$$
\int \frac{1}{1+\sqrt{2x}}dx
$$
Substituting:
$u-1=\sqrt{x}$
being used to create
$\int\frac{u-1}{u}$
which simplifies to the answer which is:
$1+\sqrt{2x}-\ln|1+\sqrt{2x}|+C$
Can I use a similar process for the first problem?
|
if $t=\sqrt{x}-3$, $x=\left(t+3\right)^2$ then $\dfrac{dt}{dx}=\frac{1}{2\sqrt{x}}\implies dt=\frac{dx}{2\sqrt{x}}\implies2\sqrt{x}dt=dx$
so$$\frac{\sqrt{x}}{\sqrt{x}-3}dx=\frac{\sqrt{x}}{\sqrt{x}-3}d2\sqrt{x}dt=2\frac{x}{t}dt=2\frac{\left(t+3\right)^2}{t}dt=2\left[\frac{t^2+6t}{t}+\frac{9}{t}\right]dt=2\left[t+6+\frac{9}{t}\right]dt=\left[2t+12+\frac{18}{t}\right]dt$$
integrate this:$$\int2t+12+\frac{18}{t}dt=\int2tdt+\int12dt+\int\frac{18}{t}dt=t^2+12t+18\ln t+c{=\left(\sqrt{x}-3\right)^2+12\left(\sqrt{x}-3\right)+18\ln\left|\sqrt{3}-x\right|}{=x-6\sqrt{x}+9+12\sqrt{x}-36+18\ln\left|\sqrt{3}-x\right|}\\{=x+6\sqrt{x}+18\ln\left|\sqrt{3}-x\right|-27+c,c_1=-27+c}\\\rightarrow x+6\sqrt{x}+18\ln\left|\sqrt{3}-x\right|+c_1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2446162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
}
|
Prove for all $x \in \mathbb{R}$ at least one of $\sqrt{3}-x$ and $\sqrt{3}+x$ is irrational I have attempted to do some work on this, but I'm not sure if I'm heading in the correct direction. If I am heading in the correct direction, then I'm not quite sure where to take it from here.
Roughly what I have so far is:
We will use proof by contradiction
Assume $\sqrt{3}-x$ and $\sqrt{3}+x$ is rational
Then $\sqrt{3}-x=\frac{a}{b}$ for some $a,b \in \mathbb{R}$ and $\sqrt{3}+x=\frac{c}{d}$ for some $c,d \in \mathbb{R}$
Then $(\sqrt{3}-x)^2=\frac{a^2}{b^2}$ and $(\sqrt{3}+x)^2=\frac{c^2}{d^2}$
Then $(\sqrt{3}-x)(\sqrt{3}-x)=\frac{a^2}{b^2}$ and $(\sqrt{3}+x)(\sqrt{3}+x)=\frac{c^2}{d^2}$
Then $x^2-2\sqrt{3}x+3=\frac{a^2}{b^2}$ and $x^2+2\sqrt{3}x+3=\frac{c^2}{d^2}$
Then $x(x-2\sqrt{3})+3=\frac{a^2}{b^2}$ and $x(x+2\sqrt{3})+3=\frac{c^2}{d^2}$
That's as far as I have got with this.
|
The sum of two rational numbers is rational, but
$$(\sqrt 3 -x)+(\sqrt 3+x)=2\sqrt 3$$
which is irrational. Therefore they cannot both be rational.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2446734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If we know range of a function , how can construct range of other function?
If we know $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$ how can obtain the range of the function below?
$$y=\frac{4 x}{9x^2+25}$$
The problem is what's the range with respect to $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$
|
$$y=\frac{4 x}{9x^2+25}\\y=\frac{4 x}{25(\frac{9}{25}x^2+1)}=\\y=\frac4{25}\frac{ x}{((\frac{3}{5}x)^2+1)}$$ now take $u=\frac{3x}{5}$
$$y=\frac4{25}\frac{ x}{((\frac{3}{5}x)^2+1)}=\\\frac4{25}\frac{ \frac{5}{3}u}{((u)^2+1)}=\\\frac{4}{15}\frac{u}{u^2+1}\\$$you know $\frac{-1}{2}\leq \frac{u}{u^2+1}\leq \frac{1}{2}$ so
$$-\frac{4}{15}\times \frac {1}{2} \leq y=\frac{4}{15}\frac{u}{u^2+1} \leq \frac{4}{15}\times \frac {1}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2451512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Find p is the prime number which $\frac{p+1}{2}$ and $\frac{p^2+1}{2}$ both are square number. Find p is the prime number which $\dfrac{p+1}{2}$ and $\dfrac{p^2+1}{2}$ both are square number.
I do not know how to use "p is prime" assumption given. I just know
$p=7$ is satisfied.
If $\dfrac{p+1}{2}=X^2$ and $\dfrac{p^2+1}{2}=Y^2$ then $(X^2;X^2-1;Y)$ is Pythagorean triple
|
\begin{align}
& p=2{x}^{2}-1 \\
& {{p}^{2}}=2{{y}^{2}}-1 \\
& \Rightarrow {{p}^{2}}-p=2{{y}^{2}}-2{{x}^{2}} \\
& \Rightarrow p(p-1)=2(y-x)(y+x)
\end{align}
Wlog, assume $x,y$ are positive and it is easy to see $p>2$.
Then, from $(4)$, we have $y> x$ and:
a) $p|y-x$ and $2(y+x)|p-1$
we get $p\leq y-x$ and $y+x\leq {p-1\over 2}$ so $p\leq {p-1\over 2}$ and no solution.
b) $p|y+x$ and $2(y-x)|p-1$
we get $p\leq y+x$ and $y-x\leq {p-1\over 2}$ so $p\leq 4x-1$.
But $p={2x^2-1}$, so we have ${2x^2-1}\leq 4x-1$ and thus $\boxed{x\leq 2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2453304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Basis for Kernel and Image of a linear map I have to find the basis for the kernel and image of the following linear map.
$ \phi: R^3 → R^2, ϕ \begin{pmatrix} \begin{pmatrix} x \\y \\z \end{pmatrix}\end{pmatrix}= \begin{pmatrix} x -y \\z \end{pmatrix} $
For the range, I think we can express any arbitrary linear transformation as:
$ x\begin{pmatrix} 1 \\ 0 \end{pmatrix} - y\begin{pmatrix} 1 \\ 0 \end{pmatrix} + z\begin{pmatrix} 0 \\ 1 \end{pmatrix} $. So I think that a basis for the range would be $\left\{{{\begin{pmatrix} 1 \\ 0 \end{pmatrix}},{\begin{pmatrix} 0 \\ 1 \end{pmatrix}}}\right\}$
As for the kernel, we set
$\begin{pmatrix} x -y \\z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
We get $x=y$ and $z=0$. Therefore a basis for the kernel would be
$\left\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\right\}$
I am doing this right?
Thanks in advance
|
Everything looks good. A slightly more elementary (but less insightful) approach that works quite well for calculations that are a little harder to "guess" as you say is to form the matrix for the linear transformation by checking the basis:
$f(1,0,0)=(1,0), \, f(0,1,0)=(-1,0) \, , f(0,0,1)=(0,1)$
So, we get the matrix
$$f:\mathbb R^3 \to \mathbb R^2 \iff\begin{pmatrix}1 & -1 & 0\\0 &0 & 1 \end{pmatrix}:\mathbb R^3 \to \mathbb R^2$$
From this, you can determine the "column space" since the first and third column vectors are linearly independent, it has to be $\mathbb R^2$ for the image, and the kernel is generated by the "null space" which amounts to solving a homogeneous system of linear equations, i.e:
$$\begin{pmatrix}1 & -1 & 0\\0 &0 & 1 \end{pmatrix} \begin{pmatrix} x \\y \\z\end{pmatrix}=\begin{pmatrix}0\\0 \end{pmatrix}$$
although this is just a restatement of your own method, it computationally allows the use of Gauss-Jordan Elimination and identifies an isomorphic subspace in $\mathbb R^3$ orthogonal to the image that is also one dimensional.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2453647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find all n for which $2^n + 3^n$ is divisible by $7$? I am trying to do this with mods. I know that:
$2^{3k+1} \equiv 1 \pmod 7$, $2^{3k+2} \equiv 4 \pmod 7$, $2^{3k} \equiv 6 \pmod 7$ and $3^{3k+1} \equiv 3 \pmod 7$, $3^{3k+2} \equiv 2 \pmod 7$, $3^{3k} \equiv 6 \pmod 7$, so I thought that the answer would be when $n$ is a multiple of $3$ since then $2^{3k} + 3^{3k} \equiv 1 + 6 \pmod 7$, but it doesn't work for $n = 6$
|
We need $$2^n+3^n\equiv0\pmod7\iff3^n\equiv-2^n\iff-1\equiv(3\cdot4)^n\equiv(-2)^n$$ using $2\cdot4\equiv1\pmod7\iff2^{-1}\equiv4$
$\implies(-2)^n\equiv-1\equiv(-2)^3\iff(-2)^{n-3}\equiv1$
$\implies n-3\equiv0\pmod6$ as $(-2)^3\equiv-1,(-2)^6\equiv1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2454402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
}
|
Why isn't there a vertical asymptote at $x = -1$ for $f(x) = \frac{(x^3-1)(\ln(x+2))}{2x^3+3x^2-2x-3}$ Why isn't there a vertical asymptote at $x = -1$ for $f(x) = \frac{(x^3-1)(\ln(x+2))}{2x^3+3x^2-2x-3}$
I understand that there is a removable discontinuity at x = 1 since $x^3 - 1 = (x-1)(x^2+x+1)$ but I don't know how x = -1 is not a vertical asymptote.
|
hint
The denominator is
$$(2x+3)(x-1)(x+1) $$
and we use
$$\lim_{x\to-1}\frac {\ln (x+2)}{x+1}=$$
$$\lim_{X\to 0}\frac {\ln (X+1)}{X}=1$$
we find that
$$\lim_{x\to-1}f (x)=\frac {-2}{-2}=1$$
Thus, there is No vertical asymptote at $x=-1$ since the limit is finite.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2454740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove fixed point iteration of $g(x) = \frac{1}{2}x + \frac{A}{2x}$ converges to $\sqrt{A}$ If we define, for any positive number $A$:
\begin{align*}
g(x) &= \frac{1}{2}x + \frac{A}{2x} \\
g'(x) &= \frac{1}{2} - \frac{A}{2x^2} \\
\end{align*}
And we define the normal fixed point iteration sequence $\{ p_n \}_{n=0}^\infty$ based on $p_n = g(p_{n-1}), n \ge 1$
How can we prove that this will converge to $\sqrt{A}$ when $p_0 > 0$? What happens if $p_0 < 0$?
The normal textbook proof of fixed point convergence depends on $g'(x) \le k < 1$ which is clearly not the case here.
|
I am not sure that this could be an answer.
$$g(x) = \frac{1}{2}x + \frac{A}{2x}=x -\frac{x^2-A}{2x}$$
$$x_{n+1} = x_n -\frac{x_n^2-A}{2x_n}$$ is the Newton formula for solving $x^2-A=0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2455260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
What is $\lim _{ x\rightarrow 0 } \frac { f(x) -x }{ x^2 }$? Given that
$$f(x)=8x-f(3x)-\sin^2(2x),$$
find
$$\lim _{ x\rightarrow 0 } \frac { f\left( x \right) -x }{ x^2 }$$
|
Note that
$$ f(x)=f(0)+f'(0)x+\frac12f''(0)x^2+O(x^3). $$
Then from $f(x)=8x-f(3x)-\sin^2(2x)$, one has
\begin{eqnarray}
&&\bigg[f(0)+f'(0)x+\frac12f''(0)x^2+O(x^3)\bigg]+\bigg[f(0)+3f'(0)x+\frac{9}2f''(0)x^2+O(x^3)\bigg]\\
&=&8x-\sin^2(2x)
\end{eqnarray}
which gives
$$ 2f(0)+4f'(0)x+5f''(0)x^2+O(x^3)=8x-\sin^2(2x). $$
Noting that $\sin(2x)\approx 2x$ and hence
$$f(0)=0,f'(0)=2,f''(0)=-\frac{4}{5}. $$
So
$$ \lim_{x\to0}\frac{f(x)-x}{x^2}=\frac{f(0)+f'(0)x+\frac12f''(0)x^2+O(x^3)-x}{x^2}=\lim_{x\to0}\frac{x-\frac{2}{5}x^2}{x^2}=DNE.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2455880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Then find the value of ... Let $a, \: b, \: c$ be three variables which can take any value (Real or Complex). Given that $ab + bc + ca = \frac{1}{2}$; $a + b + c = 2$; $abc = 4$. Then find the value of $$\frac{1}{ab + c - 1} + \frac{1}{bc + a - 1} + \frac{1}{ac + b - 1}$$
I always get stuck in such problems. Please give a detailed solution and also give me an approach that I can use to solve other such kind of problems.
|
Another way is to find a cubic $p(x)$ with roots $ab+c, bc+a, ca+b$ (which is easier than the original problem) and then use simple transformations. Here using elementary symmetric polynomials and Vieta, we get $p(x) = x^3-\frac52x^2- \frac52x-\frac{65}4$.
Now get the polynomial $p_1(x)$ which has roots $ab+c-1, bc+a-1, ca+b-1$, as $p_1(x) = p(x+1) = x^3+\frac12x^2-\frac92x-\frac{81}4$
Finally the polynomial with reciprocal roots is got by reversing coefficients just so:
$p_2(x) = -\frac{81}4x^3-\frac92x^2+\frac12x+1$
So the sum of these roots is just $-\frac92 \div \frac{81}4 = -\frac29$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2456096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Binomial theorem relating proof There is this identity
$$1 -\frac{1}{2}\binom{n}{1}+\frac{1}{3} \binom{n}{2}- \frac{1}{4}\binom{n}{3}+....+(-1)^n \frac{1}{n+1}\binom{n}{n}$$
And we are supposed to prove it using these two identities
$$k\binom{n}{k} = n\binom{n-1}{k-1}$$
and
$$\binom{n}{0} + \binom{n}{1} + \binom{n}{2} +....+ \binom{n}{n} = 2^n$$
I have been working on this problem for a long time. Can you guys help me?
|
\begin{eqnarray}
&&1 - \frac{1}{2}\binom{n}{1}+\frac{1}{3} \binom{n}{2}- \frac{1}{4}\binom{n}{3}+....+(-1)^n \frac{1}{n+1}\binom{n}{n}\\
&=&\sum_{k=0}^n(-1)^{k}\frac{1}{k+1}\binom{n}{k}\\
&=&\sum_{k=0}^n(-1)^{k}\frac{1}{n+1}\binom{n+1}{k+1}\\
&=&\frac{1}{n+1}\sum_{k=1}^{n+1}(-1)^{k}\binom{n+1}{k}\\
&=&-\frac{1}{n+1}\left[\sum_{k=0}^{n+1}(-1)^{k}\binom{n+1}{k}-1\right]\\
&=&\frac1{n+1}
\end{eqnarray}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2456255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
If $f(1)=1$, then is it true that $f(n)=n$ for all $n \in \mathbb{N}\cup\{0\}$. Let $f:\mathbb{N}\cup\{0\}\to\mathbb{N}\cup\{0\}$ be a function which satisfies $f(x^2+y^2)=f(x)^2+f(y)^2$ for all $x,y \in\mathbb{N}\cup\{0\}$. It' easy to see that $f(0)=0$ and $f(1)=0$ or $f(1)=1$. Suppose let's assume that $f(1)=1$. Then it's very easy to see that $f(2)=2$ and since $f(2)=2$, we can see that $f(4)=4$ and $f(5)=5$ because $5=2^2+1^2$. To see $f(3)=3$, note that $$25=f(5^2)=f(3)^2+f(4)^2 \implies f(3)=3$$ One can even see that $f(6)=6$, $f(7)=7$ and so on. Is it generally true that $f(n)=n$ for all $n$?
Edit. If anyone wondering how $f(7)=7$ here is the way. Note that $f$ satisfies $f(n^2)=f(n)^2$. One can see that $$25^{2}=24^2+7^2 \implies 25^{2}=f(24)^{2}+f(7)^{2}$$ We have to show $f(24)=24$. For this note that $$26^2 = 24^2+10^2 \implies f(26)^{2} = f(24)^{2}+f(10)^{2}$$ But $f(26)=26$ because $26=5^2+1^2$ and $f(10)=10$ since $10=3^{2}+1^{2}$. In short if $n=x^2+y^2$ for some $x,y$ and $f(1)=1$, then we can see that $f(n)=n$.
|
Yes.
Suppose you already know that $f(n)=n$ for all $n < N$. You want to find $a,b,c$, all smaller than $N$, such that $N^2+a^2=b^2+c^2$. If you find such a triplet, you immediately conclude that $f(N)=N$, and you have the necessary inductive step.
For odd $N>6$, use
$N^2+\left(\frac{N-5}{2}\right)^2 = (N-2)^2+\left(\frac{N+3}{2}\right)^2$
For even $N>6$, use
$N^2+\left(\frac{N}{2}-5\right)^2 = (N-4)^2 + \left(\frac{N}{2}+3\right)^2$
Since the question details already prove that $f(n)=n$ for small values of $n$, this inductive step finishes the argument.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2456751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
}
|
How to prove that $x^y - y^x = x + y$ has only one solution in positive integers? Prompted by this question, I tried to show that $(2,5)$ is the only solution in positive integers of $x^y - y^x = x+y$ (which would show, a fortiori, that it's the only solution in primes). It's convenient to rewrite the equation as $f(x,y) = x^y - y^x - x - y = 0$.
With the aid of some trial calculations, I reasoned informally as follows:
If $x=1$ then $f(x,y) = -2y=0$, implying $y=0$.
If $x=2$ and $y \leq 5$, then a case-by-case check shows that only $(2,5)$ is a solution.
If $(x,y) = (2,6)$ then $f(x,y) = 20$, and as $y$ increases above $6$, $f(x,y)$ increases.
If $x \geq 3$ and $y \leq x$ then $f(x,y) < 0$.
If $x \geq 3$ and $y=x+1$, then $f(x,y)>0$, and as $y$ increases above $x+1$, $f(x,y)$ increases.
How can the above be made into a rigorous proof? I've included calculus as a tag since it could be useful in showing under what conditions $f(x,y)$ is an increasing function of $y$ (viewing it as a real variable).
|
If $(x,y) = (2,6)$ then $f(x,y) = 20$, and as $y$ increases above $6$, $f(x,y)$ increases.
To prove this rigorously, let us prove by induction that $$2^{X-1}\gt X+1\tag1$$ for $X\ge 6$.
The base case : $2^{6-1}=32\gt 7=6+1$.
Supposing that $(1)$ holds for some $X$ gives
$$2^{(X+1)-1}=2\cdot 2^{X-1}\gt 2(X+1)=X+X+2\gt (X+1)+1\quad\blacksquare$$
Using $(1)$, we get, for $y\ge 6$,
$$f(2,y+1)-f(2,y)=2(2^{y-1}-y-1)\gt 0$$
If $x \geq 3$ and $y \leq x$ then $f(x,y) < 0$.
*
*$f(x,1)=-2\lt 0$.
*To prove that $f(x,2)\lt 0$ for $x\ge 3$, let us prove by induction that $$2^X\gt X^2-X-2\tag2$$ for $X\ge 4$. The base case : $2^4=16\gt 10=4^2-4-2$. Supposing that $(2)$ holds for some $X$ gives $$\begin{align}2^{X+1}&=2\cdot 2^X\\&\gt 2(X^2-X-2)\\&=(X+1)^2-(X+1)-2+X(X-3)-2\\&\ge (X+1)^2-(X+1)-2+4(4-3)-2\\&\gt (X+1)^2-(X+1)-2\quad\blacksquare\end{align}$$ Also, it is easy to see that $(2)$ holds for $X=3$. Using this, we get, for $x\ge 3$, $$f(x,2)=(x^2-x-2)-2^x\lt 0$$
*To prove that $f(x,y)\lt 0$ for $3\le y\le x$, we use the fact that $y=\frac{\ln x}{x}$ is decreasing for $x\gt e$. We see that $\frac{\ln y}{y}\ge \frac{\ln x}{x}$ for $3\le y\le x$ from which $x^y-y^x\le 0$ follows. It follows from this that we have, for $3\le y\le x$, $$f(x,y)=x^y-y^x-x-y\le 0-x-y\lt 0$$
If $x \geq 3$ and $y=x+1$, then $f(x,y)>0$, and as $y$ increases above $x+1$, $f(x,y)$ increases.
This answer proves that $f(x,y)\gt 0$ for $x\ge 3$ and $y\ge x+2$.
Let $y=x+a$ where $a\ge 2$ is an integer. Then,
$$\frac{f(x,x+a)}{x^x}=x^{a}-\left(1+\frac ax\right)^x-\frac{2}{x^{x-1}}-\frac{a}{x^x}\gt x^a-e^a-\frac 29-\frac{a}{x^x}:=g(x)$$
Since $g'(x)=ax^{a-1}+\frac{a+a\ln x}{x^x}\gt 0$, $g(x)$ is increasing with $g(3)=3^a-e^a-\frac 29-\frac{a}{27}:=h(a)$.
We have $h'(a)=3^a\ln 3-e^a-\frac{1}{27}$ and $h''(a)=3^a(\ln 3)^2-e^a\gt 0$ with $h'(2)=9\ln 3-e^2-\frac{1}{27}\gt 0$ and $h(2)=9-e^2-\frac 29-\frac{2}{27}\gt 0$.
So, $h(a)\gt 0$ for $a\ge 2$. It follows from this that $g(x)\gt 0$ for $x\ge 3$.
Therefore, $f(x,x+a)\gt 0$ follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2457087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
}
|
Limit of $\sqrt{x^2+3x}+x$ when $x\to-\infty$ Limit of $ \lim_{x\to -\infty}(\sqrt{x^2+3x}+x)$, I know that the final answer is $-3/2$, my question is about Wolfram Alpha step by step solution:
$$x+\sqrt{x^2+3x}=\frac{(x+\sqrt{x^2+3x})(x-\sqrt{x^2+3x})}{x-\sqrt{x^2+3x}}$$
$$=-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\infty}-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\infty}-\frac{3x}{x-\sqrt{x^2+3x}}=-3$$
$$\lim_{x\to-\infty}\frac{x}{x-\sqrt{x^2+3x}}=-3\lim_{x\to-\infty}\frac{x}{x-\sqrt{x^2+3x}}$$
$$\frac{x}{x-\sqrt{x^2+3x}}=\frac{1}{1-\frac{\sqrt{x^2+3x}}{x}}$$
$$-3\lim_{x\to-\infty}\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$$
To prepare the product $\frac{1}{1-\frac{\sqrt{x^2+3x}}{x}}$ for solution by l'Hopital's rule, write it as $\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$
$$-3\lim_{x\to-\infty}\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$$
Is it correct to use L'Hopital here like Wolfram did?
|
Yes, Wolfram is free to do that because for all finite $x$,
$$\frac{f(x)}{g(x)}=\frac{xf(x)}{xg(x)}$$ by "unsimplification", so that the limits are the same.
You can even write a "modified L'Hospital rule" theorem saying
$$\lim\frac fg=\lim\frac{f'}{g'}=\lim\frac{f+xf'}{g+xg'},$$ if that has any use.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2457183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
How do I find details of a parabola from its general two-degree equation? This general equation in two-degree represents a parabola: $$(ax + by)^2 + 2gx + 2fy + c = 0$$
How do I find the following from this equation:
*
*Vertex
*Focus
*Axis
*Length of Latus Rectum
*Co-ordinates of end points of Latus Rectum
*Equation of Directrix
I know how to find these if a parabola has its axis parallel to either of the co-ordinate axes. But I don't know how to derive these from the general equation.
N.B.: w.r.t. a comment by @Vasya, I produce this picture from my book, where it says that the general form of the equation of a parabola is the above equation. The book gives the equation, but doesn't derive any of the above from the equation:
Hence the general form of the equation of a parabola is
$$ (ax + by)^2 + 2gx + 2fy + c = 0. $$
|
$y'=bx-ay$ is parallel to the directrix and $x'=ax+by$ parallel to the axis. This we can see from comparing to the form I mentioned in the comments. So let these be the new coordinates. The inverse is $x=\frac{ax'+by'}{a^2+b^2}$, $\frac{bx'-ay'}{a^2+b^2}$, transforming your equation into $$x'^2+2g\frac{ax'+by'}{a^2+b^2}+2f\frac{bx'-ay'}{a^2+b^2}+c=0,$$ solving for $y'$ we get $$y'=\frac{a^2+b^2}{2(fa-gb)}x'^2+\frac{ga+fb}{fa-gb}x'+c\frac{a^2+b^2}{2(fa-gb)}=Ax'^2+Bx'+C,$$ making (from the formulas given here) $x'=-\frac{B}{2A}$ or $ax+by=-\frac{ga+fb}{a^2+b^2}$ the axis and $y'=\frac{4AC-B^2-1}{4A}$ or $bx-ay=\frac{c(a^2+b^2)-g^2-f^2}{2(fa-gb)}$ the directrix. The semi latus rectum is $p=\frac1{2A}=\frac{fa-gb}{a^2+b^2}$. For the vertex and focus to undo the transformation we transform the point back. The vertex in the rotated coordinates is $(-\frac{B}{2A},\frac{4AC-B^2}{4A})$ making the vertex in the first coordinates $$(\frac12 \frac{-2 g a^3 f+g^2 a^2 b-2 f^2 b a^2+b c a^4+2 c a^2 b^3+c b^5-f^2 b^3}{(a^2+b^2)^2 (f a-g b)},-\frac12 \frac{-2 g^2 a b^2+f^2 b^2 a-2 f b^3 g+c a^5+2 c a^3 b^2+a c b^4-g^2 a^3}{(a^2+b^2)^2 (f a-g b)})$$ and the focus is $(-\frac{B}{2A},\frac{4AC-B^2+1}{4A})$, using the inverse transformation the focus is $$(\frac12 \frac{b c a^2-2 f a g+g^2 b+c b^3-f^2 b}{(a^2+b^2) (f a-g b)},-\frac12 \frac{c a^3+a f^2+a c b^2-g^2 a-2 f g b}{(a^2+b^2) (f a-g b)}).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2457587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solve $2\log_bx + 2\log_b(1-x) = 4$ I need to solve $$2\log_bx + 2\log_b(1-x) = 4.$$
I have found two ways to solve the problem. The first (and easiest) way is to divide through by $2$:
$$\log_bx + \log_b(1-x) = 2.$$ Then, combine the left side: $$\log_b[x(1-x)] = 2,$$ and convert to the equivalent exponential form, $$b^2 = x(1-x) \;\;\implies\;\; x^2 - x + b^2 = 0,$$ which by the quadratic equation, we get $$x = {1\over2}\left(1\pm\sqrt{1 - 4b^2}\right).$$
My question is: How do I know that neither solution is extraneous? These are the solutions in the back of the textbook, but I am left questioning when these are the solutions.
The alternative solution involves some more clever thinking:
$$\begin{align}2\log_bx + 2\log_b(1-x) &= 4\\\log_bx^2(1-x)^2 &= 4\\ b^4 &= x^2(x^2 - 2x + 1)\\ 0&=x^4 - 2x^3 + x^2 - b^4\\ &=x^2(x - 1)^2 - b^4\\ & = [x(x-1)]^2 - (b^2)^2 \\ &= [x(x-1) + b^2][x(x-1) - b^2],\end{align}$$ which implies that $$x = {1\over2}\left(1 \pm\sqrt{1 + 4b^2}\right) \;\;\;\text{or} \;\;\; {1\over2}\left(1 \pm\sqrt{1 - 4b^2}\right),$$
which is even worse, because now there are $4$ solutions to check.
The most I know is that since $b$ is a logarithmic base, $b>0$ and $b\ne 1$. But what happens when $b$ is something like $2$? Then you end up with a complex number under the root (of the solutions from the easier way), and that doesn't necessarily make sense if I'm trying to solve the logarithmic equation over the reals.
|
You can see from the equation itself that are are, in general, two solutions: If $x$ is a solution, then so is $1-x$, since $1-(1-x)=x$.
As for the two spurious solutions $x=(1\pm\sqrt{1+4b^2})/2$ in your second approach, note that for these two, either $x$ or $1-x$ is negative. If the equation were written in the seemingly equivalent form $\log_b(x^2)+\log_b((1-x)^2)=4$, those would indeed be solutions. But since it's written as $2\log_bx+2\log_b(1-x)=4$, they are not, because you cannot take the log of a negative number.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2458109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Proof that $(abc)^2 = \frac{(c^6-a^6-b^6)}{3}$ We are given the length of the sides of a right triangle T, where $a \leq b \leq c$. We are asked to prove if $T$ is a right triangle. then $(abc)^2 = (c^6-a^6-b^6)/3$.
What I tried:
I tried substituting with the Pythagorean theorem $a^2 + b^2 = c^2$.
Where I am stuck:
I am unable to get $\frac{1}{3}$ or $3$ anywhere in my simplification. Is there some kind of identity that I am missing?
|
Hint : Cube the equation $\color{blue}{a^2+b^2=c^2}$
\begin{eqnarray*}
a^6+b^6+\color{red}{3} a^2 b^2 (\color{blue}{a^2+b^2}) =c^6.
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2459375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How would one prove that for all positive real $a$ and $b$, $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$? I think that the best way to prove this would be to prove by contradiction. Am I right?
If so, can I just assume that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ and say $a = 1$ and $b = 1$ and show that $\sqrt{1+1} \neq \sqrt{1} + \sqrt{1}$, or is that not enough?
|
If
$\sqrt{a + b} = \sqrt a + \sqrt b, \tag 1$
then, squaring,
$a + b = a + 2\sqrt a \sqrt b + b, \tag 2$
whence
$2\sqrt a \sqrt b = 0, \tag 3$
which forces
$a = 0 \vee b = 0. \tag 4$
Contradiction!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2459873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
}
|
Finding the value of the definite integral $\int_0^2{x\int_x^2{\frac{dy}{\sqrt{1+y^3}}}}dx$ If $$f(x) = \int_x^2{\frac{dy}{\sqrt{1+y^3}}}$$
then find the value of $$\int_0^2{xf(x)}dx$$
I have no idea how to solve this question. Please help.
|
Using integration by parts
$$\int_0^2xf(x)dx=\frac{1}{2}\int_{x=0}^{x=2}f(x)d(x^2)$$
$$=\frac{1}{2}x^2f(x)\bigg|_0^2-\frac{1}{2}\int_0^2x^2f'(x)dx$$
$$=2f(2)+\frac{1}{2}\int_0^2x^2\frac{1}{\sqrt{1+x^3}}dx$$
$$=0+\frac{1}{3}\sqrt{1+x^3}\bigg|_0^2$$
$$=\frac{1}{3}(3-1)$$
$$=\frac{2}{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2460221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Finding integral solutions of $x+y=x^2-xy+y^2$
Find integral solutions of $$x+y=x^2-xy+y^2$$
I simplified the equation down to
$$(x+y)^2 = x^3 + y^3$$
And hence found out solutions $(0,1), (1,0), (1,2), (2,1), (2,2)$ but I dont think my approach is correct . Is further simplification required? Is there any other method to solve this? I am thankful to those who answer!
|
$x+y=x^2-xy+y^2\to y^2-(x+1) y+x^2-x=0$
$y=\dfrac{1}{2} \left(1+x\pm\sqrt{-3 x^2+6 x+1}\right)$
discriminant must be positive $\Delta=-3 x^2+6 x+1\geq 0\to \dfrac{1}{3} \left(3-2 \sqrt{3}\right)\leq x\leq \dfrac{1}{3} \left(3+2 \sqrt{3}\right)$
which for integer $x$ means, $0\leq x \leq 2$
For $x=0$ we get $y=0;\;1$ solutions are $\color{red}{(0,0)\;(0;\;1)}$
for $x=1$ we have $y=0;\;y=2$ so $\color{red}{(1,0)\;(1;\;2)}$
for $x=2$ finally $y=1;\;y=2$ so $\color{red}{(2,1)\;(2;\;2)}$
hope this helps
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2460474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Proving $\frac{x+1}{y+1}$ is equal to or less than $\frac{x-1}{y-1}$ I know that $\frac{x+1}{y+1}$ is equal to (when $x = y$) or less than $\frac{x-1}{y-1}$. Suppose we cross multiply, then
*
*$xy - x+ y -1$ is less than or equal to $xy + x - y -1$
So,
*$y- x$ is less than or equal to $x- y$.
But when $y$ is greater than $x$ then the inequality is wrong.
Where am I wrong?
|
You have
$$
f(x,y)=\frac{x+1}{y+1}-\frac{x-1}{y-1}=
\frac{(xy+y-x-1)-(xy+x-y-1)}{(y^2-1)}=
\frac{2(y-x)}{(y^2-1)}
$$
Then
$$
f(x,y)\begin{cases}
>0 & \text{if $y>x$ and $|y|>1$} \\[4px]
>0 & \text{if $y<x$ and $|y|<1$} \\[4px]
<0 & \text{if $y>x$ and $|y|<1$} \\[4px]
<0 & \text{if $y<x$ and $|y|>1$}
\end{cases}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2462867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Combinatorics - Hexadecimals, how many possible numbers.... with ascending order?
How many different numbers with 4 figures can you make out of the sixteen hexadecimal ’figures’ $\{0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F\}$
$16^4$ possible solutions
(i) How many numbers are ’real’ 4-figure numbers, meaning the first figure $ \neq 0$?
$16^4 - 16^3 = 61440$ numbers, because $0,16,16,16 $are the undesirable outcome
(ii) How many numbers from (i) have four different figures?
$15 \cdot 14 \cdot 13 \cdot 12 = 32760$ out of the $61440$ possible numbers
(iii) How many numbers from (ii) end with figure 0?
$15 \cdot 14 \cdot 13 \cdot 11$?
(iv) How many numbers from (ii) have figures in increasing order?
This is where I am confused the most. I think it would be $1/4! \times 4$ since there are $4!$ ways to order $4$ numbers initiating with $1$ number, but there are $4$ numbers.
|
How many $4$-character strings can be formed using the sixteen hexadecimal digits?
Your answer $16^4$ is correct?
How many $4$-digit numbers can be formed using the sixteen hexadecimal digits?
Your answer $16^4 - 16^3$ is correct.
How many $4$-digit hexadecimal numbers have distinct digits?
We have $15$ choices for the leading digit since we cannot use $0$, $15$ choices for the next digit since we can now use $0$ but cannot use the leading digit, $14$ choices for the third digit since we cannot use the first two digits, and $13$ choices for the fourth digit since we cannot use the first three digits.
$$15 \cdot 15 \cdot 14 \cdot 13$$
How many $4$-digit hexadecimal numbers with distinct digits end with $0$?
We have $1$ choice for the last digit, $15$ choices for the leading digit, $14$ choices for the second digit, and $13$ choices for the third digit.
$$15 \cdot 14 \cdot 13 \cdot 1$$
How many $4$-digit hexadecimal numbers with distinct digits have digits written in increasing order?
Observe that since the leading digit cannot be $0$ and the digits are increasing, the four digits of such a number must be selected from the set
$$\{1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F\}$$
A $4$-digit hexadecimal number with strictly increasing digits is completely determined by which $4$ of these $15$ digits are selected since once the four digits are chosen, there is only one way to arrange them in increasing order. Hence, there are
$$\binom{15}{4}$$
such numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2463404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Why does the discriminant in the Quadratic Formula reveal the number of real solutions? Why does the discriminant in the quadratic formula reveal the number of real solutions to a quadratic equation?
That is, we have one real solution if
$$b^2 -4ac = 0,$$
we have two real solutions if
$$b^2 -4ac > 0,$$
and we have no real solutions if
$$b^2 -4ac < 0.$$
|
Because for $a\neq0$ we obtain:
$$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right).$$
Now, we see that if $b^2-4ac<0$ then $\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}>0$,
which says that the equation $ax^2+bx+c=0$ has no solutions.
For $b^2-4ac=0$ we have one root only:
$$x_1=-\frac{b}{2a}$$ and for $\Delta=b^2-4ac>0$ our equation has two distinct roots:
$$x_1=\frac{-b+\sqrt{\Delta}}{2a}$$ and
$$x_2=\frac{-b-\sqrt{\Delta}}{2a}$$ because in this case we obtain:
$$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right)=$$
$$=a\left(x+\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}\right)\left(x+\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}\right)=$$
$$=a\left(x-\frac{-b+\sqrt{\Delta}}{2a}\right)\left(x-\frac{-b-\sqrt{\Delta}}{2a}\right)=a(x-x_1)(x-x_2).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2464409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "59",
"answer_count": 10,
"answer_id": 0
}
|
Find $\int\sqrt{t^{3/2}+1}\,dt$
Integrate $\int\sqrt{t^{3/2}+1}\,dt$
I tried all possible substituion but its not worked.
Wolfram alpha also gives something different answer
https://www.wolframalpha.com/input/?i=integrate+(%E2%88%9A(t%E2%88%9At%2B1))
Can someone give me just a hint please?
|
When $|t|\leq1$ ,
$\int\sqrt{t^\frac{3}{2}+1}~dt$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n}{2}}{4^n(n!)^2(1-2n)}dt$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n+2}{2}}{4^n(n!)^2(1-2n)\dfrac{3n+2}{2}}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3n+2}{2}}{2^{2n-1}(n!)^2(1-2n)(3n+2)}+C$
When $|t|\geq1$ ,
$\int\sqrt{t^\frac{3}{2}+1}~dt$
$=\int t^\frac{3}{4}\sqrt{1+\dfrac{1}{t^\frac{3}{2}}}~dt$
$=\int t^\frac{3}{4}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)t^\frac{3n}{2}}dt$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{3-6n}{4}}{4^n(n!)^2(1-2n)}dt$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t^\frac{7-6n}{4}}{4^n(n!)^2(1-2n)\dfrac{7-6n}{4}}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^{n-1}(n!)^2(2n-1)(6n-7)t^\frac{6n-7}{4}}+C$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2465695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Quadrilateral in Square
$S$ is a unit square. Four points are taken randomly, one on each side of $S$. A quadrilateral is drawn. Let the sides of this quadrilateral be $a,b,c,d$. Prove that $2\leq{}a^2+b^2+c^2+d^2\leq{}4$.
My Efforts:
Let
$\begin{align}m^2+t^2&=a^2&\mathfrak{a}\\n^2+o^2&=b^2&\mathfrak{b}\\p^2+q^2&=c^2&\mathfrak{c}\\r^2+s^2&=d^2&\mathfrak{d}\end{align}$
$\mathfrak{a+b+c+d}\text{ gives}$
$m^2+n^2+o^2+p^2+q^2+r^2+s^2+t^2=a^2+b^2+c^2+d^2$
Since $m+n=o+p=q+r=s+t=1$,
This gives
$1-2mn+1-2op+1-2qr+1-2st=a^2+b^2+c^2+d^2$
Simplifying,
$4-2{(mn+op+qr+st)}=a^2+b^2+c^2+d^2$
Since the minimum value of $2{(mn+op+qr+st)}$ is $0$, I get
$a^2+b^2+c^2+d^2\leq4$.
Is there any other way to do this? How do I get $2\leq{}a^2+b^2+c^2+d^2$?
|
For the lower bound, note by CS inequality, $$(a^2+b^2+c^2+d^2)\cdot8=(m^2+n^2+\cdots)(1+1+\cdots)\geqslant(m+n+\cdots)^2=4^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2467024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Prove identity in a triangle I want to show that if $ABC$ is a triangle then
$$\sin^2(A/2)+ \sin^2(B/2) + \sin^2(C/2) =1-2\sin(A/2) \sin(B/2) \sin(C/2)$$
Well I eventually got it after much algebra, but I am looking for a shorter solution, or maybe even a geometric one?
|
Let $\alpha=\pi-2A$, $\beta=\pi-2B$ and $\gamma=\pi-2C$.
Thus, $\alpha+\beta+\gamma=\pi$ and we need to prove that
$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1,$$
which is obvious for acute-angled triangle $ABC$ (it's just law of cosines for new triangle).
In the general case we obtain:
$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=$$
$$=\cos^2\alpha+\cos^2\beta+\cos^2(\alpha+\beta)-2\cos\alpha\cos\beta\cos(\alpha+\beta)=$$
$$=\cos^2\alpha+\cos^2\beta+\cos^2\alpha\cos^2\beta+\sin^2\alpha\sin^2\beta-2\sin\alpha\sin\beta\cos\alpha\cos\beta-$$
$$-2\cos\alpha\cos\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=$$
$$=\cos^2\alpha+\cos^2\beta-\cos^2\alpha\cos^2\beta+\sin^2\alpha\sin^2\beta=$$
$$=\cos^2\alpha\sin^2\beta+\cos^2\beta+\sin^2\alpha\sin^2\beta=\sin^2\beta+\cos^2\beta=1.$$
Done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2468209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
Solve Recurrence relation -:$T(n)=T(n-1)+6n^{2}+2n$
Solve Recurrence relation -:$T(n)=T(n-1)+6n^{2}+2n$
Base case $T(1)=8$
$T(n)=T(n-1)+6n^{2}+2n$
$T(n-1)=T(n-2)+6(n-1)^{2}+2(n-1)............(1)$
$T(n-2)=T(n-3)+6(n-2)^{2}+2(n-2)...........(2)$
Substituting $(1)\,\, \text{and} \,\,(2) \text{in our question},$
$T(n)=T(n-k)+6(n-k)^{2}+2(n-k)+......6(n-2)^{2}+2(n-2)+6(n-1)^{2}+2(n-1)+6n^{2}+2n$
$T(n)=T(n-k)+6((n-k)^{2}+....+(n-2)^{2}+(n-1)^{2}+n^{2})+2((n-k)+...(n-2)+(n-1)+2n)$
finding $k$ using base case.
$T(n-k)=8$
$\Rightarrow n-k=1$
$\Rightarrow k=n-1$
putting the value of $k$ in our question.
$T(n)=T(n-k)+6((n-n+1)^{2}+....+(n-2)^{2}+(n-1)^{2}+n^{2})+2((n-n+1)+...(n-2)+(n-1)+n)$
$T(n)=T(1)+6(1^{2}+2^{2}+3^{3}+....n^{2})+2*(1+2+3..+n)$
$$T(n)=8+6\frac{n*(n+1)(2n+1)}{6}+2* \frac{n*(n+1)}{2}$$
$$T(n)=n*(n+1)(2n+1+1)+8$$
$$T(n)=n*(2n+2)*(n+1)+8$$
But annsweris given as $$T(n)=n*(2n+2)*(n+1)$$
Am i wrong?
Please help
|
I think it should be $$T(n)=T(1)+\sum_{k=2}^n(6k^2+2k)=T(1)+\sum_{k=1}^n(6k^2+2k)-8=$$
$$=T(1)+6\cdot\frac{n(n+1)(2n+1)}{6}+2\cdot\frac{n(n+1)}{2}-8=$$
$$=6\cdot\frac{n(n+1)(2n+1)}{6}+2\cdot\frac{n(n+1)}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2468759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
If $a+b+c=0$ prove that $ (a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$(a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$$
What is a good way to do this?
This question came from answering this slightly harder question. Those answers were somewhat hard to understand for me. To get something easier to digest I made a very similar but easier (lower exponent) question.
|
Let $e_1 = a+b+c, e_2 = ab+bc+ac, e_3 = abc$.
Since $A = (a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$ is symmetric and homogeneous of degree $4$, it is a linear combination of $e_1^4, e_1^2e_2, e_1e_3, e_2^2$, and when $e_1=0$ the only nonzero one is possibly $e_2^2$ :
There is a coefficient $k$ such that if $e_1=0$ then $A = ke_2^2$.
To check that $k$ is $0$, we only need to check that the identity is true for one case where $e_1=0$ and $e_2 \neq 0$.
Picking for example $a=0,b=1,c=-1$, you get $4=4$ so things work out.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2469668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
On the way finding limit.
Find The Limit $$\lim_{x\to0}\left(\frac{\tan(x)}{x}\right)^{\displaystyle\frac{1}{x^{2}}}$$
My Approach
We know that $$\lim_{x\to0}\frac{\tan(x)}{x}
= 1$$
So $$\lim_{x\to0}
\left[ 1+\left(\frac{\tan(x)}{x}-1\right)\right]
^{\displaystyle\left(\frac{1}{\left(\frac{\tan(x)}{x}-1\right)}\right)\left(\frac{\tan(x)}{x}-1\right)\left(\frac{1}{x^{2}}\right)}
=e^{\left[\displaystyle\lim_{x\to0}\left(\frac{\tan(x)}{x}-1\right)\left(\frac{1}{x^{2}}\right)\right]}$$
From here I tried many methods but couldn’t get the solution. I also tried
the logrithmic approach but everything got messesed up.
|
$$\lim_{x\to0}\left(\frac{\tan(x)}{x}\right)^{\frac{1}{x^{2}}}=e^{\lim_{x\to0}\ln\left(\left(\frac{\tan(x)}{x}\right)^{\frac{1}{x^{2}}}\right)}=e^{\lim_{x\to0}\frac{\ln\left(\frac{\tan(x)}{x}\right)}{x^2}}=e^{\frac{0}{0}}\\\lim_{x\to0}\frac{\ln\left(\frac{\tan(x)}{x}\right)}{x^2}=\lim_{x\to0}-\dfrac{\tan\left(x\right)-x\sec^2\left(x\right)}{2x^2\tan\left(x\right)}=-\lim_{x\to0}\frac{-2x\sec^2\left(x\right)\tan\left(x\right)}{4x\tan\left(x\right)+2x^2\sec^2\left(x\right)}=\\-\lim_{x\to0}\frac{-4x\sec^2\left(x\right)\tan^2\left(x\right)-2\sec^2\left(x\right)\tan\left(x\right)-2x\sec^4\left(x\right)}{4x^2\sec^2\left(x\right)\tan\left(x\right)+4\tan\left(x\right)+8x\sec^2\left(x\right)}=\\-\lim_{x\to0}\frac{-8x\sec^2\left(x\right)\tan^3\left(x\right)-8\sec^2\left(x\right)\tan^2\left(x\right)-16x\sec^4\left(x\right)\tan\left(x\right)-4\sec^4\left(x\right)}{8x^2\sec^2\left(x\right)\tan^2\left(x\right)+24x\sec^2\left(x\right)\tan\left(x\right)+4x^2\sec^4\left(x\right)+12\sec^2\left(x\right)}\\=-\frac{-4}{12}=\frac{1}{3}\\\therefore\lim_{x\to0}\left(\frac{\tan(x)}{x}\right)^{\frac{1}{x^{2}}}=e^{\frac{1}{3}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2471556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Inductive proof for recursive formula So, I have a recursion in which $$a_0 = 5$$ $$a_1 = 1$$ $$a_{n+2} = a_{n+1} + 2a_n$$
I should then prove by induction that the formula $a_n = 2^{n+1} + 3(-1)^n$ works for every number.
Anyway, I generally know how to use induction as proof but doesn't really have a clue on how to use it when it comes to a recursive formula. I have tried to prove the base case but don't really know how to use $k+1$ to my advantage afterwards since I'm not sure how to represent $n = k$ and then that it works for every $n$ by $k + 1$.
|
Start with
$a_{n+2} = a_{n+1} + 2a_n$
and substitute
the induction assumption
$a_n = 2^{n+1} + 3(-1)^n
$
for $n$ and $n+1$.
This gives
$\begin{array}\\
a_{n+2}
&= a_{n+1} + 2a_n\\
&= (2^{n+2} + 3(-1)^{n+1}) + 2(2^{n+1} + 3(-1)^n)\\
&= 2^{n+2} + 3(-1)^{n+1} + 2^{n+2} + 6(-1)^n\\
&= 2^{n+3} + 3(-1)^{n+1} - 6(-1)^{n+1}\\
&= 2^{n+3} - 3(-1)^{n+1}\\
&= 2^{n+3} + 3(-1)^{n+2}\\
\end{array}
$
Therefore,
if
$a_n = 2^{n+1} + 3(-1)^n
$
is true
for $n$ and $n+1$,
it is true for $n+2$.
Since it is true for
$0$ and $1$
(by direct calculation),
it is true for all $n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2472103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
Prove that $\int_0^\infty \frac{x\,\log x}{(1+x^2)^2} = 0$ I have been asked to prove that $$\int_0^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx = 0$$
I tried proceeding as follows:
\begin{align}
\int_0^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx &= \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx + \int_1^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx \\
& = \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx + \int_{u=1}^0 \frac{u\,\log u}{(1+u^2)^2}\,du\quad (\text{where }u=1/x) \\
&= \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx - \int_0^1 \frac{u\,\log u}{(1+u^2)^2}\,du \\
&=^? \int_0^1 0 \,dx \\
&= 0
\end{align}
But I think that I should first show that the improper integral $\int_0^1 {x\,\log x}/{(1+x^2)^2}\,dx$ is convergent.
Do I need to show it? If yes, how to do so?
(I have been taught basic methods such as comparison, limit comparison and Dirichlet test.)
|
$I$ converges because
$$\dfrac{x\ln x}{(1+x^2)^2}\le \dfrac{x^2}{(1+x^2)^2}\le\dfrac{1}{1+x^2} ~~~x>1$$
and $$ \lim_{x\to 0}\dfrac{x\ln x}{(1+x^2)^2} = 0$$
that is $$x\mapsto \dfrac{x\ln x}{(1+x^2)^2}~~~\text{is continuous on [0,1]}$$
Now Set $x=\frac{1}{u}$ that is $dx = -\frac{1}{u^2}$
$$I=\int_0^\infty\dfrac{x\ln x}{(1+x^2)^2}dx=-\int^0_\infty\dfrac{\frac{1}{u}\ln \frac{1}{u}}{(1+\frac{1}{u^2})^2}\frac{1}{u^2}du =\int_0^\infty\dfrac{\ln \frac{1}{u}}{(\frac{u^2+1}{u^2})^2}\frac{1}{u^3}du$$
But $$\color{red}{ \ln\frac{1}{u} = -\ln u}$$
hence We have
$$I=\int_0^\infty\dfrac{x\ln x}{(1+x^2)^2}dx\\=\int_0^\infty\dfrac{\ln \frac{1}{u}}{(\frac{u^2+1}{u^2})^2}\frac{1}{u^3}du\\= -\int_0^\infty\dfrac{u^4\ln u}{(u^2+1)^2}\frac{1}{u^3}du \\=-\int_0^\infty\dfrac{u\ln u}{(u^2+1)^2}du =-I$$
That $$\color{blue}{ I = -I\implies 2I= 0 \implies I= 0}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2473234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
}
|
Find number of positive integer solutions Find number of positive integer solutions $(x,y,z)$ for the following equation:
$19x + 11y + 8z = 240$
I divided the equation by $8$ and then tried to equate remainders.
It yields that $3(x + y) = 8k$ for some constant $k$ or $x + y$ is a multiple of 8.
Can't choose which combinations of $x,y$ will do the work.
Solved!
For $ x + y = 8 $ All ordered pairs are allowed. Corresponding values of $z$ are within domain of positive integers. This gives 7 solutions.
For $ x + y = 16 $ Only 7 pairs of $ x,y $ allowed. They are $ (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9) $ Pairs $(8,8)$ and beyond yield negative values for $z$. Therefore total of 14 solutions.
|
I had a very similar method, which takes advantage of $19-11=8$ and also the factor $8$ generally. Rewrite as $$11(x+y)+8(x+z)=240=8\times 30$$
$x+y$ must be divisible by $8$ and the only possibilities are $x+y=8, 16$. In the first of these cases $x+y=8$ has seven solutions. In the second we find $x+z=8 (=30-2\times 11)$ which has seven solutions. All these solutions evidently give a positive integer value for the missing number, so there are fourteen solutions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2474698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solve identity: $\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$ $$\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$$
The only way I can see of doing this is by cross multiplying but isn't that not allowed when trying to prove something?
|
It's $$\sin^2x=1-\cos^2x$$ or
$$\sin^2x+\cos^2x=1.$$
Also, we have:
$$\frac {\sin x}{1-\cos x}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^2\frac{x}{2}}=\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{1+\cos{x}}{\sin{x}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2475362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Check solution of recurrence relation $a_0 = 3$, $a_1 = 7$, $a_n = 3a_{n-1} - 2a_{n-2}$ for $n \geq 2$ $a_0 = 3, a_1 = 7, a_n = 3a_{n-1} - 2a_{n-2}$ for $n \geq 2$
I know I've got the wrong answer because my outputs don't match. So if someone could show me where I'm going wrong I would be super grateful.
Let $ y = \sum\limits_{n = 0}^{\infty} a_nx^n$
multiplying original formula through by $x^n$ I get
$a_nx^n = 3a_{n-1}x^n - 2a_{n-2}x^n$
sum over $n \geq 2$
$\sum\limits_{n \geq 2} a_nx^n = \sum\limits_{n \geq 2} 3a_{n-1}x^n - \sum\limits_{n \geq 2}2a_{n-2}x^n$
The left hand side is equal to $y - a_1 - a_0$
The first part of the right hand side
$\sum\limits_{n \geq 2} 3a_{n-1}x^n = 3x \sum\limits_{n \geq 2} a_{n-1}x^{n-1} = 3x(y-a_0)$
The second part of the right hand side
$\sum\limits_{n \geq 2}2a_{n-2}x^n = 2x^2\sum\limits_{n \geq 2}a_{n-2}x^{n-2} = 2x^2y$
So putting it all together and substituting for $a_0$ and $a_1$
$y - 10 = 3x(y-3) - 2x^2y$
all the $y$'s to one side
$y - 3xy + 2x^2y = -9x + 10 \Rightarrow y = \cfrac{-9x + 10}{(1-2x)(1-x)}$
after fraction decomposition I get
$y = \cfrac{11}{(1-2x)}-\cfrac{1}{(1-x)}$
Which then gives me
$a_n = 11 \cdot 2^n -1$
which is definitely wrong :(
|
Alternatively:
$$a_n = 3a_{n-1} - 2a_{n-2} \Rightarrow a_n-2a_{n-1}=a_{n-1}-2a_{n-2}.$$
Change: $b_n=a_n-2a_{n-1}, b_0=1$ to get:
$$b_n=b_{n-1} \Rightarrow b_n=1.$$
Hence:
$$a_n-2a_{n-1}=1, a_0=3 \Rightarrow a_n=2^{n+2}-1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2478002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Let $r$ be a root of the polynomial $p(x) = (\sqrt{5} - 2\sqrt{3})x^3 + \sqrt{3}x - \sqrt{5} + 1$. Find another polynomial $q(x)$ with integer coefficients such that $q(r) = 0$.
I have no clue how to do this question. Can't use rational root theorem and I see no feasible way to get the roots of $p(x)$. Any help would be appreciated.
|
Start by collecting all the like-terms with a particular radical and moving them to the left-hand-side; I chose $\sqrt{3}$ first. Move all terms without $\sqrt{3}$ to the right-hand-side. Then, factor out the chosen radical. This is shown in $(1)$. Next, square both sides, as shown in $(2)$. Since all terms with $\sqrt{3}$ have been collected, we've taken $\sqrt{3}$ out of the equation. Although, we're still left with all other radicals ($\sqrt{5}$ in this case).
Next, collect all terms with the next radical ($\sqrt{5}$) on one side of the equation. As before, factor out that radical, as shown in $(3)$. And once again, square both sides, as shown in $(4)$. You're left with a polynomial with coefficients in $\mathbb{Z}$. I'll leave the simplification up to you.
$$\begin{aligned}
&(\sqrt{5}-2\sqrt{3})x^3+\sqrt{3}x-\sqrt{5}+1=0
\\
\\&\bigg(-2x^3+x\bigg)\sqrt{3}=\bigg(1-x^3\bigg)\sqrt{5}-1&(1)
\\
\\&3\bigg(-2x^3+x\bigg)^2=5\left(1-x^3\right)^2-2\sqrt{5}\left(1-x^3\right)+1&(2)
\\
\\&3\bigg(-2x^3+x\bigg)^2-5\left(1-x^3\right)^2-1=-2\sqrt{5}\left(1-x^3\right)&(3)
\\
\\&\left(3\bigg(-2x^3+x\bigg)^2-5\left(1-x^3\right)^2-1\right)^2=20\left(1-x^3\right)^2&(4)\end{aligned}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2478954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
What is the sum of $E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$ What is the right way to assess this problem?
$$E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$$
To find the value of the sum I tried to use the fact that it could be something convergent like a geometric series. However it does not seem to be the case as there is no common ratio between the terms. Since the numerator is increasing from $1,2,3,...$ makes it impossible to find a ratio. What should I do?.
|
\begin{eqnarray}
E&=&\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\ldots=\sum_{k=1}^\infty\frac{k}{3^k}=\frac{1}{3}\sum_{k=1}^\infty k\cdot\left(\frac{1}{3}\right)^{k-1}\\
&=&\frac{1}{3}\dfrac{d}{dx}\left(\sum_{k=0}^\infty x^k\right)\Big|_{x=\frac{1}{3}}=\frac{1}{3}\frac{d}{dx}\left(\frac{1}{1-x}\right)\Big|_{x=\frac{1}{3}}=\frac{1}{3}\dfrac{1}{(1-x)^2}\Big|_{x=\frac{1}{3}}\\
&=&\frac{1}{3}\cdot\dfrac{1}{\left(1-\frac{1}{3}\right)^2}=\frac{1}{3}\cdot\dfrac{1}{\frac{4}{9}}=\frac{1}{3}\cdot\frac{9}{4}=\frac{3}{4}
\end{eqnarray}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2480080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Evaluate $\oint_C \dfrac{dz}{z^3 - 1}$, where $C$ is in the circle $|z + 1| = \dfrac{3}{2}$: Discrepancy Between Solutions. I have the following complex analysis contour integration problem:
Evaluate $\oint_C \dfrac{dz}{z^3 - 1}$, where $C$ is in the circle $|z + 1| = \dfrac{3}{2}$.
There is a discrepancy between my solution and the solution provided by the instructor.
The instructor's solution is as follows:
$\oint_C \dfrac{dz}{z^3 - 1} = 2\pi i \left( \dfrac{-\sqrt{3} + 3i}{6} + \dfrac{-\sqrt{3} - 3i}{6} \right) = - \dfrac{2 \pi i}{\sqrt{3}}$
My solution is as follows:
$\dfrac{1}{z^3 - 1} = \dfrac{1}{(z - 1)\left( z + \dfrac{1}{2} - \dfrac{i \sqrt{3}}{2} \right) \left( z + \dfrac{1}{2} + \dfrac{i \sqrt{3}}{2} \right)}$
Therefore, we have poles of order $1$ at $z = 1$, $z = -\dfrac{1}{2} + \dfrac{i \sqrt{3}}{2}$, $z = -\dfrac{1}{2} - \dfrac{i \sqrt{3}}{2}$. However, the pole at $z = 1$ is outside of $C$, and so we exclude it since its integral will equal $0$.
The residue theorem states that:
$\oint_C f(z) \ dz = 2\pi i \sum_{k = 1}^m res(f, c_k)$
The residue is calculated as follows:
Suppose that $f$ has a pole of order $m$ at $c$, so that
$res(f, c) = \dfrac{1}{(m - 1)!} g^{m - 1} (c)$,
where $g(z) = (z - c)^m f(z)$
We get $res\left[ \dfrac{1}{(z - 1)\left( z + \dfrac{1}{2} - \dfrac{i \sqrt{3}}{2} \right) \left( z + \dfrac{1}{2} + \dfrac{i \sqrt{3}}{2} \right)} , \dfrac{-1}{2} + \dfrac{i \sqrt{3}}{2} \right] = \dfrac{-2}{3\sqrt{3}i + 3}$
and
$res\left[ \dfrac{1}{(z - 1)\left( z + \dfrac{1}{2} - \dfrac{i \sqrt{3}}{2} \right) \left( z + \dfrac{1}{2} + \dfrac{i \sqrt{3}}{2} \right)} , \dfrac{-1}{2} - \dfrac{i \sqrt{3}}{2} \right] = \dfrac{2}{3\sqrt{3}i - 3}$
Therefore, by the residue theorem, we get
$\oint_C \dfrac{1}{z^3 - 1} = 2\pi i \left( \dfrac{2}{3\sqrt{3}i - 3} - \dfrac{2}{3\sqrt{3}i + 3}\right)$
$= \dfrac{-2\pi i}{3} $
I would greatly appreciate it if people could please take the time to check my solution and clarify any errors.
|
First of all, what you say the instructor has makes no sense and the arithmetic in it is not correct. What is missing is a sum (with the $2\pi i$ factored out). Note that your residue at the primitive cube root of unity $\omega = -\dfrac12+i\dfrac{\sqrt3}2$ is $\dfrac 1{3\omega^2} = \dfrac13 \omega$. Adding this to its conjugate makes the arithmetic a lot easier.
EDIT: So there's no difference between the solutions, except the instructor is using a more convenient computation of the residue, as I did. The instructor probably should have included a bit more detail. (That said, I hate using the definition of residue that you are using, although I realize it is widely taught. I far prefer to use Laurent series, when needed, along with the observation that
when $h$ has a simple zero at $z=a$ and $g$ is holomorphic at $z=a$, then we have
$$\text{Res}_{z=a} \frac{g(z)}{h(z)} = \frac{g(a)}{h'(a)}.$$
This should be immediate from your approach and is certainly immediate from thinking about the Laurent series.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2481011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find $\alpha + \beta$ given that: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ , ( $\alpha , \beta \in \Bbb R)$
Find $\alpha + \beta$ given that: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ , ( $\alpha , \beta \in \Bbb R)$
I have a solution involving a variable change ($\alpha=x+2$) which is not beautiful in my opinion! I'm looking for a more beautiful solution.
|
No change of variables:
Adding the two equations and using $$\alpha^2 + \beta^2 = (\alpha+\beta)^2-2\alpha\beta\\
\alpha^3+\beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$gives the equation
$$(\alpha + \beta)^3 - 6(\alpha + \beta)^2 + (13-3\alpha\beta)(\alpha + \beta) + 12\alpha\beta - 20$$
meaning that $\alpha + \beta$ is the root of the polynomial $$x^3-6x^2+(13-3\alpha\beta)x + 12\alpha\beta - 20$$
This polynomial has only one real solution, $x=4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2481243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Prove by induction that $1^1\cdot 2^2\cdot \dots\cdot n^n\leq \left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}.$ How can I prove by induction that for every natural number $n$,
$$1^1\cdot 2^2\cdot \dots\cdot n^n\leq \left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}.$$
|
Hint. Instead of induction, I warmly recommend the use of the AGM inequality. Note that
$$\left(1^1\cdot 2^2\cdot \dots\cdot n^n\right)^{\frac{2}{n(n+1)}}$$
is the geometric mean of the numbers
$$1,2,2,3,3,3,\dots,\underbrace{n,\dots,n}_{\text{$n$ times}}.$$
What is their arithmetic mean?
$$\frac{1+2+2+3+3+3+\dots+\overbrace{n+\dots+n}^{\text{$n$ times}}}{n(n+1)/2}=\frac{1+2^2+3^2+\dots+n^2}{n(n+1)/2}=?$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2484139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Let $\alpha $ is a root of $x^2+3x+5=0$ then express Let $\alpha $ is a root of $x^2+3x+5=0$ then express $\alpha^4 + 5\alpha^3 + 7\alpha^2 + 8\alpha -9$ as a linear expression of $\alpha $.
My Attempts:
since $\alpha $ is a root of $x^2+3x+5=0$
We have,
$$\alpha^2 + 3\alpha +5=0$$.
..
|
Because $$x^4+5x^3+7x^2+8x-9=$$
$$=x^2(x^2+3x+5)+2x(x^2+3x+5)-4(x^2+3x+5)+10x+11.$$
Thus, since $\alpha^2+3\alpha+5=0$, we got that our expression it's $10\alpha+11$.
Also, since a degree of the remainder should be less that $2$ we must get a linear expression of $\alpha$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2487645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Rationalize denominator: $\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}$ $$\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}$$
So this is what I thought: the square root of 1 is obviously one, so I have $1^3 +(\sqrt[3]{5} + \sqrt[3]{6})$. In my head I see that this is the first part for the sum of cubes formula. I multiplied with the rest of the formula so I can get $1^3 +(\sqrt[3]{5} + \sqrt[3]{6})^3$ in the denominator. Now when I try to do this bracket I have a problem, what do I do with this? $$11+3\sqrt[3]{25 \cdot 6}+3\sqrt[3]{5 \cdot 36}$$
How do I get rid of the square roots?
|
Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}=\frac{\sqrt[3]{5^2}+1+\sqrt[3]{36} - \sqrt[3]{5}-\sqrt[3]{6}-\sqrt[3]{30}}{12-3\sqrt[3]{30}}$$
and since $a^3-b^3=(a-b)(a^2+ab+b^2)$$$\frac{1}{\sqrt[3]{30}-4}=\frac{\sqrt[3]{30^2}+4\sqrt[3]{30}+16}{30-64}$$
you just have to multiply them to obtain the answer, but do not forget $-\frac14$ factor.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2489731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Calculus inequality involving sine and cosine community! I saw the following inequality in a calculus assignment, which I thought was harder to prove than I expected: For every $x\in\mathbb{R},$$$(\sin x + a \cos x)(\sin x + b \cos x)\leq 1+(\frac{a+b}{2})^2.$$
By means of a graphing device, I noticed that when $a=b$, $f(x)=(\sin x + a\cos x)(\sin x + b \cos x)$ has $1+(\frac{a+b}{2})^2$ as its maximum, but in other cases the maximum of $f$ is strictly less (and sometimes far less) than this number. I'd appreciate any kind of help :)
|
For any trig0nometric expression of the form $$A\sin x+B\cos x=\sqrt{A^2+B^2}\left(\dfrac{A}{\sqrt{A^2+B^2}}\sin x+\dfrac{B}{\sqrt{A^2+B^2}}\cos x\right)$$
can be written in the form $\sqrt{A^2+B^2}\sin(x+\phi)$ with $\tan\phi=B/A$ and hence $\vert A\sin x+B\cos x|\le\sqrt{A^2+B^2}$ for any $A,B\in\Bbb{R}.$
In fact these bounds are sharp.
Here I will use above fact to give a pure trig-solution (without calculus).
First observe that
\begin{align}
f(x)& = (\sin x+a \cos x)(\sin x+b \cos x) \\
& = \sin^2x+(a+b)\sin x\cos x+ab\cos^2x\\
& =\dfrac{1}{2}\left((1-\cos 2x)+(a+b)\sin 2x+ab(1+\cos 2x)\right) \\
& =\left(\dfrac{a+b}{2}\right)\sin2x+\left(\dfrac{ab-1}{2}\right)\cos 2x+\left(\dfrac{ab+1}{2}\right).\\
\end{align}
Hence $$-\sqrt{\left(\dfrac{a+b}{2}\right)^2+\left(\dfrac{ab-1}{2}\right)^2}\le f(x)-\left(\dfrac{ab+1}{2}\right)\le\sqrt{\left(\dfrac{a+b}{2}\right)^2+\left(\dfrac{ab-1}{2}\right)^2}.$$ Therefore the function has the maximum value
$$\left(\dfrac{ab+1}{2}\right)+\sqrt{\left(\dfrac{a+b}{2}\right)^2+\left(\dfrac{ab-1}{2}\right)^2}$$
and minimum value
$$\left(\dfrac{ab+1}{2}\right)-\sqrt{\left(\dfrac{a+b}{2}\right)^2+\left(\dfrac{ab-1}{2}\right)^2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2491701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Distribute n integers into m (non distinct) bins without creating consecutive triples I have $[n] := \{0, 1, \cdots, n-1\}$ integers and $m$ bins (we don't distinguish bins). How many possible ways to distribute $m$ integers from $[n]$ into $m$ bins are there such that the distributions have no consecutive triple?
In other words, how many subsets of size $m$ are there in $[n]$ which don't contain 3 consecutive integers ($6, 4, 5$ is a consecutive triple but $1, 2, 4$ isn't)?
Some subsets for $m=7$ and $n = 20$:
*
*{0, 1, 3, 4, 6, 7, 9} has no consecutive triple
*{1, 3, 6, 18, 19, 10, 2} has a consecutive triple (1, 2, 3)
*{10, 15, 18, 19, 0, 5, 8} has no consecutive triple
|
Let $A(n,m)$ be the number of ways to choose $m$ integers from $[n]$ without a consecutive triple. Clearly $A(n,m) = \binom{n}{m}$ when $m < 3$, and $A(n,m) = 0$ when $m > \left\lceil\frac23 n\right\rceil$.
For larger $m$ it might be easier to invert the question. How many subsets $S$ of size $m$ are there which do have 3 consecutive integers?
Suppose that the lexicographically first consecutive triple is $k,k+1,k+2$.
Case $k = 0$: any selection of $m-3$ elements from the remaining $n-3$ works, so we have $\binom{n-3}{m-3}$ subsets.
Case $0 < k \le n - 3$: clearly $k-1$ is not in $S$. Suppose that $i$ elements from $[k-1]$ are in $S$: they don't contain a triple, so there are $A(k-1,i)$ ways to choose them. The remaining $m-i-3$ elements of $S$ are chosen freely from the remaining $n-k-3$ elements of $[n]$.
Putting those together, $$\binom{n}{m} - A(n,m) = \binom{n-3}{m-3} + \sum_{k=1}^{n-3} \sum_{i=0}^{\left\lceil\frac{2(k-1)}{3}\right\rceil} A(k-1,i) \binom{n-k-3}{m-i-3}$$
This allows closed forms for small $m$. E.g. $$\begin{eqnarray}
A(n,3) & = & \binom{n}{4} - \binom{n-3}{1} - \sum_{k=1}^{n-3} A(k-1,0) \binom{n-k-3}{0} \\
& = & \binom{n}{3} - n + 2 \\
A(n,4) & = & \binom{n}{4} - \binom{n-3}{1} - \sum_{k=1}^{n-3} A(k-1,0) \binom{n-k-3}{1} + A(k-1,1) \binom{n-k-3}{0}\\
& = & \binom{n}{4} - (n-3) - \sum_{k=1}^{n-3} n-4\\
& = & \binom{n}{4} - (n-3)^2
\end{eqnarray}$$
etc. I wouldn't be especially optimistic about finding a closed form for general $m$, but I haven't really tried.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2492080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Justify the recurrence relation $a_{n+2}=2a_{n+1}+a_n$ and find $a_n$ Let $a_n$ be the number of ways to color the squares of a $1$ x $n$ chessboard using the colors red, white, and blue, so that no red square is adjacent to a white square. Justify the relation $a_{n+2}=2a_{n+1}+a_n$ (for certain $n$), and then find $a_n$.
My Attempt:
.If the first square is blue, then the number of ways of coloring the other $n-1$ squares is $a_{n-1}$
.If the first square is red, then the next square must be blue (1 option), and the rest can be filled in $a_{n-2}$ ways
.If it is white, then it is also $a_{n-2}$
This means the recurrence relation is $a_n=a_{n-1}+2a_{n-2}$ for $n\ge2$
By inspection, $a_0=1$ and $a_1=3$
Not sure where to go from here.
|
Generating Function Approach
$
\begin{array}{cl}
\text{function}&\text{meaning}\\
x&\text{blue}\\
\frac{x^2}{1-x}&\text{blue followed by one or more red}\\
\frac{x^2}{1-x}&\text{blue followed by one or more white}\\
\frac1x&\text{remove initial blue}
\end{array}
$
The generating function is
$$
\begin{align}
\frac1x\sum_{k=1}^\infty\left(x+\frac{x^2}{1-x}+\frac{x^2}{1-x}\right)^k
&=\frac1x\sum_{k=1}^\infty\left(\frac{x+x^2}{1-x}\right)^k\\
&=\frac1x\frac{\frac{x+x^2}{1-x}}{1-\frac{x+x^2}{1-x}}\\[3pt]
&=\frac{1+x}{1-2x-x^2}
\end{align}
$$
This generating function says
$$
a_0=1\quad a_1=3
$$
and for $n\ge2$,
$$
a_n=2a_{n-1}+a_{n-2}
$$
Solving the linear recurrence relation for $a_n$, we get
$$
a_n=\frac{2+\sqrt2}{2\sqrt2}\left(1+\sqrt2\right)^n-\frac{2-\sqrt2}{2\sqrt2}\left(1-\sqrt2\right)^n
$$
Decomposition By The Terminal Element
Let $r_n$ be the number of strings of length $n$ that end in red.
Let $w_n$ be the number of strings of length $n$ that end in white.
Let $b_n$ be the number of strings of length $n$ that end in blue.
Then we have the recursions
$
\begin{align}
&\text{recursion}&&\text{reason}\\\hline
r_n&=r_{n-1}+b_{n-1}&&\begin{array}{l}\text{ }\\\text{remove the last from a string that ends}\\\text{in red and it ends in red or blue}\end{array}\\
w_n&=w_{n-1}+b_{n-1}&&\begin{array}{l}\text{ }\\\text{remove the last from a string that ends}\\\text{in white and it ends in white or blue}\end{array}\\
b_n&=r_{n-1}+w_{n-1}+b_{n-1}&&\begin{array}{l}\text{ }\\\text{remove the last from a string that ends}\\\text{in blue and it ends in red, white, or blue}\end{array}
\end{align}
$
Therefore, we have
$$
\begin{align}
a_n
&=r_n+w_n+b_n\\
&=2r_{n-1}+2w_{n-1}+3b_{n-1}\\
&=2r_{n-1}+2w_{n-1}+2b_{n-1}+r_{n-2}+w_{n-2}+b_{n-2}\\
&=2a_{n-1}+a_{n-2}
\end{align}
$$
Thus, we get the same recursion as before.
We can count $a_1=3$ and $a_2=7$ (because we have to leave out red-white and white-red) and solve the linear recurrence relation as above.
Method Similar To That In Natash1's Answer
$$
\begin{align}
a_n
&=\overbrace{\ \ \ a_{n-1}\ \ \ \vphantom{1}}^{\text{one blue}}+\overbrace{\ \ \ \ \ \ 1\ \ \ \ \ \ }^{\text{all red}}+\sum_{k=1}^{n-1}\overbrace{\ \ a_{n-k-1}\ \ \vphantom{1}}^{\text{$k$ red, one blue}}+\overbrace{\ \ \ \ \ \ 1\ \ \ \ \ \ }^{\text{all white}}+\sum_{k=1}^{n-1}\overbrace{\ \ a_{n-k-1}\ \ \vphantom{1}}^{\text{$k$ white, one blue}}\\
&=a_{n-1}+2+2\sum_{k=0}^{n-2}a_k\tag1
\end{align}
$$
Subtracting $(1)$ for $n-1$ from $(1)$ for $n$ gives
$$
a_n-a_{n-1}=a_{n-1}-a_{n-2}+2a_{n-2}\tag2
$$
which is equivalent to
$$
a_n=2a_{n-1}+a_{n-2}\tag3
$$
which is the same recurrence gotten above.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2494428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Find $\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$ Find $$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$$
My work so far:
$$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}=\frac{\ln3-\ln5}{\ln4-\ln10}$$
Is correct?
Add:
I used $a^x\sim 1+x\ln a$ for $x\rightarrow 0$
|
L'Hospital's rule for $\frac{0}{0}$ limits works well in this case. An alternate is to use
$$a^{x} = e^{x \, \ln(a)} = 1 + \ln(a) \, x + \frac{\ln^{2}(a)}{2!} \, x^{2} + \mathcal{O}(x^{3})$$
which yields
\begin{align}
\frac{a^{x} - b^{x}}{c^{x} - d^{x}} &= \frac{(\ln(a) - \ln(b)) \, x + \frac{1}{2} \, (\ln^{2}(a) - \ln^{2}(b)) \, x^{2} + \mathcal{O}(x^{3})}{(\ln(c) - \ln(d)) \, x + \frac{1}{2} \, (\ln^{2}(c) - \ln^{2}(d)) \, x^{2} + \mathcal{O}(x^{3})} \\
&= \frac{\ln(\frac{a}{b}) + \frac{1}{2} \, \ln(\frac{a}{b}) \, \ln(a b) \, x + \mathcal{O}(x^{2})}{\ln(\frac{c}{d}) + \frac{1}{2} \, \ln(\frac{c}{d}) \, \ln(c d) \, x + \mathcal{O}(x^{2})}
\end{align}
Taking the limit as $x \to 0$ yields
$$\lim_{x \to 0} \frac{a^{x} - b^{x}}{c^{x} - d^{x}} = \frac{\ln(\frac{a}{b})}{\ln(\frac{c}{d})} = \frac{\ln (a) - \ln(b)}{\ln(c) - \ln(d)}.$$
By L'Hospital's rule:
\begin{align}
\lim_{x \to 0} \frac{a^{x} - b^{x}}{c^{x} - d^{x}} &= \lim_{x \to 0} \frac{e^{x \, \ln(a)} - e^{x \, \ln(b)}}{e^{x \, \ln(c)} - e^{x \, \ln(d)}} \\
&= \lim_{x \to 0} \frac{\ln(a) \, a^{x} - \ln(b) \, b^{x}}{\ln(c) \, c^{x} - \ln(d) \, d^{x}} \\
&= \frac{\ln (a) - \ln(b)}{\ln(c) - \ln(d)}.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2498411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Inverse Z-Transform of $(1+2/z)^{-3}$ Working through the K.A Stroud Advanced Engineering Mathematics Textbook on my own, and have really got stuck on this question with no one to help out.
The question and my working can be found here
Basically, brought it down into 3 fractions, which I could transform up until the last one $\frac{4z}{(z+2)^3}$ , as can be seen in the top corner.
If anyone knows the inverse Z-transform of $\frac{4z}{(z+2)^3}$, but not necessarily the answer to the main question it would still be really appreciated.
I tried going with the derivative rule where if $Z(x)=F(z)$ then -$zF'(z)=Z(kx)$, and ended up with a final answer of $(-2)k * (z+1)^2$ which I know was wrong. I believe it should be $(-2)k * (z+2)^2$.
Any help would be greatly appreciated as I have no access to any professors or fellow students. Thanks so much!
|
$$\frac{z}{(z+2)^3}=\frac{z^{-2}}{(1+\frac{2}{z} )^3}=z^{-2}\sum_{k=0}^\infty \binom{k+2}{k}\left(-\frac{2}{z}\right)^k=
z^{-2}\sum_{k=0}^\infty \frac{1}{2}(k+1)(k+2)\left(-\frac{2}{z}\right)^k$$
$$\frac{z}{(z+2)^3}=\sum_{k=0}^\infty (-1)^k 2^{k-1}(k+1)(k+2)\frac{1}{z^{k+2}}$$
$k=n-2$
$$\frac{z}{(z+2)^3}=\sum_{n=-2}^\infty (-1)^{n-2} 2^{n-3}(n-1)(n)\frac{1}{z^{n}}=\sum_{n=-2}^\infty \frac{a_n}{z^{n}}$$
$$\mathcal{Z}^{-1}\left(\frac{z}{(z+2)^3}\right)_{(n)}=a_n=(-1)^{n} 2^{n-3}(n-1)n \qquad\qquad n\geq 0$$
$\mathcal{Z}^{-1}$ denotes the inverse Z-transform.
$$ $$
$$\frac{1}{(1+\frac{2}{z} )^3}=\sum_{k=0}^\infty \binom{k+2}{k}\left(-\frac{2}{z}\right)^k=
\sum_{k=0}^\infty \frac{1}{2}(k+1)(k+2)\left(-\frac{2}{z}\right)^k$$
$$\frac{1}{(1+\frac{2}{z} )^3}=\sum_{k=0}^\infty (-1)^k 2^{k-1}(k+1)(k+2)\frac{1}{z^{k}}$$
$$\mathcal{Z}^{-1}\left(\frac{1}{(1+\frac{2}{z} )^3}\right)_{(n)}=(-1)^{n} 2^{n-1}(n+1)(n+2) \qquad\qquad n\geq 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2499323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How to calculate $\lim_{n \to \infty} \sum_{k=1}^n\binom nk k!k\frac{1}{n^k}$? How to calculate this limit :
$$\lim_{n\rightarrow +\infty}u_n$$
with :
$$u_n=\sum_{k=1}^n\binom nk k!k\frac{1}{n^k}$$
We can write this :
$$u_n=\sum_{k=1}^n\frac{n}{n}\times\frac{n-1}{n}\times\cdots\times\frac{n-k+1}{n}\times k$$
But I can't find the solution.
|
Hint: The number $u_n$ has a nice telescoping property. Let's for example consider the case $n=5$.
\begin{align*}
\color{blue}{u_5}&=1+2\cdot\frac{4}{5}+3\cdot\frac{4}{5}\cdot\frac{3}{5}+4\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}
+\color{blue}{5}\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}\cdot\color{blue}{\frac{1}{5}}\\
&=1+2\cdot\frac{4}{5}+3\cdot\frac{4}{5}\cdot\frac{3}{5}
+(4+1)\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}\\
&=1+2\cdot\frac{4}{5}+3\cdot\frac{4}{5}\cdot\frac{3}{5}
+\color{blue}{5}\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\color{blue}{\frac{2}{5}}\\
&=1+2\cdot\frac{4}{5}+(3+2)\cdot\frac{4}{5}\cdot\frac{3}{5}\\
&=1+2\cdot\frac{4}{5}+\color{blue}{5}\cdot\frac{4}{5}\cdot\color{blue}{\frac{3}{5}}\\
&=1+(2+3)\cdot\frac{4}{5}\\
&=1+\color{blue}{5}\cdot\color{blue}{\frac{4}{5}}\\
&=(1+4)\\
&\color{blue}{\,=5}
\end{align*}
indicating $\color{blue}{u_n=n}$ for $n\geq 1$.
We observe we can iteratively collect the two right-most summands whereby the factor $5$ and $\frac{1}{5}$ cancel in the right-most summand.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2499476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
}
|
Why is $\sqrt x \times \sqrt y = \sqrt {xy}$ Sorry I do not know latex. Why is $\sqrt x \times \sqrt y = \sqrt {xy}$? It also applies for division, but why not addition and subtraction? i.e., why is $\sqrt x + \sqrt y$ not equal to $\sqrt {x+y}$?
Thank you for editing and answering.
|
remember what $^2$ is:
$$u^2=u\cdot u\\\text{so: }\left(\sqrt{a}\cdot\sqrt{b}\right)^2=\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{a}\cdot\sqrt{b}=\sqrt{a}\cdot\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{b}=a\cdot b=\sqrt{ab}^2\implies\sqrt x \cdot\sqrt y = \sqrt {xy}\\\text{and you can do: }\left(\sqrt{a}+\sqrt{b}\right)^2=\left(\sqrt{a}+\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)=\sqrt{a}^2+2\sqrt{a}\cdot\sqrt{b}+\sqrt{b}^2{=a+2\sqrt{a}\cdot\sqrt{b}+b}=\sqrt{a+2\sqrt{a}\cdot\sqrt{b}+b}^2\ne\sqrt{a+b}^2\implies\sqrt{a}+\sqrt{b}\ne\sqrt{a+b}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2501058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $(1+a)^7.(1+b)^7.(1+c)^7>7^7.a^4 b^4 c^4$ If a,b,c are positive prove that $(1+a)^7.(1+b)^7.(1+c)^7>7^7.a^4 b^4 c^4$
My approach
I tried (1+a)(1+b)(1+c) =1+[a+b+c+ab+bc+ac+abc]
Find the AM of Square Bracket, them AM $\ge$ GM
But if I add 1 then AM>GM. Please help me with my approach
|
You almost did it.
(1+a)(1+b)(1+c) =1+a+b+c+ab+bc+ac+abc (Now subtract 1 and multiply and divide by 7)
= $\frac{7a+7b+7c+7ab+7bc+7ac+7abc}{7}\ge \sqrt[7]{(7)(7a)(7b)(7c)(7ab)(7bc)(7ac)(7abc)}$
add 1 on LHS
=>$1+\frac{7a+7b+7c+7ab+7bc+7ac+7abc}{7} > \sqrt[7]{(7)(7a)(7b)(7c)(7ab)(7bc)(7ac)(7abc)}$
Taking power 7 on both sides
$(1+a)^7(1+b)^7(1+c)^7 > 7^7.a^4 b^4 c^4$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2501208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Proof for an identity (from Ramanujan written) I saw an identity by Ramanujan
$$\forall n \in \mathbb{N} ,n>1 :\lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n
+17}\rfloor$$ I tried to prove it by limit definition . I post my trial below . If possible check my prove (right , wrong) ?
Then Is there more Idea to proof ?
|
if $$|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - \sqrt {9n
+17}|<\epsilon \\\Rightarrow \lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n
+17}\rfloor $$
so ,I will show $|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - \sqrt {9n
+17}|<\epsilon$
$$\quad{|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - \sqrt {9n
+17}|=\\|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - 3\sqrt {n
+\frac{17}{9}}|\leq\\
|\sqrt n-\sqrt {n
+\frac{17}{9}}|+|\sqrt {n+2}-\sqrt {n
+\frac{17}{9}}|+|\sqrt{n+4} - \sqrt {n
+\frac{17}{9}}|\\\leq|\frac{0-\frac{17}{9}}{\sqrt n+\sqrt {n
+\frac{17}{9}}}|+\leq|\frac{2-\frac{17}{9}}{\sqrt {n+2}+\sqrt {n
+\frac{17}{9}}}|+\leq|\frac{4-\frac{17}{9}}{\sqrt {n+4}+\sqrt {n
+\frac{17}{9}}}|\\
\leq|\frac{\frac{17}{9}}{2\sqrt n}|+|\frac{\frac{1}{9}}{2\sqrt n}|+|\frac{\frac{19}{9}}{2\sqrt n}|\\=\frac{37}{18\sqrt n}\\\leq \frac{37}{18.5\sqrt n}\\=\frac{1}{2\sqrt{n}}\\\leq \frac{1}{2}}$$ I checked max $\{\epsilon\}<0.15$ with numerical method ,also graphing..
so $\lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n
+17}\rfloor$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2501870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$ then find $f(x)$ let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$
then find $f(x)$
My Try :
$$f(\frac{x}{3})+f(\frac{2}{x})=(\frac{2}{x})^2-1+(\frac{x}{3})^2-1$$
So we have :
$$f(x)=x^2-1$$
it is right ?Is there another answer?
|
A plausible $f(x)$ is in fact $f(x)=x^2-1$. A proof of that is as follows.
Suppose $f(x)=x^2+h$ For $x=\sqrt6$ one has $\dfrac 2x=\dfrac x3$ so
$$2f(\frac{\sqrt6}{3})=2(\frac{\sqrt6}{3})^2-2=2((\frac{\sqrt6}{3})^2+h)\Rightarrow h=-1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2502655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
}
|
Probability that length of Randomly chosen chord of a circle Find Probability that length of Randomly chosen chord of a circle lies between $\frac{2}{3}$ and $\frac{5}{6}$ of its diameter.
My try: I assumed unit circle with center origin. Let two randomly chosen distinct points be $A(\cos \alpha, \sin \alpha)$ and $B(\cos \beta, \sin \beta)$
Length of the chord is $$p=2\sin \left(\frac{\alpha-\beta}{2}\right)$$
Now we have to find Probability that
$$\frac{4}{3} \le 2\sin \left(\frac{\alpha-\beta}{2}\right) \le \frac{5}{3}$$
can i have any clue here?
|
Let $r$ be the constant radius of the circle, let $L$ be the continuous random variable that represents the chord’s length, and let $\Theta$ be the continuous random variable that represents the counterclockwise angle measured from the first point chosen to the second point chosen.
As mentioned by others, per symmetry arguments, the location of the first point does not affect the probability.
I would approach this problem by writing $L$ in terms of $\Theta$ by using the law of cosines.
$$\begin{align}
c^2 &= a^2 + b^2 -2ab\cos\gamma \\
L^2 &= r^2 + r^2 -2rr\cos\Theta \\
L^2 &= 2r^2(1-\cos\Theta) \\
\end{align}$$
Let us investigate the values of $\Theta$ that correspond the values of $L$ you specified: $\frac{4r}{3}\le L\le\frac{5r}{3}$.
$$\begin{align}
\frac{L^2}{2r^2} &= 1-\cos\Theta \\
\cos\Theta &= 1 - \frac{L^2}{2r^2} \\
\end{align}$$
Plugging in the range of $L$ gives us that
$$\begin{align}
\cos\Theta &\in \left[ -\frac{7}{18} , \frac{1}{9} \right] \\
\Theta &\in \left[\arccos\left(\frac19\right) , \arccos\left(-\frac{7}{18}\right)\right] \cup \left[2\pi-\arccos\left(-\frac{7}{18}\right) , 2\pi-\arccos\left(\frac19\right)\right] \\
\end{align}$$
For this problem, solving the other side of the interval isn’t important, because only the length of the interval matters, and the two sides are equal in length—but nevertheless, there it is for reference. Call this final interval $J$ for ease.
The probability of selecting one of the desired values of $\Theta$ boils down to the length of $J$ divided by $2\pi$. This value simplifies to $$\frac{ \arccos\left(-\frac{7}{18}\right) - \arccos\left(\frac19\right) }{\pi} \approx 16.258\,194\,85 \ \%$$
This was a pretty inventive problem, so I humbly disclaim that my math might not be correct, although I feel confident that it is.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2503686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Solve $P(z)=0$, over complex field and factorise $P(z)=0$ over real field $P(z)=3z^4+10z^3+6z^2+10z+3$
The roots are $z=-3, -1/3, i,-i$ but I couldn't find $i,-i$ as the root.
Also the factorised version is meant to be $(z+3)(3z+1)(z^2+1)$ but I got something fabulous like $z^2(z+\frac{1}{z})(3z+\frac{3}{z}-10)$
I feel like I've butchered the equation already.
|
I like to write complete solution:
\begin{align}
P(z)
&= 3z^4+10z^3+6z^2+10z+3 \\
&= 3z^4+3 + 10z^3+10z+6z^2 \\
&= 3z^2\left(z^2+\dfrac{1}{z^2}\right)+10z^2\left(z+\dfrac{1}{z}\right)+6z^2 \\
&= z^2\Big[3\left(z^2+\dfrac{1}{z^2}\right)+10\left(z+\dfrac{1}{z}\right)+6\Big] \\
&= z^2\Big[3\left(z+\dfrac{1}{z}\right)^2+10\left(z+\dfrac{1}{z}\right)\Big] \\
&= z^2\left(z+\dfrac{1}{z}\right)\Big[3\left(z+\dfrac{1}{z}\right)+10\Big] \\
&= z\left(z+\dfrac{1}{z}\right)z\Big[3\left(z+\dfrac{1}{z}\right)+10\Big] \\
&= \left(z^2+1\right)(3z^2+10z+3) \\
&= (z+i)(z-i)(3z+1)(z+3) \\
\end{align}
which gives $z=-i,i,-\dfrac13,-3$ as roots.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2505312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Inequality proof $\frac{2 \cdot a \cdot b }{a+b} < \frac {a+b}{2}$ This is very simple, but impossible to Google as you can understand.
I need to prove this inequality:
$$\frac{2 \cdot a \cdot b }{a+b} < \frac {a+b}{2}$$
where $ a < b$ and $a > 0$ and $b > 0$
I have tried to rework it to:
$$4 \cdot a \cdot b < {(a+b)}^2$$ but that doesn't bring me further.
Also I tried to substitute $ a = b-\delta$ and substitute that in the inequality.
I can see where it goes when $a=b$ and when $b >>a$ but that still is not the proof.
|
We know that $(a-b)^2 > 0$ or $a^2 + b^2 -2ab > 0$
Adding $2ab$ term on both sides of the inequality we get
$a^2 + b^2 +2ab - 2ab > 2ab$
$a^2 + b^2 + 2ab > 4ab$
$(a+b)^2 > 4ab$
Equivalently
$\frac{(a+b)}{2} > \frac{2(ab)}{a+b}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2506815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Prove directly from the definition that $({1\over2}+\frac{1}{2^2}+...+\frac{1}{2^n})_n $ is Cauchy Prove directly from the definition that $({1\over2}+\frac{1}{2^2}+...+\frac{1}{2^n})_n$ is cauchy
I know from the definition of Cauchy that |$x_n$-$x_m$|< ϵ but how do you do this with |$\frac{1}{2^n}- \frac{1}{2^m}$|
what I've tried:
if $n\gt m$ then
$$ |\frac{1}{2^n}- \frac{1}{2^m}| \le |\frac{2^m-2^n}{2^n2^m}| \le |\frac{2^m +2^n}{2^{n+m}}| \le \frac{2^n+2^n}{2^{2n}}= \frac{1}{2^{n-1}} \le \frac{1}{2^{N-1}} \le \epsilon $$
and rearrange to get N $ \ge 1+ \frac{ln(\epsilon)}{ln(2)}$
is this correct?
|
To prove that the series satisfies the Cauchy condition you have to estimate, for $n,p\in\mathbb{N}$, the difference
$$
\sum_{j=n}^{n+p} \frac{1}{2^j} = \frac{1}{2^{n-1}}\left(1 - \frac{1}{2^{p+1}}\right) < \frac{1}{2^{n-1}}\,.
$$
Given $\varepsilon > 0$, it is enough to choose $N > 1 - \log_2\varepsilon$ so that
$$
\sum_{j=n}^{n+p} \frac{1}{2^j} < \varepsilon
\qquad
\forall n> N, \ p\in\mathbb{N}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2515137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
}
|
Divisibility theorem based proof for any square mod 4 being either 0 or 1 Kindly help me in understanding the below proof for given statement:
If $n$ is a square, then leaves a remainder $0$ or $1$ when divided by 4.
Proof: The divisibility theorem states that for two integers $a,b$ with $b>0$, then there is a unique pair of integers $q$ and $r$ such that $a =qb +r$ and $0\le r <b$.
Let the number $n= a^2$, for $a$ being an integer. If $b$(divisor)$=4$, then $a = 4q +r$, where $r \in \{0, 1, 2, 3\}$, so that $n = (4q+r)^2 = 16q^2 + 8qr + r^2$.
If $r=0$, then $n=4(4q^2 +2qr) +0$; if $r=1$, then $n=4(4q^2 +2qr)+1$; if $r=2$, then $n =4(4q^2 +2qr+1) +0$; and if $r =3$, then $n = (4q^2 +2qr+2)+1$. In each case , the remainder is $0$ or $1$.
The difficulty lies in the last para. where according to me it should be:
If $r=0$, then $n=4(4q^2 +2qr) +0$; if $r=1$, then $n=4(4q^2 +2qr)+1$; if $r=2$, then $n =4(4q^2 +2qr) +2$; and if $r =3$, then $n = (4q^2 +2qr)+3$.
|
I think you missed the fact that the last $r$ is squared in the expression of $n$.
If $r=2$, then you have $$\begin{align}n=&16q^2+8qr+r^2\\=&16q^2+8q\cdot 2 + 2^2 \\=& 16q^2+16q+4\\=&4(4q^2+4q+1).\end{align}$$
For $r=3$, you missed a $4$ when writing the original solution, so you have
$$\begin{align}n=&16q^2+8qr+r^2\\=&16q^2+8q\cdot 3 + 3^2\\=&16q^2+24q+9\\=&16q^2+24q+8+1\\=&4(4q^2+6q+2)+1\end{align}$$
and since $r=3$, that's the same as $$n=4(4q^2+2qr+2)+1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2522766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
find all positive integers n with $\sigma(n)=12$ How do you find all positive integers n with $\sigma(n)=12$?
$\sigma(n)$ is the sum of the divisors of $n$.
The book gives the answer:
Each factor in the formula for $\sigma(n)$ must divide $12$. The only way to get factors, other than $1$, of $12$ for sums of this type are:
$(1+2)=3$
$(1+3)=4$
$(1+5)=6$
$(1+11)=12$
Hence the only values of $n$ for which $\sigma(n)=12$ are $n=2*3=6$ and $n=11$
I totally do not understand the method, what is the thing I must do to find all positive integers n with $\sigma(n)=12$? And in general $\sigma(n)=k$
|
Let $n = \prod p_i^{a_i}$ then
Then the factors are ... any combination of primes and powers. But the important thing to realize is: Suppose there is a prime factor $p$ and $n$ has $p^a$ power dividing it. And suppose that $m$ is factor that is not divisible by $p$. Then $m, pm, p^2m..... p^am$ are also factors. So them sum of all the factors will be:
$\sigma(n)= (1 + p + p^2 + .... + p^a) + m_1(1+p+p^2 + ..... + p^a) + m_2(1+p+p^2 + ..... + p^a)+ ....$ where $m_1$ are the factors (greater than $1$) that are not divisible $p$.
Do this for all prime factors $p_i$ and you get:
$\sigma(n)= \prod(1 + p_i + .... + p_i^{a_i})$
So $\sigma(n) = 12$ means $12 = (1 + p + p^2+ ....)(1 + q + q^2 + ...)....$
That is what they mean by "factors in the formula".
But the factors of $12$ are $1,2,3,4,6,12$ so we might have $12 = 12 = (1 + p + p^2 + .....)$. The only way that can happen is if $p=11$. So the factors of $n$ are $1$ and $11$. So $n=11$.
So we want to solve any of the following:
a) $1 = (1 + p + p^2 +....)$
b) $2 = (1 + p + p^2 + .....)$
c) $3 = (1 + p + p^2 + .....)$
d) $4 = (1 + p + p^2 + ...)$
e) $6 = (1+ p + p^2 + ...)$
f) $12 = (1 + p + p^2 + ..)$.
The solution to a) is $1 = 1$
b) has no solution.
The solution to c) is $3 = 1 + 2$
The solution to d) is $4 = 1+3$
The solution to e) is $6 = 1+ 5$
The solution to f) is $12 = 1 + 11$.
So we have either $12 = (1 + 11)$ and the factors of $n$ are $1,11$ so $n = 11$.
Or we have $12= (1+2)*(1+3)$ and the factors of $n$ are $1,2,3,2*3$. so $n = 6$.
Perhaps abetter example to get the sense of the use of exponent series, would be to find all $n$ so that $\sigma(n) = 56$.
We need $56 = \prod (1 + p_i + ... +p_i^{a_i}$.
The factors of $56$ are $1,2,4,8, 7,14, 28,56$.
$1 = 1$.
$2 \ne 1 + p + p^2 +...$. Since we can't use $2$ as a factor we can not use $\frac {56}2=28$.
$4 = 1 + 3$
$8 = 1+7$
$7 = 1 + 2 + 4$
$14 = 1 + 13$
$56 \ne 1 + p + p^2$
So we can do either
$56 = 4*14 = (1 + 3)(1 + 13)$ and $n = 3*13 = 39$
or $56 = 7*8= (1 + 2 + 4)(1+7)$ and $n = 4*7 = 28$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2523187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
General solution Find the general solution of $\sec4\theta-\sec2\theta =2$
My approach:
Converted them to cosines and then further simplified. I got $\theta = \frac{n\pi}{5}+\frac{\pi}{10}$ or $\theta = -n\pi+\frac{\pi}{2}$But my book stated the answer as $\theta = \frac{2n\pi}{5}\pm\frac{\pi}{10}$ or $\theta = 2n\pi\pm\frac{\pi}{2}$. Where am i going wrong.
My simplification:
$$\cos2\theta - \cos4\theta = 2\cos4\theta \cdot\cos2\theta$$
$$\cos2\theta - \cos4\theta = \cos6\theta + \cos2\theta$$
$$\cos4\theta=-\cos6\theta$$
$$\cos4\theta=\cos(\pi-6\theta)$$
$$4\theta=2n\pi\pm(\pi-6\theta)$$
taking the plus and minus, I got my above-stated answer. But it doesn't quite agree with the answer given in my book. Can anyone explain my mistake?
|
You got the same answer, just in a different form
$$ \pm \frac{\pi}{2} + 2n\pi = -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ \cdots = \frac{\pi}{2}+n\pi $$
$$ \pm \frac{\pi}{10} + \frac{2n\pi}{5} = -\frac{\pi}{10}, \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \ \cdots = \frac{\pi}{10} + \frac{n\pi}{5} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2524318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Laplace Transform to solve system of differential equations I have the system:
\begin{align}
2\frac{dx}{dt} + \frac{dy}{dt} - 2x &= 1 \\
\frac{dx}{dt} + \frac{dy}{dt} - 7x-7y &=2
\end{align}
$y(0)=0, x(0)=0$
Which I am attempting to solve using Laplace Transform's. I change the system to
$2sx+sy-2x=1$
$sx-sy-7x-7y=2$
Solving I get $x=-{(s+7)\over(s^2-9s+14)}$ and $y={3(s+1)\over(s^2-9s+14)}$.
Plugging in the inverse laplace transforms for these I get $x=\frac{14}{5}e^{7t}+\frac{9}{5}e^{2t}$ and $y=\frac{24}{5e}^{7t} - \frac{9}{5}e^{2t}$. Which is evidently wrong.
Does anyone see what I did wrong here?
|
Using the standard Laplace transform then the equations become
\begin{align}
2(s-1) \, \overline{x} + s \overline{y} &= \frac{1}{s} \\
(s-7) \overline{x} + (s-7) \overline{y} &= \frac{2}{s}.
\end{align}
Solving for $\overline{x}$ and $\overline{y}$ then
\begin{align}
\overline{x} &= - \frac{s+7}{s (s-2) (s-7)} \\
\overline{y} &= \frac{3 (s+1)}{s (s-2) (s-7)}
\end{align}
and lead to
\begin{align}
x(t) &= \frac{9}{5} \, e^{2 t} - \frac{2}{5} \, e^{7 t} - \frac{1}{2} \\
y(t) &= \frac{24}{35} \, e^{7 t} - \frac{9}{10} \, e^{2 t} + \frac{3}{14}.
\end{align}
Note: The Laplace transform is defined by
$$\mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{- s t} f(t) \, dt,$$
in shorthand notation $f(t) \doteqdot f(s)$, and
\begin{align}
\frac{dx}{dt} &\doteqdot s \overline{x} - x(0) \\
a &\doteqdot \frac{a}{s} \hspace{5mm} \text{$a$ is a constant} \\
e^{a t} &\doteqdot \frac{1}{s - a}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2524812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
An identity involving the Gamma function I am trying to prove the following identity. Let $k\in\mathbb{N}$, $k\geq 2$. Then
$$
\sum_{i=1}^{k-1}\binom{k}{i}\prod_{l=2}^{i}(4l-6)\prod_{l=2}^{k-i}(4l-6)=\prod_{l=2}^{k}(4l-6).
$$
There is numerical evidence that the equation holds, also I have checked the identity for small values of $k$. Dividing the equation by $k!\cdot4^{k-2}$ and using the definition of the Gamma function it is easy to see that it is equivalent to show that
$$
\sum_{i=1}^{k-1}\frac{\Gamma(i-1/2)}{\sqrt{\pi}\Gamma(i+1)}\frac{\Gamma(k-i-1/2)}{\sqrt{\pi}\Gamma(k-i+1)}=4\prod_{l=2}^{k}\frac{\Gamma(k-1/2)}{\sqrt{\pi}\Gamma(k+1)}.
$$
I have tried to show this equation by induction, but I got stuck. Also, observing that
$$
\prod_{l=2}^{k}(4l-6)=\frac{(2(k-1))!}{(k-1)!}
$$
is the highest order term of the polynomial
$$
\frac{\partial^{k-1}}{\partial x^{k-1}}x^{2(k-1)}
$$
did not help me. Do you have any ideas? Thank you in advance.
|
We start from
$$\sum_{q=1}^{n-1} {n\choose q}
\prod_{l=2}^q (4l-6) \prod_{l=2}^{n-q} (4l-6)
= \prod_{l=2}^n (4l-6).$$
This is
$$\sum_{q=1}^{n-1} {n\choose q}
2^{q-1} \prod_{l=2}^q (2l-3) 2^{n-q-1} \prod_{l=2}^{n-q} (2l-3)
= 2^{n-1} \prod_{l=2}^n (2l-3)$$
or
$$\sum_{q=1}^{n-1} {n\choose q}
\prod_{l=2}^q (2l-3) \prod_{l=2}^{n-q} (2l-3)
= 2 \prod_{l=2}^n (2l-3)$$
or
$$\sum_{q=1}^{n-1} {n\choose q}
\frac{(2q-2)!}{2^{q-1} (q-1)!}
\frac{(2(n-q)-2)!}{2^{n-q-1} (n-q-1)!}
= 2 \frac{(2n-2)!}{2^{n-1} (n-1)!}$$
or
$$\sum_{q=1}^{n-1} {n\choose q}
\frac{(2q-2)!}{(q-1)!}
\frac{(2(n-q)-2)!}{(n-q-1)!}
= \frac{(2n-2)!}{(n-1)!}$$
Continuing we get
$$n! \sum_{q=1}^{n-1} \frac{1}{q (n-q)}
{2q-2\choose q-1} {2n-2q-2\choose n-q-1} = \frac{(2n-2)!}{(n-1)!}$$
or
$$\sum_{q=1}^{n-1} \frac{1}{q (n-q)}
{2q-2\choose q-1} {2n-2q-2\choose n-q-1} =
\frac{1}{n} {2n-2\choose n-1}.$$
Now recall the generating function of the Catalan numbers which says
that
$$\sum_{n\ge 0} \frac{1}{n+1} {2n\choose n} z^n
= \frac{1-\sqrt{1-4z}}{2z}.$$
This implies that
$$\sum_{n\ge 0} \frac{1}{n+1} {2n\choose n} z^{n+1}
= \frac{1-\sqrt{1-4z}}{2}$$
or
$$\sum_{n\ge 1} \frac{1}{n} {2n-2\choose n-1} z^{n}
= \frac{1-\sqrt{1-4z}}{2}.$$
Therefore the LHS of the identity is in fact
$$\sum_{q=1}^{n-1} [z^q] \frac{1-\sqrt{1-4z}}{2}
[z^{n-q}] \frac{1-\sqrt{1-4z}}{2}
\\ = \sum_{q=0}^{n} [z^q] \frac{1-\sqrt{1-4z}}{2}
[z^{n-q}] \frac{1-\sqrt{1-4z}}{2}
\\ = [z^n] \left(\frac{1-\sqrt{1-4z}}{2}\right)^2
= [z^n] \frac{1-2\sqrt{1-4z}+1-4z}{4}
\\ = [z^n] \frac{1-2z-\sqrt{1-4z}}{2}.$$
With $\sqrt{1-4z} = 1 - 2z - \cdots$ we get zero for $n\le 1$
and for $n\ge 2$ we find
$$[z^n] \frac{1-2z-\sqrt{1-4z}}{2}
= [z^n] \frac{1-\sqrt{1-4z}}{2} =
\frac{1}{n} {2n-2\choose n-1}$$
as claimed.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2525372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding coefficient of $ x^{2017}$ in expansion of $(x +1+\frac{1}{x})(x^3 + 1 + \frac{1}{x^3})...(x^{2187} + 1 + \frac{1}{x^{2187}})$ I notice that $3^7=2187$ and this implies there are 8 terms in product.
The presence of $x^3$ and its powers gives a slight possible hint that $\omega$ might do some trick. But I don't see how.Other things I thought was ${2017}$ leaved remainder 1 when divided by 3 so maybe finding coefficient of $x$,$x^4$ may help. and I multiplied the first two terms to see that all the powers of x occurred. I am thinking many ideas but none of them are solving the problem,also it looks like the answer maybe 1.Probably the only reason why we are finding coefficient of $x^{2017}$ because this year is 2017.I can't solve it directly by observation.
Please suggest some method and see where I am lacking.
|
A nice example of telescoping products. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{2017}]\prod_{j=0}^7\left(\frac{1}{x^{3^j}}+1+x^{3^j}\right)}
&=[x^{2017}]\prod_{j=0}^7\frac{1+x^{3^j}+\left(x^{3^j}\right)^2}{x^{3^j}}\\
&=[x^{2017}]\prod_{j=0}^7\frac{1-x^{3^{j+1}}}{x^{3^j}\left(1-x^{3^j}\right)}\tag{1}\\
&=[x^{2017}]\frac{1}{x^{\sum_{j=0}^{7}3^j}}\prod_{j=0}^7\frac{1}{1-x^{3^j}}\prod_{j=1}^8\left(1-x^{3^j}\right)\tag{2}\\
&=[x^{2017}]x^{-\frac{1}{2}\left(3^8-1\right)}\frac{1-x^{3^8}}{1-x}\tag{3}\\
&=[x^{5297}]\sum_{j=0}^{3^8}x^j\tag{4}\\
&\color{blue}{=1}
\end{align*}
Comment:
*
*In (1) we use of $(1+y+y^2)(1-y)=1-y^3$ with $y=x^{3^j}$.
*In (2) we factor out $\prod_{j=0}^7x^{3^j}$, seperate numerator and denominator and shift the index of the right product by one to better see the telescoping property.
*In (3) we apply the finite geometric series formula and cancel terms thanks to telescoping.
*In (4) we note that $\frac{1}{2}\left(3^8-1\right)=3280$ and apply the rule $[x^p]x^{-q}A(x)=[x^{p+q}]A(x)$. We also expand the geometric power series and see finally the coefficient of $x^{5297}$ is equal to $1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2525752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
recurrence relation, all terms of the sequence positive Let $a_1=a$, $a_2=\frac{1}{a}-a$, $a_{n+1}=\frac{n}{a_n}-a_n-a_{n-1}$ for $n=2,3,4,...$.
Find all $a$ such that $(a_n)$ is a sequence of positive reals.
My attempt was to look at $a_3=\frac{3a^2-1}{a-a^3}$, $a_4=\frac{8a^3-4a}{3a^4-4a^2+1}$ and a few more, $a_1>0$ gives $a>0$, $a_2>0$ gives $a\in(0,1)$, $a_3>0$ gives $a\in(\frac{1}{\sqrt{3}},1)$, but this probably doesn't give important information and further terms are nasty.
|
I post it as a curiosity. Considering the difference equation
$$
\frac{n}{x_n^2} = 1 + \frac{x_{n+1}+x_{n-1}}{x_n}
$$
and considering
$$
\Delta x_n = x_n-x_{n-1}
$$
we have
$$
\frac{n}{x_n^2} = 3+\frac{\Delta x_{n+1}-\Delta x_n}{x_n}
$$
which is a difference approximation for the DE
$$
\frac{t}{x^2(t)} = 3 + \frac{\ddot x(t)}{x(t)}
$$
now calling $x(1) = a = \frac{2 \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{1}{4}\right)}$ and $x(2) = a-\frac 1a$ and integrating we have
In red the plot for $x_n = \sqrt{\frac n3 + \frac{1}{36n}+ \mathcal{O}(\frac{1}{n^3})}$ and in blue the DE solution.
and now considering instead the initial conditions $x(2) = a-\frac 1a, x(3) = \frac{1-3a^2}{a^3-3}$ we obtain
This result points in the direction of good agreement between the DE and the recurrence. Now following with the initial conditions
$$
x(3) = \frac{1-3a^2}{a^3-3}\\
x(4) = \frac{8 a^3-4 a}{3 a^4-4 a^2+1}
$$
we obtain
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2525989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 0
}
|
Solving system of linear equations with rationals Problem:$$(x+11):(x+6)=(y+12):(y+7)\land(y+1):(x-1)=y:x$$
When I tried using $\frac{a1}{b1}=\frac{a2}{b2}=\frac{k1a1+k2a2}{k1b1+k2b2}$ for $\frac{x+11}{x+6}=\frac{y+12}{y+7}$ where k1 = 1 and k2 = -1, this is what I got:$$\frac{x+11-y-12}{x+6-y-7}=\frac{x-y-1}{x-y-1}=1$$
Which would mean $\frac{x+11}{x+6}=1$ and $\frac{y+12}{y+7}=1$, which are false.
If I instead transform $(x+11):(x+6)=(y+12):(y+7)$ into $\frac{x+11}{y+12}=\frac{x+6}{y+7}$ then $$\frac{x+11-x-6}{y+12-y-7}=\frac{5}{5}=1$$
Which is possible for $\frac{x+11}{y+12}$ and $\frac{x+6}{y+7}$. The equations are easy and could be solved simply by other means. I must be missing something fundamental.
Please help. Thanks.
|
$
\left\{
\begin{array}{l}
\dfrac{x+11}{x+6}=\dfrac{y+12}{y+7} \\
\dfrac{y+1}{x-1}=\dfrac{y}{x} \\
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
(x+11)(y+7)=(x+6)(y+12) \\
x(y+1)=y(x-1) \\
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
xy+7x+11y+77=xy+12x+6y+72 \\
xy+x=xy-y\\
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
x-y=1 \\
x+y=0\\
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
x=\dfrac{1}{2} \\
y=-\dfrac{1}{2}\\
\end{array}
\right.
$
Hope this can be useful
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2526489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Help thorougly proving a limit: $a>b>0; (a^n-b^n)/(a^n+b^n)$ as $n\to \infty$ Clearly, using basic calculus, I can find that the limit is 1. This is what I had so far:
$(1)$ $\frac {a^n-b^n}{a^n+b^n} = (2)$ $\frac {a^n}{a^n+b^n} - \frac {b^n}{a^n+b^n}
= (3)$ $\frac {1}{1+ \frac{b^n}{a^n}}-\frac {1}{1+ \frac{a^n}{b^n}}$
$(4)$ $a>b \implies (5)$ $\frac{b^n}{a^n}=0$ and $(6)$ $\frac{a^n}{b^n}=\infty$ as $n \to \infty$.
$(7)$ $\frac {1}{1+ \frac{b^n}{a^n}}-\frac {1}{1+ \frac{a^n}{b^n}} = (8)$ $\frac {1}{1+0} - \frac {1}{1+\infty} = (9)$ $1-\frac{1}{\infty}=(10)$ $1-0=1$
I asked another student whether what I had was correct; he said it was correct, however, I should put an indication of a theorem or definition at each step that I make.
So far I have a theorem that says I can use the step that the sum of limits is equal to the limit of sums. Some steps are trivial, but I'm not sure which ones are. Are there any more theorems or definitions I can use to identify what steps I'm taking?
|
If you mean that $n\rightarrow+\infty$ then
$$\frac{a^n-b^n}{a^n+b^n}=\frac{1-\left(\frac{b}{a}\right)^n}{1+\left(\frac{b}{a}\right)^n}\rightarrow\frac{1-0}{1+0}=1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2531193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Evaluate $\int \frac{x+1}{x^2+2x}dx$ in two different ways I am trying to evaluate
$$\int \frac{x+1}{x^2+2x}dx$$
Sounds like $$\int \frac{x+1}{x^2+2x}dx=\int \frac{(x+1)}{(x+1)^2-1}dx\\u=x+1 \implies\int\frac{u}{u^2-1}du\\t=u^2 \implies \frac{1}{2}\int\frac{1}{t-1}dt $$ Which is also known as $$\frac{1}{2}\ln|t-1|=\frac{1}{2}\ln|x^2+2x|$$But before I tried that, I had tried $$\int \frac{x+1}{x^2+2x}dx=\int \frac{1}{x+2}dx+\int \frac{1}{x^2+2x}dx\\=\ln|x+2| + \int \frac{1}{(x+1)^2-1}dx$$ Recalling that $\int \frac{1}{u^2-1}du=-\tanh^{-1}(u)$ or $-\coth^{-1}(u)$, $$\ln|x+2| + \int \frac{1}{(x+1)^2-1}dx=\ln|x+2|-\tanh^{-1}(x+1)$$ or $\ln|x+2|-\coth^{-1}(x+1)$ on the different parts of the domain.
Why did I get two different answers for the two different methods? Thanks.
Edit: They are not different. Just from the logarithmic definition of $\coth^{-1}$ and $\tanh^{-1}$ they are the same thing.
|
These are not really differents ways – only variants. You can indeed use a partial fractions decomposition:
$$\frac{x+1}{x^2+2x}=\frac{x+1}{x(x+2)}=\frac Ax+\frac B{x+2},$$
determine $A$ and $B$, then the integrals of each term.
Or you can go faster and observe that
$$\int\frac{x+1}{x^2+2x}\mathop{}\!\mathrm d x=\int\frac12\frac{(x^2+2x)'}{x^2+2x}\mathop{}\!\mathrm d x=\frac12\ln\bigl|x^2+2x\bigr|.$$
Addendum
Your second method actually yields the same result if you remember that all inverse hyperbolic functions are logs: indeed
$$\arg\coth x=\frac12\ln\Bigl(\frac{x+1}{x-1}\Bigr), $$
so that
$$\ln|x+2|-\arg\coth(x+1)=\frac12\biggl(\ln(x+2)^2-\ln\Bigl(\frac{x+2}{x}\Bigr)\biggr)=\frac12\ln\biggl(\frac{(x+2)^2x}{x+2}\biggr).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2537366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Trigonometric equation result differs from given i've got an equation:
$$\sin^6(x) + \cos^6(x) = 0.25$$
and i'm trying to solve it using the sum of cubes formula, like this:
$$ (\sin^2(x))^3 + (\cos^2(x))^3 = 0.25 $$
$$ (\sin^2(x) + \cos^2(x))^2 (\sin^4(x) - \sin^2(x)\cos^2(x) + \cos^4(x)) = 0.25 $$
$$ 1 - 3\sin^2(x)\cos^2(x) = 0.25 $$
$$ -3 \sin^2(x)\cos^2(x) = -\frac{3}{4} $$
$$ \sin^2(2x) = 1 $$
$$ \sin^2(2x) = \sin^2(2x) + \cos^2(2x) $$
$$ \cos^2(2x) = 0 $$
and here it must be $x = \frac{\pi n}{2} \pm \frac{\pi}{4} $
but solution is $ x = \pi n \pm \frac{\pi}{4}$
What's wrong?
|
Observe that $$\dfrac{\pi n}2\pm\dfrac\pi4=\dfrac{\pi(2n\pm1)}4$$
Now set a few values of $n$ to check overlap which is evident as $2n+1=2(n+1)-1$
Actually, $\cos y=0\implies y=(2m+1)\dfrac\pi2$ where $m$ is any integer
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2539488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $2^{3n+1} + 5$ is a multiple of 7 for all n ≥ 0. As the title states I need to prove that $(2^{3n+1}+5)$ is a multiple of 7 for all $n \geq 0$.
I can do this using induction but I also want to prove it using modular arithmetic. So here's what I've got so far.
Start with Fermat's little theorem:
\begin{align}2^6 &\equiv 1 \ \ \pmod{7} \\ 2^3 \cdot 2^3 &\equiv 1 \ \ \pmod{7} \\8 \cdot 2^3 &\equiv 1 \ \ \pmod{7} \\ 2^3 &\equiv 1 \ \ \pmod{7} \\ 2^{3n} &\equiv 1^n \pmod{7} \end{align}
I feel up to this point everything is correct.
How would I add the $1$ to the exponent?
\begin{align} 2^{3n+1} &\equiv 1^n (5) \pmod{7} \\ 2^{3n+1} &\equiv 5 \ \ \ \ \ \ \ \pmod{7}
\end{align}
Now adding the $5$, I am confused as to how to do that as well. I would just subtract the $5$ remainder correct? Such that:
$2^{3n+1} -5 \equiv 0 \pmod{7}$ but this is not what I intend to do.
I need, $2^{3n+1} +5 \equiv 0 \pmod{7}$
|
When you add $1$ to the exponent on $2$ you multiply by $2$, not by $5$ or $-2$. So $2^{3n+1}\equiv 1×2\bmod 7$ and the rest is easy.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2549706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$ (using induction)
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$
Base case $n = 1\to 5 + 6 - 3 = 8 \to 8 \mid 8 $
Assume that for some $n \in \mathbb{N}\to 8 \mid 5^n + 2 \cdot 3^n - 3$
Showing $8 \mid 5^{n+1} + 2 \cdot 3^{n+1} - 3$
$$5^{n+1} + 2 \cdot 3^{n+1} - 3$$ $$5\cdot 5^n + 2\cdot 3\cdot 3^n - 3$$ $$ (5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n $$ $$ 5\cdot(5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n - 4\cdot(5^n + 2\cdot 3^n - 3)$$
$$ [5\cdot(5^n + 2\cdot 3^n - 3)] - [4\cdot 3^n - 12]$$
The first term divides by 8 but I am not sure how to get the second term to divide by 8.
|
It is $4\cdot 3^n-12=4(3^n-3)$, since $3^n-3$ is even $2\mid 3^n-3$. Hence $8\mid 4\cdot 3^n-12$.
To make it clear: $3^n-3$ is even, since for $n\geq 1$ it is $3^n$ odd, and
"odd-odd=even" Since $(2k+1)-(2l+1)=2k-2l=2(k-l)$ which is even.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2554949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
}
|
If $z=\tan(\frac{x}{2})$, show that $\sin(x)=\frac{2z}{1+z^2}$ So as the title states I've got the following problem:
If $z=\tan(\frac{x}{2})$, show that $\sin(x)=\frac{2z}{1+z^2}$
So I'd guess that this probably involves the formula for half-angles, but that is is a dead-end.
Any suggestions?
Thank you in advance!
|
$$\frac{2\tan\left(\frac{x}{2} \right)}{1+\tan^2\left(\frac{x}{2} \right)}=$$
We know that $\tan^2(x)+1=\sec^2(x)$, so:
$$\frac{2\tan\left(\frac{x}{2} \right)}{\sec^2\left(\frac{x}{2} \right)}=$$
$$\frac{2\tan\left(\frac{x}{2} \right)}{\frac{1}{\cos^2\left(\frac{x}{2} \right)}}=$$
$$2\tan\left(\frac{x}{2} \right)\cos^2\left(\frac{x}{2} \right)=$$
$$2\frac{\sin\left(\frac{x}{2} \right)}{\cos\left(\frac{x}{2} \right)}\cos^2\left(\frac{x}{2} \right)=$$
$$2\sin\left(\frac{x}{2} \right)\cos\left(\frac{x}{2} \right)=$$
We know that $\sin(2x)=2\sin(x)\cos(x)$, so:
$$\sin\left(2\frac{x}{2} \right)=\sin(x)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2556046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy
Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy.
My work :
Need to show that for every $\epsilon \gt 0$ there exist $N$ such that $n,m\ge N \implies | a_n - a_m| \lt\epsilon$.
$$|a_n-a_m| = \dfrac{1}{2}|(a_{n-1} + a_{n-2}) - ( a_{m-1} + a_{m-2})|$$
I feel triangle inequality might be helpful here, but really not sure how to link it to the $\epsilon$. Appreciate any help...
|
Note that for $n\geq 0$,
$$a_{n+2}-a_{n+1} =\frac{ (a_{n+1}+a_n)}{2}-a_{n+1}=\frac{a_n-a_{n+1}}{2}.$$
Hence
$$|a_{n+2}-a_{n+1}|= \frac{1}{2}|a_{n+1}-a_n|= \frac{1}{2^2}|a_{n}-a_{n-1}|= \frac{1}{2^n}|a_{2}-a_{1}|=\frac{1}{2^n}.$$
Now if $n>m\geq 1$ then, by the triangle inequality,
$$|a_{n}-a_{m}|\leq |a_{n}-a_{n-1}|+\dots+|a_{m+1}-a_{m}|=
\frac{1}{2^{n-2}}+\dots+\frac{1}{2^{m-1}}.$$
Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2559687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
What is the probability that this quadratic equation has real roots? (I've seen the other questions in this site similar to my problem, but they didn't help much). So, the problem is:
The numbers $B$ and $C$ are chosen at random between $−1$ and $1$,
independently of each other. What is the probability that the
quadratic equation $x^2+Bx+C=0$ has real roots?
I've actually gotten an answer, but it's wrong according to my book. Here's what I did: I know that the condition for that equation to have real roots is that $B^2-4C\ge 0\Rightarrow C\le\frac{B^2}{4}$. The sample space consists of all points $(x,y):-1\le x\le1,-1\le y\le1$. So I think the desired probability should be the ratio of the area a square of side-lenght $2$ to the area under $C\le\frac{B^2}{4}$ from $-1$ to $1$.
That makes intuitive sense to me, but the book says the answer is actually $\frac{1}{4}(2+\int_{-1}^{1}x^2dx)$. Where does that $2$ come from? Thanks very much in advance.
|
I am not sure if the answer given in the book is exactly correct or not. Because I am finding a different answer while I tried to compute the probability.
What I attempted:- The numbers $B$ and $C$ are chosen randomly from $[-1,1]$. So, we can model them as uniform random variables taking values between $-1$ and $1$. If $b$ and $c$ denote the values taken by the random variables $B$ and $C$ respectively, then \begin{equation} f_B(b)=\frac{1}{1-\left( -1\right)}=\frac{1}{2} \qquad -1\le b \le 1
\end{equation} and,\begin{equation} f_C(c)=\frac{1}{1-\left( -1\right)}=\frac{1}{2} \qquad -1\le c \le 1
\end{equation}
As you have already mentioned, for real roots we must have $C\le \frac{B^2}{4}$. So, the only constraint is imposed upon $C$ and its value should be less than$\frac{B^2}{4} $. $B$ is free to vary in the range $(-1,1)$. So, the required probability is given by
\begin{equation}
\begin{aligned}
P\left(C\le \frac{B^2}{4}\right) &=\int_{-1}^{1} \int_{-1}^{\frac{b^2}{4}} f_{BC}(b,c) dc db \\
&=\int_{-1}^{1} \int_{-1}^{\frac{b^2}{4}} f_B(b).f_C(c) dc db \qquad\mbox{(Since $B$ and $C$ are independent)} \\
&=\int_{-1}^{1} \int_{-1}^{\frac{b^2}{4}} \frac{1}{4} dc db \\
&=\frac{1}{4} \left(\int_{-1}^{1}\left(\frac{b^2}{4}+1 \right)db \right) \\
&=\frac{1}{4} \left(\frac{1}{4}\times \frac{2}{3} +2\right) \\
&=\frac{13}{24}
\end{aligned}
\end{equation}
I think you missed the factor $\frac{1}{4}$ just before the integral inside the parentheses. As you mentioned in the comment, there is a general formula when $B$ and $C$ lie in an interval $\left[-n,n\right]$. So, if you put $n=1$, the required probability is $=\frac{1}{2}+\frac{1}{24}=\frac{13}{24}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2560875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Time and Work. How much work does C do per hour?
A, B and C need a certain unique time to do a certain work. C needs 1
hours less than A to complete the work. Working together, they require
30 minutes to complete 50% of the job. The work also gets completed if
A and B start working together and A leaves after 1 hour and B works
for a further 3 hours. How much work does C do per hour?
(A)16.66%
(B)33.33%
(C)50%
(D)66.66%
My attempt:
Let the total work be 100 units.
Let the work done by A,B and C be a units/hour,b units/hour,c units/hour respectively.
Let the time taken by A alone to complete the work be t hours.
ATQ:
\begin{align*}
(a+b+c) \cdot \frac{1}{2} & =50 \tag{1}\\
(a+b) \cdot 1+b \cdot 3 & =100 \tag{2}\\
c \cdot (t-1)& =100 \tag{3}\\
at & =100 \tag{4}
\end{align*}
Please help me solve these equations. When I am solving it is getting cumbersome.
Also if someone tells us some other way of solving, that would be helpful as well. Thanks.
|
Let the total work be 1, or 100%.
Let the efficiency of A,B and C be a,b and c(percentage of work be done in an hour)
(1) $\frac{1}{c}$ + 1 = $\frac{1}{a}$
(2) $(a+b+c)\cdot\frac{1}{2}$ = $\frac{1}{2}$
(3) $(a+b)\cdot1 + b\cdot3 = 1$
from (3) you get
a + 4b = 1
b = $\frac{1-a}{4}$
put it into (2)
a + $\frac{1-a}{4}$ + c = 1
4a + 1 - a + 4c = 4
a = $\frac{3-4c}{3}$
put it into (1)
$\frac{1}{c} + 1 = \frac{3}{3-4c}$
$\frac{1+c}{c} = \frac{3}{3-4c}$
$4c^2 + 4c - 3 = 0$
$4c^2 + 4c + 1 = 4$
$(2c + 1)^2 = 4$
2c + 1 = 2 (c can not be negative)
c = 0.5
The final answer is C 50%.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2566277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
}
|
Deriving the asymptotic estimate (9.62) in Concrete Mathematics I was reading Chapter 9: Asymptotics in Graham, Knuth, Patashnik: Concrete Mathematics, and I got stuck while deriving the following asymptotic estimate on page 466:
$$
\begin{equation}
g_n = \frac{e^{\pi^2/6}}{n^2} + O(\log n / n^3) \, ,
\quad \text{for } n > 1 \, . \tag{9.62}
\end{equation}
$$
The value $g_n$ is the coefficient of $z^n$ in the generating function
$$
\begin{equation}
G(z) = \exp\left(\sum_{k \geq 1} \frac{z^k}{k^2}\right) \, . \tag{9.57}
\end{equation}
$$
To derive the estimate, one starts by differentiating $(9.57)$:
$$
G'(z) = \sum_{n \geq 0} n g_n z^{n-1} = \left( \sum_{k \geq 1} \frac{z^{k-1}}{k} \right) G(z) \, .
$$
Equating coefficients of $z^{n-1}$ on both sides leads to the following recurrence for $g_n$:
$$
\begin{equation}
n g_n = \sum_{0 \leq k < n} \frac{g_k}{n-k} \, . \tag{9.58}
\end{equation}
$$
Next, one proceeds with the following bootstrapping trick. Start with a rough initial estimate $g_n = O(1)$, obtained by showing that $0 < g_n \leq 1$ for $n \geq 0$. Plug the initial estimate into the recurrence to get a better estimate $g_n = O(\log n / n)$. By repeatedly plugging new estimates back into the recurrence and massaging the recurrence when needed, one gets successively better estimates, leading up to the following one that is one step away from $(9.62)$ (note the differently placed exponent in the $O$ term):
$$
\begin{equation}
g_n = \frac{e^{\pi^2/6}}{n^2} + O(\log n / n)^3 \, ,
\quad \text{for } n > 1 \, . \tag{9.61}
\end{equation}
$$
Another bootstrapping step is supposed to give $(9.62)$. However, I fail to carry out the calculation.
Here are some of the attempts I made.
Attempt 1
Let me denote the constant $e^{\pi^2 / 6}$ by $c$. When I plug $(9.61)$ directly into $(9.58)$, I get the following:
$$
\begin{align}
n g_n = {} & \frac{1}{n} + \sum_{0 < k < n} \frac{c}{k^2 (n-k)} + \sum_{0 < k < n} \frac{O(\log n)^3}{k^3 (n-k)} \\
= {} & \frac{1}{n} + c \sum_{0 < k < n} \left( \frac{1}{n k^2} + \frac{1}{n^2 k} + \frac{1}{n^2(n-k)} \right) \\
& {} + O(\log n)^3 \sum_{0 < k < n} \left( \frac{1}{n k^3} + \frac{1}{n^2 k^2} + \frac{1}{n^3 k} + \frac{1}{n^3 (n-k)} \right) \\
= {} & \frac{1}{n} + c \left( \frac{1}{n} H^{(2)}_{n-1} + \frac{2}{n^2} H_{n-1} \right) + O(\log n)^3 \left( \frac{1}{n} H^{(3)}_{n-1} + \frac{1}{n^2} H^{(2)}_{n-1} + \frac{2}{n^3} H_{n-1} \right) \, .
\end{align}
$$
(In the last step, $H^{(i)}_{n-1}$ stands for generalized harmonic numbers, and $H_{n-1}=H^{(1)}_{n-1}$.) This seems to be worse than what I started with, since for example
$$ \frac{1}{n} H^{(3)}_{n-1} O(\log n)^3 = O\left( \frac{(\log n)^3}{n} \right) \, , $$
so $O((\log n)^3 / n^2)$ appears in the final estimate for $g_n$.
Attempt 2
Another attempt involves the trick of "pulling out the largest part." This trick is used in Concrete Mathematics to derive $(9.61)$. We can massage the recurrence $(9.58)$ to obtain the following:
$$
\begin{align}
n g_n & = \sum_{0 \leq k < n} \frac{g_k}{n} + \sum_{0 \leq k < n} g_k \left( \frac{1}{n-k} - \frac{1}{n} \right) \\
& = \frac{1}{n} \sum_{k \geq 0} g_k - \frac{1}{n} \sum_{k \geq n} g_k + \frac{1}{n} \sum_{0 \leq k < n} \frac{k g_k}{n-k} \, .
\end{align}
$$
The first sum is $G(1)=c$. For the second sum, I have
$$
\begin{align}
\sum_{k \geq n} g_k
& = \sum_{k \geq n} \frac{c}{k^2} + O\left( \sum_{k \geq n} \frac{(\log k)^3}{k^3} \right) \\
& = c \left( \frac{\pi^2}{6} - H^{(2)}_{n-1} \right) + O\left( \frac{(\log n)^3}{n^2} \right) \, .
\end{align}
$$
It doesn't look like I'm going to end up with anything as nice as $(9.62)$, but at least the final asymptotic error is still within $O(\log n / n^3)$. However, things fail again when I analyze the third sum:
$$
\begin{align}
\sum_{0 \leq k < n} \frac{k g_k}{n-k}
& = \sum_{0 < k < n} \frac{c}{k (n-k)} + \sum_{0 < k < n} \frac{O(\log n)^3}{k^2 (n-k)} \\
& = \frac{c}{n} \sum_{0 < k < n} \left( \frac{1}{k} + \frac{1}{n-k} \right) + O(\log n)^3 \left( \frac{1}{n} H^{(2)}_{n-1} + \frac{2}{n^2} H_{n-1} \right) \\
& = \frac{2c}{n} H_{n-1} + O(\log n)^3 \left( \frac{1}{n} H^{(2)}_{n-1} + \frac{2}{n^2} H_{n-1} \right) \, .
\end{align}
$$
Similarly as in Attempt 1, there is a term
$$ \frac{1}{n} H^{(2)}_{n-1} O(\log n)^3 = O\left( \frac{(\log n)^3}{n} \right) \, , $$
leading to $O(\log n / n)^3$ in the final estimate for $g_n$.
Attempt 3
After pulling out $1/n$ in Attempt 2, we see that the first sum gives us exactly the leading term in $(9.62)$, so that sum should stay as it is. The second and the third sum are problematic. Each on its own contributes too much, so I guess the trick is to somehow fuse them together. One idea is to try pulling out another $1/n$ from the third sum:
$$
\sum_{0 \leq k < n} \frac{k g_k}{n - k}
= \frac{1}{n} \sum_{0 \leq k < n} k g_k + \frac{1}{n} \sum_{0 \leq k < n} \frac{k^2 g_k}{n-k} \, .
$$
Now the sum $\sum_{0 \leq k < n} k^2 g_k / (n-k)$ is fine: with the same kind of analysis as in previous attempts we conclude it is $O((\log n)^4 / n)$, so with all the pulled-out parts it contributes $O(\log n / n)^4$ in the final estimate.
What to do with $\sum_{0 \leq k < n} k g_k$? My idea was to decompose it as in Attempt 2:
$$
\sum_{0 \leq k < n} k g_k = \sum_{k \geq 0} k g_k - \sum_{k \geq n} k g_k \, .
$$
Unfortunately, this doesn't work, because now the first sum is $G'(1)$, and this doesn't converge.
What am I doing wrong? Do I need to massage the recurrence in some other way? There must be something obvious that I just don't see.
|
We have $\enspace\displaystyle n g_n = \frac{1}{n}\sum\limits_{k\ge 0}g_k - \frac{1}{n}\sum\limits_{k\ge n}g_k + \frac{1}{n}\sum\limits_{0<k<n}\frac{k}{n-k}g_k $
with $\enspace\displaystyle g_n = \frac{G(1)}{n^2} + O\left(\frac{\ln n}{n}\right)^3 \enspace$ and $\enspace\displaystyle G(1)=e^{\zeta(2)}\,$ .
First part: $\enspace\displaystyle \sum\limits_{k\ge 0}g_k = G(1) $
With $\enspace\displaystyle O\left(\frac{G(1)}{n^2} + O\left(\frac{\ln n}{n}\right)^3\right)= O\left(\frac{1}{n^2}\right)\enspace$ we get
for the second part
$\enspace\displaystyle \sum\limits_{k\ge n}g_k = O\left(\sum\limits_{k\ge n}\frac{1}{k^2}\right)=O\left(\frac{1}{n}\right)\,$ ,
where the last step comes from $\enspace\displaystyle \lim\limits_{n\to\infty}n^x\sum\limits_{k\ge n} \frac{1}{k^{1+x}}=\frac{1}{x}\enspace$ for $\enspace x>0\,$ .
The third part is
$\enspace\displaystyle \sum\limits_{0<k<n}\frac{k}{n-k}g_k = O\left(\sum\limits_{0<k<n}\frac{1}{k(n-k)}\right)=O\left(\frac{\ln n}{n}\right) \,$ .
It follows: $\enspace\displaystyle g_n =\frac{1}{n}\left(\frac{G(1)}{n} - \frac{1}{n} O\left(\frac{1}{n}\right) + \frac{1}{n} O\left(\frac{\ln n}{n}\right)\right) = \frac{G(1)}{n^2} + O\left(\frac{\ln n}{n^3}\right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2566961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
}
|
Probability Problem with $10$ players being put on two teams Could somebody please check my answer to this problem?
Thanks
Bob
Problem:
$10$ kids are randomly grouped into an A team with five kids and a B team with five kids. Each
grouping is equally likely. There are three kids in the group, Alex and his two best friends Jose
and Carl. What is the probability that Alex ends up on the same team with at least one of his two
best friends?
Answer:
Let $p_{a}$, $p_j$, and $p_c$ be the probabilities that Alex is on team A, Jose is on team A, and Carl is on team A. Let $p$ by the probability that we seek.
\begin{eqnarray*}
p &=& 2 ( p_a p_j + p_a * p_c - p_a p_j p_c ) \\
p_a &=& \frac{1}{2} \\
p_j &=& \frac{1}{2} \\
p_c &=& \frac{1}{2} \\
p &=& =
2 ( \frac{1}{2} \Big( \frac{1}{2} \Big) + \frac{1}{2} \Big( \frac{1}{2} \Big) -
\frac{1}{2} \Big( \frac{1}{2} \Big) \Big( \frac{1}{2} \Big) \\
p &=& 2 ( \frac{1}{4} + \frac{1}{4} - \frac{1}{8} ) = 2 ( \frac{1}{2} - \frac{1}{8} ) \\
p &=& \frac{3}{4} \\
\end{eqnarray*}
|
Visualize $5$ slots each for team A and team B, viz $\;\;\boxed A\boxed A\boxed A\boxed A \boxed A\quad \boxed B\boxed B\boxed B\boxed B\boxed B$.
Alex can be on any team, occupying one slot.
P(Jose and Carl are both on the other team ) $= \dfrac59\cdot\dfrac48 = \dfrac5{18}$
Hence P(at least one of them is on Alex's team) $= 1 - \dfrac5{18} = \dfrac{13}{18}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2567048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Finding smallest possible value of expression with x and y I'm not supposed to use calculus here. I'm trying to find the smallest possible value of the expression $x^2+4xy+5y^2-4x-6y+7$ for real numbers $x$ and $y$.
Here's my attempt:
$x^2+4xy+5y^2-4x-6y+7=[(x-2)^2+3)+5((y-3/5)^2-9/25)]+4xy$
$3\le (x-2)^2+3$ for all real $x$
$-9/5\le 5((y-3/5)^2-9/25)$ for all real $y$.
Thus $6/5=3-9/5\le x^2+5y^2-4x-6y+7$ for all real $x$ and $y$. $(A)$
Therefore $x^2+4xy+5y^2-4x-6y+7 \ge 6/5+4xy$ for all real $x$ and $y$. $(B)$
The equality in $B$ does only hold when equality in $A$ holds. This happens when $x=2$ and $y=3/5$.
The smallest value of $x^2+4xy+5y^2-4x-6y+7$ is then $6/5+4*2*(3/5)=6$
Is this a legitimate procedure or am I doing something wrong? I'm a bit shaky on the statement $B$ and the following implications. Let me know.
|
Let $u=x^2+4xy+5y^2-4x-6y+7$
$\iff x^2+4x(y-1)+5y^2-6y+7-u=0$
which is a quadratic equation in $x$
As $x$ is real, the discriminant must be $\ge0$
i.e., $$16(y-1)^2-4(5y^2-6y+7-u)\ge0$$
$$\iff u\ge y^2+2y+3=(y+1)^2+2\ge2$$
The equality occurs if $y+1=0$ and $x=-\dfrac{4(y-1)}2=?$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2569688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Slope of The Tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$ Find slope of tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$
using $$\frac{dy}{dx} = \frac{\frac{dr}{dθ}\sin \theta + r \cos \theta}{\frac{dr}{dθ}\cos \theta-r \sin \theta}$$
I got
$$\frac {14\cos \theta \sin \theta}{7 \cos^2 \theta-7\sin \theta cos \theta}$$
|
Your derivation seems to be not correct.
Indeed:
$$x=r \cos \theta$$
$$y=r \sin \theta$$
$$r=7 \sin \theta \implies dr=7 \cos \theta d\theta$$
Thus:
$$dx=\cos \theta dr - r \sin \theta d\theta=7\cos^2 \theta-7\sin^2 \theta$$
$$dy=\sin \theta dr +r \cos \theta d\theta=7\cos \theta \sin \theta+7\cos \theta \sin \theta$$
That is
$$\frac{dy}{dx}=\frac {2\cos \theta \sin \theta}{\cos^2 \theta-\sin^2 \theta}=\frac{\frac{\sqrt 3}{2}}{\frac{3}{4}-\frac{1}{4}}=\sqrt 3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2571026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Is there a solution possible for the following integral? I want to solve the following integral
$$\int_0^{\infty}e^{-a(1+x^{m})^{\frac{2}{m}}}x\,{\rm d}x$$
where $a$ is real number but is not equal to $0$ and $m>2$. Any help or upper/lower bounds on this integral will be very helpful. Thank you.
|
Using CAS like Mathematica and MellinTransform can be solve it for m even integers and $a>0$.
$$\int_0^{\infty } \exp \left(-a \left(1+x^m\right)^{2/m}\right) \, dx=\mathcal{M}_a\left[\int_0^{\infty } \exp \left(-a
\left(1+x^m\right)^{2/m}\right) \, dx\right](s)=\int_0^{\infty } \mathcal{M}_a\left[\exp \left(-a \left(1+x^m\right)^{2/m}\right)\right](s) \,
dx=\mathcal{M}_s^{-1}\left[\int_0^{\infty } \left(1+x^m\right)^{-\frac{2 s}{m}} \Gamma (s) \,
dx\right](a)=\mathcal{M}_s^{-1}\left[\frac{\Gamma \left(1+\frac{1}{m}\right) \Gamma (s) \Gamma \left(\frac{-1+2 s}{m}\right)}{\Gamma
\left(\frac{2 s}{m}\right)}\right](a)$$
for m=3 and odd m Mathematica can't inverse Mellin transform.
for m=4
$$\mathcal{M}_s^{-1}\left[\frac{\Gamma \left(\frac{5}{4}\right) \Gamma (s) \Gamma \left(\frac{1}{4} (-1+2 s)\right)}{\Gamma
\left(\frac{s}{2}\right)}\right](a)=\frac{2^{3/4} \sqrt[4]{a} K_{\frac{3}{4}}(a) \Gamma \left(\frac{5}{4}\right)}{\sqrt{\pi }}$$
where $K_{\frac{3}{4}}(a)$ is modified Bessel function of the second kind.
for m=6
$\mathcal{M}_s^{-1}\left[\frac{\Gamma \left(\frac{7}{6}\right) \Gamma (s) \Gamma \left(\frac{1}{6} (-1+2 s)\right)}{\Gamma
\left(\frac{s}{3}\right)}\right](a)=\Gamma \left(\frac{7}{6}\right) \left(\frac{3 \sqrt{\pi } \,
_0F_2\left(;\frac{1}{6},\frac{1}{2};-\frac{a^3}{27}\right)}{\sqrt{a} \Gamma \left(\frac{1}{6}\right)}-\frac{2 a \sqrt{\pi } \,
_0F_2\left(;\frac{2}{3},\frac{3}{2};-\frac{a^3}{27}\right)}{3 \Gamma \left(\frac{2}{3}\right)}+\frac{2 a^2 \pi \,
_0F_2\left(;\frac{4}{3},\frac{11}{6};-\frac{a^3}{27}\right)}{9 \Gamma \left(\frac{4}{3}\right) \Gamma \left(\frac{11}{6}\right)}\right)$
where $\, _0F_2\left(;\frac{1}{6},\frac{1}{2};-\frac{a^3}{27}\right)$ is the generalized hypergeometric function.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2571213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
system of equations with 3 variables in denominator I'm having problems with this system of equations:
$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2}$$
$$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3}$$
$$\frac{3}{x-z}+\frac{5}{y-z}=-\frac{1}{4}$$
I've tried substitution method but that took me nowhere, I'm not sure I can solve it and would appreciate if anyone showed me the proper method to solve this problem.
|
$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2}$$
$$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3}$$
$$\frac{3}{x-z}+\frac{5}{\color{red}{x-z}}=-\frac{1}{4}$$
I assume that the part in red should be $y-z$; if not, please clarify.
Hint. Letting:
$$a=\frac{1}{x+y} \;,\; b=\frac{1}{x-z} \;,\;c=\frac{1}{y-z}$$
turn the system into:
$$\left\{\begin{array}{rcr}
3a&+&2b&&&=&\tfrac{3}{2}\\
a&&&-&10c&=&\tfrac{7}{3}\\
&&3b&+&5c&=&-\tfrac{1}{4}
\end{array}\right.$$
Perhaps this looks easier?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2571761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Why is function domain of fractions inside radicals not defined for lower values than those found by searching for domain of denominator in fraction? Consider function $y = \sqrt{\frac{1-2x}{2x+3}}$. To find the domain of this function we first find the domain of denominator in fraction:
$2x+3 \neq 0$
$2x \neq -3$
$x \neq - \frac{3}{2}$
So, the domain of $x$ (for fraction to be valid) is $x \in \left(- \infty, - \frac{3}{2}\right) \cup \left(- \frac{3}{2}, + \infty\right)$.
Then we find the domain for whole fraction:
$\frac{1-2x}{2x+3} \ge 0$
$1-2x \ge 0$
$-2x \ge -1$
$x \le \frac{1}{2}$
My textbook says that the (real) domain of the whole $y$ function is $x \in \left(- \frac{3}{2}, \frac{1}{2}\right]$.
I understand why the function is not defined in values larger than $\frac{1}{2}$ (because condition is $x \le \frac{1}{2}$), but I don't understand why it can't be have values less than $- \frac{3}{2}$ (because condition says only $x \neq - \frac{3}{2}$).
I checked the domain of this function and the domain given in the textbook is correct. Function has imaginary values for $x$ values less than $- \frac{3}{2}$ or bigger than $\frac{1}{2}$. It is undefined in $- \frac{3}{2}$.
Real values only in $\left(- \frac{3}{2}, \frac{1}{2}\right]$ domain.
|
If $x<-\frac{3}{2}$, then $2x+3<0$ and $1-2x>0$, so $\frac{1-2x}{2x+3}<0$, which means you can't take the square root and get a real answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2572000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
}
|
A Junior Olympiad like system of linear equations Given a system of linear equations
$$\begin{align}\frac{x}{3}+\frac{y}{5}+\frac{z}{9}+\frac{w}{17} &=1 \\
\frac{x}{4}+\frac{y}{6}+\frac{z}{10}+\frac{w}{18} &=\frac{1}{2} \\
\frac{x}{5}+\frac{y}{7}+\frac{z}{11}+\frac{w}{19} &=\frac{1}{3} \\
\frac{x}{6}+\frac{y}{8}+\frac{z}{12}+\frac{w}{20} &=\frac{1}{4} \\
\end{align}$$
Determine $$ \frac{x}{10}+\frac{y}{12}+\frac{z}{16}+\frac{w}{24}$$
This is a problem I was asked to solve a bit long before. Since I didn’t come up with any other ideas than using Gauss Jordan elimination, I did so. The answer is $\frac{1}{36}$. Could someone provide me an elegant solution to this problem?
|
Let $G(s) = \frac{x}{s+2} + \frac{y}{s+4} + \frac{z}{s+8} + \frac{w}{s+16} - \frac 1s$, and $F(s) = s(s+2)(s+4)(s+8)(s+16)G(s)$.
Clearly $F(s)$ is a degree 4 polynomial with roots being 1, 2, 3, and 4. Also, $F(0) = - 2\times4\times8\times16 = -2^{10}$. Hence
$$
F(s) = -\frac{2^{10}}{24}(s-1)(s-2)(s-3)(s-4) = -\frac{2^7}{3}(s-1)(s-2)(s-3)(s-4).
$$
We are interested in $F(8) = -\frac{2^7}{3}\times7\times6\times5\times4 = -2^{10}\times5\times7$.
Hence
$$
\frac{x}{10}+\frac{y}{12}+\frac{z}{16}+\frac{w}{24} - \frac{1}{8} = \frac{F(8)}{8\times10\times12\times16\times24} = \frac{-2^{10}\times 5\times 7}{2^{13}\times3^2\times5} = -\frac{7}{72},
$$
and
$$
\frac{x}{10}+\frac{y}{12}+\frac{z}{16}+\frac{w}{24} = \frac18 - \frac{7}{72} = \frac{1}{36}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2573182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
Does the convergence of $\sum (a_k)^2$ imply $\sum (a_k)^3$ convergence? Does the convergence of $\sum (a_k)^2$ imply $\sum (a_k)^3$ convergence? I feel like it definitely should but can't find a solid way to prove it ....
|
It's true for real sequences (as shown in other answers), but false for complex sequences.
For example, if $\omega = e^{2\pi i/3}$ and
$$
a_{3k+m} = \frac{\omega^m}{(k+1)^{1/3}}
\qquad\text{for}\qquad
k \ge 0
,\qquad
m \in \{ 0,1,2 \}
,
$$
that is, if the sequence $(a_j)_0^\infty$ is
$$
\frac{1}{1^{1/3}},
\frac{\omega}{1^{1/3}},
\frac{\omega^2}{1^{1/3}},
\quad
\frac{1}{2^{1/3}},
\frac{\omega}{2^{1/3}},
\frac{\omega^2}{2^{1/3}},
\quad
\ldots,
\quad
\frac{1}{n^{1/3}},
\frac{\omega}{n^{1/3}},
\frac{\omega^2}{n^{1/3}},
\quad
\ldots
,
$$
then $\sum a_j^2$ converges (to zero). Indeed, each group of three terms sums to zero:
$$
\left( \frac{1}{n^{1/3}} \right)^2 +
\left( \frac{\omega}{n^{1/3}} \right)^2 +
\left( \frac{\omega^2}{n^{1/3}} \right)^2
=
\frac{1+\omega^2+\omega^4}{n^{2/3}}
=
0
,
$$
so the partial sums $S_n = \sum_{j=0}^{n-1} a_j^2$ are $S_{3k}=0$, $S_{3k+1}=a_{3k}^2$, $S_{3k+2}=a_{3k}^2+a_{3k+1}^2$,
and thus $S_n \to 0$ since $a_j^2 \to 0$.
But $\sum a_j^3$ diverges, since
$$
\left( \frac{1}{n^{1/3}} \right)^3 +
\left( \frac{\omega}{n^{1/3}} \right)^3 +
\left( \frac{\omega^2}{n^{1/3}} \right)^3
=
\frac{1+1+1}{n}
$$
and $\sum (1/n)$ diverges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2574186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Muhammad Ali pulls tiles from bag $3.$ A bag contains three tiles marked $A$, $L$, and $I$. Muhammad wants to pick the letters $ALI$, in that sucession. Randomly, he pulls one til from the bag. If the letter $A$ is drawn, he keeps it. If the letter pulled is other than $A$, he puts it back into the back. He does the same thing with the next tile. If the selected tile is $L$, he keeps it. If it is not $L$, he puts it back in the bag. The probability that Muhammad draws from the bag at most $10$ times can be written in the form $\frac{x_1}{x_2}$, where $x_1$ and $x_2$ are relatively prime numbers. Compute the remainder when $x_1+x_2$ is divided by $1000$.
$\textbf{Thoughts}$
This seems like an application of casework counting. Here is my best try.
NOTE:Some of these cases could be incorrectly counted.
$3$ draws needed: $1$ case
$4$ draws needed: $2$ cases
$5$ draws needed: $3$ cases
$6$ draws needed: $4$ cases
$7$ draws needed: $5$ cases
We have an emerging pattern. Therefore, we get $1+2+3+4+5+6+7+9=45$. However, I do not know what my denominator will be...
|
What is the probability that it takes n draws, imagine the sequence of draws $x_1, x_2, \cdots, x_n$.
For this this to be a valid draw we have some $x_i=A$ and $x_{n-1}=L$ such that $1 \le i <n-1$. And since it took $n$ draws then $x_n=I$
For a particular $i,j$ the probability is $p_n(i,j)$
$$p_n(i,j)=\left(\frac{2}{3}\right)^{i-1}\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)^{n-i-2}\left(\frac12\right)$$
We need to sum this over all $i$ $$\sum_{i=1}^{n-2}\left(\frac{2}{3}\right)^{i-1}\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)^{n-i-2}\left(\frac12\right)=\frac16\frac{1}{2^{n-2}} \sum_{i=1}^{n-2}\left(\frac{2}{3}\right)^{i-1}2^i=\frac{1}{3.2^{n-1}}.\frac{2\left((\frac{4}{3})^{n-2}-1\right)}{\frac{4}{3}-1} \\\text{Probability that n draws are needed}=p_n=\frac{1}{2^{n-2}}{\left((\frac{4}{3})^{n-2}-1\right)}$$
So probability that he draws at most 10 times $$\sum_{n=3}^{10}p_n=\sum_{n=3}^{10}\left(\left(\frac23\right)^{n-2}-\left(\frac12\right)^{n-2}\right)\\=\frac{(2/3)(1-(2/3)^8)}{(1-2/3)}-\frac{(1/2)(1-(1/2)^8)}{(1-1/2)}\\=\frac{12610}{6561}-\frac{255}{256}$$
Simplifying this we get $$\frac{1555105}{1679616}$$
The final answer $x_1=1555105, x_2=1679616$ and $x_1+x_2=3234721$, therefore that last 3 digits is $721$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2575305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Factorization of a 4th degree polynomial
Factorization of :
$$x^4-x^3+3x^2+3x+5$$
$$x^4-x^3+3x^2+3x+5=(x^4-x^3+2)+3(x^2+x+1)$$
what do i do ? please help me
|
A possible error in the question, it might explain things. If we change 5 to 54, we get
$$ x^4 - x^3 + 3x^2 + 3x + 54 = ( x^2 -4 x + 9) ( x^2 + 3 x + 6) $$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
First you look for rational roots. There are none. Then we need only check, because of GAUSS,
The second result states that if a non-constant polynomial with
integer coefficients is irreducible over the integers, then it is also
irreducible if it is considered as a polynomial over the rationals.
$$ (x^2 + a x + b) ( x^2 + c x + 5 b), $$
with $$ b = \pm 1 $$
For both the case $b=1$ and $b=-1,$ we easily get correct coefficients for $x^3$ and $x,$ but then the $x^2$ term comes out to something incorrect. So the thing is irreducible. That is really all that is needed, try two cases.
$$ ( x^2 + x + 1) ( x^2 - 2 x + 5) = x^4 - x^3 + 4x^2 + 3x + 5 $$
Next comes out non-integer, and still wrong anyway. Gauss says we did not need to continue once $a,c$ came up not being integers.
$$( x^2 - \frac{x}{2} - 1) (x^2 - \frac{x}{2} - 5)= x^4 - x^3 -\frac{23x^2}{4} + 3x + 5$$
Perhaps I should add that, although there are no real roots, there are real numbers $p,q,r,s$ such that
$$ (x^2 + px+q)(x^2 + rx + s) = x^4 - x^3 + 3 x^2 + 3 x + 5 $$
where both quadratic factors are positive for any chosen $x$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2581248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Equation of the sphere that passes through 4 points Write he equation of the sphere that passes through points
$$a(-5,4,1),b(3,4,-5),c(0,0,4),d(0,0,0)$$
I tried to use four points to draw a geometric shape and then calculate the center of this shape on the basis of the circle that passing on four points. But I did not succeed
Here is the book answer
$$x^2+y^2+z^2+54x−58y+4z=0$$
|
The equation of the sphere with a centre in $(x_0,y_0,z_0) $ and radius $r $ is: $(x-x_0)^2+(y-y^0)^2+(z-z_0)^2=r^2$. For us, the unknowns are $x_0,y_0,z_0,r $.
Plug the given four points in:
$$\begin {align}(-5-x_0)^2+(4-y_0)^2+(1-z_0)^2=r^2 \\ (3-x_0)^2+(4-y_0)^2+(-5-z_0)^2=r^2 \\ (-x_0)^2+(-y_0)^2+(4-z_0)^2=r^2 \\ (-x_0)^2+(-y_0)^2+(-z_0)^2=r^2\end {align} $$
or (after a bit of transforming):
$$\begin {align}42+10x_0-8y_0-2z_0+x_0^2+y_0^2+z_0^2=r^2 \\ 50-6x_0-8y_0+10z_0+x_0^2+y_0^2+z_0^2=r^2 \\ 16-8z_0+x_0^2+y_0^2+z_0^2=r^2 \\ x_0^2+y_0^2+z_0^2=r^2\end {align} $$
Now, take away one of the equations from the rest (in this case, the last one is the easiest one):
$$\begin {align}42+10x_0-8y_0-2z_0=0 \\ 50-6x_0-8y_0+10z_0=0\\ 16-8z_0=0\\ x_0^2+y_0^2+z_0^2=r^2\end {align} $$
Solve the system of the first three (linear) equations, to obtain: $x_0=2, y_0=\frac {29}{4}, z_0=2$. Plug that into the fourth to obtain $r=\sqrt {2^2+\left (\frac {29}{4}\right)^2+2^2}=\frac {\sqrt {969}}{4}$.
Thus, the final equation is:
$$(x-2)^2+(y-\frac {29}{4})^2+(z-2)^2=\frac {969}{16} $$
or, after transforming:
$$x^2+y^2+z^2-4x-\frac {29}{2}y-4z=0$$
as all constant terms will happen to cancel each other.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2585392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
}
|
Given $3\cos x - 4 \sin x = 2$, find $3 \sin x + 4 \cos x$ without first solving for $x$ If $$3\cos{x}-4\sin{x}=2$$
find $$3\sin{x} +4\cos{x} $$
I have solved the equation for $x$, then calculated the required value, but I think there is a direct solution without solving the equation.
|
$$a \cos(x) - b \sin(x) = \sqrt{a^2+b^2} \cos(x+y)$$ where $\cos(y) = a/\sqrt{a^2+b^2}$ and $\sin(y) = b/\sqrt{a^2 + b^2}$. Then $$a \sin(x) + b \cos(x) = \sqrt{a^2 + b^2} \sin(x+y)$$
In your case, with $a=3$ and $b=4$, $\sqrt{a^2+b^2}=5$, $\cos(x+y) = 2/5$ so $\sin(x+y) = \pm \sqrt{1-(2/5)^2} = \pm \sqrt{21}/5$, so the answer is $\pm \sqrt{21}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2588061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
}
|
A double integral into polar coordinates I have the double integral
$$\int^{10}_0 \int^0_{-\sqrt{10y-y^2}} \sqrt{x^2+y^2} \,dx\,dy$$
And I am asked to evaluate this by changing to polar coordinates.
|
Complete the square:
$$
10y-y^2 = 25 - (5-y)^2,
$$
so the graph of $x = -\sqrt{10y-y^2} = - \sqrt{5^2 - (5-y)^2}$ is the left half of the circle $x^2 + (5-y)^2 = 5^2.$
Exercises with polar coordinates will have shown you that $$\tag 1 r = 5\sin\theta$$ is that circle. If you multiply both sides of $(1)$ by $r,$ you get $$ r^2 = 5r\sin\theta $$ which becomes $$ x^2+y^2 = 5y $$ and by completing the square, then becomes $$ x^2 + (y-5)^2 = 5^2. $$
Thus you have
$$
\int_{\pi/2}^\pi \int_0^{5\sin\theta} r\, dr\,d\theta.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2588968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Sum of the infinite series The series is $$\frac{5}{1\cdot2}\cdot\frac{1}{3}+\frac{7}{2\cdot3}\cdot\frac{1}{3^2}+\frac{9}{3\cdot4}\cdot\frac{1}{3^3}+\frac{11}{4\cdot5}\cdot\frac{1}{3^4}+\cdots$$
This is my attempt:
$$T_n=\frac{2n+3}{n(n+1)}\cdot\frac{1}{3^n}$$
Assuming $$\frac{2n+3}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$$
we find $A=3,B=-1.$ Putting these values in $T_n$ we get,
$$T_n=\frac{1}{n}\cdot\frac{1}{3^{n-1}}-\frac{1}{n+1}\cdot\frac{1}{3^{n}}$$
How do I find the sum of the series from here $?$
|
$\sum_{i=1}^{n} T_i =S_n - R_n$ , where
$S_n := \sum_{i=1}^{n} \dfrac{1}{i}\dfrac{1}{3^{i-1}};$
$R_n :=\sum_{i=1}^{n} \dfrac{1}{i+1} \dfrac{1}{3^i}.$
Change the dummy index in $R_n:$
$k= i +1$, then
$R_n=\sum_{k=2}^{n+1} \dfrac{1}{k}\dfrac{1}{3^{k-1}} =$
$(1 + \sum_{k=2}^{n}\dfrac{1}{k}\dfrac{1}{3^{k-1}}) -1 +
\dfrac{1}{n+1}\dfrac{1}{3^n}$
$= S_n - 1 + \dfrac{1}{n+1}\dfrac{1}{3^n}.$
Hence:
$T_n = S_n - R_n= 1 - \dfrac{1}{ n+1}\dfrac{1}{3^n}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2590246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
What is the residue for this function How to find the residue of $\dfrac{1}{\tan(z)}$? I have calculated that it has a pole of order 2 . But I'm having trouble when using the residue theorem and I end up with a residue of $2$. However, wolfram alpha calculator says that the residue=$0$.
|
Since, $ 1 + \cos z = 2\cos^2 \frac{z}{2} $, the function has second-order poles at $\frac{z}{2} = (n + \frac12)\pi$ or $z = (2n+1)\pi$
Let $w = z - (2n+1)\pi$, we can compute the residues
$$ \begin{align}
\operatorname*{Res}_{z=(2n+1)\pi} f(z) &= -\frac12 \lim_{z\to (2n+1)\pi} \frac{d}{dz}\left(\big(z -
(2n+1)\pi\big)^2\sec^2 \frac{z}{2}\right) && w = z- (2n+1)\pi \\
&= -\frac{1}{2} \lim_{w\to0} \frac{d}{dw}\left( w^2 \sec^2 \left(\frac{w + (2n+1)\pi}{2}\right)\right) && \cos(z + n\pi) = (-1)^n\cos(z) \\
&= -\frac12 \lim_{w\to0} \frac{d}{dw} \left(w^2 \sec^2 \left( \frac{w+\pi}{2} \right)\right) && \cos(z+\pi/2) = -\sin z \\
&= -\frac12 \lim_{w\to0} \frac{d}{dw} \left(w^2 \csc^2 \frac{w}{2}\right) \\
&= -\frac12 \lim_{w\to0} \left(2w\csc^2 \frac{w}{2} - w^2\csc^2 \frac{w}{2} \cot \frac{w}{2} \right) \\
&= -\frac12 \lim_{w\to0} \frac{2w\sin \frac{w}{2} - w^2 \cos \frac{w}{2}}{\sin^3 \frac{w}{2}} \\
&= -\frac12 \lim_{w\to0} \frac{2w\left(\frac{w}{2}-\frac{w^3}{48} + O(w^5)\right) - w^2 \left(1-\frac{w^2}{8}+O(w^4)\right)}{\frac{w^3}{8} + O(w^5)} \\
&= -\frac12 \lim_{w\to0} \frac{\frac{w^4}{12} + O(w^6)}{\frac{w^3}{8} + O(w^5)} \\
&= 0
\end{align} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2594768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
lim$_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+...+\frac{1}{\sqrt{3n}+\sqrt{3n+3}}\right)$
Question
Edit:
My Approach:
$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+...+\frac{1}{\sqrt{3n}+\sqrt{3n+3}}\right)$
$S_{k}= \sum_{n=1}^{k} \frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{3n}+\sqrt{3n+3}}\right) \frac{\left(\sqrt{3n}-\sqrt{3n+3}\right)}{\left(\sqrt{3n}-\sqrt{3n+3}\right)} = \sum_{n=1}^{k}\frac{1}{\sqrt{n}}\left(\frac{\sqrt{3n}-\sqrt{3n+3}}{-3}\right) = \frac{1}{3\sqrt{k}} \left\{ \sqrt{3k+3}\right\} $
$\lim_{k\rightarrow\infty} S_{k} = \frac{1}{\sqrt{3}}$
|
More generally, by Stolz-Cesaro Theorem, if $a\geq 0$ then
$$\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{3k}+\sqrt{3k+a}}=
\lim_{n\to\infty}\left(\frac{1}{\sqrt{n}-\sqrt{n-1}}\cdot\frac{1}{\sqrt{3n}+\sqrt{3n+a}}\right)\\
=
\lim_{n\to\infty}\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{3n}+\sqrt{3n+a}}=\frac{1}{\sqrt{3}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2595404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
a four variable inequality Let $a,b,c,d$ be non-negative real numbers satisfying $a+b+c+d=3$. Show
\begin{equation*}
\frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2d^3}+\frac{d}{1+2a^3} \geqslant \frac{a^2+b^2+c^2+d^2}{3}.
\end{equation*}
I got this inequality from an IMO preparation club. I no longer do contest math but still I find this interesting. There is a weird equality case : $(2,1,0,0)$ works (whereas $(2,0,1,0)$ does not).
So far I've tried regrouping terms to disminish the number of variables or the number of non-zero variables. I've shown that the inequality is implied by the three-variable version
\begin{equation*}
\frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2a^3} \geqslant \frac{a^2+b^2+c^2}{3}
\end{equation*} if $a+b+c=3$. The three variable version itself is implied by the two variable version, but the two variable version is false.
|
By AM-GM $$\sum_{cyc}\frac{a}{1+2b^3}=3+\sum_{cyc}\left(\frac{a}{1+2b^3}-a\right)=3-\sum_{cyc}\frac{2ab^3}{1+2b^3}\geq$$
$$\geq3-\sum_{cyc}\frac{2ab^3}{3b^2}=\frac{(a+b+c+d)^2-2\sum\limits_{cyc}ab}{3}=$$
$$=\frac{a^2+b^2+c^2+d^2+ac+bd}{3}\geq\frac{a^2+b^2+c^2+d^2}{3}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2596760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$5$ divides $2ax+b$ for some integer $a$,given $5$ does not divide $x$ $x$ and $b$ are given.Let,$5$ does not divide $x$.
Then $5$ divides $2ax+b$ for some integer $a$.
How do i prove this? I have not gone through a course in number theory,so a general proof would be appreciated.
|
The slightly stronger statement holds true that $\,2ax+b\,$ is a multiple of $\,5\,$ for an $\,a \in \{0, \pm 1, \pm 2\}\,$.
If $\,b\,$ is a multiple of $\,5\,$ then $\,2 \cdot 0 \cdot x+b=b\,$ is a multiple of $\,5\,$, so $\,a=0\,$ works.
Otherwise, note first that if $\,u\,$ is not a multiple of $\,5\,$ then $\,u^4 = \mathcal{M}5+1\,$ where the notation $\,\mathcal{M}5\,$ denotes any integer multiple of $\,5\,$. This follows directly from Fermat's little theorem, but is also easy to prove in this case, as to keep the answer self-contained and largely number-theory-free. Any number $\,u\,$ which is not a multiple of $\,5\,$ must be in one of two forms:
*
*if $\,u=\mathcal{M}5 \pm 1\,$, then $\,u^2= (\mathcal{M}5)^2 \pm 2 \cdot \mathcal{M}5+(\pm 1)^2 = \mathcal{M}5+1\,$, then $\,u^4=\left(u^2\right)^2$ $=\mathcal{M}5+1\,$ by the same argument;
*if $\,u=\mathcal{M}5 \pm 2\,$, then $\,u^2= (\mathcal{M}5)^2 \pm 4 \cdot \mathcal{M}5+(\pm 2)^2 = \mathcal{M}5+4 = \mathcal{M}5 - 1\,$, then $\,u^4=\left(u^2\right)^2$ $=\mathcal{M}5+1\,$ by the same argument as above.
Then what remains to be proved is that one of the numbers obtained for $\,a \in \{\pm 1, \pm 2\}\,$ is a multiple of $\,5\,$. Those numbers are $\,b-4x, b-2x, b+2x, b+4x\,$, and their product is:
$$
(b^2-4 x^2)(b^2-16 x^2) = b^4 + 64 x^4 - 20 b^2x^2 = b^4 -x^4 + 65 x^4 - 20 x^2b^2 = b^4 - x^4 +\mathcal{M}5
$$
But neither of $\,b, x\,$ is a multiple of $\,5\,$, so $\,b^4-x^4=(\mathcal{M}5+1)-(\mathcal{M}5+1)=\mathcal{M}5\,$, and therefore one of the factors $b-4x, b-2x, b+2x, b+4x$ must be a multiple of $\,5\,$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2599153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to prove that this triangle is equilateral? Question: If $\cos A +\cos B +\cos C=\frac{3}{2}$, prove that the triangle is equilateral.
My attempt: According to cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and $\cos C=\frac{a^2+b^2-c^2}{2ab}$.
$\cos A +\cos B +\cos C=\frac{3}{2}$
$\therefore\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$
$\implies\frac{ab^2+ac^2-a^3+bc^2+ba^2-b^3+ca^2+cb^2-c^3}{2abc}=\frac{3}{2}$
I have tried simplifying the given equation using cosine rule but could not get far. Please help.
|
Or $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=0.$$
Now, let $a\geq b\geq c$.
Thus,
$$0=\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}(a^3-a^2b-ab^2+abc)=\sum_{cyc}a(a-b)(a-c)\geq$$
$$\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0,$$
which says that the equality occurs for $a=b=c$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2602051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
sum $\displaystyle \sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$ I have the following series:
$$\sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$$
I am not able to do the telescoping process in the above series. I converted it into the following partial fraction:
$$\sum _{n=2}^{\infty}\frac{(-1)^n}{(n+2)(n-1)}$$
But nothing seems to cancel (as usually happens) using the telescoping method. How can I solve the above series? Is there any other method to do the above problem?
|
Writing the $n$th term of your sum in a form that telescopes can be avoid altogether by converting the problem to a double integral as follows.
Noting that
$$\frac{1}{n - 1} = \int_0^1 x^{n - 2} \, dx \qquad \text{and} \qquad \frac{1}{n + 2} = \int_0^1 y^{n + 1} \, dy,$$
the sum can be rewritten as
\begin{align*}
\sum_{n = 2}^\infty \frac{(-1)^n}{n^2 + n - 2} &= \sum_{n = 2}^\infty \frac{(-1)^n}{(n - 1)(n + 2)}\\
&= \int_0^1 \int_0^1 \sum_{n = 2}^\infty (-1)^n x^{n - 2} y^{n + 1} \, dx dy \tag1\\
&= \int_0^1 \int_0^1 \frac{y}{x^2} \sum_{n = 2}^\infty (-xy)^n \, dx dy\\
&= \int_0^1 \int_0^1 \frac{y}{x^2} \cdot \frac{x^2 y^2}{1 + xy} \, dx dy \tag2\\
&= \int_0^1 \int_0^1 \frac{y^3}{1 + xy} \, dx dy\\
&= \int_0^1 y^2 \Big{[} \ln (1 + xy) \Big{]}_0^1 \, dy\\
&= \int_0^1 y^2 \ln (1 + y) \, dy\\
&= \frac{1}{3} \ln (2) - \frac{1}{3} \int_0^1 \frac{y^3}{1 + y} \, dy \tag3\\
&= \frac{1}{3} \ln (2) - \frac{1}{3} \int_0^1 \left (y^2 - y + 1 - \frac{1}{1 + y} \right ) \, dy \tag4\\
&= \frac{1}{3} \ln (2) - \frac{1}{3} \left [\frac{y^3}{3} - \frac{y^2}{2} + y - \ln (1 + y) \right ]_0^1\\
&= \frac{2}{3} \ln (2) - \frac{5}{18}.
\end{align*}
Explanation
*
*Interchanging the summation with the double integration.
*Summing the series which is geometric.
*Integration by parts.
*Partial fraction decomposition.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2603199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Find all local maxima and minima of $x^2+y^2$ subject to the constraint $x^2+2y=6$. Does $x^2+y^2$ have a global max/min on the same constraint? This is my method for the local max/min. Does this answer sound sensible? (Not sure how to go about checking for global max/min though...)
Method
$G(x,y)=x^2+2y-6$. Rewrite in terms of $y$, so $y=((-x^2+6)/2)^2$. Then $y$ substitute into $f(x)$ so:
$f(x)=x^2+((-x^2+6)/2)^2=-2x^2+(x^4/4)+9$.
So $f'(x)=-4x+4x^3$. And if $f'(x)=0$ then $x=0,x=1$
And $f''(x)=-4+12x^2$ so when $x=0$ then $f''(x)=-4$ so there is a maximimum when $x=0$.
There is a local maximum at (0,3) with value 9.
When $x=1$ then $f''(x)=8$ so there is a minimum when $x=1$.
There is a local minimum at (1/5/2) with value 29/4.
I also tried answering this question via an alternative method using a Lagrange function, to check my result but I got a different answer...
|
From $x^2+2y=6 $ we get $ x^2=6-2y$
Substitute in $x^2+y^2$ to get $$x^2+y^2=y^2-2y+6$$ Note that $y^2-2y+6=(y-1)^2+5$.
Thus the minimum is attained when at $y=1$ at two points where and $x=2$ or $x=-2$
There is no global for $(y-1)^2+5$, hence there is no global maximum for $ x^2+y^2$ subject to $x^2+2y=6 $.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2603889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Calculating an inverse Laplace transform I have a pain for a problem... need to find the inverse Laplace transform of:
$$\frac{s^2+s+1}{(s+1)(s+2)^2(s^2+4s+9)}$$
Now, I get that we have to expand the partial fraction, I managed to get to this:
$$\frac{A}{(s+1)}+\frac{B}{(s+2)}+\frac{C}{(s+2)^2}+\frac{Ds+E}{(s^2+4s+9)}$$
and after doing some expanding and simultaneous equations, I got:
$A=\frac{5}{6}$; $B=\frac{64}{5}$; $C=\frac{-67}{5}$; $D=\frac{-289}{30}$; $C=\frac{-347}{10}$
So now, I have
$$\mathcal{L}[f(t)]=\frac{5}{6}\frac{1}{(s+1)}+\frac{64}{5}\frac{1}{(s+2)}-\frac{67}{5}\frac{1}{(s+2)^2}-\frac{289}{30}\frac{s}{(s^2+4s+9)}-\frac{347}{10}\frac{1}{(s^2+4s+9)}$$
from here I use the table of inverses in the textbook to get:
$F(s) = \frac{5}{6}e^{-t}+\frac{64}{5}e^{-2t}-\frac{67}{5}te^{-2t}+\dots$
and then I'm stuck on that last polynomial... I have looked and looked but found nothing... and I'm not even sure I did the first part correctly.
Please help!
|
Hint. Note that
$$s^2+4s+9=(s+2)^2+(\sqrt{5})^2$$
and take a look at "exponentially decaying
sine/cosine wave" in this Table.
P.S. Check your partial fraction decomposition. It should be
$$\frac{1/6}{(s+1)}-\frac{3/5}{(s+2)^2}-\frac{s/6-1/10}{(s^2+4s+9)}
=\frac{1/6}{(s+1)}-\frac{3/5}{(s+2)^2}-\frac{(s+2)/6-13/30}{(s+2)^2+(\sqrt{5})^2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2604769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Integral of a Polynomial in Square Root
I need to solve the indefinite integral:
$$\int \frac{\sqrt{x^4+x^{-4}+2}}{x^3} dx,$$
but I can not find any technique which could solve it.
Can you help me please?
|
It is :
$$(x^2+x^{-2})^2=x^4+x^{-4}+2$$
which means that your initial integral can be substituted to :
$$\int \frac{\sqrt{x^4+x^{-4}+2}}{x^3} dx = \int\frac{\sqrt{(x^2+x^{-2})^2}}{x^3}dx = \int \frac{x^2 + x^{-2}}{x^3}dx$$
since $x^2 + x^{-2} \geq 0 \space \forall x \in \mathbb R$.
So :
$$\int \frac{x^2 + x^{-2}}{x^3}dx= \int \bigg(\frac{1}{x} + \frac{1}{x^5}\bigg)dx=\int x^{-1}dx + \int x^{-5}dx= \ln|x| +-4x^{-4}+ C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2609487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Find a number $M$, such that $|x^3-4x^2+x+1| < M$ for all $1The question states that I cannot use calculus, so finding max/min of the function is not an accepted solution.
My attempt at solving the problem:
$|(x^3-4x^2) + (x+1)|$ $\le$ $|x^3-4x^2| + |x+1|$
$= |x^2||x-4| + |x+1|$
$ = x^2|x-4| + x+1$ (because 1 < x < 3 so x+1 will always be positive)
Plugging in 1 for x:
$ = 1(3) + 2$
$ = 5$
Plugging in 3 for x:
$ = 9(1) + 4 $
$ = 13 $
Therefore the value of M is 13 as there is no value in the interval from 1 < x <3 which is greater than the value outputted by 3.
Is this the right approach to the solution? When graphing the function I can see that 13 works as a value of M but there are other values that also work which are closer to the maximum of the function in that interval.
|
For all $1<x<3$ we have $$x^3-4x^2+x+1=x^3-4x^2+3x+1-2x=-x(3-x)(x-1)-2x+1<0.$$
Thus, we need to find the maximal value of $f(x)=-x^3+4x^2-x-1$ on $(1,3).$
We have $$f'(x)=-3x^2+8x-1,$$ which gives $$x_{max}=\frac{4+\sqrt{13}}{3}$$ and all $$M>f\left(\frac{4+\sqrt{13}}{3}\right)=\frac{13(5+2\sqrt{13})}{27}$$ is valid.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2610859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
An alternative way to find the sum of this series? $\displaystyle \frac{4}{20}$+$\displaystyle \frac{4.7}{20.30}$+$\displaystyle \frac{4.7.10}{20.30.40}$+...
Now I have tried to solve this in a usual way, first find the nth term $t_n$.
$t_n$= $\displaystyle \frac{1}{10}$($\displaystyle \frac{1+3}{2}$) + $\displaystyle \frac{1}{10^2}$($\displaystyle \frac{1+3}{2}$)($\displaystyle \frac{1+6}{3}$) + ...+ $\displaystyle \frac{1}{10^n}$($\displaystyle \frac{1+3}{2}$)($\displaystyle \frac{1+6}{3}$)...($\displaystyle \frac{1+3n}{n+1}$)
=$\displaystyle \frac{1}{10^n}\prod$(1+$\displaystyle \frac{2r}{r+1}$) , $r=1,2,..,n$
=$\displaystyle \prod$($\displaystyle \frac{3}{10}-\displaystyle \frac{1}{5(r+1)}$)
thus, $t_n=$ (x-$\displaystyle \frac{a}{2}$)(x-$\displaystyle \frac{a}{3}$)...(x-$\displaystyle \frac{a}{n+1}$), x=$\displaystyle \frac{3}{10}$, a=$\displaystyle \frac{1}{5}$
Now to calculate $S_n$, I have to find the product $t_n$, and then take sum over it. But this seems to be a very tedious job. Is there any elegant method(may be using the expansions of any analytic functions) to do this?
|
Through Euler's Beta function and the reflection formula for the $\Gamma$ function:
$$\sum_{n\geq 1}\frac{\prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=\sum_{n\geq 1}\frac{3^n\Gamma\left(n+\frac{4}{3}\right)}{10^n \Gamma(n+2)\Gamma\left(\frac{4}{3}\right)}=\frac{3\sqrt{3}}{2\pi}\sum_{n\geq 1}\left(\tfrac{3}{10}\right)^n B\left(\tfrac{2}{3},n+\tfrac{4}{3}\right) $$
where
$$ \sum_{n\geq 1}\left(\tfrac{3}{10}\right)^n B\left(\tfrac{2}{3},n+\tfrac{4}{3}\right) = \int_{0}^{1}\sum_{n\geq 1}\left(\tfrac{3}{10}\right)^n(1-x)^{-1/3}x^{n+1/3}\,dx=\int_{0}^{1}\frac{3x^{4/3}\,dx}{(1-x)^{1/3}(10-3x)} $$
and the last integral can be computed in a explicit way with a bit of patience. The final outcome is
$$\sum_{n\geq 1}\frac{\prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=\color{red}{10\sqrt[3]{\frac{10}{7}}-11} $$
which can also be proved by invoking Lagrange's inversion theorem or the extended binomial theorem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2611305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things.
|
For $x:0\neq|x|<1$ we have $x^3<x^2$. Therefore if we have 3 numbers $x_1^2+x_2^2+x_3^2=1$ and $x_1x_2x_3\neq0$ we have that $x_i\neq0$ $\forall i$. Therefore for every $i$ we have the strict inequallity $x_i^3<x_i^2$ therefore $x_1^3+x_2^3+x_3^3<1$. Therefore one of the $x_i$ is $0$ .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2612380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 3
}
|
Three touching circles inscribed in a rectangle 'Three equal circles, with radius r, are inscribed in a rectangle in a way that all three of them touch each other only once. Find the area of rectangle.' The thing i can't understand is that the arrangement of circles seem limitless to me, do they need to be touching the sides of the rectangle or not ?
|
A little thought will reveal that, if the three-circle grouping is to be tangent to the sides of a rectangle, then (at least) one of those circles must be tangent to two adjacent sides of the rectangle. This gives rise to the following diagram:
Let $\bigcirc O$, of radius $r$, be tangent to two sides of the rectangle. Place $\bigcirc A$ and $\bigcirc B$, such that $\overline{OA}$ makes an angle of $\alpha$ with a "horizontal" radius of $\bigcirc O$, and such that $\overline{OB}$ makes an angle of $\beta$ with a "vertical" radius. Of course, $\alpha + \beta = 30^\circ$ (and both $\alpha$ and $\beta$ are non-negative).
We see that the sides of the rectangle must have lengths $2r+2r\cos\alpha$ and $2r+2r\cos\beta$, so that the area is
$$\begin{align}
2r(1+\cos\alpha)\cdot 2r(1+\cos\beta) &= 4r^2(1+\cos\alpha)(1+\cos\beta) \tag{1a}\\
&= 16 r^2 \cos^2\frac{\alpha}{2}\cos^2\frac{\beta}{2} \tag{1b}\\
&= 16 r^2 \left( \cos\frac{\alpha}{2} \cos\frac{\beta}{2}\right)^2 \tag{1c}
\end{align}$$
But then the final factor of $(1c)$ is the square of ...
$$\frac{1}{2}\left( \cos\frac{\alpha+\beta}{2} + \cos\frac{\alpha-\beta}{2} \right) = \frac{1}{2}\left( \cos 15^\circ + \cos\frac{\alpha-\beta}{2} \right) \tag{2}$$
... so we have:
$$\text{area} = 4 r^2 \;\left(\; \cos 15^\circ + \cos\frac{\alpha-\beta}{2} \;\right)^2 \tag{3}$$
This value depends upon $\alpha$ and $\beta$, so there is no unique area. However, we can see that the area is minimized when $|\alpha-\beta|$ is maximized, and vice-versa. So,
$$\begin{align}
|\alpha - \beta| = 30^\circ &\;\to\; \,\text{min area} = 16 r^2 \;\cos^2 15^\circ = 8 r^2 \left( 1 + \cos 30^\circ \right) = 4 r^2 \left(2 +\sqrt{3}\right) \tag{4a} \\[4pt]
|\alpha - \beta| = \phantom{3}0^\circ &\;\to\; \text{max area} = \phantom{1}4 r^2 \;\left(\; 1 + \cos 15^\circ \;\right)^2 = \text{exercise for reader}\tag{4b}
\end{align}$$
Note: The minimum area occurs when either $\alpha$ or $\beta$ vanishes; that is, when two of the circles are tangent to the same side of the rectangle (equivalently, when two circles are each tangent to two adjacent edges). This is the configuration shown in @RossMillikan's answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2612707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
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.