Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Euclidean Geometry challenge.
Can someone help me on this one?
I have found that $\frac{1}{(x+1)^2}+1=\frac{1}{x^2}$, but I can't solve the fourth degree equation that comes with it. There must be a easier way!
|
here is partial answer:
we need to solve $(1 + \frac{1}{x})^2 = 1 + (1+x)^2$ on clearing you get the quartic $$ 0= f(x) = x^4 + 2x^3 + x^2 -2x - 1 = g(x) + h(x), \text{ where } g(x) = x^4 + x^2 - 1, h(x) = 2x^3 - 2x.$$
we are only interested in the solution $0 < x < 1.$ we can find a lower bound for $x$ this way. $g(x) = 0$ at $x^2 = \dfrac{-1+ \sqrt 5}{2}$ so the positive zero of $g$ is $\sqrt{\dfrac{-1+ \sqrt 5}{2}}$ and $h(x) < 0 \text{ on } 0< x < 1.$ by inspecting the sign of $g$ and $h$ we can conclude that the only zero of $f = g+h$ is $$\sqrt{\dfrac{-1+ \sqrt 5}{2}} < x < 1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1125587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $\sqrt x$ is Lipschitz on $[1, \infty)$ Prove that $\sqrt x$ is Lipschitz on $[1, \infty)$
I want to show that $|f(x) - f(y)| \leq L |x - y|$
So $|\sqrt x - \sqrt y| = \frac{|x - y|}{\sqrt x + \sqrt y} \leq \frac{1}{2}|x - y|$.
I can see that to get $|\sqrt x - \sqrt y| = \frac{|x - y|}{\sqrt x + \sqrt y}$ I just multiply the first expression by its conjugate. But I don't really understand how $\frac{|x - y|}{\sqrt x + \sqrt y} \leq \frac{1}{2}|x - y|$ How does $\frac{1}{\sqrt x + \sqrt y} \leq \frac{1}{2}$?
|
We know $x,y \in [1, \infty)$ thus both $x,y \geq 1$.
Now we can see $$\sqrt{x} + \sqrt{y} \geq 1 + 1 =2$$
and so we know $$\frac{1}{\sqrt{x} + \sqrt{y}} \leq \frac{1}{2}$$
because both values are positive.
Hence all that remains for us is to multiply through by $|x-y|$ and we get $$\frac{|x-y|}{\sqrt{x} + \sqrt{y}} \leq \frac{|x-y|}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1125681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Defining the equation of an ellipse in the complex plane
Usually the equation for an ellipse in the complex plane is defined as $\lvert z-a\rvert + \lvert z-b\rvert = c$ where $c>\lvert a-b\rvert$. If we start with a real ellipse, can we define it in the manner below?
For $x,y,h,k,a,b\in\mathbb{R}$ such that $a,b\neq 0$, we define a real ellipse as
$$
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1.
$$
Let $z = \frac{x}{a} + i\frac{y}{b}$ and $z_0 = \frac{h}{a} + i\frac{k}{b}$. If we expand the equation for an ellipse, we have
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{h^2}{a^2} + \frac{k^2}{b^2} -
\frac{2xh}{a^2} - \frac{2yk}{b^2} = 1.
$$
Notice that $\lvert z\rvert^2 = \frac{x^2}{a^2} + \frac{y^2}{b^2}$ and $\lvert z_0\rvert^2 = \frac{h^2}{a^2} + \frac{k^2}{b^2}$. Now, let's write the ellipse as
$$
\lvert z\rvert^2 + \lvert z_0\rvert^2 - \frac{2xh}{a^2} - \frac{2yk}{b^2} +
\frac{yh}{ab}i - \frac{yh}{ab}i + \frac{xk}{ab}i - \frac{xk}{ab}i =
\lvert z\rvert^2 + \lvert z_0\rvert^2 - \bar{z}z_0 - z\bar{z}_0 = 1.
$$
Thus, the equation of an ellipse in the complex plane is
$$
(z - z_0)(\bar{z} - \bar{z}_0) = \lvert z - z_0\rvert^2 = 1\Rightarrow
\lvert z - z_0\rvert = 1
$$
where $z$ and $z_0$ are defined above.
|
Your answer is correct, in the terms you are using. But note that in the standard equation
$$|z-a|+|z-b|=c\ ,$$
it is implicitly assumed that $z=x+iy$. Since your $z$ has a different form from this, you have obtained a different answer.
If you wish to plot your equation in the complex plane, you will have to rescale the axes and so the graph of $|z-z_0|=1$, which is "usually" a circle, will be "squeezed" so that it becomes an ellipse.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1126121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
}
|
Prove that a recurrence relation (containing two recurrences) equals a given closed-form formula. Prove that $a_n = 3a_{n-1} - 2a_{n-2} = 2^n + 1$ , for all $n \in \mathbb{N}$ , and $a_1 = 3$ , $a_2 = 5$ , and $n \geq 3$
Basis:
$a_1 = 2^1 + 1 = 2 + 1 = 3$ $\checkmark$
$a_2 = 2^2 + 1 = 4 + 1 = 5$ $\checkmark$
Inductive Hypothesis:
$a_k = 2^k + 1$ , where $n = k$
$a_k = 3a_{k-1} - 2a_{k-2}$
Inductive Step:
$$\begin{align}
\ a_{k+1} & = 3a_k - 2a_{k-1} \\
& = 3(2^k + 1) - 2(2^{k-1} + 1) \\
& = \color{red}{6^k} + 3 - \color{red}{4^{k-1}} - 2 \\
& = \color{red}{6^k} - \color{red}{4^{k-1}} + 1
\end{align}$$
Now, the $ + 1$ looks very promising, but $6^k - 4^{k-1}$ makes me sick. Anyone have some good hints? That second recurrence ( $2a_{n-2}$ ) seems to be stumping me.
|
Inductive Hypothesis:
$a_k = 2^k + 1$ , where $n=k$ , to show:
$$\begin{align} \ a_{k+1} & = 2^{k+1} + 1 \\ \end{align}$$
Inductive Step:
$$\begin{align}
\ a_{k+1} & = 3a_{(k+1)-1} - 2a_{(k+1)-2}\\
& = 3a_k - 2a_{k-1} \\
& = 3(2^k + 1) - 2(2^{k-1} + 1) \\
& = 3(2^k) + 3 - 2(2^{k-1}) - 2 \\
& = (2^k + 2^k + 2^k) - (2^{k-1} + 2^{k-1}) + 3 - 2 \\
& = (2^{k+1} + 2^k) -(2^k) + 1 \\
& = 2^{k+1} + 2^k - 2^k + 1 \\
& = 2^{k+1} + 1 \\
\end{align}$$
Gee, I might be having more fun formatting the MathJax than I am solving the math. Oh well, thanks to all!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1127212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
$f(xy)=\frac{f(x)+f(y)}{x+y}$ Prove that $f$ is identically equal to $0$ For all $x,y\in\mathbb{R}$. also $f : \mathbb{R} → \mathbb{R}$ and $x+y\not=0$.
My attempt:
I restated it as
$a[x^2 y^2 (\frac{x}{y}+\frac{y}{x}-\frac{1}{y^2}-\frac{1}{x^2})] +
b[xy(x+y-\frac{1}{y}-\frac{1}{x})] + c [x+y-2]=0$
because of $f(xy) - \frac{f(x)+f(y)}{x+y}=0$
(we know that $f$ is identically equal to $0$)
and later tried to prove that
$\frac{x}{y}+\frac{y}{x}-\frac{1}{y^2}-\frac{1}{x^2}$
$x+y-\frac{1}{y}-\frac{1}{x}$
$x+y-2$
are all equal to 1. I eventually got to that
$x^2 -2x -1,5 =0$ And i don't like this
Later I tried ;
$\frac{a(x^2+y^2)+b(x+y)+2c}{x+y} = \frac{f(x+y)}{x+y}$
because if
$(x+y)^2=x^2+y^2$
$xy=0$ but that implies that $c$ is equal to $0$ too. Both of my trials are propably wrong.
|
Setting $(x, y) = (1, 0)$ gives $f(1) = f(1) + f(0)$; that is, $f(0) = 0$. For $x\not =0$, setting $y = 0$ in the given equation shows that $(f(x) + f(0))/x = f(x)/x$ vanishes; that is, $f(x) = 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1128061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
$x^2+y^2+9=3(x+y)+xy$ Find all pairs of real $x,y$ that meet this equation $\frac{(x-y)^2}{(y-3)(3-x)} = 1$
That was my attempt, I can't think of anything else here. I'd prefer a hint
|
Shift variables to eliminate the linear terms. $x=u+a$ and $y=v+b$, with $a,b$ yet-to-be-determined constants.
$$
\begin{align}
x^2+y^2+9&=3(x+y)+xy\\
(u+a)^2+(v+b)^2+9&=3(u+a+v+b)+(u+a)(v+b)\\
u^2+2au+a^2+v^2+2bv+b^2+9&=3u+3a+3v+3b+uv+av+bu+ab
\end{align}
$$
We would like to be able to cancel the the terms that are linear in $u$ and $v$, so we want:
$$
\left\{\begin{aligned}
2a&=3+b\\
2b&=3+a
\end{aligned}\right.
$$
This system of linear equations in $a,b$ has solution $a=b=3$. So now we have
$$
\begin{align}
u^2+6u+9+v^2+6v+9+9&=3u+9+3v+9+uv+3v+3u+9\\
u^2+v^2&=uv\quad\text{(next, complete the square)}\\
4u^2-4uv+4v^2&=0\quad\text{(multiplying by 4 avoids fractions)}\\
(2u-v)^2+3v^2=0
\end{align}
$$
Since we have two squares adding to $0$, the only solution to this is $u=v=0$, which leads to $x=y=3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1129124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Solving the inequality $(x^2+3)/x\le 4$ This is the inequality
$$\left(\frac {x^2 + 3}{x}\right) \le 4 $$
This is how I solve it
The $x$ in the left side is canceled and $4x$ is subtracted from both sides.
$$\not{x} \left (\frac {x^2+3} {\not{x}}\right) \le 4x $$
$$ x^2+3 - 4x \le 4x - 4x $$
$$x^2 -4x + 3 \le 0 $$
Then the trinomial is factorized
$$(x-3)(x-1) \le 0 $$
Therefore
$$1\lt x \lt 3 $$
Can someone tell me if I make a mistake, or if the process is good enough. Will you have done something different?
|
To deal away with fractions, we can multiply both sides of $\left(\frac {x^2 + 3}{x}\right) \le 4$ with $x^2>0$ (here, $x\neq 0$):
$$
(x^2+3)x\leq 4x^2\iff x^3+3x-4x^2\leq 0\iff x(x-1)(x-3)\leq 0.
$$
We want the product of 3 factors above to be nonpositive so there is an odd number of nonnegative factors among $x-3<x-1<x$. This can only happen when
$$
x-3<x-1<x<0 \quad\text{or}\quad x-3\leq 0\leq x-1<x.
$$
Thus, the answers are $x<0$ or $1\leq x\leq3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1130716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
}
|
find the length of rectangle based on area of frame A rectangular picture is $3$ cm longer than its width, $(x+3)$. A frame $1$ cm wide is placed around the picture. The area of the frame alone is $42 \text{cm}^2$. Find the length of the picture.
I have tries:
$(x+6)(x+3) = 42 \\
x^2+3x+6x+12 = 42 \\
x^2+9x+12x-42 = 0 \\
\\
x^2 + 21 -42$
|
Area $A$ of the $1$-cm-wide frame in cm$^2$ is $2\times $ height $+ 2\times $ width $+ 4$ (for the corners).
$A = 2 (x+3) + 2(x+6)+4 = 4x+22 = 42$
$\implies x=5$
The length is $(x+6) = 11$ cm
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1131118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
The inequality. Regional olympiad 2015 Let $a, b, c$ - the positive real numbers, and $ab+bc+ca=1$
Prove that $\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}+\sqrt{c+\frac{1}{c}} \geqslant 2(\sqrt{a}+\sqrt{b}+\sqrt{c})$
Probably, we should use these facts:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{abc}$
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2$
But I don't know how to use them. Please, help.
|
We have
$$
\sum_{cyc} \sqrt{a+\frac{1}{a}}=\sum_{cyc} \sqrt{a+\frac{ab+bc+ca}{a}}=\sum_{cyc} \sqrt{a+\frac{bc}{a}+b+c}\ge\sum_{cyc} \sqrt{2\sqrt{bc}+b+c}=\sum_{cyc} \sqrt{\left(\sqrt b+\sqrt c\right)^2}=\sum_{cyc} \left(\sqrt b+\sqrt c\right)=2\left(\sqrt a+\sqrt b+\sqrt c\right)
$$
as desired.
Note: The inequality is due to the well known inequality between the arithmetic mean and the geometric mean, which states that:
$$
\frac{a_1+…+a_n}{n}\ge\left(a_1\cdots a_n\right)^\frac{1}{n}
$$
Taking $n=2$, $a_1=a$ and $a_2=\frac{bc}{a}$ this yields:
$$
\frac{a+\frac{bc}{a}}{2}\ge\left(a\cdot\frac{bc}{a}\right)^\frac12=\sqrt{bc}\iff a+\frac{bc}{a}\ge2\sqrt{bc}
$$
This is, as shown above, enough to prove your inequality.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1131184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
How many three digit numbers are not divisible by 3, 5 or 11? How many three digit numbers are not divisible by 3, 5, or 11?
How can I solve this?
Should I look to the divisibility rule or should I use, for instance,
$$
\frac{999-102}{3}+1
$$
|
Assuming that you mean by either $3$ or $5$ or $11$, use inclusion/exclusion principle:
*
*Amount of numbers with at most $3$ digits that are not divisible by $3$ or $5$ or $11$:
$999-\lfloor\frac{999}{3}\rfloor-\lfloor\frac{999}{5}\rfloor-\lfloor\frac{999}{11}\rfloor+\lfloor\frac{999}{3\times5}\rfloor+\lfloor\frac{999}{3\times11}\rfloor+\lfloor\frac{999}{5\times11}\rfloor-\lfloor\frac{999}{3\times5\times11}\rfloor=485$
*Amount of numbers with at most $2$ digits that are not divisible by $3$ or $5$ or $11$:
$99-\lfloor\frac{99}{3}\rfloor-\lfloor\frac{99}{5}\rfloor-\lfloor\frac{99}{11}\rfloor+\lfloor\frac{99}{3\times5}\rfloor+\lfloor\frac{99}{3\times11}\rfloor+\lfloor\frac{99}{5\times11}\rfloor-\lfloor\frac{99}{3\times5\times11}\rfloor=48$
*Amount of numbers with exactly $3$ digits that are not divisible by $3$ or $5$ or $11$:
$485-48=437$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1133197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
}
|
exercise: ordinary differential equations I am struggling with an exercise. Can you please give me a hint?
Exercise:
Show that the solution curves of the differential equation:
$\frac{dy}{dx}=-\frac{y(2x^3-y^3)}{x(2y^3-x^3)}$, are of the form $x^3+y^3=3Cxy$.
I tried the substitution $u=y/x \rightarrow y=xu, \frac{dy}{dx}=u+x\frac{du}{dx}$.
Hence I get:
$u+x\frac{du}{dx}=-\frac{u(2-u^3)}{2u^3-1}$
This gives:
$2u^4-u+(2u^3-1)x\frac{du}{dx}=-2u+u^4$
$(2u^3-1)x\frac{du}{dx}=-(u^4+u)$
$\frac{(2u^3-1)}{(u^4+u)}\frac{du}{dx}=-\frac{1}{x}$
So I can atleast reduce the problem to a seperable differential equation, but I am not able to integrate the left side up. Do you have any tips?
|
Use partial fractions
$$\begin{align}
\frac{2u^{3} - 1}{u^{4} + u} &= \frac{2u^{3} - 1}{u(u^{3} + 1)} \\
&= \frac{A}{u} + \frac{Bu^{2} + Cu + D}{u^{3} + 1} \\
\implies 2u^{3} - 1 &= (A + B)u^{3} + Cu^{2} + Du + A \\
\end{align}$$
Equating coefficients
$$\begin{align}
A &= -1 \\
B &= 3 \\
\end{align}$$
Hence
$$\begin{align}
\frac{2u^{3} - 1}{u^{4} + u} &= \frac{3u^{2}}{u^{3} + 1} - \frac{1}{u}
\end{align}$$
Integrating
$$\begin{align}
\int \bigg(\frac{3u^{2}}{u^{3} + 1} - \frac{1}{u} \bigg)du &= - \int \frac{1}{x} dx \\
\implies \ln(u^{3} + 1) - \ln(u) &= -\ln(x) + K_1 \\
\implies \ln \bigg( \frac{u^{3} + 1}{u} \bigg) &= - \ln(x) + K_1 \\
\implies \frac{u^{3} + 1}{u} &= \frac{K_2}{x} \\
\end{align}$$
With
$$\begin{align}
u &= \frac{y}{x} \\
\implies \frac{y^{2}}{x^{2}} + \frac{x}{y} &= \frac{K_2}{x} \\
\implies y^{3} + x^{3} &= K_2xy
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1134973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
Proving that there are at least $n$ primes between $n$ and $n^2$ for $n \ge 6$ I was thinking about Paul Erdos's proof for Bertrand's Postulate and I wondered if the basic argument could be used to show that there are more than $n$ primes between $n$ and $n^2$.
Is this approach valid? Is there a better approach?
Here's the argument:
*
*Let $v_p(n)$ be the maximum power of $p$ that divides $n$.
*Using Legendre's Theorem, I calculate that $v_p\left(\dfrac{(n^2)!}{(n!)^n}\right) \le (n-1)n^2$.
*
*Let $n = qp^i + r$ where $0 \le r < p^i$
*For $r < \dfrac{p^i}{n}$, $\left(\left\lfloor\dfrac{n^2}{p^i}\right\rfloor - n\cdot\left\lfloor\dfrac{n}{p^i}\right\rfloor\right) = 0$
*For $r \ge \dfrac{p^i}{n}$, $\left(\left\lfloor\dfrac{n^2}{p^i}\right\rfloor - n\cdot\left\lfloor\dfrac{n}{p^i}\right\rfloor\right) \le n-1$
*For $p^i > n^2$, $\left(\left\lfloor\dfrac{n^2}{p^i}\right\rfloor - n\cdot\left\lfloor\dfrac{n}{p^i}\right\rfloor\right) = 0$
*Using the multinomial theorem I get: $\dfrac{n^{n^2}}{n^2+1} < \dfrac{(n^2)!}{(n!)^n}$.
*
*$n^{n^2}=(1 + 1 + \cdots + 1)^{n^2}
= \sum\limits_{k_1+k_2+\cdots+k_n=n^2} {n^2 \choose k_1, k_2, \ldots, k_n}
\prod\limits_{1\le t\le n^2}x_{t}^{k_{t}}\,,$ where ${n^2 \choose k_1, k_2, \ldots, k_n}
= \frac{(n^2)!}{k_1!\, k_2! \cdots k_n!}$
*$n^{n^2} < (n^2+1)\left(\dfrac{(n^2)!}{(n!)^n}\right)$
*$\dfrac{n^{n^2}}{n^2+1} < \dfrac{(n^2)!}{(n!)^n}$
*$\dfrac{(n^2)!}{(n!)^n} < [(n-1)(n^2)]^n\prod\limits_{n < p \le n^2} n$
*$\dfrac{n^{n^2}}{n^2+1} < [(n-1)(n^2)]^n\prod\limits_{n < p \le n^2} n$
*For $n \ge 1, \dfrac{n^{n^2}}{n^{4n}} < \dfrac{n^{n^2}}{(n^2+1)[(n-1)(n^2)]^n} < \prod\limits_{n < p \le n^2} p$
*For $n \ge 6$, $n^2 - 4n = n(n-2) \ge 2n$
*$(2\log n)(\pi(n^2) - \pi(n)) > \sum\limits_{n < p \le n^2} \log p > (n^2-4n)\log n > (2n)\log n$
*So $\pi(n^2) - \pi(n) > n$.
Thanks,
-Larry
Edit: I believe that step 4 is not correct. If I can save the argument, I will update it. Otherwise, please check out the link in the answer.
|
We have the following result1:
(Chebyshev, 1850) For any $X>30$, we have: $$\pi(X)\cdot\frac{\log
X}{X} \in (A,B),$$ where
$A=\log\left(\frac{2^{\frac{1}{2}}3^{\frac{1}{3}}4^{\frac{1}{4}}}{30^{\frac{1}{30}}}\right)\approx
0.946$ and $B=\frac{6}{5}A\approx 1.135$.
It follows that for any $X>30$ we have:
$$ \pi(X^2)-\pi(X)\geq A\frac{X^2}{2\log X}-B\frac{X}{\log X}\geq\left(\frac{A}{2}-\frac{B}{30}\right)\frac{X^2}{\log X}\geq\color{red}{\frac{2}{5}\frac{X^2}{\log X}} $$
that is way greater than $X$. So we just need to check that the our claim holds in the range $[6,30]$.
1) it is a weaker version of the PNT, but not so weaker. Its proof only relies on the properties of the central binomial coefficients, and it is rather short and ingenious.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1140314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
}
|
Is $4 \times 6$ defined as $4 + 4 + 4 + 4 + 4 + 4$ or $6 + 6 + 6 + 6$? There are long debates among Indonesian netizens about this http://www.globalindonesianvoices.com/15785/is-4x6-the-same-as-6x4-this-primary-school-math-made-controversy-in-social-media/
|
Let $A \times B$ mean the addition of $A$ repeated $B$ times.
Since multiplication is commutative in $\mathbb{R}$, that tells us that $A \times B = B \times A$. In other words, the addition of $A$ repeated $B$ times is the same as the addition of $B$ repeated $A$ times.
Thus, $4 \times 6 = 4+4+4+4+4+4 = 6 \times 4 = 6+6+6+6$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1140773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 4
}
|
finding the minimum value of $\frac{x^4+x^2+1}{x^2+x+1}$ given $f(x)=\frac{x^4+x^2+1}{x^2+x+1}$.
Need to find the min value of $f(x)$.
I know it can be easily done by polynomial division but my question is if there's another way
(more elegant maybe) to find the min?
About my way: $f(x)=\frac{x^4+x^2+1}{x^2+x+1}=x^2-x+1$. (long division)
$x_{min}=\frac{-b}{2a}=\frac{1}{2}$. (when $ax^2+bx+c=0$)
So $f(0.5)=0.5^2-0.5+1=\frac{3}{4}$
Thanks.
|
$$f(x)=\frac{x^4+x^2+1}{x^2+x+1}=x^2-x+1$$
$$f'(x) = 2x-1=0,x=\frac12$$
$$f''(\frac12)=2\gt0\text{ hence this is a local minimum by the second derivative test}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1146050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
How to factorize $n^5+n+1$? How to factorize $n^5+n+1$ ?
I think I should break $a^5$ and use a factorization formula. but how is it done?
|
By trial and error, knowing that there must be a simple solution.
*
*Factoring $n^5+1$ leads nowhere.
$$n^5+n+1=(n+1)(n^4-n^3+n^2-n+1)+n.$$
*Factoring $n^5+n$ leads nowhere.
$$n^5+n+1=n(n+1)(n^3-n^2+n-1)+1.$$
*Factoring $n^5+n^2$ with an artifice fails, but shows some hope
$$n^5+n^2-n^2+n+1=n^2(n+1)(n^2-n+1)-n^2+n+1.$$
*Factoring $n^5-n^2$ instead works !
$$n^5-n^2+n^2+n+1=n^2(n-1)(n^2+n+1)+n^2+n+1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1146569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
}
|
Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}<2-\frac{1}{n^2} $
|
Here is one way...
$$\frac1{k^2} < \frac1{k(k-1)}= \frac1{k-1} - \frac1{k}$$
Now telescope to get
$$1+\sum_{k=2}^n\frac1{k^2} < 1+1-\frac1{n}< 2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1150388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
}
|
Argument of $3+4i$ for $z = \sqrt{3 + 4i}$, I am trying to put this in Standard form, where z is complex. I let $w = 3+4i$ and find that the modulus, $|w|=r$, is 5. I am having trouble solving for arg(w).
I find that $\tan^{-1}{\theta} = \frac{4}{3}$. However, this is not an angle well known. How do I find it?
|
Note, we have $|w| = 5$. Let $\theta \in Arg(w)$ and then from your corresponding diagram of the triangle form my $w$, $\cos(\theta) = \frac{3}{5}$ and $\sin(\theta) = \frac{4}{5}$. Therefore, from $\sqrt{z} = \sqrt{z}\left( \cos(\frac{\theta}{2}) + i\sin(\frac{\theta}{2})\right )$, we essentially arrive at our answer.
Recall the half-angle identities of both cosine and sine. i.e.,
$$\cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1}{2}(1 + \cos(\theta))}$$
$$\sin \left (\frac{\theta}{2} \right) = \sqrt{\frac{1}{2}(1 - \cos(\theta))}$$
From plugging in the corresponding values into the above equations, we find that $\cos(\frac{\theta}{2}) = \frac{2}{\sqrt{5}}$ and $\sin(\frac{\theta}{2}) = \frac{1}{\sqrt{5}}$.
Then we obtain $\boxed{\sqrt{3 + 4i} = \pm (2 + i)}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1151831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
}
|
how to evaulate: $\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)} $ How do I evaluate: $\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)} $ ?
according to Taylor's series, I did like this:
$$\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)}=\lim \limits_{x \to 0} \frac{x^5 - \frac{x^{10}}{2}+ O_3(x)}{(x^3+ \frac{x^6}{2!} +O_4(x))(x^2- \frac{x^6}{3!} +\frac{x^{10}}{5!} +O_4(x))} $$
but how do I continue from here?
|
From Taylor's expansions at $x=0$, $\log(1+x^5) \approx x^5$, $e^{x^3}-1 \approx x^3$ and $\sin x^2 \approx x^2$, as $x \rightarrow 0$. Thus,
$$\lim \limits_{x \rightarrow 0} \frac{\log(1+x^5)}{(e^{x^3}-1)(\sin x^2 )} = \lim \limits_{x \rightarrow 0} \frac{x^5}{x^{3} x^{2}} =1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1153010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Evaluate $\int{\frac{\sqrt[3]{x+8}}{x}}dx$ So I've tried solving the equation below by using $u=x+8$, and I get $\int{\frac{\sqrt[3]{u}}{u-8}}du$ which doesn't seem to lead anywhere, I've also tried taking the $ln$ top and bottom, but I don't know how to proceed. Any hints?
$$\int{\frac{\sqrt[3]{x+8}}{x}}dx$$
Update: Using partial fraction decomposition, $A=2, B=-2, C= -8$ (thank you to all the helpful posts) and after some tedious calculations, the answer I obtained was: $3{\sqrt[3]{x+8}}+24(2ln|\sqrt[3]{x+8}-2|-2ln|\sqrt[3]{x+8}+2|+{\frac{4}{\sqrt[3]{x+8}+2}})+C$
|
you can start with
$$\begin{align}
u^3=x+8&\Rightarrow 3u^2du=dx\\
x=u^3-8\\
\int\frac{\sqrt[3]{x+8}}{x}dx&=\int\frac{\sqrt[3]{u^3}}{u^3-8}(3u^2du)\\
&=\int\frac{3u^3}{u^3-8}\\
&=\int\frac{3(u^3-8)+24}{u^3-8}du\\
&=\int3+\frac{24}{u^3-8}du
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1153991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
interpreting partitions of integers(generating functions) the question is: In $ f(x) = \left (\frac{1}{1-x} \right )\left (\frac{1}{1-x^2} \right )\left (\frac{1}{1-x^3} \right )$the coefficient of $x^6$ is 7. Interpret this result in terms of partitions of 6.
I checked the answer and it says the number of partitions of 6 into 1's, 2's and 3's is 7. But I'm having hard time understanding it. Can anyone explain this? An easy explanation would be very much appreciated..Thankyou
|
Writing each of the terms as power series, we have
$$\eqalign{f(x)
&=(1+x+x^2+\cdots)(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots)\cr
&=(1+x^1+x^{1+1}+\cdots)(1+x^2+x^{2+2}+\cdots)(1+x^3+x^{3+3}+\cdots)\ .\cr}$$
Now if you look carefully you will see that you can multiply terms, one from each bracket, to get an $x^6$ term in the following ways:
$$x^{3+3}\,,\ x^{1+2+3}\,,\ x^{1+1+1+3}\,,\ x^{2+2+2}\,,\ x^{1+1+2+2}\,,\ x^{1+1+1+1+2}\,,\ x^{1+1+1+1+1+1}\,.$$
These seven terms combine to give the coefficient of $7$ for $x^6$, and you can see that the exponents are all possible ways of writing $6$ as a sum of $1$s, $2$s and $3$s.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1154096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$
A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\geq3\prod_{cyc}(a+b-c),$$ or
$$\sum_{cyc}(a^3-ab^2-ac^2+2abc)\geq\sum_{cyc}(-3a^3+3a^2b+3a^2c-2abc),$$ but it becomes very ugly.
|
Set $b+c-a=x, a+c-b=y$ and $a+b-c=z$
Due to triangular inequality, $x,y,z$ are positive. Hence we can apply the AM-GM inequality.
We have $x+y=b+c-a+a+c-b=2c \implies c = \frac{x+y}{2}$
Similarly $a = \frac{y+z}{2}$ and $b = \frac{z+x}{2}$
Now we have,
$$\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}=\frac{y+z}{2x}+\frac{z+x}{2y}+\frac{x+y}{2z}$$
$$\Leftrightarrow 2\left ( \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c} \right )=\left ( \frac{x}{y}+\frac{y}{x} \right )+\left ( \frac{y}{z}+\frac{z}{y} \right )+\left ( \frac{z}{x}+\frac{x}{z} \right )\geq 6$$
Hence $\ $ $\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\geq 3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1155955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 1
}
|
Prove that $m+\frac{4}{m^2}\ge3$ How to prove that $m+\frac{4}{m^2}\geq3$ for every $m>0$? I multiplied both sides by $m^2$ and finally got $m^3+4-3m^2\geq0$, yet don't know how to move further. Hint?
|
If $m=1$, then
$$m + \frac{4}{m^2}= 1 + \frac{4}{1^2} = 1 + 4 = 5 \geq 3$$
If $m = 2$, then
$$m + \frac{4}{m^2} = 2 + \frac{4}{2^2} = 2 + 1 = 3 \geq 3$$
Consider now $m\geq 3$. Then, since $\frac{4}{m^2}$ is positive for all $m$, it is obvious that also $m + \frac{4}{m^2}$ is greater than $3$.
We can conclude that
$$m + \frac{4}{m^2}\geq 3 ~ \forall m > 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1157949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
}
|
Proof of a limit of a sequence I want to prove that $$\lim_{n\to\infty} \frac{2n^2+1}{n^2+3n} = 2.$$
Is the following proof valid?
Proof
$\left|\frac{2n^2+1}{n^2+3n} - 2\right|=\left|\frac{1-6n}{n^2+3n}\right| =\frac{6n-1}{n(n+3)} $ (because $n \in \mathbb N^+)$. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$
We have $n \ge 1 \implies 6n -1 > n + 3 \implies \frac{n+3}{n(n+3)} < \frac{6n-1}{n^2+3n}.$
Let $\epsilon > 0$ be given.
Note that $ \frac{6n-1}{n^2+3n}< \epsilon \iff \frac{n+3}{n(n+3)} < \epsilon \iff n > \frac{1}{\epsilon}.$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (**)$
By the Archimedean Property of $\mathbb R$, $\exists N \in \mathbb N^+$ such that $N > \frac{1}{\epsilon}.$
If $n \ge N$, then $n > \frac{1}{\epsilon}$, and from $(*)$ and $(**)$ it follows that $\left|\frac{2n^2+1}{n^2+3n} - 2\right| < \epsilon$.
Therefore $\lim_{n\to\infty} \frac{2n^2+1}{n^2+3n} = 2.$
|
Your “if and only if” are incorrect in
$$
\frac{6n-1}{n^2+3n}< \varepsilon \iff \frac{n+3}{n(n+3)} < \varepsilon \iff n > \frac{1}{\varepsilon}
$$
What you can do is
$$
n > \frac{6}{\varepsilon}\implies
\frac{6(n+3)}{n(n+3)}<\frac{6}{\varepsilon}\implies
\frac{6n-1}{n(n+3)}<\frac{6}{\varepsilon}
$$
Alternative proof
Rewrite your fraction as
$$
\frac{2n^2+1}{n^2+3n}=\frac{2n^2+6n-6n+1}{n^2+3n}=
2-\frac{6n-1}{n^2+3n}
$$
Now use partial fractions:
$$
\frac{6n-1}{n^2+3n}=\frac{A}{n}+\frac{B}{n+3}
$$
can be easily solved to give $A=-1/3$ and $B=19/3$. So your expression is
$$
2+\frac{1/3}{n}-\frac{19/3}{n+3}
$$
Now fix $\varepsilon>0$ and find an integer $k$ such that
$$
\frac{1/3}{k}<\frac{\varepsilon}{2},\qquad
\frac{19/3}{k+3}<\frac{\varepsilon}{2}
$$
that is,
$$
k>\max\Bigl(\frac{2}{3\varepsilon},\frac{38}{3\varepsilon}-3\Bigr)
$$
and, for $n\ge k$ you have
$$
\left|\Bigl(2+\frac{1/3}{n}-\frac{19/3}{n+3}\Bigr)-2\right|=
\left|\frac{1/3}{n}-\frac{19/3}{n+3}\right|\le
\left|\frac{1/3}{n}\right|+\left|\frac{19/3}{n+3}\right|<
\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1158558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Example of contour integration Could someone help me evaluate the following integral with contour integration ?
$$\int_{0}^{2\pi}\frac{d\theta}{(a+b\cos\theta)^2}.$$
Constraints are: $a>b>0$.
|
We have:
$$ I = \int_{0}^{\pi/2}\frac{d\theta}{a+b\cos\theta}=\int_{0}^{\pi/2}\frac{d\theta}{a+b\,\frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}}=2\int_{0}^{\pi/4}\frac{d\theta}{a+b\,\frac{1-\tan^2(\theta)}{1+\tan^2(\theta)}}$$
or just:
$$ I = 2\int_{0}^{1}\frac{dt}{(a+b)+(a-b)\,t^2}=\frac{2}{\sqrt{a^2-b^2}}\,\arctan\sqrt{\frac{a-b}{a+b}} $$
that can be proven with many techniques, not only contour integration.
That gives:
$$\int_{0}^{2\pi}\frac{d\theta}{a+b\cos\theta}=2\left(I+\int_{0}^{\pi/2}\frac{d\theta}{a-b\cos\theta}\right) = \color{red}{\frac{2\pi}{\sqrt{a^2-b^2}}}\tag{1}$$
since for any $r>0$, $\arctan r+\arctan\frac{1}{r}=\frac{\pi}{2}$.
Now, what about the modified question? We have:
$$ -\frac{d}{da}\frac{1}{a+b\cos\theta} = \frac{1}{(a+b\cos\theta)^2}, \tag{2}$$
hence, by $(1)$ and $(2)$:
$$ \int_{0}^{2\pi}\frac{d\theta}{(a+b\cos\theta)^2} = -\frac{d}{da}\frac{2\pi}{\sqrt{a^2-b^2}} = \color{red}{\frac{2\pi a}{(a^2-b^2)\sqrt{a^2-b^2}}}.\tag{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1160331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Prove that the equation $x^2+y^2=57$ has no integer solution. I am trying to prove that equation $x^2+y^2=57,\{x,y\}\in\mathbb{Z}$ has no integers solution.
In case the equation is of the form $ax^2+by^2=n,\text{where}~~ a\neq 1 ~~\text{or}~~ b\neq1
,\{a,b,n\}\in\mathbb{Z}$ .
Then applying $ax^2+by^2\pmod a=n \pmod a$. or $~ax^2+by^2\pmod b=n \pmod b$
was useful .But in this case both $a=b=1$ so how to cope with this situation
|
$$x^2+y^2\equiv 57\equiv 0\pmod {3}$$
Now, the only quadratic residues modulo $3$ are $0,1$, and since if $a,b\in\{0,1\}$, then $a+b\in\{0,1,2\}$, we know that both $a=0$ and $b=0$ must hold in order for $a+b\equiv 0\pmod{3}$ to be true, and so $$\begin{cases}x^2\equiv 0\pmod{3}\\ y^2\equiv 0\pmod {3}\end{cases}\iff \begin{cases}x\equiv 0\pmod{3}\\ y\equiv 0\pmod{3}\end{cases}\iff\begin{cases} x^2\equiv 0\pmod{9}\\y^2\equiv 0\pmod{9}\end{cases}$$
But then $x^2+y^2\equiv 0+0\equiv 0\equiv 57\equiv 3\pmod{9}$, impossible.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1160730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Find the inverse function of $y = g(x) = 6 x^3 + 7$: $g^{-1}(y) =?$ The question states,
Find the inverse function of $y = g(x) = 6 x^3 + 7$, $g^{-1}(y) =?$
I have tried setting the equation to $y$ and then solving for $x.$ This resulted in the answer $\dfrac{(x-7)^{1/3}}{6}$. This answer is incorrect. I have also tried $\dfrac{(36(x-7))^{1/3}}{6}$ and just re-writing the equation as $7+6y^3.$ None of these answers are correct either. Please help.
|
You are close. Solving for $x$ in $y=6x^3+7$ goes like this:
$$y=6x^3+7 \implies y-7=6x^3 \implies \frac{y-7}{6} = x^3 \implies \left(\frac{y-7}{6}\right)^{1/3} = x.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1162111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$ Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$
$$\begin{align}\frac{1}{n^4+n^2+1}&
=\frac{1}{n^4+2n^2+1-n^2}\\
&=\frac{1}{(n^2+1)^2-n^2}\\
&=\frac{1}{(n^2+n+1)(n^2-n+1)}\\
&=\frac{1-n}{2(n^2-n+1)}+\frac{1+n}{2(n^2+n+1)}\\
\end{align}$$
For $n={1,2,3}$ it is not giving telescooping series.
$=\frac{1}{3}+\frac{3}{14}+\frac{2}{13}+0-\frac{1}{6}-\frac{1}{7}$
|
Just completing Thomas Andrew's answer,
$$\sum_{n=1}^{N}\frac{1}{n^2+n+1}=4\sum_{n=1}^{N}\frac{1}{(2n+1)^2+3}=4\sum_{k=1}^{+\infty}\sum_{k=1}^{N}\frac{3^{k-1}}{(2n+1)^{2k}}\tag{1}$$
can be put in integral form by exploiting:
$$ \int_{0}^{+\infty}\frac{\sin(m x)}{m}\,e^{-\sqrt{3}\,x}\,dx = \frac{1}{3+m^2}.\tag{2}$$
That gives, for instance:
$$\sum_{n=1}^{+\infty}\frac{1}{n^2+n+1}=4\int_{0}^{+\infty}\left(-\sin x+\varphi(x)\right) e^{-\sqrt{3}\,x}\,dx\\=-1+4\int_{0}^{+\infty}\varphi(x)\,e^{-\sqrt{3}\,x}\,dx \tag{3}$$
where $\varphi(x) = \frac{\pi}{4}\cdot(-1)^{\left\lfloor\frac{x}{\pi}\right\rfloor}$, so that (have also a look at this other question explaining the method):
$$\sum_{n=1}^{+\infty}\frac{1}{n^2+n+1}=\color{red}{-1+\frac{\pi}{\sqrt{3}}\tanh\left(\frac{\pi}{2}\sqrt{3}\right)}.\tag{4}$$
By truncating the previous sum at $N=50$ we get something smaller than the RHS of $(4)$, but not much smaller, since the series on the LHS converges quite fast.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1163703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Long limit question If $$\lim_{x \to \infty}\frac{a(2x^3-x^2)+b(x^3+5x^2-1)-c(3x^3+x^2)}{a(5x^4-x)-bx^4+c(4x^4+1)+2x^2+5x}=1$$ then find the value of $a+b+c$. I have done such problems with $x$ tending to some finite amount. Infinity is creating some problems.
|
Consider
$$
\lim_{x \to \infty}\frac{a(2x^3-x^2)+b(x^3+5x^2-1)-c(3x^3+x^2)}{a(5x^4-x)-bx^4+c(4x^4+1)+2x^2+5x}=1.
$$
rewrite this as
$$
\lim_{x \to \infty}\frac{x^3(2a+b-3c)+x^2(-a+5b-c)-b}{x^4(5a-b+4c)+2x^2+x(-a+5)+c}=1.
$$
As long as the $x^4$ term is in the denominator the limit cannot be $1$, so the coefficient of that has to be $0$. So we get
$$5a-b+4c=0.$$
Now same thing should happen in the numerator, the coefficient of $x^3$ term has to be zero.
So we have
$$2a+b-3c=0.$$
This leads us to
$$
\lim_{x \to \infty}\frac{x^2(-a+5b-c)-b}{2x^2+x(-a+5)+c}=1.
$$
Now divide both numerator and denominator by $x^2$ and use the fact that $1/x$ and $1/x^2 \to 0$ as $x \to \infty$ to get
$$
\frac{-a+5b-c}{2}=1.
$$
Now you have three equations in $a,b$ and $c$ to solve.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1166116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
}
|
Find $\lim_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$ How to calculate the following limit?
$$\lim\limits_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$$
|
$$\lim_{x\to \infty} \Big(\frac{x^2+1}{x^2-1}\Big)^{x^2}=\lim_{x\to \infty} \Big(1+\frac{2}{x^2-1}\Big)^{\frac{x^2-1}{2}\cdot\frac{2x^2}{x^2-1}}=e^{\lim_{x\to \infty}\frac{2x^2}{x^2-1}}=e^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1168828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
}
|
How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$
|
You can use "Long Division of Polynomials"
$\frac {1-(\frac y x)^3}{1-(\frac y x)}=1+\frac y x+(\frac y x)^2$
Multiply both sides by $x^3$
$\frac {x^3-y^3}{1-(\frac y x)}=x(x^2+xy+y^2)$
Then you have
$x^3-y^3=(x-y)(x^2+xy+y^2)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1172119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 12,
"answer_id": 11
}
|
Solving for real x and y If x and y be real numbers such that, $x^3 - 3x^2 + 5x = 1$ and $y^3 - 3x^2 + 5y = 5$; Find $(x + y)$ From an old Russian olympiad.
I tried to make the equations homogenous by substituting for $1 = x^3 - 3x^2 + 5x$ in the second equation for $5 * 1$, which didn't work even by repeated substitution.
Next, equating $3x^2$ in both equations and trying to homogenise them also doesn't give me anything.
Hint for this one please!
|
I think you have somewrong,I have see a book with Russian olympiad problem
Let $x,y$ are real numbers, and
$$x^3-3x^2+5x=1,y^3-\color{#0a0}{\text{$3y^2$}}+5y=5$$
Find $x+y$
since
$$(x-1)^3+2(x-1)=-2$$
$$(y-1)^3+2(y-1)=2$$
since $f(x)=x^3+2x$ is odd function,and increaing on $R$
since
$$f(x-1)+f(y-1)=0\Longleftrightarrow f(x-1)=-f(y-1)=f(1-y)$$
so
$$x-1=1-y\Longrightarrow x+y=2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1176178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
}
|
Odd numbers expressible as a sum of 3 composite odd integers. How many positive odd integers between $0$ and $999$ inclusive can be
written as sum of $3$ odd composite positive integers?
I am trying to approach this by looking at equivalence classes $\mod 6$ but the calculations are a bit tedious.
|
Let's start with integers equivalent to $3\pmod6$. The smallest is obviously $27=9+9+9$. Every following odd multiple of $3$ also works as it can be written as $9+9+3(2n+3)$. $27=6(4)+3$ and $999=6(166)+3$, so we have $163$ odd multiples of $3$.
The first odd number not a multiple of $3$ is $5(5)=25\equiv1\pmod6$. So the first value equivalent to $1\pmod6$ is $9+9+25=43=6(7)+1$. The same trick in the first step works here, so we have $160$ values here.
The first odd number equivalent to $5\pmod6$ is $5(7)=35$. The smallest value that works equivalent to $5\pmod6$ is then $9+9+35=53<9+25+25$. $53=6(8)+5$. We have $158$ numbers here that work.
So our final tally is $163+160+158=481.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1179147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
If $\frac {1}{2+a} + \frac {1}{2+b} + \frac {1}{2+c} = 1$, prove $\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \leq 3$ Let $a,b,c$ be non-negative numbers such that $$\frac {1}{2+a} + \frac {1}{2+b} + \frac {1}{2+c} = 1.$$
Prove that $ \sqrt{ab} + \sqrt{ac} + \sqrt{bc} \leq 3 $.
|
The condition gives:
$$\frac32-1 = \sum_{cyc} \left(\frac12-\frac1{2+a} \right)$$
$$\implies 1 = \sum_{cyc} \frac{a}{2+a} \ge \frac{(\sqrt a+\sqrt b+\sqrt c)^2}{a+b+c+6} \quad \text{by Cauchy-Schwarz inequality}$$
$$\implies 3 \ge \sqrt{ab}+\sqrt{bc}+\sqrt{ca}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1180443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
}
|
2-adic expansion of (2/3) I have been asked in an assignment to compute the 2-adic expansion of (2/3). It just doesn't seem to work for me though. In our definition of a p-adic expansion we have $x= \sum_{n=0}^{\infty}a_np^n$ with $0 \leq a_0<p$.
So I use the same method as for similar questions which seems to work fine. This involves solving each congruence $\frac{2}{3} = \sum_{n=0}^{\infty}a_np^n \ (mod \ p^N)$ for $N=1,2,3,4,5$ and then we are supposed to guess the pattern and show it by summing the series.
Doing this, I get $\frac{2}{3} = 2 + 2^2 + 2^4 + 2^5 + ... \\ = (2+2^2)(1+2^3+2^6+...) \\ = 6(\frac{1}{1-2^3}) \\ = \frac{6}{-7}.$
Which is just completely wrong?
An example of me solving a congruence:
$\frac{2}{3} \equiv a_0 \ (mod \ 2)$ so $0 \equiv 2 \equiv 3a_0 \ (mod \ 2)$ so $a_0 = 0$.
|
I get $2^1+2^2+2^4+2^6+\ldots$, with no $2^5$. In the $2^4$ place, you have $3\times1+1=4=0$ carry $2$, so you need $0\times 2^5$ to give $0$ carry $1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1182127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
How to prove that $\frac{\sqrt{5(160ab+8b+13)}-5}{20}$ is not an integer
Prove that $\frac{\sqrt{5(160ab+8b+13)}-5}{20}$ is not an integer where $a$ and $b$ are positive integers.
One of the roots of the equation $10x^2+5x-1-(20a+1)b=0$ is $\frac{\sqrt{5(160ab+8b+13)}-5}{20}$. Wolframalpha says that this equation has no integer solution. How can I prove that $\frac{\sqrt{5(160ab+8b+13)}-5}{20}$ is not an integer?
|
Take $b=14$ and $a=3$. In this case, $\sqrt{5(160ab+8b+13)} = 185$.
This gives $x=\frac{185-5}{20}=9$ as a valid root. So there are integer solutions for $x$ when $a$ and $b$ are positive integers.
If Wolfram told you there were no integer solutions, either Wolfram is wrong, or perhaps you made a typo when entering your equation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1183676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Show that $\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx = \frac{8 \pi ^3}{81 \sqrt{3}}$ I have found myself faced with evaluating the following integral: $$\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx. $$
Mathematica gives a closed form of $8 \pi ^3/(81 \sqrt{3})$, but I have no idea how to arrive at this closed form. I've tried playing around with some methods from complex analysis, but I haven't had much luck (it has been a while). Does anyone have any ideas? Thanks in advance!
|
Note
$$ \int_0^1x^m(\ln x)^2dx=\frac{2}{(m+1)^3} $$
So
\begin{eqnarray}
I&=&\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1}dx\\
&=&\int_0^1\frac{(1-x)(\ln x)^2}{1-x^3}dx\\
&=&\int_0^1\sum_{n=0}^\infty(1-x)x^{3n}(\ln x)^2dx\\
&=&2\sum_{n=0}^\infty\left(\frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3}\right)\\
&=&2\sum_{n=-\infty}^\infty\frac{1}{(3n+1)^3}
\end{eqnarray}
Note
\begin{eqnarray}
\sum_{n=-\infty}^\infty\frac{1}{(3n+1)^3}=-\pi\text{Res}(\frac{1}{(z+1)^3}\cot(\pi z),-\frac{1}{3})=\frac{4\pi^3}{81\sqrt3}
\end{eqnarray}
and hence
$$ I=\frac{8\pi^3}{81\sqrt3}. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1186002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
}
|
Differentiate $\,y = 9x^2 \sin x \tan x:$ Did I Solve This Correctly? I'm posting my initial work up to this point.
Criticism welcomed!
Using the formula $(fgh)' = f'gh+fg'h + fgh'$, differentiate$$y = 9x^2\sin x \tan x$$
$$\begin{align} y' &= 9\frac d{dx}(x^2)\sin x \tan x + 9x^2 \frac d{dx} (\sin x) \tan x + 9x^2\sin x \frac d{dx}(\tan x)\\ \\
& = 9(2x\sin x \tan x + 9x^2-\cos x \tan x + 9x^2\sin x \sec^2x\\ \\
&=9\Big(2x\sin x \tan x + x^2 -\cos x \tan x + x^2\sin x \sec^2 x\Big)
\end{align}$$
Have I done it correctly up and until this point?
|
You have a parenthesis error
Line 3 should read
$$
9(2x)\sin x \tan x +9x^2(-\cos x)\tan x + 9x^2 \sin x \sec^2 x
$$
You can actually factor out $9x\sin x$ because $\cos x \tan x = \sin x$.
Continuing
$$
9x \sin x (2\tan x - x + x \sec^2 x) \\
9x \sin x (2\tan x +x (\sec^2 x - 1) )\\
$$
Then you can use some trig identities to simplify further.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1188870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Solve $\int\frac{x}{\sqrt{x^2-6x}}dx$ I need to solve the following integral
$$\int\frac{x}{\sqrt{x^2-6x}}dx$$
I started by completing the square,
$$x^2-6x=(x-3)^2-9$$
Then I defined the substitution variables..
$$(x-3)^2=9\sec^2\theta$$
$$(x-3)=3\sec\theta$$
$$dx=3\sec\theta\tan\theta$$
$$\theta=arcsec(\frac{x-3}{3})$$
Here are my solving steps
$$\int\frac{x}{\sqrt{(x-3)^2-9}}dx = 3\int\frac{(3\sec\theta+3)\sec\theta\tan\theta}{\sqrt{9(\sec^2\theta-1)}}d\theta$$
$$=\int\frac{(3\sec\theta+3)\sec\theta\tan\theta}{\tan\theta}d\theta$$
$$=\int(3\sec\theta+3)\sec\theta\tan\theta$$
$$=3\int\sec^2\theta\tan\theta d\theta + 3\int\sec\theta\tan\theta d\theta$$
$$u = \sec\theta, du=\sec\theta\tan\theta d\theta$$
$$=3\int udu + 3\sec\theta$$
$$=\frac{3\sec\theta}{2}+3\sec\theta+C$$
$$=\frac{3\sec(arcsec(\frac{x-3}{3}))}{2}+3\sec(arcsec(\frac{x-3}{3}))+C$$
$$=\frac{3(\frac{x-3}{3})}{2}+3(\frac{x-3}{3})$$
$$=\frac{x-3}{2}+x-3+C$$
However, the expected answer is
$$\int\frac{x}{\sqrt{x^2-6x}}dx=\sqrt{x^2-6x}+3\ln\bigg(\frac{x-3}{3}+\frac{\sqrt{x^2-6x}}{3}\bigg)$$
What did I misunderstood?
|
I prefer to apply hyperbolic substitution.
More precisely, if we let $x - 3 = 3\cosh(y)$, we arrive at
\begin{align*}
\int\frac{x}{\sqrt{x^{2} - 6x}}\mathrm{d}x & = \int\frac{x}{\sqrt{(x - 3)^{2} - 9}}\mathrm{d}x\\\\
& = 3\int\frac{(1 + \cosh(y))\sinh(y)}{\sqrt{\cosh^{2}(y) - 1}}\mathrm{d}y\\\\
& = 3\int\mathrm{d}y + 3\int \cosh(y)\mathrm{d}y\\\\
& = 3y + 3\sinh(y) + c\\\\
& = 3\operatorname{arccosh}\left(\frac{x - 3}{3}\right) + 3\sqrt{\left(\frac{x - 3}{3}\right)^{2} - 1} + c\\\\
& = 3\ln\left(\frac{x - 3}{3} + \frac{\sqrt{x^{2} - 6x}}{3}\right) + \sqrt{x^{2} - 6x} + c\\\\
& = 3\ln\left(x - 3 + \sqrt{x^{2} - 6x}\right) + \sqrt{x^{2} - 6x} + C
\end{align*}
which coincides with the proposed result because
\begin{align*}
\operatorname{arccosh}(z) = \ln\left(z + \sqrt{z^{2} - 1}\right)
\end{align*}
Hopefully this helps!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1189943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
How to solve $\int\sqrt{1+x\sqrt{x^2+2}}dx$ I need to solve
$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$
I've chosen the substitution variables
$$u=\sqrt{x^2+2}$$
$$du=\frac{x}{\sqrt{x^2+2}}$$
However, I am completly stuck at
$$\int\sqrt{1+xu} dx$$
Which let me believe I've chosen wrong substitution variables.
I've then tried letting $u=x^2+2$ or simply $u=x$, but it does not help me at all solving it.
Would someone please give me an hint on this ?
Thanks.
|
I thought about Wolfram and Maple too, needs a little help. But I was wrong.
$$\int {\sqrt {1 + x\sqrt {{x^2} + 2} } } dx = \int {\sqrt {1 + \frac{1}{2}\sqrt {{x^2} + 2} } } 2xdx$$
\begin{gathered}
y = {x^2} + 2 \\
dy = 2xdx \\
\end{gathered}
\begin{gathered}
\int {\sqrt {1 + \frac{1}{2}\sqrt y } } dy = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt y + 2} \right)^{3/2}}\left( {3\sqrt y - 4} \right) \\
= \frac{2}{{15}}\sqrt 2 {\left( {\sqrt {{x^2} + 2} + 2} \right)^{3/2}}\left( {3\sqrt {{x^2} + 2} - 4} \right) \\
\end{gathered}
Now Wolfram: This is an identity:
$$\int {\sqrt {1 + \frac{1}{2}\sqrt y } } dy = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt y + 2} \right)^{3/2}}\left( {3\sqrt y - 4} \right)$$
Resubstituting gives me:
$$\frac{2}{{15}}\sqrt 2 {\left( {\sqrt y + 2} \right)^{3/2}}\left( {3\sqrt y - 4} \right) = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt {{x^2} + 2} + 2} \right)^{3/2}}\left( {3\sqrt {{x^2} + 2} - 4} \right)$$
Differentiation:
$$\frac{d}{{dx}}\left( {\frac{2}{{15}}\sqrt 2 {{\left( {\sqrt {{x^2} + 2} + 2} \right)}^{3/2}}\left( {3\sqrt {{x^2} + 2} - 4} \right)} \right) = \sqrt 2 x\sqrt {2 + \sqrt {{x^2} + 2} } $$`
And:
$$\sqrt 2 x\sqrt {2 + \sqrt {{x^2} + 2} } \ne \sqrt {1 + x\sqrt {{x^2} + 2} }$$
Very poor! For me. Made a mistake by myself.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1191730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
}
|
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
My Attempt: I start with direct proof.
Let $a,b,c$ be consecutive integers and $a< b < c $, there exists a integer $k$ such that $a=k, b=k+1, c=k+2$. Then $a^3+b^3=k^3+(k+1)^3=k^3+(k^3+3k^2+3k+1)$ and $c^3=(k+2)^3=k^3+6k^2+12k+8=(k^3+3k^2+3k+1)+(3k^2+9k+7)$. Hence, we have $k^3+(k^3+3k^2+3k+1)\neq (k^3+3k^2+3k+1)+(3k^2+9k+7)$ which implies $a^3+b^3\neq c^3$.
Does my proof valid ? And, is use Contradiction easier? If not, can anyone give me a hit or suggestion ?
Thanks
|
Suppose$$(k-1)^3+k^3=(k+1)^3$$ $$2k^3-3k^2+3k-1=k^3+3k^2+3k+1$$ $$k^3-6k^2-2=0$$ $$k^2(k-6)=2 ,\,\,\,\,\text{where $k$ is an integer. }$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1191907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
$\sup$ and $\inf$ of this set These are exercises from my textbook, and I am not sure if the solutions are correct or not.
Given a set $B = \{\frac{n}{2n+1} : n \in \mathbb{N} \}$
Find the $\sup$ and $\inf$ of $B$, and maxima and minima if they exist.
The solutions give $\sup B = \frac{1}{2}$ and $\inf B = \frac{1}{3}$
I am confused by this result. The $\sup$ has to be the least upper bound, $\frac{1}{2} \in B$ and $\frac{1}{3}$ is also in $B$. So how can we claim that $\sup B = \frac{1}{2}$ when $\frac{1}{3} > \frac{1}{2}$
Is this a typo?
Additionally, shouldn't $\inf B = \frac{1}{2}$ since as $n \to \infty$, $\frac{n}{2n+1} \to \frac{1}{2}$
|
We have
$$
0.333... = \frac{1}{3} < \frac{1}{2} = 0.5
$$
so the $\sup$ is right.
Furthermore, $\frac{1}{3} < \frac{1}{2}$ also implies that the infimum is at most $\frac{1}{3}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1192944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$ How do we compute this integral ?
$$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$
I have tried partial fraction but it is quite hard to factorize the denominator. Any help is appreciated.
|
Let $I$ be the integral given by
$$I= \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$
Now, make the substitution $x=u^{-1}$. Then $dx=-u^{-2}du$ and the limits of integration over $u$ extend from $\infty$ to $1$. Thus, we may write
$$
\begin{align}
I & = \int_{\infty}^1 \frac{1+u^{-6}}{1-u^{-2}+u^{-4}-u^{-6}+u^{-8}}\left(-u^{-2}\right)du \\
& = \int_1^{\infty} \frac{1+u^{6}}{1-u^{2}+u^{4}-u^{6}+u^{8}}du
\end{align}
$$
whereby
$$I=\frac12 \int_0^{\infty} \frac{1+u^{6}}{1-u^{2}+u^{4}-u^{6}+u^{8}}du$$
Next, we exploit the even symmetry of the integrand, which reveals that
$$I=\frac14 \int_{-\infty}^{\infty} \frac{1+u^{6}}{1-u^{2}+u^{4}-u^{6}+u^{8}}du$$
Using the Residue Theorem along with Jordan's Lemma, the integral $I$ is given by
$$I=2\pi i \left(\frac14\right)\sum_{i=1}^4 Res_i \left(\frac{1+z^{6}}{1-z^{2}+z^{4}-z^{6}+z^{8}}\right)$$
where the summation is over the four residues of the integrand in the upper-half plane.
All that remains is to find the four complex roots $z_i$ of the denominator that lie in the upper-half plane. These are $z_1=e^{\frac{\pi}{10}}$, $z_2=e^{\frac{3\pi}{10}}$, $z_3=e^{\frac{7\pi}{10}}$, and $z_4=e^{\frac{9\pi}{10}}$.
To find the residue of $z_i$, take the limit
$$\lim_{z \to z_i} (z-z_i) \left(\frac{1+z^{6}}{1-z^{2}+z^{4}-z^{6}+z^{8}}\right)=\frac{1+z_i^{6}}{-2z_i+4z_i^{3}-6z_i^{5}+8z_i^{7}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1193424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1}$$
$$\cos (x)=\frac{1-u^2}{u^2+1}$$
$$dx=\frac{2du}{u^2+1}$$
$$=\int_{0}^{\infty }\frac{2(1-\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+1)}du$$
$$=\int_{0}^{\infty }\frac{2(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{(u+1)^2(u^2+1)}du $$
$$=2\int_{0}^{\infty }(\frac{2}{(u+1)^2}-\frac{1}{u^2+1})du $$
$$=-2\int_{0}^{\infty }\frac{1}{u^2+1}du+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du $$
$$\lim_{b\rightarrow \infty }\left | (-2\tan^{-1}(u)) \right |_{0}^{b}+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=(\lim_{b\rightarrow \infty}-2\tan^{-1}(b))+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=-\pi+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
Let $$s=u+1$$
$$ds=du$$
$$=-\pi+4\int_{1}^{\infty}\frac{1}{s^2}ds$$
$$=-\pi+\lim_{b\rightarrow \infty}\left | (-\frac{4}{s}) \right |_{1}^{b}$$
$$=-\pi+(\lim_{b\rightarrow \infty} -\frac{4}{b}) +4$$
$$=4-\pi$$
$$\approx 0.85841$$
|
Multiply numerator and denominator by $1 - \sin x$. So that $\displaystyle\int_0^\pi \dfrac{1-\sin x}{1+\sin x} \cdot \dfrac{1-\sin x}{1-\sin x}dx $
$$ = \displaystyle\int_0^\pi \dfrac{1-2\sin x+ \sin^2x}{\cos^2x}dx = \int_0^\pi \sec^2x - 2\dfrac{\sin x}{\cos^2x} + \tan^2x dx$$
We know that $\tan^2x + 1 = \sec^2x$. So the integral would look like:
$$= \int_0^\pi 2\sec^2x -1 - 2\dfrac{\sin x}{\cos^2x}dx$$
We know the integral of $\sec^2x$ is just $\tan x$, and the last integrand can be solved by letting $u=\cos x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1194139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 9,
"answer_id": 8
}
|
Evaluate the following limit without L'Hopital I tried to evaluate the following limits but I just couldn't succeed, basically I can't use L'Hopital to solve this...
for the second limit I tried to transform it into $e^{\frac{2n\sqrt{n+3}ln(\frac{3n-1}{2n+3})}{(n+4)\sqrt{n+1}}}$ but still with no success...
$$\lim_{n \to \infty } \frac{2n^2-3}{-n^2+7}\frac{3^n-2^{n-1}}{3^{n+2}+2^n}$$
$$\lim_{n \to \infty } \frac{3n-1}{2n+3}^{\frac{2n\sqrt{n+3}}{(n+4)\sqrt{n+1}}}$$
Any suggestions/help? :)
Thanks
|
Hints :
$$ \frac{2n^2-3}{-n^2+7} = \frac{2 - \frac{3}{n^2}}{-1+\frac{7}{n^2}},$$
and
$$ \frac{3^n-2^{n-1}}{3^{n+2}+2^n} = \frac{1-\frac{1}{2}\left( \frac{2}{3} \right)^n}{3^2+\left( \frac{2}{3} \right)^n}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1197056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
How to evaluate $\sum_{n=2}^\infty\frac{(-1)^n}{n^2-n}$ How would you go about evaluating:$$\sum_{n=2}^\infty\frac{(-1)^n}{n^2-n}$$
I split it up to $$\sum_{n=2}^\infty\left[(-1)^n\left(\frac{1}{n-1}-\frac{1}{n}\right)\right]$$
but I'm not sure what to do from here. If the $(-1)^n$ term wasn't there then it would be a simple telescoping series but the alternating bit causes trouble.
|
\begin{align}
& \sum_{n=2}^\infty (-1)^n\left(\frac{1}{n-1}-\frac{1}{n}\right) \\[10pt]
= {} & \underbrace{\left(1-\frac 1 2 \right) - \left( \frac 1 2\right.}_{=\,0} - \left.\frac 1 3 \right) + \left( \frac 1 3 - \frac 1 4 \right) - \left( \frac 1 4 - \frac 1 5\right) + \left( \frac 1 5 - \frac 1 6 \right) - \cdots \\[10pt]
= {} & \frac 2 3 - \frac 2 4 + \frac 2 5 - \frac 2 6 + \cdots \\[10pt]
= {} & 2\int_0^1 \left( u^2 - u^3 + u^4 - u^5 + \cdots \right) \,du = 2\int_0^1 \frac{u^2}{1+u} \, du \\[10pt]
= {} & 2 \int_0^1 \left( u - 1 + \frac 1 {u+1} \right)\,du = \text{etc.}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1197265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
}
|
Solution of $y''+xy=0$ The differential equation $y''+xy=0$ is given.
Find the solution of the differential equation, using the power series method.
That's what I have tried:
We are looking for a solution of the form $y(x)= \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence of the power series $R>0$.
Then:
$$y'(x)= \sum_{n=1}^{\infty} n a_n x^{n-1}= \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n$$
$$y''(x)= \sum_{n=1}^{\infty} (n+1) n a_{n+1} x^{n-1}= \sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n$$
Thus:
$$\sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n+ x \sum_{n=0}^{\infty} a_n x^n=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=0}^{\infty} a_n x^{n+1}=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=1}^{\infty} a_{n-1} x^n=0 \\ \Rightarrow 2a_2+\sum_{n=1}^{\infty} \left[ (n+2) (n+1) a_{n+2}+ a_{n-1}\right] x^n=0$$
So it has to hold:
$$a_2=0 \\ (n+2) (n+1) a_{n+2}+a_{n-1}=0, \forall n=1,2,3, \dots$$
For $n=1$: $3 \cdot 2 \cdot a_3+ a_0=0 \Rightarrow a_3=-\frac{a_0}{6}$
For $n=2$: $4 \cdot 3 \cdot a_4+a_1=0 \Rightarrow a_4=-\frac{a_1}{12}$
For $n=3$: $5 \cdot 4 \cdot a_5+a_2=0 \Rightarrow a_5=0$
For $n=4$: $6 \cdot 5 \cdot a_6+a_3=0 \Rightarrow 30 a_6-\frac{a_0}{6}=0 \Rightarrow a_6=\frac{a_0}{6 \cdot 30}=\frac{a_0}{180}$
For $n=5$: $7 \cdot 6 \cdot a_7+ a_4=0 \Rightarrow 7 \cdot 6 \cdot a_7-\frac{a_1}{12}=0 \Rightarrow a_7=\frac{a_1}{12 \cdot 42}$
Is it right so far? If so, how could we find a general formula for the coefficients $a_n$?
EDIT: Will it be as follows:
$$a_{3k+2}=0$$
$$a_{3k}=(-1)^k \frac{a_0}{(3k)!} \prod_{i=0}^{k-1} (3i+1)$$
$$a_{3k+1}=(-1)^k \frac{a_1}{(3k+1)!} \prod_{i=0}^{k-1} (3i+2)$$
If so, then do we have to write seperately the formula for the coefficients of $x^0, x^1$, because otherwhise the sum would be from $0$ to $-1$ ?
|
Follow-up from my comment, every $a_{n+3}$ can be expressed in terms of $a_n$.
$$a_{n+3} = -\frac{a_{n}}{(n+3)(n+2)}$$
Since $a_2 = 0$, all coefficients with $n = 3k + 2$ will be $0$
The rest will have to based on the initial conditions of $a_0 = y(0)$ and $a_1 = y'(0)$
For $n = 3k$
$$a_3 = -\frac{a_0}{3\cdot2}$$
$$a_6 = -\frac{a_3}{6\cdot5} = \frac{a_0}{6\cdot5\cdot3\cdot2} $$
This can be extrapolated to
$$ a_{n} = \pm\frac{a_0}{n(n-1)(n-3)(n-4)\cdots\cdot6\cdot5\cdot3\cdot2} $$
Basically, the denominator is a product of all integers from $1$ to $n$, skipping every 3rd number and alternating signs. An alternate notation is
$$a_{3k} = (-1)^k\frac{a_0}{(3k)!}\prod_{i=0}^{k-1} (3i+1) $$
The same can be done for $n = 3k+1$
$$a_{3k+1} = (-1)^k\frac{a_1}{(3k+1)!}\prod_{i=0}^{k-1} (3i+2)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1197528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
}
|
How to calculate $\int \frac{x^2 }{(x^2+1)^2} dx$? I'm trying to calculate $\int \frac{x^2 }{(x^2+1)^2} dx$ by using the formula:
$$ \int udv = uv -\int vdu $$
I supposed that $u=x$ s.t $du=dx$, and also that $dv=\frac{x}{(x^2+1)^2}dx$, but I couldn't calculate the last integral. what is the tick here?
the answer must be: $ \frac{1}{2}arctan\ x - \frac{x}{2(1+x^2)} $ + C
|
First, you used parts: with $u = x$ and $dv = \frac{x}{(x^2+1)^2}dx$, you got $$\int\frac{x^2}{(x^2+1)^2}dx = xv - \int v\,dx.\tag{1}$$
From here, I suggested the substitution $t = x^2$ in order to integrate $dv$ and get $v$.
Doing this, we see that $dt = 2x\,dx$, so $x\,dx = \frac{1}{2}dt$.
Thus $$v = \int dv = \int \frac{x}{(x^2+1)^2}dx = \frac{1}{2} \int \frac{dt}{(t + 1)^2}.$$
This latter integral is easy: if you can't guess it, let $s = t + 1$.
In any case, we get $$v = -\frac{1}{2(t+1)} = -\frac{1}{2(x^2+1)}.$$
Going back to $(1)$,
$$\int\frac{x^2}{(x^2+1)^2}dx = -\frac{x}{2(x^2+1)} + \frac{1}{2} \int \frac{dx}{x^2+1}$$
and you should recognize the final integral as $\arctan{x}$.
Putting it all together, the answer is
$$\frac{1}{2}\arctan{x} - \frac{x}{2(x^2+1)} + C.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1198686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Parabolic sine approximation Problem Find a parabola ($f(x)=ax^2+bx+c$) that approximate the function sine the best on interval [0,$\pi$].
The distance between two solutions is calculated this way (in relation to scalar product): $\langle u,v \rangle=\int_0^\pi fg$.
My (wrong) solution I thought that I would get the solution by calculating the orthogonal projection $w=a+bx+cx^2$ of $v=\sin x$ on subspace $W=\langle u_1,u_2,u_3 \rangle=\langle 1,x,x^2\rangle$ using Gramm matrix. Then I have
$$\begin{pmatrix}\langle u_1, u_1\rangle&\langle u_1, u_2\rangle&\langle u_1, u_3\rangle\\
\langle u_2, u_1\rangle&\langle u_2, u_2\rangle&\langle u_2, u_3\rangle\\
\langle u_3, u_1\rangle&\langle u_3, u_2\rangle& \langle u_3, u_3\rangle\end{pmatrix}
\begin{pmatrix} a\\b\\c\end{pmatrix}=
\begin{pmatrix} \langle u_1, v\rangle\\
\langle u_2,v\rangle\\
\langle u_3, v\rangle \end{pmatrix}$$
So then
$$\begin{pmatrix} \int_0^\pi 1&\int_0^\pi x&\int_0^\pi x^2\\
\int_0^\pi x& \int_0^\pi x^2& \int_0^\pi x^3 \\
\int_0^\pi x^2&\int_0^\pi x^3&\int_0^\pi x^4\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} \int_0^\pi \sin x\\\int_0^\pi x\sin x\\\int_0^\pi x^2 \sin x \end{pmatrix}$$
But solving these equations didn't give me any good answer. So my question is - is my way of solving it completely wrong (if so, can you give me hints how to do it otherwise)?
Thank you
Edit
Then$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
\frac{\pi^2}{2}&\frac{\pi^3}{3}&\frac{\pi^4}{4}\\
\frac{\pi^3}{3}&\frac{\pi^4}{4}&\frac{\pi^5}{5}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\\pi\\\pi^2-4 \end{pmatrix}$$
$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&\frac{\pi^4}{12}&\frac{4\pi^5}{45}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&0&\frac{\pi^5}{180}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
Edit II My way (I have found the mistake I did) will get the result
$$f(x)=\dfrac{60(\pi^2-12)}{\pi^5}x^2-\dfrac{60(\pi^2-12)}{\pi^4}x+\dfrac{12(\pi^2-10)}{\pi^3}$$
which I hope is the right answer with error approx. $0,000936$
|
The answer has already been given.
In addition, the numerical solving and the graph are shown below, in order to compare the results.
Note that the result corresponds to the mean least square absolute deviation. A small modification of the method and formulas would give the result corresponding to the mean least square relative deviation, which could be more convenient in the low range of $\sin(x)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1199798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
How can I express the ration of double factorials $\frac{(2n+1)!!}{(2n)!!}$ as a single factorial? How can I change the double factorial of $$\frac{(2n+1)!!}{(2n)!!}$$ to single factorial?
|
$$(2n+1)!!=(2n+1)(2n-1)(2n-3) \cdots (3) (1)$$
$$=\frac{(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4) \cdots (4)(3)(2) (1)}{(2n)(2n-2)(2n-4) \cdots (4)(2)}=\frac{(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4) \cdots (4)(3)(2) (1)}{(2)(n)(2)(n-1)(2)(n-2) \cdots (2)(2)(2)(1)}$$
$$=\frac{(2n+1)!}{(2^n)n!}$$
Similarly,
$$(2n)!!=(2n)(2n-2)(2n-4) \cdots (4)(2)=(2)(n)(2)(n-1)(2)(n-2) \cdots (2)(2)(2)(1)=(2^n)n!$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1200129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Derivative Chain Rule $f(x) = (3x^2+2)^2 (x^2 -5x)^3$ I'm learning chain rule in derivative,
and I don't understand this in the example.
$$f(x) = (3x^2+2)^2 (x^2 -5x)^3\\$$
\begin{align}
f'(x)&= (3x^2 +2)^2[3(x^2 -5x)^2(2x-5)] + (x^2-5x)^3[2(3x^2+2)(6x)]
\\&=3(3x^2 +2)(x^2-5x)^2[(3x^2+2)(2x-5)+4x(x^2-5x)]
\\&= 3(3x^2 +2)(x^2-5x)^2[6x^3-15x^2+4x-10+4x^3-20x^2]
\\&=3(3x^2+2)(x^2-5x)^2(10x^3-35x^2+4x-10)
\end{align}
I understand the first line, because they just applied the chain rule,
but I don't understand the second line. I think they are simplifying it, but still I don't understand.
Could anyone explain what happened there? or add extra steps so it's easier??
|
Hint:
Put $(x^2- 5x)^2$, $(3x^2 + 2)$ and $3$ in evidence. Then
$$\begin{align}f'(x)&= \color{green}{(3x^2 + 2)}^2[\color{red} 3\color {#05f}{(x^2 - 5x)^2} (2x - 5)] + \color {#05f}{(x^2 - 5x)^3}[2\color{green}{(3x^2 + 2)}\color {red} 6 x]\\&=\color{red}3\color{green}{(3x^2 + 2)}\color{#05f}{ (x^2 - 5x)^2} \,[(3x^2 + 2)(2x - 5) + 4x (x^2 - 5)]\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1200300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Integral of $\log(\sin(x)) \tan(x)$ I would like to see a direct proof of the integral $$\int_0^{\pi/2} \log(\sin(x)) \tan(x) \, \mathrm{d}x = -\frac{\pi^2}{24}.$$ I arrived at this integral while trying different ways to evaluate $\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + ...$ and found this value using the known sum $\frac{\pi^2}{6}.$ It would be better to have a direct way to evaluate this.
|
Note that
$$\begin{align}\int_0^{\pi/2} \log(\sin x) \tan x \, \mathrm{d}x &= \frac1{2}\int_0^{\pi/2} \log(\sin^2 x ) \tan x \, \mathrm{d}x \\ &= \frac1{2}\int_0^{\pi/2} \frac{\log(\sin^2 x ) \sin x}{\cos x} \, \mathrm{d}x \\ &= \frac1{2}\int_0^{\pi/2} \frac{\log(1-\cos^2 x ) \cos x \sin x}{\cos^2 x} \, \mathrm{d}x \end{align}.$$
Changing variables to $u = \cos^2 x$ we get
$$\begin{align}\int_0^{\pi/2} \log(\sin x) \tan x \, \mathrm{d}x &= \frac1{4}\int_0^{1} \frac{\log(1-u)}{u} \mathrm{d}u \end{align}.$$
Now integrate the Taylor series expansion of the integrand termwise to finish.
$$\begin{align}\int_0^{\pi/2} \log(\sin x) \tan x \, \mathrm{d}x &= -\frac1{4}\int_0^{1} \sum_{k=1}^{\infty}\frac{u^{k-1}}{k} \mathrm{d}u \\ &= -\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^2} \end{align}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1200655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational
How can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?
I know that $\sqrt{3}, \sqrt{5}$ and $\sqrt{7}$ are all irrational and that $\sqrt{3}+\sqrt{5}$, $\sqrt{3}+\sqrt{7}$, $\sqrt{5}+\sqrt{7}$ are all irrational, too. But how can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?
|
Suppose $\sqrt{3}+\sqrt{5}+\sqrt{7} = r$ for some rational $r$. Then, $$(\sqrt{3}+\sqrt{5})^2 = (r-\sqrt{7})^2 \implies 8+2\sqrt{15} = 7+r^2-2r\sqrt{7}$$
So, $$1-r^2+2\sqrt{15} =-2r\sqrt{7}$$
Let $1-r^2 = k$, which will be a rational number. So,
$$(k+2\sqrt{15})^2 = k^2+ 60+4k\sqrt{15} = 28r^2$$
The LHS is irrational while the RHS is rational. Hence, we have a contradiction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1200832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
}
|
What is the name of the following matrix "product" in which elements are not multiplied? Assume that I am given the following matrices:
$A = \begin{bmatrix}
a_{1,1} & \dots & a_{1,n_a} \\
\vdots & \ddots & \vdots \\
a_{m_a,1} & \dots & a_{m_a,n_a}
\end{bmatrix}$ and $B =
\begin{bmatrix}
b_{1,1} & \dots & b_{1,n_b} \\
\vdots & \ddots & \vdots \\
b_{m_b,1} & \dots & b_{m_b,n_b}
\end{bmatrix}$. I need to consider the following matrix and I am looking for a way to relate it to $A$ and $B$.
\begin{bmatrix}
a_{1,1} & a_{1,1} & \dots & a_{1,1} & a_{1,2} & \dots & a_{1,2} & \dots & a_{1,n_a} & a_{1,n_a} & \dots & a_{1,n_a} \\
a_{2,1} & a_{2,1} & \dots & a_{2,1} & a_{2,2} & \dots & a_{2,2} & \dots & a_{2,n_a} & a_{2,n_a} & \dots & a_{2,n_a} \\
\vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\
a_{m_a,1} & a_{m_a,1} & \dots & a_{m_a,1} & a_{m_a,2} & \dots & a_{m_a,2} & \dots & a_{m_a,n_a} & a_{m_a,n_a} & \dots & a_{m_a,n_a}\\
b_{1,1} & b_{1,2} & \dots & b_{1,n_b} & b_{1,1} & \dots & b_{1,n_b} & \dots & b_{1,1} & b_{1,2} & \dots & b_{1,n_b} \\
b_{2,1} & b_{2,2} & \dots & b_{2,n_b} & b_{2,1} & \dots & b_{2,n_b} & \dots & b_{2,1} & b_{2,2} & \dots & b_{2,n_b} \\
\vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\
b_{m_b,1} & b_{m_b,2} & \dots & b_{m_b,n_b} & b_{m_b,1} & \dots & b_{m_b,n_b} & \dots & b_{m_b,1} & b_{m_b,2} & \dots & b_{m_b,n_b}
\end{bmatrix}
What would be good name and symbol for such an operation? I am thinking about cartesian product.
|
In terms of operations that already exist, we can write this as the block matrix
$$
\pmatrix{A \otimes X\\ X \otimes B}
$$
Where $X$ is the row-vector $(1,\dots,1)$ and $\otimes$ denotes the Kronecker product.
A nice shorthand for this operation (if the above is not sufficiently compact) could be $\frac{A}{B}$, which has no common second meaning in this context.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1202273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to find the limit of the following series based on a new operation? Lets define an operation as $$a*b=a+b+ab$$ Now I need to find the following limit $$\lim_{n\to\infty}\frac{1}{1}*\frac{1}{4}*\frac{1}{9}*\frac{1}{16}\cdots*\frac{1}{n^2}$$
Now I observed the series and found that it is nothing but the sum of the magnitude of coefficients of the equation except for the first term $$\prod_{i=1}^{n}(x-\frac{1}{i^2})$$ Now to get the sum of coefficients I substituted the value of $x$ as $-1$ thus the answer will be $$\prod_{i=1}^{n}(-1-\frac{1}{i^2})-1$$ After this I could not proceed. My first doubt was that for $n$ odd value will be negative , and secondly I am not able to evaluate this limit.$$\prod_{i=1}^{\infty}(1+\frac{1}{i^2})$$
|
If
$$x_n = \frac{1}{1} * \frac{1}{4} * \frac{1}{9} * \cdots * \frac{1}{n^2}$$
then
$$x_n = x_{n-1} * \frac{1}{n^2} = (1 + x_{n-1})\left(1 + \frac{1}{n^2}\right) - 1,$$
so
\begin{align}1 + x_n &= (1 + x_{n-1})\left(1 + \frac{1}{n^2}\right)\\
& = (1 + x_{n-2})\left(1 + \frac{1}{(n-1)^2}\right)\left(1 + \frac{1}{n^2}\right)\\
&\quad\vdots\\
& = \prod_{j = 1}^n \left(1 + \frac{1}{j^2}\right).\\
\end{align}
Now
$$\prod_{j = 1}^n \left(1 + \frac{1}{j^2}\right) \to \frac{\sinh \pi}{\pi}\quad \text{as}\quad n\to \infty$$
using the product formula
$$\sin \pi z = \pi z \prod_{j = 1}^\infty \left(1 - \frac{z^2}{j^2}\right)$$
and substituting $z = i$.
It follows that
$$\lim_{n\to \infty} x_n = \frac{\sinh \pi}{\pi} - 1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1202854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$ How do I show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$? Someone says I should use the rational root test, but I don't exactly know how that applies. Thanks for any input.
|
We have that $p(x)=x^4-10x^2+1$ splits as $(x^2-2)(x^2+2)$ over $\mathbb{F}_5$, hence $p(x)$ has no linear factors over $\mathbb{Q}$ and by assuming it splits over $\mathbb{Q}$, since $p(x)=p(-x)$, it must satisfy:
$$ x^4-10x^2+1 = (x^2+ax+b)(x^2-ax+b) $$
with $b^2=1$ and $a^2-2b=10$, contradiction.
With a little reverse-engineering, I guess you just want to show that your polynomial is the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$. We have that $\mathbb{Q}[x,y]/(x^2-2,y^2-3)$ is a vector space over $\mathbb{Q}$ with dimension $4$ and a base given by $1,x,y,xy$. With respect to such a base:
$$\begin{pmatrix}1 \\ (x+y) \\ (x+y)^2 \\ (x+y)^3 \\ (x+y)^4\end{pmatrix}= \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 5 & 0 & 0 & 2 \\ 0 & 9 & 11 & 0\\49& 0 & 0 & 20 \end{pmatrix}\begin{pmatrix}1 \\ x \\ y \\ xy \end{pmatrix}$$
and by applying Gaussian elimination we get that the fifth row of the last matrix can be expressed as a linear combination of the previous four rows, so that $z^4-10z^2+1$ is a polynomial in $\mathbb{Q}[z]$ that vanishes at $z=\sqrt{2}+\sqrt{3}$. Since the rank of such a matrix is four (easy to check), the last polynomial is also the minimal polynomial of $\sqrt{2}+\sqrt{3}$, so it is irreducible.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1204279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 2
}
|
Find $p$ for which all solutions of system/equation are real There is system of $5$ equations
$$
a+b+c+d+e = p; \\
a^2+b^2+c^2+d^2+e^2 = p; \\
a^3+b^3+c^3+d^3+e^3 = p; \\
a^4+b^4+c^4+d^4+e^4 = p; \\
a^5+b^5+c^5+d^5+e^5 = p, \\
\tag{1}
$$
where $p\in\mathbb{R}$.
One can prove (like here) that $a,b,c,d,e$ are roots of equation
$$
x^5-\binom{p}{1}x^4+\binom{p}{2}x^3-\binom{p}{3}x^2 + \binom{p}{4}x - \binom{p}{5} = 0.
\tag{2}
$$
I want to create related task for my son: to find polynomial $(2)$ for given system $(1)$. But it would be great to create task, where all $a,b,c,d,e$ are real and distinct. To have possibility to check manually all the sums (without complex numbers).
Question: Which values of $p$ provide $5$ real (pairwise) distinct $a,b,c,d,e$? And if there exists such one at all (for $5$ variables)?
|
In order for the roots to transition from real to complex, they must pass through a double-root first - that is, the gradient must be zero at those points. By examining how many roots are present for one solution between each pair, we can determine the regions for which all solutions are real.
$$
5x^4-4\binom{p}{1}x^3+3\binom{p}{2}x^2-2\binom{p}{3}x + \binom{p}{4}= 0
$$
Since we need roots equal to extrema, we can examine the resultant of this polynomial with the original one, which gives us (evaluated using Maxima)
$$
955514880\,{\left( p-5\right) }^{4}\,{\left( p-4\right) }^{3}\,{\left( p-3\right) }^{3}\,{\left( p-2\right) }^{3}\,{\left( p-1\right) }^{3}\,{p}^{4}
$$
and this must equal zero for a root to be a double root (or higher), and thus this occurs at $p\in\{0,1,2,3,4,5\}$.
Now, we need only determine the number of roots for $p<0$, $0<p<1$, and so on. The easiest way to do this is to test $p=\frac{2n-1}2$ for $n\in\{0,1,2,3,4,5,6\}$. For $n=0$, we have one root (near -0.7). For $n=1$, we have one root (near 0.85). For $n=2$, we have three roots (near -0.25, 0.65, and 1.05). For $n=3$, we have one root (about 0.5). For $n=4$, we have three roots (near -0.05, 0.33, and 1.25). For $n=5$, we have one root (near 0.15). And for $n=6$, we have one root (near 1.75).
Of course, we could also check the integer $p$ values... but since we already know that $p\in\{0,1,2,3,4,5\}$ produces multi-roots, and thus the roots aren't distinct (as required), we need not consider those cases.
And therefore, it will never have five distinct, real solutions. The best it achieves in terms of distinct roots is three, which occurs when $1<p<2$, and when $3<p<4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1207678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Sum of series:$ 1+(1+2+4)+(4+6+9)+(9+12+16)+.....+(361+380+400)$ How to find the sum of series
$$ 1+(1+2+4)+(4+6+9)+(9+12+16)+.....+(361+380+400)$$
I have the problem in finding the nth term of the series. Please suggest me the procedure.
|
Your terms look something of the form:
$$a_0 = 1, a_1 = (1+2+4) = 1^2 + 1\cdot 2 + 2^2$$
$$a_2 = (4 + 6 + 9) = 2^2 + 2\cdot 3 + 3^2$$
$$a_{19} = (19^2 + 19\cdot 20 + 20^2)$$
$$\sum_{j=0}\big[j^2 + j(j+1) + (j+1)^2\big]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1208918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Alternative form of a Trigonometrics Expression Express $8\sin\theta \cos\theta - 6 \sin^2 \theta$ in the form $R \sin(2\theta + \alpha) + k$
Edit: I am sorry, I thought it was a somewhat interesting question. I shall let you know of the progress I have made. I first tried to rewrite it in the form $R \sin (\theta -\alpha)$ by rewriting it as $10 (\frac{8}{10} \sin \theta \cos \theta - \frac {6}{10} \sin ^2 \theta)$. For this to have a similar form to the sine subtraction formula, I noted that therefore $\frac{6}{10}= \cot \theta$, however, I am not sure of this assertion and do not know how to proceed. I tried the standard identities already, as someone indicated below.
|
$$\begin{align}8\sin(\theta)\cos(\theta)-6\sin^2(\theta)&=4\sin(2\theta)-3(1-\cos(2\theta))\\&=5\left(\frac{4}{5}\sin(2\theta)-\frac{3}{5}\cos(2\theta)\right)-3\end{align}$$
Notice that $\left(\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2=1$, therefore there is an angle $s=\arccos\left(\frac{4}{5}\right)$ such that $\cos(s)=\frac{4}{5}$ and $\sin(s)=\frac{3}{5}$.
Therefore our expression can be written as
$$5\left(\sin(2\theta)\cos(s)-\cos(2\theta)\sin(s)\right)-3=5\sin(2\theta-s)-3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1213270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Prove that for all naturals $n \ge 6$ there is a set of $n$ positive naturals, $a_1$ to $a_n$ such that $\sum_{i=1}^n \left(\frac{1}{a_i}\right)^2 =1$ I don't know how to prove this. I know that $\{2, 2, 2, 2\}$ is a set for $n = 4$, since $\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = 1$. So a set of $n$ naturals can have repeated elements.
|
Extending your observation, you can replace $a$ in any list with $2a, 2a, 2a, 2a$, to get a new valid list. Hence if it is possible for $n$ then it is possible for $n + 3$. Then you only need to prove it for $6, 7, 8$.
Another nice fact:
$$\frac{1}{4} = \frac{1}{9} + \frac{1}{9} + \frac{1}{36}$$
So if $2$ is in your list, you can replace it with $3, 3, 6$.
Starting with $(2,2,2,2)$, applying the first strategy gives $(2,2,2,4,4,4,4)$, solving $n = 7$.
Applying the second strategy gives $(2,2,2,3,3,6)$, solving $n = 6$. And applying the second strategy again gives $(2,2,3,3,3,3,6)$, solving $n = 8$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1213809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Solving a 2nd-order linear recurrence with variable coefficients
Question: I am seeking a general closed-form evaluation for the following definite integral.
For nonnegative real parameters $a,b,z\in\mathbb{R}^{+}\land a<z\le b$ and integer index $n\in\mathbb{Z}$, define the auxiliary function $f_{n}{\left(a,b;z\right)}$ via the integral representation
$$f_{n}{\left(a,b;z\right)}:=\int_{a}^{z}\frac{t^{n}}{\sqrt{\left(t-a\right)\left(b-t\right)}}\,\mathrm{d}t.\tag{1}$$
Specifically, I am requesting a closed-form that's purely in terms of elementary functions (the problem would be practically trivial if hypergeometric type functions were permitted here). We know in principle that every integral of the form $\int R{\left(x,\sqrt{px^{2}+qx+c}\right)}\,\mathrm{d}x$, where $R{\left(x,y\right)}$ is a rational function of two variables, possesses an elementary antiderivative as a consequence of Euler's substitutions and the partial fraction decomposition algorithm.
For $n\in\mathbb{N}\land n\ge2\land a,b,z\in\mathbb{R}^{+}\land a<z\le b$, we can derive a second-order linear non-homogeneous recurrence relation with variable coefficients for the terms $f_{n}{\left(a,b;z\right)}$:
$$\begin{cases}
&\small{f_{0}{\left(a,b;z\right)}=2\arcsin{\left(\sqrt{\frac{z-a}{b-a}}\right)},}\\
&\small{f_{1}{\left(a,b;z\right)}=-\sqrt{\left(z-a\right)\left(b-z\right)}+\left(a+b\right)\arcsin{\left(\sqrt{\frac{z-a}{b-a}}\right)},}\\
&\small{n\,f_{n}=\left(\frac{a+b}{2}\right)\left(2n-1\right)F_{n-1}-ab\left(n-1\right)F_{n-2}-z^{n-1}\sqrt{\left(z-a\right)\left(b-z\right)}.}\tag{2}\\
\end{cases}$$
A general solution to recurrence relation $(2)$ would in principle solve the integral in $(1)$, but my research into general methods for solving second-order recurrence relations didn't turn up anything of much help.
How would I go about evaluating either $(1)$ or $(2)$?
|
I think your recurrence may be solved in terms of a generating function
$$G(x) = \sum_{n=0}^{\infty} F_n x^n$$
so that
$$F_n = \frac1{n!}G^{(n)}(0) $$
Because all of the coefficients are no worse than linear, the recurrence relation turns into a first-order differential equation in $G$ by using the relation
$$\sum_{n=0}^{\infty} n F_n x^n = x G'(x)$$
We multiply the recurrence relation by $x^n$ and sum over $n \ge 2$ to get
$$\sum_{n=2}^{\infty} n F_n x^n - \frac12 (a+b) \sum_{n=2}^{\infty} (2 n-1) F_{n-1} x^n + a b \sum_{n=2}^{\infty} (n-1) F_{n-2} x^n = -\frac{\sqrt{(z-a)(b-z)}}{z} \sum_{n=2}^{\infty} z^n x^n $$
This becomes
$$x G'(x) - x F_1 - (a+b) x^2 G'(x) - \frac12 (a+b) x G(x)+\frac12 (a+b) x F_0+ a b x^3 G'(x)+a b x^2 G(x) = -z \sqrt{(z-a)(b-z)} \frac{x}{1-z x} $$
or
$$(a b x^2- (a+b) x + 1)G'(x) + \left ( a b x - \frac12 (a+b) \right )G(x) = \left (F_1 - \frac12 (a+b) F_0 \right) -z \sqrt{(z-a)(b-z)} \frac{x}{1-z x}$$
The integrating factor for this equation is relatively simple, so we may rewrite this equation as
$$a b \frac{d}{dx} \left [\sqrt{\left ( \frac1{a}-x\right )\left ( \frac1{b}-x\right )} G(x) \right ] = \frac{F_1 - \frac12 (a+b) F_0}{\sqrt{\left ( \frac1{a}-x\right )\left ( \frac1{b}-x\right )}} - z \sqrt{(z-a)(b-z)} \frac{x}{(1-z x)\sqrt{\left ( \frac1{a}-x\right )\left ( \frac1{b}-x\right )} } $$
Integrating, we get
$$a b \sqrt{\left ( \frac1{a}-x\right )\left ( \frac1{b}-x\right )} G(x) +C = \left (F_1 - \frac12 (a+b) F_0 \right ) \log \left(2 a b x+2 \sqrt{a b} \sqrt{1-a x} \sqrt{1-b x}-(a+b)\right) + \sqrt{z-a} \sqrt{b-z} \log \left(2 a b x+2 \sqrt{a b} \sqrt{1-a x} \sqrt{1-b x}-(a+b)\right)+\sqrt{a b} \left[\log (1-x z)-\log \left(x (2 a b+a z+b z)-2 \sqrt{z-a}\sqrt{b-z} \sqrt{1-a x}\sqrt{1-b x} +a+b-2 z\right)\right] $$
where $C$ is a constant of integration determined by $G(0)=F_0$.
While this way looks messy, it does not rely on hypergeometrics.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1215085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Antiderivative fraction I'm would like to find the antiderivative of
$$\frac{x^2}{(x-1)^5}$$
I tried without using partial fractions but I did not manage, so I started a lesson on partial fractions but I do not succeed in proving $$\frac{x²}{(x-1)^5}=\frac{1}{(x-1)^3}+\frac{2}{(x-1)^4}+\frac{1}{(x-1)^5}$$
Could you tell me how to proceed?
Thank you
|
Indeed you are correct that,
$$\frac{x^2}{(x-1)^5} = \frac{1}{(x-1)^3} + \frac{2}{(x-1)^4} + \frac{1}{(x-1)^5}$$
One can integrate each of these terms in turn. I will do the first to help. Let $u=x-1$ and $du=dx$ then,
$$ \int\frac{1}{(x-1)^3} dx = \int u^{-3} du = -\frac{1}{2}u^{-2} = -\frac{1}{2}\frac{1}{(x-1)^2} + c$$
where $c$ is the constant of integration. The final answer is,
$$ -\frac{1}{2}\frac{1}{(x-1)^2} -\frac{2}{3}\frac{1}{(x-1)^3} -\frac{1}{4}\frac{1}{(x-1)^4}+c $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1216377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Norm of the sum of two vectors This problem has two parts.
Part a): $x$ and $y$ are vectors. If $||x|| = 7, ||y|| = 11$, what is the smallest value possible for $||x+y||$? (Note: the || || denotes the norm of a vector).
This is what I have tried so far: I put vector $x$ equal to $\begin{pmatrix} a \\ b \end{pmatrix}$ and vector $y$ equal to $\begin{pmatrix} c \\ d \end{pmatrix}$. $||x|| = 7$ would then be, after simplification, $a^2+b^2 = 49$. Similarly, for $||y|| = 11$, after simplification, $c^2+d^2 = 121$.
Then, $||x+y|| = \sqrt{(a+c)^2 + (b+d)^2}$. Expanding gives us $\sqrt{(a^2+b^2) + (c^2+d^2) + 2(ac+bd)} = \sqrt{49+121+2(ac+bd)} = \sqrt{170+2(ac+bd)}$. That is where I was stuck--any hints for the next few steps?
Part b): $x$ and $y$ are vectors (these are not the same vectors as in part a). If $||x|| = 4, ||y|| = 5, ||x+y|| = 7,$ what is $||2x-3y||$?
Using the same approach as in part a), where vector $x$ is equal to $\begin{pmatrix} a \\ b \end{pmatrix}$ and vector $y$ is equal to $\begin{pmatrix} c \\ d \end{pmatrix}$, $a^2+b^2 = 16$ and $c^2+d^2 = 25$. Similarly, for $||x+y||$, after simplification, it equals $41+2(ac+bd) = 49$ -> $ac+bd = 4$. I'm not sure what to do next after this part too. Any hints?
|
Your problem is equivalent to finding the smallest value of $\|x-y\|$, which you can obtain by the following triangle inequality
\begin{equation}
|\|x\|-\|y\|| \leq \|x-y\| \leq \|x\|+\|y\|.
\end{equation}
This will give you that the smallest value is $4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1221766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
how to factorize $x^2+10yz-2xz-2xy-3y^2-3z^2$? How to factorize
$$x^2+10yz-2xz-2xy-3y^2-3z^2$$
It is expanded and we should make them into parts and factorize each part individually.
the last answer is $$(x+y-3z)(x-3y+z)$$
but how to get it ?
|
One way to do it is
\begin{align*}
x^2+10yz-2xz-2xy-3y^2-3z^2&=(x^2+y^2+z^2-2xz-2xy+2yz)-(4y^2+4z^2-8yz)\\
&=(x-y-z)^2-(2y-2z)^2\\
&=(x+y-3z)(x-3y+z)
\end{align*}
where going from the second to the third line, we have used $a^2-b^2=(a-b)(a+b)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1222868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Find the Laurent Expansion of $f(z)$ Find the Laurent Expansion for
$$f(z)=\frac{1}{z^4+z^2}$$
about $z=0$.
I have found the partial fraction decomposition
$$f(z)=\frac{1}{z^4+z^2}=\frac{1}{z^2}-\frac{1}{2i(z-i)}+\frac{1}{2i(z+i)}.$$
Next I wanted to expand each of the three terms separately. I have
$$\frac{1}{z^2}=\frac{1}{z^2},$$
$$\frac{1}{2i(z-i)}=-\frac{1}{2z}i\sum_{n=0}^{\infty}\left(\frac{i}{z}\right)^n,\quad |z|>1$$
$$\frac{1}{2i(z+i)}=-\frac{1}{2z}i\sum_{n=0}^{\infty}\left(-\frac{i}{z}\right)^n,\quad |z
|>1.$$
Therefore, I believe that my Laurent expansion should be
$$\frac{1}{z^4+z^2}=\frac{1}{z^2}+\frac{1}{2z}i\sum_{n=0}^{\infty}\left(\frac{i}{z}\right)^n-\frac{1}{2z}i\sum_{n=0}^{\infty}\left(-\frac{i}{z}\right)^n,\quad |z|>1.$$
I had a few questions, though.
1) What about the $z$ in the denominators outside the sums? What's that all about?
2) Does the same radius of convergence $|z|>1$ apply for $\frac{1}{z^2}$ as did for the other two series? What does it mean to expand about $z=0$, and yet the radius of convergence for those two expansions above are $|z|>0$?
3) Can I do anything to clean up this answer?
|
You may just write, as $z \to 0$:
$$\begin{align}f(z)=\frac{1}{z^4+z^2}&=\frac{1}{z^2(1+z^2)}\\\\&=\frac{1}{z^2}(1-z^2+z^4-z^6+z^8...)
\\\\&=\frac{1}{z^2}-1+z^2-z^4+z^6-...
\end{align}
$$ and this gives the Laurent expansion of $f$ near $z=0$, on $0<|z|<1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1225870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Find the sum of all odd numbers between two polynomials I was asked this question by someone I tutor and was stumped.
Find the sum of all odd numbers between $n^2 - 5n + 6$ and $n^2 + n$ for $n \ge 4.$
I wrote a few cases out and tried to find a pattern, but was unsuccessful.
Call polynomial 1, $p(n) = n^2 - 5n + 6,$ then $p(4)=2.$ Next, call polynomial 2, $q(n)=n^2 + n,$ then $q(4)=20.$ Then adding all the odd numbers between 2 and 20 gives the following sum:
$3+5+7+9+11+13+15+17+19= 99. \\$
$p(5)= 6$ and $q(5)=30.$ Then adding all the odd numbers between 6 and 30 gives the following sum: $7+9+11+13+15+17+19+21+23+25+27+29=216 \\$
$p(6)=12$ and $q(6)= 42.$ Then adding all the odd numbers between 12 and 42 give the following sum: $13+15+17+19+21+23+25+27+29+31+33+35+37+39+41 = 405.$
From here I do not see any apparent patters. This problem was given in a Pre-Calculus course, so clearly only elementary methods are expected by the students.
Any help or advice would be much appreciated. Thank you!!!!!
|
@Elaqqad gave a very elegant answer, so I'll try to give the "especially pre-calc accessible" answer, exploiting the fact that we have an arithmetic series.
Note $p(n) = n^2 - 5n + 6 = (n-3)(n-2)$, while $q(n) = n^2 + n = n(n+1)$, with $p(n) < q(n)$.
Now, $p(n)$ and $q(n)$ are both even, so that we're adding up odd numbers from $p(n)+1 = n^2 - 5n + 7$ to $q(n)-1 = n^2 + n - 1$.
Thus, we have $a_1 = n^2 - 5n + 7$ and $a_k = n^2 + n - 1$, where $a_k - a_1 = 6n - 8 = 2(3n - 4)$.
We know that the sequence $a_i$ of odd numbers is arithmetic with common difference $2$, and hence has the formula $a_i = a_1 + 2(i-1)$. Since we found $a_k - a_1 = 2(3n - 4)$, we know that our last term is achieved when $i - 1 = 3n - 4$, so that $i = 3n - 3$, and we have $3(n-1)$ terms.
Thus, using infamous formula
$$S_k = \frac{k(a_1 + a_k)}{2},$$
we have $$S_k = \dfrac{3(n-1)(2n^2 - 4n + 6)}{2} = \dfrac{6(n-1)(n^2 - 2n + 3)}{2} = 3(n-1)(n^2-2n + 3).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1226224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Proving $\lim_{x\to c}x^3=c^3$ for any $c\in\mathbb R$ using $\epsilon$-$\delta$ definition
$\lim_{x\to 3}x^3=c^3$ for any $c\in\mathbb R$
Let $\epsilon>0$. Then
$$|x^3-c^3|=|x-c||x^2+xc+c^2|.$$
Let $\delta=\min\{1,\epsilon/x^2+xc+c^2\}$.
Then if $0<|x-c|<\delta$ and therefore since $|x-c|<\epsilon/(x^2+xc+c^2)$,
$$|x^3-c^3|=|x-c||x^2+xc+c^2|<ε.$$
Does this make sense or are the steps done in the right way?
|
Suppose $\varepsilon>0$. The inequality
$$
|x^3-c^3|<\varepsilon
$$
becomes
$$
c^3-\varepsilon<x^3<c^3+\varepsilon
$$
or
$$
\sqrt[3]{c^3-\varepsilon}<x<\sqrt[3]{c^3+\varepsilon}
$$
or also
$$
\sqrt[3]{c^3-\varepsilon}-c<x-c<\sqrt[3]{c^3+\varepsilon}-c
$$
Take
$$
\delta=\min\{c-\sqrt[3]{c^3-\varepsilon},\sqrt[3]{c^3+\varepsilon}-c\}
$$
after noting that
$$
c-\sqrt[3]{c^3-\varepsilon}>0
$$
and
$$
\sqrt[3]{c^3+\varepsilon}-c>0
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1227691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Integer solutions to $x^{x-2}=y^{x-1}$
Find all $x,y \in \mathbb{Z}^+ $ such that $$x^{x-2}=y^{x-1}.$$
I can only find the following solutions: $x=1,2$. Are there any other solutions?
|
The only solutions are $x=1$ and $y=$ anything, and $x = 2$ and $y = 1$.
Start by rearranging the equation: $(\frac{x}{y})^{x-1} = x$.
If x = 1 then we have: $(\frac{1}{y})^0 = 1$ which is true for any $y$.
If y = 1 then we have: $x^{x-1} = x \implies x^{x-2} = 1$, which means either $x = 1$ which is covered by the case above, or $x = 2$.
If $x \not = 1$, $y \not = 1$ and $x$ is not a multiple of $y$, then $(\frac{x}{y})^{x-1} = x$ is in the form $\text{fraction}^{\text{positive integer}} = \text{positive integer}$, which is always false.
If $x \not = 1$, $y \not = 1$ and $x$ is a multiple of $y$, then we can write $x = ky$ for some $k \in \mathbb{Z}^+$. So we get $k^{ky-1} = ky \implies k^{ky-2} = y \implies (k^k)^y = k^2y$. If $k = 1$ then $y = 1$ and $x = 1$ (covered), or if $k = 2$ then $y = 1$ and $x = 2$ (covered), so any other solutions has $k > 2$. We see then that at $y = 1$, the $\text{LHS} > \text{RHS}$. You now want to show that it's true for every $y$. You can do that by doing the following:
Let $u_y = (k^k)^y$,
$v_y = k^2y$. Compute $\frac{u_y}{u_{y-1}} = k^k \geq 27$, $\frac{v_y}{v_{y-1}} = \frac{y}{y-1} = 1 + \frac{1}{y-1} \leq 2$. So $\frac{u_y}{u_{y-1}} > \frac{v_y}{v_{y-1}}$. So $u_y > v_y, \forall y$. No more solutions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1228440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
To find the value of a function at a point where it is continuous Find $f(0)$ so that the function $f(x)=\dfrac{1-\cos(1-\cos x)}{x^4}$ is continuous everywhere.
My attempt:
By applying sandwich theorem $-1 \le \cos(x) \le 1$.
$$1 \ge -\cos(x) \ge -1$$
$$2 \ge 1-\cos(x) \ge 0$$
$\cos(x)$ is a decreasing function from $0^\circ$ to $90^\circ$.
$$\cos(2) \le \cos(1-\cos(x)) \le 1$$
$$-\cos(2) \ge -\cos(1-\cos(x)) \ge -1$$
$$1-\cos(2) \ge 1-\cos(1-\cos(x)) \ge 0$$
$$\frac{(1-\cos(2))}{(x^4)} \ge \frac{(1-\cos(1-\cos(x)))}{(x^4)} \ge 0$$
Thus I got the answer 0 applying limits every where.
But the actual answer is 1/8 apparently. How?
|
It is enough to note that $\lim_{y \to 0} \frac{1-\cos y}{y^2}=\frac{1}{2}$.
Now you write
$$ \frac{1-\cos(1-\cos x)}{x^4}= \frac{1-\cos(1-\cos x)}{(1-\cos x)^2}\left( \frac{1-\cos x}{x^2}\right)^2 \to \frac{1}{2} \frac{1}{4}=\frac{1}{8}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1229724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Sum the infinite series How to solve this:
\begin{equation*}
\sum_{n=1}^{\infty }\left[ \frac{1\cdot 3\cdot 5\cdots \left( 2n-1\right) }{
2\cdot 4\cdot 6\cdots 2n}\right] ^{3}
\end{equation*}
I can make the bracket thing, $\left[ C(2n,n)/4^{n}\right] ^{3}$, but how to proceed now.
|
We have:
$$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\,x^n = \frac{1}{\sqrt{1-x}},\tag{1} $$
$$\frac{1}{4^n}\binom{2n}{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left(\sin x\right)^{2n}\,dx \tag{2}$$
$$ \sum_{n\geq 0}\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 x^n = \frac{2}{\pi}\,K(x)=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-x\sin^2\theta}}\tag{3} $$
hence by $(2)$ and $(3)$ it follows that:
$$\begin{eqnarray*}\sum_{n\geq 0}\left(\frac{1}{4^n}\binom{2n}{n}\right)^3 &=& \frac{1}{\pi^2}\int_{-\pi}^{\pi}K(\sin^2 x)\,dx\\&=&\frac{4}{\pi^2}\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{1}{\sqrt{1-\sin^2\varphi\sin^2\theta}}\,d\theta\,d\varphi\\&=&\frac{4}{\pi^2}\,K\left(\frac{1}{2}\right)^2=\color{red}{\frac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}},\tag{4}\end{eqnarray*}$$
so the original series equals $\displaystyle\color{purple}{-1+\frac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}=0.39320392968567685918424626\ldots}.$
Footnote: this is just a very special case of the identity $(6)$ for the square of the complete elliptic integral of the first kind, plus the fact that $K(1/2)$ can be computed through the reflection and multiplication formulas for the $\Gamma$ function.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1229897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
}
|
$x^3+ (5m+1)x+ 5n+1$ is irreducible over $\Bbb Z$ How to prove that the polynomial: $x^3+ (5m+1)x+ 5n+1$ is irreducible over the set of integers for any integers $m$ and $n$?
I was trying to put $x= y+p$ for some integer $p$ so that I could apply Eisenstein's criterion. But it is very tedious to find some suitable $p$.
|
If it is not, then it can be written as a product $(x^2 + ax + b)(x + c)$, where $a,b,c$ are integers.
Expanding: $x^3 + (5m + 1)x + 5n + 1 = x^3 + (a + c)x^2 + (ac + b)x + bc$.
Then $a = -c$ and this becomes: $x^3 + (5m+1)x + 5n+1 = x^3 + (b-c^2)x + bc$
Then $c^2 = b - (5m + 1)$, so $c^3 = bc - (5m + 1)c = 5n+1-(5m+1)c \equiv 1 - c \pmod 5 $
Then $c^3 + c \equiv 1 \pmod 5 $
But this can't happen:
$0^3 + 0 = 0 \equiv 0 \pmod 5 $
$1^3 + 1 = 2 \equiv 2 \pmod 5 $
$2^3 + 2 = 10 \equiv 0 \pmod 5 $
$3^3 + 3 = 30 \equiv 0 \pmod 5 $
$4^3 + 4 = 68 \equiv 3 \pmod 5 $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1231278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
$x\over(1-x)$ $y\over(1-y)$ $z\over(1-z)$ >= 8 when $x ,y ,z $ are positive proper fractions and $x+y+z = 2$ Q. Prove that $x\over(1-x)$ $y\over(1-y)$ $z\over(1-z)$ $\geq$ 8 when $x ,y ,z $ are positive proper fractions and $x+y+z = 2$
What I did: From A.M. G.M. inequality,
$(x+y+z)\over3$ $\geq$ $(xyz)^{1/3}$
or, $xyz \leq$ $8\over27$ (Putting $x+y+z = 2$ and cubing both sides) -----------1.
And, $(1-x)+(1-y)+(1-z)\over3 $$\geq [(1-x)(1-y)(1-z)]^{1/3}$
or, $1/3 \geq [(1-x)(1-y)(1-z)]^{1/3}$ (Putting $x+y+z = 2$)
or, $1\over(1-x)(1-y)(1-z)$ $\geq$ $1\over27$ ----------------------2.
But we cannot multiply 1. by 2. Since the inequality signes are different.
Am I missing something?
|
Note that $1 = (x+y+z)/2$. The inequality becomes
$$ \frac{xyz}{(y+z-x)(x+z-y)(x+y-z)} \geq 1$$
which is equivalent to
$$ xyz \geq (y+z-x)(x+z-y)(x+y-z)\ \ \ (1)$$
To see why this is true, apply AM-GM inequality three times:
$$ x = \frac{x+y-z+x+z-y}{2} \geq \sqrt{(x+y-z)(z+x-y)}$$
and the other two for $y,z$. Now multiply all three and you'll get $(1)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1231351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Maximum value of $ x^2 + y^2 $ given $4 x^4 + 9 y^4 = 64$ It is given that $4 x^4 + 9 y^4 = 64$.
Then what will be the maximum value of $x^2 + y^2$?
I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .
|
Without using Lagrange multipliers
It's equivalent to find $x$ that maximizes
$$f(x)=x^2+\frac23\sqrt{16-x^4}$$
You have
$$f'(x)=2x-\frac13\frac{4x^3}{\sqrt{16-x^4}}$$
Hence $f'(x)=0$ iff $x=0$ or
$$\frac{2x^2}{\sqrt{16-x^4}}=3$$
$$2x^2=3\sqrt{16-x^4}$$
$$4x^4=9(16-x^4)$$
$$13x^4=12^2$$
$$x=\pm2\frac{\sqrt{3}}{\sqrt[4]{13}}$$
Let $x_0=2\dfrac{\sqrt{3}}{\sqrt[4]{13}}$. Since $x$ appears only with even powers, sign is not important. And you have
$$f(0)=\frac83$$
$$f(x_0)=\frac{12}{\sqrt{13}}+\frac{2}{3}\sqrt{16-\frac{16\times 9}{13}}$$
$$=\frac{12}{\sqrt{13}}+\frac{2}{3}\sqrt{\frac{64}{13}}=\frac{12+\frac{16}{3}}{\sqrt{13}}=\frac{4\times13}{3\sqrt{13}}=\frac{4}{3}\sqrt{13}$$
And this is larger than $f(0)$ since
$$f(0)^2-f(x_0)^2=\frac{16\times4}{9}-\frac{16\times13\times3}{9}$$
Thus the max is found for $x=\pm x_0$ and
$$y^4=\frac{1}{9}\left(64-4x_0^4\right)=\frac{1}{9}\left(64-4\frac{16\times 9}{13}\right)=\frac{64}{9}\left(1-\frac{9}{13}\right)=\frac{64\times4}{9\times13}$$
Hence
$$y=\pm \frac{4}{\sqrt{3}\sqrt[4]{13}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1234320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
}
|
How do I integrate this expression? Need to find the following: $$\int \frac{\sqrt{x^2 - 4}}{x^3} dx$$
So far, after using integration by parts, I've arrived at $\frac{-\sqrt{x^2-4}}{2x^2} + \frac12 \int \frac{1}{x\sqrt{x^2-4}} dx$. How to proceed from here?
|
$$I=\int\dfrac{\sqrt{x^2-4}}{x^3}dx=\int\dfrac{\sqrt{x^2-4}}{x^4}x\ dx$$
Set $\sqrt{x^2-4}=u\implies x^2-4=u^2\iff x\ dx=u\ du$
$$I=\int\dfrac{u^2}{(u^2+4)^2} du=\int\dfrac1{(u^2+4)} du-4\int\dfrac{du}{(u^2+4)^2}$$
For the second part set $u=2\tan\theta$
Alternatively, set $x=2\sec\theta$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1239010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction:
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove that given inequality holds for $n+1$ case by using the given $n$ case. Ok, so this is what I've got by now:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{(n+1)^2}\overset{?}{>}1$$
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1$$
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
From the $n$ case we know that:
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}$$
So we basically have to prove that:
$$1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}\frac{1}{n}$$
Since
$$n^2+1<n^2+2<\dots<n^2+2n+1<2n^2+n$$
for $n\geq2$, then also:
$$\frac{1}{n^2+1}>\frac{1}{2n^2+n}$$
$$\frac{1}{n^2+2}>\frac{1}{2n^2+n}$$
.
.
$$\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}$$
so then we have:
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}+\dots+\frac{1}{2n^2+n}=(2n+1)\frac{1}{2n^2+n}=\frac{1}{n}$$
|
In case you want a different approach, here is one:
The red rectangles represent the sum $S_n=\frac 1n+...+\frac 1{n^2}$. Above it is for $n=2$, but the principle is general. From the figure it follows immediately that
$$
\int_n^{n^2}\frac 1x\ dx=\ln(n)<\frac 1n+\frac1{n+1}+...+\frac1{n^2}
$$
and for $n>e$ this implies $S_n>1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1239518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
}
|
Pairs of integers $(a,b)$ such that $\frac{1}{6} =\frac{1}{a} + \frac{1}{b}$ How many pairs of integers are there $(a,b)$ with $a \leq b$ such that $$\frac{1}{6} =\frac{1}{a} + \frac{1}{b}$$
My attempt:
Clearing fractions we get $$ab = 6(a+b)$$
$$ \Longrightarrow \frac{ab}{a+b} = 6$$
Which means that $ab$ is a multiple of $a+b$ since the answer is an integer. Also $ab \equiv 0\pmod{a+b}$.
Any ideas or a hint to help me continue? Thanks.
|
HINT : $$ab=6(a+b)\iff (a-6)(b-6)=36.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1241683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
}
|
Solve the system $ x \lfloor y \rfloor = 7 $ and $ y \lfloor x \rfloor = 8 $.
Solve the following system for $ x,y \in \mathbb{R} $: \begin{align} x \lfloor y \rfloor & = 7, \\ y \lfloor x \rfloor & = 8. \end{align}
It could be reducing to one variable, but it is not that simple.
Another idea is substitution, but it didnt lead me anywhere.
This was taken from a competition, and it’s tricky.
From what I noticed, systems with two or more variables involving "greatest integer" function are rarely seen, probably too difficult to analyze...
This was obtained with the help of Wolfram Alpha: (I swapped places of 7 and 8 by mistake, so it is not exactly the same system; anyway, the answer should not rely on software tools, of course)
|
Note: Here we use the following characterization of the floor function
\begin{align*}
\lfloor x\rfloor = p\qquad \text{ with } \qquad p \leq x < p+1, \quad p\in \mathbb Z
\end{align*}
We use this relationship to transform the system of equations into a system of inequalities which can be easily analysed.
Let $x,y \in \mathbb{R}$. The system of equations
\begin{align*}
x \lfloor y \rfloor & = 7 \tag{1}\\
y \lfloor x \rfloor & = 8
\end{align*}
is equivalent with
\begin{align*}
x p & = 7 \qquad \text{ and }\qquad p\leq y < p+1, \quad p\in \mathbb{Z}\\
y q & = 8 \qquad \text{ and }\qquad q\leq x < q+1, \quad q\in \mathbb{Z}
\end{align*}
We substitute $x=\frac{7}{p}$ and $y=\frac{8}{q}$ and get the following system of inequalities with $p,q\in \mathbb{Z}$
\begin{align*}
p\leq \frac{8}{q} < p+1\tag{2}\\
q\leq \frac{7}{p} < q+1\tag{3}
\end{align*}
Note that the two inequalities are equivalent with the two equations in (1).
We are now looking for integer solutions of these inequalities. First we see from (1) that $x\neq 0$ and $y\neq 0$. Then we deduce from (2) that either both $p$ and $q$ are greater zero or both are less than zero.
First step: $p,q>0$
Since $q > 0$ we conclude from (3) that $1\leq p \leq 7$ and we obtain
\begin{align*}
p=1 \qquad \rightarrow \qquad q\leq &\frac{7}{1}<q+1 \qquad \rightarrow \qquad q=7\\
p=2 \qquad \rightarrow \qquad q\leq &\frac{7}{2}<q+1 \qquad \rightarrow \qquad q=3\\
p=3 \qquad \rightarrow \qquad q\leq &\frac{7}{3}<q+1 \qquad \rightarrow \qquad q=2\\
p=4 \qquad \rightarrow \qquad q\leq &\frac{7}{4}<q+1 \qquad \rightarrow \qquad q=1\\
&\ldots\\
p=7 \qquad \rightarrow \qquad q\leq &\frac{7}{7}<q+1 \qquad \rightarrow \qquad q=1\\
\end{align*}
Checking these pairs $(p,q)$ in inequality (2) we see, that only $p=1$ and $p=2$ are valid.
\begin{align*}
(p,q)=(1,7) \qquad \rightarrow \qquad 1\leq \frac{8}{7} < 2\tag{4}\\
(p,q)=(2,3) \qquad \rightarrow \qquad 2\leq \frac{8}{3} < 3\\
\end{align*}
Since $x=\frac{7}{p}$ and $y=\frac{8}{q}$ we get from (4) the solutions with positive $x,y$
\begin{align*}
A=\left\{\left(7,\frac{8}{7}\right),\left(\frac{7}{2},\frac{8}{3}\right)\right\}
\end{align*}
$$ $$
Second step: $p,q<0$
First we note, that if $p<-7$ it follows from (3) that $q=-1$. Putting $q=-1$ into (2) we obtain $p\leq -8<p+1$ and $p=-8$ follows.
We conclude that in the case $p<-7$ only $(p,q)=(-8,-1)$ is a valid solution.
Next we consider $-7 \leq p \leq -1$ similar to above
\begin{align*}
p=-1 \qquad \rightarrow \qquad q\leq &-\frac{7}{1}<q+1 \qquad \rightarrow \qquad q=-7\\
p=-2 \qquad \rightarrow \qquad q\leq &-\frac{7}{2}<q+1 \qquad \rightarrow \qquad q=-4\\
p=-3 \qquad \rightarrow \qquad q\leq &-\frac{7}{3}<q+1 \qquad \rightarrow \qquad q=-3\\
p=-4 \qquad \rightarrow \qquad q\leq &-\frac{7}{4}<q+1 \qquad \rightarrow \qquad q=-2\\
&\ldots\\
p=-7 \qquad \rightarrow \qquad q\leq &-\frac{7}{7}<q+1 \qquad \rightarrow \qquad q=-1\\
\end{align*}
Checking these seven pairs $(p,q)$ in (2) shows that the following three pairs are valid
\begin{align*}
(p,q)=(-2,-4)\qquad\Rightarrow\qquad -2\leq- \frac{8}{4}<-1\\
(p,q)=(-3,-3)\qquad\Rightarrow\qquad -3\leq- \frac{8}{3}<-2\\
(p,q)=(-4,-2)\qquad\Rightarrow\qquad -4\leq- \frac{8}{2}<-3\\
\end{align*}
Together with $(-8,-1)$ we derive from $x=\frac{7}{p}$ and $y=\frac{8}{q}$ the following $(x,y)$ pairs with negative solutions
\begin{align*}
B=\left\{\left(-\frac{7}{2},-2\right),\left(-\frac{7}{3},-\frac{8}{3}\right),\left(-\frac{7}{4},-4\right),\left(-\frac{7}{8},-8\right)\right\}
\end{align*}
Note that $A\cup B$ coincides with the solution of @AlexR.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1242157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
}
|
Show the closed form of the sum $\sum_{i=0}^{n-1} i x^i$ Can anybody help me to show that when $x\neq 1$
$$\large \sum_{i=0}^{n-1} i\, x^i = \frac{1-n\, x^{n-1}+(n-1)\,x^n}{(1-x)^2}$$
|
The following chain of identities gives the closed form without using calculus.
$$\sum_{i=1}^n i x^i= \sum_{k=0}^{n-1} x^{n-k} \sum_{i=0}^k x^i=$$ $$=\sum_{k=0}^{n-1} x^{n-k} \frac{x^{k+1}-1}{x-1}=\frac{1}{x-1}\sum_{k=0}^{n-1}(x^{n+1}-x^{n-k})=$$ $$=\frac{x^n}{x-1}\sum_{k=0}^{n-1}(x-x^{-k})=\frac{x^n}{x-1}(n x-\frac{\frac{1}{x^n}-1}{\frac{1}{x}-1})=$$ $$=\frac{1}{x-1}(n x^{n+1}-\frac{x-x^{n+1}}{1-x})=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}$$
In response to comments: I have indexed the sum in a slightly different manner than in the question asked. To derive the formula asked for, just notice that $$\sum_{i=0}^{n-1} i x^i=\sum_{i=1}^n i x^i -n x^n=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}-nx^n=\frac{nx^{n+2}-(n+1)x^{n+1}+x-n x^n(x-1)^2}{(x-1)^2}=\frac{nx^{n+2}-(n+1)x^{n+1}+x-nx^{n+2}+2nx^{n+1}-nx^n}{(x-1)^2}=\frac{(n-1)x^{n+1}-nx^{n}+x}{(x-1)^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1244297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
}
|
Find basis so matrix is in Jordan Canonical Form $M = \left(\begin{array}{ccc}0 & -3 & -2 \\1 & 3 & 1 \\1 & 2 & 3\end{array}\right)$
I want to find a basis $B$ such that matrix for $M$ w.r.t $B$ has the form:
$\left(\begin{array}{ccc}2 & 1 & 0 \\0 & 2 & 1 \\0 & 0 & 2\end{array}\right)$
The eigenvalues for $M = 2,2,2$.
$M-2I = \left(\begin{array}{ccc}-2 & -3 & -2 \\1 & 1 & 1 \\1 & 2 & 1\end{array}\right)$.
I tried to find eigenvectors. This is what I came up with.
$(M-2I)v_1 = 0$, then $v_1$ = $(-1,0,1)^t$
And so forth but I either get a wrong answer or it's not invertible.
How do I find the generalized eigenvectors so M is in Jordan Normal Form?
|
Since $(M-2I)^2=\begin{pmatrix}-1&-1&-1\\0&0&0\\1&1&1\end{pmatrix}$, we can choose any vector $v_3=\begin{pmatrix}a\\b\\c\end{pmatrix}$
which is not in the nullspace of $(M-2I)^2$, so $a+b+c\ne0$, and
then let $v_2=(M-2I)v_3$ and $v_1=(M-2I)v_2$
to get a basis $B=\{ v_1, v_2, v_3\}$ with respect to which M will be in this form.
For example, $v_3=\begin{pmatrix}0\\0\\1\end{pmatrix}$ gives $v_2=\begin{pmatrix}-2\\1\\1\end{pmatrix}$ and $v_1=\begin{pmatrix}-1\\0\\1\end{pmatrix}$;
$\;\;\;\;$similarly, $v_3=\begin{pmatrix}-2\\1\\0\end{pmatrix}$ gives $v_2=\begin{pmatrix}1\\-1\\0\end{pmatrix}$ and $v_1=\begin{pmatrix}1\\0\\-1\end{pmatrix}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1245740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Finding maxima of a 3-variable function. Let $x,y,z$ be positive real number satisfy $x+y+z=3$
Find the maximum value of
$P=\frac{2}{3+xy+yz+zx}+(\frac{xyz}{(x+1)(y+1)(z+1)})^\frac{1}{3}$
|
We'll prove that $$\frac{2}{3+xy+xz+yz}+\sqrt[3]{\frac{xyz}{(1+x)(1+y)(1+z)}}\leq\frac{5}{6}$$
for all positives $x$, $y$ and $z$ such that $x+y+z=3$.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Since by Holder $\sqrt[3]{(1+x)(1+y)(1+z)}\geq1+\sqrt[3]{xyz}=1+w$, it's enough to prove that
$$\frac{2u^2}{3u^2+3v^2}+\frac{w}{1+w}\leq\frac{5}{6},$$
which says that it's enough to prove the last inequality for a minimal value of $v^2$.
But $x$, $y$ and $z$ are positive roots of the equation $(X-x)(X-y)(X-z)=0$ or
$3v^2X=-X^3+3uX^2+w^3$, which says that a line $Y=3v^2X$
and a graph of $Y=-X^3+3uX^2+w^3$ have three common points.
Now we see that $v^2$ gets a minimal value, when a line $Y=3v^2X$
is a tangent line to the graph of $Y=-X^3+3uX^2+w^3$,
which happens for equality case of two variables.
Since the last inequality is homogeneous, we can assume $w^3=1$.
Id est, it's enough to prove the last inequality for $y=x$ and $z=\frac{1}{x^2}$, which gives
$$(x-1)^2(38x^7+7x^6-24x^5+32x^4+10x^3-12x^2+2x+1)\geq0,$$
which is obviously true.
For $x=y=z$ we get $P=\frac{5}{6}$, which says that the answer is $\frac{5}{6}$.
Done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1246298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Integration: $\int\frac{1}{(x^2+x+1)^{1/2}} dx$ Find the value of $$\int\frac{1}{(x^2+x+1)^{1/2}} dx$$
Anyone can provide hint on how to integrate this, and how you know what method to use? (I mean, is there any general guideline to follow for solving?)
Thank you!
|
$\bf{My\; Solution:: }$ Let $\displaystyle I = \int\frac{1}{\sqrt{x^2+x+1}}dx = \int\frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}dx$
Now Let $\displaystyle \left(x+\frac{1}{2}\right) = t\;,$ Then $dx = dt$ and $\displaystyle \frac{\sqrt{3}}{2} = a>0$
So Integral $$\displaystyle I = \int\frac{1}{\sqrt{t^2+a^2}}dt$$
Now Let $t^2+a^2 = y^2\;,$ Then $$\displaystyle tdt = ydy \Rightarrow \frac{dt}{y} = \frac{dy}{t}=\frac{d(t+y)}{(t+y)}$$
So Integral $$\displaystyle I = \int\frac{dt}{y} = \int\frac{d(t+y)}{(t+y)}= \ln \left|t+y\right|+\mathcal{C}$$
So $$\displaystyle I = \int\frac{1}{\sqrt{x^2+x+1}}dx = \ln \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x+1}\right|+\mathcal{C}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1246836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
Method to integrate $\cos^4(x)$ Here my attempts for integrating $\cos^4(x)$ in few methods.
1st method.
$(\cos^2x)^2=(\frac{1}{2})^2(1+\cos2x)^2$
$=\frac{1}{4}(1+2\cos2x+\cos^22x)=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\cos^22x)$
$=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\frac{1}{2}(1+\cos4x))$
$=\frac{3}{8}+\frac{1}{2}\cos2x+\frac{1}{8}\cos4x$
So, $\int\cos^4xdx=\int(\cos^2x)^2dx$
$=\int\frac{3}{8}dx+\frac{1}{2}\int\cos2xdx+\frac{1}{8}\int\cos4xdx$
$=\frac{1}{32}(12x+8\sin2x+\sin4x)+c$
2nd method.
By using reduction formula, $\cos^4xdx=\frac{1}{4}\sin x\cos^3x+\frac{3}{4}\int\cos^2xdx$
$=\frac{1}{4}\sin x\cos^3x+\frac{3}{4}\int(\frac{1}{2}\cos2x+\frac{1}{2})dx$
$=\frac{1}{4}\sin x\cos^3x+\frac{3}{8}\int\cos2x+\frac{3}{8}\int1dx$
$=\frac{1}{4}\sin x\cos^3x+\frac{3}{16}\sin 2x+\frac{3}{8}x$
$=\frac{1}{4}\sin x \cos^3x+\frac{3}{8}\sin x \cos x+\frac{3}{8}x$
3rd method, By using De Moivre's formula.
4th method
$\int \cos^4xdx=\int (\frac{e^{ix}+e^{-ix}}{2})^4dx$
$=\frac{1}{16}\int (e^{4ix}+4e^{2ix}+6+4e^{-2ix}+e^{-4ix})dx$
$=\frac{1}{16}\int (e^{4ix}+e^{-4ix}+4(e^{2ix}+e^{-2ix})+6)dx$
$=\frac{1}{16}(\frac{1}{4i}(e^{4ix}-e^{-4ix})+\frac{2}{i}(e^{2ix}-e^{-2ix})+6x)+c$
$\frac{1}{16}(\frac{1}{2}(\frac{e^{4ix}-e^{-4ix}}{2i})+4(\frac{e^{2ix}-e^{-2ix}}{2i})+6x)+c$
$\frac{1}{16}(\frac{1}{2}\sin 4x+4\sin 2x+6x)+c$
$\frac{1}{8}(\frac{1}{4}\sin 4x+2\sin 2x+3x)+c$
I want to ask is there more method to integrate it? Thanks.
|
Let us first establish a general formula, with an obvious notation:
$$(C^cS^s)'=-cC^{c-1}S^{s+1}+sC^{c+1}S^{s-1}=-cC^{c-1}(1-C^2)S^{s-1}+sC^{c+1}S^{s-1}\\
=-cC^{c-1}S^{s-1}+(s+c)C^{c+1}S^{s-1}.$$
Then, rewriting and shifting the indexes,
$$\int C^\color{green}cS^s\,dx=\frac1{s+c}C^{c-1}S^{s+1}+\frac{c-1}{s+c}\int C^{\color{green}{c-2}}S^s\,dx.$$
Then
$$C^2\to\frac12CS+\frac12x,$$
$$C^4\to\frac14C^3S+\frac34\left(\frac12CS+\frac12x\right).$$
Deriving $C^{c-1}S$ directly leads to the recurrence
$$\int C^\color{green}c\,dx=\frac1cC^{c-1}S+\frac{c-1}c\int C^{\color{green}{c-2}}\,dx.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1247913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Is this a good proof of the binomial identity? Prove that the binomial identity ${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$ is true using the following expression: $(1+x)^n = (1+x)(1+x)^{n−1}$ and the binomial theorem.
What I have:
We know from the binomial theorem that:
$$(x+1)^n= {n\choose 0} x^0 + {n\choose 1} x^1+\cdots+{n\choose k} x^k+\cdots+ {n\choose n-1} x^{n-1}+ {n\choose n} x^n.$$
By using the property that $(1+x)^{n} = (1+x)(1+x)^{n−1}$, we can take out a factor of $x$ to get:
$$x(x+1)^{n-1}={n-1\choose 0} x^0+\cdots+{n-1\choose k-1} x^{k-1}+\cdots+{n+1\choose n-1} x^{n-1}$$
If we then divide by a factor of $x$ we get:
$$1(x+1)^{n-1}={n-1\choose 0} x^0+\cdots +{n-1\choose k-1} x^k+\cdots+{n-1\choose n-1} x^{n-1}$$
Substituting $(x+1)^{n-1}={n-1\choose 0} x^0 +\cdots+{n-1\choose k-1} x^k+\cdots+{n-1\choose n-1} x^{n-1}$ into the equation $x(x+1)^{n-1}= {n-1\choose 0} x^0+\cdots+{n-1\choose k-1} x^{k-1}+\cdots+ {n+1\choose n-1}x^{n-1}$, the equation can be reduced to:
$${n\choose k}x^k = {n-1\choose k-1} x^{k-1}x+{n-1 \choose k}x^k 1$$
Dividing by $x$ we get: ${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$
Is this a good proof? What can I do to improve it? Is there a better way to solve this problem?
|
A much shorter way would be calculating the coefficient of $x^k$ independently from each side of $$(1+x)^n=(1+x)(1+x)^{n-1}$$
Using the binomial theorem, the coefficient of $x^k$ on the LHS is $\binom{n}{k}$.
For the right hand side, $x^k$ can be formed by two ways -
*
*Selecting $1$ from $(1+x)$ and $x^k$ from $(1+x)^{n-1}$. This gives coefficient $\binom{n-1}{k}$.
*Selecting $x$ from $(1+x)$ and $x^{k-1}$ from $(1+x)^{n-1}$. This gives coefficient $\binom{n-1}{k-1}$.
So the total coefficient of $x^k$ on the RHS is $\binom{n-1}{k} + \binom{n-1}{k-1}$.
Since LHS=RHS, $$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1248296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Is $\sqrt{x^2} = (\sqrt x)^2$? Take $x=4$ for example:
$ \sqrt{(4)^2} = \sqrt{16} = \pm4 $
However:
$ (\sqrt{4})^2 = \sqrt{\pm2}$
Case 1: $ (-2)^2 = 4$
Case 2: $ (2)^2 = 4$
Solution : $+4$
How come the $ \sqrt{(4)^2} = \pm4$; but $ (\sqrt{4})^2 = 4 $ ?
What is missing?
|
In general:
$$\sqrt{x^2}=|x|\implies \sqrt{x^2}=\left(\sqrt{|x|}\right)^2=|x|$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1248801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
}
|
Number of times $2^k$ appears in factorial
For what $n$ does: $2^n | 19!18!...1!$?
I checked how many times $2^1$ appears:
It appears in, $2!, 3!, 4!... 19!$ meaning, $2^{18}$
I checked how many times $2^2 = 4$ appears:
It appears in, $4!, 5!, 6!, ..., 19!$ meaning, $4^{16} = 2^{32}$
I checked how many times $2^3 = 8$ appears:
It appears in, $8!, 9!, ..., 19!$ meaning, $8^{12} = 2^{36}$
I checked how many times $2^{4} = 16$ appears:
It appears in, $16!, 17!, 18!, 19!$ meaning, $16^{4} = 2^{16}$
In all,
$$2^{18} \cdot 2^{32} \cdot 2^{36} \cdot 2^{16} = 2^{102}$$
But that is the wrong answer, its supposed to be $2^{150}$?
|
It might be more helpful to do this recursively.
Let $T(n) = \prod_{k=1}^n k!$.
We will use the notation: $2^{r} \| m$ to mean that $2^r$ is the largest power of $2$ that divides $m$.
Then we have $2 \| 2! = T(2)$. We also know that $2 \| 3!$, so $2^2 \| T(3) = 3! T(2)$. Continuing:
$$2^3 \| 4!$$
$$2^3 \| 5!$$
$$2^4 \| 6!$$
$$2^4 \| 7!$$
$$2^7 \| 8!$$
$$2^7 \| 9!$$
$$2^8 \| 10!$$
$$2^8 \| 11!$$
$$2^{10} \| 12!$$
$$2^{10} \| 13!$$
$$2^{11} \| 14!$$
$$2^{11} \| 15!$$
$$2^{15} \| 16!$$
$$2^{15} \| 17!$$
$$2^{16} \| 18!$$
$$2^{16} \| 19!$$
If we take the sum of all of those powers, $$2\cdot 1 + 2\cdot 3 + 2\cdot 4 + 2\cdot 7 + 2\cdot8 + 2 \cdot 10 + 2 \cdot 11 + 2 \cdot 15 + 2 \cdot 16$$
$$=2(1+3+4+7+8+10+11+15+16) = 2(75) = 150.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1250112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Evaluate $ \int_{0}^{\pi/4}\left(\cos 2x \right)^{11/2}\cdot \cos x\;dx $
Evaluate the definite integral
$$
I=\int_{0}^{\pi/4}(\cos 2x)^{11/2}\cdot \cos x\;dx
$$
My Attempt:
$$
I = \int \left(1-2\sin^2 x\right)^{11/2}\cdot \cos x\;dx
$$
Now, substitute $\sin x=t$ with $\cos x \,dx = dt$:
$$
I = \int (1-2t^2)^{11/2}\;dt
$$
How can I complete the solution from this point?
|
Here's an efficient method using special functions.
Applying the substitution
$$t = 2 \sin^2 x, \qquad ds = 4 \sin x \cos x \,dx $$
gives
$$\int_0^{\pi / 4} (\cos 2x)^{(2 m - 1) / 2} \cos x \,dx = \frac{1}{2 \sqrt{2}} \int_0^1 t^{-1/2} (1 - t)^{(2 m - 1) / 2} dt .$$
By the definition of and then a standard identity for the Beta function $\textrm{B}$,
$$\int_0^1 t^{-1/2} (1 - t)^{(2 m - 1) / 2} dt = \textrm{B} \left(\tfrac{1}{2}, m + \tfrac{1}{2}\right) = \frac{\Gamma(\tfrac{1}{2}) \Gamma(m + \frac{1}{2})}{\Gamma(m + 1)} .$$
where $\Gamma$ is the Gamma function, but we can easily handle all three of the factors $\Gamma(\,\cdot\,)$:
*
*By definition, $\Gamma\left(\frac{1}{2}\right) = \int_0^{\infty} q^{-1/2} e^{-q} \,dq ,$ and the substitution $q = r^2, dq = 2 r \, dr$ gives (by symmetry, one-half of) a Gaussian integral, yielding $$\Gamma\left(\tfrac{1}{2}\right) = \sqrt{\pi} .$$
*Applying the identity $\Gamma(z) = z \Gamma(z - 1)$ a total of $m$ times gives $$\Gamma\left(m + \tfrac{1}{2}\right) = \frac{(2 m)!}{4^m m!} \Gamma\left(\tfrac{1}{2}\right) = \frac{(2 m)!}{4^m m!} \sqrt\pi .$$
*Since $m$ is an integer, $\Gamma(m + 1) = m !$.
Assembling these pieces (and remembering the factor of $\frac{1}{2 \sqrt{2}}$ in the second display equation) gives
$$\int_0^{\pi / 4} (\cos 2x)^{(2 m - 1) / 2} \cos x \,dx = \frac{1}{2 \sqrt{2}} \frac{\left(\sqrt\pi\right)\left(\tfrac{(2 m)!}{4^m m!} \sqrt\pi\right)}{m!} = \frac{\pi}{2 \sqrt{2}} \cdot \frac{1}{4^m} {{2 m} \choose m} .$$
In our case, $\frac{2 m - 1}{2} = \frac{11}{2}$, so $m = 6$, and
$$\color{#bf0000}{\boxed{\int_0^{\pi / 4} (\cos 2x)^{11 / 2} \cos x \,dx = \frac{\pi}{2 \sqrt{2}} \cdot \frac{1}{4^6} {{12} \choose 6} = \frac{231 \pi}{2048 \sqrt{2}}}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1251186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
}
|
Proving that $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{13}{24}$ by induction. Where am I going wrong? I have to prove that
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{2n}>\frac{13}{24}$$ for every positive integer $n$.
After I check the special cases $n=1,2$, I have to prove that the given inequality holds for the $n+1$ case by using the $n$ case. So, I have to prove:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}$$
what is equivavlent to
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}+\frac{1}{n}$$
By using the $n$ case, I know that
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}>\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2}$$
so I basically have to prove that
$$\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}+\frac{1}{n}$$
what is equivavlent to
$$\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{1}{n}$$
$$n(2n+2)+n(2n+1)\overset{?}{>}(2n+1)(2n+2)$$
$$4n^2+3n\overset{?}{>}4n^2+6n+2$$
$$-3n\overset{?}{>}2$$
$$n\overset{?}{<}-\frac{2}{3}$$
which can not be because $n$ is a positive integer. What am I doing wrong? Tnx!
|
As I see it, here's what might be confusing you:
If you let
$$A=\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2},$$
and
$$B=\frac{13}{24}+\frac{1}{n},$$
then you are trying to prove that $A>B$. Also, if you let
$$C=\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2},$$
then you have by assumption that $A>C$. Thus if you can prove that $C>B$, as you write yourself, then your are finished, since this will imply that $A>B$.
The problem with this however is that $C\not > B$. This is why you get a contradiction, when you try to prove it, and there's nothing wrong with this. We can have both $A>B$ and $A>C$ without $C>B$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1252576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
}
|
Evaluate the integral $\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x$
Evaluate the integral $$\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x.$$
If I transfer $(x^2 + 4)^5$ to the numerator, how do I integrate?
|
Put $u = x^2+4 \implies du = 2x\,dx\implies x\,dx = \frac 12 du$.
$$ \begin{align}\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x &= \frac 12\int u^{-5} \,du\\ & = \frac 12 \left(\frac {u^{-4}}{-4}\right) +c \\ &= -\frac{1}{8u^4} + c \\ &= -\frac {1}{8(x^2 + 4)^4} + c\end{align} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1252652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
I do not have use derivatives, so I decided to apply the definition of being a strictly increasing function, which should be:
If we pick 2 numbers $a$ and $b$ from the domain of a function $f$, where $a < b$, then $f(a) < f(b)$.
Now (if that is a correct definition), I have tried to apply it to my case:
Let $a$, $b \in [-1, 1]$ and $a < b$. We want to show that $f(a) < f(b)$ or that $f(b) - f(a) > 0$.
We know that $x^2 + 3x + 2 = (x + 2)(x + 1)$, thus we have that $f(a) = (a + 2)(a + 1)$ and $f(b) = (b + 2)(b + 1)$ , therefore we need to show that:
$$(b + 2)(b + 1) - (a + 2)(a + 1) > 0$$
We can see that $(b + 2)$ and $(a + 2)$ will always be positive, and that $(b + 2) > (a + 2)$ , since $b > a$ (by assumption).
Since $b > a$, we know that $b > -1$ (otherwise $a \geq b$), thus $(b + 1) > 0$ (so we know that $(b + 2)(b + 1) > 0$ . We also have at most $(a + 1) = 0$, and thus we have that $(a + 2)(a + 1) \ge 0$.
So, here is the proof that $f(b) - f(a) > 0$ .
Am I correct? If yes, what can I improve it? If not, where are the erros and possible solutions?
|
It looks right. Another possible approach is the following one - since:
$$ f(x)=x^2+3x+2 = \left(x+\frac{3}{2}\right)^2-\frac{1}{4} $$
we have that $f(x)$ attains its minimum in $x=-\frac{3}{2}$ (the abscissa of the vertex) and for every $r>-\frac{1}{4}$ the equation
$$ f(x)=r $$
has two solutions, symmetric with respect to $x=-\frac{3}{2}$.
It follows that $f(x)$ is increasing over $\left(-\frac{3}{2},+\infty\right)$ and $[-1,1]$ is just a subset of it.
Still another way - for every $\varepsilon>0$ we have:
$$ f(x+\varepsilon)-f(x) = \varepsilon(2x+\varepsilon)+3\varepsilon>\varepsilon(2x+3) $$
hence $x\in[-1,1]$ implies $(2x+3)>0$, then $f(x+\varepsilon)>f(x)$, i.e. $f$ increasing.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1253120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
}
|
Using implicit differentiation with a fraction How do I solve this? What steps? I have been beating my head into the wall all evening.
$$ x^2 + y^2 = \frac{x}{y} + 4 $$
|
We have:
$$ \frac{d(x^2)}{dx} + \frac{d(y^2)}{dy}\cdot \frac{dy}{dx} = \frac{d( \frac{x}{y})}{dx} + \frac{d(4)}{dx}$$
$$2x+2y\frac{dy}{dx} = \frac{y-x \frac{dy}{dx} }{ y^2} +0$$
Can you finish?
Okay, to finish it up:
$$2y\frac{dy}{dx} - \frac{y-x \frac{dy}{dx} }{ y^2} = -2x $$
Multiplying both sides by $y^2$
$$2y^3\frac{dy}{dx}-(y-x\frac{dy}{dx})=-2xy^2$$
$$2y^3\frac{dy}{dx}+x\frac{dy}{dx}=-2xy^2 +y$$
$$\frac{dy}{dx} \cdot (2y^3+x) =-2xy^2 +y$$
$$\frac{dy}{dx} = \frac{-2xy^2 +y}{2y^3+x}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1255368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
What is wrong with this limit reasoning $$\lim_{x \to 0}\left(\dfrac 1 {\sin x}-\dfrac1 x\right) $$
I solved this problem using the fact that near $x=0$ , $\sin x \cong x$),$(\sin x=x+O(x^2)$ therefore subtracting them results to $0$ which I did verify using other techniques, However, the same reasoning leads to $0$ in solving
$$\lim_{x \to 0}\left(\dfrac 1 {\log (x+1)}-\dfrac1 x\right) $$
while the correct result is $1/2$ although $\log(x+1) \cong x$ when $x$ is near zero$(\log (x+1)=x+O(x^2)$.What is getting wrong here?
|
We will be using the following three Taylor expansions: $\sin{x}=x-\frac{x^3}{6}+O(x^5)$, $\log(1+x)=x-\frac{x^2}{2}+O(x^3)$ and $\frac{1}{1-x}=1+x+O(x^2)$
$$\begin{align}\frac{1}{\sin{x}}&=\frac{1}{x-\frac{x^3}{6}+O(x^5)}\\&=\frac{1}{x(1-x^2/6+O(x^4))}\\&=\frac{1}{x}(1+\frac{x^2}{6}+O(x^4))\\&=\frac{1}{x}+\frac{x}{6}+O(x^3)\end{align}$$
And the limit in this case is zero. While
$$\begin{align}\frac{1}{\log(1+x)}&=\frac{1}{x-\frac{x^2}{2}+O(x^3)}\\&=\frac{1}{x(1-x/2+O(x^2))}\\&=\frac{1}{x}(1+\frac{x}{2}+O(x^2))\\&=\frac{1}{x}+\frac{1}{2}+O(x)\end{align}$$
And the limit is $\frac{1}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1257382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
}
|
Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.
I was thinking of using induction, but wasn't really sure how to do it.
|
Here's a proof based on the fact that among any five consecutive numbers, one of them must be divisible by $5$.
Suppose that $n^2(n^2+1)(n^2-1)$ is not divisible by $5$. Then either $n^2+2$ or $n^2-2$ is divisible by $5$. Let's write this as $n^2+2\sigma=5k$, where $|\sigma|=1$.
But if $n^2(n^2+1)(n^2-1)$ is not divisible by $5$, then neither is $n(n^2-1)=n(n+1)(n-1)$, and therefore $5$ divides either $n-2$ or $n+2$. Let's write this as $n+2\tau=5h$ with $|\tau|=1$.
Combining these cleverly, we can conclude $n^2-\sigma\tau n=5(k-\sigma\tau h)$, which implies $5$ divides $n(n-\sigma\tau)$, where $|\sigma\tau|=1$. But that implies $5$ divides $n(n-1)(n+1)$, which is a contradiction. Therefore $5$ must divide $n^2(n^2+1)(n^2-1)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1257632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 8
}
|
What is the volume inside $S$, which is the surface given by the level set $\{ (x,y,z): x^2 + xy + y^2 + z^2 =1 \}$? The solution given uses a linear algebraic argument that doesn't seem very instructive -- and may not even be correct, I think.
We notice from the equation, that the surface is a quadratic form, level set = 1.
$\bullet$ The solution rewrites it as $x^TAx$,
$\bullet$ Finds a symmetric matrix $A$ that gives the resulting quadratic form,
$\bullet$ Computes the eigenvalues of $A$,
And then (here's where the explanation doesn't really follow, I think)
$\bullet$ A change of variables is made so that the surface becomes an ellipsoid,
$\bullet$ Finally using the volume of an ellipsoid formula, the final answer is given.
Is there another / better way of finding the volume of $S$?
Thanks,
|
First look at the elliptical cross-section in xy-plane:
$\{ (x,y): x^2 + xy + y^2 = 1 \}$
It is clear that this is symmetric in $x$ and $y$ so write
$x^2 + xy + y^2 = a(x+y)^2 + b(x-y)^2$ so by equating coefficients $a=\frac{3}{4}, b=\frac{1}{4}$
Therefore $\boxed{\dfrac{3}{4}(x+y)^2 + \dfrac{1}{4}(x-y)^2 = 1}$
Changing to new coordinate system: $X = \dfrac{1}{\sqrt{2}}\Big(x+y\Big), Y = \dfrac{1}{\sqrt{2}}\Big(x-y\Big)$ (where the $\frac{1}{\sqrt{2}}$ terms ensure that the new axes are of unit length), and writing in standard form:
$\dfrac{X^2}{({\sqrt{\frac{2}{3}}})^2} + \dfrac{Y^2}{({\sqrt{2}})^2} = 1$, or for the full level set $\boxed{\dfrac{X^2}{({\sqrt{\frac{2}{3}}})^2} + \dfrac{Y^2}{({\sqrt{2}})^2} + \dfrac{z^2}{1^2} = 1}$
Hence, the level set is an ellipsoid with semi-axes of lengths $R_X = \sqrt{\frac{2}{3}}$ , $R_Y = \sqrt{2}$ and $R_Z = 1$.
So the volume is $\dfrac{4}{3}\pi \cdot R_X \cdot R_Y \cdot R_Z = \dfrac{8 \pi}{3 \sqrt{3}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1258463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
On $a^4 + b^4 = c^4 + d^4 = e^5$. Let $a, b, c, d, e$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5$. Show that $ac + bd$ is a composite number.
|
Assume that $p=ac+bd$ is prime. Without loss of generality, we may assume that $a>c,d$ from which follows that $b<c,d$. Also, $a,b,c,d<p$, and so must all be relatively prime with $p$.
Computing modulo $p$, we have $ac \equiv -bd$ which leads to $(ac)^4\equiv(bd)^4$. Exploiting $a^4+b^4=c^4+d^4$ we get
$$
(a^4+b^4)c^4=(ac)^4+(bc)^4\equiv(db)^4+(bc)^4=(c^4+d^4)b^4=(a^4+b^4)b^4
$$
which means that either $p$ divides $a^4+b^4=c^4+d^4$, or $b^4\equiv c^4$.
We can exclude the alternative $p|a^4+b^4$ since that would make $p|e$, which in turn would require $e\ge p=ac+bd>a+b$ which leads to $e^5>a^4+b^4$. Note that this is the only use for the equality with $e^5$.
So now we know that $b^4\equiv c^4$, which means that
$$
p|c^4-b^4=(c^2+b^2)(c+b)(c-b).
$$
Again, $p>c+b$, so we must have $p|c^2+b^2$. However,
$c^2+b^2<ac+bd=p$ as $c<a$ and $b<d$, which makes $p|c^2+b^2$ impossible.
Thus, no such $a,b,c,d$ exist for which $p$ is prime.
This was written late at night, so I hope I didn't make any glaring mistakes.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1258623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
}
|
Demonstrating Strassen's method using domain transformation: $T(n)=7T(n/2)+an^2$ I want to solve the recurrence for Strassen's method (for multiplying square matrices) with domain transformation and get a closed form. The equation is given below:
$T(n)=b$, at $n=2$
$T(n)=7T(n/2)+an^2$, at $n>2$
I know that since I have to cancel out n/2, I should set $n=2^k$. I'll call that $G(k)$.
$G(k)=7G(k-1)+a(2^k)^2$
Now I'm not sure what to do. The $(2^k)^2$ term is quite confusing. It prevents telescoping or characteristic equations as far as I can tell. It doesn't seem to fit any of the simple forms for generating functions.
|
At each step, we replace $n$ with $\tfrac{n}{2}$ and scale each term by $7$ as follows:
\begin{align*}
T(n) &= 7T\left(\frac{n}{2}\right) + an^2 \\
7T\left( \frac{n}{2} \right) &= 7^2T\left(\frac{n}{2^2}\right) + \frac{7}{4}an^2 \\
7^2T\left( \frac{n}{2^2} \right) &= 7^3T\left(\frac{n}{2^3}\right) + \left(\frac{7}{4} \right)^2 an^2 \\
7^3T\left( \frac{n}{2^3} \right) &= 7^4T\left(\frac{n}{2^4}\right) + \left(\frac{7}{4} \right)^3 an^2 \\
&~~\, \vdots \\
7^{\log_2 n - 2}T(4) &= 7^{\log_2 n - 1}T(2) + \left(\frac{7}{4} \right)^{\log_2 n - 2} an^2 \\
7^{\log_2 n - 1}T(2) &= 7^{\log_2 n - 1}b \\
\end{align*}
Summing everything together, we observe that the terms telescope, yielding:
\begin{align*}
T(n)
&= 7^{\log_2 n - 1}b + \left(1 + \tfrac{7}{4} + (\tfrac{7}{4})^2 + \cdots + (\tfrac{7}{4})^{\log_2 n - 2} \right)an^2 \\
&= 7^{\log_2 n - 1}b + \frac{(\tfrac{7}{4})^{\log_2 n - 1} - 1}{\tfrac{7}{4} - 1}an^2 \\
&= 7^{\log_2 n - 1}b + ((\tfrac{7}{4})^{\log_2 n - 1} - 1)\tfrac{3}{4}an^2 \\
&= \tfrac{1}{7}b7^{\log_2 n} + (\tfrac{4}{7}(\tfrac{7}{4})^{\log_2 n} - 1)\tfrac{3}{4}an^2 \\
&= \tfrac{1}{7}bn^{\log_2 7} + (\tfrac{4}{7}n^{\log_2 7/4} - 1)\tfrac{3}{4}an^2 \\
&= \tfrac{1}{7}bn^{\log_2 7} + \tfrac{3}{7}an^{2 + \log_2 7/4} - \tfrac{3}{4}an^2 \\
&\in \Theta(n^{\log_2 7})
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1258729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Compare $\sum_{k=1}^n \left\lfloor \frac{k}{\varphi}\right\rfloor$ ... Given two integer sequences
\begin{equation*}
\displaystyle A_n=\sum_{k=1}^n \left\lfloor \frac{k}{\varphi}\right\rfloor ,
\end{equation*}
\begin{equation*}
B_n=\left\lfloor\dfrac{n^2}{2\varphi}\right\rfloor-\left\lfloor \dfrac{n}{2\varphi^2}\right\rfloor
\end{equation*}
here: $\quad\varphi=\dfrac{1+\sqrt{5}}{2}\quad$ (golden ratio)
Prove that: $|A_n-B_n|\leq 1.$
I realized that the difference between $A_n$ and $B_n$ is very small but the failure in finding an exact formula for $A_n$
Could you help me?
|
Note that $$A_n = \sum_{k=1}^n\lfloor k(\phi - 1) \rfloor = S_n - \frac{1}{2}n(n+1),$$ where $S_n = \sum_{k=1}^n \lfloor k\phi \rfloor$ (appeared as A054347). In The Golden String, Zeckendorf Representations, and the Sum of a Series by Martin Griffiths, he showed that $S_n = \lfloor \frac{n(n+1)\phi}{2} - \frac{n}{2} \rfloor + \delta_1$, where $\delta_1 \in \{0,1\}$. Now, we have $$A_n = \left\lfloor \frac{n(n+1)\phi}{2} - \frac{n}{2} \right\rfloor + \delta_1 -\frac{1}{2}n(n+1) = \left\lfloor \frac{n^2}{2\phi} - \frac{n}{2\phi^2} \right\rfloor + \delta_1 = B_n - \delta_2 + \delta_1,$$ where $\delta_2\in\{0,1\}$, and so $|A_n - B_n|\le 1$.
Efficient Recursive Formula for $A_n$
The following picture gives you a way to express $A_n$ in a different way. (Indeed, it is just Fubini's theorem.)
Namely, the area of the staircase (red and orange region) is exactly $A_n$.
However, one can calculate the area of the staircase in a different way. Define $m = \lfloor n / \phi\rfloor$. The area of the staircase between $y = k-1$ and $y = k$ (where $k = 1, \dots, m$) is $n - \lfloor k\phi \rfloor$.
From this, we can get a formula for $A_m$, $$A_n = mn - \sum_{k=1}^m \lfloor k\phi \rfloor = mn - S_m.$$
On the other hand, $$A_m = \sum_{k=1}^m \lfloor k(\phi-1) \rfloor = S_m - \frac{1}{2}m(m+1).$$ Therefore, we obtain $$A_n + A_m = mn - \frac{1}{2}m(m+1).$$ Now one can compute $A_n$ in $O(\log n)$ time.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1259142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
}
|
Find conditions for $\left(\frac{-3}{p}\right)=1$ $\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right)^{\frac{p-1}{2}}$
$=\begin{cases}1,\:p\equiv 1\pmod{4}\text{ or }\begin{cases}p\equiv 3\pmod{4}\\p\equiv 2\pmod {3}\end{cases}\\-1,\:\text{otherwise}\end{cases}$
$p$ is an odd prime. Something is wrong here, since $\left(\frac{-3}{5}\right)=-1$, yet $5\equiv 1\pmod {4}$. What's wrong?
|
The case $p=2,3$ is trivial, so suppose that $p\neq3$ is an odd prime with remainder $r$, when divided by three therefore by some observations we have
$$
\left(\frac{-3}p\right)=\left(\frac{-1}p\right)\left(\frac{3}p\right)=(-1)^{(\frac{p-1}2)}(-1)^{(\frac{p-1}2)(\frac{3-1}2)}\left(\frac p3\right)=(-1)^{p-1}\left(\frac p3\right)=\left(\frac p3\right)=\left(\frac r3\right)
$$
Which means that
$$
\begin {cases}
\left(\frac{-3}p\right)=1, & \text{iff $q\equiv1\pmod3$}\\
\left(\frac{-3}p\right)=-1, & \text{iff $q\equiv -1\pmod3$}
\end{cases}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1260317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
What is the ratio of $\frac{XY}{SQ}$ In the picture,$PQRS$ is a prallelogram. $PS$ is parallel to $ZX$ and $\frac{PZ}{ZQ}=\frac{2}{3}$. Then$\frac{XY}{SQ}$ equals:
The answer is $\frac{9}{40}$
Help me with the idea to solve this problem.
Thanks in advance.
|
This can be brute-forced using analytical geometry.
We start with
$$
\frac{2}{3}
= \frac{\lVert PZ\rVert}{\lVert ZQ\rVert}
= \frac{\lVert SX\rVert}{\lVert XQ\rVert}
$$
We want the ratio
$$
\frac{\lVert XY\rVert}{\lVert SQ\rVert}
$$
and use the linear interpolation formula
$$
C = (1-t) A + t B \quad (t \in [0,1])
$$
for a point $C$ on a line between points $A$ and $B$ to observe
$$
X = (1-3/5) Q + 3/5 S = 2/5 Q + 3/5 S \\
Y = (1-\lambda) Q + \lambda S = (1-\mu) Z + \mu R
$$
If we knew the coordinates for all points this should be solvable. For this we choose lengths for a particular instance of all scalings, choose $Q$ as origin (so it drops from the above formulas) and have positive coordinates to the left (to not worry about negative values too much):
$$
Q = (0,0), \quad P = (5, 0), \quad Z = (3, 0)
$$
This gives
$$
X = (x, y), \quad S = ((5/3)x, (5/3)y)
$$
For a parallelogram we have
$$
R = S + PQ = ((5/3)x, (5/3)y) + (-5, 0) = (5/3)x - 5, (5/3) y)
$$
The equation for $Y$ reduces to:
$$
Y = \lambda S = \lambda ((5/3)x, (5/3)y) = (1 - \mu) (3,0) + \mu ((5/3)x-5,(5/3)y)
$$
which gives for the components:
$$
(5/3)x\lambda = 3 - 3\mu + (5/3) x \mu - 5\mu, \quad
(5/3)y\lambda = (5/3) y \mu
$$
thus
$$
\lambda = \mu, \quad
(5/3)x\lambda = 3 - 3\lambda + (5/3) x\lambda -5 \lambda \iff
\lambda = 3/8
$$
and
$$
Y = \lambda S = ((5/8)x,(5/8)y)
$$
which gives
$$
XY = X - Y = ((3/8)x, (3/8)y) \\
SQ = S - Q = ((5/3)x, (5/3)y)
$$
So
$$
\lVert XY\rVert^2 = (3/8)^2 (x^2 + y^2) \\
\lVert SQ \rVert^2 = ((5/3)x, (5/3)y) = (5/3)^2 (x^2 + y^2)
$$
and
$$
\frac{\lVert XY\rVert}{\lVert SQ\rVert} = \frac{3/8}{5/3} = \frac{9}{40}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1260822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Solving an infimum-inequality How does one evaluate $$\inf \{c: c^2 \geq \frac{a^2x^2+b^2y^2}{x^2+y^2} \} $$ for every $a,b,x,y \in \mathbb R$ ?
Hints and tips appreciated
Edit: I tried to write the inequality in polar coordinates, which gives me:
$$ \frac{a^2r^2\cos^2(\phi) + b^2r^2\sin^2(\phi)}{r^2\cos^2(\phi) + r^2\sin^2(\phi) } \leq c^2$$ thus
$a^2\cos^2(\phi) + b^2\sin^2(\phi) \leq c^2 $
Since $\sin$ and $\cos$ are bounded by $1$, I receive $a^2+b^2 \leq c^2 $
Is this correct $( \color{blue}{really}$ the infimum and not just some upper bound?)
Edit: Without loss of generality, let $a^2 \leq b^2 $. Then I can find a lower upper bound by estimating $a^2\cos^2(\phi) + b^2\sin^2(\phi) \leq c^2$ to $b^2 \leq c^2$, thus $c = \max\{ \lvert a\rvert\, \lvert b \rvert \}$
|
It's $c = \sqrt{\max(a^2, b^2)}$. Proof:
*
*If $c = \sqrt{\max(a^2, b^2)}$, then $a^2 \leqslant c^2$ and $b^2 \leqslant c^2$, so
$$a^2 \cos^2( \phi ) + b^2 \sin^2( \phi ) \leqslant c^2 \cos^2( \phi ) + c^2 \sin^2( \phi ) = c^2.$$
*If $\gamma^2 \geqslant a^2 \cos^2( \phi ) + b^2 \sin^2( \phi )$ for all $x, y \in \mathbb{R}$, then letting $\phi = 0$ and $\phi = \frac{\pi}{2}$ we get $\gamma^2 \geqslant a^2$ and $\gamma^2 \geqslant b^2$, so $\gamma \geqslant c$.
Note: I assume you want $c$ positive, because very large negative $c$ will always satisfy the inequality.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1261704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to determine if $\sum_{n=1}^{\infty}\left (\frac{n^2-5n+1}{n^2-4n+2}\right)^{n^2}$ converges or diverges $$\sum_{n=1}^{\infty} \left(\frac{n^2-5n+1}{n^2-4n+2}\right)^{n^2}$$
Using root test seems not a efficient way since I got stuck without knowing what to do next
$$\lim_{n\to\infty}\left(\frac{n^2-5n+1}{n^2-4n+2}\right)^{n}$$
Using ratio test is too complicated
$$\left(\frac{(n+1)^2-5(n+1)+1}{(n+1)^2-4(n+1)+2}\right)^{(n+1)^2}=\left(\frac{n^2-3n-3}{n^2-2n-1}\right)^{n^2+2n+1}$$
$$\to\left(\frac{(n+1)^2-5(n+1)+1}{(n+1)^2-4(n+1)+2}*\frac{n^2-4n+2}{n^2-5n+1}\right)^{n^2}*\left(\frac{n^2-3n-3}{n^2-2n-1}\right)^{2n}*\frac{n^2-3n-3}{n^2-2n-1}$$
|
The general term decays like $\left(1-\frac{1}{n}\right)^{n^2}\approx e^{-n}$ hence the series is convergent.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1263557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
What is the connection between the discriminant of a quadratic and the distance formula? The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$
If we look at the quadratic formula
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
we can see that it specifies two points at a certain offset from the center
$$-\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
This means that $\frac{\sqrt{b^2 - 4ac}}{2a}$ is the (horizontal) distance from the vertex to the roots. If I squint, the two squared-ish quantities being subtracted under a square root sign reminds me of the Euclidean distance formula
$$\sqrt{(x_0 - x_1)^2 + (y_0 - y_1)^2}$$
Is there a connection? If not, is there any intuitive or geometric reason why $\frac{\sqrt{b^2 - 4ac}}{2a}$ should be the horizontal distance from the vertex to the roots?
|
Consider the image below.
The roots of $f(x)=ax^2+bx+c$ will lie on the circle in the complex plane whose center is $\dfrac{-b}{2a}.$ In the case where both roots are real numbers, they will lie on the real line at $R_1$ and $R_2.$ In the case where the roots are purely imaginary, they will lie on the vertical axis at $ir_1$ and $-ir_1.$ Now we want to find $d$ the radius of the circle.
To do this, note that the product of the roots of $f(x)=ax^2+bx+c$ is $\dfrac ca.$ So in the case that the roots are purely imaginary, we have
$$\frac ca=(ir_1)(-ir_1)=r_1^2,$$
And thus $r_1=\sqrt{\dfrac ca}.$ Note these roots will only occur when $b=0$ and $ac>0.$
Now to find the distance $d,$ we use the Pythagorean Theorem to get
$$d^2=\left(\frac{-b}{2a}\right)^2+\left(\pm i\sqrt{\frac{c}{a}}\right)^2=\frac{b^2}{4a^2}+\frac{-c}{a}=\frac{b^2-4ac}{4a^2},$$
and thus
$$d=\frac{\pm\sqrt{b^2-4ac}}{2a}.$$
In the case where the roots are both real, for $b^2>4ac$ we have
$$R_1=\frac{-b-\sqrt{b^2-4ac}}{2a},\quad\text{and}\quad R_2=\frac{-b+\sqrt{b^2-4ac}}{2a}.$$
One may struggle or take issue about the above argument, since $d$ is not a distance as traditionally defined in the complex plane. See related question:
Pythagorean Theorem for imaginary numbers
It should be noted that my notion of distance is being done on the hyperbolic space. In both the Euclidean plane and the Hyperbolic plane, the distance between points on the real line is the same. But once we venture off the real line we get different versions of the Pythagorean Theorem.
In Euclidean space, we have $\cos^2t+\sin^2t=1\Longrightarrow a^2+b^2=c^2.$
In Hyperbolic space, we have $\cosh^2t{\color{red}-}\sinh^2t=1\Longrightarrow a^2-b^2=c^2\equiv a^2+(\pm ib)^2=c^2.$
You may ask:
Why do we need to represent our roots on the hyperbolic plane?
Answer:
Because it allows us to use imaginary numbers as lengths in the Euclidean plane!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1264091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 8,
"answer_id": 2
}
|
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.