Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Rational bounds for $\tan^{-1} x$ I want help with this question.
Show that for all $x>0$, $$ \frac{x}{1+x^2}<\tan^{-1}x<x.$$
Thank you.
|
$$ \frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2} < \frac{d}{dx} x =1$$
and $\tan^{-1}(0)=0$ so that the last inequality is proved.
$$ \frac{d}{dx} \frac{x}{x^2+1} = \frac{(x^2+1) - x(2x) }{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2} < \frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2} $$
and $\frac{x}{1+x^2}(0)=0=\tan^{-1}(0)$ so that the first inequality is proved.
Another way :
$$\frac{x}{x^2+1}< t\frac{1}{t^2+1}|_{t=0}^{t=x} -
\int_0^x t\frac{-2t}{(t^2+1)^2} = \int_0^x \frac{1}{t^2+1}\ dt =\tan^{-1}x < \int_0^x \frac{1}{0+1}\ dt=x$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/588930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
}
|
Difference of squares Determine if $5^{36} - 1$ is divisible by $13$ using the difference of squares. I tried splitting it up by difference of squares a couple of times but can't seem to get to a point where I can determine divisibility.
|
We have $5^{36}-1=(5^{18}+1)(5^{18}-1)$. Also, $5^{18}+1=(5^6)^3+1=(5^6+1)((5^6)^2-5^6+1)$, and $5^6+1=(5^2)^3+1=(5^2+1)((5^2)^2-5^2+1)$. We are done, because $5^2+1=26=2\cdot13$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/591558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
eigenvalues of integral matrices Is it possible that a $3$-by-$3$ matrix with integer values and determinant 1 has a real eigenvalue with algebraic multiplicity 2, that is not equal to $\pm 1$?
Doing some elementary computations one can rephrase the question as follows. Do there exist integers $k$ and $m$ such that $a=\frac{1}{3}(k\pm \sqrt{k^2-3m})$ and $b=\frac{1}{3}(k\mp 2\sqrt{k^2-3m})$ are real numbers and $a^2b=1$ but $a,b \neq \pm 1$?
|
As in Ewan's proof, write
$$x = \frac{2 a^3+1}{a^2} \quad y = \frac{a^3+2}{a}.$$
Notice that
$$a = \frac{xy-9}{2 (x^2-3y)}. \quad (\ast)$$
So, if $x$ and $y$ are integers then either $a$ is rational or else $xy=9$ and $x^2=3y$. The latter case implies $x (x^2/3) = 9$, so $x=y=3$ and $a=1$, which we already ruled out.
So we focus on the case where $a$ is rational. Write $a=p/q$ in lowest terms. So
$$x = \frac{2p^3+q^3}{p^2 q} \quad y=\frac{p^3+2 q^3}{p q^2}.$$
From the formula for $x$, any prime that divides $p$ also divides $q$; from the formula $y$, any prime that divides $q$ also divides $p$. Since $p/q$ is in lowest terms, we get $a = \pm 1$.
So, where did $(\ast)$ come from? Since the map $a \mapsto \left( \frac{2a^3+1}{a^2}, \frac{a^3+2}{a} \right)$ is generically injective, the subfield of $k(a)$ generated by $(2a^3+1)/a^2$ and $(a^3+2)/a$ should be all of $k(a)$. So such a formula should exist, and I just needed to find it. To find it, I used that $a$ is double root of $t^3-x t^2+y t - 1$, so it is also a root of $3 t^2 - 2x t + y$. Therefore, $a$ is a root of $(t^3-x t^2+y t - 1) - (t/3 - x/9) (3 t^2 - 2x t + y) = (2y/3 - 2x^2/9) t +(xy/9 -1)$. I solved this equation for $t$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/592477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
}
|
Logarithm / exponential equation, not sure what to make of this, (simple) Solve for $a:(2 \log_a x)(3 \log_{x^2} 4) = 3$
No idea how to approach this problem other than moving the 2 and the 3 into an exponent..
|
This answer is full of seemingly useless derivations, but would help you in the future problems.
There are three interesting results of the base-change property:
First property: $\log _a b = \dfrac{1}{\log _b a }$
Proof. $\log_a b = \dfrac{\log b}{\log a}$ and $\dfrac{1}{\log_b a} = (\log _b a)^{-1} = \left(\dfrac{\log a}{\log b}\right)^{-1} = \dfrac{\log b }{\log a} $
Second property: $\log _{a^m}{b^n} = \dfrac{n}{m}\log _a b$
Proof. As you may be aware of, $\log x^n = n \log x$.
Hence,$$\log _{a^m}b^n = n \log_{a^ m}b = n \cdot\dfrac{1}{\log _b a^m}= n \cdot \dfrac{1}{m \log_b a} = \dfrac{n}{m} \cdot \dfrac{1}{\log _b a} = \dfrac{n}{m}\log_a b$$
Third property: $a^{\log _x b} = b^{\log _x a}$
Proof. $a^{\frac{\log_a b}{\log_ a x}} = \left(a^{\log _a b}\right)^{\frac{1}{\log _a x}} = b^{\frac{1}{\log_a x}}= b^{\log_x a}$
As for this problem,
$(2 \log_a x)(3 \log_{x^2} 4) = 3$
$\Rightarrow \log _a x^2 \cdot 3\log_{x^2} 4 = 3$
$\Rightarrow 3\dfrac{\log_{x^2} 4}{\log_{x^2 }a} = 3 $
$\Rightarrow \dfrac{\log_{x^2} 4}{\log_{x^2} a} = 1$
$\Rightarrow \log_a 4 = 1$
$\Rightarrow a^1 = 4$
$\Rightarrow a = 4$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/593134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
}
|
calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\cos ^3 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^2 x+\cos ^2 x-\sin x \cos x\right)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)}\;dx\\
&= \frac{1}{3}\int \left(\frac{2}{\left(\sin x+\cos x\right)}+\frac{\left(\sin x+\cos x \right)}{\left(1-\sin x\cos x\right)}\right)\;dx\\
&= \frac{2}{3}\int\frac{1}{\sin x+\cos x}\;dx + \frac{1}{3}\int\frac{\sin x+\cos x}{1-\sin x\cos x}\;dx
\end{align}
$$
Using the identities
$$
\sin x = \frac{2\tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}},\;\cos x = \frac{1-\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}
$$
we can transform the integral to
$$I = \frac{1}{3}\int\frac{\left(\tan \frac{x}{2}\right)^{'}}{1-\tan^2 \frac{x}{2}+2\tan \frac{x}{2}}\;dx+\frac{2}{3}\int\frac{\left(\sin x- \cos x\right)^{'}}{1+(\sin x-\cos x)^2}\;dx
$$
The integral is easy to calculate from here.
My Attempt for $(2)$:
$$
\begin{align}
J &= \int\frac{1}{\sin^5 x+\cos ^5 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^4 x -\sin^3 x\cos x+\sin^2 x\cos^2 x-\sin x\cos^3 x+\cos^4 x\right)}\;dx\\
&= \int\frac{1}{(\sin x+\cos x)(1-2\sin^2 x\cos^2 x-\sin x\cos x+\sin^2 x\cos^2 x)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x-\left(\sin x\cos x\right)^2\right)}\;dx
\end{align}
$$
How can I solve $(2)$ from this point?
|
Given $$\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx$$
First we will simplify $$\sin^5 x+\cos^5 x = \left(\sin^2 x+\cos^2 x\right)\cdot \left(\sin^3 x+\cos^3 x\right) - \sin ^2x\cdot \cos ^2x\left(\sin x+\cos x\right)$$
$$\displaystyle \sin^5 x+\cos^5 x= (\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)$$
So Integral is $$\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx $$
$$\displaystyle = \int\frac{1}{(\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$
$$\displaystyle = \int \frac{(\sin x+\cos x)}{(\sin x+\cos x)^2\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$
$$\displaystyle = \int \frac{(\sin x+\cos x)}{(1+\sin 2x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$$
Let $$(\sin x-\cos x) = t\;,$$ Then $$(\cos +\sin x)dx = dt$$ and $$(1-\sin 2x) = t^2\Rightarrow (1+\sin 2x) = (2-t^2)$$
So Integral Convert into $$\displaystyle = 4\int\frac{1}{(2-t^2)\cdot(5-t^4)}dt = 4\int\frac{1}{(t^2-2)\cdot (t^2-\sqrt{5})\cdot (t^2+\sqrt{5})}dt$$
Now Using partial fraction, we get
$$\displaystyle = 4\int \left[\frac{1}{2-t^2}+\frac{1}{(2-\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}-t^2)}+\frac{1}{(2+\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}+t^2)}\right]dt$$
$$ = \displaystyle \sqrt{2}\ln \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\frac{1}{(2-\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \ln \left|\frac{5^{\frac{1}{4}}+t}{5^{\frac{1}{4}}-t}\right|+\frac{2}{(2+\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \tan^{-1}\left(\frac{t}{5^{\frac{1}{4}}}\right)+\mathbb{C}$$
where $$t=(\sin x-\cos x)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/595038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 0
}
|
Proving $4(a^3 + b^3) \ge (a + b)^3$ and $9(a^3 + b^3 + c^3) \ge (a + b + c)^3$
Let $a$, $b$ and $c$ be positive real numbers.
$(\mathrm{i})$ Prove that $4(a^3 + b^3) \ge (a + b)^3$.
$(\mathrm{ii})$Prove that $9(a^3 + b^3 + c^3) \ge (a + b + c)^3.$
For the first one I tried expanding to get $a^3 + b^3 \ge a^2b+ab^2$ but I'm not sure how to prove it.
|
Actually both inequalities follow directly form Holder's Inequality. Obviously $q=\frac 32$ and $p = 3$ satisfy the condition $\frac 1p + \frac 1q = 1$. Then we have:
$$(1^q + 1^q)^{\frac 1q}(a^p + b^p)^{\frac 1p} \ge (a+b)$$
Substitute and we have:
$$(1^{\frac 32} + 1^{\frac 32})^{\frac 23}(a^3 + b^3)^{\frac 13} \ge (a+b)$$
Cube both sides and we have:
$$(1 + 1)^2(a^3 + b^3) \ge (a+b)^3$$
$$4(a^3 + b^3) \ge (a+b)^3$$
I'll leave the second inequality to you, but it's as easy as the first one.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/596257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Polynomials with rational zeros Find all polynomials $F(x)={a_n}{x^n}+\cdots+{a_1}x+a_0$ satisfying
*
*$a_n \neq0$;
*$(a_0, a_1, a_2, \ldots ,a_n)$ is a permutation of $(0, 1, 2 ... n)$;
*all zeros of $F(x)$ are rational.
|
The complete list of such polynomials is
$$x, \quad 2x^2+x, \quad x^2+2x, \quad 2x^3+3x^2+x, \quad x^3+3x^2+2x.$$
We now prove this. The first observation to make is that for a polynomial satisfying your hypotheses, all rational roots are non-positive, and that $0$ occurs as a root at most once and only if the constant term is $0$. The second is that by Descartes' rule of signs, the constant term of the polynomial must be zero (otherwise there are only $n-2$ sign changes in the sequence of coefficients of $F(-x)$; not enough to account for $n-1$ negative rational roots).
So now we let $g(x)=F(x)/x$. The coefficients of $g(x)$ are a permutation of the numbers $1,2,\dots,n$, and all the roots of $g$ are negative rational numbers with denominators dividing $a_n$. If $n=1$ obviously $g(x)=1$ and there is nothing to prove. From now on we assume $n>1$. We factor
$$g(x)/a_n=(x+r_1)(x+r_2) \cdots (x+r_{n-1})$$ where $r_1,r_2,\dots,r_{n-1}$ are positive rational numbers, each of the form $r_i=m_i/a_n$ for positive integers $m_i$. In particular we have $r_i \geq 1/a_n$ and hence $$r_1+r_2+\cdots+r_{n-1} \geq \frac{n-1}{a_n}.$$ Since
$$g(x)/a_n=x^{n-1}+\frac{a_{n-1}}{a_n} x^{n-2}+\cdots $$ we obtain
$$\frac{n-1}{a_n} \leq r_1+r_2+\cdots+r_{n-1}=\frac{a_{n-1}}{a_n}$$ and hence there are only two possibilities: $a_{n-1}=n-1$ or $a_{n-1}=n$.
In case $a_{n-1}=n-1$, we find that $r_i=1/a_n$ for all $i$, and hence
$$g(x)=a_n(x+1/a_n)^{n-1}.$$ If $n>2$ the constant term of this polynomial can be an integer only if $a_n=1$, so $g(x)=(x+1)^{n-1}$. But for $n>1$ this polynomial has constant term equal to its leading coefficient, contradiction. Thus $n=2$ and $g(x)=2(x+1/2)$, so that $xg(x)$ is one of the degree two polynomials in our list above.
It remains to consider the case $a_{n-1}=n$. In this case $n-2$ of the roots $r_i$ are of the form $1/a_n$ and the other one is $2/a_n$. So
$$g(x)=a_n(x+1/a_n)^{n-2}(x+2/a_n).$$ The constant term of $g(x)$ is
$$a_1=\frac{2}{a_n^{n-2}}.$$ This is an integer only if $a_n=1$ or $a_n=2$ and $n$ is $2$ or $3$. These last two cases correspond to $g(x)=2(x+1)$, contradicting our hypothesis, and $g(x)=2(x+1/2)(x+1)$, for which $x g(x)=2x^3+3x^2+x$, a polynomial in our list.
We are finally reduced to the case $a_n=1$ and
$$g(x)=(x+1)^{n-2}(x+2).$$ Now the coefficient of $x$ in $g(x)$ is $1+2(n-2)$. For $n \geq 4$ we have $$1+2(n-2)=1+2n-4 \geq 1+n > n,$$ contradiction. Thus $n \leq 3$. The polynomials $x (x+2)$ and $x(x+1)(x+2)$ appear in our list so we are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/596589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
}
|
How can I find this integral? I want to find the integral $$\int{\dfrac{\cos^2 x}{\sin^2 x + 4\sin x \cos x}} \mathrm{d} x.$$ I tried put $t=\tan \frac{x}{2}$, I got
$$\int{-\dfrac{1}{4}\dfrac{(t^2-1)^2}{t(t^2 + 1)(2t^2-t-2)} \mathrm{d} t.}$$
But, it's too difficult for me to calculate this integral.
|
HINT:
Diving the numerator & the denominator by $\sec^4x,$
$$I=\int{\dfrac{\cos^2 x}{\sin^2 x + 4\sin x \cos x}} dx=\int\frac{\sec^2xdx}{(\tan^2x+4\tan x)(\tan^2x+1)}$$
Setting $\tan x=u,$
$$I=\int\frac{du}{(u^2+4u)(u^2+1)}$$
Using Partial Fraction Decomposition
$$\frac1{u(u+4)(u^2+1)}=\frac Au+\frac B{u+4}+\frac{Cu+D}{u^2+1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/599906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
The limit $\lim_{x\to \infty}\frac{x-\frac{1}{2}\sin x}{x+\frac{1}{2}\sin x}$ How to find the value of the limit:
$$ \lim_{x\to\infty}\frac{x-\frac{1}{2}\sin x}{x+\frac{1}{2}\sin x} $$
(l'Hopital not working here, right?)
|
Numerator and denominator are both $\sim x$, because $\lim_{x \to \infty} \frac{x \pm \frac 1 2 \sin x}{x}=1$, thus your limit is also 1.
Two functions $f$ and $g$ are equivalent at some point $a$ (possibly $\infty$), and you denote $f \sim g$, if: $f(x) = (1+\varepsilon(x))g(x)$ with $\varepsilon(x) \to 0$ for $x\to a$
Or if $g(x)$ is never 0, $\frac{f(x)}{g(x)} \to 1$.
Thus from the limit $\frac{\sin x}{x} \to 0$, you get immediately that $x-\frac 1 2 \sin x \sim x$, and similarly $x+\frac 1 2 \sin x \sim x$.
And since you can divide such "equivalents" provided the limit is not 0, you get the original limit from
$$\frac{x-\frac 1 2 \sin x}{x+\frac 1 2 \sin x} \sim \frac x x \sim 1$$
You can also write, for $x \to \infty$,
$$\frac{x-\frac 1 2\sin x}{x+\frac 1 2\sin x}=\frac{1-\frac 1 2\frac{\sin x}{x}}{1-\frac 1 2\frac{\sin x}{x}} \longrightarrow 1$$
The limit follows from the fact that $x \to \infty$ and $\sin x$ is bounded, thus $\frac{\sin x}{x} \to 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/601115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Find all solutions for a system of linear equations over a given field My Problem is: to find all Solutions for the following given System of linear equations over the Field $K = \mathbb{Z}_{/7}$
The System is given with:
$$\begin{equation}
\begin{split}
x_1\quad\quad\quad + 3x_3 &= 5 \\
5x_1 + 3x_2 + 6x_3 &= 3 \\
6x_1 + 2x_2 + 5x_3 &= 6 \\
\end{split}
\end{equation}$$
My Approach was: i can see this is inhomogenous System of linear Equations. The given Field is $$\mathbb{Z}_{/7}=\{ 0,1,2,3,4,5,6\}$$ so i have to calculate in Modulo 7, i think. My task seems to be to find the general solution of this System.
I tried the Cramer's rule.
$$(A\ \ b)=\left(\begin{array}{ccc|c}
1 & 0 & 3 & 5 \\
5 & 3 & 6 & 3 \\
6 & 2 & 5 & 6
\end{array}\right)$$
Since this is not Zero:
$$\det A = \left|\begin{array}{ccc}
1 & 0 & 3 \\
5 & 3 & 6 \\
6 & 2 & 5
\end{array}\right|=-8 \neq 0$$
there must be a solution. i think.
$$x_1 = \frac{\det A_1}{\det A}=\frac{ \left|\begin{array}{ccc}
5 & 0 & 3 \\
3 & 3 & 6 \\
6 & 2 & 5
\end{array}\right| }{ \left|\begin{array}{ccc}
1 & 0 & 3 \\
5 & 3 & 6 \\
6 & 2 & 5
\end{array}\right| } =\frac{0}{-8}=0$$
$$x_2 = \frac{\det A_2}{\det A}=\frac{ \left|\begin{array}{ccc}
1 & 5 & 3 \\
5 & 3 & 6 \\
6 & 6 & 5
\end{array}\right| }{ \left|\begin{array}{ccc}
1 & 0 & 3 \\
5 & 3 & 6 \\
6 & 2 & 5
\end{array}\right| }=\frac{0}{-8}=0$$
$$x_3 = \frac{\det A_3}{\det A}=\frac{ \left|\begin{array}{ccc}
1 & 0 & 5 \\
5 & 3 & 3 \\
6 & 2 & 6
\end{array}\right| }{ \left|\begin{array}{ccc}
1 & 0 & 3 \\
5 & 3 & 6 \\
6 & 2 & 5
\end{array}\right| }=\frac{-7}{-8}=\frac{7}{8}$$
Okay now i have a solution.
$$\left(\begin{array}{c}
x_1 \\
x_2 \\
x_3
\end{array}\right) = \left(\begin{array}{c}
0 \\
0 \\
\frac{7}{8}
\end{array}\right)$$That Looks cool. But my Task is to find ALL Solutions. But the given System has as many unknowns as equations. So i learned there should be only one solution.
My conclusion is: there is either just one solution (and this one is "all") or i made one (or more) mistakes.
My question is: am i right? And in case i am wrong, what are my mistakes? How can i solve such a Task?
|
\begin{align}
(A\ \ b)=
\left(\begin{array}{ccc|c}
1 & 0 & 3 & 5 \\
5 & 3 & 6 & 3 \\
6 & 2 & 5 & 6
\end{array}\right)
&\to
\left(\begin{array}{ccc|c}
1 & 0 & 3 & 5 \\
0 & 3 & 5 & 6 \\
0 & 2 & 1 & 4
\end{array}\right)\quad R_2-5R1, R_3-6R_1
\\
&\to
\left(\begin{array}{ccc|c}
1 & 0 & 3 & 5 \\
0 & 1 & 4 & 2 \\
0 & 2 & 1 & 4
\end{array}\right)\quad 5R_2
\\
&\to
\left(\begin{array}{ccc|c}
1 & 0 & 3 & 5 \\
0 & 1 & 3 & 2 \\
0 & 0 & 0 & 0
\end{array}\right)\quad R_3-2R_2
\end{align}
So the solution set is
$$
\begin{pmatrix}5-3h\\2-3h\\h\end{pmatrix}\quad (h\in\mathbb{Z}/7\mathbb{Z})
$$
Let's look at the determinant, if you don't trust Gauss Elimination. I'll develop with respect to the first row:
\begin{align}
\det A&=(-1)^{1+1}\det\begin{pmatrix}3 & 6 \\ 2 & 5\end{pmatrix}
+0+(-1)^{1+3}\cdot 3\det\begin{pmatrix}5 & 3 \\ 6 & 2\end{pmatrix}\\
&=(15-12)+3(10-18)\\
&=3+3(-8)=3-24=-21=0
\end{align}
so you can't apply Cramer's rule.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/601912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Proof of inequality with ratio of odds to evens. A professor gave this to the class as a challenge. It may or may not have to do with calculus, but I'm running out of ideas.
Prove that for any $n > 1$
$$ \frac{1}{2\sqrt{n}} < \frac{1}{2} \cdot \frac{3}{4}\cdot\frac{5}{6}\cdots \frac{2n -1}{2n} < \frac{1}{\sqrt{2n}} $$
When I first looked at this I though it was simply something I could prove with the mean value theorem. Where
$$ \frac{2n -1}{2n} - \frac{1}{\sqrt{2n}} < 0 $$
for any $x > 1$.
I made some progress on that but then I realized that $ \frac{2n -1}{2n} $ is not a formula for the middle term that I could treat as a function, but rather the product of the terms up to $\frac{2n -1}{2n}$, so you can imagine my disappointment.
With this new realization and as I had already seen with the previous approach, I focused on
$$ \frac{2n -1}{2n} = 1 - \frac{1}{2n}. $$ Simple enough.
Which makes the middle expression
$$ \left(1 - \frac{1}{2(1)} \right)\cdot \left(1 - \frac{1}{2(2)} \right)\cdot \left(1 - \frac{1}{2(3)} \right)\cdots $$
But I'm not sure what to make of this product. It reminded me of the golden ratio definition and maybe of the divisor function, but I might be readding too much into it.
I also see that
$$ 2n - 1 < 2n $$
For $n > 1$. So the ratio
$$ \frac{2n -1}{2n} < 1 $$
and likely the product
$$ \left(1 - \frac{1}{2(1)} \right)\cdot \left(1 - \frac{1}{2(2)} \right)\cdot \left(1 - \frac{1}{2(3)} \right)\cdots < 1 $$
But I'm not sure where to go from there.
I'd appreciate any help.
|
Let $\displaystyle \frac1A = \frac{1}{2} \cdot \frac{3}{4}\cdot\frac{5}{6}\cdots \frac{2n -1}{2n}$. Then we have
$$\displaystyle \frac{1}{2\sqrt{n}} < \frac1A < \frac{1}{\sqrt{2n}} \iff 2\sqrt{n} > A > \sqrt{2n} \iff 4n > A^2 > 2n$$
For the lower bound, we have to show
$$\frac{2^2}{1} \cdot \frac{4^2}{3^2}\cdot \frac{6^2}{5^2} \cdots \frac{(2n)^2}{(2n-1)^2} > 2n$$
$$\iff \frac{2^2}{1 \cdot 3} \cdot \frac{4^2}{3\cdot 5}\cdot \frac{6^2}{5\cdot 7} \cdots \frac{(2n)^2}{(2n-1)\cdot(2n+1)} > \frac{2n}{2n+1}$$
$$\iff \left(\prod_{k=1}^{n-1} {\frac{(2k)^2}{(2k)^2-1}}\right)\frac{2n}{2n-1} > 1$$
As we can note every factor on the LHS is $> 1$, the inequality holds.
Similarly one can work for the upper bound to get the equivalent:
$$ 2n > \frac{2\cdot 4}{3^2} \cdot \frac{4\cdot 6}{5^2}\cdot \frac{6\cdot 8}{7^2} \cdots \frac{(2n-2)(2n)}{(2n-1)^2}(2n) $$
$$\iff 1 > \prod_{k=1}^{n-1} \frac{2k (2k+2)}{(2k+1)^2}$$
which is again evident as all factors in the RHS are $< 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/604431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Prove $1 + \tan^2\theta = \sec^2\theta$ Prove the following trigonometric identity: $$1 + \tan^2\theta = \sec^2\theta$$
I'm curious to know of the different ways of proving this depending on different characterizations of tangent and secant.
|
Assuming the First Pythagorean Trigonometric Identity,
$$\sin^2\theta + \cos^2\theta = 1$$
Dividing by $\cos^2\theta$,
$$\Rightarrow \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$$
$$\Rightarrow \left(\frac{\sin\theta}{\cos\theta}\right)^2 + \left(\frac{\cos\theta}{\cos\theta}\right)^2 = \left(\frac{1}{\cos\theta}\right)^2$$
Since $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\sec\theta = \dfrac{1}{\cos\theta}$,
$$\Rightarrow \tan^2\theta + 1 = \sec^2\theta $$
Hence Proved.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/607269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Divisibility by large powers Show that $2^{111} + 1$ divides $2^{555} + 1$ but does not divide $2^{444} + 1$.
My try:
$(2^{111}+1)^5 = 2^{555}+5*2^{444}+10*2^{333}+10*2^{222}+5*2^{111}+1$
$2^{555} + 1$ = $(2^{111}+1)^5$ - $(5*2^{444}+10*2^{333}+10*2^{222}+5*2^{111})$
= $(2^{111}+1)^5$ - ($5*2^{333}(2^{111}+1)+5*2^{222}(2^{111}+1)+5*2^{111}(2^{111}+1))$
Each term is divisible by $(2^{111} + 1)$ and hence divisible by $2^{111} + 1$.
Such a nice distribution is not possible for $2^{444} + 1$
I could answer the first part and the second part of indivisibility is not as clean as I would like. Could someone let me know how I could prove the second part.
|
Note that $2^{111}\equiv -1\pmod{2^{111}+1}$. Thus $2^{555}\equiv (-1)^5=-1\pmod{2^{111}+1}$.
Similarly, $2^{444}\equiv (-1)^4=1\pmod{2^{111}+1}$, and therefore $2^{111}+1$ divides $2^{444}-1$. It follows that $2^{111}+1$ cannot divide $2^{444}+1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/610148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Parity through Series expansion By Maclaurin series; $\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-....$, and we know that period of $\sin{x}$ is $2π$ because $\sin{(x+2π)} = \sin{x}$. But if we consider only the RHS of the above series then how we can tell that this expression (series) is also of period$=2π$?
|
Expanding on the comment I have given I first give the proof for $\sin (x + y)$ formula. First we define $$\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \cdots $$ and by differentiation of power series $(\sin x)'= \cos x, (\cos x)' = -\sin x$.
Let us define $$f(z) = 1 + z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + \cdots$$ for all complex $z$. Then by using rule for multiplication of two series we can show easily that $f(z_{1} + z_{2}) = f(z_{1})f(z_{2})$. Now it can be easily seen (by putting $z = ix$ in definition of $f(z)$) that $f(ix) = \cos x + i\sin x$ so that $f(i(x + y)) = f(ix + iy) = f(ix)f(iy)$ and therefore $$\cos(x + y) + i\sin (x + y) = (\cos x + i\sin x)(\cos y + i\sin y)$$ so that $\cos(x + y) = \cos x\cos y - \sin x\sin y$ and $\sin (x + y) = \sin x\cos y + \cos x \sin y$.
Again considering $g(x) = \cos^{2}x + \sin^{2}x$ we can show via differentiation formulas that $g'(x) = 0$ so that $g(x)$ is constant. Clearly since $\sin 0 = 0, \cos 0 = 1$ we have $g(x) = g(0) = 1$. So that proves the fundamental identity $\sin^{2}x + \cos^{2}x = 1$
Using the above formulas and $\cos(\pi/2) = 0$ (note that we have defined $\pi/2$ to be the smallest positive root of $\cos x = 0$), it can be easily shown that $\sin (x + 2\pi) = \sin x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/610217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
How to integrate $\int{ dx \over \sqrt{1 + x^2}}$ How to integrate $dx \over \sqrt{1 + x^2}$?
Answer should be $\log ( x + \sqrt{1 + x^2})$
Please help as possible...
Thank you
|
Let $\displaystyle 1+x^2=y^2\Rightarrow \sqrt{1+x^2} = y,$ and $2xdx = 2ydy$
$$\displaystyle xdx = ydy\Rightarrow \frac{dx}{y} = \frac{dy}{x} = \frac{d(x+y)}{(x+y)}$$ (Using Ratio and Proportion).
Now $$\displaystyle \int \frac{1}{\sqrt{1+x^2}}dx = \int \frac{dx}{y} = \int \frac{d(x+y)}{(x+y)} = \ln \left|x+y\right|+C$$
So $$\displaystyle \int \frac{1}{\sqrt{1+x^2}}dx = \ln \left|x+\sqrt{x^2+1}\right|+\mathbb{C}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/610733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
}
|
max and min $x y - \ln(x^2 + y^2)$ Find max and min
$$x y - \ln(x^2 + y^2) , 1/4 \le x^2 + y^2 \le 4$$
Problem: With this hairy expression as my partial derivatives, I do not know what to do.
Attempt:
|
\begin{align}
x-\frac{2y}{x^2+y^2}=0\\
y-\frac{2x}{x^2+y^2}=0
\end{align}
can be combined to, as you have done, to
\begin{align}
(x+y)\left(1-\frac2{x^2+y^2}\right)=0\\
(x-y)\left(1+\frac2{x^2+y^2}\right)=0
\end{align}
and also, since $(x,y)\ne(0,0)$, to
\begin{align}
2xy-2=0
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/613864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Need Sine form of Cotangent equation $$(b^2 - c^2)\cot A + (c^2 - a^2)\cot B + (a^2 -b^2)\cot C=0$$
I want this equation to be in the Sine form.
Please help me with steps.
Thanks a lot
|
Assuming that $a,b,c$ are the sides of a Triangle
Using Law of sines
$$T=(b^2-c^2)\cot A=4R^2(\sin^2B-\sin^2C)\frac{\cos A}{\sin A}$$
Again using
Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ or Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$
$\displaystyle\sin^2B-\sin^2C=\sin(B+C)\sin(B-C)$
But as $A+B+C=\pi$ and $\displaystyle\sin(\pi-x)=\sin x,\sin(B+C)=\sin(\pi-A)=\sin A$
$T$ becomes $\displaystyle 4R^2\sin A\sin(B-C)\frac{\cos A}{\sin A}$
As $\sin(B+C)=\sin A\ne0$ and $\cos(\pi-y)=-\cos y$
$T$ reduces to $\displaystyle 4R^2\sin(B-C)\{-\cos(B+C)\}=-2R^2\{2\sin(B-C)\cos(B+C)\}$
Using $\displaystyle2\sin B\cos A=\sin(A+B)-\sin(A-B), T=-2R^2(\underbrace{\sin2B-\sin2C}) $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/617974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
The second degree polynomial $f(x)$ satisfying $f(0) = 0$, $f(1) = 1$ and $f'(x)>0$ for all $x\in[0,1]$ The second degree polynomial $f(x)$ satisfying $f(0) = 0$, $f(1) = 1$ and $f'(x)>0$ for all $x\in[0,1]$
Then which of the following is right?
Options
(a) $ax+(1-a)x^2$, $a\in \mathbb{R}$
(b) $ax+(1-a)x^2$, $a\in (0,2)$
(c) no such polynomial exists.
My try
Let $f(x) = ax^2+bx+c$. Now let $x=0$; we get $f(0)=0=c\implies c=0$
Again let $x=1$; we get $f(1) = 1= a+b+c=a+b+0\implies a+b=1$
So $f(x)=ax^2+bx=(1-b)x^2+bx\implies f(x)=(1-b)x^2+bx$.
Now $f'(x)>0\implies 2ax+b>0$ for all $x\in [0,1]$
Now I did not understand how can I solve it.
|
So you found out that $f(x) = (1 - b)x^2 + bx$
Then $f'(x) = (2 - 2b)x + b$. This we know is a linear equation.
So, if this line has a positive gradient, then $2 - 2b > 0 \Rightarrow 1 > b$ We also want the points $f'(0)$ and $f'(1)$ to both be greater than 0.
$f'(0) > 0 \Rightarrow b > 0, f'(1) > 0 \Rightarrow 2 - b > 0 \Rightarrow 2 > b$
So for this case $b \in (0,1)$
If the line has a negative gradient, then $2 - 2b < 0 \Rightarrow 1 < b$
As before $b > 0$ and $2 > b$, and for this case $b \in (1,2)$
If the gradient is $0$, then $b = 1$.
Therefore, the answer is option (b).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/618063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
show that $\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$ show that
$$I=\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$$
I find this is Nice equalition!
My try: let
$$\sqrt[3]{1-x^7}=t\Longrightarrow x=\sqrt[7]{1-t^3}$$
so
$$dx=-\dfrac{3}{7}t^2(1-t^3)^{-\dfrac{6}{7}} \, dt$$
so
$$I=\frac{3}{7}\int_0^1 \frac{t^3}{\sqrt[7]{(1-t^3)^6}} \, dt-\int_0^1 \sqrt[7]{1-x^3} \, dx$$
By parts,we have
$$\int_0^1 \sqrt[7]{1-x^3} dx=\dfrac{3}{7}\int_0^1 \frac{x^3}{\sqrt[7]{(1-x^3)^6}} \, dt$$
so
$$I=0$$
this problem maybe have more other nice methods!Thank you
|
You can write the two parts of the integrals as follows:
$$
\int_0^1 \sqrt[3]{1-x^7} \, dx = \int_0^1 \int_0^\sqrt[3]{1-x^7} \, dy\,dx
$$
and
$$
\int_0^1 \sqrt[7]{1-y^3}) \, dy = \int_0^1 \int_0^\sqrt[7]{1-y^3} \, dx\,dy.
$$
Now you can show that the regions of both these integral are the same. Namely the sets $\{(x, y): 0 \leq x \leq 1, 0 \leq \sqrt[3]{1-x^7}\}$ and $\{(x, y): 0 \leq y \leq 1, 0 \leq x \leq \sqrt[7]{1-y^3}\}$ are the same.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/619333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
}
|
how to find out the region between two surfaces Describe the region cut out of the ball $x^2+y^2+z^2\le4$ by the elliptic cylinder $2x^2+z^2=1$ i.e the region inside the cylinder and ball
I equated $4-x^2-y^2=1-2x^2$ getting $y^2-x^2=3$. I guessed the region has to be
$\sqrt{1-2x^2}\le z \le \sqrt{4-x^2-y^2}$, $-\sqrt{y^2-3}\le x \le\sqrt{y^2-3},-\sqrt{3} \le x\le \sqrt{3} $.
I am not sure if I am on the right path. I don't understand how to get theses regions analytically since it is very difficult to see these regions graphically by hand which was the case for two dimensional objects.
Is there any fool proof method for seeing the regions??
|
If we want to describe the region in $1/4$ of all space, i.e; $$x\ge0, z\ge0$$ then it becomes:
$$\sqrt{1-2x^2}\le z \le \sqrt{4-x^2-y^2},~~ 0\leq x\leq\frac{\sqrt{2}}{2},~~-\sqrt{x^2+3}\le y \le\sqrt{x^2+3}\\ 0\le z \le \sqrt{4-x^2-y^2},~~ \frac{\sqrt{2}}{2}\leq x\leq 2,~~-\sqrt{x^2+3}\le y \le\sqrt{x^2+3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/619556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Help with derivative of $y=x^2\sin^5x+x\cos^{-5}x$
Find $y^{\prime}$ of $y=x^2\sin^5x+x\cos^{-5}x$
My try:
$\dfrac{d}{dx}(x^2\sin^5x)=x^2(-5\sin^4x)+(2x\sin^5x)$
$\dfrac{d}{dx}(x\cos^{-5}x)=x(-5\cos^{-6}x)+1(\cos^{-5}x)$
This doesn't seem right. Can you please show how to do it correctly?
The answer is $y^{\prime}=x\sin^4(x) ( 2 \sin(x)+5x\cos(x)) +(5x\tan(x)+1)\sec^5x$
|
$\dfrac{d}{dx}(x^2\sin^5x)=x^25\sin^4x \cos x+(2x\sin^5x)$
$\dfrac{d}{dx}(x\cos^{-5}x)=5x\cos^{-6}x \sin {x}+1(\cos^{-5}x)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/620006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
$3$ never divides $n^2+1$ Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.
Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.
If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.
And that is where I am stuck. I try to plug in numbers for $n$ but I want a more general form of showing that $3$ can't divide $n^2+1$ when $n$ is even.
|
$n= 0 \pmod3 \implies n^2 + 1 = 1\pmod3$,
$n = 1\pmod3 \implies n^2 + 1 = 2\pmod3$,
$n = 2\pmod3 \implies n^2 + 1 = 2\pmod3$.
So $n^2 + 1$ is not a multiple of $3$ for any $n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/620153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 4
}
|
Set of integers with a particular additive property I have a set of integers $S = \{a_1,a_2,\ldots,a_n\}$. Let $K = a_1+a_2+\ldots+a_n$.
Consider the space of all $n$-tuples whose values are taken from the set $S$.
For example $(a_1,a_2,\ldots,a_n)$ is one $n$-tuple and its sum is $K$, $(a_1,a_1,a_2,a_2,\ldots,a_k)$ is another.
I need $S$ such that the only $n$-tuple that sums to $K$ among the space of all $n$-tuples is $(a_1,a_2,\ldots,a_n)$.
An example that wont work is, if $S = \{1,4,7\}$, $K = 12 (= 1+4+7)$. $\{1,4,7\}$ and $\{4,4,4\}$ both sum to $12$. Similarly if we take $S = \{1,2,3,4,5,6,7\}$, I can think of 2 $7$-tuples that add to $28$.
I tried the set $S = \{1, 2, 4, 8, \ldots, 2^n\}$ and it seems to work. Can anyone give me an example that is polynomial instead of exponential in $n$. Would the set $S = \{1, 4, 9, 16, \ldots, n^2\}$ or $\{1, 8, 27,\ldots,n^3\}$ work? Thanks for the comments and suggestions.
|
The set of squares will not work in general. Any notrivial solution to $a^2+b^2=c^2+d^2$ allows us to drop $a^2$ and $b^2$ and instead have $c^2$ and $d^2$ twice. And such solutions are easy to find since the equation is equivalent to $(a-d)(a+d)=a^2-d^2=c^2-b^2=(c-b)(c+b)$. For example $1\cdot 15 = 3\cdot 5$, which leads to $a=8, b=1, c=4, d=7$. Indeed, $8^2+1^2=65=4^2+7^2$. There is also the smaller solution $5^2+5^2=1^2+7^2$. Thus already $S=\{1,4,9,16,25,36,49\}$ does not have the desired property.
And this reminds us of Ramanujan and Hardy's taxi: $1729 = 10^3+9^3=12^3+1^3$. So the set $S=\{1,8,27,\ldots, 12^3\}$ of cubes won't work either.
Here's why exponential growth is necessary:
Let $S=\{a_1,\ldots, a_n\}$. There are $2^n$ subsets of $\{1,\ldots, n\}$ be a "good" set. If $A,B\subseteq\{1,\ldots, n\}$ are distict subsets of the same size $k$ and with $\sum_{i\in A}a_i=\sum_{i\in B}a_i$, the set $S$ is "bad" as this allows us to replace summands coming from $A$ in $a_1+\ldots +a_n$ with summands coming from $B$. Since $\sum_{i\in A}a_i$ is between $0$ and $ka_n$ and there are $n\choose k$ subsets of size $k$, we conclude that $ka_n\ge {n\choose k}$. If we pick $k=\lfloor n/2\rfloor$, then using Stirling's approximation we see that ${n\choose k}\approx\frac{2^n\sqrt {2}}{\sqrt{\pi n}}$ and obtain
$$ a_n\gtrsim\frac{2^{n+1}\sqrt 2}{n\sqrt{\pi n}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/621756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
How prove this inequality $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0$ let $a,b,c,d,e$ are positive real numbers,show that
$$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+e}+\dfrac{d-e}{e+a}+\dfrac{e-a}{a+b}\ge 0$$
My try:
I have solved follow Four-variable inequality:
let $a,b,c,d$ are positive real numbers,show that
$$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge 0$$
poof:since
\begin{align*}
&\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}=\dfrac{a+c}{b+c}+\dfrac{b+d}{c+d}+\dfrac{c+a}{d+a}+\dfrac{d+b}{a+b}-4\\
&=(a+c)\left(\dfrac{1}{b+c}+\dfrac{1}{d+a}\right)+(b+d)\left(\dfrac{1}{c+d}+\dfrac{1}{a+b}\right)-4
\end{align*}
By Cauchy-Schwarz inequality we have
$$\dfrac{1}{b+c}+\dfrac{1}{d+a}\ge\dfrac{4}{(b+c)+(d+a)},\dfrac{1}{c+d}+\dfrac{1}{a+b}\ge\dfrac{4}{(c+d)+(a+b)}$$
so we get
$$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge\dfrac{4(a+c)}{(b+c)+(d+a)}+\dfrac{4(b+d)}{(c+d)+(a+b)}-4=0$$
Equality holds for $a=c$ and $b=d$
By done!
But for Five-variable inequality,I can't prove it.Thank you
|
For $n$ variables $x_i > 0$, (with the convention $x_{n+k}=x_k$) we need to show that
$$\sum_i \dfrac{x_i - x_{i+1}}{x_{i+1} + x_{i+2}} \geqslant 0 \iff \sum_i \left( \dfrac{x_i - x_{i+1}}{x_{i+1} + x_{i+2}} +\frac12 \right) \geqslant \frac{n}2 \iff \sum_i \dfrac{2x_i - x_{i+1}+x_{i+2}}{x_{i+1} + x_{i+2}} \geqslant n $$
By Cauchy-Schwarz we have
$$\sum_i \dfrac{2x_i - x_{i+1}+x_{i+2}}{x_{i+1} + x_{i+2}} \geqslant \dfrac{ \left(\sum_i (2x_i - x_{i+1}+x_{i+2}) \right)^2}{\sum_i \left( (x_{i+1} + x_{i+2}) (2x_i - x_{i+1}+x_{i+2} ) \right)} = \dfrac{ 4\left(\sum_i x_i \right)^2}{2\sum_i (x_i x_{i+1} + x_i x_{i+2})} $$
So it is sufficient to show that
$$ \left(\sum_i x_i \right)^2 \geqslant \dfrac{n}2 \sum_i x_i (x_{i+1}+x_{i+2}) \tag{*}$$
While the general case eludes me, this helps to answer your question and the case $n=6$.
For $n=5$ we have the SOS form:
$$\left(\sum_{i=1}^5 x_i \right)^2 - \dfrac{5}2 \sum_{i=1}^5 x_i (x_{i+1}+x_{i+2}) = \frac14 \sum_{i=1}^5 (x_i - x_{i+1})^2 + \frac14 \sum_{i=1}^5 (x_i - x_{i+2})^2 \geqslant 0$$
For equality we will need all $x_i$ to be the same.
For $n=6$, we can use $a=x_1 + x_4, b = x_2 + x_5, c = x_3+x_6$ to get:
$$\left(\sum_{i=1}^6 x_i \right)^2 - \dfrac{6}2 \sum_{i=1}^6 x_i (x_{i+1}+x_{i+2}) = (a+b+c)^2 - 3(ab+bc+ca) \geqslant 0$$
In this case for equality we need all $x_i$ for odd $i$ to be the same, and similarly all the $x_i$ for even $i$ must also be the same.
Added based on Dongryul Kim's comment - this application of Cauchy Schwarz requires an additional condition like $x_i \in [\frac1{\sqrt3}, \sqrt3]$ so that the numerators in the fraction on LHS are non-negative.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/621937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Finding difference of angles in triangle . If $\sin A+\sin B+\sin C=0$, $\cos A+\cos B+\cos C=0$, then prove that $$A-B=B-C=C-A=\dfrac{2\pi}3$$
|
Adding a new answer due to the volume of the other answer and these methods are markedly different from the other.
First of all, as coffeemath has observed $A,B,C$ can not be angles of a triangle as $\displaystyle0<A,B,C<\pi\implies$ all sine ratios will positive $\implies\sin A+\sin B+\sin C>0$
Method $1:$
From the other answer we have $\displaystyle A-B=2n\pi\pm \frac{2\pi}3\iff 3(A-B)=2\pi(3n\pm1)$ where $n$ is any integer
Similarly, $\displaystyle 3(B-C)=2\pi(3m\pm1)$ and $\displaystyle 3(C-A)=2\pi(3r\pm1)$ for integers $m,r$
Adding we get $\displaystyle 0=3(m+n+r)+(\pm1\pm1\pm1)\iff \pm1\pm1\pm1=3(-m-n-r)$
As the RHS is divisible by $3,$ the signs of all the three $1$ in the LHS must be same.
So, difference will be $\displaystyle 2a\pi+r\cdot\frac{2\pi}3$ where $r=1$ or $-1$ in each case
This can be corroborated by the following observation:
If $A_1,B_1,C_1$ is one of the solutions of the given relations, $-A_1,-B_1,-C_1$ and $-A_1,-B_1,-C_1$
[more generally $\displaystyle A_1+b\frac\pi2,B_1+b\frac\pi2, C_1+b\frac\pi2$ are also solutions (where $b$ is any integer)]
Method $2:$
We have $\displaystyle \sin A+\sin B=-\sin C$ and $\displaystyle \cos A+\cos B=-\cos C$
Using Prosthaphaeresis Formulas $$\frac{2\sin\frac{A+B}2\cos\frac{A-B}2}{2\cos\frac{A+B}2\cos\frac{A-B}2}=\frac{-\sin C}{-\cos C}$$
If $\displaystyle\cos\frac{A-B}2=0$ both $\sin C,\cos C$ have to be zero which is impossible
So, we have $\displaystyle\tan\frac{A+B}2=\tan C\implies \frac{A+B}2=n\pi+C$
$\displaystyle\iff A+B=2n\pi+2C\ \ \ \ (1)$
From the other answer we have $\displaystyle A-B=2m\pi\pm \frac{2\pi}3$
$\displaystyle\implies A=(n+m)\pi+C\pm\frac\pi3$ and
$\displaystyle B=(n-m)\pi+C\mp\frac\pi3$
Observe that $n-m,n+m$ have same parity
Case $1:$ If $n+m$ is even, so is $n-m$
$\displaystyle\implies \cos A=\cos\left(C\pm\frac\pi3\right)$ and
$\displaystyle \cos B=\cos\left(C\mp\frac\pi3\right)$
$\displaystyle\implies \cos A+\cos B=2\cos C\cos\frac\pi3=\cos C$
$\displaystyle\implies \cos A+\cos B+\cos C=2\cos C\ne0$ unless $\cos C=0$
Case $2:$ If $n+m$ is odd, so is $n-m$
$\displaystyle\implies \cos A=\cos\left(\pi+C\pm\frac\pi3\right)=-\cos\left(C\pm\frac\pi3\right)$ and
$\displaystyle \cos B=\cos\left(\pi+C\mp\frac\pi3\right)=-\cos\left(C\mp\frac\pi3\right)$
$\displaystyle\implies \cos A+\cos B=-2\cos C\cos\frac\pi3=-\cos C$
$\displaystyle\implies \cos A+\cos B+\cos C=0$ for all $C$
Similarly, $\displaystyle \sin A+\sin B+\sin C=0$ for all $C$
$\displaystyle\implies A=\pi+C\pm\frac\pi3\pmod{2\pi}$ and
$\displaystyle B=\pi+C\mp\frac\pi3\pmod{2\pi}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/623765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Minimum value of: $x^7(yz-1)+y^7(zx-1)+z^7(xy-1)$
$x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of:
$$x^7(yz-1)+y^7(zx-1)+z^7(xy-1)$$
I put it in the form $x^6y +x^6z+y^6x+y^6z+z^6x +z^6y$. I tried AM-GM but it's not helping.
|
$$x^6y +x^6z+y^6x+y^6z+z^6x +z^6y \ge 6 (xyz)^{7/3}$$
Also $xyz = x+y+z \ge 3 \sqrt[3]{xyz} \implies xyz \ge 3\sqrt3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/624712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
If the coefficients of the quadratic equation $ax^2+bx +c$ are u.i.i.d ran variates in $(0,1)$ what is the probability of roots being real? If the coefficients a,b,c(taken in order ,c being the constant term) of a quadratic equation are randomly and independenly chosen in the open interval(0,1) what is the probability that both the roots are real?
|
$a,b,c$ are limited to the interval $(0,1)$, so that
$$
P(b^2 < 4ac) = \int_{a=0}^1\int_{c=0}^1\int_{b=0}^{\min(2\sqrt{ac},1)}da\,db\,dc =
\int_{a=0}^1\int_{c=0}^1 \min(2\sqrt{ac},1)\,da\,dc,
$$
so that
$$
P(b^2 < 4ac) = \int_{a=0}^1\int_{c=0}^{\min(1/(4a),1)}2\sqrt{ac}\,da\,dc
+ \int_{a=0}^1\int_{c=\min(1/(4a),1)}^1\,da\,dc.
$$
The first integral is
$$
\begin{multline}
\int_{a=0}^1 2\sqrt{a}\,da\int_{c=0}^{\min(1/(4a),1)}\sqrt{c}\,dc = \frac{4}{3}\int_{a=0}^1 \sqrt{a}\min\left(\frac{1}{8a^{3/2}},1\right)\,da\\
= \frac{4}{3}\int_{a=0}^{1/4}\sqrt{a}\,da + \frac{1}{6}\int_{a=1/4}^1\frac{1}{a}\,da = \frac{1}{9} - \frac{1}{6}\ln(1/4)\tag{1}.
\end{multline}
$$
The second integral is
$$
\int_{a=0}^1 \left[1- \min\left(\frac{1}{4a},1\right)\right]\,da = \int_{a=1/4}^1\left(1 - \frac{1}{4a}\right)\,da = \frac{3}{4} + \frac{1}{4}\ln(1/4).\tag{2}
$$
Combining $(1)$ and $(2)$, we get
$$
P(b^2 < 4ac) = \frac{31}{36} + \frac{1}{12}\ln(1/4),
$$
and
$$
P(b^2 > 4ac) = \frac{5}{36} - \frac{1}{12}\ln(1/4) = \frac{5}{36}+\frac{1}{6}\ln 2.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/624900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
}
|
Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
x-4=x-5-\sqrt{x-5}+1\\
x-4=x-4-\sqrt{x-5}\\
\text{substracting x, and adding 4 to both sides}\\
0=-\sqrt(x-5)\\
\text{switching both sides}\\
\sqrt{x-5}=0\\
\text{sqaring both sides}\\
x-5=0\\
x=5\\
\text{When I place 5 in the equation, I get:}\\
\sqrt{5-4}-\sqrt{5-5}+1=0\\
\sqrt{1}-\sqrt{0}+1=0\\
1-0+1=0\\
2=0\\
\text{this means that the equation dosent have any solution, right??}\\
$$
Any advice and suggestion is helpful.
Thanks!!!
|
As $(x-4)-(x-5)=1$
$\displaystyle(\sqrt{x-4}-\sqrt{x-5})(\sqrt{x-4}+\sqrt{x-5})=1$
As $\displaystyle\sqrt{x-4}-\sqrt{x-5}=-1, \sqrt{x-4}+\sqrt{x-5}=-1$
Adding we get $\displaystyle\sqrt{x-4}=-1$ which is impossible as $\sqrt{x-1}\ge0$ for real $x$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/624974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
}
|
Compute $\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} \right )^2+\left (\frac{1}{n} \right )^2$ Compute the value of the following expression
$$\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\left ( \frac{1}{2}+\cdots + \frac{1}{n}\right )^2+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} \right )^2+\left (\frac{1}{n} \right )^2$$
The answer is $\boxed{2n-\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )}$.
I've been trying to do it but I've been failed. Any ideas ?
Wolfram's test
|
We can solve this by recursion relations. Denoting the sum as $S_n$, we have,
$$ S_{n+1} = \left (1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1} \right )^2+\left ( \frac{1}{2}+\cdots + \frac{1}{n} +\frac{1}{n+1}\right )^2+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} +\frac{1}{n+1} \right )^2+\left (\frac{1}{n} +\frac{1}{n+1} \right )^2+\left (\frac{1}{n+1} \right)^2 = S_{n} +\frac {2}{n+1}(\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )+\left ( \frac{1}{2}+\cdots + \frac{1}{n}\right )+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} \right )+\left (\frac{1}{n} \right ))+\sum_{k=1}^{n} \left (\frac{1}{n+1} \right)^2 +\left (\frac{1}{n+1} \right)^2 = S_{n} + \frac{2n+1}{n+1} $$
So, we have $$S_{n} - S_{n-1} = \frac {2n-1}{n}=2-\frac{1}{n}$$
Where from we get,
$$S_{n}-S_{1} = 2n - 1 -\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )$$
But $S_{1} = 1$. So,
$$S_{n} = 2n -\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/625939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
}
|
prove $ b^6c^8-b^5c^7+b^8c^6-b^3c^6+b^2c^6-b^7c^5-b^6c^3+b^6c^2-bc^2+c^2-b^2c+b^2 \ge 0$ $b>0,c>0, $ prove $g(b,c)=b^6c^8-b^5c^7+b^8c^6-b^3c^6+b^2c^6-b^7c^5-b^6c^3+b^6c^2-bc^2+c^2-b^2c+b^2 \ge 0$
this is from a middle process of a inequality. (I am sure it is correct because the inequality is proved.)
edit: the inequality is : $abc=1,\dfrac{a}{b^2+c^2+a}+\dfrac{b}{c^2+a^2+b}+\dfrac{c}{a^2+b^2+c} \le 1$ , replace $a=\dfrac{1}{bc}$ ,one can get the asked inequality.
I try different approach such as $c=b+u,$ or $ c=tb \to b^{12}t^8-b^{10}t^7+b^{12}t^6-b^7t^6+b^6t^6-b^{10}t^5-b^7t^3+b^6t^2-bt^2+t^2-bt+1 \ge 0$
it doesn't work.there is no factor such as $(t-1),(b-1),(bt-1),(b^2t-1),(b-c)$
when $c=b$ we have $(b-1)^2f(b)$ and $ f(b) >0$ so it is clear that $g(b,c)=0 \iff b=c=1$, but how to prove it?
thanks in advance.
|
It seems the following.
AM-GM inequality implies the following inequalities:
$$\frac 12 b^8c^6+\frac 16 b^6c^8+\frac 13 b^6c^2\ge b^7c^5,$$
$$\frac 12 b^6c^8+\frac 16 b^8c^6+\frac 13 b^2c^6\ge b^5c^7,$$
$$\frac 13 b^8c^6+\frac 12 b^6c^2+\frac 16 b^2\ge b^6c^3,$$
$$\frac 13 b^6c^8+\frac 12 b^2c^6+\frac 16 c^2\ge b^3c^6,$$
$$\frac 16 b^6c^2+\frac 56 c^2\ge bc^2,$$
$$\frac 16 b^2c^6+\frac 56 b^2\ge b^2c.$$
When we add all of them, we obtain $g(b,c)\ge 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/626526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
How to calculate Taylor expansion of $\cos(\sin x)$ I know that Taylor expansion of $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + O(x^6)$
and that of $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - O(x^7)$.
But how do I calculte the Taylor Expansion of $\cos(\sin x)$.
|
We know that for any $x$ (close to $0$):
$$
\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - O(x^6)\\
\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - O(x^7)
$$
We can find $\cos\sin x$ by substituting $x\to \sin x$ in the expansion of $\cos x$:
$$
\cos x = 1 - \frac{(x - \frac{x^3}{3!} + \frac{x^5}{5!} - O(x^7))^2}{2!} + \frac{(x - \frac{x^3}{3!} + \frac{x^5}{5!} - O(x^7))^4}{4!} + O(x^6)
$$
We know that $O(x^a)^b=O(x^{ab})$, so the error terms because of the sine expansion will be smaller (go to zero faster) than $O(x^6)$. We may also omit any power of $x$ with exponent at least $6$, because of the $O(x^6)$. Now, we can calculate the result:
$$
\cos\sin x=1-\frac 12 x^2+\frac 5{24}x^4+O(x^6)
$$
Another way to calculate this is to repeatedly differentiate $\cos\sin x$ and evaluate the result in $x=0$, but that requires some more effort I think, because you get a lot of terms/factors due to the product and chain rule.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/627204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Calculation of $f(x) = \int_{0}^{\frac{\pi}{4}}\ln \left(1+x\cdot \tan z\right)dz$ If $\displaystyle f(x) = \int_{0}^{\frac{\pi}{4}}\ln \left(1+x\cdot \tan z\right)dz,$ where $x>-1$. Then value of $\displaystyle f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right) = $
$\bf{My\; Try::}$ Given $\displaystyle f(x) = \int_{0}^{\frac{\pi}{4}}\ln \left(1+x\cdot \tan z\right)dz$
$\displaystyle \Rightarrow f^{'}(x) = \int_{0}^{\frac{\pi}{4}}\frac{\tan z}{1+x\cdot \tan z}dz = \int_{0}^{\frac{\pi}{4}}\frac{\frac{2\tan \frac{z}{2}}{1-\tan^2 \frac{z}{2}}}{1+x\cdot \frac{2\tan \frac{z}{2}}{1-\tan^2 \frac{z}{2}}}dz = \int_{0}^{\frac{\pi}{4}}\frac{2\tan \frac{z}{2}}{1-\tan^2\frac{z}{2}+2x\cdot \tan \frac{z}{2}}dz$
Now Let $\displaystyle \tan \frac{z}{2}=t$, Then $\displaystyle dz=\frac{2}{1+t^2}dt$
So $\displaystyle f^{'}(x) = \int_{0}^{\frac{\pi}{4}}\frac{4t}{1-t^2+2x\cdot t}dt$
Now How can I solve after that
Please Help me
Thanks
|
There is a somewhat simpler way to procede: once you get to
$$f^\prime(x) = \int_0^{\pi/4} \frac{\tan z}{1+x \tan z} dz,$$ make the substitution $u = 2 z,$ so the integral becomes
$$\frac12\int_0^{\pi/2} \frac{\tan u/2}{1+x \tan u/2} du.$$ Now make the substitution $\tan u/2 = t,$ to get
$$\int_0^1 \frac{t}{(1+ x t)(1+t^2)} dt.$$
This is easily integrated by partial fractions, to give you
$$
\frac{\pi x-4 \log (x+1)+\log (4)}{4 x^2+4}.
$$
You can go on from there, though I would not describe this as super fun.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/627953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
|
Try looking at the terms in binary:
2^10 10000000000
2^9 1000000000
2^8 100000000
2^7 10000000
2^6 1000000
2^5 100000
2^4 10000
2^3 1000
2^2 100
+ 2^1 10
---------------------
= 11111111110
+ 2 10
= 100000000000 = 2^11
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/628501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
}
|
Range Of Quartic Polynomial Of Two Variables $a$,$b$ are real numbers such that $~3\leq a^{2}+ab+b^{2}\leq 6$.
I would like to find the range of $~a^{4}+b^{4}$.
Is it possible to find it with (well-known) AM-GM, CS, etc...?
|
$ab \le \dfrac{a^2+b^2}{2} \implies \dfrac{3(a^2+b^2)}{2} \ge 3 \implies a^2+b^2 \ge2 \implies a^4+b^4 \ge 2(\dfrac{a^2+b^2}{2})^2 \ge 2 $ when $a=b=1$ hold "="
$a^4+b^4=(a^2+b^2+\sqrt{2}ab)(a^2+b^2-\sqrt{2}ab)=(u+(\sqrt{2}-1)ab)(u-(\sqrt{2}+1)ab)=u^2-2abu-a^2b^2, u=a^2+ab+b^2 \le 6$
it hints that when $ab<0$, the max will be got. let $c=-b>0, 3\le a^2-ac+c^2\le6 \implies ac \le6 $ when $a=c=\sqrt{6}$ hold "="
$a^4+b^4 \le 36+12ac-(ac)^2=72-(6-ac)^2\le 72$,when $a=c=-b=\sqrt{6}$ get "="
so we get : $2\le a^4+b^4 \le 72$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/632055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $\lim\limits_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)=0$
Prove that:
$$\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}
\,\right)=0$$
Epsilon>0. According to the Archimedean Property of reals, we have n1 element N with epsilon*n1>11. (Why 11? Seems so random...)
n0:=max{3,n1} (What's the point of that? It doesn't appear anywhere in the proof...).
For each n element N with n>=n0 we have:
\begin{align}
0&\lt\sqrt{n^4+n^2+20n+7}-\sqrt{n^4+n^2+1} \\\,\\
&=\dfrac{\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}\right)\cdot\left(\sqrt{n^4+n^2+20n+7}+\sqrt{n^4+n^2+1}\right)}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}} \\\,\\
&=\dfrac{20n+6}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}}\leqslant\dfrac{20n+2n}{\sqrt{n^4}+\sqrt{n^4}}=\dfrac{11}n\leqslant\dfrac{11}{n_1}\lt\epsilon
\end{align}
(I don't get the circled part. Why 20n+2n, why "cut off" the roots? And why should it equal 11/n?)
From this we get
$$\left|\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right|\lt\epsilon$$
, thus the proof is complete.
Thanks for the clarifications!
|
Setting $\frac1n=h$
$$F=\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}
\,\right)$$
$$=\lim_{h\to0}\frac{\sqrt{1+h^2+20h^3+7h^4}-\sqrt{1+h^2+h^4}}{h^2}$$
Rationalizing the numerator,
$$F=\lim_{h\to0}\frac{(1+h^2+20h^3+7h^4)-(1+h^2+h^4)}{h^2(\sqrt{1+h^2+20h^3+7h^4}+\sqrt{1+h^2+h^4})}$$
As $h\to0,h\ne0,$
$$F=\lim_{h\to0}\frac{20h+6h^2}{(\sqrt{1+h^2+20h^3+7h^4}+\sqrt{1+h^2+h^4})}=\cdots$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/632553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
}
|
Maximum number of edges that a bipartite graph with $n,m$ vertices can have when it doesn't contain $4$-cycle Let $A_{n,m}$ be the maximum number of edges that a bipartite graph with $n,m$ vertices can have when it doesn't contain $4$-cycle. I have calculated some values:
$A_{2,2}=3$, $A_{3,3}=6$, $A_{4,4}=9$, $A_{4,5}=10$, $A_{5,5}=12$.
I am trying to find formula for $A_{n,m}$. Does anyone know it or a hint how to find?
Especially I need the value of $A_{6,6}$. I only know that $A_{6,6}\ge16$.
|
I will give simple proofs of the evaluations $A_{4,6}=12,\ A_{5,6}=14,\ A_{6,6}=16,\ A_{6,7}=18,\ $ and $A_{7,7}=21$.
1. Since the ratio $\dfrac{A_{m,n}}{mn}$ is a nonincreasing function of $m$ and $n$, we have the recursive upper bound
$$A_{m,n+1}\le\lfloor\frac{n+1}nA_{m,n}\rfloor.$$
2. It's easy to see that $A_{1,n}=n$ and $A_{2,n}=n+1$ for all $n$.
3. $$A_{3,4}\le\lfloor\frac32A_{2,4}\rfloor=\lfloor\frac{15}2\rfloor=7$$
$$A_{3,5}\le\lfloor\frac54A_{3,4}\rfloor\le\frac{35}4\rfloor=8$$
$$A_{4,5}\le\lfloor\frac43A_{3,5}\rfloor\le\lfloor\frac{32}3\rfloor=10$$
$$A_{4,6}\le\lfloor\frac65A_{4,5}\rfloor\le\lfloor\frac{60}5\rfloor=12$$
$$A_{{5,5}}\le\lfloor\frac54A_{4,5}\rfloor\le\lfloor\frac{50}4\rfloor=12$$
$$A_{5,6}\le\lfloor\frac65A_{5,5}\rfloor\le\lfloor\frac{72}5\rfloor=14$$
$$A_{6,6}\le\lfloor\frac65A_{5,6}\rfloor\le\lfloor\frac{84}5\rfloor=16$$
$$A_{6,7}\le\lfloor\frac76A_{6,6}\rfloor\le\lfloor\frac{112}6\rfloor=18$$
$$A_{7,7}\le\lfloor\frac76A_{6,7}\rfloor\le\lfloor\frac{126}6\rfloor=21$$
4. Let $G$ be the bipartite graph with vertex sets $U,V$ where $|U|=6,|V|=7$, and:
*
*The elements of $U$ are the six faces of a cube;
*$V$ has four elements corresponding to four pairwise nonadjacent vertices of the cube, each of those four elements being joined to the three faces incident with the corresponding vertex of the cube;
*$V$ has three elements coresponding to pairs of opposite faces of the cube, and joined to those two faces.
The graph $G$ witnesses the inequality $A_{6,7}\ge18$; moreover, by deleting one, two, or three of the degree $2$ vertices of $G$, we get witnesses to $A_{6,6}\ge16,\ A_{5,6}=A_{6,5}\ge14$, and $A_{4,6}=A_{6,4}\ge12$. Combined with the previously obtained upper bounds, this verifies the exact evaluations $A_{4,6}=12,\ A_{5,6}=14,\ A_{6,6}=16,$ and $A_{6,7}=18$.
5. The point-line incidence graph of a finite geometry is a $C_4$-free bipartite graph. In this way we get the lower bounds
$$A_{n^2,\ n^2+n}\ge n^3+n^2$$and
$$A_{n^2+n+1,\ n^2+n+1}\ge(n^2+n+1)(n+1)$$
whenever a projective plane of order $n$ exists, e.g., whenever $n$ is a prime power.
Setting $n=2$ (the Fano plane) we get $A_{4,6}\ge12$ (again) and $A_{7,7}\ge21$ which, combined with the previously obtained upper bound, establishes that $A_{7,7}=21$.
Update. Thanks to @David for pointing out that "the graph G in 4 is in fact the incidence graph of the Fano plane with one line (or point) removed."
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/635066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ Can some one help me out on where to go?
If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ ?
|
We are given that $ a^2 + b^2 = 1$, and $ a, b \geq 0$.
Applying AM-GM or Cauchy-Schwarz, we get that
$$ 20 = 5(a^2 + a^2 + a^2 + a^2 + 4b^2) \geq (a+a+a+a+2b)^2 = (4a+2b)^2$$
Hence $ 2a + b \leq \frac{20}{16} = \frac{5}{4} $, with equality when $ a = 2b, b = \sqrt{\frac{1}{5} }$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/636361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
How to find nth derivative of $1/(1+x+x^2+x^3)$ I was trying to solve a differentiation question but unable to understand .
My question is :
find the $n^{th}$ derivative of $1/(1+x+x^2+x^3)$
I know that if we divide the numerator by denominator then the expression would be :
$1- x(1+ x^2 + x )/(1+x+x^2+x^3)$
But now how to find the nth derivative?
Please somebody explain this..
Thanx :-)
|
Hint: Note that $(1-x)(1+x+x^2+x^3)=1-x^4$, so $$\frac1{1+x+x^2+x^3}=\frac{1-x}{1-x^4}=\frac{1-x}{(1-x^2)(1+x^2)}=\frac{1}{(1+x)(1+x^2)}.$$
(After the fact, this factorization is easy to see directly.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/636718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Find minimum value of this expression: $P=\sqrt{a^2+(1-bc)^2}+\sqrt{b^2+(1-ca)^2}+\sqrt{c^2+(1-ab)^2}$ Let $a,b,c\in R$ and satisfying $a^2+b^2+c^2=1$
Find minimum value of this expression: $P=\sqrt{a^2+(1-bc)^2}+\sqrt{b^2+(1-ca)^2}+\sqrt{c^2+(1-ab)^2}$
|
$3u=a+b+c,3v=ab+bc+ac \implies 9u^2-6v=1$
edit:
$P \ge \sqrt{(a+b+c)^2+(3-(3b+bc+ac))^2}=\sqrt{(9u^2+(3-3v)^2}=\sqrt{1+6v+9-18v+9v^2}=\sqrt{(2-3v)^2+6} $
$3v=ab+bc+ac \le a^2+b^2+c^2 =1 $
$ 2-3v \ge 1 \implies \sqrt{(2-3v)^2+6} \ge \sqrt{7}$
last "=" will hold when $a=b=c=\pm \dfrac{1}{\sqrt{3}}$
another "=" will hold when $\dfrac{a}{1-bc}=\dfrac{c}{1-ab}=\dfrac{b}{1-ac}=\dfrac{a+b+c}{3-(ab+bc+ac)}=\pm \dfrac{\sqrt{3}}{2}$
$\sqrt{3}a^2 \pm 2a -\sqrt{3}=0 \implies a=\pm \sqrt{3}$ or $\pm \dfrac{1}{\sqrt{3}} $
so $P_{min}=\sqrt{7}$ will be got when $a=b=c=\pm \dfrac{1}{\sqrt{3}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/636816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Solving recurrence relation with generating functions - Nearly got the answer I'm trying to solve the following recurrence relation (Find closed formula) using generating functions:
$f(n)=10f(n-1)-25f(n-2)$, $f(0)=0$, $f(1)=1$
I'm having a small difficulty at the end and can use a nudge in the right direction.
My solution
Define $$g(x)=\sum_{n=0}^{\infty}f(n)x^n = \sum_{n=2}^{\infty}f(n)x^n+x=10\sum_{n=2}^{\infty}f(n-1)x^n-25\sum_{n=2}^{\infty}f(n-2)x^n+x =$$
$$=10x\sum_{n=2}^{\infty}f(n-1)x^{n-1}-25x^2\sum_{n=2}^{\infty}f(n-2)x^{n-2}+x =$$
$$=10x\sum_{n=1}^{\infty}f(n)x^n-25x^2\sum_{n=0}^{\infty}f(n)x^n+x$$
But since the first element of $f(n)$ is $0$, then $$10x\sum_{n=1}^{\infty}f(n)x^n=10x\sum_{n=1}^{\infty}f(n)x^n+0=10x\sum_{n=0}^{\infty}f(n)x^n$$
So overall:
$$g(x)=10x\sum_{n=0}^{\infty}f(n)x^n-25x^2\sum_{n=0}^{\infty}f(n)x^n+x=10xg(x)-25x^2g(x)+x$$
To sum up:
$$g(x)=10xg(x)-25x^2g(x)+x$$
Solve for $g(x)$ and get:
$$g(x)=\frac{x}{(x-\frac{1}{5})^2}$$
But what do I do now? How can I extract $f(n)$ from this information?
|
An orderly way of attacking such problems is as follows: Define $F(z) = \sum_{n \ge 0} f(n) z^n$, write the recurrence so there are no substractions in indices:
$$
f(n + 2) = 10 f(n + 1) - 25 f(n)
$$
Multiply by $z^n$, sum over $n \ge 0$, recognize:
\begin{align}
\sum_{n \ge 0} f(n + 1) z^n &= \frac{F(z) - f(0)}{z} \\
\sum_{n \ge 0} f(n + 2) z^n &= \frac{F(z) - f(0) - f(1) z}{z^2} \\
\end{align}
to get:
$$
\frac{F(z) - z}{z^2} = 10 \frac{F(z)}{z} - 25 F(z)
$$
Maxima spits out the partial fraction expansion:
$$
F(z) = \frac{1}{5 (1 - 5 z)^2} - \frac{1}{5 (1 - 5 z)}
$$
Using the generalized binomial theorem, as $(-1)^n \binom{-2}{n} = \binom{n + 2 - 1}{1} = n + 1$:
$$
f(n) = \frac{(n + 1) 5^n}{5} - \frac{5^n}{5}
= n 5^{n - 1}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/638354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Complex numbers - Exponential numbers - Proof Let $z$ be a complex number, and let $n$ be a positive integer such that $z^n = (z + 1)^n = 1$. Prove that $n$ is divisible by 6.
For this problem I am stumped...how should I begin?
Also there's a hint for it:
From $z^n = 1$, prove that $|z| = 1$. What does the equation $(z + 1)^n = 1$ tell you? What do the resulting equations tell you about $z$?
Could someone give me a hint on where to begin? thanks in advance
|
We can say: $$z^n=1 \implies |z|^n=1 \implies |z^n|=1 \implies |z|=1$$
Using similar logic, we can also get: $$(z+1)^n=1 \implies |z+1|^n=1 \implies |(z+1)^n|=1 \implies |z+1|=1$$
Let's set $z=a+bi$: $$|z|=|z+1| \implies a^2 + b^2 = (a+1)^2 + b^2 \implies a^2 + b^2 = a^2 + b^2 + 2a + 1$$ $$\implies 2a + 1 = 0 \implies a = -\frac{1}{2}$$
Pluging this back into the equation, we get: $$\left(-\frac{1}{2}\right)^2 + b^2 = 1 \implies b^2 = \frac{3}{4} \implies b = \pm \frac{\sqrt{3}}{2}$$
With this information, we know that: $$z= -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i \implies z = e^{\pm 2\pi i/3}$$
Also: $$z+1= \frac{1}{2} \pm \frac{\sqrt{3}}{2}i \implies z+1 = e^{\pm \pi i/3} \implies (z+1)^n = e^{\pm \pi ni/3}$$
For $(z+1)^n$ to be $1$, $\cos \pm \pi n/3 = 1$ and $\sin \pm \pi n/3 = 0$. This can only happen if $\pm \pi n/3 = 2\pi k$ for some integer $k$
Solving for $n$, we get: $$\pm \pi n/3 = 2\pi k \implies n = \pm 6k \implies \boxed{6 \mid n}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/643024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
}
|
limit problem - can't get rid of $0$. I am trying to evaluate limit:
$$\lim_{x\rightarrow 0}\frac{\arcsin x - \arctan x}{e^x-\cos x -x^2 -x}$$
I tried to use known limit in denominator to get:
$$\lim_{x\rightarrow 0}\frac{\frac{\arcsin}{x} - \frac{\arctan x}{x}}{x( \frac{e^x-1}{x}\cdot \frac{1}{x}+\frac{1-\cos x}{x^2} -1 -\frac{1}{x})}$$
But I still get $1-1=0$ in numerator. I also tried to use L'Hôpital rule, but it didn't help.
|
Using the Taylor series
$$f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f^{(3)}(0)+o(x^3)$$
for $f$ is $\arcsin$ and $\arctan$ we find
$$\arcsin x=x+\frac{x^3}{6}+o(x^3)$$
and
$$\arctan x=x-\frac{x^3}{3}+o(x^3)$$
and we have
$$e^x-\cos x-x^2-x=1+x+\frac{x^2}{2}+\frac{x^3}{6}-1+\frac{x^2}{2}-x^2-x+o(x^3)=\frac{x^3}{6}+o(x^3)$$
hence
$$\lim_{x\rightarrow 0}\frac{\arcsin x - \arctan x}{e^x-\cos x -x^2 -x}=\lim_{x\rightarrow 0}\frac{\frac{x^3}{6}+\frac{x^3}{3}+o(x^3)}{\frac{x^3}{6}+o(x^3)}=3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/643787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Infinite Geometric Series Issue i have came across a series, i am trying to find its sum knowing the fact that, if it converges and its common ratio ex. r is: -1 < r < 1, then i can use the specified formula $\frac{a}{1-r}$ , which specifically means first term of series over 1 minus common ratio
here is the series
$\sum_{n=1}^{\infty}\frac{2n-1}{2^n}$
i manipulated it this way to prove its convergence: $\sum_{n=1}^{\infty}\frac{2n-1}{2^n}=\sum_{n=1}^{\infty}(2n-1)\frac{1}{2^n}=\sum_{n=1}^{\infty}(2n-1)\left(\frac{1}{2}\right)^n$
$\frac{a}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$
using it i get the result 1, which actually should be 3
|
Since you know it converges, let $S=\sum_{n=1}^\infty\frac{2n-1}{2^n}$. Then we have
$$\begin{align}
S &=\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\cdots&\text{so}\\
\frac{1}{2}S&=\phantom{\frac{1}{2}}+\frac{1}{4}+\frac{3}{8}+\frac{5}{16}+\cdots&\text{subtracting, we get}\\
S-\frac{1}{2}S&=\frac{1}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\cdots&\text{so}\\
S-\frac{1}{2}S+\frac{1}{2}&=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\cdots&\text{so}\\
&=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots = 2
\end{align}$$
and thus we have
$$
\frac{1}{2}S+\frac{1}{2}=2
$$
from which we can conclude that $S=3$.
This idiom, suitably manipulating a series as a linear combination leading to known sums, is quite often handy.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/645560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
How to get the man's age. I am very much in need of solution for this problem. I can't figure out the answer for this. does anybody know about this problem below?
thanks,,
Problem:
A man's boyhood lasted for 1/6 of his life, he played soccer for the next 1/12 of his life, and he married after 1/7 more of his life. a daughter was born 5 years after his marriage & the daughter lived 1/2 as many years as her father did. If the man died four years after his daughter did, how old was the man when he died?
|
We can set up this problem by adding all the fractions of his life so they equal $1$ (his whole life). Also, let's say that he lived to $x$ years old.
Now,
$\frac{1}{6}$ of his life was boyhood.
$\frac{1}{12}$ of his life was soccer.
$\frac{1}{7}$ of his life was end of soccer to marriage.
After $5$ years, his daughter was born so
$\frac{5}{x}$ of his life was marriage to daughter's birth.
His daughter lived half the amount of years that he will have lived so
$\frac{1}{2}$ of his life was his daughter's life.
Then he died $4$ years later so
$\frac{4}{x}$ of his life was from his daughter's death to his death.
And all of these should equal to $1$. Thus
$\frac{1}{6}+\frac{1}{12}+\frac{1}{7}+\frac{5}{x}+\frac{1}{2}+\frac{4}{x}=1$
Now we simply solve for $x$. We get
$\frac{9}{x}=1-\frac{1}{6}-\frac{1}{12}-\frac{1}{7}-\frac{1}{2}$
then $9=x(1-\frac{1}{6}-\frac{1}{12}-\frac{1}{7}-\frac{1}{2})$
then $x=\frac{9}{1-\frac{1}{6}-\frac{1}{12}-\frac{1}{7}-\frac{1}{2}}$.
By simplifying, we get $x=84$. So he lived to $84$ years old.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/647035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
}
|
How many zero elements are there in the inverse of the $n\times n$ matrix How many zero elements are there in the inverse of the $n\times n$ matrix
$A=\begin{bmatrix}
1&1&1&1&\cdots&1\\
1&2&2&2&\cdots&2\\
1&2&1&1&\cdots&1\\
1&2&1&2&\cdots&2\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
1&2&1&2&\cdots&\cdots
\end{bmatrix}$
My try: Denote by $a_{ij}$ and $b_{ij}$ the elements of $A$ and $A^{-1}$,respectively
then for $k\neq m$,we have
$$\sum_{i=0}^{n}a_{ki}b_{im}=0$$
Then I can't.Thank you very much
|
I am excavating this post because this post needs an affirmative answer.
After some try I feel that this can be (ironically) done fastest by brute force. Let $A^{-1} = (a_{ij})$ and we just solve column-wisely $AA^{-1} = I$.
For the first column, by subtracting equations derived from neighboring row of $A$, you will get:
$$
\begin{aligned}
a_{11} + a_{21} + a_{31} + a_{41} + \cdots &= 1\\
a_{21} + a_{31} + a_{41} + \cdots &= -1 \\
-a_{31}- a_{41} - \cdots &= 0 \\
a_{41} + \cdots &= 0\\
&\vdots
\end{aligned}
$$
You then have $a_{11} = 1$, $a_{21} = -1$ and the rest are 0.
From the second to the penultimate columns, $a_{ij}$ for $i < j$ can be derived directly from the fact that $A^{-1}$ is symmetric as $A$ is, for the rest, play the same "compare neighboring row game", you will get:
$$
\begin{aligned}
(-1)^ia_{ii} + (-1)^ia_{i+1,i} + (-1)^ia_{i+2,i} + \cdots &= 1\\
(-1)^{i+1}a_{i+1,i} + (-1)^{i+1}a_{i+2,i} + \cdots &= -1 \\
(-1)^{i+2}a_{i+2,i} \cdots &= 0 \\
&\vdots
\end{aligned}
$$
As the equations are easy enough, you will immediately see that $a_{i, i+1} = (-1)^i$ and $a_{i, i-1} = (-1)^{i-1}$.
For the last column, the only thing new is that $(-1)^na_{nn} = 1$, so $a_{nn} = (-1)^n$ and $a_{n, n-1} = (-1)^{n-1}$.
At the end you see that $A^{-1}$ has $2n$ non-zero entries, where exactly 2 of them are on each row and each column.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/648340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Am I understanding induction correctly? Here is an induction proof that I have written for my homework and I want to know if I am understanding this correctly:
Prove that for:
$ \sum\limits_{i=0}^n i^2 = \frac{n(n+1)(2n+1)}{6}$
My proof:
Check the base-case, let mine be n=1 (or do I have to choose 0..? The lowest one)
$ (1^2) = \frac{1^2(1^2+1)(2 \cdot 1^2 +1)}{6} = \frac{6}{6} = 1 $
Base case proved.
Induction step. Assume that
$ (1^2+2^2=3^2...k^2) = \frac{k(k+1)(2k+1)}{6} $
works for all k.
Add another sum, $k+1$, then:
$(1^2+2^2=3^2...+k^2+(k+1)^2) = \frac{k(k+1)(2k+1)}{6}+(k+1) $
Substitude for sum up until $(k+1)^2$
$ \frac{k(k+1)(2k+1)}{6}+(k+1)^2 = \frac{k(k+1)(2k+1)}{6}+(k+1) $
Re-arrange and tidy up:
$ \frac{(k^2+k)(2k+1)+6k+6}{6} = \frac{(k^2+k)(2k+1)(6k+6}{6} $
Therefore we have proven that it works for $i=1$ and as it works for any $k+1$ it must work for the rest.
Is this correctly executed and understood?
EDIT: As my method seems to be wrong, let me have another try:
Now with:
$ \sum\limits_{i=0}^n i^2 = \frac{n(n+1)(2n+1)}{6}$
$(1^2+2^2=3^2...+k^2+(k+1)^2) = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} $
Re arrange and fiddle:
$2n^3+9n^2+13n+6 = 2n^3+9n^2+13n+6$
Is this correct?
|
If it is to be shown by induction that $$\sum_{i=0}^{n}f\left(i\right)=g\left(n\right)$$
then $f\left(0\right)=g\left(0\right)$ must be proved (the base case) and
secondly it must be shown that: $$\sum_{i=0}^{n+1}f\left(i\right)=g\left(n+1\right)$$
on base of: $$\sum_{i=0}^{n}f\left(i\right)=g\left(n\right)$$
That amounts in showing that: $$g(n)+f\left(n+1\right)=g\left(n+1\right)$$
Note that here: $$g(n)+f\left(n+1\right)=\sum_{i=0}^{n+1}f\left(i\right)$$
In your case that comes to proving that: $$\frac{1}{6}n\left(n+1\right)\left(2n+1\right)+\left(n+1\right)^{2}=\frac{1}{6}\left(n+1\right)\left(n+2\right)\left(2n+3\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/649054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Showing $\frac{x}{1+x}<\log(1+x)0$ using the mean value theorem I want to show that $$\frac{x}{1+x}<\log(1+x)<x$$ for all $x>0$ using the mean value theorem. I tried to prove the two inequalities separately.
$$\frac{x}{1+x}<\log(1+x) \Leftrightarrow \frac{x}{1+x} -\log(1+x) <0$$
Let $$f(x) = \frac{x}{1+x} -\log(1+x).$$ Since $$f(0)=0$$ and $$f'(x)= \frac{1}{(1+x)^2}-\frac{1}{1+x}<0$$ for all $x > 0$, $f(x)<0$ for all $x>0$. Is this correct so far?
I go on with the second part:
Let $f(x) = \log(x+1)$. Choose $a=0$ and $x>0$ so that there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with
$f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow \frac{1}{x_0+1}=\frac{ \log(x+1)}{x}$.
Since $$x_0>0 \Rightarrow \frac{1}{x_0+1}<1.$$ $$\Rightarrow 1 > \frac{1}{x_0+1}= \frac{ \log(x+1)}{x} \Rightarrow x> \log(x+1)$$
|
By Definition of log(which already a kind of mean value theorem ) we have
For any $x>0$ We have
$$\frac{x}{x+1} =\int_{0}^x\frac{dt}{x+1} \le \int_{0}^x\frac{dt}{t+1} =\color{red}{\ln(x+1 )}=\int_{0}^x\frac{dt}{t+1} \le \int_{0}^x\frac{dt}{1} = x $$
Thus,
$$\frac{x}{x+1} \le \ln(x+1 ) \le x $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/652581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
}
|
Find value of unending continued fraction I am trying to find the limit if it exists for the following unending continued fraction: $$1+{1\over{2+{1\over2+{1\over{2+...}}}}}$$
I have discovered this is the continued fraction for $\sqrt2$, but I am unsure how to show this.
Also I am being asked to assume that the even-order and odd-order truncations have a limit as well and calculate each of them. How would I go about finding any of these limits?
|
Let $x = 2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}}$. Then \begin{align*}
x - 2 &= \frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}} \\
\frac{1}{x-2} &= 2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}} \\
\frac{1}{x-2} &= x
\end{align*}
We find two roots to this quadratic equation, $1+\sqrt{2}$ and $1-\sqrt{2}$. Since $x = 2+\dots > 2$, we ignore the negative root and continue with $x = 1+\sqrt{2}$. The number you have is therefore $$1+\frac{1}{x} = 1 + \frac{1}{1+\sqrt{2}} = 1+\frac{1-\sqrt{2}}{1-2} = 1+\sqrt{2}-1 = \sqrt{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/652830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How to solve $\mathrm{diag}(x) \; A \; x = \mathbf{1}$ for $x\in\mathbb{R}^n$ with $A\in\mathbb{R}^{n \times n}$? I would like to solve the following equation for $x\in\mathbb{R}^{n}$
$$\mathrm{diag}(x) \; A \; x = \mathbf{1}, \quad \text{with $A\in\mathbb{R}^{n\times n}$},$$
where $\mathrm{diag}(x)$ is a diagonal matrix whose diagonal elements are the elements of $x$ and $\mathbf{1}$ is a vector whose elements are equal to 1.
I will already be very happy to find a solution if $A$ is a positive definite and symmetric.
Ideally I would like to find a closed-form solution for this quadratic equation.
Any ideas (or solution ;-) would be greatly appreciated.
Other formulation
Another way to formulate this equation is as follows
$$A \; x = 1./x, \quad \text{with $A\in\mathbb{R}^{n\times n}$},$$
where $1./x$ denotes the "element-wise inverse of the vector $x$".
Solution for the 1-dimensional case
The solution for the 1-dimensional case is straightforward
$$ x = \frac{1}{\sqrt{A}} .$$
|
Do these least squares solutions help?
$n=2$
$$
\begin{align}
%
\mathbf{D} \mathbf{A} x &= \mathbf{1} \\
%
\left[
\begin{array}{cc}
a_{\{1,1\}} & a_{\{1,2\}} \\
a_{\{2,1\}} & a_{\{2,2\}} \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
x_{\{1\}} & 0 \\
0 & x_{\{2\}} \\
\end{array}
\right]
%
\left[
\begin{array}{c}
x_{\{1\}}\\
x_{\{2\}} \\
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
1 \\
1 \\
\end{array}
\right]
%
\end{align}
$$
The least squares solution is
$$
\begin{align}
%
x_{LS} &= \left( \mathbf{D} \mathbf{A} \right)^{+} \mathbf{1} \\
%
&= \left( \det \mathbf{DA} \right)^{-1}
\left[
\begin{array}{c}
\frac{a_{\{2,2\}}}{x_{\{1\}}}-\frac{a_{\{1,2\}}}{x_{\{2\}}} \\
\frac{a_{\{1,1\}}}{x_{\{2\}}}-\frac{a_{\{2,1\}}}{x_{\{1\}}}\end{array}
\right]
%
\end{align}
$$
$n=3$
$$
\begin{align}
%
\mathbf{D} \mathbf{A} x &= \mathbf{1} \\
%
\left[
\begin{array}{ccc}
x_{\{1\}} & 0 & 0 \\
0 & x_{\{2\}} & 0 \\
0 & 0 & x_{\{3\}} \\
\end{array}
\right]
%
\left[
\begin{array}{ccc}
a_{\{1,1\}} & a_{\{1,2\}} & a_{\{1,3\}} \\
a_{\{2,1\}} & a_{\{2,2\}} & a_{\{2,3\}} \\
a_{\{3,1\}} & a_{\{3,2\}} & a_{\{3,3\}} \\
\end{array}
\right]
%
\left[
\begin{array}{c}
x_{\{1\}} \\
x_{\{2\}} \\
x_{\{3\}} \\
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
1 \\
1 \\
\end{array}
\right]
%
\end{align}
$$
The least squares solution is
$$
\begin{align}
%
x_{LS} &= \left( \mathbf{D} \mathbf{A} \right)^{+} \mathbf{1} \\
%
&= \left( \det \mathbf{DA} \right)^{-1}
\left[
\begin{array}{c}
%
\frac{a_{\{1,3\}} a_{\{3,2\}}-a_{\{1,2\}} a_{\{3,3\}}}{x_{\{2\}}}+a_{\{2,3\}} \left(\frac{a_{\{1,2\}}}{x_{\{3\}}}-\frac{a_{\{3,2\}}}{x_{\{1\}}}\right)+a_{\{2,2\}} \left(\frac{a_{\{3,3\}}}{x_{\{1\}}}-\frac{a_{\{1,3\}}}{x_{\{3\}}}\right) \\
%
\frac{a_{\{1,1\}} a_{\{3,3\}}-a_{\{1,3\}} a_{\{3,1\}}}{x_{\{2\}}}+a_{\{2,1\}} \left(\frac{a_{\{1,3\}}}{x_{\{3\}}}-\frac{a_{\{3,3\}}}{x_{\{1\}}}\right)+a_{\{2,3\}} \left(\frac{a_{\{3,1\}}}{x_{\{1\}}}-\frac{a_{\{1,1\}}}{x_{\{3\}}}\right) \\
%
\frac{a_{\{1,2\}} a_{\{3,1\}}-a_{\{1,1\}} a_{\{3,2\}}}{x_{\{2\}}}+a_{\{2,2\}} \left(\frac{a_{\{1,1\}}}{x_{\{3\}}}-\frac{a_{\{3,1\}}}{x_{\{1\}}}\right)+a_{\{2,1\}} \left(\frac{a_{\{3,2\}}}{x_{\{1\}}}-\frac{a_{\{1,2\}}}{x_{\{3\}}}\right)
%
\end{array}
\right]
%
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/655280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Proving a property of about the Fermat numbers
Show that the last digit in the decimal expansion of $F_n=2^{2^n}+1$ is $7$ for $n \geq 2$.
For our base step we let $n=2$. Now we have $2^{2^2}=16$. So the assertion holds for our base case. Then we assume it holds for $n$. For the $n+1$ case, is there a way to demonstrate this without resorting to this: $$(2^{2^n})^{{2^{n+1}}-2^n}.$$
That is the solution available in the back of the book. My attempt was to look at $2^{2^{n+1}}=2^{2^n}2^2.$
|
Observe that $\displaystyle2^1\equiv2,2^2\equiv4,2^3\equiv8,2^4=16\equiv6,2^5=32\equiv2\pmod{10}$
So, $\displaystyle2^m\equiv6\pmod{10}\iff 4|m$
$\displaystyle\implies2^{2^n}\equiv6\pmod{10}\iff 4|2^n\iff n\ge2$
Alternatively, as $\displaystyle6^{m+1}-6=6(6^m-1)\equiv0\pmod{30}$ as $6^m-1$ is divisible by $6-1=5$
$\displaystyle\implies 6^{m+1}\equiv6\pmod{30}\equiv6\pmod{10}$ for integer $m\ge0$
Again, $\displaystyle2^4=16\equiv6\pmod{10}$
So, $\displaystyle2^{2^n}=(2^4)^{2^{n-2}}=16^{(2^{n-2})}\equiv6^{(2^{n-2})}\pmod{10}$ for $\displaystyle 2^{n-2}\ge1\iff n\ge2$
Again, $\displaystyle6^{(2^{n-2})}\equiv6\pmod{10}$ for $\displaystyle 2^{n-2}\ge1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/655528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
$8$ cards are drawn from a deck of cards without replacement I draw $8$ cards randomly from a shuffled deck without replacement. What is the expected value of the sum of the largest $3$ cards? The ace is given a value of 1 and the jack, queen and king are all given a value of $10$. I can generate a simulation answer but that is all. How can this be done, I am stuck?
|
For each of the possible values for the sum, $S$, of the three highest cards -- integers on $[6,30]$ -- figure out the number of combinations of eight-card hands that would give the three highest cards that sum.
For $S=30$, we can have anywhere from three to eight $10$s, so we consider those separately:
$$P(S=30, 10, 10, 10) = \frac{{16 \choose 3}{36 \choose 5} + {16 \choose 4}{36 \choose 4} + {16 \choose 5}{36 \choose 3} + {16 \choose 6}{36 \choose 2} + {16 \choose 7}{36 \choose 1} + {16 \choose 8}{36 \choose 0}}{{52 \choose 8}}.$$
For something a little more complicated, you'll need to add up more combinations. For $S=20$, for example, there are twelve possible triples for the three highest cards:
$$(7, 7, 6), \\ (8, 8, 4), (8, 7, 5), (8, 6, 6), \\ (9, 9, 2), (9, 8, 3), (9, 7, 4), (9, 6, 5), \\ (10, 8, 2), (10, 7, 3), (10, 6, 4), (10, 5, 5).$$
Let's calculate $P(S=20, 9, 8, 3).$ We draw one $9$: ${4 \choose 1}$. We draw one $8$: also ${4 \choose 1}.$ Then there needs to be at least one $3$ (but due to the constraints on the number of $2$s and $3$s, there can't be less than two!):
$$P(S=20, 9, 8, 3) = \frac{{4 \choose 1}{4 \choose 1}\left[{4 \choose 4}{4 \choose 2} + {4 \choose 3}{4 \choose 3} + {4 \choose 2}{4 \choose 4}\right]}{{52 \choose 8}}.$$
(Note that there are little subtleties like the fact that $(10, 9, 1)$ isn't possible. If you draw a $10$ and a $9$, at least two of the remaining six cards must be something higher than an ace.)
Once you have the probability of getting each sum individually, then getting the expected value is easy. Counting up the number of combinations of cards for each sum is the laborious part.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/656310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
How do I solve $y''+4y=0$? This problem is in Penney's Elementary Differential Equations, listed as a reducible, 2nd-order DE. The chapter has taught two techniques to be used, which are for when either $x$ or $y(x)$ is missing. It didn't show how to solve when $y'(x)$ is missing.
I checked with WolframAlpha, and it suggests starting by assuming that $y$ is proportional to $e^{\lambda x}$.
If I didn't know to assume this, how could I solve this otherwise?
|
$y''+4y=0$ is a second order differential equation.
First, change the equation to
$r^2+4 = 0$
This equation will will have complex conjugate roots, so the final answer would be in the form of $e^{\alpha x}(c_1\sin(\beta x) + c_2 \cos (\beta x))$
where $\alpha$ equals the real part of the complex roots and $\beta$ equals the imaginary part of (one of) the complex roots.
We need to use the quadratic formula
\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}
In this equation $a =1$, $b=0$, and $c =4$
\begin{array}{*{20}c} {x = \frac{{ -0 \pm \sqrt {0^2 - 4(1)(4)} }}{{2}}} \\ \end{array}
\begin{array}{*{20}c} {x = \frac{{ -0 \pm \sqrt {-16} }}{{2}}} \\ \end{array}
\begin{array}{*{20}c} {x = \frac{{\pm \sqrt {-16} }}{{2}}} \\ \end{array}
Since we can't have negative signs in the square root, we have an imaginary number $i$.
\begin{array}{*{20}c} {x = \frac{{\pm \sqrt {16}i }}{{2}}} \\ \end{array}
Take the square root
\begin{array}{*{20}c} {x = \frac{{\pm 4i }}{{2}}} \\ \end{array}
\begin{array}{*{20}c} {x = \pm 2i} \\ \end{array}
Now let's bring this equation back.
$y = e^{\alpha x}(c_1 \sin(\beta x) + c_2 \cos (\beta x))$
$\alpha = 0$ and $\beta = 2$
So your answer would be
$y = e^{0x}(c_1\sin(2x) + c_2\cos (2x))$
Since $e^{0x}$ is $1$ because $0$ multiplied by $x$ is just $0$ and $e^{0}$ is $1$.
The final answer is $y = (c_1\sin(2x) + c_2\cos (2x))$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/657796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
}
|
Rationalize the denominator of the surd, giving your answer in the simplest form. Rationalize the denominator of the surd, giving your answer in the simplest form.
$\frac {3}{\sqrt2+5} $
Please help me...
It must be like this right?
$\frac {3}{\sqrt2+5} * \frac{\sqrt2-5}{\sqrt2-5}$
|
Yes. in general we do not leave the denominator irrational. So we multiply and divide the fraction by the conjugate of denominator.
In this case the conjugate of $\sqrt{2}+5$ is $\sqrt{2}-5$.
$$\frac{3(\sqrt{2}-5)}{(\sqrt{2}+5)(\sqrt{2}-5)}$$
$$\frac{3\sqrt 2 - 15}{2-25} = \frac{3\sqrt 2 - 15}{-23}$$
$$\frac{15 - 3\sqrt{2}}{23}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/659739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined I appreciate the help.
My attempt:
$$
\begin{align}
\tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\
&= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\
&= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin(x)}\\
&= \frac{1}{\cos(x) \sin(x)}\\
&= \frac{1}{\frac{1}{\sec(x)}\frac{1}{\csc(x)}}\\
&=\frac{1}{\frac{1}{\sec \csc}}\\
&=\frac{1}{1}\cdot \frac{\sec(x) \csc(x)}{1}\\
&= \sec(x) \csc(x)
\end{align}
$$
|
Your steps are correct, but just keep in mind that robotically converting everying into $\sin$s and $\cos$s isn't the only option available to you.
Note that
$$\cot\theta = \frac{\cos\theta}{\sin\theta}=\frac{\frac{1}{\sin\theta}}{\frac{1}{\cos\theta}}=\frac{\csc\theta}{\sec\theta}$$
that
$$\cot\theta\tan\theta=\frac{1}{\tan\theta}\cdot\tan\theta=1$$
and that
$$\sec^2\theta=\tan^2+1$$
then
$$\begin{array}{lll}
\tan\theta+\cot\theta&=&1\cdot(\tan\theta+\cot\theta)\\
&=&(\cot\theta\tan\theta)(\tan\theta+\cot\theta)\\
&=&(\cot\theta)(\tan\theta(\tan\theta+\cot\theta))\\
&=&\frac{\csc\theta}{\sec\theta}(\tan^2\theta+1)\\
&=&\frac{\csc\theta}{\sec\theta}\sec^2\theta\\
&=&\sec\theta\csc\theta
\end{array}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/661439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
}
|
How prove binomial cofficients $\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$ How prove this $$\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$$
This equation How prove it? Thank you
I want take this $$f(x)=(1-x)^{n+1}?$$
But I can't deal this $[n/3]$,
Thank you for you help
|
By way of enrichment here is another algebraic proof using basic
complex variables, quite similar to the accepted answer.
Note that the second binomial coefficient in both sums controls the
range of the sum, so we can write our claim like this:
$$\sum_{k=0}^{n+1} {n+1\choose k} (-1)^k {2n-3k\choose n-3k}
= \sum_{k=0}^{n+1} {n+1\choose k} {k\choose n-k}.$$
To evaluate the LHS introduce the integral representation
$${2n-3k\choose n-3k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n-3k}}{z^{n-3k+1}} \; dz.$$
We can check that this really is zero when $k>\lfloor n/3\rfloor.$
This gives for the sum the representation
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n}}{z^{n+1}}
\sum_{k=0}^{n+1} {n+1\choose k}
(-1)^k \left(\frac{z^3}{(1+z)^3}\right)^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n}}{z^{n+1}}
\left(1-\frac{z^3}{(1+z)^3}\right)^{n+1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} \frac{1}{(1+z)^{n+3}}
\left(3z^2+3z+1\right)^{n+1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} \frac{1}{(1+z)^{n+3}}
\sum_{q=0}^{n+1} {n+1\choose q} 3^q z^q (1+z)^q \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{q=0}^{n+1} {n+1\choose q} 3^q z^{q-n-1} (1+z)^{q-n-3} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{q=0}^{n+1} {n+1\choose q} 3^q
\frac{1}{z^{n+1-q}} \frac{1}{(1+z)^{n+3-q}} \; dz.$$
Computing the residue we find
$$\sum_{q=0}^{n+1} {n+1\choose q} 3^q (-1)^{n-q}
{n-q+n+2-q\choose n+2-q}
= \sum_{q=0}^{n+1} {n+1\choose q} 3^q (-1)^{n-q}
{2n-2q+2\choose n-q+2}.$$
Now introduce the integral representation
$${2n-2q+2\choose n-q+2}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n-2q+2}}{z^{n-q+3}} \; dz$$
which gives for the sum the integral
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n+2}}{z^{n+3}}
\sum_{q=0}^{n+1} {n+1\choose q} 3^q (-1)^{n-q}
\left(\frac{z}{(1+z)^2}\right)^q
\; dz
\\ = - \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n+2}}{z^{n+3}}
\left(\frac{3z}{(1+z)^2}-1\right)^{n+1}
\; dz
\\ = - \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+3}} (-1+z-z^2)^{n+1}
\; dz.$$
Put $w=-z$ which just rotates the small circle to get
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{(-w)^{n+3}} (-1-w-w^2)^{n+1}
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+3}} (1+w+w^2)^{n+1} \; dw.$$
We get for the final answer
$$[w^{n+2}] (1+w+w^2)^{n+1}$$
but we have $2n+2-n-2 = n$ and thus exploiting the symmetry
of $1+w+w^2$ we get
$$[w^n] (1+w+w^2)^{n+1}.$$
To evaluate the RHS introduce the integral representation
$${k\choose n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^k}{z^{n-k+1}} \; dz.$$
This gives for the sum the representation
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\sum_{k=0}^{n+1} {n+1\choose k} \left((1+z)z\right)^k \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z(1+z))^{n+1} \; dz
.$$
The answer is
$$[z^n] (1+z+z^2)^{n+1},$$
the same as the LHS, and we are done.
We have not made use of the properties of complex integrals here so
this computation can also be presented using just algebra of
generating functions.
This MSE link has another computation in the same spirit.
Apparently this method is due to Egorychev although some of it is
probably folklore.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/664823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $\left(\frac12(x+y)\right)^2 \le \frac12(x^2 + y^2)$ Prove that $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
I've gotten that
$$\left(\frac12(x+y)\right)^2 \ge 0 $$
but stumped on where to go from here...
|
Expand both sides and we have:
$$\frac{x^2 + y^2}{2} \ge \frac{x^2 + 2xy + y^2}{4}$$
$$\iff x^2 + y^2 \ge {2xy}$$
$$\iff (x-y)^2 \ge 0$$
Which is well-know fact.
Also there are lot of other ways to prove it, here's "fancy" one using Cauchy-Scwarz inequality:
$$(1 + 1)(x^2 + y^2) \ge (x + y)^2$$
Multiply both sides by $\frac 14$:
$$\frac 12 (x^2 + y^2) \ge \left(\frac 12 (x+y)\right)^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/665206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
}
|
Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
|
Yet another method, inspired by looking at the problem geometrically (try drawing the region $a^2+b^2\leq2$ and the line $a+b=2$): let $s=a+b$, $t=a-b$. Then $a=\frac12(s+t)$ and $b=\frac12(s-t)$, so $a^2+b^2=\frac14\bigl((s^2+2st+t^2)+(s^2-2st+t^2)\bigr) = \frac12(s^2+t^2)$ and $a+b=s$, so the problem becomes:
if $s^2+t^2\leq 4$ then show that $s\leq 2$.
But this is manifestly true; $t^2\geq 0$, so if $s^2+t^2\leq 4$ then $s^2\leq 4$ and so $s\leq 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/666217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
}
|
Generating functions - difficulty with question
Find the number of non-negative integers solutions to the equation $$x_1+x_2+x_3+x_4=12$$
when $x_1=2x_2+2$ and $x_3 \le x_4$.
My try:
Iv'e substituted $x_1$, thus the equation is $3x_2+x_3+x_4=10$. Second, we may define $x_4=x_3+y$ where $y \ge 0$ and the final equation is $3x_2+2x_3+y=10$.
Now we can write $$g(x)=\sum_{k=0}^{\infty} \left(x^{3k} \right) \sum_{i=0}^{\infty} \left(x^{2i} \right) \sum_{j=0}^{\infty} \left(x^j \right)=\frac{1}{(1-x^3)(1-x^2)(1-x)}$$
I know how to continue, but it's far too complicated and messy... I'm looking for an easier way.
Please help, thank you!
|
One way to simplify the computation would be to take out the $\frac{1}{1-x^3}$, so that you only have to do partial fractions on $\frac{1}{(1-x^2)(1-x)}$. Then you get
\begin{align*}
\frac{1}{(1-x^2)(1-x)}
&= \frac{1}{4(1+x)} + \frac{1}{4(1-x)} + \frac{1}{2(1-x)^2} \\
&= \frac{1}{4} \sum_{i=0}^\infty (-x)^i
+ \frac{1}{4} \sum_{i=0}^\infty x^i + \frac{1}{2} \sum_{i=0}^\infty (i+1)x^i \\
&= \sum_{i=0}^\infty \left[ \frac{(-1)^i + 1 + 2(i+1)}{4} x^i\right]
\end{align*}
Then,
$$
g(x) = (1 + x^3 + x^6 + x^9 + \cdots) \sum_{i=0}^\infty \left[ \frac{(-1)^i + 1 + 2(i+1)}{4} x^i\right]
$$
Your answer is the coefficient of $x^{10}$ in $g(x)$, which is, using the above,
\begin{align*}
\quad \frac{(-1)^{10} + 1 + 2(10 + 1)}{4}
&+ \frac{(-1)^{7} + 1 + 2(7 + 1)}{4} \\
+ \frac{(-1)^{4} + 1 + 2(4 + 1)}{4}
&+ \frac{(-1)^{1} + 1 + 2(1 + 1)}{4} \\
&= \frac{24}{4} + \frac{16}{4} + \frac{12}{4} + \frac{4}{4} \\
&= 6 + 4 + 3 + 1 \\
&= 14.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/666558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Does $b^2 = 4a + 2$ have integer solutions? I got the following question on an exam I had today: $$b^2 = 4a+2$$
I said that it didn't since $b^2 -2$ was not a multiple of $4a$ or in other words, was not divisible by $4$. Is this correct?
|
We assume, on the contrary, that $b^2 = 4a + 2$ does indeed have integer solutions. Then, $b^2 \equiv 2 \mod 4 \implies 2$ is a quadratic residue modulo $4$.
We now aim to obtain a contradiction by proving (as demanded by the rigorous standards of mathematics) that $2$ cannot be a quadratic residue modulo $4$, i.e. $b^2 \not\equiv{2} \mod 4$ for all $b\in\mathbb{Z}$.
Note that the integer $b$ can be written as either one of $4k, 4k + 1, 4k + 2, 4k + 3$ for some $k \in \mathbb{Z}$. Taking the residues of the square of each modulo $4$, we get :
$$\begin{align}b^2 &= (4k)^2, (4k + 1)^2, (4k + 2)^2, (4k + 3)^2\\
&\equiv 0, 1^2,2^2, 3^2 \mod 4\\
&\equiv 0,1 \mod4\end{align}$$
From the above, we see that $b^2 \not\equiv{2} \mod 4$ for all $b\in\mathbb{Z}$, and hence $2$ cannot be a quadratic residue modulo 4, leading to a contradiction. Thus, there cannot be any integer solutions in $a,b$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/669042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Optimization with a few Variables (AMC 12 question) In the 2013 AMC 12B, question 17 says:
Let $a$,$b$, and $c$ be real numbers such that
$a+b+c=2$, and $a^2+b^2+c^2=12$
What is the difference between the maximum and minimum possible values of $c$?
I was wondering if there is a quick and easy solution using multivariable calculus for this problem. (I've only taken single variable)
The only solution I've seen uses the Cauchy-Schwarz inequality.
|
Both extremes occur when $a=b.$
This is a geometric observation, the plane $a=b$ is a plane of symmetry for both the sphere and the (orthogonal) plane $a+b+c=2.$ Put another way, switching $(a,b)$ does not change anything, you still get a point in the intersection.
So, there is an argument that makes smaller demands on background. The intersection of a plane and sphere is a circle. For any point $(a,b,c),$ whenever $a \neq b,$ there are two points with the same $c$ value, $(a,b,c)$ and $(b,a,c).$ Such points are not along the circle diameter along which the maximum and minimum values of $c$ occur. So, the diameter has $a=b.$ Substitution then gives max $c = 10/3,$ min $c=-2.$
Oh, well. By carefully plugging in, one may confirm that this gives your circle as a function of an independent variable $t$
$$ a = \frac{2}{3} - \frac{4}{\sqrt 3} \cos t - \frac{4}{ 3} \sin t, $$
$$ b = \frac{2}{3} + \frac{4}{\sqrt 3} \cos t - \frac{4}{ 3} \sin t, $$
$$ c = \frac{2}{3} + \frac{8}{ 3} \sin t, $$
as $a+b+c=2$ and $a^2 + b^2 + c^2 = 12.$ As $\sin t$ varies between $1$ and $-1,$ the largest $c$ gets is $10/3,$ the smallest $-6/3 = -2.$ The extremes occur when $\cos t = 0,$ so $a=b.$
The information used to construct the parametrization above comes from the upper left three rows by three columns of
$$
\left( \begin{array}{rrrrrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9
\end{array}
\right),
$$
with columns that are orthogonal, but not orthonormal, eigenvectors for the matrix with all entries equal to $1.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/669257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
For given positive integers $n,k$ prove that there always exists some positive integer $x$ for which $2^n\mid \dfrac{x(x+1)}{2}-k$ For given positive integers $n,k$ prove that there always exists some $x$ for which $2^n \mid \dfrac{x(x+1)}{2}-k.$
My work:
$\dfrac{x(x+1)}{2}$ is the sum of all positive integers upto $x$.
Now, if we can show,
$\dfrac{x(x+1)}{2}\equiv 0,1,\ldots,2^n-1 \mod2^n$ then we are done.
Now, two cases arise,
(i) $x=2j \implies j(2j+1)\equiv 0,1,\ldots,2^n-1$
(ii) $x+1=2j \implies j(2j-1)\equiv 0,1,\ldots,2^n-1$
Now, I have reached a deadlock. Please help.
|
$$P_{n,k}(x) \equiv 2^n | \frac{x^2 + x}2 - k$$
By mechanical inspection we can see that the $x$ which satisfy $P_{n,k}$ this are all of the form $x_z = z\cdot 2^{n+1} + B$. So it suggests that we should consider this equation this equation in modulo $2^{n+1}$.
Consider possible solutions of the form $x = B \pmod {2^{n+1}}$:
$$2^n = \frac {B^2 + B} 2 - k \pmod {2^{n + 1}}$$
$$B^2 + B - 2k = 0 \pmod {2^{n+1}}$$
So we only have to show that this polynomial always has a solution in $B$. Induct on increasing $n$.
The base case for $n = 1$, $B^2 + B - 2k = 0 \pmod {4}$ finds solutions by a checking the 4 possible values of $k$.
The inductive case follows from Hensel's Lemma.
More detailed information is available from this XKCD thread.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/670893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How prove this inequality $\sum\limits_{cyc}\frac{x+y}{\sqrt{x^2+xy+y^2+yz}}\ge 2+\sqrt{\frac{xy+yz+xz}{x^2+y^2+z^2}}$ let $x,y,z$ are postive numbers,show that
$$\dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}+\dfrac{y+z}{\sqrt{y^2+yz+z^2+zx}}+\dfrac{z+x}{\sqrt{z^2+zx+x^2+xy}}\ge 2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}$$
My try: Without loss of generality，we assume that $$x+y+z=1$$
and use Holder inequality,we have
$$\left(\sum_{cyc}\dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}\right)^2\left(\sum_{cyc}((x+y)(x^2+xy+y^2+yz)\right)\ge\left(\sum_{cyc}(x+y)\right)^3$$
then we only prove this
$$\dfrac{8}{\sum_{cyc}(x+y)(x^2+xy+y^2+yz)}\ge 2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}$$
then I can't
Thank you
|
We can use here the Ji Chen's lemma:
https://artofproblemsolving.com/community/c6h194103
for $p=\frac{1}{2}.$
For which it's enough to prove that:
1.$$\sum_{cyc}\frac{(x+y)^2}{x^2+xy+y^2+yz}\geq2+\frac{xy+xz+yz}{x^2+y^2+z^2},$$
2.$$\sum_{cyc}\frac{(x+y)^2(y+z)^2}{(x^2+xy+y^2+yz)(y^2+yz+z^2+zx)}\geq1+\frac{2(xy+xz+yz)}{x^2+y^2+z^2},$$
3. $$\prod_{cyc}\frac{(x+y)^2}{x^2+xy+y^2+yz}\geq\frac{xy+xz+yz}{x^2+y^2+z^2}.$$
A proof of the first inequality.
By C-S $$\sum_{cyc}\frac{(x+y)^2}{x^2+xy+y^2+yz}\geq\frac{\left(\sum\limits_{cyc}(x+y)\right)^2}{\sum\limits_{cyc}(x^2+xy+y^2+yz)}=\frac{2\sum\limits_{cyc}(x^2+2xy)}{\sum\limits_{cyc}(x^2+xy)}.$$
Now, let $x^2+y^2+z^2=k(xy+xz+yz).$
Thus, $k\geq1$ and it's enough to prove that $$\frac{2(k+2)}{k+1}\geq2+\frac{1}{k}$$ or $$k\geq1,$$ which is true;
A proof of the second inequality.
By C-S and Rearrangement $$\sum_{cyc}\frac{(x+y)^2(y+z)^2}{(x^2+xy+y^2+yz)(y^2+yz+z^2+zx)}\geq\frac{\left(\sum\limits_{cyc}(x+y)(y+z)\right)^2}{\sum\limits_{cyc}(x^2+xy+y^2+yz)(y^2+yz+z^2+zx)}=$$
$$=\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+2x^3y+3x^3z+4x^2y^2+6x^2yz)}=$$
$$=\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+3x^3y+3x^3z+4x^2y^2+5x^2yz)-xyz\sum\limits_{cyc}\left(\frac{x^2}{z}-x\right)}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+3x^3y+3x^3z+4x^2y^2+5x^2yz)}.$$
Id est, it's enough to prove that
$$\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+3x^3y+3x^3z+4x^2y^2+5x^2yz)}\geq\frac{(x+y+z)^2}{x^2+y^2+z^2}.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, it's obvious that the last inequality is a linear inequality of $w^3$, which says that it's enough to prove the last inequality for an extreme value of $w^3$, which happens in the following cases.
a) $w^3\rightarrow0^+$.
Let $z\rightarrow0^+$ and $y=1$.
We need to prove that $$\frac{(x^2+3x+1)^2}{x^4+1+3x^3+3x+4x^2}\geq\frac{(x+1)^2}{x^2+1}$$ or
$$x(x^4+x^3-2x^2+x+1)\geq0,$$ which is obvious;
b) Two variables are equal.
Let $y=z=1$.
Here we obtain:
$$(x-1)^2(2x^3+11x^2+12x+2)\geq0.$$
A proof of the third inequality.
We need to prove that
$$(x^2+y^2+z^2)\prod_{cyc}(x+y)^2\geq(xy+xz+yz)\prod_{cyc}(x^2+xy+y^2+yz),$$
which is obviously true after full expanding:
$$\sum_{cyc}(x^6y^2+x^5y^3+x^6yz+2x^5y^2z-x^4z^3y-x^4y^2z^2-3x^3y^3z^2)\geq0.$$
Done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/673411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
}
|
Solve exponential-polynomial equation Solve the equation in $\mathbb{R}$
$$10^{-3}x^{\log_{10}x} + x(\log_{10}^2x - 2\log_{10} x) = x^2 + 3x$$
To be fair I wasn't able to make any progress. I tried using substitution for the logarithms, but it didn't help at all.
This is a contest problem, so there should be a nice solution.
Any help? Clue?
|
Clearly $x > 0$. Let $x = 10^y$. Then we have $10^{-3}x^y + x(y^2 - 2y) = x^2 + 3x$.
$$\iff 10^{-3}x^{y-1} + (y-3)(y+1) = x$$
$$\iff 10^{y^2-y-3} + (y-3)(y+1) = 10^y$$
$$\iff \underbrace{10^y(10^{(y-3)(y+1)}-1)} + \underbrace{(y-3)(y+1)} = 0$$
Now note that if $(y-3)(y+1) > 0$, both terms on the left are positive, hence the equation cannot have a solution. Similarly, if $(y-3)(y+1) < 0$, both terms are negative and again we cannot have a solution. Hence the only solutions are when $(y-3)(y+1) = 0 \implies y \in \{-1, 3\} \implies x \in \{\frac1{10}, 1000\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/675676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Real number multiplicative inverses expressed in another form I've been asked to express the multiplicative inverse of $3 + \sqrt{5}$ in the form $c + d\sqrt{5}$, where $c,d$ are rational numbers.
I understand that for some rational numbers $c,d$ we must have:
$$1 = (3 + \sqrt{5})(c + d\sqrt{5}).$$
I was able to answer for the multiplicative inverse of $2 +\sqrt{3}$. We find that $1 = (2 +\sqrt{3})(c + d\sqrt{3})$ where $c = 2$ and $d = -1$. However this seems to be related to the original $2 +\sqrt{3}$; and in the problem at hand this is not the case. So I'm rather confused.
How could i go about solving this? I'd like steps without the answer IF possible; moreover, how could i prepare for more general questions in this form?
|
We have $a+b\sqrt c$, where $a,b,c$ are known, and $c$ is not a perfect square, and we want to find $x$ and $y$ to make $$(x +y\sqrt c)(a + b\sqrt c) = 1.$$
Multiplying out, we get $$(ax + bcy) + (ay+bx)\sqrt c = 1.$$
The first term, $ax+bcy$, is an integer and its sum with $(ay+bx)\sqrt c$ is $1$, another integer, so $(ay+bx)\sqrt c$ must be an integer as well. The only way this can happen is if $ay+bx=0$, in which case we must have $ax+bcy = 1$. So we have two equations in $x$ and $y$:
$$\begin{align}
ay&+bx&=0 \\
ax&+bcy&=1
\end{align}$$
We can solve these by any of the usual methods and get $$\begin{align}
x&=\frac{a}{a^2-b^2c} \\
y&=\frac{-b}{a^2-b^2c}
\end{align}$$
so the solution is $$\frac{a}{a^2-b^2c} + \frac{-b}{a^2-b^2c}\sqrt c.$$
Checking with your $2+\sqrt3$ example, we have $a=2, b=1, c=3$, so $a^2-b^2c = 1$ and the inverse should be simply $a-b\sqrt 3 = 2 -\sqrt 3$, as you said.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/677938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Summation - relatively simple? I have a question which might be too simple for this site but I really tried many ideas without coming to a solution. This is assignment from elementary school in which I am trying to help and the solution should be relatively simple but somehow I cannot figure out the correct approach.
The assignment is as follows i.e. calculate the sum:
$$\frac{1}{1+2} + \frac{1}{2+3} + \frac{1}{3+4} + … + \frac{1}{98+99}+\frac{1}{99+100}$$
|
I think there isn't a nice closed form for this sum. But if we let $H_n$ denote $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$, then
$\frac{1}{1 + 2} + \frac{1}{2 + 3} + ... + \frac{1}{99 + 100} = \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{199} = \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{200} - 1 - (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{200}) = H_{200} - 1 - \frac{1}{2}(1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{100}) = H_{200} - \frac{1}{2}H_{100} - 1$
But the harmonic sum does not have a nice closed form as far as I knew.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/679430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
}
|
Eliminate the parameter from the parametric equations $$x=\frac{3t}{1+t^3} , y=\frac{3t^2}{1+t^3} , t \neq -1,$$
and hence find an ordinary equation in x and y for this curve, The parameter t can be interpreted as the slope of the line joining the general point $(x,y)$ to the origin. Sketch the curve and show that the line $x+y=-1$ is an asymptote.
|
Note that
$$
t=\frac yx
$$
Then
$$
x=\frac{3\frac yx}{1+\frac{y^3}{x^3}}=\frac{3x^2y}{x^3+y^3}\implies x^3+y^3-3xy=0
$$
As $x$ gets big, note that the major terms are $x^3$ and $y^3$, so if we scale $x$ and $y$ by $\lambda$, we get
$$
x^3+y^3-\frac3\lambda xy=0\to0=x^3+y^3=(x+y)(x^2-xy+y^2)
$$
Since there is no solution to $x^2-xy+y^2=0$ we are left with the scaled asymptote
$$
x+y=0
$$
If we scale back, and check with the original formulae, we get
$$
x+y\to\lim_{t\to-1}\frac{3t+3t^2}{1+t^3}=\lim_{t\to-1}\frac{3+6t}{3t^2}=-1
$$
Thus, we get the unscaled asymptote
$$
x+y+1=0
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/680610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Integer solutions of $ x^3+y^3+z^3=(x+y+z)^3 $ Consider the equation
$$ x^3+y^3+z^3=(x+y+z)^3 $$
for triples of integers $(x, y, z) $.
I noticed that this has infinitely many solutions: $ x, y $ arbitrary and $ z=-y $.
Are there more solutions?
|
For such equations :
$$(x+y)^3+x^3+y^3+z^3=(x+y+z)^3$$
You can write the formula.
$$x=3p^2+18ps-s^2$$
$$y=15p^2-6ps-5s^2$$
$$z=3p^2-6ps+7s^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/682666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
}
|
How find this integral $I=\int_{-\infty}^{+\infty}\frac{x^3\sin{x}}{x^4+x^2+1}dx$ Find this integral
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
my idea:
$$I=2\int_{0}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
because
$$\dfrac{x^3\sin{x}}{x^4+x^2+1}\approx\dfrac{\sin{x}}{x},x\to\infty$$
so
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
converges
then I can't,Thank you
|
Note that
$$I(a) = \int_{-\infty}^\infty \frac{t\sin at} {t^2+1}dt
= \pi e^{-a}
$$
(derived from $I’’(a)=I(a)$, $I(0)= -I’(0) =\pi$). Then
\begin{align}
\int_{-\infty}^{\infty}\frac{x^3\sin x}{x^4+x^2+1}\> dx
= &\frac12\int_{-\infty}^{\infty}
\overset{x=\frac{\sqrt3t -1}2}{\frac{(x+1)\sin x}{x^2+x+1}} +\overset{x= \frac{\sqrt3t +1}2}{\frac{(x-1)\sin x}{x^2-x+1}} d x\\
=&\cos\frac1{2} \int_{-\infty}^{\infty} \frac{t\sin \frac {\sqrt3}{2}t}{t^2+1} dt - \frac1{\sqrt3}\sin\frac1{2} \int_{-\infty}^{\infty} \frac{\cos\frac {\sqrt3}{2}t}{t^2+1} dt \\
=& \cos\frac1{2} I(\frac {\sqrt3}{2})
+\frac1{\sqrt3}\sin\frac1{2} I’(\frac {\sqrt3}{2})\\
=&\pi \>e^{-\frac {\sqrt3}{2} }\left( \cos\frac12-\frac1{\sqrt3} \sin\frac12\right)\\
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/683000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}=1$? I'm working on an assignment where part of it is showing that $S_k=0$ for even $k$ and $S_k=1$ for odd $k$, where
$$S_k:=\sum_{j=0}^{n}\cos(k\pi x_j)= \frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}}+e^{-ik\pi x_{j}}) $$
Here $x_j=j/(n+1)$.
So, working through the algebra:
$$\frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}} +e^{-ik\pi x_{j}}) =\dots
=\frac{1}{2}\cdot\frac{1-e^{ik\pi}}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{2}\cdot\frac{1-e^{-ik\pi}}{1-e^{-\frac{ik\pi}{n+1}}}
$$
Obviously $S_k=0$ for even $k$'s, since $e^{i\pi\cdot\text{even integer}}=1$. But when $k$ is odd we get $$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}$$
which isn't obviously one to me, at least. Wolfram alpha confirms it equals 1.
My question: How does one see that it equals 1?
|
Using trigonometric identities
\begin{align}
\cos (k \pi x_j) &= \frac{\sin (k \pi x_{j+1}) - \sin (k \pi x_{j-1})}{2 \sin (\frac{k \pi}{n+1})} \\
\sum_{j=0}^{n} \cos (k \pi x_j) &= 1+ \frac{1}{2 \sin (\frac{k \pi}{n+1})} \sum_{j=1}^{n} {\sin (k \pi x_{j+1}) - \sin (k \pi x_{j-1})} \\
\end{align}
In the summation on the RHS, the terms that remain are
$-\sin (\frac{k \pi}{n+1}) + \sin (\frac{k \pi n}{n+1}) + \sin (k \pi)$ which clearly equals $0$ when $k$ is even, and $-2 \sin(\frac{k \pi}{n+1})$ when $k$ is odd.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/686167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
}
|
How to find the sum of this power series $\sum\limits_{n=0}^\infty \frac {x^{5n}} {(5n)!}$ How to prove that
$$
\sum\limits_{n=0}^\infty \frac {x^{5n}} {(5n)!}=
\frac{2}{5} e^{-\cos \left( 1/5\,\pi \right) x}\cos \left( \sin
\left( 1/5\,\pi \right) x \right) +\frac{2}{5}\, e^{\cos \left( 2/5\,
\pi \right) x}\cos \left( \sin \left( 2/5\,\pi \right) x \right) +\frac{1}{5} e^{x}?
$$
|
Hint. Clearly
$$
\frac{1}{5}\sum_{j=1}^5\mathrm{e}^{\omega^j x}=\sum_{n=0}^\infty\frac{x^{5n}}{(5n)!}
$$
where $\omega=\mathrm{e}^{2\pi i/5}$, since
$$
\sum_{j=1}^5 \omega^{jn}=\left\{\begin{array}{ccc} 5&\text{if}& 5\mid n, \\ 0&\text{if} &5\not\mid n. \end{array}\right.
$$
But
$$
\omega=\cos (2\pi/5)+i\sin (2\pi/5), \,\,\omega^2=\cos (4\pi/5)+i\sin (4\pi/5),
\,\,\omega^3=\overline{\omega^2},\,\,\omega^4=\overline{\omega},
$$
and so
$$
\sum_{n=0}^\infty\frac{x^{5n}}{(5n)!}=\frac{1}{5}\sum_{j=1}^5\mathrm{e}^{\omega^j x}=\frac{1}{5}\left(\mathrm{e}^x+2\mathrm{e}^{x\cos(2\pi/5)}\cos\big(\sin(2\pi/5)\big)+2\mathrm{e}^{x\cos(4\pi/5)}\cos\big(\sin(4\pi/5)\big)\right).
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/686423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Prove that a pair of irrational numbers is the solution to a quadratic polynomial. Suppose a, b are two irrational numbers such that ab is rational and a+b is rational. Then a, b are the solution to a quadratic polynomial with integer coeffecients.
|
Suppose $a$, $b$ are two irrational numbers such that $ab$ is rational and $a+b$ is rational. Then $a$, $b$ are the solution to a quadratic polynomial with integer coefficients.
Proof:
Because $a+b$ and $ab$ are rational, from the definition
of a rational number we can write
$$a+b = \frac{m}{n}, \quad\mbox{and}\quad ab = \frac{p}{q},$$
where $m$, $n$, $p$ and $q$ are integers. Now we can
write
\begin{eqnarray*}
(x-a)(x-b) &=& x^2 -(a+b)x + ab\\
&=& x^2 - \frac{m}{n} x + \frac{p}{q} \\
&=& \frac{1}{nq}\left(nq\, x^2 - mq\, x + np\right).
\end{eqnarray*}
So we have a quadratic
polynomial with integer
coefficients ($nq$, $mq$ and $np$), and obviously
the roots of this polynomial
are $a$ and $b$. Therefore
we have the proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/686587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Factoring a difference of 2 cubes I am trying to factorize the expression $(a - 2)^3 - (a + 1)^3$ and obviously I would want to put it in the form of $(a - b)(a^2 + ab + b^2)$
So I start off with the first $(a - b)$ and I get $(a - 2) - (a + 1)$ which I simplify from $(a^2 + a -2a -2)$ to $(a^2 -3a -2)$
Now I'm up to $(a^2 + ab + b^2)$ and I $a^2$ would equal to $(a^2 - 4)$, $ab$ would be $(a - 2) * (a + 1)$ which is $(a^2 + a -2a -2)$ and $b^2$ would be $(a^2 + 1)$..
Then we get $(a^2 - 3a - 2)((a^2 - 4) (a^2 + a - 4a)(a^2 + 1)) $
At this point I get confused because I'm not sure if I did anything correct, and I don't know how to continue this. Any help is much appreciated
Regards,
|
Brute force algebra gives us
$$
(a-2)^3 - (a+1)^3
= (a^3 -6a^2 + 12a - 8) - (a^3 + 3a^2 + 3a +1)
= -9(a^2 - a + 1)
$$
But $a^2 - a + 1$ doesn't have any factors (not real ones, anyway).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/689671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Find all triples $(p; q; r)$ of primes such that $pq = r+ 1$ and $2(p ^ 2+q ^ 2) =r ^ 2 + 1$. We have to find all triples $(p; q; r)$ of primes such that $pq = r+ 1$ and $2(p ^ 2+q ^ 2) =r ^ 2 + 1$. This question was asked in the 2013 mumbai region RMO but i could not find a solution to it. Can you please help me out with this?
|
First, we square the first equation. We have: $p^2q^2=r^2+2r+1$. If we plug $r^2+1$ we would have: $p^2q^2=2(p^2+q^2+r)$. Since $p$ and $q$ are prime numbers and the RHS is even, at least one of them should be $2$. Without loss of generality we take $p=2$. We have: $4q^2=2(4+q^2+r)$ so $q^2=r+4$. If we plug this into the second equation, we have: $2(4+r+4)=r^2+1$ so $r^2-2r+15=0$. The solutions are $r=-3,5$ but $-3$ is not a prime number so $r=5$. Plugging this into the first equation, we get $2q=6$ so $q=3$. Thus, the only triplets would be: $(p,q,r)=(2,3,5),(3,2,5)$P.S.: As Singhal suggested, Technically −3 is a prime. But in here it does not lead to any solutions
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/689884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
}
|
How to solve simple trigonometry equation. So we are learning trigonometry in school and I would like to ask for a little help with these. I would really appreciate if somebody can explain me how I can solve such equations :)
*
*$\sin 3x \cdot \cos 3x = \sin 2x$
*$2( 1 + \sin^6 x + \cos^6 x ) - 3(\sin^4 x + \cos^4 x) - \cos x = 0$
*$3 \sin^2 x - 4 \sin x \cdot \cos x + 5 \cos^2 x = 2$
*$\sin^2 x - \sin^4 x + \cos^4 x = 1$
In our students book they're poorly explained with 2 pages, I tried to find a solution on the web, but still couldn't find similar examples. All we got from our teacher was paper with few formulas and we basically have no idea when to use them. I would show what I've tried, but the problem is that I have no idea to even start solving such equations.
|
For the second:
$$
2\left(1+\sin^6 x+\cos^6 x\right)-3\left( \sin^4 x + \cos^4 x\right) - \cos x= 0 \\
2\left(1+\left( \sin^2 x + \cos^2 x \right) \left(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x\right)\right)-3\left( \sin^4 x + \cos^4 x\right) - \cos x = 0 \\
2\left(1+\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x\right)-3\left( \sin^4 x + \cos^4 x\right) - \cos x= 0 \\
2+2\sin^4 x - 2\sin^2 x \cos^2 x + 2\cos^4 x-3\sin^4 x -3\cos^4 x - \cos x= 0 \\
2-\sin^4 x - 2\sin^2 x \cos^2 x - \cos^4 x - \cos x= 0 \\
2-\left(\sin^4 x + 2\sin^2 x \cos^2 x + \cos^4 x \right)- \cos x= 0 \\
2-\left(\sin^2 x + \cos^2 x \right)^2- \cos x= 0 \\
2-1- \cos x= 0 \\
\cos x=1 \\
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/690465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Please explain where the $2x$ came from in this cubic identity: The question is to factorize the difference of cube identity $(x + 1)^3 - y^3$
I obviously want to put it in the form of $(a - b)(a^2 + ab + b^2)$
My working out:
$(a - b) = (x + 1) - (y) = (x + 1 - y)$
$(a^2 + ab + b^2) = (x^2 + 1) + (x + 1)(y) + (y^2) = (x^2 + 1 + xy + y + y^2) $
Therefore,
$=(x + 1 - y)(x^2 + 1 + xy + y + y^2)$
Much to my dismay, the correct answer is..
$=(x + 1 - y) (x^2 + 2x + 1 + xy + y + y2)$
Any help would be much appreciated, regards.
|
Note that
$$a^2+ab+b^2={\mathbf{\color{red}{(x+1)^2}}}+(x+1)y+y^2$$
and $(x+1)^2\neq x^2+1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/690882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Find the coefficients such that all four roots of $(x^2-px+q)(x^2-qx+p)$ are natural numbers
Find all ordered pairs $(p,q)$ of natural numbers such that all $4$ the roots of $$f(x)=(x^2-px+q)(x^2-qx+p)$$ are natural numbers.
I got a solution of the problem (see below) but I want some alternatives.
My solution
Let $f(x)=(x^2-px+q)(x^2-qx+p)$. Since both quadratics must be nonnegative at $x=1$, $f(1)\ge 0$. Putting $x=1$, we get $(1-p+q)(1-q+p)\ge 0$. After simplifying this we get $(p-q)^2 \le1$, which implies that either $p-q=1$ or $p=q$ or $q-p=1$.
Let the roots of $x^2-px+q$ be $m,n$ and of $x^2-qx+p$ be $r,s$. Consider $p=q$, so $m+n=mn$ therefore $m=n=2$ which implies $p=q=4$. Now consider $p-q=1$, so $m+n-mn=1$, leading to infinite solutions, so looking at $x^2-qx+p$, $r+s=q$ and $rs=p$, so $rs-(r+s)=1$ giving $r=3 $ and $s=2$. So $p=6$ and $q=5$.
Similarly if we proceed with $q-p=1$ we get $q=6$ and $p=5$. Hence the pairs are $(4,4)$, $(5,6)$, $(6,5)$.
|
Here's my take at the problem. Let $a\leq b$ be the roots of the first equation and $c\leq d$ of the second one. Vieta's formulas then tell us that
$$\begin{array}{ccccc}
a+b & = & p & = & cd \\
c+d & = & q & = & ab \\
\end{array}$$
Adding the two and moving everything to one side yields
$$ab + cd - (a+b) - (c+d) = 0$$
or, equivalently,
$$(a-1)(b-1) + (c-1)(d-1) = 2$$
There are three possibilities:
*
*Either $(a-1)(b-1) = (c-1)(d-1) = 1$, which implies $a=b=c=d=2$ and $(p,q)=(4,4)$,
*or $(a-1)(b-1)=2$ and $(c-1)(d-1)=0$, yielding $(a,b,c)=(2,3,1)$ which produces $(p,q)=(5,6)$ (and $d=6$),
*or the pairs $(a,b)$ and $(c,d)$ are swapped, so $(p,q)=(6,5)$ and $(a,b,c,d)=(1,6,2,3)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/692212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that any palindrome with an even number of digits is divisible by 11 Confusing myself here, need some clarification..
First, we consider the palindrome $abccba$. We can see this can be written as
$$a(10^5 + 10^0) + b(10^4 + 10^1) + c(10^3 + 10^2) = a(10^5 + 1) + 10b(10^3 + 1) + 100c(10 + 1)$$ So essentially we see that all palindromes of even digits can be written in the form $x(10^{2k+1} + 1)$, i.e. we must show that any number of the form $(10^{2k+1} + 1)$ is divisible by $11$.
Base case: $10^{2(0)+1} + 1 = 10 + 1 = 11$, which is clearly divisible by $11$.
Induction hypothesis: Assume that $(10^{2k+1} + 1)$ is divisible by $11$, we work to show that $(10^{2(k+1)+1} + 1)$ is divisible by $11$. That is, $(10^{2k+3} + 1) = (10^2\cdot10^{2k+1} + 1) = \dots$
Where am I going wrong?
|
Also, notice that in your inductive proof:
$10^{2k+3} + 1$ = $10^{(2k+2)+1}+1$ = $10^{2\overline{k} +1}+1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/697096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
}
|
Reducing the System of linear equations \begin{align*}
x+2y-3z&=4 \\
3x-y+5z&=2 \\
4x+y+(k^2-14)z&=k+2
\end{align*}
I started doing the matrix of the system:
$$ \begin{pmatrix} 1 & 2 & -3 & 4 \\ 3 & -1 & 5 & 2 \\ 4 & 1 & k^2-14 & k+2 \end{pmatrix} $$
But I still don't know how can I reduce this, could you explain me what steps I should follow?
Which values of $k$ make the system inconsistent, determined, undetermined?
|
Let our augmented matrix be $$A=\begin{pmatrix} 1 & 2 & -3 & 4 \\ 3 & -1 & 5 & 2 \\ 4 & 1 & k^2-14 & k+2 \end{pmatrix}$$ as you pointed out. So, if we do $$-3R_1+R_2\to R_2~(I),~~~ -4R_1+R_3\to R_3~(II),~~~R_2-R_3\to R_3~(III)$$ then we get $$A\to B=\begin{pmatrix} 1 & 2 & -3 & 4 \\ 0 & -7 & 14 & -10 \\ 0 & 0 & 16-k^2 & -k+4 \end{pmatrix}$$ Now think about
*
*$k=4$ which makes your system consistent with infinitely many solutions,
*$k=-4$ which makes your system inconsistent,
and...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/697288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
What will be the solution of this equation? What will be the solution of the equation.
$(x^2+m^2)\frac{\partial^2y}{\partial(x^2+m^2)}+(x+m)\frac{\partial y}{\partial (x+m)}+(x^2+m^2-n^2)=0$ where $m$ may be a constant
|
Hint:
$(x^2+m^2)\dfrac{\partial^2y}{\partial^2(x^2+m^2)}+(x+m)\dfrac{\partial y}{\partial(x+m)}+x^2+m^2-n^2=0$
$(x^2+m^2)\dfrac{\partial}{\partial(x^2+m^2)}\left(\dfrac{\partial y}{\partial(x^2+m^2)}\right)+(x+m)\dfrac{\partial y}{\partial x}+x^2+m^2-n^2=0$
$\dfrac{x^2+m^2}{2x}\dfrac{\partial}{\partial x}\left(\dfrac{1}{2x}\dfrac{\partial y}{\partial x}\right)+(x+m)\dfrac{\partial y}{\partial x}+x^2+m^2-n^2=0$
$\dfrac{x^2+m^2}{2x}\left(\dfrac{1}{2x}\dfrac{\partial^2y}{\partial x^2}-\dfrac{1}{2x^2}\dfrac{\partial y}{\partial x}\right)+(x+m)\dfrac{\partial y}{\partial x}+x^2+m^2-n^2=0$
$\dfrac{x^2+m^2}{4x^2}\dfrac{\partial^2y}{\partial x^2}+\left(x+m-\dfrac{x^2+m^2}{4x^3}\right)\dfrac{\partial y}{\partial x}=n^2-m^2-x^2$
$\dfrac{\partial^2y}{\partial x^2}+\left(\dfrac{4x^2(x+m)}{x^2+m^2}-\dfrac{1}{x}\right)\dfrac{\partial y}{\partial x}=\dfrac{4x^2(n^2-m^2-x^2)}{x^2+m^2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/699101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
EXERCISE VERIFICATION: Find where $f(x):=|x|+|x+1|$ is differentiable and calculate its derivative Could someone verify my exercise?
a) $f(x):=|x|+|x+1|$
First, analyse the roots of each absolute value, where they go to zero:
$$|x|:=\left\{\begin{matrix}
& x& x>0 \\
& x- & x<0\\
& 0 & x=0
\end{matrix}\right.$$
$$|x+1|:=\left\{\begin{matrix}
& x+1& x>-1 \\
& -(x+1) & x<-1\\
& 0 & x=-1
\end{matrix}\right.$$
So the principal function will be:
$$f(x):=|x|+|x+1|:=\left\{\begin{matrix}
& x+x+1 &=2x+1& x\ge0 \\
& -x+x+1&=1 & -1\le x<0\\
& -x-x-1 &=-(2x+1) & x<-1
\end{matrix}\right.$$
So, the derivatives wil be:
for $x\ge0$:
$$\lim_{x\to c} \dfrac{f(x)-f(c)}{x-c}=\lim_{x\to c} \dfrac{2x+1-2c-1}{x-c}=\lim_{x\to c} \dfrac{2(x-c)}{x-c}=\lim_{x\to c}2= 2$$
for $-1\le x<0$:
$$\lim_{x\to c} \dfrac{f(x)-f(c)}{x-c}=\lim_{x\to c} \dfrac{1-1}{x-c}=\lim_{x\to c} \dfrac{0}{x-c}=\infty$$ so, $f(x)$ isn´t differentiable at $.-1\le x<0$
for $x<-10$:
$$\lim_{x\to c} \dfrac{f(x)-f(c)}{x-c}=\lim_{x\to c} \dfrac{-(2x+1)+(2c+1)}{x-c}=\lim_{x\to c} \dfrac{-2(x-c)}{x-c}=\lim_{x\to c}-2= -2$$
So, $f(x)$ isn´t differentiable at $-1\le x<0$
|
Since $|x|=\sqrt{x^2}$, we have $|x|'=\dfrac{x}{|x|}$ for $x\neq0$, hence $f'(x)=\dfrac{x}{|x|}+\dfrac{x+1}{|x+1|}$ for $x\notin\{-1,0\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/699992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Having problems using Hensel Lifting for Prime Power Moduli We have $P(x) = x^2 + 3x + 9 ≡ 0 \mod(3^i)$, for $i=1,2,3,4$
for $\mod(3): x^2 + 3x + 9 ≡ x^2$, so $x ≡ 0 \mod(3)$ is a solution.
From Hensel lifting, we have $s = x + a\times 7,\ a = [-P(0)\times (1/3)] \times (P'(0))^{-1}$.
Computing for $a$, we get $a = -3 \times (3)^{-1}$. But, $(3)^{-1}$ is when $3a ≡ 1 \mod(3)$, but this has no solutions since $\gcd(3,3)$ is not $1$. Where am I going wrong? Can somebody help me out please? :(
|
We go through the lifting process. The case $i=1$ has been dealt with. We now deal with $i=2$. Let $p=3$. Instead of $i$ we will use $k$.
We have $P'(x)=2x+3$. This is congruent to $0$ modulo $p$. In that case, the Hensel lifting goes as follows:
1) If we have a root $a$ modulo $p^k$, and $p^{k+1}$ divides $P(a)$, then the root $a$ modulo $p^k$ lifts to $p$ roots modulo $p^{k+1}$;
2) If $p^{k+1}$ does not divide $P(a)$, then the root $a$ does not lift to a root modulo $p^{k+1}$.
Currently we are lifting from $k=1$, with $a=0$. Note that $p^2$ divides $P(a)$, so by 1) the root $a=0$ lifts to $3$ roots modulo $p^2$, namely $a\equiv 0$, $3$, or $6$ modulo $9$.
Now we lift from $k=2$ to $k=3$. First look at $a\equiv 0\pmod{9}$. We have $P(0)\equiv 9\not\equiv 0\pmod{3^3}$, so the root $a\equiv 0\pmod{9}$ does not lift. A similar argument shows that $a\equiv 6\pmod{9}$ does not lift.
Now look at the root $a\equiv 3\pmod{9}$. The number $P(3)$ is divisible by $3^3$, so the root $a\equiv 3\pmod{9}$ lifts to $3$ solutions modulo $3^3$, namely $a\equiv 3$, $12$, and $21$ modulo $3^3$.
We now have $3$ roots modulo $3^3$. For each of them, we investigate liftability to a solution modulo $3^4$. So calculate $P(3)$, $P(12)$, and $P(21)$.
I expect you can now continue, and finish the analysis for $k=4$. As a start, the root $a\equiv 3\pmod{3^3}$ does not lift.
Remark: Since we are only asked to go to the small number $3^4$, there are less tedious ways to proceed than Hensel lifting. We went through the process on the assumption that this is the tool you were asked to use.
Since our polynomial is a quadratic, another way to proceed for general $k$ is to multiply by $4$. We get the equivalent congruence $4x^2+12x+36\equiv 0\pmod{3^k}$. Complete the square, and rewrite as $(2x+3)^2\equiv -27\pmod{3^k}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/700640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Cousin of the Vandermonde binomial identity The Vandermonde binomial identity can be expressed as
\begin{align*}
\sum_{i+j=r} \binom{m}{i} \binom{n}{j} = \binom{m+n}{r} && r \leq m +n.
\end{align*}
While working on an algebra problem, I stumbled on a formally similar, but distinct identity:
\begin{align*}
\sum_{i+j=r} \binom{i}{m}\binom{j}{n} = \binom{r+1}{m+n+1}
&& m+n \leq r.
\end{align*}
This isn't hard to prove or anything. The left-hand side enumerates the subsets $S \subseteq \{1,2,\ldots,r+1\}$ with $|S| = m+n+1$ according to the position of the $(m+1)$st largest element of $S$. But, I found the similarity striking enough to ask the following
Question: Are the parallels between these two formulas just a coincidence? Or, is there something else going on here?
|
It can be adjusted to be a convolution by shifting the upper and lower indices and then there must be a formal power series $f(x)$ for which the second identity
equates the $x^t$ terms in $f(x)^p f(x)^q = f(x)^{p+q}$,
just as the first identity compares the $x^r$ coefficients of $(1+x)^m (1+x)^n = (1+x)^{m+n}$.
Taking $(I,J,R,M,N) = (i+1, j+1, r+2, m+1, n+1) $ the formula becomes
\begin{align*}
\sum_{I+J=R} \binom{I-1}{M-1}\binom{J-1}{N-1} = \binom{R-1}{M+N-1}
&& M+N \leq R.
\end{align*}
which now has the correct form $\sum_{I+J=R} c(M,I)c(N,J) =c(M+N,R)$
for $c(p,q)={{q-1} \choose {p-1}}$.
We can read off $f(x)$ as $\sum c(1,v)x^v = 0 + x + x^2 + x^3 + \dots = \frac{x}{1-x}$ , with $(t,p,q)=(R,M,N)$ and write everything in the original variables. The identity
equates the $x^{r+2}$ coefficients of $\hskip10pt (\frac{x}{1-x})^{m+1}(\frac{x}{1-x})^{n+1} = (\frac{x}{1-x})^{m+n+2}.$
Cancelling the powers of $x$ gives a proof using the binomial theorem with negative exponents, by
equating the $x^{r-m-n}$ coefficients of $\hskip10pt (\frac{1}{1-x})^{m+1}(\frac{1}{1-x})^{n+1} = (\frac{1}{1-x})^{m+n+2}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/706190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
}
|
Prove that if $(u,v)$ is chosen randomly from $S$, there is at least a $50\%$ chance that $u\ne\pm v \bmod N$ I need serious help with this problem.
Suppose $N$ is an odd composite number and $S=\{(x,y) \in \mathbb Z^2 : x^2 \equiv y^2 \mod N\}$
*
*Prove that if $(u,v)$ is chosen randomly from $S$, there is at least a $50\%$ chance that $u \not \equiv \pm v \bmod N$
Attempt:
since $(x,y)$ satisfy $x^2 \equiv y^2 \mod N$
then $x^2 - y^2 \equiv 0 \mod N$
so,
$(x-y)(x+y)=0 \mod n$
and we know that n is a composite number, so there are prime factor p and q such that n=pq
so
$(x-y)(x+y)=0 \bmod pq$
and by Chinese Remainder Theorem we deduce
$(x-y)(x+y)=0 \bmod p$
and
$(x-y)(x+y)=0 \bmod q$
so
$(x-y)(x+y)=kpq$
$(x-y)(x+y)/(pq)=k$
for some constant $k$
that's pretty much all i got, I'm not sure how to finish the proof. any help would be greatly appreciated.
|
Let $n=15$, and fix $y$. We find the probability that if $0\le x\lt 15$, and $x^2\equiv y^2\pmod{15}$, then $x\equiv \pm y\pmod{15}$.
It is not hard to verify that if $y$ is relatively prime to $15$, then there are precisely $4$ values of $x$ such that $x^2\equiv y^2\pmod{15}$. Thus for such $y$ the probability is exactly $50\%$.
If $y^2\equiv 0\pmod{15}$, then $x\equiv y \pmod{15}$. For such $y$, the probability that $x\not\equiv \pm y\pmod{15}$ is $0$.
We cannot have $y^2\equiv 3\pmod{15}$, or $y^2\equiv 5$, or $y^2\equiv 12$.
Now look at $y^2\equiv 6\pmod{15}$. The only $x$ such that $x^2\equiv y^2\pmod{15}$ are $x\equiv \pm 6$. Thus if $y^2\equiv 6\pmod{15}$, the probability that $x\not\equiv \pm y\pmod{15}$ is $0$.
We can deal similarly with $y^2\equiv 9\pmod{15}$ and $y^2\equiv 10\pmod{15}$.
Thus the assertion we were asked to prove is not true.
Remark: If we specify in addition that $x$ and $y$ are relatively prime to $n$, then it is indeed true that the probability that $x\not\equiv \pm y\pmod{n}$ is $\ge \frac{1}{2}$, unless $n$ is a prime power. This can be done by a Chinese Remainder Theorem argument.
If $n$ is not a prime power, let $n=ab$ where $a$ and $b$ are relatively prime. Fix $y$. Then the congruence $x^2\equiv y^2\pmod{a}$ has at least two solutions, $x\equiv y$ and $x\equiv -y$.
The same is true modulo $b$.
By the CRT, we can find $x$ such that $x\equiv y\pmod{a}$ and $x\equiv -y\pmod{b}$. This $x$ is not congruent to $y$ or $-y$ modulo $ab$. Neither is $-x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/707067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How to show $\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$?
Show that $\,\displaystyle\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$.
I'm thinking right now (though not getting anywhere with it) that I want to expand out the summation portion to $i!/2!(i-2)!$ and simplify from there? Not sure if that will help, not to mention if I put $1$ in for $i$ I get $1/-2$ which I don't think is right.
Anyone care to shed some light on the subject?
Thanks.
|
Well, if $n-1 =2k$ $$\displaystyle\sum_{i=1}^{n} \binom{i}{2} = \displaystyle\sum_{i=1}^{n}\frac{i\cdot (i-1)}{2}=\frac{1\cdot 0}{2}+\frac{2\cdot 1}{2}+\frac{3\cdot 2}{2}+\frac{4\cdot 3}{2} + \ldots +\frac{n\cdot (n-1)}{2} = \frac{2\cdot 1+3\cdot 2+4\cdot 3+ \ldots +n\cdot (n-1)}{2} = \frac{2^2\cdot2+4^2\cdot2+6^2\cdot2+\ldots +(n-1)^2\cdot2}{2}=2^2+4^2+6^2+\ldots+(n -1)^2=4(1+2^2+3^2+\ldots+k^2)=\frac{4k\cdot(k+1)\cdot(2k+1)}{6}=\frac{4\frac{n-1}{2}\cdot\frac{n+1}{2}\cdot n}{6} = \frac{(n+1)\cdot n\cdot(n-1)}{3!}=\binom{n+1}{3}$$
Similarly if $n-1 =2k +1$ you have $$\displaystyle\sum_{i=1}^{n} \binom{i}{2} = \displaystyle\sum_{i=1}^{n}\frac{i\cdot (i-1)}{2}=\frac{1\cdot 0}{2}+\frac{2\cdot 1}{2}+\frac{3\cdot 2}{2}+\frac{4\cdot 3}{2} + \ldots +\frac{n\cdot (n-1)}{2} = \frac{2\cdot 1+3\cdot 2+4\cdot 3+ \ldots +n\cdot (n-1)}{2} = \frac{2^2\cdot2+4^2\cdot2+6^2\cdot2+\ldots +(n-2)^2\cdot2+n\cdot(n-1)}{2}=2^2+4^2+6^2+\ldots+(n-2)^2+n\cdot (n-1)=4(1+2^2+3^2+4^2+\ldots+(n-2)^2)+\frac{n\cdot(n-1)}{2}=\frac{4k\cdot(k+1)\cdot(2k+1)}{6}+\frac{n\cdot(n-1)}{2}=\frac{4\frac{n-2}{2}\cdot\frac{n}{2}\cdot (n-1)}{6}+\frac{n\cdot(n-1)}{2}=\frac{(n-2)\cdot n\cdot(n-1)}{3!}+\frac{n\cdot(n-1)}{2}=\frac{n\cdot(n-1)\cdot(n-2)+3n\cdot(n-1)}{3!}=\frac{n\cdot(n-1)\cdot(n-2+3)}{3!}=\frac{n\cdot(n-1)\cdot(n+1)}{3!}=\binom{n+1}{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/707256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
}
|
For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ according as $n$ is odd or even.
Since for any $n\ge3$,$2^{2^n}\equiv2^{2^{n-2}}\pmod9$, if $n=2k+3$
then
$$2^{2^n}=2^{2^{2k+3}}\equiv2^{2^{2k+1}}=(2^{2^{2k}})^2\pmod9?$$
if $n=2k+4$
then
$$2^{2^n}=2^{2^{2k+4}}\equiv2^{2^{2k+2}}=(2^{2^{2k}})^4\pmod9?$$
but I don't know how continue?
|
Note that $2^6 \equiv 1 \pmod 9$.
Also $2^n \equiv 2,4 \pmod 6$ depending on the parity of $n$. If $n$ is odd then $2^n \equiv 2 \pmod 6$, otherwise $2^n \equiv 4 \pmod 6$. This means that $2^n = 6k + 2$ or $2^n = 6k + 4$. Now just substitute and use the first conclusion.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/708110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Coin probability example I have a bag of 10 coins and 1 of them has heads on both sides you choose 1 of the coins from the bag at random and then flip it 100 times observing heads every time and then tell me what the probability you chose a fair coin is.
|
TL;DR The answer is $\frac{9}{2^{100}+9}$
Explanation
The a priori probability that you got a fair coin is $\frac{9}{10}$. The probability of 100 heads in a row with a fair coin is $\frac{1}{2^{100}}$. So, the total probability that you got a fair coin and flipped 100 heads in a row is $\frac{9}{10\cdot2^{100}}$.
The a priori probability that you got the two-heads coin is $\frac{1}{10}$. The probability of 100 heads in a row with that coin is 1. So, the total probability that you got that coin and flipped 100 heads in a row is $\frac{1}{10}$.
The overall probability is
$\frac{P_{fair}}{P_{fair} + P_{DoubleHeads}}$ = $\frac{9\cdot\frac{1}{10\cdot2^{100}}}{\frac{1}{10}+9\cdot\frac{1}{10\cdot2^{100}}}$
Simplifying the fraction gives us $\frac{9}{2^{100}+9}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/713907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Integral $\int_0^1 \left(\arctan x \right)^2\,dx$ Evaluate
$$\int_0^1 \left(\arctan x \right)^2\,dx$$
The answer should be
$${\pi^2\over16} + \frac{\pi\ln(2)}{4} -C$$
where $C$ is Catalan's constant.
How do I proceed?
I tried doing integration by parts twice and got stuck at $$\int_0^1{\frac{\log\left(\frac1{x}+x\right)}{1+x^2}}\,dx$$
|
Integrating by parts twice,
$$ \begin{align} \int_{0}^{1} (\arctan x)^{2} \ dx &= x (\arctan x)^{2} \Big|^{1}_{0} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - \arctan(x) \ln(1+x^{2}) \Big|^{1}_{0} + \int_{0}^{1} \frac{\ln (1+x^{2})}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 + \int_{0}^{1} \frac{\ln(1+x^{2})}{1+x^{2}} \ dx \end{align}$$
Let $x = \tan t $.
Then
$$\begin{align}\int_{0}^{1} (\arctan x)^{2} \ dx &=\frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 - 2 \int_{0}^{\pi /4} \ln (\cos t) \ dt \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 -2 \int_{0}^{\pi /4} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (2nt) - \ln 2 \right) \ dt \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 - 2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\int_{0}^{\pi /4} \cos (2nt) \ dt + \frac{\pi}{2} \ln 2 \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=1}^{\infty} \frac{\sin \left(\frac{\pi n}{2} \right)}{n^{2}} \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - C \end{align}$$
Fourier series of Log sine and Log cos
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/713971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 8,
"answer_id": 4
}
|
How to divide polynomials? The title says it all : How to divide polynomials?
I don't understand the method below taught in my school.
Would any of you mind explaining it or even better, suggest an alternative way to solve this?
|
Here is another method that you asked for. Its main advantage is that you can easily do this in a text editor, which helps immensely in accurately tracking your work. It is completely done with algebraic manipulation.
So, with $A$ as the highest term in numerator and $B$ as the sum of all remaining terms. And $C$ as the highest term in denominator, and $D$ is the remaining terms. The goal is to eliminate $A$ and get an answer in terms of $m + \frac{r}{C + D}$. In other words, $m$ is a multiple of the denominator, and $r$ is the remainder. So you start with two things being divided.
Here we justify an algebraic procedure. Imagine some example:
$$
\color{red}{A = 4x^2}
$$
$$
\color{green}{B = 3x + 5}
$$
$$
\color{blue}{C = 2x^2}
$$
$$
\color{purple}{D = 9x - 7}
$$
So the polynomial being divided is like this, at a high level.
So ignore their values and see what happens when we perform one step:
$$
\frac{A + B}{C + D} =
$$
And note that you can add and subtract a ratio of the highest terms $\frac{A}{C}$ without changing the answer.
$$
\frac{A}{C} - \frac{A}{C} + \frac{A + B}{C + D} =
$$
And you can multiply and divide by the denominator $C + D$ without changing the answer as well. Specifically, you can do this to the subtracted term. It just happens that $A$, which represents the highest term will always cancel out when you simplify this. This is actually the only thing you need to remember. You can derive everything else with this.
$$
\frac{A}{C} - \frac{A}{C}*\frac{C+D}{C+D} + \frac{A + B}{C + D} =
$$
If you start there, you can usually jump right to that step in your head.
So now, we can figure out what it simplifies to, just to see what happens with $A$:
$$
\frac{A}{C} + \frac{-AC -AD}{CC + CD} + \frac{AC + BC}{CC + DC} =
$$
$$
\frac{A}{C} + \frac{BC - AD}{C(C + D)}
$$
So, let's apply it to the example values:
$$
\frac{\color{red}{4x^2} + \color{green}{(3x + 5)}}{\color{blue}{2x^2}+ \color{purple}{(9x - 7)}} =
$$
$$
\frac{\color{red}{4x^2}}{\color{blue}{2x^2}} + \frac{(\color{green}{3x + 5})(\color{blue}{2x^2}) - (\color{red}{4x^2})(\color{purple}{9x-7})}{(\color{blue}{2x^2})(\color{blue}{2x^2}+ \color{purple}{(9x - 7)})} =
$$
$$
\frac{4x^2}{2x^2} + \frac{6x^3 + 10x^2 - 36x^3 + 28x^2}{(2x^2)(2x^2+ (9x - 7))} =
$$
$$
\frac{4x^2}{2x^2} + \frac{3x + 5 - 18x + 14}{2x^2+ (9x - 7)} =
$$
$$
2 + \frac{-15x + 19}{2x^2+ (9x - 7)}
$$
So, the current answer is $2$ plus remainder $(-15x+19)$. If $\frac{-15x}{2x^2} $ has a positive $x$ exponent, then we apply this method to the remainder to get more terms. Let's check the answer:
$$
2(2x^2+ (9x - 7)) + (-15x + 19) =
$$
$$
(4x^2+ 18x - 14) + (-15x + 19) =
$$
$$
\color{red}{4x^2} + \color{green}{3x +5}
$$
Which is our numerator $\color{red}{A} + \color{green}{B}$.
Let's apply this to a different polynomial that requires more than one reduction:
$$
\frac{\color{red}{5x^2} + \color{green}{2x + 3}}{\color{blue}{x} + \color{purple}{1}} =
$$
$$
\frac{5x^2}{x} + \frac{(2x + 3)x - 5x^2}{x(x + 1)} =
$$
$$
5x + \frac{2x + 3 - 5x}{x + 1} =
$$
$$
5x + \frac{\color{red}{-3x} + \color{green}{3}}{\color{blue}{x} + \color{purple}{1}} =
$$
$$
5x - 3 + \frac{3x + 3x}{x(x + 1)} =
$$
$$
5x - 3 + \frac{6}{x + 1} =
$$
So, let's check that the answer is $5x - 3$ with remainder $6$.
$$
(5x - 3)(x + 1) + 6 = 5x^2 + 5x - 3x - 3 + 6 = 5x^2 + 2x + 3
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/717179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Find the integral : $\int\frac{dx}{x^\frac{1}{2}+x^\frac{1}{3}}$ Find the integral : $\int\dfrac{dx}{x^\frac{1}{2}+x^\frac{1}{3}}$
Please guide which substitution fits in this I am not getting any clue on this .. thanks..
|
If $x=m^6$, then $dx=6m^5 dm$ and $x^{\frac{1}{2}}=(m^6)^{\frac{1}{2}}=m^3$ and $x^{\frac{1}{3}}=(m^6)^{\frac{1}{3}}=m^2$, then
$$\int \frac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\int \frac{6m^5dn}{m^3+m^2}=6\int\frac{m^5dm}{m^2(m+1)}=6\int \frac{m^3dm}{m+1}. $$
Now, $$\frac{m^3}{m+1}=m^2-m+\frac{m}{m+1}, $$ then
$$\int \frac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}= 6\int \frac{m^3dm}{m+1}=6\left[\int m^2dm-\int m dm+\int \frac{mdm}{m+1}\right]=$$
$$=6\left[\frac{m^3}{3}-\frac{m^2}{2}+ \int \frac{m+1-1}{m+1}dm \right]= $$
$$=2m^3-3m^2+6\int dm-6\int \frac{dm}{m}=2m^3-3m^2+6m-6\ln |m|+c= $$
$$=2(x^{\frac{1}{6}})^3-3(x^{\frac{1}{6}})^2+6(x^{\frac{1}{6}})-6\ln |x^{\frac{1}{6}}| +c=$$
$$=2x^{\frac{1}{2}}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6\ln |x^{\frac{1}{6}}| +c. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/717902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\frac{\pi}{4}+x\right)
$$
This changes the integral to
$$
\int \cos (2x)\cdot \ln \left(\tan \left(\frac{\pi}{4}+x\right)\right)\,dx
$$
Now let $t=\left(\frac{\pi}{4}+x\right)$ such that $dx = dt$. Then the integral with changed variables becomes
$$
\begin{align}
\int \cos \left(2t-\frac{\pi}{2}\right)\cdot \ln (\tan t)dt &= \int \sin (2t)\cdot \ln (\tan t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \frac{\sec^2(t)}{\tan t}\cdot \cos (2t)\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \cot (2t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\ln \left|\sin (2t)\right|
\end{align}
$$
where $t=\displaystyle \left(\frac{\pi}{4}+x\right)$.
Is this solution correct? Is there another method for finding the solution?
|
$$
\begin{aligned}
I&=\frac{1}{2} \int \cos 2 x \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)^{2} d x \\
&=\frac{1}{2} \int \cos 2 x \ln \left(\frac{1+\sin 2 x}{1-\sin 2 x}\right) d x \\
& =\frac{1}{4} \int \ln \left(\frac{1+y}{1-y}\right) d y, \text { where } y=\sin 2 x\\
&\stackrel{y=\sin 2x}{=} \frac{1}{4}\left[y \ln \left(\frac{1+y}{1-y}\right)+\int \frac{2 y}{y^{2}-1} d y\right] \\
&=\frac{1}{4}\left[y \ln \left(\frac{1+y}{1-y}\right)+\ln \left|y^{2}-1\right|\right]+C
\end{aligned}$$
Now we can conclude that
$$
I=\frac{1}{2} \sin 2 x \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)+\frac{1}{2} \ln |\cos 2 x|+C
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/718719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 3
}
|
Need help with Proof by Strong Induction question So, here is the question:
For any position integer $n$, let $T(n)$ be the number 1 if $n<4$ and the number $T(n-1) + T(n-2) + T(n-3)$ if $n \geq 4$.
We have $T(1)=1, T(2)=2, T(3)=3$ $$T(4)=T(3)+T(2)+T(1) = 1+1+1+1 = 3$$
$$T(5) = T(4)+T(3)+T(2) = 3+1+1 = 5$$
Prove that: (Universal n is an element of all positive integers), $T(n)<2^n$
Any suggestions?
|
Suppose that for all $m\lt n$ we have $T(m)\lt 2^m$. We want to show T(n)\lt 2^n$.
This is certainly true up to $n=3$. Past that, we have $T(n)=T(n-1)+T(n-2)+T(n-3)$. So by the induction assumption we have
$$T(n-1)\lt 2^{n-1},\qquad T(n-2)\lt 2^{n-2},\qquad\text{and}\qquad T(n-3)\lt 2^{n-3}.$$
It follows that
$$T(n)\lt 2^{n-1}+2^{n-2}+2^{n-3}.$$
We will be finished if we can show that $2^{n-1}+2^{n-2}+2^{n-3}\lt 2^n$. Calculate. We have $2^{n-1}+2^{n-2}+2^{n-3}=\frac{1}{2}\cdot 2^n+\frac{1}{4}\cdot 2^n+\frac{1}{8}\cdot 2^n=\frac{7}{8}\cdot 2^n\lt 2^n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/719590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Question about the number of primes greater than $3$ in a sequence of consecutive integers. I recently noticed that for any $x > 16$, it follows that there are at least $2$ integers in the any sequence of 3 consecutive integers that are divisible by a prime greater than $3$.
For example, for $10,11,12$, we have $5|10$ and $11$. For $18,19,20$, we have $5|20$ and $19$. (Note: For the details on my reasoning, see below the line).
For any $n$, does it follow that there is a $y$ such that for any $x \ge y$, there are at least $n-1$ integers divisible by a prime greater than $3$ in the sequence $x, x+1, \cdots x+n-1$?
Here's my reasoning for any sequence of 3 consecutive integers greater than $9$ contain $2$ integers divisible by a prime greater than $3$:
Case 1: $6 | x$
$x+1$ is clearly divisible by a prime greater than $3$. Assume that no prime greater than $3$ divides $x$. We can assume that $x=3^u2$ and $3^u2+2 = 2^v$ where $v > 4$ and $u > 1$. Then, $2^{v-1} - 3^u = 1$ which is impossible by the proof of Catalan's Conjecture.
Case 2: $6 | (x+1)$
$x$ and $x+2$ are clearly divisible by a prime greater than $3$.
Case 3: $6 | (x+2)$
$x+1$ is clearly divisible by a prime greater than $3$. Assume that no prime greater than $3$ divides $x+2$. We can assume that $x+2=3^u2$ and $3^u-2 = 2^v$ where $v > 4$ and $u > 1$. Then, $3^u - 2^{v-1}=1$ which is impossible by the proof of Catalan's Conjecture.
Case 4: $3 | x$ and $2 | x+1$
$x+2$ is clearly divisible by a prime greater than $3$. Assume both $x$ and $x+1$ are not divisible by a prime greater than $3$. Then, $2^v - 3^u = 1$ where $v > 3$ and $u > 2$ but this is impossible by the proof of Catalan's Conjecture.
Case 5: $2 | x$ and $3 | x+1$
$x$ and $x+2$ cannot both be powers of $2$ so one must be divisible by a prime greater than $3$. Further, it is not possible that $x$ is a power of $2$ and $x+1$ is a power of $3$ since by the proof of Catalan's conjecture: $3^u - 2^v \ne 1$ where $u > 2$ and $v > 3$ It is also not possible that $x+2$ is a power of $2$ and $x+1$ is a power of $3$ since $2^m - 3^n \ne 1$ where $m > 3$ and $n > 2$.
Case 6: $2 | x+1$ and $3 | x+2$
$x$ is clearly divisible by a prime greater than $3$. It is not possible for both $x+1$ to be a power of $2$ and $x+2$ to be a power of $3$ since by the proof of Catalan's Conjecture $3^v - 2^u \ne 1$ where $v > 2$ and $u > 3$.
Edit: $x > 9$ is not correct since $18-16=2$ doesn't violate Catalan's conjecture. So, I've corrected it to $x > 16$. Thanks very much to J.B. King for pointing this out.
|
I believe that I've worked out the answer for a sequence of $n$ consecutive integers: $x+1, x+2, \cdots, x+n$
We can assume at least one integer $x+c$ where $c \le n$ is not divisible by a prime greater than $3$ so that $x+c = 2^m3^n$ where $m,n \ge 0$
The question is under what circumstances will $2^m3^n \pm d = 2^u3^v$ where $u,v \ge 0$.
We know that either $2 | d$ or $3 | d$ so we just need to solve for three cases.
Case 1: $2\mid d$ and $3\nmid d$
We can assume $d>0$ and either $d=2^u3^v-2^w$ or $d=2^w - 2^u3^v$.
If $u > w$, then $\frac{d}{2^w} = 2^{u-w}3^v - 1$ and $\frac{d}{2^w}=1$ only if $v=0$ and $u-w=1$. So, it is not true if $d < 2^{u-1}$ or if $x > 2n$
If $w > u$, then $\frac{d}{2^u} = 3^v - 2^{w-u}$ or $\frac{d}{2^u} = 2^{w-u} - 3^v$. So, $\frac{d}{2^u} = 1$ only if $v=2$ and $w-u=3$ or $w-u=2$ and $v=1$ by the proof behind Catalan's Conjecture. So, it is not true if $d < \frac{x}{72}$ or if $x > 72n$ for the first condition or $d < \frac{x}{12}$ or if $x > 12n$ for the second condition.
If $w=u$, then $\frac{d}{2^w} = 3^v - 1$. If this is a power of $2$, then $3^v - \frac{d}{2^w} = 1$. So, that $v=2$ and $\frac{d}{2^w} = 8$ So, it is not true if $d < \frac{8x}{9}$ or $x > \frac{9}{8}n$
Case 2: $3\mid d$ and $2\nmid d$
We can assume $d>0$ and either $d=2^u3^v-3^w$ or $d=3^w - 2^u3^v$.
If $v > w$, then $\frac{d}{3^w} = 2^u3^{v-w} - 1$ and $\frac{d}{3^w}\ne1$ since $v-w>0$
If $w > v$, then $\frac{d}{3^v} = 3^{w-v} - 2^u$ or $\frac{d}{3^v} = 2^u - 3^{w-v}$. So, $\frac{d}{3^v} = 1$ only if $w-v=2$ and $u=3$ or $u=2$ and $w-v=1$ by the proof behind Catalan's Conjecture.
If $w=v$, then $\frac{d}{3^w} = 2^u - 1$. If this is a power of $3$, then $2^u - \frac{d}{3^w} = 1$. So, that $u=2$ and $\frac{d}{2^w} = 3$
Case 3: $6\mid d$
We can assume $d>0$ and $d=2^u3^v - 2^s3^t$ where $u,v > 1$ or $s,t > 1$
If $u = s$, then $\frac{d}{2^u} = 3^v - 3^t \ne 1$ or $\frac{d}{2^u} = 3^t - 3^v \ne 1$. So we can assume that $u \ne s$.
If $v=t$, then $\frac{d}{3^v} = 2^u - 2^s$ or $\frac{d}{3^v} - 2^s - 2^u$. In this case $\frac{d}{3^v} = 1$ only if $u=1$ and $s=0$ or $s=1$ and $u=0$.
If $u > s$ and $v > t$, then $\frac{d}{2^s3^t} = 2^{u-s}3^{v-t} - 1$ and $\frac{d}{2^s3^t} \ne 1$ since $v > t$.
If $s > u$ and $t > v$, then $\frac{d}{2^u3^v} = 2^{s-u}3^{t-v} -1$ and $\frac{d}{2^u3^v} \ne 1$ since $t > v$
If $u > s$ and $v < t$, then $\frac{d}{2^s3^v} = 2^{u-s} - 3^{t-v}$ or $\frac{d}{2^s3^v} = 3^{t-v} - 2^{u-s}$ so the only solutions are $u-s=2$ and $t-v=1$ or $t-v=2$ and $u-s=3$ by the proof of Catalan's Conjecture.
If $s > u$ and $t < v$, then $\frac{d}{2^u3^t} = 3^{v-t} - 2^{s-u}$ or $\frac{d}{2^u3^t} = 2^{s-u} - 3^{v-t}$ so the only solutions are $v-t=2$ and $s-u=3$ or $s-u=2$ and $v-t=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/721737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Does $7$ divide $2 x^2 - 4y^2$ for all $x,y$? Does $7$ divide $2 x^2 - 4y^2$ for all $x,y \in \mathbb{Z}$?
|
...suppose the result holds true, then,
$2x^2-4y^2=0 \pmod7$
since $2$ does not divide $7$, we have $x^2-2y^2=0 \pmod7$, but $2$ is congruent to $9$ modulo $7$, so, $x^2-9y^2=0 \pmod7$ or
$x=3y \pmod7$, so the result holds true $iff$ $x=3y+7k$, for some integer $k$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/722476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Is this proof by induction that $13 \mid 4^{2n+1}+3^{n+2}$ correct/sufficient? To show: $13\, |\, 4^{2n+1}+3^{n+2}$
I used induction beginning successfully with $n=0$ $($or $n=1),$
then making the step to $n+1:$
An $x$ exists so that $13x = 4^{2n+3}+3^{n+3}$
$$\begin{align*}13x &= 16\cdot4^{2n+1}+3\cdot3^{n+2}\\
\frac{13}{3}x &= \frac{16}{3}\cdot4^{2n+1}+3^{n+3}\\
\frac{13}{3}x &= \frac{13}{3}\cdot4^{2n+1}+4^{2n+1}+3^{n+3}\end{align*}$$
Here comes the actual question: Can I say that because of the premise the last two summands can be expressed as $13m$ with $m$ being a natural number? Like this:
$$\begin{align*}\frac{13}{3}x &= \frac{13}{3}\cdot4^{2n+1}+13m\\
x &= 4^{2n+1}+3m,\end{align*}$$ which is a natural number.
Is that a valid proof?
|
Yes, it seems to be correct. However, the person who would read the proof might be confused a little. Instead, you can re-write it differently, like
$$
\frac{4^{2n+3} + 3^{n+3}}{13} = \frac{13 \cdot 4^{2n+1} + 3\left(4^{2n+1} + 3^{n+1} \right)}{13} = \frac{13\cdot 4^{2n+1} + 3 \cdot 13m}{13} \in \mathbb{Z}
$$
which is the same, except it's going from 'bottom up'.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/724054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Combinatorics for integer solutions How many integer solutions are there to the following equation?
$x_1 + x_2 + x_3 = 17$
a) if $x_1 > 1, x_2 > 2, x_3 > 3$
b) if $x_1 < 6, x_3 > 5$ and $x_2$ can be any integer.
c) if $x_1 < 4, x_2 < 3, x_3 < 5$
What is a systematic way to approach and solve these types of problems?
|
Use generating functions. I.e., the value of $x_1$ is represented by $z^2 + z^3 + \cdots = z^2 (1 - z)^{-1}$, and similarly the others. For all three variables it is in case (a):
$$
[z^{17}] z^2 \cdot z^3 \cdot z^4 \cdot (1 - z)^{-3} = (-1)^{17} \binom{-3}{17}
= [z^8] (1 - z)^{-3}
= \binom{8 + 3 - 1}{3 - 1}
= 45
$$
Case (b) gives $1 + \cdots + z^5 = (1 - z^6) / (1 - z)$ for $x_1$, $z^6 + z^7 + \cdots = z^6 (1 - z)^{-1}$ for $x_2$ and $(1 - z)^{-1}$ for $x_3$ (if we restrict $0 \le x_i$, that is; otherwise there are infinite solutions). Similar to before:
$$
[z^{17}] z^6 \cdot (1 - z^6) \cdot (1 - z)^{-2}
= [z^{11}] (1 - z^6) \sum_{r \ge 0} (-1)^r \binom{-2}{r} z^r
= [z^{11}] (1 - z^6) \sum_{r \ge 0} \binom{r + 1}{1} z^r
= \binom{11 + 1}{1} - \binom{6 + 1}{1}
= 5
$$
Case (c) is the most interesting one. As before:
\begin{align}
[z^{17}] \frac{1 - z^4}{1 - z} \cdot &\frac{1 - z^3}{1 - z} \cdot \frac{1 - z^5}{1 - z}
= (1 - z^3 - z^4 - z^5 + z^7 + z^8 + z^9 - z^{12})
\cdot \sum_{r \ge 0} (-1)^r \binom{-3}{r} z^r \\
&= (1 - z^3 - z^4 - z^5 + z^7 + z^8 + z^9 - z^{12})
\cdot \sum_{r \ge 0} \binom{r + 2}{2} z^r \\
&= \binom{19}{2} - \binom{16}{2} - \binom{15}{2} - \binom{14}{2}
+ \binom{12}{2} + \binom{11}{2} + \binom{10}{2} - \binom{7}{2}
\end{align}
The result is obviously 0, as $3 + 2 + 4 < 17$...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/724822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Irreducibility of $X^6+X^3+1$ in $\mathbb{Q}[X]$
Could anyone advise me on how to prove $X^6+X^3+1$ is irreducible in $\mathbb{Q}[X] \ ?$
I'm thinking of substituting $X=Y+1$ into the equation, do some tedious computations to simplify and use Eisenstein's criterion. May I know if that is the correct approach?
Thank you.
|
If you don't want to do the multiplication, write $X^9-1=(X^6+X^3+1)\cdot(X^3-1)$. Now use your $X=Y+1$ trick and reduce mod 3:
$$\begin{align}
(Y+1)^9-1&=\left((Y+1)^6+(Y+1)^3+1\right)\cdot((Y+1)^3-1)\\
Y^9+1^9-1&=\left((Y+1)^6+(Y+1)^3+1\right)\cdot(Y^3+1^3-1)\\
Y^9&=\left((Y+1)^6+(Y+1)^3+1\right)\cdot Y^3\\
Y^6&=(Y+1)^6+(Y+1)^3+1\\
\end{align}$$
which tells you that all terms besides $Y^6$ have coefficient divisible by 3 (note that the Frobenius automorphism was used for the second line). Now plug $X=1$ into your original to find that the constant term is exactly three and then Eisenstein is good to go. The upshot to this method is that it generalizes to when you need to show, say, that $X^{78}+X^{75}+\cdots + X^3+1$ is irreducible.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/725694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Solve $5\sin^2(x) + \sin(2x) - \cos^2(x) = 1$ I tried
$$5\sin^2(x) + 2\sin(x)\cos(x)- (1-\sin^2(x)) = 1.$$
Simplifying,
$$6\sin^2(x) + 2\sin(x)\cos(x) -2 = 0$$
Then I'm stuck!
|
Using the twice-angle formulas, this becomes
$$ -3 \cos(2x) + \sin(2x) = -1 $$
Let $\alpha = \arctan(1/3)$ so $\cos(\alpha) = 3/\sqrt{10}$ and
$\sin(\alpha) = 1/\sqrt{10}$. The equation can be written as
$$ - \cos(\alpha) \cos(2x) + \sin(\alpha) \sin(2x) = - 1/\sqrt{10}$$
or $$\cos(\alpha + 2 x) = 1/\sqrt{10}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/727276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Proof of convergence of $s_n = \frac{1}{2\cdot 1} + \frac{1}{3\cdot 2} + \cdots + \frac{1}{n(n+1)}$ I am really bad at this, but I am trying my best to understand.
If I am given a sequence of partial sums such that
$$s_n = \frac{1}{2\cdot1} + \frac{1}{3\cdot2} + \cdots + \frac{1}{n(n+1)}$$
Prove that $s_n \to 1$.
I always have trouble tackling this. I know I have to find an $N$ s.t. $|s_n - 1 | < \epsilon$. But to what extent can I go to select this $N$? Any tips on selecting $N$ or is it just intuition based?
Or perhaps could I do the following:
I know
$$\begin{align}
s_1 & = \dfrac{1}{2} \\
s_2 & = \dfrac{2}{3} \\
s_3 & = \dfrac{3}{4} \\
& \vdots \\
s_n & = \dfrac{n}{n+1}
\end{align}
$$
and it is trivial to show $\left|\dfrac{n}{n+1} - 1\right|$ converges by doing the following:
$$\left|\frac{n}{n+1} - 1\right| = \left|\frac{n}{n+1} - \frac{n+1}{n+1}\right| = \left|\frac{1}{n+1}\right| \le \frac{1}{N+1} < \epsilon$$
But I am not sure if I have to prove why $s_n = \dfrac{n}{n+1}$. If so I think I would have to do so by induction? Do I have to show how I got $\dfrac{n}{n+1}$ formally and rigorously or can I assume my audience follows?
|
Can you show by induction that
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}. $$
Indeed, for $n=1$ is true because
$$\frac{1}{1\cdot 2}=\frac{1}{2}=\frac{1}{1+1}. $$
Now, suppose that the equality is true for $n=k$, i.e.,
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{k(k+1)}=\frac{k}{k+1}, $$
then, adding $\frac{1}{(k+1)(k+2)}$ we have
$$ \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=$$
$$=\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{(k^2+2k+1)}{(k+1)(k+2)}=\frac{k+1}{k+2}=\frac{k+1}{(k+1)+1},$$
then the equality follow for induction. I have helped this part.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/729724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How prove this sum $\sum_{n=1}^{\infty}\binom{2n}{n}\frac{(-1)^{n-1}H_{n+1}}{4^n(n+1)}$ show that
$$\sum_{n=1}^{\infty}\binom{2n}{n}\dfrac{(-1)^{n-1}H_{n+1}}{4^n(n+1)}=5+4\sqrt{2}\left(\log{\dfrac{2\sqrt{2}}{1+\sqrt{2}}}-1\right)$$
where $H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$
My try: we let
$$s(x)=\sum_{n=1}^{\infty}\binom{2n}{n}\dfrac{(-1)^{n-1}H_{n+1}}{(n+1)}x^{n+1}$$
then
$$s'(x)=\sum_{n=1}^{\infty}(-1)^{n-1}\binom{2n}{n}H_{n+1}x^n$$
then I can't.Thank you
|
There's a transformation formula (might have been derived from generalized Euler series transformation):
$$
f(x)=\sum_{n=0}^\infty \, \binom{2n}{n}\, a_n \, x^n = \dfrac{1}{\sqrt{1+4\, x}}\, g\left(\dfrac{x}{1+4\, x}\right)
$$
where
$$a_n=\sum_{k=0}^n\, \binom{n}{k}\, (-1)^{n-k}\, b_k$$
and
$$g(x)=\sum_{n=0}^\infty \, \binom{2n}{n}\, b_n \, x^n$$
For $a_n=(-1)^{n-1}\, H_n$ and $b_n=\dfrac{1}{n}$, we have the identity
$$
(-1)^{n-1}\, H_n = \sum_{k=1}^n \binom{n}{k}\, \dfrac{(-1)^{n-k}}{n}
$$
Also, from the generating function for central binomial coefficients, the following g.f. can be obtained:
$$
G_1(x)=\sum_{n=1}^\infty \, \binom{2n}{n}\, \dfrac{1}{n}\, x^n = 2\, \log{\left( \dfrac{2}{1+\sqrt{1-4\, x}}\right)}
$$
Therefore,
$\displaystyle \begin{aligned}
f(x)&=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, H_n \, x^n = \dfrac{1}{\sqrt{1+4\, x}}\, \sum_{n=1}^\infty \,
\binom{2n}{n} \, \dfrac{1}{n}\, \left( \dfrac{x}{1+4\, x} \right)^n\\\\
&= \dfrac{1}{\sqrt{1+4\, x}}\, G_1\left(\dfrac{x}{1+4\, x}\right)\\\\
&= \dfrac{1}{\sqrt{1+4\, x}}\, 2\, \log{\left( \dfrac{2}{1+\sqrt{1-4\, \left(\dfrac{x}{1+4\, x}\right)}}\right)}
\end{aligned}
$
Then, consider the g.f. we need in order to get the required sum:
$\displaystyle \begin{aligned}
h(x) &=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{H_{n+1}}{n+1} \, x^n\\\\
&=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{H_{n}+\dfrac{1}{n+1}}{n+1} \, x^n\\\\
&=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{H_{n}}{n+1} \, x^n+\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{1}{(n+1)^2} \, x^n\\\\
&=h_1(x)+h_2(x)
\end{aligned}
$
Both $h_1(x)$ and $h_2(x)$ can be found by integrating the g.f.'s we have already seen:
$\displaystyle \begin{aligned}
h_1(x)&=\dfrac{1}{x}\int f(x) \, dx+\dfrac{4 \, \log\left(2\right) + 1}{4 \, x}\\\\
&=\dfrac{4 \, \sqrt{4 \, x + 1} \log\left(\dfrac{2}{\sqrt{\dfrac{1}{4 \, x + 1}} + 1}\right) - 4 \, \log\left(\sqrt{4 \, x + 1} + 1\right) - 1}{4 \, x} + \dfrac{4 \, \log\left(2\right) + 1}{4 \, x}
\end{aligned}$
$\displaystyle
\begin{aligned}
h_2(x)&= \dfrac{1}{x}\, \int \left(\dfrac{1}{x}\, \int -\dfrac{1}{\sqrt{1+4\, x}}\, dx \right)\, dx + \dfrac{\log\left(x\right) + 2}{2 \, x} + 1\\\\
&= -\dfrac{2 \, \sqrt{4 \, x + 1} - \log\left(\sqrt{4 \, x + 1} + 1\right) + \log\left(\sqrt{4 \, x + 1} - 1\right)}{2 \, x} + \dfrac{\log\left(x\right) + 2}{2 \, x} + 1
\end{aligned}
$
and the required sum is:
$\displaystyle \begin{aligned} h\left(\dfrac{1}{4}\right)&=4 \, \sqrt{2} \log\left(\dfrac{2}{\sqrt{\dfrac{1}{2}} + 1}\right) - 4 \, \sqrt{2} + 4 \, \log\left(2\right) + 2 \, \log\left(\dfrac{1}{4}\right) - 2 \, \log\left(\sqrt{2} + 1\right) - 2 \, \log\left(\sqrt{2} - 1\right) + 5\\\\
&=4 \, \sqrt{2} {\left(\log\left(\dfrac{2 \, \sqrt{2}}{\sqrt{2} + 1}\right) - 1\right)} + 5\\\\
&\approx 0.238892690197059
\end{aligned}$
References:
[1]
[2]
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/730885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 1
}
|
How find this $\frac{3x^3+125y^3}{x-y}$ minimum value
let $x>y>0$,and such $xy=1$, find follow minimum of the value
$$\dfrac{3x^3+125y^3}{x-y}$$
My idea: let $x=y+t,t>0$
then
$$\dfrac{3x^3+125y^3}{x-y}=\dfrac{3(y+t)^3+125y^3}{t}=3t^2+3yt+3y^2+\dfrac{128y^3}{t}$$
and $$(y+t)y=1$$
I think this can use AM-GM inequality.But I can't.
Thank you very much
|
$\dfrac{3x^3+125y^3}{x-y}=\dfrac{3x^6+125x^3y^3}{x^4-x^3y}=\dfrac{3x^6+125}{x^2(x^2-1)}=\dfrac{3(p+1)^3+125}{p(p+1)},p=x^2-1>0$
$\dfrac{3(p+1)^3+125}{p(p+1)}=3p+6+\dfrac{3}{p}+\dfrac{125}{p(p+1)}$
$3p$ is mono increasing function, $\dfrac{3}{p}+\dfrac{125}{p(p+1)}$ is mono decreasing function,so there must be only a min point.with try $p=1,2,3,4,5,6$,we can guess $p=4$ is the point. now we need $\dfrac{1}{p}=\dfrac{k}{p(p+1)}$when we taking AM-GM, so $k=5 $ if $p=4$ is right one, the last step is to verify:
$3p+6+\dfrac{3}{p}+\dfrac{125}{p(p+1)}=25\times\dfrac{(p+1)}{20}+28\times\dfrac{p}{16}+3\times\dfrac{1}{p}+25\times\dfrac{5}{p(p+1)}+\dfrac{9}{4} \ge 91\left(\left(\dfrac{p+1}{20}\right)^{25}\left(\dfrac{p}{16}\right)^{28}\left(\dfrac{1}{p}\right)^3\left(\dfrac{5}{p(p+1)}\right)^{25}\right)^{\frac{1}{91}}+\dfrac{9}{4}=\dfrac{91}{4}+\dfrac{9}{4}=25$
when $\dfrac{(p+1)}{20}=\dfrac{p}{16}=\dfrac{1}{p}=\dfrac{5}{p(p+1)}$ get "=" ie. $p=4$
so min is $25$ when $x=\sqrt{p+1}=\sqrt{5}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/733770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.