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How to prove that for $a_{n+1}=\frac{a_n}{n} + \frac{n}{a_n}$ , we have $\lfloor a_n^2 \rfloor = n$? Let $(a_n)_{n\ge 1}$ be the sequence defined as the following :
$$a_1=1 ,\ a_{n+1}=\dfrac{a_n}{n} + \dfrac{n}{a_n} ,\ n\ge1$$
Show that for every $n\ge4,\ \lfloor a_n^2 \rfloor = n$.
My approach to this problem was trying induction and using the function $f_n(x)=\dfrac{x}{n} + \dfrac{n}{x}$ :
Proving the base case for $n= 4$ and then by the inductive hypothesis $\lfloor a_n^2 \rfloor = n$ implies that $$\sqrt{n} \le a_n \lt \sqrt{n+1}$$
We then apply $f_n$ knowing that it is decreasing in that interval following it up with the floor function and some polishing, all leads to this inequality : $$n+1\le \lfloor a_{n+1}^2 \rfloor \le n+2$$
So I can't exactly get $n+1$ since $n+2$ is a possibility, this problem is a product of the fact that if $a\lt b$ then $\lfloor a \rfloor \le \lfloor b \rfloor$.
Any insights would be greatly appreciated! I wonder if my result is correct because it seems like the only way.
|
We will prove the following result instead.
$$n+{2\over n}<{a_n}^2<n+1$$
First we check it's true for $n=4$.
Now the induction step, first notice that $x>y>1$ implies $x+{1\over x}>y+{1\over y} \equiv 1>{1\over xy}$
Therefore for RHS $${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2<{n+{2\over n}\over n^2}+{n^2 \over n+{2\over n}}+2=2+{n^4+n^2+4+{4\over n^2}\over n^3+2n}$$$$= 2+n-{n^2-4-{4\over n^2}\over n^3+2n}<n+2$$
For LHS,
$${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2>{n+1\over n^2}+{n^2\over n+1}+2={n^4+n^2+2n+1\over n^3+n^2}+2$$ $$=2+(n-1)+{2n^2+2n+1\over n^3+n^2}>n+1+{2\over n}>n+1+{2\over n+1}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $x$ is a nonnegative real number, find the minimum value of $\sqrt{x^2 +4} + \sqrt{x^2 -24x+153}$ If $x$ is a nonnegative real number, then find the minimum value of
$$\sqrt{x^2 +4} + \sqrt{x^2 -24x+153}$$
How can I approach this? Thanks
|
Process 1: Derivate it to zero
Process 2: Using triangle inequality,
Given,$$\sqrt{x^2 +4} + \sqrt{x^2 -24x+153}$$
$$\rightarrow\sqrt{x^2+2^2}+\sqrt{\left(-x+12\right)^2+3^2}\geq \sqrt{\left(x-x+12\right)^2+(2+3)^2} = \sqrt{{12}^2+{5}^2}$$
$$=\sqrt{169}=13$$
I.e, So,
The minimum value is $$\bbox[5px,border:2px solid red]{13}$$
|
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|
What is the meaning of "due to the symmetry of the coefficients, if $x=r$ is a zero of $x^4+x^3+x^2+x+1$ then $x=\frac1r$ is also a zero" I was studying this answer about factoring $x^4+x^3+x^2+x+1$:
https://socratic.org/questions/how-do-you-factor-x-4-x-3-x-2-x-1
The author says: "A cleaner algebraic approach is to notice that due to the symmetry of the coefficients, if $x=r$ is a zero of $x^4+x^3+x^2+x+1$ then $x= {1\over r}$ is also a zero"
And eventually he writes $x^4+x^3+x^2+x+1=(x^2+ax+1)(x^2+bx+1)$
Question $1$: What is the meaning of symmetry of the coefficients?
Question $2$: Can we do the same approach for $x^4-x^3+x^2-x+1$? ( I ask because it is relevant to my other question: Problem with factoring $x^4-x^3+x^2-x+1$)
|
To answer the original question, the thinking process comes as follows:
(1) If $r$ is a solution to $x^4-x^3+x^2-x+1=0$, then $r^4-r^3+r^2-r+1=0$.
(2) Divide both sides by $r^4$ you get $({1\over r})^4-({1\over r})^3+({1\over r})^2-({1\over r})+1=0$. Therefore $1\over r$ is also a solution.
(3) Hence if $(x-r)$ is a factor of the polynomial then $(x-{1\over r})$ is also a factor.
(4) Therefore the equation can be written as $(x-r)(x-{1\over r})(x-s)(x-{1\over s})$
(5) Therefore it can be written as $(x+ax+1)(x+bx+1)$
|
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|
How to prove this equality of the determinant of matrix?
Prove that
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=(a^2-bc)(b^2-ca)(c^2-ab)\end{equation*}
My attempt:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
a(b-a) & b(c-b) & c(a-c) \\
(b+a)(b-a) & (c+b)(c-b) & (a+c)(a-c)
\end{pmatrix}
\end{equation*}
But I think my direction is incorrect. Can anyone give me some hints or the solution of this question?
|
My steps to calculate your determinant. With $R$ I indicate the row and with $C$ the column.
$$C_2=C_2-\left(\frac{b^{2}}{a^{2}}\right)C_1 $$
$$\left| \begin{array}{ccc} a^{2} & 0 & c^{2} \\\\ a b & b c - \frac{b^{3}}{a} & a c \\\\ b^{2} & c^{2} - \frac{b^{4}}{a^{2}} & a^{2} \end{array} \right| \tag{first step}$$
Subtract column $1$ multiplied by $c^2/a^2$ from column $3$; you will have
$$\left| \begin{array}{ccc} a^{2} & 0 & 0 \\\\ a b & b c - \frac{b^{3}}{a} & a c - \frac{b c^{2}}{a} \\\\ b^{2} & c^{2} - \frac{b^{4}}{a^{2}} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right| \tag{second step}$$
Using the first rule of Laplace:
$$\left| \begin{array}{ccc} a^{2} & 0 & 0 \\\\ a b & b c - \frac{b^{3}}{a} & a c - \frac{b c^{2}}{a} \\\\ b^{2} & c^{2} - \frac{b^{4}}{a^{2}} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right|=\left(a^{2}\right) \cdot(-1)^{1+1}\cdot \left| \begin{array}{cc} b c - \frac{b^{3}}{a} & a c - \frac{b c^{2}}{a} \\\\ c^{2} - \frac{b^{4}}{a^{2}} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right|+0 \cdot(-1)^{1+2}\cdot \left| \begin{array}{cc} a b & a c - \frac{b c^{2}}{a} \\\\ b^{2} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right|+0 \cdot(-1)^{1+3}\cdot \left| \begin{array}{cc} a b & b c - \frac{b^{3}}{a} \\\\ b^{2} & c^{2} - \frac{b^{4}}{a^{2}} \end{array} \right|=a^{2} \left| \begin{array}{cc} b c - \frac{b^{3}}{a} & a c - \frac{b c^{2}}{a} \\\\ c^{2} - \frac{b^{4}}{a^{2}} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right|$$
Considering the last determinant you have:
$$\left| \begin{array}{cc} b c - \frac{b^{3}}{a} & a c - \frac{b c^{2}}{a} \\\\ c^{2} - \frac{b^{4}}{a^{2}} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right|=\left(b c - \frac{b^{3}}{a}\right)\cdot\left(\frac{a^{4} - b^{2} c^{2}}{a^{2}}\right)-\left(a c - \frac{b c^{2}}{a}\right)\cdot\left(c^{2} - \frac{b^{4}}{a^{2}}\right)=\frac{\left(a^{2} - b c\right) \left(a b - c^{2}\right) \left(a c - b^{2}\right)}{a^{2}}$$
Hence:
$$\left| \begin{array}{ccc} a^{2} & b^{2} & c^{2} \\\\ a b & b c & a c \\\\ b^{2} & c^{2} & a^{2} \end{array} \right|=a^{2} \cdot \left(\frac{\left(a^{2} - b c\right) \left(a b - c^{2}\right) \left(a c - b^{2}\right)}{a^{2}}\right)=(a^2-bc)(b^2-ca)(c^2-ab)$$
|
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|
Factorize: $x^3 + x + 2$. How do I factorize the term $x^3 + x +2$?
What I have previously tried is the middle term factor method but it didn't work...
$x^3 + x + 2$
$\Rightarrow x^3 + 2x - x + 2$
$\Rightarrow x(x^2 + x) - 2(x - 1)$
This doesn't work.
What should I do?
|
Note
\begin{align}
x^3+x+2& = (x^3+x^2)-(x^2-x-2)\\
&= x^2(x+1)-(x-2)(x+1)\\
&=(x+1)(x^2-x+2)
\end{align}
|
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|
Given positive $x,y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $, find minimum $(x+y)$ I am given positive numbers $x, y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $. I need to find the minimum value of $(x+y)$. Here is my try. Using AM-GM inequality for nonnegative numbers, I have
$$
\frac{(x+y)}{2} \geqslant \sqrt{x} \sqrt{y}
$$
$$
\sqrt{x} \sqrt{y}(x-y) \geqslant 2 \sqrt{x} \sqrt{y}
\\ \therefore (x-y) \geqslant 2
$$
So, I have been able to arrive at this conclusion. But I am stuck here. Any help ?
Thanks
|
By AM-GM
$$(x+y)^2=xy(x-y)^2=\frac{1}{4}\cdot4xy(x-y)^2\leq\frac{1}{4}\left(\frac{4xy+(x-y)^2}{2}\right)^2=\frac{(x+y)^4}{16},$$
which gives $$x+y\geq4.$$
The equality occurs for $(x-y)\sqrt{xy}=x+y$ and $4xy=(x-y)^2,$ which gives $$(x,y)=(2+\sqrt2,2-\sqrt2),$$ which says that we got a minimal value.
|
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|
Find the smallest possible value of an equation, where $a+b+c=3$ We have the positive real numbers $a, b, c$ such that $a+b+c=3$. Find the minimum value of the equation:
$$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$$
I solved it in the following fashion:
$$
\begin{align}
A&=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-(a^2+b^2+c^2)\\
&\ge 2\cdot\frac{3^2}{a+b+c}-(a^2+b^2+c^2) \quad \textrm{(Andreescu inequality)}\\
&=2\cdot3-(a^2+b^2+c^2)\\
&=6-(a^2+b^2+c^2).
\end{align}$$
However, $9=(a+b+c)^2\ge 3(ab+bc+ac)$, so
$$3\ge ab+bc+ac.$$
From this we have that $A \ge 6-3=3$, where equality is true for $a=b=c=1$.
Due to the total simplicity of my solution, I have difficulties believing that it is correct, despite having checked it thoroughly many times. Could you please tell me if my solution is correct and suggest some alternative solutions?
|
We have
$$A = 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\left[(a+b+c)^2-2(ab+bc+ca)\right]$$
$$=2(ab+bc+ca)\left(\frac{1}{abc}+1\right)-9.$$
Using know inequality $(ab+bc+ca)^2 \geqslant 3abc(a+b+c)$ and the AM-GM inequality, we get
$$A \geqslant 2\sqrt{3abc(a+b+c)}\cdot \frac{2}{\sqrt{abc}}-9=3.$$
Equality occur when $a=b=c=1.$ So $A_{\min} = 3.$
|
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|
Evaluate $\int \frac{11xe^{2x}}{(1+2x)^2}dx$ Evaluate $\int \frac{11xe^{2x}}{(1+2x)^2}dx$
Can somebody check my solution? Thanks!!
Let $u=11xe^{2x}$
then $\frac{du}{dx} = \frac{d}{dx}(11xe^{2x})$
$=11(e^{2x}+2xe^{2x})$
and so $du=11e^{2x}(2x+1)dx$
Now, let $dv=\frac{1}{(2x+1)^2}dx$
Then $\int dv = \int \frac{1}{(2x+1)^2}dx$
To evaluate this, let $r=2x+1$ and so $dr = 2dx$
Then we have $\int dv = \int \frac{1}{r^2}\frac{dr}{2} = \frac{-1}{2r}=\frac{-1}{2(2x+1)}$
So $v=\frac{-1}{2(2x+1)}$
Wow... So that was already a lot of work to find $du$ and $v$ once we chose $u$ and $dv$... Integration by parts totally sucks!!
Anyway, we are now ready to use our VOODOO formula!!
$\int \frac{11xe^{2x}}{(1+2x)^2}dx = \int udv = uv - \int vdu$
$=(11xe^{2x})(\frac{-1}{2(2x+1)}) - \int \frac{-1}{2(2x+1)}(11e^{2x}(2x+1))dx$
Now, there is a $2x+1$ in the denominator and the numerator so the cancel. We also pull the coefficient $\frac{-11}{2}$ outside of the integral, which we can do.
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{2}\int e^{2x}dx$
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{2} \frac{e^{2x}}{2}+C$
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{4} e^{2x}+C$
|
Your answer is correct, but it can be further simplified:
$$\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{4}e^{2x}$$
$$=\frac{11}{4}e^{2x} \left(\frac{-2x}{2x+1} + 1 \right) =\frac{11}{4}e^{2x} \left(\frac{-2x-1}{2x+1} + \frac{1}{2x+1} + 1 \right)$$
$$=\frac{11}{4}e^{2x} \cdot \frac{1}{2x+1} = \frac{11e^{2x}}{4(2x+1)}$$
which is the same as the answer in Venkat Amith's comment.
|
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|
What is the second derivative of the absolute function $\left|\frac{x+1}{x+2}\right|$? I calculated the derivative of $\left|\frac{x+1}{x+2}\right|$ in the same way that I would do with $ \frac{x+1}{x+2}$ in order to study the function.
But when I verified on wolfram, I noticed it is all wrong. Wolfram uses the chain rule as you can see here.
I don't get it. The only rule I've been taught as far as absolute function derivatives are concerned, is $|x|' = \frac{x}{|x|}$. Does a similar rule apply for $f(x)$? And why does wolfram uses chain rule?
Edit
I calculated the derivatives as there is no absolute and then, at the result, I applied the absolute.
My answers are $|(\frac{x+1}{x+2})|' = |(\frac{x+1}{x+2})'| = \frac{1}{\left(x+2\right)^2}$ and $|(\frac{x+1}{x+2})|'' = |(\frac{x+1}{x+2})''| = \frac{2}{\left(x+2\right)^3}$
Wolfram's answer is
$\left(\left|\frac{x+1}{x+2}\right|\right)'\:=\frac{\left|x+2\right|\left(x+1\right)}{\left|x+1\right|\left(x+2\right)^3}$
|
As an alternative, using sign function we have that for $x\neq -1,-2$
$$\left|\frac{x+1}{x+2}\right|=\frac{x+1}{x+2}\cdot \frac{\left|\frac{x+1}{x+2}\right|}{\frac{x+1}{x+2}}=\frac{x+1}{x+2} \operatorname{sign}\left(\frac{x+1}{x+2}\right)$$
therefore by chain rule, since $(\operatorname{sign}(x))'=0 $ for $x\neq 0$, we obtain
$$\frac d{dx}\left|\frac{x+1}{x+2}\right|=\left(\frac d{dx}\frac{x+1}{x+2}\right)\operatorname{sign}\left(\frac{x+1}{x+2}\right)=\frac1{(x+2)^2}\operatorname{sign}\left(\frac{x+1}{x+2}\right)=\frac{\left|\frac{x+1}{x+2}\right|}{(x+1)(x+2)}$$
which is an equivalent form for the derivative.
|
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|
$\cos\left(\frac{\pi}{5}\right)$ using De Moivre's Theorem This is an Exercise 3.2.5 from Beardon's Algebra and Geometry:
Show that $\cos\left(\frac{\pi}{5}\right)=\frac{\lambda}{2}$, where $\lambda$ = $\frac{1+\sqrt{5}}{2}$ (the Golden Ratio).
[Hint: As $\cos(5\theta) = 1$, where $\theta = \frac{2\pi}{5}$, we see from De Moivre's theorem that $P(\cos\theta) = 0$ for some polynomial $P$ of degree five. Now observe that $P(z)=(1-z)Q(z)^2$ for some quadratic polynomial $Q$.]
So, by De Moivre's theorem:
$$\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5=\cos(2\pi)+i\sin(2\pi)=\cos(2\pi)=1$$
And so:
$$\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5-1=0$$
Thus, $\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5-1$ is our polynomial $P$ of degree five. But how can I get from here to $Q$ and from $Q$ to $\cos\left(\frac{\pi}{5}\right)$?
|
Here is another way to prove it, we'll be using De moivre's theorem but without having to get to the $ 5^{\mathrm{th}} $ degree.
We have : $$ \cos{\left(\pi-\theta\right)}=-\cos{\theta} $$
With $ \theta $ being equal to $ \frac{2\pi}{5} $, we get : $$ \cos{\left(\frac{3\pi}{5}\right)}=-\cos{\left(\frac{2\pi}{5}\right)} \ \ \ \ \ \ \left(*\right) $$
Now using De moivre, for any $ \theta\in\mathbb{R} $, we have : \begin{aligned} \cos{\left(3\theta\right)}=\mathcal{Re}\left(\left(\cos{\theta}+\mathrm{i}\sin{\theta}\right)^{3}\right)&=\mathcal{Re}\left(\cos^{3}{\theta}+3\mathrm{i}\cos^{2}{\theta}\sin{\theta}-3\cos{\theta}\sin^{2}{\theta}-\mathrm{i}\sin^{3}{\theta}\right)\\ &=\cos^{3}{\theta}-3\cos{\theta}\left(1-\cos^{2}{\theta}\right)\\ \cos{\left(3\theta\right)}&=4\cos^{3}{\theta}-3\cos{\theta} \end{aligned}
\begin{aligned} \cos{\left(2\theta\right)}=\mathcal{Re}\left(\left(\cos{\theta}+\mathrm{i}\sin{\theta}\right)^{2}\right)&=\mathcal{Re}\left(\cos^{2}{\theta}+2\mathrm{i}\cos{\theta}\sin{\theta}-\sin^{2}{\theta}\right)\\ &=\cos^{2}{\theta}-\left(1-\cos^{2}{\theta}\right)\\ \cos{\left(2\theta\right)}&=2\cos^{2}{\theta}-1 \end{aligned}
Applying those two formulas, and setting $ X=\cos{\left(\frac{\pi}{5}\right)} $, the equation $ \left(*\right) $ becomes : \begin{aligned} 4X^{3}+2X^{2}-3X-1&=0\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(X+1\right)\left(4X^{2}-2X-1\right)&=0\\ \iff 4\left(X+1\right)\left(X-\frac{1+\sqrt{5}}{4}\right)\left(X-\frac{1-\sqrt{5}}{4}\right)&=0\end{aligned}
Since $ X $ cannot be negative, we get : $$ X=\frac{1+\sqrt{5}}{4} $$
|
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|
Find all $x\in \mathbb{R}$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$
Find all $x\in \mathbb{R}$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$
Letting $a=2^x$ and $b=3^x$ we get $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$
from the numerator we have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=7$$
since $7$ is a prime we can say that $$a+b=1, a^2-ab+b^2=7.$$
It follows that $$a=1-b$$
from where $$(1-b)^2-(1-b)b+b²=7$$
this quadratic has solutions $b=1, b=0.$
what I now did was consider cases. Firstly $a=b=0$ which has no solutions for $x$.
case $a=b=1$ has the solution $x=0$.
However the actual solutions for this were $x=1, x=-1$ which I don't see how they came up with. What is wrong with my approach?
|
You properly wrote $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$ Tak into account the homegeneity and let $b=k a$ to get
$$k+\frac 1k-1=\frac{7}{6}\implies k=\frac 23 \qquad \text{and} \qquad k=\frac 32$$ and then the slutions.
|
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|
Is my calculation of the integral $\int \tan^{-1} x \, dx$ correct? Compute $\int \tan^{-1}x \,dx$.
First, set $u = \arctan(x)$ and $dv = dx$. We want to find $du$ and we already have $v = x$.
We start by taking the tangent of both sides, leaving us with
$$\tan(u) = x.$$
Next, using implicit differentiation, we get $\frac{du}{dx}\sec^2(u) = 1$, or $$\frac{du}{dx} = \frac{1}{\sec^2(u)} = \frac{1}{1+ \tan^2(u)} = \frac{1}{1+ \tan^2(\arctan(x))} = \frac{1}{x^2+1}.$$
Therefore, $du = \frac{1}{x^2 +1}\, dx.$
We know that $\int u\, dv = uv - \int v\, du$ by Integration By Parts.
So, $\int \arctan(x)\, dx = x\arctan(x) - \int \frac{x}{x^2+1}\, dx.$
We can simplify $\int \frac{x}{x^2+1}\, dx$ by using u-substitution, making a different $u = x^2+1$ and $\frac{du}{2x} = dx.$
Subbing these in, we get
$$\int \frac{x}{u} \cdot \frac{du}{2x} = \frac{1}{2}\int \frac{1}{u}\, du = \frac{1}{2}|\log u| = \frac{1}{2}|\log(x^2+1)| + C.$$
Plugging this back into $x\arctan(x) - \int \frac{x}{x^2+1}\, dx,$ we get
$\boxed{x\arctan(x) - \frac{1}{2} |\log(x^2+1)| + C}$, where C is a constant.
|
Yes, but you can simply do like below
$$\int \arctan x~\mathrm{d}x= x\cdot \arctan x - \int \frac{x}{x^2+1}\mathrm{d}x = x\cdot \arctan x - \frac{1}{2}\log (x^2+1) +C$$
|
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|
Solving $\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right)$ for $x\in(0,\pi/2)$
Solve this equation for $x\in (0 , \frac{\pi}{2})$
$$\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right)$$
I gave a try using the $\cos(a-b)$ and $\sin(a-b)$ formulas, but it seems the problem complicated a little bit more.
Is any other elegant solution for this?
|
As an alternative, since $x=k\frac \pi 4$ and $x = \frac{\pi}{6}+k\frac \pi 2$ are not solutions, we have that
$$\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right) \iff \frac{\sin(2x)}{\cos(2x)}=\frac{\cos\left(\frac{\pi}{6} - x\right)}{\sin\left(\frac{\pi}{6} - x\right)}$$
$$\iff \tan (2x)=\cot\left(\frac{\pi}{6} - x\right)$$
and since
$$\tan A=\cot B \iff A=\frac \pi 2-(B+k\pi)$$
we obtain
$$2x=\frac \pi 2-\frac{\pi}{6} + x+k\pi \iff x=\frac \pi 3 +k\pi $$
|
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|
Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512 \equiv -6 \equiv 5 \pmod{11}\\
x_2 &= 123^{456} \equiv 123^0 \equiv 1 \pmod{8}\\
y_2 &= 1^{-1} \equiv 1 \pmod{8} \\
123^{456}
&\equiv \sum_{i=1}^2 x_i\times\frac{M}{m_i} \times y_i
\equiv 9\times\frac{88}{11}\times5 + 1\times\frac{88}{8} \times1 \equiv 371
\equiv 19 \pmod{88}
\end{split}
$$
|
$123^{456}\equiv 2^6=64\equiv9\bmod 11$.
$123^{456}\equiv 3^0=1\equiv9\bmod 8$.
Therefore, by the constant case of the Chinese Remainder Theorem, $123^{456}\equiv9\bmod88$.
|
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|
Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$
$a,b,c$ are three real numbers such that $a+b+c=5$ and $a^2+b^2+c^2=11$,
what's the maximum value of $abc$?
I thought of a way, $ab+bc+ca$ is not hard to find, $a,b,c$ satisfy the cubic equation $x^3 - 5 x^2 + 7 x - abc = 0$ , then use the discriminant of the cubic equation non-negative.
The discriminant of $x^3 + A x^2 + B x + C=0$ is
$A^2 B^2 - 4 B^3 - 4 A^3 C + 18 A B C - 27 C^2$
Is there an easier way?
|
Yes, we need to find $ab+ac+bc=7$ before.
Now, $a$, $b$ and $c$ are roots of the equation:
$$(x-a)(x-b)(x-c)=0$$ or
$$abc=x^3-5x^2+7x.$$
Now, $$(x^3-5x^2+7x)'=(x-1)(3x-7),$$ which gives that a maximal value of $abc$ for which the equation $$abc=x^3-5x^2+7x$$ has three real roots holds for $x=1$, which gives:
$$\max_{a+b+c=7,a^2+b^2+c^2=11}{abc}=3.$$
The equality occurs for example, for $(a,b,c)=(1,1,3).$
|
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|
If $a > b > 1$ and $\frac{1}{log_a(b)} + \frac{1}{log_b(a)} = \sqrt{1229}$ , find the value of :- $\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$
If $a > b > 1$ and $\frac{1}{log_a(b)} + \frac{1}{log_b(a)} = \sqrt{1229}$ , find the value of :- $\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$ .
What I Tried :- I tried the problem this way :-
As $log_a(b) = \frac{log_b(b)}{log_b(a)}$ , we have $\frac{1}{log_a(b)} = \frac{log_b(a)}{log_b(b)} = log_b(a).$ So :-
$$log_b(a) + \frac{1}{log_b(a)} = \sqrt{1229}$$
$$\rightarrow \frac{log(a)}{log(b)} + \frac{log(b)}{log(a)} = \sqrt{1229}$$
$$\rightarrow \frac{(log(a))^2+(log(b))^2}{log(a)log(b)} = \sqrt{1229}$$
Now :-
$$\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$$
$$\rightarrow \frac{log(a) + log(b)}{log(b)} - \frac{log(a) + log(b)}{log(a)}$$
$$\rightarrow \frac{(log(a))^2 + log(a)log(b) - log(a)log(b) - (log(b))^2}{log(a)log(b)}$$
$$\rightarrow \frac{(log(a))^2 - (log(b))^2}{log(a)log(b)}$$
$$\rightarrow \sqrt{1229} - \frac{2(log(b))^2}{log(a)log(b)}$$
I could conclude only upto this, other than that I have no idea . Now can anyone help me?
|
$$\dfrac{1}{\log_{ab}a}-\dfrac{1}{\log_{ab}b}=\log_ab\,-\log_ba \tag{1}$$
Let $\log_b a=x$, we get the first expression to be $$x +\frac{1}{x}=\sqrt{1229} \tag{2}$$,
Now, squaring (1) and (2), we see (1) becomes:
$$x^2+\frac{1}{x^2} + 2 = 1229\tag{3}$$
and expression (2) becomes:
$$x^2-2+\frac{1}{x^2} = A^2\tag{4}$$
Subtracting (3) and (4), we get:
$$1229-A^2=4$$
Therefore, A=$\sqrt{1225}$
|
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|
Primality test for specific class of $N=k \cdot 2^n+1$ Can you prove or disprove the following claim:
Let $N=k \cdot 2^n+1$ be a natural number that is not a perfect square such that $ 2 \nmid k$ , $n>2$ . Let $c$ be the smallest odd prime number such that $\left(\frac{c}{N}\right)=-1$ , where $\left(\frac{}{}\right)$ denotes a Jacobi symbol . Let $z$ be a real number of the form $a+b\sqrt{c}$ equal to the modular $(1-\sqrt{c})^{(N-1)/2} \operatorname{mod} N$ , then $N$ is prime iff $a=b$ .
You can run this test here. I have verified this claim for all $k \in [1,1000]$ with $n \in [3,1000]$ .
|
This fails for $N=22577=1411\cdot 2^4+1=107\cdot 211$. The value of $c$ that is taken is $3$, but
$$(1-\sqrt3)^{11288}\equiv 11502+11502\sqrt 3\bmod 22577.$$
It is true, however, if $N$ is prime. Setting $x=\sqrt c$, we have that $x^N$ is the Galois conjugate of $x$, so $x^N=-x$, whence
$$(1+x)^{N+1}+(1+x)^N(1+x)=(1+x)(1+x^N)=(1+x)(1-x)=1-x^2=1-c;$$
this is a product of $-1$ and primes strictly less than $c$ (including $2$) and thus a quadratic residue (using that $c\equiv 1\bmod 8$), so
$$(1+x)^{\frac{N+1}{2}}\in\mathbb F_N;$$
this implies that
$$(1+x)^{\frac{N+1}{2}}=(1-x)^{\frac{N+1}{2}},$$
which gives that, if $(1-x)^{\frac{N-1}{2}}=a+bx$ with $a,b\in\mathbb F_N$,
$$(a-bx)(1+x)=(a+bx)(1-x)\implies x(a-b)=x(b-a)\implies a=b.$$
|
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|
Evaluating $\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$ without a calculator? Is there a way to get this value without calculator?
$$\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$$
I'm currently studying AIME.
|
Consider
$$S_n=\sum_{a=1}^n\sum_{b=1}^n\sum_{c=1}^n\frac{ab(3a+c)}{2^a\,2^b\,2^c\,(a+b)(b+c)(c+a)}$$ and compute $S_n$ for the very first values of $n$. This gives the sequence
$$\left\{\frac{1}{16},\frac{27}{128},\frac{343}{1024},\frac{3375}{8192}\right\}$$ and you can notice that the numerators are cubes and that the denominators are powers of $2$.
So, you can conjecture that
$$S_n=\frac {\left(2^n-1\right)^3 } {2^{3 n+1} }=\frac{1}{2} \left(1-2^{-n}\right)^3$$
|
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|
Computing a finite sum involving binomial coefficients I would like to prove the following: given the sequence $$a_m = \begin{cases}1 & \text{if $m = 0, 1$} \\\frac{(\alpha+2)(\alpha+4)\cdots(\alpha+2(m-1))}{(\alpha+1)(\alpha+2)\cdots(\alpha+m-1)} & \text{otherwise}\end{cases}$$ then, $$\sum_{m=0}^{n}(-1)^m\binom{n}{m}a_m = \begin{cases}\frac{(n-1)!!}{(\alpha+1)(\alpha+3)\cdots(\alpha+n-1)}& \text{if $n$ is even and $\ge 2$} \\0 & \text{if $n$ is odd}\end{cases}$$
I understand that the left hand side in the last equation is the $n$-th difference $\Delta^na_0$, so basically I tried to compute each difference but it seems that a general formula is hard to find for $\Delta^na_k$. Can you help me? I have seen often that such problems can be attacked with special functions but unfortunately I don't have any experience :(. What do you suggest? Any hint would be greatly appreciated :)
EDIT: So here is the original problem: show that $$e^{-x}\left[1+x+\frac{\alpha+2}{\alpha+1}\cdot\frac{x^2}{2!}+\frac{(\alpha+2)(\alpha+4)}{(\alpha+1)(\alpha+2)}\cdot\frac{x^3}{3!}+\ldots\right]=$$$$1+\frac1{\alpha+1}\frac{x^2}{2^1\cdot1!}+\frac1{(\alpha+1)(\alpha+3)}\frac{x^4}{2^2\cdot2!}+\ldots$$. Doing the product one just compares the two series and obtains my equation above (if I interpreted correctly the pattern of the two series).
|
From the definition, of the Pochhammer symbol, $$(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$$we can express
\begin{equation}
a_m=2^m\frac{ \left(\frac{\alpha }{2}\right)_{m}}{(\alpha )_{m}}
\end{equation}
with $a_0=a_1=1$. (The expression given by @ClaudeLeibovici can be retrieved by remarking that
\begin{equation}
(s+1)_{m-1}=\frac{\Gamma(s+m)}{\Gamma(s+1)}=\frac{1}{s}\frac{\Gamma(s+m)}{\Gamma(s)}=\frac{1}{s}(s)_m
\end{equation}
The series proposed in the edit can be expressed as a confluent hypergeometric function
\begin{align}
\sum_{k=0}^\infty a_k\frac{x^k}{k!}&= 1+x+\frac{\alpha+2}{\alpha+1}\cdot\frac{x^2}{2!}+\frac{(\alpha+2)(\alpha+4)}{(\alpha+1)(\alpha+2)}\cdot\frac{x^3}{3!}+\ldots\\
&=\sum_{k=0}^\infty\frac{ \left(\frac{\alpha }{2}\right)_{k}}{(\alpha )_{k}}\frac{(2x)^k}{k!}\\
&=\,_1F_1\left( \frac{\alpha}{2},\alpha;2x \right)
\end{align}
It can be recognized as related to a modified Bessel function
\begin{equation}
I_\nu(z)=\frac{2^{-\nu}z^\nu e^{-z}}{\Gamma(\nu+1)}\,_1F_1\left( \nu+\frac{1}{2},2\nu+1;2z \right)
\end{equation}
Here, taking $\nu=(\alpha-1)/2$, we find
\begin{equation}
\sum_{k=0}^\infty a_k\frac{x^k}{k!}=2^{\frac{\alpha-1}{2}}\Gamma\left( \frac{\alpha+1}{2} \right)x^{\frac{1-\alpha}{2}}e^{x}I_{\frac{\alpha-1}{2}}(x)
\end{equation}
Then
\begin{equation}
e^{-x}\sum_{k=0}^\infty a_k\frac{x^k}{k!}=\Gamma\left( \frac{\alpha+1}{2} \right)\left( \frac{x}{2} \right)^{\frac{1-\alpha}{2}}e^{x}I_{\frac{\alpha-1}{2}}(x)
\end{equation}
From the series expansion
\begin{align}
I_\nu(z)&=\sum_{k=0}^\infty \frac{1}{\Gamma(k+\nu+1)k!}\left( \frac{z}{2} \right)^{2k+\nu}\\
&=\frac{1}{\Gamma(\nu+1)}\left( \frac{z}{2} \right)^\nu\left( 1+\frac{z^2}{4(\nu+1)}+\frac{z^4}{32(\nu+1)(\nu+2)}+\ldots \right)
\end{align}
we obtain
\begin{equation}
e^{-x}\sum_{k=0}^\infty a_k\frac{x^k}{k!}=1+\frac1{\alpha+1}\frac{x^2}{2^1\cdot1!}+\frac1{(\alpha+1)(\alpha+3)}\frac{x^4}{2^2\cdot2!}+\ldots
\end{equation}
as expected
|
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Can we find $ \sum_{n=1}^{\infty}\frac{1+2+\cdots +n}{n!} $? Consider the sequence $$ a_{n} = \sum_{r=1}^{n}\frac{1+2+\cdots +r}{r!} $$
Then we have, $$ a_{n} = \sum_{r=1}^{n}\frac{1}{r!} \ + 2\sum_{r=2}^{n}\frac{1}{r!} \ + 3\sum_{r=3}^{n}\frac{1}{r!} \ + \cdots + n\sum_{r=n}^{n}\frac{1}{r!} \ \geq \ 1 + \sum_{r=1}^{n}\frac{1}{r!}$$ for all $n \geq 2$.
$\implies \displaystyle \lim_{n \to \infty}a_{n} \ \geq \ e $
Now, in my book it says $\displaystyle \lim_{n \to \infty}a_{n} \ = \frac{3}{2}e $
How can I attack this problem? Anyone please?
|
Note that
$$ \sum_{k = 1}^\infty \frac{1 + 2 + \cdots + k}{k!} = \frac{1}{2} \sum_{k = 1}^{\infty} \frac{k + 1}{(k - 1)!} = \frac{1}{2} \left( \sum_{k = 1}^\infty \frac{(k - 1) + 1}{(k - 1)!} + e \right) = \frac{1}{2} \left( \sum_{k =2 }^\infty \frac{1}{(k - 2)!} + \sum_{k = 1}^{\infty} \frac{1}{(k-1)!} \right) + \frac{e}{2} = \frac{3e}{2} $$
|
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|
How do i form the inequality $ -5\le 3\cos x - 4\sin x\le5$? We know that
$-1\le \cos x\le1$ so we if we multiply 3 so we get $-3\le 3\cos x\le 3$ and we also know that $-1\le \sin x\le 1$ and and again we multiply 4 we get $-4\le 4\sin x\le 4$ and we can add these two inequality we get $-7\le 3\cos x - 4\sin x\le 7$ but how in the above inequality is forming i m not able to understand at all.
|
Hint :
\begin{eqnarray*}
3 \cos \theta -4 \sin \theta = R \cos ( \theta +\phi) .
\end{eqnarray*}
$R=?$
|
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|
Probability of taking white balls until a black shows up. (No replacement) I have an urn with $n$ white balls, and 1 black ball.
What's the probability of taking $x-1 \in \{0,n\}$ white balls, and the last extraction being a black ball? (No replacement)
My reasoning:
$$P(ww...b)=\frac{n}{n+1}\frac{n-1}{n}\cdots\frac{n-(x-2)}{n-(x-3)}\frac{1}{n-(x-2)}=\frac{1}{n+1}$$
However, when I try to think in terms of combinations:
I reason $$\frac{C^n_{x-1}C^1_{1}}{C^{n+1}_{x}}=\frac{x}{n+1}$$
Where is my flaw in the combinatorics reasoning?
|
In order to select the black ball on the $k$th draw, we must first select $k - 1$ of the $n$ white balls while selecting $k - 1$ of the $n + 1$ balls, then select the only black ball from the remaining $n + 1 - (k - 1) = n + 2 - k$ balls. Hence, the probability of selecting the black ball on the $k$th draw is
\begin{align*}
\frac{\dbinom{n}{k - 1}}{\dbinom{n + 1}{k - 1}} \cdot \frac{1}{n + 2 - k} & = \frac{\dfrac{n!}{(k - 1)![n - (k - 1)]!}}{\dfrac{(n + 1)!}{(k - 1)![n + 1 - (k - 1)]!}} \cdot \frac{1}{n + 2 - k}\\
& = \frac{\dfrac{n!}{(k - 1)!(n + 1 - k)!}}{\dfrac{(n + 1)!}{(k - 1)!(n + 2 - k)!}} \cdot \frac{1}{n + 2 - k}\\
& = \frac{n!}{(k - 1)!(n + 1 - k)!} \cdot \frac{(k - 1)!(n + 2 - k)(n + 1 - k)!}{(n + 1)n!} \cdot \frac{1}{n + 2 - k}\\
& = \frac{1}{n + 1}
\end{align*}
which agrees with your first answer.
However, there is a simpler way to see this. There are $n + 1$ balls. Since the lone black ball is equally likely to be any position in the sequence, the probability that it is in the $k$th position is
$$\frac{1}{n + 1}$$
for $1 \leq k \leq n + 1$.
|
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|
Proving $4\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \Big)+\frac{81}{(a+b+c)^2}\geqslant{\frac {7(a+b+c)}{abc}}$ For $a,b,c>0.$ Prove$:$ $$4\Big(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \Big)+\dfrac{81}{(a+b+c)^2}\geqslant{\dfrac {7(a+b+c)}{abc}}$$
My proof is using SOS$:$
$${c}^{2}{a}^{2} {b}^{2}\Big( \sum a\Big)^2 \sum a^2 \Big\{ 4\Big(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \Big)+\dfrac{81}{(a+b+c)^2}-{\dfrac {7(a+b+c)}{abc}} \Big\}$$
$$=\dfrac{1}{2} \sum {a}^{2}{b}^{2} \left( {a}^{2}+{b}^{2}-2\,{c}^{2} +5bc-10ab+5\, ac \right) ^{2} +\dfrac{1}{2} \prod (a-b)^2 \left( 7\sum a^2 +50\sum bc \right) \geqslant 0.$$
From this we see that the inequality is true for all $a,b,c \in \mathbb{R};ab+bc+ca\geqslant 0.$
But we also have this inequality for $a,b,c \in \mathbb{R}.$ Which verify by Maple.
I try and I found a proof but I'm not sure$:$
If replace $(a,b,c)$ by $(-a,-b,-c)$ we get the same inequality.
So we may assume $a+b+c\geqslant 0$ (because if $a+b+c<0$ we can let $a=-x,b=-y,c=-z$ where $x+y+z \geqslant 0$ and the inequality is same!)
Let $a+b+c=1,ab+bc+ca=\dfrac{1-t^2}{3} \quad (t\geqslant 0), r=abc.$ Need to prove$:$
$$f(r) =81\,{r}^{2}-15\,r+\dfrac{4}{9} \left( t-1 \right) ^{2} \left( t+1 \right) ^{2
}\geqslant 0.$$
It's easy to see, when $r$ increase then $f(r)$ decrease. Since $r\leqslant \dfrac{1}{27} \left( 2\,t+1 \right) \left( t-1\right) ^{2} \quad$(see here). We get$:$
$$f(r)\geqslant f\Big(\dfrac{1}{27} \left( 2\,t+1 \right) \left( t-1\right) ^{2}\Big)=\dfrac{1}{9} {t}^{2} \left( 2\,t-1 \right) ^{2} \left( t-1 \right) ^{2} \geqslant 0.$$
Done.
Could you check it for me? Who have a proof for $a,b,c \in \mathbb{R}$?
|
After using the nguyenhuyen_ag's reasoning it's enough to prove our inequality for positive variables,
and we can end the proof by $uvw$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$\frac{4(9v^4-6uw^3)}{w^6}+\frac{81}{9u^2}\geq\frac{21u}{w^3}$$ or $f(w^3)\geq0,$ where
$$f(w^3)=w^6-5u^3w^3+4u^2v^4.$$
But $$f'(w^3)=2w^2-5u^3<0,$$ which says that $f$ decreases and it's enough to prove our inequality for a maximal value of $w^3$,
which by $uvw$ happens for equality case of two variables.
Since our inequality is homogeneous and symmetric, it's enough to assume $b=c=1,$ which gives
$$(a-1)^2(a-4)^2\geq0$$ and we are done!
|
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|
Rational Roots (with Lots of Square Roots!) Find the smallest positive integer $a,$ greater than 1000, such that the equation $$\sqrt{a - \sqrt{a + x}} = x$$ has a rational root.
Squaring both sides we have $a-\sqrt{a+x}=x^2.$ We shall not square again as that gives a quartic in $x.$ Rearranging, we have $$-x^2+a-\sqrt{a+x}=0 \implies x^2-a+\sqrt{a+x}=0.$$ How would we continue? Solutions? (I don't have one in my book.)
|
Since the quartic equation is monic, i.e. it has the form
$$x^4+ \dots =0$$
by the rational root theorem, $x$ must be an integer. Moreoever
$$x=\sqrt{a-\sqrt{a+x}}>0$$
so that $x$ must be a positive integer.
Now, consider the equation
$$\sqrt{a+x}=a-x^2$$
and call $$y=\sqrt{a+x}=a-x^2$$
Note that $y=\sqrt{a+x}>0$ is a positive integer. Then you have the two equations
$$y^2=a+x$$
$$y=a-x^2$$
Subtracting the two equations side by side you get
$$y^2-y=x+x^2$$
which can be written as
$$y(y-1)=(x+1)x$$
which has a unique positive solution: $y=x+1$.
Thus
$$a=x^2+y=x^2+x+1$$
So you have to find the smallest number $a$ which can be represented as $a=x^2+x+1$ where $x$ is a positive integer, and $x^2+x+1>1000$.
Now, $x=32$ is the smallest integer solution, and $$a=32^2+32+1=1057$$ is the number you were looking for.
|
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proof that $ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $
Proof that
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $$
by induction.
Proof
Base case: Statement clearly holds for $n = 1$. Now assume that statement holds for some $n = k$ and lets show that it implies $n = k + 1$ holds. The proof:
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} + \frac{1}{n(n+1)} = \frac{3}{2} - \frac{1}{n} + \frac{1}{n(n+1)} = \frac{3}{2} - \frac{1}{n} + \frac{1}{n} - \frac{1}{n + 1} $$
$$ = \frac{3}{2} - \frac{1}{n+1} $$
Now the problem is I can't find the error. The statement doesn't clearly work for $ n = 2 $. However, the assumption seems to be correct since if I assume it's true for some $n = k$ and it is true for $ n = 1$? It shouldn't be possible to show that $p(n) \implies p(n+1)$ when $p(n)$ is true and $p(n+1)$ is false. This means that $p(n)$ has to be false in this case since when $p(n)$ is false then $p(n) \implies p(n+1)$ is tautology. The problem is I don't really see how? Isn't the whole point of induction to show that $p(n)$ is true for some specific $n = k$ (not all $n$) and then show $p(n+1)$ by assuming $p(n)$. Now when $p(n)$ is false you can show anything since it's tautology but how can you be sure $p(n)$ is true if you don't show it for all $n$? And wouldn't that defeat the purpose of induction (if you have already shown it's true for all $n$)?.
|
As $\frac 1{n(n-1)}$ is not defined for $n =1$ and also because the first term is $\frac 1{2\cdot 1} = \frac 1{2(2-1)}$ and so the first term is for $n = 2 > 1$, then it clearly does NOT work for $n= 1$.
If the statement were true for any $n$ it would be true for any subsequent natural numbers but it is not true for any $n$.
that statement is $\sum_{k=2}^n \frac 1{k(k-1)} = \frac 32 -\frac 1n$ and that just isn't true.
But $\sum_{k=2}^n \frac 1{k(k-1)} = 1 -\frac 1n$ is.
Note the first case is for $n = 2$ and not $n =1$.
Proof:
For $n=2$ then $\frac 1{2} = 1-\frac 12$.
And if $\sum_{k=2}^n \frac 1{k(k-1)}= 1-\frac 1n$ then
$\sum_{k=2}^{n+1} \frac 1{k(k-1)} = 1-\frac 1n + \frac 1{n(n+1)} = 1-\frac {(n+1) - 1}{n(n+1)} =1-\frac n{n(n+1)} = 1-\frac 1{n+1}$.
|
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How to evaluate below expression of modular arithmetic? How to evaluate (10^18)%(10^9 + 7) using modular arithmetic? How to proceed and what will be the steps?
|
Building on dave's answer above but in a way that I can understand -
Given
10^18 = (10^9 + 7) * (10^9 - 7) + 49
10^18 % (10^9 + 7) = 49
....
Generalising ...
===> a^2 = (a + b) * (a - b) + b^2
===> a^2 % (a + b) = { (a + b) * (a - b) + b^2 } % (a + b)
Consider : { (a + b) * (a - b) + b^2 } % (a + b)
===> [ { (a + b) * (a - b) % (a + b) } + { b^2 % ( a + b) } ] % (a+b) ... using properties of modular arithmetic (refer below link)...
===> { b^2 % ( a + b) ... using {(x*y + z) % x = z % x} from my own understanding but also relates to the link below.
a^2 % (a + b) = b^2 % ( a + b)
a^2 % (a + b) = b^2 ... where (a+b) > b^2
substituting a=10^9 and b=7
10^18 % (10^9 + 7) = 49
Interestingly, above works where b = -7 as well.
The given problem can be solved using a modern computer based calculator but knowing this is still useful!
References -
https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/modular-addition-and-subtraction
|
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Is there a simpler expression for this piecewise-defined function? As a math-for-fun exercise, I challenged myself to find a globally-defined, everywhere-differentiable antiderivative of $\sqrt{1-\sin(x)}$. Because of the fundamental theorem of calculus, the problem boils down to evaluating the integral
$$f(x)=\int_{-\frac{\pi}{2}}^{x}\sqrt{1-\sin(t)}\text{ }dt$$
After lots of thinking, many laborious calculations and a series of roadblocks, I arrived at this piecewise-defined expression:
$$f(x)=\begin{cases}
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + 2\sqrt{1+\sin(x)} & \text{if } 2\pi \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor - \frac{\pi}{2}\leq x \leq 2\pi \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + \frac{\pi}{2}\\
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor +4\sqrt{2} - 2\sqrt{1+\sin(x)} & \text{if } 2\pi \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + \frac{\pi}{2}\leq x \leq 2\pi \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + \frac{3\pi}{2}
\end{cases}$$
Using knowledge obtained during the solution process, I obtained the (perhaps) simpler expression
$$f(x)=\begin{cases}
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + 2\sqrt{1+\sin(x)} & \text{if } \cos(x)\geq 0\\
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor +4\sqrt{2} - 2\sqrt{1+\sin(x)} & \text{if } \cos(x)\leq 0
\end{cases}$$
Naturally, I wondered whether this expression could be simplified further. Maybe there's a non-piecewise expression for $f(x)$? I originally thought about writing
$$4\sqrt{2}\left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + 4\sqrt{2}=4\sqrt{2} \left( \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor +1 \right)=4\sqrt{2}\left \lceil \frac{x}{2\pi}+\frac{1}{4} \right \rceil$$
but quickly recognized that this is not valid when $\frac{x}{2\pi}+\frac{1}{4}$ is an integer. I also tried writing
$$f(x)=\begin{cases}
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + 2\text{sgn}(\cos x)\sqrt{1+\sin(x)} & \text{if } \cos(x)\geq 0\\
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor +4\sqrt{2} + 2\text{sgn}(\cos x)\sqrt{1+\sin(x)} & \text{if } \cos(x)\leq 0
\end{cases}$$
but this fails when $\cos(x)=0$. I don't have any other tricks up my sleeve. Could I get some assistance?
|
You can verify that
$$ \sqrt{1 - \sin t} =
\sqrt2 \left\lvert \sin \left(\frac12t - \frac\pi4\right)\right\rvert . $$
This suggests the substitution $u = \frac12t - \frac\pi4,$
$$ \int \sqrt{1 - \sin t} \,\mathrm dt
= 2\sqrt2 \int \lvert\sin u\rvert \,\mathrm du
\DeclareMathOperator{\sgn}{sgn}
= 2\sqrt2 \int \sgn(\sin u) \sin u\,\mathrm du. $$
Within any interval over which $\sin u$ does not change sign
(that is, within the interval $(n\pi, (n+1)\pi)$ where $n$ is an integer),
$$ \int \sgn(\sin u) \sin u\,\mathrm du
= \sgn(\sin u) \int \sin u\,\mathrm du
= -\sgn(\sin u) \cos u + C. $$
This does not quite work for the closed interval $[n\pi, (n+1)\pi],$
however, because there is a problem with the definition of $\sgn(0).$
Usually we would define $\sgn(0) = 0,$ which would cause
$-\sgn(\sin u) \cos u$ to be $0$ at either end of the interval while the
limit at the left end is $-1$ and the limit at the right is $1.$
We cannot fix this over the entire closed interval $[n\pi, (n+1)\pi],$
but we can at least achieve a continuous antiderivative on the half-open interval
$[n\pi, (n+1)\pi)$ by moving the rest of the curve upward to meet the left endpoint:
$$ \int \sgn(\sin u) \sin u\,\mathrm du
= -\sgn(\sin u) (1 + \cos u) + C. $$
This gives us an antiderivative that is piecewise continuous over each interval
$[n\pi, (n+1)\pi)$.
In order to have a continuous antiderivative over a larger interval,
you need to compensate for the fact that $-\sgn(\sin u) \cos u$
jumps between $1$ and $-1$ at every multiple of $\pi$.
We need to add $2$ every time we cross a multiple of $\pi$ in the increasing direction, and not otherwise.
So the missing part of the integral is a step function.
It's practically impossible to get a continuous antiderivative without introducing one of these or inventing some other piecewise definition.
Fortunately, the step function we need is just the floor function scaled as needed along both axes.
We get
$$ \int \sgn(\sin u) \sin u\,\mathrm du
= -\sgn(\sin u) (1 + \cos u) + 4 \left\lfloor \frac u{2\pi} \right\rfloor + C. $$
Plug in $u = \frac12t - \frac\pi4$ and multiply the whole thing by $2\sqrt2$
and you have your continuous antiderivative.
|
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Evaluating the limit $\lim_{h\to 0} \frac{\sqrt {1 - (x+h) ^2} - \sqrt {1-x^2}} {h}$ Find the limit
$\lim_{h\to 0} \frac{\sqrt {1 - (x+h) ^2} - \sqrt {1-x^2}} {h}$
I tried substituting "h" and also multiplying $\frac{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}$
|
Your idea is fine, indeed we obtain
$$\frac{\sqrt {1 - (x+h) ^2} - \sqrt {1-x^2}} {h}\cdot \frac{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}=$$
$$=\frac{{1 - (x+h) ^2} - ({1-x^2})}{h(\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2})}=$$
$$=\frac{-2xh-h^2}{h(\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2})}$$
and since $h$ cancels out we can conclude.
|
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What to do with exponent with different base in denominator? I‘m struggling right now with a equation. The equation is
$$\frac{2^{(5x)}}{7^{(x+2)}} = 10.$$
The solution should be $4.076$ but I don‘t know how to solve this equation.
I came to the conclusion that I can simplify the equation so it‘s $\frac{32^x}{7^{(x+2)}}$. My problem is that I don‘t know how I can get rid of the denominator.
Thank you for your help.
|
You simplified $2^{5x}$ to $\left(2^5\right)^x=32^x$.
You can also simplify $7^{x+2}$ to $7^x\cdot7^2=49\cdot7^x$, using the exponent rule $a^{b+c}=a^b\cdot a^c$.
Thus, your equation simplifies to
$$
\begin{align*}
\frac{32^x}{49\cdot7^x}&=10\\
\frac{1}{49}\cdot\frac{32^x}{7^x}&=10\\
\frac{32^x}{7^x}&=490\\
\left(\frac{32}{7}\right)^x&=490\\
\ln\left(\left(\frac{32}{7}\right)^x\right)&=\ln(490)\\
x\ln\left(\frac{32}{7}\right)&=\ln(490)\\
x&=\frac{\ln(490)}{\ln\left(\frac{32}{7}\right)}\\
x&\approx4.076
\end{align*}
$$
|
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|
Calculate $\lim\limits_{(x,y)\to (0,0)} \dfrac{x^4}{(x^2+y^4)\sqrt{x^2+y^2}}$ Calculate, $$\lim\limits_{(x,y)\to (0,0)} \dfrac{x^4}{(x^2+y^4)\sqrt{x^2+y^2}},$$ if there exist.
My attempt:
I have tried several paths, for instance: $x=0$, $y=0$, $y=x^m$. In all the cases I got that the limit is $0$. But I couldn't figure out how to prove it. Any suggestion?
|
Assuming $y<1$ and using polar coordinates we have
$$\dfrac{x^4}{(x^2+y^4)\sqrt{x^2+y^2}} \le \dfrac{x^4}{(x^2+y^2)\sqrt{x^2+y^2}} =r \sin^4\theta \to 0$$
|
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Prove that $\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}dx=\frac{1}{2}.$
Question: Prove that $$\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}dx=\frac{1}{2}.$$
Solution: Let $$I_n:=n^2\int_0^{\frac{1}{n}}x^{x+1}dx, \forall \in\mathbb{N}.$$ Substituting $nx=t$ in $I_n$, we have $$I_n=n\int_0^1\left(\frac{t}{n}\right)^{1+\frac{t}{n}}dt.$$
Now for all $0\le t\le 1$ and for all $n\in\mathbb{N}, n+t\le n+1\implies 1+\frac{t}{n}\le1+\frac{1}{n}.$ This implies that for all $0\le t\le 1$ and for all $n\in\mathbb{N}$, we have $$\left(\frac{t}{n}\right)^{1+\frac{t}{n}}\ge \left(\frac{t}{n}\right)^{1+\frac{1}{n}}.$$
Therefore, for all $n\in\mathbb{N},$ $$\int_0^1\left(\frac{t}{n}\right)^{1+\frac{t}{n}}dt\ge \int_0^1\left(\frac{t}{n}\right)^{1+\frac{1}{n}}dt=n^{-\left(1+\frac{1}{n}\right)}\frac{n}{2n+1}.$$ This implies that $$I_n\ge n^{-\frac{1}{n}}\frac{n}{2n+1},\forall n\in\mathbb{N}.$$
Next note that for all $0\le t\le 1$ and for all $n\in\mathbb{N}$, $1+\frac{t}{n}>1$, which implies that $\left(\frac{t}{n}\right)^{1+\frac{t}{n}}<\frac{t}{n}.$ Therefore, $$\int_0^1\left(\frac{t}{n}\right)^{1+\frac{t}{n}}dt<\int_0^1\left(\frac{t}{n}\right)dt=\frac{1}{2n}.$$ This implies that $$I_n<\frac{1}{2},\forall n\in\mathbb{N}.$$ Thus, for all $n\in\mathbb{N}$, we have $$n^{-\frac{1}{n}}\frac{n}{2n+1}\le I_n<\frac{1}{2}.$$ Now since $$\lim_{n\to\infty}n^{-\frac{1}{n}}\frac{n}{2n+1}=\frac{1}{2},$$ therefore by Sandwich theorem we can conclude that $$\lim_{n\to\infty}I_n=\frac{1}{2}.$$
Is this solution correct and rigorous enough and is there any other way to solve the problem?
|
$$ \lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}\mathrm{d}x = \lim_{n \to \infty} n^2 I_n =\frac{1}{2}. $$
My approach is quite similar to @Sanket. Sandwiching is the basic idea.
We have that $$0 \leqslant x \leqslant \frac{1}{n} \implies 1 \leqslant x + 1 \leqslant 1 + \frac{1}{n} \implies x \geqslant x^{x + 1} \geqslant x^{1 + \frac{1}{n}} $$
$$ \implies n^2 \int_0^{\frac{1}{n}} x~\mathrm{d}x \geqslant n^2 I_n \geqslant
n^2 \int_{0}^{\frac{1}{n}} x^{1 + \frac{1}{n}}~\mathrm{d}x \implies \frac{1}{2} \geqslant n^2 I_n \geqslant \frac{1}{n^{1/n}\left(2 + \frac{1}{n}\right)} $$
$$ \implies \frac{1}{2} \geqslant \lim_{n \to \infty} n^2 I_n \geqslant \lim_{n \to \infty} \frac{1}{n^{1/n}\left(2 + \frac{1}{n}\right)} $$
We'll basically use the standard limit $n^{1/n} \to 1$ as $n \to \infty$. Thus:
$$ \frac{1}{2} \geqslant \lim_{n \to \infty} n^2 I_n \geqslant \frac{1}{2} $$
Thus by sandwich theorem we have that $\lim\limits_{n \to \infty} n^2 I_n $ exists and equals $\frac{1}{2}$.
|
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How to factor $ 1 - 3x + x^2 + x^3$? By inspection, I can see that one of the roots is $1$.
So we can write
$$
1 - 3x + x^2 + x^3 = (x - 1)f_2(x)
$$
where $f_n(x)$ is an n-th order polynomial. I haven't used long division for polynomials in ages, but I feel like that might be overcomplicating things and there might be an easier way to determine $f_2(x)$. Is there an obvious approach to getting $f_2(x)$ here?
I tried some guess and checking to obtain it. I know that the quadratic term in $f_2(x)$ must have a coefficient of $1$, since the coefficient of $x^3$ is $1$. So $f_2(x) = x^2 + f_{1}(x)$. Now $f_{1}(x)$ is some affine equation. I know that $f_{1}(x)$ must have the constant $-1$ since we have a constant $1$ in the cubic and $(x-1)$, so we know that $f_2(x) = x^2 - 1 + f_a(n-1)$, where $f_a(x)$ is some linear equation that goes through the origin.
Now I checked $(x-1)(x^2 - 1) = x^3 - x^2 - x + 1$. When we compare this with the original cubic, we see that we're off by $2x^2 - 2x$. So this prompted me to use $f_a(x) = 2x$. So we have
$f_2(x) = x^2 + 2x - 1$ and this checks out.
This procedure that I used was just kind of just guess and checking.
|
One method to factor it is to check whether it can be separated into several parts that have a common factor. In this case, we can separate $x^3+x^2-3x+1$ into $x^3-x$ and $x^2-2x+1$. Since $x^3-x=x(x^2-1)=x(x-1)(x+1)$ and $x^2-2x+1=(x-1)^2$, the 2 parts have a common factor of $(x-1)$ and can be factored out. Thus, $x^3+x^2-3x+1=x^3-x+x^2-2x+1=x(x-1)(x+1)+(x-1)^2=(x-1)(x(x+1)+(x-1))=(x-1)(x^2+2x-1)$.
|
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|
How many $ax^2+bx+c=0$ with distinct pairs of rational roots can be made from integers $N=|ac|=|\alpha\beta|$ and $b=\alpha+\beta$? Only for illustration, let's choose an integer $N=6$. I want to create a collection of all possible quadratic equations in the form of $ax^2+bx+c=0$ where
*
*$|ac|=|\alpha\beta|=N=6$, and
*$\alpha+\beta=b$.
Note that all constants are integers but $x$ is not necessarily integer.
Here is my attempt.
How many ways to specify $a$ and $c$
The prime factors of $N$ is
$$
P=\{2,3\}
$$
The all possible pairs of $(a,c)$ are
$$
\{(1,\pm 6), (6,\pm 1), (2,\pm 3), (3,\pm 2)\}
$$
Here I don't need to consider the case in which $a<0$ because, for example,
$6x^2+5x-1=0$ is actually identical to $-6x^2-5x+1=0$. Only the sign of $b$ and $c$ do matter here.
From $2^2=4$ subsets of $P$, we can create $\frac{2^2}{2}=2$ "partitions", each "partition" can be assigned to $(a,c)$ in $2!=2$ ways, and $c$ can have two choice of signs. Thus there are
$$
\frac{2^2}{2}\times 2! \times 2 = 8
$$
ways to create $(a,c)$.
How many ways to specify $b$
The all possible pairs of $(\alpha,\beta)$ are
$$
\{(\pm 1,\pm 6), (\pm 2,\pm 3)\}
$$
Therefore the all possible values of $b$ are
$$
\{\pm |1+6|, \pm |1-6| = \pm |2+3|, \pm |2-3|\} = \{\pm 7, \pm 5, \pm 1 \}
$$
I cannot find an easier way to calculate how many possible ways to determine $b$ because the same $b$ can be obtained from two (or possibly more) "partitions". For example, partition $\{1,6\}$ and $\{2,3\}$ can produce $b=\pm 5$ as follows.
$$
\pm |1-6| = \pm |2+3|
$$
As we can see, there are $3\times 2=6$ ways to assign $b$.
Final
Thus in total, we have $8\times 6=48$ quadratic equations. I have not checked programmatically whether all of these equations have distinct pair of roots.
Question
Generally speaking, for any positive integer $N$, how many quadratic equations (with the constraints given above) are possible to make?
|
This is a partial answer.
This answer deals with $N$ such that $N\not\equiv 0\pmod 3$.
This answer proves that the number of such quadratic equations is
$$\begin{cases}2\sigma_0(N)^2&\text{if $N$ is not a square number with $N\not\equiv 0\pmod 3$}
\\\\\sigma_0(N)(2\sigma_0(N)-1)&\text{if $N$ is a square number with $N\equiv 1\pmod 3$}\end{cases}$$
where $\sigma_0(N)$ is the number of the positive divisors of $N$.
Proof :
We may suppose that $a\gt 0$.
Note that $ax^2+bx+c=0$ has two distinct rational roots if and only if $b^2-4ac$ is a non-zero square number.
Also, let $\sigma_0(N)$ be the number of the positive divisors of $N$.
($\sigma_0(N)=\displaystyle\prod_{k=1}^{d}(e_k+1)$ when $N=\displaystyle\prod_{k=1}^{d}p_k^{e_k}$ where $p_1,p_2,\cdots,p_k$ are distinct prime numbers.)
*
*Case 1 : $ac=\alpha\beta$Since we have $b^2-4ac=(\alpha+\beta)^2-4\alpha\beta=(\alpha-\beta)^2$, we see that $ax^2+bx+c=0$ has two distinct rational roots if and only if $\alpha\not=\beta$.If $\alpha\beta=\alpha'\beta',\alpha'\not=\alpha$ and $\alpha'\not=\beta$, then $$\small\alpha+\beta-(\alpha'+\beta')=\alpha+\beta-\alpha'-\frac{\alpha\beta}{\alpha'}=\frac{(\alpha-\alpha')(\alpha'-\beta)}{\alpha'}\not=0\implies \alpha+\beta\not=\alpha'+\beta'\tag1$$Case 1-1 : $ac=\alpha\beta=N$The number of $(a,c)$ is given by $\sigma_0(N)$ since $a$ is positive.If $N$ is not a square number, then the number of $(\alpha,\beta)$ is given by $2\sigma_0(N)$. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)$. So, the number of such quadratic equations is $\sigma_0(N)^2$.If $N$ is a square number, then the number of $(\alpha,\beta)$ is given by $2(\sigma_0(N)-1)$ since the case where $\alpha=\beta$ has to be excluded. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)-1$. So, the number of such quadratic equations is $\sigma_0(N)(\sigma_0(N)-1)$.Case 1-2 : $ac=\alpha\beta=-N$The number of $(a,c)$ is given by $\sigma_0(N)$ since $a$ is positive.If $N$ is not a square number, then the number of $(\alpha,\beta)$ is given by $2\sigma_0(N)$. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)$. So, the number of such quadratic equations is $\sigma_0(N)^2$.If $N$ is a square number, then the number of $(\alpha,\beta)$ is given by $2\sigma_0(N)$ since the case where $\alpha=\beta$ does not happen. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)$. So, the number of such quadratic equations is $\sigma_0(N)^2$.
*Case 2 : $ac=-\alpha\beta=\pm N$For $N$ such that $N\not\equiv 0\pmod 3$, we have
$$\begin{align}-\alpha\beta=\pm N&\implies (\alpha,\beta)\equiv (1,1),(1,2),(2,1),(2,2)\pmod 3
\\\\&\implies b^2-4ac=(\alpha+\beta)^2+4\alpha\beta\equiv 2\pmod 3\end{align}$$So, $b^2-4ac$ cannot be a square number.
Hence, it follows from the above cases that the number of such quadratic equations is
$$\begin{cases}2\sigma_0(N)^2&\text{if $N$ is not a square number with $N\not\equiv 0\pmod 3$}
\\\\\sigma_0(N)(2\sigma_0(N)-1)&\text{if $N$ is a square number with $N\equiv 1\pmod 3$}\end{cases}$$
where $\sigma_0(N)$ is the number of the positive divisors of $N$.
|
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|
finding a relation in $p:p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$
if $$p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$$ and $$p^2+ap+c=0.$$ Find $a,c$ also $|c|=2$
My progress:The general term $$T_{m+1}=\frac{(1)(3)\cdots(2m+1)}{(3)(6)\cdots(3m+3)}$$ from here i tried to make it in terms of factorial but it does not help.
Is there any algorithm to solve these problems?
(This question came in a maths magazine
|
It turns out that $p = \sqrt{3} - 1$. To see this, you need to recognize $p$ as the power series of the function $1/\sqrt{1-x} - 1$, evaluated at $2/3$. More specifically, the $n$-th derivative of $(1-x)^{-1/2}$ is
$$
\begin{align}
\frac{1}{2}\cdot \frac{3}{2}\cdots \frac{2n-1}{2} (1-x)^{-\frac{1}{2} - n} & = (1-x)^{-\frac{1}{2} - n} \prod_{i=0}^{n-1}(-1)\left(-\frac{1}{2}-i\right)\\
& = (1-x)^{-\frac{1}{2} - n} \prod_{i=1}^{n}\frac{2i-1}{2}
\end{align}
$$
Evaluating $ (1-x)^{-\frac{1}{2} - n} $ at zero just gives us $1$ for all $n$, so the power series should be
$$
1 + \frac{1}{2} x + \frac{1}{2!} \left(\frac{1}{2}\cdot\frac{3}{2}\right) x^2 + \frac{1}{3!}\left(\frac{1}{2}\cdot \frac{3}{2}\cdot\frac{5}{2} \right)x^3 + \cdots
$$
or
$$
\frac{1}{\sqrt{1-x}} = \sum_{n = 0}^\infty \frac{x^n}{n!} \prod_{i = 1}^{n} \frac{2i-1}{2} = \sum_{n=0}^\infty \prod_{i=1}^n\frac{2i-1}{2i} x
$$
Plugging in $x = 2/3$ then gives us
$$
\frac{1}{\sqrt{1-2/3}} = \sqrt{3} = \sum_{n=0}^\infty \prod_{i=1}^{n} \frac{2i-1}{2 i} \frac{2}{3} = \sum_{n=0}^\infty \prod_{i=1}^n \frac{2i-1}{3i}\ .
$$
The right hand side can now be recognized as $p+1$.
Finally, to construct a quadratic equation that has $p$ as one of its roots, you can make use of the Vieta's formulas. Under the constraint $|c| = 2$, there are really only two possibilities. If $c = 2$, then the other root $p'$ should be $c/p = 1+\sqrt{3}$, which tells us that $a = -p-p' = 2\sqrt{3}$. The $c = -2$ case is similar and I'll leave it to you to verify that $a = 2$ there.
The key recognition is inspired by the form of the general term. Its denominator contains a factorial, and its numerator contains a product of an arithmetic progression. The former suggests that we might be able to rewrite $p$ as a power series and the latter suggests that it might be the series of some negative power of $x$. The lack of alternating signs then suggests that we need to compose it with $1-x$ to get an extra minus sign each time we take the derivative.
|
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|
Finding the number of digits in repunit Find the number of digits in the smallest repunit divisible by $19$.
I believe a repunit number, with $N$ digits, is of the form $ \sum_{k=0}^{N-1} 10^k = \frac{10^N -1}{10-1} = \frac{10^N - 1}{9}, $ and is divisible by $n$ if $10^N - 1 \equiv 0 \pmod{9n}.$ The solution should be the smallest number N that satisfies $10^N -1 \equiv 0 \pmod{9 \cdot 19}.$ Can somebody finish this off?
|
$10\equiv1\pmod9$, so $10^N\equiv 1 \pmod9$ for all $N\in\mathbb N$,
so your question becomes what is the smallest number $N$ satisfying $10^N\equiv1\pmod{19}$.
By Fermat's little theorem, we know $10^{18}\equiv1\pmod{19}$;
we just have to show that $10^6\not\equiv1\pmod{19}$ and $10^{9}\not\equiv1\pmod{19}$.
Method 1
$10^2\equiv5\bmod19$, so $10^3\equiv50\equiv12\bmod19$, so $10^6\equiv144\equiv11$, and $10^9\equiv132\equiv18\bmod19$.
Method 2
$10^6-1=(10^3+1)(10^3-1)=(7\times11\times13)(27\times37)$ is not divisible by $19$.
By Euler's criterion, $10^9-1\equiv\left(\dfrac{10}{19}\right)=\left(\dfrac{2}{19}\right)\left(\dfrac{5}{19}\right)=(-1)\left(\dfrac45\right)=-1\pmod{19}$.
|
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|
$(ab + bc + ca) \left(\frac {1}{(a + pb)(a + qb)} + \frac {1}{(b + pc)(b + qc)} + \frac{1}{(c + pa)(c + qa)}\right)\ge \frac {9}{(p + 1)(q + 1)}$ One of my friends showed me this inequality.
$$(ab + bc + ca) \left(\frac {1}{(a + pb)(a + qb)} + \frac {1}{(b + pc)(b + qc)} + \frac{1}{(c + pa)(c + qa)}\right)\ge \frac {9}{(p + 1)(q + 1)}$$ for every positive numbers $a, b, c, p$ and $q.$
My first idea was to use Bergstrom's Inequality, so I would have $$\frac {1}{(a + pb)(a + qb)} + \frac {1}{(b + pc)(b + qc)} + \frac{1}{(c + pa)(c + qa)} \ge \frac {9}{(a ^ 2 + b ^ 2 + c ^ 2) (pq + 1) + pq (ab + bc + ca)} \ge \frac {9}{(a ^ 2 + b ^ 2 + c ^ 2)(pq + p + q + 1)} = \frac {9}{(a ^ 2 + b ^ 2 + c ^ 2)(p + 1)(q + 1)},$$ but in the end we get to $$\frac {ab + bc + ca}{a ^ 2 + b ^ 2 + c ^ 2} \ge 1,$$ which is obviously false.
I am eager to hear your advice.
|
The AM-GM:
$\dfrac{(q+1)(a+pb)+(p+1)(a+qb)}2 \geqslant \sqrt{(p+1)(q+1)(a+pb)(a+qb)}$
$$\implies \frac1{(a+pb)(a+qb)}\geqslant \frac{4(p+1)(q+1)}{[(q+1)(a+pb)+(p+1)(a+qb)]^2}=\frac{4(p+1)(q+1)}{[(p+q+2)a+(2pq+p+q)b]^2}$$
Defining $k$ using $2pq+p+q=k(p+q+2)$, the above can be rewritten as
$$\frac1{(a+pb)(a+qb)}\geqslant \frac{4(p+1)(q+1)}{(p+q+2)^2(a+kb)^2}$$
Using this, it is enough to show that (with $\sum$ denoting cyclic sums):
$$\left(\sum ab\right)\cdot \sum \frac1{\left(a+kb \right)^2} \geqslant \frac{9(p+q+2)^2}{4(p+1)^2(q+1)^2} = \frac9{(k+1)^2} \tag{1}$$
We can make this symmetric using the substitution $x = a+kb, y = b+kc, z = c + ka$, to instead prove:
$$\sum \frac1{x^2}\geqslant \frac{9(k^2-k+1)}{(k^2+1)\sum xy - k\sum x^2}$$
For this, it may be expedient to use the so called "uvw" method - let $3u = \sum x, 3v^2=\sum xy, w^3=xyz$, so that $u\geqslant v\geqslant w$. Note as the RHS is positive, we also have $(k^2+1)\sum xy > k\sum x^2 \implies (k+1)^2v^2>3ku^2$. Now the inequality is
$$\frac{(3v^2)^2-6uw^3}{w^6}\geqslant \frac{3(k^2-k+1)}{(k+1)^2v^2-3ku^2}$$
$$(k^2-k+1)w^6 +2u[(k+1)^2v^2-3ku^2]w^3-3v^4[(k+1)^2v^2-3ku^2]\leqslant 0$$
As the LHS is a quadratic in $w^3$ with both leading coefficients positive, it is maximised when $w$ attains its maximum, viz. $w=v$, which implies $w=v=u \implies x=y=z \implies a=b=c$, hence it is enough to check inequality $(1)$ for this case, which is obviously true.
|
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|
Generating function coefficient I have a generating function $\frac{1}{(1-x-x^{3}+x^{4})^{2}}$.
I have to find the $x^{n}$ coefficient. So far it seems that the next step is to convert this function back to the sequence, but I am not sure how to proceed.
|
First note that $$1-x-x^3+x^4=(1-x)(1-x^3)=(1-x)^2(1+x+x^2)\,,$$
so if nothing else, you can decompose
$$\frac1{(1-x)^4(1+x+x^2)^2}$$
into partial fractions and go from there. Alternatively, you can work with
$$\frac1{(1-x)^2}\cdot\frac1{(1-x^3)^2}\,.$$
It’s standard that
$$\frac1{(1-x)^2}=\sum_{n\ge 0}(n+1)x^n\,,$$
so
$$\begin{align*}
\frac1{(1-x^3)^2}&=1+2x^3+3x^6+4x^9+\ldots\\
&=\sum_{n\ge 0}a_nx^n\,,
\end{align*}$$
where
$$a_n=\begin{cases}
k+1,&\text{if }n=3k\\
0,&\text{otherwise.}
\end{cases}$$
Now use the fact that
$$\frac1{1-x}\sum_{n\ge 0}b_nx^n=\sum_{n\ge 0}\left(\sum_{k=0}^nb_k\right)x^n$$
for any power series $\sum_nb_nx^n$ to see that
$$\begin{align*}
\frac1{(1-x)(1-x^3)^2}&=1+x+x^2+3x^3+3x^4+3x^5+6x^6+\ldots\\
&=\sum_{n\ge 0}\binom{\lfloor n/3\rfloor+2}2x^n
\end{align*}$$
and hence that
$$\frac1{(1-x)^2(1-x^3)^2}=\sum_{n\ge 0}\sum_{k=0}^n\binom{\lfloor k/3\rfloor+2}2x^n\,.$$
Finally, if $n=3m+r$ for $\in\{0,1,2\}$, then
$$\begin{align*}
\sum_{k=0}^n\binom{\lfloor k/3\rfloor+2}2&=3\sum_{k=0}^{m-1}\binom{k+2}2+(r+1)\binom{m+2}2\\
&=3\binom{m+2}3+(r+1)\binom{m+2}2\\
&=3\binom{m+3}3-(2-r)\binom{m+2}2\,.
\end{align*}$$
Expressed directly in terms of $n$ this is
$$3\binom{\lfloor n/3\rfloor+3}3-\big(2-(n\bmod 3)\big)\binom{\lfloor n/3\rfloor+2}2\,.$$
For example, the coefficient of $x^{10}$ is
$$3\binom63-(2-1)\binom52=60-10=50\,.$$
|
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|
What is the solution to $\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$ $$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$$
I've already tried this many times but still not get the answer, this was one of my solution which I can't go further anymore.
$\frac{{x}^{2}}{(x+1)({\sqrt{x+1}}-1)^{2}}< \frac{{x}^{2}+3x+18}{({x+1})^{2}}$
$\frac{{x}^{2}(x+1)}{({\sqrt{x+1}}-1)^{2}}< {x}^{2}+3x+18$
$\frac{{x}^{3}+{x}^{2}}{({\sqrt{x+1}}-1)^{2}}< \frac{({x}^{2}+3x+18)({\sqrt{x+1}-1})^{2}}{({\sqrt{x+1}-1})^{2}}$
$\frac{{x}^{3}+{x}^{2}}{({\sqrt{x+1}}-1)^{2}}< \frac{{x}^{3}+5{x}^{2}+24x+36-(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}-1})^{2}}$
$\frac{{x}^{3}+{x}^{2}-{x}^{3}-5{x}^{2}-24x-36+(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
$\frac{-4{x}^{2}-24x-36+(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
$\frac{-4(x+3)^{2}+2({x}^2+3x+18)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
Does anyone have a hint for the solution?
|
We have that
$$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}
\iff \frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2}-\frac{{x}^{2}+3x+18}{({x+1})^{2}}<0 $$
$$\iff \frac{({x}^{2}(x+1)^2-({x}^{2}+3x+18)(({x+1-\sqrt{x+1}})^2) }{({x+1-\sqrt{x+1}})^2(x+1)^2} < 0$$
$$\iff ({x}^{2}(x+1)^2-({x}^{2}+3x+18)(({x+1-\sqrt{x+1}})^2)<0$$
with the condition
$$x+1>0 \implies {x+1-\sqrt{x+1}}\neq 0$$
then by $x+1=y>0$
$$\iff ((y-1)^2y^2-((y-1)^2+3(y-1)+18)(({y-\sqrt{y}})^2)<0$$
$$\iff 2y(\sqrt y-1)^2(y+2\sqrt y+4)(\sqrt y-2)<0$$
|
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|
How to Evaluate $ \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1} $ How can I evaluate
$$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1} \approx - 0.198909 $$
The Sum can be given also as
$$ \frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)\sqrt[4]{(-x)^{3}}}\,\left(\,\tan^{-1}\left(\sqrt[4]{-x}\right)-\tanh^{-1}\left(\sqrt[4]{-x}\right)\,\right) $$
Unfortunately i have not been able to evaluate either the Sum or the Integral using methods I know. Mathematica gives really weird results for the integral.
Is there a closed form for this Sum/Integral?
Thank you kindly for your help and time.
EDIT
For those of you that still care about the question i was able to find the following closed form. I will let the above $ sum = S $
and as such
$$ S = C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} $$
Where $C$ denotes Catalan's constant.
Thank you very much once again to those who provided answers!
EDIT #2 (Proof as Requested )
I will not show this one (too much typing) but ,
$$S= \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1} = 4 \sum_{n=1}^{\infty} (-1)^n \sum_{k=0}^{\infty} \frac{1}{(4k+3)} \frac{1}{(4k+(4n+3))} $$
next expand the terms on the RHS into a Matrix as such :
$$
\begin{matrix}
\color{red}{+(\frac13\times\frac13)} & -(\frac13\times\frac17)& +(\frac13\times\frac1{11})& -(\frac13\times\frac1{15}) \\
\color{blue}{-(\frac17\times\frac13)} & \color{red}{+(\frac17\times\frac17)} & -(\frac17\times\frac1{11}) & +(\frac17\times\frac1{15})\\
\color{blue}{+(\frac1{11}\times\frac13)} & \color{blue}{-(\frac1{11}\times\frac17)}&\color{red}{+(\frac1{11}\times\frac1{11})}&-(\frac1{11}\times\frac1{15})\\
\end{matrix}
$$
The black terms x 4 are our desired sum
I then added the red and blue terms to "complete" the
matrix
One can then see that the matrix (complete) may be given as
$$ \left(\frac13-\frac17+\frac1{11}...\right)\left(\frac13-\frac17+\frac1{11}...\right) $$
which is just
$$P= \left(\sum_{n=0}^{\infty} \frac{(-1)^n}{4n+3}\right)^{2} = \left(\frac{\pi}{4 \sqrt{2}}+\frac{\ln(\sqrt{2}-1)}{2 \sqrt{2}}\right)^2 $$
So
$$ P = \color{red}{\sum_{n=1}^{\infty} \frac{1}{(4n-1)^2}} + \color{blue}{\text{Blue terms}} + \text{Black terms} $$
but one can see that $ \color{blue}{\text{Blue terms}} = \text{Black terms} $
Therefore :
$$ P = \frac{\pi^2}{16}-\frac{C}{2}+\frac{S}{2} $$
Solve for S to find :
$$ S = C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} $$
where $C$ denotes Catalan's Constant.
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The result is not so bad. If
$$S=\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1}$$ For more legibility, I shall write $S$ as
$$S=\frac {A}{96}-i\frac B 4$$ where $A$ and $B$ contain real and complex parts.
$$A=24 C-5 \pi ^2+9 \log ^2\left(3-2 \sqrt{2}\right)+(24-6 i) \pi \log \left(3-2
\sqrt{2}\right)$$
$$B=\text{Li}_2\left(\frac{1+i}{2+\sqrt{2}}\right)-\text{Li}_2\left(\frac{1-i}{2+\sqrt{
2}}\right)+\text{Li}_2\left(-\frac{1+i}{-2+\sqrt{2}}\right)-\text{Li}_2\left(-\frac{1-i}{-2+\sqrt{2}}\right)+
i \left(\text{Li}_2\left(i \left(-1+\sqrt{2}\right)\right)+\text{Li}_2\left(-i
\left(1+\sqrt{2}\right)\right)\right)$$
$$S=-0.19890902742911208266537143997251410413430136724348\cdots$$
It is amazing that this number is very close to
$$\frac{1}{100} \left(\psi \left(\frac{1}{15}\right)+\psi
\left(\frac{3}{16}\right)-\psi \left(\frac{7}{10}\right)\right)$$ which is
$ -0.1989090283$
|
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|
Limit related to $ f(x) = \prod_{i = 1}^x \left( \sin\left( i \frac{\pi}{n}\right) + \frac{5}{4}\right) $? Let $n$ be a positive integer.
Let $b = 2 n - 1$.
Let $x$ be a positive integer.
Define $f(x)$ as :
$$ f(x) = \prod_{i = 1}^x \left( \sin\left( i \frac{\pi}{n}\right) + \frac{5}{4}\right) $$
Then it appears that
$$ f(b) = \frac{4}{5} + C(n)$$
And $C(n)$ is close to zero.
In fact
$$ \lim_{n \to \infty} C(n) = 0 $$
How do we prove this ??
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If you are allowed to use the complex representation of the sine and Pochhammer symbols,
$$f(b) = \prod_{i = 1}^{2n-1} \Big[\sin \left(\frac{\pi i}{n}\right)+\frac{5}{4}\Big]=-\frac{4}{5} \frac{ \left(\frac{i}{2};e^{\frac{i \pi
}{n}}\right){}_{2 n} \left(2 i;e^{\frac{i \pi }{n}}\right){}_{2 n}}{4^n\,e^{i \pi n}}$$which makes
$$C_n=-\frac{4}{5} \Big[1+\frac{ \left(\frac{i}{2};e^{\frac{i \pi
}{n}}\right){}_{2 n} \left(2 i;e^{\frac{i \pi }{n}}\right){}_{2 n}}{4^n\,e^{i \pi n}}\Big]$$
For the angles we know the exact trignometric functions, this generates for $C_n$'s the sequence
$$\left\{\frac{9}{20},-\frac{31}{320},\frac{129}{5120},-\frac{511}{81920},?,-\frac{8191}{20971520},?,-\frac{131071}{5368709120},?,?,?,-\frac{33554431}{351843720888320}\right\}$$ and
$$\log(|C_n|) \sim \frac 12 - b\,n \qquad \text{where} \qquad b=1.38861 \quad \left(\sigma_b=0.00053\right)$$
|
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|
Dependent variable substitution of a differential equation. I am attempting to answer a question from a textbook. The question is as follows:
"Use the substitution $y = x^2$ to turn the differential equation $x\frac{d^2x}{(dt)^2} + (\frac{dx}{dt})^2 + x\frac{dx}{dt} = 0$ into a second order differential equation with constant coefficients involving y and t."
The answer according to the textbook is $\frac{d^2y}{(dt)^2} + \frac{dy}{dt} = 0$
I am unsure what to do with the $(\frac{dx}{dt})^2$ term in the original equation. Is it equivalent to $\frac{d}{dt}(x^2)$? If this is the case then my working gets as far as the following before I get stuck:
$x\frac{d^2x}{(dt)^2} + (\frac{dx}{dt})^2 + x\frac{dx}{dt} = 0$
$\Rightarrow x\frac{d^2x}{(dt)^2} + \frac{d}{dt}(x^2) + x\frac{dx}{dt} = 0$
$\frac{dy}{dt} =\frac{dy}{dx}\frac{dx}{dt}=2x\frac{dx}{dt} \Rightarrow \frac{d^2x}{(dt)^2}=\frac{d}{dx}(\frac{1}{2x}\frac{dy}{dt})$
$\Rightarrow x\frac{d}{dx}(\frac{1}{2x}\frac{dy}{dt}) + 2x + x\frac{1}{2x}\frac{dy}{dt} = 0$
$\Rightarrow \frac{1}{2x}\frac{dy}{dt} + \frac{1}{2}\frac{d^2y}{(dt)^2} + 2x + \frac{1}{2}\frac{dy}{dt} = 0$
Please can someone show me where I have gone wrong?
|
As noticed $$\left(\frac{dx}{dt}\right)^2=\frac{dx}{dt}\cdot \frac{dx}{dt}\neq \frac{d(x^2)}{dt}$$
By the suggested substitution we have
$$y=x^2 \implies \frac{dy}{dt}=2x \frac{dx}{dt} \implies \frac{d^2y}{dt^2}=2 \left(\frac{dx}{dt}\right)^2+2x \frac{d^2x}{dt^2}$$
that is
*
*$\frac{dx}{dt}=\frac1{2x}\frac{dy}{dt} $
*$\frac{d^2x}{dt^2}=\frac1{2x}\frac{d^2y}{dt^2}-\frac1{4x^3}\left(\frac{dy}{dt}\right)^2 $
then
$$x\frac{d^2x}{dt^2} + \left(\frac{dx}{dt}\right)^2 + x\frac{dx}{dt}=0$$
$$\frac1{2}\frac{d^2y}{dt^2}\color{red}{-\frac1{4x^2}\left(\frac{dy}{dt}\right)^2 +\frac1{4x^2}\left(\frac{dy}{dt}\right)^2}+ \frac1{2}\frac{dy}{dt}=0$$
$$\frac1{2}\frac{d^2y}{dt^2}+ \frac1{2}\frac{dy}{dt}=0$$
|
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|
If $a^2x^4+b^2y^4=c^6$, then the maximum value of $xy$ Simply subsisting the value of $y$ in $xy$ will not work, but I don’t know any other method to solve it. Can I get a hint?
|
Method 1
You could use the parametric equation of ellipse to solve it.
Let $p=x^2,q=y^2$ then $x^4 = p^2,y^4=y^2$, the equation turns to
$\frac{p^2}{c^6/a^2} + \frac{q^2}{c^6/b^2}=1$, which is the ellipse function. Thus
$p = \frac{c^3}{a} \cos\theta$
$q = \frac{c^3}{b} \sin\theta,\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$
Since $xy = \sqrt{pq}$, $\max xy = \sqrt{\frac{c^6}{2ab}}$
the maximum is taken when $\theta=2k\pi+ \frac{\pi}{2},k\in\mathbb{Z}$
Method 2
Use the AM-GM inequality:
$c^6 = (ax^2)^2 + (by^2)^2 \ge 2ab(xy)^2$
thus
$xy \le \sqrt{\frac{c^6}{2ab}}$
equal is taken when $a x^2 = b y^2$.
|
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|
Integrate curve given in polar coords solution verification My solution to the below problem is $4\pi$ but the answer sheet says it's $8\pi$. Please verify my calculations:
$$
\rho=4\sin2\phi \implies 4\sin2\phi \ge0 \implies \sin t\ge 0
$$
Where $t = 2\phi$
Now, I'm trying to find the integration bounds like this:
$$
t \in [0 + 2k\pi, \pi+2k\pi], k\in Z \\
0+2k\pi \le t \le\pi+2k\pi \\
0+2k\pi \le 2\phi \le\pi+2k\pi \\
k\pi \le \phi \le \frac{\pi}{2}+k\pi \\
0 \le \phi \le \frac{\pi}{2} \vee \pi \le \phi \le \frac{3\pi}{2}\\
$$
Therefore the integrals should be:
$$
\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 16\sin^22\phi\ d\phi + \frac{1}{2} \int_{\pi}^{\frac{3\pi}{2}} 16\sin^22\phi\ d\phi = \\
8 \int_{0}^{\frac{\pi}{2}} \sin^22\phi\ d\phi + 8 \int_{\pi}^{\frac{3\pi}{2}} \sin^22\phi\ d\phi
$$
The indefinite integral is:
$$
\int \sin^22\phi\ d\phi = \frac{1}{2}\phi - \frac{1}{8}\sin4\phi + C
$$
And so I continue the calculations:
$$
8\left[\frac{1}{2}\phi-\frac{1}{8}\sin4\phi\right]_{0}^{\frac{\pi}{2}} + 8\left[\frac{1}{2}\phi-\frac{1}{8}\sin4\phi\right]_{\pi}^{\frac{3\pi}{2}} = 4\pi
$$
This got to be something simple...
EDIT
The problem I'm trying to solve is to calculate the area of a figure bounded by a curve given by the first equation in this question.
|
This is the bound -
i) For $0 \le \phi \le \frac{\pi}{2}, 0 \le 2\phi \le \pi$.
ii) For $\frac{\pi}{2} \le \phi \le \pi, \pi \le 2\phi \le 2\pi$.
iii) For $\pi \le \phi \le \frac{3\pi}{2}, 2\pi \le 2\phi \le 3\pi$.
iv) For $\frac{3\pi}{2} \le \phi \le 2\pi, 3\pi \le 2\phi \le 4\pi$.
This is rose curve with $4$ petals.
(i) gives you the petal in first quadrant, (ii) gives you the petal in the fourth quadrant and so on...
Given we have to find area bound by this curve, let's find area of one of the petals and multiply by $4$.
Area of the petal in the first quadrant $= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 16\sin^22\phi\ d\phi $
$= 8 \int_{0}^{\frac{\pi}{2}} \sin^22\phi\ d\phi = 8\left[\frac{1}{2}\phi-\frac{1}{8}\sin4\phi\right]_{0}^{\frac{\pi}{2}} = 2\pi$
Total area $= 4 \times 2\pi = 8\pi$
Or you can apply the bound (ii), (iii), (iv) in the integral to get to total area $8\pi$.
|
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|
Limit of sum of exponential functions under root How to solve this limit?
$$\underset{x\to \infty }{\text{lim}}\left(4*6^x-3*10^x+8*15^x\right)^{1/x}$$
It is equal $15$ and it seems obvious that it is so. I just can not write it mathematically.
I tried to get rid of $1/x$ in exponent:
$$\underset{x\to \infty }{\text{lim}}\left(4*6^x-3*10^x+8*15^x\right)^{1/x}=\exp \left(\underset{x\to \infty }{\text{lim}}\frac{\log \left(4*6^x-3*10^x+8*15^x\right)}{x}\right)$$
Then applied L'Hôpital's rule:
$$\frac{\partial \log \left(4*6^x-3*10^x+8*15^x\right)}{\partial x}=\frac{4*6^x (\log 6)+8*15^x (\log 15)-3*10^x (\log 10)}{4*6^x-3*10^x+8*15^x}$$
So we have:
$$\underset{x\to \infty }{\text{lim}}\left(4*6^x-3*10^x+8*15^x\right)^{1/x}=\\\exp \left(\underset{x\to \infty }{\text{lim}}\frac{4*6^x \log (6)-3*10^x \log (10)+8*15^x \log (15)}{4*6^x-3*10^x+8*15^x}\right)$$
I can apply the rule again but it only gets more complicated.
I was thinking also about some substitution but can not figure out what substitution to use.
|
$$\lim_{x\to \infty}\left(4\times6^x-3\times10^x+8\times15^x\right)^{1/x}=\lim_{x\to \infty}(15^x)^{1/x}\left(4\times\frac{6^x}{15^x}-3\times \frac{10^x}{15^x}+8\right)^{1/x}$$
and we have $$(15^x)^{1/x}\left(4\times\frac{6^x}{15^x}-3\times \frac{10^x}{15^x}+8\right)^{1/x}<15(4\times 1-3\times 0+8)^{1/x}\to 15$$
$$(15^x)^{1/x}\left(4\times\frac{6^x}{15^x}-3\times \frac{10^x}{15^x}+8\right)^{1/x}>15(4\times 0-3\times 1+8)^{1/x}\to 15$$
Now apply squeeze theorem.
|
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|
maclaurin series $\ln \left(\frac{1+x^2}{1-x^2}\right)$ I know that the Maclaurin seriess is: $\sum_{n=1}^\infty c_n(x)^n$ and $c_n = \frac{f^{n}(0)}{n!}$
$\ln \left(\frac{1+x^2}{1-x^2}\right)$
$f^0 = \ln (1+0) - \ln (1-0) = 0$
$f^{(1)} = \frac{4x}{(1+x^2)(1-x^2)} = 0$
$f^{(2)} = \frac{12x^4 + 4}{(1+x^2)^2(1-x^2)^2} = 4$
$\frac{0x^1}{1} + \frac{0x^2}{2!} + \frac{4x^2}{3!} ...$
I must have done a mistake, because I can't find the answer with that and $f^{(3)}$ seems insane.
|
Since the function is defined for $|x|<1$, you can write it as
$$
f(x)=\ln(1+x^2)-\ln(1-x^2)
$$
You surely know the Maclaurin series
$$
\ln(1+t)=\sum_{n\ge1}\frac{(-1)^{n+1}t^n}{n}
\qquad
\ln(1-t)=-\sum_{n\ge1}\frac{t^n}{n}
$$
and so
$$
\ln(1+x^2)=\sum_{n\ge1}\frac{(-1)^{n+1}x^{2n}}{n}
$$
and
$$
\ln(1-x^2)=-\sum_{n\ge1}\frac{x^{2n}}{n}
$$
Hence
$$
f(x)=
\sum_{n\ge1}\frac{(-1)^{n+1}x^{2n}}{n}+
\sum_{n\ge1}\frac{x^{2n}}{n}
$$
What are the terms that “survive”?
Only those with odd $n$, so you get $$\sum_{\substack{n\ge1\\n\text{ odd}}}\frac{2x^{2n}}{n}=\sum_{n\ge1}\frac{2x^{4n-2}}{2n-1}=\sum_{n\ge0}\frac{2x^{4n+2}}{2n+1}$$
|
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|
How to solve $A\sin{\theta} + B\cos{\theta} = C$? I've stumbled upon a equation in the form
$$A\sin{\theta} + B\cos{\theta} = C$$
What would be the steps necessary to solving it?
Thank you.
|
$$A\sin{\theta} + B\cos{\theta} = C$$
use the identity
$$\cos\theta= \pm \sqrt{1-\sin^2\theta}$$
so
$$A\sin{\theta} \pm B\sqrt{1-\sin^2\theta} = C$$
$$ \pm B\sqrt{1-\sin^2\theta} = C-A\sin{\theta}$$
$$ B^2(1-\sin^2\theta) = (C-A\sin{\theta})^2$$
let $x=\sin{\theta}$
$$ B^2(1-x^2) = (C-Ax)^2$$
$$B^2-B^2x^2=C^2-2ACx+A^2x^2$$
$$(A^2+B^2)x^2-(2AC)x+(C^2-B^2)=0$$
then use the quadratic formula to solve it and then complete the solution to find the $\theta$ values
|
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|
Power series of $\frac{x^{3}-2}{x^{2}+1}$ I have to find the power series of the function
$$f(x)=\frac{x^3-2}{x^{2}+1}$$
centered at $a=1$. I tried to write $f$ as
$$f(x)=(x^3-2)\cdot\frac{1}{x^{2}+1}$$
and then, find the power series of $\displaystyle\frac{1}{x^{2}+1}$ centered at $a=1$. I want to know if this is the best way to do it.
|
If you make $x=y+1$
$$\frac{x^3-2}{x^{2}+1}=\frac{y^3+3 y^2+3 y-1}{y^2+2 y+2}=1+y-\frac{y+3}{y^2+2 y+2}$$ If you want a truncated series, continue with the long division.
Looking at the coefficient of the first terms, I suppose that their definition is not the simplest.
Edit
Working
$$\frac{y+3}{y^2+2 y+2}=\sum_{n=0}^\infty a_n\,y^n$$ the coefficients are given by
$$a_n=2^{-(n+2)} \left((3-i) (-1-i)^n+(3+i) (-1+i)^n\right)$$ which are all real. Replacing $y$ by $(x-1)$, we then have
$$\frac{x^3-2}{x^{2}+1}=1+(x-1)-\sum_{n=0}^\infty a_n\,(x-1)^n=-\frac 12+2(x-1)-\sum_{n=2}^\infty a_n\,(x-1)^n$$
|
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|
Expressing $\frac{1+x+x^2}{1-x^7}$ as a closed form for the $n^{th}$ term of the associated sequence First of all I'm a little confused about the wording, am I suppose to find the sequence associated, lets say for example $(1,1,1,1,\ldots)$, and come up with an expression like $a_n=1^n$? I've gotten this far with this question
\begin{align}
\frac{1+x+x^2}{1-x^7}=&\ \frac{1}{1-x^7}+\frac{x}{1-x^7}+\frac{x^2}{1-x^7}\\
=&\ \sum^{\infty}_{k=0}x^{7k} +x\sum^{\infty}_{k=0}x^{7k}+x^2\sum^{\infty}_{k=0}x^{7k}\\
=& \sum^{\infty}_{k=0}x^{7k} +\sum^{\infty}_{k=0}x^{7k+1}+\sum^{\infty}_{k=0}x^{7k+2}
\end{align}
But I'm unsure what to do next, am I supposed to reindex each sum s.t. the exponents are all the same?
eg: $k=k-\frac{1}{7}$ for the second sum, and $k=k-\frac{2}{7}$ for the third sum? But if I do that, then the lower limit of the sums wouldn't be integers and I'm assuming that isn't allowed. Or am I overcomplicated it and I can just do:
\begin{align}
\sum^{\infty}_{k=0}x^{7k} +\sum^{\infty}_{k=0}x^{7k+1}+\sum^{\infty}_{k=0}x^{7k+2}=& \sum^{\infty}_{k=0}x^{7k} +\sum^{\infty}_{k=0}x\cdot x^{7k}+\sum^{\infty}_{k=0}x^2\cdot x^{7k}\\
=&\ \sum^{\infty}_{k=0}(1+x+x^2)x^{7k}
\end{align}
If so, would the closed form of the associated sequence be $a_n = 1+n+n^2$?
|
From expanding $$\sum_{k=0}^\infty (1+x+x^2)x^{7k} = 1 + x + x^2+x^7+x^8+x^9+x^{14}+x^{15}+x^{16}+\dots$$
the sequence associated with this G.F. is $(1,1,1,0,0,0,0,1,1,1,0,0,0,0, \dots)$, which can be written as:
$$a_n=\begin{cases}1 &\text{for } n \equiv 0,1,2 &\pmod 7\\0 &\text{for } n \equiv 3,4,5,6 &\pmod 7\end{cases}$$
where $1+x+x^2$ contributes to the $n\equiv 0,1,2$ and $x^{7k}$ contributes to$\pmod 7$.
We can also see this from the original $$\sum_{k=0}^\infty x^{7k} + \sum_{k=0}^\infty x^{7k+1} + \sum_{k=0}^\infty x^{7k+2}$$
so the coefficient of $x^n$ is nonzero and equal to $1$ if and only if $n = 7k,7k+1$ or $7k+2$.
As a side note, the G.F. for $a_n=1+n+n^2$ would look like:
$$1 + 3x + 7x^2 + 13x^3 + 21 x^4 + 31 x^5 + 43 x^6 + \cdots=\frac{1+x^2}{(1-x)^3}.$$
|
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|
Better proof of inequality $x - (1 + x) \log(1+x) \leq -\frac{x^2}{2(1+x)}$ for $x > 0$ The following inequality is valid for all positive real $x$,
$$
x - (1+x)\log(1+x) \leq \frac{-x^2}{2(1+x)}.
$$
It is possible to show that this is true by considering the function
$$
f(x) := x - (1+x)\log(1+x)+ \frac{x^2}{2(1+x)}.
$$
By differentiation it is possible to check that $f(x)$ attains it maximum on the nonnegative reals at $x = 0$.
However, is there a cleaner, more obvious way to see that this is true? Specifically, I would like a solution that does not require me to analyze the monotonicity of $f$ via differentiation if possible. I tried Taylor expansion, but do not get this inequality.
|
Better proof:
$\log(1+y)\le y-\dfrac{1}{2}y^2\quad$ for all $\;y\in\left]-1,0\right]\;.\quad\color{blue}{(*)}$
For all $\;x\ge0\;,\;$ it results that $\;y=\dfrac{1}{1+x}-1\in\left]-1,0\right],$ hence, by applying $(*)$, we get that
$\log\left(1+\dfrac{1}{1+x}-1\right)\le\dfrac{1}{1+x}-1-\dfrac{1}{2}\left(\dfrac{1}{1+x}-1\right)^2,$
$-\log\left(1+x\right)\le\dfrac{1}{1+x}-1-\dfrac{1}{2}\dfrac{x^2}{(1+x)^2}\;,$
$-(1+x)\log\left(1+x\right)\le 1-(1+x)-\dfrac{x^2}{2(1+x)}\;,$
$-(1+x)\log\left(1+x\right)\le -x-\dfrac{x^2}{2(1+x)}\;,$
$x-(1+x)\log(1+x)+\dfrac{x^2}{2(1+x)}\le0\;,\;$ for all $\;x\ge0\;.$
|
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|
$\frac{\tan8°}{1-3\tan^2 8°}+\frac{3\tan24°}{1-3\tan^2 24°} + \frac{9\tan72°}{1-3\tan^2 72°} + \frac{27\tan216°}{1-3\tan^2 216°}=x\tan108°+y\tan8°$ $$\frac {\tan 8°}{1-3\tan^2 8°}+\frac {3\tan 24°}{1-3\tan^2 24°} + \frac{9\tan 72°}{1-3\tan^2 72°} + \frac{27\tan 216°}{1-3\tan^2 216°} =x\tan108°+y\tan8° .$$
Find the value of $x$ and $y$.
I tried different approaches like using $$ \tan 3x=\frac{3\tan x-\tan^3 x}{1-3\tan^2 x}$$
and converting into $\sin$ and $\cos$, but I was unable to simplify.
Can anybody help?
|
Hint
$$f(p)=\dfrac{\tan p}{1-3\tan^2p}+z\tan p=\dfrac{(1+z)\tan p-3z\tan^3p}{1-3\tan^2p}$$
Now comparing with $\tan3p=?$ formula,
we need $$\dfrac{1+z}{3z}=\dfrac31 \iff z=\dfrac18$$
$$\implies f(p)=\dfrac{3\tan3p}8$$
Put $p=8,24,72,216^\circ$ to find the LHS to be
$$-\dfrac{\tan8^\circ}8+\dfrac{3^4\tan648^\circ}8$$
Finally $648\equiv108\pmod{180}$
|
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|
If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$? From the Pre-Regional Mathematics Olympiad, 2019:
If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$?
I have provided one answer below, and would be interested in alternative solutions.
|
We can rewrite $x =\sqrt{2} + \sqrt{3} + \sqrt6$ as
$$\begin{aligned}(x-\sqrt2)^2 &= (\sqrt2 + \sqrt6)^2\\
x^2 - 2\sqrt2x+2 &= 9 + 6\sqrt2\\
x^2-7&=2\sqrt2\space(x+3)\end{aligned}$$
On squaring both sides of the equation,
$$x^4-22x^2-48x-23=0$$
Therefore, $a = 0, b = -22, c = -48, d = -23$, implying $|a+b+c+d| = 93$
|
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|
What is $\tan \alpha$, if $(a+2)\sin\alpha +(2a - 1)\cos\alpha =2a + 1$? I tried the following:
$$\begin{aligned}a\sin\alpha +2\sin\alpha + 2a\cos\alpha - \cos\alpha &= 2a+1\\
a(\sin\alpha +2\cos\alpha)+(2\sin\alpha-\cos\alpha)&=2a+1\end{aligned}$$
Therefore,
$$\sin\alpha +2 \cos\alpha=2$$
$$2\sin\alpha - \cos\alpha=1$$
From these two equations, we get
$$\sin\alpha=\frac{4}{5},\cos\alpha=\frac{3}{5}$$
Therefore,
$$\tan\alpha = \frac{\sin\alpha} {\cos\alpha} = \frac{4} {3}$$
Is this a correct method to solve the question? Since $a$ is a constant, it does not seem necessary to me that its coefficients on the two sides of the equation be equal. Should I find $\sin\alpha$ and $\cos\alpha$ using some other method? Are there specific cases where this method of equating the coefficients will break?
|
The OP finds one constant root that applies for all $a$. But there is a second root for most specific values of $a$, which is a function of $a$. The full answer is $\tan\alpha\in\{4/3,2a/(a^2-1)\}$.
Properly, the given equation should be combined with the identity $\sin^2\alpha+\cos^2\alpha=1$. There are two ways to do this:
Method 1
Isolate one of the trigonometric function, square the resulting equations and substitute to get a quadratic equation for the remaining function. Choosing to isolate the cosine we then have
$(2a-1)\cos\alpha=(2a+1)-(a+2)\sin\alpha$
$(2a-1)^2\cos^2\alpha=(2a+1)^2-2(2a+1)(a+2)\sin\alpha+(a+2)^2\sin^2\alpha$
$(4a^2-4a+1)-(4a^2-4a+1)\sin^2\alpha=(4a^2+4a+1)-(4a^2+10a+4)\sin\alpha+(a^2+4a+4)\sin^2\alpha$
$(5a^2+5)\sin^2\alpha-(4a^2+10a+4)\sin\alpha+8a=0$
The quadratic equation looks like a mouthful, but its discriminant is a squared quantity, to wit $(4a^2-10a+4)^2$, thus we get the two roots
$\sin\alpha=\dfrac{(4a^2+10a+4)\pm(4a^2-10a+4)}{2(5a^2+5)}\in\{4/5,2a/(a^2+1)\}$
For each root of $\sin\alpha$ the previous equation with $\cos\alpha$ isolated is used to assure the proper sign of that function:
$(2a-1)\cos\alpha=(2a+1)-(a+2)(4/5); \cos\alpha=3/5$
$(2a-1)\cos\alpha=(2a+1)-(a+2)(2a/(a^2+1)); \cos\alpha=(a^2-1)/(a^2+1)$
Correspondingly $\tan\alpha\in\{4/3,2a/(a^2-1)\}$.
Thus the answer given by the OP is correct for one root that applies for all $a$, but in most cases there will be a second root for any specific value of $a$ (the only exceptions being $a=2$ where there is one doubly degenerate root instead, and $a=\pm 1$ where the second root fails to give a defined value for $\tan\alpha$; also $a=-1/2$ gives the second root with $\tan\alpha=4/3$ but different values for the sine and cosine). The existence of two roots ultimately comes from the fact that except for $\pm 1$, any value of the sine or cosine corresponds to two different arguments within any fundamental period.
Method 2
In this method, we use a trick by combining the original equation with one involving the orthogonal linear combination. To get the orthogonal combination, switch the coefficients $a+2$ and $2a-1$ and then reverse the sign before the second term.
$(a+2)\sin\alpha+(2a-1)\cos\alpha=(2a+1)$
$(2a-1)\sin\alpha-(a+2)\cos\alpha=x$
Square both sides and add them together getting:
$(5a^2+5)(\sin^2\alpha+\cos^2\alpha)=5a^2+5=(2a+1)^2+x^2$
Thus $x=\pm\sqrt{5a^2+5-(2a+1)^2}=\pm(a-2)$. We then have a linear system to solve for $\sin\alpha$ and $\cos\alpha$ for each root of $x$. For example, $x=+(a-2)$ gives
$(a+2)\sin\alpha+(2a-1)\cos\alpha=(2a+1)$
$(2a-1)\sin\alpha-(a+2)\cos\alpha=a-2$
whose solution is $\sin\alpha=4/5,\cos\alpha=3/5$. Putting $-(a-2)$ for $x$ similarly gives a linear system with solution $\sin\alpha=2a/(a^2+1),\cos\alpha=(a^2-1)/(a^2+1)$.
The remainder of the solution, and the comments that follow, are identical to Method 1.
|
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|
Prove $\forall z\in\mathbb C-\{-1\},\ \left|(z-1)/(z+1)\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$ I'm trying to prove
$$\forall z\in\mathbb C-\{-1\},\ \left|\frac{z-1}{z+1}\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$$
thus showing that the solutions to $\left|(z-1)/(z+1)\right|=\sqrt2$ form the circle of center $-3$ and radius $\sqrt8$. But my memories of algebra in $\mathbb C$ fail me. The simplest I get is writing $z=x+i\,y$ with $(x,y)\in\mathbb R^2-\{(-1,0)\}$ and doing the rather inelegant
$$\begin{align}\left|\frac{z-1}{z+1}\right|=\sqrt2&\iff\left|z-1\right|=\sqrt2\,\left|z+1\right|\\
&\iff\left|z-1\right|^2=2\,\left|z+1\right|^2\\
&\iff(x-1)^2+y^2=2\,((x+1)^2+y^2)\\
&\iff0=x^2+6\,x+y^2+1\\
&\iff(x+3)^2+y^2=8\\
&\iff\left|z+3\right|^2=8\\
&\iff\left|z+3\right|=\sqrt8\\
\end{align}$$
How can I avoid the steps with $x$ and $y$ ?
|
Using thrice that $\forall c\in\mathbb C,\, \left|c\right|^2=c\,\bar c$, I got it down to
$$\begin{align}\left|\frac{z-1}{z+1}\right|=\sqrt2&\iff\left|z-1\right|=\sqrt2\,\left|z+1\right|\\
&\iff\left|z-1\right|^2=2\,\left|z+1\right|^2\\
&\iff(z-1)\,(\bar z-1)=2\,(z+1)\,(\bar z+1)\\
&\iff0=z\,\bar z+3\,z+3\,\bar z+1\\
&\iff(z+3)(\bar z+3)=8\\
&\iff\left|z+3\right|^2=8\\
&\iff\left|z+3\right|=\sqrt8\\
\end{align}$$
|
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|
prove the convergence of the following series If $0 < d_n < 1$ with $\sum d_n$ divergent, then the two series $$\sum d_{n+1}\left[(1-d_0)\cdots(1-d_n)\right]^p$$
$$\sum\frac{d_{n+1}}{\left[(1+d_0)\cdots(1+d_n)\right]^p}$$
are convergent, for every $p>0$. I think it's enough to show the convergence of the first one, because by substituting $1-d_n' = \frac{1}{1+d_n}$ one can reduce the second series to the first one (almost, except for a bounded factor). I don't really have any ideas on how to prove this. Can you help me?
|
Let $S_0 = d_0$ and $S_n = \sum_{j=0}^n d_j $.
We have $(1+d_0)\cdots(1+d_n)\geqslant 1 + d_0 + \cdots + d_n \geqslant d_0 + \cdots + d_n + d_{n+1} = S_{n+1}$.
When $p > 1$ there exists a positive integer $m$ such that $\frac{1}{m} < p-1$ and
$$\sum_{n=0}^N\frac{d_{n+1}}{\left[(1+d_0)\cdots(1+d_n)\right]^p} \leqslant \sum_{n=0}^N\frac{S_{n+1}- S_n}{S_{n+1} ^p} \leqslant \sum_{n=0}^N\frac{S_{n+1}- S_n}{S_{n+1}S_n ^{p-1}} \\\leqslant \sum_{n=0}^N\frac{S_{n+1}- S_n}{S_{n+1}S_n ^{1/m}}$$
Note that
$$\frac{S_{n+1}- S_n}{S_{n+1}S_n ^{1/m}}= \frac{1- \frac{S_{n}}{S_{n+1}}}{S_n^{1/m}}= \frac{1- \frac{S_{n}}{S_{n+1}}}{1- \frac{S_{n}^{1/m}}{S_{n+1}^{1/m}}}\left(\frac{1}{S_{n}^{1/m}} - \frac{1}{S_{n+1}^{1/m}} \right), $$
where the term $x = \frac{S_n^{1/m}}{S_{n+1}^{1/m}}= \left(\frac{S_n}{S_{n+1}}\right)^{1/m} \in (0,1) $ since the sequence $S_n$ is increasing .
By Bernoullis' inequality, we have $x^m = [1- (1-x)]^m \geqslant 1 - m(1-x)$ which implies that $1- x^m \leqslant m(1-x)$ and, substituting for $x$,
$$1- \frac{S_{n}}{S_{n+1}} \leqslant m\left(1- \frac{S_{n}^{1/m}}{S_{n+1}^{1/m}}\right)$$
Thus,
$$\sum_{n=0}^N\frac{d_{n+1}}{\left[(1+d_0)\cdots(1+d_n)\right]^p} \leqslant m\sum_{n=0}^N\left(\frac{1}{S_{n}^{1/m}} - \frac{1}{S_{n+1}^{1/m}} \right) = \frac{m}{d_0^{1/m}} - \frac{m}{S_{N+1}^{1/m}}$$
The series $\sum d_n$ diverges to $+\infty$ which implies that $m/S_{N+1}^{1/m} \to 0$ as $N \to \infty$ and the sum on the LHS converges, with
$$\sum_{n=0}^\infty\frac{d_{n+1}}{\left[(1+d_0)\cdots(1+d_n)\right]^p} \leqslant \frac{m}{d_0^{1/m}}$$
For the first series, taking $d_n' = d_n/(1-d_n)$, we have
$$1- d_n = \frac{1}{1+d_n'}, \quad d_{n+1} = \frac{d_{n+1}'}{1+d_{n+1}'},$$
and
$$d_{n+1}[(1- d_0) \cdots (1-d_n)]^p = \frac{d_{n+1}'}{\left[(1+d_0')\cdots(1+d_n')\right]^p(1+d_{n+1}')} \leqslant \frac{d_{n+1}'}{\left[(1+d_0')\cdots(1+d_n')\right]^p}$$
Now you can prove convergence using the result for the second series, after showing that divergence of $\sum d_n$ implies divergence of $\sum d_n'$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Derivative Integral $$\frac{d}{dx} \int_{x}^{\sqrt{x}} \frac{e^{xy^2}}{y}dy$$
Where am I going wrong?
$$\begin{align*}
\frac{d}{dx} \int_{x}^{\sqrt{x}} \frac{e^{xy^2}}{y}dy = & \frac{d}{dx} ( F(\sqrt{x} ) - F(x))\\
= & \frac{F'(\sqrt{x} )}{2\sqrt{x}} - F'(x))\\
= & \frac{e^{x(\sqrt{x})^2}}{2\sqrt{x}\sqrt{x}} - \frac{e^{xx^2}}{x}\\
= & \frac{e^{x^2} - 2e^{x^3}}{2x}
\end{align*}$$
The alternatives to this question were
a) $\frac{e^{x^2}}{2x}$
b) $\frac{3e^{x^3}}{2x}$
c) $2e^{x^2} - 3e^{x^3}$
d) $\frac{ 2e^{x^2} - 3e^{x^3} }{2x}$
e) $\frac{ 2e^{x^2} + 3e^{x^3} }{2x}$
Thanks in advance!
|
We have from Leibniz's Rule for Differentiating Under the Integral
$$\begin{align}
\frac{d}{dx}\int_x^{\sqrt x}\frac{e^{xy^2}}{y}\,dy&=\color{blue}{\underbrace{\int_x^{\sqrt x}\frac{\partial}{\partial x}\left(\frac{e^{xy^2}}{y}\right)\,dy}_{\text{The Ommitted Term}}}+\color{red}{\underbrace{\left(\frac1{2\sqrt x}\frac{e^{x^2}}{\sqrt x}-\frac{e^{x^3}}{x}\right)}_{\text{This part was correct}}}\\\\
&=\color{blue}{\left(\frac{e^{x^2}}{2x}-\frac{e^{x^3}}{2x}\right)}+\color{red}{\left(\frac{e^{x^2}}{2 x}-\frac{e^{x^3}}{x}\right)}\\\\
&=\frac{2e^{x^2}-3e^{x^3}}{2x}
\end{align}$$
|
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|
Show that $\sum_{k=0}^{n} \binom{n}{k} ka^k = an(a+1)^{n-1}$ Problem:
Show that $\sum_{k=0}^{N} \binom{N}{k} ka^k = aN(a+1)^{N-1}$
Attempt:
I tried using induction but got stuck.
$N=0$ implies $\sum_{k=0}^{0} \binom{0}{k} ka^k = 0$
$N=1$ implies $\sum_{k=0}^{1} \binom{1}{k} ka^k = a = a(1)(a+1)^{1-1}$
$N=2$ implies $\sum_{k=0}^{2} \binom{2}{k} ka^k = 2a(a+1)^{2-1}$
Assume that for $N=n$, $\sum_{k=0}^{n} \binom{n}{k} ka^k = na(a+1)^{n-1}$. Then, for $N=n+1$, we find
$$\sum_{k=0}^{n+1} \binom{n+1}{k} ka^k = \sum_{k=0}^{n+1} \left(\binom{n}{k-1}+\binom{n}{k}\right) ka^k$$
Splitting up the sum on the right gives
$$\sum_{k=0}^{n+1} \binom{n}{k-1}ka^k + \sum_{k=0}^{n} \binom{n}{k}ka^k + \binom{n}{n+1}ka^k$$
and since $\binom{n}{n+1}=0$, we have
$$\sum_{k=0}^{n+1} \binom{n+1}{k} ka^k = \sum_{k=0}^{n+1} \binom{n}{k-1}ka^k + an(a+1)^{n-1}$$
and this is where I'm stuck. If we look at it from the other direction, we have
$$(n+1)a(a+1)^{n+1-1} = (n+1)a(a+1)^n = na(a+1)^n + a(a+1)$$
But I'm not sure where to go from here.
|
$$S=\sum_{k=0}^{n} k {n \choose k} a^k =\sum_{k=0}^{n} \frac{n!}{k! (n-k)!} a^k= \sum_{k=0}^{n} \frac{n (n-1)!}{(k-1)! (n-k)!}=n\sum_{k=0}^{n} {n-1 \choose k-1} a^k$$
Let $k-1=p$, then
$$S=n\sum_{p=-1}^{n-1} {n-1 \choose p} a^{p+1}=na\sum_{p=0}^{n-1} {n-1 \choose p} a^p=na(1+a)^{n-1}.$$
Note that ${m \choose -n}=0$, where $n,m$ are positive integers.
|
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|
Solving limit - $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$ $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$
Since $x$ approaches $0$ and $y$ also approaches $0$ we can suspect that $0<x^2 + y^2<1$. For every $x,y\in\Bbb R$, we have that $\frac{1}{4}(x^2 + y^2)^2\geq x^2y^2$.
Now, $1\geq (x^2+y^2)^{x^2y^2}\geq (x^2+y^2)^{\frac{1}{4}(x^2 + y^2)^2}$, then substitute $(x^2 + y^2)=t$
$1\geq \lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}\geq \lim_{t\to0}t^{\frac{1}{4}t^2}=\lim_{t\to0}e^{\frac{1}{4}t^2\ln t}=e^0=1$
This is how my professor solved this limit. What I don't understand is this part:
$\frac{1}{4}(x^2 + y^2)^2\geq x^2y^2$
How can I prove it? And this would never come to my mind, is there maybe some other way to solve the limit? Grateful in advance.
|
hint
The difference gives
$$\frac 14(x^2+y^2)^2-x^2y^2=$$
$$\frac 14\Bigl(x^4+y^4+2x^2y^2-4x^2y^2\Bigr)=$$
$$\frac 14\Bigl(x^4+y^4-2x^2y^2\Bigr)=$$
$$\frac 14(x^2-y^2)^2\ge 0$$
Other proof
Putting
$$x=r\cos(t)\;,\;y=r\sin(t)$$
We know that
$$\sin^2(2t)\le 1 \iff $$
$$4\sin^2(t)\cos^2(t)\le (\cos^2(t)+\sin^2(t))^2\iff $$
$$4r^4\sin^2(t)\cos^2(t)\le r^4(\cos^2(t)+\sin^2(t))^2\iff$$
$$4x^2y^2\le (x^2+y^2)^2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $\cos\theta-3\cos2\theta+\cos3\theta=\sin\theta-3\sin2\theta+\sin3\theta$ My attempt:
\begin{align*}
\cos\theta-3\cos2\theta+\cos3\theta&=\sin\theta-3\sin2\theta+\sin3\theta\\
\cos\theta-3\cos2\theta+4\cos^3\theta-3\cos\theta&=\sin\theta-3\sin2\theta+3\sin\theta-4\sin^3\theta\\
-2\cos\theta-3\cos2\theta+4\cos^3\theta&=4\sin\theta-3\sin2\theta-4\sin^3\theta
\end{align*}
I have faced several symmetric trigonometry problems most of them need to use product to sum identities, but this one I can't continue.
|
It's $$2\cos2\theta\cos\theta-3\cos2\theta=2\sin2\theta\cos\theta-3\sin2\theta$$ or
$$(2\cos\theta-3)(\cos2\theta-\sin2\theta)=0$$ or
$$\tan2\theta=1.$$
Can you end it now?
|
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|
Solve for closed form $a_j$ given $a_j = 1 + pa_{j+1} + (1-p)a_{j-1}$ where $a_1 = 0$.
Solve for a closed form expression for $a_j$ given constant $p \in (0,1)$ and $q=1-p$ and:
\begin{align*}
a_j &= 1 + pa_{j+1} + qa_{j-1} \\
\end{align*}
Given that $a_1 = 0$.
The text says to use the hint that $a_j = c(1-j)$, but where do we get that hint? If we plug in that hint, we get $a_j = \frac{1-j}{p-q} = \frac{1-j}{2p-1}$ which correctly satisfies the original equation. But how would I solve this without the given hint?
Is this matrix version of the original equation helpful in any way?
\begin{align*}
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ -\frac{1}{p} & \left(1 - \frac{1}{p}\right) & \frac{1}{p} \end{pmatrix} \begin{pmatrix} 1 \\ a_{j-1} \\ a_j \end{pmatrix} &= \begin{pmatrix} 1 \\ a_j \\ a_{j+1} \end{pmatrix} \\
\end{align*}
|
The intuition of the hint is that from state $j$ you need to "move to the left" a net total of $j-1$ times to reach the absorbing state $1$, and the expected number of steps to move one unit to the left is independent of $j$ because the transition probabilities are.
For an order-2 recurrence relation, you should be given two initial conditions, so I assume $a_2=1/(1-2p)$ in addition to $a_1=0$.
Without being given or recognizing the hint, one approach is to use generating functions. Let $A(z)=\sum_{j=1}^\infty a_j z^j$. The recurrence relation for $j \ge 2$ implies that
\begin{align}
A(z) - a_1 z^1
&= \sum_{j=2}^\infty \left(1 + p a_{j+1} + q a_{j-1}\right) z^j \\
&= \sum_{j=2}^\infty z^j + \frac{p}{z} \sum_{j=2}^\infty a_{j+1} z^{j+1} + q z \sum_{j=2}^\infty a_{j-1} z^{j-1} \\
&= \frac{z^2}{1-z} + \frac{p}{z} (A(z) - a_1 z^1 - a_2 z^2) + q z A(z) \\
&= \frac{z^2}{1-z} - p a_2 z + \left(\frac{p}{z} + q z\right) A(z).
\end{align}
So
\begin{align}
A(z) &= \frac{z^2/(1-z)-p a_2 z}{1-p/z-qz} \\
&= \frac{z^3-p a_2 z^2(1-z)}{(1-z)(z-p-qz^2)} \\
&= \frac{(1-2p)z^3-p z^2(1-z)}{(1-2p)(1-z)(z-p-(1-p)z^2)} \\
&= \frac{z^2}{(1-2p)(1-z)^2} \\
&= \frac{z^2}{1-2p}\sum_{j=0}\binom{j+1}{1}z^j \\
&= \frac{1}{1-2p}\sum_{j=2}\binom{j-1}{1}z^j,
\end{align}
which implies that $$a_j=\frac{j-1}{1-2p}.$$
|
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|
Solving the equation $X^2 + (3 +i)X + 1 +i =0$ I have to find the roots of the equation
$$X^2 + (3 +i)X + 1 +i =0.$$
The first step is to find the discrimant which is $4 + 2i$. Then, I assume that the square root of the discriminant is in the form of $a+bi$, so we have to solve the below system
$$a^2-b^2=4\qquad\text{ and }\qquad ab=1.$$
How can I solve the above system ? Please do not provide another solution (e.g using trigonometric form)
|
Hint:
Multiplying by $a^2$,
$$a^2-b^2=4\land ab=1\implies a^4-a^2b^2=a^4-1=4a^2$$ and you get a biquadratic equation in $a$.
More generally, the square root of $u+iv$ is obtained by solving
$$\begin{cases}a^2-b^2&=u,\\2ab&=v\end{cases}$$ and
$$a^4-ua^2-\frac{v^2}4=0.$$
So,
$$a^2=\frac{\sqrt{u^2+v^2}+u}2$$ (the negative root must be rejected) and similarly
$$b^2=\frac{\sqrt{u^2+v^2}-u}2.$$
After taking the square roots, there are two solutions such that the sign of $ab$ matches that of $v$.
|
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|
Let $f(n)=\sum_{k=0}^{\left\lfloor n/2\right\rfloor} {2k \choose k}{n \choose 2k}$ . Show that $\sum_{n\geq 0}^{} f(n)x^n=\frac{1}{\sqrt{1-2x-3x^2}}$ Using the multinomial theorem, one can show that $f(n)$ is the coefficient of $x^n$ of the polynomial $(1+x+x^2)^n$. There are $3$ obvious ways to show the equation in the title:
*
*First, you can square the $2$ sides of the equation and then multiply with $1-2x-3x^2$. In that case, the coefficients of the left formal power series seems too much.
*Second, we can factorize the polynomial,
$$
1-2x-3x^2=(-3)(x+1)(x-1/3)=(1+x)(1-3x)
$$
$$\mbox{and use the formula ,}\quad
\sqrt{1+F(x)}=\sum_{n\geq 0}^{}(-1)^n \frac{1}{4^n}{2n \choose
n}F(x)^n
$$
for $F(x)=x , G(x)=-3x$, then multiply the formal power
series and see if the equation holds.
*The third way is to use the
previous method for $F(x)=-(2x+3x^2)$. In the 2 last methods the
numbers arent very far from those we want, but i cant prove it. I
think that i am missing some identity with binomial coefficients and
thats why i cant solve it.
|
Here's a way that doesn't require knowing the result ahead of time.
\begin{align}
\sum_{n \ge 0} f(n) x^n
&= \sum_{n \ge 0}\sum_{k=0}^{\left\lfloor n/2\right\rfloor} \binom{2k}{k}\binom{n}{2k} x^n \\
&= \sum_{k \ge 0}\binom{2k}{k} \sum_{n\ge 2k} \binom{n}{2k} x^n \\
&= \sum_{k \ge 0}\binom{2k}{k} \frac{x^{2k}}{(1-x)^{2k+1}} \\
&= \frac{1}{1-x}\sum_{k \ge 0}\binom{2k}{k} \left(\left(\frac{x}{1-x}\right)^2\right)^k \\
&= \frac{1}{1-x}\cdot\frac{1}{\sqrt{1-4\left(\frac{x}{1-x}\right)^2}} \\
&= \frac{1}{\sqrt{(1-x)^2-4x^2}} \\
&= \frac{1}{\sqrt{1-2x-3x^2}}
\end{align}
|
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|
Writing a function in terms of its power series with some tricky reindexing steps Represent the following function as a power series and find it's radius of convergence:
$$\frac{x^2}{(8+x)^3}$$
By Using differentiation to find the power series of a fairly tricky function!! We know that
$\frac{1}{(8+x)^3} = \frac{1}{2} (\sum_{n=0}^\infty (-1)^{n+2} (n+2)(n+1) x^n (\frac{1}{8})^{n+3})$
Therefore we have:
$\frac{x^2}{(8+x)^3} = x^2 (\frac{1}{2} \sum_{n=0}^\infty (-1)^{n+2} (n+2)(n+1) x^n (\frac{1}{8})^{n+3})$
Multiplying the $x^2$ through yields
$= \frac{1}{2} (\sum_{n=0}^\infty (-1)^{n+2} (n+2)(n+1) x^{n+2} (\frac{1}{8})^{n+3})$
But we can't have a power series be expressed in terms of $x^{n+2}$. It needs to be in terms of $x^n$. So we subtract a $2$ from everywhere an $n$ appears in our expressoin and make the summation start at $n=2$
$= \frac{1}{2} (\sum_{n=2}^\infty (-1)^{n} (n)(n-1) x^{n} (\frac{1}{8})^{n+1})$
|
Yes it's correct. Since for $n=0$ and $n=1$ the terms vanish, if you want to start at $n=0$ we can write $$\frac{x^2}{(x+3)^3}= \frac{1}{2} \sum_{n=2}^\infty (-1)^{n} (n)(n-1) x^{n} (\frac{1}{8})^{n+1}=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n(n)(n-1)x^n(\frac{1}{8})^{n+1}$$
|
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|
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
What I Tried: Here is a picture :-
I know the centroid divides each of the medians in the ratio $2:1$ . So $AD = 3\sqrt{3}$ , $BE = 3\sqrt{2}$ , $CF = 3$ .
From this site :- https://mathworld.wolfram.com/TriangleMedian.html, I find that the area of the triangle will be :- $$\frac{4}{3}\sqrt{s_m(s_m - m_1)(s_m - m_2)(s_m - m_3)}$$
Where $m_1,m_2,m_3$ are the medians of the triangle and $s_m = \frac{m_1 + m_2 + m_3}{2}$ .
After putting the respective values for the medians I get that $[\Delta ABC]$ is :-
$$\frac{4}{3}\sqrt{\Bigg(\frac{3(\sqrt{3} + \sqrt{2} + 1)}{2}\Bigg)\Bigg(\frac{3(\sqrt{2} + 1 - \sqrt{3})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + 1 - \sqrt{2})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + \sqrt{2} - 1)}{2}\Bigg)}$$
$$\rightarrow \frac{4}{3}\sqrt{\frac{81(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3})(\sqrt{3} + 1 - \sqrt{2})(\sqrt{3} + \sqrt{2} - 1)}{16}}$$
I am almost to the answer (assuming I made no mistake), but I think this simplification is getting complicated. How do I proceed next?
Can anyone help me?
|
It is indeed a theorem that can be generalized, although Mathworlddoes n't say so explicitly.
The area of a triangle formed by medians ( computed for example from Brahmagupta/Heron formula ) is three-fourths the area formed by the corresponding sides of the given triangle.
It can also be proved by projective geometry.
Linear scale $k=\sin \frac{\pi}{3}$ can be established from relative proportions of the simplest equilateral triangle.
In our case hypothetical medians after scaling up full sides from centroid by $\text{50%} : 3(\sqrt 3, \sqrt 2,1)$ calculates to $\dfrac{9}{\sqrt 2};$
So the circumscribed triangle area would be:
$$ \dfrac{{\dfrac{9}{\sqrt 2}}} {\sin^2\dfrac{\pi}{3}} = 6 \sqrt 2.$$
|
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|
Prove by induction that $\frac{1}{2n}\leq\frac{1\text{·}3\cdot5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}$ What would be the right way to solve this by induction proof?
$$\frac{1}{2n}\leq\frac{1\text{·}3\cdot5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}$$
This is what I've done (reference https://www.slader.com/discussion/question/prove-that-12n-1-3-5-2n-12-4-2n-whenever-n-is-a-positive-integer/#):
*
*Show that $S\left(n+1\right)$ by induction proof. This is
$$\frac{1}{2(n+1)}\leq\frac{1\text{·}3\text{·}5\text{·}\ldots\text{·}(2n+1)}{2+4+6+\ldots+2(n+2)}$$
Multiplying both sides of the equation $\frac{2n\text{·}(2(n+1)-1)}{2n\text{·}(2(n+1)-1)}=\frac{2n\text{·}(2n+1)}{2n\text{·}(2n+1)}$
$$\frac{1}{2n}\leq\frac{1\text{·}3\text{·}5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}\times\frac{2n\text{·}(2(n+1)-1}{2n\text{·}(2(n+1)-1}$$
$$\frac{1}{2n}\times\frac{2n\text{·}(2n+1)}{2n\text{·}(2n+1)}$$
Rewriting we have the following
\begin{array}{c}
\frac{2n+1}{2n(2n+1)}\\
\frac{1}{2n+1}+\frac{1}{2n(2n+1)}
\end{array}
$$\frac{1}{2n}\leq\frac{1}{2n+1}+\frac{1}{2n(2n+1)}$$
|
$$\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2+4+6+\ldots+2n}\ge \frac{1}{2n}\tag{1} $$
can be simplified as
$$\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{n(n+1)}\ge \frac{1}{2n} $$
and then
$$1\cdot3\cdot5\cdot\ldots\cdot(2n-1)\ge\frac{n+1}{2}\tag{2}$$
for $n=1$ we have $1\ge 1$ true.
Now suppose $(2)$ is true and let us prove it for $(n+1)$.
$$[1\cdot3\cdot5\cdot\ldots\cdot(2n-1)](2n+1)\ge \frac{n+1}{2}\cdot(2n+1)\ge\frac{n+1+1}{2}=n+1$$
|
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|
Why using fewer terms of Taylor series doesn't give $0/0$ but gives a wrong answer? I was reading a Calculus book and I saw this problem which looks easy:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} = ?$$
It's a 0/0 limit and it's using some of the Taylor series of $\sin$ and $\cos$ expressions to solve the problem.
I know that the First and Second way should be correct because it's using more expressions of the Taylor series around 0.
What I can't figure out is WHY using fewer expressions of the Taylor series in the Third way doesn't give 0/0 but gives a wrong answer?
First way:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x-2 \sin x \cos x}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x(x-\sin x)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(x-x+\frac{x^3}{6}\right)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(\frac{x^3}{6}\right)}{x^3}=\frac{1}{3}$$
Second way:
$$\lim _{x \rightarrow 0} \frac{2x(1-\frac{x^2}{2})-(2x-\frac{8x^3}{6})}{x^3}=\lim _{x \rightarrow 0} \frac{2x-x^3-2x+\frac{8x^3}{6}}{x^3}=\lim _{x \rightarrow 0} \frac{\frac{x^3}{3}}{x^3}=\frac{1}{3}$$
Third way:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} =\lim _{x \rightarrow 0} \frac{2 x \cos x-2x}{x^3}=\lim _{x \rightarrow 0} \frac{2x(\cos x -1)}{x^3}=\lim _{x \rightarrow 0} \frac{2x(-\frac{x^2}{2})}{x^3}=-1$$
|
You are dividing by $x^3$ at the end, so you need all possible terms at least of degree $3$ in the numerator to be present, otherwise you're basically guaranteed to change the value of the limit.
Let's keep the error term in the third way, and see what happens. I will do that the following way: we have
$$
\sin(2x) = 2x + x^3\cdot g(x)\\
\cos(x) = 1-\frac{x^2}2 + x^4\cdot h(x)
$$
for some functions $g$ and $h$ where $g(x)$ and $h(x)$ are bounded as $x\to 0$. (It is more common to use $O(x^3)$ instead of $x^3\cdot g(x)$ and $O(x^4)$ rather than $x^4\cdot h(x)$. But the $O$ terms can be a bit unintuitive to arithmetize with, so if you are unaccustomed to working with error terms, I think that my approach here is closer to what you are already used to.)
Then we follow the steps in your third way and see what we get:
$$
\frac{2 x \cos x- \sin 2x}{x^3} =\frac{2 x \cos x-(2x + x^3\cdot g(x))}{x^3}\\
=\frac{2x(\cos x -1) - x^3\cdot g(x)}{x^3}\\
=\frac{2x(-\frac{x^2}{2} + x^4\cdot h(x)) + x^3\cdot g(x)}{x^3}\\
=-1 + x\cdot h(x) - g(x)
$$
and we see that in order to assess the limit as $x\to 0$, we don't need to know more about $h$ (it is bounded, so $x\cdot h(x)\to 0$), but we do need to know more about $g(x)$. Of course, it is easy to go back and check that $g(x) = -\frac8{3!} + x^2\cdot g_1(x)$ for some function $g_1$ that is bounded for $x\to 0$. Which is enough to conclude that the limit is indeed $\frac13$.
|
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|
Why is $\sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}$? I am trying to prove an inequality, but that itself is not the question here. Looking around for ideas how to proceed I found solutions for the same problem.
But in several occasions I couldn't follow the answers because of this step.
In both of these excerpts (taken from answered questions on this site, the links are below) two different square roots are combined like this.
\begin{align}
\sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}\tag{Now it's an inequality}
\end{align}
Why is this possible even though the square roots are not identical?
Here the excerpts:
\begin{align}
= & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{\sqrt{n + 2} + \sqrt{n + 1}}\right] \quad \text{multiply conjugate}\\
> & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{2\sqrt{n + 1}}\right] \\
\end{align}
taken from here
or
\begin{align}
\sqrt{k+1}-\sqrt{k}&=\frac{1}{\sqrt{k+1}+\sqrt{k}}\\
&\le\frac{1}{2\sqrt{k}}
\end{align}
taken from here
I hope someone can explain this to me or point me in the right direction
|
No, the linked examples use $\sqrt{n}+\sqrt{n+1}>2\sqrt{n}$, which is equivalent to the trivial $\sqrt{n+1}>\sqrt{n}$.
|
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|
Area of triangle inside a $30-60-90$ triangle
In a right angle triangle ABC with hypotenuse BC and C=60 degrees, M and N are the middles of AB and AC respectively. Draw ND perpendicular to BC (D is point on the side BC). If MD=7cm calculate the area of triangle DNM.
I solved this question in the following way:
$NM//=\frac{CB}{2}$ (Thales theorem)
ABC=30 degrees hence $CA=\frac{CB}{2}$ and hene $NM=CA$
$sinNCD=\frac{ND}{CN}$
$\frac{\sqrt{3}}{2}*CN=ND$
$CN=\frac{2}{\sqrt{3}}*ND$
$AC=\frac{4}{\sqrt{3}}*ND$
$MN=\frac{4}{\sqrt{3}}*ND$
since NM//CB then DNM=90 degrees.
Hence, $DN^2+DN^2*\frac{16}{9}=49$ (Pythagoras), so $DN^2*\frac{25}{9}=49$, so $DN*\frac{5}{3}=7$, so $DN=\frac{21}{5}$
From this we have that $MN=\frac{4}{\sqrt{3}}*\frac{21}{5}=\frac{84}{5\sqrt{3}}$.
Hence $Area=\frac{MN*ND}{2}=\frac{42}{5\sqrt{3}}*\frac{21}{5}=\frac{21*42}{25\sqrt{3}}=\frac{882}{25\sqrt{3}}$
Could you please tell me if I am correct and if I am wrong how much would it be reasonable to assume I would get in an olympiad for my work?
|
You have done a mistake here
$$ DN^2+DN^2*\frac{16}{\color{red}{3}}=49$$
You wrote $9$.
With correct value, I got $$ [DNM] = \dfrac{98\sqrt{3}}{19}$$
|
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|
How can I found 3rd and 4th central moments of gumbel distribution with characteristic function? I tried many ways(using digamma function, partial integration etc.) to find moments using by characteristic function but I couldn't. Is there any trick or suggestion? (I couldn't calculate integral of 2nd, 3nd and 4th derivatives of gamma function.)
Thanks in advance..
|
The characteristic function for the Gumbel distribution with location parameter $\alpha$ and scale parameter $\beta$ is
$$ \varphi_X(t) = \mathrm{e}^{\mathrm{i} t \alpha} \Gamma(1+\mathrm{i}t \beta) \text{.} $$
The $n^\text{th}$ (noncentral) moment can be found from the characteristic function using
$$ E[X^n] = \mathrm{i}^{-n} \left. \frac{\mathrm{d}^n}{\mathrm{d}t^n} \varphi_X(t) \right|_{t = 0} \text{.} $$
So, \begin{align*}
E[X^0] &= 1 \cdot \mathrm{e}^0 \Gamma(1) = 1 \\
E[X^1] &= -\mathrm{i} \cdot \left( \mathrm{e}^{\mathrm{i} \cdot 0 \cdot \alpha} \Gamma'(1 + 0)(\mathrm{i}\beta) + \mathrm{i}\alpha \mathrm{e}^{\mathrm{i}\cdot 0 \cdot \alpha} \Gamma(1+0)\right) \\
&= - \gamma \beta + \alpha \\
E[X^2] &= \cdots \\
&= \alpha ^2-2 \gamma \alpha \beta +\frac{1}{6} \left(6 \gamma ^2+\pi ^2\right) \beta ^2 \\
E[X^3] &= \cdots \\
&= \alpha ^3-3 \gamma \alpha ^2 \beta +\frac{1}{2} \left(6 \gamma ^2+\pi ^2\right) \alpha \beta ^2-\frac{1}{2} \beta ^3 \left(2 \gamma ^3+\gamma \pi ^2-2 \psi^{(2)}(1)\right) \\
E[X^4] &= \cdots \\
&= \alpha ^4-4 \gamma \alpha ^3 \beta +\left(6 \gamma ^2+\pi ^2\right) \alpha ^2 \beta ^2-2 \alpha \beta ^3 \left(2 \gamma ^3+\gamma \pi ^2-2 \psi^{(2)}(1)\right)+\frac{1}{20} \beta ^4 \left(20 \gamma ^4+20 \gamma^2 \pi^2 + 3 \pi^4 - 80 \gamma \psi^{(2)}(1)\right) \text{,}
\end{align*}
where
*
*$\gamma$ is the Euler-Mascheroni constant and
*$\psi^{(2)}$ is the polygamma function of order $2$.
Now we compute the central moments from the noncentral moments. The $n^\text{th}$ central moment, $\mu_n$ is
$$ \mu_n = \begin{cases}
1 ,& n = 0 \\
E[X^1] ,& n = 1 \\
\sum_{j=0}^n \binom{n}{j} (-1)^{n-j} E[X^j] E[X^1]^{n-j} ,& n \geq 2
\end{cases} \text{.} $$
So,
\begin{align*}
\mu_1 &= E[X^1] \\
&= -\gamma \beta + \alpha \\
\mu_2 &= \alpha ^2-2 \gamma \alpha \beta +(i \alpha -i \gamma \beta )^2+\frac{\pi^2 \beta ^2}{6}+\gamma ^2 \beta ^2 \\
&= \frac{\pi^2 \beta^2}{6} \\
\mu_3 &= 3 i \left(\alpha ^2-2 \gamma \alpha \beta +\frac{\pi ^2 \beta ^2}{6}+\gamma ^2 \beta ^2\right) (i \alpha -i \gamma \beta )+i \left(-i \alpha ^3+3 i \gamma \alpha ^2 \beta +3 i \alpha \left(-\frac{1}{6} \pi ^2 \beta ^2-\gamma ^2 \beta ^2\right)+\frac{1}{2} i \gamma \pi ^2 \beta ^3+i \gamma ^3 \beta ^3-i \beta ^3 \psi ^{(2)}(1)\right)+2 i (i \alpha -i \gamma \beta )^3 \\
&= \beta ^3 \psi ^{(2)}(1)
\end{align*}
and you should see how to compute $\mu_4$.
|
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|
Showing $\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2}$ for $0I was trying to solve the following exercise:
Show that for $0<x<1$ the following inequalities hold: $$\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2}$$
I think the Mean Value Theorem might be useful, but I don't know how to apply it.
|
$$
\frac{\pi}{4}-\frac{1-x}{1+x^2}<\arctan (x)<\frac{\pi}{4}-\frac{1-x}{2} \\
-\frac{1-x}{1+x^2}<\arctan (x) - \frac{\pi}{4}<-\frac{1-x}{2} \\
\frac{1-x}{1+x^2}>\frac{\pi}{4} - \arctan (x)>\frac{1-x}{2} \\
\frac{1}{1+x^2}>\frac{\frac{\pi}{4} - \arctan (x)}{1-x}>\frac{1}{2} \\
$$
where the last step assumes $x<1$.
could you write down Taylor series for the arc-tangent around $1$?
|
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|
How to solve this equation with matrices can you please give me some hints to solve the following? I really don't know how to start.
$$X^2= \begin{pmatrix}
6 & 2 \\ 3 & 7
\end{pmatrix}.$$
I tried to express this matrix as $4\cdot I + \begin{pmatrix}
2 & 2 \\ 3 & 3
\end{pmatrix}$ And somehow solve it, but I really have no clue. Please some help.
|
We have
$$
\begin{pmatrix}
6 & 2 \\ 3 & 7
\end{pmatrix} = QDQ^{-1}$$
where
$$
Q = \begin{pmatrix} -1 & 2 \\ 1 & 3 \end{pmatrix},\ D = \begin{pmatrix} 4& 0 \\ 0 & 9 \end{pmatrix}.
$$
It follows that we may take any of the four matrices
$$
X = Q\begin{pmatrix} \pm 2 & 0 \\ 0 & \pm 3 \end{pmatrix} Q^{-1}.
$$
|
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|
Is my $\epsilon$-$\delta$ calculation correct? I have to show that $\lim_{x \to 1} x^4-1 =0$. Here is how i have done it:
$\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < \epsilon \qquad$ and since we are close to 1, we can assume that the $\delta$-neighborhood of $c=1$ must be havea radius of max $\delta =1$ which implies that : $\mid x+1 \mid \le 2 \quad and \mid x^2+1 \mid \le 2 \quad \forall x \in V_{\delta}(c) \quad$
We now choose $\delta=min \left \{1,\frac{\epsilon}{4}\right\} \quad$ and we can conclude that if $\mid x-1 \mid < \delta$, it follows that $\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < 4\frac{\epsilon}{4} =\epsilon$. Is this calculation correct? Do I miss something? Or some details?
|
You can also express $x^4-1$ in powers of $x-1$ as in
$$|x^4-1| = |((x-1)+1)^4-1| = |(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)|.$$
If we assume $|x-1| < \delta$, the triangle inequality gives
$$|x^4-1| \le |x-1|^4+4|x-1|^3+6|x-1|^2+4|x-1| < \delta^4+4\delta^3+6\delta^2+4\delta.$$
Now we want $\delta^4+4\delta^3+6\delta^2+4\delta \le \varepsilon$. Since $$\delta^4+4\delta^3+6\delta^2+4\delta =\delta(\delta^3+4\delta^2+6\delta+4)$$
we only have to bound the second factor since the first is merely $\delta$. If we set $\delta = \min\left\{1, \frac\varepsilon{15}\right\}$, we have
$$|x^4-1| \le \delta(\delta^3+4\delta^2+6\delta+4) \le \delta \cdot (1+4+6+4) = 15\delta \le 15\cdot \frac\varepsilon{15} = \varepsilon.$$
|
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|
Perfect square involving the exponential law If $n$ is a natural number, and $2^{10} + 2^{13} + 2^n$ is a perfect square, what is the value of $n$?
I've attempted to factor out $2^{10}$ and got $2^{10}(1 + 2^3 + 2^{n-10})$. How can I move further?
|
$$(a+b)^2=a^2+2ab+b^2$$
$$2^{10}+2^{13}+2^n=(2^5)^2+2\times2^{12}+(2^7)^2=(2^5)^2+2\times2^{5+7}+(2^7)^2=(2^5+2^7)^2$$
Hence $n=14$
|
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|
Show that the sequence $X_n$ converges to a limit $Y$ Let $A$ be a positive definite $n\times n$ matrix. We use the iteration of the mapping $$f(X)=0.5(X^2+B), \ X\in \mathbb{R}^{n\times n}$$ where $B=I-A$.
Show that the sequence $X_0:=0$ (zero matrix), $X_1=f(X_0), X_2=f(f(X_0)), \ldots $ converges to a limit $Y$ if $\|B\|<1$ and show that $(I-Y)(I-Y)=A$.
We want to show that for every $\epsilon>0$ there is a natural number $N$ such that $|X_n-Y|<\epsilon$, right? How could we do that?
|
We have $$f(X) = \frac 12X^2 + \frac 12 B$$
Let's try to see what initial terms of $(X_n)$ look like (with $X_0=0$):
$$\begin{align}X_1 &= f(0) = \frac 12 0^2 + \frac 12 B = \frac 12 B\\[2mm]
X_2 &= f(f(0)) = \frac12 \bigg(\frac 12 B\bigg)^2 + \frac 12 B = \frac{1}{2^3}B^2 + \frac{1}{2}B \\[2mm]
X_3 &=f(f(f(0)))= \frac 12\bigg( \frac{1}{2^3}B^2 + \frac{1}{2}B \bigg)^2 + \frac 12 B = \frac1{2^7}B^4 + \frac{1}{2^5}B^3 + \frac {1}{2^3}B^2 + \frac{1}{2} B\end{align}$$
We now notice that (which can be proven by induction):
$$X_n = \frac{1}{2} B + \frac {1}{2^3}B^2 + \dots + \frac{1}{2^{2a_n - 1}}B^{a_n} = \sum_{k=1}^{a_n}\frac{1}{2^{2k - 1}}B^{k}$$
where $a_n = a_{n-1}^\text{th}~ \text{triangular number} + 1$, or explicitly
$$\quad a_{n+1} = \frac{a_n(a_n+1)}{2} + 1$$
for all $n \in \mathbb N$ and $a_0 = 1$.
Why exactly this number? This is because the expression $(c_1 + \dots + c_n)^2$ has $\frac{n(n+1)}{2}$ (which is $n^\text{th}$ triangular number) summands, and we add $1$ to this number since the function $f$ is adding one extra "constant" term.
Now, since $\|B\| < 1$, the following series
$$\lim_{n\to\infty}X_n = \sum_{n=0}^\infty \frac{1}{2^{2k - 1}}B^{k}$$
is convergent.
Finally, let the limit be $Y$. So, we have
$$X_{n+1} = f(X_n)$$
Now "limiting" both sides, we get
$$Y = f(Y) \\
Y = \frac 12 Y^2 + \frac 12 B \\
Y^2 - 2Y + B = 0$$
By the original substitution $B = I-A$:
$$Y^2 - 2Y + I = A \\
(Y - I)^2 = A \\
(I-Y)(I-Y) = A$$
as desired.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\sum_1^n 1/i^2 \le 2 - 1/n$ for all natural $n$. I'm trying to do this by induction. It works for $n=1$ because we get $1 \le 2 - 1 = 1$. Now suppose for some natural $k \ge 1$, we have $\sum_{i=1}^k \le 2 - 1/k$. I must show $\sum_{i=1}^{k+1} \le 2 - 1/(k+1)$. I started out from the hypothesis and added $1/(k+1)^2$ to both sides of the inequality. I get
\begin{align*}
1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{k^2} + \frac{1}{(k+1)^2} \le 2 - \frac{1}{k} + \frac{1}{(k+1)^2}
\end{align*}
and it's not obvious what to do from there. Any hints?
|
$$
{1\over(k+1)^2}<{1\over k(k+1)}={k+1-k\over k(k+1)}=\frac1k-{1\over k+1}
$$
As a result, we have
$$
\sum_{i=1}^{k+1}{1\over i^2}\le2-\frac1k+\frac1k-{1\over k+1}=2-{1\over k+1}
$$
Hence, by the principle of mathematical induction, we conclude that the proposition holds for all positive integer $n$.
|
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|
Given $T_1 = 0$ and for $i > 1$: $T_i = 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j$ Prove via induction that $T_i = \sum\limits_{j=1}^{i-1} \frac{1}{j}$
Given $T_1 = 0$ and for $i \in \mathbb{N}, i > 1$:
\begin{align*}
T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j \\
\end{align*}
Manually computing terms $2,3,4$:
\begin{align*}
T_2 &= 1 \\
T_3 &= 1 + \frac{1}{2} \\
T_4 &= 1 + \frac{1}{3} \left(1 + 1 + \frac{1}{2} \right) = 1 + \frac{1}{2} + \frac{1}{3} \\
\end{align*}
Prove by induction that this is:
\begin{align*}
T_i &= \sum\limits_{j=1}^{i-1} \frac{1}{j} \\
\end{align*}
The given equation is clearly satisfied for terms $1,2,3,4$. We need to show the inductive step. Assuming it's true for all $j < i$, show that it is also true for $i$.
\begin{align*}
T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} \sum\limits_{k=1}^{j-1} \frac{1}{k} \\
T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-2} \frac{i-1-j}{j} \\
\end{align*}
|
$$\begin{align*}
T_{i+1}&=1+\frac1i\sum_{j=1}^iT_j\\
&=1+\frac1i\sum_{j=1}^i\sum_{k=1}^{j-1}\frac1k\\
&=1+\frac1i\sum_{k=1}^{i-1}\sum_{j=k+1}^i\frac1k\\
&=1+\frac1i\sum_{k=1}^{i-1}\frac1k\sum_{j=k+1}^i1\\
&=1+\sum_{k=1}^{i-1}\frac{i-k}{ik}\\
&=1+\sum_{k=1}^{i-1}\left(\frac1k-\frac1i\right)\\
&=1+\sum_{k=1}^{i-1}\frac1k-\frac{i-1}i\\
&=1+\sum_{k=1}^{i-1}\frac1k-1+\frac1i\\
&=\sum_{k=1}^i\frac1k
\end{align*}$$
|
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|
Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$ The question is:
Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$
I did the question without using the Hint, but I don't know how to do it using the hint.
Quick working out of what I've done:
\begin{aligned}
\text { If } \theta &=\tan ^{-1} 2 \\
\tan \theta &=2 \\
0 & < \theta < \frac{\pi}{2}
\end{aligned}
\begin{aligned}
\cos 2 \theta &=2 \cos ^{2} \theta-1 \\
&=2 \times\left(\frac{2}{\sqrt{5}}\right)^{2}-1 \\
&=\frac{3}{5} \\
2 \theta =& \cos ^{-1} \frac{3}{5}, \quad \text { since } 0 < 2\theta < \pi
\end{aligned}
\begin{array}{l}
2 \tan ^{-1} 2=\cos ^{-1} \frac{3}{5} \text { . } \\
\text { Note: } \cos ^{-1} x \text { has point symmetry } \\
\text { in }\left(0, \frac{\pi}{2}\right) \text { . }
\end{array}
$$
\begin{array}{l}
\cos ^{-1} x+\cos ^{-1}(-x)=\pi \\
\cos ^{-1} \frac{3}{5}=\pi-\cos ^{-1}\left(-\frac{3}{5}\right) \\
\therefore \quad 2 \tan ^{-1} 2=\pi-\cos ^{-1}\left(-\frac{3}{5}\right)
\end{array}
$$
But I didn't use the Hint given in the question for this working out. How do I use the hint? Thank you !
|
Note
\begin{align}
& \tan(2\tan^{-1}2)-\tan(\pi-\cos^{-1}\frac35)\\
=&\frac{2\tan(\tan^{-1}2)}{1-\tan^2(\tan^{-1}2)}+\tan(\cos^{-1}\frac35)
=\frac{2\cdot2}{1-2^2}+ \frac43=0
\end{align}
which leads to
$$2\tan^{-1}2=\pi-\cos^{-1}\frac35$$
|
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|
Find all $z$ such that $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$. Find all $z$ such that $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$.
After much trial and error, I was able to rearrange the above equation
as $80z^2 - 112z-4 = 0$.
$z = \frac{7\pm3\sqrt{6}}{10}$.
I substituted $\frac{7+3\sqrt{6}}{10}$ as $z$ in the $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$. Things worked out.
Am I still not done? Must I check for extraneous solutions in this situation?
I wonder if it might not require it because if ever we know one of the roots of a quadratic with rational coefficients is irrational, we know the other.
I suppose I'm also trying to avoid going through another round of substitution and test with these irrational conjugates. I'm not very trusting of my algebra yet. So I end up rewriting the same equations time and time again.
|
$$\sqrt{5 z+5}=2 \sqrt{z}+\sqrt{3-3 z}\tag{1}$$
Square both sides
$$5z+5=4z+3-3z+4\sqrt{3z-3z^2}$$
$$2+4z=4\sqrt{3z-3z^2}\to 1+2z=2\sqrt{3z-3z^2}$$
Square again both sides
$$1+4z+4z^2=12z-12z^2$$
$$16 z^2-8 z+1=0\to (4z-1)^2=0\to z=\frac14$$
Which is actually a solution of $(1)$ (verify!)
|
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|
How to find the local maximum and minimum of $f(x)=x+\sin(2x)$ bounded by the interval $-\frac{2\pi}{3}$ and $\frac{2\pi}{3}$?
Find the local maximum and minimum of $y=x+\sin(2x)$ between
$-\frac{2\pi}{3}$ and $\frac{2\pi}{3}$.
What I have done so far:
$f'(x) = 1+2\cos(2x)$ and $f''(x)=-4\sin(2x)$.
|
First, find where the derivative is zero ($y'=1+2\cos(2x)=0$).
$$1+2\cos(2x)=0$$
$$2\cos(2x)=-1$$
$$\cos(2x)=-\frac{1}{2}$$
$$2x=-\frac{2\pi}{3}, \frac{2\pi}{3}$$
(within the bounds)
$$x=-\frac{\pi}{3}, \frac{\pi}{3}$$
Checking the second derivative, you can find that $x=-\frac{\pi}{3}$ is local minimum and $x=\frac{\pi}{3}$ is local maximum.
|
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|
Writing Fourier transform integral in terms of real-valued functions I am interested in the following integral.
$$ f(x,t) = \int \limits_{-\infty}^{\infty} \mathrm{d}k e^{-k^2}\left( e^{i(kx-pt-qk^2t)}+e^{i(kx+pt+qk^2t)} \right)$$
One way (for example, by using Mathematica) to solve it is in the following way.
$$ f(x,t) = \frac{\sqrt{\pi}}{2} \frac{(e^{-ipt}e^{\frac{ix^2}{-4i+4qt}}\sqrt{1-iqt}+e^{ipt}e^{\frac{-ix^2}{4i+4qt}}\sqrt{1+iqt})}{\sqrt{1+q^2t^2}} $$
I believe that $f(x,t)$ should be real valued, but it is not clear to me how to simplify given expression so that it consists only of real valued functions.
|
The two parts of your answer are obviously a complex conjugate pair.
Let's just take one of the terms and double its real part:
$$\begin{align*}f(x,t) &= \frac{\sqrt{\pi}}{2} \frac{(e^{-ipt}e^{\frac{ix^2}{-4i+4qt}}\sqrt{1-iqt}+e^{ipt}e^{\frac{-ix^2}{4i+4qt}}\sqrt{1+iqt})}{\sqrt{1+q^2t^2}}\\
\\
&= \sqrt{\pi}\;\Re\left[\frac{e^{-ipt}e^{-\frac{x^2}{4}\frac{1}{1+iqt}}}{\sqrt{1+iqt}}\right]\\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}\;\Re\left[\sqrt{1-iqt}\cdot e^{-ipt}e^{-\frac{x^2}{4}\frac{1-iqt}{1+q^2t^2}}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[\sqrt{1-iqt}\cdot e^{-ipt}e^{-\frac{x^2}{4}\frac{-iqt}{1+q^2t^2}}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[\sqrt{re^{i(\theta +2n\pi)}}\cdot e^{i\left(\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right)}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[{\sqrt[4]{1+q^2t^2}e^{i(\theta/2 +n\pi)}}\cdot e^{i\left(\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right)}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt[4]{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\Re\left[e^{i\left(\frac{\theta}{2}+n\pi+\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right)}\right] \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt[4]{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\cos\left(\frac{\theta}{2}+n\pi+\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right) \\
\\
&= \dfrac{\sqrt{\pi}}{\sqrt[4]{1+q^2t^2}}e^{-\frac{x^2}{4}\frac{1}{1+q^2t^2}}\;\cos\left(\frac{\mathrm{arctan2}(-qt,1)}{2}+n\pi+\frac{x^2}{4}\frac{qt}{1+q^2t^2}-pt\right) \\
\\
\end{align*}$$
You can take $n=0$ for the principal value of the square root. Or you can take $n=1$ and effectively just change the sign of the answer.
|
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|
Given $f(x) = \frac{x^3}{3} - \sqrt{x^2+1}$, proof that $f(\alpha) = \frac{\alpha^4 - 3}{3\alpha}$ I evaluated the function on $\alpha$ and removed the radical by multiplying by $\sqrt{\alpha^2+1}$ as follows:
$$ \frac{\alpha}{3} - \frac{\alpha^2 + 1}{\sqrt{\alpha^2+1}} $$
After some algebraic manipulations, I got to this step:
$$ \frac{\alpha^4 \sqrt{1 + 1/\alpha^2}}{3\alpha\sqrt{1 + 1/\alpha^2}} - \alpha(1 + 1/\alpha^2)^{-1/2} $$
Does anyone knows if I am on the good track? or is there any hints that could help me? Thanks
|
$$\frac{x^3}{3}-\sqrt{x^2+1}=\frac{x^4-3}{3 x}$$
$$\frac{x^3}{3}-\frac{x^4-3}{3 x}=\sqrt{x^2+1}$$
$$\frac{1}{x^2}=x^2+1$$
Substitute $x^2=w$
$$\frac{1}{w}=w+1\to w^2+w-1=0\to w=\frac{1}{2}\left(\sqrt{5}-1\right)$$
the root $w_2=\frac{1}{2}\left(-\sqrt{5}-1\right)$ is discarded because is negative. The solution is:
$$x=\sqrt{\frac{1}{2}\left(\sqrt{5}-1\right)}$$
|
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|
Evaluating $\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}$ Since we have $\frac00$,I Applied L'Hopital rule :
$$\lim_{x\to 0} (1+x)^{\tfrac1x}\times\left(\cfrac{-\ln(1+x)}{x^2}+\cfrac{1}{x(x+1)}\right)$$$$=\lim_{x\to 0}\cfrac{x^2(x+1)(1+x)^{\tfrac1x}-(x+1)\ln(1+x)+x}{x^2(x+1)}$$
But as you can see it is getting very ugly.
|
Without series, only L'Hospital
$$
\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}
$$
we get
$$
\lim_{x\to 0}\left[\left(-\frac{\ln(1+x)}{x^2}+\frac{1}{x}(1+x)^{-1}\right)(1+x)^{\frac{1}{x}}\right] \\
=\lim_{x\to 0}\left(-\frac{\ln(1+x)}{x^2}+\frac{\frac{1}{1+x}}{x}\right)\lim_{x\to 0}(1+x)^{\frac{1}{x}} \\
=\lim_{x\to 0}\left(\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}\right)e
$$
another L'Hospital
$$
= \lim_{x\to 0}\left(\frac{\frac{1}{1+x}+\frac{-x}{(1+x)^2}-\frac{1}{1+x}}{2x}\right)e
= \lim_{x\to 0}\frac{-1}{2(1+x)^2}e = -\frac{e}{2}\\
$$
|
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|
An estimate on number of solutions $e^z=a$ in $|z|\leq r.$ Let $n(r,a)$ be the number of solutions of $e^z=a$ in $|z|\leq r, $ where $a\in\mathbb{C}\setminus\left\{0\right\}.$
I have to estimate this number.
Attempt: Writing $a=|a|e^{i\theta}.$ Then $e^{z}=a$ gives $$z=\log|a| + i\left(\theta + 2k\pi\right),$$ where $k\in\mathbb{Z}.$
For these solutions to lie inside $|z|\leq r,$ we must have
$$|\log|a| + i\left(\theta + 2k\pi\right)|\leq r$$ which means that $$(\log|a|)^2 + (\theta + 2k\pi)^2\leq r^2$$
This finally gives $$k\leq\frac{\sqrt{r^2-(\log|a|)^2}}{2\pi}-\theta/2\pi.$$
It has been proved that $$n(r,a)=\frac{\sqrt{r^2-(\log|a|)^2}}{\pi} + O(1).$$ However, I am unable to understand how to arrive at this estimate.
|
Your calculation is correct up to
$$
(\log|a|)^2 + (\theta + 2k\pi)^2\leq r^2 \, .
$$
That is equivalent to
$$
|\theta + 2k\pi| \le \sqrt{r^2 - (\log|a|)^2} \\
\iff - \sqrt{r^2 - (\log|a|)^2} \le \theta + 2k\pi \le \sqrt{r^2 - (\log|a|)^2} \\
\iff - \frac{\sqrt{r^2 - (\log|a|)^2}}{2 \pi} - \frac{\theta}{2 \pi}
\le k \le \frac{\sqrt{r^2 - (\log|a|)^2}}{2 \pi} - \frac{\theta}{2 \pi} \, .
$$
So $k$ lies in an interval of length $\frac{\sqrt{r^2 - (\log|a|)^2}}{\pi}$. It follows that
$$
\frac{\sqrt{r^2 - (\log|a|)^2}}{\pi} \le n(r, a) \le \frac{\sqrt{r^2 - (\log|a|)^2}}{\pi} + 1
$$
because a closed interval of length $L$ contains between $L$ and $L+1$ integral points.
|
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|
Inverse function of a polynomial $x(x-1)(x+1) = 0 $ How to get the interval to get the inverse function of the polynomial $$x(x-1)(x+1) = f(x) $$, to show that the function is surjective.
The interval where the function is surjective is $x > 1$ and $x < -1$, how to get inverse on this interval ?
|
Consider $$x(x-1)(x+1) =x^3-x= y$$ and then th cubic equation
$$x^3-x-y=0$$ for which the discriminant $\Delta=4-27y^2$ must be negative in order to have only one real root. This defines
$$-\frac{2}{3 \sqrt{3}} <y <\frac{2}{3 \sqrt{3}}$$
Using the hyperboic method for only one real root, we then have
$$x=\frac{2}{\sqrt{3} }\frac{ |y|}y \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{3 \sqrt{3}
}{2}|y|\right)\right)$$
|
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|
Prove the following sequence is convergent: $a(1) = 1$ and $a(n+1) = 1/2 \cdot (a(n) + 3/a(n))$ I know the limit is $\sqrt 3$. And I tried proving that the sequence is decreasing from $a(2)$, yet when I got to this point, I have been stuck:
$a(n+1) - a(n) = (3 - a(n)^2) / 2a(n)$ ?
Thank you very much!
|
Observe that $a_n > 0, \forall n \ge 1$, and by using the AM-GM inequality: $\forall n \ge 1 \implies (a_{n+1})^2 \ge \left(\dfrac{1}{2}\cdot 2\sqrt{a_n\cdot \dfrac{3}{a_n}}\right)^2 = \left(\sqrt{3}\right)^2 = 3 \implies a_{n+1} \ge \sqrt{3}, \forall n \ge 1$. Thus: $a_{n+1} - a_n = \dfrac{1}{2}\left(a_n+\dfrac{3}{a_n}\right)-a_n= \dfrac{1}{2}\left(\dfrac{3}{a_n}-a_n\right)= \dfrac{3-a_n^2}{2a_n}< 0\implies a_{n+1} < a_n$. The sequence is decreasing and bounded below by $\sqrt{3}$ hence converges to $L$ that satisfies the equation: $L = \dfrac{1}{2}\left(L+\dfrac{3}{L}\right)\implies L^2 = 3\implies L = \sqrt{3}$ since $L \ge \sqrt{3}$.
|
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|
Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$ I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.
The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$
So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so that'd be a go-to. Also, since the degree of th enumerator is one less than the denominator we can maybe treat this as $du/u$ which would imply a natural log, though we don't quite know that yet.
So we decompose: I know that $x^2+3x-4$ can be factored as $(x-1)(x+4)$.
That gets me
$$\int{\frac{x}{(x-1)(x+4)}dx}$$
so I can do this:
$$\frac{Ax}{x-1}+\frac{B}{x+4} = \frac{x}{(x-1)(x+4)}$$
Which implies
$$A(x+4)+B(x-1) = x$$
and since the roots are at $x=1$ and $x=-4$, I can set it up like this:
$$5A = 1;-5B=1 \text{ and } A = \frac{1}{5}, B=-\frac{1}{5}$$
Leading to:
$$\frac{x}{5(x-1)}-\frac{1}{5(x+4)}$$
Which I can set up the integral like so:
$$\int{\frac{x}{5(x-1)}-\frac{1}{5(x+4)}dx}=\int{\frac{x}{5(x-1)}dx-\int{\frac{1}{5(x+4)}dx}}$$
I can now integrate by addition here $$\int{\frac{x}{x-1}}dx=\int{\frac{x-1+1}{x-1}}dx=x+\int{\frac{1}{x-1}}dx=x-\ln(x-1)$$
And doing the same thing for the second term and bringing back my $\frac{1}{5}$
$$\frac{1}{5}(x-\ln(x-1)+\ln(x+4))$$
I suspect there is a further simplification I could do. On a problem like this I also saw it integrated as an arctangent, but that seemed needlessly complex? In any case I was curious if I did this correctly.
|
HINT
Here it is an alternative way to solve it:
\begin{align*}
\frac{x}{(x-1)(x+4)} & = \frac{(x-1) + 1}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{5}\times\frac{(x+4) - (x-1)}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{5}\times\left[\frac{1}{x-1} - \frac{1}{x+4}\right]\\\\
& = \frac{1}{5(x-1)} + \frac{4}{5(x+4)}
\end{align*}
Can you take it from here?
|
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|
Show that for natural $n \ge 2$ :$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$ Show that for natural $n \ge 2$ the following does hold:
$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$
First solution:
$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$
$$\iff$$
$$\left(1-\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)^{n}>1-\frac{1}{n}$$
$$\left(1-\frac{1}{n}\right)^{n-1}\left(1+\frac{1}{n}\right)^{n}>1$$
By Bernoulli inequality:
$$\left(1-\frac{1}{n}\right)^{n-1}\left(1+\frac{1}{n}\right)^{n}\ge2\left(1-\frac{n-1}{n}\right)=\frac{2}{n}$$
Which is not useful.
Second solution:
$$\left(1-\frac{1}{n^{2}}\right)^{n}=\left(1-\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)^{n}\ge0\cdot2=0$$
Which is not useful.
Third solution:
$$\left(1-\frac{1}{n^{2}}\right)^{n}=\left(1-\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)^{n}\ge2\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{n-1}=\left(1+\left(-\frac{1}{2}\right)\right)^{n-1}$$$$\ge1-\frac{n-1}{2}=\frac{3-n}{2}$$
Final solution:
$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$
$$\iff$$
$$\left(1-\frac{1}{n^{2}}\right)^{n-1}\left(1+\frac{1}{n}\right)>1$$
By Bernoulli inequality:
$$\left(1-\frac{1}{n^{2}}\right)^{n-1}\left(1+\frac{1}{n}\right)>1-\frac{n-1}{n^{2}}=1-\frac{1}{n}+\frac{1}{n^{2}}\ge\frac{1}{2}+\frac{1}{n^{2}}$$
Which is true to claim that $$\frac{1}{2}+\frac{1}{n^{2}}>1-\frac{1}{n}$$
Since for natural $n \ge 2$ we have that $$\frac{1}{n}\left(\frac{1}{n}+1\right)>\frac{1}{2}$$
|
Show that for natural $n \ge 2$ the following does hold:
$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$
By using the Bernoulli’s inequality
$(1+x)^{n}>1+nx$
for any natural $\;n\ge2\;$ and for any $\;x>-1\;,$
with $\;x=-\dfrac1{n^2}>-1\;,\;$
we get that
$\left(1-\dfrac{1}{n^{2}}\right)^{n}>1+n\left(-\dfrac{1}{n^2}\right)=1-\dfrac1n\;,$
for any natural $\;n\ge2\;.$
|
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|
Find the best constant $c < 0$ for $c a_n = a_{n-1} + 2 a_{n-2} + 3 a_{n-3} + ... + n a_0$ Let's define the following recurrence relation:
$$
\begin{align}
a_0 &= 1 \\
c a_n &= a_{n-1} + 2 a_{n-2} + 3 a_{n-3} + ... + n a_0
\end{align}
$$
Find the best constant $c < 0, \forall n \in \mathbb{N}$ which is farest from zero such that
$$
\begin{align}
a_n &> 0, \quad \text{if n is even} \\
a_n &< 0, \quad \text{if n is odd} \\
\end{align}
$$
My guessed answer would be $-\frac{1}{4}$. But how to prove it?
My attempt so far is to reduce the recurrence relation:
$$
\begin{align}
a_n &= \sum\limits_{k=1}^n \frac{k a_{n-k}}{c} = \frac{a_{n-1}}{c} + \sum\limits_{k=2}^n \frac{(k-1) a_{n-k}}{c} + \sum\limits_{k=2}^n \frac{a_{n-k}}{c} \\
&= \frac{c + 1}{c} a_{n-1} + \sum\limits_{k=2}^n \frac{a_{n-k}}{c} \\
&= \frac{c + 1}{c} a_{n-1} + \frac{a_{n-2}}{c} + \sum\limits_{k=3}^n \frac{a_{n-k}}{c} \\
&= \frac{c + 1}{c} a_{n-1} + \frac{a_{n-2}}{c} + a_{n-1} - \frac{c + 1}{c} a_{n-2} \\
&= \frac{2c + 1}{c} a_{n-1} - a_{n-2} \\
\end{align}
$$
If I set my guess $c = -\frac{1}{4}$ in this reduced recurrence relation then:
$$
\begin{align}
a_n = -2 a_{n-1} - a_{n-2}
\end{align}
$$
How to proceed? Or is there any other way to prove it?
|
From the recurrent relation $a_n = \frac{2c + 1}{c} a_{n-1} - a_{n-2}$ with $a_0 = 1$ and $a_1 = 1/c$, you can obtain an explicit formula for $a_n$:
$$a_n = p_1x_1^n+p_2x_2^n$$
with $x_1, x_2$ are the roots (can be non real) of the equation : $x^2 = \frac{2c + 1}{c} x - 1$ and $p_1,p_2$ are the solution of the system of 2 linear equations:
$$p_1 + p_2 = a_0 = 1$$
$$p_1 x_1 + p_2 x_2 = a_1 = 1/c$$
After obtaining the analytic formula of $a_n$, the problem becomes easy. You just need to study the sign of $a_n$ in case $n$ is odd or even.
|
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|
Apparently in this form the limit is "trivially" 1/2 I'll admit, I'm struggling to keep up with the material in my analysis classes, but it only demoralizes me more when the questions in my textbooks have explained solutions, but I don't even understand how you go from one step to another.
This is the final from, that's supposed to be close to the simplest form (so close that it's "trivial" for the student to find it):
$\lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+\left(n-1\right)^{c+1}}{\left(c+1\right)\left(n^c-\left(n-1\right)^c\right)}\right)=\frac{1}{2}$
Why is this true? I don't see it at all.
|
Compute the leading terms of binomial expansions in numerator and denominator and simplify
$$
\begin{align}
& \lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+\left(n-1\right)^{c+1}}{\left(c+1\right)\left(n^c-\left(n-1\right)^c\right)}\right)= \\
& = \lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+n^{c+1}-\binom{c+1}1 n^c + \binom{c+1}2 n^{c-1} + O(n^{c-2})}{\left(c+1\right)\left(n^c-n^c + \binom{c}1 n^{c-1}+O(n^{c-2})\right)}\right)\\
& = \lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+n^{c+1}-(c+1) n^c + \frac{c(c+1)}{2} n^{c-1} + O(n^{c-2})}{\left(c+1\right)\left(n^c-n^c + c n^{c-1}+O(n^{c-2})\right)}\right)\\
& = \lim _{n\to \infty }\left(\frac{\frac{c(c+1)}{2} n^{c-1} + O(n^{c-2})}{c\left(c+1\right)n^{c-1}+O(n^{c-2})}\right)\\
& = \frac{c(c+1)}{2}\frac{1}{c(c+1)} \\
& = \frac{1}{2}.
\end{align}
$$
|
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|
$f^n(x) = \frac{5-10n}{2}*(7-5x)^{-1} * f^{n-1}(x)$. Prove valid for all $n \in \mathbb{N}$ $f^n$(x)$ = \frac{5-10n}{2}*(7-5x)^{-1} * f^{n-1}(x)$. Prove valid for all $n \in \mathbb{N}$
Where $f^n(x)$ is the $n^{th}$ derivative of $f(x)$. And $f(x) = (7-5x)^{\frac{1}{2}}$.
So far I have made my base case of $n=1$ and showed that it equals the derivative of $f(x)$ and then I assumed n=k. I then state that: $f^{k+1} = \frac{d}{dx}(f^k)$. I know that my end goal is to get: $f^{k+1} = \frac{5-10(k+1)}{2}*(7-5x)^{-1} * f^{(k+1)-1}(x)$ which equals $f^{k+1} = \frac{-5-10k}{2}*(7-5x)^{-1} * f^{k}(x)$.
I've tried plugging the equation for $f^{k+1}$ into $f^{k+1} = \frac{d}{dx}(f^k)$ but couldn't seem to get the result I wanted, as I ended up with:
$-25*(7-5x)^{-2}*f^k(x)$
Help would be much appreciated.
|
The constant term in $f^n(x)$ should be $\frac{10n-15}{2}$, not $\frac{5-10n}{2}$.
It's best to rewrite the equation as
$$f^n(x) = a_n \frac{f^{n-1}(x)}{(f(x))^2}$$
This will greatly simplify our calculation.
Induction step from $n$ to $n+1$: First notice that
$$f^{n-1} (x) = \frac{1}{a_n} (f(x))^2 f^n(x)\tag1$$
and
$$f'(x) = a_1 \frac{f(x)}{(f(x))^2} \implies f(x)f'(x)= a_1\tag2$$
with $a_1=-\frac{5}{2}$.
Then
$$f^{n+1}(x) = \frac{d}{dx} f^n(x) = \frac{d}{dx} \left(a_n \cdot \frac{f^{n-1}(x)}{(f(x))^2}\right)\\
=a_n \cdot \frac{f^n(x) (f(x))^2 - f^{n-1}(x) \cdot 2f(x)f'(x)}{(f(x))^4}\\
=a_n \cdot \frac{f^n(x) (f(x))^2 - \frac{1}{a_n} (f(x))^2 f^n(x) \cdot (2a_1)}{(f(x))^4}\\
= a_n \left(1-\frac{2a_1}{a_n}\right) \frac{f^n(x)}{(f(x))^2}\\
= (a_n-2a_1) \frac{f^n(x)}{(f(x))^2}
$$
Then we have $a_{n+1}=a_n-2a_1$, an arithmetic sequence. Therefore $a_n=(n-1)(-2a_1)+a_1 = \frac{10n-15}{2}.\blacksquare$
|
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|
$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$ I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$
My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ and $c$, then adding these 3, we get, $a ^ 2 + b ^ 2 + c ^ 2 + 6 \ge 2\sqrt 2(a + b+ c)$, but then we get to $2\sqrt2 > 3$, which is false.
Edited: I found some variants of the original problem.
Problem 1: Let $a, b, c > 0$. Prove that $a^2 + b^2 + c^2 + 6 + (abc - 1) \ge 3(a+b+c)$.
Problem 2: Let $a, b, c$ be reals with $abc \le 1$. Prove that $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.
|
By using Schur's inequality and the identity
\begin{align}
&(a+b+c)^3 + 9abc - 4(a+b+c)(ab + bc + ca)\\
=\ & a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b),
\end{align}
we have $(a+b+c)^3 + 9abc - 4(a+b+c)(ab + bc + ca) \ge 0$ which results in
$$\frac{(a+b+c)^3 + 9abc }{4(a+b+c)} \ge ab + bc + ca. \tag{1}$$
We need to prove that $(a+b+c)^2 - 2(ab + bc + ca) + 6 \ge 3(a+b+c)$.
By using (1), it suffices to prove that
$$(a+b+c)^2 - 2\cdot \frac{(a+b+c)^3 + 9abc }{4(a+b+c)} + 6 \ge 3(a+b+c)$$
that is (using $abc = 1$)
$$\frac{(a+b+c - 3)[(a+b+c)^2 - 3(a+b+c) + 3]}{2(a+b+c)} \ge 0$$
which is true (using $a+b + c \ge 3\sqrt[3]{abc} = 3$).
We are done.
Remark: Actually, it is the pqr method.
|
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|
Need help to check alternating series criterion I know this is simple calculation based question but I got stuck:
I want to check convergence of the following series:
$$1-\frac{1}{2}(1+\frac{1}{3})+\frac{1}{3}(1+\frac{1}{3}+\frac{1}{5})-\cdots$$
Clearly this is alternating series. So I need satisfy alternating series criterion. That means I need to show that its terms are decreasing monotonically.
Let $a_n=\frac{1}{n}(1+ \frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1})$ and $a_{n+1}=\frac{1}{n+1}(1+ \frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1})$.
I just have to show $a_n-a_{n+1}>0$ but right here I got stuck because of the presence of the first factor. I think i am missing a simple trick here.
Anyway, I got $$a_n-a_{n+1}=(\frac{1}{n}-\frac{1}{n+1})+\frac{1}{3}(\frac{1}{n}-\frac{1}{n+1})+\cdots+\frac{1}{2n-1}(\frac{1}{n}-\frac{1}{n+1}) {-\color{red}{\frac{1}{(2n+1)(n+1)}}}.$$
The last red color term making problem.
May be ratio test help as well ?
Any help please
|
Doesn't answer your question, since I think you've already got the answer to it. It ads something to it though.
Let's evaluate the sum of the series $ \sum\limits_{n\geq 0}{\frac{\left(-1\right)^{n}}{n+1}\sum\limits_{k=0}^{n}{\frac{1}{2k+1}}} $.
Denoting $\left(\forall n\in\mathbb{N}\right),\ a_{n}=\frac{\left(-1\right)^{n}}{n+1}\sum\limits_{k=0}^{n}{\frac{1}{2k+1}} $, and $ b_{n}=\frac{\left(-1\right)^{n}}{2n+1} $.
Let $ n\in\mathbb{N} $, we have the following : $$ \sum_{k=0}^{n}{\frac{1}{2k+1}}=\frac{1}{2}\left(\sum_{k=0}^{n}{\frac{1}{2k+1}}+\sum_{k=0}^{n}{\frac{1}{2\left(n-k\right)+1}}\right) $$
We added the sum to itsefl, but by shifting the terms, then we divided by $ 2 $.
Thus : $$ \sum_{k=0}^{n}{\frac{1}{2k+1}}=\left(n+1\right)\sum_{k=0}^{n}{\frac{1}{\left(2k+1\right)\left(2\left(n-k\right)+1\right)}} $$
Hence : $$ a_{n}=\sum_{k=0}^{n}{\left(\frac{\left(-1\right)^{k}}{2k+1}\times\frac{\left(-1\right)^{n-k}}{2\left(n-k\right)+1}\right)} $$
Now, using a Cauchy product, we have : \begin{aligned}\sum_{n=0}^{+\infty}{\frac{\left(-1\right)^{n}}{n+1}\sum_{k=0}^{n}{\frac{1}{2k+1}}}&=\sum_{n=0}^{+\infty}{a_{n}}\\&=\sum_{n=0}^{+\infty}{\sum_{k=0}^{n}{b_{k}b_{n-k}}}\\ &=\left(\sum_{n=0}^{+\infty}{b_{n}}\right)^{2}\\ &=\left(\sum_{k=0}^{n}{\frac{\left(-1\right)^{n}}{2n+1}}\right)^{2}\\ \sum_{n=0}^{+\infty}{\frac{\left(-1\right)^{n}}{n+1}\sum_{k=0}^{n}{\frac{1}{2k+1}}}&=\frac{\pi^{2}}{4}\end{aligned}
|
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|
Find the minimum value of $7x-24y$ LMNAS $25^{th}$ UGM, Indonesian
Suppose that $x,y\in\mathbb{R}$, so that :
$x^2+y^2+Ax+By+C=0$
with $A,B,C>2014$. Find the minimum value of $7x-24y$
$x^2+y^2+Ax+By+C=0$
can be written $\rightarrow$ $(x+\frac{A}{2})^2+(y+\frac{B}{2})^2+C-(\frac{A}{2})^2-(\frac{B}{2})^2=0$
Stuck,:>
|
Since
$$
x^2+Ax+\left( \frac{A}{2} \right) ^2+y^2+By+\left( \frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C\\
\left( x+\frac{A}{2} \right) ^2+\left( y+\frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C
$$
Let $\displaystyle r=\sqrt{\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C}$, we can set $x=r\cos \theta -\dfrac{A}{2},y=r\sin \theta -\dfrac{B}{2}$, then
$$
\begin{aligned}
7x-24y&=\frac{1}{2} \left(\sqrt{A^2+B^2-4 C} (7 \cos \theta-24 \sin \theta)-7 A+24 B\right)\\
&=\frac{1}{2} \left(\sqrt{A^2+B^2-4 C} \sqrt{7^2+24^2}\sin(\theta+\phi)-7 A+24 B\right)\\
&\geq \frac{1}{2} \left(25\sqrt{A^2+B^2-4 C} -7 A+24 B\right)
\end{aligned}
$$
The second step uses The auxiliary Angle formula.
|
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|
Proof that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}$ Kazakstan 2012
Suppose that $a, b \in\mathbb{R}$, and $a,b>0$. If $\frac{1}{a}+\frac{1}{b}=2$
prove that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}.$
My idea :
$a+b+\frac{1}{1+\sqrt{ab}}$ can be written as $ab(\frac{a+b}{ab}+\frac{1-\sqrt{ab}}{1-ab})=2ab+\frac{1-\sqrt{ab}}{ab-a^2b^2}$, but where can I use the fact that $\frac{1}{a}+\frac{1}{b}=2?$
|
We have: $4(ab)^2 = (a+b)^2 \ge 4ab \implies ab \ge 1\implies LHS \ge 2\sqrt{ab} + \dfrac{1}{1+\sqrt{ab}}= f(t), t = \sqrt{ab} \ge 1\implies f'(t) = 2 - \dfrac{1}{(1+t)^2}> 0\implies f(t) \ge f(1) = 2+\dfrac{1}{2} = \dfrac{5}{2} \implies LHS \ge \dfrac{5}{2} = RHS $. Equality occurs when $a = b = 1$.
|
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|
$\lim_{x→0^+} \frac{\sin(6x)}{\sqrt{\sin(2x)}}$ I'm trying to prove and compute the limit of this function.
$\lim_{x→0^+} \frac{\sin(6x)}{\sqrt{\sin(2x)}}$
I've tried converting it into different functions like $\cos(\pi/2-2x)$ or multiplying by the inverse function and so on, but it keep getting back to $0/0$.
I'm sure that the limit does in fact exist because using L'Hôpital's rule it is fairly easy to prove it, but I can't use it because the course I'm attending still hasn't defined the rule's properties, but I couldn't find a different way aside from that.
|
You can do this with regular old trig identities. Note that:
$$\sin\left(6x\right) = 6 \sin x \cos^5 x - 20 \sin^3 x \cos^3 x + 6 \sin^5 x \cos x$$
$$\sqrt{\sin\left(2x\right)} = \sqrt{2 \sin\left(x\right) \cos\left(x\right)}$$
For shorthand, let me write: $$\sin\left(x\right) = S \hspace{2.54cm} \cos\left(x\right) = C$$
So your ratio is: \begin{align*}\lim_{x \rightarrow 0^+} \frac{\sin\left(6x\right)}{\sqrt{\sin\left(2x\right)}} &= \frac{6SC^5 - 20S^3C^3 + 6S^5C}{\sqrt{2SC}} \\ & = \frac{6\sqrt{S}C^5 - 20S^2\sqrt{S}C^3 + 6S^4\sqrt{S}C}{\sqrt{2C}}\end{align*}
Finally, we have a form where the denominator won't blow up. Note that $$\lim_{x \rightarrow 0^+} \sin\left(x\right) = 0 \hspace{2.54cm} \lim_{x \rightarrow 0^+} \cos\left(x\right) = 1$$ to seal the deal: $$\lim_{x \rightarrow 0^+} \frac{\sin\left(6x\right)}{\sqrt{\sin\left(2x\right)}} = \frac{6\sqrt{0}C^5 - 20\cdot 0^2\sqrt{0}C^3 + 6\cdot 0^4\sqrt{0}C}{\sqrt{2C}} = 0$$
|
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|
$\frac{d}{dx}(\sin(x^{\frac{1}{3}}))$ from first principle The question contains a hint: at the appropriate point use the result $a^{3} - b^{3} =
\left(a - b\right)\left(a^{2} + b^{2} + ab\right)$.
$$
\frac{{\rm d}}{{\rm d}x}\sin\left(x^{1/3}\,\right)
$$
My attempt:
I did not use the hint as it was not immediately obvious to me what the simplification was. Instead I wrote with binomial expansion that $(x+h)^{\frac{1}{3}}=x^{\frac{1}{3}}(1+\frac{h}{x})^{\frac{1}{3}} \approx x^{\frac{1}{3}}(1+(\frac{1}{3})\frac{h}{x})$, used $\sin(()+())=\sin()\cos()+\sin()\cos()$ and applied $\cos() \approx 1$, $\sin() \approx ()$ for small $h$ which gave the correct answer.
But I do not see the point of using the cubes difference formula given in the question? Could this be a typo because it would make more sense to apply the identity to something like e.g. $\sin({x})^{\frac{1}{3}} (\text{not }\sin(x^{\frac{1}{3}}))\to \sin(x+h)^{\frac{1}{3}}-\sin(x)^{\frac{1}{3}} \iff (\sin(x+h)-\sin(x)=(\sin(x+h)^{\frac{1}{3}}-\sin(x)^{\frac{1}{3}})(\sin(x+h)^2+\sin(x)^2+\sin(x+h)\sin(x))$
Or am I missing something here?
|
You have
$$\begin{aligned}
\frac{\sin(x^{\frac{1}{3}}) - \sin(c^{\frac{1}{3}}) }{x-c} &= \frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)\cos\left(\frac{x^{\frac{1}{3}}+c^{\frac{1}{3}}}{2}\right)}{x-c}\\
&=\frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)\cos\left(\frac{x^{\frac{1}{3}}+c^{\frac{1}{3}}}{2}\right)}{(x^{\frac{1}{3}} - c^{\frac{1}{3}})(x^{\frac{2}{3}} + c^{\frac{2}{3}} + x^{\frac{1}{3}}c^{\frac{1}{3}})}
\end{aligned}$$
Now as $$\lim\limits_{x \to c }\frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)}{x^{\frac{1}{3}} - c^{\frac{1}{3}}} = 1$$
we get $$\frac{d}{dx}(\sin(x^{\frac{1}{3}})) = \frac{\cos(x^{\frac{1}{3}})}{3 x^{\frac{2}{3}}}$$
|
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|
The number of pairs of integers $(x,y)$ satisfying $x ≥ y ≥ -20$ and $2x + 5y = 99$ is I tried to solve this question but I was unable to think how to get the number of integer pairs satisfying these conditions. Till now I have broken down this into :-
$2x+5y = 99$
==>$y=(99-2x)/5$
$y\geq-20$
==> $(99-2x)/5 \geq20$
==> $x\leq 99.5$
$x\geq y$
==> $x \geq (99-2x)/5$
==> $x\geq 99/7$
From here I am not able to process further to think of the way that will give me the solution.
|
$2x + 5y = 99$ is very rare and a strong restriction.
If $2a + 5b=99$ is a solution to find another solution $2(a\pm d) +5(b\mp e) =99=2a+5b$ we must have $2d - 5e = 0$ and that requires that $d$ be divisible by $5$ and $e$ be divisble by $2$.
so if $2a + 5b =99 $ is one answer than any $2(a + 5k) + 5(b-2k) =99$ is also a solution and these are the only solutions.
Now lets find our first solution.
$2x +5y = 99$ so $5y = 99-2x$ and we need the RHS to be divisible by $5$. We can do that by letting $x = 2$ and $5y = 95$ so $y = 19$.
So if $2x + 5y = 99$ then $x = 2+5k$ and $y = 19 - 2k$ for some integer $k$ and for any integer $k$ $x = 2+5k$ and $y =19-2k$ will be a solution.
So we must have $2+5k \ge 19-2k\ge -20$. How many integers key can that be true for?
We must have $2+5k \ge 19-2k$ so $7k \ge 17$ so $k \ge 3$. And we have $19-2k \ge -20$ so $2k \le 39$ so $k \le 19$. So that's it.
$3\le k \le 19$ and $x = 2+5k$ and $y = 19 - 2k$ are all the solution pairs. There are $17$ of them:
$(x,y) = (17, 13),(22,11),(27,9).....,(97, -19)$.
All other solutions will either have $y > x$ (such as $(12,15)$) of have $y < -20$ (such as $(102, -21)$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/3966548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Elementary proof (involving no series expansion) that $\tan x \approx x+\frac{x^3}{3}$ for small $x$ For small values of $x$, the following widely used approximations follow immediately by Taylor expansion:
*
*$\sin x \approx x$
*$\cos x \approx 1-\frac{x^2}{2}$
*$\tan x \approx x +\frac{x^3}{3}$
I am looking for a justification of these approximations without the use of series expansions.
By purely geometric considerations, it is easy to see that for small values of $x$, we have
$$ \sin x < x < \tan x.$$
Division by $\cos x$ and an application of the squeeze lemma yield
$$\frac{\sin x}{x}\xrightarrow{x\to0}1$$
and hence the approximation (1.). Using the identity $\cos x = 1-2 \sin^2 \frac{x}{2}$, the approximation (2.) also follows.
Can one justify the approximation (3.) by a similarly elementary argument without using Taylor expansion?
I tried around using the angle addition theorems, but I did not really get anywhere, mainly because I could not make the factor $\frac13$ appear anywhere.
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Proof that $\boldsymbol{\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}}=\frac16$
As shown in $(2)$ from this answer, since $\cos(x)$ is continuous, given any $\epsilon\gt0$, we can find a $\delta\gt0$ so that if $0\lt|x|\le\delta$, we have
$$
1-\epsilon\le\cos(x)\le\frac{\sin(x)}x\le1\tag1
$$
Thus, assuming $0\lt|x|\le\delta$,
$$
\begin{align}
\frac{x-\sin(x)}{x^3}
&=\frac1{x^3}\sum_{n=0}^\infty\left(2^{n+1}\sin\left(x/2^{n+1}\right)-2^n\sin\left(x/2^n\right)\right)\tag{2a}\\
&=\frac1{x^3}\sum_{n=0}^\infty\left(2^{n+1}\sin\left(x/2^{n+1}\right)-2^{n+1}\sin\left(x/2^{n+1}\right)\cos\left(x/2^{n+1}\right)\right)\tag{2b}\\
&=\frac1{x^3}\sum_{n=0}^\infty2^{n+1}\sin\left(x/2^{n+1}\right)\left(1-\cos\left(x/2^{n+1}\right)\right)\tag{2c}\\
&=\frac1{x^3}\sum_{n=0}^\infty\frac{2^{n+1}\sin^3\left(x/2^{n+1}\right)}{1+\cos\left(x/2^{n+1}\right)}\tag{2d}\\
&=\frac12\sum_{n=0}^\infty\frac1{2^{2n+2}}\left[(1-\epsilon)^3,\frac1{1-\epsilon/2}\right]\tag{2e}\\
&=\frac16\left[(1-\epsilon)^3,\frac1{1-\epsilon/2}\right]\tag{2f}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply $\lim\limits_{x\to0}\frac{\sin(x)}x=1$ and the telescoping sum
$\text{(2b)}$: $\sin(2x)=2\sin(x)\cos(x)$
$\text{(2c)}$: factor
$\text{(2d)}$: $\sin^2(x)=(1-\cos(x))(1+\cos(x))$
$\text{(2e)}$: apply $(1)$ with $[a,b]$ representing a number in that range
$\text{(2f)}$: evaluate the sum
Therefore, by $(2)$ and the Squeeze Theorem,
$$
\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag3
$$
The Approximations
The limit $(3)$ gives the following approximation
$$
\sin(x)\approx x-\frac{x^3}6\tag4
$$
Since the Binomial Theorem gives $(1-x)^2=1-2x+x^2\approx1-2x$, we have $\sqrt{1-x}\approx1-\frac x2$.
$$
\begin{align}
\cos(x)
&=\sqrt{1-\sin^2(x)}\tag{5a}\\
&\approx\sqrt{1-x^2}\tag{5b}\\
&\approx1-\frac{x^2}2\tag{5c}
\end{align}
$$
Finally, since the Geometric Series gives $\frac1{1-x}=1+x+x^2+\dots\approx1+x$,
$$
\begin{align}
\tan(x)
&=\frac{\sin(x)}{\cos(x)}\tag{6a}\\
&\approx\left(x-\frac{x^3}6\right)\left(1+\frac{x^2}2\right)\tag{6b}\\
&\approx x+\frac{x^3}3\tag{6c}
\end{align}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3969365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Ordered pairs $(m,n)$ such that ${\frac{a^{m+n}+b^{m+n}+c^{m+n}}{m+n}=\frac{(a^m+b^m+c^m)}{m}\frac{(a^n+b^n+c^n)}{n}}$. I am unable to proceed with the following problem, : If $a+b+c=0$ then find all the ordered pairs $(m,n)$ where $m,n\in\mathbb{N}$ such that
$$\boxed{\frac{a^{m+n}+b^{m+n}+c^{m+n}}{m+n}=\frac{(a^m+b^m+c^m)}{m}\frac{(a^n+b^n+c^n)}{n}}$$
I tried but I think I have no idea regarding this kind of problem although I tried a bit hard with all possible methods (which I can do/know) but that doesn't result any substantial thing which I can add and also it will not give a complete set of solutions, this trial approach provides a few solution(and quite laborious and too many limitations) for instance this well known stuff if $a+b+c=0$ then, $\frac{a^5+b^5+c^5}{5}=(\frac{a^3+b^3+c^3}{3})(\frac{a^2+b^2+c^2}{2})$ is one of the solution. Thanks for your attention.
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Let us first consider the case $m\le n$.
*
*If $m=n$, then taking $(a,b,c)=(1,-1,0)$, we get $m=(1+(-1)^m)^2$ from which $m=4$ follows. However, when $m=n=4$, the equation does not hold for $(a,b,c)=(2,-1,-1)$.
*If $m=1$, then taking $(a,b,c)=(2,-1,-1)$, we get $2^{n}=(-1)^{n}$ which is impossible.
So, in the following, $2\le m\lt n$.
*
*If both $m$ and $n$ are odd, then taking $(a,b,c)=(2,-1,-1)$, we get
$$mn(2^{m+n-1}+1)=2(m+n)(2^{m-1}+(-1)^m)(2^{n-1}+(-1)^n)$$The LHS is odd while the RHS is even, which is impossible.
*If both $m$ and $n$ are even, then taking $(a,b,c)=(1,-1,0)$, we get $(m-2)(n-2)=4$, but there are no solutions satisfying $2\le m\lt n$.
*If exactly one of $m,n$ is odd, then taking $(a,b,c)=(2,-1,-1)$, we get
$$mn(2^{m+n}-2)=(m+n)(2^m+2(-1)^m)(2^n+2(-1)^n)\tag1$$Suppose here that $m$ is odd. Then, since we have $mn\gt m+n\ (\gt 0)$ (which is equivalent to $(m-1)(n-1)\gt 1$ which is true) and $$2^{m+n}-2\gt (2^m+2(-1)^m)(2^n+2(-1)^n)\ \ (\gt 0)$$ (which is equivalent to $2\gt 2^{m+1}-2^{n+1}$ which is true), the LHS of $(1)$ is larger than the RHS of $(1)$. So, $m$ has to be even to have
$$2^n\bigg(\underbrace{n(m2^{m}-2^{m}-2)-2^mm-2m}_{A}\bigg)+\underbrace{n(2^{m+1}-2m+4)+4m(2^{m-1}+1)}_{\gt 0}=0\tag2$$Now, suppose that $m\ge 4$. Then, we have
$$\begin{align}A=n\underbrace{(m2^{m}-2^{m}-2)}_{\gt 0}-2^mm-2m&\gt m(m2^{m}-2^{m}-2)-2^mm-2m
\\&=4m\bigg((m-2)2^{m-2}-1\bigg)
\\\\&\gt 0\end{align}$$implying that the LHS of $(2)$ is positive. So, we have to have $m=2$. Taking $(a,b,c)=(2,-1,-1)$, we have$$(n-6)2^{n-1}+2n+6=0\tag3$$Suppose here that $n\ge 7$. Then, the LHS of $(3)$ is positive. So, we get $n=3,5$ which are sufficient (see here, here, here).
In conclusion, considering the case $m\gt n$, we see that the answer is
$$\color{red}{(m,n)=(2,3),(3,2),(2,5),(5,2)}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3972109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Integral of a rational expression involving quadratics. I want to solve this integral but I have some problems...
$$\int_2^3 \frac{(x^2-2x+1)}{(x^2+2x+1)}$$
I transformed both in $(x-1)^2$ and $(x+1)^2$ respectively but didn't find any answer. I tried as well to transform the rational expression into $1 -\frac{4x}{(x^2+2x+1)}$, but I wasn't capable of finding any antiderivative.
Thanks for responding!
|
\begin{gather*}
I=\int ^{3}_{2}\frac{x^{2} -2x+1}{x^{2} +2x+1} dx\\
\text{Decompose $x^{2} -2x+1$ into partial fractions of $( x+1)$}\\
I=\int ^{3}_{2} 1-\frac{4x+4-4}{x^{2} +2x+1}\\
I=\int ^{3}_{2} 1-\frac{4}{x+1} +\frac{4}{x^{2} +2x+1}\\
I=\int ^{3}_{2} 1-\frac{4}{x+1} +\frac{4}{( x+1)^{2}} dx\\
\end{gather*}Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3972228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$? The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both side and drawing triangles. However, the worked solutions does a simplification from LHS to RHS instead, namely $$\cos\arcsin\left(\frac{3}{5}\right)\cos\arctan\left(\frac{7}{24}\right)-\sin\arcsin\left(\frac{3}{5}\right)\sin\arctan\left(\frac{7}{24}\right)=\frac{3}{5}$$which I don't understand. Can someone please explain?
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As the inverse trigonometric functions are multi-valued with a respective principle value
or
as sine/cosine/tangent functions are periodic, it is incorrect to apply cosines in both sides.
For example,
*
*$x$ in general $\ne -x$ but $\cos(-x)=\cos(x)$
*$x$ in general $\ne \pi-x$ but $\sin(\pi-x)=\sin(x)$
*$x\ne\pi+x$ but $\tan(\pi+x)=\tan(x)$
Using the ranges of principle values of $\arctan,$ if $\arctan\dfrac7{24}=y$
$\implies0<y<\dfrac\pi2$
and $\tan y=\dfrac7{24}$
and $\sin y,\cos y>0$
Consequently, $\dfrac{\sin y}7=\dfrac{\cos y}{24}=+\sqrt{\dfrac{\sin^2y+\cos^2y}{7^2+24^2}}$
$\implies\arctan\dfrac7{24}=\arcsin\dfrac7{25}=\arccos\dfrac{24}{25}$
Now use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ in the left hand side
For the right hand side, if $\arccos\dfrac35=u$
$\implies0<u<\dfrac\pi2$ and $\cos u=\dfrac35$ and $\sin u>0$
$\implies\sin u=\sqrt{1-\cos^2u}=\dfrac45$
$\implies\arcsin\dfrac45=u=\arccos\dfrac35$
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"language": "en",
"url": "https://math.stackexchange.com/questions/3973282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.