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Find the value of $x^5 + \frac{1}{x^5}$ - question about correctness of method The task is: if $x+ \frac{1}{x}= 1$ find $x^5 + \frac{1}{x^5} $.
I used the binomial formula and proved that $x^5 + \frac{1}{x^5} = 1$, but I have a question about following method, I am not sure if it's correct. If I take square of the the first equality, I get:
$x^2 +2 + \frac{1}{x^2} = 1$ so $x^2 +\frac{1}{x^2} = -1$. Now, the sum of 2 squares is nonnegative and the right side is negative, so when I come to this part does it mean that this method is wrong?
In general, when proving such equalities, when are we allowed to take square (and we don't know if one side of equality is positive as in this task)? Thanks in advance.
|
$x+\frac{1}{x}=1$ gives $x^2-x+1=0,$ which gives $x^3+1=0$ or $x^3=-1$.
Thus, by your work $$x^5+\frac{1}{x^5}=-x^2-\frac{1}{x^2}=1.$$
Now, about your last question.
You used the following statement.
If $x+\frac{1}{x}=1$ so $\left(x+\frac{1}{x}\right)^2=1$.
It's true because if $a=b$ so $a-b=0$ and from here
$$a^2-b^2=(a-b)(a+b)=0,$$ which gives $$a^2=b^2.$$
Id est, we proved the following claim:
$$a=b\Rightarrow a^2=b^2$$ is true.
|
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|
Let $x, y \in \mathbb Z$. Prove that if $7 | (x^2 + y^2 ),$ then $7 | x$ and $7 | y$
Let $x, y \in \mathbb{Z}$. Prove that if $7 | (x^2 + y^2 ),$ then $7 | x$ and $7 | y$.
How do I prove this. Sorry for not providing how I approached the problem because I have no idea how to approach it.
|
Well, you could if worst comes to worst just try every option:
$x \equiv i \pmod 7$ where $i=0,....6$ and $y\equiv j\pmod 7$ where $0..., 6$.
The $x^2\equiv i^2 \equiv 0, 1, 4, 2\pmod 7$. (Might be worthing noting that as $6,5,3 \equiv -1,-2,-3$ then $6^2,5^2, 3^2\equiv 1^2,2^2,3^2$). So $y^2\equiv j^2\equiv 0,1,4,2\pmod 7$.
The possible combinations of $x^2 + y^2$ are $0+0, 0+1,0+4,0+2,1+1,1+4,1+2,4+4,4+2,2+2$ and the only time you have $x^2 + y^2 \equiv 0$ are if $x \equiv 0; y\equiv 0$.
So $7|x^2 + y^2\implies x^2 + y^2 \equiv 0 \pmod 7 \implies x,y\equiv 0 \pmod 7\implies 7|x, 7|y$.
That was pretty clunky and tedious but it's valid.
.....
Alternatively: $x^2 + y^2 \equiv 0\pmod 7$
$x^2 \equiv -y^2 \pmod 7$.
If $A = \{[x^2]_7|$ where $[x^2]_7$ is an equivalence class of a perfect square $\mod 7\} = \{[0]_7,[1]_7,[4]_7,[2]_7\}$ and if $B =\{-[y^2]_7\}=\{[0]_7,[6]_7,[3]_7,[5]_7\}$ then as $x^2 \equiv -y^2 \pmod 7$ we must have $x^2, y^2$ both in the intersection of $A$ and $B$ and the intersection of $A$ and $B$ is $[0]_7$ so $x^2\equiv -y^2 \equiv 0$ and therefore $x \equiv y \equiv 0\pmod 7$.
.....
I thought maybe there'd be something clever with $(x+y)^2 \equiv x^2 + 2xy + y^2 \equiv 2xy\pmod 7$ and if we assume $2xy\not \equiv 0\pmod 7$ but.... That doesn't seem to get anywhere.
......
It might or might not be worth noting a lemma that if $n|x^2 + y^2$ then $[x^2],-[y^2] \in \{[a^2]_n\}\cap \{-[a^2]_n\}$. Or $x,y \in \{a| \exists b: a^2 \equiv-b^2\pmod n\}$.
If $p$ is an odd prime the intersection is $0$ so if odd prime $p|x^2 + y^2$ then $p|x$ and $p|y$. Even if $n$ isn't prime the intersection of those sets of classes is pretty exclusive.
|
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Hard limit $\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$ Prove that :
$$\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$$
I can prove that :
$$\lim_{x\to\infty}\frac{(x(x+1)(x+2))^{\frac{1}{3}}-2(x+1)}{-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}}=1$$
Using the Hospital rule but it doesn't help here .Moreover I have tried power series without success .
Any helps is welcome .
Thanks in advance
|
first of all, notice that the substitution $x+1 = t, t \to \infty$ simplifies things quite a bit:
$$L = \lim_{t \to \infty} \left( (t(t^2-1))^{1/3} + ((t-1)^{t-1}\cdot t^t \cdot (t+1)^{t+1})^{1/3t} - 2t\right)$$
Factor out $t$ from the expression in the bracket and do some manipulations to get
$$\begin{align}
L &= \lim_{t \to \infty}\ t\left( \left( 1 - \frac{1}{t^2}\right)^\frac13 + \left( \frac{t+1}{t-1}\right)^\frac{1}{3t} \cdot \left(1 - \frac{1}{t^2}\right)^\frac13 - 2\right) \\
&= \lim_{t \to \infty}\ t\left( \left( 1 + \left( \frac{t+1}{t-1}\right)^\frac{1}{3t} \right)\left( 1 - \frac{1}{t^2}\right)^\frac13 - 2\right)
\end{align}$$
The term in the first bracket tends to 2 as $t$ tends to infinity, so we're finally getting somewhere. All we have to do is prove that the order of the infinitesimal within the bracket is $\frac{1}{t^2}$ or lesser. To do this, notice that
$$\left( \frac{t+1}{t-1}\right)^\frac{1}{3t} = \left( 1 + \frac{2}{t-1}\right)^\frac{1}{3t} \approx 1 + \frac{2}{3t(t-1)}$$
(By using binomial expansion). We can also assume that the second term tends to 1, i.e.
$$\left( 1 - \frac{1}{t^2}\right)^\frac13 \approx 1$$
(I'll justify this in a minute, but let's carry on for now)
once we have done these simplifications, the limit becomes
$$\begin{align}
L &= \lim_{t \to \infty}\ t\left( 2 + \frac{2}{3t(t-1)} - 2\right) \\
&= \lim_{t \to \infty}\ \frac{2t}{3t(t+1)} \\
&= \boxed{0}
\end{align}$$
Hence proved.
For justifying the approximation(s) I did, let's look at the expression and try to simplify via binomial theorem to the first order:
$$\begin{align}
f(t) &= \left( 2 + \frac{2}{3t(t-1)}\right)\left( 1 - \frac{1}{t^2}\right)^\frac13 \\
&= \left( 2 + \frac{2}{3t(t-1)}\right)\left( 1 - \frac{1}{3t^2}\right)\\
&= 2 - \frac{2}{3t^2} + \frac{2}{3t(t-1)} - \frac{2}{9t^3(t-1)} \\
&= 2 + \frac{2}{3t^2(t-1)} - \frac{2}{9t^3(t-1)}
\end{align}$$
You notice that we get a different expression here, one that has a $t^2$ instead of a $t$ in the denominator. This doesn't matter as the power of $t$ in the denominator should be more than 1 to obtain the result as 0. While this expansion gives a more accurate answer, the approximation that I did in the answer is justified.
|
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|
prove the trigonometry identity: prove the trigonometry identity:
$$\tan^4(A) = \frac{\tan^3(A) + \frac{1 - \tan(A)}{\cot(A)}}{\frac{1 - \cot(A)}{\tan(A)} + \cot^3(A)}$$
of course i started from the complicated side the RHS and i wrote them all into tangents but then it all messy up and I'm stuck there.
|
$1. \ \cot^3A = \dfrac{1}{\tan^3A}\\$
$2. \ \dfrac{1-\cot A}{\tan A } = \dfrac1{\tan A}-\dfrac{1}{\tan^2A}\\$
$3. \ \dfrac{1-\tan A}{\cot A} = \tan A - \tan^2A$
So, the RHS becomes,
$\begin{align}\dfrac{\tan^3A + \tan A - \tan^2A}{\dfrac1{\tan A}-\dfrac{1}{\tan^2A} +\dfrac{1}{\tan^3A}} &= \dfrac{\tan^3A + \tan A - \tan^2A}{\dfrac{1}{\tan^3A}\left(\tan^2A-\tan A+1\right)} \\&= \tan^3A\dfrac{\tan A(\tan^2A+1-\tan A)}{\tan^2A+1-\tan A} \\&= \tan^4A\end{align}$
|
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An integer $n \geq 2$ is called square-positive- proof? An integer $n \geq 2$ is called square-positive if there are $n$ consecutive positive integers whose sum is a square. Determine the first four square-positive integers.
So I have found the first four square-positive numbers, but I need to prove that why it $4$ is not a square-positive number and I also need to write a general formula for determining whether a number is square-positive or not. I have tried to write the sum of consecutive positive integers like this $a + a +1 + a + 2 + a+3 \dots a - 1$ and I wrote it like this for all numbers, and part of the proof for why $4$ isn't a square-positive number is that $4a + 6$ is not divisible with $4$. But I haven't got so far.
Here is my answer:
2 : 4 + 5 = 9 which is 3^2
3 : 2 + 3 + 4 = 9 which is 3^2
5: 18 + 19 + 20 + 21 + 22 = 100 which is 10^2
6: 35 + 36 + 37 + 38 + 39 + 40 = 225 which is 15^2
Interesting fact is that for all odd numbers and some even numbers like 6 and 10, you can find out which number is the first (the one you start with and then go forward here like 3, 2, 18 and 35) using this formula :
(I show it in an example because I still can't write it algebraically):
For example: the sum of 95 subsequent numbers is 5n + 10
(10^2 - 10) /5 = 18
So your first number is 18
And if you keep adding, 18 + 19 + 20 + 21 + 22 you get 100 which is 10^2, the same number you squared.
|
Let $b-2, b-1, b, b+1$ be four consecutive numbers. The sum is $4b -2$. Let $m^2 = 4b-2$ Then $\frac {m^2}2 = 2b-1$. And $2|m^2$. So $2|m$. Let $m=2k$ then $4k^2 = 4b -1$ and $2k^2 = 2b-1$. That's impossible as the LHS is even and the RHS is odd. So $4$ is not square positive.
=====
In general:
........
If the first integer is $a$ and there are $n$ consecutive integers the numbers are $a, a+ 1,......, a+(n-1)$ and the sum of the the integers is $a + (a+ 1) + ..... + (a + (n-1)) = n\cdot a + (1+ 2 + ...... + (n-1)) = n\cdot a + \frac {(n-1)n}2$.
If $n\cdot a + \frac {(n-1)n}2 = M^2$ then by completing the square
$\frac 12n^2 - (a-\frac 12)n -M^2 = 0$ and .....
well, we can keep going but.... let's try to be clever....
If $n=2k+1$ is an odd number and the middle number is $b$ then the first number is $b-k$ and the last number is $b+k$ and the sum of the number is $$(b-k) + (b-k+1) + ..... + (b-1) + b+ (b+1) + ..... + (b+k) =\\ nb + (-k+(-k+1)+...... + (-1) + 0 + 1 + 2 + ..... k) =\\nb + 0= nb$$. ANd $nb$ can be square positive whenever $b= n$. (and other times). For intsance. $1=1^2$ and $2+3+4=3^2$ and $3+4+5+6+7 = 5^2$ and $(n-\frac {n-1}2) + ...... + (n + \frac {n-1}2) = n^2$.
So all odd numbers are square positive.
And if $n= 2k$ is even the there is no middle number but if the "middlish" number (off by one half) is $b$:Then we can notice that first $k$ of these numbers are $b-k ,b-(k-1) , b-(k-2),........, b-1$; and the second $k$ of these numbers are $b, b+1, b+2, ......, b+(k-1)$; and the sum of these numbers are $$nb + (-k,-(k-1),-(k-2),......., (k-2),(k-1)) =\\ nb - k =\\ 2kb -k = k(2b-1)$$. And that can be a square positive every time we have $2b-1 = k=\frac n2$.
But this is possible only if $\frac n2$ is odd.
But if $k=2m$ is even we must have $2m(2b-1)$ is a perfect square which requires $m$ to be even.
By induction will prove $n$ is square positive if and only if $n = 2^{odd}\cdot odd$.
So the list of all square positive numbers are $2,3,5,6,7,8,9,10,11,13,14,15,17,18,19,21,22,32,24,......$ etc.
|
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Non-linear system of two equations (Method of Moments for Generalized Pareto) In a Method of Moments Estimation problem involving the generalized Pareto Distribution, the following system of 2 non-linear equations arise
\begin{align*}
\bar{X} &= \frac{\alpha \beta}{\alpha - 1}. \\
\frac{1}{n}\sum_{i=1}^{n} X^2_i &= \frac{\alpha \beta^2}{\alpha - 2}.
\end{align*}
The problem is to solve the system for $\alpha$ and $\beta$ in terms of $\bar{X} $ and $\frac{1}{n}\sum_{i=1}^{n} X^2_i $. The thing is, it's easy to isolate $\alpha$ or $\beta$ from any of the two questions, but after this has been done you cannot isolate the other parameter in the other equation. Using Mathematica you can get two solutions but I cannot just say "According to Mathematica, the solution is ..."
|
$$
\begin{cases}
\frac{a b}{a-1}=x\\
\frac{a b^2}{a-2}=y\\
\end{cases}
$$
multiply the first equation by $b$
$$
\begin{cases}
a b^2=(a-1)bx\\
a b^2=(a-2)y\\
\end{cases}
$$
thus
$$(a-1)bx=(a-2)y\to b=\frac{(a-2) y}{(a-1) x} $$
Now plug this in the first, the very first equation
$$ab=x(a-1)\to a\cdot \frac{(a-2) y}{(a-1) x}=x(a-1)$$
You get a quadratic equation in $a$ which gives
$$a=\frac{x^2-y\pm\sqrt{y^2-x^2 y}}{x^2-y}$$
plug back to get $b$
$$b_1=\frac{y \left(\frac{x^2-y-\sqrt{y^2-x^2 y}}{x^2-y}-2\right)}{x \left(\frac{x^2-y-\sqrt{y^2-x^2 y}}{x^2-y}-1\right)}=\frac{y}{x}\cdot \frac{x^2-y-\sqrt{y \left(y-x^2\right)}}{\sqrt{y \left(y-x^2\right)}}=\frac{y-\sqrt{y \left(y-x^2\right)}}{x}$$
and
$$b_2=\frac{y+\sqrt{y \left(y-x^2\right)}}{x}$$
|
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How to calculate the angle between matrices? I understand the formula:
$cos\left(θ\right)=\frac{tr\left(AB^T\right)}{\sqrt{tr\left(AA^T\right)\cdot tr\left(BB^T\right)}}$
Only one thing is not clear to me,the number I get is the angle itself?
Or the number should be obtained by the cos function?
I will give an example with simple numbers to make the question clearer.
$A=\begin{pmatrix}1&2\\ 0&1\end{pmatrix} B=\begin{pmatrix}2&0\\ 0&2\end{pmatrix}$
$\frac{tr\left(\begin{pmatrix}1&2\\ 0&1\end{pmatrix}\cdot \begin{pmatrix}2&0\\ 0&2\end{pmatrix}^t\right)}{\sqrt{tr\left(\begin{pmatrix}1&2\\ 0&1\end{pmatrix}\cdot \begin{pmatrix}1&2\\ 0&1\end{pmatrix}^t\right)\cdot tr\left(\begin{pmatrix}2&0\\ 0&2\end{pmatrix}\cdot \begin{pmatrix}2&0\\ 0&2\end{pmatrix}^t\right)}}=\frac{1}{\sqrt{3}}$
So the angle is $\frac{1}{\sqrt{3}}(\approx 0.577)$ or $cos (\frac{1}{\sqrt{3}})\approx 0.837$?
|
Neither. With $\theta$ your angle, you have $\cos \theta = \frac{1}{\sqrt{3}}$, and so $\theta = 2 \pi n \pm \arccos\left(\frac{1}{\sqrt{3}}\right)$, where $\arccos\left(\frac{1}{\sqrt{3}}\right) \approx 0.955$ (radians).
|
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$(ab+bc+ca)^3=abc(a+b+c)^3$, prove that $a,b,c$ are in $G.P.$ Suppose $a,b,c$ are non-zero real numbers such that
$$(ab+bc+ca)^3=abc.(a+b+c)^3$$
Prove that $a,b,c$ must be terms of a $G.P.$
I simplified this equation too
$$(ab)^3+(bc)^3+(ca)^3=abc.(a^3+b^3+c^3)$$
I tried to subtract $3(abc)^2$ from both sides and it gave a factorised form.
$$(ab+bc+ca).((ab)^2+(bc)^2+(ca)^2-abc.(a+b+c))=abc.(a+b+c).(a^2+b^2+c^2-ab-bc-ca)$$
How can I proceed after that?
|
There was a good answer posted by @Albusdumbledore that was deleted. I am just reproducing that answer and will delete mine if the other is undeleted.
The suggestion was to write $a = q, b = q r$ and $c = q r k$. Naturally, since $a,b,c \ne 0$, we also have $q,r,k \ne 0$. Plugging this into the relation, one concludes that $r, k$ must satisfy the relation
$$
(k-r) \left(k^2 r-1\right) \left(k r^2-1\right)=0
$$
All solutions lead to GP's. More specifically,
Case 1. If $k=r$, $a=q$, $b = r q$ and $c= r^2 q$ are consecutive terms in a GP with ratio $r$.
Case 2. If $k=\frac{1}{r^2}$ then $c=\frac{1}{r^2} q$, $a = q$ and $b=rq$ are (nonconsecutive) terms in a GP with ratio $r$
Case 3. If $k=\frac{1}{\sqrt{r}}$ then $c= \frac 1r q$, $a=q$ and $b= rq$ are (nonconsecutive) terms in a GP with ratio $\sqrt{r}$.
Case 4. If $k=-\frac{1}{\sqrt{r}}$ then $a=q$, $c=-\sqrt{r} q$, $b = rq$ are consecutive terms in a GP with ratio $-\sqrt{r}$
|
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|
Prove by induction $\left(1-\frac{1}{2}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$ Prove by induction $\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$
What would be the best way to solve this by the induction method?
|
Hint: Prove that
$$
\left(\frac{1}{4}+\frac{1}{2^{n+1}}\right)
\left(1-\frac{1}{2^{n+1}}\right)
\ge
\frac{1}{4}+\frac{1}{2^{n+2}}
$$
|
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|
Number of sequences of the form $x_1x_3x_5$
Find the number of sequences containing $5$ distinct terms $x_1,x_2,x_3,x_4,x_5$ from the set $A=\left\{1,2,3,4....20\right\}$ such that $x_1<x_2>x_3<x_4>x_5$.
My try:
The total number of sequences with $5$ distinct terms is $5!\times \binom{20}{5}$, among which there are $\binom{20}{5}$ strictly increasing sequences and $\binom{20}{5}$ strictly decreasing sequences. But now I am stuck in finding the sequences with $x_1<x_2>x_3<x_4>x_5$ among $5!\times \binom{20}{5}-2\times \binom{20}{5}$
|
Note that for any set of five distinct numbers, the number of ways of arranging them so that they satisfy the given condition must be the same as for any other five distinct numbers. Thus we really only need to count the number of ways of ordering $\{1,2,3,4,5\}$ to satisfy the given condition.
Note that $5$ has to be either $x_2$ or $x_4$; note also that whichever of $x_2$ or $x_4$ is not $5$, it will have to be either $3$ or $4$.
Suppose $(x_2, x_4) \in \{(4,5),(5,4)\}$; in each case, there are $3!$ ways of arranging the other three numbers. Now suppose $(x_2, x_4) \in \{(3,5),(5,3)\}$; then the $4$ has to go in the outside position next to the position containing the $5$, which leaves $2!$ ways of arranging the other numbers.
Thus the number of arrangements of $\{1,2,3,4,5\}$ satisfying the given condition is $2 \cdot 3! + 2 \cdot 2! = 16$. But then the number of ways of sequences from $\{1,2,3,4,...,20\}$ satisfying the condition will just be
$$16 \binom{20}{5} = \boxed{248,064}$$
|
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The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
What I Tried: Here is the diagram :-
You can see I marked the angles as required. Now let $AB = x$ . We then have :-
$$[\Delta ADE] = \frac{\sqrt{3}}{4}x^2$$
Now from here :- https://www.quora.com/What-is-the-ratio-of-sides-of-a-30-75-75-angle-triangle-without-sine-rule , I could understand and show that :- $$ED : DC : CE = \bigg(\frac{\sqrt{3} + 1}{2} : \frac{\sqrt{3} + 1}{2} : 1\bigg)$$
So let $EC = k$ , $CD = DE = \frac{(\sqrt{3} + 1)k}{2}$ .
From here :- $$x = \frac{(\sqrt{3} + 1)k}{2}$$
$$\rightarrow k = EC = \frac{2x}{(\sqrt{3} + 1)}$$
Now, we can find area by Heron's Formula. We have :- $$s = x + \frac{x}{(\sqrt{3} + 1)}$$
$$\rightarrow s = \frac{x\sqrt{3} + 2x}{(\sqrt{3} + 1)}$$
So :- $[\Delta DEC] = \sqrt{s(s-a)(s-b)(s-c)}$
$$\rightarrow \sqrt{\Bigg(\frac{(x\sqrt{3} + 2x)}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{(x\sqrt{3})}{(\sqrt{3} + 1)}\Bigg)}$$
This looks like really a complicated expression, and I really am not going to attempt to simplify this. So can anyone give me a different solution?
Thank You.
|
If $a,b$ are sides of a triangle and $x$ the mesure of angle between them, then the area of it is $${a\cdot b \cdot \sin x\over 2}$$
We use that formula here.
We have $AD = AE =DE =DC=a $ so$$\frac{[\Delta ADE]}{[\Delta DEC]} = {{a^2\sin 60 \over 2}\over {a^2 \sin 30 \over 2}} = \sqrt{3}$$
|
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"url": "https://math.stackexchange.com/questions/3896680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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|
Prove the identity $\frac{1}{1+x^{1}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$ Show that for all non-negatives integers $n$, it is true that
$$\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$$
For $x \ne 1$, $x \ne -1$
I tried to solved it by appliying geometric series but I got the following
after doing the $2f\left(x\right)-f\left(x\right)$ step, where $f\left(x\right)=\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}$
$$f\left(x\right)=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n}}}$$
Not sure if Fermat's theorem is needed in order to solve it
|
Subtract $1/(x-1)$ from LHS and note that,
$$\frac 1{x+1}-\frac 1{x-1}=-\frac 2{x^2-1} \\ \frac 2{x^2+1}-\frac 2{x^2-1}=-\frac 4{x^4-1}\\ \frac 4{x^4+1}-\frac 4{x^4-1}=-\frac 8{x^8-1}$$
and so on. Do you see the telescoping pattern?
To finish, note that,
$$\frac {2^n}{x^{2^n}+1}-\frac {2^n}{x^{2^n}-1}=-\frac{2^{n+1}}{x^{2^{n+1}}-1}=\frac{2^{n+1}}{1-x^{2^{n+1}}}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3897133",
"timestamp": "2023-03-29T00:00:00",
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|
Determine all pairs of perfect powers $2^m$, $3^n$ such that their difference is a perfect square. Determine all pairs of perfect powers $2^m$, $3^n$ such that their difference is a perfect square.
I found these pairs work by mod 3 and mode 4:
Case 1, $2^m \ge 3^n$, $(0,0), (2,1), (1,0)$
Case 2, $2^m \le 3^n$, $(0,0), (1,1), (1,3), (3, 2), (5, 4)$
How to prove there's no others?
I also found this post: Pairs forming perfect square. There's good discussions. Do we have any complete solution?
|
This is a partial solution concerning $m\ne 1$.
First we consider the cases $m\ge 2$.
For case 1 ($2^m \ge 3^n$), write $2^m = 3^n + k^2$.
Taking modulo $3$, we have $(-1)^m \equiv k^2 \pmod 3$. This gives $m$ is even.
Write $m = 2s$, then $3^n = 2^m-k^2 = (2^s-k)(2^s+k)$, and both $2^s \pm k$ are powers of $3$.
Summing them up, however, gives $2^{s+1}$ which is not a multiple of $3$. Hence $2^s-k = 1$.
$2^s+k = 2^{s+1}-1$ is a power of $3$. [This gives $s=1, m=2, n=1$.]
For case 2 ($2^m \le 3^n$), write $3^n = 2^m + k^2$.
Taking modulo $4$, we have $(-1)^n\equiv k^2 \pmod 4$. This gives $n$ is even.
Write $n=2t$, then $2^m = 3^n-k^2 = (3^t-k)(3^t+k)$, and both $3^t \pm k$ are powers of $2$.
If we write $3^t-k = 2^x, 3^t+k =2^y$, we have $2^x+2^y = 2^x(1+2^{y-x})$.
Summing them up, however, gives $2 \times 3^t$. Since $3^t$ is odd, we must have $2^x=2$ and $1+2^{y-1} = 3^t$.
[This equation has $2$ solutions: $y=2, t=1$ and $y=4, t=2$.]
Both cases above end with solving an equation of the form $1+a^b= c^d$, which by Catalan's Conjecture/Mihăilescu's theorem only has the solution $a=2,b=3,c=3,d=2$, for integers $a,b,c,d >1$. However there are elementary proofs for this special case $a,c \in \{2,3\}$, which I choose to omit (unless requested)
For the case $m=0$, write $3^n = 1+k^2$. By modulo $3$, no solutions exist for $n \ge 1$. This forces $n=0$.
For the case $m=1$, write $3^n = 2+k^2$. By modulo $3$ we have $k^2 \equiv 1\pmod 3$, and by modulo $4$ we have $(-1)^n \equiv 2+k^2$, forcing $n$ to be odd and $k^2\equiv 1 \pmod 4$. $k$ must be an odd number not divisible by $3$, but as of right now I am not sure how to proceed.
|
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|
if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then...
if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then find $$\sin (d/2)+2\sin (c/2)+3\sin (b/2)+4\sin (a/2)$$.$(a<b<c<d)$
Attempts
$$\sin x=1/4$$,$$4\sin^2(x/2)(1-\sin^2(x/2))=\frac{1}{16}$$
Let $\sin^2 (x/2)=t$:it remains to solve $$t-t^2=\frac{1}{64}$$ .The rest is just brute force i.e we calculate the 4 values $\sin(x/2)$ can take , make sure the angles are arranged in ascending order and substitute the roots.As far as i can see i a getting irrational roots.
My question is does anyone spot a trick, that can avoid all this work.!
|
So a is in the first quadrant
$b = \pi-a$
$c = a + 2\pi$
$d = 3\pi-a$
$\sin(\frac{b}{2}) = \sin(\frac{\pi}{2}-\frac{a}{2}) = \sin(\frac{\pi}{2})\cos(-\frac{a}{2})+\cos(\frac{\pi}{2})\sin(-\frac{a}{2}) = \cos(\frac{a}{2})$
$\sin(\frac{c}{2}) = \sin(\frac{a}{2}+\pi) = \sin(\frac{a}{2})\cos(\pi)+\cos(\frac{a}{2})\sin(\pi) = -\sin(\frac{a}{2})$
$\sin(\frac{d}{2}) = \sin(\frac{3\pi}{2}-\frac{a}{2}) = \sin(\frac{3\pi}{2})\cos(-\frac{a}{2})+\cos(\frac{3\pi}{2})\sin(-\frac{a}{2}) = -\cos(\frac{a}{2})$
So $\sin(\frac{d}{2}) + 2\sin(\frac{c}{2}) + 3\sin(\frac{b}{2}) + 4\sin(\frac{a}{2})$
$=2\sin(\frac{a}{2}) + 2\cos(\frac{a}{2})$
We know $\cos(a) = \sqrt{1-\frac{1}{16}} = \frac{\sqrt{15}}{4}$
Using the half-angle formla
$\sin(\frac{a}{2}) = \sqrt{0.5(1-\cos(a))}$
$\sin(\frac{a}{2}) = \sqrt{\frac{4-\sqrt{15}}{8}}$
similarly
$\cos(\frac{a}{2}) = \sqrt{\frac{4+\sqrt{15}}{8}}$
So
$2\sin(\frac{a}{2}) + 2\cos(\frac{a}{2}) = \sqrt{\frac{4-\sqrt{15}}{2}}+\sqrt{\frac{4+\sqrt{15}}{2}}$
ok. simplifying the right side by squaring (comes out to 5) and then taking the square root, I get $\sqrt{5}$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If $f$ is continuous on $[0,1]$, prove that $\lim_{n\to\infty}\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx=\frac{\pi}{2}f(0)$. The solution given:
$$\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx=\int_{0}^{n^{-\frac{1}{3}}} \frac{nf(x)}{1+n^2x^2}dx+\int_{n^{-\frac{1}{3}}}^{1} \frac{nf(x)}{1+n^2x^2}dx$$
okay, my first question here: why the $n^{-1/3}$?
$$|\int_{n^{-\frac{1}{3}}}^{1} \frac{nf(x)}{1+n^2x^2}dx|\leq \int_{n^{-\frac{1}{3}}}^{1} |\frac{nf(x)}{1+n^2x^2}|dx \leq \frac{nM}{1+n^{1+1/3}}$$
where M is the global maximum of $|f|$. how did the middle term turn into the rightmost one? How is the mean value theorem of integrals used here? what's with the $1+1/3$?
Since the rightmost term tends to $0$ as $n$ tends to infinity, by squeeze theorem we have the leftmost term tends to $0$ as well.
Since $\frac{n}{1+n^2x^2}$ does not change sign on $[0,1]$, then there exists $c\in[0,n^{-1/3}]$ such that
$$\int_{0}^{n^{-\frac{1}{3}}} \frac{nf(x)}{1+n^2x^2}dx=f(c)\int_{0}^{n^{-\frac{1}{3}}} \frac{n}{1+n^2x^2}dx=f(c)\tan^{-1}n^{2/3}$$
Since $c\in[0,n^{-1/3}]$, $n\to\infty$ implies $c\to0$ and $\tan^{-1}n^{2/3}\to \frac{\pi}{2}$ and so $\int_{0}^{n^{-\frac{1}{3}}} \frac{nf(x)}{1+n^2x^2}dx\to\frac{\pi}{2}f(0)$
So the final answer is $0+\frac{\pi}{2}f(0)=\frac{\pi}{2}f(0)$
I've highlighted my questions! Thanks for any advice given!
|
First for your highlighted question: note that
$$
\newcommand{\abs}[1]{\left\vert #1 \right\vert}
\newcommand\rme{\mathrm e}
\newcommand\imu{\mathrm i}
\newcommand\diff{\,\mathrm d}
\DeclareMathOperator\sgn{sgn}
\renewcommand \epsilon \varepsilon
\newcommand\trans{^{\mathsf T}}
\newcommand\F {\mathbb F}
\newcommand\Z{\mathbb Z}
\newcommand\R{\Bbb R}
\newcommand \N {\Bbb N}
\renewcommand\geq\geqslant
\renewcommand\leq\leqslant
\newcommand\bm\boldsymbol
\newcommand\stpf\blacktriangleleft
\newcommand\qed\blacktriangleright
\newcommand\upint[2][a]{\bar {\phantom \int} \mspace{-21mu}{\int_{#1}^{#2}}}
\frac {n \abs f(x)} {1 + n^2 x^2} \leq \frac {nM} {1 + n^2 n^{-1/3 \times 2}} = \frac {nM} {1 + n^{4/3}} = \frac {nM} {1 + n^{1 + 1/3}},
$$
where $x \in [n^{-1/3}, 1]$. Now integrate over this interval,
$$
\int_{n^{-1/3}}^ 1 \frac {n \abs f(x)} {1 + n^2 x^2} \diff x\leqslant \int_{n^{-1/3}}^1 \frac {nM} {1 + n^{1 +1/3}} \diff x = (1 - n^{-1/3}) \frac {nM} {1 + n^{1+ 1/3}} \color{red}{\leq} 1 \cdot \frac {nM} {1 + n^{1+ 1/3}},
$$
where $\color{red}\leq$ comes from $(1 - n^{-1/3}) \leq 1$.
For the solution, $n^{-1/3}$ seems tricky. So we try another explanation, which unfortunately may require some knowledge about superior/inferior limits.
At the first glimpse, we may want to take limit under $\int$, but in general we can't. But intuitively, we might feel that for $x$ close enough to $1$, the $n^2$ part in the denominator would dominate, since $f$ is bounded, and $n f$ is of course "weaker" than $n^2$. The problematic point is $0$. So we could pick any temporarily fixed $\delta > 0$ and break the interval into two parts. One part could be estimated like this:
\begin{align*}
&\quad \abs {\int_\delta^1 \frac {nf (x) \diff x }{1 + n^2 x^2} } \\
&\leq \int_\delta ^1 \frac {n \abs f (x)}{ 1+ n^2x^2 }\diff x \\
&\leq \int_\delta^1 \frac {nM}{1 + n^2 \delta^2} \\
&= (1 - \delta) \frac {nM} {1 + n^2 \delta^2}\leq \frac {nM}{1 + n^2 \delta^2}.
\end{align*}
For the other, $[0, \delta]$, we use the continuity, and note that
$$
\abs {\int_0^{\delta} \frac {n (f(x) - f(0))}{1 + n^2 x^2} \diff x} \leq \max_{0 \leq x \leq \delta} \abs {f(x) - f(0)}\int_0^\delta \frac {\diff (nx)}{ 1 + (nx)^2} =: N(\delta) \arctan (n \delta).
$$
If the limit of both part exists, then
$$
\lim_{n \to \infty} \abs {\int_\delta^1 \frac {nf (x) \diff x }{1 + n^2 x^2} } \leq \lim_{n \to \infty} \frac {nM}{1 + n^2 \delta^2} = 0,
$$
and
$$
\lim_{n \to \infty} \abs {\int_0^{\delta} \frac {n (f(x) - f(0))}{1 + n^2 x^2} \diff x} \leq \lim_{n \to \infty} N(\delta) \arctan (n \delta) = \frac \pi 2 N(\delta),
$$
and if we let $\delta \to 0^+$, then we can expect that the original limit to be $\lim_n \int_0^1 n f(0)\diff x /(1 +n^2 x^2) = \pi f(0)/2 $, since $N(\delta) \to 0$ according to the continuity of $f$ at $0$. To make these argument work, we shall find some $\delta(n)$ that variates as $n \to \infty$ such that
*
*$\delta (n) \to 0$ as $n \to \infty$;
*$1 + n ^2 \delta (n)^2$ "dominates" as $n \to \infty$, i.e. $nM / (1 + n ^2 \delta(n)^2)\to 0$.
Easy to see a qualified candidate could be $\bm {n^{-1/3}}$. Hence the solution works. Of course we could pick other forms, but for the efficiency, we might pick simple ones.
In fact we could write the following using superior limits:
\begin{align*}
&\quad \varlimsup_n \abs {\int_0^1 \frac {n (f(x) - f(0))}{1 + n^2 x^2} \diff x }\\
&\leq \varlimsup_n \int_0^\delta \abs {\frac {n (f(x) - f(0))}{1 + n^2 x^2} \diff x}+ \varlimsup_n \int_\delta^1 \abs { \frac {n (f(x) - f(0))}{1 + n^2 x^2} \diff x}\\
&\leq \varlimsup_n \frac{n \cdot 2M} {1 + n^2 \delta^2} + \varlimsup_n N(\delta) \arctan (n \delta) \\
&= 0 + \frac {\pi}2 N(\delta)\\
&\xrightarrow {\delta\to 0^+} 0,
\end{align*}
which is a standard proof of the limit.
|
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|
Recurring Sequence with Exponent Working with recurring sequences and generating functions, I'm generally lost on solving a general expression of $a_n$ for any $n$ when the next part of the sequence, that is $a_{n+1}$, is in the form of an exponent, such that $a_n = a_{n-1} +k^{n-1}$, where k is some constant. I have no clue on how to approach this problem.
I've solved the Fibonacci sequence by subtracting the two pervious terms and shifting the sequence, but it does not seem to work here.
I'm particularly working with $a_n = 2a_{n-1} + 5^{n-1}$, but the sequence expands extremely fast. The base case, $a_{0} = 1$.
Any help would be appreciated!
|
I love to telescope.
If
$a_n = ua_{n-1} + vc^{n}
$,
then
$\dfrac{a_n}{u^n}
= \dfrac{a_{n-1}}{u^{n-1}} + v(c/u)^{n}
$.
Let
$b_n = \dfrac{a_n}{u^n}$.
Then
$b_n
=b_{n-1}+vd^n
$
where
$d = c/u$.
Then
$b_n-b_{n-1}
=vd^n
$.
Summing,
$\begin{array}\\
b_m-b_0
&=\sum_{n=1}^m (b_n-b_{n-1})\\
&=\sum_{n=1}^m vd^n\\
&=v\dfrac{d-d^{m+1}}{1-d}\\
&=vd\dfrac{1-d^{m}}{1-d}\\
\end{array}
$
so
$\begin{array}\\
\dfrac{a_m}{u^m}
&=a_0+vd\dfrac{1-d^m}{1-d}\\
\text{or}\\
a_m
&=a_0u^m+\dfrac{vc}{u}u^m\dfrac{1-(c/u)^m}{1-c/u}\\
&=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\
&=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\
\end{array}
$
In this case,
$u=2, c=5, v = \frac15, a_0 = 1
$
so
$a_m
= 2^m + \dfrac{2^m-5^m}{2-5}
= 2^m + \dfrac{5^m-2^m}{3}
$.
This can be rewritten as
$\begin{array}\\
a_m
&=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\
&=\dfrac{(u-c)a_0u^m+vc(u^m-c^m)}{u-c}\\
&=\dfrac{(a_0(u-c)+vc)u^m-vc^{m+1}}{u-c}\\
\end{array}
$
Again, we get
$=\dfrac{(a_0(u-c)+vc)u^m-vc^{m+1}}{u-c}
=\dfrac{(-3+1)2^m-5^{m}}{-3}
=\dfrac{2\cdot 2^m+5^{m}}{3}
$.
|
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|
$16$ people, $8$ men, $8$ women, divide the group to $8$ couples, what is the chance for exactly $3$ male couples? We have a group of $16$ people, $8$ men, $8$ women.
We divide them to $8$ couples.
Let $X$ Be the number of couples make of $2$ men.
Calculate: $P(X = 3)$
I am not sure how to approach this.
I thought that this has an hypergeometric distribution.
Therefore i chose:
$$
N = 16, n = 16, K = 8, k = 6
$$
For:
$$
\frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}
$$
But its wont gonna work.
Another way i thought is to choose the couples like this:
$$
\frac{\binom{8}{2} \binom{6}{2} \binom{4}{2} \binom{8}{1} \binom{7}{1} \binom{6}{2} \binom{4}{2}}{\frac{16!}{2^8}}
$$
The logic wass:
But again its not working.
I would like some help
|
Complicated problem.
Answer will be
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$
For the denominator, not only do you have to consider how many ways there are of choosing the 1st couple, then the 2nd, ...
You then have to adjust for over-counting, re each grouping into 8 couples will be counted $8!$ ways.
Therefore,
$$D = \frac{\binom{16}{2}\binom{14}{2}\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}}{8!}.$$
For the numerator, first you have to create exactly 3 male couples.
So
$$N_1 = \frac{\binom{8}{2}\binom{6}{2}\binom{4}{2}}{3!}.$$
Then, you have to select among 8 women for the 1st odd man, and then among 7 women for the 2nd odd man.
$$N_2 = 8 \times 7.$$
Then, the 6 remaining women must pair up.
$$N_3 = \frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}.$$
Final answer is
$$\frac{N_1 \times N_2 \times N_3}{D}.$$
|
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|
Which matrix belongs to the stabiliser of a vector? We consider the action of $GL_2(\mathbb{R})$ on the plane $\mathbb{R}^2$ by linear transformation. Which of these matrices belong to the stabiliser of the vector $\begin{pmatrix} 1\\3\end{pmatrix}$?
|
$\begin{pmatrix} -1 & 1 \\ 4 & -1 \end{pmatrix}$
B: $\begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix}$
C: $\begin{pmatrix} 1 & 0 \\ -3 & 2 \end{pmatrix}$
D: $\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$
What I have tried so far:
Given a group action of a group $G$ on a set $X$. For $x \in X$, the stabiliser of $x$, denoted $\text{Stab}(x)$ is such that $g \cdot x=x $. Hence I think it is D Since $\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1\\3\end{pmatrix}= \begin{pmatrix} 1\\3\end{pmatrix}$
Is this method correct?
|
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|
How to evaluate $\sum _{n=1}^{\infty }\left(\frac{H_n^2+H_n^{\left(2\right)}}{n}\right)^2$ in a particular way. How to evaluate:
$$\sum _{n=1}^{\infty }\left(\frac{H_n^2+H_n^{\left(2\right)}}{n}\right)^2,$$
without splitting the expression into more sums.
Here $H_n^{\left(m\right)}=\sum _{k=1}^n\frac{1}{k^m}$ is the harmonic number of order $m$.
If one just wants to evaluate it if we split we have,
$$2\sum _{n=1}^{\infty }\frac{H_n^2H_n^{\left(2\right)}}{n^2}+\sum _{n=1}^{\infty }\frac{H_n^4}{n^2}+\sum _{n=1}^{\infty }\frac{\left(H_n^{\left(2\right)}\right)^2}{n^2},$$
Then making use of this results one only has to compute
$$\sum _{n=1}^{\infty }\frac{\left(H_n^{\left(2\right)}\right)^2}{n^2}$$
But I'd like to know if its possible to evaluate the series without splitting or expanding the terms.
|
To find the desired series we must first consider the following integral.
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx$$
To evaluate it one can make use of the following trilogarithm identity.
$$\operatorname{Li}_3\left(\frac{x}{x-1}\right)=-\operatorname{Li}_3\left(x\right)-\operatorname{Li}_3\left(1-x\right)+\zeta \left(3\right)+\frac{1}{6}\ln ^3\left(1-x\right)$$
$$+\zeta \left(2\right)\ln \left(1-x\right)-\frac{1}{2}\ln \left(x\right)\ln ^2\left(1-x\right)$$
Using it on the previous integral yields:
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(x\right)}{x}\:dx-\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(1-x\right)}{x}\:dx$$
$$+\zeta \left(3\right)\int _0^1\frac{\ln ^2\left(1-x\right)}{x}\:dx+\frac{1}{6}\int _0^1\frac{\ln ^5\left(1-x\right)}{x}\:dx+\zeta \left(2\right)\int _0^1\frac{\ln ^3\left(1-x\right)}{x}\:dx$$
$$-\frac{1}{2}\int _0^1\frac{\ln \left(x\right)\ln ^4\left(1-x\right)}{x}\:dx$$
$$=-\frac{81}{2}\zeta \left(6\right)+2\zeta ^2\left(3\right)+12\sum _{k=1}^{\infty }\frac{H_k}{k^5}-\sum _{k=1}^{\infty }\frac{H_k^2}{k^4}-\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{k^4}-2\sum _{k=1}^{\infty }\frac{H_k^{\left(3\right)}}{k^3}$$
The series remaining can be calculated quite easily and a nice thing to know is that to evaluate them one does not have to cross paths with the series in the body of the OP.
Thus:
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\frac{581}{24}\zeta \left(6\right)-4\zeta ^2\left(3\right)$$
Now what is left to do is to consider the following generating function.
$$\sum _{k=1}^{\infty }\frac{x^{k-1}}{k}\left(H_k^2+H_k^{\left(2\right)}\right)=-2\frac{\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}$$
Which can be found along other generating functions in the book (Almost) Impossible Integrals, Sums, and Series, page $\#285$.
Using it on the previously found integral means that,
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\frac{1}{2}\sum _{k=1}^{\infty }\left(\frac{H_k^2+H_k^{\left(2\right)}}{k}\right)^2$$
Thus:
$$\sum _{k=1}^{\infty }\left(\frac{H_k^2+H_k^{\left(2\right)}}{k}\right)^2=\frac{581}{12}\zeta \left(6\right)+8\zeta ^2\left(3\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$.........? Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$ are both divisible by $2^{n+1}$. Is this the highest power of $2$ dividing either of the numbers?
I am not well versed with binomial theorem hence I am not able to proceed. I am also interested in the different ways this problem can be solved.
Any help would be appreciated.
|
Let $a_n = (1+\sqrt{3})^n + (1 - \sqrt{3})^n$ for $n \ge 0$.
Since $1 \pm \sqrt{3}$ are roots of $(\lambda - 1)^2 - 3 = \lambda^2 - 2\lambda - 2$, $a_n$ satisfies a recurrence relation:
$$a_{n+2} = 2(a_{n+1} + a_n)$$
Define two auxiliary sequences $b_n, c_n$ by
$\begin{cases}
b_n &= 2^{-(n+1)} a_{2n}\\
c_n &= 2^{-(n+1)} a_{2n+1}
\end{cases}$, we have
$$\begin{align}
b_{n} &= 2^{-(n+1)}a_{2n} = 2^{-n}(a_{2n-1} + a_{2n-2}) = c_{n-1} + b_{n-1}\\
c_{n} &= 2^{-(n+1)}a_{2n+1} = 2^{-n}(a_{2n} + a_{2n-1}) = 2b_n + c_{n-1} = 2b_{n-1} + 3c_{n-1}
\end{align}$$
or in matrix form:
$$\begin{bmatrix}b_n \\ c_n\end{bmatrix} =
\begin{bmatrix} 1 & 1\\ 2 & 3\end{bmatrix} \begin{bmatrix}b_{n-1} \\ c_{n-1}\end{bmatrix}$$
Since $a_0 = a_1 = 2 \implies b_0 = c_0 = 1$, this leads to following representation of $(a_{2n}, a_{2n+1})$:
$$\begin{bmatrix}a_{2n} \\ a_{2n+1}\end{bmatrix}
= 2^{n+1}\begin{bmatrix}b_n \\ c_n\end{bmatrix}
= 2^{n+1}\begin{bmatrix} 1 & 1\\ 2 & 3\end{bmatrix}^n \begin{bmatrix}1 \\ 1\end{bmatrix}$$
From above expression, it is easy to see $(b_n, c_n)$ are integers. This implies $2^{n+1}$ divides both $a_{2n}$ and $a_{2n+1}$.
In fact, since $c_0 = 1$ is odd and
$$c_{n} = 2b_{n-1} + 3c_{n-1} \equiv c_{n-1} \pmod 2$$
all $c_n$ are odd. From this, we can deduce $2^{n+1}$ is the highest factor of $2$ that divides both $a_{2n}$ and $a_{2n+1}$.
|
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|
System of congruences with moduli not coprime and chinese remainder theorem application Find all solutions, if any, to the system of congruences
$$\Bigg\{
\begin{array}{c}
x \equiv 1 (\mod 6)\\
x \equiv 7 (\mod 9)\\
x \equiv 4 (\mod 15)\\
\end{array}$$
we can see that
$ x \equiv 1 \pmod 6 \implies
\left\{
\begin{array}{l}
x\equiv 1 \pmod 2 \\
x \equiv 1 \pmod 3
\end{array}
\right.$
as well as
$ x \equiv 7 \pmod 9\\
\qquad \Rightarrow \quad x=7+9k\\
\qquad \Rightarrow \quad x=1+3(2+3k)\\
\qquad \Rightarrow \quad x\equiv 1 \pmod 3\\
$
and
$ x \equiv 4 \pmod{15} \implies
\left\{
\begin{array}{l}
x\equiv 4 \pmod 5 \\
x \equiv 1 \pmod 3
\end{array}
\right.
$
After removal of redundant congruences and picking the one with the highest power of 3 we are left with:
$
\Bigg\{
\begin{array}{c}
x \equiv 1 \pmod 2\\
x \equiv 7 \pmod {3^2}\\
x \equiv 4 \pmod 5\\
\end{array}
$
Now we reduced it to standard CRT problem.
$ x=2k+1 \\
2k+1 \equiv 7\pmod{9}\\
2k \equiv 6\pmod{9}\quad \gcd{(2,9)}=1\\
k\equiv 3\pmod{9}\\
k=9\cdot l+3$
$x=2\cdot(9\cdot l+3)+1=18l+7$
$
18l+7\equiv 4\pmod{5}\\
18l\equiv -3\pmod{5}\\
18l\equiv 2\pmod{5}\\
3l\equiv 2\pmod{5}\\
\quad \Rightarrow \quad 7\pmod{5}\\
\quad \Rightarrow \quad 12\pmod{5} \quad \gcd{(3,5)}=1\\
l\equiv 4\pmod{5}\\
l=5\cdot m +4\\
$
The solution to the original system:
$
x=2\cdot(9\cdot (5\cdot m +4)+3)+1=90\cdot m+79\\
x\equiv 79\pmod{90}
$
with the smallest x = 79
Is it a correct approach?
|
It is correct, but you can make the solution a bit shorter using the explicit inverse isomorphism in the C.R.T. for two coprime moduli $a$ and $b$, given a Bézout's relation $ua+vb=1$:
$$x\equiv \begin{cases}\alpha\mod a,\\\beta\mod b,\end{cases}\iff x\equiv \beta ua+\alpha vb\mod ab$$
As $4\cdot 3^2-7\cdot 5=1$, the last two final congruences yield as a first step $$x\equiv 4\cdot 4\cdot 9-7\cdot 7\cdot 5=144-245=-101\mod 45.$$
Next, from $45-22\cdot 2=1$, you deduce from this congruence and the first above that
$$x\equiv 45+101\cdot 22\cdot 2=4489\equiv 79\mod 90.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\sum_{k=2}^\infty \frac{1}{1+(-1)^k \sqrt k}$ converges or diverges? $\frac{1}{1+\sqrt k (-1)^k}= \frac{1-(-1)^k \sqrt k}{1-k}$. Not quite sure how to move on from here.
|
Rewrite
\begin{align}\sum_{k=2}^{2N+1}\frac{1}{1+(-1)^k \sqrt k}&= \sum_{k=1}^N\left(\frac{1}{1+(-1)^{2k} \sqrt {2k}}+\frac{1}{1+(-1)^{2k+1} \sqrt {2k+1}}\right)\\&=\sum_{k=1}^N\left(\frac{1}{1+\sqrt {2k}}+\frac{1}{1- \sqrt {2k+1}}\right)\\&=\sum_{k=1}^N\left(\frac{1- \sqrt {2k+1}+1+\sqrt {2k}}{(1+\sqrt {2k})(1- \sqrt {2k+1})}\right)
\\&<\sum_{k=1}^N\left(\frac{2}{(1+\sqrt {2k+1})(1- \sqrt {2k+1})}\right)
\\&=\sum_{k=1}^N\left(\frac{2}{-2k}\right)\to-\infty
\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 !
|
Let $\cos^{-1}\dfrac35=y,\cos y=?$
Using Principal values, as $1>\dfrac35>0; 0<y<\dfrac\pi2$
$\tan y=\dfrac{+\sqrt{1-\left(\dfrac35\right)^2}}{\dfrac35}=?, y=\tan^{-1}\dfrac43$
$$\pi-2\tan^{-1}2=2\left(\dfrac\pi2-\tan^{-1}2\right)=2\cot^{-1}2$$
Use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?
and
Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
|
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|
How to prove that $n!>2^n$ holds for all $n>3$? I suspect something with sets since one with cardinality $n$ has $n!$ permutations and its powerset contains $2^n$ elements. It could also involve binomial coefficients because of$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{pmatrix}+\cdot\cdot\cdot+\begin{pmatrix}n\\n\end{pmatrix}=2^n$$
|
Since $n\ge 4$ we have $n! = 1\cdot 2\cdot 3\cdot 4 \cdot 5\cdots n = 24 \cdot 5 \cdots n > 16 \cdot 5 \cdots n\ge 2^4\cdot 2 \cdots 2 = 2^42^{n-4}=2^n$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Number theory in sequence $x_{n+1}=x_n^3-2x_n^2+2$ $x_1=5, x_{n+1}=x_n^3-2x_n^2+2$
Prove that, there is no prime $p=4k+3(k>1)$ and $p\mid x_n^2-3x_n+3$
I think I can have $p\mid t^2+1$ and then I have QED.
But $p\mid x_n^2-3x_n+3$ means that $p\mid (2x_n-3)^2+3=t^2+3$
What should I do next?
|
I found this to be a quite interesting (and challenging to solve) question. It's asking about the primes $p$ where
$$p \mid x_n^2 - 3x_n + 3 \tag{1}\label{eq1A}$$
As you showed, multiplying by $4$ gives
$$p \mid 4x_n^2 - 12x_n + 12 = (2x_n - 3)^2 + 3 \tag{2}\label{eq2A}$$
Since $p \neq 3$ (note $x_n \equiv 5 \pmod{72}$ for all $n \ge 1$, with this giving $x_n^2 - 3x_n + 3 \equiv 13 \pmod{72}$), this shows $-3$ is a quadratic residue modulo $p$.
Next, multiplying \eqref{eq1A} by $x_n - 3$ gives
$$\begin{equation}\begin{aligned}
(x_n - 3)(x_n^2 - 3x_n + 3) & = x_n^3 - 3x_n^2 + 3x_n - 3x_n^2 + 9x_n - 9 \\
& = x_n^3 - 6x_n^2 + 12x_n - 8 - 1 \\
& = (x_n - 2)^3 - 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This means
$$(x_n - 2)^3 \equiv 1 \pmod{p} \tag{4}\label{eq4A}$$
If $n \gt 1$, substituting $x_n = x_{n-1}^3 - 2x_{n-1}^2 + 2$ gives
$$\begin{equation}\begin{aligned}
(x_{n-1}^3 - 2x_{n-1}^2)^3 & \equiv (x_{n-1}^2(x_{n-1} - 2))^3 \\
& \equiv x_{n-1}^6(x_{n-1} - 2)^3 \\
& \equiv 1 \pmod{p}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Repeating the substitutions for each $x_i - 2 = x_{i-1}^2(x_{i-1} - 2)$ for $i$ from $n - 1$ down to $2$ gives a product of $0$ or more $x_{i-1}^6$ values multiplied by $(x_1 - 2)^3 = 3^3$, i.e.,
$$\left(\prod_{i=1}^{n-1}x_i\right)^6(3^3) \equiv 1 \pmod{p} \tag{6}\label{eq6A}$$
Multiplying both sides by $3$ gives
$$\left(\left(\prod_{i=1}^{n-1}x_i\right)^3 3^2\right)^2 \equiv 3 \pmod{p} \tag{7}\label{eq7A}$$
This shows $3$ is also a quadratic residue. Thus, $3^{-1}(-3) \equiv -1 \pmod{p}$ is a quadratic residue, so there's an integer $x$ with $x^2 \equiv -1 \pmod{p} \implies x^4 \equiv 1 \pmod{p}$. Therefore, $4$ is the multiplicative order of $x$ modulo $p$, so $4 \mid p -1 \implies p \equiv 1 \pmod{4}$. This proves that $p$ cannot be of the form $4k + 3$.
|
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|
Typesetting Multiline Inequalities If we write
$$\begin{align}
a&<b\\
&=c\end{align}$$
does this mean
$$a<b=c$$
or
$$a<b; a=c$$
?
How do we write the other in multiline format?
(Edited to Add)
What about
$$\begin{align}
a&<b\\
&=c\\
&=d\\
&=f\end{align}$$
?
|
Generally,
\begin{align*}
a &< b \\
&= c
\end{align*}
is to be interpreted as $a < b = c$. This can be convenient for a series of derivations or inequalities, when putting them on the same line could be ugly or hard to follow. To pull an example from a homework of mine:
\begin{align*}
f(z)
&= \frac{1}{(z-ia)^2} \frac{1}{(2ia)^2} \frac{1}{(z+ia)^2} \\
&= \frac{1}{(z-ia)^2} \frac{1}{(2ia)^2} \ \sum_{k=0}^\infty \binom{2+k-1}{k} \left( - \frac{z-ia}{2ia} \right)^k \\
&= \frac{1}{(z-ia)^2} \frac{1}{(2ia)^2} \sum_{k=0}^\infty \frac{(-1)^k (k+1)}{(2ia)^k} (z-ia)^k \\
&= \sum_{k=0}^\infty \frac{(-1)^k (k+1)}{(2ia)^{k+2}} (z-ia)^{k-2} \\
&= \sum_{n=-2}^\infty \frac{(-1)^n (n+3)}{(2ia)^{n+4}} (z-ia)^{n}
\end{align*}
(Usually I accompany these with minor notes on the right side, or number the equal signs and then explain the equalities afterwards, it simply depends. The former looks ugly on MSE most of the time so I left it out.)
If you want to write $a < b; a = c$, then you could just use
\begin{align*}
a &< b \\
a &= c
\end{align*}
which yields
\begin{align*}
a &< b \\
a &= c
\end{align*}
Perhaps throw a curly brace on the left-hand side to indicate they should considered together or a group?
|
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|
Limit of $\sqrt[10]{n^{10} + 8n^9} - n$ as $n \rightarrow \infty$ using two "standard" limits I must use the two standard limits $\lim\limits_{n \rightarrow \infty} \frac{e^{\alpha_n}-1}{\alpha_n} = 1$ and $\lim\limits_{n \rightarrow \infty} \frac{\ln(1+\beta_n)}{\beta_n} = 1$ if $\alpha_n, \beta_n \rightarrow 0$ so multiplying by conjugate approach won't work.
My attempt:
$\lim\limits_{n \rightarrow \infty} (\sqrt[10]{n^{10} + 8n^9} - n) = \lim\limits_{n \rightarrow \infty} e^{\ln(\sqrt[10]{n^{10} + 8n^9} - n)} = e^{\lim\limits_{n \rightarrow \infty} (\ln(\sqrt[10]{n^{10} + 8n^9} - n))}$
I am not sure how to proceed further. Any help/hints will be appreciated.
|
$$(\sqrt[10]{n^{10} + 8n^9} - n)=n\left(\sqrt[10]{1+\frac{8}{n}}-1\right) $$
$$\sqrt[10]{1+\frac{8}{n}}=\sum_{k=0}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}=1+\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}$$
$$n\left(\sqrt[10]{1+\frac{8}{n}}-1\right)=\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^{k-1}}=\frac{4}{5}-\frac{72}{25 n}+\cdots $$
|
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|
show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $ We want to show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $
I've tried to use AM-GM directly on the LHS but that obviously failed.
Simplifying the question gives $a^2b + b^2c + c^2a \geq ab + bc + ac$
The fact that this is $a(ab) + b(bc) + c(ca) \geq ab + bc + ac$ makes me think that there's a way to proceed from here, but I'm not quite sure how.
Note : a,b,c are real positive numbers
|
Case 1: Let $ab+bc+ca\ge a+b+c$, then the AM-HM of $a,b,c$ with weights as $1/b,1/c,1/a$ gives
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge \frac{1/a+1/b+1/c)^2}{1/(ab)+1/(bc)+1/ca)}=\frac{(ab+bc+ca)(1/a+1/b+1/c)}{a+b+c} \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Case 2: Let $(ab+bc+ca) \le a+b+c$, then AM-HM of $1/b,1/c,1/a$ with weights as $a,b,c$, we get
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{(a+b+c)^2}{ab+bc+ca} \ge ab+bc+ca=\frac{1}{b}+\frac{1}{c}+\frac{1}{a}$$
|
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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.
|
Its wrong ! the step $$A(x+4)+B(x-1) = x$$ is not true instead you should have $$Ax(x+4)+B(x-1)=x$$ As you can see we would then have to take $A=0$ and the purpose of partial fractions is gone
I prefer $$\frac{x}{x^2-3x+4}=\frac{1}{5}(\frac{1}{x-1})+\frac{4}{5}(\frac{1}{x+4})$$
another way is to write $x=l(2x-3)+m$ (because derivative of $x^2-3x+4=2x-3$) then we have $$\int \frac{x}{x^2-3x+4}=\int \frac{2x-3}{2(x^2-3x+4)}+\int \frac{3}{2(x^2-3x+4)}$$ which is easy
|
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|
Proving $ \sum_{cyc} \frac{1}{a^6 + b^6 + 3c^3 + 4} \leq \frac{3}{3 + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})}$ Problem proposed for JBMO practice in symmetrical inequalities (Chebyshev, rearrangement):
For every positive real numbers $a, b, c$, for which $a + b + c = 3$ we have:
$$\sum_{cyc} \frac{1}{a^6 + b^6 + 3c^3 + 4} \leq \frac{3}{3 + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})}$$
My attempt:
Step 1
AM-GM states that $$a^6 + b^6 \geq 2a^3b^3$$
Therefore, $$ LHS \leq \sum_{cyc} \frac{1}{2a^3b^3 + 2c^3 + 2 + c^3 + 2}$$
Step 2
AM-GM states that $2a^3b^3 + 2c^3 + 2 \geq 6abc$ and that $c^3 + 1 + 1 \geq 3c$
This means that $$ LHS \leq \sum_{cyc} \frac{1}{6abc + 3a} = \frac{1}{3} \sum_{cyc} \frac{1}{2abc + a}$$
Step 3
Rearrangement inequality states that $$\sqrt{ab} + \sqrt{bc} + \sqrt{ca} \leq a + b + c = 3$$
This means: $$ RHS \geq \frac{1}{3}$$
Remaining of the proof and further ideas
We should prove $$ \sum_{cyc} \frac{1}{2abc + a} \leq 1$$
I should probably apply Chebyshev but I could get no good result after applying its variants...
Thanks in advance!
|
By AM-GM, we have $\sqrt{ab} + \sqrt{bc} + \sqrt{ca} \le a + b + c = 3$. Thus,
$\mathrm{RHS} \ge \frac{1}{3}$.
By Chebyshev's sum inequality, we have
\begin{align}
a^3 + b^3 + c^3 - (a+b+c) &= (a-1)(a^2 + a) + (b-1)(b^2+b) + (c-1)(c^2+c)\\
&\ge \frac{1}{3}(a-1 + b-1 + c-1)(a^2+a + b^2+b + c^2 + c)\\
& = 0.
\end{align}
Thus, we have $a^3 + b^3 + c^3 \ge a + b + c = 3$. Thus, we have
\begin{align}
a^6 + b^6 + 3c^3 + 4 &= (a^6 + 1) + (b^6 + 1) + 3c^3 + 2\\
&\ge 2a^3 + 2b^3 + 3c^3 + 2\\
&= 2(a^3 + b^3 + c^3) + c^3 + 2\\
&\ge 8 + c^3.
\end{align}
Thus, $\mathrm{LHS} \le \sum_{\mathrm{cyc}} \frac{1}{8 + c^3}$.
It suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{1}{8 + c^3} \le \frac{1}{3}$$
or (tangent line method)
$$\sum_{\mathrm{cyc}} \left(\frac{1}{8 + c^3} - \frac{1}{9} + \frac{1}{27}(c-1) \right) \le 0$$
or
$$\sum_{\mathrm{cyc}} \frac{(c^2-2c-5)(c-1)^2}{27(8 + c^3)} \le 0$$
which is true since $x^2-2x-5 < 0$ for all $0 \le x \le 3$.
|
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|
Solving $(\sin x-\cos x)^2+\tan x=2\sin^2x$ From this post (Solving $\frac{\cos^23t}{\tan t}+\frac{\cos^2t}{\tan3t}=0$), which is my post, I have tried solving the last equation. I have gotten a solution(s). Would like to see if I did correctly.
$$\bigl(\sin\left(x\right)-\cos\left(x\right)\bigr)^2+\tan\left(x\right)=2\sin^2\left(x\right)$$
Expanding:
$$sin^2\left(x\right)-2sin\left(x\right)\cos\left(x\right)+\cos^2\left(x\right)+tan\left(x\right)=2\sin^2\left(x\right)$$
Using $\sin^2\left(x\right)+\cos^2\left(x\right)=1$, and bringing the RHS to the LHS, the expansion becomes:
$$-2sin\left(x\right)\cos\left(x\right)-2\sin^2\left(x\right)+1+\tan\left(x\right)=0$$
Then using $\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$ and multiplying the LHS by cos(x), I get the following:
$$\frac{-2\sin\left(x\right)\cos^2\left(x\right)-2\sin^2\left(x\right)\cos\left(x\right)+\cos\left(x\right)+\sin\left(x\right)}{\cos\left(x\right)}=0$$
Now strictly dealing with the numerator, making two separate terms, one being $-2\sin\left(x\right)\cos^2\left(x\right)-2\sin^2\left(x\right)\cos\left(x\right)$ and the other being $\cos\left(x\right)+\sin\left(x\right)$, then factoring out $-2\sin(x)\cos(x)$, I get
$$\color{blue}{-2\sin\left(x\right)\cos\left(x\right)+1}\color{red}{\bigl(\cos\left(x\right)+\sin\left(x\right)\bigr)}+\color{red}{\bigl(\cos\left(x\right)+\sin\left(x\right)\bigr)}=0$$
This makes two terms
$$\biggl(\color{blue}{-2\sin\left(x\right)\cos\left(x\right)+1}\biggr)\biggl(\color{red}{\cos\left(x\right)+\sin\left(x\right)}\biggr)=0$$
For the first term, aka the red term, I got $$\tan(x)=-1$$
where $$x=\frac{3\pi}{4}+n\pi $$
and the second term, aka the blue term.
using the identity $\sin\left(2x\right)=2\sin\left(x\right)\cos\left(x\right)$
$$\sin(2x)=1$$
where $$x=\frac{\pi}{4}+n\pi$$
Thanks for reading!
|
Alternatively
\begin{align}
& (\sin x-\cos x)^2+\tan x-2\sin^2x\\
=& (1-\sin2x )+\tan x -(1-\cos 2x) \\
=&- \frac{2\tan x}{1+\tan^2 x}+ \tan x+\frac{1-\tan^2x}{1+\tan^2 x}\\
= &\frac1{1+\tan^2x}(\tan x-1)^2(\tan x+1)=0
\end{align}
Thus, $\tan x= \pm1$, leading to the solutions
$$x =\frac\pi4 +\frac{\pi n}2$$
|
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|
Prove $\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$ I have to prove that :
$$\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$$
I already know that the Maclaurin series of $\ln(1+x)$ is $x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5+\ldots$ But how can I use it to prove the first claim? Should I just multiply $\frac{1}{1+x}$ to the Maclaurin series of $\ln(1+x)$? Still, I don't know how to do that.
I tried to use the derivatives of $\frac{\ln(1+x)}{1+x}$ then substitute that into the Maclaurin formula. But I don't think that's the right way to do this efficiently.
|
Multiply the two series $\ln(1+x) = x - \frac{1}{2} x^2 + \frac{1}{3} x^3 - \cdots$ and $\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots$.
For instance, the coefficient of $x^2$ in the product is $(-\frac{1}{2} x^2) \cdot 1 + x \cdot (-x) = -(1 - \frac{1}{2}) x^2$.
|
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|
Are functions: $1$; $\cos(x)$; $\cos^2(\frac{x}{2})$ linearly independent? I used the definition.
$1$; $\cos(x)$; $\cos^2(\frac{x}{2})$
$c_1\cdot1+c_2\cdot\cos(x)+c_3\cdot\cos^2(\frac{x}{2}) = 0$
I tried converting $\cos^2(\frac{x}{2})$ into something better: $\frac{1+\cos(x)}{2}$
$c_1+c_2\cdot\cos(x)+c_3\cdot\frac{1+\cos(x)}{2} = 0$
$c_1+\cos(x)\cdot\left(c_2+c_3\cdot\frac{1+\cos(x)}{2\cos(x)}\right) = 0+0\cdot\cos(x)$
$c_1=0$
$c_2+c_3\cdot\frac{1+\cos(x)}{2\cos(x)}=0 \to c_2=-c_3\cdot\frac{1+\cos(x)}{2\cos(x)} \to \frac{c_2}{c_3}=\frac{-1-\cos(x)}{2\cos(x)}$
$c_2=-1-\cos(x)$ and $c_3=2\cos(x)$
There exist $c_i$-s outside of $c_i=0$ which can be solutions. So these functions are linearly dependent.
Is this correct?
Thanks.
|
Once you know that $$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2},$$ you already know that $$(1/2, 1/2, -1) \cdot (1, \cos x, \cos^2 \tfrac{x}{2}) = 0.$$
|
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|
matrix and linear transformations With regards to this (Matrix associated to a linear transformation with respect to a given basis)
I saw Ivo Terek solution which goes:
\begin{align}L(1,1,1) &= (4,6,6) = 6\cdot (1,1,1) + 0\cdot (1,1,0) -
2\cdot (1,0,0) \\ L(1,1,0) &= (0,8,2) = 2\cdot (1,1,1) + 6\cdot (1,1,0) -
8\cdot (1,0,0) \\ L(1,0,0) &= (0,3,1) = 1\cdot (1,1,1) + 2\cdot (1,1,0) -
3\cdot (1,0,0),\end{align}
However, I was not able to understand how:
$ 6\cdot (1,1,1) + 0\cdot (1,1,0) - 2\cdot (1,0,0) \\2\cdot (1,1,1) + 6\cdot (1,1,0) - 8\cdot (1,0,0) \\1\cdot (1,1,1) + 2\cdot (1,1,0) - 3\cdot (1,0,0) $
was derived. I only know half of where the equation came from.
|
I suppose you understand the first equality in
$$L(1,1,1) = (4,6,6) = 6\cdot (1,1,1) + 0\cdot (1,1,0) -
2\cdot (1,0,0)$$
well, as $L$ is defined by $L \left( \begin{bmatrix}x_{1}\\x_{2}\\x_{3} \end{bmatrix} \right) = \begin{bmatrix}4x_{3}\\3x_{1}+5x_{2}-2x_{3}\\x_{1}+x_{2}+4x_{3} \end{bmatrix}$ in the linked question.
Observe that for each vector $v$ expressed as a linear combination of basic vectors in the basis $B = \left(\begin{bmatrix}1\\1\\1 \end{bmatrix} , \begin{bmatrix}1\\1\\0 \end{bmatrix}, \begin{bmatrix}1\\0\\0 \end{bmatrix} \right)$, the first basic vector $(1,1,1)^T$ determines the third component of $v$, the second basic vector $(1,1,0)^T$ determines the second component of $v$.
In particular, you have $(4,6,6)^T$, so you need $6$ as the coefficient for $(1,1,1)^T$, as the basic vectors $(1,1,0)^T$ and $(1,0,0)^T$ can't contribute to the third component $z$ in the vector $(x,y,z)^T$. Use a similar reasoning to get $0$ for the coefficient of $(1,1,0)^T$. The rest is a simple arithmetic exercise.
|
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|
Surface integral over ellipsoid $ax^2+by^2+cz^2=1$ I'm not sure how to compute the integral
$$\int_{s} (a^2x^2+b^2y^2+c^2z^2)^{-1/2} d\vec{S}\cdot \vec{n}$$
over the surface of the ellipsoid
$$ax^2+by^2+cz^2=1$$, $$z>0$$
Where $\vec{n}$ is the unitary normal vector to the surface.
Any help is appreciated. Thank you.
|
We propose to attack the problem with a blunt instrument: on the surface
$$\vec{r}=\langle x,y,z\rangle=\left\langle x,y,\pm\sqrt{\frac{1-ax^2-by^2}{c}}\right\rangle$$
so
$$\begin{align}d\vec{r}&=\left\langle1,0,\mp\frac{\frac{ax}c}{\sqrt{\frac{1-ax^2-by^2}{c}}}\right\rangle\,dx+\left\langle0,1,\mp\frac{\frac{by}c}{\sqrt{\frac{1-ax^2-by^2}{c}}}\right\rangle\,dy\\
&=\left\langle1,0,-\frac{ax}{cz}\right\rangle\,dx+\left\langle0,1,-\frac{by}{cz}\right\rangle\,dy\end{align}$$
So that the vector areal element is
$$\begin{align}d^2\vec{A}&=\pm\left\langle1,0,-\frac{ax}{cz}\right\rangle\,dx\times\left\langle0,1,-\frac{by}{cz}\right\rangle\,dy\\
&=\pm\left\langle\frac{ax}{cz},\frac{by}{cz},1\right\rangle\,dx\,dy=\frac{\langle ax,by,cz\rangle}{c|z|}dx\,dy\end{align}$$
Then
$$\hat{n}=\frac{d^2\vec{A}}{\left\lVert d^2\vec{A}\right\rVert}=\frac{\langle ax,by,cz\rangle}{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}$$
$$\begin{align}I&={\large\bigcirc}\kern-1.55em\iint_S\frac1{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\frac{\langle ax,by,cz\rangle}{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\cdot d^2\vec{A}\\
&=2\int_{-\frac1{\sqrt a}}^{\frac1{\sqrt a}}\int_{-\sqrt{\frac{1-ax^2}b}}^{\sqrt{\frac{1-ax^2}b}}\frac1{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\frac{\langle ax,by,cz\rangle}{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\cdot\frac{\langle ax,by,cz\rangle}{c|z|}dy\,dx\\
&=2\int_{-\frac1{\sqrt a}}^{\frac1{\sqrt a}}\int_{-\sqrt{\frac{1-ax^2}b}}^{\sqrt{\frac{1-ax^2}b}}\frac1{\sqrt c\sqrt{1-ax^2-by^2}}\,dy\,dx=2\int_{-\frac1{\sqrt a}}^{\frac1{\sqrt a}}\int_{-\frac{\pi}2}^{\frac{\pi}2}\frac1{\sqrt{bc}}\,d\theta\,dx=\frac{4\pi}{\sqrt{abc}}\end{align}$$
Where we have used the substitution
$$y=\sqrt{\frac{1-ax^2}b}\sin\theta$$
to solve the $y$-integral. Note that if $a=b=c$ then $a^2x^2+b^2y^2+c^2z^2=a$ on the surface and $d^2\vec{A}\cdot\hat{n}=d^2A$ and the integral is over the surface of a sphere of radius $1/\sqrt a$ so we should get
$$I=\frac1{\sqrt a}\frac{4\pi}{a}$$
so this checks.
|
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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 $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.
|
My third solution:
Actually $a^2 + b^2 + c^2 + 6 - 3(a+b+c) + (abc - 1) \ge 0$ for all $a, b, c \ge 0$.
(pqr method)
Let $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$.
By Schur's inequality $a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) \ge 0$
which is written as $p^3 - 4pq + 9r \ge 0$, we have $\frac{p^3 + 9r}{4p}\ge q$.
We need to prove that $p^2 - 2q + 6 - 3p + r - 1 \ge 0$.
It suffices to prove that
$p^2 - 2 \cdot \frac{p^3 + 9r}{4p} + 6 - 3p + r - 1 \ge 0$ or
$$\frac{(2p-9)r}{2p} + \frac{1}{2}p^2 - 3p + 5 \ge 0. \tag{1}$$
If $2p- 9 \ge 0$, clearly (1) is true.
If $2p - 9 < 0$, since $\frac{p^3}{27} \ge r$, it suffices to prove that
$$\frac{(2p-9)\frac{p^3}{27}}{2p} + \frac{1}{2}p^2 - 3p + 5 \ge 0$$
that is
$$\frac{1}{27}(p+15)(p-3)^2 \ge 0.$$
We are done.
|
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|
Change of variable to show $\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$ Which change of variable can be used to show
$$\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$$
(direct integration isn't allowed, must use change of variable)
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$$I=\int_0^\infty\frac{1}{1+x^2}dx=\int_0^1\frac{1}{1+x^2}dx+\int_1^\infty\frac{1}{1+x^2}dx$$
now:
$$I_1=\int_1^\infty\frac{1}{1+x^2}dx$$
$$u=\frac1x,du=-1/x^2dx\Rightarrow dx=-\frac{1}{u^2}du$$
$$I_1=-\int_1^0\frac{1}{1+(1/u)^2}\frac{du}{u^2}=\int_0^1\frac{1}{1+u^2}$$
$$\therefore I=2\int_0^1\frac{1}{1+x^2}dx$$
|
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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,:>
|
Hint:
Let $7x-24y=z\iff x=?$
Replace the value of $x$ in $$x^2+y^2+Ax+By+C=0$$ to form a Quadratic Equation in $y$
As $y$ is real, the discriminant must be $\ge0$
|
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|
partial fractions when the fraction cannot be decomposed I am trying to find partial fractions of $\frac {1}{(x^2+1)^2}$. All the coefficients I get are zeros except the coefficient for the constant term which is 1, leaving me with the fraction I started with, so it seems like the fraction cannot be decomposed. How can I then go about writing this fraction as the sum $\frac {1}{2}\left [\frac {1}{x^2+1}-\frac{x^2-1}{(x^2+1)^2}\right]$? Is it just manipulation and trial and error?
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$\frac{1}{(x^2+1)^2}$ is already in "partial fractions form", so...
"Is it just manipulation and trial and error?"
Yes.
It's not super trivial, but the result is true.
I did:
\begin{align} \frac{1}{(x^2+1)^2}
&= \frac{1 + x^2 - x^2}{(x^2+1)^2} \\\\
&= \frac{1}{(x^2+1)}\ - \frac{x^2}{(x^2+1)^2} \\\\
&= \frac12\left[\frac{2}{(x^2+1)}\ - \frac{2x^2}{(x^2+1)^2}\right] \\\\
&= \frac12\left[\frac{1}{(x^2+1)} + \frac{1}{(x^2+1)}\ - \frac{2x^2}{(x^2+1)^2}\right]\\\\
&= \frac12\left[\frac{1}{(x^2+1)}\ + \frac{x^2+1-2x^2}{(x^2+1)^2}\right] \\\\
&= \frac12\left[\frac{1}{(x^2+1)}\ - \frac{x^2-1}{(x^2+1)^2}\right]. \\\\
\end{align}
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|
Prove that integral converges How can I prove that this integral converges?
$$\int\limits_1^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}\ dx$$
I tried to show that function $0 < \frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ is decreasing and is continious for $x$ in $(1,\infty)$. Furthermore the above integral converges when series $\sum_{1}^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ converges.
Are there better methods?
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First of all $ \frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}=\frac{\ln{x}}{\left(x-1\right)\sqrt{x+1}}\cdot\frac{1}{\sqrt{x-1}}\underset{x\to 1}{\sim}\frac{1}{\sqrt{2}\sqrt{x-1}} $, and we know that $ \int_{1}^{a}{\frac{\mathrm{d}x}{\sqrt{x-1}}} $ converges for any $ a>1 $.
We also have that $ \frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}=\underset{\overset{x\to +\infty}{}}{\scriptsize\mathcal{O}\normalsize}\left(\frac{\sqrt{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\right)=\underset{\overset{x\to +\infty}{}}{\scriptsize\mathcal{O}\normalsize}\left(\frac{1}{x\sqrt{x}}\right) $, and $ \int_{b}^{+\infty}{\frac{\mathrm{d}x}{x\sqrt{x}}} $ converges for any $ b>1 $.
Thus our integral $ \int_{1}^{+\infty}{\frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\,\mathrm{d}x} $ converges.
Let's use a substitution $ \left\lbrace\begin{matrix}x=\sec{y}\\ \sec{y}\,\mathrm{d}y=\frac{\mathrm{d}x}{\sqrt{x^{2}-1}}\ \ \ \ \ \ \ \end{matrix}\right. $, we get : \begin{aligned}\int_{1}^{+\infty}{\frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\,\mathrm{d}x}&=\int_{0}^{\frac{\pi}{2}}{\frac{\sec{y}\ln{\left(\sec{y}\right)}}{\sec{y}-1}\,\mathrm{d}y}\\ &=-\int_{0}^{\frac{\pi}{2}}{\frac{\ln{\left(\cos{y}\right)}}{1-\cos{y}}\,\mathrm{d}y}\\ &=\left[\left(\cot{\left(\frac{y}{2}\right)}-1\right)\ln{\left(\cos{y}\right)}\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}{\left(\cot{\left(\frac{x}{2}\right)}-1\right)\tan{x}\,\mathrm{d}x}\\ &=\int_{0}^{\frac{\pi}{2}}{\left(\cot{\left(\frac{x}{2}\right)}-1\right)\tan{x}\,\mathrm{d}x}\\ &=4\int_{0}^{\frac{\pi}{4}}{\frac{\cos{x}}{\sin{x}+\cos{x}}\,\mathrm{d}x}\\ &=4\int_{0}^{\frac{\pi}{4}}{\frac{\cos{\left(\frac{\pi}{4}-x\right)}}{\cos{\left(\frac{\pi}{4}-x\right)}+\sin{\left(\frac{\pi}{4}-x\right)}}\,\mathrm{d}y}\\ &=2\int_{0}^{\frac{\pi}{4}}{\left(1+\tan{x}\right)\mathrm{d}x}\\ &=2\left[x-\ln{\left(\cos{x}\right)}\right]_{0}^{\frac{\pi}{4}}\\ \int_{1}^{+\infty}{\frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\,\mathrm{d}x}&=\frac{\pi}{2}+\ln{2}\end{aligned}
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|
Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$
Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3)(x-4)=0$$
So, the answer is $\boxed{4}$.
Is correct my solution?
Can you show other ways for to solve this problem?
Can you suggest me any textbooks with similar problems?
Thank you so much!
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Edit
First see the comment of NinadMunshi immediately following this answer. I have edited the answer accordingly.
Continuing Infinity_hunter's answer, and letting
$$x_n~ \text{denote}~
\sqrt{12 + \sqrt{12 + \cdots \text{n times} }}
$$
To show convergence, all that is necessary is to show that
*
*the sequence is bounded
*the sequence is strictly increasing.
Clearly, $0 < x_1 < 4.$ Assume that $0 < x_n < 4.$ Then
$$x_{(n+1)} = \sqrt{12 + x_n} < \sqrt{12 + 4} < 4$$
and
$$x_{(n+1)} = \sqrt{12 + x_n} > \sqrt{12 + 0} > 0.$$
Therefore,
$$0 < x_n < 4 \implies 0 < x_{(n+1)} < 4.$$
To show that the sequence is (therefore) strictly increasing:
Since
$$ 0 < x_n < 4$$
and
$$[(x_n)^2 - x_n - 12] = (x_n - 4)(x_n + 3)$$
I conclude that
$$[(x_n)^2 - x_n - 12] < 0 \implies (x_n)^2 < x_n + 12.$$
However, by the definition of the sequence
$$\left[x_{(n+1)}\right]^2 = x_n + 12.$$
Therefore
$$\left[x_{(n+1)}\right]^2 > (x_n)^2.$$
Therefore, since each element in the sequence is positive,
$$x_{(n+1)} > (x_n).$$
Therefore, the sequence is strictly increasing.
|
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|
Evaluate the integral $\int_0^{\pi/2}\frac{\tan x}{\tan(\frac{x}{2})} \, dx$ Evaluate the integral.
$$\int_0^{\frac{\pi}{2}}\frac{\tan{x}}{\tan(\frac{x}{2})}\,dx$$
I tried to solve it with $u = \tan{x/2}$, but i got divergent part of the solution. How can I integrate it, such that, when boundaries are plugged in, the result will be able to be calculated.
|
Write $$\frac{\tan x}{\tan \frac{x}{2}} = \frac{\sin x}{\cos x} \frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} = \frac{2\sin\frac{x}{2} \cos\frac{x}{2}}{\cos x}\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} = \frac{2\cos^2\frac{x}{2}}{\cos x} = \frac{1 + \cos x}{\cos x}$$Then your integral reduces to $$\int_0^{\frac{\pi}{2}} (\sec x + 1)\, dx$$
|
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|
Can each root equation be rearranged to a corresponding polynomial equation? Let $n\in\mathbb{N}_+$.
Can each root equation of $n$ unknowns be rearranged to a polynomial equation of $n$ unknowns whose solution set contains the solution set of the root equation?
If not, is this true at least for root equations of one or two unknowns?
I already know the solution method with raising both sides of the root equation to the same power. But I guess this method is limited to simple root equations because raising to a power of a sum of at least three summands on one side of the equation does not reduce the number of roots on this side of the equation. Take e.g. the equation $\sqrt{A(x)}+\sqrt{B(x)}+\sqrt{C(x)}=\sqrt{D(x)}+\sqrt{E(x)}$.
I already know the method of introducing new unknowns. Does this method answer my question?
|
By irrational equation, I assume you meant one with one or more variable under a radical as assigned here. The technique of raising both sides to the same power may be invoked repeatedly if needed until there are no radicals. Take, for example.
$$A=\sqrt{x}+\sqrt{y}\implies A^2=\big(\sqrt{x}+\sqrt{y}\big)^2 = 2 \sqrt(x) \sqrt(y) + x + y$$
$$A^2 = 2 \sqrt(x) \sqrt(y) + x + y\implies A^2-x-y=2\sqrt{x}\sqrt{y}$$
$$\big(A^2-x-y\big)^2=\big(2\sqrt{x}\sqrt{y}\big)^2\implies
A^4 - 2 A^2 x - 2 A^2 y + x^2 + 2 x y + y^2 = 4 x y$$
$$A^4 - 2 A^2 x - 2 A^2 y + x^2 - 2 x y + y^2 = A^4-2A^2(x+y)+(x-y)^2=0$$
The last is an algebraic equation with no radicals an no loss of the "solutions" of the original. The example has been limited to two unknowns and of the same $[(1/2)]$ power. We can, to an extent, also mix the powers of the radicals as in:
$$A=\sqrt{x}+\sqrt[3]{y}\implies (\sqrt[3]{y})^3=\big(A-\sqrt{x}\big)^3
=A^3 - 3 A^2 (x)^{1/2} + 3 A x - x^{3/2}$$
$$\implies y = A^3-3Ax-\sqrt{x}(3A^2+x)$$
$$\big(\sqrt{x}(3A^2+x)\big)^2=\big(A^3-3Ax-y\big)^2$$
$$\implies 9 A^4 x + 6 A^2 x^2 + x^3=A^6 - 6 A^4 x - 2 A^3 y + 9 A^2 x^2 + 6 A x y + y^2$$
$$\implies A^6 - 15 A^4 x - 2 A^3 y + 3 A^2 x^2 + 6 A x y - x^3 + y^2 = 0$$
|
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|
Determinant of $n\times n$ Matrix Linear Algebra So, I have a matrix
$$
A = \begin{pmatrix}
0 & 1 & 1 & ... & 1 \\
1 & 0 & x & ... & x \\
1 & x & 0 & ... & x \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & x & x & ... & 0
\end{pmatrix}
$$
I need to evaluate it's determinant. At first I calculated $\det A$ for $n=2,3,4$. And I got a pattern $\det A=(-1)^{n-1}(n-1)x^{n-2}$ for every $n\ge2$. But I need to solve it differently. I added all rows to the first, then multiplied every row (except 1) by $(1+x(n-2))$ and subtracted first row multiplied by $x$. This is what I got:
$$
\begin{pmatrix}
n-1 & 1+x(n-2) & 1+x(n-2) & ... & 1+x(n-2) \\
1-x & -x(1+x(n-2)) & 0 & ... & 0 \\
1-x & 0 & -x(1+x(n-2)) & ... & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1-x & 0 & 0 & ... & -x(1+x(n-2))
\end{pmatrix}
$$
Now I can multiply diagonal elements, but I don't know what can I do with the rest of it. Any hints will be helful.
|
First factor out $1+x(n-2)$ from all columns except the first to get
$$\det A(x)=\begin{vmatrix}
n-1 & 1 & 1 & \cdots & 1 \\
1-x & -x & 0 & \cdots & 0 \\
1-x & 0 & -x & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1-x & 0 & 0 & \cdots & -x
\end{vmatrix}$$
and note that this cancels the multplication by $(1+x(n-2))^{n-1}$ from before. Now multiply the first column by $x$ and the rest of them by $(1-x)$ and then add them all to the first one:
$$\det A(x)=\frac1{x(1-x)^{n-1}}\begin{vmatrix}
n-1 & 1-x & 1-x & \cdots & 1-x \\
0 & -x(1-x) & 0 & \cdots & 0 \\
0 & 0 & -x(1-x) & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & -x(1-x)
\end{vmatrix}$$
which yields
$$\det A = \frac{(n-1) (-x(1-x))^{n-1}}{x(1-x)^{n-1}} = (-1)^n(n-1)x^{n-2}.$$
Note that this calculation was assuming that $x \notin \left\{0,1,-\frac1{n-2}\right\}$ so these cases should be considered separately. Alternatively, $\det A(x)$ is a continuous function in $x$ equal to $(-1)^n(n-1)x^{n-2}$ on the dense set $\Bbb{R}\setminus \left\{0,1,-\frac1{n-2}\right\}$ so it must be $\det A(x) = (-1)^n(n-1)x^{n-2}$ everywhere.
|
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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 $\frac{1}{a}+\frac{1}{b}=2?$
|
Here is a solution that does not involve derivatives.
By AM-GM, we have that $a+b \geq 2\sqrt{ab}$. Hence it suffices for us to prove that $2\sqrt{ab}+\dfrac{1}{1+\sqrt{ab}} \geq \dfrac{5}{2}.$
From the given condition, $\dfrac{1}{a} + \dfrac{1}{b} =2 \Rightarrow a+b=2ab \Rightarrow 2ab \geq 2\sqrt{ab} \Rightarrow \sqrt{ab} \geq 1$. Hence, letting $x=\sqrt{ab}, x \geq 1$, we have to prove that:
\begin{align}
& 2x+\dfrac{1}{1+x} \geq \dfrac{5}{2} \\
& \iff \dfrac{2x(1+x)+1}{1+x} \geq \dfrac{5}{2} \\
& \iff \dfrac{2x^2+2x+1}{1+x} \geq \dfrac{5}{2} \\
& \iff 2(2x^2+2x+1) \geq 5+5x \\
& \iff 4x^2-x \geq 3 \\
& \iff x(4x-1) \geq 3 \\
\end{align}
But the last inequality is obvious since $x \geq 1$ and $4x-1 \geq 4-1=3$; hence we are done.
|
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|
Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$ Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how to deal with the equation. Thank you in advance!
|
Denote $t:=\sqrt{45x^2-30x+1}$. Then we observe that
$$t^2=45x^2-30x+1=-5(7+6x-9x^2)+36.$$
As a result, it follows that $$t=-\frac{t^2-36}{5}\implies t^2+5t-36=0\implies(t+9)(t-4)=0.$$
Since $t\geq 0$, it follows that $t=4$. Therefore, $$45x^2-30x+1=16\implies45x^2-30x-15=0\implies 3x^2-2x-1=0.$$
Hence $(3x+1)(x-1)=0$. Either $x=1$ or $x=-1/3$.
|
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|
Summation of Cosine Series Find the summation of the series $\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} $
My approach is as follow
$\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} \Rightarrow \sum\limits_{k = 0}^n {\left( {1 - \cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} \right)} $
$ \Rightarrow \sum\limits_{k = 0}^n {\left( 1 \right)} - \sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} \Rightarrow \left( {n + 1} \right) - \sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} $
$\sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} = \sum\limits_{k = 0}^n {\cos \left( {\frac{{2\pi }}{{n + 2}} + \frac{2}{{n + 2}}k} \right)} ;a = \frac{{2\pi }}{{n + 2}};d = \frac{2}{{n + 2}}$
From the website I got the following formula but the summation of the series is from $0$ to $n-1$.
$\sum\limits_{k = 0}^{n - 1} {\cos \left( {a + kd} \right)} = \frac{{\sin \left( {\frac{{nd}}{2}} \right)}}{{\sin \left( {\frac{d}{2}} \right)}} \times \cos \left( {a + \frac{{\left( {n - 1} \right)d}}{2}} \right)$
Where as in the question it is from $0$ to $n$, a total of $n+1$ terms. How do I proceed
|
Move the last term of the sum out of the sum. The remaining sum will match your reference.
$$\sum_{k=0}^n f(k) = \left( \sum_{k=0}^{n-1} f(k) \right) + f(n) \text{.}$$
|
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|
Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$ As stated in the title.
My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \infty}\frac{(\frac{x+1}{x-2})^{\frac{2}{3}}-(\frac{x-1}{x-2})^{\frac{2}{3}}}{(\frac{x+2}{x-2})^{\frac{2}{3}}-1}$$
L'Hopital
$$L=\lim_{x \to \infty}\frac{\frac{2}{3}(\frac{x+1}{x_2})^{-\frac{1}{3}}(\frac{x-2-(x+1)}{(x-2)^2})-\frac{2}{3}(\frac{x-1}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x-1)}{(x-2)^2})}{\frac{2}{3}(\frac{x+2}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x+2)}{(x-2)^2})}=\lim_{x\ \to \infty}{\frac{3(x+1)^{-\frac{1}{3}}-(-1)(x-1)^{-\frac{1}{3}}}{(x+2)^{-\frac{1}{3}}(-4)}}$$
$$=\lim_{x \to \infty}{\frac{3(1+\frac{1}{x})^{-\frac{1}{3}}-(1-\frac{1}{x})^{-\frac{1}{3}}}{4(1+\frac{2}{x})^{-\frac{1}{3}}}}=\frac{3-1}{(4)(1)}=\frac{1}{2}.$$
Is this correct and is there a more elegant way of doing it?
EDIT: strictly speaking L'Hopital is not applicable with $x \to \infty$ so just got lucky here...
|
From the binomial theorem, we know $(1+y)^n\approx1+ny$ for $y\approx0$. Divide both numerator and denominator by $x^{2/3}$. The limit is$$\lim_{x \to \infty}{\frac{\left(1+\frac1x\right)^{\frac{2}{3}}-\left(1-\frac1x\right)^{\frac{2}{3}}}{\left(1+\frac2x\right)^{\frac{2}{3}}-\left(1-\frac2x\right)^{\frac{2}{3}}}}=\lim_{x \to \infty}\frac{1+\frac{2/3}x-1+\frac{2/3}x}{1+\frac{4/3}x-1+\frac{4/3}x}=1/2.$$
|
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|
Solve $ \int\frac{1}{\sin(x)-\cos(2x)}dx $ ... Weierstrass substitution :
$$\tan(\frac{x}{2})=t$$ $$\sin(x)=(\frac{2t}{1+t^2})$$ $$\cos(x)=(\frac{1-t^2}{1+t^2})$$ $$dx=(\frac{2\,dt}{1+t^2})$$
Than : $$\cos(2x)=\cos^2(x)-\sin^2(x)$$
P.s I tried using these substitutions, but I couldn't get so far .
Need a bit help to solve this problem, if it's possible by using these substitutions.
Thank you in advance :)
|
Since $\cos2x=2\cos^2x-1=\frac{1-6t^2+t^4}{(1+t^2)^2}$, $\sin x-\cos 2x=\frac{-1+2t+6t^2+2t^3-t^4}{(1+t^2)^2}$, so your integral is$$\int\frac{2(1+t^2)}{-1+2t+6t^2+2t^3-t^4}dt.$$Sincce $-1+2t+6t^2+2t^3-t^4=-(t^2-4t+1)(t+1)^2$ (the repeated root is easily guessed with the rational root theorem), we seek a partial fraction decomposition:$$\begin{align}\frac{-2(1+t^2)}{(t^2-4t+1)(t+1)^2}&=\frac{At+B}{t^2-4t+1}+\frac{Ct+D}{t^2-4t+1},\\-2(1+t^2)&=(At+B)(t+1)^2+(Ct+D)(t^2-4t+1)\\&=(A+C)t^3+(2A+B-4C+D)t^2\\&+(A+2B+C-4D)t+B+D.\end{align}$$Since $A+C=0,\,2A+B-4C+D=-2,\,A+2B+C-4D=0,\,B+D=-2$, the solution is$$\int\frac{-2(1+t^2)dt}{(t^2-4t+1)(t+1)^2}=\int\left(\frac{-4/3}{t^2-4t+1}+\frac{-2/3}{(t+1)^2}\right)dt,$$which you can finish.
|
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|
Solve for $x$ in $ \sin x + \sin x \cos x - (\frac{1}{\sqrt{2}} + \frac{1}{2})= 0 $ How can to solve for $x$ in $$ \sin x + \sin x \cos x - \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right)= 0.$$
My try:
$$ \sin x + \sin x \cos x = \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right) $$
$$ (\sin x + \sin x \cos x)^2 = (\frac{1}{\sqrt{2}} + \frac{1}{2})^2
$$
$$ \sin^2x + 2\sin^2 x \cos x + \sin^2x \cos^2x - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
$$ \sin^2 x( 1 + 2 \cos x + \cos^2 x ) - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
$$ (1 - \cos^2x)( 1 + 2 \cos x + \cos^2 x ) - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
$$ 1 + 2 \cos x + \cos^2 x - \cos^2 x - 2 \cos^3 x - \cos^4 x- (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
$$ 1 + 2 \cos x - 2 \cos^3 x - \cos^4x -(\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
thx for all ur help
|
You may continue with
$$ 1 + 2 \cos x - 2 \cos^3 x - \cos^4x -(\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
knowing that $ \cos x = \frac1{\sqrt2} $ is a solution from inspection, as commented. So, factorize the equation as
$$ (\cos x - \frac1{\sqrt2})(\cos^3x +(2+\frac1{\sqrt2})\cos^2x+ (\frac12+\sqrt2)\cos x-1+\frac1{2\sqrt2})=0
$$
where the cubic polynomial has one real root that can be calculated analytically with the Cardano formula.
|
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|
Find all positive integer solutions for the following equation: Find all positive integer solutions for the following equation:
$(x^2+1)(y^2+1)+2(x-y)(1-xy)=4(1+xy)$
I've tried simplifying the equation and then refactoring but I can't find any solutions.
|
We write the given equation equivalently:
$$
\begin{aligned}
0 &= x^2y^2 - 2x^2y + 2xy^2 + x^2 - 4xy + y^2 + 2x - 2y - 3\ ,\\
0 &= x^2(y-1)^2 + 2x(y-1)^2 + (y-1)^2-4\ ,\\
4 &= (x+1)^2(y-1)^2\ .
\end{aligned}
$$
Now consider all possible ways to write $4$ as a product of two perfect squares.
|
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|
Series representation of the reciprocal of prime In the article on unique primes in Wikipedia. It is stated that the representation of the reciprocal of a prime $p$ in the numeral base $b$ is periodic of period $n$ if
$$\displaystyle {\frac {1}{p}}=\sum _{i=1}^{\infty }{\frac {q}{(b^{n})^{i}}}$$ where $q$ is a positive integer smaller than $b^n$
I have no idea how this particular representation is obtained. Kindly provide a proof or point to some resources where I can find this representation of $\frac{1}{p}$.
Thanks in advance !
Update:
Let $p$ be a prime and $n$ be the period of $\displaystyle \frac{1}{p}$ in base $b$ and $b \not\mid p$.
If, $\displaystyle \frac{1}{p} = (0.\overline{a_1a_2a_3\dots a_n})_b$ where $a_1, a_2, a_3, \dots \in \mathbb{Z}_b$
Then, $\displaystyle \frac{1}{p} = a_1a_2a_3 \dots a_n\left (\frac{1}{b^n} + \frac{1}{b^{2n}}+\frac{1}{b^{3n}}+\dots \right) = \sum_{i=1}^{\infty}\frac{q}{(b^n)^i}$, where $ q = a_1a_2a_3 \dots a_n$.
I am not sure how $(0.\overline{a_1a_2a_3\dots a_n})_b = a_1a_2a_3 \dots a_n\left (\frac{1}{b^n} + \frac{1}{b^{2n}}+\frac{1}{b^{3n}}+\dots \right)$ holds
|
I am not sure how $(0.\overline{a_1a_2a_3\dots a_n})_b = a_1a_2a_3 \dots a_n\left (\frac{1}{b^n} + \frac{1}{b^{2n}}+\frac{1}{b^{3n}}+\dots \right)$ holds
Well, $(0.\overline{a_1a_2a_3\dots a_n})_b=0.a_1a_2a_3\dots a_na_1a_2a_3\dots a_na_1a_2a_3\dots a_n\dots$
$=\dfrac{a_1}{b}+\dfrac{a_2}{b^2}+\cdots\dfrac{a_n}{b^n}+\dfrac{a_1}{b^{n+1}}+\dfrac{a_2}{b^{n+2}}+\cdots\dfrac{a_n}{b^{2n}}+\dfrac{a_1}{b^{2n+1}}+\dfrac{a_2}{b^{2n+2}}+\cdots+\dfrac{a_n}{b^{3n}}+\cdots$
$=\dfrac{a_1b^{n-1}+a_2b^{n-2}\cdots+a_n}{b^n}+\dfrac{a_1b^{n-1}+a_2b^{n-2}+\cdots+a_n}{b^{2n}}+\dfrac{a_1b^{n-1}+a_2b^{n-2}+\cdots+a_n}{b^{3n}}+\cdots$
$=a_1a_2\dots a_n\left(\dfrac1{b^n}+\dfrac1{b^{2n}}+\dfrac1{b^{3n}}+\cdots\right).$
|
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Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1} + \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.
Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1}
+ \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.
What I Tried: I thought of substituting $\sqrt{a - 1} = x$ , $\sqrt{a} = y$ .
This gives :- $$\rightarrow \sqrt[3]\frac{(x - y)^5}{(x + y)} + \sqrt[3]\frac{(x + y)^5}{(y - x)}$$
But I was not able to find any good factorisation for this. I even took some help from Wolfram Alpha and it gives me this :-
Another thing I thought of was to substitute only $\sqrt{a} = x$. This would give :-
$$\rightarrow \sqrt[3]\frac{(\sqrt{(x + 1)(x - 1)} - x)^5}{(2x^2 - 1)} + \sqrt[3]\frac{(\sqrt{(x + 1)(x - 1)} + x)^5}{1}$$
This looks more or less simpler to work with, but unfortunately I could not get any ideas.
Can anyone help me?
|
You can first get rid of the denominator in each fraction by multiplying above and below by the conjugate expression of the denominator. Then you will have 6th powers above, and you can get rid of the cubic roots.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
What I Tried: I checked similar questions and answers in the Art of Problem Solving here and here and tried to get some ideas.
First thing which I did is thinking of pairing the values, I took for example, $f(-1)$ and $f(1)$.
We have :-
$$\rightarrow f(-1) = \frac{1}{\frac{1}{3} + \sqrt{3}} = \frac{3\sqrt{3} + 1}{3}$$
$$\rightarrow f(1) = \frac{1}{3 + \sqrt{3}}$$
Adding both gives $\frac{7 + 6\sqrt{3}}{12 + 10\sqrt{3}}$, which more or less looks like a random sum.
So my idea of pairing did not work, or at least I couldn't pair them nicely or missed a pattern. So how would I start solving it?
Can anyone help?
|
You're ever so slightly off. Notice the median of $(-5, -4, ..., 5, 6)$ is $\frac{-5+6}{2}=\frac 12$ which hints at trying $$f(\frac12+x)+f(\frac 12-x)\overbrace{=}^{y=x+\frac 12}f(y)+f(1-y)$$
We see that: $$\frac{1}{3^x+\sqrt 3}+\frac{1}{3^{1-x}+\sqrt 3}=\frac{3^x+3^{1-x}+2\cdot 3^\frac 12}{3^{x+\frac 12}+3^{\frac32-x}+2\cdot 3^1}=\frac{\alpha}{3^\frac 12 \cdot \alpha}=\frac{1}{\sqrt 3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_{2}^{7} \frac{x}{1-\sqrt{2+x}} d x$ We have the following integral:
$$
\int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\, dx
$$
And this is my solution, which seems to be wrong, and I am failing to see where exactly I failed at:
We have $u=1-\sqrt{2+x}, x=u^2-2u-1, dx=-2\sqrt{2+x}\, du$, and we know that $x\geq -2$ and thus $u\leq 1$:
\begin{align}
\int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\, dx
&=-2\int_{-1}^{-2} \frac{(u^2+2u-1)(u-1)}{u}\, du\\
&= -2 \left( \int_{-1}^{-2} u^2 d u + \int_{-1}^{-2} u\, du +\int_{-1}^{-2} -3\, du + \int_{-1}^{-2} \frac{1}{u}\, du \right) \\
&= -2\left[\frac{u^3}{3}+\frac{u^2}{2}-3u+\ln{|u|}\right]_{-1}^{-2}\approx -18
\end{align}
Can someone please help me pinpoint the issue?
|
I would have calculated it like this:
Let $u=1-\sqrt{2+x} \Leftrightarrow x=(1-u)^2-2$ where $x\ge-2$. We then have
$$ \mathrm dx=2(1-u)(-1)\,\mathrm du$$
so
\begin{align*}
\int_2^7\!\frac{x}{1-\sqrt{2+x}}\,\mathrm dx
&=\int_{-1}^{-2}\!\frac{(1-u)^2-2}{u}\cdot2(1-u)(-1)\,\mathrm du
\\&=2\int_{-2}^{-1}\!\frac{((1-u)^2-2)(1-u)}{u}\,\mathrm du
\\&=2\int_{-2}^{-1}\!\frac{(1-2u+u^2-2)(1-u)}{u}\,\mathrm du
\\&=2\int_{-2}^{-1}\!\frac{(u^2-2u-1)(1-u)}{u}\,\mathrm du
\\&=2\int_{-2}^{-1}\!\frac{-u^3+3 u^2-u-1}{u}\,\mathrm du
\\&=2\int_{-2}^{-1}\!\Bigl(-u^2+3u-1-\frac{1}{u}\Bigr)\,\mathrm du
\\&=2\Bigl[-\frac13u^3+\frac32u^2-u-\ln(|u|)\Bigr]_{-2}^{-1}
\\&=-\frac{47}{3}+2\ln(2).
\end{align*}
|
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|
How to solve this equation $(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$? $(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$
$Domain: x\in (-\infty;-2]\cup(1;+\infty)$
$x = 2\longrightarrow$ Done
How to prove $x = \dfrac{-1-3\sqrt{17}}{2}$ is the last solution? With raise both side by power of two?
Any better way such as factoring, grouping,...?
Please help!!
|
Hint:
*
*Note that $$(x-1)\sqrt{\frac{x+2}{x-1}} = \sqrt{(x-1)(x+2)}$$ when $x>1$ and
$$\begin{align}(x-1)\sqrt{\frac{x+2}{x-1}} &= -(1-x)\sqrt{\frac{x+2}{x-1}} \\[1mm]&= -\sqrt{(1-x)^2\frac{x+2}{x-1}}\\[1mm]&=-\sqrt{(x-1)(x+2)}\end{align}$$ when $x \le -2$.
*Substitute $$t = \sqrt{(x-1)(x+2)}$$ Then, how will the equation look like?
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find n where $(2n+1)2^{4n+5} = 3 \pmod{7}$ For $n$ normal number, the book solved it like this:
If $n$ can be divided by $3$ (which is $n = 3k$) then $n = 21L + 9$.
If $n$ can't be divided by 3(Which is either $n = 3k +1$ or $n = 3k + 2$) then $n = 21L + 1$ or $n = 21L + 2$ .
But I didn't solve it like this.
My logic is that since 3 is a prime number then $(2n+1)2^{4n+5} = 3 \pmod{7}$ means either $2^{4n+5} = 3\pmod{7}$ and $ 2n + 1 = 1\pmod{7} $ or the other way around.
But since there is no $n$ value that can make $2^{4n+5} = 3\pmod{7}$ then it means:
$ 2n + 1 = 3 \pmod{7}$ and by calculating we find in the end $n = 7k' + 1$
And $ 2^{4n+5} = 1\pmod{7}$ and by calculating we find that $n = 3k''$
So it means that $n = 7k' + 1$ AND $n = 3k''$
So, I know that my solution is faulty but can anyone explains to me the right solution or point out where I went wrong?
|
Update:
I found that this solution provides a neat way in general to solve this kind of problems. I will apply the method here:
$$(2n+1)2^{4n+5} \equiv 3 \pmod 7 \\ \iff g(n)=(2n+1)2^n\equiv (2n+1) 2^{4n+6} \equiv 3\cdot 2 \equiv -1 \pmod 7$$
Suppose $n=3k+i, i=0,1,2$, then
$$-1 \equiv g(3k+i) = (6k+2i+1)2^{3k+i} \equiv (-k+2i+1)2^i=g(i)-k\cdot 2^i \pmod 7\\
\iff k\equiv 2^{3-i}(g(i)+1)\equiv (2i+1) 2^3 + 2^{3-i} \equiv 2^{3-i}+2i+1 \pmod 7 \\
(\text{ now write } k = 2^{3-i}+2i+1 + 7j)\\
\iff n=3k+i = 3(2^{3-i}+2i+1+7j)+i \equiv 3\cdot 2^{3-i}+7i+3 \pmod {21}\\
\iff n \equiv \begin{cases}3 \cdot 8 + 0 + 3 \equiv 6 \\ 3 \cdot 4 + 7 +3\equiv 1 \\ 3 \cdot 2 + 14+3\equiv 2 \\ \end{cases} \pmod{21}$$
Just some comments:
$f(n)=(2n+1)2^{4n+5}$, then $f(n+21)\equiv f(n) \pmod 7$ (notice that $2^3\equiv 1 \pmod 7$). So the most direct way is to plug in $n=1, 2, \ldots, 21$ into $f(n)$ and see which ones yield $3 \pmod 7$. No fancy theorems needed.
To save time, do as zwim showed by reusing partial results (you can make a $3 \times 7$ table).
|
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"timestamp": "2023-03-29T00:00:00",
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|
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$
I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.
Please give some idea/hint for the other part.
|
Let $f(a,b)=a^3+b^3-6ab+11$. At the extrema, the two curves $f(a,b)=0$ and $k=a+b$ are tangential to each other, i.e.
$$\frac{f_a’}{f_b‘}=\frac{3a^2- 6b}{3b^2-6a}=1\implies (a-b)(a+b+2)=0
$$
which leads to the tangential points $a=b$ and $a+b=-2$. It is straightforward to verify the maximum $a+b<-2$ and the minimum is determined by $f(a,a)=0$, or $2a^3-6a^2+11=0$, whose sole real root is given by the Cardano’s formula and is greater than $-\frac76$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Show that $\left\lceil x-\frac{1}{2} \right\rceil$ is the closest integer to the number $x$ Show that $\left\lceil x-\frac{1}{2} \right\rceil$ is the closest integer to the number $x$, except when $x$ is midway between two integers $n$ and $n+1$, when it is the smaller of these two integers.
We want to prove 2 cases here:
*
*When $x$ is midway between the two integers, $\left\lceil x-\frac{1}{2} \right\rceil$ is equal to the smaller of the two integers.
*When $x$ is not midway between the two integers, $\left\lceil x-\frac{1}{2} \right\rceil$ is equal to the closest integer to the number $x$.
If we let $x$ be midway between the two integers $n$ and $n+1$, then
$$x = n + \frac{(n+1)-n}{2} = n+ \frac{1}{2}$$
Now, substitute back in $x$ in $\left\lceil x-\frac{1}{2} \right\rceil$,
$$\left\lceil n+ \frac{1}{2} - \frac{1}{2}\right\rceil = \left\lfloor n \right\rfloor$$
The problem is why the ceiling function was changed to the floor function?
|
The problem is why the ceiling function was changed to floor function?
Because $n$ is an integer, $\lfloor n\rfloor= \lceil n \rceil = n.$
|
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|
Is there an other way for it? Consider two equations:
$x+y=2$$y+z=4$
Find the value of $(x+z)$.($x,y,z$ all are positive real numbers)
My Approach:
$\because x+y=2 \Longrightarrow y=0$ or $y=1$.
Case 1:
taking $y=0$$ \Longrightarrow$ $x=2$ ; $z=4$.$\therefore x+z = 2+4 = 6.$Case 2:
taking $y=1 \Longrightarrow$
$x=1$ ; $z=3$.
$\therefore x+z = 1+3 = 4.$
$x+z=6$or$x+z=4$
I want to know if my approach is correct or is there more better way to evaluate this ?
|
Given
$\qquad x+y=2\space\land\space y+z=4\qquad$
we can start by eliminating $y$.
$$
y+z=4\implies z=4-y\\
(x+y=2)\implies x=2-y\\
z-x=(4-y)-(2-y)=2\qquad\longrightarrow (z-x=2)\\
(x+z=4)-(z-x=2)\implies 2x=2\implies x=1\\
(x+z=4)+(z-x=2)\implies 2z=6\implies z=3\\
\therefore\quad x+z=1+3=4
$$
At this point we can also see that $y=1$ but who cares? Ha ha
|
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|
Solving state-space function with using of Runge-Kutta method I need to implement my own integration routine that will take state space function $f$, free variable $t$, and initial state $x(0)$ as input and produce the solution $x(t)$ as output. I thought that using Runge-Kutta method will be great. But I cannot understand how to apply it to the state-space function matrix.
$$\dot{x}=f(t,x)$$
$$\dot{x}=A\cdot{x}$$
I have only $A$ matrix. How can I apply Runge-Kutta method?
For example, could you provide step-by-step solution for:
$$
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
-10 & -5 & -2 \\
\end{bmatrix}
\quad
$$
Thanks in advance!
|
You have a system $\dot{x} = f(x)$ with
$$
f(x) = A x = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
-10 & -5 & -2
\end{pmatrix} \begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix} = \begin{pmatrix}
x_2 \\
x_3 \\
-10 x_1 - 5 x_2 - 2 x_3
\end{pmatrix}
$$
The formula for Runge Kutta with step size $h$ is:
$$
\begin{align}
k_1 &= f(x(t)) \\
k_2 &= f(x(t) + \frac{h}{2}k_1) \\
k_3 &= f(x(t) + \frac{h}{2}k_2) \\
k_4 &= f(x(t) + h k_3) \\
x(t + h) &= x(t) + \frac{h}{6}(k_1 + 2 k_2 + 2 k_3 + k_4)
\end{align}
$$
So all you need to do is choose a step size $h$ and initial condition $x(0) = x_0$.
For example use $h = 0.01$ and
$$
x_0 = \begin{pmatrix}
5 \\
-2 \\
3
\end{pmatrix}
$$
Insert this:
$$
\begin{align}
k_1 &= f(x_0) = \begin{pmatrix}
-2 \\
3 \\
-46
\end{pmatrix} \\
k_2 &= f(x_0 + \frac{h}{2}k_1) = \begin{pmatrix}
-1.985 \\
2.77 \\
-45.515
\end{pmatrix} \\
k_3 &= f(x_0 + \frac{h}{2}k_2) = \begin{pmatrix}
-1.98615 \\
2.772425 \\
-45.51485
\end{pmatrix} \\
k_4 &= f(x_0 + h k_3) = \begin{pmatrix}
-1.97227575 \\
2.5448515 \\
-45.02970925
\end{pmatrix} \\
\end{align}
$$
And so:
$$
x(0.01) = x(0) + \frac{0.01}{6}(k_1 + 2 k_2 + 2 k_3 + k_4) = \begin{pmatrix}
4.98014237375 \\
-1.972283830833333 \\
2.544850984583333
\end{pmatrix}
$$
If you repeat this 500 times you get:
If you simulate even longer you can see this is an undamped oscillation which makes sense because the three eigenvalues of $A$ are $-2, \sqrt{5}i, -\sqrt{5}i$.
|
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|
Showing that the function $f(x) = \frac{x}{1+x^2}$ is continuous using the $\epsilon \delta$ definition I am attempting to prove that $f(x) = \frac{x}{1+x^2}$ is continuous. Here's my attempt so far.
In order to prove that $f(x)$ is continuous, it is necessary to show that for every $\epsilon>0$ there exists a $\delta>0$ such that $|\frac{x}{1+x^2}-\frac{c}{1+c^2}|<\epsilon$ when $|x-c|<\delta$.
I first start by working with the left side of the inequality. We can say that $$|\frac{x}{1+x^2}-\frac{c}{1+c^2}| = \frac{|x-c+c^2x-cx^2|}{|(1+x^2)(1+c^2)|}$$ After this we can use the triangle inequality to say, $$\frac{|x-c+c^2x-cx^2|}{|1+x^2||1+c^2|}\leq \frac{|x-c|+|c^2x|+|-cx^2|}{|1+x^2||1+c^2|}$$
Now we can break the function into 3 terms $$\frac{|x-c|}{|1+x^2||1+c^2|}+\frac{|c^2x|}{|1+x^2||1+c^2|}+\frac{|x^2c|}{|1+x^2||1+c^2|}$$
Then we use the fact that $$\frac{c^2x}{|1+x^2||1+c^2|}<1$$ which can be applied to the last 2 terms using the same logic. This gives $$\frac{|x-c|}{|1+x^2||1+c^2|}+2$$
Now we want to isloate $|x-c|$ in order to solve for $\delta$.
I have
$$\frac{|x-c|}{|1+x^2||1+c^2|}<\epsilon-2$$. This is where I get stuck. Any help would be appreciated, thanks!
|
When $c \neq 0$, use the bound
$$\tag{*}\left|\frac{x}{1+x^2}-\frac{c}{1+c^2}\right| = \frac{|x-c+c^2x-cx^2|}{(1+x^2)(1+c^2)} =\frac{|x-c+c^2x-cx^2+c^3 - c^3|}{(1+x^2)(1+c^2)} \\ \leqslant\frac{|x-c|+c^2|x-c|+|c||x^2-c^2|}{(1+x^2)(1+c^2)}= \frac{1+c^2+|c||x+c|}{(1+x^2)(1+c^2)}|x-c|$$
If $|x-c| < \frac{|c|}{2}$, then, since $|x| - |c| \leqslant |x-c|$, we have $\frac{|c|}{2} < |x| < \frac{3|c|}{2}$ and
$$\tag{**}|x+c| \leqslant |x|+|c| <\frac{5|c|}{2}, \quad 1+x^2 > 1+ \frac{c^2}{2}$$
Bounding on the RHS of (*) with (**), we get
$$\left|\frac{x}{1+x^2}-\frac{c}{1+c^2}\right| < \frac{1+c^2 + \frac{5c^2}{2}}{\left(1 + \frac{c^2}{2}\right)(1+c^2)}|x-c|$$
You should be able to finish from here by finding $\delta > 0$ such that if $|x-c| < \min(\delta, \frac{|c|^2}{2})$, then $|f(x) - f(c)| < \epsilon$.
Handle the case where $c=0$ separately.
|
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|
Find all positive integers $x$ such that $x(x+2021)$ is a perfect square I completed the square as to get $(x+1010.5)^2-102110.25 = k^2$ but I don't know where to go from here.
Please help, thank you
I then got $(2x+2021)^2-4084441=4k^2$ then $(2k-2x-2021)(2k+2x+2021)=43^2*47^2$
|
Start from $$x (x+2021)=y^2$$
multiply both sides by $4$ and add $2021^2$ to both sides
$$4 x^2+8084 x+2021^2=4 y^2+2021^2\to (2x+2021)^2=4y^2+2021^2$$
set $2x+2021=z$ and remember that $2021^2=43^2\times 47^2$. The equation becomes
$$z^2-4y^2=43^2\times 47^2\to (z+2y)(z-2y)=43^2\times 47^2$$
we get the following possibilities
$$
\begin{array}{ll}
& \begin{cases}
z-2 y=1\\
2 y+z=2021^2\\
\end{cases}\to (y= 1021110,z= 2042221)\to x=1020100\\
&\begin{cases}
z-2 y=43\\
2 y+z=43\times 47^2\\
\end{cases}\to (y= 23736,z= 47515)\to x=22747\\
& \begin{cases}
z-2 y=47\\
2 y+z=43^2\times 47\\
\end{cases}\to (y= 21714,z= 43475)\to x=20727\\
& \begin{cases}
z-2 y=43^2\\
2 y+z= 47^2\\
\end{cases}\to (y= 90,z= 2029)\to x=4\\
\end{array}
$$
The values are only four:
$$x=4,20727,22747,1020100$$
|
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|
Show that there is only one real root of the equation $1 + a^2 + ax - x^3$ I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?
Thanks.
|
A cubic has only one real root iff its discriminant is negative. (See Wikipedia.)
The discriminant of $1 + a^2 + ax - x^3$ is $-27 a^4 + 4 a^3 - 54 a^2 - 27$.
Now, $-27 a^4 + 4 a^3 - 54 a^2 = a^2 (-27 a^2 + 4 a - 54)$ and $-27 a^2 + 4 a - 54=-27 \left(a - \frac{2}{27}\right)^2 - \frac{1454}{27} < 0$ for all $a$.
Therefore, $-27 a^4 + 4 a^3 - 54 a^2 - 27 < 0$ for all $a$.
|
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|
Complex Integral : How can I evaluate $\int_{C_R} f(z) dz$? I have to calculate $I$ ,using complex integral.
\begin{equation}
I:=\displaystyle\int_{-\infty}^{\infty} \dfrac{\text{log}{\sqrt{x^2+a^2}}}{1+x^2}.(a>0)
\end{equation}
Let $f(z)=\dfrac{\text{log}(z+ia)}{1+z^2}$.
$
C_R : z=Re^{i\theta}, \theta : 0 \to \pi.$
$C_1 : z=t, t : -R \to R.$
From Residue Theorem,
\begin{equation}
\displaystyle\int_{C_1} f(z) dz + \displaystyle\int_{C_R} f(z) dz = 2\pi i \text{Res}(f, i).
\end{equation}
If $R \to \infty$, $\displaystyle\int_{C_1} f(z) dz \to \displaystyle\int_{-\infty}^{\infty} f(x) dx. $
And my mathematics book says that if $R \to \infty$, $\displaystyle\int_{C_R} f(z) dz \to 0.$ I cannot understand why this holds.
My attempt is following :
\begin{align}
\Bigg|\displaystyle\int_{C_R} f(z) dz \Bigg|
&=\Bigg|\displaystyle\int_0^{\pi} f(Re^{i\theta}) i Re^{i\theta} d\theta \Bigg| \\
&\leqq \displaystyle\int_0^{\pi} \Bigg| R \dfrac{\text{log} (Re^{i\theta} + ia)}{1+R^2e^{2i\theta}} \Bigg| d\theta \\
&\leqq \displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (Re^{i\theta}+ia) \Bigg| d\theta \\
&= \displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (R\cos \theta+i(R\sin \theta +a)) \Bigg| d\theta
\end{align}
I expect that
$ \displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (R\cos \theta+i(R\sin \theta +a)) \Bigg| d\theta \to 0$,
but I cannot prove this.
I would like to give me some ideas.
|
$$I(a):=\int_{-\infty}^{\infty} \frac{\ln\sqrt{x^2+a^2}}{1+x^2}dx=
\int_0^{\infty} \frac{\ln(x^2+a^2)}{1+x^2}dx$$
$$\frac{dI}{da}=\int_0^{\infty} \frac{2a}{(1+x^2)(x^2+a^2)}dx$$
$$\int \frac{2a}{(1+x^2)(x^2+a^2)}dx=\frac{2}{a^2-1}\tan^{-1}(\frac{x}{a})-\frac{2a}{a^2-1}\tan^{-1}(x)$$
$$\frac{dI}{da}=\frac{2}{a^2-1}\frac{\pi}{2}-\frac{2a}{a^2-1}\frac{\pi}{2}$$
After simplification :
$$\frac{dI}{da}=\frac{\pi}{a+1}$$
$$I(a)=\pi\ln(a+1)+C$$
In order to determine $C$ we compute the integral for a particular value of $a$, for example $a=0$ :
$$I(0)=\int_0^{\infty} \frac{\ln(x^2)}{1+x^2}dx$$
Change of variable $\quad x=\frac{1}{t}$ :
$$I(0)=\int_{\infty}^0 \frac{\ln(\frac{1}{t^2})}{1+\frac{1}{t^2}}(-\frac{dt}{t^2})=
-\int_0^{\infty} \frac{-\ln(t^2)}{t^2+1}(-dt)=-\int_0^{\infty} \frac{\ln(t^2)}{t^2+1}dt$$
This implies
$$\int_0^{\infty} \frac{\ln(x^2)}{1+x^2}dx=-\int_0^{\infty} \frac{\ln(t^2)}{1+t^2}dt=0\quad\text{thus}\quad I(0)=0.$$
$$I(0)=\pi\ln(0+1)+C\quad\implies\quad C=0$$
$$\boxed{I(a):=\int_{-\infty}^{\infty} \frac{\ln\sqrt{x^2+a^2}}{1+x^2}dx=\pi\ln(a+1)}$$
|
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|
A sequence $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and find $x_{2020}$ The sequence is given by the formula $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and it is known that $x_{2017} + x_{2023} = 990$, then what is $x_{2020}$ = ?
My little approch:
It is given that $x_{2017} + x_{2023} = 990$ ----- (1) and $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$
So,
$x_{2017} = x_{2016+1} = 3 x_{2016} + \sqrt{8x^2_{2016} + 2}$ -------(2)
$x_{2023} = x_{2022+1} = 3 x_{2022} + \sqrt{8x^2_{2022} + 2}$ -------(3)
Then from (1),(2),(3) =>
$3 x_{2016} + \sqrt{8x^2_{2016} + 2} + 3 x_{2022} + \sqrt{8x^2_{2022} + 2} = 990$
$3 (x_{2016}+x_{2022} ) + \sqrt{8x^2_{2016} + 2} + \sqrt{8x^2_{2022} + 2} = 990$
$3 (x_{2016}+x_{2020} + x_{2021}) + \sqrt{8x^2_{2016} + 2} + \sqrt{8x^2_{2020} + 2}+ \sqrt{8x^2_{2021} + 2}+ \sqrt{8x^2_{2022} + 2} = 990$
I stuck here and can't go further. Please help me with it.
|
After you get $x_{n+1} - 6x_n + x_{n-1}=0$ as in the other two answers, a conceptually easy way to establish the relationship between $x_n$ and $x_{n-3}+x_{n+3}$ is as follows:
Rewrite the recurrence equation as
$$(\mathbb E^2-6\mathbb E+1)x_n=0$$
where $\mathbb E$ is the forward shift operator $\mathbb E^{i} x_n = x_{n+i}, \forall i \in \mathbb N$. Then the characteristic equation is $$\lambda^2 - 6\lambda + 1 =0$$ with two roots $a, b$ such that $a+b=6, ab=1$. Now
$$a^3\cdot b^3 =1, a^3+b^3 = (a+b)((a+b)^2-3ab)=6\cdot (6^2-3)=198$$
Therefore $$\lambda^6 - 198 \lambda^3+1=(\lambda^3-a^3)(\lambda^3-b^3)=g(\lambda)(\lambda-a)(\lambda-b)=g(\lambda)(\lambda^2-6\lambda+1)$$
where $g(\lambda)=(\lambda^2 + a\lambda +a^2)(\lambda^2 + b\lambda +b^2)$ (we don't need to know the coefficients, we only need the fact that $g$ is a polynomial)
Then
$$(\mathbb E^6 - 198 \mathbb E^3 + 1) x_n = g(\mathbb E)(\mathbb E^2 - 6\mathbb E+1)x_n = 0\\
\implies x_{n+6} - 198 x_{n+3} + x_n=0 \implies x_{2020} = 990/198=5.$$
|
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|
$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$ without using L'Hospital or Taylor's series Find limit without using L'Hospital or Taylor's series:
$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$$
$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = \displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} \cdot \frac{\sin(2x^5)}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = $$
$$= \displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} \cdot \frac{\sin(2x^5)}{x^5} \cdot \frac{x^2}{5^{x^{2}} - 1} \cdot \frac{ x^3}{\tan^3(3x)}$$
*
*$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} = 1$
*$\displaystyle \lim_{x \to 0} \frac{\sin(2x^5)}{x^5} = 2$
*$\displaystyle \lim_{x \to 0} \frac{x^2}{5^{x^{2}} - 1} = \frac{1}{\ln5}$
*$\displaystyle \lim_{x \to 0} \frac{ x^3}{\tan^3(3x)} = \displaystyle \lim_{x \to 0} \cos^3(3x) \cdot \displaystyle \lim_{x \to 0} \frac{x^3}{\sin^3(3x)} = 1 \cdot \left( \frac{1}{3} \right)^3 = \frac{1}{27}$
We get:
$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = 1 * 2 * \frac{1}{\ln5} * \frac{1}{27} = \frac{2}{27 \ln5}$$
All credits to @Bernard. Thank you all, question closed.
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Just a slight variant on Bernard's answer: rewrite the fraction as
$${\ln(1+\sin(2x^5))\over\sin(2x^5)}\cdot{\sin(2x^5)\over2x^5}\cdot\left(3x\over\sin(3x)\right)^3\cdot{x^2\ln5\over e^{x^2\ln5}-1}\cdot{2\cos^3(3x)\over3^3\ln5}$$
then use the familiar limits
$$\lim_{u\to0}{\ln(1+u)\over u}=\lim_{u\to0}{\sin u\over u}=\lim_{u\to0}{e^u-1\over u}=1$$
to get the limit
$$1\cdot1\cdot1^3\cdot1\cdot{2\cdot1^3\over27\ln5}={2\over27\ln5}$$
|
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|
A problem on inverse trigonometric function How to sum the following series: $$S=\cot^{-1}2+\cot^{-1}8+\cot^{-1}18+\cot^{-1}32+\cdots+\cot^{-1}∞$$ My attempt better to say a solution was,\begin{align}S &=\sum_{n=1}^{∞}\cot^{-1}2n^2\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{1}{2n^2}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{2}{4n^2}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{2}{1+4n^2-1}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\\
&=\sum_{n=1}^{∞}[\tan^{-1}(2n+1)-\tan^{-1}(2n-1)]\\
&=\tan^{-1}∞-\tan^{-1}1\\
&=\frac{π}{2}-\frac{π}{4}\\
&=\frac{π}{4}\end{align} Now one thing I have to confess that I knew this method previously,so I did and I don't understand from where this idea suddenly comes about telescoping? Can someone help me explain with any different approach is added as an answer. Thanks for your attention.
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The series transformations might seem at first glance rather arbitrary. But there are some nice aspects behind the curtain which can be sometimes conveniently used.
Let's assume we want to calculate
\begin{align*}
\color{blue}{\sum_{n=1}^\infty \tan^{-1}\left(\frac{1}{2n^2}\right)}\tag{1}
\end{align*}
We know that telescoping is one of the techniques which can be sometimes used to tackle such problems. But is it plausible in this case?
Addition theorem: Here we might recall the addition theorem for the tangent function and its inverse
\begin{align*}
\tan\left(\alpha\pm\beta\right)&=\frac{\tan \alpha \pm\tan \beta}{1\mp\tan\alpha \tan\beta}\tag{2.a}\\
\alpha\pm\beta &=\tan^{-1}\left(\frac{\tan \alpha \pm\tan \beta}{1\mp\tan\alpha \tan\beta}\right)\tag{2.b}
\end{align*}
Letting $x:=\tan \alpha$ and $y:=\tan\beta$ we obtain from (2.b) the identity
\begin{align*}
\color{blue}{\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x- y}{1+ xy}\right)}\tag{2.c}
\end{align*}
The point is, if we recall one of (2.a), (2.b) or (2.c) the telescoping approach becomes interesting, especially in it's most convenient representation (2.c).
Transformation: When looking at what we have in (1) the challenge is to find a transformation from
\begin{align*}
\tan^{-1}\frac{1}{2n^2}\quad\to\quad (2.c) \quad\to \quad\tan^{-1} x - \tan^{-1} y
\end{align*}
We put the focus on the fraction $\frac{x-y}{1+xy}$ at (2.c) and obtain
\begin{align*}
\color{blue}{\frac{1}{2n^2}}&=\frac{1}{1+\left(2n^2-1\right)}\tag{3}\\
&=\frac{2}{1+\left(4n^2-1\right)}\tag{4}\\
&=\frac{2}{1+\left(2n+1\right)\left(2n-1\right)}\tag{5}\\
&\,\,\color{blue}{=\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}}\tag{6}
\end{align*}
Comment:
*
*In (3) we write the denominator as $1+\left(2n^2-1\right)$ since we want as in (2.c) something like $1+xy$. We see we have $1+\left(a^2-b^2\right)=1+(a+b)(a-b)$. The factor $2$ in $\left(2n^2+1\right)$ is not that convenient, but we can cope with it by expanding by a factor $2$ to get $4n^2-1=(2n+1)(2n-1)$ which is done in (4).
*In (5) we have the denominator according to that in (2.c) and we are lucky, since the numerator $2$ can be written as $2=(2n+1)-(2n-1)$ to obtain in (6) the wanted representation as in (2.c).
Series calculation: Now we are well prepared to tackle the series by a telescoping approach. We obtain
\begin{align*}
\color{blue}{\sum_{n=1}^\infty}\color{blue}{\tan^{-1}\frac{1}{2n^2}}
&=\sum_{n=1}^{\infty}\tan^{-1}\left(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right)\tag{$\to (6)$}\\
&=\sum_{n=1}^{\infty}\left[\tan^{-1}\left(2n+1\right)-\tan^{-1}\left(2n-1\right)\right]\tag{$\to (2.c)$}\\
&=\lim_{N\to\infty}\sum_{n=1}^{N}\left[\tan^{-1}\left(2n+1\right)-\tan^{-1}\left(2n-1\right)\right]\\
&=\lim_{N\to\infty}\left[\tan^{-1}\left(2N+1\right)-\tan^{-1}\left(1\right)\right]\\
&=\lim_{N\to\infty}\tan^{-1}\left(2N+1\right)-\lim_{N\to\infty}\tan^{-1}\left(1\right)\\
&=\frac{\pi}{2}-\frac{\pi}{4}\\
&\,\,\color{blue}{=\frac{\pi}{4}}
\end{align*}
|
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|
Nested equilateral triangles Let triangle $ABC$ is an equilateral triangle. Triangle $DEF$ is also an equilateral triangle and it is inscribed in triangle $ABC \left(D\in BC,E\in AC,F\in AB\right)$. Find $\cos\measuredangle DEC$ if $AB:DF=8:5$.
Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.
Let $\measuredangle DEC=\alpha$. We can note that $\triangle AEF \cong \triangle BFD \cong CDE$. This is something we can always use in such configuration. So $$AE=BF=CD, $$ $$AF=BD=CE.$$ Let $AB=BC=AC=8x$ and $DF=DE=EF=5x$. If we denote $CD=y,$ then $CE=AC-AE=AC-CD=8x-y$. Cosine rule on $CED$ gives $$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$ which is a homogenous equation. I got that $\dfrac{y}{x}=4\pm\sqrt{3}.$ Now using the sine rule on $CED$ $$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$ Now we can use the trig identity $\sin^2x+\cos^2x=1$ but it doesn't seem very rational. Can you give me a hint? I was able to find $\sin\measuredangle DEC$ in acceptable way, but I can't find $\cos\measuredangle DEC$...
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As others have done, I'll assume $AB=8,DF=5$ without loss of generality. Then equilateral triangles $\triangle ABC$ and $\triangle DEF$ have area $(\sqrt{3}/4)AF^2 = 16\sqrt{3}$ and $(\sqrt{3}/4)DF^2=(25/4)\sqrt{3}$ respectively. Since the triangles $\triangle FAE,\triangle DBF, \triangle ECD$ are all congruent, they must each have area $$\frac{1}{3}\left(16\sqrt{3}-\frac{25}{4}\sqrt{3}\right)=\frac{13\sqrt{3}}{4}.$$ But each triangle has a 60-degree base angle, so we can also write the area of $\triangle FAE$ as
$$\frac12 AF\cdot AE \cdot \sin 60^\circ$$
Since $AE=AC-EC=8-AF$, this becomes $$\frac12 AF(8-AF)\frac{\sqrt{3}}{2}= \frac{13\sqrt{3}}{4}\implies AF(8-AF)=13$$
with $x=4\pm \sqrt{3}$ as solutions. As a minor variation on the other approaches, we may then use the law of cosines on triangle $\triangle CED$ to write
$$CD^2=CE^2+ED^2-2CE\cdot ED\cos\measuredangle DEC$$
and therefore
\begin{align}
\cos\measuredangle DEC
&=\frac{CE^2+ED^2-CD^2}{2CE\cdot ED}\\
&=\frac{(4\pm \sqrt{3})^2+5^2-(4\mp \sqrt{3})^2}{2(5)(4\mp \sqrt{3})}\cdot \frac{4\mp \sqrt{3}}{4\mp \sqrt{3}}\\
&=\frac{(25\pm 16\sqrt{3})(4\mp \sqrt{3})}{130}\\
&=\frac{52\pm 39\sqrt{3}}{130}\\
&=\frac{4\pm 3\sqrt{3}}{10}
\end{align}
as others have obtained.
|
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|
Legendre 3 Square problem I got a little help on the way here yesterday but it seems like my question is dead.
Problem:
If $ n \in \mathbb{N} $ can be represented as $ n = n_1^2 + n_2^2 + n_3^2 ,\quad n_1, n_2, n_3 \in \mathbb{Z}, $ show that then $ n \neq 4^a(8m+7)$ for $a,m \in \mathbb{Z}.$
Solution:
$ 4^a(8m+7) \equiv 0 \mod 4, \forall a \geq 1.$ If $ n_i $ is an odd square, $ (n_i)^2 = (2k +1)^2 = 4(k^2+1) + 1 \equiv 1 \mod 4.$
So $n_1, n_2, n_3 $ must be even squares. So we can write $ (\frac{n_1}{2})^2 +(\frac{n_2}{2})^2 + (\frac{n_3}{2})^2 = \frac{n}{4} $, which yields to
$$ \frac{n}{4} = 4^{a-1}(8m+7). $$
Now I want to apply $ \mod 8 $ somehow I believe.
We got the easier parts: $ (8m+7) \equiv 7 \mod 8 ,\quad 4^{a-1} \equiv \begin{cases} 4 \quad a =2\\ 0 \quad a > 2\end{cases} $.
So $ 4^{a-1}(8m+7) \equiv 4 \quad \text{or} \quad 0 \mod 8. $
How do I proceed? $n$ is even, $ \frac{n}{4} \equiv \quad ? \mod8$. Any help is muy appriciated! <3
Edit;
The squares in $ \mathbb{Z_8} $:
*
*$ 0^2 = 0$
*$ 1^2 = 1 $
*$ 2^2 = 4 $
*$3^2 = 1 $
*$4^2 = 0 $
*$5^2 = 1 $
*$ 6^2 = 4 $
*$7^2 = 1 $
With the sum of three squares, we can never reach $ 7 \mod 8$?
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What you have done so far is to prove that:
Let $a \geq 1$. If $n = 4^a(8m+7)$ can be written as a sum of three squares, then so can $\frac{n}4 = 4^{a-1}(8m+7)$.
Applying this observation $a$ times, one notices that if $n = 4^a(8m+7)$ can be written as a sum of three squares, then so can $\frac{n}{4^a} = 8m+7$. Thus, if we prove that no number of the form $8m+7$ can be written as a sum of three squares, then no number of the form $4^a(8m+7)$ can be written as a sum of three squares either.
So consider a number of the form $8m+7$. Modulo $8$, it is congruent to $7$.
As you've noticed, the squares modulo $8$ are $0$, $1$ and $4$. So if $8m+7 = x_1^2+x_2^2+x_3^2$ with $x_j \in \mathbb Z$, then
$$8m + 7 \equiv 7 \equiv x_1^2 + x_2^2 + x_3^2 \pmod 8,$$
which is not a solvable equation, since $7$ cannot be written as the sum of $0$, $1$ and $4$, using three summands (repetitions allowed).
|
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|
Calculate integral $\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx$.
Calculate integral $$\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx.$$
My direction: Since this integral can't calculate normally, I tried to use the property following:$$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx.$$
Then, I have
$$I=\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx = \int_0^{\pi/2} \frac{\sin^3x}{\cos^2x + \sin^3x}dx.$$
Therefore
$$2I = \int_0^{\pi/2} \left(\frac{\cos^3x}{\sin^2x + \cos^3x} + \frac{\sin^3x}{\cos^2x + \sin^3x}\right) dx.$$
I stucked here.
|
This integral does have a closed form despite being very messy. We have \begin{align}I&=\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}\,dx\\&=\frac14\int_{-\pi}^\pi\frac{\cos^3(u/2)}{\sin^2(u/2)+\cos^3(u/2)}\,du\\&=\frac14\oint_{|z|=1}\frac{\frac18(z^{1/2}+z^{-1/2})^3}{-\frac14(z^{1/2}-z^{-1/2})^2+\frac18(z^{1/2}+z^{-1/2})^3}\frac{dz}{iz}\\&=-\frac i4\oint_{|z|=1}\frac{(z+1)^3}{z((z+1)^3-2z^{1/2}(z-1)^2)}\,dz.\end{align} Note that $(z+1)^3-2z^{1/2}(z-1)^2=(w^2+1)-2w(w^2-1)^2=\prod\limits_{i=1}^6(w-z_i)$ where $z_1=\overline{z_2}$, $z_3=\overline{z_4}$ and $z_5=\overline{z_6}$. Hence $$I=-\frac i4\cdot2\pi i\left(\operatorname{Res}(f,0)+\sum_{i=1}^6\operatorname{Res}(f,z_i^2)\right)=\frac\pi2\left(1+\sum_{i=1}^6\operatorname{Res}(f,z_i^2)\right).$$ An interesting property about this polynomial is that it is symmetric about its coefficients \begin{align}(w^2+1)-2w(w^2-1)^2&=w^6-2w^5+3w^4+4w^3+3w^2-2w+1\\&=w^3(w^3-2w^2+3w+4+3w^{-1}-2w^{-2}+w^{-3}).\end{align} This means we can make the substitution $w=v+1/v$ to obtain a cubic equation which is solvable in radicals, as in @Quanto's answer.
|
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|
Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$ Calculate:
$$\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$
The problem with that case is that the roots are in different powers so multiplication in nominator and denominator by conjugate is not an option (at least I think it's not).
|
$$\frac{\sqrt{19-x}-2 \sqrt[4]{x+13}}{\sqrt[3]{11-x}-x+1}$$
plug $x=y+3$ so we get
$$\underset{y\to 0}{\text{lim}}\frac{\sqrt{16-y}-2 \sqrt[4]{y+16}}{\sqrt[3]{8-y}-y-2}$$
we have the following expansions as $y\to 0$
$$
\begin{array}{rll}
\sqrt{16-y} &\sim &4-\frac{y}{8} \\
2 \sqrt[4]{y+16} &\sim& \frac{y}{16}+4 \\
\sqrt[3]{8-y} &\sim& 2-\frac{y}{12} \\
\end{array}
$$
So the limit can be written as
$$\underset{y\to 0}{\text{lim}}\frac{4-\frac{y}{8}-\left(\frac{y}{16}+4\right)}{2-\frac{y}{12}-y-2}=\lim_{y\to0}\frac{-\frac{3}{16}y}{-\frac{13}{12}y}=\frac{9}{52}$$
|
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|
Prove that $\left \lfloor \frac{1+\lfloor na+1/a\rfloor}{a} \right \rfloor=n$ If $a \geq \frac{1+\sqrt{5}}{2}$ and $n \in \Bbb W$, prove that
$$\left \lfloor \frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a} \right \rfloor=n.$$
I could prove only when $a$ is an integer, that is $a \geq 2$. If $a \in \Bbb Z$ we have: $$\left \lfloor \frac{1+na^2}{a} \right \rfloor=na+\left\lfloor \frac{1}{a} \right\rfloor=na$$ so we get: $$\left \lfloor \frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a} \right \rfloor=\left \lfloor \frac{1+na}{a} \right \rfloor=n+\left\lfloor \frac{1}{a} \right\rfloor=n.$$ But what if $a \notin \Bbb Z$?
|
From the well known
$$x-1<\lfloor x\rfloor \leq x \tag{1}$$
Thus
$$\frac{1+na^2}{a}-1<
\left\lfloor \frac{1+na^2}{a}\right\rfloor\leq
\frac{1+na^2}{a}$$
and
$$n<n+\frac{1}{a^2}<
\frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a}\leq
n+\frac{1}{a^2}+\frac{1}{a}\tag{2}$$
Finally, for $\color{red}{a>\phi}$
$$\frac{1}{a}+\frac{1}{a^2} <\frac{2}{1+\sqrt{5}}+\frac{4}{(1+\sqrt{5})^2}=
\frac{2+2\sqrt{5}+4}{(1+\sqrt{5})^2}\\
=\frac{6+2\sqrt{5}}{6+2\sqrt{5}}=1$$
and from $(2)$
$$n<
\frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a}<
n+1\tag{3}$$
the result follows from $(3)$.
The corner case for $\color{red}{a=\phi}$ leads to
$$\left\lfloor \frac{1+n\phi^2}{\phi}\right\rfloor=
\left\lfloor \frac{\sqrt{5}-1}{2}+n\cdot\frac{\sqrt{5}+1}{2}\right\rfloor=\\
\left\lfloor (n+1)\cdot\frac{\sqrt{5}}{2}+\frac{n+1}{2}-1\right\rfloor=
\left\lfloor (n+1)\cdot\frac{\sqrt{5}+1}{2}-1\right\rfloor=\\
\left\lfloor (n+1)\cdot\phi-1\right\rfloor < ...$$
because $\phi$ is irrational, thus $(n+1)\cdot\phi-1$ can never be an integer
$$...< (n+1)\cdot\phi-1$$
Then $(2)$ becomes
$$n<
\frac{1+\left\lfloor \frac{1+n\phi^2}{\phi}\right\rfloor}{\phi}<
n+1$$
I wanted to make it a short answer, but the corner case and the explanatory notes spoiled the effort.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Mistake proving that $3 = 0$ I was watching this video in which they proved that $3 = 0$ and the objective is to find the mistake in the proof. They did the following:
*
*Let $x$ be a solution for the equation: $x^2 + x + 1 = 0$ $(1)$.
*Because $x \neq 0$ we can devide both sides by $x$: $x + 1 + \frac{1}{x} = 0$ $(2)$
*From eq. $(1)$ we conclude that $x +1 = -x^2$ and if we substitude that in eq. $(2)$ we get: $-x^2 + \frac{1}{x} = 0$ $(3)$
*Simplifying eq $(3)$ we get: $x^2 = \frac{1}{x} \iff x^3 = 1$, so we get that $x = 1$ is a solution.
*If we substitude the solution $x = 1$ back into eq. $(1)$ we get $1^2 + 1 + 1 = 0 \iff 3 = 0$
I'll leave the answer and the rest of my question as a spoiler if you first want to try this yourself:
He explains that the mistake is when we substitute $x + 1=-x^2$ into eq. $(2)$, because when we do so we are actually adding the solution $x = 1$. The thing is that in the video he only says that and that got my wondering, why is that so?
|
Call $f(x)=x^2 + x + 1$, then his equation is $f(x)=0$.
From there, he derives the equation $(2)$, $\frac{f(x)}{x}=0$.
To reach equation $(3)$, he substracts the second equation to the first one, obtaining the equation $\left(\frac{1}{x}-1\right)f(x)=0$.
Thus, the equation $(3)$ will be true when either $f(x)=0$, which are the solutions of the original problem, or when $\frac{1}{x}-1=0$, that is, $x=1$. That´s what he means when he says he is adding the solution $x=1$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Integral of $\sqrt{4-x^2}$ by Riemann sum I was trying to compute $\int_{0}^{2}\sqrt{4-x^2}\,dx$ by Riemann sum like this
$$\sum_{i=1}^{n}\sqrt{4-x_i^2}\,\Delta x_i$$
where $x_i = \frac{2i}{n}$ and $\Delta x_i = \frac{2}{n}$, but I couldn't get something useful for the answer. I think it should be with another kind of partition, but I don't know which kind of partition to use.
|
The right Riemann sum using the partition with points $x_k = 2 \sin \frac{\pi k}{2n}\,\,(k = 0,1,2,\ldots, n)$ is
$$\tag{*}S_n = \sum_{k=1}^n \sqrt{4 - 4 \sin^2 \frac{\pi k}{2n}}\left(2 \sin \frac{\pi k}{2n}- 2\sin \frac{\pi (k-1)}{2n}\right)\\ = 4\sum_{k=1}^n \cos \frac{\pi k}{2n}\left( \sin \frac{\pi k}{2n}- \sin \frac{\pi k}{2n}\cos \frac{\pi}{2n}+ \sin \frac{\pi }{2n}\cos \frac{\pi k}{2n}\right)\\ = 4\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \cos \frac{\pi k}{2n}\sin \frac{\pi k}{2n} + 4 \sin \frac{\pi}{2n}\sum_{k=1}^n \cos^2 \frac{\pi k}{2n}$$
Note that
$$\sum_{k=1}^n \cos \frac{\pi k}{2n}\sin \frac{\pi k}{2n}= \frac{1}{2}\sum_{k=1}^n\sin \frac{2\pi k}{2n} =\frac{1}{2}\sum_{k=1}^n\sin \frac{\pi k}{n},\\\sum_{k=1}^n \cos^2 \frac{\pi k}{2n}= \sum_{k=1}^n \left(\frac{1}{2} - \frac{1}{2} \cos \frac{2\pi k}{2n} \right) = \frac{n}{2}- \frac{1}{2}\sum_{k=1}^n\cos \frac{\pi k}{n}$$
Substituting into (*), we get
$$S_n = \underbrace{2\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \sin \frac{\pi k}{n}}_{A_n} + \underbrace{2n \sin \frac{\pi}{2n}}_{B_n}- \underbrace{\sin \frac{\pi}{2n}\sum_{k=1}^n \cos\frac{\pi k}{n}}_{C_n} $$
It is shown below that $A_n - C_n \to 0$ as $n \to \infty$, and, therefore,
$$\int_0^2 \sqrt{4-x^2}\, dx= \lim_{n \to \infty}S_n = \lim_{n \to \infty}2n \sin \frac{\pi}{2n} = \lim_{n \to \infty}\pi \frac{\sin \frac{\pi}{2n}}{\frac{\pi}{2n}} = \pi$$
Limits of $A_n$ and $C_n$:
We have, since $\frac{1-\cos x}{\sin x} \to 0$ as $x \to 0$,
$$A_n = 2\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \sin \frac{\pi k}{n}= 2 \left(1 -\cos\frac{\pi}{2n} \right) \frac{\sin \frac{n}{2}\frac{\pi}{n}\sin \frac{n+1}{2} \frac{\pi}{n}}{\sin \frac{\pi}{2n}}\\ = 2 \frac{1 -\cos\frac{\pi}{2n} }{\sin \frac{\pi}{2n}}\sin \left[\left(1 + \frac{1}{n}\right)\frac{\pi}{2}\right]\underset{n\to \infty}\longrightarrow 0$$
Also,
$$C_n = \sin\frac{\pi}{2n} \sum_{k=1}^n \cos \frac{\pi k}{n} = \sin\frac{\pi}{2n}\frac{\sin \frac{n}{2}\frac{\pi}{n}\cos \frac{n+1}{2} \frac{\pi}{n}}{\sin \frac{\pi}{2n}}\\ = \cos \left[\left(1+\frac{1}{n} \right)\frac{\pi}{2}\right]\underset{n\to \infty}\longrightarrow 0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the Fourier Coefficient to this discrete time signal? I'm having trouble finding the fourier coefficients to this discrete time signal:
My approach:
I start by first finding the discrete period, $N$, and discrete frequency, $\Omega_0$
From the problem, we can easily identify that $N = 10$ and so $\Omega_{0} = \frac{2\pi}{10} = \frac{\pi}{5}$
$$x[n] = \sum_{k = 0}^{9}F[K]e^{j\left( \frac{\pi}{5}\right)kn} $$
where $F[K]$ are the Fourier coefficients we want to find.
$$F[K] = \frac{1}{N}\sum_{n = 0}^{9}x[n]e^{-j\left( \frac{\pi}{5}\right)kn}$$
$$F[K] = \frac{1}{10}\left( \sum_{n = 0}^{4}\frac{7}{2}e^{-j\left( \frac{\pi}{5}\right)kn} + \sum_{n = 4}^{9}\frac{-7}{2}e^{-j\left( \frac{\pi}{5}\right)kn}
\right)$$
$$F[K] = \frac{7}{20}\left( \sum_{n = 0}^{4}e^{-j\left( \frac{\pi}{5}\right)kn} - \sum_{n = 4}^{9}e^{-j\left( \frac{\pi}{5}\right)kn}
\right)$$
$$F[K] = \frac{7}{20} \left( 1 + \cos(\frac{\pi}{5}k) + j\sin(\frac{\pi}{5}k) + \cos(\frac{2\pi}{5}k) + j\sin(2\frac{\pi}{5}k) + \cos(\frac{3\pi}{5}k) + j\sin(3\frac{\pi}{5}k) + \cos(\frac{4\pi}{5}k) + j\sin(\frac{4\pi}{5}k) - \cos(\pi k) - j\sin(\pi k) - \cos(\frac{6\pi}{5}k) - j\sin(\frac{6\pi}{5}k) - \cos(\frac{7\pi}{5}k) - j\sin(\frac{7\pi}{5}k) - \cos(\frac{8\pi}{5}k) - j\sin(\frac{8\pi}{5}k) - \cos(\frac{9\pi}{5}k) - j\sin(\frac{9\pi}{5}k)
\right)$$
This looks like such a mess to reduce... or is there something elegant that I'm missing?
According to the solutions
$
X[K]=
\begin{cases}
\frac{7}{10} + \frac{7j \pi}{5}\left( \sin(\frac{\pi}{5}) + \sin( \frac{2\pi}{5}) \right) , &\text{if}\, k = - 1\\
\frac{7}{10} + \frac{7j \pi}{5}\left( \sin(\frac{\pi}{5}) + \sin( \frac{2\pi}{5}) \right) , &\text{if}\ k = + 1\\
\frac{7}{10} - \frac{7j \pi}{5}\left( \sin(\frac{2 \pi}{5}) - \sin( \frac{\pi}{5}) \right) , &\text{if}\ k = + 3\\
\frac{7}{10} + \frac{7j \pi}{5}\left( \sin(\frac{2 \pi}{5}) - \sin( \frac{\pi}{5}) \right) , &\text{if}\ k = - 3\\
0 &\text{otherwise}
\end{cases}
$
HOW DO YOU FIGURE THAT OUT!!? Is there some super clever technique I'm not employing? Or is all my math wrong?
|
Your second sum should index from $n=5$. Then here is a small simplification:
\begin{align}
F[k] &= \frac{7}{20} \Big( \sum_{n=0}^4 e^{-ink\pi/5}-\sum_{n=5}^9e^{-ink\pi/5} \Big) \\
&=\frac{7}{20} \Big( \sum_{n=0}^4 e^{-ink\pi/5}-\sum_{n=0}^4 e^{-5ik \pi/5} e^{-ink\pi/5} \Big) \\
&= \frac{7(1 - (-1)^k)}{20} \sum_{n=0}^4 e^{-ink\pi/5} \\
&=\left\{ \begin{array} ~0 & \text{when } k \text{ is even} \\
\frac{7}{10} \sum_{n=0}^4 e^{-ink\pi/5} & \text{otherwise} \\
\end{array}\right .
\end{align}
Thus for even $k$, $F[k] = 0$. Otherwise, pair the terms in the sum using the fact that $k$ is odd,
\begin{align}
e^{-ik\pi/5}+e^{-i4k\pi/5}&=e^{-ik\pi/5}-e^{ik\pi/5}=-2i\sin(k\pi/5), \quad\text{ pairing } n = 1 \text{ and } n=4\\
e^{-i2k\pi/5}+e^{-i3k\pi/5}&=e^{-2ik\pi/5}-e^{i2k\pi/5}=-2i\sin(2k\pi/5), \quad\text{ pairing } n = 2 \text{ and } n = 3 .
\end{align}
So taking the real and imaginary, the $\cos$ terms in the sum cancel for all values of $k$ and we are left only with two $\sin$ terms in the imaginary part,
\begin{align}
F[k]=\left\{
\begin{array}
~0 & \text{when } k \text{ is even} \\
~\frac{7}{10} -\frac{7}{5}i \Big(\sin\big(\frac{1}{5}k\pi\big)+\sin\big(\frac{2}{5}k\pi\big) \Big)&\text{when } k \text{ is odd},
\end{array} \right.
\end{align}
which is close to the answer you wanted; the stated answer has a $\pi$ factor in the imaginary part which I think must be an error, and we differ in some signs.
PS. I have used $i$ instead of $j$ out of habit.
|
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|
A number N in base 10, is 503 in base b and 305 in base b + 2. What is the product of the digits of N? Pls help me with A number N in base 10, is 503 in base b and 305 in base b + 2. What is the product of the digits of N? I amen't solve it. Help needed
|
A number $\overline{xyz}$ in base $b$ is equal to $x\cdot b^2 + y\cdot b + z$ and so on for larger numbers continuing the pattern increasing the power of $b$ as we continue to the left. For instance, in base 10 we have $35013$ is equal to $3\cdot 10^4 + 5\cdot 10^3 + 0\cdot 10^2 + 1\cdot 10 + 3$.
So, we are told that $N = 5\cdot b^2 + 0\cdot b + 3$ and also that $N = 3\cdot (b+2)^2 + 0\cdot (b+2) + 5$
Setting these equal, we can find $b$.
Now, armed with the knowledge of what $b$ is we can find $N$.
Now, armed with the knowledge of what $N$ is in decimal, we can find the product of the digits of $N$.
|
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|
Computation of density function of $Y=\frac1X-X$ where $X\stackrel{\mathrm d}= U[0,1]$
Calculate the density function of $Y=\frac1X-X$, where $X\stackrel{\mathrm d}= U[0,1]$.
I am confused about the technique to deal with the problem like this when after the transformation, I can not express the density of $Y$ in terms of $X$
|
$1/X-X$ is a decreasing function of $X$ so $Y\in(0,\infty)$ for $X\in(0,1)$.
$$P(Y\le y)=\begin{cases}0,&y\le0\\P(1-X^2\le yX),&y>0\end{cases}$$
Now $X^2+yX-1$ is an upward opening parabola with zeroes at $\frac{-y\pm\sqrt{y^2+4}}2$, and thus $$\begin{align*}P(X^2+yX-1\ge0)&=\underbrace{P\left(X\le\frac{-y-\sqrt{y^2+4}}2\right)}_{0}+P\left(X\ge \frac{-y+\sqrt{y^2+4}}2\right)\\&=1-P\left(X\le\frac{-y+\sqrt{y^2+4}}2\right)\\&=1-\min\left\{1,\frac{-y+\sqrt{y^2+4}}2\right\}\end{align*}$$
Can you simplify this?
Rationalizing$$\frac{-y+\sqrt{y^2+4}}2\left[\frac{\sqrt{y^2+4}+y}{\sqrt{y^2+4}+y}\right]=\frac2{\sqrt{y^2+4}+y}$$Call it $g(y)$. It is decreasing for $y>0$ so $g(y)<g(0)=1$, giving $\min\{1,g(y)\}=g(y)$.
|
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|
Prove that sin A/2 * sin B/2 * sin C/2 = r/4R The other day I came across an identity in the book "Problems from the Book" and it was presented as well known:
$$\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} = \frac{r}{4R}$$
However I wasn't familiar with thee identity so I tried proving it.
Here is part of my attempt in solving this problem:
I first rewrote $\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}$ as $\frac{r^3}{AI\cdot BI \cdot CI}$ (where I is the incenter of $\bigtriangleup ABC$). Then I tried using the fact that [ABC] =abc/4R (where [ABC] represents the area of $\bigtriangleup ABC$)which allowed me to rewrite the equation as $\frac{r^2}{AI\cdot BI \cdot CI} = \frac{[ABC]}{abc}$. I tried various things at this point but none of my attempts were successful...
Does anyone have a proof of this identity?
|
$$LHS = \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} $$
$$= \sqrt{\frac{(s-b)(s-c)}{bc}} \cdot \sqrt{\frac{(s-c)(s-a)}{ca}} \cdot \sqrt{\frac{(s-a)(s-b)}{ab}}$$
$$= \frac{(s-a)(s-b)(s-c)}{abc}$$
Note that $$r = \frac{\Delta}{s}$$
$$R = \frac{abc}{4 \Delta}$$
$$\Delta^2 = s(s-a)(s-b)(s-c)$$
Now $$ RHS = \frac{r}{4R} = \frac{\Delta}{s} \cdot \frac{\Delta}{abc} = \frac{(s-a)(s-b)(s-c)}{abc}$$
Hence LHS = RHS
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to simply this radical expression $\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$ $$\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$$
I could not multiply by the conjugate since it is a cube root. Can you show me a way to simplify it?
Thanks!
|
Note $(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)=(\sqrt[3]{2})^3-1=1$ and
\begin{align}
\sqrt[3]{\sqrt[3]{2}-1}
=& \sqrt[3]{\frac1{\sqrt[3]{4}+\sqrt[3]{2}+1}}
=\sqrt[3]{\frac3{2+3\sqrt[3]{4}+3\sqrt[3]{2}+1}}
=\sqrt[3]{\frac3{(\sqrt[3]{2}+1)^3}}=
\frac{\sqrt[3]{3}}{\sqrt[3]{2}+1}
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Determine if $x^3+y^3+z^3+t^3 = 10^{2021}$ has a solution I want to know if the equation $x^3+y^3+z^3+t^3=10^{2021}$ has distinct positive integer solutions
PowersRepresentations[10^2021, 4, 3]
return
PowersRepresentations::ovfl: Overflow occurred in computation.
FindInstance[{x^3 + y^3 + z^3 + t^3 == 10^2021, 0 < x < y < z < t}, {x,y,z,t}, Integers]
My computer runs too long. How can I reduce timing to solve this equation?
|
Easy, notice that $10^{2021}=100\times 10^{3\times 673}$. Next use your code, but for the factor 100.
FindInstance[{x^3 + y^3 + z^3 + t^3 == 100, 0<x<y<z<t}, {x,y,z,t}, Integers]
yielding a single result
(*{{x -> 1, y -> 2, z -> 3, t -> 4}}*)
Now verify the solution
(x^3 + y^3 + z^3 + t^3 /. {x -> 1 10^673,y -> 2 10^673,z -> 3 10^673,t -> 4 10^673}) == 10^2021
(* True*)
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
given $a+b+c=3$ prove that $abc(a^2+b^2+c^2) \leq 3$ I am preparing for inmo and I came accross this problem while solving a worksheet, but couldn't solve it, pl help me...
Problem-
Prove that if a,b,c are non negative real numbers such that a+b+c=3, then $abc(a^2+b^2+c^2)\leq3$ ... (0)
Developments-
Firstly applying am-gm on $a+b+c$ to get $abc\leq1$
Then applying quadratic mean arthematic mean to get
$a^2+b^2+c^2 \ge 3$
Then writing the expansion of $(a+b+c)^2$ we get
$a^2+b^2+c^2 = 9-2(ab+bc+ca)$ ...(I)
Thn applying am-hm on $ab+bc+ca$ to get
$ab+bc+ca \ge 3abc$
Then substituting in (I) we get
$a^2+b^2+c^2 \leq 9-6(abc)$
Then we substituting in (0)
$LHS \leq abc(9-6abc)$
Whose max value is $27/8$ but is not less than three, so I am stuck for this point onwards
|
I will use a well known inequality for my solution -
$a^2b^2+b^2c^2+c^2a^2 \geq abc(a+b+c)$
$\implies (ab+bc+ca)^2 \geq 3abc(a+b+c) = 9abc$
Now using AM-GM,
$a^2+b^2+c^2+2 (ab + bc + ca) \geq 3 [(a^2+b^2+c^2)(ab+bc+ca)^2]^{1/3}$
$9 \geq 3[(a^2+b^2+c^2)(ab+bc+ca)^2]^{1/3}$
$3^3 \geq 9 abc(a^2+b^2+c^2)$
$3 \geq abc(a^2+b^2+c^2)$
On the inequality I used to answer comes from $(ab-bc)^2+(bc-ca)^2+(ca-ab)^2 \geq 0$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving that $f$ verifies $f(1-x)=-f(x)$ I'm trying to show an identity verified by a function. I have this function which is defined for all real numbers as
$ f(x) = \sum_{k=0}^{p-1} \frac{(x+k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!}$
I would like to show that $f(1-x) = -f(x)$.
First thing that tried:
$
\begin{align*}
f(1-x)
&= \sum_{k=0}^{p-1} \frac{(1-x+k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!} \\
&= \sum_{k=0}^{p-1} \frac{(-1)^{2m-1}(x-1-k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!}
\quad \text{factoring the numerator by }(-1)^{2m-1}\\
&= -\sum_{k=0}^{p-1} \frac{(x-1-k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!}
\quad \text{reducing }(-1)^{2m-1}=-1\\
&= -\sum_{i=-p}^{-1} \frac{(x+i)^{2m-1}}{(-1)^{p-i-1} (p+i)!(p-i-1)!}
\quad \text{doing a change of variable } i = -k-1 \\
\end{align*}
$
I'm kind of stuck at this point and I don't see how to move forward with it. I think I'm not going to the optimal direction to prove that relation.
Second thing that I am investigating:
Since $f(x)$ is a polynomial of degree $2m-1$, I could write it as a finite power series centered at $0$. So with Taylor's theorem I would have
$f(x) = \sum_{i=0}^{2m-1} \frac{f^{(i)}(0) }{i!} x^{i}$.
and
$
\begin{align*}
f(1-x)
&= \sum_{i=0}^{2m-1} \frac{ f^{(i)}(0) }{i!} (1-x)^{i} \\
&= \sum_{i=0}^{2m-1} \frac{ f^{(i)}(0) }{i!} \Bigg( \sum_{j=0}^{i} \binom{i}{j} (-x)^{j} \Bigg)
\quad \text{binomial expansion applied to } (1-x)^{i} \\
&= \sum_{j=0}^{2m-1} \Bigg((-1)^{j}\sum_{i=j}^{2m-1} \binom{i}{j} \frac{ f^{(i)}(0) }{i!} \Bigg) x^{j}
\quad \text{changing the summation order} \sum_{i=0}^{2m-1}\sum_{j=0}^{i} \text{ to } \sum_{j=0}^{2m-1}\sum_{i=j}^{2m-1} \\
\end{align*}
$
So showing that $f(1-x) = -f(x)$ is equivalent to showing that
$(-1)^{j}\sum_{i=j}^{2m-1} \binom{i}{j} \frac{ f^{(i)}(0) }{i!} = - \frac{f^{(j)}(0) }{j!} $.
I think this is the more elegant way to go but I don't know if it is going to make things worse.
Any feedback ?
EDIT : clarification of the parameters m and p :
I forgot to mention hypothesis on the parameters $m$ and $p$. So we consider that $ 1 \leq m < p$
|
We can rewrite the sum as
$$
\eqalign{
& f\left( x \right) = \sum\limits_{k = 0}^{p - 1} {{{\left( {x + k} \right)^{\,2m - 1} }
\over {\left( { - 1} \right)^{\,p + k} \left( {p - 1 - k} \right)!\left( {p + k} \right)!}}} = \cr
& = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k} {{\left( {x + k} \right)^{\,2m - 1} }
\over {\left( {p - 1 - k} \right)!\left( {p + k} \right)!}}} \quad \Rightarrow \cr
& \Rightarrow \quad g\left( x \right) = \left( {2p - 1} \right)!f\left( x \right) = \cr
& = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
\left( \matrix{ 2p - 1 \cr p + k \cr} \right)\left( {x + k} \right)^{\,2m - 1} } \cr}
$$
Then for the sum $g(1-x)+g(x)$ we get
$$
\eqalign{
& g\left( {1 - x} \right) + g(x) = \cr
& = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
\left( \matrix{2p - 1 \cr p + k \cr} \right)\left( {1 - x + k} \right)^{\,2m - 1} } + \cr
& + \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
\left( \matrix{2p - 1 \cr p + k \cr} \right)\left( {x + k} \right)^{\,2m - 1} } = \cr
& = \left( { - 1} \right)^{\,2m - 1} \sum\limits_{k = 0}^{p - 1}
{\left( { - 1} \right)^{\,p + k} \left( \matrix{ 2p - 1 \cr p + k \cr} \right)
\left( {x - k - 1} \right)^{\,2m - 1} } + \cr
& + \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
\left( \matrix{ 2p - 1 \cr p + k \cr} \right)\left( {x + k} \right)^{\,2m - 1} } = \cr
& = - \left( { - 1} \right)^{\,2m - 1} \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p - 1 - k}
\left( \matrix{2p - 1 \cr p - 1 - k \cr} \right)\left( {x - p + p - 1 - k} \right)^{\,2m - 1} } + \cr
& + \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
\left( \matrix{2p - 1 \cr p + k \cr} \right)\left( {x - p + p + k} \right)^{\,2m - 1} } = \cr
& = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,k}
\left( \matrix{2p - 1 \cr k \cr} \right)\left( {x - p + k} \right)^{\,2m - 1} } + \cr
& + \sum\limits_{k = p}^{2p - 1} {\left( { - 1} \right)^{\,k}
\left( \matrix{2p - 1 \cr k \cr} \right)\left( {x - p + k} \right)^{\,2m - 1} } = \cr
& = \sum\limits_{k = 0}^{2p - 1} {\left( { - 1} \right)^{\,k}
\left( \matrix{2p - 1 \cr k \cr} \right)\left( {x - p + k} \right)^{\,2m - 1} } = \cr
& = \left( { - 1} \right)^{\,2p - 1} \Delta ^{\,2p - 1} \left( {x - p} \right)^{\,2m - 1} = 0
\quad \;\left| {\,1 \le m < p} \right. \cr}
$$
and that is the $2p-1$-th finite difference of a polynomial of degree $2m-1$
which is null if $ m < p$
|
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|
if $f(x,y)=\frac{xy^2}{x^2+y^6}$ and $f(0,0)=0$ show that $f$ is ubounded at any neighborhood of (0,0)
if $f(x,y)=\frac{xy^2}{x^2+y^6}$ and $f(0,0)=0$ show that $f$ is ubounded at any neighborhood of (0,0) but the restritions to straight line in $\mathbb{R}^2$ is cont.
My attempt:
**Proving continuity **
if $f(x,y)=\frac{xy^2}{x^2+y^6}$ then $y=mx+c$ is equation of any line then we can get $f(x,mx+c)= \frac{x.(mx)^2+c^2.x+2mxc}{x^2+(mx+c)^6}$ then the function is continuous at any point $(x,y)$ which is on the line$y=mx+c$,
We pick any point $(a,ma+c)$ on the line $y=mx+c$,then we try to show that the function is continuous at the point
We know that $f(a,ma+c)= \frac{ma^3+c^2.a+2mac}{a^2+(ma+c)^6}$ .
Now, $\lim_{x \to a}(f(x,mx+c))=\lim_{x \to a}\frac{x^3m^2+c^2x+2mxc}{x^2+(mx+c)^6}=f(a,ma+c)$(by the property of limits).
Proving uboundedness
In this part I did some back calculation . I showed that the function will not be bounded if the numerator becomes infinity or when the denominator becomes infinity.
For the denominator to become $0$ we see that $y^6+x^2=0$ then $(y^2)^3= -x^2$ then $y^2=(-x)^{\frac{2}{3}}$ . So I have shown that if $(x,y) \to (0,0) $ along the given curve $y^2=(-x)^{\frac{2}{3}}$ the limit of the function $f$ doesnt exist.
I am not sure whether this is correct but this has been essentially my attempt.
|
Equation $y^6+x^2=0$ has only one solution in #\mathbb{R}^2$, and that is $(0,0)$.
To prove that the function is unbounded in the neighbourhood of $(0,0)$ consider (for $y\neq 0$)
$$ f(y^3,y) = \frac{y^3 \cdot y^2}{(y^3)^2+ y^6} = \frac{1}{2y}$$
|
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|
An Elegant New Proof of Herons Formula? A triangle with side lengths $a, b, c$ an altitude($h$), where the height($h_a$) intercepts the hypotenuse($a$) such that it is the sum of two side lengths, $a = u +v$ and height($h_b$) intercepts hypotenuse($b$) such that it is also the sum of two side lengths $b = x + y$, we can find a simple proof of herons formula.
Image here:
(Note that $h = \sqrt{au}$ and $h = \sqrt{bx}$, giving $au = bx$)
First the equations we have are:
$u + v = a$
$x + y = b$
$au = bx$
$\sqrt{av} + \sqrt{by} = c$
Substituting $v$ and $y$ with $u$ and $x$ for the $4$th equation:
$\sqrt{a^2 - au} + \sqrt{b^2 - bx} = c$
Then using $au = bx$ we eliminate $x$ to find $u$:
$\sqrt{a^2-au}+\sqrt{b^2-b\left(\frac{au}{b}\right)}=c$
$u = \frac{1}{4ac^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)$
Finally, substituting this into $h = \sqrt{au}$:
$h = \sqrt{\frac{1}{4c^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)}$
Plugging this into the area formula ($A = \frac{1}{2}ch$) gives:
$A = \frac{1}{2}c\sqrt{ \frac{1}{4c^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)} $
$A = \sqrt{\frac{1}{16}(c^2 - (a - b)^2)(( a + b)^2 - c^2)} $
$A = \sqrt{\frac{1}{16}(a + b - c)( a + b + c)( b + c - a)(a + c - b)} $
$A = \sqrt{s(s - a)(s- b)(s- c)}$
Q.E.D.
Thoughts?
|
It is certainly a correct proof of Heron's formula that I have not seen before. I don't believe that it is any simpler than the other proofs that I have seen, but I am still entertained by it. However, I do not know if this is an appropriate posting since you do not really have a question other than possibly 'is my proof correct?'. I believe that the stack exchange guidelines suggest we only post questions and answers, not discussions.
|
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|
Show that $a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}$ converges to zero According to WolframAlpha, the following sequence
\begin{align*}
a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}
\end{align*}
seems to converges to zero. However, how can I prove this ? The difficulty I encounter is this "one quarter of an integer" in the Gamma functions.
|
First approach. By the beta integral
\begin{align*}
\frac{{\Gamma (x)}}{{\Gamma \left( {x + \tfrac{1}{2}} \right)}} & = \frac{1}{{\Gamma \left( {\frac{1}{2}} \right)}}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}\int_0^1 {\frac{{t^{x - 1} }}{{\sqrt {1 - t} }}dt} \\ & = \frac{1}{{\sqrt \pi }}\int_0^{ + \infty } {\frac{{e^{ - xs} }}{{\sqrt {1 - e^{ - s} } }}ds} = \frac{1}{{\sqrt \pi }}\int_0^{ + \infty } {e^{ - (x - 1/4)s} s^{ - 1/2} \sqrt {\frac{{s/2}}{{\sinh (s/2)}}} ds}
\end{align*}
for all $x>0$. It is well known that $w < \sinh w$ for all $w>0$. Thus, for $x>\frac{1}{4}$,
$$
\!\! \frac{{\Gamma (x)}}{{\Gamma \left( {x + \frac{1}{2}} \right)}} \le \frac{1}{{\sqrt \pi }}\int_0^{ + \infty }\! {e^{ - (x - 1/4)s} s^{ - 1/2} ds} = \frac{1}{{\sqrt {\pi \!\left( {x - \frac{1}{4}} \right)} }}\int_0^{ + \infty }\! {e^{ - t} t^{ - 1/2} dt} = \frac{1}{{\sqrt {x - \frac{1}{4}} }}.
$$
Consequently,
$$
\frac{{\Gamma \left( {\frac{n}{2} + \frac{1}{4}} \right)}}{{\Gamma \left( {\frac{n}{2} + \frac{3}{4}} \right)}} \le \sqrt {\frac{2}{n}}
$$
for all $n\geq 1$. From the asymptotics I gave in the comments, it is seen that this bound is sharp.
Second approach. By the log-convexity and the functional equation of the gamma function, we have
$$
\log \Gamma \left( {x + \tfrac{1}{2}} \right) \le \frac{1}{2}\log \Gamma (x) + \frac{1}{2}\log \Gamma (x + 1) = \log \Gamma (x + 1) - \frac{1}{2}\log x,
$$
i.e.,
$$
\frac{{\Gamma \left( {x + \frac{1}{2}} \right)}}{{\Gamma (x + 1)}} \le \frac{1}{{\sqrt x }}
$$
for all $x>0$. Consequently,
$$
\frac{{\Gamma \left( {\frac{n}{2} + \frac{1}{4}} \right)}}{{\Gamma \left( {\frac{n}{2} + \frac{3}{4}} \right)}} \le \frac{2}{{\sqrt {2n - 1} }}
$$
for all $n\geq 1$.
|
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|
Find the last two digit of $5^{121}*3^{312}$ The answers key says by using
$$5^k\equiv 25 \pmod{100}, k>=2$$
and $$3^{40}\equiv 1 \pmod{100}$$
can have $$ 5^2 *3^4 \equiv25 \pmod{100},$$
and it follows that the last two digits of $5^{143}*3^{312}$ are $25$.
Don't know why $ 5^2 *3^4 \equiv25 \pmod{100}$ can applies that the last two digits of $5^{143}*3^{312}$ are $25$.
Also, the question is asking about $5^{121}$, why the last step relates to $5^{143}$
|
The first sentence says that whatever power greater than $1$ you pick, $5$ to that power ends in $25$. Note that the powers of $5$ are $5,25,125,625,3125\ldots $ The end digits are always $25$ after the first one because when you multiply some number of hundreds by $5$ you still have some number of hundreds and when you multiply $25$ by $5$ you get $125$. The $1$ goes into the hundreds and you are left with $25$.
Then from $3^{40} \equiv 1 \pmod {100}$ you get $3^{280} \equiv 1 \pmod {100}$ by taking the seventh power. We now want $5^{121} \cdot 3^{32}$ or $5^{143} \cdot 3^{32} \pmod {100}$. Noting that $5^2 \cdot 3^4 \equiv 1 \pmod {100}$ is new information, it is not derived from the preceding. Now we can say $(5^2 \cdot 3^4)^8=5^{16} \cdot 3^{32} \equiv 1 \pmod {100}$ We multiply by the remaining factors of $5$, whether $105$ or $127$ and get $25 \pmod {100}$
|
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|
Find all $m,n \in \mathbb N$ Such that $3(2^m)+4=n^2$.
Find All $n,m\in \mathbb N$ Such that $3(2^m)+4=n^2$
I’ve tried to plug some values of $m$, and it turns up that the only valid values is $$m\in \{2,5,6\}.$$
So once i saw that i tried to prove that it doesn’t exist a solution to the equation $\forall m\geq7$. But i have no idea how to prove this.
Something i’m not sure whether it’s true:
$$3(2^m)+4=n^2 \iff3(2^m+1)=n^2-1 \\ \iff 3(2^m+1)=(n+1)(n-1)$$
Thus $3\mid (n+1)(n-1) $, But since $3$ is a prime and $\gcd(n+1,n-1)=1$ or $2 \implies$
$3\mid n+1$ or $3\mid n-1$ What’s next?
|
(Here's a suggested approach. If you're stuck, explain where you're stuck and what you've tried.)
*
*Factor it as $ 3 \times 2^m = (n-2) ( n+2)$.
*Hence $ \{ n-2, n+2 \} = \{ 3 \times 2^a, 2^b \}$.
*Hence $ 3 \times 2^a = 2^b \pm 4 $.
*If $ b = 0, 1, 2$, what can we say? What are the solutions with these small cases?
$3 \times 2^a = 1+4, 2+4, 4+4$ (cannot take negative branch).
So we get $ b=1, a = 1 \Rightarrow 2 \times 6 = 3\times 2^2 $.
*
*If $ b \geq 3$, what can we say?
RHS is a multiple of 4 but not 8, so $ a = 2$.
$3 \times 4 = 2^b \pm 4$, which gives us $b=3, 4$.
$a = 2, b = 3 \Rightarrow 8\times 12 = 3\times 2^5 .$
$a = 2, b = 4 \Rightarrow 12 \times 16 = 3\times 2^6 . $
In conclusion, we have solutions only when $ m = 2, 5, 6$.
|
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|
A very small disc winds around a larger circular disc of radius $R$ connected to it by a string. How long is the spiral it travels? A very small disc winds around a larger circular disc of radius $R$. It is connected to it by a string of length l that remains tight. What distance does it travel before it hits the larger disc of radius $R$ ? I've estimated that it is $2\pi(l-R)$ based on assuming a spiral is a little like a series of concentric circles.
|
Another way to add up your diminishing circle circumferences
is to use the formulas $2\pi R = \frac LN$ and $N\pi = \frac{L}{2R},$
both of which are derived from $N = \frac{L}{2\pi R}.$
\begin{align}
\left(L - \frac LN\right)2\pi &+ \left(L - 2\frac LN\right)2\pi
+ \left(L - 3\frac LN\right)2\pi + \cdots \\
&= \left(1 - \frac 1N\right)2\pi L + \left(1 - \frac 2N\right)2\pi L
+ \left(1 - \frac 3N\right)2\pi L + \cdots \\
&= \left(N - 1\right)\frac{2\pi L}{N}
+ \left(N - 2\right)\frac{2\pi L}{N}
+ \left(N - 3\right)\frac{2\pi L}{N} + \cdots \\
&= \left((N - 1) + (N - 2) + (N - 3) + \cdots + 1\right)\frac{2\pi L}{N} \\
&= \left(\frac{N(N - 1)}{2}\right)\frac{2\pi L}{N} \\
&= (N - 1)\pi L \\
&= N\pi L - \pi L \\
&= \frac{L^2}{2R} - \pi L. \\
\end{align}
But this is an underestimate, because at the beginning the end of the string is tracing a path of length approximately $L \theta$, where $\theta$ is the change in the angle of the string, although you have assumed it is only
$\left(1 - \frac 1N\right)L \theta.$
You could get a better approximation by wrapping the string around a regular polygon of $k$ sides and then letting $k$ increase to infinity
(effectively computing the integral of the path length as the string wraps around the circle);
the $\pi L$ term then drops out.
Your answer would also come up with the same sum except for a couple of errors.
The first error is that $2\pi \times 2\pi = 4\pi^2$ but you wrote $4\pi.$
The second error is on the very last step. Note that $2\pi N = \frac LR$
and $2\pi RN = L$ but $\pi(N + 1) = \frac{L}{2R} + \pi,$
so the result actually should be
\begin{align}
2\pi NL - 2\pi^2 RN(N+1)
&= \frac{L^2}{R} - L\left(\frac{L}{2R} + \pi\right) \\
&= \frac{L^2}{R} - \frac{L^2}{2R} - \pi L \\
&= \frac{L^2}{2R} - \pi L.
\end{align}
|
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|
Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$ Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
I attempted to solve this question as follows:
$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
$a^2(b+c)+b^2(c+a)+c^2(a+b)-(a^3+b^3+c^3)\le 3abc$
$(a+b)(b+c)(c+a)-2abc-3abc\le 3abc$
$(a+b)(b+c)(c+a)-5abc\le 3abc$
$(a+b)(b+c)(c+a)\le 8abc$
And I got stuck here. Could you please explain to me how to finish it off?
|
It is just Schur’s $^{[1]}$ Inequality of degree 3. You can search the web and read about it. It is pretty well known and someone named Aritra12 posted a handout on it recently on Aops.
|
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|
Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $ How to determine the range of the function
$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$
It it is straightforward to verify that $f(x)$ is even and
$$f(0)= \frac13, \>\>\>\>\> \lim_{x\to\pm \infty} f(x) \to -\infty$$
which implies $f(x) \in (-\infty,\frac13]$, i.e.
$$\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le \frac13
$$
and is visually confirmed below
However, it is not obvious algebraically that $f(x)$ monotonically decreases away from $x=0$. The standard derivative tests are not viable due to their rather complicated functional forms.
So, the question is how to prove the inequality $f(x) \le \frac13$ with rigor.
Note that it is equivalent to proving
$$\cot \left(\frac{\pi(1+x)}{3+x^2}\right)
\cot \left(\frac{\pi(1-x)}{3+x^2}\right)\le \frac13
$$
|
Partial answer :
Well we can use Jensen's inequality remarking that :
$$f(x)=\ln\Big(\tan\Big(\frac{\pi}{2}\frac{(1+x)^2}{3+x^2}\Big)\Big)$$
Is concave on $[\frac{-17}{100},\frac{17}{100}]$
So we have :
$$f(x)+f(-x)\leq 2f(0)$$
A bit of algebra and we get the result !
We can improve the reasoning for that we use a substitution:
$$x=\frac{a}{a+1}$$
And follow the same reasoning !
The concavity (an attempt)for $-\frac{17}{100}<x<0$:
We start by introduce a function called $j$ :
$$j(x)=\ln\left(\tan\left(\frac{\left(\frac{x}{x+1}-1\right)^2}{3+\left(\frac{x}{x+1}\right)^2}\right)\right)$$
The derivative is equal to :
$$j'(x)=-\frac{\pi(4x+3)\csc\left(\frac{\pi}{2(4x^2+6x+3)}\right)\sec\left(\frac{\pi}{2(4x^2+6x+3)}\right)}{(4x^2+6x+3)^2}$$
Using the substitution $y=4x^2+6x+3$
The function :
$$h(y)=-\frac{\csc\left(\frac{\pi}{2(y)}\right)\sec\left(\frac{\pi}{2(y)}\right)}{(y)}$$
Is increasing and negative .On the other hand the function :
$$g(x)=\frac{\pi(4x+3)}{(4x^2+6x+3)}$$ is decreasing positive .
So as a multiplication we deduce that the function is negative increasing so the second derivative is positive . $j$ is convex .
We have $j\left(\frac{x}{1-x}\right)=f(-x)$
We Differentiate and using the fact that :
$k(x)=\frac{-1}{(1-x)^2}$ is negative decreasing and $j$ is negative decreasing we deduce by multiplication that $-f'(-x)$ is decreasing so $f(-x)$ is concave on the interval .
Hope it helps you !
|
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|
Find constant $c$ such that all intersection points of two spheres have perpendicular tangent planes I have come to a problem in a multivariable calculus book that I'm having trouble with.
The problem statement is :
"Find a constant $c$ such that for any point of intersection of the two spheres
$(x-c)^{2} + y^{2} + z^{2} = 3$ and $x^{2} + (y-1)^{2} + z^{2} = 1$, the corresponding
tangent planes will be perpendicular to each other."
So I define two functions :
\begin{align}
f(x,y,z) & = (x-c)^{2} + y^{2} + z^{2} - 3 \\
g(x,y,z) & = x^{2} + (y-1)^{2} + z^{2} - 1
\end{align}
We denote the partial derivative with respect to $x$ as $f_{x}$ and so on...
We see :
\begin{align}
f_{x}(x,y,z) & = 2(x-c) = 2x - 2c\\
f_{y}(x,y,z) & = 2y\\
f_{z}(x,y,z) & = 2z
\end{align}
and :
\begin{align}
g_{x}(x,y,z) & = 2x \\
g_{y}(x,y,z) & = 2(y-1) = 2y - 2 \\
g_{z}(x,y,z) & = 2z
\end{align}
I assume that if the tangent planes of the two spheres are perpendicular at an intersection point, then the normals are also perpendicular. So for every intersection point (x,y,z)
we have :
\begin{equation}
\bigtriangledown f(x,y,z) \cdot \bigtriangledown g(x,y,z) = 0
\end{equation}
So :
\begin{align}
\require{cancel}
(2x - 2c, 2y, 2z) \cdot (2x, 2y-2, 2z) & = 0\\
(2x-2c)2x + (2y)(2y) - 2(2y) + 4z^{2} & = 0\\
4x^{2} - 4xc + 4y^{2} - 4y + 4z^{2} & = 0\\
4(x^{2} + y^{2} + z^{2}) - 4(xc + y) & = 0\\
\cancel{4} \left[ (x^{2}+y^{2}+z^{2}) - (xc + y) \right] & = 0\\
(x^{2}+y^{2}+z^{2}) - (xc + y) & = 0 \\
x^{2} + y^{2} + z^{2} & = xc + y
\end{align}
We also see :
\begin{align}
\require{cancel}
f(x,y,z) & = x^{2} - 2xc + c^{2} + y^{2} + z^{2} - 3 \\
& = (x^{2} + y^{2} + z^{2}) + (c^{2} - 2xc - 3) \\
g(x,y,z) & = x^{2} + y^{2} - 2y + \cancel{1} + z^{2} - \cancel{1} \\
& = (x^{2}+y^{2}+z^{2}) - 2y
\end{align}
and :
\begin{align}
f(x,y,z) = 0 & \Rightarrow (x^{2}+y^{2}+z^{2}) = -(c^{2}-2xc-3) = -c^{2} + 2xc + 3\\
g(x,y,z) = 0 & \Rightarrow (x^{2}+y^{2}+z^{2}) = 2y
\end{align}
So we have :
\begin{equation}
xc + y = -c^{2} + 2xc + 3 = 2y
\end{equation}
It is here that I am stuck. It seems that the goal here would be to obtain an expression for $c$ that doesn't include $x$ or $y$, but I do not know how to obtain it.
Can someone help with this ?
|
Because of the symmetry of spheres with respect to rotation, the statement “for any point of intersection between two spheres $A$ and $B$, the corresponding tangent planes are perpendicular” is equivalent to “between two spheres $A$ and $B$, there is a point of intersection at which the corresponding tangent planes are perpendicular.”
This means that we can reduce this problem to two dimensions by looking at the intersection of the spheres with a plane that passes through the center of both. Doing so we get two intersecting circles, one with radius $1$, the other with radius $\sqrt{3}$.
Rather than look at the two circles actually created by this, for now we will look at the semicircles $y=\sqrt{1-x^2}$ and $y=\sqrt{3-(x-d)^2}$. These intersect at $x=\frac{d^2-2}{2d}$, and for their tangent lines to be perpendicular, their derivatives must be perpindicular, so we have
$$0=\frac{\sqrt{3-(x-d)^2}}{x-d}+\frac{x}{\sqrt{1-x^2}}$$
Substituting for $x$ and solving, we get $d=\pm 2$. Because of this, we know that the centers of the spheres in the problem must be a distance of 2 apart, which is equivalent to $c^2+1^2=2^2 \to c=\pm \sqrt{3}$.
|
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|
Find $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ exactly by hand Show the exact value of $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ is $\frac{-(1-\sqrt{17})}{4}$ (no calculator).
By and large these things require considerations of Euler's formula so I rewrote the expression as $$\frac{1}{2}(a+a^2+a^4+a^8)+\frac{1}{2}(a^{-1}+a^{-2}+a^{-4}+a^{-8})$$ where $a=e^\frac{2\pi i}{17}$. This looks really nice like you could sum it or something but it really just doesn't play nice when you try to. It is not geometric, so pretty much the only way forward is to find a function whose taylor series starts with $f=x+x^2+x^4+x^8+...$ or something workable to look like that, but this is not obvious at all either, and the next problem would be whether that function can be used to evaluate $f(e^\frac{2\pi i}{17})$ easily which at this point is hopelessly improbable. By now I have no where else to turn, it seems likely that the solution doesn't involve complex numbers.
|
Let
$$s_1=\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}
$$
$$s_2=\cos\frac{6\pi}{17} + \cos\frac{10\pi}{17} + \cos\frac{12\pi}{17} + \cos\frac{14\pi}{17}
$$
and note
$$s_1s_2=2(s_1+s_2) =2\cdot (-\frac12)=-1
$$
Thus, $s_1$ and $s_2$ satisfy $s^2+\frac12 s-1=0$, which yields $s_1= \frac{-1+\sqrt{17}}4$ and, as a by-product, $s_2= \frac{-1-\sqrt{17}}4$.
|
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|
Easiest way to solve $\frac{2}{9}x=\sin(\frac{\pi}{9}x)$ We wanted to find the points of intersection between 2 curves, so we have to solve this equation $\frac{2}{9} x= \sin(\frac{\pi}{9}x)$.
I couldn't figure out an easy way to find the points of intersection, they are $-\frac{9}{2}$, $0$, and $\frac{9}{2}$.
Any ideas please on how to solve such equations?
|
uhm the following way isn't that easy... you have to consider several properties of functions...
first off: $\sin(\dfrac{\pi x}{9}) = \dfrac{2x}{9} \iff \sin^{-1}(\dfrac{2x}{9}) = \dfrac{\pi x}{9}$
Let: $f(x) =\sin^{-1}(\dfrac{2x}{9}) - \dfrac{\pi x}{9} $
Our goal is to find where $f(x) = 0$
Notice $f(x) = -f(-x) \implies f(x)$ is an odd function. If an odd function is defined at zero, then its graph must pass through the origin. Proof. This means $x=0$ is a solution.
Now lets examine the domain of $f(x)$. Note $\sin^{-1}(x)$ is defined only when $-1\leq x\leq1$
This means domain of $f(x) =[-\dfrac{9}{2},\dfrac{9}{2}]$
Now lets examine the derivative of f(x):
$f'(x) = \dfrac{2}{9\sqrt{1-\dfrac{4x^2}{81}}} - \dfrac{\pi}{9}$
Now $f'(x) =0 \implies x = \pm \dfrac{9\sqrt{1-\dfrac{4}{\pi^2}}}{2} = \pm a, a>0$.
After drawing a sign diagram one can see:
f(x) is increasing $-\dfrac{9}{2}\leq x\leq -a$
f(x) is decreasing $-a\leq x\leq a$
f(x) is increasing $a\leq x\leq \dfrac{9}{2}$
Since f(x) is strictly decreasing in the interval $[-a,a]$ where $a>0$ and $f(-a) >0$ and $f(a)<0$, then there must be only one root from $[-a,a]$ and we already saw that $x = 0$ is the solution and now we can confirm that this is the only root in $[-a,a]$ Also notice the boundary points are $\pm \dfrac{9}{2}$ where the function is zero. With the same increasing/decreasing function logic we can confirm these two are also the only roots in their respective interval.
|
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|
Proving that $(X-1,Y^2-4Y-XY+Y+4)=(X-1,Y-2) $ We consider the ideal $I=(X-1,Y^2-4Y-XY+Y+4)$ of the polynomial ring $k[X,Y]$. I need to prove that $(X-1,Y^2-4Y-XY+Y+4)=(X-1,Y-2) $. Since $Y^2-4Y-XY+Y+4=(Y-2)^2-Y(X-1)$, it is easy to see that $I \subseteq (X-1,Y-2) $, but I can't prove that $(X-1,Y-2) \subseteq I$.
Any help would be appreciated.
|
The claim is false since
$$\begin{align}
(x-1,f(x)) &= (x-1,\,f(x)\bmod x-1)\\
&= (x-1,\,f(1))\\
&= (x-1,\,y^2-4y+4)\ \ {\rm in\ OP}\\
&= (x-1,\,(y-2)^2)
\end{align}\qquad$$
but $\,(x\!-\!1,(y\!-\!2)^2)\neq (x\!-\!1,y\!-\!2),\,$ else $\,y\!-\!2 = (x\!-\!1)f + (y\!-\!2)^2g,\,$ so $\,y\!-\!2\mid f,\,$ so cancelling $y\!-\!2$ yields $\,1 = (x\!-\!1)f'+(y\!-\!2)g,\,$ so $\,1 = 0\,$ by eval at $\,x=1,y=2$.
|
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|
Proof for The Limit $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4},\space n\in\Bbb{N}$ I am revising my knowledge on the topic of real analysis by attempting some simple proofs. The question requires for the proof of the Limit for $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4}$ using the definition of convergence: $$\forall\epsilon>0\space\space\exists N\in\Bbb{R}\space s.t\space\space \forall n>N,|s_n-L|<\epsilon$$ Below is my working: $$|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\epsilon$$ $$|\frac{3n^2-n}{4n^2+1}-\frac{3n^2+\frac{3}{4}}{4n^2+1}|=|\frac{-n-\frac{3}{4}}{4n^2+1}|=\frac{|-n-\frac{3}{4}|}{4n^2+1}<\frac{4n}{4n^2}=\frac{1}{n}<\epsilon $$ $$\frac{1}{n}<\epsilon\rightarrow\frac{1}{\epsilon}<n$$ $$\forall n>N=\frac{1}{\epsilon},\space|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\frac{1}{n}<\epsilon$$ By the definition of convergence,$\space\frac{3n^2-n}{4n^2+1}\rightarrow\frac{3}{4}$ as $n\rightarrow\infty.$
Is this correct? Any feedback would be greatly appeciated.
|
Given $\epsilon>0$ be any number.
Now
\begin{align}
&\left|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}\right|\\
&=\left|\frac{-4n-3}{4(4n^2+1)}\right|\\
&=\frac{4n+3}{4(4n^2+1)}\\
&<\frac{n+1}{(4n^2+1)}\\
&<\frac{n+1}{(n^2+1)}\\
&<\frac{n+1}{n^2}<\epsilon\quad &\text{whenever}\quad n^2\epsilon-n-1>0\\
&&\text{i.e whenever}\quad n>\dfrac{1+\sqrt {1+4\epsilon}}{2\epsilon}\\
\end{align}
Now you can take $N=\left[{\dfrac{1+\sqrt {1+4\epsilon}}{2\epsilon}}\right]+1$ and this also allows $N$ to be a positive integer rather than just a real number which, in cases, can be difficult to find out.
Therefore
$\left|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}\right|<\epsilon$ whenever $n>N$
|
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|
Prove that $\tan(x)\tan(x+r) = \frac{\tan(x+r) - \tan(x)}{\tan r} -1$ $\tan(x)\cdot \tan(x+r) = \dfrac{\tan(x+r) - \tan(x)}{\tan r} -1$
I tried L.H.S = $\dfrac{\sin(x)\cdot \sin(x+r)}{\cos(x)\cdot \cos(x+r)}$
= $\dfrac{\sin(x+r- x)}{\sin(r)}\dfrac{\sin(x)\cdot\sin(x+r)}{\cos(x)\cdot \cos(x+r)}$
= $\dfrac{\sin(x+r- x)}{\sin(r)}\dfrac{\sin(x)\cdot\sin(x+r)}{\cos(x)\cdot\cos(x+r)}$
= $\dfrac{\sin(x)\cdot\sin(x+r)}{\sin(r)}\dfrac{\sin(x+r)\cos(x) - \cos(x+r)\cos(x)}{\cos(x)\cdot \cos(x+r)}$
= $\dfrac{\sin(x)\cdot\sin(x+r)}{\sin(r)}\left(\tan(x+r) -\tan(x)\right)$
I get stuck here; what else is there to do?
Edit: using Tavish's hint, we can get it directly.
$$\tan(x+r -x) = \dfrac{\tan(x+r) - \tan(x)}{1 + \tan(x+r)\tan(x)}$$
$$1 + \tan(x+r)\tan(x)= \dfrac{\tan(x+r) - \tan(x)}{\tan(r)}$$
$$\tan(x+r)\tan(x) = \dfrac{\tan(x+r) - \tan(x)}{\tan(r)} -1$$
|
Hint:
Use $$\tan(a-b)=\frac{\tan a -\tan b}{1+\tan a \tan b} $$ with $a=x+r, b=x$.
|
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|
It is given that $x.y.z=12^3$ where $x,y,z$ are positive integers.Then , how many different solutions are there for $x+y+z$?
It is given that $x.y.z=12^3$ where $x,y,z$ are positive integers.Then , how many different solutions are there for $x+y+z$?
Its answer is easy if it were $x.y=12^3$ , because there are $28$ possible $(a,b)$ pairs by $x.y=2^6 \times 3^3$ and $C(8,2) \times C(4,1) =28$.
Moreover, half of $28$ will give us the number of $x+y$ thanks to symmetry property.
Let's come to original question:
Firtly , i found the number of how many possible $x,y,x$ there exist by combination with repetition such that $C(6+3-1,6) \times C(3+3-1,3)=280$.
However , when i want to find the number of different values of $x+y+z$ , i stuck in.Because , i thought that if i divide $280$ by $3$ , i can obtain the result.However , it did not work and i dont know what i should do.
Therefore, i hope to find tricks or solutions for my problem. Moreover , what can i do for the expanding versions of this question such as $x+y+z+t$
|
Let's define an order that $a < b$ if the power of $3$ that divides $a$ is less than the power of $3$ that divide $a$ or those powers are equal then if power of $2$ that divides $a$ is larger then the power than divides $b$
Then wolog $x \le y \le z$ and we need to consider the cases
$x=y=z$
$x=y < z$
$x < y = z$
$x < y < z$.
Case 1: $x = y = z$ so $3^1|x,y,z$ and $2^2|x,y,z$ and $x=y=z=12$ and $x+y+z = 36$.
Case 2: $x=y < z$ then either $3|x,y,z$ and $2^3|x,y$ or $3^3|z$ and $1,2,4|x,y$
That gives us $4$ more options $x=y=24;z=3$ or $x=y=1,2,4; z = 27\cdot(2^6,2^3,1)$
So that's $5$ options so far.
Case 3: $x< y = z$ then $3|x,y,z$ and either $2^6|x$ or $2^4|x, 2|y,z$
So that's $2$ more options. $7$ so far.
Case 4: $x < y < z$.
either $3|y; 3^2|z$ or $3|x,y,z$ and $2^4|x; 2^2|y$ or $2^3|x; 2^2|y; 2|z$
The latter two give us $2$ more for $9$.
And finally The first case we can have the powers of $2$ distributed any way.
That $3$ times $1,1,2^6$.
$1$ times $1,2^2,2^4$ and $2$ times $1,2^3,2^3$.
And $2$ times $2,2,2^4$. And $6$ times $2,2^2,2^3$.
And $1$ times $2^2,2^2,2^2$ so that is $3+1+2+2+6+1=15$.
So a total of $24$.
I'm taking it on faith, perhaps incorrectly, that we can show these solutions are pairwise relatively prime so distinct.
And as Robert Isreal got $49$ sums I'm obviously making an error. Probably missing some options.
Still I'll leave this up as a strategy that I think is valid.
|
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|
I do not understand this particular step in a proof using the Well Ordering Principle Below is a proof using the Well Ordering Principle. I get lost starting at $(13)$...
$$
\begin{aligned}
P(c): &:=c^{3} \leq 3^{c} \\
& \equiv c^{3} \leq 3(c-1)^{3} \\
& \equiv c \leq \sqrt[3]{3} \times(c-1)
\end{aligned}
$$
I don't understand how we get $c^{3} \leq 3(c-1)^{3}$ from $c^{3} \leq 3^{c}$.. why does the righthand side of the inequality change like that?
|
You want to show that
$c^3 \le 3^c$
by induction.
This means that you want to show that
$(c-1)^3 \le 3^{c-1}$
implies that
$c^3 \le 3^c$.
By the induction hypothesis,
$3^c = 3\cdot 3^{c-1}
\ge 3(c-1)^3
$.
Therefore,
if you can show that
$3(c-1)^3
\ge c^3
$
then
$3^c \ge c^3
$.
Now we can proceed as above.
Taking cube roots,
$3(c-1)^3
\ge c^3
$
becomes
$3^{1/3}(c-1)
\ge c
$
or
$c(3^{1/3}-1)
\ge 3^{1/3}
$
or
$c
\ge \dfrac{3^{1/3}}{3^{1/3}-1}
= \dfrac{1}{1-3^{-1/3}}
\approx 3.26...
$
so the induction will work for
$c \ge 4$.
Since
$4^3 (64) < 3^4 (81)
$,
$c^3 \le 3^c$
for $c \ge 4$.
(With induction,
you always have to check
the base case.)
|
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|
Find the least $n$ such that $f(n)/g(n)=4/7$
Let $f(n)$ and $g(n)$ be functions satisfying $$f(n)=\cases{\sqrt{n} \hspace{50pt} \sqrt{n} \in \mathbb{N} \\ 1+f(n+1) \hspace{14pt} \text{otherwise}} $$
$$g(n)=\cases{\sqrt{n} \hspace{50pt} \sqrt{n} \in \mathbb{N} \\ 2+g(n+2) \hspace{14pt} \text{otherwise}}$$
for positive integers $n$. Find the least positive value of $n$ such that $\dfrac{f(n)}{g(n)}=\dfrac47$.
I found the values for $f$ and $g$ till $n=9$, I noticed that $f(n)=g(n)$ happened at $n=1,2,4,5,7,9$ and for some values it also happened that if $f(a)=b$, then $f(b)=a$. I don't know if that is useful in anyway. I can't see how to approach the problem.
Please give some hints.
|
I can not believe the number of arithmetic errors I made. All inexcusable. but.
Let $n > 0$ so there is a $k$ so that $(k-1)^2 < k \le n^2$ and let $j = k^2 - n$.
Then $$f(n) = f(k^2-j)= 1 + f(k^2-j+1) = 2+f(k^2 - j + 2) =.... =j + f(k^2 -j + j) = j + f(k^2) = j + k$$. (That holds if $j=0$.)
Now similarly if $j$ is even then $g(n) = j +k$ as well. But if $j$ is odd then.
$$g(k^2 - j) = (j-1) + g(k^2 - 1) = (j+1) + g(k^2 + 1) = (j + 2k + 1) + g(k^2+ 2k + 1) = (j+2k +1) + g((k+1)^2) = (j+2k +1) + (k+1) = j + 3k + 2$$.
So if $\frac {f(n)}{g(n)} \ne 1$ we must have $j$ is odd and $\frac {f(n)}{g(n)} = \frac {j+k}{j + 3k + 2}$.
So we need $\frac {j+k}{j + 3k + 2} = \frac {4M}{7M}$ and
$j + k = 4M; j + 3k + 2 = 7M$ and $j$ is odd. And ... arithmetic the bane of mathematicians.
$j= 4M - k$ (so $k$ is odd)
$4M - k + 3k + 2 = 4m + 2k + 2 = 7M$ so $k = \frac {3M-2}2$.
A bit of manipulation we can see than $M$ is divisible by $4$ as $k$ is odd. ($\frac {3M-2}2 = 2w -1\implies 3M=4m$).
Plugging in $M=4w$ we get
$k = \frac {12w-2}2 = 6w -1$ and $j = 16M -k = 10w + 1$.
But we must also have that $j < k^2 - (k+1)^2 = 2k-1$ so $10w + 1 < 12w -3$ so $w > 2$.
Now $w = 3$ and we have $k=17$ and $j = 31$ and $n = 17^2 - 31 > 17^2 -34 +1 = 16^2$.
And $f(17^2 - 31) = 31 + f(17^2) = 48 = 4\cdot 12$.
And $$g(17^2 -31) = 30 + g(17^2 -1)=32 + g(17^2+1)=32+34 + g(17^2 + 34 + 1)=66+g(18^2) = 84 = 7\cdot 12$$.
So $n = 17^2 - 31 = 258$ is the smallest such $n$.
(To minimimize $(k-1)^2 < n=k^2 -j < k^2$ where $n = k^2 -j=(6w-1)^2-(10w + 1)=36w^2 -20w=4w(6w-5)$ we must minimize $w$.)
|
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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.