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Prove by induction that $\sum _{r=1}^n \cos((2r-1)\theta) = \frac{\sin(2n\theta)}{2\sin\theta}$ is true $\forall \ n \in \mathbb{Z^+}$
Prove by induction that $$\sum _{r=1}^n \cos((2r-1)\theta) = \frac{\sin(2n\theta)}{2\sin\theta}$$ is true $\forall \ n \in \mathbb{Z^+}$
So my attempt as this is as follows, I've started the inductive step, but I don't know where to continue from now, any help would be great.
If $n=1$
LHS = $\cos\theta$, RHS = $\frac{\sin(2\theta)}{2\sin\theta} = \cos\theta$ so $\therefore$ true when $n=1$.
Assume true for $n=k$,
$$\sum _{r=1}^k \cos((2r-1)\theta) = \frac{\sin(2k\theta)}{2\sin\theta}$$
If $n=k+1$
$$\sum _{r=1}^{k+1} \cos((2r-1)\theta) = \sum _{r=1}^k \cos((2r-1)\theta) + \cos((2k+1)\theta)$$
$$ = \frac{\sin(2k\theta)}{2\sin\theta} + cos((2k+1)\theta)$$
$$= \frac{\sin(k\theta)\cos(k\theta)}{\sin\theta} + \cos(2k\theta)\cos\theta - \sin(2k\theta)\sin\theta$$
I'm not really sure my last step has lead me anywhere, but i wasn't sure on what else to do than apply the compound angle formulae.
| Starting from where you left off:
$$\frac{\sin (2k\theta)}{2\sin \theta} + \cos ((2k+1)\theta)$$
Let $x=2k\theta$. Then the expression is
$$\frac{\sin x}{2\sin\theta} + \cos (x+\theta)$$
Angle sum identity gives
$$\frac{\sin x}{2\sin\theta} + \cos x\cos\theta - \sin x\sin\theta$$
$$=\frac{\sin x + 2\cos x\cos\theta\sin\theta - 2\sin x\sin^2 \theta}{2\sin\theta}.$$
Factoring, we get
$$=\frac{(1-2\sin^2\theta)\sin x + (2\cos\theta\sin\theta)\cos x}{2\sin\theta}.$$
We use the double angle formula and angle sum formula in reverse:
$$=\frac{\cos 2\theta \sin x + \sin 2\theta \cos x}{2\sin \theta}$$
$$=\frac{\sin(x+2\theta)}{2\sin\theta}$$
$$=\frac{\sin(2k\theta + 2\theta)}{2\sin \theta}$$
$$=\frac{\sin(2(k+1)\theta)}{2\sin \theta}.$$
This completes the inductive step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Given $f(x)=ax^3-ax^2+bx+4$ Find the Value of $a+b$ Let $f(x)=ax^3-ax^2+bx+4$. If $f(x)$ divided by $x^2+1$ then the remainder is $0$. If $f(x)$ divided by $x-4$ then the remainder is $51$. What is the value of $a+b$?
From the problem I know that $f(4)=51$.
Using long division, I found that remainder of $\frac{ax^3-ax^2+bx+4}{x^2+1}$ is $a+b+x(b-a)$.
Then
$$a+b+x(b-a)=0$$
I can't proceed any further so I'm guessing the other factor of $f(x)$ is $ax+4$.
Then
$$f(x)=(ax+4)(x^2+1)=ax^3+4x^2+ax+4=ax^3-ax^2+bx+4$$
I found that $a=-4$ and $b=a=-4$. Then $f(x)=-4x^3+4x^2-4x+4$. But I doesn't satisfy $f(4)=51$
| This question is very interesting......
Only by this statement, the question can be solved
$f(x)= ax^3-ax^2+bx+4$ is a multiple of $x^2+1$;
Because the roots of the equation $x^2+1$ would be the roots of $f(x)$.
Therefore if $j,k$ are roots of the equation then $j+k=0$ and $j.k =1$.
Let the roots of $f(x)$ be $j,k,l$.
Then,
$j+k+l = -\frac{-a}{a}$
$l=1$
now, $j.k.l= -\frac{4}{a}$
so, $a= -4$
now $j.k + l(k+j) = \frac{b}{-4}$
Therefore $b=-4$
Finally $a+b=-8$
But $f(4)= 4(-4^3+4^2-4+1)$
Therefore $f(4)= -204$
So the question may be incorrect.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Maclaurin series for $f(x) = \frac{1}{1+x+x^2}$ I need to get the Maclaurin series and its radius of the convergence for $f(x) = \dfrac{1}{1+x+x^2}$. I tried to do it manually, with getting the derivatives, but I gave up after some time, because I thought there had to be a better way to solve this. Could anyone help? Thanks in advance!
| Alternatively:
$$\begin{align}\frac{1}{1+x+x^2}&=1-(x+x^2)+(x+x^2)^2-(x+x^2)^3+(x+x^2)^4-(x+x^2)^5+\cdots\\
&=1-x+x^2-x^3+x^4-x^5+x^6-x^7+\cdots\\
&\qquad -x^2+2x^3-3x^4+4x^5-5x^6+\cdots\\
&\qquad \ \ \ \ \ x^4-3x^5+6x^6-10x^7+\cdots\\
&\qquad -x^6+4x^7-10x^8+\cdots\\
&\qquad \ \ \ \ \ x^8-5x^9+\cdots\\
&\qquad -x^{10}+\cdots=\\
&=(1-x+0\cdot x^2)+(x^3-x^4+0\cdot x^5)+(x^6-x^7+0\cdot x^8)+\cdots.\end{align}$$
Do you see Pascal's triangle?
And also the pattern of coefficients $(1,-1,0)$ inside brackets at the end?
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does Brent/Salamin algorithm double the digits of $\pi$ with each iteration? The Brent-Salamin-Formula uses the arithmetic-geometric mean to calculate $\pi$.
There are many sophisticated proofs proving very sharp error bounds, for example in Salamin's 1976 paper or in this 2018 paper by Richard Brent, equation (20), p. 9.
Is there an easy way to prove the (much weaker) error bound $|\pi-p_{n+1}|<|\pi-p_n|^2$?
A partial solution proving the quadratic convergence of the AGM is this:
\begin{align*}
d_{n+1}&=c_{n+1}^2 =a_{n+1}^2-b_{n+1}^2\\
&= \left(\frac{a_n+b_n}{2}\right)^2-a_n\cdot b_n\\
&= \frac{a_n^2+2a_nb_n+b_n^2-4a_nb_n}{4}\\
&= \left(\frac{a_n-b_n}{2}\right)^2\\
&=\frac{(a_n-b_n)^2\cdot(a_n+b_n)^2}{4\cdot(a_n+b_n)^2}\\
&= \frac{(a_n^2-b_n^2)^2}{4\cdot(a_n+b_n)^2}\\
&< \frac{\left(c_n^2\right)^2}{4\cdot\left(1/\sqrt{2}+1/\sqrt{2}\right)^2}\\
&=\frac{1}{8}\cdot d_n^2
\end{align*}
| An easy proof of the quadratic convergence can be found on p. 13-14 of this preprint.
| {
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Solving a limit of integral with L'Hopital I have this limit:
$\displaystyle \lim\limits_{x\to \infty}\dfrac{1}{x}\int_{0}^x \dfrac{1}{2+\cos(\mathrm t)}\, \mathrm{dt}$
I said that
Edit:
$-1\leq \cos\mathrm{t}\leq 1 \Rightarrow \dfrac{1}{2+\cos\mathrm{t}} > 0 \text{ and because that expression has no limit as x -> inf}\\\Rightarrow \displaystyle \int_{0}^x \dfrac{1}{2+\cos\mathrm{t}}\, dt \to \infty \\ \\ \Rightarrow \lim\limits_{x\to \infty} \dfrac{\displaystyle\int_{0}^x \dfrac{1}{2+\cos\mathrm{t}}\,\mathrm{dt}}{x} = \lim\limits_{x\to \infty} \dfrac{\Big( \displaystyle\int_{0}^x \dfrac{1}{2+\cos \mathrm{t}}\, \mathrm{dt}\Big)'}{x'} = \lim\limits_{x\to \infty} \dfrac{\dfrac{1}{2+\cos \mathrm{x}}}{1} = \lim\limits_{x\to \infty} \dfrac{1}{2+\cos \mathrm{x}}$
But here I got stuck because that limit does not exist, because cos x has no limit when x goes to infinity.
But the correct answer is $\dfrac{1}{\sqrt 3}$, what did I do wrong?
| Note that
\begin{align}
\int^{2\pi n}_0 \frac{dt}{2+\cos t} = n \int^{2\pi}_0 \frac{dt}{2+\cos t}
\end{align}
which means
\begin{align}
\frac{1}{2\pi n}\int^{2\pi n}_0 \frac{dt}{2+\cos t} = \frac{1}{2\pi}\int^{2\pi }_0 \frac{dt}{2+\cos t}.
\end{align}
Hence
\begin{align}
\lim_{n\rightarrow \infty} \frac{1}{2\pi n}\int^{2\pi n}_0 \frac{dt}{2+\cos t}= \frac{1}{2\pi}\int^{2\pi }_0 \frac{dt}{2+\cos t}= \frac{1}{\sqrt{3}}.
\end{align}
This suggests that
\begin{align}
\lim_{x\rightarrow \infty} \frac{1}{x}\int^x_0 \frac{dt}{2+\cos t} = \frac{1}{\sqrt{3}}
\end{align}
if the limit exists.
To show that the limit exists, observe we have that
\begin{align}
\left|\frac{1}{x}\int^{x}_0 \frac{dt}{2+\cos t}- \frac{1}{2\pi} \int^{2\pi}_0 \frac{dt}{2+\cos t}\right| =&\ \left|\frac{1}{2\pi n+r}\int^{2\pi n+r}_0 \frac{dt}{2+\cos t}- \frac{1}{2\pi} \int^{2\pi}_0 \frac{dt}{2+\cos t}\right|\\
=&\ \left|\frac{n}{2\pi n+r}-\frac{1}{2\pi}\right|\int^{2\pi}_0\frac{dt}{2+\cos t}+ \frac{1}{2\pi n+r}\int^{r}_{0} \frac{dt}{2+\cos t}
\end{align}
where $0\le r< 2\pi$. Then as $n\rightarrow \infty$ we get the desired result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$ The following trig identities have shown up in various questions on MSE:
$$-\tan\frac{\pi}{5}+4\sin\frac{2\pi}{5}=\tan\frac{\pi}{5}+4\sin\frac{\pi}{5}=\sqrt{5+2\sqrt{5}}$$
$$-\tan\frac{2\pi}{7}+4\sin\frac{3\pi}{7}=-\tan\frac{\pi}{7}+4\sin\frac{2\pi}{7}=\sqrt{7}$$
$$\tan\frac{\pi}{11}+4\sin\frac{3\pi}{11}=\tan\frac{3\pi}{11}+4\sin\frac{2\pi}{11}=\sqrt{11}$$
$$\tan\frac{6\pi}{13}-4\sin\frac{5\pi}{13}=\tan\frac{2\pi}{13}+4\sin\frac{6\pi}{13}=\sqrt{13+2\sqrt{13}}$$
Does anyone know of any analogous identities for larger primes? I have not been able to find anything similar for $p=17$ or $p=19$.
(I am not asking for proofs of the above equations.)
| There is rule to satisfy either to find the identities or to find any particular value of a trigonometric function with that prime in the denominator and a multiple of $ \pi$ in the numerator.
*
*Considering the case of $5$, it is a Fermat prime and hence it can be constructed using straight edges and compass. Because, $F_n = 2^{2^n}+1$ and $F_1=5$. And since the divisor neglecting the $1$ is $ k=2$, it can be manipulated with simple bisectors or straight edges and compass.
*Considering the number $7$, it is not a Fermat prime. But it is a Pierpont prime i.e. of the form $2^u 3^v+1$. for $7$, $u=v=1$. For those numbers which are Pierpont primes they are either constructible using angle trisectors (due to the factor $3$) or through neusis construction.
*Considering the number $11$, it neither Fermat not Pierpont prime. Hence, it can neither be constructed using the bisectors (straight edges and a compass) nor using an angle trisector. It is only possible through neusis construction.
*For $13$-gon, $13$ is a Pierpont prime with $u=2, v=1$ and hence it can be constructed using angle trisectors or neusis but it can't be constructed using straight edges since it is not a Fermat prime.
*Considering the number $15$, it is a product of distinct Fermat primes $3$ ($ F_0$) and $5$ ($F_1$) and hence it is constructible using straight edges and a compass.
*Considering the number $17$, it is a Fermat prime ($F_2$) and hence it is constructible using straight edges and compass.
*But $19$ is not a Fermat prime but a Piermont prime and hence it is constructible using angle trisectors or neusis. This holds for any prime number. Only the Fermat prime-sided polygons will have a defined value for all the trigonometric functions but Pierpont primes will not have a clearly defined value for any trigonometric function. Instead, they give the identity for a sum or product (or mixed) of different amplitudes of sine and cosine of angles of the format $\frac{n\pi}{p}$ where $p$ is the prime under consideration and the number $n$ need not be the same throughout the identity.
For example, consider the pentagon. It gives the value for $\cos{\frac{\pi}{10}}$ which may be used to give the value for $\cos{\frac{\pi}{5}}$ using trigonometric identity for $\cos{2x}$. The value of is $\cos{\frac{\pi}{10}}=\sqrt{\frac{5+\sqrt{5}}{8}}$. Similarly for the next Fermat prime 17, it is given as:
\begin{equation}
16\cos{\frac{2\pi}{17}}=-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}
\end{equation}
For the Pierpont number 13, it can't be given a defined value for a trigonometric function as mentioned above and hence the identity holds as:
\begin{equation}
\cos^2 {\frac{\pi}{13}}+\cos^2 {\frac{3\pi}{13}}+\cos^2 {\frac{4\pi}{13}}=\frac{11+\sqrt{13}}{8}
\end{equation}
\begin{equation}
\sin{\frac{\pi}{13}}+\sin{\frac{3\pi}{13}}+\sin{\frac{4\pi}{13}}=\sqrt{\frac{13+3\sqrt{13}}{8}}
\end{equation}
But Wolfram has revealed about the number $23$ as stated here. Also, its general solution is as follows (i.e. any trigonometric function can be written in terms of 1 and fractional powers of -1):
$-\frac{1}{2}(-1)^{\frac{22}{23}}\big[1+(-1)^{\frac{2}{23}}\big]$
It is the same case for $7$, a Pierpont prime as:
\begin{equation}
\prod_{k=1}^{3} \sin{\frac{k\pi}{7}} =\frac{\sqrt{7}}{8}
\end{equation}
\begin{equation}
\prod_{k=1}^{3} \cos{\frac{k\pi}{7}} =\frac{1}{8}
\end{equation}
\begin{equation}
\cos^2 {\frac{\pi}{7}}-\cos{\frac{\pi}{7}}\cos{\frac{2\pi}{7}}=\frac{1}{4}
\end{equation}
And another identity for the same is:
\begin{equation}
\cos^{\frac{1}{3}} {\frac{2\pi}{7}}-\Bigg[-\cos{\frac{4\pi}{7}}\Bigg]^{\frac{1}{3}}-\Bigg[-\cos{\frac{6\pi}{7}}\Bigg]^{\frac{1}{3}}=-\Bigg[\frac{1}{2} \bigg(3\times7^{\frac{1}{3}}-5\bigg)\Bigg]^{\frac{1}{3}}
\end{equation}
The reason for saying that a number of sides, $n$ which is not a Fermat prime but a Pierpont prime for a $n-$gon being constructible only by trisectors or neusis and not by bisectors is that Fermat primes allows to factor out 2 i.e. bisect since it is in that form of power of 2 and since $n$ is not a Fermat prime but a Pierpont prime, the only factor that distinguishes Pierpont from Fermat prime is 3. Hence, it is constructible using trisectors. That is, consider Fermat prime to be of the form $F_n = 2^{k(n)}+1; k(n)=2^n$ i.e. $k(n)=k$ in short. Pierpont prime adds another factor of 3 i.e. $2^u3^v+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
We have that $xy^2 + y + 1 \mid x^2y + x + y \implies xy^2 + y + 1 \mid y(x^2y + x + y) - x(xy^2 + y + 1)$
$\iff xy^2 + y + 1 \mid y^2 - x \implies xy^2 + y + 1 \mid xy^2 + y + 1 - x(y^2 - x)$
$\iff xy^2 + y + 1 \mid x^2 + y + 1 \implies xy^2 + y + 1 \le x^2 + y + 1$
$\iff xy^2 \le x^2 \iff y^2 \le x$. And I don't know what to do next.
This problem is adapted from a recent competition.
| We know that $(xy^2+y+1) \mid (x^2y+x+y)$. Let $k \in \mathbb{N}$ s.t. $$(x^2y+x+y)=k(xy^2+y+1)$$
We can rewrite this expression as a quadratic in $x$:
$$yx^2+(1-ky^2)x+(-ky-k+y)=0$$
Now, we can see that the discriminant is:
$$\Delta = (ky^2-1)^2+4y(ky+k-y)=(ky^2+1)^2+4y(k-y)$$
We can see that the discriminant must be a perfect square. However:
$$(ky^2)^2 \geqslant (ky^2+1)^2+4y(k-y) \implies 4y(y-k) \geqslant 2ky^2+1 \tag{1}$$
$$(ky^2+2)^2 \leqslant (ky^2+1)^2+4y(k-y) \implies 4y(k-y) \geqslant 2ky^2+3 \tag{2}$$
However, for $k \geqslant 2$ and $y \geqslant 2$, we have:
$$2ky^2 \geqslant 4y^2 \tag{3}$$
$$2ky^2 \geqslant 4ky$$
Thus, we have:
$$2ky^2+3 > 2ky^2+1 > 2y^2 \geqslant 4y \cdot \mathrm{max}(k,y) \geqslant 4y|k-y|$$
Thus, we either have $k=1$, $y=1$ or $(ky^2)^2 < \Delta (ky^2+2)^2 \implies \Delta = (ky^2+1)^2 \implies y=k$.
$(i)$ $k=1$
We have: $$x^2y+x+y=xy^2+y+1 \implies x^2y+x=xy^2+1 \implies x \mid 1 \implies x=1$$
Thus, we have:
$$y+1=y^2+1 \implies y=1$$
This gives us the solution $(x,y)=(1,1)$ for $k=1$.
$(ii)$ $y=1$
We have: $$x^2+x+1=x+2 \implies x=1$$
This gives us the same solution as before.
$(iii)$ $k=y$
From our previous quadratic:
$$x=\frac{ky^2-1 \pm \sqrt{\Delta}}{2y}=\frac{(y^3-1) \pm (y^3+1)}{2y}>0 \implies x=y^2$$
This gives us the solution $(x,y)=(t^2,t)$ for any $t \in \mathbb{N}$. This also includes $(x,y)=(1,1)$, which is our other solution.
Thus, the only possible pairs of positive integers $(x,y)$ such that $(xy^2+y+1) \mid (x^2y+x+y)$ are all pairs: $$(x,y)=(t^2,t) \space \forall \space t \in \mathbb{N}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate without expansion? I want to evaluate
$$ \int_0^1 ( 1 - x^2)^{10} dx $$
One way I can do this is by expanding out $(1 - x^2)^{10}$ term by term, but is there a better way to do this?
| When you encounter this type of problem, the first thing to do is to come up with a reduction formula. Let us find the reduction formula for $\int(k^2 - x^2)^n dx $.
Using integration by parts:
$u = (k-x^2)^n $ and $dv = dx$ then $du = n(k^2 - x^2)^{n-1}(-2x)dx $ and $v =x$
Thus,
$$\int(k^2 - x^2)^n dx = x(k^2 - x^2)^{n-1} + 2n \int x^2(k^2 - x^2)^{n-1} \\
= x(k^2 - x^2)^{n-1} + 2n \int \bigg[ k^2(k^2 - x^2)^{n-1} - (k^2 - x^2)^{n} \bigg] dx \\
= x(k^2 - x^2)^{n-1} + 2n \int k^2(k^2 - x^2)^{n-1} dx - 2n \int (k^2 - x^2)^{n} dx
$$
From here you can see that:
$$2n\int(k^2 - x^2)^n dx + \int(k^2 - x^2)^n dx = x(k^2 - x^2)^{n-1} + 2n \int k^2(k^2 - x^2)^{n-1} dx - 2n \int (k^2 - x^2)^{n} dx $$
Thus,
$$\int(k^2 - x^2)^n dx = \frac{x(k^2 - x^2)^{n-1}}{2n+1} + \frac{2nk^2}{2n+1} \int (k^2 - x^2)^{n-1} dx$$
For your problem, $k=1$ and $n=10$. Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $7^x -5x^3 \equiv 0 \quad \pmod{11}?$ I have to study the solvability of the equation $$ 7^x -5x^3 \equiv 0 \quad \pmod{33} $$ and determine its integer solutions $ x $ with $ 0 \le x \le 110 $.
I started dividing this equation into two equations $$\cases {7^x -5x^3 \equiv 0 \quad \pmod{3} \\ 7^x -5x^3 \equiv 0 \quad \pmod{11}}.$$ For the first one I tried to substitute the values $0,1,2 $ and found that $2$ is the only possibile solution.
Then I tried to solve the second one with the method of indices:
$$ x \cdot\mbox{ind}_{11}(7)-\mbox{ind}_{11}(5)-3\cdot\mbox{ind}_{11}(x) \equiv 0 \pmod{\phi(11)}.$$ I noticed that $2$ is a primitive root $\pmod{11}$ and computed its powers which lead to $$2^4 \equiv 5 \pmod{11} \\2^7 \equiv 7 \pmod{11}.$$
The equation became $$ x \cdot 7 - 4- 3 \cdot \mbox{ind}_{11}(x) \equiv 0 \pmod{\phi(11)} \quad \longrightarrow \quad 7 x - 3 \cdot \mbox{ind}_{11}(x) \equiv 4 \quad \pmod{\phi(11)}.$$
Then I stopped because I had no clues on how to continue.
Have you any ideas?
Thank you in advance.
| Instead of using the primitive root $2$, note that $7$ is a primitive root mod 11, because $7^2\equiv 5$ and $7^5\equiv -1$. So using 7-indices is a sensible move.
We list the values:
$$
\begin{array}{c|c|c}
\hline
x\pmod{11} & 5x^3\equiv 7^x\pmod{11} & x\pmod{10}
\\\hline
1 & 5&2\\
2 & 7&1\\
3 & 3&4\\
4 & 1&0\\
5 & 9&8\\
6 & 2&3\\
7 &10&5\\
8 & 8&9\\
9 & 4&6\\
10& 6&7\\\hline
\end{array}
$$
So combining the outer columns, you know ten solutions $x\pmod{110}$ and you know $x\equiv 2\pmod{3}$. Hence you can write down the ten solutions mod 330, and determine the solutions between 0 and 110.
| {
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Proving $\arctan (-\sqrt3) = \frac{2\pi}{3}$ using basic methods Background
A recent exam I'm looking at, has the following trigonometric equation:
$$\sin(\pi x) + \sqrt3 \cos(\pi x) = 0, \quad x \in [0, 2]$$
After a simple re-write, we get $$\tan(\pi x) = -\sqrt 3$$
Note, on this exam, the student doesn't have any tools or sheets available. They may or may not remember the inverse tangent here, and let's assume they do not.
Question
What are some solid ways that a student at this level, knowing basic trigonometry, could reason their way to the right answer here?
My thoughts
In a $30-60-90$ triangle, we soon discover that $\tan(60^\circ) = \sqrt3$, but this requires memorizing the ratios of the sides in a $30-60-90$. From here, we can use the unit circle to look at where the tangent function is positive and negative, and we would find the answer.
Not entirely unreasonable, but if we assume they remember this, they may remember the inverse tangent just as well.
Are there any better, more reasonable ways a student could be expected to find that $$\displaystyle\arctan(-\sqrt3) = -\frac{\pi}{3}$$?
| The fact that $\cos \frac{\pi}{3} = \frac{1}{2}$ is not something that requires memorization, but like many identities in mathematics, it is convenient and efficient to memorize because the proof is more sophisticated than the result.
In an equilateral triangle $\triangle ABC$, draw the altitude $\overline{AD}$ from $A$ to $\overline{BC}$. Since $\angle A = \angle B = \angle C$ and their sum is $\pi$, it easily follows that $\angle B = \angle C = \pi/3$. Since the altitude $\overline{AD}$ is by construction perpendicular to $\overline{BC}$, we conclude $\angle DAC = \angle DAB = \pi/6$, since $\angle ADC = \angle ADB = \pi/2$. Consequently $\triangle ADC \cong \triangle ADB$ as they share $\overline{AD}$, thus $BD = CD = AC/2 = AB/2$. It follows that $$\cos \frac{\pi}{3} = \cos \angle C = \frac{CD}{CA} = \frac{1}{2}.$$
Then $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\tan \frac{\pi}{3} = \sqrt{3}$ follow from the Pythagorean theorem.
It is worth mentioning that the original equation $$\sin \pi x + \sqrt{3} \cos \pi x = 0$$ has another method of solution that uses the angle addition identity $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.$$ By dividing both sides by $2$, we obtain $$\frac{1}{2} \sin \pi x + \frac{\sqrt{3}}{2} \cos \pi x = 0.$$ Then recognizing the established identities above, we can write this as $$\cos \frac{\pi}{3} \sin \pi x + \sin \frac{\pi}{3} \cos \pi x = 0.$$ Applying the angle addition identity, we get $$\sin \left(\frac{\pi}{3} + \pi x\right) = 0,$$ from which it follows that $$\frac{\pi}{3} + \pi x = k \pi$$ for some integer $k$. Therefore, $x = k - \frac{1}{3}$ for some integer $k$. Since $x \in [0,2]$, the only such $k$ corresponding to $x$ in this interval are $k \in \{1, 2\}$. This gives us $x \in \{2/3, 5/3\}$ as the complete solution set.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$x^2 + 4x - \lambda +2 \equiv 0 \pmod 7$ I solved this problem but I don't know if my solution is right:
Find the values of $\lambda, 0 \le \lambda \le 6$, such that the
congruence $x^2 + 4x - \lambda +2 \equiv 0 \pmod 7$ has a solution.
Find all the solutions for the minimum value of $\lambda$ for which the
congruence is solvable.
I have to solve the system of congruences $$\cases{y^2 \equiv b^2-4ac \pmod p\\ 2ax \equiv y-b \pmod p} $$
$y^2 \equiv 4^2-4(2-\lambda) \equiv 1+4\lambda \pmod 7$, so I need to know when $1+4\lambda$ is a quadratic residue $\pmod 7$.
For the Euler's criterion I know that $(1+4\lambda |7) \equiv (1+4\lambda)^3 \equiv 1+64 \lambda^3 +12\lambda+12\lambda^2 \equiv 1+ \lambda(\lambda^2+5\lambda+5) \pmod 7$.
But $5$ is not a quadratic residue $\pmod 7$ because $(5|7) = (7|5)(-1)^6=(2|5)=(-1)^3=-1$, so the quadratic equation $\lambda^2+5\lambda+5 \equiv 0 \pmod 7$ is not solvable.
I conclude that the unique value for which $x^2 + 4x - \lambda +2 \equiv 0 \pmod 7$ has a solution is $\lambda =0$.
My solutions are therefore $$\cases{y^2 \equiv 1 \pmod 7 \longrightarrow y \equiv \pm 1 \pmod 7 \\ 2x \equiv -5 \pmod 7 \longrightarrow x \equiv 1 \pmod 7\\ 2x \equiv -3 \pmod 7 \longrightarrow x \equiv 2 \pmod 7}$$
Is my solution right?
Thanks you for all your suggestions.
| No! There are four values of $\lambda$ such that $x^2+4x-\lambda+2\equiv 0\pmod 7$ has a solution $x\in\mathbb{Z}$.
Note that $x^2+4x-\lambda+2=(x+2)^2-(\lambda+2)$, so you want $\lambda+2$ to be a square, i.e., $\lambda+2\equiv 0,1,2,4\pmod{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int_{-π/2}^{π/2} \frac{28\cos^2(θ)+10\cos(θ)\sin(θ)-28\sin^2(θ)}{2\cos^4(θ)+3\cos^2(θ)\sin^2(θ)+m\sin^4(θ)}\ dθ$ I am studying the integral
$$I=\int_{-\pi/2}^{\pi/2}
\frac{28\cos^2(\theta)+10\cos(\theta)\sin(\theta)-28\sin^2(\theta)}{2\cos^4(\theta)+3\cos^2(\theta)\sin^2(\theta)+m\sin^4(\theta)}d\theta,$$
where $m>0$. In the problem I am working, it is very important to know what value of $m$ makes this integral positive, negative or equal to zero. I found that if $m=2$, then the integral is zero (introducing it in Wolfram Alpha). However, I don't know how to prove it formally.
On the other hand, I tried some values of $m$ in Wolfram and it seems that if $m<2$, then the integral is negative, and if $m>2$, the integral is positive. But again, I have no proof of this.
Any ideas of how to approach this problem?
Just in case, I found this alternative representation of the integral
$$I=\int_{-\pi/2}^{\pi/2}
\frac{8[5\sin(2\theta)+28\cos(2\theta)]}{\cos(4\theta)+15+8(m-2)\sin^4(\theta)}d\theta.$$
Any help would be appreciated.
| note that since the function part of the function is odd i.e:
$$f(x)=\frac{28\cos^2x+10\cos x\sin x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}$$
$$f(-x)=\frac{28\cos^2x-10\cos x\sin x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}$$
you could notice that the integral can be simplified to:
$$\int_{-\pi/2}^{\pi/2}\frac{28\cos^2x+10\cos x\sin x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$
$$=\int_{-\pi/2}^{\pi/2}\frac{28\cos^2x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$
$$=2\int_0^{\pi/2}\frac{28\cos^2x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$
$$=56\int_0^{\pi/2}\frac{\cos^2x-\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$
One route you could try to take is Tangent half-angle substitution, which yields:
$$112\int_0^1(1+t^2)\frac{(1-t^2)^2-(2t)^2}{2(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4}dt$$
the bottom of this fraction can be expanded to:
$$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$
this may be factorisable for certain values of $m$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Second order ODE with piecewise term I have the differential equation $\frac{d^2 n}{d z^2} + 2 \frac{dn}{dz} + f(n) = 0$, where $f(n) = \begin{cases} n, \quad 0 \leq n \leq \frac{1}{2} \\ 1 - n, \quad \frac{1}{2} \leq n \leq 1 \end{cases}$.
I am confused about how to handle the $f(n)$ term. I believe it should be solved in cases, say $0 \leq n \leq \frac{1}{2}$ and $\frac{1}{2} \leq n \leq 1$, like this:
If $0 \leq n \leq \frac{1}{2}$, then $\frac{d^2 n}{d z^2} + 2 \frac{dn}{dz} + n = 0$, which has solution $n(z) = c_1 e^{-z} + c_2 z e^{-z}$.
If $\frac{1}{2} \leq n \leq 1$, then $\frac{d^2 n}{d z^2} + 2 \frac{dn}{dz} + 1-n = 0$ which has solution $n(z) = c_3e^{(-1+\sqrt{2})z} + c_4 e^{(-1 - \sqrt{2})z} + 1$.
Would I write the final answer as $$n(z) =
\begin{cases}
C_1e^{-z} + C_2 z e^{-z}, \quad & 0 \leq n \leq \frac{1}{2} \\
C_3e^{(-1+\sqrt{2})z} + C_4 e^{(-1 - \sqrt{2})z} + 1, & \frac{1}{2} \leq n \leq 1
\end{cases}?
$$
I am uncomfortable with this because the answer seems "circular" to me because the name of the solution is $n$ but the conditions are on $n$ and not $z$.
| Hint:
Let $u=\dfrac{dn}{dz}$ ,
Then $\dfrac{d^2n}{dz^2}=\dfrac{du}{dz}=\dfrac{du}{dn}\dfrac{dn}{dz}=u\dfrac{du}{dn}$
$\therefore\begin{cases}u\dfrac{du}{dn}+2u+n=0,\quad0\leq n\leq\dfrac{1}{2}\\u\dfrac{du}{dn}+2u+1-n=0,\quad\dfrac{1}{2}\leq n\leq1\end{cases}$
$\begin{cases}\dfrac{du}{dn}=-2-\dfrac{n}{u},\quad0\leq n\leq\dfrac{1}{2}\\\dfrac{du}{dn}=\dfrac{n-1}{u}-2,\quad\dfrac{1}{2}\leq n\leq1\end{cases}$
$\begin{cases}\dfrac{n}{u+n}+\ln(u+n)=c_1,\quad0\leq n\leq\dfrac{1}{2}\\(2-\sqrt2)\ln((\sqrt2-1)(n-1)-u)+(2+\sqrt2)\ln((\sqrt2+1)(n-1)+u)=c_2,\quad\dfrac{1}{2}\leq n\leq1\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
To show that $o(o(x^3)-\frac{1}{2}x^2)-o(x^3)=o(x^2) \: as \: x \rightarrow 0.$ Apostol Calculus Example 1 pg.288 I encounter this problem when I was trying to show that
$\sec x=1+\frac{1}{2}x^2+o(x^2) \: as \: x \rightarrow 0.$
We know that $\cos x=1-\frac{1}{2}x^2+o(x^3).$
So $\sec x=\frac{1}{1-\frac{1}{2}x^2+o(x^3)} = 1+\frac{1}{2}x^2-o(x^3)+o(o(x^3)-\frac{1}{2}x^2)$
In the the book Calculus Apostol, it states that the the two little o's behind is equal to $o(x^2)$ without an explanation.
I could prove that they equal to $o(x)$ since
$-o(x^3)+o(o(x^3)-\frac{1}{2}x^2)$
$= -o(x^3)+o(o(x^3)-o(x))$
$= -o(x^3)+o(o(x^3)+o(x)) = -o(x^3)+o(o(x))=o(x)+o(x^3)=o(x).$
But is this correct, if then how can it equal $o(x^2)$?
| You are trying to understand in Example 1 that
$$
\frac{1}{\cos x}=\frac{1}{1-\frac12x^2+o(x^3)}=1+\frac12 x^2+o(x^2)
\quad\textrm{as }x\to 0.
$$
Apostol says that this is "from part (e) of Theorem 7.8", i.e.,
As $x\to a$, if $g(x)\to 0$, then
$$
\frac{1}{1+g(x)}={1-g(x)+o(g(x))}.
$$
So if you let $g(x)=-\frac12x^2+o(x^3)$, then
$$
\sec x
= 1-(-\frac12x^2+o(x^3))+o(-\frac12x^2+o(x^3))
=1+\frac12x^2-o(x^3)+o(-\frac12x^2+o(x^3)).
$$
Note that $-o(x^3)=o(-x^3)=o(x^3)$ by Theorem 7.8 (c) and (b). Also,
$$
o(-\frac12x^2+o(x^3))=o(x^2).
$$
Hence,
$$
\sec x = 1+\frac12 x^2+o(x^3)+o(x^2)
=1+\frac12 x^2+o(x^2).
$$
[Added.] To see why $o(-\frac12x^2+o(x^3))=o(x^2)$, suppose $f(x)=o(-\frac12x^2+h(x))$ for some $h$ with $\lim_{x\to 0}\frac{h(x)}{x^3}=0$. Then by definition,
$$
\lim_{x\to 0}\frac{f(x)}{-\frac12x^2+h(x)}=0.\tag{1}
$$
We want to show that $f(x)=o(x^2)$, i.e.,
$$
\lim_{x\to 0}\frac{f(x)}{x^2}=0.\tag{2}
$$
But
$$
\lim_{x\to 0}\frac{f(x)}{x^2}
=\lim_{x\to 0}\frac{f(x)}{-\frac12x^2+h(x)}\cdot
\frac{-\frac12x^2+h(x)}{x^2}
=\lim_{x\to 0}\frac{f(x)}{-\frac12x^2+h(x)}\cdot
\lim_{x\to 0}\frac{-\frac12x^2+h(x)}{x^2}=0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Determinant with rows $a_1$ to $a_n$ with $-x$ on the diagonal $\left|\begin{matrix}
-x&a_2&\cdots&a_{n}\\
a_1&-x&\cdots&a_{n}\\
a_1&a_2&\cdots&a_{n}\\
\vdots&\vdots&\ddots&\vdots\\
a_1&a_2&\cdots&-x
\end{matrix}\right|$
so i tried to find and expression $D_{n}=kD_{n-1}+tD_{n-2}$
by doing this $\left|\begin{matrix}
-x&a_2&\cdots&a_{n}\\
a_1&-x&\cdots&a_{n}\\
a_1&a_2&\cdots&a_{n}\\
\vdots&\vdots&\ddots&\vdots\\
a_1&a_2&\cdots&a_n-x-a_n
\end{matrix}\right|$ so we split it and get
$\left|\begin{matrix}
-x&a_2&\cdots&a_{n}\\
a_1&-x&\cdots&a_{n}\\
a_1&a_2&\cdots&a_{n}\\
\vdots&\vdots&\ddots&\vdots\\
a_1&a_2&\cdots&a_n
\end{matrix}\right|$ which is an upper trianglular matrix hence the determinant is $\prod_{1\le i\le n}(-x-a_i)$
$\left|\begin{matrix}
-x&a_2&\cdots&0\\
a_1&-x&\cdots&0\\
a_1&a_2&\cdots&0\\
\vdots&\vdots&\ddots&\vdots\\
a_1&a_2&\cdots&-x-a_n
\end{matrix}\right|$ now i expand it along the last column and get $(-x-a_n)D_{n-1}$ but the expression gets very complicated , can i please get a hint on a good way to solve it , i didnt see a better way by manipulating rows/columns(e.g. adding/subtracting all columns to the first but since i miss the $a_i$ in the $i$-th row i cant manipulate the determinant better).
| Subtract the last row from each preceding row to obtain
$$D_n(a_1, \ldots, a_n;x) = \left|\begin{matrix}
-x&a_2&\cdots&a_{n}\\
a_1&-x&\cdots&a_{n}\\
\vdots&\vdots&\ddots&\vdots\\
a_1&a_2&\cdots&-x
\end{matrix}\right| = \left|\begin{matrix}
-x-a_1&0&\cdots&0&a_{n}+x\\
0&-x-a_2&\cdots&0&a_{n}+x\\
\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&\cdots&-x-a_{n-1}&a_n+x\\
a_1&a_2&\cdots&a_{n-1}&-x
\end{matrix}\right|$$
Now expand along the first column and iterate the procedure:
$$= (-x-a_1)\left|\begin{matrix}
-x-a_2&\cdots&0&a_{n}+x\\
\vdots&\ddots&\vdots&\vdots\\
0&\cdots&-x-a_{n-1}&a_n+x\\
a_2&\cdots&a_{n-1}&-x
\end{matrix}\right|+(-1)^{n+1}a_1\left|\begin{matrix}
0&\cdots&0&a_{n}+x\\
-x-a_2&\cdots&0&a_{n}+x\\
\vdots&\ddots&\vdots&\vdots\\
0&\cdots&-x-a_{n-1}&a_n+x\\
a_2&\cdots&a_{n-1}&-x
\end{matrix}\right|$$
\begin{align}
&=(-x-a_1)D_{n-1}(a_2, \ldots, a_n;x)+a_1(-x-a_2)\cdots(-x-a_{n-1})\\
&=(-x-a_1)(-x-a_2)D_{n-2}(a_3, \ldots, a_n;x)+a_1(-x-a_2)\cdots(-x-a_{n-1})+a_2(-x-a_1)(-x-a_3)\cdots (-x-a_n)\\
&=\cdots\\
&=(-x-a_1)\cdots(-x-a_n)+\sum_{i=1}^n a_i(-x-a_1)\cdots(-x-a_{i-1})(-x-a_{i+1})\cdots (-x-a_n)
\end{align}
which is the same as @user1551's result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+7=(1-\sqrt{x+2})^2$ # square both sides
(Use perfect square formula on right hand side $a^2-2ab+b^2$)
$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$ # lhs radical is removed, rhs use perfect square formula
$3x+7=1+2(\sqrt{x+2})+x+2$ # simplify
$3x+7=x+3+2\sqrt{x+2}$ # keep simplifying
$2x+4=2\sqrt{x+2}$ # simplify across both sides
$(2x+4)^2=(2\sqrt{x+2})^2$
$4x^2+16x+16=4(x+2)$ # now that radical on rhs is isolated, square both sides again
$4x^2+12x+14=0$ # a quadratic formula I can use to solve for x
For use int he quadratic function, my parameters are: a=4, b=12 and c=14:
$x=\frac{-12\pm\sqrt{12^2-(4)(4)(14)}}{2(4)}$
$x=\frac{-12\pm{\sqrt{(144-224)}}}{8}$
$x=\frac{-12\pm{\sqrt{-80}}}{8}$
$x=\frac{-12\pm{i\sqrt{16}*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{4i*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{-4\sqrt{5}}}{8}$ #since $4i*i\sqrt{5}$ and i^2 is -1
This is as far as I get:
$\frac{-12}{8}\pm\frac{4\sqrt{5}}{8}$
I must have gone of course somewhere further up since the solution is provided as x=-2.
How can I arrive at -2?
| This line
$$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$$
should be
$$3x+7=1^2-2(1)(\sqrt{x+2})+x+2$$
You put in one too many minuses.
This is my solution.
\begin{align}
\sqrt{3x+7} - \sqrt{x+2} &= 1 \\
(\sqrt{3x+7} + \sqrt{x+2})(\sqrt{3x+7} - \sqrt{x+2})
&= (\sqrt{3x+7} + \sqrt{x+2}) \\
(3x+7) - (x+2) &= \sqrt{3x+7} + \sqrt{x+2} \\
\sqrt{3x+7} + \sqrt{x+2} &= 2x+5 \\
\hline
(\sqrt{3x+7} + \sqrt{x+2}) - (\sqrt{3x+7} - \sqrt{x+2})
&= (2x+5) - 1 \\
2\sqrt{x+2} = 2x+4 \\
\sqrt{x+2} = x+2 \\
x+2 &= x^2+4x+4 \\
x^2 + 3x + 2 &= 0 \\
(x+1)(x+2) &= 0 \\
x &\in \{-1, -2\}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 8
} |
Show that $\int_{0}^{\pi/2} \frac{\sin x~dx}{1+\sin 2x}=\frac{\coth^{-1} \sqrt{2}}{\sqrt{2}}$ The integral $$\int_{0}^{\pi/2} \frac{\sin x~dx}{1+\sin 2x}=\frac{\coth^{-1} \sqrt{2}}{\sqrt{2}}?$$ can be checked at Mathematica. The question is how to do it by hand?
| $$I=\int_{0}^ {\pi/2}\frac{\sin x~dx}{1+\sin 2x} = \frac{1}{2} \int_{0}^{\pi/2} \frac{(\sin x+\cos x)-(\cos x-\sin x)}{(\sin x+ \cos x)^2} dx=\frac{1}{2}\left (\int_{0}^{\pi/2} \frac{dx}{\sin x+\cos x}-\int_{1}^{1} \frac{dt}{t^2} \right).$$
$$\Rightarrow I=\frac{1}{2\sqrt{2}} \int_{0}^{\pi/2} \mbox{cosec} (x+\pi/4) dx=\frac{1}{2\sqrt{2}}\left . \ln(\tan(x/2+\pi/8)\right |_{0}^{\pi/2}
=\frac{1}{2\sqrt{2}} \ln\left( \frac{\tan (3\pi/8)}{\tan (\pi/8)} \right).$$
Note that $\tan(\pi/8)=\sqrt{2}-1=t$, then
$$ I= \frac{1}{2\sqrt{2}} \ln \left (\frac{3-t^2}{1-3t^2} \right )= \frac{1}{2\sqrt{2}} \ln (3+2\sqrt{2})=\frac{1}{\sqrt{2}} \ln (1+\sqrt{2})=
\frac{\mbox{coth}^{-1}\sqrt{2}}{\sqrt{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Prove $\lim_{n\to \infty} \sum_{k=0}^{n} 2^{\frac{-kn}{k+n}}=2$ I'm asked to prove
$$\displaystyle\lim_{n\to \infty} \sum_{k=0}^{n} 2^{\frac{-kn}{k+n}}=2$$
Define $$b_{k,n} = \begin{cases}
0,& n \le k\\
2^{\frac{-kn}{k+n}},& n \ge k\\
\end{cases}$$
Our limit is equivalent to finding $\displaystyle\lim_{n\to \infty} \sum_{k=0}^{\infty} b_{k,n}$. It's appealing to write
$$\displaystyle\lim_{n\to \infty} \sum_{k=0}^{\infty} b_{k,n}=^?\sum_{k=0}^{\infty}\lim_{n\to \infty} b_{k,n}=\sum_{k=0}^{\infty} 2^{-k}=2$$
Question 1:
Can we justify the swapping of sum and limit here? (solved)
Question 2:
Is there any alternative approach?
| The series can be dominated by $$ b_{k,n} \le 2^{-\frac{k}{2}}$$
For $n < k$ this is obivously true, as we have $b_{k,n} = 0$. For $n\ge k$ we have
$$ n \ge k$$
$$ \frac1n \le \frac 1k$$
$$ \frac1k + \frac1n \le \frac2k$$
$$ \frac{kn}{k+n} = \frac{1}{\frac1k + \frac1n} \ge \frac{k}{2}$$
$$ b_{k,n} = 2^{-\frac{kn}{k+n}} \le 2^{-\frac{k}{2}}$$
Series $ \sum_{k=0}^\infty 2^{-\frac{k}{2}} = \sum_{k=0}^\infty \frac{1}{(\sqrt{2})^k}$ is convergent, so from the dominated convergence theorem, you can justify the swapping of the sum and the limit.
More fundamental proof:
Let us decompose $$ \sum_{k=0}^n 2^{-\frac{kn}{k+n}} = \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{k+n}} + \sum_{k={\lfloor\sqrt{n}\rfloor+1}}^n 2^{-\frac{kn}{k+n}}$$
For $k \in [\lfloor\sqrt{n}\rfloor + 1,n] $ we have $$ \frac{kn}{k+n} > \frac{\sqrt{n} n}{n+ n}= \frac{\sqrt{n}}{2}$$ $$ 2^{-\frac{kn}{k+n}} < 2^{-\frac{\sqrt{n}}{2}}$$
We have then $$ \sum_{k={\lfloor\sqrt{n}\rfloor+1}}^n 2^{-\frac{kn}{k+n}} < (n-\lfloor\sqrt{n}\rfloor) 2^{-\frac{\sqrt{n}}{2}} < n2^{-\frac{\sqrt{n}}{2}} \rightarrow 0$$
so
$$ \lim_{n\rightarrow\infty} \sum_{k={\lfloor\sqrt{n}\rfloor+1}}^n 2^{-\frac{kn}{k+n}} = 0$$
For $k \in [0,\lfloor\sqrt{n}\rfloor]$ we have $$ \frac{kn}{\sqrt{n}+n}\le \frac{kn}{k+n} \le k $$
$$ 2^{-k} \le 2^{-\frac{kn}{k+n}} \le 2^{-\frac{kn}{\sqrt{n}+n}} $$
$$ \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-k} \le \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{k+n}} \le \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{\sqrt{n}+n}} $$
$$ \frac{1-2^{-\lfloor\sqrt{n}\rfloor-1}}{1-2^{-1}} \le \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{k+n}} \le \frac{1-2^{-\frac{(\lfloor\sqrt{n}\rfloor+1)n}{\sqrt{n}+n}}}{1-2^{-\frac{n}{\sqrt{n}+n}}} $$
We have $$ \lim_{n\rightarrow\infty} \frac{1-2^{-\lfloor\sqrt{n}\rfloor-1}}{1-2^{-1}} = \lim_{n\rightarrow\infty} \frac{1-2^{-\frac{(\lfloor\sqrt{n}\rfloor+1)n}{\sqrt{n}+n}}}{1-2^{-\frac{n}{\sqrt{n}+n}}} = 2 $$
which means that also $$ \lim_{n\rightarrow\infty} \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{k+n}} = 2 $$
Altogether we have $$ \lim_{n\rightarrow\infty} \sum_{k=0}^n 2^{-\frac{kn}{k+n}} = \lim_{n\rightarrow\infty} \Big(\sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{k+n}} + \sum_{k={\lfloor\sqrt{n}\rfloor+1}}^n 2^{-\frac{kn}{k+n}}\Big) = 2 + 0 =2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the least common factor in equation $\frac{5}{x+4}=4+\frac{3}{x-2}$ I am attempting to solve for x $\frac{5}{x+4}=4+\frac{3}{x-2}$
I know that I need to find the least common denominator. In this case, since I cannot see a clear relationship among them all I think it's just the product of all 3 denominators:
$\frac{5}{x+4}=4+\frac{3}{x-2}$ = $\frac{5}{x+4}=\frac{4}{1}+\frac{3}{x-2}$
LCD: $(x+4)(1)(x-2)$ = $(x+4)(x-2)$
Is this the LCD?
Because I tried to use this in solving my equation but I arrived at a quadratic. I don't think that my textbook wants me to use quadratics in this section but I'm not sure. Here's how I arrived at that:
$\frac{5}{x+4}=\frac{4}{1}+\frac{3}{x-2}$
$(x+4)(x-2)\frac{5}{x+4}=(x+4)(x-2)\frac{4}{1}+(x+4)(x-2)\frac{3}{x-2}$
Then cancel out common factors:
$(x-2)5=(x+4)(x-2)(4)+(x+4)(3)$
$5x-10=(x+4)(x-2)(4)+3x+12$
And if I multiple out the middle term I'll get a polynomial, which is unexpected so I'm not sure I'm on the right path here... am I?
$5x-10=4x^2-4x-8+3x+12$
$5x-10=4x^2-x+4$
| You have a mistake in this step:
$$5x - 10 = (x+4)(x-2)(4) + 3x+12$$
$$5x-10 = (x^2+2x-8)(4) + 3x+12$$
$$5x-10 = 4x^2+\color{red}{8x}-\color{red}{32}+3x+12$$
$$5x-10 = 4x^2+11x-20$$
$$4x^2-6x-10 = 0$$
Alternatively, you can save a step by grouping under the $x+4$ term:
$$(x-2)(5)=(x+4)(x-2)(4)+(x+4)(3)$$
$$5x-10 = (x+4) \left(4(x-2) + 3 \right)$$
and you should get the same answer as before.
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed form for the integral $\int_{-1}^{1} \frac{\ln[(1+x^2)~ -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}\mathrm dx$
Somewhere, the integral
$$\int_{-1}^{1} \frac{\ln[(1+x^2)~ -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}\mathrm dx$$
has been found numerically to be real and finite, if $k \in (-\infty,3].$
Can one find a closed form for this integral?
| $$I=\int_{-1}^{1} \frac{\ln[(1+x^2) -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}dx$$
Split in two parts and let $x\to-x$ for the $(-1,0)$ interval to see that:
$$ I=\int_{0}^{1} \frac{\ln[(1+x^2) +x\sqrt{k+x^2}]}{\sqrt{1-x^2}}dx+\int_{0}^{1} \frac{\ln[(1+x^2) -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}dx$$
$$=\int_0^1 \frac{\ln(1+(2-k)x^2)}{\sqrt{1-x^2}}dx\overset{x=\sin t}=\int_0^\frac{\pi}{2} \ln(1+(2-k)\sin^2 t)dt\overset{t=\operatorname{arccot x}}=\int_0^\infty \frac{\ln\left(\frac{3-k+x^2}{1+x^2}\right)}{1+x^2}dx$$
For simplicity let's put $3-k=a$, then we have:
$$I(a)=\int_0^\infty \frac{\ln(a+x^2)-\ln(1+x^2)}{1+x^2}dx\Rightarrow I'(a)=\int_0^\infty \frac{1}{(1+x^2)(a+x^2)}dx$$
$$=\frac{1}{1-a}\left(\int_0^\infty \frac{1}{a+x^2}dx-\int_0^\infty \frac{1}{1+x^2}dx\right)=\frac{1}{1-a}\left(\frac{1}{\sqrt a}\cdot \frac{\pi}{2}-\frac{\pi}{2}\right)=\frac{\pi}{2}\frac{1}{\sqrt a(1+\sqrt a)}$$
Now we have to integrate back, but notice that $I(1)=0$ then:
$$I(a)=\frac{\pi}{2} \int_1^a\frac{1}{\sqrt x(1+\sqrt x)}dx\overset{\sqrt x=t}=\pi \int_1^{\sqrt a}\frac{1}{1+t}dt=\pi \ln(1+\sqrt a)-\pi \ln 2 $$
Finally, returning back to $k$ we can write the closed form as:
$$\boxed{\int_{-1}^{1} \frac{\ln[(1+x^2) -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}dx=\pi\ln\left(\frac{1+\sqrt{3-k}}{2}\right)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the value of $\cos\left (\frac{\pi }{28} \right )-\cos\left (\frac{3\pi }{28} \right )+\sin\left ( \frac{5\pi }{28} \right )$? Once , I do a problem $ \sum_{n=1}^{14} \cos\left ( \frac{n^{2}\pi }{14} \right )$
when angle modular by $28$ we've got
$2\left ( \cos\left (\frac{\pi }{14} \right ) - \cos\left (\frac{2\pi }{14} \right ) + \cos\left (\frac{3\pi }{14} \right ) + \cos\left (\frac{4\pi }{14} \right ) - \cos\left (\frac{5\pi }{14} \right ) - \cos\left (\frac{6\pi }{14} \right ) \right ) + 1$
and then using identity $\cos\left ( a \right ) - \cos\left ( b \right )$ and $\cos\left ( a \right ) + \cos\left ( b \right )$
leads to $2\sqrt{2} \left (\cos\left ( \frac{\pi }{28} \right ) -\cos\left ( \frac{3\pi }{28} \right )+\sin\left ( \frac{5\pi }{28} \right )\right ) + 1$
the final answer is $\sqrt{7}$ but I don't know how to compute $\left ( \cos\left ( \frac{\pi }{28} \right ) - \cos\left ( \frac{3\pi }{28} \right )+\sin\left ( \frac{5\pi }{28} \right )\right )$ by hands.
I appreciate for your helps.
| Your sum can be written in terms of complex exponentials as
$$ 1+\exp(\pi i/14) + \exp(3 \pi i/14) + \exp(4 \pi i/14) + \exp(8 \pi i/14) + \exp(9 \pi i/14) + \exp(12 \pi i/14) + \exp(16 \pi i/14) + \exp(19 \pi i/14) + \exp(20 \pi i/14) + \exp(24 \pi i/14) + \exp(25 \pi i/14) + \exp(27 \pi i/14)
$$
If $w = e^{i\pi/14}$, this is $$Q(w) = {w}^{27}+{w}^{25}+{w}^{24}+{w}^{20}+{w}^{19}+{w}^{16}+{w}^{12}+{w}^{9}
+{w}^{8}+{w}^{4}+{w}^{3}+w+1
$$
The claim is that $Q(w)^2 - 7 = 0$
which must mean that $Q(w)^2-7$ is divisible by the minimal polynomial of $w$. Since
$w$ is a primitive $28$'th root of $1$, that minimal polynomial is the $28$'th cyclotomic polynomial $C_{28}(w)$, namely ${w}^{12}-{w}^{10}+{w}^{8}-{w}^{6}+{w}^{4}-{w}^{2}+1$.
And indeed it turns out that $(Q(w)^2-7)/C_{28}(w)$ is a polynomial in $w$ of degree
$42$ (but I'd hate to have to verify that by hand).
| {
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"url": "https://math.stackexchange.com/questions/3280171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Moment of inertia of hemisphere I want to find the moment of inertia of the hemisphere shown in the picture about the $O$ axis.
I am getting different results depending on the approach.
*
*The moment of inertia of the hemisphere about an axis parallel to $O$ passing through its Center of Mass (CoM) is
$
I_{CoM} = \frac{83}{320}ma^2
$
wherein $m$ is the mass of the hemisphere.
The distance from the $O$ axis to its CoM is $\frac{5}{8}a$.
Therefore, from the parallel axis theorem,
$
I_O = \frac{83}{320}ma^2 + m(\frac{5}{8}a)^2=\frac{13}{20}ma^2
$
*The moment of inertia of the hemisphere about an axis parallel to $O$ passing through $a$ (the "base" of the hemisphere) is
$
I_{CoM} = \frac{2}{5}ma^2
$
Therefore, from the parallel axis theorem,
$
I_O = \frac{2}{5}ma^2 + ma^2=\frac{7}{5}ma^2
$
*The general expression for the moment of inertia is:
$
I_O = \int_m{r^2dm} = \int_m{(x^2+y^2)\cdot\rho dV} = \int_m{(x^2+y^2)\cdot\rho \cdot (\pi \cdot y^2 \cdot dx)}
$
Substituting $y^2=a^2-(x-a)^2$, we get
$
I_O = \int_0^a{(x^2+(a^2-(x-a)^2))\cdot\rho \cdot (\pi \cdot (a^2-(x-a)^2) \cdot dx)} = \frac{5}{6} \pi \rho a^5
$
The mass of the hemisphere is $m=\frac{2}{3}\pi \rho a^3$, so the expression above can be written in terms of the mass $m$ as
$
I_O = \frac{5}{6} \frac{3}{2} \left( \frac{2}{3} \pi \rho a^3 \right) a^2 = \frac{15}{12}ma^2 = \frac{5}{4}ma^2
$
*Since the differential volume in (3) was $dV=\pi y^2 dx$, maybe the general expression for the moment of inertia reduces to:
$
I_O = \int_m{r^2dm} = \int_m{x^2 \cdot\rho dV} = \int_m{x^2 \cdot\rho \cdot (\pi \cdot y^2 \cdot dx)}
$
Which, after substituting $y^2=a^2-(x-a)^2$, results in:
$
I_O = \int_0^a{x^2 \cdot\rho \cdot (\pi \cdot (a^2-(x-a)^2) \cdot dx)} = \frac{3}{10} \pi \rho a^5 = \frac{9}{20}ma^2
$
In summary, four different approaches and four different results. What am I missing here?
| *
*is correct, $\dfrac{83}{320}+\dfrac{5^2}{8^2}=\dfrac{13}{20}.$
*is incorrect, the parallel axis theorem works with an axis through the gravity center;
*is incorrect, there is no account for the $z$ coordinate.
| {
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"timestamp": "2023-03-29T00:00:00",
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Maclaurin series of $\tan(x+x^2)$ to order 3 I know that the maclaurin series of $\tan(x)$ is $\tan(x)=x+\frac{1}{3}x^3+\frac{2}{15}x^5+...$, then shouldn't be $\tan(x+x^2)=(x+x^2)+\frac{1}{3}(x+x^2)^3+\frac{2}{15}(x+x^2)^5+...$?
Mathematica actually gives me $\tan(x+x^2)=x+x^2+\frac{x^3}{3}+o(x^4)$ to order 3.
| Expanding the monomial terms gives Mathematica's result – and you can stop at $(x+x^2)^3$:
$$(x+x^2)+\frac13(x+x^2)^3=x+x^2+\frac13(x^3\color{lightgrey}{+3x^4+3x^5+x^6})=x+x^2+\frac{x^3}3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx=\frac{2}{3}G$ Discovered the integral below
$$I=\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx= \frac23G$$
which looks clean, yet challenging. Have not seen it before. Post it here in case anyone is interested.
Edit:
Here is a solution. Let $J(a)=\int_0^{\frac\pi{3}}\tanh^{-1}\frac{2a\sin x}{1+a^2}dx$, with $I(0)=0$
$$ J’(a) = \int_0^{\frac\pi{3}}\frac{2(1-a^2)\sin x}{4a^2\cos^2x+(1-a^2)^2}dx
=\frac{\tan^{-1}\frac {a(1-a^2)}{1+a^4}}{a}
$$
Then
\begin{align}
I
& =J(1) =J(0)+\int_0^1 J’(a)da = \int_0^1\frac{\tan^{-1}\frac {a(1-a^2)}{1+a^4}}{a} da\\
&=\int_0^1\left(\frac{\tan^{-1}a}{a}\right.
-\underset{a^3\to a}{\left.\frac{\tan^{-1}a^3}{a}\right)}da=\left(1-\frac13\right) \int_0^1\frac{\tan^{-1}a}{a}da=\frac23G
\end{align}
| Use that
$$\operatorname{arctanh}(\sin(x))=2\sum_{n=1}^{\infty} (-1)^{n-1}\frac{\sin((2n-1)x)}{2n-1}, \ 0 <x<2\pi$$and after integration, employ the result in $(3.238)$, page $215$, from the book (Almost) Impossible Integrals, Sums, and Series$\displaystyle \left(\sum_{n=1}^{\infty} (-1)^{n-1} \frac{\sin(\pi/6(4n+1))}{(2n-1)^2}=\sum_{n=1}^{\infty} \frac{\sin(\pi/6(2n-1))}{(2n-1)^2}=\frac{2}{3}G\right)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the orthogonal complement The question is, considering
*
*The vector space $\mathcal{P}_3(\mathbb{R})$ of the polynomials with real coefficients of degree $\leq$ 3
*The inner product defined by $\left<p,q\right>=\int_{-1}^{1}pq$
how to find a basis for the orthogonal complement of the space spanned by $\{x-1, x^2+3\}$?
I tried making $$\int_{-1}^{1}(x-1)p(x)=0$$ and $$\int_{-1}^{1}(x^2+3)p(x)=0,$$ which gave me the spanning sets $\{x^3+\frac{1}{5}, x^2-\frac{1}{3}, x+\frac{1}{3}\}$ and $\{x^3, x^2-\frac{9}{25}, x\}$ respectively. Then I think we may determine a basis for the intersection os these subspaces, but didn't see how.
| Let $\textsf{W}=\operatorname{span}(\{x-1,x^2+3\})$ be the subspace of $\textsf{P}_3(\mathbb R)$. Then $p(x)\in \textsf{W}^\perp$ if and only if
$$\langle p(x),x-1\rangle=\int_{-1}^1 p(x)(x-1)dx=0$$
$$\langle p(x),x^2+3\rangle=\int_{-1}^1 p(x)(x^2+3)dx=0$$
at the same time, since $\{x-1,x^2+3\}$ is a basis for $\textsf{W}$. Let $p(x)=a_3x^3+a_2x^2+a_1x+a_0$ for some scalars $a_0,a_1,a_2,a_3$. Then, the two above equations are equivalent to
$$\left\{\begin{align}
-2a_0+\frac{2}{3}a_1-\frac{2}{3}a_2+\frac{2}{5}a_3=0 \\
\frac{20}{3}a_0+\frac{12}{5}a_2=0
\end{align}\right.$$
(after perform every integral) and solving this last system gives us
$$a_2=-\frac{25}{9}a_0, \qquad a_3=\frac{10}{27}a_0-\frac{5}{3}a_1$$
Making $a_0=t$ and $a_1=s$, we can guarantee that
$$\begin{align}
\textsf{W}^\perp&=\left\{ \left(\frac{10}{27}t-\frac{5}{3}s\right)x^3-\frac{25}{9}tx^2+sx+t:\, s,t\in \mathbb R\right\} \\
&=\operatorname{span}\left(\left\{ \frac{10}{27}x^3-\frac{25}{9}x^2+1,-\frac{5}{3}x^3+x \right\}\right)
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Factoring $-3x^2 - 2xy + 8y^2 + 11x - 8y - 6$ This is quite easy to do if we do the middle term splitting method, but here's another method given in the book:
Only considering terms with powers of $x$ and constants:
$-3x^2 + 11x - 6 = (3x - 2)(-x + 3)$
Only considering terms with powers of y and constants:
$8y^2 - 8y - 6 = (-4y - 2)(-2y + 3)$
Merging both we get $-3x^2 - 2xy + 8y^2 + 11x - 8y - 6 = (3x - 4y - 2)(-x - 2y + 3)$
Why does this method work, and is there a name to this method?
| So you want: $$-3x^2 - 2xy + 8y^2 + 11x - 8y - 6= (ax+by+c)(dx+ey+f)$$
If we set $y=0$ we get $$\boxed{-3x^2+11x-6 = (ax+c)(dx+f)}$$
and since $$-3x^2+11x-6 = (-3x+2)(x-3)$$ so $a=-3$, $c=2$, $d=1$ and $f=-3$ (or $a=3$, $c=-2$, $d=-1$ and $f=3$.)
Seting $x= 0$ we get $$\boxed{ 8y^2-8y-6 = (by+c)(ey+f)}$$ so ...
| {
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"timestamp": "2023-03-29T00:00:00",
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Factorization in a proof of induction I have to prove the following:
$1+3^3+ ... + (2n+1)^3=(n+1)^2(2n^2+4n+1)$ by induction.
My try:
Base case, $n=1$:
$1+3^3=(2)^2(2\cdot1^2 + 4\cdot1 + 1)$, which is true.
By inductive hypothesis, assume $n=k$:
$1 + 3^3 + ... + (2k+1)^3=(k+1)^2 (2k^2 + 4k + 1)$
For $n=k+1$
$1 + 3^3 + ... + (2k+1)^3 + (2k+3)^3 = (k+2)^2 ( 2(k+1)^2 + 4(k+1) + 1)$
Using the inductive hypothesis we need to prove that
$(k+1)^2(2k^2+4k+1) + (2k+3)^3= (k+2)^2 ( 2(k+1)^2 + 4(k+1) + 1)$
and here I have the problem, because I don't know how to manipulate any of the sides of the equation to prove this. I tried in both sides but I can't find the way. One of my last attempts ended here:
$(k+1)^2(2k^2+4k+1) + (2k+3)^3= (k+2)^2 ( 2(k+1)(k+3) + 1 )$
Which is true by wolfram alpha.
PS: I'm aware that there are a lot of papers and information about this proof but I'm not searching another way or something like that, because I saw a lot of posts about proving this statement via induction, but all of them used the last term as (2n-1)^3, and I need to prove it when it is (2n+1)^3, and the final expression is a little bit different. I just need help factoring my last step.
Any corrections on the inductive steps that i followed are appreciated too.
| You can get the factor $(k+2)^2$ by the following way.
$$(k+1)^2(2k^2+1)+(2k+3)^2=2k^4+16k^3+47k^2+60k+28=$$
$$=2k^4+8k^3+8k^2+8k^3+32k^2+32k+7k^2+28k+28=$$
$$=(k+2)^2(2k^2+8k+7)=(k+2)^2(2(k+1)^2+4(k+1)+1).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence bound for $p\left(n\right) = \text{arctan}\left(\frac{2-\sqrt{4-\left(\frac{2}{n}+\frac{1}{n^2}\right)}}{(2+\frac{1}{n})}\right)$ For my engineering-related work, I ended up with the following expression
$p\left(n\right) = \text{arctan}\left(\frac{2-\sqrt{4-\left(\frac{2}{n}+\frac{1}{n^2}\right)}}{(2+\frac{1}{n})}\right)$,
where $n\in \left[1, \infty\right)$. Now I am aware that for large $n$, the nominator and thus also the $\text{arctan}$ "quickly" converges to zero.
I can of course visualize that "quick" convergence.
What I am actually interested in is a more comprehensible statement on the order of convergence like "an exponential/quadratic convergence".
I have looked into the Taylor expansion of the $\text{arctan}$ but got lost after that.
| $$ p(n) = \arctan(\frac{2 - \sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})}}{(2+\frac{1}{n})} ) = \arctan(\frac{(2 - \sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})})(2+\sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})})}{(2+\frac{1}{n})(2+\sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})})}) $$
After using $(a-b)(a+b) = a^2 - b^2 $, we have:
$$ p(n) = \arctan(\frac{\frac{1}{n}\cdot(2+\frac{1}{n})}{(2+\frac{1}{n})(2+\sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})})}) = \arctan(\frac{1}{n} \cdot \frac{1}{2 + \sqrt{4+\frac{1}{n}\cdot(2+\frac{1}{n})}})$$
Now using the taylor expansion of $\arctan$ function (near zero) : $\arctan(x) = x - \frac{x^3}{3} + o(x^3) = x + o(x^2)$
we have: as $n\to \infty$, then $p(n) = \frac{1}{n} \cdot \frac{1}{2 + \sqrt{4+\frac{1}{n}\cdot(2+\frac{1}{n})}} + o(\frac{1}{n^2}) $
And after a little thinking, as $n \to \infty $, then $\frac{1}{n} \cdot \frac{1}{2 + \sqrt{4+\frac{1}{n}\cdot(2+\frac{1}{n})}}$ isn't that different from $\frac{1}{4n}$
Formally ($E$ is error)$$E(n) = \frac{1}{n} \cdot \frac{1}{2 + \sqrt{4+\frac{1}{n}\cdot(2+\frac{1}{n})}} - \frac{1}{4n} = \frac{1}{n}(\frac{2 -\sqrt{4+\frac{1}{n}(2+\frac{1}{n})} }{8+4\sqrt{4+\frac{1}{n}(2+\frac{1}{n})}})$$
So similarly (multiplying both sides by $2 +\sqrt{4+\frac{1}{n}(2+\frac{1}{n})}$, we get:
$$ E(n) = -\frac{1}{4n^2}\cdot (2 + \frac{1}{n})\cdot \frac{1}{(2 +\sqrt{4+\frac{1}{n}(2+\frac{1}{n})})^2 }, $$ so after ellaborating a little bit: $E(n) = -\frac{1}{4n^2} + o(\frac{1}{n^2}) $ ( just tackle the error function once again)
SO:
$$ p(n) = \frac{1}{4n} - \frac{1}{4n^2} + o(\frac{1}{n^2})$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/3289798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generating prime numbers of the form $\lfloor \sqrt{3} \cdot n \rfloor $ How to prove the following claims ?
Let $b_n=b_{n-1}+\operatorname{lcm}(\lfloor \sqrt{3} \cdot n \rfloor , b_{n-1})$ with $b_1=3$ and $n>1$ . Let $a_n=b_{n+1}/b_n-1$ .
*
*Every term of this sequence $a_i$ is either prime or $1$ .
*Every odd prime of the form $\left\lfloor \sqrt{3}\cdot n \right\rfloor$ greater than $3$ is a term of this sequence.
*At the first appearance of each prime of the form $\left\lfloor \sqrt{3}\cdot n \right\rfloor$ greater than $5$, it is the next prime of the given form after the largest prime that has already appeared.
A few first terms of this sequence can be found at A323388 .
Implementation of this generator in PARI/GP can be found here.
| We prove that the second and the third claims are true.
The second claim is true.
If $a=d\alpha$, $b=d\beta$ and $(a,b)=d$, we have $\mathrm{lcm}(a,b) / b = \alpha = a / d.$
We may rewrite the sequence $a_n$ using above.
$$
a_1=1,$$
$$
a_n=\frac{\lfloor(n+1)\sqrt 3\rfloor}{\left(\lfloor(n+1)\sqrt 3\rfloor, (a_{n-1}+1)\cdots (a_1+1) 3 \right)}, \ \ n\geq 2.
$$
Thus, if $n\geq 2$ and $\lfloor(n+1)\sqrt 3\rfloor=p>3$ is prime, then $p$ cannot divide $a_i+1$ for all $1\leq i\leq n-1$.
It is clear that $p$ cannot divide $a_i+1$ for $1\leq i\leq n-2$. Otherwise, $p | a_i+1 \leq \lfloor (n-1)\sqrt 3 \rfloor +1<\lfloor(n+1)\sqrt 3\rfloor=p$ is a contradiction.
To see that $p$ cannot divide $a_{n-1}+1$, assume otherwise. Then $p|a_{n-1}+1 \leq \lfloor n\sqrt 3\rfloor +1\leq \lfloor(n+1)\sqrt 3\rfloor=p$. This gives the equality
$$
p=a_{n-1}+1=\lfloor n\sqrt 3\rfloor +1=\lfloor(n+1)\sqrt 3\rfloor.
$$
Then $\lfloor n\sqrt 3 \rfloor = p-1$ is an even number. Moreover, $\lfloor \sqrt 3\rfloor = 1$, $\lfloor 2\sqrt 3\rfloor=5$ gives $n\geq 3$. But, $2=a_1+1$ gives
$$
a_{n-1}=\frac{\lfloor n\sqrt 3 \rfloor}{\left(\lfloor n\sqrt 3\rfloor, (a_{n-2}+1)\cdots (a_1+1) 3 \right)}\leq \frac{p-1}2 <p-1.
$$
This is also a contradiction.
Therefore, we have $a_n=p$ in such case.
The third claim is true.
Recall that
$$
a_1=1,$$
$$
a_n=\frac{\lfloor(n+1)\sqrt 3\rfloor}{\left(\lfloor(n+1)\sqrt 3\rfloor, (a_{n-1}+1)\cdots (a_1+1) 3 \right)}, \ \ n\geq 2.
$$
Suppose that $n\geq 2$ and $\lfloor(n+1)\sqrt 3\rfloor = p>3$ is prime. Then we have for any $1\leq i\leq n-1$,
$$
a_i\leq \lfloor (i+1)\sqrt 3 \rfloor \leq \lfloor n\sqrt 3\rfloor < \lfloor (n+1)\sqrt 3 \rfloor =p,
$$
Therefore, $p$ is the next prime of the given form after the largest prime that has already appeared.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3290665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving the diophantine equation $x^3+y^3 = z^6+3$ I've the following problem:
Show that the congruence $x^3+y^3 \equiv z^6+3\pmod{7}$ has no solutions. Hence find all integer solutions if any to $x^3+y^3-z^6-3 = 0.$
We can rearrange the first equation to $z^6 \equiv (x^3+y^3-3) \mod{7}$. But $z^6\equiv 1\mod{7}$ so this is only possible when $x^3+y^3=4$ which no $x,y \in \mathbb{Z}$ satisfy.
Now I don't know if that helps at all solve the second bit. We have $z^6 = x^3+y^3-3$. I know that $x^3+y^3 = (x+y)(x^2-xy+y^2)$.
| Render any cube $\in\{-1,0,1\}\bmod 9$. Then $z$ can't be a multiple of $3$ because that makes $ x^6+3\equiv 3\bmod 9$. Try $z$ not a multiple of $3$, then $z^6=(z^2)^3$ where $z^2\equiv 1\bmod9$ forcing $z^6\equiv 1\bmod9$. Then $x^3+y^3 =x^6+3\equiv 4\bmod9$. Well, today is not the day Serena Williams ties that tennis record either.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Value of $\prod_{n>1} \frac{1}{1-\frac{1}{n^s}}$ or $-\sum_{n=2}^{\infty} \log(1-\frac{1}{n^s} )$ I know
\begin{align}
\prod_{p~is~ prime} \frac{1}{1-\frac{1}{p^2}} = \zeta(2) = \frac{\pi^2}{6}
\end{align}
which has a convergent number.
actually I can even generalized this to
\begin{align}
\prod_{p ~is~ prime} \frac{1}{1-\frac{1}{p^s}} = \zeta(s)
\end{align}
For $s>1$ [consider $s\in \mathbb{R}$] we know zeta function converges, so this has a convergent number.
How about generalization to arbitrary integers? [i.e., I want to replace $p$ with arbitrary integer $n$.]
For example $s=2$, we have
\begin{align}
\prod_{n>1} \frac{1}{1-\frac{1}{n^2}}
\end{align}
taking log we need to show
\begin{align}
- \sum_{n=2} \log\left(1-\frac{1}{n^2} \right)
\end{align}
is convergent or not.
simply by telescope method I can see this value converges to $\log(2)$, that means $ \prod_{n>1} \frac{1}{1-\frac{1}{n^2}} = 2$.
Now consider $s>1$.
\begin{align}
-\sum_{n>1} \log\left(1-\frac{1}{n^s}\right)
\end{align}
This is convergent from comparison test.
Simply take $a_n = -\log(1-\frac{1}{n^s})$ and $b_n = \frac{1}{n^s}$, then
\begin{align}
\lim_{n\rightarrow \infty} \frac{a_n}{b_n} = \lim_{x\rightarrow 0} \frac{-\log(1-x)}{x} = 1 >0
\end{align}
and since $\sum_{n=1}^{\infty} b_n = \zeta(s)$ is convergent for $s>1$, $\sum_{n=2}^{\infty} a_n$ also converges.
What I want to obtain is the value of such convergent series, first i tried telescope method, but it seems difficult even for $s=3$.
Is there a way to compute exact value of those products?
How and what is the values of those products?
| For $2k\ge 2$ $$\prod_{n=2}^\infty (1-\frac{1}{n^{2k}})=\prod_{m=1}^{2k} \prod_{n=2}^\infty (1-\frac{e^{2i \pi m/(2k)}}{n}) = \prod_{m=1}^{k}\prod_{n =-\infty, |n|\ge 2}^\infty (1-\frac{e^{2i \pi m/(2k)}}{n}) = \prod_{m=1}^{k} f(e^{2i \pi m/(2k)})$$
where $f(x) =\frac{\sin(\pi x)}{\pi x(1-x^2)} $ and in those products the order of summation is meant to be $\lim_{N \to \infty} \prod_{|n| \le N} $
For $k \ge 2$ $$\prod_{n=2}^\infty (1-\frac{1}{n^k})=\prod_{m=1}^kg(e^{2i \pi m/k}), \qquad g(x) =\frac{1}{(-x)(1-x)\Gamma(-x)}= \frac{1}{\Gamma(2-x)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many primes are there of the form $x^6+y^6$, where $x,y\in\mathbb{Z}$? I'm trying to find such primes that are of the form $x^6+y^6$; $x,y\in\mathbb{Z}$.
If one of $x$ and $y$ is $0$, say $y=0$ then $x^6+y^6=x^6$ which is not a prime. So assume that $(x,y)\ne(0,0).$
I factored $x^6+y^6$ as follows:
$x^6+y^6=(x^2)^3+(y^2)^3=(x^2+y^2)(x^4+y^4-x^2y^2)=(x^2+y^2)[(x^2+y^2)^2-3x^2y^2]=(x^2+y^2)(x^2+y^2+\sqrt{3}x^2y^2)(x^2+y^2-\sqrt{3}x^2y^2)$
If $x^4+y^4-x^2y^2=1$ then $(x^2+y^2+\sqrt{3}x^2y^2)(x^2+y^2-\sqrt{3}x^2y^2)=1$ this implies that the two numbers $x^2+y^2+\sqrt{3}x^2y^2$ and $x^2+y^2-\sqrt{3}x^2y^2$ are reciprocal to each other. So if $x^2+y^2+\sqrt{3}x^2y^2\in\mathbb{Z}$ then $x^2+y^2-\sqrt{3}x^2y^2\not\in\mathbb{Z}$ or vise-versa.
But $x^2+y^2\pm\sqrt{3}x^2y^2\not\in\mathbb{Z}$ for $(x,y)\ne(0,0)$.
So that $x^4+y^4-x^2y^2$ can't be factored out in $\mathbb{Z}$. That is we can conclude that this is a prime.
To $x^6+y^6$ be a prime, we must have $x^2+y^2=1$, which is possible only for $(x,y)=(+1,0);(-1,0);(0,+1);(0,-1)$.
But in all the above cases, $x^6+y^6=1$, which is not a prime.
So I can conclude that there is no prime of the form $x^6+y^6.$
My question is my argument ok? And is there any other way to think about this problem?
| So you know that $x^6 + y^6 = (x^2 + y^2)(x^4 - x^2y^2 + y^4)$, and both are not factorizable over $\mathbb Q$.
If $a, b, c \in \mathbb Z$ and $a = bc$, then if $a$ is prime, either $b = 1$ or $c = 1$.
You already know that $x^2 + y^2 = 1$ if and only if $(x, y) \in \{(0, \pm 1), (\pm 1, 0)\}$. Let us briefly look at the other term.
Actually, $x^4 - x^2y^2 + y^4$ will not be small too often, since $x^4 - x^2y^2 + y^4 = (x^2-y^2)^2 + x^2y^2 \geq x^2y^2$. In particular, it is $1$ only if $(x, y) = (\pm1, \pm 1)$.
Combining all these, the only prime $x^6+y^6$ hits is $2$.
For your other question in your comment, actually one can prove more:
For any multivariate non-constant polynomial $f \in \mathbb Z[x_1, \cdots, x_n]$ with integral coefficients, there exists $(a_1, \cdots, a_n) \in \mathbb Z^n$ such that $(a_i)$ are pairwise relatively prime and $|f(a_1, \cdots, a_n)|$ is composite.
If you care about why, here is a sketch of the construction. The construction can roughly be taken in the following steps:
*
*Fix temporarily $a_2 = \cdots = a_n = 1$, and consider the univariate polynomial $f_1(x_1) = f(x_1, 1, \cdots, 1)$. There exists a $a_1 \neq 0$ such that $|f_1(a_1)| \geq 2$. If it is composite, we are done. If not, let $p$ be a prime divisor of $|f_1(a_1)|$.
*The iteration: For $i = 2, \cdots n$, determine the final value $a_i$ satisfying the following three conditions:
(a) $a_i \equiv 1 \pmod p$.
(b) $a_i$ is relatively prime to $a_1, \cdots, a_{i-1}$.
(c) $|f(a_1, \cdots, a_i, 1, \cdots, 1)| > p$
This is always possible. For the correctness, (a) guarantees us that $p \ | \ |f(a_1, \cdots, a_i, 1, \cdots, 1)|$; (b) guarantees everything are still relatively prime, and (c) guarantees that it is composite.
After the iteration we should get $(a_1, \cdots, a_n)$ satisfy the condition we requested.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find basis for the image and the kernel of a linear map How do I calculate a basis for $\ker (\varphi)$ and $\operatorname{im}(\varphi)$?
Where, given
$$ A = \begin{pmatrix}
1 & -1\\
-1 & 1 \\
\end{pmatrix}$$
we define
$$ \begin{matrix}
\varphi: \mathbb{R}^{2 \times 2} \to \mathbb{R}^{2 \times 2} \\
X \mapsto XA+A^t X^t
\end{matrix}$$
Let $B$ be the standard basis for $\mathbb{R}^{2 \times 2}$ :
$$B =\left\{ \begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}, \begin{pmatrix}
0 & 1\\
0 & 0\\
\end{pmatrix},\begin{pmatrix}
0 & 0\\
1 & 0\\
\end{pmatrix},\begin{pmatrix}
0 & 0\\
0 & 1\\
\end{pmatrix} \right\}$$
Calculate $\textsf{M}_B(\varphi)$ we come to $$\textsf{M}_B(\varphi) = \begin{pmatrix}
0 & 0 & 0 & 0\\
-1 & 1 & -1 & 1 \\
1 & -1 & 1 & -1\\
0 & 0 & 0 & 0\\
\end{pmatrix}$$
We calculate a basis for the kernel like this:
If
$$X:= \begin{pmatrix}
a & b\\
c & d\\
\end{pmatrix}$$
then $$\varphi(X) = \begin{pmatrix}
a-b & -a+b\\
c-d & -c+d\\
\end{pmatrix}+\begin{pmatrix}
a-b & c-d\\
-a+b & -c+d\\
\end{pmatrix} = \begin{pmatrix}
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\
\end{pmatrix}$$
Now we have to look, for what values $$\begin{pmatrix}
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\
\end{pmatrix}$$
by definition, the kernel of a linear transformation is $\varphi(X) = 0$, therefore our basis for $\ker(\varphi)$ should be
$$\left\{ \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} \right\}$$
Now here comes the part where I'm confused. How do I calculate a basis for $\operatorname{im}(\varphi)$ ?
$\textsf{M}_B(\varphi)$ is the transformation matrix. I've read that you'd just transpose the matrix $\textsf{M}_B(\varphi)$ and row reduce to calculate a basis. I just don't get it.
The solution according to the solution it should be the basis $$ \left\{\begin{pmatrix}
0 & 1 \\
1 & -2 \end{pmatrix}, \begin{pmatrix}
1 & 0 \\
0 & -1 \end{pmatrix} \right\}$$
Also little side questions. I do know about
$$\dim(A) = \dim(\operatorname{im} A) + \dim(\ker A)$$
but how exactly do you know the dimension of the Kernel/Image?
| EDIT: This answer assumes that $\varphi(X) = XA - (XA)^T$, so it only illustrates the method.
Your matrix $M_\varphi$ should be
$$M_B(\varphi) = \begin{pmatrix}
0 & 0 & 0 & 0\\
-1 & 1 & -1 & 1 \\
1 & -1 & 1 & -1\\
0 & 0 & 0 & 0\\
\end{pmatrix}$$
which implies the basis for $\ker\varphi$ is $b_1+b_2, b_1-b_3, b_1+b_4$, or
$$\left\{\begin{pmatrix} 1 & 1 \\ 0 & 0\end{pmatrix}, \begin{pmatrix} 1 & 0 \\ -1 & 0\end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}\right\}$$
where $B = \{b_1, b_2, b_3, b_4\}$.
$\operatorname{Im}\varphi$ should be spanned by images of basis elements:
$$\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$$
so the basis is $$\left\{\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Evaluate $\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx$ Find:
$$\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx$$
I don't know how I starte & evaluate this integral
Wolfram alpha give $=1,28553$
My problem whene I use $t=\cos x$ I get $\arccos x$
Same problem with $t=\sin x$
If any one have idea please help me
| $$\mathcal{J}=\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx=\int_0^{\frac{\pi}{4}} \color{red}{\frac{1+x\sin{x}}{\cos{x}\sqrt{1-x^2}}} \cdot \color{blue}{\frac{\cos{x} \left(1+\cos x-x^{2}\right)}{(1+x\sin x)^2}\,dx}$$
If $\displaystyle t=\frac{\sin{x}+x}{1+x\sin{x}} \text{ then }\color{blue}{dt=\frac{\cos{x}\left(1+\cos{x}-x^2\right)}{\left(1+x\sin{x}\right)^2} dx}$, and:
\begin{align*}
\color{red}{\frac{1+x\sin{x}}{\cos{x}\sqrt{1-x^2}}} &= \frac{1+x\sin{x}}{\sqrt{(1-\sin^2{x})(1-x^2)}}\\
&= \frac{1+x\sin{x}}{\sqrt{\left(1+x\sin{x}\right)^2-\left(\sin{x}+x\right)^2}}\\
&= \frac{1}{\sqrt{1-\left(\frac{\sin{x}+x}{1+x\sin{x}}\right)^2}} \\
&= \color{red}{\frac{1}{\sqrt{1-t^2}} }
\end{align*}
Putting this together,
\begin{align*}
\mathcal{J}&=\int_0^{\frac{\pi}{4}} \color{red}{\frac{1+x\sin{x}}{\cos{x}\sqrt{1-x^2}}} \cdot \color{blue}{\frac{\cos{x} \left(1+\cos x-x^{2}\right)}{(1+x\sin x)^2}\,dx} \\&=\int_0^{\frac{4+\pi \sqrt{2}}{\pi + 4\sqrt{2}}} \frac{dt}{\sqrt{1-t^2}} \\
&= \color{green}{\boxed{\arcsin{\left(\frac{4+\pi \sqrt{2}}{\pi + 4\sqrt{2}}\right)}}}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\sin(x) - \sin(y) = -\frac{1}{3}$, $\cos(x) - \cos(y) = \frac{1}{2}$, what is $\sin(x+y)$? If $\sin(x) - \sin(y) = -\frac{1}{3}$ and $\cos(x) - \cos(y) = \frac{1}{2}$, then what is $\sin(x+y)$?
Attempt:
$$ \sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y) $$
If we multiply the two "substraction identities" we get
$$ (\sin(x) - \sin(y))(\cos(x) - \cos(y)) = -\frac{1}{6} $$
$$ \sin(x)\cos(x) + \sin(y) \cos(y) - \sin(x+y) = \frac{1}{6}$$
thus
$$ \sin(x+y) =\sin(x)\cos(x) + \sin(y) \cos(y) - \frac{1}{6} $$
Next if I do a sum and organize and squaring we get:
$$ (\sin(x) + \cos(x))^{2} + (\sin(y) + \cos(y))^{2} - 2 (\sin(x) + \cos(x)) (\sin(y) + \cos(y)) = \frac{1}{36} $$
$$ 2 + 2 ( \sin(x) \cos(x) + \sin(y) \cos(y)) - 2 \left( \sin(x+y) + \cos(x-y) \right) = \frac{1}{36} $$
I have no idea after this.
Another method is I suspect that we have to solve for $\sin(x), \cos(x), \sin(y), \cos(y)$ instead of directly finding $\sin(x+y)$ algebraically.
| We have
$$\cos \frac{x+y}{2} \sin \frac{x-y}{2}=-\frac{1}{6}$$
$$\sin \frac{x+y}{2} \sin \frac{x-y}{2}=-\frac{1}{4}$$
Thus
$\tan \frac{x+y}{2}=\frac{3}{2}$.
And hence
$$\sin(x+y)=\frac{2\tan\frac{x+y}{2}}{1+\tan^2\frac{x+y}{2}}=\frac{12}{13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the constant $k$ from the determinant
Given: $$\begin{vmatrix}(b+c)^2 &a^2&a^2\\b^2 &(c+a)^2&b^2 \\c^2&c^2& (a+b)^2\end{vmatrix}=k(abc)(a+b+c)^3$$ Find $k$.
If I directly open the determinant it will go to long I can't apply most of the row or column operation as they keep making it more complex.
| It's not long:
$$\prod_{cyc}(a+b)^2+2a^2b^2c^2-\sum_{cyc}a^2b^2(a+b)^2=$$
$$\sum_{cyc}\left(a^4b^2+a^4c^2+2a^3b^3+2a^4bc+2a^3b^2c+2a^3c^2b+2a^2b^2c^2+4a^3b^2c+4a^3c^2b+\tfrac{4}{3}a^2b^2c^2\right)+$$
$$+\sum_{cyc}\left(\frac{2}{3}a^2b^2c^2-a^4b^2-a^4c^2-2a^3b^3\right)=$$
$$=\sum_{cyc}(2a^4bc+6a^3b^2c+6a^3c^2b+4a^2b^2c^2)=2abc(a+b+c)^3.$$
Two minutes of computations! By hand, of course.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\int \frac{2x^2-4x+3}{(x-1)^2}\, dx$? My book says the answer to $\int \frac{2x^2-4x+3}{(x-1)^2} \, dx$ is $2x-\frac{1}{x-1}+C$ but symbolab says it is $2x-\frac{2}{x-1}+\frac{4}{x-1}-\frac{3}{x-1}-2+C$. Who is correct and how would I get to the answer? I tried to use u-substitution but that doesn't work because the derivative of $2x^2-4x+3$ is not a multiple of $(x-1)^{-2}$ and the derivative of $x-1$ is not a multiple of $2x^2-4x+3$.
| Note that $ \int \frac{2x^2-4x+3}{(x-1)^2}\, dx = \int \frac{2\left(x-1\right)^2+1}{\left(x-1\right)^2}\, dx = \int 2+\frac{1}{\left(x-1\right)^2}\, dx = 2x - \frac{1}{\left(x-1\right)} + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
| If you want to change $\sqrt{2+ \sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$, then go through the following calculation steps from bottom to top. If you want to change $\dfrac{\sqrt{6}+\sqrt{2}}{2}$ into $\sqrt{2+ \sqrt{3}}$, then go through the following calculation steps from top to bottom. Either way we get equality $\sqrt{2+ \sqrt{3}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}$.
\begin{align*}
\dfrac{\sqrt{6}+\sqrt{2}}{2} &= \sqrt{\left(\dfrac{\sqrt{6}+\sqrt{2}}{2}\right)^2}
&& a = \sqrt{a^2} \text{ for } a \geq 0
\\&= \sqrt{\dfrac{(\sqrt{6}+\sqrt{2})^2}{2^2}}
&& \left(\dfrac{a}{b}\right)^2 = \dfrac{a^2}{b^2}
\\&= \sqrt{\dfrac{(\sqrt{6})^2+2\cdot\sqrt{2}\cdot\sqrt{6}+(\sqrt{2})^2}{4}}
&& (a+b)^2 = a^2+2\cdot a\cdot b+b^2
\\&= \sqrt{\dfrac{6+2\cdot\sqrt{2}\cdot\sqrt{6}+2}{4}}
&& a = (\sqrt{a}) \text{ for } a \geq 0
\\&= \sqrt{\dfrac{6+\sqrt{4}\cdot\sqrt{2}\cdot\sqrt{6}+2}{4}}
&& 2 = \sqrt{4}
\\&= \sqrt{\dfrac{6+\sqrt{4\cdot 2\cdot 6}+2}{4}}
&& \sqrt{a}\cdot \sqrt{b}\cdot \sqrt{c} = \sqrt{a\cdot b\cdot c}
\\&= \sqrt{\dfrac{6+\sqrt{48}+2}{4}}
&& 4\cdot 2\cdot 6=8\cdot6=48
\\&= \sqrt{\dfrac{6+\sqrt{3\cdot 16}+2}{4}}
&& 48 = 3 \cdot 16
\\&= \sqrt{\dfrac{6+\sqrt{3}\cdot \sqrt{16}+2}{4}}
&& \sqrt{a\cdot b} = \sqrt{a}\cdot \sqrt{b}
\\&= \sqrt{\dfrac{6+\sqrt{3}\cdot 4+2}{4}}
&& \sqrt{16} = \sqrt{4^2} = 4
\\&= \sqrt{\dfrac{6+4 \cdot \sqrt{3}+2}{4}}
&& a \cdot b = b\cdot a
\\&= \sqrt{\dfrac{8+4 \cdot \sqrt{3}}{4}}
&& 6+2=8
\\&= \sqrt{\dfrac{8}{4}+\dfrac{4 \cdot \sqrt{3}}{4}}
&& \dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}
\\&= \sqrt{2+ \sqrt{3}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Proof by epsilon-delta of $\lim \limits_{x \to 1} 3x^2+1=4$ I have some doubts proving the following:
$$\lim \limits_{x \to 1} 3x^2+1=4$$
My try
$(\forall \varepsilon > 0)(\exists \space \delta > 0): (0<|x-1|< \delta \implies |3x^2+1-4| < \varepsilon)$
$\implies 0<|x-1|< \delta \iff -\delta \lt x -1 \lt \delta \iff -\delta +2 \lt x+1 \lt \delta + 2 $
By transitivity:
$\implies -(\delta + 2) \lt x+1 \lt \delta + 2 \iff 0 \lt |x+1| \lt \delta + 2$
Now, working with the epsilon part:
$|3x^2+1-4| < \varepsilon \iff 3 |x-1||x+1| \lt e$
Using the fact that $|x+1| \lt \delta + 2 \iff 3|x+1| \lt 3(\delta + 2)$, and $|x-1| \lt \delta$ by transitivity:
$3|x-1||x+1| \lt 3(\delta + 2)\delta$
Then, $\varepsilon = 3(\delta + 2) \delta$ should be enough, but, if i solve the quadratic for delta:
$\delta = \frac{-6 \pm \sqrt{36 + 12\varepsilon}}{6}$, a contradiction, because in one of the solutions $\delta \lt 0 \space \forall \varepsilon \gt 0$.
After that i tried using $\delta = 1$:
$\iff |x-1| \lt 1 \implies |3x^2-3| \lt \varepsilon$
$|x-1| \lt 1 \iff -1 \lt x-1 \lt 1 \iff 1 \lt x+1 \lt 3$
$\implies -3 \lt x+1 \lt 3 \iff |x+1| \lt 3$
With $|x-1| \lt 1$ and $|x+1| \lt 3 \iff 3|x+1| \lt 9: $
$3 |x-1||x+1| \lt \varepsilon \implies 9 |x-1| \lt \varepsilon \iff |x-1| \lt \frac{\varepsilon}{9}$.
So $\delta = \frac{\varepsilon}{9}$ should satisfy, but i used the fact that $\delta = 1.$ How can i proceed here?. I saw that i have to use $\delta = \min\{1, \frac{\varepsilon}{9}\}$, but i don't know why. Any hints?. Are my steps correct or i did something wrong?. I'm not familiarized with epsilon-delta proofs.
| Given $\varepsilon>0$, your formula $\delta:= \min\{1, \frac{\varepsilon}{9}\}$ works! Let's verify it.
If $|x-1|<\delta$ with $\delta>0$, then
$$|3x^2+1-4|=|3(x-1)(x-1+2)|\leq 3|x-1|(|x-1|+2)<3\delta(\delta+2).$$
Now, by the above definition, $\delta\leq 1$ AND $\delta\leq \frac{\varepsilon}{9}$ which implies
$$3\delta(\delta+2)<3\cdot \frac{\varepsilon}{9}\cdot (1+2)=\varepsilon$$
and we are done.
P.S. If we take just $\delta:=\frac{\varepsilon}{9}$, then we have that $$|3x^2+1-4|<3\cdot\frac{\varepsilon}{9}\cdot (\frac{\varepsilon}{9}+2).$$
Is it true that for all $\varepsilon>0$, $$3\cdot\frac{\varepsilon}{9}\cdot (\frac{\varepsilon}{9}+2)<\varepsilon\;?$$
No, because the LHS is quadratic in $\varepsilon$! Therefore we have to impose a bound on $\delta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding the adjoint of a differentiation map The integral inner product is defined as
$$\langle p(x),q(x) \rangle = \int_{-1}^1 p(t)q(t)dt$$
on both $\textsf{P}_2(\mathbb{R})$ and $\textsf{P}_1(\mathbb{R})$.
Find the adjoint of the differentiation map
$$\begin{align}
\textsf{T} : \textsf{P}_2(\mathbb{R}) & \to \textsf{P}_1(\mathbb{R}) \\
p(x) & \mapsto p'(x)
\end{align}$$
Any help on finding the adjoint of $\textsf T$ above is appreciated.
| Let $p\in P_2(\Bbb R)$ and $q\in P_1(\Bbb R)$ so that
\begin{align*}
p(t) &= a_{2} t^{2} + a_{1} t + a_{0} & q(t) &= b_{1} t + b_{0}
\end{align*}
Then
$$
\langle Tp, q\rangle
= \int_{-1}^1 p^\prime(t)\cdot q(t)\,dt
= 2 \, a_{1} b_{0} + \frac{4}{3} \, a_{2} b_{1}
$$
We may also write $(T^\ast q)(t)=c_{2} t^{2} + c_{1} t + c_{0}$ so
$$
\langle p, T^\ast q\rangle
= \int_{-1}^1 p(t)\cdot\{c_{1} t + c_{0}\}\,dt
= 2 \, a_{0} c_{0} + \frac{2}{3} \, a_{2} c_{0} + \frac{2}{3} \, a_{1} c_{1} + \frac{2}{3} \, a_{0} c_{2} + \frac{2}{5} \, a_{2} c_{2}
$$
Since $\langle Tp, q\rangle = \langle p, T^\ast q\rangle$, we can compare our two expressions and obtain
\begin{align*}
0
&= \langle Tp, q\rangle - \langle p, T^\ast q\rangle \\
&= 2 \, a_{1} b_{0} + \frac{4}{3} \, a_{2} b_{1} - 2 \, a_{0} c_{0} - \frac{2}{3} \, a_{2} c_{0} - \frac{2}{3} \, a_{1} c_{1} - \frac{2}{3} \, a_{0} c_{2} - \frac{2}{5} \, a_{2} c_{2} \\
&= a_0\cdot\left\{-2\,c_0-\frac{2}{3}\,c_2\right\}+a_1\cdot\left\{2\,b_0-\frac{2}{3}\,c_1\right\}+a_2\cdot\left\{\frac{4}{3}\,b_1-\frac{2}{3}\,c_0-\frac{2}{5}\,c_2\right\}
\end{align*}
This gives the system of equations $A\vec{x}=\vec{b}$ where
\begin{align*}
A &= \left[\begin{array}{rrr}
-2 & 0 & -\frac{2}{3} \\
0 & -\frac{2}{3} & 0 \\
-\frac{2}{3} & 0 & -\frac{2}{5}
\end{array}\right] & \vec{x} &= \left[\begin{array}{r}
c_{0} \\
c_{1} \\
c_{2}
\end{array}\right] & \vec{b} &= \left[\begin{array}{r}
0 \\
-2 \, b_{0} \\
-\frac{4}{3} \, b_{1}
\end{array}\right]
\end{align*}
Solving this system gives
\begin{align*}
c_0 &= -\frac{5}{2} \, b_{1} & c_1 &= 3 \, b_{0} & c_2 &= \frac{15}{2} \, b_{1}
\end{align*}
So, our formula for $T^\ast$ is
$$
T^\ast(b_{1} t + b_{0})
= -\frac{5}{2} \, b_{1}
+3 \, b_{0}\,t
+\frac{15}{2} \, b_{1}\,t^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3307059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac{2a}{a} } = 4^{ b } + 3^{ a }$$
Firstly is this right? Secondly how to complete?
| Here a short way:
*
*$3^a = 4^b \Leftrightarrow 3 = 4^{\frac{b}{a}}$ and $4= 3^{\frac{a}{b}}$
*$\Rightarrow 9^{\frac{a}{b}} = \left(3^{\frac{a}{b}}\right)^2 = 4^2$ and
*$\Rightarrow 16^{\frac{b}{a}} = \left(4^{\frac{b}{a}}\right)^2 = 3^2$
Now, just sum up.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Interesting series involving factorials and logarithms Not sure if the following is easy to prove or not, I obtained the result using probabilistic arguments exclusively.
Define $x_1 = 2; x_2 = 4/3; x_3 = (8\cdot 6) / (7\cdot 5); x_4 = (16\cdot 14 \cdot 12 \cdot 10) / (15\cdot 13 \cdot 11 \cdot 9)$, and so on. Prove that
*
*$x_k \rightarrow \sqrt{2}$.
*$\sum_{k=1}^\infty (\log x_k)\cdot 2^{-k}= 1/2.$
A more challenging (unsolved problem): for which values of $k$ do we have $\log x_k < (\log 2)/2$ ? Maybe one can use the Stirling formula to solve this last problem.
| Note that, for $k>1, x_k=\dfrac{y_k/y_{k-1}}{z_k/z_{k-1}}$ where $y_k=2^k(2^k-2)\cdot\cdot\cdot2=2^{2^{k-1}}(2^{k-1})!$
and $z_k=(2^k-1)(2^k-3)\cdot\cdot\cdot1=(2^k)!/y_k$.
Thus $x_k=\dfrac{y_k/y_{k-1}}{\dfrac{(2^k)!}{(2^{k-1})!}/\dfrac{y_{k-1}}{y_k}}=
\dfrac{y_k^2}{y_{k-1}^2}\cdot\dfrac{(2^{k-1})!}{(2^k)!}$.
Substituting $y_k=2^{2^{k-1}}(2^{k-1})!$ and the corresponding expression for $y_{k-1}$ and simplifying yields $$x_k=\dfrac{2^{2^{k-1}}(2^{k-1})!^3}{(2^{k-2})!^2(2^k)!}.$$
Applying Stirling's formula ${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}}$ then shows $x_k\sim\sqrt2$ after a lot of cancellation
[for example, $\sqrt{2\pi}^3$ in the numerator cancels $\sqrt{2\pi}^2\sqrt{2\pi}$ in the denominator, and
$\left(\dfrac1e\right)^{2^{k-1}3}$ in the numerator cancels $\left(\dfrac1e\right)^{2^{k-2}2}$$\left(\dfrac1e\right)^{2^{k}} $ in the denominator since $2^{k-1}3=2^{k-1}+2^k$].
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $ \sin a +\sin c =2 \sin b $, show that $ \tan\frac{a+b}{2}+\tan\frac{b+c}{2} = 2 \tan\frac{c+a}{2}$ Found the question in the textbook.
I tried many methods of manipulating the identity to be proved but I did not even see a clue of using the given condition(of different forms like
$$
\sin\frac{a+c}{2} \cos\frac{a-c}{2} = \sin \frac{b}{2} \cos \frac{b}{2}
$$
and
$$
\cos \frac{a+b}{2} \sin \frac{a-b}{2} = \cos \frac{b+c}{2} \sin \frac{b-c}{2}
$$
| $$\tan\dfrac{c+a}2-\tan\dfrac{a+b}2$$
$$=\dfrac{\sin\dfrac{c+a-(a+b)}2}{\cos\dfrac{c+a}2\cos\dfrac{a+b}2}$$
$$=\dfrac{2\sin\dfrac{c-b}2\cos\dfrac{b+c}2}{2\cos\dfrac{c+a}2\cos\dfrac{a+b}2\cos\dfrac{b+c}2}$$
$$=\dfrac{\sin c-\sin b}{2\cos\dfrac{c+a}2\cos\dfrac{a+b}2\cos\dfrac{b+c}2}$$
Similarly,
$$\tan\dfrac{b+c}2-\tan\dfrac{c+a}2=\dfrac{\sin b-\sin a}{2\cos\dfrac{c+a}2\cos\dfrac{a+b}2\cos\dfrac{b+c}2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$$ holds.
My work. I proved this inequality, but my proof is ugly (it is necessary to check by brute force whether the inequality holds for $n=1,2,3,...,15$). I hope that there is nice proof of this inequality. Michael Rozenberg wrote a very nice solution to a similar problem ( Prove the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ ). I think this inequality has a similar proof, but I can’t prove in a similar way. I will write as I proved the inequality. Let $S_n= \sum \limits_{k=1}^n \frac{1}{k} $. Then $$ \sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}=n^2+2n-S_n $$ and $$\sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}=n^2-2n-3+\frac{3}{n+1}+3S_n$$ We need to prove that $$3S_n^2-S_n \left( 2n^2+8n+3-\frac{3}{n+1}\right)+\frac{n^4}{8}+7n^2+3n-3+\frac{3}{n+1} \ge 0$$ To prove this inequality, I found discriminant of the quadratic polynomial and used the fact that $S_n \le n$. It was possible to prove that the inequality holds for all $n \ge 16$.
| We can use also the Cassel's inequality:
Let $a$, $b$ and $w$ be sequences of $n$ positive numbers such that $1<m\leq\frac{a_k}{b_k}\leq M$ for any $k.$ Prove that:
$$\sum_{k=1}^nw_ka_k^2\sum_{k=1}^nw_kb_k^2\leq\frac{(M+m)^2}{4Mm}\left(\sum_{k=1}^nw_ka_kb_k\right)^2.$$
This inequality was here:
G.S. WATSON, Serial Correlation in Regression Analysis, Ph.D. Thesis, Dept. of Experimental
Statistics, North Carolina State College, Raleigh; Univ. of North Carolina, Mimograph Ser., No.
49, 1951, appendix 1.
In our case $w_k=2k-1$, $a_k=\sqrt{\frac{k+1}{k}}$, $b_k=\sqrt{\frac{k}{k+1}},$ $M=2$ and $m=1$, which gives:
$$\sum_{k=1}^n(2k-1)\frac{k+1}{k}\sum_{k=1}^n(2k-1)\frac{k}{k+1}\leq\frac{(2+1)^2}{4\cdot2\cdot1}\left(\sum_{k=1}^n(2k-1)\right)^2=\frac{9n^4}{8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
The natural numbers $a, b, c$, formed by the same $n$ digits $x$, $n$ digits $y$, and $2n$ digits $z$ satisfy $a^2 + b = c$
Given that the natural numbers $a, b, c$ are formed by the same $n$
digits $x$, $n$ digits $y$, and $2n$ digits $z$ respectively. For any
$n \geq 2$, find the digits $x, y, z$ such that $a^2 + b = c$
Greetings, I was doing this question above and I couldn't figure out how to do.
Here's my progress so far:
The relation $a^2 + b = c$ means $$(xxx...x)^2 + (yy..yy) = (zzz...zz)$$ where the $x\text{ and }y$ are $n$ in number and the $z$ is in $2n$ in number. This simplifies to $$x^2(11···1)^2 +y(11···1) = z(11···1)$$
and I couldn't figure out further. :(
By some trial and error, I found $(333)^2 + 222 = 111,111$.
Any help would be appreciated. I tried but couldn't find out even a similar question somewhere else. Please inform if this question has been asked before.
Thank You
| If we let $1_n= \underbrace{1111....1}_{n}$
$a^2 + b = c$
$\frac {a^2}{1_n} + \frac {b}{1_n} = \frac {c}{1_n}$
$x^2*1_n + y = z*(10^n + 1)$
Let's think about what that means:
If $x^2 = 10j + k$ we have
$10j + k + y \equiv z$. And we carry $j$ or $j+1$
Then we have $j +k (+1)\equiv 0$. Which means $j+k(+1) = 10$.
This can occur if:
$x=1$ so $k=1$ and $j=0$, $j+k(+1) \equiv 1,2 \not \equiv 0$.
$x = 2$ so $k=4$ and $j=0$, $j+k(+1) \equiv 4,5\not equiv 0$.
$x=3$ so $k=9$ and $j=0$ and $j+k(+1)\equiv 0$ if we carry.
$x=4$ so $k=6$ and $j=1$ and $j+k(+1)\not \equiv 0$.
$x=5$ so $k=5$ and $j=2$ and $j+k(+1)\not \equiv 0$.
$x=6$ so $k=6$ and $j=3$ and $j+k+1\equiv 0$.
$x=7$ so $k=9$ and $j=4$ and $j+k(+1) \not \equiv 0$.
$x=8$ so $k =4$ and $j=6$ and $j+k\equiv 0$.
$x=9$ so $k =1$ and $j = 8$ and $j+k(+1)\equiv 0$.
Now we repeat this and carry again to get the third digit. Assume $n > 2$ As $j+k(+1) = 10$ we must carry $1$ and get $j+k+1$ so this can only happen if $j+k+1 = 10$; not $j+k=10$.
So this can occur if:
$x=3; k=9;j=0$ and $z= j+1=1$ and $9+y \equiv 1$ so $y = 2$.
$x=6; k=9;j=3$ and $z = j+1 = 4$ and $36+y\equiv 4$ so $y=8$.
$x=9; k=1; j=8$ and $z = j+1=9$ and $81+y\equiv 9$ so $y=8$ but.. then we don't carry the $1$. we must have $y=18$ but that's not a single digit.
But if $n = 2$ we can have:
$x^2 = 10j + k$ and $x^2*11 = 100j + 10(k+j) + k$ and $x^2*11 + y= 100j + 10(k+j) + k+y$ where $k+y = z$ and $k+j = 10$ and $j+1 = z$
Then we can have $x=8;k=4;j=6;z=7$ and $y = 3$. (i.e $8^2*11 + 3= 707$ or $(88)^2 + 33 = 7777$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Coordinates of centres of central triangles with respect to the reference triangle In Kimberling's Encyclopedia of Triangle Centers, a lot of centres are described as the centres of certain central triangles of the reference triangle, whether as a main or alternate definition. For example, $X_{164}$ is defined as the incentre of the excentral triangle (whose vertices are the triangle's excentres) and the Kiepert antipode $X_{114}$ can be defined as the Tarry point $X_{98}$ of the medial triangle (whose vertices are the triangle's medians).
For a programmatic implementation of triangle centres these indirect definitions, when used judiciously, reduce the workload and error bounds associated with the centre's computation as compared to simply using trilinear/barycentric coordinates (which may be very complicated).
However, I cannot figure out how to combine the pieces together to generate the centre's coordinates in either system with respect to the reference triangle, which is the question I am asking here. For example, I can easily look up the trilinear coordinates of the vertices of the excentral triangle ($-1:1:1$ and permutations) and the trilinear coordinates of the incentre ($1:1:1$), but how do I combine these two parts to obtain trilinear coordinates for $X_{164}$ ($\sin B/2+\sin C/2-\sin A/2$ and cyclic permutations of $A,B,C$)?
| This answer is an answer for the very specific question:
How to obtain or verify $X(164)$ knowing it is the $X(1)$-point of the excentral triangle?
Explicitly:
How to obtain the barycentric coordinates of $X(164)$, knowing it is the incenter of the circle through the points with (inhomogeneous) barycentric coordinates
$$\begin{aligned}
S&(-a:b:c)\ ,\\
T&(a:-b:c)\ ,\\
U&(a:-b:c)\ ?
\end{aligned}
$$
As a reference, the following collects a lot of information in a beautiful presentation:
Max Schindler, Evan Chen, Barycentric Coordinates for the Impatient
We fix the frame, we will work usually with baricentric normalized coordinates $(x,y,z)$ of a point $P=xA+yB+zC$, considered w.r.t. the triangle $ABC$ with sides $a=BC$, $b=CA$, $c=AB$. The word normalized refers to the relation $x+y+z=1$. While working with points in barycentric coordinates, often expressions get complicated, so it is natural to show the point $(x,y,z)$ only up to the common denominator, which is omitted. In such a case we write $(x:y:z)$, and this point stays for $(\ x/(x+y+z),\ y/(x+y+z),\ z/(x+y+z)\ )$.
The formula for the squared distance $OP^2$ between two points with barycentric normalized coordinates $P(x,y,z)$ and $O(x_0,y_0,z_0)$ is algebraic, explicitly
$$
PO^2
=
-a^2(y-y_0)(z-z_0)
-b^2(x-x_0)(z-z_0)
-c^2(x-x_0)(y-y_0) \ ,
$$
and it factorizes through building the "displacement" (vector) $(x-x_0,y-y_0,z-z_0)$.
In our case, let $S,T,U$ be the ex-centers of $\Delta ABC$. Let us compute first $TU^2$. To have an easy typing, let me please type sage code:
sage: var('a,b,c');
sage: def d2(P, Q):
....: x1, y1, z1 = P
....: x2, y2, z2 = Q
....: x12, y12, z12 = x2-x1, y2-y1, z2-z1
....: return -a^2*y12*z12 -b^2*x12*z12 -c^2*x12*y12
....:
sage: S, T, U = vector([-a,b,c]), vector([a,-b,c]), vector([a,b,-c])
sage: S, T, U = S/sum(S), T/sum(T), U/sum(U)
sage: S, T, U
((a/(a - b - c), -b/(a - b - c), -c/(a - b - c)),
(a/(a - b + c), -b/(a - b + c), c/(a - b + c)),
(a/(a + b - c), b/(a + b - c), -c/(a + b - c)))
sage: d2(T, U).factor()
4*a^2*b*c/((a + b - c)*(a - b + c))
So we can write:
$$
\begin{aligned}
TU^2 &= a^2\cdot \frac{bc}{(p-b)(p-c)}
&&=a^2\cdot\sin^{-2}\frac A2\ ,\text{ and similarly}
\\
US^2 &= b^2\cdot \frac{ac}{(p-a)(p-c)}
&&=b^2\cdot\sin^{-2}\frac B2\ ,
\\
ST^2 &= c^2\cdot \frac{ab}{(p-a)(p-b)}
&&=c^2\cdot\sin^{-2}\frac C2\ .
\end{aligned}
$$
(Above $p=(a+b+c)/2$, since i need $s$ for an other purpose. This is a rare case where i do not use $p$ for a prime number.)
Above it is important that we could write the expressions $TU^2$, $US^2$, $ST^2$ also as squares of "elements" directly connected to the original triangle $\Delta ABC$. (Taking the sine of half an angle brings us slightly out of the world of algebraic expressions in $a,b,c$.) Else i could not continue. The continuation is now clear. Let $s,t,u$ be the sides of $\Delta STU$, so
$$
\begin{aligned}
s & = TU = a\cdot\sin^{-1}\frac A2\ ,
\\
t & = US = b\cdot\sin^{-1}\frac B2\ ,
\\
u & = ST = c\cdot\sin^{-1}\frac C2\ .
\\[3mm]
&\qquad\text{ Then we use the formula for $X(1)$ in $\Delta STU$:}
\\[3mm]
X(1)_{\Delta STU}
&=
\frac 1{s+t+u}(sS+tT+uU)\ ,\text{ and here we plug in }
\\
S &=
\frac{-a}{-a+b+c}A +
\frac{b}{-a+b+c}B +
\frac{c}{-a+b+c}C \ ,\\
\\
T &=
\frac{a}{a-b+c}A +
\frac{-b}{a-b+c}B +
\frac{c}{a-b+c}C \ ,\\
\\
U &=
\frac{a}{a+b-c}A +
\frac{b}{a+b-c}B +
\frac{-c}{a+b-c}C \ ,
\\[3mm]
&\qquad\text{ and we compute of the $A$-coefficient in $X(1)_{\Delta STU}$,}
\\[3mm]
\text{($A$-coefficient)}
&=\frac a{s+t+u}
\left(
\frac{-s}{2(p-a)} +
\frac{t}{2(p-b)} +
\frac{u}{2(p-c)}
\right)
\\
&\sim
a
\left(
-
\frac{a}{(p-a)\sin\frac A2} +
\frac{b}{(p-b)\sin\frac B2} +
\frac{c}{(p-c)\sin\frac B2}
\right)
\\
&=
a
\left(
-
\frac{a}{r\cot\frac A2\sin\frac A2} +
\frac{b}{r\cot\frac B2\sin\frac B2} +
\frac{c}{r\cot\frac C2\sin\frac C2}
\right)
\\
&\sim
a
\left(
-
\frac{\sin A}{\cos\frac A2} +
\frac{\sin B}{\cos\frac B2} +
\frac{\sin C}{\cos\frac C2}
\right)
\\
&\sim
a
\left(
-
\sin\frac A2 +
\sin\frac B2 +
\sin\frac C2
\right)\ .
\end{aligned}
$$
The corresponding coefficients of $A,B,C$ are the needed barycentric coefficients, and they match the one in the ETC.
Above, the symbol $\sim$ denotes equality up to a factor, which is a symmetric polynomial in $a,b,c$.
Conclusion:
The OP reopens implicitly the following door. Given the triangle $ABC$, we associate other triangles $STU$, where $S\cong S(a,b,c)$ is an asymmetric expression, a linear combination in $A,B,C$ with polynomial (or slightly more general) weights in $a,b,c$, maybe symmetric w.r.t. $b\leftrightarrow c$, and $T,U$ are obtained correspondingly using cyclic permutations of $(a,b,c)$ (and implicitly $A,B,C$), so $T\cong S(b,c,a)$, $U\cong S(c,a,b)$.
We compute the squared sides of $STU$ as above, and if the explicit expressions for $TU^2$, $US^2$, $ST^2$ admit a radical, then we can proceed as above to write all centers $X(k)_{STU}$ in terms of $S,T,U$, then using the expressions for $S,T,U$ in terms of $A,B,C$ we obtain weights w.r.t. the initial triangle, and we may want to match them with existing centers.
Application:
Let $STU$ be constructed based on $S=(0,\frac 12,\frac 12)=\frac 12(B+C)$, so $S$ is the mid point of $BC$, and we chose $T,U$ similarly. The sides of $STU$ are $s=a/2$, $t=b/2$, $u=c/2$, so
$$
\begin{aligned}
X(98)_{\Delta STU}
&=
\left(\
\frac 1{t^4+u^4-s^2(t^2+u^2)}\ :\
\dots
\ \right)
\\
&=
\left(\
\frac 1{b^4+c^4-a^2(b^2+c^2)}\ :\
\dots
\ \right)
\\
&\sim \frac 1{b^4+c^4-a^2(b^2+c^2)}S+\dots
\\
&\sim \frac 1{b^4+c^4-a^2(b^2+c^2)}(B+C)+\dots
\\
&\sim
\left(
\frac 1{a^4+b^4-c^2(a^2+b^2)}+
\frac 1{a^4+c^4-b^2(a^2+c^2)}
\right)A+\dots
\\
&\sim
(b^4+c^4-a^2(b^2+c^2))
\cdot\Big(\
(a^4+b^4-c^2(a^2+b^2))+
(a^4+c^4-b^2(a^2+c^2))
\ \Big)A+\dots
\\
&\qquad\text{ matching again the ETC, and giving}
\\
&=X(114)_{\Delta ABC}\ .
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Chord $x+y=1$ Subtends $45^{\circ}$ at the centre If the Chord $x+y=1$ Subtends $45^{\circ}$ at the centre of the circle $x^2+y^2=a^2$, Find $a$
Method $1$:
We have radius=$a$
If $O$ is the centre and chord be $AB$ we have
$$\angle AOB =45$$
If $OM$ is drawn perpendicular to the chord we have:
$$\angle AOM=22.5$$
In $\Delta AOM$ we have
$$cos(22.5)=\frac{OM}{a}$$
$OM$ is perpendicular distance from $(0,0)$ to $x+y=1$ which is
$$OM=\left|\frac{0+0-1}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}}$$
Hence we get $$a=\sqrt{2-\sqrt{2}}$$
Method $2.$
By Homogenization of line and curve we have:
$$x^+y^2=a^2(x+y)^2$$
$$(a^2-1)x^2+2xya^2+(a^2-1)y^2=0$$
This pair of straight lines subtends angle of $45^{\circ}$
We have formula that:
$$\tan (\theta)=\left|\frac{2\sqrt{H^2-AB}}{A+B}\right|$$
We have
$A=a^2-1=B$ and $H=a^2$
We get
$$1=\left|\frac{2\sqrt{a^4-(a^2-1)^2}}{2(a^2-1)}\right|$$
Simplifying we get
$$a^4-4a^2+2=0$$
Giving
$$a=\sqrt{2\pm \sqrt{2}}$$
Why in first Method, i could not get second answer?
| When $a=1$, the angle subtended is $90º$. This angle will continually decrease until the circle is tangent to the chord, giving a subtended angle of $0º$. Similarly, increasing $a$ will cause the subtended angle to be larger than $90º$, eventually approaching $180º$.
Therefore there is only one solution for $a$. If you get two solutions, you should substitute $a$ into the original equation to verify which solution makes sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do I solve this exponential-surdic equation? I have been trying to solve a problem today, but have had very little to no progress. The problem is:
Find the real values of $x$ satisfying $$(26 + 15\sqrt{3})^x + 6(2 + \sqrt{3})^x + (2 - \sqrt{3})^x - 5(7 + 4\sqrt{3})^x = 5$$
Clear explanations would surely be appreciated.
| Notice, that $26+15\sqrt3=(2+\sqrt3)^3$ and $7+4\sqrt3=(2+\sqrt3)^2$. Now the equation reads $$(2 + \sqrt{3})^{3x} + 6(2 + \sqrt{3})^x + (2 - \sqrt{3})^x - 5(2 + \sqrt{3})^{2x} = 5$$
Let's multiply both parts of it by a nonzero $(2 + \sqrt{3})^x$. We receive $$(2 + \sqrt{3})^{4x} + 6(2 + \sqrt{3})^{2x} + ((2 - \sqrt{3})(2 + \sqrt{3}))^x - 5(2 + \sqrt{3})^{3x} = 5(2 + \sqrt{3})^x$$
The term $((2 - \sqrt{3})(2 + \sqrt{3}))^x$ equals 1. Let's introduce $a=(2+\sqrt3)^x$. Then we get $$a^4-5a^3+6a^2-5a+1=0$$ This is $$(a^2-4a+1)(a^2-a+1)=0$$ Roots are $2-\sqrt3$ and $2+\sqrt3$, which means that $x=1$ is the solution and another one is $\log_{2+\sqrt3}(2-\sqrt3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Derive formula for the area of triangle provided that you know 2 angles and side between them. I am trying to derive the formula for finding the area of the triangle that uses 2 angles and and 1 side between them.
Formula provided by the book:
If we know two angles (call them $A$ and $B$) and the side between
them call it $b$, then Area = $\frac{b^2 \sin B \sin C}{2 \sin A}$
However, I keep getting different result.
My attempt as follows:
First we draw following picture:
$\angle YPX = 90^{\circ}$ and $\angle YPZ = 90^{\circ}$
We know that $\angle YXZ = A$ and $\angle YZX = C$ and XZ = $b$
Let $XP = d$ and $PZ = b-d$
$\tan A = \frac{YP}{XP} = \frac{YP}{d} \implies YP = \tan A \cdot d$
$\tan C = \frac{YP}{PZ} = \frac{YP}{b-d} \implies YP = \tan C \cdot (b-d)$
It follows that $$\tan A \cdot d = \tan C \cdot (b-d)$$
$$d = \frac{\tan C \cdot (b-d)}{\tan A}$$
$$\frac{d}{b-d} = \frac{\tan C}{\tan A}$$
$$\frac{b-d}{d} = \frac{\tan A}{\tan C}$$
$$\frac{b}{d} - 1 = \frac{\tan A}{\tan C}$$
$$\frac{b}{d} = \frac{\tan A}{\tan C} + 1$$
$$\frac{1}{d} = \frac{\tan A}{\tan C \cdot b} + \frac{1}{b}$$
$$\frac{1}{d} = \frac{\tan A}{\tan C \cdot b} + \frac{\tan C}{\tan C \cdot b}$$
$$\frac{1}{d} = \frac{\tan A + \tan C}{\tan C \cdot b} $$
$$ d= \frac{\tan C \cdot b}{\tan A + \tan C} $$
$$ d= \frac{\frac{\sin C}{\cos C} \cdot b}{\frac{\sin A}{\cos A} + \frac{\sin C}{\cos C}} $$
$$ d= \frac{\frac{\sin C}{\cos C} \cdot b}{\frac{\sin A \cos C + \sin C \cos A}{\cos C \cos A} }$$
$$ d = \frac{\frac{\sin C}{\cos C} \cdot b \cdot \cos C \cos A}{\sin A \cos C + \sin C \cos A}$$
$$ d = \frac{\sin C \cos A\cdot b}{\sin A \cos C + \sin C \cos A}$$
We know that $$YP = \tan A \cdot d$$
$$YP = \frac{\sin A}{\cos A} \frac{\sin C \cos A\cdot b}{\sin A \cos C + \sin C \cos A}$$
$$YP = \frac{\sin A}{\cos A} \frac{\sin C \cos A\cdot b}{\sin(A+C)}$$
$$YP = \frac{\sin A \sin C \cdot b}{\sin(A+C)}$$
Because
$$\text{Area} =\frac{1}{2} \cdot YP \cdot b$$
Then
$$\text{Area} =\frac{1}{2} \cdot \frac{\sin A \sin C \cdot b}{\sin(A+C)} \cdot b$$
$$\text{Area} = \frac{\sin A \sin C \cdot b^2}{2 \sin(A+C)} $$
Which, as can be seen, is different from the one provided by the textbook. I've tried to find areas of several triangles using formula above and this calculator, both give the same results. However, I'm still unsure whether it is correct. Thus I would like to ask you, is the the derivation (and the formula) above correct?
| It is simply use are of triangle is ∆=(1/2)cbSinA=(1/2)abSinB
Therefore ∆^2 = (1/4) acb^2 SinA.SinB
Also ∆=(1/2)acSinC
Dividing both we get Area ∆ =( c^2 SinA.SinB)/2 Sin C
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve this geometric series question? A geometric series has first term 4 and common ratio r, where 0< r<1. The first, second, and fourth terms of this geometric series form three successive terms of an arithmetic series. Find the sum to infinity of the geometric series.
$S = \frac{t_1}{1-r}$ is the sum to infinity where $t_1$ is the first term in the geometric series.
The second term of the arithmetic series is $t_1*r$, the third term is $t_1*r^3$.
From there we can deduce that $(r-1)(r^2+r-1)=0.$ So r must be either $\frac{-1 - \sqrt{5}}{2}$ or $\frac{-1 + \sqrt{5}}{2}$.
From there how can I find $t_1$ and the sum to infinity?
Thank you very much!
| The question says the first term is $4$, so $t_1 = 4$. Thus, since $0 \lt r \lt 1$, you have $r = \frac{-1 + \sqrt{5}}{2}$ giving
$$S = \frac{t_1}{1-r} = \frac{4}{1 - \frac{-1 + \sqrt{5}}{2}} = \frac{8}{2 - (-1 + \sqrt{5})} = \frac{8}{3 - \sqrt{5}} = 2\left(3 + \sqrt{5}\right) \tag{1}\label{eq1}$$
Thanks to J. W. Tanner for suggesting providing the normalized value at the end.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find equations of the two lines through the origin that make an angle $\tan^{-1}(1/2)$ with $3y=2x$
If two straight lines pass through the origin and makes an angle $\tan^{-1}(1/2)$ with $3y=2x$, then find its equations.
Let $m$ be the gradient of the line then, $$\frac{1}{2}=\frac{m-2/3}{1-2m/3}$$ I don't know whether my approach is correct or not.
| $tan^{-1}(\frac{1}{2}) = \theta$
$tan(\theta) = \frac{1}{2}$
Line $3y = 2x$ have the slope of $\frac{2}{3}$, suppose this line is called a and the two lines in question is b and c (with $m_{b} > m_{a}, m_{c} < m_{a})$.
So $m_{a} = \frac{2}{3}$ $=>$ $tan(A) = \frac{2}{3}$, $m_{b} = tan(B)$, and $m_{c} = tan{C}$.
Hint :
$B - A = tan^{-1}(\frac{1}{2})$
$A-C = tan^{-1}(\frac{1}{2}) $
$tan(B-A) = \frac{1}{2}$
$tan(A-C) = \frac{1}{2}$
$tan(B)$ and $tan(C)$ will be the slope of the line you want to find, with $tan(A) = \frac{2}{3}$.
Rather complicated but i think the easiest one you can do with logical thinking rather than looking up formula
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Simplify $ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $ Simplify
$$ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $$
Attempt:
$$ \frac{ \sqrt[3]{16} - 1}{3 + \sqrt[3]{4} + \sqrt[3]{2}} = \frac{ \sqrt[3]{16} - 1}{ (3 + \sqrt[3]{4}) + \sqrt[3]{2}} \times \frac{ (3 + \sqrt[3]{4}) - \sqrt[3]{2}}{ (3 + \sqrt[3]{4}) - \sqrt[3]{2}} $$
$$ = \frac{ (\sqrt[3]{16} - 1) [(3 + \sqrt[3]{4}) - \sqrt[3]{2}]}{ (3 + \sqrt[3]{4})^{2} - 2^{2/3}} $$
$$ = \frac{ 3 \sqrt[3]{16} - 3\sqrt[3]{4} + \sqrt[3]{2} + 1}{ (9 + 5 \sqrt[3]{4} + \sqrt[3]{16}) } $$
From here on I don't know how to continue.
I can let $a = \sqrt[3]{2}$, but still cannot do anything.
| Let $x=\sqrt[3]{2}$ then we have $\frac{x^4-1}{x^2+x+3}$
Multiply by $x-1$
$\frac{x^4-1}{x^2+x+3}\cdot\frac{x-1}{x-1}=\frac{(x^4-1)(x-1)}{x^3+2x-3}=\frac{(x^4-1)(x-1)}{2+2x-3}=\frac{(x^4-1)(x-1)}{2x-1}$
Multiply by $4x^2+2x+1$
$\frac{(x^4-1)(x-1)(4x^2+2x+1)}{(2x-1)(4x^2+2x+1)}=\frac{(x^4-1)(x-1)(4x^2+2x+1)}{8x^3-1}=\frac{(x^4-1)(x-1)(4x^2+2x+1)}{15}=\frac{1 + x + 2 x^2 - 4 x^3 - x^4 - x^5 - 2 x^6 + 4 x^7}{15}=\frac{15\sqrt[3]{2}-15}{15}=\sqrt[3]{2}-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Convert standard form to factor form Im so confused now
I have no idea how to factor a standard form
i have example shown below
Standard form
$$x^4+6x^3-x^2-6x$$
Factor form
$$x(x+6)(x+1)(x-1)$$
Can you explain a little bit
I have no idea
Thank you
| $$x^4+6x^3-x^2-6x=x^4-x^2+6x^3-6x=x^2(x^2-1)+6x(x^2-1)=$$ $$(x^2-1)x(x+6)=(x-1)(x+1) x(x+6)=x(x-1)(x+1)(x+6).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Solution verification on homework problem. Separable first order ODE IVP. The answer is supposedly $y^2 = 1 + \sqrt{x^2 - 16}$ I don't know where I went wrong cause I know for a fact that my substitution of $x = 4 \sec(\theta)$ is correct. I know for a fact that after substitution the integral becomes $\int \sec^2(\theta)= \tan(\theta)+C$. I am pretty sure that my substitution of $\tan(\theta)$ is correct. Am I missing something?
$2y \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 16}} \space \space \space y(5)=2$
$\int 2y \space dy = \int \frac{4\sec(\theta)}{\sqrt{(4\sec(\theta))^2-16}}\space d(\theta)$
$y^2 = \int \frac{4\sec(\theta)}{\sqrt{(\sec^2(\theta)-1)16}}\sec(\theta) \ tan(\theta)$
$y^2 = \frac{4\sec(\theta)}{4\tan(\theta)}\sec(\theta)\tan(\theta)$
$y^2 = \int sec^2(\theta)$
$y^2 = \tan(\theta) + c$
Using the reference triangle: Tangent is equal to $\frac{\sqrt{x^2-16}}{4}$
$y^2= \frac{\sqrt{x^2-16}}{4} + C$
$4 = \frac{3}{4}+ C$
$C = \frac{15}{4}$
$y^2= \frac{\sqrt{x^2-16}}{4} + \frac{15}{4}$
| In integrating the RHS of in$$2y \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 16}} \space \space \space y(5)=2$$
it is advised to let $$u=x^2-16$$ and you have $$du=2xdx$$
Then you do not have to worry about trig substitution at all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculate how many ways you can give $7$ children $7$ identical candies My try:
Calculate how many ways you can give $7$ children $7$ identical candies if each child got at most 2 candies.
$$x_1+x_2+x_3+x_4+x_5+x_6+x_7=7 \text{ for } x_i \in \left\{ 0,1,2\right\}$$
$$[t^7](1+t+t^2)^7[t^7](\frac{1-t^3}{1-t})^7=[t^7](1-t^3)^7 \sum {n+6 \choose 6}t^n$$
$$\begin{array}{|c|c|c|c||c|c|}
\hline
\text{first parenthesis} & \text{ways in the first} & \text{ways in the second }\\ \hline
\text{1} & 1 & { 13 \choose 6} \\ \hline
{t^3} & { 1 \choose 1} & { 10 \choose 6} \\ \hline
{t^6} & { 7 \choose 2} & { 7 \choose 6}\\ \hline
\end{array}$$
Sollution:$${ 7 \choose 2}{ 7 \choose 6}+{ 7 \choose 1}{ 10 \choose 6}+{ 13 \choose 6}=3333$$
But I checked it in Mathematica and I get $393$. So can you check where the error is?
| The solution will be confficent of $\,t^7 $ in the
expension of $${(1\, - \ t^3)}^7{(1 - t)}^{-7}$$
I.e coefficient of $t^7$ in $( \,1 - \, {7 \choose 1} t^3\,+{7 \choose 2} t^6)(1\,-t)^{-7}$ $$
= \,{13 \choose 7} \, - 7\,{10 \choose 4}\,+21{7 \choose 1} $$
= 393
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Angles between vectors of center of two incircles I have two two incircle between rectangle and two
quadrilateral circlein. It's possible to determine exact value of $\phi,$ angles between vectors of center of two circles.
|
Let $|AB|=|CD|=a$,$|BC|=|AD|=b$,
$|O_1E|=r_1$,
$|O_2F|=r_2$,
$\angle O_1AO_2=\phi$,
$\angle O_1AF=\alpha$
$\angle O_2AF=\beta$.
Then
\begin{align}
\tan\alpha&=\frac{b-r_1}{a/2}
\tag{1}\label{1}
,\\
\sin\alpha&=\frac{b-r_1}{b+r_1}
,\\
\tan\alpha&=\frac{\sin\alpha}{\sqrt{1-\sin^2\alpha}}
=\frac{b-r_1}{b+r_1}
\left/
\sqrt{1-\Big( \frac{b-r_1}{b+r_1} \Big)^2} \right.
=\tfrac12\,\frac{b-r_1}{\sqrt{b\,r_1}}
\tag{2}\label{2}
.
\end{align}
From \eqref{1}$=$\eqref{2} it follows
\begin{align}
a&=4\,\sqrt{b\,r_1}
\tag{3}\label{3}
.
\end{align}
Similarly,
\begin{align}
\tan\beta&=\frac{2r_2}a
\tag{4}\label{4}
,\\
\sin\beta&=\frac{r_2}{b-r_2}
,\\
\tan\beta&=
\frac{\sin\beta}{\sqrt{1-\sin^2\beta}}
=\frac{r_2}{\sqrt{b\,(b-2\,r_2)}}
\tag{5}\label{5}
.
\end{align}
From \eqref{4}$=$\eqref{5}:
\begin{align}
a&=2\,\sqrt{b\,(b-2\,r_2)}
\tag{6}\label{6}
,
\end{align}
and from \eqref{3}$=$\eqref{6}
we have
\begin{align}
b&=4r_1+2r_2
,\\
a&=4\,\sqrt{r_1\,(4\,r_1+2\,r_2)}
.
\end{align}
\begin{align}
\tan\phi&=\tan(\alpha-\beta)
=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}
=\frac{2(3r_1+r_2)\,\sqrt{2\,r_1\,(2\,r_1+r_2)}}{16\,r_1^2+11\,r_1\,r_2+2\,r_2^2}
\end{align}
For $r_1=1$, $r_2=2$ we have
\begin{align}
a&=8\,\sqrt2\approx 11.31370850
,\\
b&=8
,\\
\phi&=\arctan\Big(\frac{10\,\sqrt2}{23} \Big)
\approx 31.586338^\circ
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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} |
Providing divisibility condition given fraction identity If $x,y,z$ are positive integers satisfying
$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$$
prove that $20{\,\mid\,}xy$.
My work:
Expanding, we find
$$(xz)^2+(yz)^2=(xy)^2$$
I know the Pythagorean triple formula and I tried applying that, but I couldn't find a way to get $20$.
I've found a lot of questions on this website relating to similar questions, but none of them seem to refer to a divisibility condition.
If someone could help me find one, that'd also be greatly appreciated.
| Suppose $x,y,z$ are positive integers such that
$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$$
Equivalently, $x,y,z$ are positive integers such that
$$\qquad\qquad\; x^2y^2=z^2(x^2+y^2)\qquad(\textbf{eq})$$
Aqua has already shown that $5{\,\mid\,}(xy)$.
To show that $4{\,\mid\,}(xy)$, we can argue as follows . . .
If $x,y$ are both even, then $4{\,|\,}(xy)$, and we're done.
If $x,y$ are both odd, then $x^2+y^2$ is even, hence the RHS of $(\textbf{eq})$ is even, contradiction, since the LHS is odd.
It remains to resolve the case where exactly one of $x,y$ is even.
Without loss of generality, assume $x$ is even and $y$ is odd.
Let $2^k$ be the largest power of $2$ which divides $x$.
Since $y^2$ and $x^2+y^2$ are both odd, it follows from $(\textbf{eq})$ that $2^k$ is also the highest power of $2$ which divides $z$.
Thus we can write $x=2^kx_1$ and $z=2^kz_1$, where $x_1,z_1$ are both odd.
\begin{align*}
\text{Then}\;\;&x^2y^2=z^2(x^2+y^2)\\[4pt]
\implies\;&x_1^2y^2=z_1^2(x^2+y^2)\\[4pt]
\implies\;&x_1^2y^2\equiv z_1^2(x^2+y^2)\;(\text{mod}\; 8)\\[4pt]
\implies\;&(1)(1)\equiv (1)(x^2+1)\;(\text{mod}\; 8)\;\;\;\text{[since $x_1,z_1,y$ are odd]}\\[4pt]
\implies\;&x^2\equiv 0\;(\text{mod}\; 8)\\[4pt]
\implies\;&8{\,\mid\,}x^2\\[4pt]
\implies\;&16{\,\mid\,}x^2\\[4pt]
\implies\;&4{\,\mid\,}x\\[4pt]
\implies\;&4{\,\mid\,}(xy)\\[4pt]
\end{align*}
as required.
This completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Limit of a function in which square roots are involved $$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}$$
Could someone please help me solve this problem.
I tried multiplying by a unity factor but I end up stuck.
| You can simplify it:
$$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}=\lim_{x \to 0}\frac{x+2-\sqrt{2}\sqrt{x+2}}{3(x+1)+\sqrt{x+1}-4}=\\
\lim_{x \to 0}\frac{\sqrt{x+2}(\sqrt{x+2}-\sqrt{2})}{3(\sqrt{x+1}-1)(\sqrt{x+1}+\frac43)}=\\
\lim_{x \to 0}\frac{\sqrt{x+2}-\sqrt{2}}{\sqrt{x+1}-1}\cdot \lim_\limits{x\to 0}\frac{\sqrt{x+2}}{3\sqrt{x+1}+4}=\\
\frac{\sqrt{2}}{7}\cdot \lim_{x \to 0}\frac{\color{red}{\sqrt{x+2}-\sqrt{2}}}{\color{blue}{\sqrt{x+1}-1}}=\cdots$$
Can you continue?
Answer:
Multiply both top and bottom by conjugates:$$\frac{\sqrt{2}}{7}\cdot\lim_{x \to 0}\frac{\require{cancel}\cancel{(\color{red}{(x+2)-2})}(\sqrt{x+1}+1)}{\cancel{(\color{blue}{(x+1)-1})}(\sqrt{x+2}+\sqrt{2})}=\frac{\sqrt{2}}{7}\cdot \frac1{\sqrt{2}}=\frac17.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Use properties of determinant and show
Let $n$ be a positive integer and
\begin{align} M =
\begin{pmatrix}
n! & (n+1)! & (n+2)! \\
(n+1)! &(n+2)! & (n+3)! \\
(n+2)! & (n+3)! & (n+4)! \\
\end{pmatrix}
\end{align}
Use properties of determinant to show that \begin{align}\left(\frac{|M|}{(n!)^3}- 4\right)\end{align} is divisible by $n$.
I took an $n!$ factor out of the matrix, getting a new matrix $B$ such that $\det A = n^3 \det B$, since $A$ and $B$ are row equivalents. But what is $\det B$ ?
| After you take the common factor $n!$ out of all columns, take $(n+1)$ and $(n+1)(n+2)$ out of 2nd and 3rd columns, respectively:
$$\frac{|M|}{(n!)^3}=\left|\begin{array}{ccc}
1 & \color{red}{n+1} & \color{blue}{(n+1)(n+2)}\\
n+1 & (\color{red}{n+1})(n+2) & \small{\color{blue}{(n+1)(n+2)}(n+3)}\\
(n+1)(n+2) & \small{(\color{red}{n+1})(n+2)(n+3)} & \scriptsize{\color{blue}{(n+1)(n+2)}(n+3)(n+4)}
\end{array}\right|=\\
\color{red}{(n+1)}\color{blue}{(n+1)(n+2)}\left|\begin{array}{ccc}
1 & 1 & 1\\
n+1 & n+2 & \small{n+3}\\
(n+1)(n+2) & \small{(n+2)(n+3)} & \scriptsize{(n+3)(n+4)}
\end{array}\right|\stackrel{C_3-C_2\to C_3\\ C_2-C_1\to C_2}{=}\\
(n+1)^2(n+2)\left|\begin{array}{ccc}
1 & 0 & 0\\
n+1 & 1 & 1\\
(n+1)(n+2) & 2n & 2n+2
\end{array}\right|\stackrel{C_3-C_2\to C_3}=\\
(n+1)^2(n+2)\left|\begin{array}{ccc}
1 & 0 & 0\\
n+1 & 1 & 0\\
(n+1)(n+2) & 2n & \color{red}2
\end{array}\right|=\color{red}2(n+1)^2(n+2)$$
Hence:
$$\begin{align}\left(\frac{|M|}{(n!)^3}- 4\right)\end{align}=2n^3+8n^2+10n\equiv 0 \pmod{n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the minimum $n$ such that $x^2+7=\sum_{k=1}^n f_k(x)^2$ where $f_k(x)\in \mathbb{Q}[x]$ Recently, I have found this problem:
Find the minimum $n \in N$ such that $x^2+7=f_1(x)^2+f_2(x)^2+\cdots+f_n(x)^2$ where $f_1(x),+f_2(x),+\cdots+f_n(x)$ are polynomials with rational coefficients.
I have tried to solve this problem when $n=2$ using $f_1(x)=(a_1x+b_1)^2$ and $f_2(x)=-(a_2x+b_2)^2$, but I can't go on. Any idea?
| The answer of @Mindlack contains a full solution to the problem. This thread just collects some partial results I got while studying the problem, using only the idea from @Mindlack that $7$ is not the sum of $3$ or less squares of rationals.
Partial results
Notice we can only use polynomials with degree at most $1$, since the coeficient of the leading term of $f_i(x)^2$ is always positive.
Easy upper bound. Taking $f_1(x) = x$ and using the idea on the comments that we can write $7 = 1^2 + 1^2 + 1^2 + 2^2$. This gives a way of using $5$ polynomials.
The argument of @Mindlack. Looking only at the constant coeficients modulo $8$, if we used only $3$ polynomials we would have $7d^2 = a^2 + b^2 + c^2$ with $a,b,c,d$ integers and $\mathrm{gcd}(a,b,c,d)=1$. Since squares modulo $8$ are only $0, 1, 4$, this is impossible. This shows that $n \geq 4$. Also, it will be used again on the arguments below.
Using only one linear polynomial. Using $f_1(x) = ax + b$ implies that $f_1(x)^2 = a^2 x^2 + 2ab x + b^2$ and we need that $a^2 = 1$, $ab=0$, thus $a = \pm 1$ and $b=0$. Since the other $f_i$ are rationals, the argument of @Mindlack implies we need at least $4$ other polynomials in this case.
Using only two linear polynomials. Take $f_1(x) = a_1x + b_1$ and $f_2(x) = a_2x + b_2$, implying that
$$
f_1(x)^2 + f_2(x)^2 = (a_1^2 + a_2^2) x^2 + 2(a_1b_1 + a_2b_2)x + (b_1^2 + b_2^2).
$$
Since the coeficient of $x^2$ needs to be $1$, we should study the solutions of
$$
a_1^2 + a_2^2 = 1 \quad \text{for rational $a_i$}.
$$
Writing $a_i = \tfrac{p_i}{q_i}$ in lowest terms, we have that
$$
p_1^2 q_2^2 + p_2^2 q_1^2 = q_1^2 q_2^2
$$
Notice that in this case we must have $q_1 \mid q_2$, since $p_1$ and $q_1$ do not have common prime factors. Analogously, $q_2 \mid q_1$ and thus they are equal. Now, we are looking at integer solutions for
$$
p_1^2 + p_2^2 = q^2,
$$
that is, Pythagorean triples. Considering
$$
f_1(x) = \frac{p_1}{q} x + b_1 \quad \text{and} \quad f_2(x) = \frac{p_2}{q} x + b_2
$$
we have
$$
f_1(x)^2 + f_2(x)^2 = x^2 + \frac2q (p_1 b_1 + p_2 b_2) + b_1^2 + b_2^2.
$$
Since the term in $x$ should be zero, we have $b_2 = - b_1 \frac{p_1}{p_2}$ and then
$$
f_1(x)^2 + f_2(x)^2 = x^2 + b_1^2 (1 + \frac{p_1^2}{p_2^2}) = x^2 + b_1^2 \frac{q^2}{p_2^2}.
$$
Notice that the independent term is the square of a rational. Then, using the argument of @Mindlack once again, we cannot use only $4$ polynomials. Indeed, this would imply writing $7$ as a sum of $3$ rational squares.
Partial conclusion. If there is an example with only $4$ polynomials, it must used either $3$ or $4$ linear polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding all real solutions of $x-8\sqrt{x}+7=0$ Finding all real solutions of $x-8\sqrt{x}+7=0$.
Man, I tried subtituting $x=y^2$ but IDK things got complicated. What is the best way to figure this out? Thanks!
| If
$x - 8\sqrt x + 7 = 0, \tag 1$
then
$x + 7 = 8\sqrt x; \tag 2$
then
$x^2 + 14 x + 49 = (x + 7)^2 = (8\sqrt x)^2 = 64x; \tag 3$
thus
$x^2 - 50x + 49 = 0, \tag 4$
which factors as
$(x - 1)(x - 49) =
x^2 - 50x + 49 = 0; \tag 5$
thus
$x = 1 \; \text{or} \; x = 49; \tag 6$
it is now a simple matter to check that $1$, $49$ obey (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
} |
If $x>4$, what is the minimum value of $\frac{x^4}{(x-4)^2}$. If $x>4$, what is the minimum value of $\frac {x^4}{(x-4)^2}$ ?
I have tried using AM-GM Inequality here by letting $y=x-4$ and ended up getting $224$ but that does not seem to be the correct answer. I find out by trial and error that the minimum value is when $x=8$ which is $256$. Are there any better ways to solve for the minimum value other than trial and error?
| Let $p=x-4>0$. By AM-GM inequality,
$$
\frac{x^4}{(x-4)^2}
=\frac{(p+4)^4}{p^2}
\ge\frac{\left(2\sqrt{4p\,}\right)^4}{p^2}
=256
$$
and equality holds when $p=4$ or $x=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means the maximum must be $4$, but the answer key says it's $16$. How come?
| Let $5x+6y=k$.
Thus, $y=\frac{k-5x}{6}$ and the following equation has real roots:
$$4x^2+\left(\frac{k-5x}{6}\right)^2=7+4x-2\cdot\frac{k-5x}{6}$$ or
$$165x^2-2(5k+102)x+k^2+12k-252=0,$$ which gives
$$(5k+102)^2-169(k^2+12k-252)\geq0$$ or
$$(k-16)(k+23)\leq0$$ or
$$-23\leq k\leq16.$$
For $k=16$ the equality occurs for
$$x=\frac{5\cdot16+102}{169}=\frac{14}{13}$$ and from here $$y=\frac{23}{13},$$ which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ A question asks
Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ for positive integers $x$ and $y$.
I tried factoring the LHS by adding $1$ to both sides so we get $(y+1)^3$ in the LHS. But I couldn't get any factorisation for the RHS, neither could think of any other ways to proceed.
How to proceed?
Thank you.
| Given:
$$y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$$
Let $K(x,y)$ be:
$$K(x,y)=y^3 + 3y^2 + 3y - (x^3 + 5x^2 - 19x + 20)$$
We could write this as:
$$K(x,y)=f(y)-f(x)+g(x)$$
If a root $r$ exists for $g$, then, we would have:
$$K(r,r)=f(r)-f(r)+g(r) $$
We can find g(x) such that:
$$g(x)=K(x,y)-f(y)+f(x) \tag1$$
By inspecting K(x,y), we need to choose the function $f$ first, then calculate $g$ after that.
Let $f(y)$ be:
$$f(y)=y^3 + 3y^2 + 3y$$
This implies $f(x)$ be:
$$f(x)= x^3 + 3x^2 + 3x$$
Using (1) above:
$$g(x)=[y^3 + 3y^2 + 3y - (x^3 + 5x^2 - 19x + 20)] - (y^3 + 3y^2 + 3y) - (x^3 + 3x^2 + 3x)$$
Simplifying, we get:
$$g(x)=2x^2-22x+20$$.
The above equation has 2 real roots $r=1,r2=10$.
Testing these values, $K(1,1)=1+3+3-(1+5-19+20)=0$ and $K(10,10)=1000+300+30-(1000+500-190+20)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Calculate $\int\sqrt{(x^2+1)}dx$ Calculate $I$ =$\int\sqrt{(x^2+1)}dx$
I have tried calculating it using integration by parts:
$$f'(x) = 1, f(x) = x$$
$$g(x) = \sqrt{x^2+1}, g'(x) = \frac{x}{\sqrt{x^2+1}}$$
$$\int\sqrt{x^2+1}dx = x\sqrt{x^2+1} - \int\frac{x}{\sqrt{x^2+1}}$$
Then I make the substitution:
$$x^2+1 = u$$
$$2xdx = du$$
$$ I = \frac{1}{2}\int{\frac{1}{\sqrt{u}}}du = \sqrt u$$
Substituting back $x$ I get:
$$\int\sqrt{(x^2+1)}dx = x\sqrt{x^2+1} - (x^2+1)^{\frac{1}{4}} $$
What am I doing wrong? Because I plugged this into an integral calculator and the answer is $\ln(|\sqrt{x^2+1}+x|) +x\sqrt{x^2+1}$.
| Your IBP should have given $$\int\sqrt{x^2+1}dx=x\sqrt{x^2+1}-\int\frac{x^2dx}{\sqrt{x^2+1}}=x\sqrt{x^2+1}-\int\sqrt{x^2+1}dx+\int\frac{dx}{\sqrt{x^2+1}}.$$Rearranging,$$\int\sqrt{x^2+1}dx=\frac{x\sqrt{x^2+1}+\int\frac{dx}{\sqrt{x^2+1}}}{2}.$$For the last integral, use $x=\tan t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the sum of the first term and common difference when given sum of first 5 and sum of first 10 the sum of the first 5 terms of an arithmetic series is 110 and the sum of the first 10 terms is 320. How do i go about finding the first term and common difference.
Sn = n/2 [2a+d(n−1)] is the equation for working out the sum of an arithmetic series, but how can i rearrange to find for the first term and common difference. I belive it would be using simulatenous equations.
| $\begin{array}{c| cccccccccccccc}
\text{index} & 1 & 2 & 3 & 4 & 5\\
\text{term} & a & a+d & a+2d & a+3d & a+4d \\
\text{sum} & a & 2a+d & 3a+3d & 4a+6d & \color{red}{5a+10d=110} \\
\text{formula} &&&&& 5\dfrac{(a)+(a+4d)}{2}
\end{array}$
$\begin{array}{c| cccccccccccccc}
\text{index} & 6 & 7 & 8 & 9 & 10\\
\text{term} & a+ 5d & a+ 6d & a+ 7d & a+ 8d & a+9d \\
\text{sum} & 6a+15d & 7a+21d & 8a+28d & 9a+36d & \color{red}{10a+45d=320}\\
\text{formula} &&&&& 10\dfrac{(a)+(a+9d)}{2}
\end{array}$
\begin{align}
5a+10d &=110 \\
10a+45d &=320 \\
\hline
10a+20d &=220 \\
10a+45d &=320 \\
\hline
25d &= 100 \\
\hline
d &= 4 \\
a &= 14
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum of the number of all continuous runs of all possible sequences with $2019$ ones and $2019$ zeros I recently had a test that is quite difficult because it is for selecting people to participate important mathematics competition. It has 6 questions like IMO, and the last question is pretty hard. Although I have done it, it takes me a long time. I hope someone would give me a better solution. The question is that:
For a sequence with some ones and zeros, we count the number of continuous runs of equal digits in it. For example, the sequence $011001010$ has $7$ continuous runs; $0,11,00,1,0,1,0$. Find the sum of the number of all continuous runs of all possible sequences with $2019$ ones and $2019$ zeros
That's the question. I did an answer $$4040 \begin{pmatrix} 4037 \\ 2018 \end{pmatrix}$$
I think it is true that I have checked the answer in different ways. However, I hope there are faster or more elegant ways to solve this question because I use a lot of algebraic-combinatorics. I hope you guys will help me. Thank you!
Spoiler(my answer)
Let $a_n$ be the number of the sequences with $n$ continuous runs. Obviously, $a_n \ne 0$ if and only if $n$ is a positive integer between $2$ and $4038$ inclusive. After some simple calculation, we know that: $$\text{For any positive integer }n\text{ between }2\text{ and }4038\text{ inclusive, }a_n=\begin{cases} 2{\begin{pmatrix} 2018 \\ \frac{n}{2}-1 \end{pmatrix}}^2 & \text{if }n\text{ is even}\\ 2\begin{pmatrix} 2018 \\ \frac{n-1}{2} \end{pmatrix} \begin{pmatrix} 2018 \\ \frac{n-3}{2} \end{pmatrix}& \text{if }n\text{ is odd}\end{cases}$$ However, our answer is to find the sum of all continuous runs of all possible sequence, so the answer is $\sum_{k=2}^{2018} ka_k$. After some horrible calculation, you'll get $4040 \begin{pmatrix} 4037 \\ 2018 \end{pmatrix}$.
| Number of Arrangements with $\boldsymbol{k}$ Runs
Using Stars and Bars, The number of ways to get a sum of $n$ with $k$ positive numbers is $\binom{n-1}{k-1}$.
The number of arrangements with $k$ runs is twice the number of ways (one starting with $0$ and one starting with $1$) to get a sum of $n$ with $\left\lfloor\frac{k+1}2\right\rfloor$ positive numbers times the number of ways to get a sum of $n$ with $\left\lfloor\frac{k}2\right\rfloor$ positive numbers.
My Interpretation of "The Sum of All Continuous Runs"
The question explicitly states that "the sequence $011001010$ has $7$ continuous runs". Here we sum the number of continuous runs for all sequences consisting of $n$ zeros and $n$ ones.
$$
\begin{align}
&\sum_{k=1}^n2(2k)\binom{n-1}{k-1}\binom{n-1}{k-1}+\sum_{k=1}^n2(2k+1)\binom{n-1}{k}\binom{n-1}{k-1}\tag1\\
&=\sum_{k=1}^n\left[4(k-1)\binom{n-1}{k-1}\binom{n-1}{k-1}+4\binom{n-1}{k-1}\binom{n-1}{k-1}\right]\tag{2a}\\
&+\sum_{k=1}^n\left[4k\binom{n-1}{k}\binom{n-1}{k-1}+2\binom{n-1}{k}\binom{n-1}{k-1}\right]\tag{2b}\\
&=\sum_{k=1}^n4(n-1)\left[\binom{n-2}{k-2}\binom{n-1}{n-k}+4\binom{n-1}{k-1}\binom{n-1}{n-k}\right]\tag{3a}\\
&+\sum_{k=1}^n\left[4(n-1)\binom{n-2}{k-1}\binom{n-1}{n-k}+2\binom{n-1}{k}\binom{n-1}{n-k}\right]\tag{3b}\\
&=4(n-1)\binom{2n-3}{n-2}+4\binom{2n-2}{n-1}+4(n-1)\binom{2n-3}{n-1}+2\binom{2n-2}{n}\tag4\\
&=2(n+1)\binom{2n-1}{n}\tag5
\end{align}
$$
Explanation:
$\phantom{\text{a}}\text{(1)}$: separate the even and odd cases
$\text{(2a)}$: $2(2k)=4(k-1)+4$
$\text{(2b)}$: $2(2k+1)=4k+2$
$\text{(3a)}$: $(k-1)\binom{n-1}{k-1}=(n-1)\binom{n-2}{k-2}$
$\text{(3b)}$: $k\binom{n-1}{k}=(n-1)\binom{n-2}{k-1}$
$\phantom{\text{a}}\text{(4)}$: Vandermonde's Identity
$\phantom{\text{a}}\text{(5)}$: put everything over $n!(n-1)!$ and simplify
Plug in $n=2019$ and we get $4040\binom{4037}{2019}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the 94th term of this sequence? $1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$
What is the 94th term of the following sequence?
$$1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$$
*
*8
*9
*10
*11
My Attempt: I found that the answer is 3rd option i.e. 94th term is 10. As every number is written 2n: n is natural number. Here 94 = 2(47) so sum of first few natural numbers should be greater than or equal to 47. Since $$1+2+3+4+5+6+7+8+9 = 45 < 47$$ so options 1,2 are not possible and $$1+2+3+4+5+6+7+8+9+10 = 55 >47$$ But this is a lengthy process.
Please tell me easiest way to approach the answer.
| Look at the last numbers of the repeating numbers. Notice the arithmetic sequence: $2,4,6,...,2+2(n-1)$, whose sum of $n$ terms is: $S_n=(n+1)n$. So, the general formula is: $a_{S_n}=n$. For example:
$$a_{S_\color{red}1}=a_2=\color{red}1\\
a_{S_\color{red}2}=a_6=\color{red}2\\
\vdots\\
a_{S_n}=a_{94}=?$$
We make up the equation:
$$(n+1)n=94 \Rightarrow n^2+n-94=0 \Rightarrow n\approx 9.2>9 \Rightarrow n=10.$$
Verify:
$$S_9=(9+1)\cdot 9=90 \Rightarrow a_{90}=9\\
S_{10}=(10+1)\cdot 9=99 \Rightarrow a_{99}=10$$
So, $a_{91}=a_{92}=\cdots=a_{99}=10$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $\sin x + \cos x = \sqrt{1+k}$ for $\sin 2x$, $\sin x-\cos x$, and $\tan x$ in terms of $k$ Given that $\sin x + \cos x = \sqrt{1+k}$, $-1 \le k \le 1$
*
*Find the value of $\sin 2x$ in terms of k
*Given that $x \in (45^{\circ}, 90^{\circ})$ deduce that $\sin x - \cos x = \sqrt{1-k}$
*Hence, show that $\tan x = \dfrac{1 + \sqrt{1-k^2}}{k}$
Okay, I have figured out that
$$ \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1+k \implies \sin 2x = k $$
Now, I am not sure what approach I should use to deduce the second one.
And can somebody give me a hint on how to start with the third one?
| Try this for the second one,
$$ (\sin x- \cos x)^2= (\sin x+\cos x)^2 -2\sin 2x$$
For the third one, solve the equations below for $\sin x$ and $\cos x$,
$$ \sin x + \cos x = \sqrt{1+k}$$
$$ \sin x- \cos x = \sqrt{1-k}$$
and then plug them into
$$\tan x= \frac{\sin x}{\cos x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A matrix of order 8 over $\mathbb{F}_3$ What is an example of an invertible matrix of size 2x2 with coefficients in $\mathbb{F}_3$ that has exact order 8?
I have found by computation that the condition that the 8th power of a matrix
$\begin{bmatrix}a & b\\c & d\end{bmatrix}$ is the identity is
$$
b c (a + d)^2 (a^2 + 2 b c + d^2)^2 + ((a^2 + b c)^2 +
b c (a + d)^2)^2=1, \qquad b (a + d) (a^2 + 2 b c + d^2) (a^4 + 4 a^2 b c + 2 b^2 c^2 +
4 a b c d + 4 b c d^2 + d^4)=0, \qquad c (a + d) (a^2 + 2 b c + d^2) (a^4 + 4 a^2 b c + 2 b^2 c^2 +
4 a b c d + 4 b c d^2 + d^4)=0, \qquad b c (a + d)^2 (a^2 + 2 b c +
d^2)^2 + (b c (a + d)^2 + (b c + d^2)^2)^2=1
$$
and the condition for invertibility is $ad\neq bc$. If the 4th power is not the identity, then no power that is not a multiple of 8 is not the identity (because we could cancel out to either get that the first power is the identity or that the second power is the identity, both lead to contradiction). That is another cumbersome condition to write out.
I hope somebody can suggest a nicer way.
| If $A$ satisfies $A^8 = I$, then all of the eigenvalues $\lambda$ of $A$ satisfy $\lambda^8 = 1$. Rearranging and factoring gives
$$0 = \lambda^8 - 1 = (\lambda^4 + 1)(\lambda^4 - 1) .$$ Since we want $A^4 \neq I$ (otherwise the order of $A$ divides $4$), at least one eigenvalue of $A$ must be a solution of $\lambda^4 + 1$.
But $$0 = \lambda^4 + 1 = (\lambda^2 - \lambda - 1) (\lambda^2 + \lambda - 1) ,$$ and both of the quadratic factors are irreducible, so the eigenvalues of $A$ must be the roots of $\lambda^2 \pm \lambda - 1$ for some choice of $\pm$, and hence that quantity is the characteristic polynomial of $A$. Any monic polynomial is the characteristic polynomial of its companion matrix, giving the solutions $$\pmatrix{0 & 1 \\ 1 & \mp 1} .$$ In fact, the uniqueness of the Jordan normal form implies that all matrices of order $8$ in $GL(2, \Bbb F_3)$ are conjugates of these.
| {
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Difficult diophantine equation Solve for integers:
$4n^4+7n^2+3n+6=m^3.$
Hi this is a problem from an Bulgarian olympiad for which I have no idea how to solve.
I figured out using wolfram alpha that $16\cdot m^3-47$ must be a square number.
I would appreciate any solutions. Thank you in advance!
| The idea is going modulo $9$.
Indeed, we will prove that $4n^4 + 7n^2+3n+6$ leaves only remainders $2,5,6$ modulo $9$. None of these are cubes modulo $9$(only $0,1,8$ are), completing the proof that no such integers $n,m$ exist.
For this, we note that if $n \equiv 0 \pmod{3}$ then $4n^4 + 7n^2+3n+6 \equiv 6\pmod{9}$.
If $n \equiv 1 \pmod{3}$ then $4n^4 + 7n^2+3n+6 \equiv 2 \pmod{9}$.
Finally, if $n \equiv - 1 \pmod{3}$ then $4n^4+7n^2+3n+6 \equiv 5 \pmod{9}$.
| {
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Solve the following system of equations: $(x-1)^3+2016(x-1)=-1$ and $(y-1)^3+2016(y-1)=1$ This problem came up in my homework for an online class and try as I might, I can't find a solution.
$(x-1)^3+2016(x-1)=-1$
$(y-1)^3+2016(y-1)=1$
Find $x+y$
So far, I've tried letting $(x-1)=a$ and $(y-1)=b$ and then adding the two equations and factoring out $(a+b)$ to get:
$(a+b)(a^2-ab+b^2+2016)=0$
But I don't know where to go from here.
| From your point you have that or $a+b=0$ or $a^2-ab+b^2+2016=0$. But $a^2-ab+b^2+2016=(a-\frac b2)^2 +b^2-\frac{b^2}4+2016>0\forall a,b\in\Bbb R$. So you have $a+b=0\implies x+y=2$
| {
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For each positive integer $k$, there is a positive integer $m\neq 3k$ such that $\varphi(3k)=\varphi(m)$ (with $\varphi(n)$ being Euler's totient)
Prove that for every positive integer $k$, $\exists m\neq 3k, m\in\mathbb{Z}^+$ such that $\varphi(3k)=\varphi(m)$. Here $\varphi(n)$ is Euler's totient function.
I am interested in the problem above. The above is my observation, but I have to idea how to prove it. Any help will be highly appreciated.
| Partial answer. Euler's totient function is multiplicative, i.e. if $\gcd(m,n)=1$ then $\varphi(m\cdot n)=\varphi(m)\cdot \varphi(n)$.
Let's consider $\gcd(6, k)=1$, then $\gcd(2^2,k)=1$ and $\gcd(3,k)=1$. And:
$$\varphi(3\cdot k)=
\varphi(3)\cdot \varphi(k)=
2\cdot \varphi(k)$$
$$\varphi(2^2\cdot k)=
\varphi(2^2)\cdot \varphi(k)=
2\cdot \varphi(k)$$
So, we may consider $m=2^2\cdot k \ne 3\cdot k$.
Let's consider $\gcd(3, k)=1$ and $k=2^n\cdot k_1$, where $\gcd(2,k_1)=1$ (or $n$ is the maximum power of $2$ dividing $k$). Then:
$$\varphi(3\cdot k)=
\varphi(3)\cdot \varphi(k)=
2\cdot \varphi(2^{n}\cdot k_1)=
2\cdot 2^{n-1}\varphi(k_1)=
2^n\cdot\varphi(k_1)$$
$$\varphi(2\cdot k)=\varphi(2^{n+1}\cdot k_1)=\varphi(2^{n+1})\cdot \varphi(k_1)=2^{n}\cdot\varphi(k_1)$$
So, we may consider $m=2\cdot k \ne 3\cdot k$.
Let's consider $\gcd(2, k)=1$ and $k=3^n\cdot k_1$, where $\gcd(3,k_1)=1$ (or $n$ is the maximum power of $3$ dividing $k$). Then:
$$\varphi(2\cdot3\cdot k)=\varphi(2)\cdot \varphi(3\cdot k)=\varphi(3^{n+1}\cdot k_1)=\varphi(3^{n+1})\cdot \varphi(k_1)=3^{n}\cdot 2 \cdot \varphi(k_1)$$
$$\varphi(3\cdot k)=\varphi(3^{n+1}\cdot k_1)=\varphi(3^{n+1})\cdot \varphi(k_1)=3^{n}\cdot 2 \cdot \varphi(k_1)$$
So, we may consider $m=6\cdot k \ne 3\cdot k$.
The remaining part is $6 \mid k$ or $k=2^{n_1}\cdot 3^{n_2}\cdot k_1$ where $n_1>0$, $n_2>0$, $\gcd(k_1,6)=1$.
$$\varphi(3\cdot k)=
\varphi(2^{n_1}\cdot 3^{n_2+1}\cdot k_1)=
2^{n_1-1}\cdot 3^{n_2}\cdot 2\cdot \varphi(k_1)=
2^{n_1}\cdot 3^{n_2}\cdot \varphi(k_1)$$
*
*if $n_1=1$ then for $m=3^{n_2+1}\cdot k_1\ne 3k$ we have $\varphi(m)=2\cdot 3^{n_2}\cdot \varphi(k_1)$
*for $n_1 \geq 2$, we can use $2^2=\varphi(5)$ (assuming $k_1$ is not a multiple of $5$) or $2\cdot 3=\varphi(7)$ (assuming $k_1$ is not a multiple of $7$)
From here onwards, just observations ...
The last part of the previous section leads to $k=2^{n_1}\cdot 3^{n_2}\cdot 5^{n_3}\cdot 7^{n_4}\cdot k_1$ where $n_i>0$, $\gcd(k_1,2\cdot3\cdot5\cdot7)=1$. In this case we have $2\cdot5=\varphi(11)$.
The previous section leads to $k=2^{n_1}\cdot 3^{n_2}\cdot 5^{n_3}\cdot 7^{n_4}\cdot 11^{n_5}\cdot k_1$ where $n_i>0$, $\gcd(k_1,2\cdot3\cdot5\cdot7\cdot 11)=1$. In this case we have $2\cdot11=\varphi(23)$ or $2\cdot3\cdot5=\varphi(31)$
Then $2\cdot23=\varphi(47)$ or $2\cdot3\cdot7=\varphi(43)$.
How far did you go?
| {
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Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f(-1)$.
Now since $x^{3}-x = x(x^{2}-1)=x(x-1)(x+1)$ is there any relation between the remainder of $f(x)$ divided by $x^{3}-x$ and remainder when $f(x)$ divided by $(x-1)$ and $(x+1)$?
| Hint
$$x^{81} + x^{49} + x^{25} + x^{9} + x=(x^{3}-x)Q(x)+R(x)$$where $$R(x)=ax^2+bx+c$$and $$R(0)=0\\R(1)=5\\R(-1)=-5$$
| {
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Limit of a sequence depending on a parameter The problem given by our teacher during the class goes as follows: Determine the limit of a sequence $a_n$ depending on the parameter $p$ if $a_n=\sqrt{4n^2+3n+5}-pn-1$.
My attempt of solution:
$$\lim{(\sqrt{4n^2+3n+5}-pn-1)}=\lim{n\left(\sqrt{4+\frac{3}{n}+\frac{5}{n^2}}-p-\frac{1}{n}\right)}=\lim{n} \cdot \lim{\left(\sqrt{4+\frac{3}{n}+\frac{5}{n^2}}-p-\frac{1}{n}\right)}.$$
Now it's easy to see that all fractions under the second limit vanish and we are left with
$\lim{n} \cdot \lim{(2-p)}$. $2-p$ is a constant value, and hence $\lim{(2-p)}=2-p$. The whole expression then becomes $$(2-p)\lim{n}.$$ Thus, for $p>2$ we get $-\infty$, and for $p<2$ the result is $+\infty$. The only unknown sitution happens for $p=2$, because we get an undefined "$0 \cdot \infty$". We've got to investigate this situation further. By plugging in $p=2$ to the original form and using a well-known technique, i.e. multiplying $\sqrt{4n^2+3n+5}-(2n+1)$ by $$\frac{\sqrt{4n^2+3n+5}+(2n+1)}{\sqrt{4n^2+3n+5}+(2n+1)},$$
we get $\lim{\sqrt{4n^2+3n+5}-(2n+1)}=-\frac{1}{4}$. And these are correct, according to the answer sheet in the textbook. My teacher, however, claims that the method I used is invalid. The problem is, I don't see any mistake here... Can you point it out for me?
| Let's see: $$\lim{(\sqrt{4n^2+3n+5}-pn-1)}=\lim{\frac{(\sqrt{4n^2+3n+5}-pn-1)(\sqrt{4n^2+3n+5}+pn+1)}{\sqrt{4n^2+3n+5}+pn+1}}=\lim{\frac{4n^2+3n+5-(pn+1)^2}{\sqrt{4n^2+3n+5}+pn+1}}=\lim{\frac{n^2(4-p^2)+n(3-2p)+4}{\sqrt{4n^2+3n+5}+pn+1}}=\lim{\frac{n(4-p^2)+3-2p+\frac{4}{n}}{\sqrt{4+\frac{3}{n}+\frac{5}{n^2}}+p+\frac{1}{n}}}$$ From here it's clear than the limit won't exist if $4-p^2 \ne 0$. If $p=2$, the limit exists ant it's equal to $-\frac{1}{2}$if $p=-2$ the limit does not exist because the denominator becomes zero.
| {
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Solve $(x+1)(y+1)(z+1)=144$ in primes "Solve $(x+1)(y+1)(z+1)=144$ in primes".
So far, I have concluded that the solutions are $(x,y,z)=(2,3,11)$ or $(2,5,7)$ and their permutations. I worked like this:
*
*$x \equiv 0\mod 2\Rightarrow x+1=3, 144=2^4*3^2 \Leftrightarrow (y+1)(z+1)=48=2^4*3$
*
*$y \equiv 0\mod 2\Rightarrow y+1=3 \Leftrightarrow z=15$, contradiction since $15$ is not prime
*$y \equiv 1\mod 2\Rightarrow y+1\geq 2^2 \Leftrightarrow y+1=2*3, z+1=2^3$ or $y+1=2^2*3, z+1=2^2 \Leftrightarrow y=5, z=7$ or $y=11, z=3$
And since the equation is symmetric the solutions are the permutations of the latter ones. Similarly, through casework, we find the same solutions if $x \equiv 1\mod 2$ (if I haven't made a mistake). My question is if there exists a more simple way to solve this problem besides lots of casework (and if there exists another triplet that satisfies the given equation).
| Assume x <= y <= z. The factors can be 3, 4, 6, 8, 12, 18, 24 and 72 (primes where p+1 divides 144). The smallest factor cannot be 6 or larger since 6^3 >= 216, so we get 3x48 or 4x36. 48 = 4x12 or 6x8, 36 = 6x6. So the solutions (x, y, z) are (2, 3, 11), (2, 5, 7) and (3, 5, 5).
And of course all the permutations.
You can also see that the product of the two smallest factors is at least 9, so the last one cannot be more than 16. That could be useful if you solve the same problem for much larger numbers.
| {
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Homotopy in $U_{2}$ Let $\mathbb{S}^{1}$ be the unit circle and $U_{2}\subset M_{2\times 2}(\mathbb{C})$ be the unitary group. Let $f,g:\mathbb{S}^{1}\rightarrow U_{2}$ be maps defined by
\begin{align}
f(x)=
\begin{pmatrix}
x & 0\\ 0 & 1
\end{pmatrix}
\quad &\text{and} \quad
g(x)=
\begin{pmatrix}
1 & 0\\ 0 & x
\end{pmatrix}.
\end{align}
Show that $f$ and $g$ are homotopic.
I was wondering if somebody could help me with this question. I tried to define a homotopy by using
\begin{align}
H(x,t)=
\begin{pmatrix}
(1-t)x+t & 0\\ 0 & (1-t)+tx
\end{pmatrix}.
\end{align}
But this is not in $U_{2}$ for all $t\in [0,1]$. Also, I want to somehow use the fact that
\begin{align}
\begin{pmatrix}
x & 0\\ 0 & 1
\end{pmatrix}
=
\begin{pmatrix}
0 & 1\\ 1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0\\ 0 & x
\end{pmatrix}
\begin{pmatrix}
0 & 1\\ 1 & 0
\end{pmatrix}.
\end{align}
Thanks!
| Your homotopy will not work. Instead, follow the hint: $U_2$ is path-connected and the identity matrix can be connected to the matrix
$$
\begin{pmatrix}
0 & 1\\ 1 & 0
\end{pmatrix}
$$
by a certain path $M_t, t\in [0,1]$, in $U_2$.
| {
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Find the projection matrix that projects every vector of $\mathbb{R}^3$ on to the plane of 2 vectors The two vectors that are mentioned above are $$\begin{pmatrix}1 \\1 \\0\end{pmatrix}$$ $$\begin{pmatrix}0 \\ 1 \\1 \end{pmatrix}$$
I am not really sure what i am doing but i tried finding the projection matrix by putting the 2 vectors into a different matrix A =$$\begin{pmatrix}1 & 0 \\1 &1 \\0&1\end{pmatrix}$$ and then calclulating the projection matrix $$ P = A (A^TA)^{-1} A^T$$
.My result is that the P matrix is $$ 1/2\begin{pmatrix}1/2 & 1 & -1\\1&2&1\\-1 & 1 &2\end{pmatrix}$$
Is my answer correct? If not what am i doing wrong? Thanks in advance
| You must have made a mistake, as can be seen by checking $P\begin{pmatrix}1\\1\\0\end{pmatrix}$.
Using your approach, $ A =\begin{pmatrix}1 & 0 \\1 &1 \\0&1\end{pmatrix},$ $A^T=\begin{pmatrix}1 &1 &0 \\0 &1 &1 \end{pmatrix},$
$ A^TA=\begin{pmatrix}2&1\\1&2\end{pmatrix}$, $(A^TA)^{-1}=\dfrac13\begin{pmatrix}2 & -1 \\-1 & 2\end{pmatrix},$ $A(A^TA)^{-1}=\dfrac13\begin{pmatrix}2 &-1 \\ 1& 1 \\ -1 &2\end{pmatrix}$, and $P=A(A^TA)^{-1}A^T=\dfrac13\begin{pmatrix}2 &1 &-1\\1 &2 &1 \\-1 &1& 2\end{pmatrix}$. Can you check that $P\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix}?$
| {
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What did I get wrong when solving $\int\frac{\sqrt{x^2-1}}{x^4}dx$? I'm not sure that this is the problem, but I think I may not know how to find the $\theta$ value when solving an integral problem with trigonometric substitution.
I got $\frac{\sin^3(\sec^{-1}(x))}{3}+C$ for the answer, but the answer should be, $\frac{1}{3}\frac{(x^2-1)^{3/2}}{x^3}+C$
$$\int\frac{\sqrt{x^2-1}}{x^4}dx$$
Let $x=\sec\theta$
Then $dx=\sec\theta\tan\theta d\theta$
$$\int\frac{\sqrt{\sec^2\theta-1}}{\sec^4\theta}\sec\theta\tan\theta d\theta$$
$$=\int\frac{\sec\theta}{\sec^4\theta}\sqrt{\tan^2\theta}\tan\theta d\theta$$
$$=\int\frac{1}{\sec^3\theta} \tan^2\theta d\theta$$
$$=\int\frac{1}{\sec^3\theta}\frac{\sec^2\theta}{\csc^2\theta}d\theta$$
$$=\int\frac{1}{\sec\theta}\frac{1}{\csc^2\theta}d\theta$$
$$=\int \cos\theta\sin^2\theta d \theta$$
Using $u$-substition, let $u=\sin\theta$
Then $du=\cos\theta d\theta$ and $dx = \frac{1}{\cos\theta}du$
$$\int\cos\theta u^2 \frac{1}{\cos\theta}du$$
$$=\int u^2 du$$
$$=\frac{u^3}{3}+C$$
$$=\frac{\sin^3\theta}{3}+C$$
Since $x=\sec\theta$, $\sec^{-1}(x)=\theta$
$$=\frac{\sin^3(\sec^{-1}(x))}{3}+C$$
What am I doing wrong?
| You've got the right answer, you've just missed that $$\sin(\sec^{-1}(x))=\sin(\cos^{-1}(1/x))=\sqrt{1-1/x^2}=\frac{1}{x}\sqrt{x^2-1}.$$
| {
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Expansion $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Expand $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Right now, I am able to expand this expression by simplifying it to:
$\frac{(x^2+1)^4 (x^2-1)^2}{x^6}$
I used the formula $(a+b)^2$ and $(a-b)^2$ a bunch of times to arrive at the answer. But, is there any simpler/smarter way to do this?
| The following procedure, I believe, is somewhat simpler
$$\begin{equation}\begin{aligned}
\left(x+{\frac{1}{x}}\right)^4 \left(x-{\frac{1}{x}}\right)^2 & = \left(x+{\frac{1}{x}}\right)^2\left(x+{\frac{1}{x}}\right)^2\left(x-{\frac{1}{x}}\right)^2 \\
& = \left(x+{\frac{1}{x}}\right)^2\left(\left(x+{\frac{1}{x}}\right)\left(x-{\frac{1}{x}}\right)\right)^2 \\
& = \left(x+{\frac{1}{x}}\right)^2\left(x^2 - \frac{1}{x^2}\right)^2 \\
& = \left(\left(x+{\frac{1}{x}}\right)\left(x^2 - \frac{1}{x^2}\right)\right)^2 \\
& = \left(x^3 - \frac{1}{x} + x - \frac{1}{x^3}\right)^2 \\
& = \left(\left(x^3 - \frac{1}{x^3}\right) + \left(x - \frac{1}{x}\right)\right)^2 \\
& = \left(x^3 - \frac{1}{x^3}\right)^2 + 2\left(x^3 - \frac{1}{x^3}\right)\left(x - \frac{1}{x}\right) + \left(x - \frac{1}{x}\right)^2 \\
& = x^6 - 2 + \frac{1}{x^6} + 2\left(x^4 - x^2 - \frac{1}{x^2} + \frac{1}{x^4}\right) + x^2 - 2 + \frac{1}{x^2} \\
& = x^6 + 2x^4 - x^2 - 4 -\frac{1}{x^2} + \frac{2}{x^4} + \frac{1}{x^6}
\end{aligned}\end{equation}\tag{1}\label{eq1}$$
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How to calculate this line integral?(complex plane)
There are counter-clockwise two curves, $C_1$ and $C_2$ on complex plane.
$C_1$ ; ${1 \over 3} \leq \vert z \vert \leq 2$ (on the 1st Quadrant.)
$C_2$ ; ${1 \over 3} \leq \vert z \vert \leq 2$ (on the 4th Quadrant)
Calculate $\int_{C_1} {z^3+1 \over z^4 + 4z -1} dz$ - $\int_{C_2} {z^3+1 \over z^4+4z-1} dz$
My attempt) By Rouche thm, $z^4 + 4z -1$ have 3 roots on the curve, $C$: ${1 \over 3} \leq \vert z \vert \leq 2$
Since, $\int_{C_1} {z^3+1 \over z^4 + 4z -1} dz$ - $\int_{C_2} {z^3+1 \over z^4+4z-1} dz$ = ${1 \over 2} {\int_{C} {z^3+1 \over z^4 + 4z -1} dz} $
Therefore, by argumented principle
${1 \over 2} {\int_{C} {z^3+1 \over z^4 + 4z -1} dz} $ = ${1 \over 8} {\int_{C} {4(z^3+1) \over z^4 + 4z -1} dz} = {1 \over 8} \times 2\pi i \times (3-0) = {3 \over 4}\pi i $
But the answer was $0$
What point do I have a mistake?
Any help would be thanksful.
| The denominator has one root close to $\frac14$ outside the region and a triple of roots close to the roots of $z^3+4=0$, where the conjugate pair $\frac{1\pm i\sqrt3}2\sqrt[3]4$ has one root inside $C_1$ and $C_2$, respectively.
Let $\zeta$ be the root inside $C_1$. Then the integral of $\frac{P(z)}{Q(z)}$ over the curve $C_1$ is equal to the residuum $2\pi i \frac{P(ζ)}{Q'(ζ)}=2\pi i \frac{ζ^3+1}{4ζ^3+4}=\frac{\pi i}2$.
Likewise the integral over $C_2$ is equal to the residual value $2\pi i \frac{P(\bar ζ)}{Q'(\bar ζ)}=2\pi i \frac{\bar ζ^3+1}{4\bar ζ^3+4}=\frac{\pi i}2$.
Their difference is clearly zero.
As $P(z)=\frac14Q'(z)$, the integral over any closed curve $C$ is clearly $\frac{\pi i}2$ times the winding number of the image under $Q$ of the curve around zero, which is the number of roots of $Q$ inside the curve. As $Q$ has real coefficients, the roots come in conjugate pairs, thus giving equal integrals over $C_1$ and $C_2$ independent of the number of roots contained within.
| {
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"url": "https://math.stackexchange.com/questions/3363386",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Help with unsolved indefinite integral ! $\int \!r\ln \left( r \right) \sqrt {ar+{b}^{2}+{r}^{2}}\,{\rm d}r$ I really need to solve this indefinite integral:
$\int \!r\ln \left( r \right) \sqrt {ar+{b}^{2}+{r}^{2}}\,{\rm d}r$
It seems much more complicated than it looks. I have found a integral table with
a integral resembles my integral, and the solution involves Polylogs. The link of the intgral table is below:
http://www-elsa.physik.uni-bonn.de/~dieckman/IntegralsIndefinite/IndefInt.html
Anyone willing to take this challenge and help with this integral ?
Thank you very much in advance
| The integral can be constructed from the following generic cases
$$
\int x \log(x)\sqrt{a^2+x^2} dx=
\frac{1}{3} \left(-\frac{1}{3} \sqrt{a^2+x^2} \left(4 a^2+x^2\right)-a^3 \log(x)+\left(a^2+x^2\right)^{3/2} \log(x)+a^3 \log\left[a
\left(a+\sqrt{a^2+x^2}\right)\right]\right)
$$
$$
\int x \log(x+b)\sqrt{a^2+x^2}dx
=\frac{1}{6} \left(-\frac{1}{3} \sqrt{a^2+x^2} \left(8 a^2+6 b^2-3 b x+2 x^2\right)-2 \left(a^2+b^2\right)^{3/2} \log(b+x)+2 \left(a^2+x^2\right)^{3/2}
\log(b+x)+b \left(3 a^2+2 b^2\right) \log\left[x+\sqrt{a^2+x^2}\right]+2 \left(a^2+b^2\right)^{3/2} \log\left[a^2-b x+\sqrt{a^2+b^2}
\sqrt{a^2+x^2}\right]\right)
$$
and
$$
\int x \log(c x+b)\sqrt{a^2+x^2}dx=\frac{1}{18 c^3}\left(-c \sqrt{a^2+x^2} \left(6 b^2-3 b c x+2 c^2 \left(4 a^2+x^2\right)\right)-6 \left(b^2+a^2 c^2\right)^{3/2} \log(b+cx)+6 c^3 \left(a^2+x^2\right)^{3/2} \log(b+c x)+3 b \left(2 b^2+3 a^2 c^2\right) \log\left[x+\sqrt{a^2+x^2}\right]+6 \left(b^2+a^2 c^2\right)^{3/2}
\log\left[a^2 c-b x+\sqrt{b^2+a^2 c^2} \sqrt{a^2+x^2}\right]\right)
$$
Finally, we consider a form that yields result in terms of dilogarithm function (setting $y(x)=\sqrt{1+x^2}$)
$$
\int \log(x)\sqrt{1+x^2}dx=\frac{1}{24} (-\pi^2-6 x y+12 x y \log(x)+6 \text{arcsinh}(x) \Big(\text{arcsinh}(x)+2 \log(x)-2 \log(1+x+y)-1-12 \text{Li}_2(-x-y)+12 \text{Li}_2(1-x-y)\Big)
$$
| {
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"url": "https://math.stackexchange.com/questions/3367062",
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"source": "stackexchange",
"question_score": "1",
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Why is $\operatorname{Var}(X^2) \ge\operatorname{Var}(X − \mu)^2 + \mu^2)$? I wonder why $\operatorname{Var}(X^2) \ge \operatorname{Var}((X − \mu)^2 + \mu^2)$.
IN ADDITION
What factor is the variance of $(X − \mu)^2 + \mu^2$ smaller than the variance of $X^2$?
What I've done so far:
$$\operatorname{Var}((X − \mu)^2 + \mu^2) =\operatorname{Var}(X^2 -2X\mu+2\mu^2) = \operatorname{Var}(X^2 -2X\mu).$$
From here, I wonder if you have any ideas?
| The result is only conditionally true (I assume you mean $\mu = E(X)$). It is valid assuming that either $E(X) = 0$, $X\ge 0$ or $X\le 0$. If no condition is given, it does not always hold, and we prove this by a simple counterexample below.
$\operatorname{Var}X = E(X^2) - E(X)^2$. Sub in
\begin{align} \mathbf \Delta(X) &:=\operatorname{Var}(X^2) - \operatorname{Var}((X-\mu)^2 + \mu^2)\\ &\phantom{:}= E(X^4) - E(X^2)^2 - E((X^2-2X\mu )^2) + (E(X^2-2X\mu))^2.\end{align}
We have by direct computation,
\begin{align} E((X^2-2X\mu )^2) &= E(X^4) - 4\mu E(X^3) + 4\mu^2E(X^2),\\
(E(X^2-2X\mu))^2 &= E(X^2)^2 - 4\mu E(X)E(X^2) + 4\mu^2E(X)^2.\end{align}
thus
$$ \mathbf \Delta(X) = 4\mu E(X^3) - 4\mu^2 E(X^2) - 4 \mu E(X) E(X^2) + 4\mu^2 E(X)^2.$$
Setting $\mu = E(X)$,
\begin{align} \mathbf \Delta(X) &= 4 \mu E(X^3) - 8 \mu^2 E(X^2) + 4\mu^4 \\
&= 4\mu [ E(X^3) - 2 \mu E(X^2) + \mu^2E(X)] \\
&=4\mu E(X^3 - 2 \mu X^2 + \mu^2 X) \\
& = 4\mu E\Big (X ( X-\mu)^2 \Big) = 4E(X) E\Big (X ( X-E(X))^2 \Big)
\end{align}
Suppose that $X \ge 0$. Then $E(X)\ge 0$. Also, $X ( X-E(X))^2\ge 0$, so $E(X ( X-E(X))^2) \ge 0$, and therefore the inequality $\mathbf\Delta(X) \ge 0$ holds for $X\ge 0$. But if $X\le 0$, then setting $Y=-X\ge 0$, since $X ( X-E(X))^2 = -Y(Y-E(Y))^2$, we also have
$$ \mathbf \Delta (X) = \mathbf \Delta (Y) \ge 0.$$
Now we give a simple counterexample that shows it cannot hold in general. Let $X$ be the following random variable that only takes the values $\pm 1$,
$$ P(X=1) = 1-P(X=-1) = p_1 \notin \{0,1/2,1\}.$$
Then as $X^2 \equiv 1$ is almost surely constant, $\operatorname{Var}(X^2) = 0$. But the right-hand side is randomly varying since $p_1 \neq 0,1$, and we avoided the sneaky cancellation from $E(X)$ by choosing $p_1 \neq 1/2$. Thus $\operatorname{Var}((X-\mu)^2 + \mu^2)>0$, and therefore
$$ \mathbf \Delta(X) < 0.$$
Concretely, with $p_1 \approx 0.853$, we have $\mathbf \Delta(X) \approx -0.99999021$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Definite Integral $\int{x^2+1 \over x^4+1}$ Evaluate
$$\int_0^{\infty}{x^2+1 \over x^4+1}$$
I tried using Integration by parts ,
$$\frac{{x^3 \over 3 }+x}{x^4+1}+\int\frac{{x^3 \over 3 }+x}{(x^4+1)^2}.4x^3.dx$$
First term is zero
But it got me no where.
Any hints.
| Another approach: Separate out the integral into
$$
I = \int_0^\infty \frac{x^2}{x^4 + 1}\:dx + \int_0^\infty \frac{1}{x^4 + 1}\:dx
$$
Both take the form of known integral
$$
\int_0^\infty \frac{x^k}{\left(x^n + a\right)^m}\:dx = a^{\frac{k +1}{n} -m}\frac{\Gamma\left(m - \frac{k + 1}{n}\right)\Gamma\left(\frac{k + 1}{n}\right)}{n\Gamma(m)}
$$
if you wish to re-position your integrals into this form, take the following sequences of actions
(1) Substitute $x \mapsto x^4$
Then
(2) Substitute $x \mapsto \frac{1}{1 + x}$
This will put the integral into the form of the Beta Function
(3) Use the relationship between the Beta and Gamma Function:
$$
B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)}
$$
And you're done.
For your specific case this then yields
\begin{align*}
I &= \int_0^\infty \frac{x^2}{x^4 + 1}\:dx + \int_0^\infty \frac{1}{x^4 + 1}\:dx \\
&= 1^{\frac{2 +1}{4} -1}\frac{\Gamma\left(1 - \frac{2 + 1}{4}\right)\Gamma\left(\frac{2 + 1}{4}\right)}{4\Gamma(1)} + 1^{\frac{0 +1}{4} -1}\frac{\Gamma\left(1 - \frac{0 + 1}{4}\right)\Gamma\left(\frac{0 + 1}{4}\right)}{4\Gamma(1)} \\
&= \frac{\Gamma\left(1 - \frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)}{4} + \frac{\Gamma\left(1 - \frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{4}
\end{align*}
Here we employ Euler's Reflection Formula
\begin{equation*}
\Gamma(z)\Gamma(1 - z) = \frac{\pi}{\sin\left(\pi z\right)} \qquad z \not \in \mathbb{Z}
\end{equation*}
Here we have $z = \frac{3}{4}, \frac{1}{4}$ which are both not integers and thus we can use the identity:
\begin{align*}
I &= \frac{\Gamma\left(1 - \frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)}{4} + \frac{\Gamma\left(1 - \frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{4} = \frac{1}{4} \cdot \frac{\pi}{\sin\left(\pi \cdot \frac{3}{4} \right)} + \frac{1}{4} \cdot \frac{\pi}{\sin\left(\pi \cdot \frac{1}{4} \right)} \\
&= \frac{1}{4} \cdot \frac{\pi}{\frac{1}{\sqrt{2}}} + \frac{1}{4} \cdot \frac{\pi}{\frac{1}{\sqrt{2}}} = \frac{\pi}{\sqrt{2}}
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/3370526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Indefinite integral of rational expression involving cubic polynomials I was given the following exercise:
$$\int \frac{x^3}{(x^2+1)^3}dx$$
As a tip, my professor suggested using the following substitution: $t=x^2+1$.
Notice that if $t=x^2+1$, then $x^2=t-1$ and therefore $x=(t-1)^\frac{1}{2}$. Then
$$x^3=x^2\cdot x=(t-1)(t-1)^\frac{1}{2}=(t-1)^\frac{3}{2}$$
and we have that
$$\int \frac{x^3}{(x^2+1)^3}dx=\int \frac{(t-1)^\frac{3}{2}}{t^3}dt.$$
But now that I have applied the suggested substitution, I don't really see how to continue with this integral. Integrating by parts doesn't seem to get me nowhere, and I can't seem to find any way to apply the substitution method. Am I missing something, or perhaps I made a mistake changing my integration variable from $x$ to $t=x^2+1$?
Thanks in advance.
| You forgot that, since $x=(t-1)^{\frac{1}{2}}$, we have also
$$dx=D((t-1)^{\frac{1}{2}})\, dt=\frac{1}{2}(t-1)^{-\frac{1}{2}}\, dt.$$
Therefore, we find
$$\int \frac{x^3}{(x^2+1)^3}dx=\int \frac{(t-1)^\frac{3}{2}}{t^3}\cdot
\color{blue}{\frac{1}{2}(t-1)^{-\frac{1}{2}}}\, dt=
\frac{1}{2}\int \frac{t-1}{t^3}\, dt=\frac{1}{2}\int \frac{dt}{t^2}-\frac{1}{2}\int \frac{dt}{t^3}.$$
Now it should be easy to finish the job.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all real matrices such that $X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$ The following question come from the 1998 Romanian Mathematical Competition:
Find all matrices in $M_2(\mathbb R)$ such that $$X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$$
Can you guy please help me? Thanks a lot!
| dan_fulea mentioned in another answer that the contestants are not expected to know diagonalisation or Jordan form. So, I will give a more elementary solution below. Let
$$
A=uv^T=\pmatrix{2\\ 1}\pmatrix{1&2}.
$$
The equation in question is equivalent to
$$
X^3-4X^2+5X=5A.\tag{1}
$$
One can easily verify that $A^2=4A$ and $X=A$ is a solution to $(1)$. In general, if $X$ satisfies $(1)$, we must have $XA=AX$, i.e. $Xuv^T=uv^TX$. Therefore $Xu=ku$ and $v^TX=kv^T$ for some common real factor $k$, and $XA=AX=kA$. It follows from $(1)$ that
\begin{aligned}
X^3A-4X^2A+5XA&=5A^2,\\
k^3A-4k^2A+5kA&=20A,\\
k^3-4k^2+5k-20&=0,\\
(k-4)(k^2+5)&=0.
\end{aligned}
Therefore $k=4$ and $XA=AX=4A$. Since $A^2=4A$, if we put $Y=X-A$, we get
$YA=AY=0$ or $Yuv^T=uv^TY=0$. Hence $Y$ must be a real scalar multiple of
$$
B=\pmatrix{2\\ -1}\pmatrix{1&-2}
$$
and $X=A+bB$ for some real scalar $b$. As $X=A$ is a solution to $(1)$, $AB=BA=0$ and $B^2=4B$, if we substitute $X=A+bB$ into $(1)$, we get
\begin{aligned}
b^3B^3-4b^2B^2+5bB&=0,\\
16b^3-16b^2+5b&=0,\\
b(16^2-16b+5)&=0,\\
b&=0.
\end{aligned}
Hence the only solution to $(1)$ is given by $X=A$.
| {
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If $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, maximise $x+y$. The question is: if $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, what is the greatest possible value of $x+y$?
The book says "the greatest value of $4$ is divided by $6$, which produces a remainder of $4$. The greatest value of $y$ is when $2$ is divided by, which produces a remainder of $2$. Therefore, the greatest value of $x+y$ is $6$."
I think what's throwing me off is the phrase "multiple of $4$" bc it makes me think that any multiple of $4$ can be divisible by $6$ (i.e. $24/6 = 4$). The books answer doesn't use multiples, just the $4$ and $2$, respectively. I don't understand how this works. Can someone please clarify?
| Take any integer $m$ and divide it by $6$. The possible remainders are $0,1,2,3,4,5$. Therefore, if we take $4m$ and divide it by $6$ then the possible remainders are $0,4,2,0,4,2$. The maximum remainder when a multiple of $4$ is divided by $6$ is $4$.
Similarly, if you divide an integer $n$ by $3$ the possible remainders are $0,1,2$. Therefore, if we take $2n$ and divide it by $3$ then the possible remainders are $0,2,1$. The maximum remainder when a multiple of $2$ is divided by $3$ is $2$.
Hence the maximum of the sum of the two remainders is $4+2=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof that $3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2$ This should be rather straightforward, but the goal is to prove that
$$3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2.$$
A possibility is to use
$$\begin{align*}3^{10^{n+1}}-1&=\left(3^{10^n}-1\right)\left(1+\sum_{k=1}^9 3^{10^n k}\right)\\&=\left(3^{10^n}-1\right)\left(3^{9\cdot 10^n}-1+3^{8\cdot 10^n}-1+\cdots +3^{10^n}-1+10\right),\end{align*}$$
but I don't see how this proves the equality in the question. I got only
$$3^{10^{n}}\equiv 1\pmod{3^{10^{n-1}}-1}.$$
Perhaps someone here can explain.
| Hint:
To get to your goal $3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2,$
can you prove using binomial expansion with $9=10-1$ that $9^{10^{n-1}}\equiv1\mod 10^n $?
| {
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"source": "stackexchange",
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Prove $\frac{1}{n} \sum^{n}_{k = 1} \left(1 + \frac{1}{k} \right) \geq (n + 1)^{\frac{1}{n}}$ Prove for every $n \in \mathbb{N}$ with $n \geq 1$ that
$$\frac{1}{n} \sum^{n}_{k = 1} \left(1 + \frac{1}{k} \right) \geq (n + 1)^{\frac{1}{n}}$$
What I've done so far:
Since $n \in \mathbb{N}_{\geq 1}$, it follows that $1 + \frac{1}{n} \in \mathbb{R}_{\geq 1}$. Then, by the AM-GM inequality we have
$$
\frac{1}{n} \sum^{n}_{k = 1} (1 + \frac{1}{k}) = \frac{1}{n}\left(1 + \frac{1}{1} + 1 + \frac{1}{2} + \cdot \cdot \cdot + 1 + \frac{1}{n} \right)
$$
$$
\geq \left[1 \cdot \frac{1}{1} \cdot 1 \cdot \frac{1}{2} \cdot \cdot \cdot 1 \cdot \frac{1}{n} \right]^{\frac{1}{n}}
$$
$$
= \left[\frac{1}{1} \cdot \frac{1}{2} \cdot \cdot \cdot \frac{1}{n} \right]^{\frac{1}{n}}.
$$
However, I'm not sure how to continue after this.
| $$\frac{1}{n} \sum^{n}_{k = 1} (1 + \frac{1}{k})\geq [(1 + \frac{1}{1})(1 + \frac{1}{2}) \cdot \cdot \cdot (1 + \frac{1}{n})]^{\frac{1}{n}}$$ $$= [\frac{2 \cdot 3 \cdot \cdot \cdot (n+1)}{n!}]^{\frac{1}{n}}=(n+1)^{\frac{1}{n}}$$
| {
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Inequality with a rational function Let $0< a_1\leq a_2\leq a_3\leq a_4\leq a_5$. Define the function $$f(x,y,z)=\cfrac{(a_1(x+y+z)^3+a_2(x)^3+a_3(x+y)^3+a_4(y+z)^3+a_5(z)^3 )^2}{(a_1(x+y+z)^2+a_2(x)^2+a_3(x+y)^2+a_4(y+z)^2+a_5(z)^2 )^3}.$$
Show that for all $x,y,z\in \mathbb R$ we have $f(x,y,z)\leq \cfrac{1}{a_1+a_2+a_3}.$
Information about the problem:
*
*The function forms an asymptote at $\cfrac{1}{a_1+a_2+a_3}$.
*We have $f(1,0,0)=\cfrac{1}{a_1+a_2+a_3}$.
*As a function of $x$, it seems there are only three critical points. Same for $y$ and $z$.
*Perhaps it helps to show that
$$\cfrac{(a_1(x+y+z)^3+a_2(x)^3+a_3(x+y)^3+a_4(y+z)^3+a_5(z)^3 )^2}{(a_1(x+y+z)^2+a_2(x)^2+a_3(x+y)^2+a_4(y+z)^2+a_5(z)^2 )^3}\leq \cfrac{(a_1(x+y+z)^3+a_2(x)^3+a_3(x+y)^3+a_3(y+z)^3+a_3(z)^3 )^2}{(a_1(x+y+z)^2+a_2(x)^2+a_3(x+y)^2+a_3(y+z)^2+a_3(z)^2 )^3}.$$
| For convenience, to simplify, take the sixth root of $f$ to give $$g(x,y,z)=\frac{(a_1(x+y+z)^3+a_2x^3+a_3(x+y)^3+a_4(y+z)^3+a_5z^3)^{1/3}}{(a_1(x+y+z)^2+a_2x^2+a_3(x+y)^2+a_4(y+z)^2+a_5z^2)^{1/2}}$$ which is the ratio of a weighted 3-norm to its 2-norm. Notice also that the function is homogeneous in $x,y,z$, so one can divide the numerator and denominator by any scalar, $$g(x,y,z)=\frac{(a_1+a_2X^3+a_3(X+Y)^3+a_4(1-X)^3+a_5(1-X-Y)^3)^{1/3}}{(a_1+a_2X^2+a_3(X+Y)^2+a_4(1-X)^2+a_5(1-X-Y)^2)^{1/2}}$$ where $X=x/(x+y+z)$, $Y=y/(x+y+z)$.
A complete solution will take up too much space but here's the main idea. Suppose you want to maximize $(\alpha U^3+\beta V^3)^{1/3}/(\alpha U^2+\beta V^2)^{1/2}$, with $\alpha<\beta$, then that is the same as maximizing $$\alpha U^3+\beta V^3\qquad \mathrm{such\ that} \qquad \alpha U^2+\beta V^2=1.$$ If you use Lagrange multipliers, we find $$3\alpha U^2=2\lambda\alpha U,\qquad 3\beta V^2=2\lambda\beta V$$ so either $U=0,V=\beta^{-1/2}$, or $U=\alpha^{-1/2},V=0$, or $U=V=(\alpha+\beta)^{-1/2}$. The function evaluated at each of these points is equal to $\beta^{-1/6}$, $\alpha^{-1/6}$, $(\alpha+\beta)^{-1/6}$ respectively. The maximum is the middle one, with $U\ne0,V=0$. In general, the components associated with the largest coefficients need to be made $0$.
Back to the main problem, we need to make $1-X-Y=0$, so $X+Y=1$; then also $1-X=0$, so $X=1$. Hence $X=1$, $Y=0$, which is the same as $y=0$, $x=x+y+z$, so $z=0$. Any vector of the form $(x,0,0)$ will achieve the maximum value.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\iint_R\big(x^2+y^2\big)\,dA$
Evaluate the following double integral:
$$\iint_R\big(x^2+y^2\big)\,dA,$$
where $R$ is the region given by plane $x^2+y^2\leq a^2$.
My attempts:
\begin{align}
\iint_{R}\big(x^2+y^2\big)\,dA
&=\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\big(x^2+y^2\big)\,dy\,dx\\
&=\int_{-a}^{a}\left(x^2y+\dfrac{y^3}{3}\right)_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dx\\
&=\dfrac{2}{3}\int_{-a}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx.
\end{align}
I can't go further from here, please help.
| Hint: Try $x=a\sin(t)$, but really polar coordinates would be the move here.
| {
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What are the non-negative integers of this form? What are $( \frac{x^2-1-my}{x}, \frac{y^2-1-nx}{y} ) \in \mathbb{Z}^2_{\ge 0}$ for $(m,n) \in \mathbb{Z}^2_{\ge 0}$, $(x,y) \in \mathbb{R}^2_{\ge 1}$ and $mx+ny=xy$?
Some investigations:
*
*According to this answer of Ivan Neretin, assuming $x,y$ integers, the only possibility is $(0,1)$ with $(m,n) = (0,1)$ and $(x,y) = (1,2)$, up to permutation.
*If $x=y$, then $(m+n)x = x^2$, so $x=m+n$ positive integer.
In general $y=mx/(x-n)$ with $x>n$. Thus we need to consider
$$ ( \frac{x^2-1-m\frac{mx}{x-n}}{x}, \frac{(mx/(x-n))^2-1-nx}{mx/(x-n)} ) $$ $$ = (\frac{(x^2-1)(x-n)-m^2x}{x(x-n)},\frac{m^2x^2-(x-n)^2(1+nx)}{mx(x-n)}) \in \mathbb{Z}^2_{\ge 0} $$
Let $\alpha=\alpha_{n,m}$ be the root of $$P_{n,m}(x):=(x^2-1)(x-n)-m^2x$$ when $x>n$ (show unicity). Then $$m^2\alpha^2-(\alpha-n)^2(1+n\alpha) = (\alpha^2-1)(\alpha-n)\alpha -(\alpha-n)^2(1+n\alpha) $$ $$ = (\alpha-n)((\alpha^2-1)(\alpha-n)-\alpha+n^2\alpha) = (m^2+n^2-1)\alpha(\alpha-n)$$
Let $\beta = \beta_{n,m}$ be the root of $$Q_{n,m}(x):=m^2x^2-(x-n)^2(1+nx)$$ when $x>n$ (show unicity). Note that if $n=m$ then $\beta = n \alpha/(\alpha-n)$.
We can assume (by symmetry) that $\alpha \le \beta$.
Let $(a,b) = ( \frac{x^2-1-my}{x}, \frac{y^2-1-nx}{y} )$. Then for $x=\alpha$ we get $$(a,b) = (0,\frac{m^2+n^2-1}{m}),$$ and by symmetry, for $x=\beta$, we get $$(a,b) = (\frac{m^2+n^2-1}{n},0).$$
The problem reduces to find all the $x \in [\alpha, \beta]$ such that $P_{n,m}(x)$ an $Q_{n,m}(x)$ are integers.
Remark: If $n |(m^2-1)$ or $m |(n^2-1)$, then above is at least one solution.
Example: $m=n=1$, then $\alpha=1.80193773580484...$ and $\beta=2.24697960371747...$.
Let us plot $P_{1,1}$ and $Q_{1,1}$:
It follows that $(0,1)$ and $(1,0)$ are the only solutions if $m=n=1$.
What are all the others solutions in general?
An other approach
Take $(p,q) = ( \frac{x^2-1-my}{x}, \frac{y^2-1-nx}{y} )$. Then:
$$\left\{
\begin{array}{ll}
x^2 - px-1-my = 0 \\
y^2 -qy-1-nx = 0
\end{array}
\right.$$
Because $(x,y) \in \mathbb{R}^2_{>1}$, it follows that:
$$\left\{
\begin{array}{ll}
2x = p + (p^2+4(1+my))^{1/2}\\
2y = q + (q^2+4(1+nx))^{1/2}
\end{array}
\right.$$
So $$ 2x = p + [p^2+4+2m(q + [q^2+4(1+nx)]^{1/2})]^{1/2} $$
Then
$$ [(2x-p)^2-p^2-4-2mq]^2 = q^2+4(1+nx) $$
Thus
$$ (4x^2-4xp-4-2mq)^2 = q^2+4(1+nx) $$
We got a polynomial of degree $4$, allowing us to get at most $4$ roots $x_i(m,n,p,q)$ with $i \le 4$. Then $$2y_i(m,n,p,q) = q + (q^2+4(1+nx_i(m,n,p,q)))^{1/2}$$
It remains to solve the Diophantine equation $$mx_i(m,n,p,q)+ny_i(m,n,p,q)=x_i(m,n,p,q)y_i(m,n,p,q).$$
| A pair $(p,q) = ( \frac{x^2-1-my}{x}, \frac{y^2-1-nx}{y} )$ is achievable if and only if $p^2+q^2+4$ is the sum of two squares.
Proof. In one direction, we assume that the following three polynomials have a common zero:
\begin{split}
f(x,y)&:=x^2-1-my-px,\\
g(x,y)&:=y^2-1-nx-qy, \\
h(x,y)&:=xy-mx-ny.
\end{split} Then eliminating $x,y$ via computing resultants, we have with necessity:
$$0 = \mathrm{Res}_y(\mathrm{Res}_x(f,g),\mathrm{Res}_x(g,h)) = -(m^2-mq+n^2-np-1)^3 n^4,$$
implying that $m^2-mq+n^2-np-1=0$, which is equivalent to
$$(\star)\qquad p^2 + q^2 + 4 = (2n-p)^2 + (2m-q)^2.$$
That is, $p^2+q^2+4$ is the sum of two squares.
In the reverse direction, we assume that $p^2 + q^2 + 4 = a^2 + b^2$ for some nonnegative integers $a,b$. Without loss of generality we have $a\equiv p\pmod{2}$ and $b\equiv q\pmod{2}$. We set $n:=\frac{a+p}2$ and $m:=\frac{b+q}2$, which makes them satisfy $(\star)$.
Then we set $x_0>n$ be a zero of the polynomial:
$$F(x):=x^3-(p+n)x^2+(pn-1-m^2)x+n.$$
Such zero does exist since $F(n) = -m^2n<0$ while $F(x)\to+\infty$ as $x$ grow. Then, we set $y_0:=\tfrac{mx_0}{x_0-n}$ to get $h(x_0,y_0)=0$. Since $F(x) = (x-n)f(x,y)+mh(x,y)$, we also have $f(x_0,y_0)=0$, which together with $(\star)$ further implies that $g(x_0,y_0)=0$. That is, for $(x,y)=(x_0,y_0)$ we get $( \tfrac{x^2-1-my}{x}, \tfrac{y^2-1-nx}{y} ) = (p,q)$. QED
P.S. Similarly to $x_0$, we also have that $y_0$ is a zero of the polynomial:
$$G(y):= y^3-(q+m)y^2+(qm-1-n^2)y+m.$$
which satisfies $G(y)=(y-m)g(x,y) + nh(x,y)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3380735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}$ How do I ind the limit of
$$\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}$$
by L'Hopital's Rule?
Using Desmos, I get the answer that this limit evaluates to $\frac{3}{15}$, but I can't get that answer.
This is what I've done so far:
\begin{align}
&\lim_{x \rightarrow 0}\frac{\tan x -x}{\sin^5x}-\lim_{x \rightarrow 0}\frac{x^3/3}{\sin^5x}\\
&=\lim_{x \rightarrow 0}\frac{\sec^2x-1}{5\sin^4x\cos x}-\lim_{x \rightarrow 0}\frac{x^2}{5\sin^4x\cos x}\\
&=\lim_{x \rightarrow 0}\frac{\tan^2x}{5\sin^4x\cos x}-\lim_{x \rightarrow 0}\frac{x^2}{\sin^2x}\lim_{x \rightarrow 0}\frac{1}{5\sin^2x\cos x}\\
&=\lim_{x \rightarrow 0}\frac{1}{5\sin^2x\cos^3x}-\lim_{x \rightarrow 0}\frac{\cos^2x}{5\sin^2x\cos^3 x}\\
&=\frac{1}{5}
\end{align}
I checked by answer with Desmos, and the 4th line is where the difference occurs. In the 4th line, I did this.
$$
\lim_{x \rightarrow 0}\frac{x^2}{\sin^2x}=\lim_{x \rightarrow 0}\frac{1}{(\sin x/x)^2}=1
$$
Shouldn't this be a valid operation? I'm just using the product law on limits. If anyone could tell me where I've gone wrong, I will be grateful.
| As noted by А.Г., you can not split the limit unless both exist.
You can use L'Hospital right away and see how splittings are done:
$$\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}\stackrel{LR}=
\lim_{x \rightarrow 0}\frac{\sec ^2x- 1-x^2}{5\sin^4x\cdot \cos x}=\\
\lim_{x \rightarrow 0}\frac{1-\cos^2x-x^2\cos^2x}{5\sin^4x}\cdot \underbrace{\lim_{x\to 0}\frac1{\cos^3x}}_{=1}\stackrel{LR}=\\
\lim_{x \rightarrow 0}\frac{\sin 2x-2x\cos^2x+x^2\sin 2x}{20\sin^3x}\cdot \underbrace{\lim_{x\to 0}\frac{1}{\cos x}}_{=1}\stackrel{LR}=\\
\lim_{x \rightarrow 0}\frac{2\cos 2x-2\cos^2x+4x\sin 2x+2x^2\cos 2x}{60\sin^2x}\cdot \underbrace{\lim_{x\to 0}\frac{1}{\cos x}}_{=1}=\\
\lim_{x \rightarrow 0}\frac{-2\sin^2x}{60\sin^2x}+\lim_{x \rightarrow 0}\frac{4x\sin 2x}{60\sin^2x}+\lim_{x \rightarrow 0}\frac{2x^2\cos 2x}{60\sin^2x}=\\
-\frac1{30}+\frac2{15}+\frac1{30}=\frac2{15}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I made an inequality to solve myself, but someone pointed out my solution is wrong. I made the following inequality for myself to solve, but my friend found out an mistake:
If $a,b,c\in\mathbb{R^{+}}$ and $abc=1$, prove that $$\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}\ge\frac32$$
I tried to substitute $a=\frac xy, b= \frac yz, c= \frac zx$, but that friend found out a mistake. He wrote a solution using Muirhead’s inequality, but that is not what I want. I also tried trigonometric substitution, but it turned out that there is a mistake also. Is there any method which doesn’t use Muirhead’s inequality? Thanks for any help!
| The friend he told in the problem was me, yeah. And I will post my solution by using the Muirhead Inequality (and also AM-GM Inequality).
$$\dfrac{a}{b+1}+\dfrac{b}{c+1}+\dfrac{c}{a+1}\ge\dfrac{3}{2} \\ \Updownarrow \\ a\left(a+1\right)\left(c+1\right)+b\left(b+1\right)\left(a+1\right)+c\left(c+1\right)\left(b+1\right)\ge\dfrac{3}{2}\left(a+1\right)\left(b+1\right)\left(c+1\right)\\ \Updownarrow \\ a^2c+b^2a+c^2b+a^2+b^2+c^2+ab+bc+ca+a+b+c\ge\\\dfrac{3}{2}abc+\dfrac{3}{2}ab+\dfrac{3}{2}bc+\dfrac{3}{2}ca+\dfrac{3}{2}a+\dfrac{3}{2}b+\dfrac{3}{2}c+\dfrac{3}{2} \\ \Updownarrow (\text{Mutiply by 2 both side and rearranging})\\ 2a^2c+2b^2a+2c^2b+2a^2+2b^2+2c^2\ge3abc+ab+bc+ca+\left(a+b+c\right)a^\frac{1}{3}b^\frac{1}{3}c^\frac{1}{3}+3abc \\ \Updownarrow \\ \left(2a^2c+2b^2a+2c^2b-6abc\right)+\sum_{sym}\left(a^2-\dfrac{1}{2}ab-\dfrac{1}{2}a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}\right)\ge0 $$ By AM-GM Inequality, $$2a^2c+2b^2a+2c^2b\ge 3\sqrt[3]{8a^3b^3c^3}=6abc \\ \therefore 2a^2c+2b^2a+2c^2b-6abc\ge 0$$
By Muirhead Inequality,
$$\because \left(2,0,0\right)\succ\left(1,1,0\right),\left(2,0,0\right)\succ\left(\dfrac{4}{3},\dfrac{1}{3},\dfrac{1}{3}\right) \\ \therefore \sum_{sym}\left(a^2-\dfrac{1}{2}ab-\dfrac{1}{2}a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}\right)=\dfrac{1}{2}\sum_{sym}\left(a^2-ab\right)+\dfrac{1}{2}\sum_{sym}\left(a^2-a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}\right)\ge0$$
Adding the two inequality we prove, we get the answer.
| {
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"url": "https://math.stackexchange.com/questions/3383742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2$ Prove that $$\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2 \tag{1} $$ $\forall$ $n \gt 1$
I tried using Induction:
For the Base Step $n=2$ we have:
$$x=\sqrt{1+\frac{1}{\sqrt{2}}}+\sqrt{1-\frac{1}{\sqrt{2}}}$$
Then we get:
$$x^2=2+\sqrt{2}\lt 4$$
So $x \lt 2$
Now Let $P(n)$ is True, We shall need to prove $P(n+1)$ is also True
We have $P(n+1)$ as:
$$\left ( 1+\frac{(n+1)^{\frac{1}{n+1}}}{n+1} \right )^\frac{1}{n+1}+\left ( 1-\frac{(n+1)^{\frac{1}{n+1}}}{n+1} \right )^\frac{1}{n+1}$$
Now i tried to use the fact that:
$$f(x)=x^{\frac{1}{x}}$$ is a Monotone Decreasing $\forall x \ge e$
Hence $\forall n \ge 3$ we have:
$$(n+1)^{\frac{1}{n+1}} \lt n^{\frac{1}{n}} \tag{2}$$ and also
$$\frac{1}{n+1} \lt \frac{1}{n} \tag{3}$$
Multiplying $(2),(3)$ We get:
$$1+\frac{(n+1)^{\frac{1}{n+1}}}{n+1}\lt 1+\frac{n^{\frac{1}{n}}}{n}$$
Can we proceed from here?
| Since $f(x)=x^{\frac{1}{n}}$ is a concave function for $n>1$, by Jensen we obtain:
$$\left (1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}\leq2\left (\frac{1+\frac{n^{\frac{1}{n}}}{n}+ 1-\frac{n^{\frac{1}{n}}}{n}}{2}\right )^\frac{1}{n}=2.$$
The equality occurs for $$1+\frac{n^{\frac{1}{n}}}{n}=1-\frac{n^{\frac{1}{n}}}{n},$$ which says that the equality does not occur.
| {
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"url": "https://math.stackexchange.com/questions/3383865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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In how many ways can 10 blankets be given to 3 beggars such that each recieves at least one blanket? The question was to find the number of ways in which 10 identical blankets can be given to 3 beggars such that each receives at least 1 blanket. So I thought about trying the multinomial theorem...this is the first time I've tried it so I'm stuck at a point...
So $$x_1+x_2+x_3 = 10$$
Subject to the condition that :
$$1\leq x_1 \leq8$$
$$1\leq x_2 \leq8$$
$$1\leq x_3 \leq8$$
As each beggar can get at maximum 8 blankets and at minimum, 1.
So the number of ways must correspond to the coefficient of $x^{10}$ in:
$$(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)$$
= coeff of $x^{10}$ in $x^3(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)$
= coeff of $x^{10}$ in $x^3(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)^3$
= coeff of $x^{10}$ in $x^3(1-x^7)^3(1-x)^{-3}$
= coeff of $x^{10}$ in $x^3(1-x^{21}-3x^7(1-x^7))(1-x)^{-3}$
= coeff of $x^{10}$ in $(x^3-3x^{10})(1+\binom{3}{1}x + \binom{4}{2}x^2+...+ \binom{12}{10}x^{10})$
= $\binom{9}{7} - 3 = 33$
Is this right? From here I get the answer as $\binom{9}{7} - 3 = 33$ but the answer is stated as $36$. I don't understand where I'm making a mistake
| First give one blanket to each recipient.
Now distribute the remaining $7$ blankets to $3$ recipients without constraints: $36$
If one person gets $7$ (additional) and the others get none, there are $3$ ways to do that. And likewise for the following distributions:
*
*$700$, $3$ ways
*$610$, $6$ ways
*$520$, $6$ ways
*$511$, $3$ ways
*$430$, $6$ ways
*$421$, $6$ ways
*$331$, $3$ ways
*$322$, $3$ ways (thanks @JMoravitz)
Total = $36$ ways
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $x^4-20200y^2=1$ has no solution in postive integers Show that $x^4-20200y^2=1$ has no solution in postive integers.
This topic is a question for the Chinese middle school students' mathematics competition today, so I think this problem has a simple solution.
| We have to solve
$$
x^4-20200y^2=1\tag 1
$$
Observe that $20200=6^2+142^2$. Then (1) becomes
$$
x^4=(6y)^2+(142y)^2+1.\tag 2
$$
Clearly $x=2k+1$ is odd. Hence $x^2=8T+1$,$T=\frac{k(k+1)}{2}$. Hence $x^4=8T_1+1=16T(4T+1)+1\Rightarrow T_1=2T(4T+1)=\frac{4T(4T+1)}{2}$. In general holds the following
THEOREM.
If $x$ is odd then $x^2=8T+1$ iff $T$ is triangular number.
PROOF.
Easy.
But then (since $6=2\cdot3$ and $142=2\cdot 71$), we have $x^4=4(3y)^2+4(71y)^2+1$. Hence $\frac{x^4-1}{4}=(3y)^2+(71y)^2$. Thus $\frac{8T_1+1-1}{4}=2T_1=5050y^2$ and we arrive to
$$
T_1=2T(4T+1)=2525y^2.\tag 2
$$
Hence $4T^2+T=2\cdot2525y_1^2$, since $2|y\Rightarrow y=2y_1$. Hence $2|T\Rightarrow T=2T'\Rightarrow 8T'^2+T'=2525y_1^2$. But exists $y_2$ such that $y_1^2=8y_2+1$. Hence $2525 y_1^2\equiv 5(mod 8)$. Hence $T'\equiv 5(mod 8)$. Hence
$$
T\equiv 2(mod 8).\tag 3
$$
Lastly from (2): $2T(4T+1)=2525y^2$ we have $8T^2+2T=2525y^2$ and since $y^2\equiv 1(mod 8)$ and $2525\equiv 5(mod 8)$, we get $2T=5(mod 8)$, which is contradiction to (3).
Hence equation (1) is imposible.
| {
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"url": "https://math.stackexchange.com/questions/3390449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Find the range of values of $k$ for which $kx^2 + 8x + k <6$ for all real values of $k$
Find the range of values of $k$ for which $kx^2 + 8x + k <6 $ for all real values of $k$.
I'm unsure if the discriminant must be greater than zero or less than zero.
My working steps: \begin{align}b^2 - 4ac = (8)^2 - 4(-2)(17-k) &> 0\\64 - 4(-2)(17-k) &> 0\\64 + 136 -8k &> 0\\200 &> 8k\end{align} so my answer is $$k < 200/8.$$
| In order for a quadratic function to always be less than a constant, the leading coefficient must be negative. We can then attempt to complete the square in such a way that the maximum value of the function is $6$, then apply the appropriate inequalities to $k$.
\begin{align}
kx^2+8x+k&=k(x^2+\frac8kx)+k\\
&=k(x^2+\frac8kx+\frac{16}{k^2}-\frac{16}{k^2})+k\\
&=k(x^2+\frac8kx+\frac{16}{k^2})-\frac{16}k+k\\
&=k(x+\frac4k)^2+k-\frac{16}k\end{align}
Therefore, we want $$k-\frac{16}k<6 \quad\text{AND}\quad k<0.$$
Solving the inequality, we get
\begin{align}
k-\frac{16}k&<6\\
k-6-\frac{16}k&<0\\
\frac{k^2-6k-16}{k}&<0\\
\frac{(k+2)(k-8)}{k}&<0\\
k&\in(-\infty,-2)\cup(0,8)
\end{align}
The largest intersection of the two inequations is $\boxed{k\in(-\infty,-2)}$, and that is the answer we want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
For $a,b,c \in \mathbb{R}$ and $|a| \geq |b+c|, |b| \geq |c+a|, |c| \geq |a+b|$ prove then $a+b+c = 0$
For $a,b,c \in \mathbb{R}$ and $|a| \geq |b+c|, |b| \geq |c+a|, |c|
\geq |a+b|$ prove then $a+b+c = 0$
Solution:
$$
|a| + |b| + |c| \geq |b+c| + |c+a| + |a+b| \geq 0
$$
and
$$0 \leq |a+b+c|\leq |a+b| + |c| \leq |a| + |b| + |c|$$
so we have
$$0 \leq |a+b+c| \geq 0 $$
$$ |a+b+c|= 0 \iff a+b+c =0 $$
I need check my solution.
Many thanks!
| hint: Square each of the $| ..|$ inequality and consider $f(a) = a^2 + (2b+2c)a + (b+c)^2$. Show that $f(a) \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3392565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find limit of sequence $\lim_{n\to\infty} \frac{n}{\sqrt{n^2+n+1}} $ I need help to find the limit:
$$\lim_{n\to\infty} \frac{n}{\sqrt{n^2+n+1}} $$
I've tried this:
$$\lim_{n\to\infty} \frac{\sqrt{n^2}}{\sqrt{n^2+n+1}} =\lim_{n\to\infty} \sqrt{\frac{n^2}{n^2+n+1}} = \lim_{n\to\infty} \sqrt{\frac{1}{1+\frac{1}{n}+\frac{1}{n^2}}}$$
but what's next?
| Since
$n^2
\lt n^2+n+1
\lt (n+1)^2
=n^2+2n+1
$,
$n
\lt \sqrt{n^2+n+1}
\lt n+1
$
so
$1
\gt \dfrac{n}{\sqrt{n^2+n+1}}
\gt \dfrac{n}{n+1}
= 1-\dfrac1{n+1}
$.
Therefore
$\lim_{n\to\infty} \dfrac{n}{\sqrt{n^2+n+1}}
=1
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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