Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Is $\frac{\lim_{x \to y} a}{\lim_{x \to y} b} \ne \lim_{x\to y}\frac{a}{b}$? So I had this question
$$\lim_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx) + ax}$$
I did it this way:
$$\lim_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx) + ax}$$
$$= \frac{\lim_{x \to 0}\sin(ax) + \lim_{x \to 0}bx}{\lim_{x \to 0}\sin(bx) + \lim_{x \to 0}ax}$$
$$= \frac{\lim_{x \to 0}\sin(ax)}{\lim_{x \to 0}\sin(bx)}$$
$$= \frac{\displaystyle\frac{\lim_{x \to 0}\sin(ax) \cdot ax}{ax}}{\displaystyle\frac{\lim_{x \to 0}\sin(bx) \cdot bx}{bx}}$$
$$=\frac{\lim_{x \to 0} ax}{\lim_{x \to 0} bx}$$
$$=\lim_{x \to 0}\frac{ax}{bx}$$
$$=\lim_{x \to 0}\frac{a}{b} = \frac{a}{b}$$
Which is incorrect.
I did it again and with a different approach :
$$\lim_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx) + ax}$$
$$= \lim_{x \to 0} \frac{\displaystyle\frac{\sin(ax) + bx}{bx}}{\displaystyle\frac{\sin(bx) + ax}{bx}}$$
$$= \lim_{x \to 0} \frac{\displaystyle\frac{\sin(ax)}{bx} +1}{\displaystyle\frac{\sin(bx) + ax}{bx}}$$
$$= \frac{\lim_{x \to 0} \left(\displaystyle\frac{\sin(ax)}{bx} +1\right)}{\lim_{x \to 0} \left( \displaystyle\frac{\sin(bx) + ax}{bx} \right) }$$
$$= \frac{\lim_{x \to 0} \left(\displaystyle\frac{\sin(ax)}{bx}\right) +1}{\displaystyle\frac{a}{b} + 1}$$
$$= \frac{\lim_{x \to 0} \left(\displaystyle\frac{\sin(ax) \cdot ax}{bx \cdot ax}\right) +1}{\displaystyle\frac{a}{b} + 1}$$
$$= \frac{\displaystyle\frac{a}{b} +1}{\displaystyle\frac{a}{b} + 1} = 1$$
Which is the correct answer.
Can anyone tell me where was my first approach incorrect ?
I am guessing that this line
$$\frac{\lim_{x \to 0} ax}{\lim_{x \to 0} bx} = \lim_{x \to 0}\frac{ax}{bx}$$
is incorrect.
Thanks in advance.
| In short, you cannot naively "split" limits. A first reason for that is that they don't necessarily exist. For example, the sequence $0$ has a limit in infinity (here, 0) and $(-1)^n$ has no limit when $n\to\infty$, but $0=(-1)^n-(-1)^n$.
Also, both the numerator and denominator may have a limit, but what if the denominator's limit is 0? Then the ratio is not defined.
An efficient way to deal with such limits is to use Taylor's expansion. Here,
$\sin(ax)+bx=ax-(ax)^3/6+bx+o(x^3)$ and $\sin(bx)+ax=bx-(bx)^3+ax+o(x^3)$, so, (assuming $a\neq -b$, otherwise, it's easy):
\begin{align}\dfrac{\sin(ax)+bx}{\sin(bx)+ax}&=\dfrac{ax+bx-a^3x^3/6}{bx+ax-b^3x^3/6}+o(x^2)=\dfrac{\frac{1}{b+a}(a+b-a^3x^2/6)}{\frac{1}{b+a}(b+a-b^3x^2/6)}+o(x^2) \\ &=1+\dfrac{b^3-a^3}{6(a+b)}x^2+o(x^2) & \\\end{align}
This proves that the limit in $0$ is $1$, and also that the way the fraction approaches 1 is parabolic is the neighbourhood of 0.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why can't we integrate $\int\frac{1}{1+x^2}$ to $\frac{\ln(1+x^2)}{2x}$? $\int\frac{1}{1+2x}$ is integrated to $\frac{\ln(1+2x)}{2}$. As we are multiplying and dividing by the derivative of the function inside, why is it not possible to do the same with non-linear polynomials. For example, why can't we do the following:
$\int\frac{1}{1+x^2}$ = $\frac{\ln(1+x^2)}{2x}$
| Because
$\frac{d}{dx}\frac{\ln(1+x^2)}{2x}=
\frac{1}{x^2+1}-\frac{\ln(x^2+1)}{2 x^2}$
And as you know:
$\frac{1}{x^2+1}-\frac{\ln(x^2+1)}{2 x^2}\neq\frac{1}{x^2+1}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to determine the remainder of $ \frac{7^{369}}{350} $? How to determine the remainder of the following :
$$ \frac{7^{369}}{350} $$
Is there any tricky way to determine this?
I start as $$\frac{7^{368}}{50} \ \ , $$
This type of problem are highly asked in aptitude exam.
| One thing you can sometimes do with problems like this is to find the remainder modulo the different prime powers and then piece the information together to find the original remainder. Essentially this amounts to using the Chinese Remainder Theorem which you may not have learned yet but one doesn't really need to know it here.
Since $350 = 2\cdot 5^2 \cdot 7$ we compute
$$ 7^{369} \equiv 1^{369}=1 \pmod{2} $$
and
$$ 7^{369} = 7\cdot (7^2)^{184} \equiv 7\cdot(-1)^{184} = 7\cdot 1 = 7\pmod{5^2}$$
and
$$ 7^{369} \equiv 0^{369}=0\pmod{7}.$$
Now only possible remainders $r< 350$ sastisfying $r\equiv 7 \pmod{5^2}$ are $7, 25+7, \ldots, 325+7$. Since we must also have $r\equiv 0 \pmod{7}$ we are only left with $r=7$ or $r=7\cdot 25+7=182$. Finally using $r\equiv 1\pmod{2}$, we see that $r=7$.
| {
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location of roots of quadratic with natural coefficients the quadratic equation $ ax^2-bx+c=0 $ ; $a,b,c \in \mathbb{N}$, has two distinct real roots belonging to the interval $(1,2)$ , then what would be least value of $a$ and $b$?
I tried to solve these four inequalities with iterations,which seems to be a lengthy process
as roots are real and distinct $b^2-4ac>0$
abscissa of vertex of parabola will lie between $(1,2)$ so $ 1<\frac{b}{2a} <2$
also quadratic will have positive values at $x=1$ and $x=2$ so $a-b+c> 0 $ and $4a-2b+c>0$
please help me out .
| You have correctly derived the conditions $2a<b<4a$, $a-b+c>0$, $4a-2b+c>0$, $c<\frac{b^2}{4a}$. So if we take $a=1$, then $b=3$. But now the conditions on $c$ cannot be satisfied.
For $a=2$ we must have $b=5,6,7$. If $b=5$ we require $c>3$ and $c\le3$. If $b=6$ we require $c>4$ and $c\le4$. If $b=7$ we require $c>6$ and $c\le6$.
For $a=3$ we must have $b=7,8,9,10,11$. If $b=7$ we require $c>4$ and $c\le4$. If $b=8$ we require $c>5$ and $c\le5$. If $b=9$ we require $c>6$ and $c\le6$. If $b=10$ we require $c>8$ and $c\le8$. If $b=11$ we require $c>10$ and $c\le10$.
For $a=4$ we must have $b=9,10,11,12,13,14,15$. If $b=9$ we require $c>5$ and $c\le5$. If $b=10$ we require $c>6$ and $c\le6$. If $b=11$ we require $c>7$ and $c\le7$. If $b=12$ we require $c>8$ and $c<9$ ($c=9$ gives equal real roots). For $b=13$ we require $c>10$ and $c<11$. For $b=14$ we require $c>12$ and $c\le12$. For $b=15$ we require $c>14$ and $c\le14$.
For $a=5$ we must have $b=11,12,13,14,15,16,17,18,19$. For $b=11$ we require $c>6$ and $c\le6$. For $b=12$ we require $c>7$ and $c\le7$. For $b=13$ we require $c>8$ and $c\le8$. For $b=14$ we require $c>9$ and $c\le9$. For $b=15$ we find $c=11$ gives two roots in $(1,2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trig substitution in the definite integral $\int_1^2 x \sqrt{1 - x^2} dx$. I have the following integral
$$\int_1^2 x \sqrt{1 - x^2}\, \mathrm dx$$
I want to use the trig substitution $x = \sin \theta$ but I can't because the limits of integration do not permit such a substitution.
So I tried $x = \sin\theta + 1$ but this does not seem to help. So I am left wondering: Is there a smart way to use a trig substitution here or should I try an other path?
| Simple way
There is a simple way to calculate this integral:
$$
F(x) = \int x\sqrt{1-x^2}dx = \frac{1}{2}\int(1-x^2)^{1/2}dx^2 = -\frac{1}{3}(1-x^2)^{3/2} + C.
$$
Then as $\int_1^2x\sqrt{1-x^2}dx = F(2) - F(1)$ we have
$$
\int_1^2 x\sqrt{1-x^2}dx = -\frac{1}{3}(1 - 2^2)^{3/2} = i\sqrt{3}.
$$
Using trig substitution
If you want to use exactly trig substitution $x = \sin\vartheta$, you will have
$$
F(x) = \int\sin\vartheta\cos\vartheta \;d\sin\vartheta = -\int \cos^2\vartheta\;d\cos\vartheta = -\frac{1}{3}\cos^3\vartheta + C = -\frac{1}{3}\left(\sqrt{1-x^2}\right)^3 + C,
$$
which is the same, of cource. But if you want to use trig substitution in definite integral you will need to calculate $\arcsin(2)$. To do this you need to represent $\arcsin$ function as $\arcsin(x) = i\ln(ix + \sqrt{1 - x^2})$ (using inverse hyperbolic sine function). In our case we have
$$
\arcsin(2) = i\ln\left((2+\sqrt{3})i\right).
$$
To calculate the answer we need to represent $\cos$ function using hyperbolic cosine function as $\cos x = (e^{ix} + e^{-ix})/2$.
Summing up, we get the same result
$$
\int_1^2 x\sqrt{1-x^2}dx = -\int_{\pi/2}^{i\ln\left((2+\sqrt{3})i\right)}\cos^2\vartheta\;d\cos\vartheta = -\frac{1}{3}\cos^3\left(i\ln\left((2+\sqrt{3})i\right)\right) = i\sqrt{3}
$$
as
$$
\cos\left(i\ln\left((2+\sqrt{3})i\right)\right) = \frac{1}{2}\left(e^{\ln((2 + \sqrt{3})i)} + e^{-\ln((2 + \sqrt{3})i)}\right) = \frac{i}{2}\left(2 + \sqrt{3} - \frac{1}{2 + \sqrt{3}}\right) = i\sqrt{3}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $f(1-x) + 2f(x) = 3x$, what is $f(0)?$ I did the following:
$$f(1-0) + 2f(0) = 3\cdot 0$$
$$f(1) + 2f(0) = 0$$
This reminds me of the equation of the straight line in the plane, then:
$$\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\begin{pmatrix}
{f(1)}\\
{f(0)}
\end{pmatrix} \right> =0$$
$(1,2)$ is a normal vector to $(f(1),f(0))$, then It's possible that: $(f(1)=-2,f(0)=1)$ because:
$$\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\begin{pmatrix}
{-2}\\
{1}
\end{pmatrix}\right> =-2+2=0$$
With generality, It is possible that for all $\alpha$:
$$\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\begin{pmatrix}
{\alpha\cdot (-2)}\\
{\alpha \cdot 1}
\end{pmatrix}\right> =
\left< \begin{pmatrix}
{1}\\
{2}
\end{pmatrix}
,
\alpha \begin{pmatrix}
{ -2}\\
{ 1}
\end{pmatrix}\right> =1\cdot -2\alpha+2\alpha \cdot 1=-2\alpha + 2 \alpha =0$$
I tagged with "functional equations" because it seems to be related. I don't know if my solution is correct, what I'm saying with this is that there are infinite solutions but I don't know if assuming arbitrary values for $f(1),f(0)$ can be made because of $3x$.
EDIT: I know that It is possible to solve:
$$f(1) + 2f(0) = 0$$
$$f(0) + 2f(1) = 3$$
And get the solution. But why do I have to use the latter instead of the former?
| Another approach could be this:
$f(1-x) + 2f(x) = 3x$
$f(1-(1-x))+2f(1-x) = 3(1-x) \implies 2f(1-x) + f(x) = 3-3x$
thus we have a system
$$\left(\begin{array}{c c}
1 & 2 \\
2 & 1
\end{array}\right)\left(\begin{array}{c}
f(1-x) \\
f(x)
\end{array}\right) = \left(\begin{array}{c}
3x \\
3-3x
\end{array}\right)$$
which has unique solution $f(x) = 3x - 1$, $f(1-x) = 2-3x$, and indeed, quick check shows that this is consistent.
| {
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Five white balls, five black balls There are 5 white balls and 5 black balls in a box. A visually-challenged man takes 5 balls from the box... What's the probability that all the balls which this individual picked are white if we know for sure that there 3 white balls among those 5 balls he took?
Using the formula $P(A | B) = P(A \cap B)/P(B)$, I get a probability of $1/100$.
| The probability that all five balls are white is $\frac{5\cdot 4 \cdot 3 \cdot 2 \cdot 1}{10\cdot 9 \cdot 8 \cdot 7 \cdot 6}$. [One way to think of this is $5/10$ chance to choose the first white ball, $4/9$ for the second ball, and so on. Another way is to consider all the balls to be distinct, and count the number of orderings of five balls that are white ($5!$) and divide by the total number of orderings of five balls ($10\cdot 9 \cdots 6$).]
The probability that exactly four are white can be computed similarly. The denominator is still $10 \cdot 9 \cdots 6$, but the numerator is now $5 \cdot 5 \cdot 5!$: five ways to choose a black ball, five ways to choose the white ball that remains in the box, and $5!$ ways to order them.
Similarly, the numerator for the probability of exactly three white balls is $\binom{5}{2} \binom{5}{3} 5!$.
So, the probability of at least three white balls is $\frac{5!}{10\cdot 9 \cdots 6} (1+25+100)=126 \frac{5!}{10\cdot 9 \cdots 6}$.
Using the formula for conditional probability that you wrote, we get $$\frac{\frac{5!}{10\cdot 9 \cdots 6}}{126 \frac{5!}{10\cdot 9 \cdots 6}} = \frac{1}{126}.$$
| {
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Prove that $\lfloor\frac{n+1}{2}\rfloor+\lfloor\frac{n+2}{4}\rfloor+\lfloor\frac{n+4}{8}\rfloor+\lfloor\frac{n+8}{16}\rfloor+ \dots=n$
Prove $$\left[\dfrac{n+1}{2}\right]+\left[\dfrac{n+2}{4}\right]+\left[\dfrac{n+4}{8}\right]+\left[\dfrac{n+8}{16}\right] + \dots=n$$
where $[x]=\lfloor x\rfloor$
$$$$
It was suggested that somehow I use the identity $[x]=\left[\dfrac x2\right]+\left[\dfrac{x+1}{2}\right]$$$$$After struggling for a while, I realised I wasn't getting anywhere using his hint, probably because I couldn't really understand how I was to use it. Instead I tried to use the Squeeze Theorem by rewriting the $nth$ term of the series (referred to later as S) as$$$$ $$t_n=\left[\dfrac{n+2^k}{2^{k+1}}\right] \text{ where } 0\le k<\infty$$
$$\Rightarrow \dfrac{n+2^k}{2^{k+1}}-1<\left[\dfrac{n+2^k}{2^{k+1}}\right]\le \dfrac{n+2^k}{2^{k+1}}$$
$$$$
$$ \lim_{k\to \infty}(k+1)\left(\dfrac{n+2^k}{2^{k+1}}-1\right)<S\le \lim_{k\to \infty}(k+1)\left( \dfrac{n+2^k}{2^{k+1}}\right)$$
However these bounds are too loose as the limits diverge to $-\infty$ and $\infty$ respectively.
$$$$
Could somebody please show me how to prove the series is equal to $n$, either through the given hint, or through the selection of tighter bounds for the Squeeze Theorem? Many thanks!
| Let's start with the observation that, for any integers $a,b\gt0$,
$$\left\lfloor{a\over b}+{1\over2b}\right\rfloor=\left\lfloor{a\over b}\right\rfloor\quad(*)$$
Now define
$$f(n)=n-\left(\left\lfloor n+1\over2 \right\rfloor+\left\lfloor n+2\over4 \right\rfloor+\left\lfloor n+4\over8 \right\rfloor+\left\lfloor n+8\over16 \right\rfloor+\cdots \right)$$
Then
$$\begin{align}
f(2n)
&=2n-\left(\left\lfloor 2n+1\over2 \right\rfloor+\left\lfloor 2n+2\over4 \right\rfloor+\left\lfloor 2n+4\over8 \right\rfloor+\left\lfloor 2n+8\over16 \right\rfloor+\cdots \right)\\
&=2n-\left(\left\lfloor n+{1\over2} \right\rfloor+\left\lfloor n+1\over2 \right\rfloor+\left\lfloor n+2\over4 \right\rfloor+\left\lfloor n+4\over8 \right\rfloor+\cdots \right)\\
&=2n-\left(n+\left\lfloor n+1\over2 \right\rfloor+\left\lfloor n+2\over4 \right\rfloor+\left\lfloor n+4\over8 \right\rfloor+\cdots \right)\\
&=f(n)
\end{align}$$
and
$$\begin{align}
f(2n+1)
&=2n+1-\left(\left\lfloor 2n+2\over2 \right\rfloor+\left\lfloor 2n+3\over4 \right\rfloor+\left\lfloor 2n+5\over8 \right\rfloor+\left\lfloor 2n+9\over16 \right\rfloor+\cdots \right)\\
&=2n+1-\left(\left\lfloor n+1 \right\rfloor+\left\lfloor {n+1\over2}+{1\over4} \right\rfloor+\left\lfloor {n+2\over4}+{1\over8} \right\rfloor+\left\lfloor {n+4\over8}+{1\over16} \right\rfloor+\cdots \right)\\
&=2n+1-\left(n+1+\left\lfloor n+1\over2 \right\rfloor+\left\lfloor n+2\over4 \right\rfloor+\left\lfloor n+4\over8 \right\rfloor+\cdots \right)\quad\text{using (*)}\\
&=f(n)
\end{align}$$
Consequently $f(n)$ is a constant function for all $n\gt0$, so it suffices to evaluate
$$\begin{align}
f(1)&=1-\left(\left\lfloor 1+1\over2 \right\rfloor+\left\lfloor 1+2\over4 \right\rfloor+\left\lfloor 1+4\over8 \right\rfloor+\left\lfloor 1+8\over16 \right\rfloor+\cdots \right)\\
&=1-(1+0+0+0+\cdots)\\
&=0
\end{align}$$
| {
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Prove that the determinant is $(a-b)(b-c)(c-a)(a+b+c)$ I have the determinant :
\begin{vmatrix}
1 &1 &1 \\
a &b &c \\
a^3 &b^3 &c^3 \\
\end{vmatrix}
How do I prove that this determinant is equal to
$$ (a-b)(b-c)(c-a)(a+b+c) $$
| Expand this
$$\begin{vmatrix}
1 &1 &1 \\
a &b &c \\
a^3 &b^3 &c^3 \\
\end{vmatrix}$$
to an expanded polynomial expression. Now expand this $$ (a-b)(b-c)(c-a)(a+b+c) $$
to an expanded polynomial expression.
Now check that they're both equal.
| {
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Trigonometry-circumcircle and sides of triangle. How to prove that
$$4R\sin A\sin B\sin C=a \cos A+b \cos B+c\cos C$$ where R is the radius of the circumcircle and $a$,$b$ and $c$ the respective sides of the triangle.
I wrote $R=a/2\sin A$ and then got $\sin A$ cancelled then changed $\sin B \sin C$ to the sums of cosine but that didn't work.
| We have $4R=\frac{abc}{\Delta}$ and $\sin A=\frac{2\Delta}{bc}$, hence the LHS equals
$$ \frac{abc}{\Delta}\cdot\frac{8\Delta^3}{a^2 b^2 c^2}= \frac{8\Delta^2}{abc}.$$
Now exploit $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$ and Heron's formula.
| {
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Finding partial fractions expansions mentally On a problem on a test, my students were asked to find $\displaystyle\int\frac{6x^4-7x^3-13x-6}{x^3-2x^2} dx$,
and one student began by writing $\displaystyle\int\frac{6x^4-7x^3-13x-6}{x^2(x-2)} dx=\int\left(\frac{3}{x^2}+6x+\frac{2}{x-2}+\frac{8}{x}+5\right)dx$.
My question is how someone can get this result without doing any division or partial fraction decomposition -- what techniques can be used to get this?
| The residue theorem, for instance. If $f(x)$ is a polynomial with degree $4$,
$$ g(x)=\frac{f(x)}{x^2(x-2)} = \frac{A}{x^2}+\frac{B}{x}+\frac{C}{x-2}+D+Ex \tag{1}$$
is granted, and
$$ A=\text{Res}\left(x\cdot g(x),x=0\right),\quad B = \text{Res}(g(x),x=0),\quad C=\text{Res}(g(x),x=2)\tag{2} $$
are straightforward to compute through simple limits. Then, by computing
$$ g(x)-\frac{A}{x^2}-\frac{B}{x}-\frac{C}{x-2}, \tag{3} $$
that we know in advance to be a polynomial with degree $\leq 1$, we recover $D$ and $E$, too.
As an alternative, it is enough to notice that:
$$\begin{eqnarray*}6x^4-7x^3-13x-6 &=& 6x\cdot x^2(x-2)+5x^3-13x-6\\&=&6x\cdot x^2(x-2)+5\cdot x^2(x-2)+10x^2-13x-6 \end{eqnarray*}$$
and:
$$ 10x^2-13x-6 = 10\cdot x^2 + 3(x-2)-10 x,\quad \frac{1}{x(x-2)}=\frac{1}{2}\left(\frac{1}{x-2}-\frac{1}{x}\right).$$
| {
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How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$? I came around this expression when solving a problem.
$$\sqrt{7+4\sqrt{3}}$$
WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this
$$\left(2+\sqrt{3}\right)^2 \;=\; 4+4\sqrt{3} + 3 \;=\; 7 + 4\sqrt{3}.$$
However, the only way I can think of how to simplify that expression in hand is guessing. Is there a better way of calculating square root of a sum like that one?
| If you want $\sqrt{7+4\sqrt{3}}=a+b\sqrt d$ (where $a$ and $b$ are rational numbers, and $d$ is a square-free integer) then $7+4\sqrt{3} = a^2+db^2+2ab\sqrt d$, which yields the natural choice $d=3$.
Then $a^2+3b^2=7$ and $2ab=4$ can be solved in the rational numbers, and you find $a=2,b=1$ (because $a^2+3 \cdot (2/a)^2=7$ has solutions $±2$ and $±\sqrt 3$).
| {
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Solving the given differential equation. We need to solve : $$ ( \sqrt{x+y} + \sqrt{x-y}) \,dx + ( \sqrt{x-y} - \sqrt{x+y})\,dy=0$$
I tried as follows :
$$ \frac{dy}{dx} = \frac{ \sqrt{x+y} + \sqrt{x-y}}{\sqrt{x-y} - \sqrt{x+y}}$$
And hence letting $ y = vx$ yields :
$$ v + x \: \frac{dv}{dx} = \frac{\sqrt{1+v} + \sqrt{1-v}}{\sqrt{1-v} - \sqrt{1+v}}$$
Continuing from here yields a very complicated integral , is there a simpler way to proceed ?
| This is so close to an exact differential equation. But I'm going to make the substitutions $u=x+y$ and $v=x-y$ so that $x=\frac12(u+v)$ and $y=\frac12(u-v)$. Then
$$\left(\sqrt u+\sqrt v\right)\cdot\frac12(du+dv)+\left(\sqrt v-\sqrt u\right)\cdot\frac12(du-dv)=\sqrt v\,du+\sqrt u\,dv=0$$
This is now separable and
$$u^{-1/2}du=-v^{-1/2}dv$$
$$2u^{1/2}=-2v^{1/2}+C_1$$
$$u+v=\sqrt{x+y}+\sqrt{x-y}=\frac12C_1=C\tag{1}$$
Let's check this solution:
$$\left(\frac1{2\sqrt{x+y}}+\frac1{2\sqrt{x-y}}\right)dx+\left(\frac1{2\sqrt{x+y}}-\frac1{2\sqrt{x-y}}\right)dy=0$$
And on multiplying by $2\sqrt{x^2-y^2}$ which ends up being the reciprocal of the integrating factor that makes it exact we do in fact reproduce our original differential equation.
EDIT: I was so busy congratulating myself on seeing the key to the problem that I didn't stop to see that there were several interesting subtleties in the solution. First off, I divided by $\sqrt{uv}$ without checking to see what happens if $u=0$ or $v=0$. In fact, these lead to the singular solutions to the differential equation, $x=y\ge0$ and $x=-y\ge0$. Then it's possible to simplify the general solution by squaring to
$$x+y+2\sqrt{x^2-y^2}+x-y=C^2$$
$$\sqrt{x^2-y^2}=\frac12C^2-x$$
$$x^2-y^2=\frac14C^4-C^2x+x^2$$
$$x=\frac{\frac14C^4+y^2}{C^2}\tag{2}$$
If $C=0$, that implies $y=0$ and so $x=0$ from eq. $(1)$, which was already covered under the singular solutions. Noting that eq. $(1)$ also implies $C\ge0$, we find that
$$\sqrt{x+y}=\frac{\left|\frac12C^2+y\right|}C$$
And
$$\sqrt{x-y}=\frac{\left|\frac12C^2-y\right|}C$$
Then for eq. $(1)$ to be satisfied,
$$-\frac12C^2\le y\le\frac12C^2$$
Thus if $x=f(y)$ we can only follow eq. $(2)$ forward until $y=\frac12C^2$, after which $x=y$. Similarly we can only follow eq. $(2)$ backwards until $y=-\frac12C^2$, before which $x=-y$. Here is a plot of some solution curves along with the two singular solutions.
| {
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Quadratic inequality proof. I recently encountered a quadratic equations property that
$ax^2+bx+c>0$ $ \forall$ $x\in \Re \Rightarrow D<0$ $and$ $a>0$
and $ax^2+bx+c<0$ $ \forall$ $x\in \Re \Rightarrow D<0$ $and$ $a<0$.
Now, i tried to prove it algebraically and through graphs. Its clear that the equation seems much simple by just making a parabola and as $D<0$ the parabola never touches the $x$ axis and hence the two equations follow.
Can there be any algebraic proof the relation mentioned above.
| For $a>0$ we have:
$$
ax^2+bx+c>0 \quad \iff \quad x^2+\frac{b}{a}+\frac{c}{a}>0
$$
so:
$$
x^2+\frac{b}{a}>-\frac{c}{a}
$$
$$
x^2+\frac{b}{a} +\left(\frac{b}{2a} \right)^2>-\frac{c}{a}+\left(\frac{b}{2a} \right)^2
$$
$$
\left(x+\frac{b}{2a} \right)^2>-\frac{c}{a}+\left(\frac{b}{2a} \right)^2
$$
that is always verified ( since The LHS is a square) if
$$
-\frac{c}{a}+\left(\frac{b}{2a} \right)^2<0
$$
i.e.
$$
\frac{b^2-4ac}{4a^2}=\frac{\Delta}{4a^2} <0
$$
and since $4a^2>0$ we have $\Delta<0$.
You can do the same for the other case.
| {
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Finding equation of tangent lines to hyperbola $xy=1$ I am working through some calculus problems (this is in a section on implicit differentiation) and this one is giving me trouble.
I am trying to find the equations of the tangent lines to the hyperbola $$xy=1$$
That pass through the point $(-1,1)$.
As in the other problems of this type, I implicitly differentiated the relation between $x$ and $y$
$$\frac{dy}{dx}x+y=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$$
Then we have:
$$\frac{dy}{dx}\vert_{(-1,1)}=-(1/-1)=1$$
So I would think lines passing through this point would have the equation:
$$y-1=x+1\Rightarrow y=x+2$$
But this is definitely not true from plotting on wolfram alpha. The solutions in the book are
$$y=\pm(2\sqrt{2}-3)x-2\pm2\sqrt{2} $$
| Using homogenous coordinates, your hyperbola can be expressed as the solution set to
$P(X) = X^TQX = X^T\begin{pmatrix}0&1&0\\1&0&0\\0&0&-2\end{pmatrix}X = 0$
What makes this form neat is that the tangent line at any point of this parabola can be calculated simply by
$QX \equiv Q\begin{pmatrix}x\\y\\1\end{pmatrix}$ where $x,y$ denote the coordinates in the Euclidean plane.
Since the line must also pass through (-1,1) we need only solve
$\begin{pmatrix}-1&1&1\end{pmatrix}Q\begin{pmatrix}x\\y\\1\end{pmatrix}=0\\
\begin{pmatrix}-1&1&1\end{pmatrix}\begin{pmatrix}y\\x\\-2\end{pmatrix}=0\\$
which gives $x-y-2=0$
We also have the condition $y = 1/x$, combined we have
$x^2-2x-1=0$ which can easily be solved.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$
$$|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$$
I have tried: $$|a\sin x+b \cos x|\leq |a+b|\leq \sqrt{a^2+b^2}$$
enough to prove: $$|a+b|\leq \sqrt{a^2+b^2}$$
But I can find how to continue from here
| Find the extrema of
$$a\cos(x)+b\sin(x).$$
By canceling the derivative,
$$-a\sin(z)+b\cos(z)=0$$ or$$\tan(z)=\frac ba.$$
Then, with
$$\cos(z)=\pm\frac1{\sqrt{\tan^2(z)+1}}=\pm\frac a{\sqrt{a^2+b^2}},\\\sin(z)=\pm\frac{\tan(z)}{\sqrt{\tan^2(z)+1}}=\pm\frac b{\sqrt{a^2+b^2}},$$
you obtain
$$a\cos(z)+b\sin(z)=\pm\sqrt{a^2+b^2}.$$
Then for all $x$,
$$-\sqrt{a^2+b^2}\le a\cos(x)+b\sin(x)\le\sqrt{a^2+b^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve without solving by inspection? $\frac{x+5}{x+k}=\frac{-kx+5}{x-1}$ Background: This is from a test review on functions. The original problem was
Find the value of $k$ so that the function $f(x) = \frac{x+5}{x+k}$ will be its own inverse.
I found the answer by inspection, and then tried to solve it through more rigorous means. Continuing from the question title, I cross multiplied and used the quadratic formula on $k.$
$x^2+4x-5=5x+5k-kx^2-k^2x$
$k^2x+k(x^2-5)+(x^2-x-5)=0$
$k = \frac{-x+5\pm\sqrt{x^4-10x^2+25-4x(x^2-x-5)}}{2x}$
$k = \frac{-x+5\pm\sqrt{x^4-4x^3-6x^2+20x+25}}{2x}$
This is where I got stuck. Any ideas?
| For the original problem, you can also use the fact that a self-inverse function requires that $ \ f \ \circ \ f (x) \ = \ f( \ f(x) \ ) \ = \ x \ $ . (A rational function of linear polynomials is easily shown to be one-to-one.) Hence,
$$ \ \frac{\left(\frac{x+5}{x+k}\right) + 5 }{ \left(\frac{x+5}{x+k}\right) + k } \ \ = \ \ x \ \ \Rightarrow \ \ \frac{\left(\frac{x+5}{x+k}\right) + 5 }{ \left(\frac{x+5}{x+k}\right) + k } \ \cdot \ \frac{x+k}{x+k} \ \ = \ \ x $$
$$\Rightarrow \ \ \frac{ ( x+5) + 5 \ (x + k) }{ (x+5) + k \ (x+k) } \ \ = \ \ x \ \ \Rightarrow \ \ \frac{ 6x + (5 k+5) }{ (k+1)x + (k ^2 + 5) } \ \ = \ \ x \ \ . $$
For this last equation to hold, the "constant term" $ \ (5k \ + \ 5) \ $ in the numerator and the "linear term" $ \ (k \ + \ 1) \ x \ $ must vanish. The one value of $ \ k \ $ for which this occurs also makes the "constant term" $ \ (k^2 \ + \ 5) \ $ in the denominator match the coefficient of the "linear term" $ \ 6x \ $ in the numerator, thereby satisfying the equation.
Alternatively, in the spirit of Dave L. Renfro's answer, the domain(range) of a one-to-one function is the same as the range(domain) of its inverse. Finding the asymptotes of $ \ f(x) \ $ tells us that the domain of the function is $ \ x \ \neq \ -k \ $ and the range is $ \ y \ \neq \ 1 \ $ (in $ \ \mathbb{R} $ ). But for a self-inverse function, its domain and range are identical. Thus, $ \ -k \ = \ 1 \ $ .
| {
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Find positive values such that $xy = 32$ and the sum $4x+y$ is as small as possible.
Find positive values such that $xy = 32$ and the sum $4x+y$ is as small as possible.
How can I solve this? I know that the answer is $2\sqrt{2} = x$ and $8\sqrt{2} = y$, but I can't seem to figure out a concrete method for solving this type of problem.
An explanation of how to solve the question would be greatly appreciated!
| We can write: $4x+y = \left(2\sqrt{x}-\sqrt{y}\right)^2+ 4\sqrt{xy}= \left(2\sqrt{x}-\sqrt{y}\right)^2+ 16\sqrt{2}\geq 16\sqrt{2}\implies \text{min} = 16\sqrt{2},$ it occurs when $\sqrt{y}=2\sqrt{x}\implies x = 2\sqrt{2}, y = 8\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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real values of $x$ which satisfy the equation $\sqrt{1+\sqrt{1+\sqrt{1+x}}}=x$
All real values of $x$ which satisfy the equation $\sqrt{1+\sqrt{1+\sqrt{1+x}}}=x$
$\bf{My\; Try::}$ Here $\sqrt{1+\sqrt{1+\sqrt{1+x}}} = x>0$
Now Let $f(x)=\sqrt{1+x}\;,$ Then equation convert into $f(f(f(x)))=x$
Now Here $f(x)=x$ be one function which satisfy above equation.
My question is how can we calculate other function which satisfy above functional equation.
Help required, Thanks
| \begin{align}\sqrt{1+{\sqrt{1+\sqrt{1+x}}}}=x&\implies \sqrt{1+{\sqrt{1+\sqrt{1+x}}}}=x\\&\implies \left(1+{\sqrt{1+\sqrt{1+x}}}\right)=x^2\\&\implies\sqrt{1+\sqrt{1+x}}=x^2-1\\&\implies 1+\sqrt{1+x}=x^4-2x^2+1\\&\implies \sqrt{1+x}=x^4-2x^2\\&\implies 1+x=x^4(x^2-2)^2\\&\implies 1+x=x^4(x^4-4x^2+4)\\&\implies x^8-4x^6+4x^4-x-1=0\end{align}Now observe that , $$x^8-4x^6+4x^4-x-1=x^8\color{red}{-x^7}-x^6\color{red}{+x^7}-x^6\color{blue}{-x^5}-2x^6\color{blue}{+2x^5}+2x^4\color{blue}{-x^5}+x^4\color{green}{+x^3}+x^4\color{green}{-x^3}\color{violet}{-x^2}\color{violet}{+x^2}-x-1$$Consequently, $$x^8-4x^6+4x^4-x-1=0\implies (x^2-x-1)(x^6+x^5-2x^4-x^3+x^2+1)$$Now observe that the function $$g(x)=x^6+x^5-2x^4-x^3+x^2+1=\left(x^3-1\right)^2+\left(x-1\right)^2\,x^3+x^2 >0$$ for all $x\in \mathbb{R}^+$. So $(x^2-x-1)=0$ and hence $x=\dfrac{1+\sqrt{5}}{2}$ is the only positive solution.
| {
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Sum of $1^2+3^2+\cdots+(2n+1)^2$ Have trouble with proof. I've been working through a question on Courant's What is Mathematics?
This is the question: Prove $1^2+3^2+\cdots+(2n+1)^2=\frac{n(n+1)(2n+1)(2n+3)}{3}$. I called this $S_{(2n+1)^2}$.
What I've been given as a hint is that I can use the sum $1+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ (I called this $S_{n^2}$) in some way.
So here's my line of thought:
What do I need to do to $S_{n^2}=1^2+2^2+3^2+\cdots+n^2$ to get $S_{(2n+1)^2}=1^2+3^2+\cdots+(2n+1)^2$?
When I compared them, I thought that $S_{n^2}-(2^2+4^2+\cdots+(2n)^2) = S_{(2n+1)^2}$.
I defined $2^2+4^2+\cdots+(2n)^2$ as $S_{(2n)^2}$.
After a bit of head-scratching I managed to figure out $S_{(2n)^2}=2^2(0)+2^2(1)^2+2^2(2)^2+\cdots+2^2(n)^2=4(1^2+2^2+3^2+\cdots+n^2)=4S_{n^2} =\frac{2}{3}n(n+1)(2n+1)$
Now is where my confusion starts. So from my logic above I went forward with $S_{n^2}-S_{(2n)^2}=S_{(2n+1)^2}$ which gave me $-\frac{1}{2}(n(n+1)(2n+1))$ which isn't $S_{(2n+1)^2}$.
I don't know where I really went wrong. One of my main suspicions is that $S_{n^2}-S_{(2n)^2}=S_{n^2}-4S_{n^2}=-3S_{n^2}\neq S_{(2n+1)^2}$. But I don't see why.
| Very easy proof
$$\sum_{k=1}^n(2k+1)^2=4\sum_{k=1}^n k^2+4\sum_{k=1}^nk+\sum_{k=1}^n1.$$
Using your hint, you can very easily conclude.
| {
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$\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are all even numbers.
Consider the binomial coefficients $\binom{n}{k}=\frac{n!}{k!(n-k)!}\ (k=1,2,\ldots n-1)$. Determine all positive integers $n$ for which $\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are all even numbers.
The answer is $n\in\{2^k|k\in\mathbb{N}\}$, but we have to show two things. First, if $n = 2^k$ then $\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are all even numbers and second if $n \neq 2^k$ then $\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are not all even numbers. I think we can use this for the first part but what about the second part?
| As someone mentioned in the comments we'll use Lucas' Theorem and we'll show that if $n=2^km$ for some odd integer $m>1$, then we'll prove that $\binom{2^km}{2^k}$ is odd. We have:
$$2^km = a_t2^t + a_{t-1}2^{t-1} + ... + 1\cdot2^{k} + 0 \cdot 2^{k-1} +... + 0\cdot2^0$$
$$2^k = 1\cdot2^{k} + 0 \cdot 2^{k-1} +... + 0\cdot2^0$$
Now we have by Lucas' Theorem we have:
$$\binom{2^km}{2^k} \equiv \prod_{n=0}^{t} \binom{a_n}{b_n} \equiv 1 \pmod 2$$
This is true as all factors are equal to $1$, as they are either $\binom{0}{0}$, $\binom{1}{1}$ or $\binom{1}{0}$. Hence $\binom{2^km}{2^k}$ is odd.
Now if $n=2^k$, then we have one of the binomial coefficient zero, as $b_s= 1$ and $a_s=0$, for some $0 \le s \le k-1$. And as by definition $\binom{0}{1} = 0$ we have that:
$$\binom{2^k}{q} \equiv \prod_{n=0}^{t} \binom{a_n}{b_n} \equiv 0 \pmod 2$$
for all $1 \le q \le 2^k - 1$. Hence all of them are even.
| {
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Number of solutions to this nice equation $\varphi(n)+\tau(n^2)=n$
How many natural numbers $n$ satisfy the equation$$\varphi(n)+\tau(n^2)=n$$where $\varphi$ is the Euler's totient function and $\tau$ is the divisor function i.e. number of divisors of an integer.
I made this equation and I think it is not hard. I haven't solved this completely yet, so I want you to work on this along with me. I'd love to see your solutions!
| Hint: You can show that $n$ is odd and has at most two (distinct) prime factors. Then, it follows that $n=21$ and $n=25$ are the only solutions.
Further Hint: Suppose $n=p^aq^br^cm$ where $p,q,r,a,b,c\in\mathbb{N}$ with $p,q,r$ being pairwise distinct primes, and $m\in\mathbb{N}$ is not divisible by $p$, $q$, or $r$. Prove that
$$n-\phi(n)> n\left(\frac{1}{2p}+\frac{1}{q}+\frac{1}{r}\right)>\tau\left(n^2\right)\,,$$
provided that $p,q,r$ are the smallest primes dividing $n$ with $2<p<q<r$.
Remark: I posted weaker inequalities (now removed for being redundant), and realized that I could make a significant improvement from what I had gotten in my scratch work. A complete solution is in the hidden portion below.
Let $u_1,u_2,\ldots,u_k$ be distinct prime numbers dividing $m$. Then, $$\begin{align}n-\phi(n)&=n\left(1-\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)\left(1-\frac{1}{r}\right)\,\prod_{i=1}^k\,\left(1-\frac{1}{u_i}\right)\right)\\&\geq n\Biggl(1-\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)\left(1-\frac{1}{r}\right)\Biggr)\\&>n\Biggl(\left(\frac{1}{p}-\frac{1}{pq}-\frac{1}{pr}-\frac{1}{qr}\right)+\frac{1}{q}+\frac{1}{r}\Biggr)\\&>n\Biggl(\left(\frac{1}{p}-\frac{1}{5p}-\frac{1}{7p}-\frac{1}{7p}\right)+\frac{1}{q}+\frac{1}{r}\Biggr)>n\left(\frac{1}{2p}+\frac{1}{q}+\frac{1}{r}\right)\,.\end{align}$$ First, note that $m\geq\tau\left(m^2\right)$. If $b>1$, then $$\frac{n}{q}= p^aq^{b-1}r^cm> (2a+1)(2b+1)(2c+1)\,\tau\left(m^2\right)\geq \tau\left(n^2\right)\,.$$ Similarly, if $c>1$, then $$\frac{n}{r}> \tau\left(n^2\right)\,.$$ If $a>1$ and $b=c=1$, then $$\frac{n}{2p}=\frac{1}{2}p^{a-1}qrm> 9(2a+1)\,\tau\left(m^2\right)=\tau\left(n^2\right)\,.$$ If $a=b=c=1$, then $$n\left(\frac{1}{q}+\frac{1}{r}\right)=p(q+r)m> 27\,\tau\left(m^2\right)=\tau\left(n^2\right)\,.$$ This proves that $n-\phi(n)>\tau\left(n^2\right)$ for any odd natural number $n$ with at least three distinct prime divisors. Now, we shall prove that $n=21$ and $n=25$ are the only solutions in $\mathbb{N}$ to $n=\phi(n)+\tau\left(n^2\right)$. It is clear that $n\neq 1$ and that $n$ is odd. From the paragraph above, $n$ has at most two distinct prime divisors. If $n$ has one prime divisor, say $n=p^a$, then the required condition gives $p^{a-1}=2a+1$, which leads to $p=5$ and $a=2$, whence $n=25$. If $n$ has two prime divisors, say $n=p^aq^b$ with $2<p<q$, then we have $$2(2a+1)q^{b-1}\leq 2p^aq^{b-1}<p^{a-1}(q-1)q^{b-1}+p^aq^{b-1}=(2a+1)(2b+1)\,.$$ That is, $q^{b-1}<\frac{2b+1}{2}$, which means $b=1$. Hence, we have $$2p^a=2p^aq^{b-1}<(2a+1)(2b+1)=3(2a+1)\,.$$ This gives $p=3$ and $a=1$. Therefore, $n=3q$. Ergo, $$3q-2(q-1)=n-\phi(n)=\tau\left(n^2\right)=9\,,$$ leading to $q=7$, whence $n=21$.
| {
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How to prove that $\int_{0}^{1}\ln{(x/(1-x))}\ln{(1+x-x^2)}\frac{dx}{x}=-\frac{2}{5}\zeta{(3)}$
$$\int_{0}^{1}\ln{\big(\frac{x}{1-x}\big)}\ln{(1+x-x^2)}\frac{dx}{x}=-\frac{2}{5}\zeta{(3)}$$
Put $$\frac{x}{1-x}=y$$
$$I=\int_{0}^{\infty}\ln{y}\ln{(1+3y+y^2)}\frac{dy}{y(y+1)}=\frac{8}{5}\zeta{(3)}$$
Simple integral at first sight, however I cannot prove that. I would appreciate your help.
| I'll prove the second integral in the post.
Let $I(a)=\int^{\infty}_{0}\frac{\log y\log(1+ay)}{y(y+1)}dy$, we have $I'(a)=\int^{\infty}_{0}\frac{\log y\,dy}{(1+y)(1+ay)}$, and $I=I(\frac{3+\sqrt{5}}{2})+I(\frac{3-\sqrt{5}}{2})$.
Now $I(0)=0$, and $I'(a)=\frac{(\log a)^2}{2(1-a)}$(Proof: Substitute $y\rightarrow\frac{1}{ay}$), so for any $0<a<1$ we have
$\begin{align*}
I(a)&=\int^a_0\frac{(\log b)^2}{2(1-b)}db\\
&=\frac12(\log a)^2\log(1-a)+\log a Li_2(a)-Li_3(a)
\end{align*}$
and
$\begin{align*}
I(1/a)&=\int^{1/a}_0\frac{(\log b)^2}{2(1-b)}db\\
&=\int^{a}_{+\infty}\frac{(\log b)^2}{2b(1-b)}db\\
&=-\frac12(\log a)^2\log(1-a)+\frac16(\log a)^3-\log a Li_2(a)+Li_3(a)
\end{align*}$
therefore
$\begin{align*}
I(a)+I(1/a)&=-(\log a)^2\log(1-a)+\frac16(\log a)^3-2\log a Li_2(a)+2Li_3(a).
\end{align*}$
and we put $a=\frac{3-\sqrt{5}}{2}$ to get
$\begin{align*}
I&=-(\log \frac{3-\sqrt{5}}{2})^2\log(\frac{1+\sqrt{5}}{2})+\frac16(\log\frac{3-\sqrt{5}}{2})^3-2\log \frac{3-\sqrt{5}}{2} Li_2(\frac{3-\sqrt{5}}{2})+2Li_3(\frac{3-\sqrt{5}}{2})\\
&=\frac{8}{3}(\log\frac{1+\sqrt{5}}{2})^3+4\log \frac{1+\sqrt{5}}{2} Li_2(\frac{3-\sqrt{5}}{2})+2Li_3(\frac{3-\sqrt{5}}{2}).
\end{align*}$
The claim follows from the specific values $Li_2(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}-(\log\frac{1+\sqrt{5}}{2})^2$ and $Li_3(\frac{3-\sqrt{5}}{2})=-\frac{2\pi^2}{15}\log\frac{1+\sqrt{5}}{2}+\frac23(\log\frac{1+\sqrt{5}}{2})^3+\frac{4}{5}\zeta(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1839178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
A fraction problem $$a = x + \frac1x \\b =y + \frac1y \\ c = xy +\frac1{xy} $$
Express $c$ in terms of $a$ and $b$
| Multiply all things by the LCD to remove fractions.
$$ax=x^2+1$$
$$by=y^2+1$$
$$cxy=x^2y^2+1$$
Multiply the first two equations to get
$$abxy=x^2y^2+1+x^2+y^2=c+x^2+y^2$$
Adding the first two equations, we get
$$ax+by=x^2+y^2+2$$
Combining these two lines,
$$abxy=c+ax+by-2$$
At this point, I'd just go back and solve for $x,y$ using the quadratic formula,
$$c+a\left(\frac{a\pm\sqrt{a^2-4}}2\right)+b\left(\frac{b\pm\sqrt{b^2-4}}2\right)-2=ab\left(\frac{a\pm\sqrt{a^2-4}}2\right)\left(\frac{b\pm\sqrt{b^2-4}}2\right)$$
Not the most beautiful thing in the world, but at least you don't have to square the thing or anything messy like that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
If $x,y,z\gt 0$ and $xyz=1$ Then minimum value of $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$
If $x,y,z\gt 0$ and $xyz=1$ Then find the minimum value of $\displaystyle \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$
$\bf{My\; Try::}$Using Titu's Lemma $$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2}\ge 3\frac{\sqrt[3]{xyz}}{2} = \frac{3}{2}$$
and equality holds when $$x=y=z=1$$
My question is how can we solve it without the above lemma, like using Jensen's Inequality or other inequality.
Please explain me.
Thanks
| Just find the minimum value of $e=(x+y)(y+z)(z+x)$ because give expression E satisfies $E\ge \frac{3}{[(x+y)(y+z)(z+x)]^{\frac{1}{3}}}$ (using AM-GM inequality). Now you can find the minimum value of $e$ by lagrange multiplier method with constraint $xyz=1$
Hope this helps !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$\frac{x^2+y^2}{x+y}$ is a divisor of $1978$
Two nonzero integers $x,y$ (not necessarily positive) are such that $x+y$ is a divisor of $x^2+y^2$, and the quotient $\dfrac{x^2+y^2}{x+y}$ is a divisor of $1978$. Prove that $x = y$.
Let $A = \dfrac{x^2+y^2}{x+y}$ where $A$ is an integer divisor of $1978$. Then $A \mid x^2+y^2$ and if $p$ is a prime divisor of $A$ we must have $x^2+y^2 \equiv 0 \pmod{p}$.
We first deal with the case $x,y \equiv 0 \pmod{q}$. Otherwise $x,y \not \equiv 0 \pmod{q}$ where $q \equiv 3 \pmod{4}$ and is a prime divisor of $A$.
I didn't find an easy way of dealing with the case that $x,y \equiv 0 \pmod{q}$. Is there an easier way of solving this?
| If a prime $p$ divides $x^2+y^2$ then $x^2\equiv -y^2$ modulo $p$ and thus, if $p$ does not divide both $x$ and $y$ we find that $-1$ is a quadratic residue modulo $p$. This cannot happen unless $p$ is of the form $4k+1$.
First let's suppose that $x,y$ are coprime. If they are not, then simply divide by their $gcd$ and find some $x',y'$ which are coprime and still satisfy the hypothesis.
Now you know that $\frac{x^2+y^2}{x+y}$ divides $2\cdot 23\cdot 43$. If $23$ or $43$ divides $x^2+y^2$ then bot $x$ and $y$ have $23$ or $43$ as a factor, which contradicts the fact that they are coprime. This leaves us with
$x^2+y^2=2(x+y)$ which is equivalent to $(x-1)^2+(y-1)^2=2$. From here we see that $x=y=2$.
There is another case: $x^2+y^2 = x+y$ which implies $(2x-1)^2+(2y-1)^2 = 2$. This implies $x=y=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
How to integrate $\int \frac{x^{13}\ dx}{x^5 + 1}$ We get this problem from our teacher today. I only wish that it was $x^{14}$ in the numerator, so we can use substitution method:
$$\int \dfrac{x^{13}\ dx}{x^5 + 1}$$
I can't find way to integrate this. Please help me out to solve this seeming simple integral.
| $$
\int\frac{x^{13}}{x^5+1}\,dx
=\int\left(x^8-x^3+\frac{x^{3}}{x^5+1}\right)\,dx
=\frac{x^9}{9}-\frac{x^4}{4}+\int\frac{x^{3}}{x^5+1}\,dx
$$
Note that
$$
x^5+1=(x+1)(x^2+\phi x+1)(x^2-(\phi-1) x+1),
$$
where $\phi=\frac{\sqrt{5}+1}{2}$ is the "Golden Ratio". It can be shown that
$$
\frac{x^3}{x^5+1}=\frac{1}{5}\left(\frac{\phi x+1-\phi}{x^2-(\phi-1)x+1}+
\frac{(1-\phi)x+\phi}{x^2+\phi x+1}-\frac{1}{x+1}\right)
$$
You can take it from here...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable
Prove that if $b^2-4ac=k^2$ (for some positive integer $k$ ) then $ax^2+bx+c$ is factorizable
I related this to the roots of equation: $ax^2+bx+c=0$ and using roots $x_1,x_2$:
$(x-x_1)(x-x_2)=0\implies(2ax+k-b)(2ax+k+b)=0\implies 4a^2x^2+4akx+k^2-b^2=0$ but this seems has nothing to do with original equation.Any better way??!!
Under what conditions can we be sure the factorization involves just integers??
| $$ax^2+bx+c=a\left(\left(x-\frac{b}{a}\right)^2-\frac{b^2-4ac}{a^2}\right)=a\left(\left(x-\frac{b}{a}\right)^2-\frac{k^2}{a^2}\right)$$
which is of the form $$\alpha ^2-\beta^2=(\alpha -\beta )(\alpha +\beta ).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to solve this inequality with absolute value: $ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 $ Good morning to everyone. I have an inequality that I don't know how to solve: $$ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 $$ I tried to solve it in this way: $$ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 \rightarrow \frac{\left|x-3\right|}{\left|x+2\right|} - 3 \le 0 \rightarrow \frac{\left|x-3\right|-3\left|x+2\right|}{\left|x+2\right|}\le 0 \rightarrow $$
Case 1:
$$ \left|x+2\right| > 0 \wedge \left|x-3\right|-3\left|x+2\right| \le 0 $$
Case 1 a) $$ x+2 >0 \rightarrow x>-2$$
Case 1 b) $$ -x-2 <0 \rightarrow x>-2$$
Case 1 c) $$ \left|x-3\right|-3\left|x+2\right| \le 0 \rightarrow \left|x-3\right| \le 3\left|x+2\right| \rightarrow x-3 \le x+2 \rightarrow $$
The solution are all real numbers
Case 1 d)$$ -x+3 < x+2 \rightarrow x>\frac{1}{2} $$
Case 1 e) $$ x-3 \le -x-2 \rightarrow x \le \frac {1}{2} $$
Case 1 f) $$ x-3 < x+2 $$
The solution: all real numbers
Case 2: $$ \left|x+2\right| \le 0 \wedge \left|x-3\right|-3\left|x+2\right| > 0 $$
It'll have the same solutions because are the same inequalities. Therefore $x$ belongs to $(\frac{1}{2}, \infty)$ But my answer sheet shows that it belongs to $(-\infty, -\frac{9}{4}) \wedge (-\frac{3}{4},\infty)$
| If $|y|\le a$ for real $y, a\ge0,$
we need $-a\le y\le a$
So, $-3\le\dfrac{x-3}{x+2}\le3$
Now $\dfrac{x-3}{x+2}\le3\iff\dfrac{x-3}{x+2}-3\le0\iff\dfrac{2x+9}{x+3}\ge0$
$\iff x\le-\dfrac92$ or $x>-3$
can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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First one to roll an even number wins the game
Bill, George, and Ross, in order, roll a die. The first one to roll an even number wins and the game is ended. What is the probability that Bill will win the game?
As far as I know, for the first roll Bill has a chance of $\frac{1}{2}$, then followed by George with a less than $\frac{1}{2}$ chance and Ross has a lesser than the second roll from George. Why is the answer $\frac{4}{7}$?
| Call the winning probabilities $b,g,r$ respectively. Then clearly $b + g + r = 1$. Also, call the probabilities that they each win in the first round $b_1, g_1, r_1$. Then $b_1 = \frac{1}{2}, g_1 = \frac{1}{4}$ and $r_1 = \frac{1}{8}$. So we have no decision in the first round with chance $\frac{1}{8}$.
Now, note that $b = b_1 + \frac{1}{8}b = \frac{1}{2} + \frac{1}{8}b$, because $b$ wins with chance $b_1$ in the first round and if gets a second round (which happens with chance $\frac{1}{8}$) he has chance $b$ again, as if he started anew.
This gives us $b = \frac{8}{7} \cdot \frac{1}{2} = \frac{4}{7}$.
Similarly $g = g_1 + \frac{7}{8}p = \frac{1}{4} + \frac{7}{8}p$, so $g = \frac{8}{7}\cdot \frac{1}{4} = \frac{2}{7}$.
So $r = \frac{1}{7}$, by summing to 1, or doing a similar calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the Prime Factorization of $\varphi(11!)$
Find the Prime Factorization of $\varphi(11!)$
What I did:
$\varphi(11!)=\varphi(11)\cdot\varphi(10)...\varphi(1)$
$$\varphi(11)=2\cdot 5\\
\varphi(10)=2^2\\
\varphi(9)=3\cdot 2\\
\varphi(8)=2^2\\
\varphi(7)=2\cdot 3\\
\varphi(6)=2\\
\varphi(5)=2^2\\
\varphi(4)=2\\
\varphi(3)=2\\
$$
$$\Longrightarrow\text{ the answer is } 2^{12}\cdot 5\cdot 3^2$$
$11!$
$\varphi(11!)=\varphi(39916800)=8294400$
$8294400=2^{12}3^45^2$
Where am I wrong?
Is there a simpler way to solve this?
| \begin{align}
& \bigg\lfloor \frac{11}{2} \bigg\rfloor+\bigg\lfloor \frac{11}{4} \bigg\rfloor+\bigg\lfloor \frac{11}{8} \bigg\rfloor=8 \\
& \bigg\lfloor \frac{11}{3} \bigg\rfloor+\bigg\lfloor \frac{11}{9} \bigg\rfloor=4 \\
& \bigg\lfloor \frac{11}{5} \bigg\rfloor=2 \\
& \bigg\lfloor \frac{11}{7} \bigg\rfloor=1 \\
&\bigg\lfloor \frac{11}{11} \bigg\rfloor=1 \\
\end{align}
$$\varphi(11!)=\varphi ({{2}^{8}}\times {{3}^{4}}\times {{5}^{2}}\times {{7}^{1}}\times {{11}^{1}})=\varphi ({{2}^{8}})\varphi ({{3}^{4}})\varphi ({{5}^{2}})\varphi ({{7}^{1}})\varphi ({{11}^{1}})$$
Note if $P$ be prime then
$$\varphi(P^n)=P^n-P^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove: $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive
$$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$
$$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$
Is the proof valid?
| There is a simpler way, by eliminating at once the square roots.
Setting $x=X^2$ and $y=Y^2$ with $X>0$ and $Y>0$, the issue is equivalent to:
$$\text{show that} \ \ \ \dfrac{X^2}{Y}+\dfrac{Y^2}{X} \geq X+Y$$
which amounts to prove that:
$$X^3+Y^3 \geq XY(X+Y)$$
knowing factorization $X^3+Y^3=(X+Y)(X^2-XY+Y^2)$ and simplifying by $X+Y$, one is brought back to show that $X^2-2XY+Y^2 \geq 0$ i.e. $(X-Y)^2\geq 0$ which is eviedently true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Determining about roots of quadratics from linear equation system Let $a,b,c,d$ be real numbers in G.P.If $u,v,w$ satisfy the system of equations
$u+2v+3w=6$,
$4u+5v+6w=12$,
$6u+9v=4$
then show that the roots of the equation $$\left(\frac{1}{u}+\frac{1}{v} + \frac{1}{w} \right)x^2 + \left[(b-c)^2+(c-a)^2+(d-b)^2\right]x+u+v+w=0$$
and $20x^2+10(a-d)^2x-9=0$ are reciprocals of each other.
I am stuck in how to implement the linear equations in determining about the roots of the quadratic.Any hints or suggestion will be helpful.
Source:JEE 1999
| Subracting first $u,v,w$ equation from second we get $u+v+w=2$.
We can easily solve to get $u=-\frac{1}{3},v=\frac{2}{3},w=\frac{5}{3}$. So $\frac{1}{u}+\frac{1}{v}+\frac{1}{w}=-\frac{9}{10}$.
Now suppose the roots of $$\left(\frac{1}{u}+\frac{1}{v}+\frac{1}{w}\right)x^2+[(b-c)^2+(c-a)^2+(d-b)^2]x+u+v+w=0\ \ (*)$$ are $\alpha,\beta$. Then we have
$$\frac{1}{\alpha\beta}=\frac{\frac{1}{u}+\frac{1}{v}+\frac{1}{w}}{u+v+w}=-\frac{9}{10}\frac{1}{2}=-\frac{9}{20}$$ Which is indeed the product of the roots of $$20x^2+10(a-d)^2x-9=0$$
We also have $$\alpha+\beta=-\frac{(b-c)^2+(c-a)^2+(d-b)^2}{\frac{1}{u}+\frac{1}{v}+\frac{1}{w}}=\frac{10}{9}\left((b-c)^2+(c-a)^2+(d-b)^2\right)$$
so $$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=-\frac{1}{2}\left((b-c)^2+(c-a)^2+(d-b)^2\right)$$
Thus we are done provided we can show that the last expression is $$-\frac{1}{2}(a-d)^2$$ or that $$(b-c)^2+(c-a)^2+(d-b)^2-(a-d)^2=0\ \ (**)$$ In general that is not true, but we are also given that $a,b,c,d$ is a geometric progression. So we have $b^2=ac,c^2=bd,bc=ad$. But expanding $(**)$ we get $(2b^2-2ac)+(2c^2-2bd)+(2ad-2bc)$ which is 0.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a nicer way to show that the series is convergent? I'd like to show that for a fixed $z\in\mathbb C\setminus\mathbb Z$ the series $$\sum_{n=1}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right|$$ is convergent.
I think, one can do it as follows. Fix some $n_0> |z|$. Then
\begin{align*}
\sum_{n=1}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| & = \underbrace{\sum_{n=1}^{n_0-1} \left| \frac{1}{z-n} + \frac{1}{n} \right|}_{=:C} + \sum_{n=n_0}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| \\
& = C + \sum_{n=n_0}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| \\
& = C + \sum_{n=n_0}^\infty \left| \frac{z}{(z-n)n} \right| \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{|z-n|n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{||z|-|n||n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{(n-|z|)n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{n^2-|z|n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{n^2-|z|n^2} \\
& \leq C + \frac{|z|}{1-|z|} \underbrace{\sum_{n=n_0}^\infty \frac{1}{n^2}}_{<\infty} \\
\end{align*}
That strikes me as somewhat cumbersome. Is there a nicer way? E.g. without separating the series into before and after $n_0$?
| Your answer seems to show that the series converges for all $z\in\mathbb{C}$. However, $z$ cannot be a natural number (positive integer).
$$
\begin{align}
\sum_{n=1}^\infty\left|\frac1{z-n}+\frac1n\right|
&=\sum_{n=1}^\infty\left|\frac z{n(z-n)}\right|\\
&=\sum_{n=1}^{\lfloor2|z|\rfloor}\left|\frac z{n(z-n)}\right|
+\sum_{n=\lfloor2|z|\rfloor+1}\left|\frac z{n(z-n)}\right|\\
&=\underbrace{\sum_{n=1}^{\lfloor2|z|\rfloor}\left|\frac z{n(z-n)}\right|}_{\text{finite for $z\not\in\mathbb{N}$}}
+\left|2z\right|\underbrace{\sum_{n=\lfloor2|z|\rfloor+1}^\infty\frac1{n^2}}_{\le\frac{\pi^2}6}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the smallest number $\alpha$, such that for all $x,y,z$ $\alpha(x^2-x+1)(y^2-y+1)(z^2-z+1)\ge(xyz)^2-xyz+1$
Find a smallest number $\alpha$, such that for all $x,y,z$ (not all of which are positive) inequality
$$\alpha(x^2-x+1)(y^2-y+1)(z^2-z+1)\ge(xyz)^2-xyz+1$$
My work so far:
Let $f(t)=t^2-t+1$. Then $f(t) \ge \frac34$.
If $x=0, y=z=\frac12$, then $$\alpha\ge \frac{16}9$$
| In the starting formulation the answer is $\frac{16}{9}$.
If $x$, $y$ and $z$ are non-positives so after replacing $x\rightarrow-x$...
we need to prove that
$$\frac{16}{9}(x^2+x+1)(y^2+y+1)(z^2+z+1)\geq x^2y^2z^2-xyz+1$$
which is true for all non-negatives $x$, $y$ and $z$ because we'll prove now that even
$$\frac{16}{9}(x^2+x+1)(y^2-y+1)(z^2-z+1)\geq x^2y^2z^2+xyz+1$$.
If $x\leq0$, $y\leq0$ and $z\geq0$ we need to prove that
$$\frac{16}{9}(x^2+x+1)(y^2+y+1)(z^2-z+1)\geq x^2y^2z^2-xyz+1$$
for non-negatives $x$, $y$ and $z$, which follows from
$$\frac{16}{9}(x^2+x+1)(y^2-y+1)(z^2-z+1)\geq x^2y^2z^2+xyz+1$$
again.
Now we'll prove it:
$$\begin{align}
&16(x^2+x+1)(y^2-y+1)(z^2-z+1)-9(x^2y^2z^2+xyz+1)
\\&\phantom{aaa}=(16(y^2-y+1)(z^2-z+1)-9y^2z^2)x^2+(16(y^2-y+1)(z^2-z+1)-9yz)x
\\
&\phantom{aaaaa}+16(y^2-y+1)(z^2-z+1)-9
\\&\phantom{aaa}
\geq\left(16\left(\frac{3}{4}y^2+\left(\frac{y}{2}-1\right)^2\right)\left(\frac{3}{4}z^2+\left(\frac{z}{2}-1\right)^2\right)-9y^2z^2\right)x^2+
+(3y\cdot3z-9yz)x
\\&\phantom{aaaaa}+
16\left(\frac{3}{4}+\left(\frac{1}{2}-y\right)^2\right)\left(\frac{3}{4}+\left(\frac{1}{2}-z\right)^2\right)-9
\\&\phantom{aaa}\geq0
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the orthogonal projection of a vector I can't figure this out. How the book explains to do this problem doesn't make sense to me. I'm hoping someone can simplify this so I can understand how to do this step by step.
Let $L$ be the line in $\mathbb{R}^3$ that consists of all scalar multiples of $$ \left[
\begin{array}{ccc}
2\\
1\\
2\\
\end{array}
\right] $$ Find the reflection of the vector
$$ \left[
\begin{array}{ccc}
1\\
4\\
1\\
\end{array}
\right] $$ onto $L$
| Let $L$ be the line spanned by some non-zero vector $\mathbf{v} \in \mathbb{R}^3$, so that $L = \{a\mathbf{v} \mid a \in \mathbb{R}\}$ is the space of all scalar multiples of $\mathbf{v}$. Then the orthogonal projection of a vector $\mathbf{x} \in \mathbb{R}^3$ onto the line $L$ can be computed as
$$
\operatorname{Proj}_L(\mathbf{x}) = \frac{\mathbf{v} \cdot \mathbf{x}}{\mathbf{v}\cdot\mathbf{v}} \mathbf{v}.
$$
So, in this case, we have
$$
\mathbf{v} = \begin{pmatrix}2\\1\\2\end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix}1\\4\\1\end{pmatrix},
$$
so that
$$
\mathbf{v} \cdot \mathbf{x} = 2 \cdot 1 + 1 \cdot 4 + 2 \cdot 1 = 8, \quad \mathbf{v} \cdot \mathbf{v} = 2^2 + 1^2 + 2^2 = 9,
$$
and hence
$$
\operatorname{Proj}_L(\mathbf{x}) = \frac{8}{9}\begin{pmatrix}2\\1\\2\end{pmatrix}.
$$
Now, you probably wanted to compute the orthogonal projection of some vector $\mathbf{x}$ onto the line $L$ spanned by some non-zero vector $\mathbf{v}$. But what if you did want to compute the reflection of $\mathbf{x}$ in the line $L$? What would this mean? Well, in general, suppose that $S$ is a subspace of $\mathbb{R}^3$ (e.g., a line through the origin or a plane through the origin), so that for any vector $\mathbf{x} \in \mathbb{R}^3$, we have
$$
\mathbf{x} = \operatorname{Proj}_S(\mathbf{x}) + \operatorname{Proj}_{S^\perp}(\mathbf{x}),
$$
where $\operatorname{Proj}_S(\mathbf{x})$ is the orthogonal projection of $\mathbf{x}$ onto $S$ and $\operatorname{Proj}_{S^\perp}(\mathbf{x})$ is the orthogonal projection of $\mathbf{x}$ onto the orthogonal complement
$$
S^\perp = \{\mathbf{y} \in \mathbb{R}^3 \mid \text{for every $\mathbf{z} \in S$}, \; \mathbf{y} \cdot \mathbf{z} = 0\}
$$
of $S$; in particular, observe that
$$
\operatorname{Proj}_{S^\perp}(\mathbf{x}) = \mathbf{x} - \operatorname{Proj}_S(\mathbf{x}).
$$
Then, geometrically, the reflection $\operatorname{Refl}_S(\mathbf{x})$ of $\mathbf{x}$ in $S$ is given by fixing the component of $\mathbf{x}$ in $S$ and flipping the direction of the component of $\mathbf{x}$ in $S^\perp$, i.e.,
$$
\operatorname{Refl}_S(\mathbf{x}) = \operatorname{Proj}_S(\mathbf{x}) - \operatorname{Proj}_{S^\perp}(\mathbf{x}) = \operatorname{Proj}_S(\mathbf{x}) - \left( \mathbf{x} - \operatorname{Proj}_S(\mathbf{x})\right) = 2\operatorname{Proj}_S(\mathbf{x}) - \mathbf{x}.
$$
So, suppose that $L$ is the line spanned by some non-zero vector $\mathbf{v} \in \mathbb{R}^3$. On the one hand, the reflection of $\mathbf{x}$ in $L$ is given by
$$
\operatorname{Refl}_L(\mathbf{x}) = 2\operatorname{Proj}_L(\mathbf{x}) - \mathbf{x} = 2\frac{\mathbf{v} \cdot \mathbf{x}}{\mathbf{v}\cdot\mathbf{v}} \mathbf{v} - \mathbf{x},
$$
which in your case yields
$$
\operatorname{Refl}_L(\mathbf{x}) = 2 \cdot \frac{8}{9}\begin{pmatrix}2\\1\\2\end{pmatrix} - \begin{pmatrix}1\\4\\1\end{pmatrix} = \frac{1}{9}\begin{pmatrix}23\\-20\\23\end{pmatrix}.
$$
On the other hand, the reflection of $\mathbf{x}$ in the plane
$$
L^\perp = \{\mathbf{y} \in \mathbb{R}^3 \mid \mathbf{v} \cdot \mathbf{y} = 0\}
$$
with normal vector $\mathbf{v}$ is given by
$$
\operatorname{Refl}_{L^\perp}(\mathbf{x}) = 2\operatorname{Proj}_{L^\perp}(\mathbf{x}) - \mathbf{x} = 2\left(\mathbf{x} - \operatorname{Proj}_L(\mathbf{x})\right) - \mathbf{x} = \mathbf{x} - 2\operatorname{Proj}_L(\mathbf{x}) = \mathbf{x} - 2\frac{\mathbf{v} \cdot \mathbf{x}}{\mathbf{v}\cdot\mathbf{v}} \mathbf{v},
$$
which in your case yields
$$
\operatorname{Refl}_{L^\perp}(\mathbf{x}) = \begin{pmatrix}1\\4\\1\end{pmatrix} - 2 \cdot \frac{8}{9}\begin{pmatrix}2\\1\\2\end{pmatrix} = -\frac{1}{9}\begin{pmatrix}23\\-20\\23\end{pmatrix}.
$$
One last cultural note: the reflection $\operatorname{Refl}_{L^\perp}(\mathbf{x})$ of $\mathbf{x}$ in the plane $L^\perp$ with normal vector $\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image
$$
H_{\mathbf{v}}(\mathbf{x}) := \operatorname{Refl}_{L^\perp}(\mathbf{x}) = \mathbf{x} - 2\frac{\mathbf{v} \cdot \mathbf{x}}{\mathbf{v}\cdot\mathbf{v}} \mathbf{v}
$$
of $\mathbf{x}$ under the Householder transformation $H_\mathbf{v}$ corresponding to $\mathbf{v}$. So, in your case, as we just saw,
$$
H_{\mathbf{v}}(\mathbf{x}) = -\frac{1}{9}\begin{pmatrix}23\\-20\\23\end{pmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Existence and uniqueness of $A\in M_2(\mathbb R)$ such that $(A^{-1} - 3 I_2)^t = 2 \begin{pmatrix} -1 & -2 \\ 1 & -5 \end{pmatrix}$ I recently thought about this exercise that states:
Does there exist a matrix $A \in M_2(\mathbb R)$ s.t.
$$(A^{-1} - 3 I_2)^t = 2 \left( \begin{array}{ccc}
-1 & -2 \\
1 & -5 \\
\end{array} \right)$$
But such a matrix $A$ is not uniquely defined?
I thought that the determinant of $A$ must have determinant = 3 to have a matrix that is not uniquely defined. But I am a bit lost on how to actually find it.
Anyone mind lending a hand?
| $$(A^{-1} - 3 I_2)^t = 2 \left( \begin{array}{ccc}
-1 & -2 \\
1 & -5 \\
\end{array} \right)$$
$$(A^{-1} - 3 I_2) = 2 \left( \begin{array}{ccc}
-1 & 1 \\
-2 & -5 \\
\end{array} \right)=\begin{pmatrix} -2 & 2 \\ -4 & -10\end{pmatrix}$$
$$A^{-1} =\begin{pmatrix} 1 & 2 \\ -4 & -7\end{pmatrix}$$
$$A=\begin{pmatrix} 1 & 2 \\ -4 & -7\end{pmatrix}^{-1}$$
Are you able to compute $A$ explicitly?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Rational root coefficient I saw this question in my exam recently,
If a, b, c are distinct rational roots of $x^3+ax^2+bx+c=0$, find the values of a, b, c.
Can someone give me a hint or answer? I tried factoring it and then expand it all out, but it doesn't seem very useful as they are all "rational". Can someone help me with it?
| Apologies for the length of the argument. I tried to format it to be more comprehensible.
Multiplying out $(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$, so we have the equations:
$1)$ $a=-a-b-c$
$2)$ $b=ab+ac+bc$
$3)$ $c=-abc$
Notice that if $a = 0$ or $b = 0$, then by the $(3)$, $c=0$ contradicting the fact that they must be distinct. Therefore $a,b\neq 0$.
Now if $c=0$, then by $(1)$ and $(2)$, $2a=-b$ and $b=ab$, which has the solution $a=1, b=-2, c=0$. The goal now is to show no other solution exists.
Consider when $c \neq 0$. Cancelling the $c$'s in $(3)$ gives us $1=-ab \implies a=-\frac{1}{b}$. Then we can simplify $(1)$ into $b^2+cb-2=0$
$\implies b=\frac{-c \pm \sqrt{c^2+8}}{2}$
and $(2)$ simplifies to $(c-1)b^2-b-c=0$
$\implies b=\frac{1\pm\sqrt{4c^2-4c+1}}{2(c-1)}=\frac{1\pm(2c-1)}{2(c-1)}$
for $c\neq 1$
(if $c=1$, then $(c-1)b^2-b-c=0\implies b=-1 \implies a=1$, contradicting distinct values again).
Setting the equations for $b$ equal yields
$\frac{-c \pm \sqrt{c^2+8}}{2}=\frac{1\pm(2c-1)}{2(c-1)}$
or
$\frac{-c \pm \sqrt{c^2+8}}{2}=\frac{1\mp(2c-1)}{2(c-1)}$
Changing these two equations to the one
$c^2+8=\left(\frac{1\pm(2c-1)}{c-1}+c \right)^2$
makes it slightly easier.
Taking the $-$ gives us $c=-1 \implies b=-1$ contradicting distinct again, and taking the $+$ gives us the more ugly cubic equation
$c^3-2c^2+4c-2=0$
By the Rational Root Theorem, one need only test the values $\pm 1, \pm 2$, none of which yield a solution to this cubic, so this cubic has no rational solutions.
Therefore, the only solution was the first one we found; namely $\boxed{a=1, b=-2,c=0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Summation of a convergent series I have the following problem:
So I start as follows:
$B_{2}M_{2}=\frac{1}{\sqrt{3}}$ and I realize that $B_{2}M_{2} = A_{2}B_{2}$, so $B_{3}M_{3} = \frac{1}{\sqrt{3}}^{2}$. Next, I compute $A_{1}B_{2}=A_{1}B_{1}-B_{2}M_{2} = 1-\frac{1}{\sqrt{3}}$. For $A_{2}B_{3}$, we can do something similar, namely: $A_{2}B_{3} = A_{2}B_{2}-B_{3}M_{3}=\frac{1}{\sqrt{3}}-(\frac{1}{\sqrt{3}})^{2}$.
We can now derive the formula for computing the area of both triangles:
$$\frac{1}{2}\sqrt{(A_{1}B_{1})^{2}+(B_{1}M_{1})^{2}}\cdot A_{1}B_{2} \sin(30)$$
Now since $A_{1}B_{1}=B_{1}M_{1}$, we can simplify it to:
$$\frac{1}{\sqrt{2}}\sqrt{(A_{1}B_{1})^{2}}\cdot A_{1}B_{2}$$
This can be written as:
$$S_{n} = \frac{1}{\sqrt{2}}\sqrt{(A_{n}B_{n})^{2}}\cdot A_{n}B_{n+1}$$
I fill in the missing pieces and I get:
$$S_{n} = \frac{1}{\sqrt{2}} \cdot \sqrt{\left(\frac{1}{3}\right)^{n-1}} \cdot \left( \left(\frac{1}{3} \right) ^{n-1} - \left( \frac{1}{3} \right)^{n} \right) $$
Now I simplify it to:
$$S_n = \sqrt{6} \cdot 3^{-1.5n}$$
Now I have a small problem, if I try to solve $\sum_{n=1}^{\infty} \sqrt{6} \cdot 3^{-1.5n}$, I know we can rewrite it to $\sqrt{6} \sum_{n=1}^{\infty} 3^{-1.5n}$, but I have no idea how to continue to get the closed form.
In addition, I was wondering, is there a faster way to get the solution?
| where does this come from?
$\frac{1}{2}\sqrt{(A_{1}B_{1})^{2}+(B_{1}M_{1})^{2}}\cdot A_{1}B_{2} \sin(30)$
I think you want $\sin 45$. But that is overly complicated.
$\frac 12 bh$ will suffice with $b = A_1B_2$ and $h = A_1B_1$ and there are 2 of these triangles, so you can double that.
$S_1 = 1 - \frac 1{\sqrt{3}}\\
S_n = (\frac 1{\sqrt 3}) S_{n-1}\\
S_n = (\frac 1{\sqrt 3})^{n-1} S_{1}\\
\sum_\limits{n=1}^{\infty} S_n = S_1\sum_\limits{n=0}^{\infty} (\frac 1{\sqrt 3})^n = \frac {S_1}{(1-\frac 1{\sqrt 3})} = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$
Prove the given trigonometric identity
$$\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$$
My attempt
L.H.S.$=\cos5A$
$$\cos(A+4A)$$
$$\cos A\cos4A-\sin A\sin4A$$
Now how should I move further?
| Good job. Just keep applying double angle formula.
Hint:
$$\cos 4A= \cos(2(2A))=2\cos^2(2A)-1$$
$$\cos 2A = 2 \cos^2A-1$$
$$\sin 4A = 2 \sin 2A \cos 2A$$
$$\sin 2A = 2\sin A \cos A$$
$$ \sin^2A +\cos^2A=1$$
Edit:
\begin{align}
\cos(5A)&=\cos A \cos 4A-\sin A \sin 4A \\
&= \cos A (2 \cos^2(2A)-1)-\sin A (2 \sin 2A \cos 2A) \\
&= \cos A (2 (2 \cos^2 A-1)^2)-\cos A-2 \sin A (2 \sin A \cos A) (2 \cos^2 A -1) \\
&= 2 \cos A (4 \cos^4A-4 \cos^2A+1)-\cos A-4\sin^2A \cos A (2 \cos^2A-1)\\
&= 8\cos^5A-8 \cos^3A+\cos A-4(1-\cos^2A) (2 \cos^3A-\cos A)
\end{align}
Are you able to complete it?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Trouble Finding $f \circ g \; \text{ and } \; g\circ f$ for this function?
$f(x) =
\begin{cases}
2x+3, & \text{if x $\lt$ 3} \\[2ex]
x^2, & \text{if $x \ge 3$ }
\end{cases}$ $,\qquad$ $g(x) =
\begin{cases}
7-2x, & \text{if x $\le$ 2 } \\[2ex]
x+1, & \text{if $x \gt 2$ }
\end{cases}$
Finding $f \circ g \; \text{ and } \; g\circ f$ for this function?
The first step to solve this is to show the definitions
$\left( f\circ g \right)(x) = f(g(x)) =\begin{cases}
2(7-2x)+3, & \text{if x $\lt$ 3} \\[2ex]
(x+1)^2, & \text{if $x \ge 3$ }
\end{cases}$
$\left( g\circ f \right)(x) = g(f(x)) =\begin{cases}
7-2(2x+3), & \text{if x $\le$ 2 } \\[2ex]
(x)^2+1, & \text{if $x \gt 2$ }
\end{cases} $
However I do not think this is correct way of going through this. I think that when $x<3 \text{ and } x\ge 3 $ have something to do with this problem but I just do not know what it is. Any insight on how I went wrong would be very superb.
| A neat way to do these types of questions would be through graphs.
It is given that $$f(x) =
\begin{cases}
2x+3, & \text{if $x \lt$ 3} \\[2ex]
x^2, & \text{if $x \ge 3$ }
\end{cases}$$
$$g(x) =
\begin{cases}
7-2x, & \text{if $x \le$ 2 } \\[2ex]
x+1, & \text{if $x \gt 2$ }
\end{cases}$$
To simplify the question lets first plot a graph of the given functions
Now lets begin with $\left( f\circ g \right)(x)$
So, we have
$$\left( f\circ g \right)(x) = f(g(x)) =\begin{cases}
2g(x)+3, & \text{if $g(x) \lt$ 3} \\[2ex]
(g(x))^2, & \text{if $g(x) \ge 3$ }
\end{cases}$$
Now from the graph itself we see that $g(x) \ge 3$, so it saves us the work of checking the other case.So, we get
$$\left( f\circ g \right)(x) = f(g(x))=(g(x))^2, \text{if $g(x) \ge 3$ }= \begin{cases}
(7-2x)^2, & \text{if $x\le$ 2} \\[2ex]
(x+1)^2, & \text{if $x \gt 2$ }
\end{cases}
$$
Lastly come to the case of $\left( g\circ f \right)(x)$, this one is a bit longer.
$$\left( g\circ f \right)(x) = g(f(x)) =\begin{cases}
7-2f(x), & \text{if $f(x) \le$ 2 } \\[2ex]
f(x)+1, & \text{if $f(x) \gt 2$ }
\end{cases}$$
From the graph we can see that $f(x) \le 2$ for $x \le -\dfrac{1}{2}$ and $f(x)\gt 2$ for $x \gt -\dfrac{1}{2}$. So we can redefine $\left( g\circ f \right)(x)$ as follows
$$\left( g\circ f \right)(x) = g(f(x)) =\begin{cases}
7-2f(x), & \text{if $x \le -\dfrac{1}{2}$} \\[2ex]
f(x)+1, & \text{if $x \gt -\dfrac{1}{2}$}
\end{cases}$$
Now we break the defining of $\left( g\circ f \right)(x)$ as per the definition of $f(x)$ to further berak down the intervals of $x$.
So, at last we get
$$\left( g\circ f \right)(x) = g(f(x)) =\begin{cases}
7-2(2x+3), & \text{if $x \le -\dfrac{1}{2}$} \\[2ex]
(2x+3)+1, & \text{if $-\dfrac{1}{2} \lt x \lt 3$} \\[3ex]
x^2+1, & \text{if $x \ge 3$}
\end{cases}=\begin{cases}
1-4x, & \text{if $x \le -\dfrac{1}{2}$} \\[2ex]
2x+4, & \text{if $-\dfrac{1}{2} \lt x \lt 3$} \\[3ex]
x^2+1, & \text{if $x \ge 3$}
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How do I prove that $\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$ How do I prove that
$$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$$
without using the calculator?
| If we consider an infinite chain.
Suppose $x = \sqrt{20 +\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{\cdots}}}}}$
$x = \sqrt{20 +x}\\
x^2 = 20 + x\\
x^2 - x - 20 = 0\\
(x-5)(x+4) = 0$
$x$ must be greater than $0, x = 5$
and $y = \sqrt{20 -\sqrt{20-\sqrt{20-\sqrt{20-\sqrt{\cdots}}}}}$
$y = \sqrt{20 - y}\\
y^2 + y - 20=0\\
y = 4$
$x-y = 1$
As we add more terms under those square roots
$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}}$
converges toward 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 1
} |
Find $\int\frac1{\sqrt{1+x^4}}\,dx$ $$ \text{Find } \int \frac 1 {\sqrt{1+x^4}} \, dx$$
Let $x^2=\tan u$
$\implies 2x \,dx=\sec^2 u \,du$
$\implies dx=\dfrac{\sec^2 u}{2\sqrt{\tan u}}\,du$
$$= \int \frac{\sec^2 u}{2\sec u\sqrt{\tan u}} \, du $$
$$= \int \frac{\sec u}{2\sqrt{\tan u}} \, du $$
I am unsure how to continue..
| Let $$I = \int\frac{1}{\sqrt{1+x^4}}dx\;,$$ Now put $x^2=\tan t\;,$ Then $2xdx = \sec^2 tdt$
So $$I = \frac{1}{2}\int\frac{\sec^2 t}{\sec t \sqrt{\tan t}}dt = \frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{\sin 2 t}}dt$$
Now put $\displaystyle 2t=\frac{\pi}{2}-2\theta\;,$ Then $dt = -d\theta\;,$ So we get $$I = -\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{\cos 2\theta}}d\theta$$
So $$I = -\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{1-2\sin^2 \theta}}d\theta = -\frac{1}{\sqrt{2}}\int^{\theta}_{0}\frac{1}{\sqrt{1-2\sin^2 u}}du = -\frac{1}{\sqrt{2}}F\left(\theta\mid 2\right)+\mathcal{C}$$
$$ = -\frac{1}{\sqrt{2}}F\left(\frac{\frac{\pi}{4}-t}{2}\mid 2\right) +\mathcal{C}= -\frac{1}{\sqrt{2}}F\left(\frac{\frac{\pi}{4}-\tan^{-1}x^2}{2}\mid 2\right)+\mathcal{C}$$
Using elliptical integral of first kind, $$F(\phi\mid k^2) = \int^{\phi}_{0}\frac{1}{\sqrt{1-k^2\sin^2\theta}}d\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Help with the integral $\int x\sqrt{\frac{1-x^2}{1+x^2}}dx$ I would like to know what is $$\int x\sqrt{\frac{1-x^2}{1+x^2}}dx.$$ I put $x=\tan(y)$ to get integral of $\displaystyle \int \frac{\sin(y)}{\cos^3(y)}.\sqrt{\cos(2y)}dy$ I don't know whether $\sin(x)=t$ is a good substitution.
| HINT:
Metheod$\#1:$
Let $\sqrt{\dfrac{1-x^2}{1+x^2}}=y\implies \dfrac{1-x^2}{1+x^2}=y^2\implies x^2=\dfrac{1-y^2}{1+y^2}=\dfrac{2-(1+y^2)}{1+y^2}=\cdots$
$$2x\ dx=-2\dfrac{2y\ dy}{(1+y^2)^2}$$
$$\dfrac{d(y/1+y^2)}{dy}=?$$
Metheod$\#2:$
Let $x^2=\cos2y\implies0\le2y\le\pi$ and $\sqrt{\dfrac{1-x^2}{1+x^2}}=+\tan y$
$x\ dx=-2\sin2y\ dy=-4\sin y\cos y\ dy$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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ODE system solving by sequence of functions Let $y' = Ay$ where $A = \begin{pmatrix} 0&1 \\ -1& 0 \end{pmatrix}$ and $y( 0 ) = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. Consider the map $$G: C(\mathbb{R},\mathbb{R}^2) \to C(\mathbb{R},\mathbb{R}^2), G(\phi)(x) = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \int_0^x A\phi(t) dt$$.
With $\phi_0(x) = \begin{pmatrix} 1\\0\end{pmatrix}$ and $\phi_{n+1} = G(\phi_n)$, how to find and prove the general formula for $(\phi_n)_{n\in\mathbb{N}}$?
| Note that
\begin{equation}
\begin{split}
& \phi_{n+1}' = A\phi_n \\
\implies & \phi_{n}^{(n)} = A^n\phi_0,
\end{split}
\end{equation}
which is a constant vector. This shows $\phi_n$ is a vector of polynomials with degree at most $n$. The $2(n+1)$ coefficients can be found by a system of linear equations: for $s = 0, 1, \ldots k$,
\begin{equation}
\begin{split}
\phi_n^{(s)} = A^{s}\phi_{n-s} \implies \phi_n^{(s)}(0) = A^s \begin{pmatrix} 1 \\ 0 \end{pmatrix}.
\end{split}
\end{equation}
For each $s$, the above formula contributes to $2$ linear equations in the coefficients.
More precisely, if we let
$$p_s(x) = \dfrac{1}{s!}x^s, s = 0, 1, 2, \ldots, n,$$
and write
$$\phi_n(x) = \begin{pmatrix} \sum_{s=0}^n C_s p_s(x) \\ \sum_{s=0}^n K_s p_s(x) \end{pmatrix}.$$
Then the coefficients $C_s, K_s$ are given by
$$ \phi_n^{(s)}(0) = \begin{pmatrix} C_s \\ K_s \end{pmatrix} = A^s \begin{pmatrix} 1 \\ 0 \end{pmatrix}. $$
It is now reduced to find a general formula for $A^s \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Compute the sum of the series.
I just see the equality in my textbook, but I really have no idea how it arises (maybe it is obvious to the author), and it seems Fourier methods are not applicable. I would appreciate if someone could show me some hints or background of it.
| Let $f(x)=\cosh{\alpha x}$, its Fourier series on $\left[-\pi,\pi\right]$
$$f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos nx + b_n \sin nx$$
has $b_n =0$ for all $n$ since $f(x)$ is symmetric, so
$$a_n=\frac{2}{\pi}\int_0^\pi\cosh \alpha x \cos nx \;\mathrm{d}x$$
This integral can be computed via per partes :
$$\frac{\pi a_n}{2}=\frac{1}{\alpha}\sinh\alpha x\cos nx\bigg{|}_0^\pi+\frac{n}{\alpha^2}\cosh\alpha x\sin nx\bigg{|}_0^\pi-\frac{n^2}{\alpha^2}\int_0^\pi\cosh\alpha x\cos nx\;\mathrm{d}x=\\ \frac{1}{\alpha}\sinh\alpha\pi\cos\pi n-\frac{n^2}{\alpha^2}\frac{\pi a_n}{2}$$
So
$$a_n=\frac{2\alpha}{\pi}\frac{ \sinh\pi \alpha}{\alpha^2+n^2}\cos{\pi n}$$
hence
$$\cosh \alpha x= \frac{\sinh\pi \alpha}{\pi\alpha}+\frac{2\alpha\sinh\pi \alpha}{\pi}\sum_{n=1}^\infty\frac{\cos{\pi n}\cos n x}{\alpha^2+n^2}$$
For $x=\pi-y$ we get :
$$\cosh \alpha y\cosh \pi \alpha-\sinh \alpha y\sinh \pi \alpha= \frac{\sinh\pi \alpha}{\pi\alpha}+\frac{2\alpha\sinh\pi \alpha}{\pi}\sum_{n=1}^\infty\frac{\cos n y}{\alpha^2+n^2}$$
Rearanging :
$$\frac{2}{\pi}\sum_{n=1}^\infty\frac{\alpha \cos n x}{\alpha^2+n^2}= \cosh \alpha x\coth \pi\alpha-\sinh \alpha x-\frac{1}{\pi \alpha} \quad ;\text{for}\quad x\in\left[0,2\pi\right]\,\tag{1}$$
Denote
$$S(x)= \frac{2}{\pi}\sum_{n=1}^\infty\frac{\alpha\cos n x}{\alpha^2+n^2}$$
splitting into odd and even terms :
$$S(\alpha,x)= \frac{2}{\pi}\sum_{n=1}^\infty\frac{\alpha\cos (2n) x}{\alpha^2\!+\!(2n)^2}\!+\!\frac{2}{\pi}\sum_{n=0}^\infty\frac{\alpha\cos (2n\!+\!1) x}{\alpha^2\!+\!(2n\!+\!1)^2}=\frac{1}{2}S\left(\frac{\alpha}{2},2x\right)\!+\!\frac{2}{\pi}\sum_{n=0}^\infty\frac{\alpha\cos (2n\!+\!1) x}{\alpha^2\!+\!(2n\!+\!1)^2}$$
ergo, by $(1)$
$$\begin{align}W(\alpha,x) &=\frac{2}{\pi}\sum_{n=0}^\infty\frac{\alpha\cos (2n+1) x}{\alpha^2+(2n+1)^2}=S(\alpha,x)-\frac{1}{2}S\left(\frac{\alpha}{2},2x\right) = \\
&=\cosh \alpha x \left(\coth \pi\alpha-\frac{1}{2}\coth \frac{\pi\alpha}{2}\right)-\frac{1}{2}\sinh \alpha x \quad ;\text{for}\quad x\in\left[0,\pi\right]\end{align}$$
this can be further simplified to :
$$W(\alpha,x)=\frac{1}{2}\frac{\sinh \alpha(\pi-2x)/2}{\cosh \alpha\pi/2}$$
Therefore
$$\sum_{n=0}^\infty\frac{\cos\left[(2n+1) x\right]}{\alpha^2+(2n+1)^2}= \frac{\pi}{4\alpha}\frac{\sinh \alpha(\pi-2x)/2}{\cosh \alpha\pi/2} \quad\quad\quad ;\text{for}\quad x\in\left[0,\pi\right] $$
Tracing back to the original series is done by taking $\alpha=\sqrt{24y}$ and $x=\frac{\pi}{6}$
Ergo :
$$\frac{48}{\sqrt3}\sum_{n=0}^\infty \frac{\cos\left[(2n+1)\pi/6\right]}{(2n+1)^2+24y}= \pi\sqrt{\frac{2}{y}}\frac{\sinh \pi\sqrt{8y/3}}{\cosh \pi\sqrt{6}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Area of circle (double integral and cartesian coordinates)? I know that the area of a circle, $x^2+y^2=a^2$, in cylindrical coordinates is
$$
\int\limits_{0}^{2\pi} \int\limits_{0}^{a} r \, dr \, d\theta = \pi a^2
$$
But how can find the same result with a double integral and only cartesian coordinates?
| Just as an alternative solution to the qbert answer.
Note that $|x|$, $|y|$ are not strictly less than $r$, but instead $|x| \leq r$, $|y| \leq r$. This can still insure that $r^2 \geq y^2$ and so $\sqrt{r^2-y^2}$ is real.
After the first integral
$$
\int_{-r}^r\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\mathrm dx\mathrm dy=
\int_{-r}^r2\sqrt{r^2-y^2}\mathrm dy
$$
Instead of the change of variable, you can remember the known integral:
$$\int \sqrt{r^2 - y^2} = \frac{1}{2} \left( r^2 \arcsin \frac{y}{r} + y \sqrt{r^2 - y^2} \right) + C$$
whose condition $|y| \leq |r|$ has just been mentioned and it is satisfied. The result follows almost immediately:
$$\int_{-r}^r 2 \sqrt{r^2 - y^2} \mathrm{d} y = \left[ r^2 \arcsin \frac{y}{r} + y \sqrt{r^2 - y^2} \ \right]_{-r}^r = r^2 \arcsin (1) - r^2 \arcsin (-1) = \\
= r^2 \frac{\pi}{2} + r^2 \frac{\pi}{2} = \pi r^2$$
So, using the cartesian coordinates, the only important observation is simply to consider the right extreme values for each variable.
Using the circumference equation $x^2 + y^2 = r^2$, you can choose $x$ as a function of $y$, obtaining $x = \pm \sqrt{r^2 - y^2}$, which will be the extreme values for $x$. Then, you let $y$ sweep from $-r$ to $r$, which are the extreme values for $y$. Of course, you can alternatively do vice-versa, with $y$ as function of $x$.
Unlike the cylindrical coordinates, there is no angular variation here, but only an horizontal variation between $- \sqrt{r^2 - y^2}$ and $\sqrt{r^2 - y^2}$ as regards $x$, and a vertical variation between $-r$ and $r$ as regards $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 1
} |
Fully simplify $\sqrt {14 + 8\sqrt {3}}$. Please help me with this problem: Fully simplify $\sqrt{14 + 8\sqrt {3}}$.
I've tried to assume that the radical is in the form of $a+b\sqrt{3}$, so $a+b\sqrt{3}=\sqrt {14 + 8\sqrt {3}}$, then I squared both sides and got $$a^2+3b^2+2ab\sqrt{3}=14+8\sqrt{3}.$$
So I assumed that $a^2+3b^2$ must be equalto $14$, and $2ab$ must be equal to $-8$. So $ab$ must equal to $-4$. So now I know that $a$ and $b$ must be factors of $-4$. But now I am stuck because the cases $a=-1$, $b=4$, $a=1$, $b=-4$, $a=4$, $b=-1$, and $a=-4$, $b=1$ do not work because of the fact that $a^2+3b^2$ must equal to $14$. (We have that $(-1)^2+3\cdot4^2=1+3\cdot16$ which is clearly not $14$, same goes with $a=1$, $b=-4$; $4^2+3(-1)^2=16+3=19$ which is also not $14$.)
So now I am out of ideas to simplify this expression, or maybe this is already fully simplified? Please help!
| This can be computed by a very simple $ $ Square Root Denesting Rule:
Simple Denesting Rule $\rm\ \ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $
$\ 14+8\sqrt 3\ $ has norm $= 4.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 2\,\ $ yields $\,\ 12+8\sqrt 3\:$
with $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{24}\, =\, 2\sqrt 6.\ \ \ \rm \color{brown}{Dividing\ it\ out}\ $ of above yields $\,\ \sqrt 8 + \sqrt 6$
Remark $\ $ Many more worked examples are in prior posts on this denesting rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove: $\log _{c+b}\left(a\right)+\log _{c-b}\left(a\right)=2\log _{c+b}\left(a\right)\cdot \log _{c-b}\left(a\right)$, where $a^2 + b^2 = c^2$ Could you help me proving this?
$$\log _{c+b}\left(a\right)+\log _{c-b}\left(a\right)=2\log _{c+b}\left(a\right)\cdot \log _{c-b}\left(a\right)$$
where $c$ is the length of the hypotenuse of a right triangle, and $a$ and $b$ are the lengths of the other sides.
I tried that but not sure if it helps and what to do next:
$$\frac{\log\left(a\right)}{\log\left(c+b\right)}+\frac{\log\left(a\right)}{\log\left(c-b\right)}=\frac{2\log^2\left(a\right)}{\log\left(c+b\right)\log\left(c-b\right)}$$
Thank you in advance!
| We have $$a^2+b^2=c^2$$ so that $$(c+b)(c-b)=c^2-b^2=a^2.$$ Take the logarithm of both sides $$\log(c+b)+\log(c-b)=2\log(a).$$
Divide by $\log(c+b)$:
$$\frac{\log(c+b)}{\log(c+b)}+\frac{\log(c-b)}{\log(c+b)}=2\frac{\log(a)}{\log(c+b)}$$
Simplify, using the fact that $\frac{\log(x)}{\log(y)}=\log_y(x)$:
$$1+\log_{c+b}(c-b)=2\log_{c+b}(a).$$
Multiply by $\log_{c-b}(a)$:
$$\log_{c-b}(a)+\log_{c-b}(a)\log_{c+b}(c-b)=2\log_{c+b}(a)\log_{c-b}(a).$$
Finally, $$\log_{c-b}(a)\log_{c+b}(c-b)=\frac{\log(a)}{\log(c-b)}\frac{\log(c-b)}{\log(c+b)}=\frac{\log(a)}{\log(c+b)}=\log_{c+b}(a).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Proving that ${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $ How can I prove that $${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $$
I tried the following:
We use the falling factorial power:
$$y^{\underline k}=\underbrace{y(y-1)(y-2)\ldots(y-k+1)}_{k\text{ factors}},$$
so that $\binom{y}k=\frac{y^{\underline k}}{k!} .$
Then
$${x +y+n- 1 \choose n} = \frac{(x +y+n- 1)!}{n! ((x +y+n- 1) - n)!} =
\frac{1}{n!}. (x +y+n \color{#f00}{-1})^{\underline n} $$
And
$$ {x+n-k-1 \choose n-k}{y+k-1 \choose k}$$
$$\frac{1}{(n-k)!}.(x+n-k-1)^{\underline{n-k}}.\frac{1}{k!}.(y+k-1)^{\underline{k}}$$
$$\frac{1}{k!.(n-k)!}.(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$
$$\sum_{k=0}^n{n \choose k}(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$
According to the Binomial-coefficients:
$$ ((x+n-k-1) + (y+k-1))^{\underline{n}}$$
$$ (x+y+n\color{#f00}{- 2})^{\underline{n}}$$
What is wrong ? und How can I continue? :/
| Using Negative Binomial Coefficients and Vandermonde's Identity, we get
$$
\begin{align}
\sum_{k=0}^n\binom{x+n-k-1}{n-k}\binom{y+k-1}{k}
&=\sum_{k=0}^n(-1)^{n-k}\binom{-x}{n-k}(-1)^k\binom{-y}{k}\tag{1}\\
&=(-1)^n\sum_{k=0}^n\binom{-x}{n-k}\binom{-y}{k}\tag{2}\\
&=(-1)^n\binom{-x-y}{n}\tag{3}\\
&=\binom{n+x+y-1}{n}\tag{4}
\end{align}
$$
Explanation:
$(1)$: Negative Binomial Coefficient conversion
$(2)$: algebra
$(3)$: Vandermonde's Identity
$(4)$: Negative Binomial Coefficient conversion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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} |
Expansion of $(1+\sqrt{2})^n$ I was asked to show that $\forall n\in \mathbb N$ there exist a $p\in \mathbb N^\ast$ such that $$(1+\sqrt{2})^n = \sqrt{p} + \sqrt{p-1}$$
I used induction but it wasn't fruitful, so I tried to use the binomial expansion of $(1+\sqrt{2})^n$ but it seems I lack some insight to go further.
Any hint is welcomed.
| Note:
I added some lines to
show this explicit formula:
$p
=1+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k-1}
$.
If
$(1+\sqrt{2})^n = \sqrt{p} + \sqrt{p-1}
$,
then
$(1+\sqrt{2})^{2n} = 2p-1+2\sqrt{p(p-1)}
$
so that
$\begin{array}\\
4p(p-1)
&=((1+\sqrt{2})^{2n} - (2p-1))^2\\
&=(1+\sqrt{2})^{4n}-2(1+\sqrt{2})^{2n} (2p-1)+(2p-1)^2\\
&=(1+\sqrt{2})^{4n}-2(1+\sqrt{2})^{2n} (2p-1)+4p^2-4p+1\\
\text{so that}\\
-1
&=(1+\sqrt{2})^{4n}-2(1+\sqrt{2})^{2n} (2p-1)\\
\implies\\
2p-1
&=\frac{(1+\sqrt{2})^{4n}+1}{2(1+\sqrt{2})^{2n} }\\
&=\frac12(1+\sqrt{2})^{2n}+\frac{1}{2(1+\sqrt{2})^{2n} }\\
&=\frac12(1+\sqrt{2})^{2n}+\frac12(-1+\sqrt{2})^{2n}
\qquad\text{ since} (1+\sqrt{2})(-1+\sqrt{2})=1\\
&=\frac12\sum_{k=0}^{2n} \binom{2n}{k}(\sqrt{2})^{2n-k}(1+(-1)^k)\\
&=\sum_{k=0}^{n} \binom{2n}{2k}(\sqrt{2})^{2n-2k}\\
&=\sum_{k=0}^{n} \binom{2n}{2k}2^{n-k}\\
&=\binom{2n}{2n}2^{n-n}+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k}\\
&=1+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k}\\
&=1+2\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k-1}\\
\text{so}\\
p
&=1+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k-1}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 4
} |
Find the $\sum\limits_{sym}ab$ maximum of the value Let $a,b,c,d,e\in (0,1)$ and such $$a+b+c+d+e=1$$
find the maximun of the value
$$S=ab+ac+ad+ae+bc+bd+be+cd+ce+de$$
I Conjecture the maximun is $\dfrac{2}{5}?$,such $a=b=c=d=e=\dfrac{1}{5}$,so $$S\le\dfrac{10}{25}=\dfrac{2}{5}?$$
right?Thanks,
| Since
$$4(a+b+c+d+e)^2-10(ab+ac+ad+ae+bc+bd+ce+cd+ce+de)=$$
$$=(a-b)^2+(a-c)^2+(a-d)^2+(a-e)^2+(b-c)^2+(b-d)^2+(b-e)^2+(c-d)^2+(c-e)^2+(d-e)^2\geq0$$
so the answer is $\frac{2}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$\epsilon$-$\delta$ proof, $\lim\limits_{x \to a}$ $\frac{1}{x}$ = $\frac{1}{a}.$ It's been a while since I've studied $\epsilon$-$\delta$ proofs so I'm trying to get a good understanding of how to go about solving this.
Working backwards using this as a reference: $\epsilon$-$\delta$ proof that $\lim\limits_{x \to 1} \frac{1}{x} = 1$.
Solve:
$\lim\limits_{x \to a}$ $\frac{1}{x}$ = $\frac{1}{a}.$
We assume:
$\lim_{x\to a}f(x)=\ell \neq 0$.
And have:
$\left|\frac{1}{f(x)}-\frac 1{\ell}\right|=\frac{|f(x)-\ell|}{|f(x)||\ell|}$
Now suppose that a = 4.
So:
$\left|\frac{1}{(\frac{1}{x})}-\frac 1{(\frac{1}{4})}\right|=\frac{|\frac{1}{x}-\frac{1}{4}|}{|\frac{1}{x}||\frac{1}{4}|}$
Which yields:
$\left|x-4\right|={|\frac{1}{x}-\frac{1}{4}|}*|{x|*|4|}$
I am confused as to how to go about solving for epsilon or delta at this point.
| Let $\epsilon>0$ be given. We have to find a number $\delta>0$ such that $\left|\frac{1}{x}-\frac{1}{a}\right|<\epsilon$ whenever $|x-a|<\delta$. But
$$
\left|\frac{1}{x}-\frac{1}{a}\right|=\left|\frac{a-x}{xa}\right|=\frac{|x-a|}{|xa|}.
$$
We find a positive constant $C$ such that $\frac{1}{|xa|}<C\Rightarrow\frac{|x-a|}{|xa|}<C|x-a|$, and we can make $C|x-a|<\epsilon$ by taking $|x-a|<\frac{\epsilon}{C}=\delta$. We restrict $x$ to lie in the interval $|x-a|<1$ which implies that $a-1<x<a+1$ so that
$$
\frac{1}{a-1}>\frac{1}{x}>\frac{1}{a+1}\Rightarrow\frac{1}{a(a-1)}<\frac{1}{xa}<\frac{1}{a(a+1)}\Rightarrow\frac{1}{|xa|}<\frac{1}{a(a+1)}.
$$
So $C=\frac{1}{a(a+1)}$ is suitable. Thus, we should choose $\delta=\min\left\{1,\epsilon\cdot a(a+1)\right\}$.
To verify this choice of $\delta$, consider the following:
Given $\epsilon>0$, we let $\delta=\min\left\{1,\epsilon\cdot a(a+1)\right\}$. If $|x-a|<\delta$, then $|x-a|<1\Rightarrow\frac{1}{|xa|}<\frac{1}{a(a+1)}$ and also $|x-a|<\epsilon\cdot a(a+1)$. Hence,
$$
\left|\frac{1}{x}-\frac{1}{a}\right|=\frac{|x-a|}{|xa|}<\frac{1}{a(a+1)}\cdot[\epsilon\cdot a(a+1)]=\epsilon,
$$
as desired. $\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
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When $abc=1$ find the minimum of $a^2+b^2+c^2$. When $abc=1$ find the minimum of $a^2+b^2+c^2$.
My attempt:I put $\frac{1}{bc}$ for $a$ and then I used AM-GM inequality.
$(\frac{1}{bc})^2+b^2+c^2\ge 3$
But finally we should get the answer $a+b+c$.
Where am I wrong?
| AM-GM says:
$$\dfrac{a^2+b^2+c^2}3\ge\sqrt[3]{a^2b^2c^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878862",
"timestamp": "2023-03-29T00:00:00",
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How do I determine the general solution of $L=\frac{\partial ^2}{\partial t^2}$? How do I determine the general solution of $L=\frac{\partial ^2}{\partial t^2}$? $L\begin{pmatrix}x\\ y\\ z\end{pmatrix}=A\begin{pmatrix}x\\ y\\ z\end{pmatrix}$. $A=\begin{pmatrix} 4 &0 &2 \\ 0&4 &6 \\ 2& 6 &1 \end{pmatrix}$. I calculated eigenvalues and those are -4,4,9, but what is next that I should do?
| Let $X(t) = \pmatrix{x(t) \\y(t)\\z(t)} = \pmatrix{x_1(t) \\ x_2(t) \\ x_3(t)} $. The equation is equivalent to $X''(t) = AX(t)$.
As you calculated, the eigenvalues of $A$ are $-4, 4$ and $9$. Three associated eigenvectors are
$$ \begin{pmatrix}-\frac{1}{4}\\ -\frac{3}{4}\\ 1\end{pmatrix},\:\begin{pmatrix}-3\\ 1\\ 0\end{pmatrix},\:\begin{pmatrix}\frac{2}{5}\\ \frac{6}{5}\\ 1\end{pmatrix}. $$
Thus, one has
$$A = \underbrace{\begin{pmatrix}-\dfrac{1}{4}&-3&\dfrac{2}{5}\\ -\dfrac{3}{4}&1&\dfrac{6}{5}\\ 1&0&1\end{pmatrix}}_{P} \underbrace{\begin{pmatrix}-4&0&0\\ 0&4&0\\ 0&0&9\end{pmatrix}}_D P^{-1}.$$
Let $Y(t) = \pmatrix{y_1(t)\\y_2(t)\\y_3(t)} = P^{-1}X(t) \Leftrightarrow X = PY$, hence $Y'' = P^{-1}X''$ and $X'' = PY''$.
$$ \begin{align} X'' & = AX \\ PY'' & = APY \\ Y'' & = P^{-1}APY \\ Y'' &= DY \tag{*}\end{align}$$
The new system (*) is equivalent to
$$ \cases{y_1''(t) = -4y(t) \\ y_2''(t) = 4y_2(t) \\ y_3''(t) = 9y_3(t)} $$
Using for instance each equation's characteristic polynomial, one can easily find its solution:
$$ \cases{y_1(t) = c_1 \cos(2 t) + c_2 \sin(2 t) \\ y_2(t) = c_3 e^{2t}+c_4 e^{-2t} \\ y_3(t) = c_5 e^{3t}+c_6 e^{-3t} } \qquad \text{with} \: c_1, c_2, c_3, c_4, c_5, c_6 \in \mathbb{R}.$$
Using $X = PY$, one can eventually get to the solution of the initial system:
$$\pmatrix{x_1(t) \\ x_2(t) \\ x_3(t)} = \begin{pmatrix}-\dfrac{1}{4}&-3&\dfrac{2}{5}\\ -\dfrac{3}{4}&1&\dfrac{6}{5}\\ 1&0&1\end{pmatrix} \pmatrix{y_1(t)\\y_2(t)\\y_3(t)},$$
which yields, using your initial notation,
$$\bbox[5px,border:2px solid green,lightgreen]{\cases{x(t) = -\frac{1}{4}\left(c_1 \cos(2t) + c_2 \sin(2t)\right) - 3 \left(c_3 e^{2t} + c_4 e^{-2t}\right) + \frac{2}{5} \left(c_5e^{3t} + c_6 e^{-3t}\right) \\ y(t) = -\frac{3}{4}\left(c_1 \cos(2t) + c_2 \sin(2t)\right) + c_3 e^{2t} + c_4 e^{-2t} + \frac{6}{5} \left(c_5e^{3t} + c_6 e^{-3t}\right) \\ z(t) = c_1 \cos(2t) + c_2 \sin(2t) + c_5e^{3t} + c_6 e^{-3t}}}.$$
Of course, to determine the constants $c_1, c_2, c_3, c_4, c_5$ and $c_6$, you'll need initial conditions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$ Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$
My attempt:First solve when $n$ is not infinity then put infinity in.
$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}{2^n}}$$
$$=2^{\frac{1}{2}}\cdot 2^{\frac{2}{4}}\cdot \dots\cdot 2^{\frac{n}{2^n}}$$
Now calculate the sum of the powers:
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}$$
$$=\frac{2^{n-1}+2\cdot2^{n-2}+3\cdot2^{n-3}+\dots+n\cdot2^0}{2^n}$$
Now calculate the numerator:
$$2^0+2^1+2^2+\dots+2^{n-1}=2^n-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-2}=2^{n-1}-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-3}=2^{n-2}-1$$
$$+$$
$$\vdots$$
$$+$$
$$2^0=2^1-1$$
$$=2^1+2^2+2^3+\dots+2^n-n=2^{n+1}-n-1$$
Now put the numerator on the fraction:
$$\frac{2^{n+1}-n-1}{2^n}=2-\frac{n}{2^n}-\frac{1}{2^n}$$
Now we can easily find $\lim_{n \to \infty}\frac{1}{2^n}=0$
Then we just have to find $\lim_{n \to \infty }\frac{n}{2^n}$, that by graphing will easily give us the answer zero.
That gives the total answer is $4$.
But now they are two problems:
1.I cannot find $\lim_{n \to \infty }\frac{n}{2^n}$ without graghing.
2.My answer is too long.
Now I want you to help me with these problems.Thanks.
| A standard thing to do to, is too use the commonly known geometric series formula, and set:
$$f(x)=\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}$$
The formula is not hard to prove, and it holds for $|x|<1$ (which can be seen by the ratio test for convergence) we take $x \in (-1,1)$.
Then by term by term differentiation we get:
$$f'(x)=\sum_{n=1}^{\infty} nx^{n-1}=\frac{1}{x}\sum_{n=1}^{\infty} nx^n=\frac{1}{(1-x)^2}$$
Hence we have:
$$\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}$$
And the case your interested about is $x=\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 5
} |
What is the square root of $3 + 2\sqrt{10}i$? I need to compute the square root of $3 + 2\sqrt{10}i$.
I know how to solve it, but for some reason I'm not getting the correct answer. I attempted to solve it like this:
$$
\sqrt{3 + 2\sqrt{10}i} = x + iy
\quad \longrightarrow \quad
3 + 2\sqrt{10}i = x^2 - y^2 +2xyi
$$
and so forth, but my answer isn't correct.
| Your "and so on" could go like this:
$$\begin{cases}x^2-y^2=3,\\2xy=2\sqrt{10}.\end{cases}$$
Then squaring and adding both,
$$x^4-2x^2y^2+y^4+4x^2y^2=(x^2+y^2)^2=49,$$ so that
$$x^2+y^2=\pm7.$$
Solving with the help of the first,
$$x^2=5,y^2=2\text{ or }x^2=-2,y=-5.$$
This leaves the possibilities
$$x=\pm\sqrt5,y=\pm\sqrt2.$$
By the second equation, we know the signs are synchronized, hence
$$\sqrt5+i\sqrt2\text{ or }-\sqrt5-i\sqrt2.$$
More generally,
$$\begin{cases}x^2-y^2=u,\\2xy=v.\end{cases}$$
yields
$$u^2=\frac12(\sqrt{v^2+u^2}+u),v^2=\frac12(\sqrt{v^2+u^2}-u),$$
and
$$x=\pm\sqrt{\frac12(\sqrt{v^2+u^2}+u)},
y=\pm\sqrt{\frac12(\sqrt{v^2+u^2}-u)},$$
where the sign of $xy$ must match the sign of $v$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
$a_1 + a_2 + \dots a_n = 1$ find min of $a_1^2 +\frac{a_2^2}{2} + \dots + \frac{a_n^2}n.$ Given $n$ numbers $a_1$, $\cdots$ and such that $a_1$, $a_2$, $\cdots$, $a_n > 0$ and their sum is $1$, I want to find the minimum value of
$$a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}.$$
I have tried using weighted AM-GM inequality, like this:
$$\frac{a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}}{a_1 + a_2 + \cdots + a_n } \geqslant \frac{a_1^{a_1} \cdots a_n^{a_n}}{2^{a_2} \cdots n^{a_n}}$$
but was unable to make progress on the right hand side.
Is there a better way to apply AM-GM inequality? Or is there some different way altogether to solve this?
| we have $$a_1^2+\frac{a_2^2}{2}+...\frac{a_n^2}{n}\geq \frac{(a_1+a_2+...+a_n)^2}{1+2+...+n}\geq \frac{1}{1+2+...+n}=\frac{1}{\frac{n(n+1)}{2}}=\frac{2}{n(n+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Is $\ 7!=5040\ $ the largest highly composite factorial? A highly composite number is a natural number $\ n\ge 1\ $ that has more divisors than any smaller natural number $\ m\ge 1$.
Checking the $\ 1000\ $ entries in the OEIS-sequence, I noticed that only the following factorials are highly composite :
$$[1, 2, 6, 24, 120, 720, 5040]$$
Is it known whether $\ 7!=5040\ $ is the largest highly composite factorial ?
The factorials have some conditions necessary for a number to be highly composite :
*
*They contain the first $k$ prime numbers as prime factors.
*The exponents are non-increasing.
*The exponent corresponding to the largest prime factor is $\ 1$.
So, it is not obvious whether the above list is complete.
| Suppose $n \ge 20$. Trivially, $n!$ is then divisible by $16$, so $\dfrac{13}{16}n!$ is an integer smaller than $n!$.
Let the prime factorization of $n!$ be $n! = 2^{e_1}3^{e_2}\cdots11^{e_5}13^{e_6}17^{e_7}\cdots p_r^{e_r}$.
Then the prime factorization of $\dfrac{13}{16}n!$ is $\dfrac{13}{16}n! = 2^{e_1-4}3^{e_2} \cdots11^{e_5}13^{e_6+1}17^{e_6}\cdots p_r^{e_r}$.
Then, the number of divisors of $n!$ is $\sigma_0(n!) = (e_1+1)(e_6+1)\displaystyle\prod_{k \neq 1,6}(e_k+1)$ and the number of divisors of $\dfrac{13}{16}n!$ is $\sigma_0(\dfrac{13}{16}n!) = (e_1-3)(e_6+2)\displaystyle\prod_{k \neq 1,6}(e_k+1)$.
Hence, $\sigma_0(\dfrac{13}{16}n!) > \sigma_0(n!)$ iff $(e_1-3)(e_6+2) > (e_1+1)(e_6+1)$ iff $e_1 > 4e_6+7$.
Using the well known formula for the largest power of a prime dividing a factorial, we have
$e_1 = \displaystyle\sum_{k = 1}^{\left\lfloor \log_2 n\right\rfloor}\left\lfloor \dfrac{n}{2^k} \right\rfloor > \displaystyle\sum_{k = 1}^{\left\lfloor \log_2 n\right\rfloor} \left(\dfrac{n}{2^k} - 1\right) \ge n - \dfrac{n}{2^{\left\lfloor \log_2 n\right\rfloor}} - \left\lfloor \log_2 n\right\rfloor \ge n - 2 - \log_2 n$.
Similarly, $e_6 = \displaystyle\sum_{k = 1}^{\left\lfloor \log_{13} n\right\rfloor}\left\lfloor \dfrac{n}{13^k} \right\rfloor \le \displaystyle\sum_{k = 1}^{\left\lfloor \log_{13} n\right\rfloor}\dfrac{n}{13^k} \le \dfrac{n}{12}$. Hence, $4e_6+7 \le \dfrac{n}{3}+7$.
I'll leave it as an exercise to show that $\dfrac{n}{3}+7 < n - 2 - \log_2 n$ holds for all integers $n \ge 20$.
Therefore, for all integers $n \ge 20$, we have $4e_6+7 \le \dfrac{n}{3}+7 < n - 2 - \log_2 n < e_1$, and thus, $\sigma_0(\dfrac{13}{16}n!) > \sigma_0(n!)$.
So, $n!$ is not highly composite for $n \ge 20$. Now, it remains to check if any factorials between $7!$ and $20!$ are highly composite.
EDIT: From the list of the first 1000 highly composite numbers that OP mentioned, it appears that the $149$-th largest highly composite number is $\approx 1.49 \times 10^{17}$, while $19! \approx 1.22 \times 10^{17}$. None of the numbers $8!, 9!, \ldots, 19!$ appear in that list, so $7!$ is indeed the largest highly composite factorial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
help determining $a,b$ such that limit will exist and calculating it So i have a limit:
$$\lim_{x\rightarrow 0} \frac{e^{ax}+b\ln(2-x)-1}{x-\sin{x}}$$
And i need to determine values for $a,b$ so that the limit will exist, and calculate the limit then.
There is also a $hint$, that i need to use Taylor series for this, so it might be helpful, but i don't know if i should look for the whole function or just a part of fraction.
Any help would really be appreciated.
Thank you in advance.
| It is sure that Taylor series help a lot. Let us try
$$e^{ax}=1+a x+\frac{a^2 x^2}{2}+\frac{a^3 x^3}{6}+O\left(x^4\right)$$ $$\log(x-2)=\log (2)-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{24}+O\left(x^4\right)$$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ Replacing $$\frac{e^{ax}+b\ln(2-x)-1}{x-\sin{x}}=\frac{b \log (2)+x \left(a-\frac{b}{2}\right)+x^2
\left(\frac{a^2}{2}-\frac{b}{8}\right)+x^3
\left(\frac{a^3}{6}-\frac{b}{24}\right)+O\left(x^4\right)}{\frac{x^3}{6}+O\left(x^4\right)}$$ In order the limit exists, the constant term and the coefficients of the first and second powers of $x$ in numerator must be $0$. Then $???$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How can I get from one factored form to another? This came from trying to factor $x^6 + 1$. We know that
$x^4 - x^2 + 1 = (x^4 +2 x^2 + 1) -3x^2 = (x^2+1)^2 - 3x^2 = (x^2 - \sqrt 3 x + 1)(x^2 + \sqrt 3 x + 1)$
I expected that if I factored the $x^4 - x^2 + 1$ completely into $4$ factors and then regrouped them $2$ by $2$, I could arrive at $(x^2 - \sqrt 3 x + 1)(x^2 + \sqrt 3 x + 1)$
$$\left(x+ \sqrt {\dfrac {1 + \sqrt 3 i}{2}}\right) \left(x+ \sqrt {\dfrac {1 - \sqrt 3 i}{2}}\right) \left(x- \sqrt {\dfrac {1 + \sqrt 3 i}{2}}\right) \left(x- \sqrt {\dfrac {1 - \sqrt 3 i}{2}}\right)$$
But that didn't happen. No matter how I group them into quadratics, I always end up with a complex coefficient. How can I get $(x^2 - \sqrt 3 x + 1)(x^2 + \sqrt 3 x + 1)$ from the $4$ factors?
| Solve the equation
$$
x^6=-1
$$
using polar coordinates: $-1=e^{\pi i}$, so
$$
x_k=e^{\frac{\pi+2\pi k}{6}i}=e^{\frac{(2k+1)\pi}{6}i},
$$
for $k\in\{0,1,2,3,4,5\}$. Therefore
$$
\begin{split}
x^6+1 &=\prod_{k=0}^{5}\left(x-e^{\frac{(2k+1)\pi}{6}i}\right)\\
&=\textstyle(x-\frac{\sqrt{3}}{2}-\frac{1}{2}i)(x-i)(x+\frac{\sqrt{3}}{2}-\frac{1}{2}i)
(x+\frac{\sqrt{3}}{2}+\frac{1}{2}i)(x+i)(x-\frac{\sqrt{3}}{2}+\frac{1}{2}i)
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sqrt{2}+\sqrt{3}$ is irrational. Problem: Prove that $\sqrt{2}+\sqrt{3}$ is irrational.
The book where I encountered this problem had the following hint:
We make a polynomial with integer coefficients called $f(x)$ that $f(\sqrt{2}+\sqrt{3})=0$. (Why?)
Accepting this I solved the problem like this:
If $x=\sqrt{2}+\sqrt{3}$ then $x^2=5+2\sqrt{6}$ and so $(x^2-5)^2=24$ thus:
$$x^4-10x^2-1=0$$
But I want to know the reason that we should do this.
| Directly:
If
$r = \sqrt{2}+\sqrt{3}$
is rational,
$\begin{array}\\
\dfrac1{r}
&=\dfrac1{\sqrt{2}+\sqrt{3}}\\
&=\dfrac1{\sqrt{2}+\sqrt{3}}\dfrac{-\sqrt{2}+\sqrt{3}}{-\sqrt{2}+\sqrt{3}}\\
&=-\sqrt{2}+\sqrt{3}\\
\end{array}
$
is also rational.
Adding and subtracting these,
$\sqrt{2}$ and $\sqrt{3}$
are rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
The number of all 3-digit numbers abc for which abc + ab+bc + ac+a+b+c = 29 The number of all 3-digit numbers abc (in base 10) for which $\ abc+ab+bc+ac+a+b+c = 29$ is
(A) 6
(B) 10
(C) 14
(D) 18
My working:
$\ ab (c + 1) +b (c + 1) + a(c + 1) + c + 1 = 30$
$\ (a + 1) (b +1) (c+1) = 30 $
$\ 9>a>0$
$\ 0\le b,c<9$
The problem I am facing:
I don't know how to figure out the no. of solutions for which this is true.
| The prime fctorisation of $30$ is $5\times3\times2$. The only ordered 3-uples for $a+1,b+1$ and $c+1$ where $a+1 \in [2,10]$ and $b+1,c+1 \in [1,10]$ are : $$(2,3,5),(2,5,3),(3,1,10),(3,2,5),(3,5,2),(3,10,1),(5,1,6),(5,2,3),(5,3,2),(5,6,1),(6,1,5),(6,5,1),(10,1,3),(10,3,1)$$
So the solutions are : $$124,142,209,214,241,290,405,412,421,450,504,540,902,920$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
solve $\tan{x} = \tan{3x}$ I'm asked to solve $\tan{x} = \tan{3x}$
Here's my attempt:
$$\tan{x} = \tan{3x}$$
$$\tan{x} = \tan{(x + 2x)}$$
$$\tan{x} = \frac{\tan{x} + \tan{2x}}{1-\tan{x}\tan{2x}}$$
Recall the identity:
$$\tan{2x} = \frac{2\tan{x}}{1-\tan^2{x}}$$
So then we have:
$$\tan{x} = \frac{\tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}}{1-\tan{x}\frac{2\tan{x}}{1-\tan^2{x}}}$$
$$\tan{x} - \tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$
$$\tan^2{x} = -1$$
This does obviously not compute. Why is my way wrong and how can I go about solving it?
| Why do you try to make things more complex than they are? Formally,
\begin{align*}\tan x=\tan 3x&\iff 3x\equiv x\mod \pi \\
&\iff 2x\equiv0\mod \pi
\iff x\equiv 0\mod\frac\pi2.
\end{align*}
However, $\tan x$ is defined if and only if $x\not\equiv\dfrac\pi2\mod\pi$, so the effective solutions are
$$x\equiv 0\mod \pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Expected position of a random walk with a biased coin directly after two consecutive heads are observed Suppose a coin is tossed repeatedly and independently, and the probability of observing a heads on any toss is 0.6. Now suppose that a one-dimensional simple random walk $\{\sigma_1,\sigma_2,\cdots\}$ is designed based on the coin tossing. For $i=1,2,\cdots,$ a coin is tossed and
$$
\sigma_i =
\left\{
\begin{array}{ll}
1, & \text{if a head is observed} \\
-1, & \text{if a tails is observed} \\
\end{array}
\right.
$$
The random walk starts at the origin. Compute the expected position of the random walk when a string of 2 consecutive heads is observed for the first time. That is compute
$$ E\left(\sum_{i=1}^T \sigma_i\right) $$
where $\{T=i \in \mathbb{N}: \text{A string of 2 consecutive heads is observed for the first time after toss i}\} $.
So I figured I should first compute T, the expected number of flips to get two consecutive heads.
Basically, I did this by conditioning as shown below:
$$E(T) = (1+E(T))\frac{2}{5}+(2+E(T)\frac{3}{5}\frac{2}{5}+2\frac{3}{5}\frac{3}{5}$$
Solving for E(T) I then get E(T) = $\frac{40}{9}$.
Thus, I'm assuming when I calculate $ E\left(\sum_{i=1}^T \sigma_i\right) $, have T=5 since it is the next natural number.
Now even with this assumption I am stuck on what to do next. I guess I could just brute force it by writing out the terms of the sum from the standard expectation formula, but I think there might be an easier way. Any insight would be appreciated.
| Let $E_{\text{prev}=H}$ be the expected number of total flips to get two consecutive heads, given that the previous flip was a head. Let $E_{\text{prev}\neq H}$ be the expected number of total flips to get two consecutive heads, given that the previous flip was not a head.
$E_{\text{prev}=H} = \frac{6}{10}(1 + 0) + \frac{4}{10}(1 + E_{\text{prev}\neq H})$
$E_{\text{prev}\neq H} = \frac{6}{10}(1 + E_{\text{prev}=H}) + \frac{4}{10}(1 + E_{\text{prev}\neq H})$
We use $E_{\text{prev}\neq H}$ to get the total number of flips. Labeling $E_{\text{total}} = E_{\text{prev}\neq H}$ and combining equations, we have:
$E_{\text{total}} = \frac{6}{10}(1 + \frac{6}{10}(1 + 0) + \frac{4}{10}(1 + E_{\text{total}})) + \frac{4}{10}(1 + E_{\text{total}})$
Solving for $E_{\text{total}}$ we get $\frac{40}{9}$.
However, what we really need is the expected number of heads and the expected number of tails during this time, as this will tell us the expected value of the walk. Since the equation above counts both heads and tails, we can modify the equation to exclude what we don't want to count.
The expected number of heads:
$E_{\text{heads}} = \frac{6}{10}(\color{blue}1 + \frac{6}{10}(\color{blue}1 + 0) + \frac{4}{10}(\color{red}0 + E_{\text{heads}})) + \frac{4}{10}(\color{red}0 + E_{\text{heads}})$
The expected number of tails:
$E_{\text{tails}} = \frac{6}{10}(\color{red}0 + \frac{6}{10}(\color{red}0 + 0) + \frac{4}{10}(\color{blue}1 + E_{\text{tails}})) + \frac{4}{10}(\color{blue}1 + E_{\text{tails}})$
The red designates coin flips that have been removed from the equations, and blue designates the flips we want to count. We get $\frac{8}{3}$ heads and $\frac{16}{9}$ tails on average.
Then the expected value of the walk is $\frac{8}{3}- \frac{16}{9} = \frac{8}{9}$.
Another way to look at it: Since the probability distribution of a single flip also extends to the whole (due to the flips being independent and identically distributed), then if we know that the total number of flips is $\frac{40}{9}$ on average, then $\frac{6}{10}$ of these flips will be heads, with the rest tails. Therefore we get $\frac{40}{9}(\frac{6}{10} - \frac{4}{10}) = \frac{8}{9}$ again for the expected value of the walk.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trying to find an explicit sum of an infinite series: $\sum_{n=1}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n }$ I have the following series
$$\sum_{n=1}^{\infty}\frac{(-1)^n }{(2n+1)3^n}$$
I am trying to find an explicit sum. I know this looks like $\arctan x = \sum_{n=0}^{\infty} \frac{ (-1)^n x^{2n+1}}{2n+1} $. I do the following
$$ \sum_{n=1}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n } = \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n } - 1 =\sqrt{3} \sum_{n=1}^{\infty} \frac{ (-1)^n \sqrt{1/3}^{2n+1}}{(2n+1) } - 1$$
Since $(1/3)^n = (\sqrt{1/3})^2n = \sqrt{3} (\sqrt{1/3})^{2n+1}$. Thus, the sum is
$$ \sqrt{3} \sum_{n=1}^{\infty} \frac{ (-1)^n \sqrt{1/3}^{2n+1}}{(2n+1) } - 1 = \sqrt{3} \arctan(1/\sqrt{3}) = 1 = \boxed{\frac{ \sqrt{3} \pi }{6} - 1 }$$
Is this a correct solution? Do you guys a differenti method?
| It is correct. From
$$
\sum _{n=1}^{\infty } \frac{(-1)^n x^{2 n+1}}{2 n+1}=-x+\arctan x, \quad |x|<1,
$$ you get, with $x=\dfrac1{\sqrt{3}}$:
$$
\sum _{n=1}^{\infty } \frac{(-1)^n }{(2n+1)3^n}=\sqrt{3}\left(-\dfrac1{\sqrt{3}}+\arctan \dfrac1{\sqrt{3}}\right)=-1+\frac{\sqrt{3}\pi}6,
$$ using
$$
\arctan \dfrac1{\sqrt{3}}=\frac{\pi}6.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find all multiplicative continuous functions on $(0,\infty)$
If $x>0,y>0$ and if $f(xy)=f(x)f(y)$, then $f=\, ?$
I tried the problem. And got it as $f(x)^n=f(x^n)$.
But answer is $f(x)=x^n$. How?
$f$ is a continuous function.
| 4 obvious solutions spring to mind.
$f(x) = 0$ then $f(xy) = 0 = 0*0 =f(x)f(y) $.
$f(x) = 1$ then $f(xy) = 1 = 1*1 = f(x)f(y) $.
$f(x) = x$ then $f(xy) = xy = f(x)f(y)$.
And $f(x) = x^n$ for some $n \in \mathbb R$ then $f(xy) = x^ny^n = f(x)f(y)$.
These are all really the same class of answers. If $n = 1; f(x) = x = x^1$. If $n = 0; f(x) =1 = x^0$. $f(x) = 0$ can be considered an exception. However if we were to allowed $n$ in the extended reals. $n = -\infty; f(x) = 0 = x^{-\infty}$.
But are is this the only class of solutions?
let $b \ne 1; f(b) = c$. Then $c = b^{\log_b c}$. Let $n = \log_b c$. So $f(b) = b^n$.
(I am assuming $c > 0$.[footnote])
Let $x > 0$. Let $k = \log_b x$. So $x = b^k$.
$f(x) = f(b^k) = f(b)^k = b^{nk} = x^n$. And we are done
(I am assuming $f(x^k) = f(x)^k$ for all $k \in \mathbb R$. [footnote 2])
====
[footnote]
If $f(b) = 0$ then $f(1) = f(b)f(1/b) = 0$. S0 $f(x) = f(x)f(1) = 0$ for all $x$ and $f(x) = 0$ is one of the cases we discussed and we can assume we are trying for other cases. So we may assume $f(b) \ne 0$. Furthermore no $x > 0$ is such $f(x) = 0$.
$f(x) = f(x)f(1)$ so $f(1) = 1 > 0$. If $f(b) < 0$ then by intermediate value theorem there exist a value $c$ between $b$ and $x$ where $f(c) = 0$ but we've shown that is not the case. So $f(x) >0$ for all $x$.
[footnote 2]
$f(x^n) = f(x*x*...) = f(x)^n$ and $f({x}^{1/n})^n = f(x)$ so $f({x}^{1/n}) = f(x)^{1/n}$. So $f(x^r)= f(x)^r; \forall r \in \mathbb Q$. As $f$ is continuous, $f(x^y) = f(x)^y$ for all $y \in \mathbb R$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving the Diophantine Equation $2 \cdot 5^n = 3^{2m} + 1$ over $\mathbb{Z}^+$.
Prove that $(m, n) = (1, 1)$ is the only solution for the Diophantine Equation $$2 \cdot 5^n = 3^{2m} + 1$$ where $(m, n) \in (\mathbb{Z}^+)^2$.
I've managed to prove that both $m$ and $n$ are odd seeing $\bmod 3\text{ and } 10$ respectively. Also, $\forall n \ge 1$, $10$ divides the LHS. I am not able to proceed from here. Any help would be appreciated.
| Here is an elementary solution. $n = 1$ gives the solution above. Now, suppose $ n > 1$. Thus, $9^m + 1 \equiv 0 \pmod {25}$. Looking at all the possible cases, we have $m = 10k + 5$ for integer $k$. We can check that there are no solutions for $m = 2, \ldots, 5$. Thus, $k > 0$. Theefore,
$$ 2 \cdot 5^n = 3^{2(10k + 5)} + 1 = 3^{10} \cdot 3^{20k} + 1.$$
Now $1181$ divides $3^{20} - 1$. Thus, $3^{20} \equiv 1 \pmod {1181}$ and $3^{10} \equiv - 1 \pmod{1181}$. Now taking the equation $\pmod{1181}$ gives
$$2 \cdot 5^n \equiv -1 \cdot 1 + 1 \equiv 0 \pmod {1181}$$ which has no solutions. Thus $(m,n) = (1,1)$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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System of Equations involving exponents I would like to solve the system:
$$2^{x+y^2}+2^{x^2+y}=8 $$
$$\sqrt x+ \sqrt y =2 $$
I have come up with a solution that is not rigorous, but somehow it gives the right solution:
Expanding the second equation:
$$x+y +2\sqrt {xy}=4$$
Squaring again
$$(x+y-4)^2 =4xy$$
$$x^2+y^2-2xy-8x-8y+16=0$$
$$2^{ x^2+y^2-2xy-8x-8y+16}=1$$
Rewriting the expression in the form:
$$2^{x^2+y^2+x+y} =2^{9x+9y+2xy-16}$$
Now I let $a(x,y)=2^{x^2+y}$ and $b(x,y)=2^{y^2 +x}$, and let $f(x,y)$ be the right hand side of the above equation, then we have:
$$a+b=8$$
$$ab=f(x,y)$$
Without justification, I use quadratic formula by treating $f(x,y)$ like a coefficient, I get:
$$a(x,y)=4 \pm \sqrt{16-f(x,y)}$$
$$b(x,y)=4 \mp \sqrt{16-f(x,y)}$$
Since $a(x,y)=a(y,x)$ based on the expression given from quadratic formula,
$$2^{x+y^2}=2^{x^2+y}$$
Factorize:
$$(x-y)(x+y-1)=0$$
$x=y$ or $x+y=1$
At this point it is easy to verify that $x=y=1$ is the only solution.
Question: Is my method of applying quadratic formula to a function valid? What are the only ways to solve this question?
| By quadratic-arithmetic inequality, we have $$(x+y)\geq \frac{(\sqrt{x}+\sqrt{y})^2}{2}=2.$$
Similarly, $$(x^2+y^2)\geq\frac{(x+y)^2}{2}\geq 2.$$
Also, by arithmetic-geometric inequality, we have $$2^{x+y^2}+2^{x^2+y}\geq 2\sqrt{2^{x+y^2}2^{x^2+y^2}}\geq 2\sqrt{2^{4}}=8.$$ Here equality holds, which is possible only if $\sqrt{x}=\sqrt{y}$.
Therefore, $x=y=1$ is the only solution.
| {
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Identity for $f(x) = \sum_{k=1}^{n}(x+k)^2$ I have a function $$f(x)=(x+1)^2+(x+2)^2 + \dots + (x+n)^2 = \sum_{k=1}^{n}(x+k)^2$$
for some positive integer $n$. I started wondering if there is an equivalent expression for $f(x)$ that can be calculated more directly (efficiently).
I began by expanding some terms to look for a pattern.
$$
(x^2+2x+1) + (x^2+4x+4) + (x^2+6x+9) + (x^2+8x+16) + \dots
$$
By regrouping the $x^2$, $x$, and constant terms, I can see that
$$
f(x) = \sum_{k=1}^{n}x^2 + \sum_{k=1}^{n}2kx + \sum_{k=1}^{n} k^2
$$
for which I've found some identities to get
$$
f(x) = n x^2 + n(n+1)x + \frac{1}{6}n(n+1)(2n+1)
$$
and simplifying some (attempting to make it computationally efficient)
$$
f(x) = n \left[ x^2 + (n+1)x + \frac{1}{6}(n+1)(2n+1) \right]
$$
$$
f(x) = n \left[ x^2 + (n+1) \left( x + \frac{2n+1}{6} \right) \right]
$$
Is this a particular type of summation (maybe just exponential?), and if so is there a standard way to write it?
Along those lines, is there a more direct derivation than what I've attempted here, perhaps using an identity I don't know?
| Given the identity:
$$ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
Write
$$ \sum_{k=1}^{n}(x+k)^2 = \sum_{k=x+1}^{n+x}k^2 = \sum_{k=1}^{n+x}k^2 - \sum_{k=1}^{x}k^2$$
$$ \sum_{k=1}^{n}(x+k)^2 = \frac{(n+x)(n+x+1)(2n+2x+1)}{6} - \frac{x(x+1)(2x+1)}{6}$$
Which reduces to:
$$ \sum_{k=1}^{n}(x+k)^2 = n \left[ x^2 + (n+1) \left( x + \frac{2n+1}{6} \right) \right]$$
| {
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If $a$ and $b$ are prime to each other and $n$ is prime then$\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor. Question is:
If $a$ and $b$ are prime to each other and $n$ is prime then prove that $\frac{a^n+b^n}{a+b}$ and
$a+b$ have no common factor unless $a+b$ is a multiple of $n$.
This is what I thought so far:
$a^n$ is also prime to $b^n$ because of the theorem (If $a$ is prime to $b$ then $a^n$ is also prime to $b^n$), Also $a^n$ and $b^n$ is a divisor of $(a+b)$ therefore by theorem (If $a$ is prime to $b$, and each of these numbers is a divisors of $N$, then $ab$ is a divisor of $N$) $a^nb^n$ is a divisor of $(a+b)$.
Thereafter no idea, How should I proceed further if I'm thinking in a right way and If not, How should I prove this?
Source: Higher Algebra by Barnard and Child
| First, when is $\frac {a^n + b^n}{a+b}$ an integer.
Or when does $(a+b)|(a^n + b^n)$
$n$ is odd
$\frac {a^n + b^n}{a+b} = a^{n-1} - a^{n-2}b + a^{n-3}b^2 -\cdots ab^{n-2} + b^{n-1}$
$n$ is even:
Apply the euclidean algorithm.
$(a+b)|(a^n + b^n) \iff (a+b)| (a^n + b^n)+k(a+b)$
$(a^n + b^n) - a^{n-1} (a+b) + a^{n-2}b (a+b)\cdots$
$(a^n+b^n) - (a+b)\sum_\limits{i=0}^{n-1} (-1)^i a^{n-1-i} b^i = 2b^n$
$(a+b)|(a^n + b^n) \iff (a+b)|2b^n$
$(a+b) = 2,$ or $a = 0$ or $b = 0$ or $a|b^n$
since $a,b$ are co-prime.
$a\ne0, b\ne0$ and $a$ does not divide $b$
$a = b = 1$ is a degenerate case.
In this case, 2 does not divide 1.
n is odd:
again apply the Euclidean algorithm.
$\sum_\limits{i=0}^{n-1} (-1)^i a^{n-1-i}b^i + (a+b)\sum_\limits{i=1}^{n-1} (-1)^i ia^{n-1-i}b^{i-1} = nb^{n-1}$
$(a+b)|\frac {a^n + b^n}{a+b} \implies (a+b)|nb^{n-1}$
as dicussed above $(a+b)$ does not divide $b^{n-1}$
$(a+b)|\frac {a^n + b^n}{a+b} \implies (a+b)|n$
| {
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Prove that $\sum\limits_{j=k}^n\,(-1)^{j-k}\,\binom{j}{k}\,\binom{2n-j}{j}\,2^{2(n-j)}=\binom{2n+1}{2k+1}$. In an attempt to answer this thread, I discovered an identity involving binomial coefficients. However, I am not able to find a proof. All tricks are welcome.
Let $n$ and $k$ be nonnegative integers with $k\leq n$. Prove that $$\sum\limits_{j=k}^n\,(-1)^{j-k}\,\binom{j}{k}\,\binom{2n-j}{j}\,2^{2(n-j)}=\binom{2n+1}{2k+1}\,.$$
| Suppose we seek to show that
$$\sum_{q=k}^n (-1)^{q-k} {q\choose k} {2n-q\choose q}
2^{2(n-q)} = {2n+1\choose 2k+1}.$$
Re-index to get
$$\sum_{q=0}^{n-k} (-1)^{q} {q+k\choose k} {2n-k-q\choose k+q}
2^{2(n-k-q)}.$$
Introduce
$${2n-k-q\choose k+q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k-2q+1}} \frac{1}{(1-z)^{k+q+1}}
\; dz.$$
Observe that this vanishes when $q\gt n-k$ so it provides range
control and we may extend $q$ to infinity, getting for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}} \frac{1}{(1-z)^{k+1}}
\sum_{q\ge 0} (-1)^q {q+k\choose q}
\frac{z^{2q}}{(1-z)^q} 2^{2(n-k-q)}
\; dz
\\ = \frac{2^{2(n-k)}}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}} \frac{1}{(1-z)^{k+1}}
\frac{1}{(1+z^2/(1-z)/2^2)^{k+1}}
\; dz
\\ = \frac{2^{2(n-k)}}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}}
\frac{1}{(1-z+z^2/2^2)^{k+1}}
\; dz
\\ = \frac{2^{2n+2}}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}}
\frac{1}{(4-4z+z^2)^{k+1}}
\; dz
\\ = \frac{2^{2n+2}}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}}
\frac{1}{(z-2)^{2k+2}}
\; dz
\\ = \frac{2^{2n-2k}}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}}
\frac{1}{(z/2-1)^{2k+2}}
\; dz
\\ = \frac{(-1)^{2k+2} 2^{2n-2k}}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}}
\frac{1}{(1-z/2)^{2k+2}}
\; dz.$$
This is
$$2^{2n-2k} [z^{2n-2k}] \frac{1}{(1-z/2)^{2k+2}}
= 2^{2n-2k} {2n-2k+2k+1\choose 2k+1} \frac{1}{2^{2n-2k}}
\\ = {2n+1\choose 2k+1}$$
which is the claim.
Remark. We discover having reached the end of the computation
that we never needed to substitute the variable in the integral, which
means it could be done using formal power series and the
coefficient-of method.
Remark II. For the geometric series to converge we need with $\epsilon\ll 1$ (pole at $z=1$) that $|z^2/(1-z)/4| \lt 1$ which requires $\epsilon^2/(1-\epsilon)/4\lt 1.$ With $\epsilon = 1/Q$ this is $1/Q/(Q-1)/4\lt 1$ which is seen to hold for $Q\ge 2.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find n-th derivative of $f(x)=\sqrt{1+\cosh{x}}$ $f(x)=\sqrt{1+\cosh{x}}$
$f'(x)=\frac{\sinh{x}}{2\sqrt{\cosh{x}+1}}$
$f''(x)=\frac{(\cosh{\frac{x}{2}})^4}{(\cosh{x}+1)^\frac{3}{2}}$
I wanted to use Leibnitz rule, but i can't seem to find any similarities between derivatives. I don't know how to expand $\cosh{x}$ as Taylor series. These are the only two general methods i know for finding upper derivatives. I think there must be some trick. Anyone has any ideas?
This is what i got so far. $$f(x)=\frac{\sum_{n=0}^\infty ( \frac{(\frac{x}{2})^n}{n!}+\frac{(\frac{-x}{2})^n}{n!})}{\sqrt{2}}=\frac{\sum_{k=0}^\infty \frac{2(\frac{x}{2})^{2k}}{(2k)!}}{\sqrt{2}}$$
Dunno what to do now
New approach.
$f(x)=\sqrt{2}\cosh{\frac{x}{2}}$
$f'(x)=\frac{\sqrt{2}}{2}\sinh{\frac{x}{2}}$
$f''(x)=\frac{\sqrt{2}}{4}\cosh{\frac{x}{2}}=\frac{1}{4}f(x)$
For $n$ even
$f^{(n)}=\frac{1}{2^n}f(x)=\frac{\sqrt{2}}{2^n}\cosh{\frac{x}{2}}$
For $n$ odd
$f^{(n)}=\frac{\sqrt{2}}{2^n}\sinh{\frac{x}{2}}$
| $\cosh x = \frac 12 (e^x+ e^{-x})\\
\cosh^2 x = \frac 14 (e^{2x}+ e^{-2x}+1) = \frac 12 (\cosh 2x + 1)\\
2cosh^2 \frac x2 = \cosh x + 1\\
\sqrt 2 \cosh \frac x2 = \sqrt {\cosh x + 1}\\
f(x) = \sqrt 2 \cosh \frac x2$
Just wanted to verify.
The derivatives:
$f(x) = \sqrt 2 \cosh \frac x2\\
f'(x) = \frac {\sqrt 2}{2} \sinh \frac x2\\
f''(x) = \frac {\sqrt 2}{4} \cosh \frac x2\\
f^{(2k)}(x) =\frac {\sqrt 2}{2^{2k}} \cosh \frac x2\\
f^{(2k+1)}(x) =\frac {\sqrt 2}{2^{2k+1}} \sinh \frac x2$
| {
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Show that $n(2n + 1)(7n + 1)$ is divisible by 6 for all integers $n$ and find all integers $n$ such that $n(2n + 1)(7n + 1)$ is divisible by 12. a) Show that $n(2n + 1)(7n + 1)$ is divisible by 6 for all integers $n$.
b) Find all integers $n$ such that $n(2n + 1)(7n + 1)$ is divisible by 12.
I tried expanding but it doesn't really help much. Thanks in advance!
| $$n(2n+1)(7n+1)\equiv n(2n+1)(n+1)\equiv 0\pmod{2},$$
because exactly one of $n$, $n+1$ is even.
$$n(2n+1)(7n+1)\equiv n(-n+1)(n+1)\pmod{3}$$
$$\equiv n(-1)(n-1)(n+1)\equiv 0\pmod{3},$$
because exactly one of $n$, $n-1$, $n+1$ is divisible by $3$.
Therefore $2\cdot 3\mid n(2n+1)(7n+1)$ for all $n\in\mathbb Z$.
The condition $2^2\cdot 3\mid n(2n+1)(7n+1)$ is therefore equivalent to $2^2\mid n(2n+1)(7n+1)$.
Since $2n+1$ is odd for all $n\in\mathbb Z$, this is equivalent to $2^2\mid n(7n+1)$ (by common sense, or you can see Generalized Euclid's lemma. We have $\gcd\left(2^2,2n+1\right)=1$ for all $n\in\mathbb Z$).
$$n(7n+1)\equiv n(-n+1)\equiv -n(n-1)\pmod{2^2}$$
is equivalent to $0$ if and only if either $2^2\mid n$ or $2^2\mid n-1$ (because $n$, $n-1$ are coprime).
The answer is: all $n\in\mathbb Z$ such that $n\equiv \{0,1\}\pmod{4}$.
| {
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Why we use $2\cos\theta=\sqrt3$ to find $x^{72}+x^{66}$ given that $x+\frac1x=\sqrt3$? $x + \frac{1}{x} = \sqrt 3$ then find the value of $x^{72}+x^{66}$.
My teacher said that in this case, where $x + \frac{1}{x}$ is equal to any value which is less than $2$, like $1$ or $\sqrt{3}$, just put this value in equal of $2 \cos \theta$ and after you will get value of $\theta$ after that divide the value $180$ by that theta value and after getting the result put that result in equation $x^n + 1=0$, as "$n$".
Please explain what is basic concept behind this.
| We have to solve the equation $\ x+\frac{1}{x}=a$. This leads to the equation
$$x^2-ax+1=0$$
If $\ 0\le a<2$, the solutions must be complex.
If $u+vi$ is one of the solutions, the other one is $u-vi$. So, the absolute values of the solutions are equal and multiply to $1$. Therefore, every solution of this equation has absolute value $1$.
If the angle that is formed by a solution is a divisor of $180$ , we have $k\cdot\theta=180$. That means, every solution of the equation satisfies $x^k=-1$.
In the given example, the angle is $30°$, that means $k=6$, so every solution satisfies $x^6=-1$
This immediately shows $x^{72}+x^{66}=0$
What remains is : Why do we take $\ 2cos\theta=a$ ?
The sum of the solutions must be $a$, so the real part of both solutions is $u=\frac{a}{2}$. The absolute value of the solutions is $1$, so we have $u=cos(\theta)$ and therefore $a=2cos(\theta)$.
| {
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Absolute Extrema Multivariable Help Please I been struggling with question for quite a while now and wanted some help. The question is to find the absolute max and min of the function $f(x,y) = x^3+y^3-3x-3y+2$, bounded by the region $x^2+y^2 \le 1$. I first found the partial derivatives and solved them by setting them zero. However, the values are inconclusive since the coordinates lie outside the region. This is where I get stuck. I tried by substituting $x^2=1-y^2$, but I keep getting the wrong answer. Then I tried using parameterization by setting $x=\cos t$ and $y=\sin t$, but I continue to get the wrong answer. Can you please help me.
Thank you
P.S. Sorry for the equations I am new to the site and this is my first question.
| Let $x = r\cos \theta, y = r\sin \theta$ with $0 \le r \le 1, 0 \le \theta \le 2\pi$. Also, let $u = \cos \theta + \sin \theta, v = \sin \theta \cos \theta$. Then:$ |u| \le \sqrt{2}, x^3 + y^3 = r^3(\cos^3 \theta + \sin^3 \theta)=r^3(\cos \theta+\sin \theta)(1-\sin \theta \cos \theta)= r^3u(1-v)$. And we have: $u^2 - 2v = 1\implies v = \dfrac{u^2-1}{2}\implies x^3+y^3 = \dfrac{r^3u(3-u^2)}{2}\implies f(x,y) = \dfrac{1}{2}r^3u(3-u^2) - 3ru+2= g(r,u) + 2$. We can find the min and max values of $g$, and then add $2$ to get the min and max values of $f$. We use the traditional "critical points" method to find the relative extrema for $g$. We have: $\dfrac{\partial g}{\partial r} = 0 \iff \dfrac{3r^2u(3-u^2)}{2} - 3u = 0$, and $\dfrac{\partial g}{\partial u} = 0 \iff \dfrac{r^3(3-3u^2)}{2}-3r = 0$. Solving this system of equations in $r,u$ we have the solutions: $(u,r) = (0,0), (0,\sqrt{2}), (0,-\sqrt{2})$ of these only the point $(0,0)$ is a valid solution since $0 \le r \le 1$. Thus the only critical point for $g$ is the origin $(0,0)$, and $g(0,0) = 0$. We now check the boundary points that are points on the $4$ sides of the rectangular in the $u$-axis (horizontal), and $r$-axis (vertical) whose vertices are: $A = (-\sqrt{2}, 0), B = (-\sqrt{2}, 1), C = (1,\sqrt{2}), D = (\sqrt{2}, 0)$. On the segment $AB$, we have $g(r,u) = g(r,-\sqrt{2})= 3\sqrt{2}r - \dfrac{r^3\sqrt{2}}{2}$, has $g'(r) = 3\sqrt{2}-\dfrac{3r^2\sqrt{2}}{2} > 0$. Thus on $AB$, $g_{\text{min}} = g(0,-\sqrt{2}) = 0, g_{\text{max}} = g(1,-\sqrt{2}) = \dfrac{5\sqrt{2}}{2}$, you can do the same for $BC$, $CD$, and $DA$, and the min of these values of $g$ together with $0$ is the global min, and the max values of these values together with $0$ is the global max of $g$, then add $2$ to them to get the global min and max for $f$.
Note: The method that you chose still works since if the critical point does not lie in the domain, then you just consider the boundary points which is points on the unit circle $x^2+y^2 = 1$, and you can use Lagrange Multiplier or the traditional critical points methods that both work !.
| {
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Rectifiability of space curve given rectifiability of projections Let $\gamma:[0,1] \to \mathbb{R}^3$ be rectifiable. Then the projections of $\gamma$ onto the planes perpendicular to the coordinate axes are rectifiable:
$$\int_{0}^{1} \left \| \dot{\gamma}(s) \right \| ds=\int_{0}^{1} \sqrt{x^2+y^2+z^2}ds<\infty$$
$$\sqrt{x^2+y^2+z^2} > \sqrt{x^2+y^2}= \left \| \dot{\gamma}_{xy} \right \|$$
$$\sqrt{x^2+y^2+z^2} > \sqrt{y^2+z^2}= \left \| \dot{\gamma}_{yz} \right \|$$
$$\sqrt{x^2+y^2+z^2} > \sqrt{z^2+x^2}= \left \| \dot{\gamma}_{zx} \right \|$$
with $\gamma_{ij}$ the projection on the $ij$-plane.
Does the converse hold?
| $\gamma:[a,b]\to\mathbb{R}$ is rectifiable if $$\sup_{P} \sum_{i=1}^n |\gamma(t_i)-\gamma(t_{i-1})|<\infty,$$ where $P=\{t_i\}$ is a partition of $[a,b].$
If the projection over each coordinate plane is rectifiable then the curve is rectifiable. It is enough to have in mind the inequality
$$\sqrt{x^2+y^2+z^2}\le \sqrt{2}(\sqrt{x^2+y^2}+\sqrt{x^2+z^2}+\sqrt{y^2+z^2}).$$ Assume $x^2\le y^2\le z^2.$ Then
$$\sqrt{x^2+y^2+z^2}\le \sqrt{2y^2+2z^2}=\sqrt{2} \sqrt{y^2+z^2},$$ from where the inequality follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Give an upper bound for $x^2 + y^2$ for $(x, y)$ on the ellipse $(x - y - a)^2 + y^2 = 1$ without using Lagrange multipliers Let $a \in \mathbb{R}$ and consider the ellipse $(x - y - a)^2 + y^2 = 1$. My goal is to maximize (or at least give an upper bound for) $x^2 + y^2$ on this ellipse. This can be solved with Lagrange multipliers however the calculations are quite messy and hard to simplify (mostly due to $a$ being any real number).
Are there any easier alternative ways that avoid using Lagrange multipliers, perhaps at the cost of showing instead there is a $C$ depending on $a$ such that $x^2 + y^2 \leq C$ for all $(x, y)$ on the ellipse?
| This doesn't avoid Lagrange multipliers, but you can let $z=x-y$ and then
try to find an upper bound for $(z+y)^2+y^2$ on the circle $(z-a)^2+y^2=1$:
1) If we maximize $(z+y)^2$ on the circle, we get
$\hspace{.3 in}2(z+y)=\lambda\cdot2(z-a)$ and $2(z+y)=\lambda\cdot2y$, so the maximum occurs for $z=a+y$,
and then $(z-a)^2+y^2=1\implies y=\pm\frac{1}{\sqrt{2}}$ and $z=a\pm\frac{1}{\sqrt{2}}$.
Therefore $(z+y)^2+y^2\le (a+\sqrt{2})^2+y^2\le a^2+2\sqrt{2}a+2+1=\color{green}{a^2+2\sqrt{2}a+3}$
| {
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Proof for an inequality problem using the rearrangement inequality I have found this problem in a book and would like someone to give me a good and elegant proof for it.
The problem is:
If $a_1,a_2,...,a_n$ are positive real numbers and $s=a_1+a_2+...+a_n$, then
$$\frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+...+\frac{a_n}{s-a_n} \ge \frac{n}{n-1}$$
The book does not provide solutions but only gives hints, which is that it can be solved like another problem using n variables
The problem is:
If $a,b,c$ are the lengths of the sides of a triangle, prove that
$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \ge 3$$
Again only hints are given and this is the solution I am proposing:
let $(b_1,b_2,b_3)=\frac{a}{b+c-a},\frac{b}{c+a-b},\frac{c}{a+b-c}$
Let $(a_1,a_2,a_3)=a,b,c$
By the rearrangement inequality we have,
$$a_1b_1+a_2b_2+a_3b_3 \ge a_2b_1 +a_3b_2+a_1b_3$$
$$a_1b_1+a_2b_2+a_3b_3 \ge a_3b_1 +a_1b_2+a_2b_3$$
$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \ge \frac{b}{b+c-a}+\frac{c}{c+a-b}+\frac{a}{a+b-c}$$
$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \ge \frac{c}{b+c-a}+\frac{a}{c+a-b}+\frac{b}{a+b-c}$$
Let $$Q = \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}$$
Now by adding the two inequalities we have,
$$2Q \ge \frac{b}{b}+\frac{b}{c}-\frac{b}{a}+\frac{c}{c}+\frac{c}{a}-\frac{c}{b}+\frac{a}{a}+\frac{a}{b}-\frac{a}{c}+\frac{c}{b}+\frac{c}{c}-\frac{c}{a}+\frac{a}{c}+\frac{a}{a}-\frac{a}{b}+\frac{b}{a}+\frac{b}{b}-\frac{b}{c}=6+\frac{a-b-a+b}{a}+\frac{a-c-a+c}{b}+\frac{b-a-b+a}{c}=6$$
$$Q \ge 3$$
I do understand the general idea but what is needed in the problem above is a generalization of the proof I provided. However I fail at producing one each time I try. Can someone please provide me with one that is simple and easy to read. Thanks.
UPDATE:
Thanks for the answers but I am looking for a proof using the rearrangement inequality in a similar way to the example.
| We need to prove that $\sum\limits_{i=1}^n\left(\frac{a_i}{s-a_i}-\frac{1}{n-1}\right)\geq0$ or $\sum\limits_{i=1}^n\frac{na_i-s}{s-a_i}\geq0$.
It's obvious that $\left(na_1-s,na_2-s,...,na_n-s\right)$ and $\left(\frac{1}{s-a_1},\frac{1}{s-a_2},...,\frac{1}{s-a_n}\right)$ are same ordered.
Thus, by Rearrangement (Chebyshov) we obtain:
$$n\sum\limits_{i=1}^n\frac{na_i-s}{s-a_i}\geq\sum\limits_{i=1}^n\left(na_i-s\right)\sum\limits_{i=1}^n\frac{1}{s-a_i}=0$$
Done!
| {
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Maxima and minima of $f(x)=\sin x+\cos x+\tan x+\arcsin x+\arccos x+\arctan x$
Let $f(x)=\sin x+\cos x+\tan x+\arcsin x+\arccos x+\arctan x$ . If $M$ and $m$ are maximum and minimum values of $f(x)$ then their arithmetic mean is equal to?
My Approach:
$\arcsin x + \arccos x =\frac{\pi}{2}$, so the equation becomes $\sin x + \cos x + \tan x + \frac{\pi}{2} + \arctan x$ . After this I differentiated this equation but did not came to any conclusion . Also I tried to solve the equation to make something relevant but the equation did not simplified . Please help...
| Hint: We have $f(x) = \sin x + \cos x + \tan x + \frac{\pi}{2} + \arctan x$ on $[-1,1]$. Now
$$ f'(x) = \cos x - \sin x + \sec^2 x + \frac{1}{1+x^2}. $$
On $[-1,1]$, we have $\cos x > 0$, $-\sin x > -1$, $\sec^2x\ge 1$, and $\frac{1}{1+x^2}>0$, so for all $x\in[-1,1]$ we have
$$ f'(x) > 0 - 1 + 1 + 0 = 0. $$
Thus $f$ is strictly increasing. What can you say about where it attains its maximum and minimum?
| {
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"timestamp": "2023-03-29T00:00:00",
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show that $\cos(\theta_{2}-\theta_{3})+\cos(\theta_{3}-\theta_{1})+\cos(\theta_{1}-\theta_{3})+1=0$ If $$a = \cos\theta_{1} + i\sin\theta_{1}, \\ b = \cos\theta_{2} + i\sin\theta_{2}, \\ c = \cos\theta_{3} + i\sin\theta_{3}$$ and $a+b+c=abc$,
then show that
$$\cos(\theta_{2}-\theta_{3})+\cos(\theta_{3}-\theta_{1})+\cos(\theta_{1}-\theta_{2})+1=0$$
I am confused with where to start. I did try to proceed with the De Moivre's Theorem but was not able to get hold of the required equation.
Kindly provide a hint.
Thanks in advance !
| Let $\bar{z}$ denote the complex conjugate of $z$. Multiplying $a+b+c = abc$ by $\bar{a}$, we get
\begin{align*}
1 + \bar{a}b + \bar{a}c = bc
\end{align*}
Similarly,
\begin{align*}
1+\bar{b}c+\bar{b}a &= ca\\
1+\bar{c}a + \bar{c}b & = ab
\end{align*}
Adding we get,
\begin{align*}
3 + (\bar{a}b +a\bar{b}+ \bar{b}c +b\bar{c}+ \bar{c}a + c\bar{a}) = ab+bc+ca
\end{align*}
Also,
from $a+b+c = abc$, we get $\bar{a} +\bar{b}+\bar{c} = \bar{a}\bar{b}\bar{c}$ and hence dividing throughout by $\bar{a}\bar{b}\bar{c}$ we get
\begin{align*}
bc+ca+ab = 1
\end{align*}
Thus,
\begin{align*}
(\bar{a}b +a\bar{b}+ \bar{b}c +b\bar{c}+ \bar{c}a + c\bar{a}) = -2
\end{align*}
and hence
\begin{align*}
\cos(\theta_2-\theta_3) + \cos(\theta_3 - \theta_1) + \cos(\theta_1 - \theta_2) + 1 = 0
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a quicker proof to show that $2^{10^k} \equiv 7 \pmod{9}$ for all positive integers $k$? I noticed this pattern while playing with digit sums and noticed that the recursive digit sums (until you arrive at a single digit number) of numbers like, $2^{10}$, $2^{100}$, $2^{1000}$ and so on is always $7$. So, I decided to find out if it is true that for all positive integers $k$,
$$ 2^{10^k} \equiv 7 \pmod{9} $$
My proof is as follows:
Lemma 1. $\, 10^k \equiv 4 \pmod{6}$ for all integers $k \geq 1$
Proof. For all integers $k \geq 1$, the number $10^k + 2$ must be divisible by $6$ since it is even (implying divisibility by $2$), and its digit sum is $3$ (implying divisibility by $3$). Therefore, you can show that
\begin{align*}
10^k + 2 &\equiv 0 \pmod{6} \\
10^k &\equiv 4 \pmod{6} \,\, \text{ for all integers } k \geq 1
\end{align*}
Lemma 2. $\, 2^{4 + 6k} \equiv 7 \pmod{9}$, for all integers $k \geq 0$
Proof. If $a \equiv c \pmod{n}$ and $b \equiv d \pmod{n}$, then $a\,b \equiv c\,d \pmod{n}$.
And, by extension, $a\,b^k \equiv c\,d^k \pmod{n}$ (for integers $k \geq 0$). Therefore, with
\begin{align*}
2^4 \equiv 16 \equiv 7 \pmod{9}
\end{align*}
and
\begin{align*}
2^6 \equiv 64 \equiv 1 \pmod{9}
\end{align*}
we can show that,
\begin{align*}
2^{4 + 6k} \equiv 7 \cdot 1^k \equiv 7 \pmod{9} \,\, \text{ for all integers } k \geq 0
\end{align*}
Theorem. $\, 2^{10^k} \equiv 7 \pmod{9} $ for all integers $k \geq 1$
Lemma 1 implies that for all integers $k \geq 1$, $10^k = 4 + 6n$ where $n$ is some positive integer. Therefore,
\begin{align*}
2^{10^k} &= 2^{4 + 6n} \\
2^{10^k} \!\!\!\! \mod a &= 2^{4 + 6n} \!\!\!\! \mod a
\end{align*}
for any positive integer $a$. Using this result along with Lemma 2,
\begin{align*}
2^{10^k} \equiv 7 \pmod{9} \,\, \text{ for all integers } k \geq 1
\end{align*}
$$\tag*{$\blacksquare$}$$
I think the proof is correct, but I am not a fan of it. Mainly because, I think proving Lemma 2 is a much too long a way to prove this theorem - since it proves a generalization of the theorem first. Also, I discovered that lemma numerically, which feels a bit like cheating.
Either way, is there a quicker, more elegant proof which does not:
*
*Use Lemma 2, or prove some generalization of the theorem first.
*Uses Euler's Theorem. I feel that using Euler's Theorem is using a needlessly complicated theorem to prove something as simple as this.
I am sure some of the people here can come up with single line proofs. I am curious to see if there's such a proof.
| Don't fear Euler's theorem. You have it. Use it. It's not like it costs a lot of gas money or you have to pay tolls.
$2^6 \equiv 1 \mod 9$ so $2^{10^k} \equiv 2^{(10^k \mod 6 = 4^k \mod 6)} \mod 9$.
One little observation. If $4^n \equiv 4 \mod 6$ then $4^{n+1} \equiv 16 \equiv 4 \mod 6$ so inductively $4^n \equiv 4 \mod 6$ for all $n$.
So $2^{10^k} \equiv 2^{4^k} \equiv 2^4 =16 \equiv 7 \mod 9$.
| {
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"timestamp": "2023-03-29T00:00:00",
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RMO inequality problem If $a,b,c,d,e>1$, then prove that:
$$\frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 20. $$
I don't know how to begin. What should be the approach?
| Note that we can rewrite your expression as
$$ \left( \frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{c^2}{a-1} \right) + \left( \frac{d^2}{e-1} + \frac{e^2}{d-1} \right) \geq 20. $$
Let us consider first the baby problem of finding the minimum value of $f(x) = \frac{x^2}{x-1}$ assuming $x > 1$. The minimum is readily seen to be attained at $x = 2$ with value $4$.
Next, consider the problem of finding the minimum value of
$$ \frac{d^2}{e-1} + \frac{e^2}{d-1} = f(d) \frac{d-1}{e-1} + f(e) \frac{e-1}{d-1} $$
assuming $d,e > 1$. The arithmetic-geometric inequality shows that
$$ f(d) \frac{d-1}{e-1} + f(e) \frac{e-1}{d-1} \geq 2 \sqrt{f(d)f(e)} \geq 2f(2) = 8 $$
where the minimum value of $8$ is indeed attained when $d = e = 2$.
Finally, consider the problem of finding the minimum value of
$$ \frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{c^2}{a-1} = f(a)\frac{a-1}{b-1} + f(b)\frac{b-1}{c-1} + f(c)\frac{c-1}{a-1} $$
assuming $a,b,c > 1$. Again, the arithmetic-geometric inequality shows that
$$ f(a)\frac{a-1}{b-1} + f(b)\frac{b-1}{c-1} + f(c)\frac{c-1}{a-1} \geq 3 (f(a)f(b)f(c))^{\frac{1}{3}} \geq 3f(2) = 12 $$
where the minimum indeed attained at $a = b = c = 2$.
Combining everything above, we obtain the required result.
| {
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The equations $ax^2+by^2+cz^2=0$ and $ax^2+by^2+cz^2=1$
Let $a,b,c -$ a non-zero integers. It is known that the equation $ax^2+by^2+cz^2=0$ it has a non-zero integer solution. Prove that the equation $ax^2+by^2+cz^2=1$ has solution in rational numbers.
My work so far:
Let $a>0,b>0,c<0$ and $z<0$.
Let $(x,y,z) -$ solution of $ax^2+by^2+cz^2=0$. Then:
If $c=-n^2$ that
$$ax^2+by^2+cz^2=0 \Leftrightarrow$$
$$ax^2+by^2-n^2z^2=0 \Leftrightarrow$$
$$25ax^2+25by^2-25n^2z^2=0 \Leftrightarrow$$
$$a(5x)^2+b(5y)^2-16n^2z^2=9n^2z^2 \Leftrightarrow$$
$$a\left(\frac{5x}{3nz}\right)^2+b\left(\frac{5y}{3nz}\right)^2+c \left(\frac{4}{3n}\right)^2=1$$
Hence, $\left(\frac{5x}{3nz};\frac{5y}{3nz};\frac{4}{3n}\right)$ solution of $ax^2+by^2+cz^2=0$.
Let $c\not=-n^2$.
I need help here...
| By the hypothesis, there exists rational $u,v$ such that $au^2+b v^2+c=0$. This means that $a(-\frac{u^2}{c})+b(-\frac{v^2}{c})=1$. Now we want a solution of the equation $ax^2+b y^2+c z^2=1$ in rational numbers $x,y,z$. Replacing $c$ by $-au^2-bv^2$ Give $a(x^2-u^2z^2)+b(y^2-v^2z^2)=1$. Hence, if we can find rational $x,y,z$ such that $x^2-u^2z^2=-\frac{u^2}{c}$ and $y^2-v^2z^2=-\frac{v^2}{c}$, we are done. This imply that $(\frac{x}{u})^2=z^2-\frac{1}{c}= (\frac{y}{v})^2$. So we put $x=ku$ and $y=kv$, with the property that $k^2=z^2-1/c$. There is a lot of solutions, we can take $2k=1-1/c$ and $2z=(1+1/c)$, this gives $2x=(1-1/c)u$ and $2y=(1-1/c)v$.
It remain to show that these values gives a solution, that is easy.
| {
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The inequality $p<\frac{a+m_b}{b+m_a}
Let in triangle $ABC$ $BC=a$ and $AC=b$, $m_a, m_b -$ the medians drawn from the vertex $A$ and $B$. Find the greatest real number $p$ and the lowest real $q$, such that inequality
$$p<\frac{a+m_b}{b+m_a}<q$$
hold.
My attempt:
I used formula:
$$m_a=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$$
| If $a\rightarrow0^+$ and $b=c=1$ so $\frac{a+m_b}{b+m_a}\rightarrow\frac{1}{4}$.
We'll prove that $\frac{a+m_b}{b+m_a}>\frac{1}{4}$.
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that
$$\frac{2(y+z)+\sqrt{4y(x+y+z)+(x-z)^2}}{2(x+z)+\sqrt{4x(x+y+z)+(y-z)^2}}>\frac{1}{4}$$ or
$$8y+6z-2x+4\sqrt{4y(x+y+z)+(x-z)^2}>\sqrt{4x(x+y+z)+(y-z)^2}$$
1. $4y+3z-x\geq0$.
After squaring of the both sides we'll obtain something obvious;
*$4y+3z-x<0$.
We need to prove that
$$4\sqrt{4y(x+y+z)+(x-z)^2}>\sqrt{4x(x+y+z)+(y-z)^2}+2(x-4y-3z)$$
In this case we need to square twice and we'll obtain something obvious.
Id est, $p=\frac{1}{4}$.
By the same way we can obtain $q=4$.
Done!
| {
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How to prove $\left\lceil\frac{x}y\right\rceil=\left\lfloor\frac{x-1}y\right\rfloor+1$ for positive integers $x,y$? I have to prove that
$$\left\lceil\frac{x}y\right\rceil=\left\lfloor\frac{x-1}y\right\rfloor+1\;?$$
For any positive integers $x, y$.
Can anyone help me?
| As @6005 clearly said, you need to agree that the $x,y$ are positive integers with the constraint $y \gt 0$.
This is a hint regarding the right side of the inequality:
$$\left\lceil\frac{x}y\right\rceil=\left\lfloor\frac{x-1}y\right\rfloor+1=\left\lfloor\frac{x-1}y+1\right\rfloor=\left\lfloor\frac{x-1+y}y\right\rfloor=\left\lfloor\frac{x}{y} + \frac{y-1}y\right\rfloor=\left\lfloor\frac{x}{y} + 1 - \frac{1}y\right\rfloor$$
| {
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if $n^2 + 2016n$ is equal to a square number, how would i find the largest value of n? Q: $n^2 + 2016n$ is equal to a square number. What is the largest value of n for which this is true, giving the answer as a remainder when divided by 1000.The answer is a whole number and is a number from 1-999.
I'm not sure if my answer is correct. I factorised n out giving $n(n+2016)$ and then I said that $(n+2016)=na^2$ where a is a number.
Then $n(1-a^2)=2^5.3^2.7$
$n(1-a)(1+a)=2^5.3^2.7$
So then wouldn't the largest number for n be when $(1-a)(1+a)=2^0.3$
then the answer = $2^5.3.7$ Is that correct?
| Here's a start:
$n^2 + 2016n = M^2$
$n^2 + 2016n + 1008^2 - M^2 = 1008^2 = 2^8*3^4*7^2$
$(n-M + 1008)(n+M + 1008) = 2^8*3^4*7^2$
Let $(n-M + 1008) = 2^a3^b7^c$ and $(n+M + 1008) = 2^{8-a}3^{4-b}7^{2-c}$
So $2n + 2016 = 2^a3^b7^c + 2^{8-a}3^{4-b}7^{2-c}$
So we need to maximize $n = \frac{2^a3^b7^c + 2^{8-a}3^{4-b}7^{2-c}}2 - 1008 \le 999$
Or $2^a3^b7^c + 2^{8-a}3^{4-b}7^{2-c} \le 4014$
Not entirely sure how to maximize to such a range. Picking values at random I get $16*81*7 + 16*7 = 16*7(82)=9184$
So $n = 3584$
And $(3584)^2 + 2016(3584) = 4480^2$.
But that's too big.
I'm not sure why they are restricting to $n < 1000$. The true largest would be $n = (2^7*3^4*7^2 + 2)/2 - 1008 = 253009$
With $(253009) + 2016(253009) = 254015^2$.
| {
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Number of ways to get from $(0, 0)$ to $(2, 2)$ in coordinates system I'm in point $\left(0, 0\right)$ of coordinates system:
*
*I want to konw how many ways are there to the point $(2, 2)$ in at most 6 steps.
*One step is moving up, down, left or right by one cell, so from
$\left(0, 0\right)$ in one step I can get to
$\left(0, 1\right), \left(1, 0\right), \left(-1, 0\right), \left(0, -1\right)$.
*Of course, the minimum number of steps is $4$ and there are $6$ ways of length $4$. You can't get there in $5$ steps.
*What about ways of length exactly $6$ ?. How can I count them ?.
|
We can do the job algebraically. We encode steps in direction $E$ and $W$ by $x^1$ and $x^{-1}$ and similarly we encode steps in direction $N$ and $S$ by $y^1$ and $y^{-1}$.
We obtain the expression
\begin{align*}
x^1+x^{-1}+y^{1}+y^{-1}\tag{1}
\end{align*}
with the $+$ sign indicating the alternatives we have for each step.
Let's denote with $[x^j]$ the coefficient of $x^j$ in an expression. This way we can write e.g.
\begin{align*}
[x^2](1+x)^6=\binom{6}{2}
\end{align*}
Since one step is encoded by (1) and we are looking for the number of possibilities to go from $(0,0)$ to $(2,2)$ in $6$ steps, we have to look for the coefficient of
\begin{align*}
[x^2y^2](x^1+x^{-1}+y^{1}+y^{-1})^6
\end{align*}
For example one of the paths of length $6$ from $(0,0)$ to $(2,2)$ is
\begin{align*}
E.N.S.N.N.E \qquad \longleftrightarrow\qquad x^1y^1y^{-1}y^{1}y^{1}x^{1}=x^2y^2
\end{align*}
We obtain
\begin{align*}
[x^2y^2]&(x^1+x^{-1}+y^{1}+y^{-1})^6\\
&=[x^2y^2]\sum_{j=0}^6\binom{6}{j}\left(x+\frac{1}{x}\right)^j\left(y+\frac{1}{y}\right)^{6-j}\tag{2}\\
&=[x^2y^2]\binom{6}{2}\left(x+\frac{1}{x}\right)^2\left(y+\frac{1}{y}\right)^4\\
&\qquad+[x^2y^2]\binom{6}{4}\left(x+\frac{1}{x}\right)^4\left(y+\frac{1}{y}\right)^2\tag{3}\\
&=[x^2y^2]\left(\binom{6}{2}\binom{2}{0}x^2\binom{4}{1}y^2+\binom{6}{4}\binom{4}{1}x^2\binom{2}{0}y^2\right)\tag{4}\\
&=\binom{6}{2}\cdot 4+\binom{6}{4}\cdot 4\\
&=120
\end{align*}
and conclude there are $120$ different paths to go from $(0,0)$ to $(2,2)$ in $6$ steps using steps $(1,0),(-1,0),(0,1)$ and $(0,-1)$.
Comment:
*
*In (2) we group the polynomial $(x^1+x^{-1}+y^{1}+y^{-1})^6=\left(\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)\right)^6$ and apply the binomial theorem.
*In (3) we observe only $j=2$ and $j=4$ provide a contribution to $[x^2y^2]$.
*In (4) we select the coefficients of $[x^2y^2]$.
| {
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"url": "https://math.stackexchange.com/questions/1920574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Determine the coefficient of $xy$ in the expansion of $(x+y+2)^7$ The problem: Determine the coefficient of $xy$ in the expansion of $(x+y+2)^7$
My approach:
We can rewrite the equation substituting $x+y =j$
$$(x+y+2)^7=(j+2)^7$$
This is simpler because we know the coefficients thanks to the formula:
$$(a+b)^{n}=\sum _{{k=0}}^{n}{n \choose k}a^{{n-k}}b^{{k}} $$
With $n=7$ we have $1,7,21,35,35,21,7,1$ as coefficients and the expansion looks like this:
$$j^7+14\cdot j^6+(21\cdot 2^2) j^5+(35\cdot 2^3) j^4+(35\cdot 2^4) j^3+(21\cdot 2^5) j^2+(7\cdot 26)j+2^7$$
Now the only time that $xy$ appears is in the expansion of $j^2$ therefore we have:
$$(21\cdot 2^5) j^2=(21\cdot 2^5) (x+y)^2= (21\cdot 2^5)(x^2+2xy+y^2)=21\cdot 2^5x^2+21\cdot 2^6xy+21\cdot 2^5y^2$$
The coefficient is $21\cdot 2^6$
Is this correct? Is there a simpler proof?
| (x+y+2)(x+y+2)(x+y+2)*(4 more times).
To get xy as a coefficient you can choose x from any block and y from any other block in 7c2 ways and since xy can be permuted in two ways you get 2*7c2 ways and for the remaining you must choose 2 from each block so you get an extra 2^5 . so coefficient is 2* 7c2 *2^5 = 21*2^6.
so for any general case (ax+by+cz+dw)^n . Coefficient of x^r1*y*r2*z^r3*w^r4 . where
r1+r2+r3+r4 = n. can be written as >>
n!*(a^r1*b^r2*c^r3*d^r4)/(r1!*r2!*r3!*r4!)
| {
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"url": "https://math.stackexchange.com/questions/1920638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How can I prove the this inequality? $$ a^6+b^6+c^6+3a^2 b^2 c^2 \geq 2(a^3 b^3 + b^3 c^3 +c^3 a^3)$$
$\forall a,b,c \in \mathbb{R}$
Can this be done with just weighted AM-GM?
| By Schur's and followed by AM-GM inequalites: $x^3+y^3+z^3 + 3xyz \ge xy(x+y) + yz(y+z) + zx(z+x) \ge 2xy\sqrt{xy} + 2yz\sqrt{yz} + 2zx\sqrt{zx}$, with $x = a^2,y=b^2,z=c^2$, the inequality follows.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of $\sqrt[n]{2^n + 3^n} + \sqrt[n]{n^2 + 1}$ I'm learning calculus, specifically limit of sequences, and need with with the following exercise:
Find the limit of the following sequences
$$(a) \ \ a_n = \sqrt{2^n + 5^n} \quad \quad (b) \ \ b_n = \sqrt[n]{2^n + 3^n} + \sqrt[n]{n^2 + 1}.$$
Since I'm having difficulties for $(b)$, I'm going to show my work for $(a)$.
$(a)$ Using the Squeeze Theorem, we observe that
$$\sqrt{2}(2)^{\frac n2} = \sqrt{2^n + 2^n} \leq a_n \leq \sqrt{5^n + 5^n} = \sqrt{2}(5)^{\frac n2}.$$
Obviously
$$\sqrt{2}\lim_{n \to \infty} (2)^{\frac n2} = +\infty = \sqrt{5}\lim_{n \to \infty} (2)^{\frac n2}$$
so for the original sequence we have
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \sqrt{2^n + 5^n} = +\infty.$$
Is my work/method correct for $(a)$. I wanted to use a similar argument for $(b)$ but I was not able to do it. Is there a clever way to solve $(b)$? I also thought of calculating the limit by multiplying by the conjugate by I think it's not the right method here.
| $$\sqrt[n]{2^n + 3^n} + \sqrt[n]{n^2 + 1}=3\sqrt[n]{1 + (\frac{2}{3})^n} + n^{\frac{1}{n}}\sqrt[n]{1 + \frac{1}{n^2}}=3\left(1 + (\frac{2}{3})^n\right)^{\frac{1}{n}}+n^{\frac{1}{n}}\left(1 + \frac{1}{n^2}\right)^{\frac{1}{n}}$$
we know that limit of
$$\lim_{n\rightarrow \infty }n^{\frac{1}{n}}=1$$
see
$$1 + (\frac{2}{3})^n<n$$
and
$$1 + \frac{1}{n^2}<n$$
so the
$$3\lim_{n\rightarrow \infty }\left(1 + (\frac{2}{3})^n\right)^{\frac{1}{n}}+\lim_{n\rightarrow \infty }n^{\frac{1}{n}}\lim_{n\rightarrow \infty }\left(1 + \frac{1}{n^2}\right)^{\frac{1}{n}}=3(1)+1(1)=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x$ and $y$ are real and $x^2 + y^2 = 1$, compute the maximum value of $(x+y)^2.$ If $x$ and $y$ are real and $x^2 + y^2 = 1$, compute the maximum value of $(x+y)^2.$
Should I manipulate the equation? I'm stuck on this problem. Answers are greatly appreciated.
| Use the Cauchy-Schwarz inequality:
$$(x\times 1+y\times 1 )^2 \leq (x^2+y^2)(1^2+1^2), x^2+y^2=1$$
$$ (x+y)^2 \leq (1)(2)$$
$$-\sqrt{2} \leq x+y \leq \sqrt{2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does this series expansion for $\frac{1}{\cos(x)}$ fail using $\frac{1}{1-x}$? I'm trying to do (2nd order) Taylor expansion for
$$\frac{1}{\cos(x)}$$ using $$\frac{1}{1-x}$$
What I do is write
$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+ \cdot\cdot\cdot, \text{ at } x_0=0$$
and
$$\frac{1}{1-x}=1+x+x^2+\cdot\cdot\cdot$$
so
$$\frac{1}{\cos(x)}=\frac{1}{1-(1-\cos(x))}=1+(1-\cos(x))+(1-\cos(x))^2$$
$$=1+(1-1+\frac{x^2}{2!})+(1-1+\frac{x^2}{2!})^2$$
$$=1+\frac{1}{2}x^2+\frac{1}{4}x^4$$
which is correct up to order 2 (or order 1?), but the coefficient of $x^4$ (is that 3rd or 4th order?) is wrong.
Why? How can I get the correct coefficients for higher orders?
And if I do want only 2nd order Taylor polynomial, then what is the term $\frac{1}{4}x^4$?
| You did not calculate the 3rd and 4th order coefficients like this, only the second (you have some $o(x^2)$ remaining from the first term, of which the $\frac{x^4}{4}$ is a part of).
In order to get the fourth (the third is null by parity), you need all terms that will give a fourth order coefficient, so that you are only left with some $o(x^4)$
In your case, for the fourth coefficient, it would be
$\frac{1}{cos(x)} = 1 + (1-1+\frac{x^2}{2!}-\frac{x^4}{4!} + o(x^4))+(1-1+\frac{x^2}{2!}-\frac{x^4}{4!} + o(x^4))^2$
$= 1 + \frac{x^2}{2!}-\frac{x^4}{4!} + \frac{x^4}{4} + o(x^4) = 1 + \frac{x^2}{2!}+\frac{5x^4}{24} + o(x^4)$
| {
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Prove $\sqrt{2}$ is between $\dfrac{a}{b}$ and $\dfrac{a+2b}{a+b}$ Let $a$ and $b$ be positive integers. Show that $\sqrt{2}$ always lies between $\dfrac{a}{b}$ and $\dfrac{a+2b}{a+b}$.
Please give the easy solution as possible.
| Denote $x=\frac{a}{b}$ and $y=\frac{a+2b}{a+b}=\frac{x+2}{x+1}=1+\frac{1}{1+x}$.
If $x< \sqrt{2}$ ($x$ cannot be $\sqrt{2}$ because $x$ is a rational), then $1+x< \sqrt{2}+1$, $\frac{1}{1+x}> \frac{1}{\sqrt{2}+1}=\sqrt{2}-1$. Thus $y> \sqrt{2}$.
The other case can be proved similarly.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int_0^{2\pi}\frac{dx}{(2+\cos(x))}$ using complex integral theorem Would someone help me to integrate
$$\int_0^{2\pi}\frac{dx}{(2+\cos(x))}$$
by means of residue theorem?
| Passing into the complex plane with the substitution
$$\cos x \to \cos z = \frac{\left(z + z^{-1}\right)}{2} ~~~~~~~ \text{d}x = \frac{\text{d}z}{iz}$$
an you get:
$$\int_{|z| = 1} -i \frac{\text{d}z}{z\left(2 + \frac{z + z^{-1}}{2}\right)} = \int_{|z| = 1} -2i \frac{\text{d}z}{z\left(4 + \frac{z^2 + 1}{z}\right)}$$
Arrange the terms:
$$\int_{|z| = 1} -2i \frac{\text{d}z}{4z + z^2 + 1}$$
Now calculate the poles, solving $z^2 + 4z +1 = 0$
The first pole (solution) is $z = -2+\sqrt{3}$ inside the unit circle and it's ok.
The second one is $z = -2-\sqrt{3}$ outside the unit circle so it's not ok.
Result:
$$2\pi i\ \text{Res}_{z = -2+\sqrt{3}}\ (z - 2-\sqrt{3})\frac{-2i}{(z-2-\sqrt{3})(z-2+\sqrt{3})} = \frac{2\pi}{\sqrt{3}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factorize the polynomial $ 6x^4 - 11x^3 - 30x^2 + 29x - 6 $ Using an online service I can see that the polynomial $$ 6x^4 - 11x^3 - 30x^2 + 29x - 6 $$ can be factorized as $$ (3 x-1) (2 x-1) (x+2) (x-3) $$ using the Rational Root Theorem, but how is it applied?
| From Wiki, if $\frac{p}{q}$ is a factor of $P(x)$, then $p$ is a factor of, in your case, $-6$, and $q$ is a factor of $6$. So, the possibilities are:
$$
1, -1, \frac{1}{2}, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{3}, \frac{1}{6}, -\frac{1}{6} \\
2, -2, \frac{2}{2}, -\frac{2}{2}, \frac{2}{3}, -\frac{2}{3}, \frac{2}{6}, -\frac{2}{6},
\\ 3, -3, \frac{3}{2}, -\frac{3}{2}, \frac{3}{3}, -\frac{3}{3}, \frac{3}{6}, -\frac{3}{6},
\\ 6, -6, \frac{6}{2}, -\frac{6}{2}, \frac{6}{3}, -\frac{6}{3}, \frac{6}{6}, -\frac{6}{6}
$$
Now it is just a matter of checking each one of them, it is not that hard because many of the numbers that appear in the list are not unique.
| {
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Prove that $\sum_{k=0}^x {n \choose k}p^k(1-p)^{n-k}=(n-x){n \choose x}\int_0^{1-p}t^{n-x-1}(1-t)^xdt $ For $0\le p \le 1$ and $x=0,1,2,...,n$, I need to prove that $\sum_{k=0}^x {n \choose k}p^k(1-p)^{n-k}=(n-x){n \choose x}\int_0^{1-p}t^{n-x-1}(1-t)^xdt $ and the two hints are either differentiate both sides with respect to $p$ or integrate by parts.
First since this is a finite sum
$$\frac{d}{dp}\sum_{k=0}^x {n \choose k}p^k(1-p)^{n-k} =\sum_{k=0}^x \frac{d}{dp} {n \choose k}p^k(1-p)^{n-k} =\sum_{k=0}^x {n \choose k}\left(kp^{k-1}(1-p)^{n-k}-(n-k) p^{k}(1-p)^{n-k-1}\right)$$
and $$ \frac{d}{dp} \int_0^{1-p}t^{n-x-1}(1-t)^xdt = -(1-p)^{n-x-1}(p)^{x} $$
and assuming I didn't make a mistake I don't see how this helps. I was also having trouble trying to make integration by parts to work. Could someone please help put me in the right direction?
| \begin{align*}
(n-x){n \choose x}&\int_0^{1-p}t^{n-x-1}(1-t)^xdt \\
&= (n-x)\frac{n!}{x!(n-x)!}\left\{ \left.\frac{t^{n-x}}{n-x}(1-t)^x \right|_0^{1-p} + \int_0^{1-p} x(1-t)^{x-1}\frac{t^{n-x}}{n-x}dt\right\}\\
&= (n-x)\frac{n!}{x!(n-x)!}\left\{\frac{(1-p)^{n-x}p^x}{n-x} + \int_0^{1-p} x(1-t)^{x-1}\frac{t^{n-x}}{n-x}dt\right\} \\
&= \binom{n}{x}(1-p)^{n-x}p^x \\
& + \frac{n!}{x!(n-x)!}\int_0^{1-p} x(1-t)^{x-1}t^{n-x}dt \\
&= \binom{n}{x}(1-p)^{n-x}p^x \\
& + \frac{n!}{(x-1)!(n-x)!}\left\{\left.\frac{t^{n-x+1}}{n-x+1}(1-t)^{x-1}\right|_0^{1-p} + \int_0^{1-p}(x-1)(1-t)^{x-2}\frac{t^{n-x+1}}{n-x+1}\right\} \\
&=\binom{n}{x}(1-p)^{n-x}p^x +\binom{n}{x-1}(1-p)^{n-x+1}p^{x-1} \\
& + \frac{n!}{(x-2)!(n-x+1)!}\int_0^{1-p}(1-t)^{x-2}t^{n-x+1}dt
\end{align*}
and we see the pattern.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the number of ways to select 2n balls from n identical blue balls, n identical red balls and n identical white balls, where n $\in$ $\mathbb N$ Q: Find the number of ways to select 2n balls from n identical blue balls, n identical red balls and n identical white balls, where n $\in$ $\mathbb N$.
My working:
$(x+x^2+x^3+...)^3$
$=x^3(1+x+x^2+...)^3$
$= x^3 \sum_{r=0}^\infty \begin{pmatrix} {r+3-1}\\{r}\end{pmatrix}x^r$
$= \sum_{r=0}^\infty \begin{pmatrix} {r+2}\\{r}\end{pmatrix}x^{3+r}$
Hence the number of ways is $\begin{pmatrix} {2n-1}\\{2n-3}\end{pmatrix}$=$\begin{pmatrix} {2n-1}\\{2}\end{pmatrix}$.
However, the actual answer is $\begin{pmatrix} {2n+2}\\{2}\end{pmatrix}$$-3$$\begin{pmatrix} {n+1}\\{2}\end{pmatrix}$. Did i go wrong somewhere in my proof? Or is it some conceptual understanding gone wrong? Thanks!
| I think it's easiest to count this directly, without any fancy tricks. We actually end up with a simpler formula: $$\binom{n+2}{2}$$ or, equivalently, $$\frac{(n+1)(n+2)}{2}$$ (which is equal to the longer formula in @Muralidharan's answer).
Here's how it works:
The number $r$ of red balls is in the range from $0$ to $n,$ and the number $b$ of blue balls is also in the range from $0$ to $n.$
Given values for $r$ and $b,$ there's at most one possible value for the number of white balls: $w = 2n-r-b.$ But this potential number of white balls also has to be in the range from $0$ to $n.$ If $2n-r-b$ is in the range from $0$ to $n,$ then $\langle r, b, 2n-r-b\rangle$ is a triple we have to count. If $2n-r-b$ is not in that range, then we should skip that triple and not count it.
Since each ordered pair $\langle r, b\rangle$ has at most one corresponding $w,$ this is the same as counting the number of ordered pairs $\langle r, b\rangle$ such that $0 \le r \le n, 0 \le b \le n,$ and $0 \le 2n-r-b \le n.$
So our count is as follows, using the notation $\#X$ for the cardinality of a set $X\!\!:$
\begin{align}
&\#\lbrace \langle r, b\rangle \mid 0 \le r \le n \;\land\; 0 \le b \le n \;\land\; 0 \le 2n-r-b \le n\rbrace
\\= &\#\lbrace \langle r, b\rangle \mid 0 \le r \le n \;\land\; 0 \le b \le n \;\land\; r + b \le 2n \;\land\; n \le r+b\rbrace
\\= &\#\lbrace \langle r, b\rangle \mid 0 \le r \le n \;\land\; 0 \le b \le n \;\land\; n-r \le b \le 2n-r\rbrace
\\= &\#\lbrace \langle r, b\rangle \mid 0 \le r \le n \;\land\; \max(0,n-r) \le b \le \min(n,2n-r)\rbrace.
\end{align}
For any $r \le n,$ we have $\max(0,n-r)=n-r$ and $\min(n,2n-r)=n,$ so the count above equals
\begin{align}
&\#\lbrace \langle r, b\rangle \mid 0 \le r \le n \;\land\; n-r \le b \le n\rbrace.
\end{align}
You can see that for each value of $r$ from $0$ to $n,$ we have $r+1$ possible values of $b$, namely $n-0, n-1, n-2, \dots, n-r.$
It follows that the total count equals
\begin{align}
\sum_{r=0}^n (r+1) &= \sum_{k=1}^{n+1} k
\\&= \frac{(n+1)(n+2)}{2},
\end{align}
and this equals $$\binom{n+2}{2}$$ if you want to phrase the answer as a binomial coefficient.
| {
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What comes before the cycles in sequences of repeated modular multiplication? I was working on this problem and found a solution that involved sequences generated by modular exponentiation with an incrementing exponent:
$$x_{m,b,i} = b^{i} \mod m$$
For example, in modular base 10:
*
*$0^1 \equiv 0$, $0^2\equiv 0$, $0^3\equiv 0$, …
*$1^1 \equiv 1$, $1^2\equiv 1$, $1^3\equiv 1$, …
*$2^1 \equiv 2$, $2^2\equiv 4$, $2^3\equiv 8$, $2^4\equiv 6$, $2^5\equiv 2$, $2^6\equiv 4$, $2^7\equiv 8$, $2^6\equiv 6$, …
*$3^1\equiv 3$, $3^2\equiv 9$, $3^3\equiv 7$, $3^4\equiv 1$, $3^5\equiv 3$, $3^6\equiv 9$, $3^7\equiv 7$, $3^6\equiv 1$, …
*$4^1 \equiv 4$, $4^2\equiv 6$, $4^3\equiv 4$, $4^4\equiv 6$, …
*$5^1 \equiv 5$, $5^2\equiv 5$, $5^3\equiv 5$, …
*$6^1 \equiv 6$, $6^2\equiv 6$, $5^3\equiv 6$, …
*$7^1\equiv 7$, $7^2\equiv 9$, $7^3\equiv 3$, $7^4\equiv 1$, $7^5\equiv 7$, $7^6\equiv 9$, $3^7\equiv 3$, $7^6\equiv 1$, …
*$8^1 \equiv 8$, $8^2\equiv 4$, $8^3\equiv 2$, $8^4\equiv 6$, $8^5 \equiv 8$, $8^6\equiv 4$, $8^7\equiv 2$, $8^8\equiv 6$, …
*$9^1 \equiv 9$, $9^2\equiv 1$, $9^3\equiv 9$, $9^4\equiv 1$, …
This property of forming cycles is quite nice, and can be proven trivially to happen in all sequences for every exponential base and every modular base (every element depends only on the previous one, and using the pigeon hole principle we can exhaust the congruence classes until an element is produced that appeared before).
But these cycles do not always start at the beginning of the sequences. Exceptional examples I found:
*
*$x_{b,m,0} = b^0 \mod m \equiv 1$ for all sequences
*$x_{2,4} = 1, 2, 0, 0, 0$…
*$x_{3,9} = 1, 3, 0, 0, 0$…
*$x_{2,8} = 1, 2, 4, 0, 0$…
Now my question: What's special about these bases? How long are those irregular beginnings? Do they ever prepend a cycle of more than one element?
I did some research and modular multiplicative inverses seem to be related, as all my examples did not have one. I also found modular multiplicative groups (Wolfram article) which seem to describe the cycles, but I didn't completely understand them and didn't notice anything about to the sequence beginnings.
| Your examples of $x_{2,4},x_{3,9},x_{2,8}$ sequences are cases where $b$ is a power of $m$, and will therefore eventually produce zeros when you take a sufficiently large power of $m$, but will not initially when $1 \le m\lt b$
You can go further than this: if $m$ and $b$ have the same distinct prime factors then the same sort of thing will happen, for example $12=2^2\times 3^1$ and $54=2^1\times 3^3$ so they both have the distinct prime factors $2$ and $3$, while $12^1\equiv 12 \pmod{54}$, $12^2\equiv 36 \pmod{54}$, and $12^i\equiv 0 \pmod{54}$ for $i \ge 3$
On the neater patterns you spotted, some are related to Fermat's little theorem $$a^p \equiv a \pmod{p}$$ for prime $p$ and $1\le a \lt p$ and to its generalisation, Euler's theorem $$a^{\varphi (n)} \equiv 1 \pmod{n}$$ for $a$ coprime to $n$ where $\varphi (n)$ is Euler's totient function
| {
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How is this property of real numbers proven?
Prove that $\dfrac{x^2}{(x − 1)^2} + \dfrac{y^2}{(y − 1)^2} + \dfrac{z^2}{(z − 1)^2} ≥ 1$ for all real numbers $x, y, z$, each different from $1$ and satisfying $xyz = 1$.
How do I prove this?
| Substitute $a=\frac{x}{x-1}, b=\frac{y}{y-1}, c=\frac{z}{z-1}$.
Then we have $x=\frac{a}{a-1}$ and the similar identities so that the condition implies $abc=(a-1)(b-1)(c-1)$ and hence $ab+ac+bc=a+b+c-1$.
We want to prove $a^2+b^2+c^2 \ge 1$ which is equivalent to $(a+b+c)^2 - 2(ab+ac+bc)-1 \ge 0$ or, using the condition, $(a+b+c)^2 -2(a+b+c)+1 \ge 0$.
But the LHS is $(a+b+c-1)^2$ which is clearly non-negative. Hence the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.