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The right way of proving LHS = RHS I'm currently brushing up on my math before heading back to school for my bachelor's and am practicing on some questions.
Given the following equation :
$$\frac{x^2 - 3x}{x^2 - 9} = 1-\frac{3x-9}{x^2-9}.$$
and that I have to prove that the LHS = RHS; How would I go about doing so?
What I've done:
I've simply brought $(3x-9)/(x^2-9)$ over to the LHS, resulting in the equation $(x^2 - 3x)/(x^2-9) + (3x-9)/(x^2-9) = 1$ , which gives me $(x^2-9)/(x^2-9) = 1$.(LHS = RHS)
However, I do not think that what I've done is the correct method of proving. Should I have instead transformed the LHS into RHS without bringing anything from LHS to RHS and vice versa?
Furthermore , after which I'm integrating $(x^2-3x)/(x^2-9)$ by doing the following:
What I first did was factorise the denominator into $(x+3)(x-3)$.
I then equated $(x^2 - 3x) = A(x + 3) + B(x - 3)$.
Given that $x = 3 \implies A = 0$ and given that $x = -3 \implies B = -3$.
This would then give me the following equation to integrate:
$[ (-3)/(x+3) + 0(x-3) ]dx$
which led me to the answer $-3ln|x+3| + C$.
However, when I checked my answer, the correct answer should be
$$x -3ln|x+3| + C.$$
Where did the missing $x$ come from?
Any assistance would be greatly appreciated
Thanks!
|
First, to show the desired equality, we have: $$\frac{x^2-3x}{x^2-9}=\frac{x^2-9-3x+9}{x^2-9}=\frac{x^2-9}{x^2-9}-\frac{3x-9}{x^2-9}=1-\frac{3x-9}{x^2-9}.$$
Now, assuming that you are trying to find the integral of $\dfrac{x^2-3x}{x^2-9}$, we have:
\begin{align}
\int{\dfrac{x^2-3x}{x^2-9}dx}&=\int\left({1-\dfrac{3x-9}{x^2-9}}\right)dx \\
&=\int{\left(1-3\cdot\frac{x-3}{(x-3)(x+3)}\right)dx} \\
&=\int{dx}-3\int{\frac{1}{x+3}dx} \\
&=x-3\ln|x+3|+C.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Indefinite integrals with absolute values Which is the right way to solve indefinite integrals which contain absolute values? For example if I have
$\int |2x+3| e^x dx$
Can I consider the sign function and integrate separetly? I mean doing:
$ Sign(2x+3) \int (2x+3)e^x dx$
Or maybe I should use the definition of absolute value and divide the two possibilities
$\int (2x+3)e^x dx$ if $ (2x+3)>0$ and
$\int (-2x-3)e^x dx$ if $ (2x+3)<0$
But I think that's more suitable for definite rather than indefinite integrals
How can I solve this type of integrals? Thanks a lot in advice
|
$\int|2x+3|e^x~dx$
$=\text{sgn}(2x+3)\int(2x+3)e^x~dx$
$=\text{sgn}(2x+3)\int_{-\frac{3}{2}}^x(2x+3)e^x~dx+C$
$=\text{sgn}(2x+3)\int_{-\frac{3}{2}}^x(2x+3)~d(e^x)+C$
$=\text{sgn}(2x+3)[(2x+3)e^x]_{-\frac{3}{2}}^x-\text{sgn}(2x+3)\int_{-\frac{3}{2}}^xe^x~d(2x+3)+C$
$=\text{sgn}(2x+3)(2x+3)e^x-2~\text{sgn}(2x+3)\int_{-\frac{3}{2}}^xe^x~dx+C$
$=|2x+3|e^x-2~\text{sgn}(2x+3)[e^x]_{-\frac{3}{2}}^x+C$
$=|2x+3|e^x-2~\text{sgn}(2x+3)\left(e^x-e^{-\frac{3}{2}}\right)+C$
|
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|
Evaluating $ \int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x$. $$
\int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x.
$$
What I have tried:
I tried writing denominator as
$ \sin^4x+\cos^4x = 1-2\sin^2x\cos^2x $ and $ 2\sin^2x\cos^2x = \frac{1}{2}\sin^2(2x) $
so the integral becomes,
$$
\int\frac{\sin x+\cos x}{1-\frac{\sin^2(2x)}{2}}\,\text{d}x.
$$
Anyone, how do I solve this further?
|
I suggest that you instead split the integral as
$$
\int\frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,dx=\int\frac{\sin x}{(1-\cos^2x)^2+\cos^4x}\,dx+\int\frac{\cos x}{\sin^4x+(1-\sin^2x)^2}\,dx,
$$
and then let $u=\cos x$ and $u=\sin x$ in the respective integral. You will get a (somewhat nasty) integral of a rational function, but it is standard.
Edit
According to your comment (and I agree), you found out that one get integrals like (after expanding the square)
$$
\int\frac{1}{2u^4-2u^2+1}\,du.
$$
Now, you need to factor the denominator, and this is where things get somewhat nasty. I write down the result (in real terms),
$$
2u^4-2u^2+1=2\Bigl(u^2+\sqrt{1+\sqrt{2}}u+\sqrt{2}/2\Bigr)\Bigl(u^2-\sqrt{1+\sqrt{2}}u+\sqrt{2}/2\Bigr).
$$
Thus, there are constants $A$, $B$, $C$ and $D$ such that
$$
\frac{1}{2u^4-2u^2+1}=\frac{Au+B}{u^2+\sqrt{1+\sqrt{2}}u+\sqrt{2}/2}+\frac{Cu+D}{u^2-\sqrt{1+\sqrt{2}}u+\sqrt{2}/2}.
$$
When that is done, you integrate "as usual". You will typically get logarithms and arctans, depending on what the constants $A$, $B$, $C$ and $D$ turn out to be.
The solution by juantheron is (as often) a short cut.
|
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|
What is the sum of the series $x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n}$? $x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n}$
I'm not sure if this can even be summed. Any help is appreciated.
|
It is
$S=x^{n-1}+2x^{n-2}+3x^{n-3}+\ldots+(n-2)\cdot x^2+(n-1)\cdot x^1+n\cdot x^0 \quad (1)$
$x\cdot S=x^{n}+2x^{n-1}+3x^{n-2}+\ldots+(n-2)\cdot x^3+(n-1)\cdot x^2+n\cdot x^1 \quad (2)$
Subtracting (1) from (2):
$(x-1)S=\color{blue}{x^n+x^{n-1}+x^{n-2}+x^{n-3}+\ldots + x^2+x}-n$
The formula for the blue expression is $x\cdot \frac{1-x^{n}}{1-x}$ (geometric series)
Now solve for $S$.
|
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|
Limit of a fraction with a square root
Find $$\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}$$ (without L'Hopital)
Where is the following wrong? (The limit is 6.)
\begin{align}\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{9-x^2-5}}= \\
& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{-x^2+4}}=\\
& = \lim_{x \to 2}\sqrt{(2-x)(2+x)}=0.
\end{align}
|
$$\lim_{x\to 2}\frac{4-x^2}{3-\sqrt{x^2+5}}=\lim_{x\to 2}\frac{\left(4-x^2\right)\left(3+\sqrt{x^2+5}\right)}{\left(3-\sqrt{x^2+5}\right)\left(3+\sqrt{x^2+5}\right)}$$
$$=\lim_{x\to 2}\frac{\left(4-x^2\right)\left(3+\sqrt{x^2+5}\right)}{3^2-\left(\sqrt{x^2+5}\right)^2}=\lim_{x\to 2}\frac{\left(4-x^2\right)\left(3+\sqrt{x^2+5}\right)}{4-x^2}$$
$$=\lim_{x\to 2}\left(3+\sqrt{x^2+5}\right)=3+\sqrt{2^2+5}=6$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$ Without using L'Hopital's rule, prove that
$$\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$$
|
Don't know if that is rigorous enough for you, but this is one way using $e^t = \sum_{k=0}^{\infty} \frac{t^k}{k!}$, $e^{\ln x} = x$ and some results about absolute convergent series:
\begin{align*}
\frac{a^x - 1}{x} & = (e^{x \cdot \ln a} -1)/x = \left(\sum_{k=0}^{\infty} \frac{(x \cdot \ln a)^k}{k!} - 1 \right)/x\\
& = \left( x \cdot \ln a + \frac{(x \cdot \ln a)^2}{2} + \ldots \right)/x = \ln a + (\ln a)^2 \cdot \frac{x}{2} + (\ln a)^3 \cdot \frac{x^2}{6} +\ldots\\
\end{align*}
Hence, $\lim_{x \to 0} \frac{a^x -1}{x} = \ln a$.
|
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|
System of congruence relations Solve the system of congruence relations:
$2x+3y\equiv 1\pmod {11}$
$x+4y\equiv 4\pmod {11}$
Could someone give a hint how to solve this system.
I know that Chinese remainder theorem can't be used because modulo numbers ($11$) are not relatively prime.
|
Write your system in matrix form: $\left( \begin{matrix} 2 & 3 \\ 1 & 4 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} 1 \\ 4 \end{matrix} \right) \pmod {11}$. The matrix has determinant $5$ which is invertible $\mod 11$, therefore the matrix itself is invertible and its inverse is $\left( \begin{matrix} 3 & 6 \\ 2 & 7 \end{matrix} \right)$, therefore $\left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} 3 & 6 \\ 2 & 7 \end{matrix} \right) \left( \begin{matrix} 1 \\ 4 \end{matrix} \right) \pmod {11} = \left( \begin{matrix} 5 \\ 8 \end{matrix} \right)$.
Alternatively, if you haven't studied matrices, multiply the second equation by $2$ in order to get $2x + 8y = 8 \pmod {11}$, then subtract the first equation from it and obtain $5y = 7 \pmod {11}$, and thus $y = 5^{-1} \cdot 7 = 9 \cdot 7 = 63 = 8 \pmod {11}$, and now use this to find $x$ from the second original equation $x = 4 - 4y = -28 = 5 \pmod {11}$.
|
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|
Finding $\binom n0+\binom n3+\binom n6+\cdots $ How to get
$$\binom n0 + \binom n3 + \binom n6 + \cdots$$
MY ATTEMPT
$$(1+\omega)^n = \binom n0 + \binom n1 \omega^1 + \binom n2 \omega^2 + \cdots$$
$$(1+\omega^2)^n = \binom n0 + \binom n1 \omega^2 + \binom n2 \omega^4 + \cdots $$
$$(1 + 1)^n = 2^n = \binom n0 + \binom n1 + \binom n2 + \cdots$$
$$(1+\omega)^n + (1+\omega^2)^n + (1 + 1)^n = 3 \left(\binom n0 + \binom n3 + \binom n6 + \cdots\right)$$
But how to solve LHS? I got the required equation in RHS
|
You have
$$(1+\omega)^n+(1+\omega^2)^n+2^n=3\left(\binom n0+\binom n3+\binom n6+\cdots\right)$$
Now note that
$$(1+\omega)^n+(1+\omega^2)^n=(-\omega^2)^n+(-\omega)^n=(-1)^n(\omega^n+\omega^{2n})$$
This is equal to $(-1)^n\cdot 2$ if $n\equiv 0\pmod 3$ or $(-1)^n\cdot (-1)=(-1)^{n+1}$ if $n\not\equiv 0\pmod 3.$
|
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|
Can you solve $y'+x+e^y=0$ by series expansion?
Find an approximate solution by series expansion of $y(x)$ around $x =
0$ up to fourth order in $x$ given the inital conditions $y(0)=0$
Let
$$
y=\sum_0^{\infty}a_nx^n \implies y'=\sum_0^{\infty}a_n nx^{n-1} \\
\implies \sum_0^{\infty}a_n nx^{n-1}+x+e^{\sum_0^{\infty}a_nx^n}=0
$$
Usually I would try to combine my terms but I can't do it here. Is this even the right approach?
|
Here is another way of solving the equation:
$$y'(x)+x+e^{y(x)}=0\Longleftrightarrow$$
$$\frac{\text{d}y(x)}{\text{d}x}+x+e^{y(x)}=0\Longleftrightarrow$$
Let $y(x)=\ln(v(x))$, which gives $\frac{\text{d}y(x)}{\text{d}x}=\frac{\frac{\text{d}v(x)}{\text{d}x}}{v(x)}$:
$$x+v(x)+\frac{\frac{\text{d}v(x)}{\text{d}x}}{v(x)}=0\Longleftrightarrow$$
$$\frac{\frac{\text{d}v(x)}{\text{d}x}}{v(x)}+x=-v(x)\Longleftrightarrow$$
$$-\frac{\frac{\text{d}v(x)}{\text{d}x}}{v^2(x)}-\frac{x}{v(x)}=1\Longleftrightarrow$$
Let $u(x)=\frac{1}{v(x)}$, which gives $\frac{\text{d}u(x)}{\text{d}x}=-\frac{\frac{\text{d}v(x)}{\text{d}x}}{v^2(x)}$:
$$\frac{\text{d}u(x)}{\text{d}x}-xu(x)=1\Longleftrightarrow$$
Let $\mu(x)=e^{\int -x\space\text{d}x}=e^{-\frac{x^2}{2}}$:
$$e^{-\frac{x^2}{2}}\frac{\text{d}u(x)}{\text{d}x}-\left(e^{-\frac{x^2}{2}}x\right)u(x)=e^{-\frac{x^2}{2}}\Longleftrightarrow$$
Substitute $e^{-\frac{x^2}{2}}x=\frac{\text{d}}{\text{d}x}\left(e^{-\frac{x^2}{2}}\right)$:
$$e^{-\frac{x^2}{2}}\frac{\text{d}u(x)}{\text{d}x}+\frac{\text{d}}{\text{d}x}\left(e^{-\frac{x^2}{2}}\right)u(x)=e^{-\frac{x^2}{2}}\Longleftrightarrow$$
$$\frac{\text{d}}{\text{d}x}\left(e^{-\frac{x^2}{2}}u(x)\right)=e^{-\frac{x^2}{2}}\Longleftrightarrow$$
$$\int\frac{\text{d}}{\text{d}x}\left(e^{-\frac{x^2}{2}}u(x)\right)\space\text{d}x=\int e^{-\frac{x^2}{2}}\space\text{d}x\Longleftrightarrow$$
$$e^{-\frac{x^2}{2}}u(x)=\sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{x}{\sqrt{2}}\right)+\text{C}\Longleftrightarrow$$
$$u(x)=e^{\frac{x^2}{2}}\left(\sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{x}{\sqrt{2}}\right)+\text{C}\right)\Longleftrightarrow$$
$$v(x)=\frac{2e^{-\frac{x^2}{2}}}{\sqrt{2\pi}\text{erf}\left(\frac{x}{\sqrt{2}}\right)+2\text{C}}\Longleftrightarrow$$
$$v(x)=\frac{2e^{-\frac{x^2}{2}}}{\sqrt{2\pi}\text{erf}\left(\frac{x}{\sqrt{2}}\right)+\text{C}}\Longleftrightarrow$$
$$e^{y(x)}=\frac{2e^{-\frac{x^2}{2}}}{\sqrt{2\pi}\text{erf}\left(\frac{x}{\sqrt{2}}\right)+\text{C}}\Longleftrightarrow$$
$$y(x)=\frac{1}{2}\left(2\ln(2)-x^2-2\ln\left(\sqrt{2\pi}\text{erf}\left(\frac{x}{\sqrt{2}}\right)+\text{C}\right)\right)$$
With $\text{C}$ is an arbitrary constant.
|
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|
Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
I'm not sure how to do this integration. It looks like partial fractions but I'm unsure.
|
We have
$$I = \int_0^{\infty} \dfrac{dx}{(1+x)^3+1} = \int_1^{\infty} \dfrac{dx}{x^3+1} = \int_1^0 \dfrac{-dx/x^2}{1/x^3+1} = \int_0^1 \dfrac{xdx}{1+x^3}$$
Hence,
$$I = \int_0^1\dfrac{x+1}{3(x^2-x+1)}dx - \int_0^1\dfrac{dx}{3(x+1)} = \dfrac{\pi}{3\sqrt{3}}-\dfrac{\log2}3$$
|
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|
Finding $\sqrt{(14+6\sqrt 5)^3}+\sqrt{(14-6\sqrt 5)^3}$ Find $$\sqrt{(14+6\sqrt{5})^3}+ \sqrt{(14-6\sqrt{5})^3}$$
A.$72$
B.$144$
C.$64\sqrt{5}$
D.$32\sqrt{5}$
How to cancel out the square root?
|
Note that $$14\pm 6\sqrt{5}=14\pm 2\sqrt{45}=9+5\pm 2\sqrt{9\times 5}=(\sqrt{9}\pm\sqrt{5})^2$$
|
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|
Given the conditions above, find when $x$, $y$, $z$ satisfy below: $ (x^2-1)(y+1)=\dfrac{z^2+1}{y-1}$ Let $x,y,z \in \mathbb{Z^+}$ and $x \neq y \neq z$.
Given the conditions above, find when $x$, $y$, $z$ satisfy below:
$$ (x^2-1)(y+1)=\frac{z^2+1}{y-1}\,.$$
What I did was I factored the numerator to
$$(x+1)(x-1)(y+1)=\frac{z^2+1}{y-1}\,,$$
but I am having trouble figuring out how to isolate the variables. I tried some values with trial and error and wasn't able to get any.
|
The only solution is $x = y = z = 0$.
Suppose there were a solution with x, y, and z not all zero. Choose the maximum $k$ with $2^k | gcf(x, y, z)$. So $x = 2^k m$, $y = 2^k n$, and $z = 2^k p$, and $m$, $n$, and $p$ are not all even.
$(xy)^2 = x^2 + y^2 + z^2$
$(2^k m 2^k n)^2 = (2^k m)^2 + (2^k n)^2 + (2^k p)^2$
$(4^k mn)^2 = 4^k m^2 + 4^k n^2 + 4^k p^2$
$16^k (mn)^2 = 4^k m^2 + 4^k n^2 + 4^k p^2$
$4^k (mn)^2 = m^2 + n^2 + p^2$
But the last equation only has solutions (mod 4) when m, n, and p are all even, contradicting the choice of k.
|
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|
Integrate $\int{ \frac{r}{(h^2 + r^2 - 2^{1/2}hr)^{1/2}} dr }$ This integral comes from a physics book when calculating potential difference between vertex and center of cone. I'm not good at integration. Please help.
update:
This is the required answer.
The second integral is in ln form. I have reached there by using the required formula but equation inside ln is less by factor of $2$. What am i doing wrong?
Also how formula for ln is derived?
|
Let me write your integral as $$ \int \frac{x}{\sqrt{x^2-\sqrt{2}ax+a^2}}dx$$
Then first consider the change of variable $u={x^2-\sqrt{2}ax+a^2} $, the $du=(2x-\sqrt{2}a )dx$, thus the integral becomes: $$ \int \frac{x}{\sqrt{x^2-\sqrt{2}ax+a^2}}dx = \frac{1}{2} \int \frac{2x-\sqrt{2}a+\sqrt{2}a}{\sqrt{x^2-\sqrt{2}ax+a^2}}dx=$$
$$\frac{1}{2} \int \frac{2x-\sqrt{2}a}{\sqrt{x^2-\sqrt{2}ax+a^2}}dx + \frac{1}{2} \int \frac{\sqrt{2}a}{\sqrt{x^2-\sqrt{2}ax+a^2}}dx = $$
$$ \frac{1}{2} \int \frac{du}{\sqrt{u}} + \frac{a}{\sqrt{2}} \int \frac{1}{\sqrt{x^2-\sqrt{2}ax+a^2}}dx $$
The first integral is simply $\sqrt{u}$, again to the initial variable, the first integral is $\sqrt{x^2-\sqrt{2}ax+a^2}$.
Now, consider the second inetgral, and use complete square in the denomenator $$\int \frac{1}{\sqrt{x^2-\sqrt{2}ax+a^2}} = \int \frac{dx}{\sqrt{x^2-\sqrt{2}ax+ \frac{a^2}{2}+\frac{a^2}{2}}}= \int \frac{dx}{\sqrt{(x-\frac{a}{\sqrt{2}})^2+ \frac{a^2}{2}}}= \int \frac{dx}{\frac{a}{\sqrt{2}}\sqrt{\frac{2}{a^2}(x-\frac{a}{\sqrt{2}})^2+ 1}} =\frac{\sqrt{2}}{a}\int \frac{dx}{\sqrt{(\frac{\sqrt{2}}{a}x-1)^2+ 1}} $$
let $v= \frac{\sqrt{2}}{a}x-1$, then $dv=\frac{\sqrt{2}}{a} dx $, and so this last integral can be written as $$ \int \frac{dv}{\sqrt{v^2+1}}= arsinh(v) $$ thus your final integral is
$$ \int \frac{x}{\sqrt{x^2-\sqrt{2}ax+a^2}}dx=\sqrt{x^2-\sqrt{2}ax+a^2} +\frac{a}{\sqrt{2}} arcsinh\Big(\frac{\sqrt{2}}{a}x-1 \Big) +C$$
|
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Convergence of $\sum_{n = 1}^\infty 1/n^2$. I know that $\sum_{n=1}^\infty 1/n$ diverges whereas $\sum_{n=1}^\infty 1/n^2$ converges.
Intuitively, I do not see the difference. If $n \to \infty$, the denominators in both fractions will be so big that the fraction approaches zero. So why doesn't both the series converge against zero?
I have read something about it, but I do not see the logic in it, and I guess it's quite obvious, so I hope someone can bring a good explanation.
|
To see that $\displaystyle\sum_{n=1}^\infty \frac{1}{n}$ diverges, you can group the terms,
$$\begin{alignat*}{2}
\frac{1}{2} + \frac{1}{3} &> \frac{1}{2}+\frac{1}{2} &= 1\\
\frac{1}{4} + \dots +\frac{1}{7} &> \frac{1}{4} + \dots + \frac{1}{4} &= 1\\
&\;\;\vdots&\vdots\;\;\;\,\\
\frac{1}{2^n}+\dots+\frac{1}{2^{n+1}-1} &> \frac{1}{2^n}+\dots +\frac{1}{2^n} &= 1
\end{alignat*}$$
And $\displaystyle\sum_{n=2}^\infty \frac{1}{n^2}$ converges by comparison:
$$
\sum_{n=2}^N\frac{1}{n^2} < \sum_{n=2}^N \frac{1}{n(n-1)} = \sum_{n=2}^N\frac{1}{n-1}-\frac{1}{n} = 1 -\frac{1}{N}\to 1 \quad(\text{telescoping}).
$$
These are the most intuitive explanations I've got.
|
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|
A fence of $y$ ft is $x$ ft from a wall, find shortest ladder using trigonometry
A fence $y$ ft high is $x$ ft from a wall. Find the length of the
shortest ladder that will rest with one end on the ground and the
other end on the wall.
This is a classic problem that is expected to be solved with trigonometry but I can't seem to get the right answer.
First, let the angle between the ground and the ladder be $\theta$, I express the total length of the ladder $L$ as the sum of the length between the ground and the fence $l_1$ and the length between the fence and the wall $l_2$:
$$
l_1 = \frac{y}{\sin\theta} \\
l_2 = \frac{x}{\cos\theta} \\
L = l_1 + l_2 = \frac{y}{\sin\theta} + \frac{x}{\cos\theta} = y\csc\theta + x\sec\theta \\
$$
Then, since this is an optimization problem, I differentiate the length:
$$
L' = -y\csc\theta\cot\theta + x\sec\theta\tan\theta = \frac{-y\cos\theta}{\sin^2\theta} + \frac{x\sin\theta}{\cos^2\theta} \\
$$
With this equation, I find the critical point where the derivative is zero:
$$
\frac{y\cos\theta}{\sin^2\theta} = \frac{x\sin\theta}{\cos^2\theta} \\
\frac{y}{x} = \frac{\sin^3\theta}{\cos^3\theta} = \tan^3\theta \\
$$
Now I make the assumption that $tan\theta = (y/x)^{1/3}$ which I then proceed to replace in the equation for the length of the ladder:
$$
L = \frac{y\cos\theta}{\sin\theta\cos\theta} + \frac{x\sin\theta}{\cos\theta\sin\theta} = \frac{y}{\cos\theta\tan\theta} + \frac{x\tan\theta}{sin\theta} \\
L = \frac{y}{\cos\theta(y/x)^{1/3}} + \frac{x(y/x)^{1/3}}{\sin\theta} = \frac{y^{2/3}x^{1/3}}{\cos\theta} + \frac{x^{2/3}y^{1/3}}{\sin\theta}
$$
Finally I make another assumption that $sin\theta = y^{1/3}$ and $cos\theta = x^{1/3}$ because $tan\theta = (\sin\theta/\cos\theta) = (y/x)^{1/3}$:
$$
L = \frac{y^{2/3}x^{1/3}}{x^{1/3}} + \frac{x^{2/3}y^{1/3}}{y^{1/3}} = y^{2/3} + x^{2/3}
$$
But the actual answer is:
$$
L = (y^{2/3} + x^{2/3})^{3/2}
$$
So where did I miss the $3/2$ exponent?
|
Your last assumption is the problem:
$$
\tan\theta = \frac{\sin\theta}{\cos\theta} = (\frac{y}{x})^{1/3} \\
\frac{\sin^2\theta}{\cos^2\theta} = (\frac{y}{x})^{2/3} \\
x^{2/3}\sin^2\theta = y^{2/3}\cos^2\theta \\
\sin^2\theta + \cos^2\theta = \frac{y^{2/3}\cos^2\theta}{x^{2/3}} + \cos^2\theta \\
\cos^2\theta(\frac{y^{2/3}}{x^{2/3}} + 1) = 1 \\
\cos^2\theta = \frac{x^{2/3}}{y^{2/3} + x^{2/3}} \\
\cos\theta = \frac{x^{1/3}}{(y^{2/3} + x^{2/3})^{1/2}} \\
$$
If you do the same for $\sin\theta$, you get:
$$
\sin\theta = \frac{y^{1/3}}{(x^{2/3} + y^{2/3})^{1/2}} \\
$$
If you replace $\cos\theta$ and $\sin\theta$ with those expressions, you then get the right answer.
|
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|
Evaluate the integral $\int \sqrt{(x-a)(b-x)}$ I'm trying to figure out how to evaluate the following integral:
$$\int \sqrt{(x-a)(b-x)} \, dx $$
I've tried various trig substitutions, but can't seem to get anywhere. This is an exercise in Apostol's Calculus Volume 1 (Section 6.22, Exercise 46). The solution provided in the text is
$$\frac{1}{4} |b-a|(b-a) \arcsin \sqrt{\frac{x-a}{b-a}} + \frac{1}{4} \sqrt{(x-a)(b-x)} (2x-(a+b)) + C.$$
|
Notice, $$\int \sqrt{(x-a)(b-x)}\ dx$$ $$=\int\sqrt{(a+b)x-x^2-ab}\ dx$$
$$=\int \sqrt{\left(\frac{a+b}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2-ab}\ dx$$
$$=\int \sqrt{\left(\frac{a+b}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2-ab}\ dx$$
$$=\int \sqrt{\left(\frac{\sqrt{a^2+b^2}}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2}\ dx$$
substituting $\left(x-\frac{a+b}{2}\right)=t$, one should get
$$=\frac{1}{2}\left[\left(x-\frac{a+b}{2}\right)\sqrt{\left(\frac{\sqrt{a^2+b^2}}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2}+\frac{a^2+b^2}{4}\sin^{-1}\left(\frac{\left(x-\frac{a+b}{2}\right)}{\frac{\sqrt{a^2+b^2}}{2}}\right)\right]$$
$$=\frac{1}{2}\left[\left(x-\frac{a+b}{2}\right)\sqrt{(x-a)(b-x)}+\frac{a^2+b^2}{4}\sin^{-1}\left(\frac{2x-a-b}{\sqrt{a^2+b^2}}\right)\right]$$
|
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|
If $a,b$ are the roots of the equation $x^2-2x+3=0$ obtain the equation whose roots are $a^3-3a^2+5a-2$, $b^3-b^2+b+5$ I have been trying this using sum of roots and product of roots but it gets too lengthy. So I found the roots of the given equation which are imaginary and tried to replace the values in the two given roots. Still I am not able to solve this.
|
We have
$$ a^3-3a^2+5a-2=(a-1)(a^2-2a+3)+1=1$$
and
$$b^3-b^2+b+5=(b+1)(b^2-2b+3)+2=2. $$
So the desired polynomial is
$$ (X-1)(X-2)=X^2-3X+2.$$
|
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|
Find the Values of n and k for which the determinant of the Matrix M(n,k) is Singular I have been stuck on this problem for a couple of days, I don't want the answer, but I would appreciate some help in finding it! Thanks in advance!
Consider a Symmetric Square Matrix $M(n,k)$ such that there is $n$ lines and $n$ columns, in which there is $-1$ everywhere expect on the diagonal, where there is $k$ everywhere.
i.e. $M(4,7) = \begin{bmatrix}7&-1&-1&-1\\-1&7&-1&-1\\-1&-1&7&-1\\-1&-1&-1&7\end{bmatrix}$
For which values of $n$ and of $k$ $M(n,k)$ is $Singular$. There is also a related Question where I initially need to prove that $Det(M(n,k)) =(k-n+1)*[(k+1)^{n-1}]$
|
Let me demonstrate the process for $M(4, 7)$. By adding rows $2-4$ to the first row, we move from $M(4,7)$ to the matrix
$$ B = \begin{pmatrix} 4 & 4 & 4 & 4 \\ -1 & 7 & -1 & -1 \\ -1 & -1 & 7 & -1 \\ -1 & -1 & -1 & 7 \end{pmatrix} $$
and we have $\det M(4,7) = \det B$. Multiplying the first row by $\frac{1}{4}$ we move from $B$ to the matrix
$$ C = \begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & 7 & -1 & -1 \\ -1 & -1 & 7 & -1 \\ -1 & -1 & -1 & 7 \end{pmatrix} $$
and we have $4 \det C = \det B$. Finally, by adding the first row to rows $2-4$, we move from $C$ to the matrix
$$ D = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 8 & 0 & 0 \\ 0 & 0 & 8 & 0 \\ 0 & 0 & 0 & 8 \end{pmatrix}. $$
Calculating $\det D$ recursively using the first column, we see that $\det D = 8^3$ and so
$$ \det M(4,7) = \det B = 4 \det C = 4 \det D = 4 \cdot 8^3. $$
Applying this process to a general matrix $M(n,k)$ will show you that $$ \det M(n,k) = (k - (n - 1)) \cdot (k + 1)^{n-1} $$
and so $M(n,k)$ will be singular if and only if $\det M(n,k) = 0$ if and only if $k = -1$ or $k = (n - 1)$.
|
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|
Power series representation of $\frac{1+x}{1-x}$ explanation? I don't really get how any of this works. I tried looking at this power function example already to figure it out. But I got lost when this part comes up. $$\sum_{n=0}^{\infty}x^n=x^0+\sum_{n=1}^{\infty}x^n=1+\sum_{n=1}^{\infty}x^n$$
Why is this entire step happening and why is the $n$ index moving? And do you just distribute the $\sum_{n=1}^{\infty}x^n$ into $(1+x)$?
|
You have a sum of two geometric series.
\begin{align}
\frac 1 {1-x} & = 1 + x + x^2 + x^3 + \cdots \\[10pt]
\frac x {1-x} & = x + x^2 + x^3 + x^4 + \cdots
\end{align}
You're looking for the sum of the two above.
\begin{align}
\sum_{n=0}^\infty x^n & = x^0 + x^1 + x^2 + x^3 + x^4 + \cdots \\[10pt]
& = x^0 + \Big( x^1 + x^2 + x^3 + x^4 + \cdots \Big) \\[10pt]
& = x_0 + \sum_{n=1}^\infty x^n \\[10pt]
& = 1+\sum_{n=1}^{\infty}x^n
\end{align}
|
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|
Evaluation of $\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$ Evaluate
$$\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$$
My main concern is finding the indefinite integral as once i have that the rest is fairly straight forward. Please give a detailed answer with reference to why you made each substitution (what indicated that said substitution would work etc.)
My initial substitution was $x= \sin \theta$ which tidies it up a bit
$$\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \space dx$$
Also the answer from Wolfram Alpha is;
$$ = \frac{1}{4}\bigg( log(1-2 x^2)+2 \tanh^{-1}\bigg(\frac{x}{\sqrt{1-x^2}}\bigg)+2 \sin^{-1}x\bigg)+constant$$
I need to be able to answer questions like this in under 15 mins so any help is appreciated
|
Since no one used it, let me consider the case of the antiderivative $$I=\int\frac{\cos x}{\sin x + \cos x} \space dx$$ Now, use the tangent half-angle substitution (Weierstrass substitution) $t=\tan(\frac x 2)$.
We so obtain $$I=\int \frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}\space dt$$ But $$t^4-2 t^3-2 t-1=(t^2+1)(t-r_1)(t-r_2)\qquad r_{1,2}=1\pm \sqrt 2$$ Using partial fraction decomposition, we have $$\frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}=\frac{1-t}{t^2+1}+\frac{1}{2 t-2 \sqrt{2}-2}+\frac{1}{2 t+2 \sqrt{2}-2}$$ that is to say $$\frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}=\frac{1}{1+t^2}-\frac{2t}{1+t^2}+\frac{1}{2 t-2 \sqrt{2}-2}+\frac{1}{2 t+2 \sqrt{2}-2}$$ and each term is easy to integrate. After simplifications, this leads to $$I=\tan ^{-1}(t)+\frac{1}{2} \log \left(\frac{-t^2+2 t+1}{t^2+1}\right)$$ Now, if integration is from $t=0$ to $t=1$, the logarithms disappear and youare just let with $\tan ^{-1}(1)=\frac \pi 4$.
|
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|
Prove for all reals: $a^2+b^2+c^2\geq 2(a+b+c)-3$ Problem:
Prove this for all reals $a,b,c$:
$$a^2+b^2+c^2\geq 2(a+b+c)-3$$
Attempt:
I am trying to work backwards.
$$a^2+b^2+c^2\geq 2(a+b+c)-3$$
$$(a+b+c)^2\geq 2(a+b+c)-3 + 2(ab+bc+ca)$$
$$(a+b+c)^2-2(a+b+c)\geq 2(ab+bc+ca)-3 $$
$$(a+b+c)(a+b+c-2)\geq 2(ab+bc+ca)-3 $$
I am stuck here. And need hints.
|
it is equivalent to $$(a-1)^2+(b-1)^2+(c-1)^2\geq 0%$$
|
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|
Solve $y^3-3y-\sqrt{2}=0$ using trigonometry This is a part of a larger question.
I had to show that for $4x^3-3x-\cos 3\alpha=0$ one of the solutions is $\cos \alpha$ and then find the other two solutions. Here they are:
$$4x^3-3x-\cos 3\alpha = (x-\cos \alpha)(2x+\cos \alpha + \sqrt{3} \sin \alpha)(2x+\cos \alpha - \sqrt{3} \sin \alpha)$$
I have to use the above and the following results: $\cos 15^{\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin 15^{\circ}=\frac{\sqrt{3}-1}{2\sqrt{2}}$ to find the solutions of the following:
$$y^3-3y-\sqrt{2}=0$$
I assumed that the constant term must be the equivalent of the cosine term and tried to find alpha so that I have one solution and then can derive the other. But since $\arccos {\sqrt{2}}$ is not trivially defined, this is not a correct approach. Or at least it is not correct the way I am doing it. Also I would have to do a bit more trigs since the second polynomial is not equivalent to the first one. There must be an easier, neater solution.
|
Recall the triplication formula:
$$
\cos3\alpha=4\cos^3\alpha-3\cos\alpha
$$
so, if your equation is $y^3-3y=\sqrt{2}$, you can first set $y=at$, so
$$
a^3t^3-3at=\sqrt{2}
$$
and you'd like that $a^3/3a=4/3$, so you can take $a=2$: $8t^3-6t=\sqrt{2}$; setting $t=\cos\alpha$, we get
$$
\cos3\alpha=\frac{\sqrt{2}}{2}
$$
so
$$
3\alpha=\frac{\pi}{4}+2k\pi
$$
where $k=0$, $k=1$ or $k=2$. Hence
$$
\frac{\pi}{12},\quad\frac{3\pi}{4},\quad\frac{17\pi}{12}
$$
are the solutions for $\alpha$.
Now,
$$
2\cos\frac{\pi}{12}=2\sqrt{\frac{1+\cos(\pi/6)}{2}}=
\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}
$$
then
$$
2\cos\frac{3\pi}{4}=-\sqrt{2}
$$
and
$$
2\cos\frac{17\pi}{12}=-2\cos\frac{5\pi}{12}=
-2\sqrt{\frac{1+\cos(5\pi/6)}{2}}=
-\sqrt{2-\sqrt{3}}=\frac{\sqrt{6}-\sqrt{2}}{2}
$$
|
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|
Derivative of $10^x\cdot\log_{10}(x)$
Derive $10^x\cdot\log_{10}(x)$
$$10^x\cdot \ln(10)\cdot \log_{10}(x)+\frac{1}{x\cdot \ln(10)}\cdot 10^x$$
But WolframAlpha gives another solution. Where am I wrong?
|
$$\frac{\text{d}}{\text{d}x}\left(10^x\cdot\log_{10}(x)\right)=\frac{\text{d}}{\text{d}x}\left(\frac{10^x\ln(x)}{\ln(10)}\right)=$$
$$\frac{\frac{\text{d}}{\text{d}x}\left(10^x\ln(x)\right)}{\ln(10)}=\frac{\ln(x)\frac{\text{d}}{\text{d}x}(10^x)+10^x\frac{\text{d}}{\text{d}x}(\ln(x))}{\ln(10)}=$$
$$\frac{10^x\ln(10)\ln(x)+10^x\frac{\text{d}}{\text{d}x}(\ln(x))}{\ln(10)}=\frac{10^x\ln(10)\ln(x)+\frac{10^x}{x}}{\ln(10)}=$$
$$10^x\left(\ln(x)+\frac{1}{x\ln(10)}\right)$$
|
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|
Find all irreducible polynomials of degrees 1,2 and 4 over $\mathbb{F_2}$. Find all irreducible polynomials of degrees 1,2 and 4 over $\mathbb{F_2}$.
Attempt: Suppose $f(x) = x^4 + a_3x^3 + a_2x^2 + a_1x + a_0 \in \mathbb{F_2}[x]$. Then since $\mathbb{F_2} =${$0,1$}, then we have either $0$ or $1$ for each $a_i$. Then we have two choices for the $4$ coefficients, hence there are 16 polynomials of degree $4$ in $\mathbb{F_2}[x]$.
Recall $f(x)$ is irreducible if and only if it has not roots. Then
$f_1 = x$ is irreducible because it has not roots
$f_2 = x + 1$ is also another irreducible polynomial.
$f_3 = x^4 + x^2 + x = x ( x^3 + x + 1) $ is reducible.
$f_4 = x^4 = x^3* x$ is reducible
$f_5 = x^4 + x + 1$
Can someone please help me? Is there a way I can save time in finding the irreducible polynomials, other than just trying to come up with polynomials. Any better approach or hint would really help! Thank you !
|
Degree $1$, clearly $x$ and $x+1$.
Degree $2$, notice the last coefficient must be one, so there are only two options, $x^2+x+1$ and $x^2+1$. Clearly only $x^2+x+1$ is irreducible.
Degree $4$. There are $8$ polynomials to consider, again, because the last coefficient is $1$, now notice a polynomial is divisible by $x+1$ if and only if the sum of its coefficients is even. So the only polynomials without factors of degree $1$ are four:
$x^4+x^3+x^2+x+1$
$x^4+x^3+1$
$x^4+x^2+1$
$x^4+x+1$.
Of course, we are missing the possibility it is the product of two irreducibles of degree $2$, but the only combination is $(x^2+x+1)(x^2+x+1)=x^4+x^2+1$.
Hence the irreducible ones are:
$x^4+x^3+x^2+x+1,x^4+x^3+1,x^4+x+1$
|
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|
$a^2-b^2$ will be a perfect square if the ratio of $a$ and $b$ is
$a^2-b^2$ will be a perfect square if the ratio of $a$ and $b$ is
Options are: A) $1:2$, B) $3:4$, C) $5:4$, D) $4:5$, E) None of these
So my first observation was that $|a|$ has to be greater than $|b|$. So only options that suffice this condition are $C)$ and $E)$. Also, if $C)$ was correct, then we get $a=5k$; $b=4k$, $25k^2-16k^2=9k^2$. Which is giving us a perfect square. But isn't there any rigorous way also?
|
A) $1^2-2^2=-3$,
B) $3^2-4^2=-7$,
C) $5^2-4^2=9$,
D) $4^2-5^2=-9$.
This rules out A), B) and D).
C) implies $(5k)^2-(4k)^2=(3k)^2$, so that the property holds for all $(a,b)=k(5,4)$.
If you don't want to use the multiple choices, let
$$a^2-b^2=c^2.$$
You can factor as
$$(a+b)(a-b)=c^2=pq$$
and solve with
$$a=\frac{p+q}2,b=\frac{p-q}2.$$
The ratio is
$$\frac ab=\frac{p+q}{p-q},$$ where $pq$ is a perfect square.
For instance, $p=c^2,q=1$ for any $c$ gives
$$\frac ab=\frac{c^2+1}{c^2-1}$$among which
$$\frac ab=\frac{9+1}{9-1}=\frac54.$$
And there are other possibilities...
|
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|
Show $\sin(\frac{\pi}{3})=\frac{1}{2}\sqrt{3}$ I have to show that
$$\sin\left(\frac{\pi}{3}\right)=\frac{1}{2}\sqrt{3}$$
and
$$\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$$
Should I use the exponential function?
|
Not to confuse you. But..
$$x=\dfrac{\pi}{3}$$
$$3x=\pi$$
$$\sin 3x=\sin \pi$$
$$3\sin x-4\sin^3x=0$$
$$\sin x=0 \text{ or } \sin x = \dfrac{\sqrt{3}}{2} \text{ or } \sin x = \dfrac{-\sqrt{3}}{2}$$
$$\text{as } 0<\dfrac{\pi}{3}<\dfrac{\pi}{2} \text{, } \sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$$
|
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Find all functions $f$ such that $f(x-f(y)) = f(f(x)) - f(y) - 1$ Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that $f(x-f(y)) = f(f(x)) - f(y) - 1$.
So far, I've managed to prove that if $f$ is linear, then either $f(x) = x + 1$ or $f(x) = -1$ must be true. I did this by plugging in $x=0$ to the above equation, which yields
$$ f(-f(y)) = f(f(0)) - f(y) - 1$$
and plugging in $x$ instead of $y$ and subtracting, this becomes
$$ f(-f(y)) - f(-f(x)) = f(x) - f(y)$$
Assuming $f(x) = ax + b$ then gives
\begin{align*}
f(-ay-b) - f(-ax-b) = ax - ay &\Rightarrow -a^2y - ab + a^2x + ab = a(x-y) \\&\Rightarrow a^2(x-y) = a(x-y) .
\end{align*}
Thus $a=1$ or $a=0$. If $a=0$, then the original equation becomes $b = b - b - 1$, thus $b=-1$. If $a=1$, the original equation becomes
$$ x-y-b+b = x+2b-y-b-1 \Longrightarrow b=1.$$
I briefly tried finding a quadratic function that works but didn't find anything. So my question is: how can I either show that $f$ must be linear or find all other representations?
|
Let $f\colon \Bbb Z\to \Bbb Z$ be a any function function with $$\tag0f(x-f(y)) = f(f(x)) - f(y) - 1$$
for all $x,y\in\Bbb Z$.
Letting $y=f(x)$ we find $$f(x-f(f(x)))=-1. $$
So for $a=-f(f(0))$ we have $f(a)=-1$. Then with $y=a$, $$\tag1f(x+1)=f(f(x)) $$
Then $(0)$ becomes
$$\tag2 f(x-f(y))=f(x+1)-f(y)-1 $$
Or with $g(x):=f(x)+1$ (and $x\leftarrow x-1$)
$$\tag3g(x-g(y))=g(x)-g(y)$$
From $(3)$ we see that the image of $g$ is a subgroup of $\Bbb Z$, hence it is either $\{0\}$ (in which case $f(x)=-1$), or $c\Bbb Z$ for some $c\ge 1$.
In that case, for $n\in\Bbb Z$ we find $y$ with $g(y)=nc$ and so have $g(x+ nc)=g(x)+nc$. Thus $g$ is determined by the values $g(0),\ldots, g(c-1)$. On the other hand, these values can indeed be chosen freely. In other words:
Claim 1. Let $c\in\Bbb N$ and $b_0,\ldots, b_{c-1}\in\Bbb Z$. Then the function $g$ given by
$$ g(x)= (n+b_r)c$$
where $x=nc+r$, $0\le r<c$
is a solution to $(3)$, and all non-zero solutions of $(3)$ are obtained this way.
Proof.
Let $x=nc+r$, $y=mc+s$ with $0\le r,s<c$. Then
$$\begin{align}g(x-g(y))&=g(nc+r-(m+b_s)c)\\
&=g((n-m-b_s)c+r)\\
&=(n-m-b_s+b_{r})c\\
&=(n+b_r)c-(m+b_s)c\\&=g(x)-g(y)\end{align}$$
That all non-zero solutions are of this form has been shown above. $\square$
Then the solutions $f$ of $(2)$ (apart from $f(x)=-1$) are precisely those of the form $f(x)=g(x)-1$ with $g$ as in Claim 1.
Such $f$ is a solution to the original $(0)$ if and only if we additionally have $(1)$ for all $x$.
Note that for $x=nc+r$, $0\le r<c$, we have $f(x)=g(x)-1=(n+b_r)c-1=(n+b_r-1)c+c-1$ so that $$f(f(x))=(n+b_r-1+b_{c-1})c-1.$$
On the other hand,
$$f(x+1)=g(x+1)-1=\begin{cases}(n+b_{r+1})c-1&\text{if }r<c-1\\(n+1+b_0)c-1&\text{if }r=c-1\end{cases}$$
We conclude that $b_{r+1}=b_r+b_{c-1}-1$ for $0\le r<c-1$, and that $2b_{c-1}-1=b_0+1$. From the first we see that $b_r=b_0+rb_{c-1}-r$, so
$$\begin{align}b_{c-1}&=b_0+1+(c-1)b_{c-1}-c\\
&=2b_{c-1}-1+(c-1)b_{c-1}-c\\
&=(c+1)b_{c-1}-c-1\end{align}$$ and finally
$$b_{c-1} = \frac {c+1}c.$$
This is an integer only for $c=1$ and in that case we arrive at $b_0=2$
Thus the only solutions to $(0)$ apart from $f(x)=-1$ is
$$f(x)=x+1.$$
|
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|
How to see this inequality Consider
$$
\frac{dx}{dt}=\sqrt{x^2+1}+t^2,\qquad x\in\mathbb{R}.
$$
Then I do not see why it is
$$
\lvert\frac{dx}{dt}\rvert\leq 2\lvert x\rvert+t^2.
$$
|
The steps that come to mind are
$$\begin{array}{}
\frac{dx}{dt}=\sqrt{x^2+1}+t^2 \\
|\frac{dx}{dt}|=|\sqrt{x^2+1}+t^2| \\
|\frac{dx}{dt}|\le|\sqrt{x^2+1}|+|t^2| & \qquad \text{Triangle Inequality} \\
|\frac{dx}{dt}|\le \sqrt{x^2+1} + t^2 \\
\end{array}$$
If we had $\sqrt{x^2+1} \le 2|x|$, then we could get your final result but
$$\begin{array}{}
\sqrt{x^2+1} \le 2|x| \\
x^2+1 \le 4x^2 \\
3x^2 \ge 1 \\
x^2 \ge \frac{1}{3} \\
|x| \ge \frac{1}{\sqrt3}
\end{array}$$
So we can say that
$$|\frac{dx}{dt}|\le \ 2|x| + t^2, \qquad \quad |x| \ge \frac{1}{\sqrt3}$$
|
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|
Prove that $n^{n^{n^{n}}}- n^{n^{n}}$ is divisible by $1989$ Problem
Let $n$ be a positive integer with $n \geq 3$. Show that $$n^{n^{n^{n}}}- n^{n^{n}}$$ is divisible by $1989$.
I don't really know where to begin with this question. Maybe I could do some case work on $n$ being even or odd, but I am not sure if that would work or not.
|
$1989$ factors as $3^2\cdot 13\cdot 17$, so it is enough to show that $n\uparrow\uparrow 4 - n\uparrow\uparrow 3 \equiv 0 \pmod a$ for $a=9,13,17$.
ANALYSIS. Suppose $n$ is coprime to $a$ -- otherwise the congruence clearly holds. Then,
$$ n\uparrow\uparrow 4 - n\uparrow\uparrow 3 = n^{n\uparrow\uparrow 2} (n^{n\uparrow\uparrow 3-n\uparrow\uparrow 2} - 1) $$
This will be divisible by $a$ if
$$ n^{n\uparrow\uparrow 3-n\uparrow\uparrow 2} \equiv 1 \pmod a $$
which is the case if
$$ n\uparrow\uparrow 3 - n\uparrow\uparrow 2 \equiv 0 \pmod{\lambda(a)} $$
When $a=9,13,17$ we have $\lambda(a)=6,12,16$, so it will be enough to show
$$ n\uparrow\uparrow 3 - n\uparrow\uparrow 2 \equiv 0 \pmod b \qquad\text{for } b=3, 16$$
Again, we can suppose without loss of generality that $n$ is coprime to $b$ (this wouldn't work if $n=2$, but fortunately we're assuming $n\ge 3$; if $n$ is an even number that is at least $4$, both $n\uparrow\uparrow 3$ and $n\uparrow\uparrow 2$ will be divisible by 16). As before we have
$$ n\uparrow\uparrow 3 - n\uparrow\uparrow 2 = n^{n^n}- n^n = n^n (n^{n^n-n} - 1) $$
so we look for a proof that
$$ n^n - n \equiv 0 \pmod{\lambda(b)} $$
and $\lambda(b)=2,4$ when $b=3,16$, so we look for
$$ n^n-n \equiv 0 \pmod 4 $$
This is not actually true when $n$, is an odd multiple of $2$, but luckily for the $b=16$ case we have already assumed that $n$ is odd, and for $b=3$ we only need $n^n-n$ to be even which is certainly always the case.
Thus we can start SYNTHESIZING a proof from the bottom up:
Lemma 1. $n^n-n$ is always even. Proof. Trivial.
Lemma 2. If $n$ is odd, then $n^n-n \equiv 0 \pmod 4$.
Proof. $n^n-n=n(n^{n-1}-1)$, and since $n-1$ is even, it is a multiple of $\lambda(4)=2$, so $n^{n-1} \equiv 1 \pmod 4$.
Lemma 3. $n^{n^n}-n^n \equiv 0 \pmod 3$.
Proof. This is immediate if $3\mid n$. Otherwise, we have $n^{n^n}-n^n = n^n(n^{n^n-n}-1)$, and $n^n-n$ is even by Lemma 1; since $\lambda(3)=2$ we have $n^{n^n-n}\equiv 1\pmod 3$.
Lemma 4. If $n\ge 3$, then $n^{n^n}-n^n \equiv 0 \pmod{16}$.
Proof. If $n$ is even and $\ge 4$, then this is immediate. Otherwise $n$ is coprime to 16, and we have $n^{n^n}-n^n = n^n(n^{n^n-n}-1)$. By Lemma 2, $n^n-n$ is a multiple of $\lambda(16)=4$, so $n^{n^n-n}\equiv 1\pmod{16}$.
Lemma 5. If $n\ge 3$. then $n^{n^n}-n^n \equiv 0$ modulo $6$, $12$, or $16$.
Proof. By Lemmas 3 and 4.
Lemma 6. If $n\ge 3$, then $n^{n^{n^n}}-n^{n^n} \equiv 0$ modulo $p\in\{13,17\}$.
Proof. This is immediate if $p\mid n$. Otherwise $n^{n^{n^n}}-n^{n^n} = n^{n^n}(n^{(n^{n^n}-n^n)}-1)$, and the exponent $n^{n^n}-n^n$ is a multiple of $\lambda(p)$ by Lemma 5, so $n^{(n^{n^n}-n^n)}\equiv 1\pmod p$.
Lemma 7. If $n\ge 3$, then $n^{n^{n^n}}-n^{n^n} \equiv 0 \bmod9$.
Proof. Immediate if $3\mid n$, since $n^{n^n}$ and $n^n$ are both $\ge 2$. Otherwise, $n$ is coprime to $9$, and the same reasoning as in Lemma 6 applies.
Corollary. If $n\ge 3$, then $n^{n^{n^n}}-n^{n^n} \equiv 0 \bmod{1989}$. Proof. By Lemmas 6 and 7.
|
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|
Simplify $2\cos(6\pi/7) + 2\cos(2\pi/7) + 2\cos(4\pi/7) + 1$ . I am trying to simplify
$$2\cos(6\pi/7) + 2\cos(2\pi/7) + 2\cos(4\pi/7) + 1$$
However if I plug this in the calculator the answer is zero. Is there a way to keep on simplifying without the calculator?
I know the identity $2\cos(\theta) = (e^{i\theta} +e^{-i\theta}) $ but I think that might make it worst.
Thanks!
|
HINT:
Your sum equals
$$\sum_{k=0}^6 \cos\frac{2 k \pi}{7}$$
because $\cos (\theta) = \cos (2\pi - \theta)$, so for instance $\cos\frac{4 \pi}{7} = \cos\frac{10 \pi}{7}$. Now this sum equals the real part of
$$\sum_{k=0}^6 (\cos\frac{2 k \pi}{7}+ i \sin \frac{2 k \pi}{7}) = \sum_{k=0}^6 (\cos\frac{2 \pi}{7}+ i \sin \frac{2 \pi}{7})^k$$
and the last sum equals $0$ ( sum of a geometric progression, and use $(\cos\frac{2 \pi}{7}+ i \sin \frac{2 \pi}{7})^7 = 1$)
|
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|
Calculus - limit of a function: $\lim\limits_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$ How do you compute the following limit without using the l'Hopital rule?
If you were allowed to use it, it becomes easy and the result is $\sqrt{3}\over 3$ but without it, I am not sure how to proceed. $$\lim_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$$
|
$\frac {\sin x\cos \frac {\pi}{3}- \cos x\sin\frac {\pi}{3}}{1+2\cos x}\\
\frac {\frac 12\sin x- \frac {\sqrt 3}{2} \cos x}{1-2\cos x}\\
\frac {\sin x- \sqrt 3\cos x}{2-4\cos x}\\
\frac {\sin^2 x- 3\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\
\frac {\sin^2 x + \cos^2 x- 4\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\
\frac {1- 4\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\
\frac {(1- 2\cos x)(1+2\cos x)}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\
\frac {(1+2\cos x)}{2(\sin x + \sqrt 3 \cos x)}$
And now we can plug $x = \frac {\pi}{3}$
$\frac {2}{2(\frac {\sqrt 3}{2} + \frac {\sqrt 3}{2})}$
$\frac {1}{\sqrt 3}$
|
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|
On $p^2 + nq^2 = z^2,\;p^2 - nq^2 = t^2$ and the "congruent number problem" (Much revised for brevity.) An integer $n$ is a congruent number if there are rationals $a,b,c$ such that,
$$a^2+b^2 = c^2\\
\tfrac{1}{2}ab = n$$
or, alternatively, the elliptic curve,
$$y = x^3-n^2x = x(x-n)(x+n)\tag1$$
is solvable in the rationals. Assume $x=(p/q)^2$. Then $(1)$ becomes,
$$\frac{p^2}{q^6}(p^4-n^2q^4) = y^2$$
or simply,
$$(p^2+nq^2)(p^2-nq^2) = w^2\tag2$$
Assuming $w=z\,t$ and equating factors, then $(2)$ becomes,
$$p^2 + nq^2 = z^2\\
p^2 - nq^2 = t^2\tag3$$
This is the implication given by Mathworld and OEIS.
Questions:
*
*Is it true that if $p^4-n^2q^4 = w^2$ is solvable, then is
$$p^2 + nq^2 = m z_1^2\\
p^2 - nq^2 = m z_2^2\tag4$$
necessarily also solvable for $m=1$?
*Is the solution to $(4)$ for $m=1$ the smallest for $p^4-n^2q^4 = w^2$?
P.S. This post made me re-visit congruent numbers.
|
Probably for the system.
$$\left\{\begin{aligned}&x^2+qy^2=a^2\\&x^2-qy^2=b^2\end{aligned}\right.$$
Write a simple formula as these numbers just to find.
$$x=p^4+s^4$$
$$y=2ps\sqrt{\frac{(p^2+s^2)(p^2-s^2)}{q}}$$
$$a=p^4+2p^2s^2-s^4$$
$$b=s^4+2p^2s^2-p^4$$
Or so.
$$x=p^4+6p^2s^2+s^4$$
$$y=2(p^2-s^2)\sqrt{\frac{2ps(p^2+s^2)}{q}}$$
$$a=p^4+4sp^3-2p^2s^2+4ps^3+s^4$$
$$b=p^4-4sp^3-2p^2s^2-4ps^3+s^4$$
It is clear that the number $q$ must be chosen so that the root was rational. Of course it's not really a solution, but it can generate and try all possible options. For finding solutions to the desired number - the task can be facilitated by considering the possible options - factorization.
|
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|
Find the length of AB In the diagram 4 circles of equal radius stand in a row in such a way that each circle touches the next one. $P$ is a point on the circumference of the first circle. The center of the fourth circle is point $Q$. The line $PQ$ goes through the centers of all four circles. $PC$ is a tangent on the fourth circle such that it intersects the second circle at points $A$ and $B$. Radius of the circles is $7$ and the length of $AB$ is $a\sqrt b$.
Find the length.
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Let $\angle CPQ=\alpha$. We have : $\sin \alpha=\dfrac{CQ}{QP}=\dfrac{7}{7^2}= \dfrac{1}{7}$.
The distance from PC tho the center of the second circle is $d=3\cdot 7 \cdot \sin \alpha=3$ so the chord $AB$ is $AB=2\sqrt{7^2-3^2}=4\sqrt{10}$
|
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Integral $\int_0^\infty\frac{\tanh^2(x)}{x^2}dx$ It appears that
$$\int_0^\infty\frac{\tanh^2(x)}{x^2}dx\stackrel{\color{gray}?}=\frac{14\,\zeta(3)}{\pi^2}.\tag1$$
(so far I have about $1000$ decimal digits to confirm that).
After changing variable $x=-\tfrac12\ln z$, it takes an equivalent form
$$\int_0^1\frac{(1-z)^2}{z\,(1+z)^2 \ln^2z}dz\stackrel{\color{gray}?}=\frac{7\,\zeta(3)}{\pi^2}.\tag2$$
Quick lookup in Gradshteyn—Ryzhik and Prudnikov et al. did not find this integral, and it also is returned unevaluated by Mathematica and Maple. How can we prove this result? Am I overlooking anything trivial?
Further questions: Is it possible to generalize it and find a closed form of
$$\mathcal A(a)=\int_0^\infty\frac{\tanh(x)\tanh(ax)}{x^2}dx,\tag3$$
or at least of a particular case with $a=2$?
Can we generalize it to higher powers
$$\mathcal B(n)=\int_0^\infty\left(\frac{\tanh(x)}x\right)^ndx?\tag4$$
Thanks to nospoon's comment below, we know that
$$\mathcal B(3)=\frac{186\,\zeta(5)}{\pi^4}-\frac{7\,\zeta(3)}{\pi^2}\tag5$$
I checked higher powers for this pattern, and, indeed, it appears that
$$\begin{align}&\mathcal B(4)\stackrel{\color{gray}?}=-\frac{496\,\zeta(5)}{3\,\pi^4}+\frac{2540\,\zeta(7)}{\pi^6}\\
&\mathcal B(5)\stackrel{\color{gray}?}=\frac{31\,\zeta(5)}{\pi^4}-\frac{3175\,\zeta(7)}{\pi^6}+\frac{35770\,\zeta(9)}{\pi^8}\\
&\mathcal B(6)\stackrel{\color{gray}?}=\frac{5842\,\zeta(7)}{5\,\pi^6}-\frac{57232\,\zeta(9)}{\pi^8}+\frac{515844\,\zeta(11)}{\pi^{10}}\end{align}\tag6$$
|
$$\mathcal{A}(m,n)=\int_{0}^{\infty}\frac{\tanh(z)^m}{z^n}\text{d}z\qquad m+1>n>1$$
We have
$$\begin{aligned}
&\mathcal{A}\left(1,p+1\right)
=\frac{(2^{p+1}-1)}{\pi^p\cos\frac{p\pi}{2} } \zeta(p+1)\\
&\mathcal{A}\left(2,p+1\right)=
\frac{(p+1)(2^{p+2}-1)}{\pi^{p+1}\sin\frac{p\pi}{2} } \zeta(p+2)\\
&\mathcal{A}\left(3,p+1\right)=
\frac{(2^{p+1}-1)}{\pi^p\cos\frac{p\pi}{2} } \zeta(p+1)
-\frac{(p+1)(p+2)(2^{p+3}-1)}{2\pi^{p+2}\cos\frac{p\pi}{2} }\zeta(p+3)\\
&\mathcal{A}\left(4,p+1\right)=
\frac{4(p+1)(2^{p+2}-1)}{3\cdot\pi^{p+1}\sin\frac{p\pi}{2} } \zeta(p+2)
-\frac{(p+1)(p+2)(p+3)(2^{p+4}-1)}{6\cdot\pi^{p+3}\sin\frac{p\pi}{2} }\zeta(p+4)\\
&\mathcal{A}\left(5,p+1\right)=
\frac{(2^{p+1}-1)}{\pi^p\cos\frac{p\pi}{2} } \zeta(p+1)
-\frac{5(p+1)(p+2)(2^{p+3}-1)}{6\cdot\pi^{p+2}\cos\frac{p\pi}{2} } \zeta(p+3)
+\frac{(p+1)(p+2)(p+3)(p+4)(2^{p+5}-1)}{24\cdot\pi^{p+4}\cos\frac{p\pi}{2} }\zeta(p+5)\\
&\mathcal{A}\left(6,p+1\right)=
\frac{23(p+1)(2^{p+2}-1)}{15\cdot\pi^{p+1}\sin\frac{p\pi}{2} } \zeta(p+2)
-\frac{(p+1)(p+2)(p+3)(2^{p+4}-1)}{3\cdot\pi^{p+3}\sin\frac{p\pi}{2} } \zeta(p+4)
+\frac{(p+1)(p+2)(p+3)(p+4)(p+5)(2^{p+6}-1)}{120\cdot\pi^{p+5}\sin\frac{p\pi}{2} }\zeta(p+6)
\end{aligned}$$
And substitute different p(can be a non-integer),your cases will done.
Proof:
For $n+m=2k$,consider the function
$$
f(z)=\frac{\tanh(z)^m}{z^n}
$$
And I get
$$
\int_{0}^{\infty}\frac{\tanh(z)^m}{z^n} \text{d}z
=-\pi i\sum_{k=-\infty}^{\infty} \text{Res}(f,\frac{\pi}{2}(2k+1)i )
$$
For $n+m=2k+1$,consider
$$
f(z)=\frac{\tanh(z)^m}{z^n}\ln z
$$
And I get
$$
\int_{0}^{\infty}\frac{\tanh(z)^m}{z^n} \text{d}z
=-\sum_{k=-\infty}^{\infty} \text{Res}(f,\frac{\pi}{2}(2k+1)i )
$$
I dont have time to write a full answer.Thanks.
(I will have exams for three days tomorrow.)
If we consider the function
$$
f(z)=\frac{\tanh z}{z^{1+p}}
$$
And use the keyhole contour,we can obtain
$$
(1-e^{-2p\pi i})\int_{0}^{\infty} \frac{\tanh z}{z^{1+p}}\text{d}z
= 2\pi i\sum_{k=-\infty}^{\infty}\text{Res}\left (\frac{\tanh z}{z^{1+p}}, \frac{\pi}{2}(2k+1)i \right )
$$
A few examples
$$
\begin{aligned}
&\int_{0}^{\infty} \frac{\tanh z}{z^{3/2}} \text{d}z
=\frac{4-\sqrt{2} }{\sqrt{\pi} } \zeta\left ( \frac{3}{2} \right )\\
&\int_{0}^{\infty} \frac{\tanh z}{z^{5/3}} \text{d}z
=\frac{2(2^{5/3}-1)}{\pi^{2/3}}\zeta\left ( \frac{5}{3} \right )\\
&\int_{0}^{\infty} \frac{\tanh^2 z}{z^{2}} \text{d}z
= \frac{14}{\pi^2}\zeta(3)\\
&\int_{0}^{\infty} \frac{\tanh^2 z}{z^{3/2}} \text{d}z
= \frac{24-3\sqrt{2} }{2\pi^{3/2}} \zeta\left ( \frac{5}{2} \right )\\
&\int_{0}^{\infty} \frac{\tanh^2 z}{z^{5/2}} \text{d}z
= \frac{80-5\sqrt{2} }{2\pi^{5/2}} \zeta\left ( \frac{7}{2} \right )\\
&
\end{aligned}
$$
|
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|
What are all the concordant forms $n$ such that $a^2+b^2 = c^2,\,a^2+nb^2=d^2$ for $n<1000$? Part I. The list of congruent numbers $n<10^4$ such that the system,
$$a^2-nb^2 = c^2$$
$$a^2+nb^2 = d^2$$
has a solution in the positive integers is known (A003273)
$$n = 5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34,\dots$$
Part II. Strangely, for the similar concordant forms/numbers $n$ such that,
$$a^2+b^2 = c^2$$
$$a^2+nb^2 = d^2$$
is not even in the OEIS,
$$n = 1,7,10,11,17,20,22,23,24,27,30,31,34,\dots$$
The list of $104$ prime $n<10^3$ is known (by Kevin Brown, David Einstein (hm?), and Allan MacLeod) though several troublesome primes were excluded assuming the Birch/Swinnerton-Dyer conjecture.
Question: Anybody knows how to generate the list of all concordant forms $n<1000$? (Elkies describes a method here.)
P.S. Incidentally, the special case $n=52$ appears in equal sums of like powers. Let,
$$a^2+b^2 = c^2$$
$$a^2+52b^2 = d^2$$
Then for $k=1,2,4,6,8,10,$
$$(8b)^k + (5a-4b)^k + (-a-2d)^k + (a-2d)^k + (-5a-4b)^k + (-12b+4c)^k + (12b+4c)^k =\\
(4a+8b)^k + (3a-2d)^k + (-3a-2d)^k + (-4a+8b)^k + (-16b)^k + (a+4c)^k + (-a+4c)^k$$
found by J. Wroblewski and yours truly. An initial primitive solution is $a,b,c,d = 3,4,5,29$ and an infinite more.
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While individ in his answer has shown there is an infinite number of concordant forms $n$, it can also be shown there is an infinite number of the special case $n = \pm N^2$. Given the system,
$$a^2+b^2 = c^2\tag1$$
$$a^2+nb^2 = d^2\tag2$$
First, let $a,\,b,\,c = p^2-q^2,\,2pq,\,p^2+q^2$ to satisfy $(1)$. The second becomes,
$$(p^2-q^2)^2+n(2pq)^2 =d^2\tag3$$
Two parametric solutions to $(3)$ are,
$$n = -(2t^2-2)^2,\quad p =t(2t^2-1),\quad q=1$$
found by Will Jagy and,
$$n = (4t^2+3)^2,\quad p =t(4t^2+1),\quad q=1$$
by this OP.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1584485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Find dimension and a basis of a subspaces $U+V$, $U \cap V$ in terms of the parameter $\alpha$ Let $U=span((1,1,1,2),(1,2,2,\alpha))$ and $V=span((1,3,4,\alpha+2),(1,4,\alpha,\alpha+1))$ are the subspaces of $\mathbb{R^4}$. ($\alpha\in\mathbb{R}$).
Find dimension and one basis of $U+V$, $U \cap V$ in terms of $\alpha$.
For values $\alpha$, which satisfies that $\dim(U+V)$ is the smallest, check if $v=(1,2,3,4)$ is an element of $U+V$.
Attempt:
From the definition of vector space span $\Rightarrow$
$U=\{(x+y,x+2y,x+2y,2x+\alpha y):x,y\in\mathbb{R}\}$
$V=\{(t+p,3t+4p,4t+\alpha p,(\alpha+2)t+(\alpha+1)p):t,p\in\mathbb{R}\}$
$U\cap V\Rightarrow$
$$x+y=t+p$$
$$x+2y=3t+4p$$
$$x+2y=4t+\alpha p$$
$$2x+\alpha y=(\alpha+2)t+(\alpha+1)p$$
We can use Kronecker-Capelli's theorem to find for which $\alpha$ the system has a unique solution.
When will $\dim(U+V)$ be the smallest?
$\dim(U+V)=\dim(U)+\dim(V)-\dim(U\cap V)$
|
First, note that $\DeclareMathOperator{rref}{rref}$
\begin{align*}
\rref
\begin{bmatrix}
1&1&1&2\\
1&2&2&\alpha
\end{bmatrix}
&=
\begin{bmatrix}
1&0&0&4-\alpha\\
0&1&1&\alpha-2
\end{bmatrix}
&
\rref
\begin{bmatrix}
1&3&4&2+\alpha\\
1&4&\alpha&1+\alpha
\end{bmatrix}
&=
\begin{bmatrix}
1&0&16-3\,\alpha&\alpha+5\\
0&1&\alpha-4&-1
\end{bmatrix}
\end{align*}
This proves that $\dim(U)=2$ and that $\dim(V)=2$.
Next, note that
$$
\det
\begin{bmatrix}
1&1&1&2\\
1&2&2&\alpha\\
1&3&4&\alpha+2\\
1&4&\alpha&\alpha+1
\end{bmatrix}
=(\alpha-7)(\alpha-3)
$$
This proves that $\dim(U+V)=4$ provided that $\alpha\neq7$ and $\alpha\neq 3$. Consequently
$$
\dim(U\cap V)=\dim(U)+\dim(V)-\dim(U+V)=2+2-4=0
$$
provided that $\alpha\neq7$ and $\alpha\neq 3$.
Now, note that
\begin{align*}
\rref
\begin{bmatrix}
1&1&1&2\\
1&2&2&7\\
1&3&4&7+2\\
1&4&7&7+1
\end{bmatrix}
&=
\begin{bmatrix}
1&0&0&-3\\
0&1&0&8\\
0&0&1&-3\\
0&0&0&0
\end{bmatrix}
&
\rref
\begin{bmatrix}
1&1&1&2 \\
1&2&2&3 \\
1&3&4&3+2\\
1&4&3&3+1
\end{bmatrix}
&=
\begin{bmatrix}
1&0&0&1\\
0&1&0&0\\
0&0&1&1\\
0&0&0&0
\end{bmatrix}
\end{align*}
This proves that $\dim(U+V)=3$ provided that $\alpha=7$ or $\alpha=3$. Consequently
$$
\dim(U\cap V)=\dim(U)+\dim(V)-\dim(U+V)=2+2-3=1
$$
provided that $\alpha=7$ or $\alpha=3$.
This covers the questions about dimension. Can you use the above to answer the question about bases?
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do you evaluate this summation: $S=\sum\limits_{r=0}^{15} (-1)^r \frac{\binom{15}{r}}{\binom{r+3}{r}}$ Find S:
$$S=\sum_{r=0}^{15} (-1)^r \frac{\binom{15}{r}}{\binom{r+3}{r}}$$
My attempt:
I tried writing the summation as:
$$S=3!(15!)\sum_{r=0}^{15} (-1)^r \frac{1}{(15-r)!(r+3)!}$$
and tried to convert it to a telescoping form, but to no avail.
I don't know how to go about solving this question and any help will be appreciated.
|
$$S=\sum_{r=0}^{15} (-1)^r \frac{\binom{15}{r}}{\binom{r+3}{r}} = \sum_{r=0}^{15} (-1)^r \frac{15!}{r! (15-r)!} \frac{3!\cdot r!}{(r+3)!}$$
$$= 15!\cdot3!\cdot\sum_{m=3}^{18} (-1)^{m+1} \frac{1}{(18-m)!} \frac{1}{(m)!}$$
$$= \frac{3!}{16 \cdot 17 \cdot 18} \sum_{m=3}^{18} (-1)^{m+1} \binom{18}{m}$$
$$= \frac{3!}{16 \cdot 17 \cdot 18} \left( \left(- \sum_{m=0}^{18} (-1)^{m} \binom{18}{m} \right) + 1 - \binom{18}{1} + \binom{18}{2} \right) $$
Now since
$$\sum_{i=0}^n (-1)^i \binom{n}{i} = 0 $$
it becomes
$$= \frac{3!}{16 \cdot 17 \cdot 18} \left( 1 - \binom{18}{1} + \binom{18}{2} \right) = \frac{1}{6}$$
|
{
"language": "en",
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|
Nesbitt's Inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ I'm reading a book which focus in inequality.
I'm stuck in this question. Let $a,b,c$ be positive real numbers. (Nesbitt's inequality) Prove the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$
So the first step of solution given is $\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{a+c}{c+b}+\frac{c+b}{a+c}+\frac{b+a}{a+c}+\frac{a+c}{b+a}\geq2+2+2=6$
I don't know how to proceed from the question to the first step of solution. Can anyone explain?
|
HINT: set
$$b+c=x$$
$$c+a=y$$
$$a+b=z$$
adding we get
$$a+b+c=\frac{x+y+z}{2}$$ and we can compute $$a+x=\frac{x+y+z}{2}$$ thus $$a=\frac{-x+y+z}{2}$$ etc
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$ It's required to prove that
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
I managed to go about out it two ways:
*
*Show it is equivalent to $\mathsf{true}$:
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
$$\Longleftrightarrow\sin x(1+\cos x+\sin x)\equiv(1+\cos x)(1-\cos x+\sin x)$$
$$\Longleftrightarrow\sin x+\cos x\sin x+\sin^2 x\equiv1-\cos x+\sin x+\cos x-\cos^2 x+\sin x \cos x$$
$$\Longleftrightarrow\sin^2 x\equiv1-\cos^2 x$$
$$\Longleftrightarrow\cos^2 x +\sin^2 x\equiv1$$
$$\Longleftrightarrow \mathsf{true}$$
*Multiplying through by the "conjugate" of the denominator:
$${\rm\small LHS}\equiv\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} $$
$$\equiv\frac{1+\cos x + \sin x}{1 - (\cos x - \sin x)} ~~\cdot ~~\frac{1+(\cos x - \sin x)}{1 +(\cos x - \sin x)}$$
$$\equiv\frac{(1+\cos x + \sin x)(1+\cos x - \sin x)}{1 - (\cos x - \sin x)^2}$$
$$\equiv\frac{1+\cos x - \sin x+\cos x + \cos^2 x - \sin x \cos x+\sin x + \sin x \cos x - \sin^2 x}{1 - \cos^2 x - \sin^2 x + 2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- \sin^2 x}{2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- 1 + \cos^2 x}{2\sin x \cos x}$$
$$\equiv\frac{2\cos x (1+\cos x)}{2\cos x(\sin x)}$$
$$\equiv\frac{1+\cos x}{\sin x}$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\equiv {\rm\small RHS}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$$
Both methods of proof feel either inelegant or unnecessarily complicated. Is there a simpler more intuitive way to go about this? Thanks.
|
if
$$
\frac{a}{b}=\frac{c}{d}=k
$$
then
$$
\frac{a+c}{b+d} = \frac{kb+kd}{b+d} =k =\frac{a}{b}
$$
since $$1-\cos^2 x =(1+\cos x)(1-\cos x) =\sin^2 x$$ we have
$$
\frac{1+\cos x}{\sin x} = \frac{\sin x}{1-\cos x} =\frac{1+\cos x +\sin x}{1 -\cos x +\sin x}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1590059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Prove that $xy+yz+zx+\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge 2x^2+2y^2+2z^2$ for every $x,y,z$ strictly positive I checked the inequality for $y=z$. In this particular case, after some simplifications, the inequality becomes:
$$
2x^3+y^3\ge 3x^2y,
$$
which is true, according to the arithmetic-geometric mean inequality applied to the numbers $x^3,\ x^3$ and $y^3$. I have no idea, at least for now, on how to proceed in the general case. Please give me a hint.
|
Let $\frac{xy}{z}=a^2$, $\frac{yz}{x}=b^2$, and $\frac{zx}{y}=c^2$.
Now, we have $y=ab$, $z=bc$, and $x=ca$.
It now suffices to prove $$a^4+b^4+c^4+a^2bc+b^2ca+c^2ab \ge 2(a^2b^2+b^2c^2+c^2a^2)$$
From Schur's Inequality, we have $$a^2(a-b)(a-c)+b^2(b-c)(b-a)+c^2(c-a)(c-b)=(a^4+b^4+c^4)-(a^3b+a^3c+b^3c+b^3a+c^3a+c^3b)+(a^2bc+b^2ca+c^2ab) \ge 0$$
This gives us $$a^4+b^4+c^4+a^2bc+b^2ca+c^2ab \ge (a^3b+b^3a)+(b^3c+c^3b)+(c^3a+a^3c) \ge 2(a^2b^2+b^2c^2+c^2a^2)$$
as desired.
|
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"url": "https://math.stackexchange.com/questions/1590461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Convergence of sequence of ratio of consecutive terms. I would like to prove that the following sequence of ratios $(x_n)$ where $x_n = c_n/c_{n-1}$ converges to a finite limit $L$:
$$\lim_{n\to\infty} x_n = \lim_{n\to\infty}\frac{c_n}{c_{n-1}} = L$$
Where $c_n$ is defined by the recurrence relation:
$c_n = c_{n-1} + \lfloor c_{n-4}/2 \rfloor$ for $n \geq 4$, with initial conditions $c_0=3, c_1=4, c_2=5, c_3=6$.
I feel there should be a similar approach to proving limit of consecutive Fibonacci numbers converges, which I think might require me to solve the recurrence, but the floor function doesn't help.
Some observations, verification by computer software has led me to predict that the limit tends towards $L = 1.25372495821...$, which appears to be very close to a solution to the equation:
$$x^4 - x^3 - \frac{1}{2}=0$$
This would be the characteristic equation to the recurrence relation $\alpha_n = \alpha_{n-1} + \alpha_{n-4}/2$, which is very similar to the way how $c_n$ is defined.
Any ideas? I exhausted most of the standard undergraduate analysis techniques for proving sequences converge, no luck though.
|
It is clear that $c_n\rightarrow\infty$, and $x_n\ge 1$ for all $n$. Note that
$$ x_n = \frac{c_n}{c_{n-1}} = \frac{c_{n-1} + \lfloor\frac{c_{n-4}}{2}\rfloor}{c_{n-1}} = 1 + \frac{c_{n-2}}{c_{n-1}}\frac{c_{n-3}}{c_{n-2}}\frac{c_{n-4}}{c_{n-3}}\frac{\lfloor\frac{c_{n-4}}{2}\rfloor}{c_{n-4}} = 1 + \frac{1}{x_{n-1}x_{n-2}x_{n-3}}\frac{\lfloor\frac{c_{n-4}}{2}\rfloor}{c_{n-4}}. $$
Since $x_n\ge 1$ and $\frac{\lfloor\frac{c_{n-4}}{2}\rfloor}{c_{n-4}}\le\frac{1}{2}$ for all $n$, we have
$$ x_n\le 1 + \frac{1}{1\cdot1\cdot1}\frac{1}{2} = \frac{3}{2} $$
for all $n$. Since $c_n\rightarrow\infty$, for every $\epsilon>0$ there exists $N$ such that if $n>N$, then $\frac{\lfloor\frac{c_{n-4}}{2}\rfloor}{c_{n-4}}\ge\frac{1}{2}-\epsilon$, and hence for $n>N$ we have
$$ x_n\ge\frac{1}{\frac{3}{2}\frac{3}{2}\frac{3}{2}}\left(\frac{1}{2}-\epsilon\right) = 1 + \left(\frac{2}{3}\right)^3\left(\frac{1}{2}-\epsilon\right). $$
Letting $\alpha$ be the constant on the right-hand side, we again have
$$ x_n \le 1 + \frac{1}{\alpha\alpha\alpha}\frac{1}{2} = 1 + \frac{1}{2\alpha^3} $$
for $n>N$. We continue with the bounds similarly, hoping that the lower and upper bounds converge.
Specifically, we make the following claim:
Claim: Let $\alpha_0 = 1$, and let $\{\epsilon_m\}$ be a decreasing sequence converging to $0$. Define the sequences $\{\alpha_m\}$ and $\{\beta_m\}$ as follows:
\begin{align*}
\beta_m &= 1 + \frac{1}{2a_m^3} \\
\alpha_{m+1} &= 1 + \frac{1}{\beta_m^3}\left(\frac{1}{2}-\epsilon_m\right)
\end{align*}
Then, for each $m$, there exists $N_m$ such that if $n>N_m$, then
$$ \alpha_m\le x_n\le \beta_m. $$
Proof: Suppose $\alpha_m\le x_n\le\beta_m$ for $n>N_m$. Let $N$ satisfy $\frac{\lfloor\frac{c_{n-4}}{2}\rfloor}{c_{n-4}}\ge\frac{1}{2}-\epsilon_m$ for $n>N$ and $N\ge N_m+3$. Then, for $n>N$, we have
$$ x_n = 1 +\frac{1}{x_{n-1}x_{n-2}x_{n-3}}\frac{\lfloor\frac{c_{n-4}}{2}\rfloor}{c_{n-4}} \ge 1+\frac{1}{\beta_m^3}\left(\frac{1}{2}-\epsilon_m\right) = \alpha_{m+1}. $$
If we let $N_{m+1} = N+3$, then for $n>N_{m+1}$ we also have
$$ x_n = 1 +\frac{1}{x_{n-1}x_{n-2}x_{n-3}}\frac{\lfloor\frac{c_{n-4}}{2}\rfloor}{c_{n-4}}\le 1 + \frac{1}{\alpha_{m+1}^3}\frac{1}{2} = \beta_{m+1} $$
as desired. $\square$
Now, we wish to show that $\{\alpha_m\}$ and $\{\beta_m\}$ are increasing and decreasing, respectively. Suppose $\alpha_m\ge\alpha_{m-1}$ and $\beta_m\le\beta_{m-1}$ (this is easily to be seen to be true for $m=1$). Then
$$\alpha_{m+1} = 1 + \frac{1}{\beta_m^3}\left(\frac{1}{2}-\epsilon_m\right) \ge 1 + \frac{1}{\beta_{m-1}^3}\left(\frac{1}{2}-\epsilon_{m-1}\right) = \alpha_{m}$$
since $\beta_m\le\beta_{m-1}$ and $\epsilon_m\le\epsilon_{m-1}$. Similarly, we have
$$\beta_{m+1} = 1 + \frac{1}{2\alpha_{m+1}^3}\le 1+\frac{1}{2\alpha_m^3} = \beta_m.$$
It follows that $\{\alpha_m\}$ and $\{\beta_m\}$ converge to the limits $\alpha$ and $\beta$, respectively, and these limits satisfy
\begin{align*} \beta &= 1 + \frac{1}{2\alpha^3} \\
\alpha &= 1 + \frac{1}{2\beta^3}
\end{align*}
(the second equality follows from the fact that $\epsilon_m\rightarrow 0$). This implies that
$$ \alpha\left(1+\frac{1}{2\alpha^3}\right) = \left(1 + \frac{1}{2\beta^3}\right)\beta\implies \alpha+\frac{1}{2\alpha^2} = \beta + \frac{1}{2\beta^2}. $$
Since $x\mapsto x + \frac{1}{2x^2}$ is strictly increasing on $(1,\infty)$, and $\beta\ge\alpha>1$, it follows that $\alpha = \beta$. Thus, by the sandwich property, we have
$$\lim\limits_{n\rightarrow\infty}{x_n} = \alpha$$
where $\alpha$ satisfies
$$\alpha = 1 + \frac{1}{2\alpha^3}\iff \alpha^4 - \alpha^3 - \frac{1}{2} = 0.$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Integral $\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$ It's a follow-up to my previous question.
Can we find an anti-derivative
$$\int\arcsin x\cdot\ln^3x\,dx$$
or, at least, evaluate the definite integral
$$\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$$
in a closed form (ideally, as a combination of elementary functions and polylogarithms)?
|
The indefinite integral can be expressed in terms of hypergeometrics.
We start with integration by parts with $dv=\sin^{-1}(x)$ and $u=\log^3x$.
$$\begin{align}
I&=\int \arcsin x \cdot \ln^3x\,dx\\
&=\ln^3x\cdot\left(\sqrt{1-x^2}+x\arcsin x\right)-\int \left(\sqrt{1-x^2}+x\arcsin x\right)\cdot \frac{3\ln^2x}{x}\,dx
\end{align}$$
Expanding the integrand we get
$$\frac{3\sqrt{1-x^2}\cdot\ln^2x}{x}+3\arcsin(x)\ln^2(x)$$
The integration of the second term is addressed in your previous question so I will focus on just the first term.
$$K=\int\frac{3\sqrt{1-x^2}\ln^2x}{x}\,dx$$
Mathematica gives the result $K=$
$$\frac{3\sqrt{1-x^2}}{x\sqrt{1-\frac{1}{x^2}}} \left[2 x \, _4F_3\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)-2 x \log (x) \, _3F_2\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)+x \sqrt{1-\frac{1}{x^2}} \log ^2(x)+\log ^2(x) \csc ^{-1}(x)\right]$$
Therefore
$$I=\ln^3x\cdot\left(\sqrt{1-x^2}+x\arcsin x\right)-K-3\int\arcsin x\ln^2x\,dx$$
Here are the sum representations of the hypergeometrics
$$\, _3F_2\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)=-\frac{1}{\sqrt \pi}\sum_{k=0}^\infty\frac{x^{-2k}}{(2k-1)^3}\cdot\frac{\Gamma\left(k+\frac 1 2\right)}{\Gamma(k+1)}$$
$$\, _4F_3\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)=\frac{1}{\sqrt \pi}\sum_{k=0}^\infty \frac{x^{-2k}}{(2k-1)^4}\cdot\frac{\Gamma\left(k+\frac 1 2\right)}{\Gamma(k+1)}$$
|
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|
Uniform convergence of $\sum_{n=0}^{\infty} \frac{\int_{\sin nx}^{\sin(n+1)x}\sin t^2dt \int_{nx}^{\infty}\frac{dt}{\sqrt{t^4+1}}}{1+n^3x^2}$ I am preparing for the exam. Please help me to solve the following problem:
Given a series
$$\sum_{n=0}^{\infty} \frac{\int_{\sin nx}^{\sin(n+1)x}\sin t^2dt \int_{nx}^{\infty}\frac{dt}{\sqrt{t^4+1}}}{1+n^3x^2}$$
determine, if it is uniformly convergent on 1) $(0, \infty)$ ? 2) $(0, c)$, where $c < \infty$ ?
Thanks a lot for your help!
|
First
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}t}{\sqrt{t^4+1}}
&=\frac14\int_0^\infty\frac{t^{-3/4}\,\mathrm{d}t}{(t+1)^{1/2}}\\
&=\frac14\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac14\right)}{\Gamma\left(\frac12\right)}\\
&\le\int_0^1\mathrm{d}t+\int_1^\infty\frac{\mathrm{d}t}{t^2}\\[9pt]
&=1+1=2
\end{align}
$$
Furthermore,
$$
\begin{align}
\int_{nx}^\infty\frac{\mathrm{d}t}{\sqrt{t^4+1}}
&\le\int_{nx}^\infty\frac{\mathrm{d}t}{t^2}\\
&=\frac1{nx}
\end{align}
$$
Therefore, using the harmonic mean of the bounds,
$$
\bbox[5px,border:2px solid #C0A000]{\int_{nx}^\infty\frac{\mathrm{d}t}{\sqrt{t^4+1}}\le\frac4{1+2nx}}
$$
Since
$$
\left|\sin((n+1)x)-\sin(nx)\right|\le2
$$
and
$$
\left|\sin((n+1)x)-\sin(nx)\right|\le(n+1)x-nx=x
$$
using the harmonic mean of the bounds, we have
$$
\bbox[5px,border:2px solid #C0A000]{\int_{\sin(nx)}^{\sin((n+1)x)}\sin\left(t^2\right)\,\mathrm{d}t\le\frac{4x}{2+x}}
$$
Thus, when $x\le2$, the sum is less than
$$
\begin{align}
\sum_{n=N}^\infty\frac{16x}{(1+2nx)(2+x)(1+n^3x^2)}
&\le\sum_{n=N}^\infty\frac{8x}{\left(1+2nx\right)\left(1+n^3x^2\right)}\\
&\le\sum_{n=N}^\infty\frac{8x}{\left(1+2nx\right)\left(1+Nn^2x^2\right)}\\
&\le\int_0^\infty\frac{8\mathrm{d}t}{\left(1+2t\right)\left(1+Nt^2\right)}\\
&\le\frac{4\pi}{\sqrt{N}}
\end{align}
$$
When $x\gt2$, the sum is less than
$$
\begin{align}
\sum_{n=N}^\infty\frac{16x}{(1+2nx)(2+x)(1+n^3x^2)}
&\le\sum_{n=N}^\infty\frac{16}{\left(1+4n\right)\left(1+4n^3\right)}\\
&\le\sum_{n=N}^\infty\frac1{n^4}\\
&\le\frac1{3(N-1)^3}
\end{align}
$$
Therefore, the convergence is uniform over $(0,\infty)$.
|
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|
Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?
|
We have $(a+b)^2=a^2+b^2 + 2ab.$ Thus $$(1+2+...+(k+1))^2=(1+2+...+k)^2+(k+1)^2+ 2(1+2+...+k)(k+1)=1^3+...+k^3+(k+1)^2 + 2(1+2+...+k)(k+1)$$
Using $1+...+k=\dfrac{k(k+1)}{2}$ we get $$(1+2+...+k+1)^2=1^3+...+k^3+ (k+1)^2 + k(k+1)^2=1^3+...+k^3+ (k+1)^2(k+1)=1^3+...+(k+1)^3$$ and the inductive step is proven.
|
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|
Find the real ordered pairs $(x,y)$ satisfying $x^2+x=y^4+y^3+y^2+y.$ Find the real ordered pairs $(x,y)$ satisfying $x^2+x=y^4+y^3+y^2+y.$
$x^2+x=y^4+y^3+y^2+y$
$4x^2+4x=4y^4+4y^3+4y^2+4y$
$4x^2+4x+1=4y^4+4y^3+4y^2+4y+1$
$(2x+1)^2=4y^4+4y^3+4y^2+4y+1$
I am stuck here.Is there a general method to solve such type of equations?Please help me.
|
HINT: $$x^2+x=y^4+y^3+y^2+y \Rightarrow \left(x+\frac{1}{2}\right)^2=y^3(y+1)+\left(y+\frac{1}{2}\right)^2$$
$$\Rightarrow \left(x+\frac{1}{2}\right)^2-\left(y+\frac{1}{2}\right)^2=y^3(y+1)$$
$$\Rightarrow (x+y+1)(x-y)=y^3(y+1)$$
Now consider the $4$ cases where $x$ is even or odd and $y$ is even or odd and use divisibility of $2$.
Can you proceed now?
|
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|
Prove that $\sum_{n=1}^{\infty}(\frac{2}{9})^na_n=\frac{1}{3}$ Let $a_n(n\geq1)$ be the sequence of numbers defined by the recurrence relation$\hspace{1cm}a_1=1,\hspace{1cm}a_n=a_{n-1}a_1+a_{n-2}a_2+...+a_2a_{n-2}+a_1a_{n-1}$
Prove that $\sum_{n=1}^{\infty}(\frac{2}{9})^na_n=\frac{1}{3}$
Let $\alpha=\sum_{n=1}^{\infty}a_nt^n$
$\alpha^2=\sum_{n=1}^{\infty}a_nt^n\sum_{n=1}^{\infty}a_nt^n$
$a_{n-1}a_1+a_{n-2}a_2+...+a_2a_{n-2}+a_1a_{n-1}$ is the coefficient of $t^n$ in the expansionn of $\alpha^2$
I could not further solve this problem.How should i solve it further?
|
Let $f(x)$ be $\sum_{i=1}^\infty a_ix^i$. Note that $$f(x) \cdot f(x) = \sum_{i=1}^\infty a_ix^i \cdot \sum_{i=1} a_ix^i = \sum_{i=2}^\infty \sum_{j=1}^{n-1} a_ja_{n-j} x^i = \sum_{i=2}^\infty a_ix^i = f(x)-a_1x=f(x)-x$$
Now we have $f(x)^2-f(x)+x=0$, or $f(x)=\frac{1 \pm \sqrt{1-4x}}{2}$.
Since $f(0)=0$, we have $f(x)=\frac{1-\sqrt{1-4x}}{2}$.
This gives $f(\frac{2}{9})=\frac{1-\sqrt{\frac{1}{9}}}{2}=\frac{1}{3}$ as desired.
For similar recursion trick see Catalan Numbers.
|
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|
Find all real numbers $x,y > 1$ such that $\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$
Find all real numbers $x,y > 1$ such that $$\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$$
Attempt
We can rewrite this as $x^2(x-1)+y^2(y-1) = 8(x-1)(y-1)$. Then I get a multivariate cubic, which I find hard to find all solutions to.
|
Note $a=x-1$ and $b=y-1$ and apply $r^2+s^2\geqslant2rs$. The solution is $x=y=2$.
|
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|
Evaluate the following integral I got stuck with the integral below. I have tried to make it look like the derivative of arctan.
$$\int \frac{2-x}{x^2-x+1}\,dx$$
Thank you!
|
$$\int \frac{2-x}{x^2-x+1}\ dx=\frac{1}{2}\int \frac{3-(2x-1)}{x^2-x+1}dx\\=\frac{1}{2}\int \frac{3}{x^2-x+1}\ dx-\frac 12\int \frac{
2x-1}{x^2-x+1}\ dx\\=\frac{3}{2}\int \frac{1}{\left(x-\frac 12\right)^2+\frac{3}{4}} dx-\frac 12\int \frac{x^2-x+1}{x^2-x+1}dx$$
Substitute $u:=x^2-x+1$ and $du=(2x-1)dx$
Substitute $\varphi:=x-\frac 1 2$ and $d\varphi=dx$
$$-\frac 1 2\int \frac 1 u du+\frac 3 2\int \frac {1} {\varphi ^2+3/4} d\varphi$$
From here it should be easy, in the end you should get $$\boxed{
\color{blue}{=\sqrt 3 \arctan\bigg(\frac{2x-1}{\sqrt 3}\bigg)-\frac 1 2 \ln|x^2-x+1|+C}}$$
|
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|
Area of the shadow of a regular polygon inscribed in a sphere. Consider the situation given below:
Let a regular polygon be inscribed in a sphere such that its circumcentre is at a distance $r$ from the centre of the sphere of radius $R$. A point source of light is kept at the centre of the sphere. How can we calculate the area of the shadow made on the surface of the sphere.
I tried to use the relation: $ \Omega = \frac{S}{R^2} $
But of course that is the case when a circle would be inscribed. So can I somehow relate it for any general polygon?
|
Let $\theta = \cos^{-1}\frac{r}{R}$ and $\phi = \frac{2\pi}{n}$ where
$n$ is the number of sides of the regular polygon.
Choose the coordinate system so that the vertices of the regular polygon are located at
$$\vec{v}_k = (R\sin\theta\cos(k\phi), R\sin\theta\sin(k\phi), R\cos\theta) \quad\text{ for } k = 0, 1, \ldots, n - 1.$$
Let $\hat{v}_k = \frac{\vec{v}_k}{|\vec{v}_k|}$ be the corresponding unit vectors
and $\hat{z} = (1,0,0)$.
Recall following formula by Oosterom and Strackee on solid angle,
Given any 3 unit vectors, $\vec{a}, \vec{b}, \vec{c}$, the solid angle $\Omega$ spanned by these 3 vectors satisfy
$$\tan\frac{\Omega}{2} = \frac{ \vec{a} \cdot (\vec{b} \times \vec{c}) }{1 + \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}}$$
Apply this to the unit vectors $\hat{v}_0, \hat{v}_1$ and $\hat{z}$, we find the corresponding solid angle $\Omega$ satisfy
$$\tan\frac{\Omega}{2} = \frac{\sin^2\theta\sin\phi}{1 + 2\cos\theta + \cos^2\theta + \sin^2\theta\cos\phi}
= \frac{\sin\phi}{
\frac{1+\cos\theta}{1-\cos\theta} + \cos\phi}
= \frac{\sin\phi}{\frac{R+r}{R-r}+\cos\phi}
$$
So the desired area is
$$\verb/Area/ = nR^2\Omega = 2nR^2\tan^{-1}\left[\frac{\sin\left(\frac{2\pi}{n}\right)}{\frac{R+r}{R-r}+\cos\left(\frac{2\pi}{n}\right)}\right]$$
Update
Let $s = \sin\frac{\pi}{n}$, $c = \cos\frac{\pi}{n}$ and $D = \frac{R+r}{R-r}$, we
can simplify above expression as
$$\begin{align}\tan^{-1}\left[\frac{2sc}{D+c^2-s^2}\right]
&= \frac{\pi}{n} - \tan^{-1}\left[\frac{\frac{s}{c} - \frac{2sc}{D+c^2-s^2}}{1 + \frac{s}{c}\frac{2sc}{D + c^2 - s^2}}\right]
= \frac{\pi}{n} - \tan^{-1}\left[\frac{s(D-1)}{c(D+1)}\right]\\
&= \frac{\pi}{n} - \tan^{-1}\left[\frac{sr}{cR}\right]
\end{align}$$
So a simpler version of the area is
$$\verb/Area/ = 2R^2\left(\pi - n\tan^{-1}\left(\frac{r}{R}\tan\frac{\pi}{n}\right)\right)$$
It is easy to see this equals to the expression in Harish Chandra Rajpoot's answer.
|
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|
Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$ Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$
My try
My book gives as a hint to move everything to the left hand side of the inequality and then factor and see what I get in the long factorization process and to lookout for squares.
So that's what I have tried:
\begin{array}
((a^7+b^7)(a^2+b^2) &\ge (a^5+b^5)(a^4+b^4) \\\\
(a^7+b^7)(a^2+b^2)-(a^5+b^5)(a^4+b^4) &\ge 0 \\\\
(a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)(a^2+b^2)-(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)(a^4+b^4) &\ge 0 \\\\
(a+b)\left[(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)((a+b)^2-2ab)-(a^4+b^4)(a^4-a^3b+a^2b^2-ab^3+b^4)\right] &\ge 0
\end{array}
Now it's not clear what I have to do next.I am stuck.
Note: My book doesn't teach any advanced technique for solving inequality as AM-GM ,Cauchy inequality etc..
|
You can simplify considerably by multiplying out both sides $$a^9+a^7b^2+a^2b^7+b^9\ge a^9+a^5b^4+a^4b^5+b^9$$
Cancelling the common terms from each side and dividing through by $a^2b^2$ gives $$a^5+b^5\ge a^3b^2+a^2b^3$$
Now move everything to the LHS and factorise.
Note if $a$ or $b$ is zero, equality is obvious.
|
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|
Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$
It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?
|
Let $a=x^3$, $b=y^3$, $c=z^3$, then it can be rewritten as:
$$
x^6+y^6+z^6+3 x^2 y^2 z^2-2 \left(x^3 y^3+x^3 z^3+y^3 z^3\right)\geq 0
$$
Use the following notations:
$$S_{3}:=xyz\qquad S_2:=xy+yz+xz\qquad S_1=x+y+z$$
Then:
$$
x^6+y^6+z^6=S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-12 S_2 S_3 S_1-2 S_2^3+3 S_3^2
$$
$$
x^3 y^3+x^3 z^3+y^3 z^3=S_2^3-3 S_1 S_3 S_2+3 S_3^2
$$
$$
3x^2y^2z^2=3S_3^2
$$
Then we only have to prove:
$$
S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-6 S_2 S_3 S_1-4 S_2^3\geq 0
$$
Now put $S_2=S_1^2$, and notice that with this:
$$
\left.S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-6 S_2 S_3 S_1-4 S_2^3\right|_{S_2=S_1^2}=0
$$
Thus this can be factorised as:
$$
\left(S_1^2-S_2\right) \left(S_1^4-5 S_2 S_1^2+6 S_3 S_1+4 S_2^2\right)\geq0
$$
Since: $(x+y+z)^2\geq 3(xy+yz+xz)\Rightarrow S_1^2\geq 3S_2$ by rearrangement, it is enough to prove that the second factor is non-negative. Return to our previous notations, enough to show:
$$
x^4+y^4+z^4+(x+y+z)xyz-x^3y-y^3x-y^3z-z^3y-x^3z-xz^3=
$$
$$
=x^2(x-y)(x-z)+y^2(y-x)(y-z)+z^2(z-x)(z-y)\geq 0
$$
Which is trivially true by applying Schur's inequality
|
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|
prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$
If $x<y<z$ are real numbers, prove that $$(x-y)^3+(y-z)^3+(z-x)^3 > 0.$$
Attempt
We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of using rearrangement but then I would get $\geq$ instead of $>$.
|
Notice that the polynomial
$$S = (x-y)^3 + (y-z)^3 + (z-x)^3$$
is zero whenever two of the variables are equal so we must have
$$S = C(y-x) (z-x) (z-y)$$
for some constant $C$. A direct computation gives $C=3$ and the result follows since all factors are positive when $x < y < z$.
|
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|
Prove $\sqrt{n}+\sqrt{n+k}$ is irrational For what $k\in\mathbb N$, $\sqrt{n}+\sqrt{n+k}$ is irrational? ($\forall n\in\mathbb N$)
|
This result is not true in general for example, suppose that $n$ is a square $n=l^2$ and $k=0$ $\sqrt{l^2+0}+\sqrt{l^2}$ is rational, suppose that $n=9, k=16$, $\sqrt{16+9}+\sqrt{9}$ is rational.
The question is given $n$ for what values of $k$, $\sqrt{n+k}+\sqrt{n}$ is irrrational?
Proposition
Suppose that $k,n\in N$ and $n(n+k)$ is not a square then $\sqrt{n+k}+\sqrt{n}$ is irrational.
Proof:
$(\sqrt{n}+\sqrt{n+k})^2=n+n+k+2\sqrt{n}\sqrt{n+k}$
$((\sqrt{n}+\sqrt{n+k})^2-2n-k)^2=4n(n+k)$
Consider $P(X)= (X^2-2n-k)^2-4n(n+k)=X^4-2(2n+k)X^2+(2n+k)^2-4n(n+k)=X^4-2(2n+k)X^2+k^2$ $\sqrt{n+k}+\sqrt{n}$ is a root of $P(X)$.
Consider $Q(X)=U^2-2(2n+k)U+k^2$. The discriminant of $Q(X)$ is $4(2n+k)^2-4k^2=4(4n^2+4nk+k^2)-4k^2 =16n(n+k)$, thus if $n(n+k)$ is not a square, $Q(X)$ and hence $P(X)$ is irreducible.
|
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|
Prove that if $a,b,$ and $c$ are positive real numbers then $(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}.$
Prove that if $a,b,$ and $c$ are positive real numbers then $$(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}.$$
This looks like a simple question. We can apply AM-GM twice to get $(a+b)(a+c) \geq 4a\sqrt{bc}$. Then how do I use that fact to get $(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}$?
|
We have
$$(a+b)(a+c) = a(a+b+c) +bc.$$
We can now apply AM-GM.
|
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|
Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$ I've worked out a proof, but I was wondering about alternate, possibly more elegant ways to prove the statement. This is my (hopefully correct) proof:
Starting from the identity $2^m = \sum_{k=0}^m \binom{m}{k}$ (easily derived from the binomial theorem), with $m = 2n$:
$2^{2n} = 4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{2n-1} + \binom{2n}{2n}$
Applying the property $\binom{m}{k} = \binom{m}{m-k}$ to the second half of the list of summands in RHS above:
$4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n-1} + \binom{2n}{n} + \underbrace{\binom{2n}{n-1} +\cdots \binom{2n}{1} + \binom{2n}{0}}_{\binom{m}{k} = \binom{m}{m-k} \text{ has been applied}}$
Rearranging the above sum by alternately taking terms from the front and end of the summand list in RHS above (and introducing the term $\binom{2n}{-1} = 0$ at the beginning just to make explicit the pattern being developed):
$4^n = (\binom{2n}{-1} + \binom{2n}{0}) + (\binom{2n}{0} + \binom{2n}{1}) + \cdots + (\binom{2n}{n-1} + \binom{2n}{n})$
Finally, using the property $\binom{m}{k} + \binom{m}{k-1} = \binom{m+1}{k}$ on the paired summands, we get the desired result:
$4^n = \binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n}$
|
Applying what you mention on (in this context nicer) odd $2n+1$ instead of even $2n$ we find:$$2\times4^n=2^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}=2\sum_{k=0}^{n}\binom{2n+1}{k}$$
Now divide by $2$.
|
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|
How to evaluate this Integral $\int { {\sqrt{5^2+K^2}}dK \over {\sqrt{10^2+K^2}K}} $ While working on an Exact Differential Equation, I encounter the following Integral.
$$\int { {\sqrt{5^2+K^2}} \over {K\sqrt{10^2+K^2}}} dK$$
I have tried substitution and all the other elementary methods, but the Integral simply refuses to yield to all my attempts. I reckon that this requires some concept that I've yet not studied. Maybe elliptical Integrals? But that's just speculation. Please Help....
Thank You.
|
This is not a trivial one $$I=\int\frac{\sqrt{x^2+25}}{x \sqrt{x^2+100}}\,dx$$ Let us try using $$\frac{\sqrt{x^2+25}}{ \sqrt{x^2+100}}=u^2 \implies x=\frac{5 \sqrt{1-4 u^4}}{\sqrt{u^4-1}}\implies dx= \frac{30 u^3}{\sqrt{1-4 u^4} \left(u^4-1\right)^{3/2}}du$$ So, $$I=-\int\frac{6 u^5}{4 u^8-5 u^4+1}\,du$$ Now, since the denominator shows pretty nice roots (it is a quadratic in $u^4$), partial fraction decomposition leads to $$\frac{-6 u^5}{4 u^8-5 u^4+1}=\frac{u}{u^2+1}+\frac{u}{2 u^2-1}-\frac{u}{2 u^2+1}-\frac{1}{2 (u-1)}-\frac{1}{2
(u+1)}$$ makes the problem much more pleasant.
The result of the integration is just a sum of logarithms $$I=\frac{1}{4} \log \left(1-2 u^2\right)-\frac{1}{2} \log
\left(1-u^2\right)+\frac{1}{2} \log \left(1+u^2\right)-\frac{1}{4} \log \left(1+2
u^2\right)$$ Recombining all of that simplifies again and $$I=\tanh ^{-1}\left(u^2\right)-\frac{1}{2} \tanh ^{-1}\left(2 u^2\right)$$
Edit
Applying the same method to
$$I=\int\frac{\sqrt{x^2+a^2}}{x \sqrt{x^2+b^2}}\,dx$$ would lead to $$I=\tanh ^{-1}\left(u^2\right)-\frac{a }{b}\tanh ^{-1}\left(\frac{b }{a}u^2\right)$$
|
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|
Proving sum of product forms a pattern in n * nnnnnn.... I am consider a problem regarding numbers which are, in decimal, one digit repeated - for instance, $88888888$ is such a number. In particular, I am looking at the following problem:
The sum of the digits of the number $$8\cdot \underbrace{88\ldots 88}_{n\text{ times}}$$ is $1000$. How many $8$'s are there? (i.e. what is $n$?)
I know that the correct answer is $991$ and I can observe a pattern that lets me solve it; in particular, the sum of digits of $8\cdot 8=64$ is $10$ and the sum of digits of $8\cdot 88=704$ is $11$ and the sum of digits of $8\cdot 888=7104$ is $12$ - so it appears that the sum of the digits is exactly $n+9$. One can note that if one replaces all the eights with other digits, similar patterns exist - for instance, with sevens, the sum increases by $4$ for each digit.
How can this pattern be proven?
|
$a*aaaa....aaa = a^2*1111111....111$ (m a's and m 1's)
Three cases to consider:
$a^2$ has one digit. (i.e. $a = 1,2,3$)
$a^2$ has two digits and the sum of the digits is less than 10. (i.e. $a = 4,5,6, 9$)
$a^2$ has two digits and the sum of the digits is 10 or more. (i.e. $a = 7,8$)
====
If $a^2 = b$ has one digit,$b$, then $a*aaa... = a^2*111... = bbbbb$ and the sum of the digits is $m*b$. That was easy.
===
If $a^2 = 10b + c$ has two digits, $b$ and $c$, and $b + c < 10$ then:
$a*aaa... = a^2*1,1,1... = (10b+c),(10b + c),...,(10b + c) = b,(b+c),(b+c),...,(b + c), c$.
[Here I use notation x,x,x ... as shorthand for $\sum x*10^i$. In the $b,(b+c),...,(b+c), c$ each term is a single digit although it wasn't for the $(10b+c),(10b+c)...$ representation. But from that one we "carried the b's" to get the final representation.]
The sum of the digits $b,(b+c),(b+c),...,(b + c), c$ is $b + (m-1)(b+c) + c= m(b+c)$.
===
If $a^2 = 10b + c$ is two digits ($b$ and $c$) and $b + c = 10 + d$ for some digit d...
First note that $d < 9$ so $d + 1 < 10$.
So $a*a*1111... = a^2*1111.... = (10b +c),(10b + c), .... (10b + c),(10b + c) = b,(b + c),....(b+c),c = b,(10 + d),(10 + d), .....(10 + d),c = (b+1),(d+1),(d+1),....(d+1),d,c$.
And the sum of the digits is $b + 1 + (m-2)(d+1) + d + c = b + c + d+ (m-2)(d+1) = 11 + 2d + (m-2)(d +1) = m(d+1) + 9$.
====
In summary: Sum of digits = $m*e \{+ 9 \}$ where $m$ is the number of digits and $e$ is the $a^2$, sum of digits of $a^2$, or sum of sum of digits of $a^2$ and $" + 9"$ applies only if sum of digits of $a^2 \ge 10$.
And hence, each increase in $m$ the sums increase by $e$.
===
And, BTW, back to the original problem: Su of digits of $a*aaaa....$ = 1000- how many digits are there:
Sum of digits is $m*e \{+9\} = 1000$. $e$ can be: $a=1 \implies e =1;a=2 \implies e =4;a=3 \implies e =9;a=4 \implies e =7;a=5 \implies e =7;a=6 \implies e =9;a=7 \implies e =4;a=8 \implies e =1; a=9 \implies e =9;$
$m*e = \{1000, 991\}$. $991$ is primes so if $m*e = 991$ then $e = 1; a=8; m=991$. If $m*e = 1000$ then $e$ is 1 or a multiple of 2 or 5 only, so possibilities:
$e = 1;a = 1; m = 1000; $
$e = 4; a = 2; m = 250;$
$e = 4; a =7$ is not possible as that would imply $m = 991$.
Three answers.
|
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|
Second derivative of $x^3+y^3=1$ using implicit differentiation I need to find the $D_x^2y$ of $x^3+y^3=1$ using implicit differentiation
So,
$$
x^3 + y^3 =1 \\
3x^2+3y^2 \cdot D_xy = 0 \\
3y^2 \cdot D_xy= -3x^2 \\
D_xy = - {x^2 \over y^2}
$$
Now I need to find the $D_x^2y$.
I am pretty sure that means the second derivative.
How would I do it to find the second derivative? apparently, it is supposed to be$$-{2x \over y^5}$$
|
If $Dx$ is the first derivative and $Dx^2$ is the second derivative, than your first derivative is correct. For the second derivative we have:
$$
\frac{d}{dx}y'=\frac{d}{dx}\left(\frac{-x^2}{y^2}\right)$$
that, using fraction rule and chain rule for $y$, becomes:
$$\frac{-2xy^2+2x^2y(y')}{y^4}$$
substituting $y'=-x^2/y^2$ and wi a bit of algebra:
$$\frac{-2xy^3-2x^4}{y^5}=\frac{-2x(y^3+x^3)}{y^5}$$
finally, using $x^3+y^3=1$:
$$\frac{d}{dx}y'=y''=\frac{-2x}{y^5}
$$
|
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|
Find the value of $\frac{i}{4-\pi}\int_{|z|=4}\frac{1}{z\cos{z}}dz$ Find the value of $$\frac{i}{4-\pi}\int_{|z|=4}\frac{1}{z\cos{z}}dz$$.
My attempt:
The integrand has singularities at $z=0, \frac{\pi}{2}, \frac{-\pi}{2}$, so
$$\frac{i}{4-\pi}\int_{|z|=4}\frac{1}{z\cos{z}}dz=\frac{i}{4-\pi}2\pi i ~Res_{z=z_k}\phi(z)=\frac{-2\pi}{4-\pi}\{Res_{z=0}\frac{1}{\cos(z)}+Res_{z=\pi/2}\frac{1}{z}+Res_{z=-\pi/2}\frac{1}{z}\}=\frac{-2\pi}{4-\pi}$$
But the given answer is $2$. Where did I made mistake? Please help me out.
|
The computation of the residues is messed up. You must consider the function as a whole, but you tossed away $1/z$ and $\cos z$ where it mattered. The rule is: $${\rm Res}\left(\frac{f(z)}{g(z)},a\right) = \frac{f(a)}{g'(a)},$$if $a$ is a simple pole of $g$. Let's save the $i/(4-\pi)$ for later. We have: $$\int_{|z|=4}\frac{{\rm d}z}{z\cos z} = 2\pi i\left(\underbrace{{\rm Res}\left(\frac{1}{z \cos z}, 0\right)}_{(a)} + \underbrace{{\rm Res}\left(\frac{1}{z \cos z}, \frac{\pi}{2}\right)}_{(b)}+\underbrace{{\rm Res}\left(\frac{1}{z \cos z}, -\frac{\pi}{2}\right)}_{(c)}\right)$$
One at a time, we get:
$$(a) = {\rm Res}\left(\frac{1}{z \cos z}, 0\right) = {\rm Res}\left(\frac{1/\cos z}{z}, 0\right) = \frac{1/ \cos 0}{1} = 1;$$
$$(b) = {\rm Res}\left(\frac{1}{z\cos z},\frac{\pi}{2}\right) = {\rm Res}\left(\frac{1/z}{\cos z},\frac{\pi}{2}\right) = \frac{1/(\pi/2)}{-\sin (\pi/2)} = -\frac{2}{\pi};$$
$$ (c) = {\rm Res}\left(\frac{1}{z\cos z},-\frac{\pi}{2}\right) = {\rm Res}\left(\frac{1/z}{\cos z},-\frac{\pi}{2}\right) = \frac{1/(-\pi/2)}{-\sin (-\pi/2)} = -\frac{2}{\pi},$$so that: $$\int_{|z|=4}\frac{{\rm d}z}{z \cos z} = 2\pi i \left(1- \frac{2}{\pi} - \frac{2}{\pi}\right) = 2\pi i \frac{(\pi-4)}{\pi} = 2i(\pi - 4),$$and finally: $$\frac{i}{4-\pi}\int_{|z|=4}\frac{{\rm d}z}{z\cos z} = \frac{i}{4-\pi}(2i(\pi-4)) = 2,$$as wanted.
|
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|
Find the least value of $4\csc^{2} x+9\sin^{2} x$
Find the least value of $4\csc^{2} x+9\sin^{2} x$
$a.)\ 14 \ \ \ \ \ \ \ \ \ b.)\ 10 \\
c.)\ 11 \ \ \ \ \ \ \ \ \ \color{green}{d.)\ 12} $
$4\csc^{2} x+9\sin^{2} x \\
= \dfrac{4}{\sin^{2} x} +9\sin^{2} x \\
= \dfrac{4+9\sin^{4} x}{\sin^{2} x} \\
= 13 \ \ \ \ \ \ \ \ \ \ \ \ (0\leq \sin^{2} x\leq 1)
$
But that is not in options.
I look for a short and simple way.
I have studied maths upto $12$th grade.
|
By AM-GM inequality
$$4\csc^{2} x+9\sin^{2} x \geq 2 \sqrt{36\csc^{2} x\sin^{2} x}$$
|
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|
Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \ \ \ d.)\ 7 $
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta \\
=\sin^{2} \theta +\dfrac{1}{\sin^{2} \theta }+\cos^{2} \theta+\dfrac{1}{\cos^{2} \theta }+\tan^{2} \theta+\dfrac{1}{\tan^{2} \theta } \\
\color{blue}{\text{By using the AM-GM inequlity}} \\
\color{blue}{x+\dfrac{1}{x} \geq 2} \\
=2+2+2=6 $
Which is not in options.
But I am not sure if I can use that $ AM-GM$ inequality in this case.
I look for a short and simple way .
I have studied maths upnto $12$th grade .
|
Hint:
We can use the Pythagorean identities $\color{blue}{\sin^2\theta+\cos^2\theta=1}$, $\color{blue}{\sec^2 \theta=\tan^2 \theta+1}$ and $\color{blue}{\csc^2\theta=\cot^2 \theta+1}$, giving us
\begin{align}\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta&=3+2\tan^2\theta+2\cot^2\theta\\
&=3+2\left(\tan^2\theta+\frac{1}{\tan^2\theta}\right)
\end{align}
|
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|
Convergence of $\int_\Bbb{R^2} \frac{\log(x^2+y^2)}{x^2+y^2}$ I tried to figure out if $\int_\Bbb{R^2} \frac{\log(x^2+y^2)}{x^2+y^2}$ converges
I think I should split it into two integrals $$\lim_{k\to\infty}\int_{B_\frac{1}{k}} \frac{\log(x^2+y^2)}{x^2+y^2} + \lim_{r\to\infty}\int_{B_r} \frac{\log(x^2+y^2)}{x^2+y^2}$$
where $B_\frac{1}{k}=${$ x \in \Bbb{R^2} |$ $ \frac{1}{k}\leq \|x\|\leq \frac{1}{2} $} and $B_r=${$ x \in \Bbb{R^2} | $ $ \frac{1}{2}\leq \|x\|\leq r $} and with changing coordinates {$x^2+y^2=r^2, \arctan\frac{y}{x}=\phi$} we get $$\lim_{k \to \infty} \int^{2\pi} _0\int^{\frac{1}{2}} _\frac{1}{k} \frac{r\log(r^2)}{r^2} drd\phi + \lim_{r \to \infty} \int^{2\pi} _0\int^{r} _\frac{1}{2} \frac{r\log(r^2)}{r^2} drd\phi$$
but Im not sure if that correct or if I can use again change of variables for $\log(r^2) = t, \frac{2r}{r^2}dr=dt$
thanks ahead
|
You can form the following chain of inequalities
$$ \int_4^\infty \frac{\log r^2}{r} dr > \int_4^\infty \frac{1}{r} dr = \lim_{b \to \infty} \log b - \log 4 = \infty.$$
So your integral diverges at $\infty$. In fact, it also diverges at $0$.
|
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|
$f(6)=144$ and $f(n+3) = f(n+2)\{f(n+1)+f(n)\}$, Then $f(7) =$? Given that $f(6)=144$ and $f(n+3) = f(n+2) \cdot\Big(f(n+1)+f(n)\Big)$ $[$For $n = 1,2,3,4]$
Then find the value of $f(7)$.
The solution is not unique but all of them are positive integers.
I can't find a way out.
|
Letting $a=f(1)$, $b=f(2)$, $c=f(3)$, $d=f(4)$, $e=f(5)$, $144=f(6)$, and $g=f(7)$ for notational simplicity, we have
$$\begin{align}
d&=c(b+a)\\
e&=d(c+b)\\
144&=e(d+c)\\
g&=144(e+d)\\
\end{align}$$
If I've done the algebra correctly, we can eliminate $d$ and $e$ leaving
$$144=c^2(a+b+1)(a+b)(b+c)$$
and
$$g=144c(a+b+1)(a+b)$$
If we assume that $a$, $b$, and $c$ are positive integers, the first equation leaves only a handful of possibilities, such as $(a,b,c)=(7,1,1)$. In particular, since $a+b$ and $a+b+1$ must be consecutive divisors of $144$, $a+b$ can only be $2$, $3$, or $8$.
|
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|
Prove that $\frac{2^{122}+1}{5}$ is a composite number As in the title. It's very easy to show that $5|2^{122}+1$, but what should I do next to show that $\frac{2^{122}+1}{5}$ is a composite number? I'm looking for hints.
|
$$1+2^{122}=\left(1+2^{61}\right)^2-\left(2^{31}\right)^2$$
$$=\left(1+2^{61}+2^{31}\right)\left(1+2^{61}-2^{31}\right)$$
Both factors are larger than $5$, also $$2^{122}\equiv \left(2^4\right)^{30}\cdot 2^2\equiv (1)^{30}\cdot 2^2\equiv -1\pmod{5},$$ so your result follows.
More generally, the following is called the Sophie-Germain identity (for all $a,b\in\mathbb R$):
$$a^4+4b^4=\left(a^2+2b^2\right)^2-\left(2ab\right)^2$$
$$=\left(a^2+2b^2+2ab\right)\left(a^2+2b^2-2ab\right)$$
The following factorization is a Aurifeuillean factorization (for all $k\in\mathbb R$):
$$2^{4k+2}+1=\left(2^{2k+1}-2^{k+1}+1\right)\left(2^{2k+1}+2^{k+1}+1\right)$$
Some other Aurifeuillean factorizations (not related to the Sophie-Germain identity):
$$3^{6k+3}+1=\left(3^{2k+1}+1\right)\left(3^{2k+1}-3^{k+1}+1\right)\left(3^{2k+1}+3^{k+1}+1\right)$$
$$5^{10k+5}-1=\left(5^{2k+1}-1\right)\left(5^{4k+2}+5^{3k+2}+3\cdot 5^{2k+1}+5^{k+1}+1\right)\times$$
$$\times \left(5^{4k+2}-5^{3k+2}+3\cdot 5^{2k+1}-5^{k+1}+1\right)$$
$$6^{12k+6}+1=\left(6^{4k+2}+1\right)\left(6^{4k+2}+6^{3k+2}+3\cdot 6^{2k+1}+6^{k+1}+1\right)\times$$
$$\times \left(6^{4k+2}-6^{3k+2}+3\cdot 6^{2k+1}-6^{k+1}+1\right)$$
There are a lot more of these on the Wikipedia page about Aurifeuillean factorization.
|
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|
Compute $\lim_{n \to +\infty} n^{-\frac12 \left(1+\frac{1}{n}\right)} \left(1^1 \cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\frac{1}{n^2}}$
How to compute
$$\displaystyle \lim_{n \to +\infty} n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} \left(1^1\cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\dfrac{1}{n^2}}$$
I'm interested in more ways of computing limit for this expression
My proof:
Let $u_n$be that sequence we've:
\begin{eqnarray*}
\ln u_n &=& -\frac{n+1}{2n}\ln n + \frac{1}{n^2}\sum_{k=1}^n k\ln k\\
&=& -\frac{n+1}{2n}\ln n + \frac{1}{n^2}\sum_{k=1}^n k\ln \frac{k}{n}+\frac{1}{n^2}\sum_{k=1}^n k\ln n\\
&=& \frac{1}{n^2}\sum_{k=1}^n k\ln \frac{k}{n}\\
&=& \frac{1}{n}\sum_{k=1}^n \frac{k}{n}\ln \frac{k}{n}\\
&\to&\int_0^1 x\ln x\,dx = -1/4
\end{eqnarray*}
Therefore the limit is $e^{-\frac{1}{4}}$
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Another approach, considering $$A_n= n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} \left(\prod_{i=1}^n i^i \right)^{\dfrac{1}{n^2}}=n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} H(n)^{\dfrac{1}{n^2}}$$ where appears the hyperfactorial function. Taking logarithms $$\log(A_n)={-\dfrac12 \left(1+\dfrac{1}{n}\right)}\log(n)+\dfrac{1}{n^2}\log(H(n))$$ Now, using Stirling like approximations $$\log(H(n))=n^2 \left(\frac{1}{2} \log(n)-\frac{1}{4}\right)+\frac{1}{2} n
\log (n))+\left(\log (A)+\frac{1}{12} \log
\left(n\right)\right)+O\left(\frac{1}{n^2}\right)$$ where appears Glaisher constant.
All of that, once recombined, leads to $$\log(A_n)=-\frac{1}{4}+\frac{\log (A)+\frac{1}{12} \log
\left(n\right)}{n^2}+O\left(\frac{1}{n^{5/2}}\right)$$ which shows the limit and how it is approached.
|
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|
Integral $\int_0^{\pi/2} \sin(ax)\cos(x)\,dx$ I have to evaluate an integral $I(a) = \sin(ax)\cos(x)$ from $0$ to $\pi/2$.The variable of $a$ is not is greater than $1$:
$$\int_0^{\pi/2} \sin(ax)\cos(x)\,dx$$
I attempted to change the function to $[\sin(ax+x)+\sin(ax-x)]/2$ and then integrate, but I am left with (-)cosines with a zero in the denominator. How do I remedy this? Am I missing something conceptual.
|
Your approach is good. Integrate the following identity term-wise:
$$
\sin(a x) \cos(x) = \frac{1}{2} \sin\left(\left(1+a\right)x\right) - \frac{1}{2} \sin\left(\left(1-a\right) x\right)
$$
getting
$$
\int_0^{\pi/2} \sin(a x) \cos(x) \mathrm{d}x = \frac{1}{2} \int_0^{\pi/2} \sin\left(\left(1+a\right)x\right)\mathrm{d}x - \frac{1}{2} \int_0^{\pi/2} \sin\left(\left(1-a\right)x\right)\mathrm{d}x
$$
And using $\int \sin(\omega x) \mathrm{d}x = -\frac{1}{\omega} \cos(\omega x)$, hence
$$
\frac{1}{2} \int_0^{\pi/2} \sin\left(\left(1\pm a\right)x\right)\mathrm{d}x = \left. -\frac{1}{2(1 \pm a)} \cos\left(\left(1 \pm a\right) x \right) \right|_{x = 0}^{x = \pi/2} = \frac{1}{2 \left(1\pm a\right)} \left(1- \cos\left(\frac{\pi}{2} \left(1 \pm a\right)\right)\right) = \frac{\sin^2\left(\frac{\pi}{2} \frac{1 \pm a}{2}\right)}{ 1\pm a}
$$
Further simplifying the difference of $+$ and $-$ parts we get
$$
\int_0^{\pi/2} \sin(a x) \cos(x) \mathrm{d}x = \frac{\sin\left(\frac{\pi}{2} a \right) - a}{1-a^2}
$$
|
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|
Result of matrix $A^{2016}$ I want to find the result of $A^{2016}$ but I cannot find any pattern except for the zeros in the middle row and column.
$$A=\begin{bmatrix}1 & 0 & {-2}\\0 & 0 & {0}\\3 & 0 & {-4}\end{bmatrix}$$
|
Note that $A$ has characteristic polynomial $p(\lambda)=-\lambda^3-3\lambda^2-2\lambda$, which yields $\lambda_1=-2$, $\lambda_2=-1$ and $\lambda_3=0$. Since each eigenvalue is different and there are 3, then $A$ is diagonalizable.
It is easy to see that $A=PDP^{-1}$ with
$P=\left(\begin{array}{ccc}2&1&0\\0&0&1\\3&1&0\end{array}\right)$, $P^{-1}=\left(\begin{array}{ccc}-1&0&1\\3&0&-2\\0&1&0\end{array}\right)
$ and $D=\left(\begin{array}{ccc}-2&0&0\\0&-1&0\\0&0&0\end{array}\right)$
Now:
$\begin{eqnarray}A^m&=&(PDP^{-1})^m=(PDP^{-1})(PDP^{-1})...(PDP^{-1})\\&=&PD(P^{-1}P)D(P^{-1}P)D...(P^{-1}P)DP^{-1}\\
&=&PDD...DP^{-1}\\
&=&PD^{m}P^{-1}\end{eqnarray}$
For $m=2016$ we have
$\begin{eqnarray}
A^{2016}&=&PD^{2016}P^{-1}\\
&=&\left(\begin{array}{ccc}2&1&0\\0&0&1\\3&1&0\end{array}\right)\left(\begin{array}{ccc}-2&0&0\\0&-1&0\\0&0&0\end{array}\right)^{2016}\left(\begin{array}{ccc}-1&0&1\\3&0&-2\\0&1&0\end{array}\right)\\
&=&\left(\begin{array}{ccc}2&1&0\\0&0&1\\3&1&0\end{array}\right)\left(\begin{array}{ccc}(-2)^{2016}&0&0\\0&(-1)^{2016}&0\\0&0&0\end{array}\right)\left(\begin{array}{ccc}-1&0&1\\3&0&-2\\0&1&0\end{array}\right)\\
&=&\left(\begin{array}{ccc}2&1&0\\0&0&1\\3&1&0\end{array}\right)\left(\begin{array}{ccc}2^{2016}&0&0\\0&1&0\\0&0&0\end{array}\right)\left(\begin{array}{ccc}-1&0&1\\3&0&-2\\0&1&0\end{array}\right)\\
&=&\left(\begin{array}{ccc}-2^{2017}+3&0&2^{2017}-2\\0&0&0\\-3\cdot2^{2016}+3&0&3\cdot2^{2016}-2\end{array}\right)
\end{eqnarray}$
|
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Proving Trig Identities (Complex Numbers)
Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n + {1\over z^n} = 2\cos(n\theta) $$
Hence prove that $\cos^6(\theta)$ $=$ $\frac 1{32}$$(\cos(6\theta)$ + $6\cos(4\theta)$ + $15\cos(2\theta)$ + $10$)
I learnt to prove the first part in another post linked here.
The second part is where I am confused because there is a 'hence'
so I thought of taking 2 approaches:
either $$ z^6 + \frac 1{z^6} $$
or $$ z^6 $$ and equating real parts.
I will start with my first approach
$$ z^6 + \frac 1{z^6} $$
$$ (\cos(x)+i\sin(x))^6 + \frac 1{(\cos(x)+i\sin(x))^6} $$
$$ (\cos(x)+i\sin(x))^6 + {(\cos(x)+i\sin(x))^{-6}} $$
$$ \cos(6x) + i\sin(6x) + \cos(-6x)+ i \sin(-6x) $$
$$ 2\cos(6x) $$
Which is no where near what I am suppose to prove..
So with my second approach (expanding and equation real parts)
$$ z^6 $$
$$ (\cos(x) + i \sin(x))^6 $$
Using pascals
$$ \cos^6(x) + i*6\cos^5(x)\sin(x) + i^2*15\cos^4(x)\sin^2(x) + i^3*20\cos^3(x)\sin(x) + i^4*15\cos^2(x)\sin^4(x) + i^5*6\cos(x)\sin^5(x) + i^6 * \sin^6(x) $$
Simplifying
$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) +i(6\cos^5(x)\sin(x)-20\cos^3(x)\sin(x) + 6cos(x)\sin^5(x)$$
Now considering only real
$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) $$
At this point I'm confused , am I on the right approach?
|
By De Moivre's formula if $z = \cos (\theta) + i\sin(\theta)$ then $z^n = \cos (n\theta) + i\sin(n\theta)$.
So $$z^n = \cos (n\theta) + i\sin(n\theta)$$
and
$$z^{-n} = \cos (-n\theta) + i\sin(-n\theta)=\cos (n\theta) - i\sin(n\theta)$$
So $$z^n+z^{-n}=2\cos (n\theta)$$
Thus when $n=1$ ,
$$z+z^{-1}=2\cos (\theta) \Rightarrow \left(z+\frac{1}{z} \right)^6=2^6 \cdot \cos^6(\theta) $$
$$2^6 \cdot \cos^6(\theta)=\left(z+\frac{1}{z} \right)^6 $$
$$2^6 \cdot \cos^6(\theta)=z^6+6z^4+15z^2+20+15+\frac{1}{z^2} +6\frac{1}{z^4} +\frac{1}{z^6} $$
$$2^6 \cdot \cos^6(\theta)=\left(z^6+\frac{1}{z^6}\right)+6\left(z^4+\frac{1}{z^4}\right)+15\left(z^2+\frac{1}{z^2}\right)+20 $$
$$64 \cdot \cos^6(\theta)=\left(2\cos 6\theta \right)+6\left(2\cos 4\theta\right)+15\left(2\cos 2\theta\right)+20 $$
Therefore
$$ \cos^6(\theta)=\frac{1}{32}\{\left(\cos 6\theta \right)+6\left(\cos 4\theta\right)+15\left(\cos 2\theta\right)+10 \}$$
|
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|
Solution to the equation $\sqrt{x^2 - 2x + 1} - 5 = 0$ I had this equation on my exam :
$$\sqrt{x^2-2x+1} - 5 = 0$$
My friends said the the solution could be :
$$|x-1| = -5$$
So the solution is nothing!
But I say the solution is:
$$x^2-2x+1 = 25 $$
so $$x = 6\ |\ x = -4$$
My Question here is which solution is right, and why.
Thanks in advance.
|
It is $$\sqrt{x^2-2x+1} - 5 = 0 \\ \Rightarrow \sqrt{(x-1)^2} - 5 = 0 \\ \Rightarrow |x-1|-5=0 \\ \Rightarrow |x-1|=5 \\ \Rightarrow x-1=5 \text{ or } x-1=-5 \\ \Rightarrow x=6 \text{ or } x=-4$$ So, your solution is correct!!
$$$$
If the equation is $\sqrt{x^2-2x+1} + 5 =0$ then we have the following:
$$\sqrt{x^2-2x+1} + 5 = 0 \\ \Rightarrow \sqrt{(x-1)^2} + 5 = 0 \\ \Rightarrow |x-1|+5=0 \\ \Rightarrow |x-1|=-5$$
So, there is no solution.
In this case the solution of your friend is correct.
|
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|
Discrete Time Markov Chain question Let $\{X_n : n \ge 0 \}$ be a Markov chain with state space $ \{0, 1, 2, 3\} $ and transition matrix
$$P=\begin{pmatrix}
\frac{1}{4} & 0 & \frac{1}{2} & \frac{1}{4}\\
0 & \frac{1}{5} & 0 & \frac{4}{5}\\
0 & 1 & 0 & 0\\
\frac{1}{3} & \frac{1}{3} & 0 & \frac{1}{3}\\
\end{pmatrix}
$$
A new process $\{Z_n : n \ge 0\}$ is defined by $Z_n = 0$ if $X_n = 0$ or $1$, and $Z_n = X_n$ if $X_n = 2$ or $3$.
Find $P(Z_{n+1} = 2 \mid Z_n = 0, Z_{n−1} = 2)$.
I'm not too sure how to deal with the fact that $Z_n = 0$ implies that $X_n = 0$ or $1$. I know you're supposed to plug in the corresponding $X_n$ values into the question, but I get $P(Z_{n+1} = 2 \mid Z_n = 0, Z_{n−1} = 2)$ = $P(X_{n+1} = 2 \mid X_n = 0 \ \text{or} \ 1, X_{n−1} = 2)$.
Is this simply equal to $P(X_{n+1} = 2 \mid X_n = 0, X_{n−1} = 2)+P(X_{n+1} = 2\mid X_n = 1, X_{n−1} = 2)$?
|
First of all, notice that $Z_{n+1} = 2 \Leftrightarrow X_{n+1} = 2$ and $Z_n = 0 \Leftrightarrow X_n \in \{0,1\}$. So
$$P(Z_{n+1} = 2 \mid Z_n = 0, Z_{n-1} = 2) = P(X_{n+1} = 2 \mid X_n \in \{0,1\}, X_{n-1} = 2).$$
Write this quantity as
$$P(X_{n+1} = 2 \mid \{X_n = 0\} \cup \{X_n = 1\}, X_{n-1} = 2).$$
The strong Markov property says that you can drop the $X_{n-1}$ conditioning.
Now try to use laws of conditional probability, along with your transition matrix $P$, to get the answer. I'd use Bayes' theorem, and then you're going to have to do some slightly messy calculations.
|
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|
A series related to $\pi\approx 2\sqrt{1+\sqrt{2}}$ This question follows a suggestion by Tito Piezas in Is there an integral or series for $\frac{\pi}{3}-1-\frac{1}{15\sqrt{2}}$?
Q: Is there a series by Ramanujan that justifies the approximation $\pi\approx2\sqrt{1+\sqrt{2}}?$
|
The answer needs some context.
Part I. One may ask why,
$$\frac{1}{2\pi\sqrt{2}} - \frac{1103}{99^2} \approx 10^{-9}\tag1$$
is such a good approximation? In fact, the convergents of the continued fraction of $\displaystyle\frac{1}{2\pi\sqrt{2}}$ start as,
$$0,\;\frac{1}{8},\;\color{brown}{\frac{1}{3^2}},\;\frac{8}{71},\;\frac{9}{80},\dots\color{brown}{\frac{1103}{99^2}},\dots$$
and it turns out using the 3rd and 9th convergents,
$$\frac{1}{2\pi\sqrt{2}}-\frac{1}{3^2} = 16\sum_{k=1}^\infty \frac{(4k)!}{k!^4} \frac{1+10k}{(12^4)^{k+1/2}}$$
$$\frac{1}{2\pi\sqrt{2}}-\frac{1103}{99^2} = 16\sum_{k=1}^\infty \frac{(4k)!}{k!^4} \frac{1103+26390k}{(396^4)^{k+1/2}}\tag2$$
so truncating $(2)$ "explains" the approximation $(1)$.
Part II. The OP's post originated with the observation that,
$$\frac{\pi}{3} -\Big(1+\frac{1}{15\sqrt{2}}\Big) \approx 10^{-5}\tag3$$
The convergents of the continued fraction of $\displaystyle\sqrt{2}\big(\frac{\pi}{3}-1\big)$ are,
$$0,\;\frac{1}{14},\;\color{brown}{\frac{1}{15}},\;\frac{55}{824},\dots$$
As there are dozens and dozens of Ramanujan-Sato pi formulas that use $\sqrt{2}$, it is possible (though not certain) there is one that is directly responsible $(3)$.
Part III. Finally, the question. The modest approximation,
$$\pi \approx 2\sqrt{1+\sqrt{2}}\tag4$$
has an explanation as it is the first term (after some manipulation) of the Ramanujan-type formula,
$$\frac{1}{\pi}= \sum_{n=0}^\infty (-1)^n\frac{(2n)!^3}{n!^6} \frac{16(8+5\sqrt{2})n+8(2+\sqrt{2})}{(C)^{n+1/2}},\quad C = 2^9(1+\sqrt{2})^3\tag5$$
And there are many other similar ones.
|
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|
Bounded sequence $a_n=\sqrt{4+2 \sqrt{4+\cdots+2 \sqrt{4+2 \sqrt{4+2 \sqrt{4+4}}}}}$ Let
$$a_n=\sqrt{4+2 \sqrt{4+\cdots+2 \sqrt{4+2 \sqrt{4+2 \sqrt{4+4}}}}}$$
the sign $\sqrt{}$ occurs $n$ times.
a) Prove, that $a_n< \sqrt{5}+1$ for all $n$.
b) Find $\lim_{n\rightarrow \infty } a_n$
Author O.Kukush
|
For the case (b) when $n \to \infty$ we can write the following,
\begin{equation}
a_n^2 = 4 + 2a_n \\ a_n^2 - 2a_n - 4 = 0 \\ a_n = \cfrac{2+\sqrt{20}}{2} = 1 + \sqrt{5}
\end{equation}
So, it may be stated that for all other $n$, $a_n < 1 + \sqrt{5} $
|
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|
Limit of the sequence defined by a recurrence Given a recurrence formula for an arithmetic sequence, $$U_{n} = \frac{1}{2+U_{n-1}}$$
Show that$$\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+ ...}}}} = (SomeGivenValue)$$
How can we solve questions like this?
|
The method is to make a substitution $U_n=\frac{T_n}{T_{n+1}}$ and you would get much more tangible
$$T_{n+1}^2-2T_nT_{n+1}-T_n^2=0$$
$$(T_{n+1}-T_n-\sqrt{2}T_n)(T_{n+1}-T_n+\sqrt{2}T_n)=0$$
$$T_{n+1}-T_n-\sqrt{2}T_n=0$$
$$T_{n+1}-T_n+\sqrt{2}T_n=0$$
These two you solve classically assuming $T_n=a^n$ and when you substitute and solve you have that $a_1=1-\sqrt{2}$ and $a_2=1+\sqrt{2}$ which gives
$$T_n=c(1-\sqrt{2})^n+d(1+\sqrt{2})^n$$
and the solution follows. Set initial conditions and you have the solution.
$$U_n=\frac{c(1-\sqrt{2})^n+d(1+\sqrt{2})^n}{c(1-\sqrt{2})^{n+1}+d(1+\sqrt{2})^{n+1}}$$
Now all you need to do to get to the answer of some value is to find:
$$\lim\limits_{n \to \infty}\frac{c(1-\sqrt{2})^n+d(1+\sqrt{2})^n}{c(1-\sqrt{2})^{n+1}+d(1+\sqrt{2})^{n+1}} = \sqrt{2}-1$$
Since $c$ and $d$ depend on the initial value (in your case that would be $U_0 = \frac{1}{2}$) this proves that the initial value is irrelevant.
Three years later....
A small technical glitch
We cannot have
$$ \frac{c}{d}=\left ( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right )^n= \left (3+2\sqrt{2} \right)^n$$
as this would lead to $0$ appearing down the track of $U_n$ and the quotient is then invalid. This does not affect the final conclusion as for $U_0 = \frac{1}{2}$ we have
$$ \frac{c}{d}=\left ( \frac{\sqrt{2}-1}{\sqrt{2}+1} \right )^n= \left (3-2\sqrt{2} \right)^n$$
|
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|
Probability an integer chosen at random from 1 to 1000 is divisible by 3,5, or 7 A number is chosen at random from the first 1,000 positive integers. What is the probability that it's divisible by 3,5, or 7?
So I started off by breaking the problem up and having:
divisible by 3: p(a)
divisible by 5: p(b)
divisible by 7 p(c)
I know I'm going to apply the exclusion inclusion principle, but how do I find out how many numbers are divisible by each without going through all the numbers between 1 and 1000?
|
Use inclusion/exclusion principle:
*
*Include the amount of numbers divisible by $3$, which is $\Big\lfloor\frac{1000}{3}\Big\rfloor=333$
*Include the amount of numbers divisible by $5$, which is $\Big\lfloor\frac{1000}{5}\Big\rfloor=200$
*Include the amount of numbers divisible by $7$, which is $\Big\lfloor\frac{1000}{7}\Big\rfloor=142$
*Exclude the amount of numbers divisible by $3$ and $5$, which is $\Big\lfloor\frac{1000}{3\cdot5}\Big\rfloor=66$
*Exclude the amount of numbers divisible by $3$ and $7$, which is $\Big\lfloor\frac{1000}{3\cdot7}\Big\rfloor=47$
*Exclude the amount of numbers divisible by $5$ and $7$, which is $\Big\lfloor\frac{1000}{5\cdot7}\Big\rfloor=28$
*Include the amount of numbers divisible by $3$ and $5$ and $7$, which is $\Big\lfloor\frac{1000}{3\cdot5\cdot7}\Big\rfloor=9$
Hence the amount of numbers divisible by $3$ or $5$ or $7$ is:
$$333+200+142-66-47-28+9=543$$
|
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|
How to simplify $\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}$ The answer is 2. But I want to learn how to simplify this expression without the use of calculator.
|
More generally, if $a + b \sqrt{2}$ (with $a$, $b$ integers) has a cube root in $\mathbb Z[\sqrt{2}]$, the norm $N(a + b \sqrt{2}) = a^2 - 2 b^2 \sqrt{2}$ must be the cube of an integer, say $m^3$, and then that cube root would be
$x + y \sqrt{2}$ with $x^2 - 2 y^2 = m$. From $(x+y \sqrt{2})^3 = a + b \sqrt{2}$ we get $x^3 + 6 x y^2 = 4 x^3 - 3 m x = a$. In particular, $x$ must be a divisor of $a$, which brings you down to a finite (and hopefully not too big) list of possiblities that are easily checked.
|
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|
How do I finish this trig integral $\int_0^{\pi/4}\frac{\sin^2 \theta}{\cos \theta}d\theta$? I got up to the part where it's $$\frac{9}{125}\int_0^{\large \frac{\pi}{4}}\frac{\sin^2\theta}{\cos\theta}\,\,d\theta$$
but I can't figure out how to finish it off.
By the way the original problem was:
$$\int_0^{0.6}\frac{x^2}{\sqrt{9-25x^2}}\,\,dx$$
|
Use the trick to add and remove a $\cos^2\theta$ term in the numerator of the integral:
$$\int_0^{\pi/4}\frac{\sin^2\theta + \cos^2\theta - \cos^2\theta}{\cos\theta}\ \text{d}\theta = \int_0^{\pi/4}\frac{1}{\cos\theta} - \cos\theta\ \text{d}\theta$$
Which now you can split.
Remembering that $$\frac{1}{\cos\theta} = \text{sec}(\theta)$$
you have immediately, by remembering standard integral of the secant:
$$\int_0^{\pi/4}\text{sec}\theta\ \text{d}\theta = \left[\ln\left(\cos\frac{\theta}{2} - \sin\frac{\theta}{2}\right) + \ln\left(\cos\frac{\theta}{2} + \sin\frac{\theta}{2}\right)\right]\bigg|_0^{\pi/4} = \frac{1}{2}\ln(3 + 2\sqrt{2})$$
The other one is trivial:
$$\int_0^{\pi/4}\cos\theta\ \text{d}\theta = \frac{\sqrt{2}}{2}$$
Thence:
$$\int_0^{\pi/4}\frac{1}{\cos\theta} - \cos\theta\ \text{d}\theta = \frac{1}{2}\ln(3 + 2\sqrt{2}) - \frac{\sqrt{2}}{2}$$
|
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|
Find the integral $\int \frac{1+x}{\sqrt{1-x^2}}\mathrm dx$ The integral can be represented as
$$
\int \frac{1+x}{\sqrt{1-x^2}}\mathrm dx=
\int \left(\frac{1+x}{1-x}\right)^{1/2}\mathrm dx
$$
Substitution $$t=\frac{1+x}{1-x}\Rightarrow x=\frac{t-1}{t+1}\Rightarrow dx=\frac{2}{(t+1)^2}dt\Rightarrow \int\limits \left(\frac{1+x}{1-x}\right)^{1/2}\mathrm dx=2\int\limits \frac{\sqrt{t}}{(t+1)^2}\mathrm dt$$
What substitution to use for solving the integral $\int\limits \frac{\sqrt{t}}{(t+1)^2}\mathrm dt$?
|
No substitutions:
$$
\int\left(\frac{1}{\sqrt{1-x^2}}-\frac{-x}{\sqrt{1-x^2}}\right)\,dx
=\arcsin x-\sqrt{1-x^2}+c
$$
You can also do that way; continue with $u=\sqrt{t}$, so $t=u^2$ and $dt=2u\,du$; so you get
$$
\int\frac{4u^2}{(u^2+1)^2}\,du=
\int 2u\cdot\frac{2u}{(u^2+1)^2}\,du
$$
Noticing that $2u$ is the derivative of $u^2+1$ you can use integration by parts
$$
=2u\cdot\left(-\frac{1}{u^2+1}\right)-
\int2\left(-\frac{1}{u^2+1}\right)\,du
=2\arctan u-\frac{2u}{u^2+1}
$$
Do the back substitutions.
Alternative method: set $x=\cos4t$, so you have
$$
\sqrt{\frac{1+\cos4t}{1-\cos4t}}=\frac{\cos2t}{\sin2t}
$$
and the integral becomes
$$
-8\int\cos^22t\,dt=-4\int(1+\cos t)\,dt
$$
|
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|
Evaluation of $\lim_{n\rightarrow \infty}\sum_{k=1}^n\sin \left(\frac{n}{n^2+k^2}\right)$
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\sin \left(\frac{n}{n^2+1}\right)+\sin \left(\frac{n}{n^2+2^2}\right)+\cdots+\sin \left(\frac{n}{n^2+n^2}\right)$
$\bf{My Try::}$ We can write the Sum as $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\sin\left(\frac{n}{n^2+r^2}\right)$$
Now how can I convert into Riemann Sum, Help me
Thanks
|
We can use Riemann sums to evaluate the limit. We have
$$\sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\right)=\sum_{k=1}^n\,n\sin\left(\frac{1/n}{1+(k/n)^2}\right)\frac1n\to \int_0^1\frac{1}{1+x^2}\,dx=\pi/4$$
since we have $$\left(\frac{1}{1+(k/n)^2}\right)-\frac{1}{6n^2}\left(\frac{1}{1+(k/n)^2}\right)^3\le n\sin\left(\frac{1/n}{1+(k/n)^2}\right)\le \frac{1}{1+(k/n)^2}$$
|
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|
Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
\sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of
\begin{equation}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) without using the identities of trigonometry.
The function $\sin x+\cos x$ could be transformed using some trigonometric identities to a single function. In fact, WolframAlpha says it is equal to $\sqrt2\sin\left(x+\frac\pi4\right)$ and there also are some posts on this site about this equality. So probably in this way we could calculate $x$ from the first equation - and once we know $\sin x$ and $\cos x$, we can calculate $\dfrac{1}{\sin x}+\dfrac{1}{\cos x}$. Is there a simpler solution (perhaps avoiding explicitly finding $x$)?
|
Notice, $$\frac{1}{\sin x}+\frac{1}{\cos x}$$
$$=\frac{\sin x+\cos x}{\sin x\cos x}$$
$$=2\cdot \frac{\sin x+\cos x}{2\sin x\cos x}$$
$$=2\cdot \frac{\sin x+\cos x}{(\sin x+\cos x)^2-1}$$
setting the value of $\sin x+\cos x$,
$$=2\cdot \frac{\frac 75}{\left(\frac{7}{5}\right)^2-1}$$
$$=\frac{35}{12}$$
|
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|
Diophantine equation: $\frac1x=\frac{a}{x+y}-\frac1y$ For how many different natural values of $a$ the Diophantine equation: $\frac1x=\frac{a}{x+y}-\frac1y$ has natural roots?
I rearranged the equation as: $xy+x^2+y^2=(a-1)xy$ , hence I said we must have: $(a-1)>2$
I tried another idea: reform the equation to: $x^2+(2-a)yx+y^2=0$ , considering this as a second order equation in terms of x, the Delta must be non negative and perfect square,but now I can't go on!!
Please help.
|
Rearrange it to
$$\frac{a}{x+y} = \frac{1}{x} + \frac{1}{y} = \frac{y+x}{xy}.\tag{1}$$
Multiply $(1)$ with $x+y$ to obtain
$$a = \frac{(x+y)^2}{xy}.\tag{2}$$
So the question is how many natural numbers are values of $\frac{(x+y)^2}{xy}$ with $x,y\in \mathbb{N}$. Suppose $x,y\in \mathbb{N}$ with $\frac{(x+y)^2}{xy}\in \mathbb{N}$. Write $x = d\cdot \xi$ and $y = d\cdot\eta$ with $d = \gcd(x,y)$. Then
$$\frac{(x+y)^2}{xy} = \frac{(d\xi + d\eta)^2}{d\xi d\eta} = \frac{d^2(\xi + \eta)^2}{d^2\xi \eta} = \frac{(\xi + \eta)^2}{\xi \eta}.$$
In particular $\xi \mid (\xi + \eta)^2$, whence $\xi \mid \eta^2$. Since $\gcd(\xi,\eta) = 1$, it follows that $\xi = 1$. By symmetry, we also have $\eta = 1$, so
$$\frac{(\xi + \eta)^2}{\xi\eta} = \frac{2^2}{1^2} = 4.$$
It follows that $a = 4$ is the only natural number for which $(1)$ has natural solutions.
|
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|
Coefficents of cubic polynomial and its least root Let $x^3-(m+n+1)x^2+(m+n-3+mn)x-(m-1)(n-1)=0$,
be a cubic polynomial with positive roots, where $m,n \ge2$ are natural nos. For fixed $m+n$, say $15$, it turns out that least root of the polynomial will be smallest in case of $m=2,n=13$ i.e the case in which difference is largest.
I checked it for many values of $m+n$ and same thing is coming out. It seems that observation is right. How to prove it?
I noticed that coefficient of x and constant are also of largest modulus when difference between $m,n$ is largest. Does it help anyway?
|
Some Remarks:
There is an explicit formula to solve cubic equations here . So you can always use it and analyze the roots in a brute-force way. I will follow a somewhat simpler method. I will assume that :
in the cases you are interested in you always have three (not necessarily distinct) real roots... otherwise the question isn't saying anything useful.
The method:
Let $f(x) = x^{3} + (m+ n+1) x^{2} + (mn + m + n -3) x - (m-1)(n-1)$.
The idea is we are interested in the smallest real root of $f(x)$ and this will be somewhere in the left of the smallest root of $f^{'}(x)$ (the derivative)- which is also real. So we want to minimize the roots of $f^{'}(x)$.
This is just a simple application of Mean Value theorem.
But $f^{'}(x)$ is a quadratic function whose roots are obtained by quadratic formula. Let us use new variable : $ p = m+ n$ and $q = m-n$.
Then we have $$f^{'}(x) = 3x^{2} - 2(p+1)x +(\frac{p^{2}-q^{2}}{4} + p -3).$$
In the question $p$ is fixed so it is a constant. Using the quadratic formula you can check that the minimum value of the real roots are achieved when the discriminant $$D(q): 4\left ( (p+1)^{2} - 3(\frac{p^{2}-q^{2}}{4} + p -3)\right)$$ (as a function of $q$ since $p$ is fixed anyway) is maximized.
This is what you observe in your calculations.
EDIT
Regarding the discussion OP mentioned in the comments here is full set of solution to the equation $x^{3} - 16x^{2} = 68x -75 = 0$ using the software SAGE. It is free and can be used online so anyone can try it and it is wonderful!
The solutions are:
$$
\left[x = -\frac{1}{2} \, {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}} {\left(i \, \sqrt{3} + 1\right)} + \frac{26 i \, \sqrt{3} - 26}{9 \, {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}}} + \frac{16}{3}, \; x = -\frac{1}{2} \, {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}} {\left(-i \, \sqrt{3} + 1\right)} + \frac{-26 i \, \sqrt{3} - 26}{9 \, {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}}} + \frac{16}{3}, \; x = {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}} + \frac{52}{9 \, {\left(\frac{1}{18} i \, \sqrt{14141} \sqrt{3} + \frac{425}{54}\right)}^{\frac{1}{3}}} + \frac{16}{3}\right]
.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1656617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
If $x,y,z>0$ and $x+y+z=1$ Then prove that $xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz$
If $x,y,z$ are positive real number and $x+y+z=1\,$ Then prove that
$xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz$
Let $$f(x,y,z)=xy(x+y)^2+yz(y+z)^2+zx(z+x)^2$$
Then $$\frac{f(x,y,z)}{xyz} = \frac{(x+y)^2}{z}+\frac{(y+z)^2}{x}+\frac{(z+x)^2}{y}$$
Now Using $\bf{Cauchy-Schwarz\; }$ Inequality, We get$$\frac{f(x,y,z)}{xyz}\geq \frac{4(x+y+z)^2}{x+y+z} = 4\Rightarrow f(x,y,z)\geq 4xyz$$
My Question is How can we solve it Using $\bf{A.M\geq G.M}$ nequality,
plz explain me, Thanks
|
Brilliant answer from @deepsea but it took me a while to understand it. For anyone else struggling here are some baby steps:
$$
\frac{f(x,y,z)}{xyz}=\frac{(x+y)^2}{z}+\frac{(y+z)^2}{x}+\frac{(z+x)^2}{y}
$$
$$
=\sum_{cyc}\frac{(x+y)^2}{z}=\sum_{cyc}\frac{(y+z)^2}{x}=\sum_{cyc}\frac{(1-x)^2}{x}\ \text{, because}\ 1-x=y+z
$$
$$
\sum_{cyc}\frac{(1-x)^2}{x}=\sum_{cyc}\frac{(1-2x+x^2)}{x}=\sum_{cyc}(\frac{1}{x}-2+x)
$$
$$
\sum_{cyc}\frac{1}{x}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{(\sqrt1+\sqrt1+\sqrt1)^2}{x+y+z}\ \text{, by Titu's Lemma}
$$
$$
\frac{(\sqrt1+\sqrt1+\sqrt1)^2}{x+y+z}=\frac{3^2}{x+y+z}=\frac{9}{1}=9
$$
$$
\sum_{cyc}2=2+2+2=6\ \text{and}\ \sum_{cyc}x=x+y+z=1
$$
$$
\text{Putting it all back together,}\ \sum_{cyc}(\frac{1}{x}-2+x) \ge 9-6+1=4
$$
$$
\frac{f(x,y,z)}{xyz}\ge4\implies f(x,y,z)\ge4xyz
$$
As far as I can tell this does not depend on AM-GM. Also, I found it helpful to realise it was Titu's Lemma employed in the OP solution rather than Cauchy-Schwarz, although I appreciate one is a specialisation of the other.
|
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"url": "https://math.stackexchange.com/questions/1657914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Value of the product: $ \sqrt{2} \sqrt{2 - \sqrt{2}} \sqrt{2 - \sqrt{2 - \sqrt{2}}} \sqrt{2 - \sqrt{2 - \sqrt{2-\sqrt{2}}}} \cdots $ =? Let the recursive sequence
$$ a_0 = 0, \qquad a_{n+1} = \sqrt{2-a_n},\,\,n\in\mathbb N.
$$
T
Can we find the value of the product
$$
\prod_{n=1}^{\infty}{a_n}?
$$
Well, from here I don't seem to follow. I can understand that there would be some good simplification and the product will hopefully telescope but I'm lacking the right algebra. I also thought of finding a recurrence solution probably from the corresponding DE but that didn't follow as well.
|
Another approach is the following one: if we assume $ a_n = 2\cos(\theta_n) $ it follows that
$$ \cos(\theta_{n+1})=\sqrt{\frac{1-\cos\theta_n}{2}} = \sin\left(\frac{\theta_n}{2}\right) = \cos\left(\frac{\pi-\theta_n}{2}\right)\tag{1} $$
from which we have $\theta_{n+1}=\frac{\pi-\theta_n}{2}$ and, by induction:
$$ \theta_{n+k} = \frac{\pi}{3}-(-1)^k\frac{\pi}{3\cdot 2^k}+(-1)^k\frac{\theta_n}{2^k}.\tag{2}$$
Since $\theta_0=\frac{\pi}{2}$,
$$ \theta_k = \frac{\pi}{3}+(-1)^k \frac{\pi}{6\cdot 2^k},\qquad \color{red}{a_k = \cos\left(\frac{\pi}{6\cdot 2^k}\right)-(-1)^k\sqrt{3}\sin\left(\frac{\pi}{6\cdot 2^k}\right)}\tag{3} $$
but since $2\cos\theta_n = \frac{\sin(2\theta_n)}{\sin(\theta_n)}$ and $\sin(\pi-\theta)=\sin(\theta)$, we also have a telescopic product.
In particular:
$$ a_1\cdot a_2\cdot\ldots\cdot a_n = \frac{\sin(2\theta_1)}{\sin(\theta_n)} \tag{4}$$
hence:
$$ \prod_{n\geq 1} a_n = \frac{\sin(2\theta_1)}{\sin(\lim_{n\to +\infty}\theta_n)} = \frac{1}{\sin\frac{\pi}{3}}=\color{red}{\frac{2}{\sqrt{3}}}.\tag{5}$$
|
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"url": "https://math.stackexchange.com/questions/1658073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
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|
$\int\frac{a^2\sin^2x+b^2\cos^2x}{a^4\sin^2x+b^4\cos^2x}dx$ $\int\frac{a^2\sin^2x+b^2\cos^2x}{a^4\sin^2x+b^4\cos^2x}dx$
My Try:
$\int\frac{a^2\sin^2x+b^2\cos^2x}{a^4\sin^2x+b^4\cos^2x}dx$
$\int\frac{a^2\tan^2x+b^2}{a^4\tan^2x+b^4}dx$
$\int\frac{1}{b^2}\frac{\frac{a^2}{b^2}\tan^2x+1}{\frac{a^4}{b^4}\tan^2x+1}dx$
I do not know how to solve it further.
|
The integrand can be written as $\frac{1}{a^2+b^2}\left[1+\frac{2a^2b^2}{\left(a^4+b^4\right)+\left(b^4-a^4\right)\cos 2\theta}\right]$. It is easy from here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1658195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Basic Homogenous Differential Equation Solve the following differential equation
$$\frac{dy}{dx} = \frac{2x^2y+x^3}{y^3+2x^3}$$
I have identified that the differential equation is homogeneous because $f(rx,ry)$ = $r^3f(x,y)$ . I then used the substitution $u = y/x$ and $\frac{dy}{dx} = x\frac{du}{dx} + u$ reducing the equation to $$x\frac{du}{dx} = \frac{1}{2u^3} + \frac{1}{u^2}$$ I believe i have made a mistake somewhere in the middle of this and if not how would I go about solving this differential equation. (The only methods that have been presented in class are 1st order linear using mu substitution, separable, exact, homogeneous and Bernoulli). If it helps my profesor wrote down a solution $y = \int \frac{v+1}{v^3+2}$dv (He did not have the dv but i believe he just forgot it so I added it)
|
On substituting $u=\frac{y}{x}$ in the differential equation $$\frac{dy}{dx} = \frac{2x^2y+x^3}{y^3+2x^3}$$
we get
$$x\frac{du}{dx} + u = \frac{2u+1}{u^3+2}$$
or $$x\frac{du}{dx} = \frac{2u+1-u^4-2u}{u^3+2}$$
or $$\frac{u^3+2}{1-u^4}du = \frac{dx}{x}$$
or $$-\frac{1}{4}\cdot \frac{-4u^3}{1-u^4}du + \frac{2}{(1-u^2)(1+u^2)}du= \frac{dx}{x}$$
or $$-\frac{1}{4}\cdot \frac{d(1-u^4)}{(1-u^4)} + \frac{(1-u^2)+(1+u^2)}{(1-u^2)(1+u^2)}du= \frac{dx}{x}$$
or $$-\frac{1}{4}\cdot \frac{d(1-u^4)}{(1-u^4)} + \frac{du}{1+u^2} + \frac{du}{1-u^2}= \frac{dx}{x}$$
Integrating both sides, we get that
$$-\frac{1}{4}\cdot \ln|1-u^4| + \arctan u + \frac{1}{2}\ln|\frac{1+u}{1-u}|=\ln |x| + c$$
Hope this helps.
|
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|
integrate $\int \frac{dx}{1+cos^2x}$
$$\int \frac{dx}{1+\cos^2x}$$
I used $\cos x=\frac{1-v^2}{1+v^2}$ and $dx=\frac{2dv}{1+v^2}$
and got $$2\int \frac{dv}{v^4-v^2+1}=2\int \frac{dv}{(v^2-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\frac{4}{\sqrt{3}}arctan(\frac{2v^2-1}{\sqrt{3}})+c$$
How to continue?
|
Note that $\cos^2(x)=\frac{1+\cos(2x)}{2}$. Then,
$$\int \frac{1}{1+\cos^2(x)}\,dx=\int \frac{2}{3+\cos(2x)}\,dx$$
Enforcing the substitution $x=u/2$ yields
$$ \int \frac{2}{3+\cos(2x)}\,dx=\int \frac{1}{3+\cos(u)}\,du \tag 1$$
Now, making the Weierstrass Substitution in $(1)$, as in the OP, we find
$$\int \frac{1}{1+\cos^2(x)}\,dx=\frac{1}{\sqrt 2}\arctan\left(\frac{\tan(u/2)}{\sqrt 2}\right)+C=\frac{1}{\sqrt 2}\arctan\left(\frac{\tan(x)}{\sqrt 2}\right)+C$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1660682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac{1-v^2}{(1+v^2)(-v^2+v+1)} \, dv$$
Using partial fractions
$$-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}=\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{2v}{v^2+1}-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}$$
$$=\frac{1}{5}ln|v^2+1|-\frac{4}{5}\arctan(v)+\frac{1}{5}ln|-v^2+v+1|$$ from $\frac{\pi}{8}$ to $0$
$0.02-0+0.299-0+0.04-0=0.359$
But it should come out 0.32
|
I have done the sum u just look at it.
Here after step 2 I have used the method where u have to do like this
Af(x) +Bf'(x) =numerator. and here A and B are constant.
Here f(x)=denominator =2cosx +sinx
Hence f'(x) =-2sinx +cosx
So we get A = 2/5 B= 1/5
Thanks for asking such a nice question.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding sine of an angle in degrees without $\pi$ The following is found using a combination of: (a) a polygon with an infinite number of sides is a circle, (b) the perimeter of that polygon is the circumference of the circle that it becomes (of course), (c) the sine theorem, and (d) the ratio between the circumference of a circle and the diameter is $\pi$.
$$\pi=\lim_{n\to\infty} n\sin \left(\frac{180}{n}\right)^o$$
I have been trying to rewrite the above expression without $sin$ in it explicitly. So, I researched other expressions for $sin$, and found this:
$$\sin (x) = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\cdots$$
Attempting to rewrite this Taylor series in Sigma notation yielded this:
$$\sin(x)=\sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)!}(-1)^{n-1}$$
The issue for me is that I have an angle measure in degrees, but $x$ above is in radians. I would convert the degrees to radians and be done with it just like that, except that I feel that solving for $\pi$ with $\pi$ is like defining a word with itself, and is 'cheating'.
I have been trying to find a way to make this conversion without $\pi$ but to no avail. I also tried using trig identities, but that has just led me in circles.
Is there another way to find the sine of $x^o$ without involving $\pi$ at all?
An even more explicit and blunt form of my question: What is another way to express $sin(x^o)$ without using $\pi$?
To be clear, I am not looking for anything that involves approximations or infinite series that cannot be wholly expressed in a finite amount of space (ex. Taylor series above is okay because it can be expressed in sigma notation).
|
As noted in a comment, there are equations you can write for the
trigonometric functions of rational portions of a right angle
(and therefore for the sine of any rational number of degrees)
without using $\pi$ or any trigonometric functions.
To actually use these equations may prove somewhat cumbersome, however.
Instead of attempting a formula for an arbitrary integer or rational
number of degrees, let me just address the desire to evaluate
$$
\newcommand{\Sin}{\mathop{\mathrm{Sin}}}
\newcommand{\Cos}{\mathop{\mathrm{Cos}}}
\lim_{n\to\infty} n \Sin\left(\frac{180}{n}\right)
$$
where $\Sin$ is the sine function that takes its parameter in degrees,
for example, $\Sin(90) = 1$.
It is sufficient to examine the following limit for integer values of $k$:
$$
\lim_{k\to\infty} 2^k \Sin\left(\frac{180}{2^k}\right).
$$
The quantity $\Sin\left(\frac{180}{2^k}\right)$ is easy to compute
(at least, it is easy compared to such values as $\Sin(1)$).
Let $\Cos(x)$ be the cosine function for $x$ measured in degrees;
then apply the half-angle formula for cosines of angles in the
interval from $0$ to $180$ degrees, inclusive:
$$
\Cos\left(\frac{x}{2}\right) = \sqrt{\frac{1 + \Cos x}{2}}.
$$
If $x = \dfrac{180}{2^m}$, then $\dfrac x2 = \dfrac{180}{2^{m+1}}$
and the half-angle formula just says that
$$
\Cos\left(\frac{180}{2^{m+1}}\right)
= \sqrt{\frac{1 + \Cos \left(\frac{180}{2^m}\right)}{2}}.
$$
You can find the value for any $k$ by starting at $m=1$ and applying the
half-angle formula repeatedly for $m=2,3,\ldots, k$:
\begin{align}
\Cos\left(\frac{180}{2}\right) &= \Cos(90) = 0,\\
\Cos\left(\frac{180}{4}\right) &=
\sqrt{\frac{1 + \Cos \left(\frac{180}{2}\right)}{2}}
= \sqrt{\frac{1 + 0}{2}} = \frac{1}{\sqrt 2}, \\
\Cos\left(\frac{180}{8}\right) &=
\sqrt{\frac{1 + \Cos \left(\frac{180}{4}\right)}{2}}
= \sqrt{\frac{1 + \frac{1}{\sqrt 2}}{2}}
= \sqrt{\frac{\sqrt 2 + 1}{2\sqrt 2}}, \\
\Cos\left(\frac{180}{16}\right) &=
\sqrt{\frac{1 + \Cos \left(\frac{180}{8}\right)}{2}}
= \sqrt{\frac{1 + \sqrt{\frac{\sqrt 2 + 1}{2\sqrt 2}}}{2}}
= \sqrt{\frac{\sqrt{2\sqrt 2} + \sqrt{\sqrt 2 + 1}}{2\sqrt{2\sqrt 2}}}, \\
\end{align}
and so forth to obtain $\Cos\left(\frac{180}{2^k}\right)$
for $k$ as large as desired.
Then obtain the sine by
$$
\Sin\left(\frac{180}{2^k}\right) =
\sqrt{1 - \left(\Cos\left(\frac{180}{2^k}\right)\right)^2}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Evaluating the integral $\int \sqrt{1 + \frac{1}{x^2}} dx$
$$\int \sqrt{1 + \frac{1}{x^2}} dx$$
This is from the problem calculating the arc length of $y=\log{x}$.
I tried $x = \sinh{t}$ or $\frac{1}{x} = \tan{t}$ but all failed.
|
If $x>0$,
$$\begin{align*}
I&=\int \sqrt{1 + \frac1{x^2}} \, dx \\
&= \int \frac{\sqrt{x^2+1}}{x} \, dx\\
&= -\frac12 \int \frac{(1+t^2)^2}{t^2(1-t^2)}\,dt \\
&= \int \frac{(1+u^2)^2}{u(1-u^2)^2}\,du
\end{align*}$$
by either substituting
$$t=\sqrt{1+x^2}-x \implies x=\frac{1-t^2}{2t} \implies dx = -\frac{1+t^2}{2t^2}\,dt$$
or
$$u=\frac{\sqrt{x^2+1}-1}x \implies x=\frac{2u}{1-u^2} \implies dx = 2\frac{1+u^2}{(1-u^2)^2}\,du$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1663637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
How is Faulhaber's formula derived? I have been wanting to understand how to find the sum of this series.
$$1^p + 2^p + 3^p +{\dots} + n^p$$
I am familiar with Gauss' diagonalised adding trick for the sum of the first $n$ natural numbers.
I can prove the formulas for
$$\begin{align}
\sum_{1}^{n} k^2 &= \frac{n(n+1)(2n+1)}{6}\\
\sum_{1}^{n} k^3 &= \frac{n^2(n+1)^2}{4}\\
\sum_{1}^{n} k^4 &= \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} \\
\end{align}$$
With mathematical induction. But, beyond that even proofs with mathematical induction are difficult.
I'm interested in learning the theory and the proof behind Faulhaber's formula. What is the knowledge required to understand this proof ?
|
The following interesting derivation is from Aigner's "A Course in Enumeration" (Springer, 2007).
Remember that if we have the following exponential generating functions:
$\begin{align}
\widehat{A}(z)
&= \sum_{n \ge 0} a_n \frac{z^n}{n!} \\
\widehat{B}(z)
&= \sum_{n \ge 0} b_n \frac{z^n}{n!}
\end{align}$
then also:
$\begin{align}
\widehat{A}(z) \cdot \widehat{B}(z)
&= \sum_{n \ge 0}
\left(
\sum_{0 \le k \le n} \frac{a_k}{k!} \frac{b_{n - k}}{(n - k)!}
\right) z^n \\
&= \sum_{n \ge 0}
\left(
\sum_{0 \le k \le n} \frac{n!}{k! (n - k)!} a_k b_{n - k}
\right) \frac{z^n}{n!} \\
&= \sum_{n \ge 0}
\left(
\sum_{0 \le k \le n} \binom{n}{k} a_k b_{n - k}
\right) \frac{z^n}{n!}
\end{align}$
Let's define:
$\begin{align}
S_m(n)
= \sum_{1 \le k \le n - 1} k^m
\end{align}$
We can define the exponential generating function:
$\begin{align}
\widehat{S}_n(z)
&= \sum_{m \ge 0} S_m(n) \frac{z^m}{m!} \\
&= \sum_{1 \le k \le n - 1} \sum_{m \ge 0} \frac{k^m z^m}{m!} \\
&= \sum_{1 \le k \le n - 1} \mathrm{e}^{k z} \\
&= \frac{\mathrm{e}^{n z} - 1}{\mathrm{e}^z - 1} \\
\end{align}$
This is almost the exponential generating function of the powers of $n$:
$\begin{align}
\widehat{P}_n(z)
&= \sum_{m \ge 0} n^m \frac{z^m}{m!} \\
&= \mathrm{e}^{n z}
\end{align}$
We can write:
$\begin{align}
(\widehat{P}_n(z) - 1) \widehat{B}(z)
= z \widehat{S}_n(z) \tag{1}
\end{align}$
where we have the exponential generating function of the Bernoulli numbers:
$\begin{align}
\widehat{B}(z)
&= \frac{z}{\mathrm{e}^z - 1} \\
&= \sum_{n \ge 0} B_n \frac{z^n}{n!}
\end{align}$
Comparing coefficients of $z^{m + 1}$ in (1):
$\begin{align}
\sum_{m \ge 1} z^m
\sum_{0 \le k \le m}
\binom{m}{k} \frac{(n z)^{m - k}}{(m - k)!} B_k
= \sum_{m \ge 0} S_m(n) \frac{z^{m + 1}}{m!}
\end{align}$
we get after simpĺifying:
$\begin{align}
S_m(n)
= \frac{1}{m + 1} \,
\sum_{0 \le k \le m} \binom{m + 1}{k} B_k n^{m + 1 - k}
\end{align}$
Note: The formula given is often associated with Faulhaber, but Faulhaber's formulas where quite different (and computationally more efficient). This formula is due to Bernoulli.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $(n + 1)a^n < \frac{b^{n + 1} - a^{n + 1}}{(b-a)} < (n + 1)b^n$ $(b-a)(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n)$
$= (b^{n + 1} - ab^n) + (ab^n - a^2b^{n - 1}) + (a^2b^{n - 1} - a^3b^{n - 2}) + \ldots + (a^nb - a^{n + 1})$
$= b^{n + 1} - a^{n + 1}$, so
$(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n) = \frac{b^{n + 1} - a^{n + 1}}{(b-a)}$
What can I do from here?
|
Hint
We have that
$$\dfrac{b^{n+1}-a^{n+1}}{b-a}=b^n+b^{n-1}a+b^{n-2}a^2+\cdots +a^2b^{n-2}+ba^{n-1}+a^n$$
Is not $a^n< b^ia^{n-i}< b^n, \forall i\in \{0,\dots, n\}?$ Are not there $n+1$ summands?
|
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|
Solving recurrence relation? Consider the recurrence relation $a_1=8,a_n=6n^2+2n+a_{n−1}$. Let $a_{99}=K\times10^4$ .The value of $K$ is______ .
My attempt :
$a_n=6n^2+2n+a_{n−1}$
$=6n^2+2n+6(n−1)^2+2(n−1)+a_{n−2}$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+a_1$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+6.1^2+2.1$
$=6(n^2+(n−1)^2+...+2^2+1^2)+2(n+(n−1)+...+2+1)$
$=6×n(n+1)(2n+1)/6+2×n(n+1)/2$
$=n(n+1)(2n+1+1)$
$=2n^3+2n^2+2n^2+2n$
$=2n(n^2+n+n+1)$
$=2n(n^2+2n+1)$
$a_n=2n(n+1)^2$
for $n=99, a_{99}=2×99×(99+1)^2=198×10^4$
I'm looking for short trick or alternative way, can you explain please?
|
The idea is to find a closed form solution. We start with
$$
a_1 = 8 \\
6n^2+2n = a_n -a_{n−1}
$$
which is an inhomogenous linear recurrence relation and look for the previous
sequence elements:
$$
6(n-1)^2 + 2(n-1) = 6n^2 -12 n + 6 + 2n - 2 = 6 n^2 - 10n + 4
= a_{n-1} - a_{n-2}
$$
Substraction gives
$$
12n - 4 = a_n - 2 a_{n-1} + a_{n-2}
$$
where the inhomogenous polynomial has been reduced by one degree and the order has increased from $1$ to $2$. Repeating the procedure leads to
$$
12(n-1) - 4 = 12 n - 16 = a_{n-1} - 2 a_{n-2} + a_{n-3} \Rightarrow \\
12 = a_n - 3 a_{n-1} + 3 a_{n-2} - a_{n-3} \Rightarrow \\
12 = a_{n-1} - 3 a_{n-2} + 3 a_{n-3} - a_{n-4} \Rightarrow \\
a_n = 4 a_{n-1} - 6 a_{n-2} + 4 a_{n-3} - a_{n-4}
$$
which now is a homogenous recurrence relation of order $4$.
The characteristic polynomial is
$$
p(t) = t^4 - 4 t^3 + 6 t^2 - 4 t + 1 = (t-1)^4
$$
so the solution is
$$
a_n = k_1 1^n + k_2 n 1^n + k_3 n^2 1^n + k_4 n^3 1^n = k_1 + k_2 n + k_3 n^2 + k_4 n^3
$$
The $k_i$ follow from the $a_i$ values:
\begin{align}
a_1 &= 8 \\
a_2 &= 6 \cdot 2^2 + 2 \cdot 2 + 8 = 36 \\
a_3 &= 6 \cdot 3^2 + 2 \cdot 3 + 36 = 96 \\
a_4 &= 6 \cdot 4^2 + 2 \cdot 4 + 96 = 200
\end{align}
We get a linear system
$$
\begin{pmatrix}
1 & 1 & 1 & 1 \\
1 & 2 & 4 & 8 \\
1 & 3 & 9 & 27 \\
1 & 4 & 16 & 64
\end{pmatrix}
\begin{pmatrix}
k_1 \\
k_2 \\
k_3 \\
k_4
\end{pmatrix}
=
\begin{pmatrix}
8 \\
36 \\
96 \\
200
\end{pmatrix}
$$
with the solution
$$
k =
\begin{pmatrix}
0 \\
2 \\
4 \\
2
\end{pmatrix}
$$
so we have
$$
a_n = 2 n + 4 n^2 + 2 n^3 = 2n (1 + 2n + n^2) = 2 n (n + 1)^2
$$
Then
$$
a_{99} = 2 \cdot 99 \cdot 100^2 = 198 \cdot 10^4
$$
and $K = 198$.
|
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|
Distributing k objects among m people where each person gets i objects such that aIn an examination, the score in each of the four subjects ( say ) - A ,B,C & D can be between integers 0 and 10. Then , how many are there such that the student can secure a total of 21 ?
My attempt - My initial attempt was to find out the total number of ways in which the student could score 21 and then look to subtract the number of ways which included a student scoring above 10. But I realised that this would include a lot of cases to be considered as I will have to weed out manually the attempts for each integer great than 10 which is too computationally intensive .
Please tell me if there is any formulae ( along with justification for the formulae ) for solving the same.
|
Let $a$, $b$, $c$, $d$ denote the student's scores in subjects $A$, $B$, $C$, and $D$, respectively. Then
$$a + b + c + d = 21 \tag{1}$$
is an equation in the non-negative integers subject to the constraints $a, b, c, d \leq 10$.
A particular solution of equation 1 corresponds to the insertion of three addition signs in a row of $21$ ones. For instance,
$$1 1 1 1 1 1 + 1 1 1 1 1 1 1 + 1 1 + 1 1 1 1 1 1$$
corresponds to $a = 6$, $b = 7$, $c = 2$, and $d = 6$, while
$$1 1 1 1 1 1 1 + 1 1 1 1 1 1 + + 1 1 1 1 1 1 1 1$$
corresponds to $a = 7$, $b = 6$, $c = 0$, and $d = 8$. Thus, the number of solutions of equation 1 is the number of ways three addition signs can be inserted into a row of $21$ ones, which is
$$\binom{21 + 3}{3} = \binom{24}{3}$$
since we must select which three of the $21$ symbols (three addition signs and $21$ ones) will be addition signs.
However, we must exclude solutions in which one or more of the variables exceeds $10$. Observe that at most one of the variables can exceed $10$ since $2 \cdot 11 = 22 > 21$.
Suppose $a > 10$. Let $a' = a - 11$. Then $a'$ is a non-negative integer. Substituting $a' + 11$ for $a$ in equation 1 yields
\begin{align*}
a' + 11 + b + c + d & = 21\\
a' + b + c + d & = 10 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers with $\binom{10 + 3}{3} = \binom{13}{3}$ solutions. By symmetry, there are also $\binom{13}{3}$ solutions in which $b$, $c$, or $d$ exceeds $10$. Hence, there are
$$\binom{4}{1}\binom{13}{3}$$
solutions of equation 1 in which one of the variables exceeds $10$.
Thus, the number of solutions of equation 1 in which none of the variables exceeds $10$ is
$$\binom{24}{3} - \binom{4}{1}\binom{13}{3}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
AB-BA so that you get another matrix I want to find out what the matrices are (a) and (b) so that $ab-ba = $
$$
\begin {pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\
\end {pmatrix} $$
This is the outcome of $ab-ba$ is. Ive been struggling with this for the past hour or so. All matrices have to be $4\times,4$
|
I like $$ A = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \\
B = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} \text{.} $$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.
|
Interesting answers .Let me add another, just for filing a somewhat more "theoretical" one, using Chebyshev polynomials $U_n(x)$ of the second kind defined by
$$U_0(x)=1\\U_1(x)=2x\\U_{n+1}(x)=2x\space U_n(x)-U_{n-1}(x)$$
and for which one has the formula
$$U_n(\cos x)=\frac{\sin (n+1)x}{\sin x}\qquad (*)$$
We use the data (if you don’t want to calculate it)
$$\begin{cases}U_6(x)=64x^6-80x^4+24x^2-1\\U_4(x)=16x^4-12x^2+1\\U_2(x)=4x^2-1\\U_0(x)=1\end{cases}$$
Therefore, using $(*)$ and simplifying the given identity,we have
$$(\sin x)^6=aU_6(\cos x)+bU_4(\cos x)+cU_2(\cos x)+dU_0(\cos x)$$
It follows, putting $\cos x= t$,
$$(1-t^2)^3=64at^6+(-80a+16b)t^4+(24a-12b+4c)t^2+(-a-c+b+d)$$
Hence the easy system
$$\begin{cases}64a=-1\\-80a+16b=3\\24a-12b+4c=-3\\-a+b-c+d=1\end{cases}$$
which is resolved line by line giving in succession
$$(a,b,c,d)=\left(\frac {-1}{64},\frac {7}{64},\frac{-21}{64},\frac {35}{64}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Interval of convergence of $\sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!}$
Given the series
$$ \sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!} \quad \quad k \geq 1 $$
Find the interval of convergence.
I started by applying the Ratio test
$$
\lim_{n\to \infty}\left|\frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)(k+n)x^{n+1}}{(n+1)!}\cdot \frac{n!}{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}\right|$$
$$\lim_{n\to \infty}\left|\frac{(k+n)x}{(n+1)}\right|$$
to show that the series converges when $|x| \lt 1$.
However, when I test the end points of $(-1,1)$ for convergence, I end up with two series whose convergence I am unable to show. Namely,
$$
\sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)}{n!}
$$
and
$$
\sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)(-1)^n}{n!}
$$
How can I show that these two series converge or diverge?
|
Note that for $k\ge1$, we have
$$
\frac{k(k+1)\cdots(k+n-1)}{n!}=\frac k1\frac{k+1}2\cdots\frac{k+n-1}{n}\ge1
$$
Thus, for $|x|=1$, the terms do not go to $0$.
|
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|
integrate $\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}dxdy, D: (x-a)^2+y^2 \le a^2$ $$I=\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}dxdy, D: (x-a)^2+y^2 \le a^2(a>0)$$
The difficulty is to find a proper and easy way to solve this double integrals.
If do it like this, $ 0\le x \le 2a, -\sqrt{2ax-x^2} \le y \le \sqrt{2ax-x^2}$, wolfram calculation exceeds the standard time.
Maybe using polar coordinate is easier? $x-a=r\cos\theta, y=r\sin\theta(-\frac\pi2 \le \theta \le \frac\pi2, 0\le r \le 2a)$, then $I=\int_{-\pi/2}^{\pi/2} d\theta\int 2a/\sqrt{4a^2-(a+r\cos\theta)^2-(r\sin\theta)^2}rdr$ which looks rather complex too.
Am I doing it wrong? How to integrate this $I$ ?
|
Let $x=r\cos(\theta),y=r\sin(\theta)$. We know that $0\le x\le 2a,-a\le y\le a$, hence $0\le r\le 2a$ and $\displaystyle -\frac{\pi}{2}\le \theta\le \frac{\pi}{2}$. The Jacobian is $r$, thus \begin{align*}\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}\text{d}x\text{d}y&=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int\limits_{0}^{2a}\frac{2ar}{\sqrt{4a^2-r^2}}\text{d}r\text{d}\theta=\begin{bmatrix}4a^2-r^2=u\\-2r\text{d}r=\text{d}u\end{bmatrix}\\&=\pi a\int\limits_{0}^{4a^2}\frac{1}{\sqrt{u}}\text{d}u=\Big.2\pi a\sqrt{u}\Big\vert_{0}^{4a^2}=4\pi a^2\end{align*}
|
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|
Factorize $(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2$ I am unable to factorize this over $\mathbb{Z}:$
$$(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2$$
Since, this from an exercise of a book (E. J. Barbeau, polynomials) it must have a neat factorization.
I tried to guess some factors by putting $x-y=0$, $x+y=0$, $x=0$.
I also notice that the expression is symmetric with respect to $x, y, z$.
I am trying to factor it without expanding the whole thing though the process of expansion can be eased by observing that $(x+y+z)(x+y-z)(y+z-x)(z+x-y)$ is a well known expression (Heron's formula), so one already knows it is
$$2(x^2y^2+y^2z^2+z^2x^2)-x^4-y^4-z^4$$
But I think there should be a way to do this without expanding the thing.
So, can someone tell me the factorization (preferably with the steps) or any hints as to how should I approach this problem.
|
We write the expression as polynomial in $z^2$. This way we obtain
\begin{align*}
P(x,y,z)&=(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2\\
&=-(x^2+y^2+z^2)\left((x+y)^2-z^2\right)\left((x-y)^2-z^2\right)-8x^2y^2z^2\\
&=-z^6+(x^2+y^2)z^4+(x^2-y^2)^2z^2-(x^2-y^2)^2(x^2+y^2)
\end{align*}
Analyzing the last line and looking for a substitution for $z^2$ which makes this polynomial vanish it's not too hard to see that
\begin{align*}
z^2=x^2+y^2
\end{align*}
does the job.
We obtain
\begin{align*}
P(x,y,z)=(x^2+y^2-z^2)Q(x,y,z)
\end{align*}
with $Q(x,y,z)$ a homogeneous polynomial in $x,y$ and $z$.
Division of $P(x,y,z)$ by this factor gives
\begin{align*}
\frac{-z^6+(x^2+y^2)z^4+(x^2-y^2)^2z^2-(x^2-y^2)^2(x^2+y^2)}{-z^2+x^2+y^2}=z^4-(x^2-y^2)^2
\end{align*}
From the RHS we derive the other irreducible factors of $P(x,y,z)$
\begin{align*}
z^2-(x^2-y^2)\qquad\text{and}\qquad z^2+(x^2-y^2)
\end{align*}
and conclude
\begin{align*}
P(x,y,z)=(x^2+y^2-z^2)(y^2+z^2-x^2)(z^2+x^2-y^2)
\end{align*}
|
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|
evaluate $\sum^{\infty}_{n=2} \frac{3}{10^n}$
evaluate $$\sum^{\infty}_{n=2} \frac{3}{10^n}$$
I know I can factor out $$\sum^{\infty}_{n=2} \frac{3}{10^n}=3\sum^{\infty}_{n=2} \frac{1}{10^n}$$
And I know that the sequence converges $${{\large \frac{1}{10^{n+1}}}\over{\large \frac{1}{10^n}}}=\frac{1}{10}<1$$
But how do I find the sum?
|
You know that $\sum_{n\ge 1}q^n$ converges (you have $q=\frac{1}{10}$, similarily it works for all $q\in(-1,1)$), so let $$S=\sum_{n\ge 1}q^n=q+q^2+q^3+\dots$$ then $$qS=q^2+q^3+\dots = S-q$$ so$$S=\frac{q}{1-q}$$
Hence
$$\sum_{n=2}^{\infty}\frac{3}{10^n}=3\left(\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^n-\frac{1}{10}\right)=\frac{1}{30}$$
|
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|
Maximum value of $\frac{3x^2+9x+17}{x^2+2x+9}$ If $x$ is real, the maximum value of $\frac{3x^2+9x+17}{x^2+2x+9}$ is?
Is it necessary that this function will attain maximum when the denominator will be minimum?
|
Note that by division $\dfrac{3x^2+9x+17}{x^2+2x+9}=3+\dfrac{3x-10}{x^2+2x+9}$, so if the first expression has a maximum (or minimum) value it will occur when $\dfrac{3x-10}{x^2+2x+9}$ has its maximum (or minimum) value. The advantage of using $\dfrac{3x-10}{x^2+2x+9}$ is that the derivative is algebraically simpler.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.